Nuevo examen - 20 de Octubre de 2014 - David Delgado - DP Maths Teacher | Maths … · 2016-03-02 · Nuevo examen - 20 de Octubre de 2014 [1207 marks] 1.The equation has two distinct

Nuevo examen - 20 de Octubre de 2014 [1207 marks]

1. [6 marks]The equation has two distinct real roots. Find the possible values of k .

Markschemeevidence of rearranged quadratic equation (may be seen in working) A1

e.g. ,

evidence of discriminant (must be seen explicitly, not in quadratic formula) (M1)

e.g. ,

recognizing that discriminant is greater than zero (seen anywhere, including answer) R1

e.g. ,

correct working (accept equality) A1

e.g. , ,

both correct values (even if inequality never seen) (A1)

e.g. ,

correct interval A1 N3

e.g. ,

Note: Do not award the final mark for unfinished values, or for incorrect or reversed inequalities, including , , .

Special cases:

If working shown, and candidates attempt to rearrange the quadratic equation to equal zero, but find an incorrect value of c, awardA1M1R1A0A0A0.

If working shown, and candidates do not rearrange the quadratic equation to equal zero, but find or , awardA0M1R1A0A0A0.

[6 marks]

Examiners reportThe majority of candidates who attempted to answer this question recognized the need to use the discriminant, however very fewwere able to answer the question successfully. The majority of candidates did not recognize that the quadratic equation must first beset equal to zero. In addition, many candidates simply set their discriminant equal to zero, instead of setting it greater than zero. Evenmany of the strongest candidates, who obtained the correct numerical values for k, were unable to write their final answers as acorrect interval.

This question is a good example of candidates who reach for familiar methods, without really thinking about what the question isasking them to find. There were many candidates who attempted to solve for x using the quadratic formula or factoring, even thoughthe question did not ask them to solve for x.

− 3x+ = 4x2 k2

− 3x+ − 4 = 0x2 k2 − 4k2

− 4acb2 Δ = (−3 − 4(1)( − 4))2 k2

− 4ac > 0b2 9 + 16 − 4 > 0k2

25 − 4 > 0k2 4 < 25k2 =k2 254

± 254

−−√ ±2.5

− < k <52

52

−2.5 < k < 2.5

≤ k > −2.5 k < 2.5

c = k2 c = ±4

f(x) = − 2x− 43

2a. [1 mark]

Let . The following diagram shows part of the curve of f .

The curve crosses the x-axis at the point P.

Write down the x-coordinate of P.

Markscheme (accept ) A1 N1

[1 mark]

Examiners reportThis question was generally done well.

f(x) = − 2x− 4x3

x = 2 (2, 0)

2b. [2 marks]Write down the gradient of the curve at P.

Markschemeevidence of finding gradient of f at (M1)

e.g.

the gradient is 10 A1 N2

[2 marks]

Examiners reportMost candidates did not use their GDC in part (b), resulting in a variety of careless errors occasionally arising either in differentiatingor substituting.

x = 2

(2)f ′

2c. [3 marks]Find the equation of the normal to the curve at P, giving your equation in the form .y = ax+ b

Markschemeevidence of negative reciprocal of gradient (M1)

e.g. ,

evidence of correct substitution into equation of a line (A1)

e.g. ,

(accept , ) A1 N2

[3 marks]

Examiners reportThere were some candidates who did not know the relationship between gradients of perpendicular lines while others found theequation of the tangent rather than the normal in part (c).

−1(x)f′

− 110

y− 0 = (x− 2)−110

0 = −0.1(2) + b

y = − x+110

210

a= −0.1 b = 0.2

3a. [2 marks]

Let . The diagram below shows part of the graph of f , for .

The graph has a local maximum at P(3, 5) , a local minimum at Q(7, − 5) , and crosses the x-axis at R.

Write down the value of

(i) ;

(ii) .

Markscheme(i) (accept ) A1 N1

(ii) (accept , if ) A1 N1

Note: Accept other correct values of c, such as 11, , etc.

[2 marks]

Examiners reportPart (a) (i) was well answered in general. There were more difficulties in finding the correct value of the parameter c.

f(x) = acos(b(x− c)) 0 ≤ x ≤ 10

a

c

a= 5 −5

c = 3 c = 7 a= −5

−5

3b. [2 marks]Find the value of b .

Markschemeattempt to find period (M1)

e.g. 8 ,

(exact), , 0.785 [ ] (do not accept 45) A1 N2

[2 marks]

Examiners reportFinding the correct value of b in part (b) also proved difficult as many did not realize the period was equal to 8.

b = 2πperiod

0.785398…

b = 2π8

π

40.785, 0.786

3c. [2 marks]Find the x-coordinate of R.

Markschemevalid approach (M1)

e.g. , symmetry of curve

(accept A1 N2

[2 marks]

Examiners reportMost candidates could handle part (c) without difficulties using their GDC or working with the symmetry of the curve althoughfollow through from errors in part (b) was often not awarded because candidates failed to show any working by writing down theequations they entered into their GDC.

f(x) = 0

x = 5 (5 ,0))

4. [6 marks]

The diagram below shows part of the graph of .

(a) Write down the -intercepts of the graph of .

(b) Find the coordinates of the vertex of the graph of .

f(x) = (x− 1)(x+ 3)

x f

f

Markscheme(a) , (accept ( , ), ( , ) ) A1A1 N2

[2 marks]

(b) METHOD 1

attempt to find -coordinate (M1)

eg , ,

correct value, (may be seen as a coordinate in the answer) A1

attempt to find their -coordinate (M1)

eg , ,

A1

vertex ( , ) N3

METHOD 2

attempt to complete the square (M1)

eg

attempt to put into vertex form (M1)

eg ,

vertex ( , ) A1A1 N3

[4 marks]

Examiners reportMost candidates recognized the values of the x-intercepts from the factorized form of the function. Candidates also showed littledifficulty finding the vertex of the graph, and employed a variety of techniques: averaging -intercepts, using , completing thesquare.

x = 1 x = −3 1 0 −3 0

x

1+−32

x = −b2a

(x) = 0f ′

x = −1

y

f(−1) −2 × 2 y = −D4a

y = −4

−1 −4

+ 2x+ 1 − 1 − 3x2

(x+ 1 − 4)2 (x− 1 + 4)2

−1 −4

x x = −b2a

5a. [3 marks]

Let , for .

Find .

MarkschemeMETHOD 1

attempt to set up equation (M1)

eg ,

correct working (A1)

eg ,

A1 N2

METHOD 2

interchanging and (seen anywhere) (M1)

eg


eg ,

A1 N2

[3 marks]

f(x) = x− 5− −−−√ x ≥ 5

(2)f−1

2 = y− 5− −−−√ 2 = x− 5− −−−√

4 = y− 5 x = + 522

(2) = 9f−1

x y

x = y− 5− −−−√

= y− 5x2 y = + 5x2

(2) = 9f−1

Examiners reportCandidates often found an inverse function in which to substitute the value of . Some astute candidates set the function equal to and solved for . Occasionally a candidate misunderstood the notation as asking for a derivative, or used .

2 2x 1

f(x)

5b. [3 marks]Let be a function such that exists for all real numbers. Given that , find .

Markschemerecognizing (M1)

eg


eg ,

A1 N2

Note: Award A0 for multiple values, eg .

[3 marks]

Examiners reportFor part (b), many candidates recognized that if then , and typically completed the question successfully.Occasionally, however, a candidate incorrectly answered .

g g−1 g(30) = 3 (f ∘ )(3)g−1

(3) = 30g−1

f(30)

(f ∘ )(3) =g−1 30 − 5− −−−−√ 25−−√

(f ∘ )(3) = 5g−1

±5

g(30) = 3 (3) = 30g−1

= ±525−−√

6. [4 marks]Find the value of .

Markschemeattempt to write as a power of (seen anywhere) (M1)

eg , ,

multiplying powers (M1)

eg ,


eg , ,

A1 N3

[4 marks]

Examiners reportPart (b) proved challenging for most candidates, with few recognizing that changing to base is a helpful move. Some made it asfar as yet could not make that final leap to an integer.

8 5log2

8 2

(23) 5log2 = 823 2a

23 5log2 a 5log2

2 125log2 log253 ( )2 5log23

= 1258 5log2

8 223 5log2

7a. [2 marks]

Consider .

Find the value of .

f(x) = ln( + 1)x4

f(0)

Markschemesubstitute into (M1)

eg ,

A1 N2

[2 marks]

Examiners reportMany candidates left their answer to part (a) as . While this shows an understanding for substituting a value into a function, itleaves an unfinished answer that should be expressed as an integer.

0 f

ln(0 + 1) ln1

f(0) = 0

ln1

7b. [3 marks]

The second derivative is given by .

The equation has only three solutions, when , .

There is a point of inflexion on the graph of at .

Sketch the graph of , for .

Markscheme

A1A1A1 N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

Only if this A1 is awarded, then award the following:

A1 for curve through ( , ) , A1 for increasing throughout.

Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

Examiners reportFew candidates created a correct graph from the information given or found in the question. This included the point ( , ), the factthat the function is always increasing for , the concavity at and the change in concavity at the given point of inflexion.Many incorrect attempts showed a graph concave down to the right of , changing to concave up.

(x) =f ′′ 4 (3− )x2 x4

( +1)x4 2

(x) = 0f ′′ x = 0 ± 3√4 (±1.316…)

f x = 3√4 (x = 1.316…)

f x ≥ 0

0 0

0 0x > 0 x = 1

x = 0

−1 v = − 1sin t

8a. [3 marks]

The velocity of a particle in ms is given by , for .

On the grid below, sketch the graph of .

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape crossing x-axis with .

Only if this A1 is awarded, award the following:

A1 for maximum in circle, A1 for endpoints in circle.

[3 marks]

Examiners reportThere was a minor error on this question, where the units for velocity were given as ms rather than ms . Examiners were instructedto notify the IB assessment centre of any candidates adversely affected, and these were considered at the grade award meeting.

Candidates continue to produce sloppy graphs resulting in loss of marks. Although the shape was often correctly drawn, studentswere careless when considering the domain and other key features such as the root and the location of the maximum point.

−1 v = − 1esin t 0 ≤ t ≤ 5

v

3 < x < 3.5

-2 -1

8b. [1 mark]Find the total distance travelled by the particle in the first five seconds.

Markscheme (exact), A1 N1

[1 mark]

Examiners reportThe fact that most candidates with poorly drawn graphs correctly found the root in (b)(i), clearly emphasized the disconnect betweengeometric and algebraic approaches to problems.

t = π 3.14

9. [6 marks]

Let and be functions such that .

(a) The graph of is mapped to the graph of under the following transformations:

vertical stretch by a factor of , followed by a translation .

Write down the value of

(i) ;

(ii) ;

(iii) .

(b) Let . The point A( , ) on the graph of is mapped to the point on the graph of . Find .

Markscheme(a) (i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

(b) recognizing one transformation (M1)

eg horizontal stretch by , reflection in -axis

is ( , ) A1A1 N3

[3 marks]

Total [6 marks]

Examiners reportPart (a) was frequently done well but a lack of understanding of the notation often led to an incorrect value for . In part (b),candidates did not recognize the simplicity of the problem. Most seemed to be unable to correctly recognize the two transformationsimplied in the question and were thus unable to attempt a geometric solution. Flawed algebraic approaches to part (b) were commonand many could not interpret the notation as multiplying the -value by .

f g g(x) = 2f(x+ 1) + 5

f g

k ( )pq

k

p

q

h(x) = −g(3x) 6 5 g A′ h A′

k = 2

p = −1

q = 5

13

x

A′ 2 −5

f(x+ 1) p

g(3x) x 13

10a. [1 mark]

Let . Part of the graph of is shown below.

Write down .

f(x) = 100(1+50 )e−0.2x

f

f(0)

Markscheme (exact), A1 N1

[1 mark]

Examiners reportCandidates had little difficulty with parts (a), (b) and (c).

f(0) = 10051

1.96

10b. [2 marks]Solve .

Markschemesetting up equation (M1)

eg , sketch of graph with horizontal line at

A1 N2

[2 marks]

Examiners reportCandidates had little difficulty with parts (a), (b) and (c). Successful analytical approaches were often used in part (b) but again, thiswas not the most efficient or expected method.

f(x) = 95

95 = 1001+50e−0.2x

y = 95

x = 34.3

10c. [3 marks]Find the range of .

Markschemeupper bound of is (A1)

lower bound of is (A1)

range is A1 N3

[3 marks]

Examiners reportCandidates had little difficulty with parts (a), (b) and (c). In part (c), candidates gained marks by correctly identifying upper and lowerbounds but often did not express them properly using an appropriate notation.

f

y 100

y 0

0 < y < 100

11a. [3 marks]

Let and .

Find .

f(x) = 4x− 2 g(x) = −2 + 8x2

(x)f−1

Markschemeinterchanging and (seen anywhere) (M1)

eg

evidence of correct manipulation (A1)

eg

(accept , , A1 N2

[3 marks]

Examiners reportThe overwhelming majority of candidates answered both parts of this question correctly. There were a few who seemed unfamiliarwith the inverse notation and answered part (a) with the derivative or the reciprocal of the function.

x y

x = 4y− 2

x+ 2 = 4y

(x) =f−1 x+24

y = x+24

x+24

(x) = x+f−1 14

12

11b. [3 marks]Find .

MarkschemeMETHOD 1

attempt to substitute into (M1)

eg

(A1)

A1 N3

METHOD 2

attempt to form composite function (in any order) (M1)

eg

correct substitution

eg ,

A1 N3

[3 marks]

Examiners reportThe overwhelming majority of candidates answered both parts of this question correctly. A few candidates made arithmetic errors inpart (b) which kept them from finding the correct answer.

(f ∘ g)(1)

1 g(x)

g(1) = −2 × + 812

g(1) = 6

f(6) = 22

(f ∘ g)(x) = 4(−2 + 8) − 2x2 (= −8 + 30)x2

(f ∘ g)(1) = 4(−2 × + 8) − 212 −8 + 30

f(6) = 22

12a. [1 mark]

The diagram below shows the graph of a function , for .

Write down the value of .

Markscheme A1 N1

[1 mark]

Examiners reportIn part (a) of this question, most candidates were able to find the value of correctly, while some had trouble finding .Many candidates tried to find an equation for the function, or to make tables of values to help them find their answers. The intent ofthis question was to read the answers from the given graph. Candidates should be reminded that when the command term is "writedown", there is no need for them to do large amounts of working.

f −1 ≤ x ≤ 2

f(2)

f(2) = 3

f(2) (−1)f−1

12b. [2 marks]Write down the value of .

Markscheme A2 N2

[2 marks]

Examiners reportIn part (a) of this question, most candidates were able to find the value of correctly, while some had trouble finding .Many candidates tried to find an equation for the function, or to make tables of values to help them find their answers. The intent ofthis question was to read the answers from the given graph. Candidates should be reminded that when the command term is "writedown", there is no need for them to do large amounts of working.

(−1)f−1

(−1) = 0f−1

f(2) (−1)f−1

−1

12c. [3 marks]Sketch the graph of on the grid below.

MarkschemeEITHER

attempt to draw on grid (M1)

OR

attempt to reverse x and y coordinates (M1)

eg writing or plotting at least two of the points

, , ,

THEN

correct graph A2 N3

[3 marks]

Examiners reportIn part (b) of this question, candidates were generally successful in reversing the and coordinates of key points or reflecting in the

line to correctly sketch the graph of the inverse function. Common errors included not sketching the graph for the appropriatedomain, or sketching the graph of or the graph of .

f−1

y = x

(−2,−1) (−1,0) (0,1) (3,2)

x y

y = x

f(−x) −f(x)

13a. [3 marks]

Let , for .

Let be a quadratic function such that . The line is the axis of symmetry of the graph of .

Find .

Markschemerecognizing gives the point ( , ) (R1)

recognize symmetry (M1)

eg vertex, sketch

A1 N3

[3 marks]

Examiners reportIn part (b), many candidates did not understand the significance of the axis of symmetry and the known point ( , ), and so wereunable to find using symmetry. A few used more complicated manipulations of the function, but many algebraic errors wereseen.

f(x) = sinx+ − 2x12x2 0 ≤ x ≤ π

g g(0) = 5 x = 2 g

g(4)

g(0) = 5 0 5

g(4) = 5

0 5g(4)

13b. [4 marks]

The function can be expressed in the form .

(i) Write down the value of .

(ii) Find the value of .

g g(x) = a(x−h + 3)2

h

a

Markscheme(i) A1 N1

(ii) substituting into (not the vertex) (M1)

eg ,

working towards solution (A1)

eg ,

A1 N2

[4 marks]

Examiners reportIn part (c), a large number of candidates were able to simply write down the correct value of , as intended by the command term inthis question. A few candidates wrote down the incorrect negative value. Most candidates attempted to substitute the and valuesof the known point correctly into the function, but again many arithmetic and algebraic errors kept them from finding the correct valuefor .

h = 2

g(x) = a(x− 2 + 3)2

5 = a(0 − 2 + 3)2 5 = a(4 − 2 + 3)2

5 = 4a+ 3 4a= 2

a= 12

h

x y

a

14a. [3 marks]

Consider the functions , and . The following table gives some values associated with these functions.

Write down the value of , of , and of .

Markscheme , , A1A1A1 N3

[3 marks]

Examiners reportNearly all candidates who attempted to answer parts (a) and (c) did so correctly, as these questions simply required them tounderstand the notation being used and to read the values from the given table.

f(x) g(x) h(x)

g(3) (3)f ′ (2)h′′

g(3) = −18 (3) = 1f ′ (2) = −6h′′

′′

14b. [2 marks]

The following diagram shows parts of the graphs of and .

There is a point of inflexion on the graph of at P, when .

Given that ,

find the -coordinate of P.

Markschemewriting as a product of and A1

eg ,

A1 N1

[2 marks]

Examiners reportNearly all candidates who attempted to answer parts (a) and (c) did so correctly, as these questions simply required them tounderstand the notation being used and to read the values from the given table.

h h′′

h x = 3

h(x) = f(x) × g(x)

y

h(3) f(3) g(3)

f(3) × g(3) 3 × (−18)

h(3) = −54

15. [7 marks]

The following diagram shows a circle with centre O and radius cm.

Points A and B are on the circumference of the circle and radians .

The point C is on [OA] such that radians .

The area of the shaded region is cm . Find the value of .

r

A B = 1.4O

B O =C π

2

25 2 r

Markschemecorrect value for BC

eg , (A1)

area of A1

area of sector A1

attempt to subtract in any order (M1)

eg sector – triangle,

correct equation A1

eg

attempt to solve their equation (M1)

eg sketch, writing as quadratic,

A1 N4

[7 marks]

Note: Exception to FT rule. Award A1FT for a correct FT answer from a quadratic equation involving two trigonometric functions.

Examiners reportAs to be expected, candidates found this problem challenging. Those who used a systematic approach in part (b) were moresuccessful than those whose work was scattered about the page. While a pleasing number of candidates successfully found the area ofsector AOB, far fewer were able to find the area of triangle BOC. Candidates who took an analytic approach to solving the resultingequation were generally less successful than those who used their GDC. Candidates who converted the angle to degrees generallywere not very successful.

BC = rsin 1.4 −r2 (rcos1.4)2− −−−−−−−−−−−√

ΔOBC = rsin 1.4 × rcos1.412

(= sin 1.4 × cos1.4)12r2

OAB = × 1.412r2

sin 1.4 × cos1.4 − 0.712r2 r2

0.7 − rsin 1.4 × rcos1.4 = 25r2 12

250.616…

r = 6.37

16a. [2 marks]

Let and , where , and . Let be the region enclosed by the -axis, the graph of , and the graphof .

Let .

Sketch the graphs of and on the same axes.

f(x) = ex

4 g(x) =mx m ≥ 0 −5 ≤ x ≤ 5 R y f

g

m = 1

f g

Markscheme

A1A1 N2

Notes: Award A1 for the graph of positive, increasing and concave up.

Award A1 for graph of increasing and linear with -intercept of .

Penalize one mark if domain is not [ , ] and/or if and do not intersect in the first quadrant.

[2 marks]

Examiners reportThere was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed whatwas intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notifythe IB assessment centre of any candidates adversely affected, and these were looked at during the grade award meeting.

While some candidates sketched accurate graphs on the given domain, the majority did not. Besides the common domain error, someexponential curves were graphed with several concavity changes.

f

g y 0

−5 5 f g

16b. [8 marks]Consider all values of such that the graphs of and intersect. Find the value of that gives the greatest value for the areaof .

m f g m

R

Markschemerecognize that area of is a maximum at point of tangency (R1)

eg

equating functions (M1)

eg ,

(A1)

equating gradients (A1)

eg ,

attempt to solve system of two equations for (M1)

eg

(A1)

attempt to find (M1)

eg ,

(exact), A1 N3

[8 marks]

Examiners reportThere was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed whatwas intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notifythe IB assessment centre of any candidates adversely affected, and these were looked at during the grade award meeting.

While some candidates were able to show some good reasoning in part (b), fewer were able to find the value of which maximizedthe area of the region. In addition to the answer obtained from the restricted domain, full marks were awarded for the answer obtainedby using the point of tangency.

R

m = (x)f ′

f(x) = g(x) =mxex

4

(x) =f ′ 14

ex

4

(x) = (x)f ′ g′ =m14

ex

4

x

×x =14

ex

4 ex

4

x = 4

m

(4)f ′ 14

e44

m = e14

0.680

m

17a. [3 marks]

Consider the points A( , , ) , B( , , ) , and C( , , ) , .

Find

(i) ;

(ii) .

Markscheme(i) appropriate approach (M1)

eg ,

A1 N2

(ii) A1 N1

[3 marks]

5 2 1 6 5 3 7 6 a+ 1 a ∈ R

AB−→

AC−→−

+AO−→−

OB−→

B − A

=AB−→ ⎛

⎝⎜132

⎞⎠⎟

=AC−→− ⎛

⎝⎜24a

⎞⎠⎟

Examiners reportThe majority of candidates successfully found the vectors between the given points in part (a).

17b. [4 marks]

Let be the angle between and .

Find the value of for which .

Markschemevalid reasoning (seen anywhere) R1

eg scalar product is zero,

correct scalar product of their and (may be seen in part (c)) (A1)

eg

correct working for their and (A1)

eg ,

A1 N3

[4 marks]

Examiners reportIn part (b), while most candidates correctly found the value of , many unnecessarily worked with the magnitudes of the vectors,sometimes leading to algebra errors.

q AB−→

AC−→−

a q = π

2

cos =π

2u∙v

|u||v|

AB−→

AC−→−

1(2) + 3(4) + 2(a)

AB−→

AC−→−

2a+ 14 2a= −14

a= −7

a

17c. [4 marks]Show that .

Markschemecorrect magnitudes (may be seen in (b)) (A1)(A1)

,

substitution into formula (M1)

eg ,

simplification leading to required answer A1

eg

AG N0

[4 marks]

Examiners reportSome candidates showed a minimum of working in part (c)(i); in a “show that” question, candidates need to ensure that their workingclearly leads to the answer given. A common error was simplifying the magnitude of vector AC to instead of .

cosq = 2a+14

14 +280a2√

(= )+ +12 32 22− −−−−−−−−√ 14−−√ (= )+ +22 42 a2− −−−−−−−−√ 20 + a2

− −−−−−√

cosθ 1×2+3×4+2×a

+ +12 32 22√ + +22 42 a2√

14+2a

14√ 4+16+a2√

cosθ = 14+2a

14√ 20+a2√

cosθ = 2a+14

14 +280a2√

20a2− −−−√ 20 + a2− −−−−−

√

17d. [4 marks]

Consider the points A( , , ) , B( , , ) , and C( , , ) , .

Let be the angle between and .

Hence, find the value of a for which .

Markschemecorrect setup (A1)

eg

valid attempt to solve (M1)

eg sketch, , attempt to square

A2 N3

[4 marks]

Examiners reportIn part (c)(ii), a disappointing number of candidates embarked on a usually fruitless quest for an algebraic solution rather than simplysolving the resulting equation with their GDC. Many of these candidates showed quite weak algebra manipulation skills, with errorsinvolving the square root occurring in a myriad of ways.

5 2 1 6 5 3 7 6 a+ 1 a ∈ R

q AB−→

AC−→−

q = 1.2

cos1.2 = 2a+14

14 +280a2√

− cos1.2 = 02a+14

14 +280a2√

a= −3.25

18a. [4 marks]

A particle’s displacement, in metres, is given by , for , where t is the time in seconds.


s(t) = 2t cost 0 ≤ t ≤ 6

s

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct shape (do not accept line segments).

Only if this A1 is awarded, award the following:

A1 for maximum and minimum within circles,

A1 for x-intercepts between 1 and 2 and between 4 and 5,

A1 for left endpoint at and right endpoint within circle.

[4 marks]

Examiners reportMost candidates sketched an approximately correct shape for the displacement of a particle in the given domain, but many lost marksfor carelessness in graphing the local extrema or the right endpoint.

(0, 0)

18b. [3 marks]Find the maximum velocity of the particle.

Markschemeappropriate approach (M1)

e.g. recognizing that , finding derivative,

valid method to find maximum (M1)

e.g. sketch of , ,

A1 N2

[3 marks]

Examiners reportIn part (b), most candidates knew to differentiate displacement to find velocity, but few knew how to then find the maximum.Occasionally, a candidate would give the time value of the maximum. Others attempted to incorrectly set the first derivative equal tozero and solve analytically rather than take the maximum value from the graph of the velocity function.

v = s′ a= s′′

v (t) = 0v′ t = 5.08698…

v = 10.20025…

v = 10.2 [10.2, 10.3]

19a. [4 marks]

Consider the function .

Sketch the graph of f , for .

Markscheme

A1A1A1A1 N4

Note: The shape must be an approximately correct upwards parabola.

Only if the shape is approximately correct, award the following:

A1 for vertex , A1 for x-intercepts between 0 and 1, and 3 and 4, A1 for correct y-intercept , A1 for correct domain .

Scale not required on the axes, but approximate positions need to be clear.

[4 marks]

Examiners reportA good number of students provided a clear sketch of the quadratic function within the given domain. Some lost marks as they didnot clearly indicate the approximate positions of the most important points of the parabola either by labelling or providing a suitablescale.

f(x) = − 4x+ 1x2

−1 ≤ x ≤ 5

x ≈ 2 (0, 1)[−1, 5]

19b. [1 mark]This function can also be written as .

Write down the value of p .

Markscheme A1 N1

[1 mark]

Examiners reportThere were few difficulties in part (b).

f(x) = (x− p − 3)2

p = 2

( )

19c. [4 marks]The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of .

Show that .

Markschemecorrect vertical reflection, correct vertical translation (A1)(A1)

e.g. , , , ,

transformations in correct order (A1)

e.g. ,

simplification which clearly leads to given answer A1

e.g. ,

AG N0

Note: If working shown, award A1A1A0A0 if transformations correct, but done in reverse order, e.g. .

[4 marks]

Examiners reportIn part (c), candidates often used an insufficient number of steps to show the required result or had difficulty setting out their worklogically.

( )06

g(x) = − + 4x+ 5x2

−f(x) −((x− 2 − 3))2 −y −f(x) + 6 y+ 6

−( − 4x+ 1) + 6x2 −((x− 2 − 3) + 6)2

− + 4x− 1 + 6x2 −( − 4x+ 4 − 3) + 6x2

g(x) = − + 4x+ 5x2

−( − 4x+ 1 + 6)x2

19d. [3 marks]The graph of g is obtained by reflecting the graph of f in the x-axis, followed by a translation of .

The graphs of f and g intersect at two points.

Write down the x-coordinates of these two points.


e.g. sketch,

,

(exact), ; A1A1 N3

[3 marks]

Examiners reportPart (d) was generally done well though many candidates gave at least one answer to fewer than three significant figures, potentiallyresulting in more lost marks.

( )06

f = g

−0.449489… 4.449489…

(2 ± )6√ −0.449 [−0.450, − 0.449] 4.45 [4.44, 4.45]

19e. [3 marks]The graph of is obtained by reflecting the graph of in the x-axis, followed by a translation of .

Let R be the region enclosed by the graphs of f and g .

Find the area of R .

g f ( )06

Markschemeattempt to substitute limits or functions into area formula (accept absence of ) (M1)

e.g. , ,

approach involving subtraction of integrals/areas (accept absence of ) (M1)

e.g. ,

A1 N3

[3 marks]

Examiners reportIn part (e), many candidates were unable to connect the points of intersection found in part (d) with the limits of integration. Mistakeswere also made here either using a GDC incorrectly or not subtracting the correct functions. Other candidates tried to divide theregion into four areas and made obvious errors in the process. Very few candidates subtracted from to get a simple functionbefore integrating and there were numerous, fruitless analytical attempts to find the required integral.

dx

((− + 4x+ 5) − ( − 4x+ 1))dx∫ b

ax2 x2 (f− g)∫ −0.449

4.45 ∫ (−2 + 8x+ 4)dxx2

dx

(− + 4x+ 5) − ( − 4x+ 1)∫ b

ax2 ∫ b

ax2 ∫ (f− g)dx

area = 39.19183…

area = 39.2 [39.1, 39.2]

f(x) g(x)

20a. [3 marks]

Let and .

Find .

Markschemeinterchanging x and y (seen anywhere) (M1)

e.g.

correct manipulation (A1)

e.g.

A1 N2

[3 marks]

Examiners reportThis question was answered correctly by nearly all candidates.

f(x) = 2x− 1 g(x) = 3 + 2x2

(x)f−1

x = 2y− 1

x+ 1 = 2y

(x) =f−1 x+12

20b. [3 marks]Find . (f ∘ g)(1)

MarkschemeMETHOD 1

attempt to find or or (M1)

(A1)

A1 N2

[3 marks]

METHOD 2

attempt to form composite (in any order) (M1)

e.g. ,

(A1)

A1 N2

[3 marks]

Examiners reportThis question was answered correctly by nearly all candidates. In part (b), there were a few who seemed unfamiliar with the notationfor composition of functions, and attempted to multiply the functions rather than finding the composite, and there were a few whofound the correct composite function but failed to substitute in to find the value.

g(1) f(1)

g(1) = 5

f(5) = 9

2(3 + 2) − 1x2 3(2x− 1 + 2)2

(f ∘ g)(1) = 2(3 × + 2) − 112 (= 6 × + 3)12

(f ∘ g)(1) = 9

x = 1

f(x)

21a. [2 marks]


Sketch the graph of on the grid below.

f(x) −2 ≤ x ≤ 3

f(−x)

Markscheme

A2 N2

[2 marks]

Examiners reportIn part (a) of this question, a large number of candidates correctly sketched the graph of , as asked. A fairly common error,however, was to graph .

f(−x)−f(x)

21b. [4 marks]The graph of f is transformed to obtain the graph of g . The graph of g is shown below.

The function g can be written in the form . Write down the value of a and of b .

g(x) = af(x+ b)

Markscheme A2A2 N4

Note: Award A1 for , A1 for .

[4 marks]

Examiners reportIn part (b), many candidates seemed to recognize that the value of a was related to a vertical stretch, though some omitted the negativerequired for the vertical reflection. Similarly, some candidates gave a positive value for b.

a= −2,b = −1

a= 2 b = 1

22. [7 marks]Consider the equation , where k is a real number.

Find the values of k for which the equation has two equal real solutions.

MarkschemeMETHOD 1

evidence of valid approach (M1)

e.g. , quadratic formula

correct substitution into (may be seen in formula) (A1)

e.g. , ,

setting their discriminant equal to zero M1

e.g.

attempt to solve the quadratic (M1)

e.g. , factorizing

correct working A1

e.g. ,

, (do not accept inequalities) A1A1 N2

[7 marks]

METHOD 2

recognizing perfect square (M1)

e.g. ,

correct expansion (A1)(A1)

e.g. ,

equating coefficients of x A1A1

e.g. ,

, A1A1 N2

[7 marks]

Examiners reportMost candidates approached this question correctly by using the discriminant, and many were successful in finding both of therequired values of k. There did seem to be some confusion about the expression "two equal real solutions", as some candidatesapproached the question as though the equation had two distinct real roots, using , rather than .

There were also a good number who recognized that the quadratic must be a perfect square, although many who used this methodfound only one of the two possible values of k. In addition, there were many unsuccessful candidates who tried to use the entirequadratic formula as though they were solving for x, without ever seeming to realize the significance of the discriminant.

+ (k− 1)x+ 1 = 0x2

− 4acb2

− 4acb2

(k− 1 − 4 × 1 × 1)2 (k− 1 − 4)2 − 2k− 3k2

Δ = 0,(k− 1 − 4 = 0)2

(k− 1 = 4)2

(k− 1) = ±2 (k− 3)(k+ 1)

k = −1 k = 3

(x+ 1 = 0)2 (x− 1)2

+ 2x+ 1 = 0x2 − 2x+ 1x2

k− 1 = −2 k− 1 = 2

k = −1 k = 3

− 4ac > 0b2 − 4ac = 0b2

23a. [4 marks]

The following diagram shows the graph of a quadratic function f , for .

The graph passes through the point P(0, 13) , and its vertex is the point V(2, 1) .

The function can be written in the form .

(i) Write down the value of h and of k .

(ii) Show that .

Markscheme(i) , A1A1 N2

(ii) attempt to substitute coordinates of any point (except the vertex) on the graph into f M1

e.g.

working towards solution A1

e.g.

AG N0

[4 marks]

Examiners reportIn part (a), nearly all the candidates recognized that h and k were the coordinates of the vertex of the parabola, and most were able tosuccessfully show that . Unfortunately, a few candidates did not understand the "show that" command, and simply verifiedthat would work, rather than showing how to find .

0 ≤ x ≤ 4

f(x) = a(x−h + k)2

a= 3

h = 2 k = 1

13 = a(0 − 2 + 1)2

13 = 4a+ 1

a= 3

a= 3a= 3 a= 3

23b. [3 marks]Find , giving your answer in the form .f(x) A +Bx+Cx2

f(x) = 3( − 2 × 2x+ 4) + 1x2 (x− 2 = − 4x+ 4)2 x2

f(x) = 3 − 12x+ 12 + 1x2

f(x) = 3 − 12x+ 13x2 A = 3 B = −12 C= 13

Examiners reportIn part (b), most candidates were able to find in the required form. For a few candidates, algebraic errors kept them from findingthe correct function, even though they started with correct values for a, h and k.

f(x)

23c. [8 marks]Calculate the area enclosed by the graph of f , the x-axis, and the lines and .x = 2 x = 4

MarkschemeMETHOD 1

integral expression (A1)

e.g. ,

A1A1A1

Note: Award A1 for , A1 for , A1 for .

correct substitution of correct limits into their expression A1A1

e.g. ,

Note: Award A1 for substituting 4, A1 for substituting 2.


e.g.

A1 N3

[8 marks]

METHOD 2


e.g. ,

A2A1

Note: Award A2 for , A1 for .

correct substitution of correct limits into their expression A1A1

e.g. , ,

Note: Award A1 for substituting 4, A1 for substituting 2.


e.g.

A1 N3

[8 marks]

METHOD 3

recognizing area from 0 to 2 is same as area from 2 to 4 (R1)

e.g. sketch,


e.g. ,

A1A1A1


correct substitution of correct limits into their expression A1(A1)

e.g. ,

Note: Award A1 for substituting 2, (A1) for substituting 0.

A1 N3

[8 marks]

(3 − 12x+ 13)∫ 42 x2 ∫ fdx

Area = [ − 6 + 13xx3 x2 ]42

x3 −6x2 13x

( − 6 × + 13 × 4) − ( − 6 × + 13 × 2)43 42 23 22 64 − 96 + 52 − (8 − 24 + 26)

64 − 96 + 52 − 8 + 24 − 26,20 − 10

Area = 10

(3 + 1)∫ 42 (x− 2)2 ∫ fdx

Area = [(x− 2 +x)3 ]42

(x− 2)3 x

(4 − 2 + 4 − [(2 − 2 + 2])3 )3 + 4 − ( + 2)23 03 + 4 − 223

8 + 4 − 2

Area = 10

f = f∫ 42 ∫ 2

0

(3 − 12x+ 13)∫ 20 x2 ∫ fdx

Area = [ − 6 + 13xx3 x2 ]20

x3 −6x2 13x

( − 6 × + 13 × 2) − ( − 6 × + 13 × 0)23 22 03 02 8 − 24 + 26

Area = 10

Examiners reportIn part (c), nearly all candidates knew that they needed to integrate to find the area, but errors in integration, and algebraic andarithmetic errors prevented many from finding the correct area.

24a. [6 marks]

Let for , , . The graph of is given below.

The graph of has a local minimum at A( , ) and a local maximum at B.

Use the quotient rule to show that .

Markschemecorrect derivatives applied in quotient rule (A1)A1A1

,

Note: Award (A1) for 1, A1 for and A1 for , only if it is clear candidates are using the quotient rule.

correct substitution into quotient rule A1

e.g. ,


e.g.

expression clearly leading to the answer A1

e.g.

AG N0

[6 marks]

Examiners reportWhile most candidates answered part (a) correctly, there were some who did not show quite enough work for a "show that" question.A very small number of candidates did not follow the instruction to use the quotient rule.

f(x) = x

−2 +5x−2x2−2 ≤ x ≤ 4 x ≠ 1

2x ≠ 2 f

f 1 1

(x) =f ′ 2 −2x2

(−2 +5x−2)x2 2

1 −4x+ 5

−4x 5

1×(−2 +5x−2)−x(−4x+5)x2

(−2 +5x−2)x2 2

−2 +5x−2−x(−4x+5)x2

(−2 +5x−2)x2 2

−2 +5x−2−(−4 +5x)x2 x2

(−2 +5x−2)x2 2

−2 +5x−2+4 −5xx2 x2

(−2 +5x−2)x2 2

(x) =f ′ 2 −2x2

(−2 +5x−2)x2 2

24b. [7 marks]Hence find the coordinates of B.

Markschemeevidence of attempting to solve (M1)

e.g.

evidence of correct working A1

e.g.

correct solution to quadratic (A1)

e.g.

correct x-coordinate (may be seen in coordinate form ) A1 N2

attempt to substitute into f (do not accept any other value) (M1)

e.g.

correct working

e.g. A1

correct y-coordinate (may be seen in coordinate form ) A1 N2

[7 marks]

Examiners reportIn part (b), most candidates knew that they needed to solve the equation , and many were successful in answering thisquestion correctly. However, some candidates failed to find both values of x, or made other algebraic errors in their solutions. Onecommon error was to find only one solution for ; another was to work with the denominator equal to zero, rather than thenumerator.

(x) = 0f ′

2 − 2 = 0x2

= 1, , 2(x− 1)(x+ 1)x2 ± 16√4

x = ±1

x = −1 (−1, )19

−1

f(−1) = −1−2× +5×(−1)−2(−1)2

−1−2−5−2

y = 19

(−1, )19

(x) = 0f ′

= 1x2

24c. [3 marks]Given that the line does not meet the graph of f , find the possible values of k .

Markschemerecognizing values between max and min (R1)

A2 N3

[3 marks]

Examiners reportIn part (c), a significant number of candidates seemed to think that the line was a vertical line, and attempted to find the verticalasymptotes. Others tried looking for a horizontal asymptote. Fortunately, there were still a good number of intuitive candidates whorecognized the link with the graph and with part (b), and realized that the horizontal line must pass through the space between thegiven local minimum and the local maximum they had found in part (b).

y = k

< k < 119

y = k

25a. [4 marks]

Let for . Part of the graph of f is shown below.

The graph passes through A(6, 2) , has an x-intercept at (−2, 0) and has an asymptote at .

Find p .

Markschemeevidence of substituting the point A (M1)

e.g.

manipulating logs A1

e.g.

A2 N2

[4 marks]

Examiners reportIn part (a), many candidates successfully substituted the point A to find the base of the logarithm, although some candidates lost amark for not showing their manipulation of the logarithm equation into the exponential equation.

f(x) = (x+ 3)logp x > −3

x = −3

2 = (6 + 3)logp

= 9p2

p = 3

25b. [5 marks]The graph of f is reflected in the line to give the graph of g .

(i) Write down the y-intercept of the graph of g .

(ii) Sketch the graph of g , noting clearly any asymptotes and the image of A.

y = x

Markscheme(i) (accept A1 N1

(ii)

A1A1A1A1 N4

Note: Award A1 for asymptote at , A1 for an increasing function that is concave up, A1 for a positive x-intercept and anegative y-intercept, A1 for passing through the point .

[5 marks]

Examiners reportA number of candidates who correctly stated the y-intercept was had difficulty sketching the graph of the reflection in the line

. A number of candidates graphed directly on the question paper rather than sketching their own graph; candidates should bereminded to show all working for Section B on separate paper. Some correct sketches did not have the position of A indicated. Manycandidates had difficulty reflecting the asymptote.

y = −2 (0, − 2))

y = −3(2, 6)

−2y = x

25c. [4 marks]The graph of is reflected in the line to give the graph of .

Find .

f y = x g

g(x)

MarkschemeMETHOD 1

recognizing that (R1)


e.g. switching x and y (seen anywhere), solving for x

correct manipulation (A1)

e.g.

A1 N3

METHOD 2

recognizing that (R1)

identifying vertical translation (A1)

e.g. graph shifted down 3 units,


e.g. substituting point to identify the base

A1 N3

[4 marks]

Examiners reportPart (c) was often well done, with candidates showing clear and correct working.

The most successful candidates clearly appreciated the linkage between the question parts.

g = f−1

= y+ 33x

g(x) = − 33x

g(x) = + bax

f(x) − 3

g(x) = − 33x

26a. [4 marks]

Let .

(i) Write down the coordinates of the vertex.

(ii) Hence or otherwise, express the function in the form .

Markscheme(i) or , A1A1 N2

(ii) evidence of valid approach (M1)

e.g. graph, completing the square, equating coefficients

A1 N2

[4 marks]

Examiners reportThis question was well done by the majority of candidates.

f(x) = 2 − 8x− 9x2

f(x) = 2(x−h + k)2

(2, − 17) x = 2 y = −17

f(x) = 2(x− 2 − 17)2

26b. [3 marks]Solve the equation .f(x) = 0

Markschemeevidence of valid approach (M1)

e.g. graph, quadratic formula

,

, A1A1 N3

[3 marks]

Examiners reportThis question was well done by the majority of candidates. There were still many however who opted for an analytical approach inpart (b), which often led to errors in sign and accuracy. Some candidates used the trace feature on their GDC to find the vertex whichoften resulted in accuracy errors.

−0.9154759… 4.915475…

x = −0.915 4.92

27a. [2 marks]

The graph of , for , is shown below.

The graph has -intercepts at , , and .

Find k .


e.g. ,

A1 N2

[2 marks]

Examiners reportCandidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of courseincluded the omission of dx , using the given domain as the upper and lower bounds of integration, forgetting to square their functionand/or the omission of . There were still many who were unable to use their calculator successfully to find the required volume.

y = (x− 1)sinx 0 ≤ x ≤ 5π2

x 0 1 π k

y = 0 sinx = 0

2π = 6.283185…

k = 6.28

π

27b. [3 marks]The shaded region is rotated about the x-axis. Let V be the volume of the solid formed.

Write down an expression for V .

360∘

Markschemeattempt to substitute either limits or the function into formula (M1)

(accept absence of )

e.g. , ,

correct expression A2 N3

e.g. ,

[3 marks]

Examiners reportCandidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of courseincluded the omission of dx , using the given domain as the upper and lower bounds of integration, forgetting to square their functionand/or the omission of . There were still many who were unable to use their calculator successfully to find the required volume.

dx

V = π dx∫ k

π(f(x))2 π∫ ((x− 1)sinx)2 π dx∫ 6.28…

πy2

π xdx∫ 6.28π

(x− 1)2sin2 π dx∫ 2ππ

((x− 1)sinx)2

π

27c. [2 marks]The shaded region is rotated about the x-axis. Let V be the volume of the solid formed.

Find V .

Markscheme

A2 N2

[2 marks]

Examiners reportCandidates showed marked improvement in writing fully correct expressions for a volume of revolution. Common errors of courseincluded the omission of dx, using the given domain as the upper and lower bounds of integration, forgetting to square their functionand/or the omission of . There were still many who were unable to use their calculator successfully to find the required volume.

360∘

V = 69.60192562…

V = 69.6

π

28a. [5 marks]

The following diagram shows two ships A and B. At noon, ship A was 15 km due north of ship B. Ship A was moving south at 15 km hand ship B was moving east at 11 km h .

Find the distance between the ships

(i) at 13:00;

(ii) at 14:00.

–1

–1

Markscheme(i) evidence of valid approach (M1)

e.g. ship A where B was, B away

A1 N2

(ii) evidence of valid approach (M1)

e.g. new diagram, Pythagoras, vectors

(A1)

A1 N2

Note: Award M0A0A0 for using the formula given in part (b).

[5 marks]

Examiners reportPart (a) was generally well done although some candidates incorrectly used the function given in part (b) to find the required values.There was evidence that some candidates are not comfortable with a 24-hour clock.

11 km

distance = 11

s = +152 222− −−−−−−√

= 26.62705709−−−√

s = 26.6

28b. [6 marks]Let be the distance between the ships t hours after noon, for .

Show that .


e.g. a table, diagram, formula

distance ship A travels t hours after noon is (A2)

distance ship B travels in t hours after noon is (A1)

evidence of valid approach M1

e.g.

correct simplification A1

e.g.

AG N0

[6 marks]

Examiners reportCandidates had difficulty generalizing the problem and therefore, were unable to show how the function was obtained in part (b).

s(t) 0 ≤ t ≤ 4

s(t) = 346 − 450t+ 225t2− −−−−−−−−−−−−−

√

d = r× t

15(t− 1)

11t

s(t) = +[15(t− 1)]2 (11t)2− −−−−−−−−−−−−−−√

225( − 2t+ 1) + 121t2 t2− −−−−−−−−−−−−−−−−−√

s(t) = 346 − 450t+ 225t2− −−−−−−−−−−−−−

√

s(t)

28c. [3 marks]Sketch the graph of .s(t)

Markscheme

A1A1A1 N3

Note: Award A1 for shape, A1 for minimum at approximately , A1 for domain.

[3 marks]

Examiners reportSurprisingly, the graph in part (c) was not well done. Candidates often ignored the given domain, provided no indication of scale, anddrew "V" shapes or parabolas.

(0.7, 9)

28d. [3 marks]Due to poor weather, the captain of ship A can only see another ship if they are less than 8 km apart. Explain why the captaincannot see ship B between noon and 16:00.


e.g. , find minimum of , graph, reference to "more than 8 km"

(accept 2 or more sf) A1

since , captain cannot see ship B R1 N0

[3 marks]

Examiners reportIn part (d), candidates simply regurgitated the question without providing any significant evidence for their statements that the twoships must have been more than 8 km apart.

(t) = 0s′ s(t)

min = 8.870455…

> 8smin

29a. [2 marks]

Let , for .

Find .

f(x) = cos( )ex −2 ≤ x ≤ 2

(x)f ′

Markscheme A1A1 N2

[2 marks]

Examiners reportMany students failed in applying the chain rule to find the correct derivative, and some inappropriately used the product rule.However, many of those obtained full follow through marks in part (b) for the sketch of the function they found in part (a).

(x) = − sin( )f ′ ex ex

29b. [4 marks]On the grid below, sketch the graph of .(x)f ′

Markscheme

A1A1A1A1 N4

Note: Award A1 for shape that must have the correct domain (from to ) and correct range (from to ), A1 for minimum incircle, A1 for maximum in circle and A1 for intercepts in circles.

[4 marks]

Examiners reportMany students failed in applying the chain rule to find the correct derivative, and some inappropriately used the product rule.However, many of those obtained full follow through marks in part (b) for the sketch of the function they found in part (a).

Most candidates sketched an approximately correct shape in the given domain, though there were some that did not realize they hadto set their GDC to radians, producing a meaningless sketch.

It is very important to stress to students that although they are asked to produce a sketch, it is still necessary to show its key featuressuch as domain and range, stationary points and intercepts.

−2 +2 −6 4

30a. [2 marks]

Let , where a , b and c are real numbers. The graph of f passes through the point (2, 9) .

Show that .

Markschemeattempt to substitute coordinates in f (M1)

e.g.

correct substitution A1

e.g.

AG N0

[2 marks]

f(x) = a + b + cx3 x2

8a+ 4b+ c = 9

f(2) = 9

a× + b× + c = 923 22

8a+ 4b+ c = 9

Examiners reportPart (a) was generally well done, with a few candidates failing to show a detailed substitution. Some substituted 2 in place of x, butdidn't make it clear that they had substituted in y as well.

30b. [7 marks]The graph of f has a local minimum at .

Find two other equations in a , b and c , giving your answers in a similar form to part (a).

Markschemerecognizing that is on the graph of f (M1)

e.g.

correct equation A1

e.g.

recognizing that at minimum (seen anywhere) (M1)

e.g.

(seen anywhere) A1A1

correct substitution into derivative (A1)

e.g.

correct simplified equation A1

e.g.

[7 marks]

Examiners reportA great majority could find the two equations in part (b). However there were a significant number of candidates who failed toidentify that the gradient of the tangent is zero at a minimum point, thus getting the incorrect equation .

(1, 4)

(1, 4)

f(1) = 4

a+ b+ c = 4

= 0f ′

(1) = 0f ′

(x) = 3a + 2bxf ′ x2

3a× + 2b× 1 = 012

3a+ 2b = 0

3a+ 2b = 4

30c. [4 marks]Find the value of a , of b and of c .

Markschemevalid method for solving system of equations (M1)

e.g. inverse of a matrix, substitution

, , A1A1A1 N4

[4 marks]

Examiners reportA considerable number of candidates only had 2 equations, so that they either had a hard time trying to come up with a third equation(incorrectly combining some of the information given in the question) to solve part (c) or they completely failed to solve it.

Despite obtaining three correct equations many used long elimination methods that caused algebraic errors. Pages of calculationsleading nowhere were seen.

Those who used matrix methods were almost completely successful.

a= 2 b = −3 c = 5

31a. [4 marks]

Consider the following circle with centre O and radius r .

The points P, R and Q are on the circumference, , for .

Use the cosine rule to show that .

Markschemecorrect substitution into cosine rule A1

e.g. ,

substituting for (seen anywhere) A1

e.g.

working towards answer (A1)

e.g.

recognizing (including crossing out) (seen anywhere)

e.g. ,

AG N0

[4 marks]

Examiners reportThis exercise seemed to be challenging for the great majority of the candidates, in particular parts (b), (c) and (d).

Part (a) was generally attempted using the cosine rule, but many failed to substitute correctly into the right hand side or skippedimportant steps. A high percentage could not arrive at the given expression due to a lack of knowledge of trigonometric identities ormaking algebraic errors, and tried to force their way to the given answer.

The most common errors included taking the square root too soon, and sign errors when distributing the negative after substituting by .

P Q = 2θO 0 < θ < π

2

PQ = 2rsin θ

P = + − 2(r)(r)cos(2θ)Q2 r2 r2 P = 2 − 2 (cos(2θ))Q2 r2 r2

1 − 2 θsin2 cos2θ

P = 2 − 2 (1 − 2 θ)Q2 r2 r2 sin2

P = 2 − 2 + 4 θQ2 r2 r2 r2sin2

2 − 2 = 0r2 r2

P = 4 θQ2 r2sin2 PQ = 4 θr2sin2− −−−−−√

PQ = 2rsinθ

cos2θ 1 − 2 θsin2

31b. [5 marks]Let l be the length of the arc PRQ .

Given that , find the value of .1.3PQ − l = 0 θ

Markscheme (seen anywhere) (A1)

correct set up A1

e.g.

attempt to eliminate r (M1)

correct equation in terms of the one variable (A1)

e.g.

1.221496215

(accept (69.9)) A1 N3

[5 marks]

Examiners reportThis exercise seemed to be challenging for the great majority of the candidates, in particular parts (b), (c) and (d).

In part (b), most candidates understood what was required but could not find the correct length of the arc PRQ mainly due tosubstituting the angle by instead of .

PRQ = r× 2θ

1.3 × 2rsin θ− r× (2θ) = 0

θ

1.3 × 2 sin θ− 2θ = 0

θ = 1.22 70.0∘

θ 2θ

31c. [4 marks]Consider the function , for .

(i) Sketch the graph of f .

(ii) Write down the root of .

Markscheme(i)

A1A1A1 N3

Note: Award A1 for approximately correct shape, A1 for x-intercept in approximately correct position, A1 for domain. Do notpenalise if sketch starts at origin.

(ii)

A1 N1

[4 marks]

f(θ) = 2.6sin θ− 2θ 0 < θ < π

2

f(θ) = 0

1.221496215

θ = 1.22

Examiners reportRegarding part (c), many valid approaches were seen for the graph of f, making a good use of their GDC. A common error wasfinding a second or third solution outside the domain. A considerable amount of sketches were missing a scale.

There were candidates who achieved the correct equation but failed to realize they could use their GDC to solve it.

31d. [3 marks]Use the graph of f to find the values of for which .

Markschemeevidence of appropriate approach (may be seen earlier) M2

e.g. , , showing positive part of sketch

(accept ) A1 N1

[3 marks]

Examiners reportPart (d) was attempted by very few, and of those who achieved the correct answer not many were able to show the method they used.

θ l < 1.3PQ

2θ < 2.6sin θ 0 < f(θ)

0 < θ < 1.221496215

0 < θ = 1.22 θ < 1.22

32a. [1 mark]

Let be a quadratic function. Part of the graph of is shown below.

The vertex is at P( , ) and the y-intercept is at Q( , ) .

Write down the equation of the axis of symmetry.

Markscheme (must be an equation) A1 N1

[1 mark]

Examiners reportA surprising number of candidates missed part (a) of this question, which required them to write the equation of the axis of symmetry.Some candidates did not write their answer as an equation, while others simply wrote the formula .

f f

4 2 0 6

x = 4

x = − b

2a

32b. [2 marks]The function f can be written in the form .

Write down the value of h and of k .

f(x) = a(x−h + k)2

Markscheme , A1A1 N2

[2 marks]

Examiners reportThis was answered correctly by the large majority of candidates.

h = 4 k = 2

32c. [3 marks]The function f can be written in the form .

Find a .

Markschemeattempt to substitute coordinates of any point on the graph into f (M1)

e.g. , ,

correct equation (do not accept an equation that results from ) (A1)

e.g. ,

A1 N2

[3 marks]

Examiners reportThe rest of this question was answered correctly by the large majority of candidates. The mistakes seen in part (c) were generally dueto either incorrect substitution of a point into the equation, or substitution of the vertex coordinates, which got the candidates nowhere.

f(x) = a(x−h + k)2

f(0) = 6 6 = a(0 − 4 + 2)2 f(4) = 2

f(4) = 2

6 = a(−4 + 2)2 6 = 16a+ 2

a= (= )416

14

33a. [4 marks]

Let , where .

Find the values of k such that has two equal roots.

MarkschemeMETHOD 1

evidence of discriminant (M1)

e.g. , discriminant = 0

correct substitution into discriminant A1

e.g. ,

A1A1 N3

METHOD 2

recognizing that equal roots means perfect square (R1)

e.g. attempt to complete the square,

correct working

e.g. , A1

A1A1 N3

[4 marks]

f(x) = + kx+ 812x2 k ∈ Z

f(x) = 0

− 4acb2

− 4 × × 8k2 12

− 16 = 0k2

k = ±4

( + 2kx+ 16)12x2

(x+ k12

)2 = 812k2

k = ±4

Examiners reportA good number of candidates were successful in using the discriminant to find the correct values of k in part (a), however, there weremany who tried to use the quadratic formula without recognizing the significance of the discriminant.

33b. [4 marks]Each value of k is equally likely for . Find the probability that has no roots.

Markschemeevidence of appropriate approach (M1)

e.g.

correct working for k A1

e.g. , , list all correct values of k

A2 N3

[4 marks]

Examiners reportPart (b) was very poorly done by nearly all candidates. Common errors included finding the wrong values for k, and not realizing thatthere were 11 possible values for k.

−5 ≤ k ≤ 5 f(x) = 0

− 4ac < 0b2

−4 < k < 4 < 16k2

p = 711

34a. [3 marks]

Let and .

Find .

Markschemeinterchanging x and y (may be seen at any time) (M1)


e.g.

(accept ) A1 N2

[3 marks]

Examiners reportAll parts of this question were well answered by most of the candidates. Some misunderstood part (a) and found the derivative or thereciprocal, indicating they were not familiar with the notation for an inverse function. Occasionally, the composition symbol wasmistaken for multiplication. Additionally, some candidates composed in the incorrect order.

f(x) = 2x+ 4 g(x) = 7x2

(x)f−1

x = 2y+ 4

(x) =f−1 x−42

y = ,x−42

x−42


Markschemeattempt to form composite (in any order) (M1)

e.g.

A1 N2

(f ∘ g)(x)

f(7 ), 2(7 ) + 4, 7(2x+ 4x2 x2 )2

(f ∘ g)(x) = 14 + 4x2


34c. [2 marks]Find .

Markschemecorrect substitution (A1)

e.g. ,

(accept 176) A1 N2

[2 marks]


(f ∘ g)(3.5)

7 × 3.52 14(3.5 + 4)2

(f ∘ g)(3.5) = 175.5

35a. [1 mark]

Jose takes medication. After t minutes, the concentration of medication left in his bloodstream is given by , where A is inmilligrams per litre.

Write down .

Markscheme A1 N1

[1 mark]

Examiners reportFor a later question in Section A, a pleasing number of candidates made good progress. Some candidates believed that raising a baseto the zero power gave zero which indicated that they most likely did not begin by analysing the function with their GDC. For part(c), many candidates could set up the equation correctly and had some idea to apply logarithms but became lost in the algebra. Thosewho used their GDC to find when the function equalled 0.395 typically did so successfully. A common error for those who obtaineda correct value for time in minutes was to treat 5.55 hours as 5 hours and 55 minutes after 13:00.

A(t) = 10(0.5)0.014t

A(0)

A(0) = 10

35b. [2 marks]Find the concentration of medication left in his bloodstream after 50 minutes.

Markschemesubstitution into formula (A1)

e.g. ,

A1 N2

[2 marks]

10(0.5)0.014(50) A(50)

A(50) = 6.16


35c. [5 marks]At 13:00, when there is no medication in Jose’s bloodstream, he takes his first dose of medication. He can take his medicationagain when the concentration of medication reaches 0.395 milligrams per litre. What time will Jose be able to take his medication again?

Markschemeset up equation (M1)

e.g.

attempting to solve (M1)

e.g. graph, use of logs


e.g. sketch of intersection,

A1

correct time 18:33 or 18:34 (accept 6:33 or 6:34 but nothing else) A1 N3

[5 marks]


A(t) = 0.395

0.014t log0.5 = log0.0395

t = 333.00025…

36a. [4 marks]

Let , where . The function v is obtained when the graph of f is transformed by

a stretch by a scale factor of parallel to the y-axis,

followed by a translation by the vector .

Find , giving your answer in the form .

Markschemeapplies vertical stretch parallel to the y-axis factor of (M1)

e.g. multiply by , ,

applies horizontal shift 2 units to the right (M1)

e.g. ,

applies a vertical shift 4 units down (M1)

e.g. subtracting 4, ,

A1 N4

[4 marks]

f(t) = 2 + 7t2 t > 0

13

( )2−4

v(t) a(t− b + c)2

13

13

f(t)13

× 213

f(t− 2) t− 2

f(t) − 4 − 473

v(t) = (t− 2 −23

)2 53

Examiners reportWhile a number of candidates had an understanding of each transformation, most had difficulty applying them in the correct order,and few obtained the completely correct answer in part (a). Many earned method marks for discerning three distinct transformations.Few candidates knew to integrate to find the distance travelled. Many instead substituted time values into the velocity function or itsderivative and subtracted. A number of those who did recognize the need for integration attempted an analytic approach rather thanusing the GDC, which often proved unsuccessful.

36b. [3 marks]A particle moves along a straight line so that its velocity in ms , at time t seconds, is given by v . Find the distance theparticle travels between and .

Markschemerecognizing that distance travelled is area under the curve M1

e.g. , sketch

distance = 15.576 (accept 15.6) A2 N2

[3 marks]

Examiners reportWhile a number of candidates had an understanding of each transformation, most had difficulty applying them in the correct order,and few obtained the completely correct answer in part (a). Many earned method marks for discerning three distinct transformations.Few candidates knew to integrate to find the distance travelled. Many instead substituted time values into the velocity function or itsderivative and subtracted. A number of those who did recognize the need for integration attempted an analytic approach rather thanusing the GDC, which often proved unsuccessful.

−1

t = 5.0 t = 6.8

∫ v, (t− 2 − t29

)3 53

37a. [4 marks]Consider an infinite geometric sequence with and .

(i) Find .

(ii) Find the sum of the infinite sequence.

Markscheme(i) correct approach (A1)

e.g. , listing terms

A1 N2

(ii) correct substitution into formula for infinite sum (A1)

e.g. ,

A1 N2

[4 marks]

= 40u1 r = 12

u4

= (40)u412

(4−1)

= 5u4

=S∞40

1−0.5=S∞

400.5

= 80S∞

Examiners reportMost candidates found part (a) straightforward, although a common error in (a)(ii) was to calculate 40 divided by as 20.1

2

37b. [5 marks]Consider an arithmetic sequence with n terms, with first term ( ) and eighth term ( ) .

(i) Find the common difference.

(ii) Show that .

Markscheme(i) attempt to set up expression for (M1)

e.g.

correct working A1

e.g. ,

A1 N2

(ii) correct substitution into formula for sum (A1)

e.g.

correct working A1

e.g. ,

AG N0

[5 marks]

Examiners reportIn part (b), some candidates had difficulty with the "show that" and worked backwards from the answer given.

−36 −8

= 2 − 38nSn n2

u8

−36 + (8 − 1)d

−8 = −36 + (8 − 1)d −8−(−36)7

d = 4

= (2(−36) + (n− 1)4)Snn

2

= (4n− 76)Snn

2−36n+ 2 − 2nn2

= 2 − 38nSn n2

37c. [5 marks]The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n .

Markschememultiplying (AP) by 2 or dividing S (infinite GP) by 2 (M1)

e.g. , , 40

evidence of substituting into A1

e.g. , ( )

attempt to solve their quadratic (equation) (M1)

e.g. intersection of graphs, formula

A2 N3

[5 marks]

Examiners reportMost candidates obtained the correct equation in part (c), although some did not reject the negative value of n as impossible in thiscontext.

Sn

2SnS∞

2

2 =Sn S∞

2 − 38n = 40n2 4 − 76n− 80n2 = 0

n = 20

( ) = 20x

38a. [3 marks]

Let , for .

Sketch the graph of f .

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape with inflexion/change of curvature, A1 for maximum skewed to the left, A1 forasymptotic behaviour to the right.

[3 marks]

Examiners reportMany candidates earned the first four marks of the question in parts (a) and (b) for correctly using their GDC to graph and find themaximum value.

f(x) = 20xe0.3x

0 ≤ x ≤ 20

38b. [3 marks](i) Write down the x-coordinate of the maximum point on the graph of f .

(ii) Write down the interval where f is increasing.

Markscheme(i) A1 N1

(ii) correct interval, with right end point A1A1 N2

e.g. ,

Note: Accept any inequalities in the right direction.

[3 marks]

Examiners reportMany candidates earned the first four marks of the question in parts (a) and (b) for correctly using their GDC to graph and find themaximum value.

x = 3.33

3 13

0 < x ≤ 3.33 0 ≤ x < 3 13

38c. [5 marks]Show that .(x) =f ′ 20−6xe0.3x


e.g. quotient rule, product rule

2 correct derivatives (must be seen in product or quotient rule) (A1)(A1)

e.g. , or

correct substitution into product or quotient rule A1

e.g. ,

correct working A1

e.g. , ,

AG N0

[5 marks]

Examiners reportMost had a valid approach in part (c) using either the quotient or product rule, but many had difficulty applying the chain rule with afunction involving e and simplifying.

20 0.3e0.3x −0.3e−0.3x

20 −20x(0.3)e0.3x e0.3x

( )e0.3x 220 + 20x(−0.3)e−0.3x e−0.3x

20 −6xe0.3x e0.3x

e0.6x

(20−20x(0.3))e0.3x

( )e0.3x 2(20 + 20x(−0.3))e−0.3x

(x) =f ′ 20−6xe0.3x

38d. [4 marks]Find the interval where the rate of change of f is increasing.

Markschemeconsideration of or (M1)

valid reasoning R1

e.g. sketch of , is positive, , reference to minimum of

correct value (A1)

correct interval, with both endpoints A1 N3

e.g. ,

[4 marks]

Examiners reportPart (d) was difficult for most candidates. Although many associated rate of change with derivative, only the best-prepared studentshad valid reasoning and could find the correct interval with both endpoints.

f ′ f ′′

f ′ f ′′ = 0f ′′ f ′

6.6666666… (6 )23

6.67 < x ≤ 20 6 ≤ x < 2023

39a. [2 marks]

Let and .

Find .

Markschemeattempt to form composite (M1)

e.g. ,

A1 N2

[2 marks]

f(x) = 7 − 2x g(x) = x+ 3

(g ∘ f)(x)

g(7 − 2x) 7 − 2x+ 3

(g ∘ f)(x) = 10 − 2x

Examiners reportA majority of candidates found success in the opening question. Common errors in (a) were to give or to multiply f by g. f ∘ g

39b. [1 mark]Write down .

Markscheme A1 N1

[1 mark]

Examiners reportFor (b) some gave the inverse as the reciprocal function , or wrote .

(x)g−1

(x) = x− 3g−1

1x+3

x = y+ 3


MarkschemeMETHOD 1

valid approach (M1)

e.g. , ,

A1 N2

METHOD 2

attempt to form composite of f and (M1)

e.g. ,

A1 N2

[2 marks]

Examiners reportMost candidates chose to find a composite in (c), sometimes making simple errors when working with brackets and a negative sign.Only a handful used the more efficient . Additionally, it was not uncommon for candidates to give a correct substitution butnot complete the result. Simple expressions such as should be finished as .

(f ∘ )(5)g−1

(5)g−1 2 f(5)

f(2) = 3

g−1

(f ∘ )(x) = 7 − 2(x− 3)g−1 13 − 2x

(f ∘ )(5) = 3g−1

f(2) = 3(7 − 2x) + 3 10 − 2x

40a. [5 marks]

Consider , for . The equation has two equal roots.

Find the value of k .

f(x) = 2k − 4kx+ 1x2 k ≠ 0 f(x) = 0


e.g. , ,

correct equation A1

e.g. , ,

correct manipulation A1

e.g. ,

A2 N3

[5 marks]

Examiners reportThose who knew to set the discriminant to zero had little trouble completing part (a). Some knew that having two equal roots meansthe factors must be the same, and thus surmised that will achieve . This is a valid approach, provided thereasoning is completely communicated. Many candidates set and used the quadratic formula, which misses the approachentirely.

− 4acb2 Δ = 0 (−4k − 4(2k)(1))2

(−4k − 4(2k)(1) = 0)2 16 = 8kk2 2 − k = 0k2

8k(2k− 1) 8± 64√32

k = 12

k = 12

(x− 1)(x− 1)f = 0

40b. [2 marks]The line intersects the graph of f . Find all possible values of p .

Markschemerecognizing vertex is on the x-axis M1

e.g. (1, 0) , sketch of parabola opening upward from the x-axis

A1 N1

[2 marks]

Examiners reportPart (b) proved challenging for most, and was often left blank. Those who considered a graphical interpretation and sketched theparabola found greater success.

y = p

p ≥ 0

41a. [1 mark]

The velocity v ms of a particle at time t seconds, is given by , for .

Write down the velocity of the particle when .

Markscheme A1 N1

[1 mark]

Examiners reportMany candidates gave a correct initial velocity, although a substantial number of candidates answered that .

−1 v = 2t+ cos2t 0 ≤ t ≤ 2

t = 0

v = 1

0 + cos0 = 0

41b. [8 marks]When , the acceleration is zero.

(i) Show that .

(ii) Find the exact velocity when .

t = k

k = π

4

t = π

4

Markscheme(i) A1

A1A1

Note: Award A1 for coefficient 2 and A1 for .

evidence of considering acceleration = 0 (M1)

e.g. ,


e.g. ,

(accept ) A1

AG N0

(ii) attempt to substitute into v (M1)

e.g.

A1 N2

[8 marks]

Examiners reportFor (b), students commonly applied the chain rule correctly to achieve the derivative, and many recognized that the acceleration mustbe zero. Occasionally a student would use a double-angle identity on the velocity function before differentiating. This is not incorrect,but it usually caused problems when trying to show . At times students would reach the equation and then substitutethe , which does not satisfy the “show that” instruction.

(2t) = 2ddt

(cos2t) = −2 sin 2tddt

−sin 2t

= 0dvdt

2 − 2 sin 2t = 0

sin 2k = 1 sin 2t = 1

2k = π

22t = π

2

k = π

4

t = π

4

2 ( ) + cos( )π

42π4

v = π

2

k = π

4sin 2k = 1

π

4

41c. [4 marks]When , and when , .

Sketch a graph of v against t .

Markscheme

A1A1A2 N4

Notes: Award A1 for y-intercept at , A1 for curve having zero gradient at , A2 for shape that is concave down to the leftof and concave up to the right of . If a correct curve is drawn without indicating , do not award the second A1 for the zerogradient, but award the final A2 if appropriate. Sketch need not be drawn to scale. Only essential features need to be clear.

[4 marks]

t < π

4> 0dv

dtt > π

4> 0dv

dt

(0, 1) t = π

4π

4π

4t = π

4

Examiners reportThe challenge in this question is sketching the graph using the information achieved and provided. This requires students to makegraphical interpretations, and as typical in section B, to link the early parts of the question with later parts. Part (a) provides the y-intercept, and part (b) gives a point with a horizontal tangent. Plotting these points first was a helpful strategy. Few understood eitherthe notation or the concept that the function had to be increasing on either side of the , with most thinking that the point was either amax or min. It was the astute student who recognized that the derivatives being positive on either side of creates a point of inflexion.

Additionally, important points should be labelled in a sketch. Indicating the on the x-axis is a requirement of a clear graph.Although students were not penalized for not labelling the on the y-axis, there should be a recognition that the point is higher thanthe y-intercept.

π

4π

4

π

4π

2

41d. [3 marks]Let d be the distance travelled by the particle for .

(i) Write down an expression for d .

(ii) Represent d on your sketch.

Markscheme(i) correct expression A2

e.g. , , ,

(ii)

A1

Note: The line at needs to be clearly after .

[3 marks]

Examiners reportWhile some candidates recognized that the distance is the area under the velocity graph, surprisingly few included neither the limits ofintegration in their expression, nor the “dt”. Most unnecessarily attempted to integrate the function, often giving an answer with “+C”,and only earned marks if the limits were included with their result. Few recognized that a shaded area is an adequate representation ofdistance on the sketch, with most fruitlessly attempting to graph a new curve.

0 ≤ t ≤ 1

(2t+ cos2t)dt∫ 10 [ + ]t2 sin 2t

2

1

01 + sin 2

2vdt∫ 1

0

t = 1 t = π

4

42a. [2 marks]

Let . The graph of f is translated 1 unit to the right and 2 units down. The graph of g is the image of the graph of f after thistranslation.

Write down the coordinates of the vertex of the graph of g .

Markscheme A1A1 N2

[2 marks]

f(x) = 3x2

(1, − 2)

Examiners reportMost candidates had little difficulty with this question.

42b. [2 marks]Express g in the form .

Markscheme (accept , ) A1A1 N2

[2 marks]

Examiners reportMost candidates had little difficulty with this question.

g(x) = 3(x− p + q)2

g(x) = 3(x− 1 − 2)2 p = 1 q = −2

42c. [2 marks]The graph of h is the reflection of the graph of g in the x-axis.

Write down the coordinates of the vertex of the graph of h .

Markscheme A1A1 N2

[2 marks]

Examiners reportMost candidates had little difficulty with this question. In part (c), a few reflected the vertex in the y-axis rather than the x-axis.

(1, 2)

43. [6 marks]Let and , for .

Find the area of the region enclosed by the graphs of f and g .

Markschemeevidence of finding intersection points (M1)

e.g. , , sketch showing intersection

, (may be seen as limits in the integral) A1A1

evidence of approach involving integration and subtraction (in any order) (M1)

e.g. , ,

A2 N3

[6 marks]

Examiners reportThis question was poorly done by a great many candidates. Most seemed not to understand what was meant by the phrase "regionenclosed by" as several candidates assumed that the limits of the integral were those given in the domain. Few realized what area wasrequired, or that intersection points were needed. Candidates who used their GDCs to first draw a suitable sketch could normallyrecognize the required region and could find the intersection points correctly. However, it was disappointing to see the number ofcandidates who could not then use their GDC to find the required area or who attempted unsuccessful analytical approaches.

f(x) = cos( )x2 g(x) = ex −1.5 ≤ x ≤ 0.5

f(x) = g(x) cos =x2 ex

x = −1.11 x = 0

cos −∫ 0−1.11 x2 ex ∫ (cos − )dxx2 ex ∫ g− f

area = 0.282

( ) = − 2

44a. [2 marks]

The following diagram shows the graph of .

The points A, B, C, D and E lie on the graph of f . Two of these are points of inflexion.

Identify the two points of inflexion.

MarkschemeB, D A1A1 N2

[2 marks]

Examiners reportMost candidates were able to recognize the points of inflexion in part (a).

f(x) = e−x2

44b. [5 marks](i) Find .

(ii) Show that .

Markscheme(i) A1A1 N2

Note: Award A1 for and A1 for .

(ii) finding the derivative of , i.e. (A1)

evidence of choosing the product rule (M1)

e.g.

A1

AG N0

[5 marks]

Examiners reportMost candidates were able to recognize the points of inflexion in part (a) and had little difficulty with the first and second derivativesin part (b). A few did not recognize the application of the product rule in part (b).

(x)f ′

(x) = (4 − 2)f ′′ x2 e−x2

(x) = −2xf ′ e−x2

e−x2 −2x

−2x −2

−2e−x2 −2x× −2xe−x2

−2 + 4e−x2x2e−x2

(x) = (4 − 2)f ′′ x2 e−x2

44c. [4 marks]Find the x-coordinate of each point of inflexion.

Markschemevalid reasoning R1

e.g.

attempting to solve the equation (M1)

e.g. , sketch of

, A1A1 N3

[4 marks]

Examiners reportObtaining the x-coordinates of the inflexion points in (c) usually did not cause many problems.

(x) = 0f ′′

(4 − 2) = 0x2 (x)f ′′

p = 0.707 (= )12√

q = −0.707 (= − )12√

44d. [4 marks]Use the second derivative to show that one of these points is a point of inflexion.

Markschemeevidence of using second derivative to test values on either side of POI M1

e.g. finding values, reference to graph of , sign table

correct working A1A1

e.g. finding any two correct values either side of POI,

checking sign of on either side of POI

reference to sign change of R1 N0

[4 marks]

Examiners reportOnly the better-prepared candidates understood how to set up a second derivative test in part (d). Many of those did not show, orclearly indicate, the values of x used to test for a point of inflexion, but merely gave an indication of the sign. Some candidates simply

resorted to showing that , completely missing the point of the question. The necessary condition for a point of

inflexion, i.e. and the change of sign for , seemed not to be known by the vast majority of candidates.

f ′′

f ′′

(x)f ′′

(± ) = 0f ′′ 12√

(x) = 0f ′′ (x)f ′′

45a. [2 marks]

Let , for .

Show that .

Markschemecombining 2 terms (A1)

e.g. ,

expression which clearly leads to answer given A1

e.g. ,

AG N0

[2 marks]

f(x) = + 16 − 4log3x

2log3 log3 x > 0

f(x) = 2xlog3

8x− 4log3 log3 x+ 4log312

log3

log38x4

log34x2

f(x) = 2xlog3

Examiners reportFew candidates had difficulty with part (a) although it was often communicated using some very sloppy applications of the rules of

logarithm, writing instead of .log 16log 4

log( )164

45b. [3 marks]Find the value of and of .

Markschemeattempt to substitute either value into f (M1)

e.g. ,

, A1A1 N3

[3 marks]

Examiners reportPart (b) was generally done well.

f(0.5) f(4.5)

1log3 9log3

f(0.5) = 0 f(4.5) = 2

45c. [6 marks]The function f can also be written in the form .

(i) Write down the value of a and of b .

(ii) Hence on graph paper, sketch the graph of f , for , , using a scale of 1 cm to 1 unit on each axis.

(iii) Write down the equation of the asymptote.

Markscheme(i) , A1A1 N1N1

(ii)

A1A1A1 N3

Note: Award A1 for sketch approximately through , A1 for approximately correct shape, A1 for sketchasymptotic to the y-axis.

(iii) (must be an equation) A1 N1

[6 marks]

f(x) = ln axln b

−5 ≤ x ≤ 5 −5 ≤ y ≤ 5

a= 2 b = 3

(0.5 ± 0.1, 0 ± 0.1)

x = 0

Examiners reportPart (c) (i) was generally done well; candidates seemed quite comfortable changing bases. There were some very good sketches in (c)(ii), but there were also some very poor ones with candidates only considering shape and not the location of the x-intercept or theasymptote. A surprising number of candidates did not use the scale required by the question and/or did not use graph paper to sketchthe graph. In some cases, it was evident that students simply transposed their graphs from their GDC without any analyticalconsideration.

45d. [1 mark]Write down the value of .

Markscheme A1 N1

[1 mark]

Examiners reportPart (d) was poorly done as candidates did not consider the command term, “write down” and often proceeded to find the inversefunction before making the appropriate substitution.

(0)f−1

(0) = 0.5f−1

45e. [4 marks]The point A lies on the graph of f . At A, .

On your diagram, sketch the graph of , noting clearly the image of point A.

Markscheme

A1A1A1A1 N4

Note: Award A1 for sketch approximately through , A1 for approximately correct shape of the graph reflectedover , A1 for sketch asymptotic to x-axis, A1 for point clearly marked and on curve.

[4 marks]

x = 4.5

f−1

(0 ± 0.1, 0.5 ± 0.1)y = x (2 ± 0.1, 4.5 ± 0.1)

Examiners reportPart (e) eluded a great many candidates as most preferred to attempt to find the inverse analytically rather than simply reflecting thegraph of f in the line . This graph also suffered from the same sort of problems as the graph in (c) (ii). Some students did nothave their curve passing through nor did they clearly indicate its position as instructed. This point was often mislabelled onthe graph of f. The efforts in this question demonstrated that students often work tenuously from one question to the next, withoutconsidering the "big picture", thereby failing to make important links with earlier parts of the question.

y = x

(2, 4.5)

46a. [4 marks]

Let and .

Express in the form , where .

Markschemeattempt to apply rules of logarithms (M1)

e.g. ,

correct application of (seen anywhere) A1

e.g.

correct application of (seen anywhere) A1

e.g.

so

(accept ) A1 N1

[4 marks]

Examiners reportThis question was very poorly done by the majority of candidates. While candidates seemed to have a vague idea of how to apply therules of logarithms in part (a), very few did so successfully. The most common error in part (a) was to begin incorrectly with

. This error was often followed by other errors.

f(x) = 3 lnx g(x) = ln5x3

g(x) f(x) + lna a ∈ Z+

ln = b lnaab lnab = lna+ lnb

ln = b lnaab

3 lnx = lnx3

lnab = lna+ lnb

ln5 = ln5 + lnx3 x3

ln5 = ln5 + 3 lnxx3

g(x) = f(x) + ln5 g(x) = 3 lnx+ ln5

ln5 = 3 ln5xx3

46b. [3 marks]The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.

Markschemetransformation with correct name, direction, and value A3

e.g. translation by , shift up by , vertical translation of

[3 marks]

Examiners reportIn part (b), very few candidates were able to describe the transformation as a vertical translation (or shift). Many candidates attemptedto describe numerous incorrect transformations, and some left part (b) entirely blank.

( )0ln5

ln5 ln5

47a. [2 marks]

The following diagram shows part of the graph of a quadratic function f .

The x-intercepts are at and , and the y-intercept is at .

Write down in the form .

Markscheme A1A1 N2

[2 marks]

Examiners reportParts (a) and (c) of this question were very well done by most candidates.

(−4, 0) (6, 0) (0, 240)

f(x) f(x) = −10(x− p)(x− q)

f(x) = −10(x+ 4)(x− 6)

47b. [4 marks]Find another expression for in the form .

MarkschemeMETHOD 1

attempting to find the x-coordinate of maximum point (M1)

e.g. averaging the x-intercepts, sketch, , axis of symmetry

attempting to find the y-coordinate of maximum point (M1)

e.g.

A1A1 N4

METHOD 2

attempt to expand (M1)

e.g.

attempt to complete the square (M1)

e.g.

A1A1 N4

[4 marks]

f(x) f(x) = −10(x−h + k)2

= 0y′

k = −10(1 + 4)(1 − 6)

f(x) = −10(x− 1 + 250)2

f(x)

−10( − 2x− 24)x2

−10((x− 1 − 1 − 24))2

f(x) = −10(x− 1 + 250)2

Examiners reportIn part (b), many candidates attempted to use the method of completing the square, but were unsuccessful dealing with the coefficientof . Candidates who recognized that the x-coordinate of the vertex was 1, then substituted this value into the function from part(a), were generally able to earn full marks here.

−10

47c. [2 marks]Show that can also be written in the form .

Markschemeattempt to simplify (M1)

e.g. distributive property,


e.g. ,

AG N0

[2 marks]

Examiners reportParts (a) and (c) of this question were very well done by most candidates.

f(x) f(x) = 240 + 20x− 10x2

−10(x− 1)(x− 1) + 250

−10( − 6x+ 4x− 24)x2 −10( − 2x+ 1) + 250x2

f(x) = 240 + 20x− 10x2

47d. [7 marks]A particle moves along a straight line so that its velocity, , at time t seconds is given by , for .

(i) Find the value of t when the speed of the particle is greatest.

(ii) Find the acceleration of the particle when its speed is zero.

Markscheme(i) valid approach (M1)

e.g. vertex of parabola,

A1 N2

(ii) recognizing (M1)

A1A1

speed is zero (A1)

( ) A1 N3

[7 marks]

Examiners reportIn part (d), it was clear that many candidates were not familiar with the relationship between velocity and acceleration, and did notunderstand how those concepts were related to the graph which was given. A large number of candidates used time in part b(ii),rather than . To find the acceleration, some candidates tried to integrate the velocity function, rather than taking the derivative ofvelocity. Still others found the derivative in part b(i), but did not realize they needed to use it in part b(ii), as well.

v ms−1 v = 240 + 20t− 10t2

0 ≤ t ≤ 6

(t) = 0v′

t = 1

a(t) = (t)v′

a(t) = 20 − 20t

⇒ t = 6

a(6) = −100 ms−2

t = 1t = 6

48a. [2 marks]

Let , and .

Find .

f(x) = 3x g(x) = 2x− 5 h(x) = (f ∘ g)(x)

h(x)

Markschemeattempt to form composite (M1)

e.g.

A1 N2

[2 marks]

Examiners reportMost candidates handled this question with ease. Some were not familiar with the notation of composite functions assuming that

implied finding the composition and then multiplying this by x . Others misunderstood part (b) and found the reciprocalfunction or the derivative, indicating they were not familiar with the notation for an inverse function.

f(2x− 5)

h(x) = 6x− 15

(f ∘ g)(x)


Markschemeinterchanging x and y (M1)


e.g. ,

A1 N3

[3 marks]

Examiners reportMost candidates handled this question with ease. Some were not familiar with the notation of composite functions assuming that

implied finding the composition and then multiplying this by x . Others misunderstood part (b) and found the reciprocalfunction or the derivative, indicating they were not familiar with the notation for an inverse function.

(x)h−1

y+ 15 = 6x = y−x

652

(x) =h−1 x+156

(f ∘ g)(x)

49a. [2 marks]

Let and .

Find .

Markschemeattempt to form composition (in any order) (M1)

A1 N2

[2 marks]

Examiners reportCandidates showed good understanding of finding the composite function in part (a).

f(x) = + 4x2 g(x) = x− 1

(f ∘ g)(x)

(f ∘ g)(x) = (x− 1 + 4)2 ( − 2x+ 5)x2

49b. [3 marks]The vector translates the graph of to the graph of h .

Find the coordinates of the vertex of the graph of h .

( )3−1

(f ∘ g)

MarkschemeMETHOD 1

vertex of at (1, 4) (A1)

evidence of appropriate approach (M1)

e.g. adding to the coordinates of the vertex of

vertex of h at (4, 3) A1 N3

METHOD 2


e.g. ,

(A1)

vertex of h at (4, 3) A1 N3

[3 marks]

Examiners reportThere were some who did not seem to understand what the vector translation meant in part (b).

f ∘ g

( )3−1

f ∘ g

h(x)

((x− 3) − 1 + 4 − 1)2 h(x) = (f ∘ g)(x− 3) − 1

h(x) = (x− 4 + 3)2

49c. [2 marks]The vector translates the graph of to the graph of h .

Show that .


e.g. ,

simplifying A1

e.g. ,

AG N0

[2 marks]

Examiners reportCandidates showed good understanding of manipulating the quadratic in part (c).

( )3−1

(f ∘ g)

h(x) = − 8x+ 19x2

(x− 4 + 3)2 (x− 3 − 2(x− 3) + 5 − 1)2

h(x) = − 8x+ 16 + 3x2 − 6x+ 9 − 2x+ 6 + 4x2

h(x) = − 8x+ 19x2

49d. [5 marks]The vector translates the graph of to the graph of h .

The line is a tangent to the graph of h at the point P. Find the x-coordinate of P.

( )3−1

(f ∘ g)

y = 2x− 6

MarkschemeMETHOD 1

equating functions to find intersection point (M1)

e.g. ,

A1

evidence of appropriate approach to solve (M1)

e.g. factorizing, quadratic formula

appropriate working A1

e.g.

A1 N3

METHOD 2


A1

recognizing that the gradient of the tangent is the derivative (M1)

e.g. gradient at

A1

A1 N3

[5 marks]

Examiners reportThere was more than one method to solve for h in part (d), and a pleasing number of candidates were successful in this part of thequestion.

− 8x+ 19 = 2x− 6x2 y = h(x)

− 10x+ 25 + 0x2

(x− 5 = 0)2

x = 5 (p = 5)

(x)h′

h(x) = 2x− 8

p = 2

2x− 8 = 2 (2x = 10)

x = 5

50a. [4 marks]

Let . Part of the graph of f is shown below.

Find the x-intercepts of the graph.

f(x) = 8x− 2x2

Markschemeevidence of setting function to zero (M1)

e.g. ,


e.g. ,

x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0) , or , ) A1A1 N1N1

[4 marks]

Examiners reportThis question was answered well by most candidates.

f(x) = 0 8x = 2x2

0 = 2x(4 −x) −8± 64√−4

x = 4 x = 0

50b. [3 marks](i) Write down the equation of the axis of symmetry.

(ii) Find the y-coordinate of the vertex.

Markscheme(i) (must be equation) A1 N1

(ii) substituting into (M1)

A1 N2

[3 marks]

Examiners reportThis question was answered well by most candidates. Some did not give an equation for their axis of symmetry.

x = 2

x = 2 f(x)

y = 8

51a. [2 marks]

Let , for .

Show that .


e.g. (accept any base)

evidence of correct manipulation A1

e.g. , , ,

AG N0

[2 marks]

Examiners reportCandidates were generally skilled at finding the inverse of a logarithmic function.

f(x) = log3 x√ x > 0

(x) =f−1 32x

x = log y√

=3x y√ =3y x12 x = y1

2log3 2y = xlog3

(x) =f−1 32x

51b. [1 mark]Write down the range of .f−1

Markscheme , A1 N1

[1 mark]

Examiners reportFew correctly gave the range of this function, often stating “all real numbers” or “ ”, missing the idea that the range of aninverse is the domain of the original function.

y > 0 (x) > 0f−1

y ≥ 0

51c. [4 marks]Let , for .

Find the value of , giving your answer as an integer.

MarkschemeMETHOD 1

finding (seen anywhere) A1

attempt to substitute (M1)

e.g.

evidence of using log or index rule (A1)

e.g. ,

A1 N1

METHOD 2

attempt to form composite (in any order) (M1)

e.g.

evidence of using log or index rule (A1)

e.g. ,

A1

A1 N1

[4 marks]

Examiners reportSome candidates answered part (c) correctly, although many did not get beyond . Some attempted to form the composite in theincorrect order. Others interpreted as multiplication by 2.

g(x) = xlog3 x > 0

( ∘ g)(2)f−1

g(2) = lo 2g3

( ∘ g)(2) =f−1 32 log 23

( ∘ g)(2) =f−1 3log 43 3log322

( ∘ g)(2) = 4f−1

( ∘ g)(x) =f−1 32 xlog3

( ∘ g)(x) =f−1 3log3x2

3log3x2

( ∘ g)(x) =f−1 x2

( ∘ g)(2) = 4f−1

32 2log3

( ∘ g)(2)f−1

f(x) = − − 3x1 3 2

52a. [8 marks]


There is a maximum point at A and a minimum point at B(3, − 9) .

Find the coordinates of A.

Markscheme A1A1A1

evidence of solving (M1)

e.g.


e.g. ,

(ignore ) (A1)

evidence of substituting their negative x-value into (M1)

e.g. ,

A1

coordinates are N3

[8 marks]

Examiners reportA majority of candidates answered part (a) completely.

f(x) = − − 3x12x3 x2

f(x) = − 2x− 3x2

(x) = 0f ′

− 2x− 3 = 0x2

(x+ 1)(x− 3) 2± 16√2

x = −1 x = 3

f(x)

(−1 − (−1 − 3(−1)13

)3 )2 − − 1 + 313

y = 53

(−1, )53

52b. [6 marks]Write down the coordinates of

(i) the image of B after reflection in the y-axis;

(ii) the image of B after translation by the vector ;

(iii) the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor .

( )−25

12

Markscheme(i) A1 N1

(ii) A1A1 N2

(iii) reflection gives (A1)

stretch gives A1A1 N3

[6 marks]

Examiners reportCandidates were generally successful in finding images after single transformations in part (b). Common incorrect answers for (biii)

included , (6, 9) and (6, 18) , demonstrating difficulty with images from horizontal stretches.

(−3, − 9)

(1, − 4)

(3, 9)

( , 9)32

( , )32

92

53a. [2 marks]


The graph passes through the points (−2, 0), (0, − 4) and (4, 0) .

Write down the value of q and of r.

Markscheme , or , A1A1 N2

[2 marks]

Examiners reportThe majority of candidates were successful on some or all parts of this question, with some candidates using a mix of algebra andgraphical reasoning and others ignoring the graph and working only algebraically. Some did not recognize that p and q are the rootsof the quadratic function and hence gave the answers as 2 and .

f(x) = p(x− q)(x− r)

q = −2 r = 4 q = 4 r = −2

−4

53b. [1 mark]Write down the equation of the axis of symmetry.

Markscheme (must be an equation) A1 N1

[1 mark]

x = 1

Examiners reportA common error in part (b) was the absence of an equation. Some candidates wrote down the equation but were not able tosubstitute correctly. Those students did not realize that the axis of symmetry is always halfway between the x-intercepts.

x = −b2a

53c. [3 marks]Find the value of p.

Markschemesubstituting into the equation (M1)

e.g. ,

correct working towards solution (A1)

e.g.

A1 N2

[3 marks]

Examiners reportMore candidates had trouble with part (c) with erroneous substitutions and simplification mistakes commonplace.

(0, − 4)

−4 = p(0 − (−2))(0 − 4) −4 = p(−4)(2)

−4 = −8p

p = 48

(= )12

54a. [2 marks]

Let and .

Find .

Markscheme (A1)

A1 N2

[2 marks]

Examiners reportIn part (a), a number of candidates were not able to evaluate , either leaving it or evaluating it incorrectly.

f(x) = cos2x g(x) = 2 − 1x2

f( )π2

f( ) = cosππ

2

= −1

cosπ


Markscheme (A1)

A1 N2

[2 marks]

Examiners reportAlmost all candidates evaluated the composite function in part (b) in the given order, many earning follow-through marks forincorrect answers from part (a). On both parts (a) and (b), there were candidates who correctly used double-angle formulas to comeup with correct answers; while this is a valid method, it required unnecessary additional work.

(g ∘ f)( )π2

(g ∘ f)( ) = g(−1)π

2(= 2(−1 − 1))2

= 1

54c. [3 marks]Given that can be written as , find the value of k, .

Markscheme A1

evidence of (seen anywhere) (M1)

A1 N2

[3 marks]

Examiners reportCandidates were not as successful in part (c). Many tried to use double-angle formulas, but either used the formula incorrectly or usedit to write the expression in terms of and went no further. There were a number of cases in which the candidates "accidentally"came up with the correct answer based on errors or lucky guesses and did not earn credit for their final answer. Only a few candidatesrecognized the correct method of solution.

(g ∘ f)(x) cos(kx) k ∈ Z

(g ∘ f)(x) = 2(cos(2x) − 1)2 (= 2 (2x) − 1)cos2

2 θ− 1 = cos2θcos2

(g ∘ f)(x) = cos4x

k = 4

cosx

55a. [3 marks]

The velocity v ms of an object after t seconds is given by , for .

On the grid below, sketch the graph of v , clearly indicating the maximum point.

−1 v(t) = 15 − 3tt√ 0 ≤ t ≤ 25

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct shape, A1 for right endpoint at and A1 for maximum point in circle.

[3 marks]

Examiners reportThe graph in part (a) was well done. It was pleasing to see many candidates considering the domain as they sketched their graph.

(25, 0)

55b. [4 marks](i) Write down an expression for d .

(ii) Hence, write down the value of d .

Markscheme(i) recognizing that d is the area under the curve (M1)

e.g.

correct expression in terms of t, with correct limits A2 N3

e.g. ,

(ii) (m) (accept 149 to 3 sf) A1 N1

[4 marks]

Examiners reportPart (b) (i) asked for an expression which bewildered a great many candidates. However, few had difficulty obtaining the correctanswer in (b) (ii).

∫ v(t)

d = (15 − 3t)dt∫ 90 t√ d = vdt∫ 9

0

d = 148.5

56a. [3 marks]

Let .

There are two points of inflexion on the graph of f . Write down the x-coordinates of these points.

(x) = −24 + 9 + 3x+ 1f ′ x3 x2

Markschemevalid approach R1

e.g. , the max and min of gives the points of inflexion on f

(accept ( ) and ( ) A1A1 N1N1

[3 marks]

Examiners reportThere were mixed results in part (a). Students were required to understand the relationships between a function and its derivative andoften obtained the correct solutions with incorrect or missing reasoning.

(x) = 0f ′′ f ′

−0.114, 0.364 −0.114, 0.811 0.364, 2.13)

56b. [2 marks]Let . Explain why the graph of g has no points of inflexion.

MarkschemeMETHOD 1

graph of g is a quadratic function R1 N1

a quadratic function does not have any points of inflexion R1 N1

METHOD 2

graph of g is concave down over entire domain R1 N1

therefore no change in concavity R1 N1

METHOD 3

R1 N1

therefore no points of inflexion as R1 N1

[2 marks]

Examiners reportIn part (b), the question was worth two marks and candidates were required to make two valid points in their explanation. There weremany approaches to take here and candidates often confused their reasoning or just kept writing hoping that somewhere along theway they would say something correct to pick up the points. Many confused and .

g(x) = (x)f ′′

(x) = −144g′′

(x) ≠ 0g′′

f ′ g′

f(x) = x ln(4 − )2

57a. [3 marks]

Let , for . The graph of f is shown below.

The graph of f crosses the x-axis at , and .

Find the value of a and of b .


e.g. , graph

, A1A1 N3

[3 marks]

Examiners reportThis question was well done by many candidates. If there were problems, it was often with incorrect or inappropriate GDC use. Forexample, some candidates used the trace feature to answer parts (a) and (b), which at best, only provides an approximation.

f(x) = x ln(4 − )x2 −2 < x < 2

x = a x = 0 x = b

f(x) = 0

a= −1.73 b = 1.73 (a= − , b = )3√ 3√

57b. [2 marks]The graph of f has a maximum value when .

Find the value of c .

Markschemeattempt to find max (M1)

e.g. setting , graph

(accept (1.15, 1.13)) A1 N2

[2 marks]

Examiners reportThis question was well done by many candidates. If there were problems, it was often with incorrect or inappropriate GDC use. Forexample, some candidates used the trace feature to answer parts (a) and (b), which at best, only provides an approximation.

x = c

(x) = 0f ′

c = 1.15

57c. [3 marks]The region under the graph of f from to is rotated about the x-axis. Find the volume of the solid formed.x = 0 x = c 360∘

Markschemeattempt to substitute either limits or the function into formula M1

e.g. , ,

A2 N2

[3 marks]

Examiners reportMost candidates were able to set up correct expressions for parts (c) and (d) and if they had used their calculators, could find thecorrect answers. Some candidates omitted the important parts of the volume formula. Analytical approaches to (c) and (d) werealways futile and no marks were gained.

V = π dx∫ c

0 [f(x)]2 π∫ [x ln(4 − )]x2 2π dx∫ 1.149…

0 y2

V = 2.16

57d. [4 marks]Let R be the region enclosed by the curve, the x-axis and the line , between and .

Find the area of R .

Markschemevalid approach recognizing 2 regions (M1)

e.g. finding 2 areas


e.g. ,

area (accept 2.06) A2 N3

[4 marks]

Examiners reportMost candidates were able to set up correct expressions for parts (c) and (d) and if they had used their calculators, could find thecorrect answers. Some candidates omitted the important parts of the volume formula. Analytical approaches to (c) and (d) werealways futile and no marks were gained.

x = c x = a x = c

f(x)dx+ f(x)dx∫ −1.73…0 ∫ 1.149…

0 − f(x)dx+ f(x)dx∫ 0−1.73… ∫ 1.149…

0

= 2.07

58a. [4 marks]

Let , for .

Sketch the graph of g on the following set of axes.

g(x) = x sinx12

0 ≤ x ≤ 4

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct shape, A1 for left end point in circle, A1 for local maximum in circle, A1 for right endpoint in circle.

[4 marks]

Examiners reportThis question was well done by the majority of candidates. Most sketched an approximately correct shape in the given domain,though some candidates did not realize they had to set their GDC to radians, producing a meaningless sketch. Candidates need to beaware that unless otherwise specified, questions will expect radians to be used. The most confident candidates used a table to aid theirgraphing. Although most recognized the need of the GDC to answer part (b), some used the trace function, hence obtaining aninaccurate result, while others attempted a fruitless analytical approach. Merely stating "using GDC" is insufficient evidence ofmethod; a sketch or an equation set equal to zero are both examples of appropriate evidence.

58b. [2 marks]Hence find the value of x for which .

Markschemeattempting to solve (M1)

e.g. marking coordinate on graph,

A1 N2

[2 marks]

Examiners reportThis question was well done by the majority of candidates. Most sketched an approximately correct shape in the given domain,though some candidates did not realize they had to set their GDC to radians, producing a meaningless sketch. Candidates need to beaware that unless otherwise specified, questions will expect radians to be used. The most confident candidates used a table to aid theirgraphing. Although most recognized the need of the GDC to answer part (b), some used the trace function, hence obtaining aninaccurate result, while others attempted a fruitless analytical approach. Merely stating "using GDC" is insufficient evidence ofmethod; a sketch or an equation set equal to zero are both examples of appropriate evidence.

g(x) = −1

g(x) = −1

x sinx+ 1 = 012

x = 3.71

59a. [1 mark]

Consider the expansion of .

Write down the number of terms in this expansion.

Markscheme12 terms A1 N1

[1 mark]

(x+ 2)11

Examiners reportMost candidates attempted this question, and many made good progress. A number of candidates spent time writing out Pascal’striangle in full. Common errors included 11 for part (a) and not writing out the simplified form of the term for part (b). Anothercommon error was adding instead of multiplying the parts of the term in part (b).

59b. [4 marks]Find the term containing .

Markschemeevidence of binomial expansion (M1)

e.g. , an attempt to expand, Pascal’s triangle

evidence of choosing correct term (A1)

e.g. 10th term , , ,

correct working A1

e.g. ,

A1 N2

[4 marks]

Examiners reportMost candidates attempted this question, and many made good progress. A number of candidates spent time writing out Pascal’striangle in full. Common errors included 11 for part (a) and not writing out the simplified form of the term for part (b). Anothercommon error was adding instead of multiplying the parts of the term in part (b).

x2

( )nr

an−rbr

r = 9 ( )119

(x (2)2 )9

( ) (x (2119

)2 )9 55 × 29

28160x2

60. [7 marks]Solve , for .

Markschemerecognizing (seen anywhere) (A1)

e.g. ,

recognizing (A1)

e.g.


e.g. ,

evidence of correct approach to solve (M1)

e.g. factorizing, quadratic formula

correct working A1

e.g. ,

A2 N3

[7 marks]

x+ (x− 2) = 3log2 log2 x > 2

loga+ logb = logab

(x(x− 2))log2 − 2xx2

b = x ⇔ = bloga ax

= 823

x(x− 2) = 23 − 2x− 8x2

(x− 4)(x+ 2) 2± 36√2

x = 4

Examiners reportCandidates secure in their understanding of logarithm properties usually had success with this problem, solving the resulting quadraticeither by factoring or using the quadratic formula. The majority of successful candidates correctly rejected the solution that was not inthe domain. A number of candidates, however, were unclear on logarithm properties. Some unsuccessful candidates were able todemonstrate understanding of one property but without both were not able to make much progress. A few candidates employed a“guess and check” strategy, but this did not earn full marks.

61a. [5 marks]


The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.

Solve for

(i) ;

(ii) .

Markscheme(i) A1

, A1A1 N2

(ii) A1

A1 N1

[5 marks]

Examiners reportMany candidates again had difficulty finding the common angles in the trigonometric equations. In part (a), some did not showsufficient working in solving the equations. Others obtained a single solution in (a)(i) and did not find another. Some candidatesworked in degrees; the majority worked in radians.

f(x) = 6 + 6 sinx

0 ≤ x < 2π

6 + 6 sinx = 6

6 + 6 sinx = 0

sinx = 0

x = 0 x = π

sinx = −1

x = 3π2

61b. [1 mark]Write down the exact value of the x-intercept of f , for .

Markscheme A1 N1

[1 mark]

0 ≤ x < 2π

3π2

Examiners reportWhile some candidates appeared to use their understanding of the graph of the original function to find the x-intercept in part (b), mostused their working from part (a)(ii) sometimes with follow-through on an incorrect answer.

61c. [6 marks]The area of the shaded region is k . Find the value of k , giving your answer in terms of .

Markschemeevidence of using anti-differentiation (M1)

e.g.

correct integral (seen anywhere) A1A1

correct substitution (A1)

e.g. ,

A1A1 N3

[6 marks]

Examiners reportMost candidates recognized the need for integration in part (c) but far fewer were able to see the solution through correctly to the end.Some did not show the full substitution of the limits, having incorrectly assumed that evaluating the integral at 0 would be 0; withoutthis working, the mark for evaluating at the limits could not be earned. Again, many candidates had trouble working with thecommon trigonometric values.

π

(6 + 6 sinx)dx∫3π2

0

6x− 6 cosx

6 ( ) − 6 cos( ) − (−6 cos0)3π2

3π2

9π− 0 + 6

k = 9π+ 6

61d. [2 marks]Let . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

Markschemetranslation of A1A1 N2

[2 marks]

Examiners reportWhile there was an issue in the wording of the question with the given domains, this did not appear to bother candidates in part (d).This part was often well completed with candidates using a variety of language to describe the horizontal translation to the right by .

g(x) = 6 + 6 sin (x− )π2

( )π

2

0

π

2

61e. [3 marks]Let . The graph of f is transformed to the graph of g.

Given that and , write down the two values of p.

Markschemerecognizing that the area under g is the same as the shaded region in f (M1)

, A1A1 N3

[3 marks]

g(x) = 6 + 6 sin (x− )π2

g(x)dx = k∫ p+ 3π2

p0 ≤ p < 2π

p = π

2p = 0

Examiners reportMost candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majoritytried to integrate the function g without success. Some candidates used sketches to find one or both values for p. The problem in thewording of the question did not appear to have been noticed by candidates in this part either.

62a. [3 marks]

Let , for .

Find .

Markschemeevidence of choosing the product rule (M1)

e.g.

A1A1 N3

[3 marks]

Examiners reportThis problem was well done by most candidates. There were some candidates that struggled to apply the product rule in part (a) andoften wrote nonsense like .

f(x) = xcosx 0 ≤ x ≤ 6

(x)f ′

x× (−sinx) + 1 × cosx

(x) = cosx−x sinxf ′

−x sinx = −sinx2

62b. [4 marks]On the grid below, sketch the graph of .y = (x)f ′

Markscheme

A1A1A1A1 N4

Note: Award A1 for correct domain, with endpoints in circles, A1 for approximately correct shape, A1 for local minimumin circle, A1 for local maximum in circle.

[4 marks]

Examiners reportIn part (b), few candidates were able to sketch the function within the required domain and a large number of candidates did not havetheir calculator in the correct mode.

0 ≤ x ≤ 6

63. [6 marks]

The graph of , for , is shown below.

There is a minimum point at (0, −3) and a maximum point at (4, 7) .

Find the value of

(i) p ;

(ii) q ;

(iii) r.

y = pcosqx+ r −5 ≤ x ≤ 14

Markscheme(i) evidence of finding the amplitude (M1)

e.g. , amplitude

A1 N2

(ii) period (A1)

A1 N2

(iii) (A1)

A1 N2

[6 marks]

Examiners reportMany candidates did not recognize that the value of p was negative. The value of q was often interpreted incorrectly as the period butmost candidates could find the value of r, the vertical translation.

7+32

= 5

p = −5

= 8

q = 0.785 (= = )2π8

π

4

r = 7−32

r = 2

64a. [1 mark]

The diagram below shows a quadrilateral ABCD with obtuse angles and .

AB = 5 cm, BC = 4 cm, CD = 4 cm, AD = 4 cm , , , .

Use the cosine rule to show that .

Markschemecorrect substitution A1

e.g. ,

AG

[1 mark]

Examiners reportMany candidates worked comfortably with the sine and cosine rules in part (a) and (b).

A CB A CD

B C =A 30∘ A C =B x∘ A C =D y∘

AC = 41 − 40 cosx− −−−−−−−−−√

25 + 16 − 40 cosx + − 2 × 4 × 5 cosx52 42

AC = 41 − 40 cosx− −−−−−−−−−√

64b. [2 marks]Use the sine rule in triangle ABC to find another expression for AC.

Markschemecorrect substitution A1

e.g. ,

(accept ) A1 N1

[2 marks]

Examiners reportMany candidates worked comfortably with the sine and cosine rules in part (a) and (b). Equally as many did not take the cue from theword "hence" and used an alternate method to solve the problem and thus did not receive full marks. Those who managed to set upan equation, again did not go directly to their GDC but rather engaged in a long, laborious analytical approach that was usuallyunsuccessful.

=ACsinx

4sin 30

AC = 4 sinx12

AC = 8 sinx 4 sinxsin 30

64c. [6 marks](i) Hence, find x, giving your answer to two decimal places.

(ii) Find AC .

Markscheme(i) evidence of appropriate approach using AC M1

e.g. , sketch showing intersection

correct solution , (A1)

obtuse value (A1)

to 2 dp (do not accept the radian answer 1.94 ) A1 N2

(ii) substituting value of x into either expression for AC (M1)

e.g.

A1 N2

[6 marks]

Examiners reportEqually as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did notreceive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long,laborious analytical approach that was usually unsuccessful. No matter what values were found in (c) (i) most candidates recoveredand earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode androunded prematurely throughout this question often resulting in accuracy penalties.

8 sinx = 41 − 40 cosx− −−−−−−−−−√

8.682… 111.317…

111.317…

x = 111.32

AC = 8 sin 111.32

AC = 7.45

64d. [5 marks](i) Find y.

(ii) Hence, or otherwise, find the area of triangle ACD.

Markscheme(i) evidence of choosing cosine rule (M1)

e.g.


e.g. , ,

A1 N2

(ii) correct substitution into area formula (A1)

e.g. ,

area A1 N2

[5 marks]

Examiners reportEqually as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did notreceive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long,laborious analytical approach that was usually unsuccessful. No matter what values were found in (c) (i) most candidates recoveredand earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode androunded prematurely throughout this question often resulting in accuracy penalties.

cosB = + −a2 c2 b2

2ac

+ −42 42 7.452

2×4×4= 32 − 32 cosy7.452 cosy = −0.734…

y = 137

× 4 × 4 × sin 13712

8 sin 137

= 5.42

65a. [2 marks]


The y-intercept is at (0, 13) .

Show that .

Markschemesubstituting (0, 13) into function M1

e.g.

A1

AG N0

[2 marks]

Examiners reportThis question was quite well done by a great number of candidates indicating that calculus is a topic that is covered well by mostcentres. Parts (a) and (b) proved very accessible to many candidates.

f(x) = A + 3ekx

A = 10

13 = A + 3e0

13 = A+ 3

A = 10

65b. [3 marks]Given that (correct to 3 significant figures), find the value of k.

Markschemesubstituting into A1

e.g. ,

evidence of solving equation (M1)

e.g. sketch, using

(accept ) A1 N2

[3 marks]

Examiners reportThis question was quite well done by a great number of candidates indicating that calculus is a topic that is covered well by mostcentres. Parts (a) and (b) proved very accessible to many candidates.

f(15) = 3.49

f(15) = 3.49

3.49 = 10 + 3e15k 0.049 = e15k

ln

k = −0.201 ln 0.04915

65c. [5 marks](i) Using your value of k , find .

(ii) Hence, explain why f is a decreasing function.

(iii) Write down the equation of the horizontal asymptote of the graph f .

Markscheme(i)

A1A1A1 N3

Note: Award A1 for , A1 for , A1 for the derivative of 3 is zero.

(ii) valid reason with reference to derivative R1 N1

e.g. , derivative always negative

(iii) A1 N1

[5 marks]

Examiners reportThe chain rule in part (c) was also carried out well. Few however, recognized the command term “hence” and that guarantees a decreasing function. A common answer for the equation of the asymptote was to give or .

(x)f ′

f(x) = 10 + 3e−0.201x

f(x) = 10 × −0.201e−0.201x (= −2.01 )e−0.201x

10e−0.201x × − 0.201

(x) < 0f ′

y = 3

(x) < 0f ′

y = 0 x = 3

65d. [6 marks]Let .

Find the area enclosed by the graphs of f and g .

g(x) = − + 12x− 24x2

3.8953… 8.6940…

g(x) − f(x)dx∫ 8.693.90 [(− + 12x− 24) − (10 + 3)]dx∫ 8.69

3.90 x2 e−0.201x

= 19.5

Examiners reportIn part (d), it was again surprising and somewhat disappointing to see how few candidates were able to use their GDC effectively tofind the area between curves, often not finding correct limits, and often trying to evaluate the definite integral without the GDC,which led nowhere.

66a. [3 marks]

Consider , for and , for . The graph of f is given below.

On the diagram above, sketch the graph of g.

Markscheme

A1A1A1 N3

[3 marks]

Examiners reportThis question was answered well by a pleasing number of candidates.

For part (a), many good graphs were seen, though a significant number of candidates either used degrees or a function such as .There were students who lost marks for poor diagrams. For example, the shape was correct but the maximum and minimum were notaccurate enough.

f(x) = 2 −x2 −2 ≤ x ≤ 2 g(x) = sin ex −2 ≤ x ≤ 2

sin ex

66b. [2 marks]Solve .

Markscheme , (accept , if working in degrees) A1A1 N2

[2 marks]

f(x) = g(x)

x = −1.32 x = 1.68 x = −1.41 x = 1.39

Examiners reportThere were candidates who struggled in vain to solve the equation in part (b) algebraically instead of using a GDC. Those that diduse their GDCs to solve the equation frequently gave their answers inaccurately, suggesting that they did not know how to use the"zero" function on their calculator but found a rough solution using the "trace" function.

66c. [2 marks]Write down the set of values of x such that .

Markscheme (accept if working in degrees) A2 N2

[2 marks]

Examiners reportIn part (c) they often gave the correct solution, or obtained follow-through marks on their incorrect results to part (b).

f(x) > g(x)

−1.32 < x < 1.68 −1.41 < x < 1.39

67a. [1 mark]

Let , for . Part of the graph of f is given below.

There is an x-intercept at the point A, a local maximum point at M, where and a local minimum point at N, where .

Write down the x-coordinate of A.

Markscheme A1 N1

[1 mark]

Examiners reportParts (a) and (b) were generally well answered, the main problem being the accuracy.

f(x) = sin 2x+ 10ex 0 ≤ x ≤ 4

x = p x = q

2.31

67b. [2 marks]Find the value of

(i) p ;

(ii) q .

Markscheme(i) 1.02 A1 N1

(ii) 2.59 A1 N1

[2 marks]

Examiners reportParts (a) and (b) were generally well answered, the main problem being the accuracy.

67c. [3 marks]Find . Explain why this is not the area of the shaded region.

Markscheme A1 N1

split into two regions, make the area below the x-axis positive R1R1 N2

[3 marks]

Examiners reportMany students lacked the calculator skills to successfully complete (6)(c) in that they could not find the value of the definite integral.Some tried to find it by hand. When trying to explain why the integral was not the area, most knew the region under the x-axis wasthe cause of the integral not giving the total area, but the explanations were not sufficiently clear. It was often stated that the areabelow the axis was negative rather than the integral was negative.

f(x)dx∫ q

p

f(x)dx = 9.96∫ q

p

68a. [2 marks]

The number of bacteria, n , in a dish, after t minutes is given by .

Find the value of n when .

Markscheme (A1)

A1 N2

[2 marks]

Examiners reportThis question seemed to be challenging for the great majority of the candidates.

Part (a) was generally well answered.

n = 800e0.13t

t = 0

n = 800e0

n = 800

68b. [2 marks]Find the rate at which n is increasing when .

Markschemeevidence of using the derivative (M1)

A1 N2

[2 marks]

t = 15

(15) = 731n′


Part (a) was generally well answered but in parts (b) and (c) they did not consider that rates of change meant they needed to usedifferentiation. Most students completely missed or did not understand that the question was asking about the instantaneous rate ofchange, which resulted in the fact that most of them used the original equation. Some did attempt to find an average rate of changeover the time interval, but even fewer attempted to use the derivative.

Of those who did realize to use the derivative in (b), a vast majority calculated it by hand instead of using their GDC feature toevaluate it.

68c. [4 marks]After k minutes, the rate of increase in n is greater than bacteria per minute. Find the least value of k , where .

MarkschemeMETHOD 1

setting up inequality (accept equation or reverse inequality) A1

e.g.

evidence of appropriate approach M1

e.g. sketch, finding derivative

(A1)

least value of k is 36 A1 N2

METHOD 2

, and A2

least value of k is 36 A2 N2

[4 marks]


Part (a) was generally well answered but in parts (b) and (c) they did not consider that rates of change meant they needed to usedifferentiation. Most students completely missed or did not understand that the question was asking about the instantaneous rate ofchange, which resulted in the fact that most of them used the original equation. Some did attempt to find an average rate of changeover the time interval, but even fewer attempted to use the derivative.

Of those who did realize to use the derivative in (b), a vast majority calculated it by hand instead of using their GDC feature toevaluate it.

The inequality for part (c) was sometimes well solved using the original function but many failed to round their answers to the nearestinteger.

10000 k ∈ Z

(t) > 10000n′

k = 35.1226…

(35) = 9842n′ (36) = 11208n′

f(x) = x ln(4 − )2

69a. [5 marks]

Consider , for . The graph of f is given below.

Let P and Q be points on the curve of f where the tangent to the graph of f is parallel to the x-axis.

(i) Find the x-coordinate of P and of Q.

(ii) Consider . Write down all values of k for which there are exactly two solutions.

Markscheme(i) A1A1 N2

(ii) recognizing that it occurs at P and Q (M1)

e.g. ,

, A1A1 N3

[5 marks]

Examiners reportMany candidates correctly found the x-coordinates of P and Q in (a)(i) with their GDC. In (a)(ii) some candidates incorrectlyinterpreted the words “exactly two solutions” as an indication that the discriminant of a quadratic was required. Many failed to realisethat the values of k they were looking for in this question were the y-coordinates of the points found in (a)(i).

f(x) = x ln(4 − )x2 −2 < x < 2

f(x) = k

−1.15, 1.15

x = −1.15 x = 1.15

k = −1.13 k = 1.13

69b. [4 marks]Let , for .

Show that .

Markschemeevidence of choosing the product rule (M1)

e.g.

derivative of is (A1)

derivative of is (A1)


e.g.

AG N0

[4 marks]

g(x) = ln(4 − )x3 x2 −2 < x < 2

(x) = + 3 ln(4 − )g′ −2x4

4−x2x2 x2

u + vv′ u′

x3 3x2

ln(4 − )x2 −2x4−x2

× + ln(4 − ) × 3x3 −2x4−x2

x2 x2

(x) = + 3 ln(4 − )g′ −2x4

4−x2x2 x2

Examiners reportMany candidates were unclear in their application of the product formula in the verifying the given derivative of g. Showing that thederivative was the given expression often received full marks though it was not easy to tell in some cases if that demonstration camethrough understanding of the product and chain rules or from reasoning backwards from the given result.

69c. [2 marks]Let , for .

Sketch the graph of .

Markscheme

A1A1 N2

[2 marks]

Examiners reportSome candidates drew their graphs of the derivative in (c) on their examination papers despite clear instructions to do their work onseparate sheets. Most who tried to plot the graph in (c) did so successfully.

g(x) = ln(4 − )x3 x2 −2 < x < 2

g′

69d. [3 marks]Let , for .

Consider . Write down all values of w for which there are exactly two solutions.

Markscheme , A1A2 N2

[3 marks]

Examiners reportCorrect solutions to 10(d) were not often seen.

g(x) = ln(4 − )x3 x2 −2 < x < 2

(x) = wg′

w= 2.69 w< 0

70a. [5 marks]

Let and .

(i) Find .

(ii) Find .

f(x) = 2 + 3x3 g(x) = − 2e3x

g(0)

(f ∘ g)(0)

Markscheme(i) (A1)

A1 N2

(ii) METHOD 1

substituting answer from (i) (M1)

e.g.


A1 N3

METHOD 2


e.g.

correct expression for (A1)

e.g.

A1 N3

[5 marks]

Examiners reportThis question was generally done well, although some students consider to be 0, losing them a mark.

g(0) = − 2e0

= −1

(f ∘ g)(0) = f(−1)

f(−1) = 2(−1 + 3)3

f(−1) = 1

(f ∘ g)(x)

(f ∘ g)(x) = f( − 2)e3x = 2( − 2 + 3e3x )3

(f ∘ g)(x)

2( − 2 + 3e3x )3

(f ∘ g)(0) = 1

e0



e.g.

attempt to solve (M1)

e.g.

A1 N3

[3 marks]

Examiners reportA few candidates composed in the wrong order. Most found the formula of the inverse correctly, even if in some cases there were

errors when trying to isolate x (or y). A common incorrect solution found was to find .

(x)f−1

x = 2 + 3y3

=y3 x−32

(x) =f−1 x−32

− −−√3

y = x−32

− −−√3

f(x)

71a. [3 marks]


Let . Sketch the graph of on the grid below.

f(x) −2 ≤ x ≤ 4

h(x) = f(−x) h

Markscheme

A2 N2

[2 marks]

Examiners reportPart (a) was generally solved correctly. Students had no trouble in deciding what transformation had to be done to the graph, althoughsome confused with . f(−x) −f(x)

71b. [3 marks]Let . The point on the graph of is transformed to the point P on the graph of . Find thecoordinates of P.


e.g. reference to any horizontal shift and/or stretch factor, ,

P is (accept , ) A1A1 N3

[3 marks]

Examiners reportPart (b) was generally poorly done. They could not "read" that the transformation shifted the curve 1 unit to the right and stretched itin the -direction with a scale factor of . It was often seen that the shift was interpreted, but in the opposite direction. Also, thestretch was applied to both coordinates of the point. Those candidates who answered part (a) incorrectly often had trouble on (b) aswell, indicating a difficulty with transformations in general. However, there were also candidates who solved part (a) correctly butcould not interpret part (b). This would indicate that it is simpler for them to plot the transformation of an entire function than to findhow a particular point is transformed.

g(x) = f(x− 1)12

A(3, 2) f g

x = 3 + 1 y = × 212

(4, 1) x = 4 y = 1

y 12

72a. [3 marks]

Let .

Given that , find the value of .

f(x) = k xlog2

(1) = 8f−1 k

MarkschemeMETHOD 1

recognizing that (M1)

e.g.

recognizing that (A1)

e.g.

A1 N2

METHOD 2

attempt to find the inverse of (M1)

e.g. ,

substituting 1 and 8 (M1)

e.g. ,

A1 N2

[3 marks]

Examiners reportA very poorly done question. Most candidates attempted to find the inverse function for and used that to answer parts (a) and (b).Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able toactually get the correct inverse. Almost none recognized that if , then . Many thought that the letters "log" couldbe simply "cancelled out", leaving the and the .

f(8) = 1

1 = k 8log2

8 = 3log2

1 = 3k

k = 13

f(x) = k xlog2

x = k ylog2 y = 2x

k

1 = k 8log2 = 821k

k = 18log2

(k = )13

f

(1) = 8f−1 f(8) = 12 8


MarkschemeMETHOD 1

recognizing that (M1)

e.g.

(A1)

(accept ) A2 N3

METHOD 2

attempt to find inverse of (M1)

e.g. interchanging x and y , substituting into

correct inverse (A1)

e.g. ,

A2 N3

[4 marks]

( )f−1 23

f(x) = 23

= x23

13

log2

x = 2log2

( ) = 4f−1 23

x = 4

f(x) = x13

log2

k = 13

y = 2x

k

(x) =f−1 23x 23x

( ) = 4f−1 23

Examiners reportA very poorly done question. Most candidates attempted to find the inverse function for and used that to answer parts (a) and (b).Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able toactually get the correct inverse. Almost none recognized that if , then . Many thought that the letters "log" couldbe simply "cancelled out", leaving the and the .

f

(1) = 8f−1 f(8) = 12 8

73a. [7 marks]

Let , for . The graph of f is given below.

The y-intercept is at the point A.

(i) Find the coordinates of A.

(ii) Show that at A.

Markscheme(i) coordinates of A are A1A1 N2

(ii) derivative of (seen anywhere) (A1)

evidence of correct approach (M1)

e.g. quotient rule, chain rule

finding A2

e.g. ,

substituting into (do not accept solving ) M1

at A AG N0

[7 marks]

Examiners reportAlmost all candidates earned the first two marks in part (a) (i), although fewer were able to apply the quotient rule correctly.

f(x) = 3 + 20−4x2

x ≠ ±2

(x) = 0f ′

(0, − 2)

− 4 = 2xx2

(x)f ′

(x) = 20 × (−1) × ( − 4 × (2x)f ′ x2 )−2 ( −4)(0)−(20)(2x)x2

( −4)x2 2

x = 0 (x)f ′ (x) = 0f ′

(x) = 0f ′

73b. [2 marks]Write down the range of .f

Markschemecorrect inequalities, , , FT from (a)(i) and (c) A1A1 N2

[2 marks]

Examiners reportIn (d) even those who knew what the range was had difficulty expressing the inequalities correctly.

y ≤ −2 y > 3

74. [2 marks]

Let . Line L is the normal to the graph of f at the point (4, 2) .

Point A is the x-intercept of L . Find the x-coordinate of A.

Markschemerecognition that at A (M1)

e.g.

A1 N2

[2 marks]

Examiners reportParts (a) and (b) were well done by most candidates.

f(x) = x√

y = 0

−4x+ 18 = 0

x = 184

(= )92

75. [7 marks]

A farmer wishes to create a rectangular enclosure, ABCD, of area 525 m , as shown below.

The fencing used for side AB costs per metre. The fencing for the other three sides costs per metre. The farmer createsan enclosure so that the cost is a minimum. Find this minimum cost.

2

$11 $3

MarkschemeMETHOD 1

correct expression for second side, using area = 525 (A1)

e.g. let ,

attempt to set up cost function using $3 for three sides and $11 for one side (M1)

e.g.

correct expression for cost A2

e.g. , ,

EITHER

sketch of cost function (M1)

identifying minimum point (A1)

e.g. marking point on graph,

minimum cost is 420 (dollars) A1 N4

OR

correct derivative (may be seen in equation below) (A1)

e.g.

setting their derivative equal to 0 (seen anywhere) (M1)

e.g.


METHOD 2

correct expression for second side, using area = 525 (A1)

e.g. let ,

attempt to set up cost function using for three sides and for one side (M1)

e.g.

correct expression for cost A2

e.g. , ,

EITHER

sketch of cost function (M1)

identifying minimum point (A1)

e.g. marking point on graph,


OR

correct derivative (may be seen in equation below) (A1)

e.g.

setting their derivative equal to 0 (seen anywhere) (M1)

e.g.


[7 marks]

AB = x AD = 525x

3(AD + BC + CD) + 11AB

× 3 + × 3 + 11x+ 3x525x

525x

× 3 + × 3 + 11AB + 3AB525AB

525AB

+ 14x3150x

x = 15

(x) = + + 14C′ −1575x2

−1575x2

+ 14 = 0−3150x2

AD = x AB = 525x

$3 $11

3(AD + BC + CD) + 11AB

3 (x+x+ ) + × 11525x

525x

3 (AD + AD + ) + × 11525AD

525AD

6x+ 7350x

x = 35

(x) = 6 −C′ 7350x2

6 − = 07350x2

Examiners reportAlthough this question was a rather straight-forward optimisation question, the lack of structure caused many candidates difficulty.Some were able to calculate cost values but were unable to create an algebraic cost function. Those who were able to create a costfunction in two variables often could not use the area relationship to obtain a function in a single variable and so could make nofurther progress. Of those few who created a correct cost function, most set the derivative to zero to find that the minimum costoccurred at , leading to . Although this is a correct approach earning full marks, candidates seem not to recognise that theresult can be obtained from the GDC, without the use of calculus.

x = 15 $420

76a. [3 marks]

Let and for .

On the same diagram, sketch the graphs of f and g .

Markscheme

A1A1A1 N3

Note: Award A1 for f being of sinusoidal shape, with 2 maxima and one minimum, A1 for g being a parabola opening down, A1 fortwo intersection points in approximately correct position.

[3 marks]

Examiners reportGraph sketches were much improved over previous sessions. Most candidates graphed the two functions correctly, but many ignoredthe domain restrictions.

f(x) = 5 cos xπ4

g(x) = −0.5 + 5x− 8x2 0 ≤ x ≤ 9

76b. [3 marks]Consider the graph of g . Write down

(i) the two x-intercepts;

(ii) the equation of the axis of symmetry.

Markscheme(i) , (accept , ) A1A1 N1N1

(ii) (must be an equation) A1 N1

[3 marks]

Examiners reportMany candidates found parts (b) and (c) accessible, although quite a few did not know how to find the period of the cosine function.

(2, 0) (8, 0) x = 2 x = 8

x = 5

77a. [2 marks]

Let and .

The graph of g can be obtained from the graph of f using two transformations.

Give a full geometric description of each of the two transformations.

Markschemein any order

translated 1 unit to the right A1 N1

stretched vertically by factor 2 A1 N1

[2 marks]

Examiners reportThe translation was often described well as horizontal (or shift) one unit right. There was considerable difficulty describing thevertical stretch as it was often referred to as "stretch by 2" or "amplitude of 2". A full description should include the name (e.g.vertical stretch) and value for full marks. Candidates also had difficulty applying two consecutive transformations to a single point.Often the translations were applied directly to instead of first mapping from f to g . A good number of candidates correctlyfound , but most could not find P from this function.

f(x) = x2 g(x) = 2(x− 1)2

(−1, 1)h(x)

77b. [4 marks]The graph of g is translated by the vector to give the graph of h.

The point on the graph of f is translated to the point P on the graph of h.

Find the coordinates of P.

MarkschemeMETHOD 1

finding coordinates of image on g (A1)(A1)

e.g. , , ,

P is (3, 0) A1A1 N4

METHOD 2

(A1)(A1)

P is A1A1 N4

Examiners reportThe translation was often described well as horizontal (or shift) one unit right. There was considerable difficulty describing thevertical stretch as it was often referred to as "stretch by 2" or "amplitude of 2". A full description should include the name (e.g.vertical stretch) and value for full marks. Candidates also had difficulty applying two consecutive transformations to a single point.Often the translations were applied directly to instead of first mapping from f to g . A good number of candidates correctlyfound , but most could not find P from this function.

( )3−2

(−1, 1)

−1 + 1 = 0 1 × 2 = 2 (−1, 1) → (−1 + 1, 2 × 1) (0, 2)

h(x) = 2(x− 4 − 2)2

(3, 0)

(−1, 1)h(x)

78. [3 marks]

Let .

(i) Show that .

(ii) Write down the domain of .

f(x) = ex+3

(x) = lnx− 3f−1

f−1

Markscheme(i) interchanging x and y (seen anywhere) M1

e.g.


e.g. ,

AG N0

(ii) A1 N1

[3 marks]

Examiners reportMany candidates interchanged the and to find the inverse function, but very few could write down the correct domain of theinverse, often giving , and "all real numbers" as responses.

x = ey+3

lnx = y+ 3 lny = x+ 3

(x) = lnx− 3f−1

x > 0

x y

x ≥ 0 x > 3

79a. [2 marks]

Let , , .The graph of f is shown below.

The region between and is shaded.

Show that .

MarkschemeMETHOD 1

evidence of substituting for (M1)

A1

AG N0

METHOD 2

is reflection of in x axis

and is reflection of in y axis (M1)

sketch showing these are the same A1

AG N0

[2 marks]

f(x) = ax

+1x2−8 ≤ x ≤ 8 a ∈ R

x = 3 x = 7

f(−x) = −f(x)

−x x

f(−x) = a(−x)

+1(−x)2

f(−x) = −ax+1x2

(= −f(x))

y = −f(x) y = f(x)

y = f(−x) y = f(x)

f(−x) = −ax+1x2

(= −f(x))

Examiners reportPart (a) was achieved by some candidates, although brackets around the were commonly neglected. Some attempted to show therelationship by substituting a specific value for x . This earned no marks as a general argument is required.

−x

79b. [7 marks]It is given that .

(i) Find the area of the shaded region, giving your answer in the form .

(ii) Find the value of .

Markscheme(i) correct expression A2

e.g. , ,

area = A1A1 N2

(ii) METHOD 1

recognizing the shift that does not change the area (M1)

e.g. ,

recognizing that the factor of 2 doubles the area (M1)

e.g.

(i.e. their answer to (c)(i)) A1 N3

METHOD 2

changing variable

let , so

(M1)

substituting correct limits

e.g. , , (M1)

A1 N3

[7 marks]

Examiners reportFor those who found a correct expression in (c)(i), many finished by calculating . Few recognized that thetranslation did not change the area, although some factored the 2 from the integrand, appreciating that the area is double that in (c)(i).

∫ f(x)dx = ln( + 1) +Ca

2x2

p lnq

2f(x− 1)dx∫ 84

[ ln( + 1)]a

2x2

7

3ln50 − ln10a

2a

2(ln50 − ln10)a

2

ln5a

2

f(x− 1)dx = f(x)dx∫ 84 ∫ 7

3 ln5a

2

2f(x− 1)dx =2 f(x− 1)dx∫ 84 ∫ 8

4 (= 2 f(x)dx)∫ 73

2f(x− 1)dx = a ln5∫ 84 2 ×

w= x− 1 = 1dwdx

2 ∫ f(w)dw= ln( + 1) + c2a2

w2

[a ln[ + 1]](x− 1)28

4[a ln( + 1)]w2 7

3a ln50 − a ln10

2f(x− 1)dx = a ln5∫ 84

ln50 − ln10 = ln40

80. [3 marks]

Let , . Let .

Find an expression for .

f(x) = + 13x2

g(x) = 4 cos( ) − 1x

3h(x) = (g ∘ f)(x)

h(x)

Markschemeattempt to form any composition (even if order is reversed) (M1)

correct composition (A1)

A1 N3

[3 marks]

Examiners reportThe majority of candidates handled the composition of the two given functions well. However, a large number of candidates haddifficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidatesknew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the periodand range, but this approach was not the most efficient.

h(x) = g( + 1)3x2

h(x) = 4 cos( ) − 1+13x

2

3(4 cos( x+ ) − 1,4 cos( ) − 1)1

213

3x+26

81. [4 marks]

In a geometric series, and .

Find the smallest value of n for which .

MarkschemeMETHOD 1

setting up an inequality (accept an equation) M1

e.g. , ,

evidence of solving M1

e.g. graph, taking logs

(A1)

A1 N2

METHOD 2

if , sum ; if , sum A2

(is the smallest value) A2 N2

[4 marks]

Examiners reportIn part (b) a good number of candidates did not realize that they could use logs to solve the problem, nor did they make good use oftheir GDCs. Some students did use a trial and error approach to check various values however, in many cases, they only checked oneof the "crossover" values. Most candidates had difficulty with notation, opting to set up an equation rather than an inequality.

=u11

81=u4

13

> 40Sn

> 40( −1)1

813n

2> 40

(1− )181

3n

−2> 64813n

n > 7.9888…

∴ n = 8

n = 7 = 13.49… n = 8 = 40.49…

n = 8

82. [2 marks]

Let .

Write down the range of values for the gradient of .

Markscheme , , A2 N2

[2 marks]

f(x) = − 4x+ 1x3

f

(x) ≥ −4f ′ y ≥ −4 [−4,∞[

Examiners reportPart (e) was often not attempted and if it was, candidates were not clear on what was expected.

83a. [2 marks]

Let and .

Find .

Markschemefor interchanging x and y (may be done later) (M1)

e.g.

(accept ) A1 N2

[2 marks]

Examiners reportMany candidates performed successfully in finding the inverse function, as well as the composite at a specified value of x.

f(x) = x2 g(x) = 2x− 3

(x)g−1

x = 2y− 3

(x) =g−1 x+32

y = ,x+32

x+32


MarkschemeMETHOD 1

(A1)

evidence of composition of functions (M1)

A1 N3

METHOD 2

(M1)

(A1)

= 25 A1 N3

[3 marks]

Examiners reportMany candidates performed successfully in finding the inverse function, as well as the composite at a specified value of x. Somecandidates made arithmetical errors especially if they expanded the binomial before substituting .

(f ∘ g)(4)

g(4) = 5

f(5) = 25

f ∘ g(x) = (2x− 3)2

f ∘ g(4) = (2 × 4 − 3)2

x = 4

84a. [2 marks]

Consider the graph of shown below.

On the same grid sketch the graph of .

Markscheme

A2 N2

[2 marks]

Examiners reportThis question was reasonably solved by many students, though a good number confused with in part (a), thus reflectingthe original diagram in the x-axis. Candidates need more practice in correctly and fully describing transformations.

f

y = f(−x)

f(−x) −f(x)

84b. [2 marks]

The following four diagrams show images of f under different transformations.

Complete the following table.

Markscheme

A1A1 N2

[2 marks]

Examiners reportCandidates need more practice in correctly and fully describing transformations. There was often confusion between the descriptionof the transformation and the equation that represents it. A fairly low percentage of the candidates used the term "translation".

84c. [2 marks]Give a full geometric description of the transformation that gives the image in Diagram A.

Markschemetranslation (accept move/shift/slide etc.) with vector A1A1 N2

[2 marks]

Examiners reportCandidates need more practice in correctly and fully describing transformations. There was often confusion between the descriptionof the transformation and the equation that represents it. A fairly low percentage of the candidates used the term "translation".

( )−6−2

85. [5 marks]Solve the equation , for .

Markschemeevidence of appropriate approach M1

e.g. a sketch, writing

, A2A2 N2N2

[5 marks]

Examiners reportAlthough many students started with an analytical approach, many also realized they were not going further and successfully usedtheir GDC to find the intercepts with the x-axis if they had set the equation equal to 0, or in other cases, they found the intersection ofthe two graphs. The better candidates drew a reasonable sketch and found the two values without difficulty. A good number ofcandidates did not provide a sketch, however, and they had more trouble earning the mark for showing method. Accuracy penaltieswere relatively common on this question.

= 4 sinxex 0 ≤ x ≤ 2π

− 4 sinx = 0ex

x = 0.371 x = 1.36

86a. [5 marks]

The quadratic equation has two equal real roots.

Find the possible values of k.

Markschemeattempt to use discriminant (M1)

correct substitution, (A1)

setting their discriminant equal to zero M1

e.g. ,

, A1A1 N3

[5 marks]

Examiners reportAlthough some candidates correctly considered the discriminant to find the possible values of k , many of them did not set it equal to

, writing an inequality instead.

k + (k− 3)x+ 1 = 0x2

(k− 3 − 4 × k× 1)2

(k− 3 − 4 × k× 1 = 0)2 − 10k+ 9 = 0k2

k = 1 k = 9

0

86b. [2 marks]Write down the values of k for which has two equal real roots.

Markscheme , A2 N2

[2 marks]

Examiners reportIn part (b), some students realized that the discriminants in parts (a) and (b) were the same, earning follow through marks just bywriting the same (often incorrect) answers they got in part (a). Many, however, did not see the connection between the two parts.

+ (k− 3)x+ k = 0x2

k = 1 k = 9

87a. [3 marks]

Let , for .

Sketch the graph of f .

f(x) = 3 sinx+ 4 cosx −2π ≤ x ≤ 2π

Markscheme

A1A1A1 N3

Note: Award A1 for approximately sinusoidal shape, A1 for end points approximately correct , A1 forapproximately correct position of graph, (y-intercept , maximum to right of y-axis).

[3 marks]

Examiners reportSome graphs in part (a) were almost too detailed for just a sketch but more often, the important features were far from clear. Somegraphs lacked scales on the axes.

(−2π, 4) (2π, 4)(0, 4)

87b. [3 marks]Write down

(i) the amplitude;

(ii) the period;

(iii) the x-intercept that lies between and 0.

Markscheme(i) 5 A1 N1

(ii) (6.28) A1 N1

(iii) A1 N1

[3 marks]

Examiners reportA number of candidates had difficulty finding the period in part (b)(ii).

− π

2

2π

−0.927

87c. [5 marks]Let , for . There is a value of x, between and , for which the gradient of f is equal to thegradient of g. Find this value of x.

g(x) = ln(x+ 1) 0 ≤ x ≤ π 0 1

MarkschemeMETHOD 1

graphical approach (but must involve derivative functions) M1

e.g.

each curve A1A1

A2 N2

METHOD 2

A1

A1

evidence of attempt to solve M1

A2 N2

[5 marks]

Examiners reportIn part (f), many candidates were able to get as far as equating the two derivatives but fewer used their GDC to solve the resultingequation. Again, many had trouble demonstrating their method of solution.

x = 0.511

(x) =g′ 1x+1

(x) = 3 cosx− 4 sinxf ′ (5 cos(x+ 0.927))

(x) = (x)g′ f ′

x = 0.511

88a. [3 marks]

Let f be the function given by , . The diagram shows the graph of f .

On the same diagram, sketch the graph of .

f(x) = e0.5x 0 ≤ x ≤ 3.5

f−1

Markscheme

A1A1A1 N3

Note: Award A1 for approximately correct (reflected) shape, A1 for right end point in circle, A1 for through .

Examiners reportThere were a large number of candidates who were unaware of the geometric relationship between a function and its inverse. Thosethat had some idea of the shape of the graph often did not consider the specified domain. Many more students were able to use ananalytical approach to finding the inverse of a function and had little problem using logarithms to solve for y. Candidates were clearlymore comfortable with algebraic procedures than graphical interpretations.

(1, 0)

88b. [1 mark]Write down the range of .

Markscheme A1 N1

[1 mark]


f−1

0 ≤ y ≤ 3.5


Markschemeinterchanging x and y (seen anywhere) M1

e.g.

evidence of changing to log form A1

e.g. , (any base), (any base)

A1 N1

[3 marks]

(x)f−1

x = e0.5y

lnx = 0.5y lnx = lne0.5y lnx = 0.5y lne

(x) = 2 lnxf−1


89a. [2 marks]

Let , . Part of the graph of is given below.

When , there is a maximum value of 29, at M.

When , there is a minimum value of 15.

The transformation P is given by a horizontal stretch of a scale factor of , followed by a translation of .

Let be the image of M under P. Find the coordinates of .

Markschemestretch takes 3 to 1.5 (A1)

translation maps to (so is ) A1 N2

[2 marks]

Examiners reportFew answered part (b) correctly as most could not interpret the horizontal stretch.

f(t) = acosb(t− c) + d t ≥ 0 y = f(t)

t = 3

t = 9

12

( )3−10

M ′ M ′

(1.5, 29) (4.5, 19) M ′ (4.5, 19)

89b. [4 marks]The graph of g is the image of the graph of f under P.

Find in the form .

Markscheme A1A2A1 N4

Note: Award A1 for , A2 for 4.5, A1 for 12.

Other correct values for c can be found, , .

[4 marks]

g(t) g(t) = 7 cosB(t− c) +D

g(t) = 7 cos (t− 4.5) + 12π

3

π

3

c = 4.5 ± 6k k ∈ Z

Examiners reportFew answered part (b) correctly as most could not interpret the horizontal stretch. As a result, there were many who were unable toanswer part (c) although follow through marks were often obtained from incorrect answers in both parts (a) and (b). The link betweenthe answer in (b) and the value of C in part (c) was lost on all but the most attentive.

89c. [3 marks]The graph of g is the image of the graph of f under P.

Give a full geometric description of the transformation that maps the graph of g to the graph of f .

Markschemetranslation (A1)

horizontal stretch of a scale factor of 2 (A1)

completely correct description, in correct order A1 N3

e.g. translation then horizontal stretch of a scale factor of 2

[3 marks]

Examiners reportIn part (d), some candidates could name the transformations required, although only a handful provided the correct order of thetransformations to return the graph to its original state.

( )−310

( )−310

90a. [3 marks]

Let .

Express in the form .

Markschemeevidence of obtaining the vertex (M1)

e.g. a graph, , completing the square

A2 N3

[3 marks]

Examiners reportMany candidates answered this question with great ease. Still, some found themselves unable to correctly find the vertexalgebraically, often mixing the signs of the h and k values. Using the GDC may have been a more fruitful approach. Some candidatesdid not write the axis of symmetry as an equation.

f(x) = 2 + 4x− 6x2

f(x) f(x) = 2(x−h + k)2

x = − b

2a

f(x) = 2(x+ 1 − 8)2

90b. [1 mark]Write down the equation of the axis of symmetry of the graph of f .

Markscheme (equation must be seen) A1 N1

[1 mark]

x = −1


90c. [2 marks]Express in the form .

Markscheme A1A1 N2

[2 marks]


f(x) f(x) = 2(x− p)(x− q)

f(x) = 2(x− 1)(x+ 3)

91a. [3 marks]

Let , .

Sketch the graph of f on the following set of axes.

Markscheme

A1A2 N3

Notes: Award A1 for correct domain, . Award A2 for approximately correct shape, with local maximum in circle 1 andright endpoint in circle 2.

[3 marks]

f(x) = xcos(x− sinx) 0 ≤ x ≤ 3

0 ≤ x ≤ 3

Examiners reportMany candidates sketched a clear and smooth freehand curve with the local maximum, x-intercept and endpoints in approximatelycorrect positions. Commonly, candidates sketched a graph across , which neglects the given domain of the function. Therewere some candidates who sketched a straight line through the origin, presumably from being in the degree mode of their GDC.

[−3, 3]

91b. [1 mark]The graph of f intersects the x-axis when , . Write down the value of a.

Markscheme A1 N1

[1 mark]

Examiners reportMany candidates sketched a clear and smooth freehand curve with the local maximum, x-intercept and endpoints in approximatelycorrect positions. Commonly, candidates sketched a graph across , which neglects the given domain of the function. Therewere some candidates who sketched a straight line through the origin, presumably from being in the degree mode of their GDC.

x = a a≠ 0

a= 2.31

[−3, 3]

92a. [4 marks]

Let , for .

Find .

MarkschemeMETHOD 1

(A1)

interchanging x and y (seen anywhere) (M1)

e.g.


e.g.

A1 N2

METHOD 2

(A1)


e.g.

interchanging x and y (seen anywhere) (M1)

e.g.

A1 N2

[4 marks]

Examiners reportThis was one of the more difficult problems for the candidates. Knowledge of the laws of logarithms appeared weak as did theinverse nature of the exponential and logarithmic functions. There were a number of candidates who mistook the notation for theinverse to mean either the derivative or the reciprocal. The order of composition seemed well understood by most candidates but theywere unable to simplify by the rules of indices to obtain the correct final answer.

f(x) = ln(x+ 5) + ln2 x > −5

(x)f−1

ln(x+ 5) + ln2 = ln(2(x+ 5)) (= ln(2x+ 10))

x = ln(2y+ 10)

= 2y+ 10ex

(x) =f−1 −10ex

2

y = ln(x+ 5) + ln2

y− ln2 = ln(x+ 5)

= x+ 5ey−ln 2

= y+ 5ex−ln 2

(x) = − 5f−1 ex−ln 2

g(x) = x

92b. [3 marks]Let .

Find , giving your answer in the form , where .

MarkschemeMETHOD 1

evidence of composition in correct order (M1)

e.g.

A1A1 N2

METHOD 2

evidence of composition in correct order (M1)

e.g.

A1A1 N2

[3 marks]

Examiners reportThis was one of the more difficult problems for the candidates. Knowledge of the laws of logarithms appeared weak as did theinverse nature of the exponential and logarithmic functions. There were a number of candidates who mistook the notation for theinverse to mean either the derivative or the reciprocal. The order of composition seemed well understood by most candidates but theywere unable to simplify by the rules of indices to obtain the correct final answer.

g(x) = ex

(g ∘ f)(x) ax+ b a,b ∈ Z

(g ∘ f)(x) = g(ln(x+ 5) + ln2)

= = 2(x+ 5)eln(2(x+5))

(g ∘ f)(x) = 2x+ 10

(g ∘ f)(x) = eln(x+5)+ln 2

= × = (x+ 5)2eln(x+5) eln 2

(g ∘ f)(x) = 2x+ 10

93a. [2 marks]

Let .

Show that .

Markscheme A1

A1

AG N0

[2 marks]

Examiners reportThis problem was generally well done. The “show that” question in part (a) was done correctly by most candidates, with a fewattempting to show it by working backwards, which earned no marks.

f(x) = 3(x+ 1 − 12)2

f(x) = 3 + 6x− 9x2

f(x) = 3( + 2x+ 1) − 12x2

= 3 + 6x+ 3 − 12x2

= 3 + 6x− 9x2

93b. [8 marks]For the graph of f

(i) write down the coordinates of the vertex;

(ii) write down the equation of the axis of symmetry;

(iii) write down the y-intercept;

(iv) find both x-intercepts.

Markscheme(i) vertex is A1A1 N2

(ii) (must be an equation) A1 N1

(iii) A1 N1

(iv) evidence of solving (M1)

e.g. factorizing, formula,

correct working A1

e.g. ,

, A1A1 N1N1

[8 marks]

Examiners reportMost candidates were able to identify the vertex but were unable to write the equation for the axis of symmetry. There was a greatdeal of success with the x and y intercepts.

(−1, − 12)

x = −1

(0, − 9)

f(x) = 0

3(x+ 3)(x− 1) = 0 x = −6± 36+108√6

(−3, 0) (1, 0)

93c. [2 marks]Hence sketch the graph of f .

Markscheme

A1A1 N2

Note: Award A1 for a parabola opening upward, A1 for vertex and intercepts in approximately correct positions.

[2 marks]

Examiners reportSome of the sketches of the graph left much to be desired even if they were technically correct; many were v-shaped.

93d. [3 marks]Let . The graph of f may be obtained from the graph of g by the two transformations:

a stretch of scale factor t in the y-direction

followed by a translation of .

Find and the value of t.

g(x) = x2

( )pq

( )pq

Markscheme , (accept , , ) A1A1A1 N3

[3 marks]

Examiners reportThe final part was poorly done, indicating that defining a graph in terms of stretch and translation was unfamiliar to many candidates.

( ) = ( )p

q

−1−12

t = 3 p = −1 q = −12 t = 3

94a. [3 marks]

Let , .


Markscheme

A1A1A1 N3

Note: Award A1 for passing through , A1 for correct shape, A1 for a range of approximately to 15.

[3 marks]

Examiners reportIn part (a), some did not realize that they should copy the curve from their GDC, paying attention to domain and range.

f(x) = 4 x− 4 sinxtan2 − ≤ x ≤π

3π

3

y = f(x)

(0, 0) −1

94b. [3 marks]Solve the equation .

Markschemeevidence of attempt to solve (M1)

e.g. line on sketch, using

, A1A1 N3

[3 marks]

Examiners reportNot using their GDC, and trying to solve the equation analytically in part (b) proved to be very difficult for many. A common errorwas to substitute .

f(x) = 1

f(x) = 1

tanx = sinxcos x

x = −0.207 x = 0.772

x = 1

95. [4 marks]

The following diagram shows the graphs of and , for .

There are two values of x for which the gradient of f is equal to the gradient of g. Find both these values of x.

Markschemeevidence of using derivatives for gradients (M1)

correct approach (A1)

e.g. , points of intersection

, A1A1 N2N2

[4 marks]

Examiners reportMost candidates realized the relationship between the gradient and the first derivative and set the two derivatives equal to one another.Once again many did not realize that the intersection could be easily found on their GDC.

f(x) = ln(3x− 2) + 1 g(x) = −4 cos(0.5x) + 2 1 ≤ x ≤ 10

(x) = (x)f ′ g′

x = 1.43 x = 6.10

( ) = − − 22

96a. [4 marks]

The following diagram shows part of the graph of f , where .

Find both x-intercepts.

Markschemeevidence of attempting to solve (M1)


e.g. ,

intercepts are and (accept , ) A1A1 N1N1

[4 marks]

Examiners reportThis question was consistently the best handled one on the entire paper.

f(x) = −x− 2x2

f(x) = 0

(x+ 1)(x− 2) 1± 9√2

(−1, 0) (2, 0) x = −1 x = 2

96b. [2 marks]Find the x-coordinate of the vertex.

Markschemeevidence of appropriate method (M1)

e.g. , , reference to symmetry

A1 N2

[2 marks]

Examiners reportThis question was consistently the best handled one on the entire paper.

=xv+x1 x2

2= −xv

b

2a

= 0.5xv

97a. [2 marks]

Part of the graph of a function f is shown in the diagram below.

On the same diagram sketch the graph of .

Markscheme

M1A1 N2

Note: Award M1 for evidence of reflection in x-axis, A1 for correct vertex and all intercepts approximately correct.

Examiners reportThis question was reasonably well done. Many recognized the graph of as a reflection in a horizontal line, but fewerrecognized the x-axis as the mirror line.

y = −f(x)

−f(x)

97b. [4 marks]Let .

(i) Find .

(ii) Describe fully the transformation that maps the graph of f to the graph of g.

g(x) = f(x+ 3)

g(−3)

Markscheme(i) (A1)

A1 N2

(ii) translation (accept shift, slide, etc.) of A1A1 N2

[4 marks]

Examiners reportA fair number gave , but did not carry through to . The majority of candidates recognized that moving thegraph of by 3 units to the left results in the graph of , but the language used to describe the transformation was often farfrom precise mathematically.

g(−3) = f(0)

f(0) = −1.5

( )−30

g(−3) = f(0) f(0) = −1.5f(x) g(x)

98. [1 mark]

Let .

Part of the graph of , for , is shown below. The x-coordinates of the local minimum and maximum points are r and srespectively.

Write down the equation of the horizontal asymptote.

Markscheme A1 N1

[1 mark]

Examiners reportFor part (b), the equation of the horizontal asymptote was commonly written as .

f(x) = (1 − )ex x2

y = f(x) −6 ≤ x ≤ 2

y = 0

x = 0

99a. [6 marks]

A city is concerned about pollution, and decides to look at the number of people using taxis. At the end of the year 2000, there were 280 taxisin the city. After n years the number of taxis, T, in the city is given by

(i) Find the number of taxis in the city at the end of 2005.

(ii) Find the year in which the number of taxis is double the number of taxis there were at the end of 2000.

T = 280 × .1.12n

Markscheme(i) (A1)

A1 N2

(ii) evidence of doubling (A1)

e.g. 560

setting up equation A1

e.g. ,

(A1)

in the year 2007 A1 N3

[6 marks]

Examiners reportA number of candidates found this question very accessible. In part (a), many correctly solved for n, but often incorrectly answeredwith the year 2006, thus misinterpreting that 6.12 years after the end of 2000 is in the year 2007.

n = 5

T = 280 × 1.125

T = 493

280 × = 5601.12n = 21.12n

n = 6.116…

99b. [6 marks]At the end of 2000 there were people in the city who used taxis.

After n years the number of people, P, in the city who used taxis is given by

(i) Find the value of P at the end of 2005, giving your answer to the nearest whole number.

(ii) After seven complete years, will the value of P be double its value at the end of 2000? Justify your answer.

Markscheme(i) (A1)

(A1)

A1 N3

(ii)

A1

not doubled A1 N0

valid reason for their answer R1

e.g.

[6 marks]

Examiners reportMany found correct values in part (b) and often justified their result by simply noting the value after seven years is less than 51200. Acommon alternative was to divide 46807 by 25600 and note that this ratio is less than two. There were still a good number ofcandidates who failed to provide any justification as instructed.

25600

P = .2560000

10 + 90e−0.1n

P = 256000010+90e−0.1(5)

P = 39635.993…

P = 39636

P = 256000010+90e−0.1(7)

P = 46806.997…

P < 51200

99c. [5 marks]Let R be the ratio of the number of people using taxis in the city to the number of taxis. The city will reduce the number oftaxis if .

(i) Find the value of R at the end of 2000.

(ii) After how many complete years will the city first reduce the number of taxis?

R < 70

Markscheme(i) correct value A2 N2

e.g. , 91.4,

(ii) setting up an inequality (accept an equation, or reversed inequality) M1

e.g. ,

finding the value (A1)

after 10 years A1 N2

[5 marks]

Examiners reportPart (c) proved more challenging to candidates. Many found the correct ratio for R, however few candidates then created a properequation or inequality by dividing the function for P by the function for T and setting this equal (or less) than 70. Such a function,although unfamiliar, can be solved using the graphing or solving features of the GDC. Many candidates chose a tabular approach butoften only wrote down one value of the table, such as , . What is essential is to include the two values betweenwhich the correct answer falls. Sufficient evidence would include , so that it is clear the value of has beensurpassed.

25600280

640 : 7

< 70P

T< 702560000

(10+90 )280×e−0.1n 1.12n

9.31…

n = 10 R = 68.3n = 9 R = 70.8 R = 70

100a. [2 marks]

Let .

Show that .

Markscheme A1

A1

AG N0

[2 marks]

Examiners report[N/A]

f(x) = 3(x+ 1 − 12)2

f(x) = 3 + 6x− 9x2

f(x) = 3( + 2x+ 1) − 12x2

= 3 + 6x+ 3 − 12x2

= 3 + 6x− 9x2

100b. [7 marks]For the graph of f

(i) write down the coordinates of the vertex;

(ii) write down the y-intercept;

(iii) find both x-intercepts.

Markscheme(i) vertex is A1A1 N2

(ii) , or A1 N1

(iii) evidence of solving M1

e.g. factorizing, formula

correct working A1

e.g. ,

, , or A1A1 N2

[7 marks]


(−1,−12)

y = −9 (0,−9)

f(x) = 0

3(x+ 3)(x− 1) = 0 x = −6± 36+108√6

x = −3 x = 1 (−3, 0), (1, 0)

100c. [3 marks]Hence sketch the graph of f .

Markscheme

A1A1A1 N3

Note: Award A1 for a parabola opening upward, A1 for vertex in approximately correct position, A1 for intercepts in approximatelycorrect positions. Scale and labelling not required.

[3 marks]


100d. [3 marks]Let . The graph of f may be obtained from the graph of g by the following two transformations

a stretch of scale factor t in the y-direction,

followed by a translation of .

Write down and the value of t .

Markscheme , A1A1A1 N3

[3 marks]

g(x) = x2

( )pq

( )pq

( ) = ( )p

q

−1−12

t = 3


101a. [3 marks]

Let , for .

Find the x-intercepts of the graph of f .

Markschemeintercepts when M1

(0.827, 0) (4.78, 0) (accept , ) A1A1 N3

[3 marks]


f(x) = 4x− − 3ex−2 0 ≤ x ≤ 5

f(x) = 0

x = 0.827 x = 4.78

101b. [3 marks]On the grid below, sketch the graph of f .

Markscheme

A1A1A1 N3

Note: Award A1 for maximum point in circle, A1 for x-intercepts in circles, A1 for correct shape (y approximately greater than ).

[3 marks]


−3.14

102a. [4 marks]

Let , .

Find .

Markscheme

interchanging x and y (seen anywhere) M1

e.g.

correct working A1

e.g.

collecting terms A1

e.g. ,

A1 N2

[4 marks]


h(x) = 2x−1x+1

x ≠ −1

(x)h−1

y = 2x−1x+1

x = 2y−1y+1

xy+x = 2y− 1

x+ 1 = 2y−xy x+ 1 = y(2 −x)

(x) =h−1 x+12−x

102b. [7 marks](i) Sketch the graph of h for and , including any asymptotes.

(ii) Write down the equations of the asymptotes.

(iii) Write down the x-intercept of the graph of h .

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct intercepts, A1 for correct shape, A1 for asymptotes, A1 for approximately correct domainand range.

(ii) , A1A1 N2

(iii) A1 N1

[7 marks]


−4 ≤ x ≤ 4 −5 ≤ y ≤ 8

x = −1 y = 2

12

103a. [5 marks]

A rock falls off the top of a cliff. Let be its height above ground in metres, after seconds.

The table below gives values of and .

Jane thinks that the function is a suitable model for the data. Use Jane’s model to

(i) write down the height of the cliff;

(ii) find the height of the rock after 4.5 seconds;

(iii) find after how many seconds the height of the rock is .

Markscheme(i) A1 N1

(ii) substitute M1

A1 N2

(iii) set up suitable equation M1

e.g.

A1 N1

[5 marks]

h t

h t

f(t) = −0.25 − 2.32 + 1.93t+ 106t3 t2

30 m

106 m

t = 4.5

h = 44.9 m

f(t) = 30

t = 4.91


103b. [3 marks]Kevin thinks that the function is a better model for the data. Use Kevin’s model to find when therock hits the ground.

Markschemerecognizing that height is 0 A1

set up suitable equation M1

e.g.

A1 N2

[3 marks]


g(t) = −5.2 + 9.5t+ 100t2

g(t) = 0

t = 5.39 secs

103c. [6 marks](i) On graph paper, using a scale of 1 cm to 1 second, and 1 cm to 10 m, plot the data given in the table.

(ii) By comparing the graphs of f and g with the plotted data, explain which function is a better model for the height of the falling rock.

Markscheme

A1A2 N3

Note: Award A1 for correct scales on axes, A2 for 5 correct points, A1 for 3 or 4 correct points.

(ii) Jane’s function, with 2 valid reasons A1R1R1 N3

e.g. Jane’s passes very close to all the points, Kevin’s has the rock clearly going up initially – not possible if rock falls

Note: Although Jane’s also goes up initially, it only goes up very slightly, and so is the better model.

[6 marks]


104a. [2 marks]

The line L passes through the point and is parallel to the vector .

Write down a vector equation for line L .

Markschemeany correct equation in the form (accept any parameter for t)

where a is , and b is a scalar multiple of A2 N2

e.g.

Note: Award A1 for the form , A1 for , A0 for .

[2 marks]

Examiners reportIn part (a), the majority of candidates correctly recognized the equation that contains the position and direction vectors of a line.However, we saw a large number of candidates who continue to write their equations using " ", rather than the mathematically

correct " " or " ". r and represent vectors, whereas L is simply the name of the line. For part (b), very few

candidates recognized that a general point on the x-axis will be given by the vector . Common errors included candidates

setting their equation equal to , or , or even just the number .

(5, −4, 10)⎛⎝⎜

4−25

⎞⎠⎟

r = a+ tb

⎛⎝⎜

5−410

⎞⎠⎟

⎛⎝⎜

4−25

⎞⎠⎟

r = + t , r = 5i− 4j+ 10k+ t(−8i+ 4j− 10k)⎛⎝⎜

5−410

⎞⎠⎟

⎛⎝⎜

4−25

⎞⎠⎟a+ tb L = a+ tb r = b+ ta

L =

r = =⎛⎝⎜x

y

z

⎞⎠⎟

⎛⎝⎜x

y

z

⎞⎠⎟

⎛⎝⎜x

00

⎞⎠⎟

⎛⎝⎜

000

⎞⎠⎟

⎛⎝⎜

100

⎞⎠⎟ 0

104b. [6 marks]The line L intersects the x-axis at the point P. Find the x-coordinate of P.

Markschemerecognizing that or at x-intercept (seen anywhere) (R1)

attempt to set up equation for x-intercept (must suggest ) (M1)

e.g. , ,

one correct equation in one variable (A1)

e.g. ,

finding A1


e.g.

(accept ) A1 N3

[6 marks]

Examiners reportIn part (a), the majority of candidates correctly recognized the equation that contains the position and direction vectors of a line.However, we saw a large number of candidates who continue to write their equations using " ", rather than the mathematically

correct " " or " ". r and represent vectors, whereas L is simply the name of the line. For part (b), very few

candidates recognized that a general point on the x-axis will be given by the vector . Common errors included candidates

setting their equation equal to , or , or even just the number .

y = 0 z= 0

x ≠ 0

L =⎛⎝⎜x

00

⎞⎠⎟ 5 + 4t = x r =

⎛⎝⎜

100

⎞⎠⎟

−4 − 2t = 0 10 + 5t = 0

t = −2

x = 5 + (−2)(4)

x = −3 (−3, 0, 0)

L =

r = =⎛⎝⎜x

y

z

⎞⎠⎟

⎛⎝⎜x

y

z

⎞⎠⎟

⎛⎝⎜x

00

⎞⎠⎟

⎛⎝⎜

000

⎞⎠⎟

⎛⎝⎜

100

⎞⎠⎟ 0

105a. [1 mark]

Let A and B be points such that and .

Show that .

=OA−→− ⎛

⎝⎜521

⎞⎠⎟ =OB

−→ ⎛⎝⎜

603

⎞⎠⎟

=AB−→ ⎛

⎝⎜1

−22

⎞⎠⎟

Markschemecorrect approach A1

e.g.

AG N0

[1 mark]

Examiners reportPart (a) was answered correctly by nearly every candidate.

+ , −AO−→−

OB−→ ⎛

⎝⎜603

⎞⎠⎟

⎛⎝⎜

521

⎞⎠⎟

=AB−→ ⎛

⎝⎜1

−22

⎞⎠⎟

105b. [4 marks]Let C and D be points such that ABCD is a rectangle.

Given that , show that .

Markschemerecognizing is perpendicular to (may be seen in sketch) (R1)

e.g. adjacent sides of rectangle are perpendicular

recognizing dot product must be zero (R1)

e.g.


e.g. ,

equation which clearly leads to A1

e.g. ,

AG N0

[4 marks]

Examiners reportIn part (b), the candidates who realized that the vectors must be perpendicular were successful using the scalar product to find p .Incorrect approaches included using magnitudes, or creating vector equations of lines for the sides and setting them equal to eachother. In addition, there were a good number of candidates who worked backwards, using the given value of 3 for p to find thecoordinates of point D. Candidates who work backwards on a "show that" question will earn no marks.

=AD−→− ⎛

⎝⎜4p

1

⎞⎠⎟ p = 3

AD−→−

AB−→

∙ = 0AD−→−

AB−→−

(1 × 4) + (−2 × p) + (2 × 1) 4 − 2p+ 2 = 0

p = 3

6 − 2p = 0 2p = 6

p = 3

105c. [4 marks]Let C and D be points such that ABCD is a rectangle.

Find the coordinates of point C.

Markschemecorrect approach (seen anywhere including sketch) (A1)

e.g. ,

recognizing opposite sides are equal vectors (may be seen in sketch) (R1)

e.g. , , ,

coordinates of point C are (10, 3, 4) (accept ) A2 N4

Note: Award A1 for two correct values.

[4 marks]

Examiners reportPart (c) was more difficult for candidates, and was left blank by some. Some candidates found rather than , as required.Many candidates recognized that the opposite sides of the rectangle must be equal, but did not consider the directions of the vectorsfor those sides. There were also a good number of candidates who mislabelled the vertices of their rectangles, which led to themworking with a rectangle ABDC, rather than ABCD.

= +OC−→−

OB−→

BC−→

+OD−→−

DC−→−

=BC−→

AD−→−

=DC−→−

AB−→

+⎛⎝⎜

603

⎞⎠⎟

⎛⎝⎜

431

⎞⎠⎟ +

⎛⎝⎜

952

⎞⎠⎟

⎛⎝⎜

1−22

⎞⎠⎟

⎛⎝⎜

1034

⎞⎠⎟

AC−→−

OC−→−

105d. [5 marks]Let C and D be points such that ABCD is a rectangle.

Find the area of rectangle ABCD.

Markschemeattempt to find one side of the rectangle (M1)

e.g. substituting into magnitude formula

two correct magnitudes A1A1

e.g. , 3 ; ,

multiplying magnitudes (M1)

e.g.

(accept ) A1 N3

[5 marks]

Examiners reportThe majority of candidates who attempted part (d) were successful in multiplying the magnitudes of the sides. Unfortunately, therewere some who set up their solutions correctly, then had arithmetic errors in their working.

+ +(1)2 (−2)2 22− −−−−−−−−−−−−√ 16 + 9 + 1− −−−−−−−√ 26−−√

×26−−√ 9√

area = (= 3 )234−−−√ 26−−√ 3 × 26−−√

−→− −→− −→−

106a. [5 marks]

The following diagram shows the cuboid (rectangular solid) OABCDEFG, where O is the origin, and , , .

(i) Find .

(ii) Find .

(iii) Show that .

Markscheme(i) valid approach (M1)

e.g.

A1 N2

(ii) valid approach (M1)

e.g. ; ;

A1 N2

(iii) correct approach A1

e.g. ; ;

AG N0

[5 marks]

Examiners reportAlthough a large proportion of candidates managed to answer this question, their biggest challenge was the use of a proper notation torepresent the vectors and vector equations of lines.

In part (a), finding and was generally well done, although many lost the mark for (iii) due to poor working or not clearlyshowing the result.

= 4iOA−→−

= 3jOC−→−

= 2kOD−→−

OB−→

OF−→

= −4i+ 3j+ 2kAG−→−

OA + OB

= 4i+ 3jOB−→

+ +OA−→−

AB−→

BF−→

+OB−→

BF−→

+ +OC−→−

CG−→−

GF−→

= 4i+ 3j+ 2kOF−→

+ +AO−→−

OC−→−

CG−→−

+ +AB−→

BF−→

FG−→

+ +AB−→

BC−→

CG−→−

= −4i+ 3j+ 2kAG−→−

OB−→

OF−→

106b. [4 marks]Write down a vector equation for

(i) the line OF;

(ii) the line AG.

Markscheme(i) any correct equation for (OF) in the form A2 N2

where is 0 or , and is a scalar multiple of

e.g. , ,

(ii) any correct equation for (AG) in the form A2 N2

where is or and is a scalar multiple of

e.g. , ,

[4 marks]

Examiners reportPart (b) was very poorly done. Not all the students recognized which correct position vectors they had to use to write the equations ofthe lines. It was seen that they frequently failed to present the equations in the required format, which prevented these candidates fromachieving full marks. The notations generally seen were , or .

r = a+ tb

a 4i+ 3j+ 2k b 4i+ 3j+ 2k

r = t(4, 3, 2) r =⎛⎝⎜

4t3t2t

⎞⎠⎟ r = 4i+ 3j+ 2k+ t(4i+ 3j+ 2k)

r = a+ sb

a 4i 3j+ 2k b −4i+ 3j+ 2k

r = (4, 0, 0) + s(−4, 3, 2) r =⎛⎝⎜

4 − 4s3s2s

⎞⎠⎟ r = 3j+ 2k+ s(−4i+ 3j+ 2k)

AG = a+ bt r = 4 + t(4, 3, 2) L = a+ bt

106c. [7 marks]Find the obtuse angle between the lines OF and AG.

Markschemechoosing correct direction vectors, and (A1)(A1)

scalar product (A1)

magnitudes , , (A1)(A1)

substitution into formula M1

e.g.

,

or A1 N4

[7 marks]

Examiners reportMost achieved the correct result in part (c) with many others gaining most of the marks as follow through from choosing incorrectvectors. Some students did not state which vectors had been used, another cause for losing marks. A few showed poor notation,including i, j and k in the working.

OF−→

AG−→−

= −16 + 9 + 4 (= −3)

+ +42 32 22− −−−−−−−−√ + +(−4)2 32 22− −−−−−−−−−−−√ ( , )29−−√ 29−−√

cosθ = = (− )−16+9+4

( )×+ +42 32 22√ + +(−4)2 32 22√3

29

95.93777∘ 1.67443 radians

θ = 95.9∘ 1.67

107a. [3 marks]

A line passes though points P(−1, 6, −1) and Q(0, 4, 1) .

(i) Show that .

(ii) Hence, write down an equation for in the form .

L1

=PQ−→ ⎛

⎝⎜1

−22

⎞⎠⎟

L1 r = a+ tb

Markscheme(i) evidence of correct approach A1

e.g. ,

AG N0

(ii) any correct equation in the form A2 N2

where a is either or and b is a scalar multiple of

e.g. , ,

[3 marks]

Examiners reportA pleasing number of candidates were successful on this straightforward vector and line question. Part (a) was generally wellanswered, although a few candidates still labelled their line or used a position vector for the direction vector. Follow-throughmarking allowed full recovery from the latter error.

= −PQ−→

OQ−→−

OP−→

Q−P

=PQ−→ ⎛

⎝⎜1

−22

⎞⎠⎟

r = a+ tb

OP−→

OQ−→−

PQ−→

r = + t⎛⎝⎜

−16

−1

⎞⎠⎟

⎛⎝⎜

1−22

⎞⎠⎟ r =

⎛⎝⎜

t

4 − 2t1 + 2t

⎞⎠⎟ r = 4j+ k+ t(i− 2j+ 2k)

L =

107b. [7 marks]

A second line has equation .

Find the cosine of the angle between and .

Markschemechoosing a correct direction vector for (A1)

e.g.

finding scalar products and magnitudes (A1)(A1)(A1)

scalar product

magnitudes ,


e.g.

A2 N5

[7 marks]

Examiners reportFew candidates wrote down their direction vector in part (b) which led to lost follow-through marks, and a common error was findingan incorrect scalar product due to difficulty multiplying by zero.

L2 r = + s⎛⎝⎜

42

−1

⎞⎠⎟

⎛⎝⎜

30

−4

⎞⎠⎟

PQ−→

L2

L2

⎛⎝⎜

30

−4

⎞⎠⎟

= 1(3) − 2(0) + 2(−4) (= −5)

= + +12 (−2)2 22− −−−−−−−−−−−√ (= 3) + +32 02 (−4)2

− −−−−−−−−−−−√ (= 5)

cosθ = −5×9√ 25√

cosθ = − 13

107c. [7 marks]The lines and intersect at the point R. Find the coordinates of R.L1 L2


e.g. equating lines,

EITHER

one correct equation in one variable A2

e.g.

OR

two correct equations in two variables A1A1

e.g. ,

THEN


one correct parameter A1

e.g. ,

correct substitution of either parameter (A1)

e.g. ,

coordinates A1 N3

[7 marks]

Examiners reportPart (c) was generally well understood with some candidates realizing that the equation in just one variable led to the correctparameter more quickly than solving a system of two equations to find both parameters. Some candidates gave the answer as (s, t)instead of substituting those parameters, indicating a more rote understanding of the problem. Another common error was using thesame parameter for both lines.

There were an alarming number of misreads of negative signs from the question or from the candidate working.

=L1 L2

6 − 2t = 2

2t+ 4s = 0 t− 3s = 5

t = 2 s = −1

r = + (−1)⎛⎝⎜

42

−1

⎞⎠⎟

⎛⎝⎜

30

−4

⎞⎠⎟ r = + (+2)

⎛⎝⎜

−16

−1

⎞⎠⎟

⎛⎝⎜

1−22

⎞⎠⎟

R(1, 2, 3)

108a. [4 marks]

The line passes through the points P(2, 4, 8) and Q(4, 5, 4) .

(i) Find .

(ii) Hence write down a vector equation for in the form .

L1

PQ−→

L1 r = a+ sb

Markscheme(i) evidence of approach (M1)

e.g. ,

A1 N2

(ii) any correct equation in the form (accept any parameter for s)

where a is or , and b is a scalar multiple of A2 N2

e.g. , ,

Note: Award A1 for the form , A1 for , A0 for .

[4 marks]

Examiners reportIn part (a), nearly all the candidates correctly found the vector PQ, and the majority went onto find the correct vector equation of theline. There are still many candidates who do not write this equation in the correct form, using "r = ", and these candidates werepenalized one mark.

+PO−→

OQ−→−

P − Q

=PQ−→ ⎛

⎝⎜21

−4

⎞⎠⎟

r = a+ sb

⎛⎝⎜

248

⎞⎠⎟

⎛⎝⎜

454

⎞⎠⎟

⎛⎝⎜

21

−4

⎞⎠⎟

r = + s⎛⎝⎜

248

⎞⎠⎟

⎛⎝⎜

21

−4

⎞⎠⎟ r =

⎛⎝⎜

4 + 2s5 + 1s4 − 4s

⎞⎠⎟ r = 2i+ 4j+ 8k+ s(2i+ 1j− 4k)

a+ sb L = a+ sb r = b+ sa

108b. [7 marks]

The line is perpendicular to , and parallel to , where .

(i) Find the value of p .

(ii) Given that passes through , write down a vector equation for .

Markscheme(i) choosing correct direction vectors for and (A1) (A1)

e.g. ,

evidence of equating scalar product to 0 (M1)

correct calculation of scalar product A1

e.g. ,

A1 N3

(ii) any correct expression in the form (accept any parameter for t)

where a is , and b is a scalar multiple of A2 N2

e.g. , ,

Note: Award A1 for the form , A1 for (unless they have been penalised for in part (a)), A0 for .

[7 marks]

L2 L1

⎛⎝⎜

3p2p4

⎞⎠⎟ p ∈ Z

L2 R(10, 6, − 40) L2

L1 L2

⎛⎝⎜

21

−4

⎞⎠⎟

⎛⎝⎜

3p2p4

⎞⎠⎟

2 × 3p+ 1 × 2p+ (−4) × 4 8p− 16 = 0

p = 2

r = a+ tb

⎛⎝⎜

106

−40

⎞⎠⎟

⎛⎝⎜

644

⎞⎠⎟

r = + t⎛⎝⎜

106

−40

⎞⎠⎟

⎛⎝⎜

644

⎞⎠⎟ r =

⎛⎝⎜

10 + 6s6 + 4s

−40 + 4s

⎞⎠⎟ r = 10i+ 6j− 40k+ s(6i+ 4j+ 4k)

a+ tb L = a+ tb L = a+ sbr = b+ ta

Examiners reportIn part (b), the majority of candidates knew to set the scalar product equal to zero for the perpendicular vectors, and were able to findthe correct value of p.

108c. [7 marks]The lines and intersect at the point A. Find the x-coordinate of A.


e.g.

any two correct equations with different parameters A1A1

e.g. , ,

attempt to solve simultaneous equations (M1)


e.g. , , ,

one correct parameter , A1

(accept (22, 14, −32)) A1 N4

[7 marks]

Examiners reportA good number of candidates used the correct method to find the intersection of the two lines, though some algebraic and arithmeticerrors kept some from finding the correct final answer.

L1 L2

+ s = + t⎛⎝⎜

248

⎞⎠⎟

⎛⎝⎜

21

−4

⎞⎠⎟

⎛⎝⎜

106

−40

⎞⎠⎟

⎛⎝⎜

644

⎞⎠⎟

2 + 2s = 10 + 6t 4 + s = 6 + 4t 8 − 4s = −40 + 4t

−6 = −2 − 2t 4 = 2t −4 + 5s = 46 5s = 50

s = 10 t = 2

x = 22

109a. [2 marks]

A line L passes through and is parallel to the line .

Write down a vector equation for L in the form .

Markschemecorrect equation in the form A2 N2

[2 marks]

Examiners reportMany candidates answered this question well. Some continue to write the vector equation in (a) using "L =", which does not earn fullmarks.

A(1, − 1, 2) r = + s⎛⎝⎜

−215

⎞⎠⎟

⎛⎝⎜

13

−2

⎞⎠⎟

r = a+ tb

r = a+ tb

r = + t⎛⎝⎜

1−12

⎞⎠⎟

⎛⎝⎜

13

−2

⎞⎠⎟

109b. [4 marks]

The line L passes through point P when .

Find

(i) ;

(ii) .

Markscheme(i) attempt to substitute into the equation (M1)

e.g. ,

A1 N2

(ii) correct substitution into formula for magnitude A1

e.g. ,

A1 N1

[4 marks]

Examiners reportPart (b) proved accessible for most, although small arithmetic errors were not uncommon. Some candidates substituted into the

original equation, and a few answered . A small but surprising number of candidates left this question blank,

suggesting the topic was not given adequate attention in course preparation.

t = 2

OP−→

| |OP−→

t = 2

⎛⎝⎜

26

−4

⎞⎠⎟ + 2

⎛⎝⎜

1−12

⎞⎠⎟

⎛⎝⎜

13

−2

⎞⎠⎟

=OP−→ ⎛

⎝⎜35

−2

⎞⎠⎟

+ + −32 52 22− −−−−−−−−−√ + +32 52 22− −−−−−−−−√

| | =OP−→

38−−√

t = 2

=OP−→ ⎛

⎝⎜26

−4

⎞⎠⎟

110a. [3 marks]

The following diagram shows the obtuse-angled triangle ABC such that and .

(i) Write down .

(ii) Find .

=AB−→ ⎛

⎝⎜−30

−4

⎞⎠⎟ =AC

−→− ⎛⎝⎜

−22

−6

⎞⎠⎟

BA−→

BC−→

Markscheme

(i) A1 N1

(ii) evidence of combining vectors (M1)

e.g. , ,

A1 N2

[3 marks]

Examiners reportMany candidates answered (a) correctly, although some reversed the vectors when finding , while others miscopied the vectorsfrom the question paper.

=BA−→ ⎛

⎝⎜304

⎞⎠⎟

+ =AB−→

BC−→

AC−→−

+BA−→

AC−→−

−⎛⎝⎜

−22

−6

⎞⎠⎟

⎛⎝⎜

−30

−4

⎞⎠⎟

=BC−→ ⎛

⎝⎜12

−2

⎞⎠⎟

BC−→

110b. [7 marks](i) Find .

(ii) Hence, find .

Markscheme(i) METHOD 1

finding , ,

e.g. , ,

substituting into formula for M1

e.g. ,

A1 N3

METHOD 2

finding , , (A1)(A1)(A1)

e.g. , ,

substituting into cosine rule M1

e.g. ,

A1 N3

(ii) evidence of using Pythagoras (M1)

e.g. right-angled triangle with values,

A1 N2

[7 marks]

Examiners reportStudents had no difficulty finding the scalar product and magnitudes of the vectors used in finding the cosine. However, fewrecognized that is the vector to apply in the formula to find the cosine value. Most used to obtain a positive cosine, whichneglects that the angle is obtuse and thus has a negative cosine. Surprisingly few students could then take a value for cosine and use itto find a value for sine. Most left (bii) blank entirely.

cosA CB

sinA CB

∙BA−→

BC−→

| |BA−→

| |BC−→

∙ = 3 × 1 + 0 + 4 × −2BA−→

BC−→

| =|BA−→−

+32 42− −−−−√ | = 3|BC−→−

cosθ

3×1+0+4×−2

3 +0+32 42√

−55×3

cosA C = −B 515

(= − )13

| |AC−→−

| |BA−→

| |BC−→

| | =AC−→−

+ +22 22 62− −−−−−−−−√ | | =BA−→

+32 42− −−−−√ | | = 3BC−→

+ −52 32 ( )44√ 2

2×5×325+9−44

30

cosA C = −B 1030

(= − )13

x+ x = 1sin2 cos2

sin A C =B 8√3

(= )2 2√3

BA−→

AB−→

110c. [6 marks]The point D is such that , where .

(i) Given that , show that .

(ii) Hence, show that is perpendicular to .

Markscheme(i) attempt to find an expression for (M1)

e.g. ,

correct equation A1

e.g. ,

A1

AG N0

(ii) evidence of scalar product (M1)

e.g. ,

correct substitution

e.g. , A1

A1

is perpendicular to AG N0

[6 marks]

Examiners reportPart (c) proved accessible for many candidates. Some created an expression for and then substituted the given to obtain

, which does not satisfy the "show that" instruction. Many students recognized that the scalar product must be zero for vectors tobe perpendicular, and most provided the supporting calculations.

=CD−→− ⎛

⎝⎜−45p

⎞⎠⎟ p > 0

=|CD|− →−

50−−√ p = 3

CD−→−

BC−→

| |CD−→−

+ +(−4)2 52 p2− −−−−−−−−−−−√ | = + +CD

−→−|2 42 52 p2

=+ +(−4)2 52 p2− −−−−−−−−−−−√ 50−−√ + + = 5042 52 p2

= 9p2

p = 3

∙⎛⎝⎜

−453

⎞⎠⎟

⎛⎝⎜

12

−2

⎞⎠⎟ ∙CD

−→−BC−→

−4 × 1 + 5 × 2 + 3 × −2 −4 + 10 − 6

∙ = 0CD−→−

BC−→

CD−→−

BC−→

| |CD−→−

p = 350−−√

111a. [2 marks]

The following diagram shows quadrilateral ABCD, with , , and .

Find .

=AD−→−

BC−→

= ( )AB−→ 3

1= ( )AC

−→− 44

BC−→


e.g. ,

A1 N2

[2 marks]

Examiners reportThis question on two-dimensional vectors was generally very well done.

−AC−→−

AB−→ ( )4 − 3

4 − 1

= ( )BC−→ 1

3

111b. [2 marks]Show that .

MarkschemeMETHOD 1

(A1)

correct approach A1

e.g. ,

AG N0

METHOD 2

recognizing (A1)

correct approach A1

e.g. ,

AG N0

[2 marks]

Examiners reportThis question on two-dimensional vectors was generally very well done. A very small number of candidates had trouble with the"show that" in part (b) of the question.

= ( )BD−→ −2

2

= ( )AD−→− 1

3

−AD−→−

AB−→ ( )1 − 3

3 − 1

= ( )BD−→ −2

2

=CD−→−

BA−→

+BC−→

CD−→− ( )1 − 3

3 − 1

= ( )BD−→ −2

2

111c. [3 marks]Show that vectors and are perpendicular.BD−→

AC−→−

MarkschemeMETHOD 1

evidence of scalar product (M1)

e.g. ,


e.g. ,

A1

therefore vectors and are perpendicular AG N0

METHOD 2

attempt to find angle between two vectors (M1)

e.g.


e.g. ,

A1

therefore vectors and are perpendicular AG N0

[3 marks]

Examiners reportNearly all candidates knew to use the scalar product in part (c) to show that the vectors are perpendicular.

∙BD−→

AC−→− ( ) ∙ ( )−2

244

(−2)(4) + (2)(4) −8 + 8

∙ = 0BD−→

AC−→−

BD−→

AC−→−

a∙bab

(−2)(4)+(2)(4)

8√ 32√cosθ = 0

θ = 90∘

BD−→

AC−→−

112. [6 marks]

Consider the vectors and .

(a) Find

(i) ;

(ii) .

Let , where is the zero vector.

(b) Find .

a= ( )2−3

b = ( )14

2a+ b

|2a+ b|

2a+ b+ c = 0 0

c

Markscheme(a) (i) (A1)

correct expression for A1 N2

eg , ,

(ii) correct substitution into length formula (A1)

eg ,

A1 N2

[4 marks]

(b) valid approach (M1)

eg , ,

A1 N2

[2 marks]

Examiners reportMost candidates comfortably applied algebraic techniques to find new vectors. However, a significant number of candidatesanswered part (b) as the absolute numerical value of the vector components, which suggests a misunderstanding of the modulusnotation. Those who understood the notation easily made the calculation.

2a= ( )4−6

2a+ b

( )5−2

(5,−2) 5i− 2j

+52 22− −−−−√ + −52 22− −−−−−−√

|2a+ b| = 29−−√

c = −(2a+ b) 5 +x = 0 −2 + y = 0

c = ( )−52

113a. [2 marks]

Consider points A( , , ) , B( , , ) and C( , , ) . The line passes through C and is parallel to .

Find .


eg , ,

A1 N2

[2 marks]

Examiners reportWhile many candidates can find a vector given two points, few could write down a fully correct vector equation of a line.

1 −2 −1 7 −4 3 1 −2 3 L1 AB−→

AB−→

−⎛⎝⎜

7−43

⎞⎠⎟

⎛⎝⎜

1−2−1

⎞⎠⎟ A − B = +AB

−→AO−→−

OB−→

=AB−→ ⎛

⎝⎜6

−24

⎞⎠⎟

113b. [2 marks]Hence, write down a vector equation for .L1

Markschemeany correct equation in the form (accept any parameter for )

where and is a scalar multiple of A2 N2

eg , ,


[2 marks]

Examiners reportWhile many candidates can find a vector given two points, few could write down a fully correct vector equation of a line. Mostcandidates wrote their equation as “ ”, which misrepresents that the resulting equation must still be a vector.

r = a+ tb t

a=⎛⎝⎜

1−23

⎞⎠⎟ b AB

−→

r = + t⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

6−24

⎞⎠⎟ (x,y,z) = (1,−2,3) + t(3,−1,2) r =

⎛⎝⎜

1 + 6t−2 − 2t3 + 4t

⎞⎠⎟

a+ tb = a+ tbL1 r = b+ ta

=L1

113c. [3 marks]

A second line, , is given by .

Given that is perpendicular to , show that .

Markschemerecognizing that scalar product (seen anywhere) R1

correct calculation of scalar product (A1)

eg ,

correct working A1

eg ,

AG N0

[3 marks]

Examiners reportThose who recognized that vector perpendicularity means the scalar product is zero found little difficulty answering part (b).Occasionally a candidate would use the given to show the scalar product is zero. However, working backward from the givenanswer earns no marks in a question that requires candidates to show that this value is achieved.

L2 r = + s⎛⎝⎜

−12

15

⎞⎠⎟

⎛⎝⎜

3−3p

⎞⎠⎟

L1 L2 p = −6

= 0

6(3) − 2(−3) + 4p 18 + 6 + 4p

24 + 4p = 0 4p = −24

p = −6

p = 6

113d. [7 marks]The line intersects the line at point Q. Find the -coordinate of Q.L1 L2 x

Markschemesetting lines equal (M1)

eg ,

any two correct equations with different parameters A1A1

eg , ,

attempt to solve their simultaneous equations (M1)


eg ,

attempt to substitute parameter into vector equation (M1)

eg ,

(accept (4, -3, 5), ignore incorrect values for and ) A1 N3

[7 marks]

Examiners reportWhile many candidates knew to set the lines equal to find an intersection point, a surprising number could not carry the process tocorrect completion. Some could not solve a simultaneous pair of equations, and for those who did, some did not know what to dowith the parameter value. Another common error was to set the vector equations equal using the same parameter, from which thecandidates did not recognize a system to solve. Furthermore, it is interesting to note that while only one parameter value is needed toanswer the question, most candidates find or attempt to find both, presumably out of habit in the algorithm.

=L1 L2 + t = + s⎛⎝⎜

1−23

⎞⎠⎟

⎛⎝⎜

6−24

⎞⎠⎟

⎛⎝⎜

−12

15

⎞⎠⎟

⎛⎝⎜

3−3−6

⎞⎠⎟

1 + 6t = 1 + 3s −2 − 2t = 2 − 3s 3 + 4t = 15 − 6s

t = 12s = 5

3

+⎛⎝⎜

1−23

⎞⎠⎟ 1

2

⎛⎝⎜

6−24

⎞⎠⎟ 1 + × 61

2

x = 4 y z

114. [7 marks]Line has equation and line has equation .

Lines and intersect at point A. Find the coordinates of A.


eg ,

any two correct equations A1A1

eg , ,


eg substituting one equation into another


eg ,


eg , ,

A ( , , ) (accept column vector) A1 N4

[7 marks]

L1 = + sr1

⎛⎝⎜

106

−1

⎞⎠⎟

⎛⎝⎜

2−5−2

⎞⎠⎟ L2 = + tr2

⎛⎝⎜

21

−3

⎞⎠⎟

⎛⎝⎜

352

⎞⎠⎟

L1 L2

+ s = + t⎛⎝⎜

106

−1

⎞⎠⎟

⎛⎝⎜

2−5−2

⎞⎠⎟

⎛⎝⎜

21

−3

⎞⎠⎟

⎛⎝⎜

352

⎞⎠⎟ =L1 L2

10 + 2s = 2 + 3t 6 − 5s = 1 + 5t −1 − 2s = −3 + 2t

s = −1 t = 2

2 + 3(2) 1 + 5(2) −3 + 2(2)

= 8 11 1

Examiners reportMost students were able to set up one or more equations, but few chose to use their GDCs to solve the resulting system. Algebraicerrors prevented many of these candidates from obtaining the final three marks. Some candidates stopped after finding the value of sand/or .t

115a. [5 marks]

The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) and D(4, −1) .

(i) Show that .

(ii) Find .

(iii) Show that is perpendicular to .

Markscheme(i) correct approach A1

e.g. ,

AG N0

(ii) appropriate approach (M1)

e.g. , , move 3 to the right and 6 down

A1 N2

(iii) finding the scalar product A1

e.g. ,

valid reasoning R1

e.g. , scalar product is zero

is perpendicular to AG N0

[5 marks]

= ( )AC−→− 4

2

BD−→

AC−→−

BD−→

−OC−→−

OA−→− ( ) − ( )5

210

= ( )AC−→− 4

2

D − B ( ) − ( )4−1

15

= ( )BD−→ 3

−6

4(3) + 2(−6) 12 − 12

4(3) + 2(−6) = 0

AC−→−

BD−→

Examiners reportThe majority of candidates were successful on part (a), finding vectors between two points and using the scalar product to show twovectors to be perpendicular.

115b. [4 marks]The line (AC) has equation .

(i) Write down vector u and vector v .

(ii) Find a vector equation for the line (BD).

Markscheme(i) correct “position” vector for u; “direction” vector for v A1A1 N2

e.g. , ; ,

accept in equation e.g.

(ii) any correct equation in the form , where

, A2 N2

[4 marks]

Examiners reportAlthough a large number of candidates answered part (b) correctly, there were many who had trouble with the vector equation of aline. Most notably, there were those who confused the position vector with the direction vector, and those who wrote their equation inan incorrect form.

r = u+ sv

u = ( )52

u = ( )10

v = ( )42

v = ( )−2−1

( ) + t( )52

−4−2

r = a+ tb b = BD−→

r = ( ) + t( )15

3−6

( ) = ( ) + t( )x

y

4−1

−12

115c. [3 marks]The lines (AC) and (BD) intersect at the point .

Show that .

P(3, k)

k = 1

MarkschemeMETHOD 1

substitute (3, k) into equation for (AC) or (BD) (M1)

e.g. ,

value of t or s A1

e.g. , , ,

substituting A1

e.g.

AG N0

METHOD 2

setting up two equations (M1)

e.g. , ; setting vector equations of lines equal

value of t or s A1

e.g. , , ,

substituting A1

e.g.

AG N0

[3 marks]

Examiners reportIn part (c), most candidates seemed to know what was required, though there were many who made algebraic errors when solving forthe parameters. A few candidates worked backward, using , which is not allowed on a "show that" question.

3 = 1 + 4s 3 = 1 + 3t

s = 12

− 12t = 2

3− 1

3

k = 0 + (2)12

k = 1

1 + 4s = 4 + 3t 2s = −1 − 6t

s = 12

− 12t = 2

3− 1

3

r = ( ) − ( )4−1

13

3−6

k = 1

k = 1

115d. [5 marks]The lines (AC) and (BD) intersect at the point .

Hence find the area of triangle ACD.

Markscheme (A1)

(A1)

(A1)

area ( ) M1

A1 N4

[5 marks]

Examiners reportIn part (d), candidates attempted many different geometric and vector methods to find the area of the triangle. As the question said"hence", it was required that candidates should use answers from their previous working - i.e. and . Somegeometric approaches, while leading to the correct answer, did not use "hence" or lacked the required justification.

P(3, k)

= ( )PD−→ 1

−2

| | =PD−→

+22 12− −−−−√ (= )5√

| | =AC−→−

+42 22− −−−−√ (= )20−−√

= × | | × | |12

AC−→−

PD−→

= × ×12

20−−√ 5√

= 5

AC⊥BD P(3, 1)

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

116a. [2 marks]

The line is represented by the vector equation .

A second line is parallel to and passes through the point B( , , ) .

Write down a vector equation for in the form .

Markschemeany correct equation in the form (accept any parameter) A2 N2

e.g.


[2 marks]

Examiners reportMany candidates gave a correct vector equation for the line.

L1 r = + p⎛⎝⎜

−3−1

−25

⎞⎠⎟

⎛⎝⎜

21

−8

⎞⎠⎟

L2 L1 −8 −5 25

L2 r = a+ tb

r = a+ tb

r = + t⎛⎝⎜

−8−525

⎞⎠⎟

⎛⎝⎜

21

−8

⎞⎠⎟

a+ tb L = a+ tb r = b+ ta

116b. [5 marks]A third line is perpendicular to and is represented by .

Show that .

Markschemerecognizing scalar product must be zero (seen anywhere) R1

e.g.

evidence of choosing direction vectors (A1)(A1)

correct calculation of scalar product (A1)

e.g.

simplification that clearly leads to solution A1

e.g. ,

AG N0

[5 marks]

Examiners reportA common error was to misplace the initial position and direction vectors. Those who set the scalar product of the direction vectors tozero typically solved for k successfully. Those who substituted earned fewer marks for working backwards in a "show that"question.

L3 L1 r = + q⎛⎝⎜

503

⎞⎠⎟

⎛⎝⎜

−7−2k

⎞⎠⎟

k = −2

a ∙ b = 0

,⎛⎝⎜

21

−8

⎞⎠⎟

⎛⎝⎜

−7−2k

⎞⎠⎟

2(−7) + 1(−2) − 8k

−16 − 8k −16 − 8k = 0

k = −2

k = −2

116c. [6 marks]The lines and intersect at the point A.

Find the coordinates of A.

L1 L3

Markschemeevidence of equating vectors (M1)

e.g. ,

any two correct equations A1A1

e.g. , ,

attempting to solve equations (M1)

finding one correct parameter ( , ) A1

the coordinates of A are A1 N3

[6 marks]

Examiners reportMany went on to find the coordinates of point A, however some used the same letter, say p, for each parameter and thus could notsolve the system.

=L1 L3 + p = + q⎛⎝⎜

−3−1

−25

⎞⎠⎟

⎛⎝⎜

21

−8

⎞⎠⎟

⎛⎝⎜

503

⎞⎠⎟

⎛⎝⎜

−7−2−2

⎞⎠⎟

−3 + 2p = 5 − 7q −1 + p = −2q −25 − 8p = 3 − 2q

p = −3 q = 2

(−9,−4,−1)

116d. [5 marks]The lines and intersect at point C where .

(i) Find .

(ii) Hence, find .

Markscheme(i) evidence of appropriate approach (M1)

e.g. ,

A1 N2

(ii) finding A1

evidence of finding magnitude (M1)

e.g.

A1 N3

[5 marks]

Examiners reportPart (d) proved challenging as many candidates did not consider that . Rather, many attempted to find thecoordinates of point C, which became a more arduous and error-prone task.

L2 L3 =BC−→ ⎛

⎝⎜63

−24

⎞⎠⎟

AB−→

| |AC−→−

+ =OA−→−

AB−→

OB−→

= −AB−→ ⎛

⎝⎜−8−525

⎞⎠⎟

⎛⎝⎜

−9−4−1

⎞⎠⎟

=AB−→ ⎛

⎝⎜1

−126

⎞⎠⎟

=AC−→− ⎛

⎝⎜722

⎞⎠⎟

| | =AC−→−

+ +72 22 22− −−−−−−−−√

| | =AC−→−

57−−√

+ =AB−→

BC−→

AC−→−

⎛⎝⎜

⎞⎠⎟

117a. [2 marks]

Let and \overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}} { - 2}\\ { - 3}\\ 2 \end{array}} \right) .

Find \overrightarrow {{\rm{BC}}} .


e.g. \overrightarrow {{\rm{BC}}} {\rm{ = }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} , \left({\begin{array}{*{20}{c}} { - 2}\\ { - 3}\\ 2 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 6\\ { - 2}\\ 3 \end{array}}\right)

\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}} { - 8}\\ { - 1}\\ { - 1} \end{array}} \right) A1 N2

[2 marks]

Examiners reportPart (a) was generally done well with candidates employing different correct methods to find the vector \overrightarrow{{\rm{BC}}} . Some candidates subtracted the given vectors in the wrong order and others simply added them. Calculation errorswere seen with some frequency.

=AB−→ ⎛

⎝⎜6

−23

⎞⎠⎟

117b. [3 marks]Find a unit vector in the direction of \overrightarrow {{\rm{AB}}} .

Markschemeattempt to find the length of \overrightarrow {{\rm{AB}}} (M1)

\overrightarrow {|{\rm{AB}}} | = \sqrt {{6^2} + {{( - 2)}^2} + {3^2}} ( = \sqrt {36 + 4 + 9} = \sqrt {49} = 7) (A1)

unit vector is \frac{1}{7}\left( {\begin{array}{*{20}{c}} 6\\ { - 2}\\ 3 \end{array}} \right) \left( { = \left({\begin{array}{*{20}{c}} {\frac{6}{7}}\\ { - \frac{2}{7}}\\ {\frac{3}{7}} \end{array}} \right)} \right) A1 N2

[3 marks]

Examiners reportMany candidates did not appear to know how to find a unit vector in part (b). Some tried to write down the vector equation of a line,indicating no familiarity with the concept of unit vectors while others gave the vector (1, 1, 1) or wrote the same vector\overrightarrow {{\rm{AB}}} as a linear combination of i, j and k. A number of candidates correctly found the magnitude but didnot continue on to write the unit vector.

117c. [3 marks]Show that \overrightarrow {{\rm{AB}}} is perpendicular to \overrightarrow {{\rm{AC}}} .

Markschemerecognizing that the dot product or \cos \theta being 0 implies perpendicular (M1)

correct substitution in a scalar product formula A1

e.g. (6) \times ( - 2) + ( - 2) \times ( - 3) + (3) \times (2) , \cos \theta = \frac{{ - 12 + 6 + 6}}{{7 \times \sqrt {17} }}

correct calculation A1

e.g. \overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AC}}} = 0 , \cos \theta = 0

therefore, they are perpendicular AG N0

[3 marks]

Examiners reportCandidates were generally successful in showing that the vectors in part (c) were perpendicular. Many used the efficient approach ofshowing that the scalar product equalled zero, while others worked a little harder than necessary and used the cosine rule to find theangle between the two vectors.

118a. [3 marks]Let u = \left( {\begin{array}{*{20}{c}} 2\\ 3\\ { - 1} \end{array}} \right) and w = \left( {\begin{array}{*{20}{c}} 3\\ { -1}\\ p \end{array}} \right) . Given that u is perpendicular to w , find the value of p .

Markschemeevidence of equating scalar product to 0 (M1)

2 \times 3 + 3 \times ( - 1) + ( - 1) \times p = 0 (6 - 3 - p = 0, 3 - p = 0) A1

p = 3 A1 N2

[3 marks]

Examiners reportMost candidates knew to set the scalar product equal to zero.

118b. [3 marks]Let {\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}} 1 \\ q \\ 5 \end{array}} \right) . Given that \left|{\boldsymbol{v}} \right| = \sqrt {42} , find the possible values of q .

Markschemeevidence of substituting into magnitude formula (M1)

e.g. \sqrt {1 + {q^2} + 25} , 1 + {q^2} + 25

setting up a correct equation A1

e.g. \sqrt {1 + {q^2} + 25} = \sqrt {42} , 1 + {q^2} + 25 = 42 , {q^2} = 16

q = \pm 4 A1 N2

[3 marks]

Examiners reportMost candidates knew to set the scalar product equal to zero. A pleasing number found both answers for q, although some oftenneglected to provide both solutions.

119a. [1 mark]

Consider the points P(2, −1, 5) and Q(3, − 3, 8). Let {L_1} be the line through P and Q.

Show that \overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 3 \end{array}} \right) .

Markschemeevidence of correct approach A1

e.g. \overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{OQ}}} - \overrightarrow {{\rm{OP}}} , \left({\begin{array}{*{20}{c}} 3\\ { - 3}\\ 8 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right)

\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 3 \end{array}} \right) AG N0

[1 mark]

Examiners reportMost candidates answered part (a) easily.

119b. [3 marks]The line {L_1} may be represented by {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 3 \\ { - 3} \\ 8\end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \\ 3 \end{array}} \right) .

(i) What information does the vector \left( {\begin{array}{*{20}{c}} 3\\ { - 3}\\ 8 \end{array}} \right) give about {L_1} ?

(ii) Write down another vector representation for {L_1} using \left( {\begin{array}{*{20}{c}} 3\\ { - 3}\\ 8 \end{array}} \right) .

Markscheme(i) correct description R1 N1

e.g. reference to \left( {\begin{array}{*{20}{c}} 3\\ { - 3}\\ 8 \end{array}} \right) being the position vector of a point on the line, avector to the line, a point on the line.

(ii) any correct expression in the form {\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}} A2 N2

where {\boldsymbol{a}} is \left( {\begin{array}{*{20}{c}} 3 \\ { - 3} \\ 8 \end{array}} \right) , and {\boldsymbol{b}} is ascalar multiple of \left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \\ 3 \end{array}} \right)

e.g. {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 3 \\ { - 3} \\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ { - 3} \end{array}} \right) , {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} {3 + 2s} \\ { - 3 - 4s} \\ {8 +6s} \end{array}} \right)

[3 marks]

Examiners reportFor part (b), a number of candidates stated that the vector was a "starting point," which misses the idea that it is a position vector tosome point on the line.

119c. [3 marks]The point {\text{T}}( - 1{\text{, }}5{\text{, }}p) lies on {L_1} .

Find the value of p .

Markschemeone correct equation (A1)

e.g. 3 + s = - 1 , - 3 - 2s = 5

s = - 4 A1

p = - 4 A1 N2

[3 marks]

Examiners reportParts (c) and (d) proved accessible to many.

119d. [3 marks]The point T also lies on {L_2} with equation \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left({\begin{array}{*{20}{c}} { - 3}\\ 9\\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ q \end{array}} \right) .

Show that q = - 3 .

Markschemeone correct equation A1

e.g. - 3 + t = - 1 , 9 - 2t = 5

t = 2 A1

substituting t = 2

e.g. 2 + 2q = - 4 , 2q = - 6 A1

q = - 3 AG N0

[3 marks]

Examiners reportParts (c) and (d) proved accessible to many.

119e. [7 marks]Let \theta be the obtuse angle between {L_1} and {L_2} . Calculate the size of \theta .

Markschemechoosing correct direction vectors \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 3 \end{array}} \right) and \left({\begin{array}{*{20}{c}} 1\\ { - 2}\\ { - 3} \end{array}} \right) (A1)(A1)

finding correct scalar product and magnitudes (A1)(A1)(A1)

scalar product (1)(1) + ( - 2)( - 2) + ( - 3)(3) ( = - 4)

magnitudes \sqrt {{1^2} + {{( - 2)}^2} + {3^2}} = \sqrt {14} , \sqrt {{1^2} + {{( - 2)}^2} + {{( - 3)}^2}} = \sqrt {14}

evidence of substituting into scalar product M1

e.g. \cos \theta = \frac{{ - 4}}{{3.741 \ldots \times 3.741 \ldots }}

\theta = 1.86 radians (or 107^\circ ) A1 N4

[7 marks]

Examiners reportFor part (e), a surprising number of candidates chose incorrect vectors. Few candidates seemed to have a good conceptualunderstanding of the vector equation of a line.

120a. [3 marks]

The vertices of the triangle PQR are defined by the position vectors

\overrightarrow {{\rm{OP}}} = \left( {\begin{array}{*{20}{c}} 4\\ { - 3}\\ 1 \end{array}} \right) , \overrightarrow {{\rm{OQ}}} = \left({\begin{array}{*{20}{c}} 3\\ { - 1}\\ 2 \end{array}} \right) and \overrightarrow {{\rm{OR}}} = \left( {\begin{array}{*{20}{c}} 6\\ { -1}\\ 5 \end{array}} \right) .

Find

(i) \overrightarrow {{\rm{PQ}}} ;

(ii) \overrightarrow {{\rm{PR}}} .


e.g. \overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} , {\rm{Q}} - {\rm{P}}

\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}} { - 1}\\ 2\\ 1 \end{array}} \right) A1 N2

(ii) \overrightarrow {{\rm{PR}}} = \left( {\begin{array}{*{20}{c}} 2\\ 2\\ 4 \end{array}} \right) A1 N1

[3 marks]

Examiners reportCombining the vectors in (a) was generally well done, although some candidates reversed the subtraction, while others calculated themagnitudes.

120b. [7 marks]Show that \cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2} .

MarkschemeMETHOD 1

choosing correct vectors \overrightarrow {{\rm{PQ}}} and \overrightarrow {{\rm{PR}}} (A1)(A1)

finding \overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} , \left| {\overrightarrow {{\rm{PQ}}} } \right| , \left|{\overrightarrow {{\rm{PR}}} } \right| (A1) (A1)(A1)

\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} = - 2 + 4 + 4( = 6)

\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( - 1)}^2} + {2^2} + {1^2}} \left( { = \sqrt 6 } \right) , \left|{\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \left( { = \sqrt {24} } \right)

substituting into formula for angle between two vectors M1

e.g. \cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6 \times \sqrt {24} }}

simplifying to expression clearly leading to \frac{1}{2} A1

e.g. \frac{6}{{\sqrt 6 \times 2\sqrt 6 }} , \frac{6}{{\sqrt {144} }} , \frac{6}{{12}}

\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2} AG N0

METHOD 2

evidence of choosing cosine rule (seen anywhere) (M1)

\overrightarrow {{\rm{QR}}} = \left( {\begin{array}{*{20}{c}} 3\\ 0\\ 3 \end{array}} \right) A1

\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} , \left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 and \left|{\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} (A1)(A1)(A1)

\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} - {{\left( {\sqrt{18} } \right)}^2}}}{{2\sqrt 6 \times \sqrt {24} }} A1

\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 - 18}}{{24}} \left( { = \frac{{12}}{{24}}} \right) A1

\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2} AG N0

[7 marks]

Examiners reportMany candidates successfully used scalar product and magnitude calculations to complete part (b). Alternatively, some used thecosine rule, and often achieved correct results. Some assumed the triangle was a right-angled triangle and thus did not earn full marks.Although PQR is indeed right-angled, in a “show that” question this attribute must be directly established.

121. [7 marks]Let {\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}} 2\\ { - 3}\\ 6 \end{array}} \right) and {\boldsymbol{w}} = \left({\begin{array}{*{20}{c}} k\\ { - 2}\\ 4 \end{array}} \right) , for k > 0 . The angle between v and w is \frac{\pi }{3} .

Find the value of k .

Markschemecorrect substitutions for {\boldsymbol{v}} \bullet {\boldsymbol{w}} ; \left| {\boldsymbol{v}} \right| ; \left| {\boldsymbol{w}}\right| (A1)(A1)(A1)

e.g. 2k + ( - 3) \times ( - 2) + 6 \times 4 , 2k + 30 ; \sqrt {{2^2} + {{( - 3)}^2} + {6^2}} , \sqrt {49} ; \sqrt {{k^2} + {{( - 2)}^2} +{4^2}} , \sqrt {{k^2} + 20}

evidence of substituting into the formula for scalar product (M1)

e.g. \frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}


e.g. \cos \frac{\pi }{3} = \frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}

k = 18.8 A2 N5

[7 marks]

Examiners reportFor the most part, this question was well done and candidates had little difficulty finding the scalar product, the appropriatemagnitudes and then correctly substituting into the formula for the angle between vectors. However, few candidates were able tosolve the resulting equation using their GDCs to obtain the correct answer. Problems arose when candidates attempted to solve theresulting equation analytically.

122a. [4 marks]

In this question, distance is in metres.

Toy airplanes fly in a straight line at a constant speed. Airplane 1 passes through a point A.

Its position, p seconds after it has passed through A, is given by \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left({\begin{array}{*{20}{c}} 3\\ { - 4}\\ 0 \end{array}} \right) + p\left( {\begin{array}{*{20}{c}} { - 2}\\ 3\\ 1 \end{array}} \right) .

(i) Write down the coordinates of A.

(ii) Find the speed of the airplane in {\text{m}}{{\text{s}}^{ - 1}}.

Markscheme(i) (3, - 4, 0) A1 N1

(ii) choosing velocity vector \left( {\begin{array}{*{20}{c}} { - 2}\\ 3\\ 1 \end{array}} \right) (M1)

finding magnitude of velocity vector (A1)

e.g. \sqrt {{{( - 2)}^2} + {3^2} + {1^2}} , \sqrt {4 + 9 + 1}

speed = 3.74 \left( {\sqrt {14} } \right) A1 N2

[4 marks]

Examiners reportMany candidates demonstrated a good understanding of the vector equation of a line and its application to a kinematics problem bycorrectly answering the first two parts of this question.

122b. [5 marks]After seven seconds the airplane passes through a point B.

(i) Find the coordinates of B.

(ii) Find the distance the airplane has travelled during the seven seconds.

Markscheme(i) substituting p = 7 (M1)

{\text{B}} = ( - 11{\text{, }}17{\text{, }}7) A1 N2

(ii) METHOD 1

appropriate method to find \overrightarrow {{\rm{AB}}} or \overrightarrow {{\rm{BA}}} (M1)

e.g. \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} , {\rm{A}} - {\rm{B}}

\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} { - 14}\\ {21}\\ 7 \end{array}} \right) or \overrightarrow{{\rm{BA}}} = \left( {\begin{array}{*{20}{c}} {14}\\ { - 21}\\ { - 7} \end{array}} \right) (A1)

distance = 26.2 \left( {7\sqrt {14} } \right) A1 N3

METHOD 2

evidence of applying distance is speed × time (M2)

e.g. 3.74 \times 7

distance = 26.2 \left( {7\sqrt {14} } \right) A1 N3

METHOD 3

attempt to find AB , AB (M1)

e.g. {(3 - ( - 11))^2} + {( - 4 - 17)^2} + (0 - 7){)^2} , \sqrt {{{(3 - ( - 11))}^2} + {{( - 4 - 17)}^2} + (0 - 7){)^2}}

AB = 686, AB = \sqrt {686} (A1)

distance AB = 26.2 \left( {7\sqrt {14} } \right) A1 N3

[5 marks]

Examiners reportMany candidates demonstrated a good understanding of the vector equation of a line and its application to a kinematics problem bycorrectly answering the first two parts of this question.

Some knew that speed and distance were magnitudes of vectors but chose the wrong vectors to calculate magnitudes.

2

2

122c. [7 marks]Airplane 2 passes through a point C. Its position q seconds after it passes through C is given by \left({\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ { - 5}\\ 8 \end{array}} \right) + q\left({\begin{array}{*{20}{c}} { - 1}\\ 2\\ a \end{array}} \right),a \in \mathbb{R} .

The angle between the flight paths of Airplane 1 and Airplane 2 is {40^ \circ } . Find the two values of a.

Markschemecorrect direction vectors \left( {\begin{array}{*{20}{c}} { - 2}\\ 3\\ 1 \end{array}} \right) and \left( {\begin{array}{*{20}{c}} { -1}\\ 2\\ a \end{array}} \right) (A1)(A1)

\left| {\begin{array}{*{20}{c}} { - 1}\\ 2\\ a \end{array}} \right| = \sqrt {{a^2} + 5} , \left( {\begin{array}{*{20}{c}} { - 2}\\ 3\\ 1\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1}\\ 2\\ a \end{array}} \right) = a + 8 (A1)(A1)

substituting M1

e.g. \cos {40^ \circ } = \frac{{a + 8}}{{\sqrt {14} \sqrt {{a^2} + 5} }}

a = 3.21 , a = - 0.990 A1A1 N3

[7 marks]

Examiners reportVery few candidates were able to get the two correct answers in (c) even if they set up the equation correctly. Much contorted algebrawas seen and little evidence of using the GDC to solve the equation. Many made simple algebraic errors by combining unlike termsin working with the scalar product (often writing 8a rather than 8 + a ) or the magnitude (often writing 5{a^2} rather than 5 + {a^2}).

123a. [2 marks]

Line {L_1} passes through points {\text{A}}(1{\text{, }} - 1{\text{, }}4) and {\text{B}}(2{\text{, }} - 2{\text{, }}5) .

Find \overrightarrow {{\rm{AB}}} .


e.g. \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} , B - A

\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right) A1 N2

[2 marks]

Examiners reportFinding \overrightarrow {{\rm{AB}}} was generally well done, although some candidates reversed the subtraction.

123b. [2 marks]Find an equation for {L_1} in the form {\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}} .

Markschemeany correct equation in the form {\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}} A2 N2

where {\boldsymbol{b}} is a scalar multiple of \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right)

e.g. {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 4 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1\\ {- 1}\\ 1 \end{array}} \right) , {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} {2 + t}\\ { - 2 - t}\\ {5 + t} \end{array}} \right) ,{\boldsymbol{r}} = 2{\boldsymbol{i}} - 2{\boldsymbol{j}} + 5{\boldsymbol{k}} + t({\boldsymbol{i}} - {\boldsymbol{j}} +{\boldsymbol{k}})

[2 marks]

Examiners reportIn part (b) not all the candidates recognized that \overrightarrow {{\rm{AB}}} was the direction vector of the line, as some used theposition vector of point B as the direction vector.

123c. [7 marks]

Line {L_2} has equation {\boldsymbol{r}} = \left( \begin{array}{l} 2\\ 4\\ 7 \end{array} \right) + s\left( \begin{array}{l} 2\\ 1\\ 3\end{array} \right) .

Find the angle between {L_1} and {L_2} .

Examiners reportMany candidates successfully used scalar product and magnitudes in part (c), although a large number did choose vectors other thanthe direction vectors and many did not state clearly which vectors they were using.

123d. [6 marks]The lines {L_1} and {L_2} intersect at point C. Find the coordinates of C.

MarkschemeMETHOD 1 (from {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 4 \end{array}} \right) + t\left({\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 1 \end{array}} \right) )

appropriate approach (M1)

e.g. {\boldsymbol{p}} = {\boldsymbol{r}} , \left( {\begin{array}{*{20}{c}} 1\\ { - 1}\\ 4 \end{array}} \right) + \left({\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right)t = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ 7 \end{array}} \right) + \left({\begin{array}{*{20}{c}} 2\\ 1\\ 3 \end{array}} \right)s

two correct equations A1A1

e.g. 1 + t = 2 + 2s , - 1 - t = 4 + s , 4 + t = 7 + 3s



e.g. t = - 3 , s = - 2

C is ( - 2{\text{, }}2{\text{, }}1) A1 N3

METHOD 2 (from {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ 5 \end{array}} \right) + t\left({\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 1 \end{array}} \right) )

appropriate approach (M1)

e.g. {\boldsymbol{p}} = {\boldsymbol{r}} , \left( {\begin{array}{*{20}{c}} 2\\ { - 2}\\ 5 \end{array}} \right) + t\left({\begin{array}{*{20}{c}} 1\\ { - 1}\\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ 7 \end{array}} \right) + \left({\begin{array}{*{20}{c}} 2\\ 1\\ 3 \end{array}} \right)s


e.g. 2 + t = 2 + 2s , - 2 - t = 4 + s , 5 + t = 7 + 3s



e.g. t = - 4 , s = - 2

C is ( - 2{\text{, }}2{\text{, }}1) A1 N3

[6 marks]

Examiners reportCandidates who were comfortable on the first three parts often had little difficulty with the final part. While the resulting systems wereeasily solved algebraically, a surprising number of candidates did not check their solutions either manually or with technology. Anoccasionally seen error in the final part was using a midpoint to find C. Some candidates found the point of intersection in part (c)rather than in part (d), indicating a familiarity with the type of question but a lack of understanding of the concepts involved.

124a. [4 marks]

A particle is moving with a constant velocity along line L . Its initial position is A(6 , −2 , 10) . After one second the particle has moved to B(9, −6 , 15) .

(i) Find the velocity vector, \overrightarrow {{\rm{AB}}} .

(ii) Find the speed of the particle.


e.g. \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} , {\rm{B}} - {\rm{A}} , \left( {\begin{array}{*{20}{c}} {9 -6}\\ { - 6 + 2}\\ {15 - 10} \end{array}} \right)

\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 3\\ { - 4}\\ 5 \end{array}} \right) (accept \left( {3, - 4, 5} \right) ) A1 N2

(ii) evidence of finding the magnitude of the velocity vector M1

e.g. {\text{speed}} = \sqrt {{3^2} + {4^2} + {5^2}}

{\text{speed}} = \sqrt {50} \left( { = 5\sqrt 2 } \right) A1 N1

[4 marks]

Examiners reportThis question was quite well done. Marks were lost when candidates found the vector \overrightarrow {{\text{BA}}} instead of\overrightarrow {{\text{AB}}} in part (a) and for not writing their vector equation as an equation.

124b. [2 marks]Write down an equation of the line L .

Markschemecorrect equation (accept Cartesian and parametric forms) A2 N2

e.g. {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 6\\ { - 2}\\ {10} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}}3\\ { - 4}\\ 5 \end{array}} \right) , {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 9\\ { - 6}\\ {15} \end{array}} \right) + t\left({\begin{array}{*{20}{c}} 3\\ { - 4}\\ 5 \end{array}} \right)

[2 marks]

Examiners reportIn part (b), a few candidates switched the position and velocity vectors or used the vectors \overrightarrow {{\text{OA}}} and\overrightarrow {{\text{OB}}} to incorrectly write the vector equation.

125a. [5 marks]

The diagram shows a parallelogram ABCD.

The coordinates of A, B and D are A(1, 2, 3) , B(6, 4,4 ) and D(2, 5, 5) .

(i) Show that \overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 5\\ 2\\ 1 \end{array}} \right) .

(ii) Find \overrightarrow {{\rm{AD}}} .

(iii) Hence show that \overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}} 6\\ 5\\ 3 \end{array}} \right) .

Markscheme(i) evidence of approach M1

e.g. {\text{B}} - {\text{A}} , \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} , \left( {\begin{array}{*{20}{c}} 6\\4\\ 4 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right)

\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 5\\ 2\\ 1 \end{array}} \right) AG N0

(ii) evidence of approach (M1)

e.g. {\text{D}} - {\text{A}} , \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}} , \left( {\begin{array}{*{20}{c}} 2\\5\\ 5 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right)

\overrightarrow {{\rm{AD}}}= \left( {\begin{array}{*{20}{c}} 1\\ 3\\ 2 \end{array}} \right) A1 N2

(iii) evidence of approach (M1)

e.g. \overrightarrow {{\rm{AC}}} = \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{AD}}}


e.g. \overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}} 5\\ 2\\ 1 \end{array}} \right) + \left({\begin{array}{*{20}{c}} 1\\ 3\\ 2 \end{array}} \right)

\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}} 6\\ 5\\ 3 \end{array}} \right) AG N0

[5 marks]

Examiners reportCandidates performed very well in this question, showing a strong ability to work with the algebra and geometry of vectors.

125b. [3 marks]Find the coordinates of point C.

Markschemeevidence of combining vectors (there are at least 5 ways) (M1)

e.g. \overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AC}}} , \overrightarrow{{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{AD}}} , \overrightarrow {{\rm{AB}}} = \overrightarrow{{\rm{OC}}} - \overrightarrow {{\rm{OD}}}


\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}}6\\ 5\\ 3 \end{array}} \right)\left( { = \left( {\begin{array}{*{20}{c}} 7\\ 7\\ 6 \end{array}} \right)} \right)

e.g. coordinates of C are (7{\text{, }}7{\text{, }}6) A1 N1

[3 marks]

Examiners reportCandidates performed very well in this question, showing a strong ability to work with the algebra and geometry of vectors.

125c. [7 marks](i) Find \overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} .

(ii) Hence find angle A.

Markscheme(i) evidence of using scalar product on \overrightarrow {{\rm{AB}}} and \overrightarrow {{\rm{AD}}} (M1)

e.g. \overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 5(1) + 2(3) + 1(2)

\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 13 A1 N2

(ii) \left| {\overrightarrow {{\rm{AB}}} } \right| = 5.477 \ldots , \left| {\overrightarrow {{\rm{AD}}} } \right| = 3.741 \ldots (A1)(A1)

evidence of using \cos A = \frac{{\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} }}{{\left| {\overrightarrow{{\rm{AB}}} } \right|\left| {\overrightarrow {{\rm{AD}}} } \right|}} (M1)


e.g. \cos A = \frac{{13}}{{20.493}}

\widehat A = 0.884 (50.6^\circ ) A1 N3

[7 marks]

Examiners reportSome candidates were unable to find the scalar product in part (c), yet still managed to find the correct angle, able to use the formulain the information booklet without knowing that the scalar product is a part of that formula.

126. [7 marks]Let {\boldsymbol{v}} = 3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}} and {\boldsymbol{w}} ={\boldsymbol{i}} + 2{\boldsymbol{j}} - 3{\boldsymbol{k}} . The vector {\boldsymbol{v}} + p{\boldsymbol{w}} is perpendicular tow. Find the value of p.

Markschemep{\boldsymbol{w}} = p{\boldsymbol{i}} + 2p{\boldsymbol{j}} - 3p{\boldsymbol{k}} (seen anywhere) (A1)

attempt to find {\boldsymbol{v}} + p{\boldsymbol{w}} (M1)

e.g. 3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}} + p({\boldsymbol{i}} + 2{\boldsymbol{j}} -3{\boldsymbol{k}})

collecting terms (3 + p){\boldsymbol{i}} + (4 + 2p){\boldsymbol{j}} + (1 - 3p){\boldsymbol{k}} A1

attempt to find the dot product (M1)

e.g. 1(3 + p) + 2(4 + 2p) - 3(1 - 3p)

setting their dot product equal to 0 (M1)

e.g. 1(3 + p) + 2(4 + 2p) - 3(1 - 3p) = 0

simplifying A1

e.g. 3 + p + 8 + 4p - 3 + 9p = 0 , 14p + 8 = 0

p = - 0.571 \left( { - \frac{8}{{14}}} \right) A1 N3

[7 marks]

Examiners reportThis question was very poorly done with many leaving it blank. Of those that did attempt it, most were able to find{\boldsymbol{v}} + p{\boldsymbol{w}} but really did not know how to proceed from there. They tried many approaches, such as,finding magnitudes, using negative reciprocals, or calculating the angle between two vectors. A few had the idea that the scalarproduct should equal zero but had trouble trying to set it up.

127a. [8 marks]

The point O has coordinates (0 , 0 , 0) , point A has coordinates (1 , – 2 , 3) and point B has coordinates (– 3 , 4 , 2) .

(i) Show that \overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} { - 4}\\ 6\\ { - 1} \end{array}} \right) .

(ii) Find {\rm{B}}\widehat {\rm{A}}{\rm{O}} .


e.g. \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} = \overrightarrow {{\rm{AB}}}

\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} { - 4}\\ 6\\ { - 1} \end{array}} \right) AG N0

(ii) for choosing correct vectors, (\overrightarrow {{\rm{AO}}} with \overrightarrow {{\rm{AB}}} or \overrightarrow{{\rm{OA}}} with \overrightarrow {{\rm{BA}}} ) (A1)(A1)

Note: Using \overrightarrow {{\rm{AO}}} with \overrightarrow {{\rm{BA}}} will lead to \pi - 0.799 . If they then say{\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799 , this is a correct solution.

calculating \overrightarrow {{\rm{AO}}} \bullet \overrightarrow {{\rm{AB}}} , \left| {\overrightarrow {{\rm{AO}}} } \right| ,\left| {\overrightarrow {{\rm{AB}}} } \right| (A1)(A1)(A1)

e.g. {d_1} \bullet {d_2} = ( - 1)( - 4) + (2)(6) + ( - 3)( - 1)( = 19)

\left| {{d_1}} \right| = \sqrt {{{( - 1)}^2} + {2^2} + {{( - 3)}^2}} ( = \sqrt {14} ) , \left| {{d_2}} \right| = \sqrt {{{( - 4)}^2} +{6^2} + {{( - 1)}^2}} ( = \sqrt {53} )

evidence of using the formula to find the angle M1

e.g. \cos \theta = \frac{{( - 1)( - 4) + (2)(6) + ( - 3)( - 1)}}{{\sqrt {{{( - 1)}^2} + {2^2} + {{( - 3)}^2}} \sqrt {{{( - 4)}^2} + {6^2}+ {{( - 1)}^2}} }} , \frac{{19}}{{\sqrt {14} \sqrt {53} }} , 0.69751 \ldots

{\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799 radians (accept 45.8^\circ ) A1 N3

[8 marks]

Examiners reportPart (ai) was done well by most students. Most knew how to approach finding the angle in part (aii). The problems occurred when theincorrect vectors were chosen. If the vectors being used were stated, then follow through marks could be given.

127b. [2 marks]The line {L_1} has equation \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left({\begin{array}{*{20}{c}} { - 3}\\ 4\\ 2 \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} { - 4}\\ 6\\ { - 1} \end{array}} \right) .

Write down the coordinates of two points on {L_1} .

Markschemetwo correct answers A1A1

e.g. (1, - 2, 3) , ( - 3, 4, 2) , ( - 7, 10, 1), ( - 11, 16, 0) N2

[2 marks]

Examiners reportPart (b) was well done.

127c. [6 marks]The line {L_2} passes through A and is parallel to \overrightarrow {{\rm{OB}}} .

(i) Find a vector equation for {L_2} , giving your answer in the form {\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}} .

(ii) Point C(k, - k,5) is on {L_2} . Find the coordinates of C.

Markscheme(i) {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { -3}\\ 4\\ 2 \end{array}} \right) A2 N2

(ii) C on {L_2} , so \left( {\begin{array}{*{20}{c}} k\\ { - k}\\ 5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 3}\\ 4\\ 2 \end{array}} \right) (M1)

evidence of equating components (A1)

e.g. 1 - 3t = k , - 2 + 4t = - k , 5 = 3 + 2t

one correct value t = 1 , k = - 2 (seen anywhere) (A1)

coordinates of C are ( - 2{\text{, }}2{\text{, }}5) A1 N3

[6 marks]

Examiners reportIn part (ci), the error that occurred most often was the incorrect choice for the direction vector.

127d. [2 marks]The line {L_3} has equation \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left({\begin{array}{*{20}{c}} 3\\ { - 8}\\ 0 \end{array}} \right) + p\left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ { - 1} \end{array}}\right) and passes through the point C.

Find the value of p at C.

Markschemefor setting up one (or more) correct equation using \left( {\begin{array}{*{20}{c}} { - 2}\\ 2\\ 5 \end{array}} \right) = \left({\begin{array}{*{20}{c}} 3\\ { - 8}\\ 0 \end{array}} \right) + p\left( {\begin{array}{*{20}{c}} 1\\ { - 2}\\ { - 1} \end{array}}\right) (M1)

e.g. 3 + p = - 2 , - 8 - 2p = 2 , - p = 5

p = - 5 A1 N2

[2 marks]

Examiners reportThose that were able to find the coordinates in part (cii) were also able to be successful in part (d).

128a. [3 marks]

Consider the points A (1 , 5 , 4) , B (3 , 1 , 2) and D (3 , k , 2) , with (AD) perpendicular to (AB) .

Find

(i) \overrightarrow {{\rm{AB}}} ;

(ii) \overrightarrow {{\rm{AD}}} giving your answer in terms of k .

[3 marks]

Markscheme(i) evidence of combining vectors (M1)

e.g. \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} - \overrightarrow {{\rm{OA}}} (or \overrightarrow{{\rm{AD}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}} in part (ii))

\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 2\\ { - 4}\\ { - 2} \end{array}} \right) A1 N2

(ii) \overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}} 2\\ {k - 5}\\ { - 2} \end{array}} \right) A1 N1

[3 marks]

Examiners reportThis question was well done by many candidates. Most found \overrightarrow {{\rm{AB}}} and \overrightarrow {{\rm{AD}}} correctly.

128b. [3 marks]Show that k = 7 .

Markschemeevidence of using perpendicularity \Rightarrow scalar product = 0 (M1)

e.g. \left( {\begin{array}{*{20}{c}} 2\\ { - 4}\\ { - 2} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2\\ {k - 5}\\ { -2} \end{array}} \right) = 0

4 - 4(k - 5) + 4 = 0 A1

- 4k + 28 = 0 (accept any correct equation clearly leading to k = 7 ) A1

k = 7 AG N0

[3 marks]

Examiners reportThe majority of candidates correctly used the scalar product to show k = 7 .

128c. [4 marks]The point C is such that \overrightarrow {{\rm{BC}}} = \frac{1}{2}\overrightarrow {{\rm{AD}}} .

Find the position vector of C.

Markscheme\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}} 2\\ 2\\ { - 2} \end{array}} \right) (A1)

\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1} \end{array}} \right) A1

evidence of correct approach (M1)

e.g. \overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{BC}}} , \left({\begin{array}{*{20}{c}} 3\\ 1\\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1} \end{array}} \right) , \left({\begin{array}{*{20}{c}} {x - 3}\\ {y - 1}\\ {z - 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1}\end{array}} \right)

\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}} 4\\ 2\\ 1 \end{array}} \right) A1 N3

[4 marks]

Examiners reportSome confusion arose in substituting k = 7 into \overrightarrow {{\rm{AD}}} , but otherwise part (c) was well done, though findingthe position vector of C presented greater difficulty.

128d. [3 marks]Find \cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} .

MarkschemeMETHOD 1

choosing appropriate vectors, \overrightarrow {{\rm{BA}}} , \overrightarrow {{\rm{BC}}} (A1)

finding the scalar product M1

e.g. - 2(1) + 4(1) + 2( - 1) , 2(1) + ( - 4)(1) + ( - 2)( - 1)

\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}} A1 N1

METHOD 2

\overrightarrow {{\rm{BC}}} parallel to \overrightarrow {{\rm{AD}}} (may show this on a diagram with points labelled) R1

\overrightarrow {{\rm{BC}}} \bot \overrightarrow {{\rm{AB}}} (may show this on a diagram with points labelled) R1

{\rm{A}}\widehat {\rm{B}}{\rm{C}} = 90^\circ

\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}} A1 N1

[3 marks]

Examiners reportOwing to \overrightarrow {{\rm{AB}}} and \overrightarrow {{\rm{BC}}} being perpendicular, no problems were created by usingthese two vectors to find \cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = 0 , and the majority of candidates answering part (d) didexactly that.

129. [6 marks]The line L is represented by {{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}} 2\\ 5\\ 3 \end{array}} \right) + s\left({\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right) and the line L by {{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}} 3\\{ - 3}\\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 1}\\ 3\\ { - 4} \end{array}} \right) .

The lines L and L intersect at point T. Find the coordinates of T.

Markschemeevidence of equating vectors (M1)

e.g. {L_1} = {L_2}

for any two correct equations A1A1

e.g. 2 + s = 3 - t , 5 + 2s = - 3 + 3t , 3 + 3s = 8 - 4t

attempting to solve the equations (M1)

finding one correct parameter (2 = - 1{\text{, }}t = 2) A1

the coordinates of T are (1{\text{, }}3{\text{, }}0) A1 N3

[6 marks]

Examiners reportThose candidates prepared in this topic area answered the question particularly well, often only making some calculation error whensolving the resulting system of equations. Curiously, a few candidates found correct values for s and t, but when substituting back intoone of the vector equations, neglected to find the z-coordinate of T.

1

2

1 2

130. [6 marks]Two lines with equations {{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}} 2\\ 3\\ { - 1} \end{array}} \right) +s\left( {\begin{array}{*{20}{c}} 5\\ { - 3}\\ 2 \end{array}} \right) and {{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}} 9\\ 2\\2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 3}\\ 5\\ { - 1} \end{array}} \right) intersect at the point P. Find thecoordinates of P.


e.g. \left( {\begin{array}{*{20}{c}} 2\\ 3\\ { - 1} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 5\\ { - 3}\\ 2 \end{array}}\right) = \left( {\begin{array}{*{20}{c}} 9\\ 2\\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 3}\\ 5\\ { - 1}\end{array}} \right)


e.g. 2 + 5s = 9 - 3t , 3 - 3s = 2 + 5t , - 1 + 2s = 2 - t

attempting to solve the equations (M1)

one correct parameter s = 2 , t = - 1 A1

P is (12, - 3,3) (accept \left( {\begin{array}{*{20}{c}} {12}\\ { - 3}\\ 3 \end{array}} \right)) A1 N3

[6 marks]

Examiners reportIf this topic had been taught well then the candidates scored highly. The question was either well answered or not at all. Manycandidates did not understand what was needed and tried to find the length of vectors or mid-points of lines. The other most commonmistake was to use the values of the parameters to write the coordinates as {\text{P}}(2{\text{, }} - 1).

131. [6 marks]Find the cosine of the angle between the two vectors 3{\boldsymbol{i}} + 4{\boldsymbol{j}} + 5{\boldsymbol{k}} and4{\boldsymbol{i}} - 5{\boldsymbol{j}} - 3{\boldsymbol{k}} .

Markschemefinding scalar product and magnitudes (A1)(A1)(A1)

scalar product = 12 - 20 - 15 ( = - 23)

magnitudes = \sqrt {{3^2} + {4^2} + {5^2}} , = \sqrt {{4^2} + {{( - 5)}^2} + {{( - 3)}^2}} , \left( {\sqrt {50} ,\sqrt {50} } \right)


e.g. \cos \theta = \frac{{12 - 20 - 15}}{{\left( {\sqrt {{3^2} + {4^2} + {5^2}} } \right) \times \left( {\sqrt {{4^2} + {{( - 5)}^2} +{{( - 3)}^2}} } \right)}}

\cos \theta = - \frac{{23}}{{50}} ( = - 0.46) A2 N4

[6 marks]

Examiners reportMany candidates performed well in finding the magnitudes and scalar product to use the formula for angle between vectors. Someexperienced trouble with the arithmetic to obtain the required result. A significant number of candidates isolated the theta answeringwith {\text{arccos}}\left( {\frac{{ - 23}}{{50}}} \right) .

132a. [2 marks]

The line {L_1} is parallel to the z-axis. The point P has position vector \left( {\begin{array}{*{20}{c}} 8\\ 1\\ 0 \end{array}} \right) and lieson {L_1}.

Write down the equation of {L_1} in the form {\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}.

Markscheme{L_1}:{\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 8\\ 1\\ 0 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 0\\ 0\\1 \end{array}} \right) A2 N2

[2 marks]

Examiners reportVery few candidates gave a correct direction vector parallel to the z-axis. Provided they wrote down an equation here they were ableto earn most of subsequent marks on follow through.

132b. [4 marks]The line {L_2} has equation {\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ { - 1} \end{array}} \right) + s\left({\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) . The point A has position vector \left( {\begin{array}{*{20}{c}} 6\\ 2\\ 9\end{array}} \right) .

Show that A lies on {L_2} .

Markschemeevidence of equating {\boldsymbol{r}} and \overrightarrow {{\rm{OA}}} (M1)

e.g. \left( {\begin{array}{*{20}{c}} 6\\ 2\\ 9 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ { - 1} \end{array}}\right) + s\left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) , A = r

one correct equation A1

e.g. 6 = 2 + 2s , 2 = 4 - s , 9 = - 1 + 5s

s = 2 A1

evidence of confirming for other two equations A1

e.g. 6 = 2 + 4 , 2 = 4 - 2 , 9 = - 1 + 10

so A lies on {L_2} AG N0

[4 marks]

Examiners reportFor (b), many found the correct parameter but neglected to confirm it in the other two equations.

132c. [7 marks]Let B be the point of intersection of lines {L_1} and {L_2} .

(i) Show that \overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}} 8\\ 1\\ {14} \end{array}} \right) .

(ii) Find \overrightarrow {{\rm{AB}}} .


e.g. \left( {\begin{array}{*{20}{c}} 2\\ 4\\ { - 1} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}}\right) = \left( {\begin{array}{*{20}{c}} 8\\ 1\\ 0 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 0\\ 0\\ 1 \end{array}}\right) {L_1} = {L_2}

one correct equation A1

e.g. 2 + 2s = 8 , 4 - s = 1 , - 1 + 5s = t


finding s = 3 A1

substituting M1

e.g. \overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}} 2\\ 4\\ { - 1} \end{array}} \right) + 3\left({\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right)

\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}} 8\\ 1\\ {14} \end{array}} \right) AG N0

(ii) evidence of appropriate approach (M1)

e.g. \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} , \overrightarrow{{\rm{AB}}} = \overrightarrow {{\rm{OB}}} - \overrightarrow {{\rm{OA}}}

\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) A1 N2

[7 marks]

Examiners reportIn (c) some performed a trial and error approach to obtaining an integer parameter and thus did not "show" the mathematical origin ofthe result. Finding vector \overrightarrow {{\rm{AB}}} proved accessible.

132d. [3 marks]The point C is at (2, 1, − 4). Let D be the point such that ABCD is a parallelogram.

Find \overrightarrow {{\rm{OD}}} .


e.g. \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{DC}}}

correct values A1

e.g. \overrightarrow {{\rm{OD}}} + \left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) = \left({\begin{array}{*{20}{c}} 2\\ 1\\ { - 4} \end{array}} \right) , \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) + \left({\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2\\ 1\\ { - 4} \end{array}} \right) ,\left( {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {2 - x}\\ {1 - y}\\ { - 4 - z}\end{array}} \right)

\overrightarrow {{\rm{OD}}} = \left( {\begin{array}{*{20}{c}} 0\\ 2\\ { - 9} \end{array}} \right) A1 N2

[3 marks]

Examiners reportA good number of candidates had an appropriate approach to (d), although surprisingly many subtracted \overrightarrow{{\rm{OC}}} from \overrightarrow {{\rm{AB}}} in finding \overrightarrow {{\rm{OD}}} .

133a. [3 marks]

In the following diagram, \boldsymbol{u} = \overrightarrow {{\rm{AB}}} and \boldsymbol{v} = \overrightarrow {{\rm{BD}}} .

The midpoint of \overrightarrow {{\rm{AD}}} is E and \frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{1}{3} .

Express each of the following vectors in terms of \boldsymbol{u} and \boldsymbol{v} .

\overrightarrow {{\rm{AE}}}

Markscheme\overrightarrow {{\rm{AE}}} = \frac{1}{2}\overrightarrow {{\rm{AD}}} A1

attempt to find \overrightarrow {{\rm{AD}}} M1

e.g. \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BD}}} , {\boldsymbol{u}} + {\boldsymbol{v}}

\overrightarrow {{\rm{AE}}} = \frac{1}{2}(u + v) \left( { = \frac{1}{2}{\boldsymbol{u}} + \frac{1}{2}{\boldsymbol{v}}}\right) A1 N2

[3 marks]


133b. [4 marks]\overrightarrow {{\rm{EC}}}

Markscheme\overrightarrow {{\rm{EC}}} = \overrightarrow {{\rm{AE}}} = \frac{1}{2}({\boldsymbol{u}} + {\boldsymbol{v}}) A1

\overrightarrow {{\rm{DC}}} = 3{\boldsymbol{v}} A1

attempt to find \overrightarrow {{\rm{EC}}} M1

e.g. \overrightarrow {{\rm{ED}}} + \overrightarrow {{\rm{DC}}} , \frac{1}{2}({\boldsymbol{u}} + {\boldsymbol{v}}) +3{\boldsymbol{v}}

\overrightarrow {{\rm{EC}}} = \frac{1}{2}{\boldsymbol{u}} + \frac{7}{2}{\boldsymbol{v}} \left( { =\frac{1}{2}({\boldsymbol{u}} + 7{\boldsymbol{v}})} \right) A1 N2

[4 marks]


© International Baccalaureate Organization 2014

134a. [1 mark]

Consider the lines {L_1} , {L_2} , {L_2} , and {L_4} , with respective equations.

{L_1} : \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 3 \end{array}} \right) + t\left({\begin{array}{*{20}{c}} 3\\ { - 2}\\ 1 \end{array}} \right)

{L_2} : \left( \begin{array}{l} x\\ y\\ z \end{array} \right) = \left( \begin{array}{l} 1\\ 2\\ 3 \end{array} \right) + p\left( \begin{array}{l} 3\\2\\ 1 \end{array} \right)

{L_3} : \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ 1\\ 0 \end{array}} \right) + s\left({\begin{array}{*{20}{c}} { - 1}\\ 2\\ { - a} \end{array}} \right)

{L_4} : \left( {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right) = q\left( {\begin{array}{*{20}{c}} { - 6}\\ 4\\ { - 2} \end{array}}\right)

Write down the line that is parallel to {L_4} .

Markscheme{L_1} A1 N1

[1 mark]


134b. [1 mark]Write down the position vector of the point of intersection of {L_1} and {L_2} .

Markscheme\left( \begin{array}{l} 1\\ 2\\ 3 \end{array} \right) A1 N1

[1 mark]


134c. [5 marks]Given that {L_1} is perpendicular to {L_3} , find the value of a .

Markschemechoosing correct direction vectors A1A1

e.g. \left( {\begin{array}{*{20}{c}} 3\\ { - 2}\\ 1 \end{array}} \right) , \left( {\begin{array}{*{20}{c}} { - 1}\\ 2\\ { - a}\end{array}} \right)

recognizing that {\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0 M1


e.g. - 3 - 4 - a = 0

a = - 7 A1 N3

[5 marks]


Printed for sek-catalunya

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Typesetting math: 82%

Documents

Nuevo examen - 20 de Octubre de 2014 - David Delgado - DP Maths Teacher | Maths … · 2016-03-02 · Nuevo examen - 20 de Octubre de 2014 [1207 marks] 1.The equation has two distinct