IB Questionbank Test

2.2 SL

1a. [1 mark]

Markscheme

  (must be an equation)                       A1   N1

[1 mark]

Examiners report

[N/A]

1b. [3 marks]

Markscheme

interchanging and (seen anywhere)      (M1)

eg   ,  

evidence of correct manipulation      (A1)

eg  ,  ,  

  (accept )                 A1   N3

[3 marks]

Examiners report

[N/A]

1c. [2 marks]

Markscheme

valid approach to find horizontal asymptote     (M1)

eg  , vert.asymp of  becomes horiz.asymp of ,  ,  

 (must be an equation)                A1   N2

[2 marks]

Examiners report

[N/A]

2a. [1 mark]

Markscheme

  (must be an equation)        A1 N1

[1 mark]

Examiners report

[N/A]

2b. [2 marks]

Markscheme

valid approach      (M1)

eg   ,  ,  ,  ,  ,  

  (must be an equation)        A1 N2

[2 marks]

Examiners report

[N/A]

2c. [3 marks]

Markscheme

METHOD 1

attempt to substitute 1 into  or        (M1)

eg   ,  

      (A1)

        A1 N2

 

METHOD 2

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

eg   ,  

correct substitution       (A1)

eg   

        A1 N2

 

[3 marks]

Examiners report

[N/A]

3a. [4 marks]

Markscheme

evidence of valid approach       (M1)

eg   sketch of triangle with sides 3 and 5, 

correct working       (A1)

eg  missing side is 4 (may be seen in sketch), ,  

       A2 N4

[4 marks]

Examiners report

[N/A]

3b. [2 marks]

Markscheme

correct substitution of either gradient or origin into equation of line        (A1)

(do not accept )

eg   ,   ,   

     A2 N4

Note: Award A1A0 for .

[2 marks]

Examiners report

[N/A]

3c. [5 marks]

Markscheme

  (seen anywhere, including answer)       A1

choosing product rule       (M1)

eg   

correct derivatives (must be seen in a correct product rule)       A1A1

eg   ,  

       A1 N5

[5 marks]

Examiners report

[N/A]

3d. [4 marks]

Markscheme

valid approach to equate their gradients       (M1)

eg   ,   ,  ,   

correct equation without         (A1)

eg   ,  ,  

correct working       (A1)

eg   ,  

(do not accept )       A1 N1

Note: Do not award the final A1 if additional answers are given.

[4 marks]

 

Examiners report

[N/A]

4a. [3 marks]

Markscheme

  A1A1A1 N3

Note: Only if the shape is approximately correct with exactly 2 maximums and 1 minimum on the interval 0 ≤ ≤ 0, award the following:A1 for correct domain with both endpoints within circle and oval.A1 for passing through the other -intercepts within the circles.A1 for passing through the three turning points within circles (ignore -intercepts and extrema outside of the domain).

[3 marks]

Examiners report

[N/A]

4b. [3 marks]

Markscheme

evidence of reasoning (may be seen on graph)      (M1)

eg  ,  (0.524, 0),  (0.785, 0)

0.523598,  0.785398

,       A1A1  N3

Note: Award M1A1A0 if any solution outside domain (eg ) is also included.

[3 marks]

Examiners report

[N/A]

4c. [2 marks]

Markscheme

     A2  N2

Note: Award A1 if any correct interval outside domain also included, unless additional solutions already penalized in (b).Award A0 if any incorrect intervals are also included.

[2 marks]

Examiners report

[N/A]

5a. [2 marks]

Markscheme

attempt to find      (M1)

eg   ,   ,  

−0.25 (exact)     A1 N2

[2 marks]

Examiners report

[N/A]

5b. [2 marks]

Markscheme

u   or any scalar multiple    A2 N2

[2 marks]

Examiners report

[N/A]

5c. [5 marks]

Markscheme

correct scalar product and magnitudes           (A1)(A1)(A1)

scalar product 

magnitudes ,    

substitution of their values into correct formula           (M1)

eg  ,  ,  2.1112,  120.96° 

1.03037 ,  59.0362°

angle = 1.03 ,  59.0°    A1 N4

[5 marks]

Examiners report

[N/A]

5d. [3 marks]

Markscheme

attempt to form composite      (M1)

eg    ,   ,  

correct working     (A1)

eg   ,  

     A1 N2

[3 marks]

Examiners report

[N/A]

5e. [1 mark]

Markscheme

  (accept , )    A1 N1

Note: Award A0 in part (ii) if part (i) is incorrect.Award A0 in part (ii) if the candidate has found by interchanging and .

[1 mark]

Examiners report

[N/A]

5f. [3 marks]

Markscheme

METHOD 1

recognition of symmetry about     (M1)

eg   (2, 8) ⇔ (8, 2) 

evidence of doubling their angle        (M1)

eg   ,  

2.06075, 118.072°

2.06 (radians)  (118 degrees)     A1  N2

 

METHOD 2

finding direction vector for tangent line at       (A1)

eg   ,  

substitution of their values into correct formula (must be from vectors)      (M1)

eg   ,  

2.06075, 118.072°

2.06 (radians)  (118 degrees)     A1  N2

 

METHOD 3

using trigonometry to find an angle with the horizontal      (M1)

eg   ,  

finding both angles of rotation      (A1)

eg   ,  

2.06075, 118.072°

2.06 (radians)  (118 degrees)     A1  N2

[3 marks]

Examiners report

[N/A]

6a. [2 marks]

Markscheme

valid approach          (M1)

eg   ,   

0.693147

= ln 2 (exact), 0.693      A1 N2

[2 marks]

Examiners report

[N/A]

6b. [3 marks]

Markscheme

attempt to substitute either their correct limits or the function into formula         (M1)

involving 

eg    ,   ,   

3.42545

volume = 3.43     A2 N3

[3 marks]

Examiners report

[N/A]

7a. [2 marks]

Markscheme

valid approach      (M1)

eg   ,  ,  ,  

(0, 6)  (accept  = 0 and  = 6)     A1 N2

 

[2 marks]

Examiners report

[N/A]

7b. [2 marks]

Markscheme

     A2 N2

 

[2 marks]

Examiners report

[N/A]

7c. [4 marks]

Markscheme

valid approach      (M1)

eg   

correct working      (A1)

eg   ,  slope = ,  

attempt to substitute gradient and coordinates into linear equation      (M1)

eg   ,  ,  ,  L 

correct equation      A1 N3

eg  ,  ,  

 

[4 marks]

Examiners report

[N/A]

7d. [8 marks]

Markscheme

valid approach to find intersection      (M1)

eg   

correct equation      (A1)

eg   

correct working      (A1)

eg   ,  

at Q      (A1)

 

valid approach to find minimum      (M1)

eg   

correct equation      (A1)

eg   

substitution of their value of at Q into their equation      (M1)

eg   ,  

= −4     A1 N0

 

[8 marks]

Examiners report

[N/A]

8a. [2 marks]

Markscheme

valid method      (M1)

eg   (0),  sketch of graph

-intercept is   (exact),  −0.333,        A1 N2

 

[2 marks]

Examiners report

[N/A]

8b. [1 mark]

Markscheme

      A1 N1

 

[1 mark]

Examiners report

[N/A]

8c. [2 marks]

Markscheme

valid method      (M1)

eg     ,  , sketch of graph

   (must be an equation)      A1 N2

 

[2 marks]

Examiners report

[N/A]

8d. [2 marks]

Markscheme

valid approach      (M1)

eg   recognizing that  is related to the horizontal asymptote, 

table with large values of , their value from (a)(iii), L’Hopital’s rule .

      A1 N2

 

[2 marks]

Examiners report

[N/A]

9a. [1 mark]

Markscheme

     A1 N1

[1 mark]

Examiners report

[N/A]

9b. [1 mark]

Markscheme

f  (1) = 2     A1 N1

[1 mark]

Examiners report

[N/A]

9c. [1 mark]

Markscheme

−2 ≤ y ≤ 2, y∈ [−2, 2]  (accept −2 ≤ x ≤ 2)     A1 N1

[1 mark]

Examiners report

[N/A]

9d. [4 marks]

Markscheme

A1A1A1A1  N4

Note: Award A1 for evidence of approximately correct reflection in y = x with correct curvature.

(y = x does not need to be explicitly seen)

Only if this mark is awarded, award marks as follows:

A1 for both correct invariant points in circles,

A1 for the three other points in circles,

A1 for correct domain.

[4 marks]

Examiners report

[N/A]

10a. [2 marks]

Markscheme

recognize that  is the gradient of the tangent at      (M1)

eg   

  (accept m = 3)     A1 N2

[2 marks]

Examiners report

[N/A]

10b. [2 marks]

Markscheme

recognize that      (M1)

eg 

     A1 N2

[2 marks]

Examiners report

[N/A]

10c. [5 marks]

Markscheme

recognize that the gradient of the graph of g is       (M1)

choosing chain rule to find       (M1)

eg  

     A2

     A1

     AG N0

[5 marks]

 

 

Examiners report

[N/A]

10d. [7 marks]

Markscheme

 at Q, L = L (seen anywhere)      (M1)

recognize that the gradient of L is g'(1)  (seen anywhere)     (M1)eg  m = 6

finding g (1)  (seen anywhere)      (A1)eg  

attempt to substitute gradient and/or coordinates into equation of a straight line      M1eg  

correct equation for L 

eg       A1

correct working to find Q       (A1)eg   same y-intercept, 

     A1 N2

[7 marks]

 

Examiners report

[N/A]

11a. [2 marks]

Markscheme

A2 N2[2 marks]

Examiners report

[N/A]

11b. [4 marks]

Markscheme

recognizing horizontal shift/translation of 1 unit      (M1)

eg  b = 1, moved 1 right

recognizing vertical stretch/dilation with scale factor 2      (M1)

eg   a = 2,  y ×(−2)

a = −2,  b = −1     A1A1 N2N2

[4 marks]

Examiners report

[N/A]

12a. [1 mark]

Markscheme

(1,5) (exact)      A1 N1

[1 mark]

Examiners report

[N/A]

12b. [3 marks]

Markscheme

      A1A1A1  N3

Note: The shape must be a concave-down parabola.Only if the shape is correct, award the following for points in circles:A1 for vertex,A1 for correct intersection points,A1 for correct endpoints.

[3 marks]

Examiners report

[N/A]

12c. [3 marks]

Markscheme

integrating and subtracting functions (in any order)      (M1)eg  

correct substitution of limits or functions (accept missing dx, but do not accept any errors, including extra bits)     (A1)eg 

area = 9  (exact)      A1 N2

[3 marks]

Examiners report

[N/A]

13a. [2 marks]

Markscheme

valid approach     (M1)eg  or 0…

1.14472

   (exact), 1.14      A1 N2

[2 marks]

Examiners report

[N/A]

13b. [3 marks]

Markscheme

attempt to substitute either their limits or the function into formula involving .     (M1)

eg  

2.49799

volume = 2.50      A2 N3

[3 marks]

Examiners report

[N/A]

14a. [2 marks]

Markscheme

valid approach       (M1)eg  

c = −2      A1 N2

[2 marks]

Examiners report

[N/A]

14b. [2 marks]

Markscheme

valid approach (M1)eg  

y = −4 (must be an equation)      A1 N2

[2 marks]

Examiners report

[N/A]

14c. [3 marks]

Markscheme

valid approach to analyze modulus function      (M1)eg   sketch, horizontal asymptote at y = 4, y = 0

k = 4, k = 0     A2 N3

[3 marks]

Examiners report

[N/A]

15a. [1 mark]

Markscheme

correct range (do not accept )     A1     N1

eg

[1 mark]

Examiners report

[N/A]

15b. [1 mark]

Markscheme

    A1     N1

[1 mark]

Examiners report

[N/A]

15c. [1 mark]

Markscheme

    A1     N1

[1 mark]

Examiners report

[N/A]

15d. [3 marks]

Markscheme

     A1A1A1     N3

 

Notes:     Award A1 for both end points within circles,

A1 for images of and within circles,

A1 for approximately correct reflection in , concave up then concave down shape (do not accept line segments).

 

[3 marks]

Examiners report

[N/A]

16a. [2 marks]

Markscheme

valid approach     (M1)

eg

0.816496

(exact), 0.816     A1     N2

[2 marks]

Examiners report

[N/A]

16b. [2 marks]

Markscheme

    A1A1     N2

[2 marks]

Examiners report

[N/A]

16c. [3 marks]

Markscheme

     A1A1A1     N3

 

Notes:     Award A1 for correct domain and endpoints at and in circles,

A1 for maximum in square,

A1 for approximately correct shape that passes through their -intercept in circle and has changed from concave down to concave up between 2.29 and 7.

 

[3 marks]

Examiners report

[N/A]

17a. [3 marks]

Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an -intercept     (M1)

eg,

valid approach     (M1)

eg

    A1     N2

METHOD 2 (expanding f (x)) 

correct expansion (accept absence of )     (A1)

eg

valid approach involving equation of axis of symmetry     (M1)

eg

    A1     N2

METHOD 3 (using derivative)

correct derivative (accept absence of )     (A1)

eg

valid approach     (M1)

eg

    A1     N2

[3 marks]

Examiners report

[N/A]

17b. [3 marks]

Markscheme

attempt to substitute     (M1)

eg

correct working     (A1)

eg

    A1     N2

[3 marks]

Examiners report

[N/A]

17c. [8 marks]

Markscheme

METHOD 1 (using discriminant)

recognizing tangent intersects curve once     (M1)

recognizing one solution when discriminant = 0     M1

attempt to set up equation     (M1)

eg

rearranging their equation to equal zero     (M1)

eg

correct discriminant (if seen explicitly, not just in quadratic formula)     A1

eg

correct working     (A1)

eg

    A1A1     N0

METHOD 2 (using derivatives)

attempt to set up equation     (M1)

eg

recognizing derivative/slope are equal     (M1)

eg

correct derivative of     (A1)

eg

attempt to set up equation in terms of either or     M1

eg

rearranging their equation to equal zero     (M1)

eg

correct working     (A1)

eg

    A1A1     N0

[8 marks]

Examiners report

[N/A]

18a. [1 mark]

Markscheme

    A1     N1

[1 mark]

Examiners report

[N/A]

18b. [2 marks]

Markscheme

attempt to substitute into their derivative     (M1)

gradient of is     A1     N2

[2 marks]

Examiners report

[N/A]

18c. [5 marks]

Markscheme

METHOD 1 

attempt to substitute coordinates of A and their gradient into equation of a line     (M1)

eg

correct equation of in any form     (A1)

eg

valid approach     (M1)

eg

correct substitution into equation     A1

eg

correct working     A1

eg

    AG     N0

METHOD 2

valid approach     (M1)

eg

recognizing at B     (A1)

attempt to substitute coordinates of A and B into slope formula     (M1)

eg

correct equation     A1

eg

correct working     A1

eg

    AG     N0

[5 marks]

Examiners report

[N/A]

18d. [2 marks]

Markscheme

valid approach to find area of triangle     (M1)

eg

area of     A1     N2

[2 marks]

Examiners report

[N/A]

18e. [7 marks]

Markscheme

METHOD 1 ()

valid approach to find area from to 0     (M1)

eg

correct integration (seen anywhere, even if M0 awarded)     A1

eg

substituting their limits into their integrated function and subtracting     (M1)

eg, area from to 0 is

 

Note:     Award M0 for substituting into original or differentiated function.

 

attempt to find area of     (M1)

eg

correct working for     (A1)

eg

correct substitution into     (A1)

eg

    A1     N2

METHOD 2 ()

valid approach to find area of     (M1)

eg

correct integration (seen anywhere, even if M0 awarded)     A2

eg

substituting their limits into their integrated function and subtracting     (M1)

eg

 

Note:     Award M0 for substituting into original or differentiated function.

 

correct working for     (A1)

eg

correct substitution into     (A1)

eg

    A1     N2

[7 marks]

Examiners report

[N/A]

19a. [2 marks]

Markscheme

valid approach     (M1)

eg,

-intercept is 2.9     A1     N2

[2 marks]

Examiners report

[N/A]

19b. [2 marks]

Markscheme

valid approach involving equation or inequality     (M1)

eg

(must be an equation)     A1     N2

[2 marks]

Examiners report

[N/A]

19c. [2 marks]

Markscheme

7.01710

    A2     N2

 

Note:     If candidate gives the minimum point as their final answer, award A1 for .

 

[2 marks]

Examiners report

[N/A]

20a. [2 marks]

Markscheme

attempt to form composite in either order     (M1)

eg

    A1

    AG     N0

[2 marks]

Examiners report

[N/A]

20b. [3 marks]

Markscheme

   A1

A1A1     N3

 

Note:     Award A1 for approximately correct shape which changes from concave down to concave up. Only if this A1 is awarded, award the following:

A1 for left hand endpoint in circle and right hand endpoint in oval,

A1 for minimum in oval.

 

[3 marks]

Examiners report

[N/A]

20c. [3 marks]

Markscheme

evidence of identifying max/min as relevant points     (M1)

eg

correct interval (inclusion/exclusion of endpoints must be correct)     A2     N3

eg

[3 marks]

Examiners report

[N/A]

21a. [2 marks]

Markscheme

evidence of valid approach     (M1)

eg

2.73205

    A1     N2

[2 marks]

Examiners report

[N/A]

21b. [2 marks]

Markscheme

1.87938, 8.11721

    A2     N2

[2 marks]

Examiners report

[N/A]

21c. [1 mark]

Markscheme

rate of change is 0 (do not accept decimals)     A1     N1

[1 marks]

Examiners report

[N/A]

21d. [4 marks]

Markscheme

METHOD 1 (using GDC)

valid approach     M1

eg, max/min on

sketch of either or , with max/min or root (respectively)     (A1)

    A1     N1

Substituting their value into     (M1)

eg

    A1     N1

METHOD 2 (analytical)

    A1

setting     (M1)

    A1     N1

substituting their value into     (M1)

eg

    A1     N1

[4 marks]

Examiners report

[N/A]

21e. [3 marks]

Markscheme

recognizing rate of change is     (M1)

eg

rate of change is 6     A1     N2

[3 marks]

Examiners report

[N/A]

21f. [3 marks]

Markscheme

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

involving (accept absence of and/or )

eg

128.890

    A2     N3

[3 marks]

Examiners report

[N/A]

22a. [2 marks]

Markscheme

     A1A1     N2

[2 marks]

Examiners report

[N/A]

22b. [5 marks]

Markscheme

(i) (ii)          A1

A1A1A1     N4

A1     N1

 

Notes: (i) Award A1 for correct cubic shape with correct curvature.

Only if this A1 is awarded, award the following:

A1 for passing through their point A and the origin,

A1 for endpoints,

A1 for maximum.

(ii) Award A1 for horizontal line through their A.

 

[5 marks]

Examiners report

[N/A]

23a. [3 marks]

Markscheme

(i)     3     A1     N1

(ii)     valid attempt to find the period     (M1)

eg

period      A1     N2

[3 marks]

Examiners report

Almost all candidates correctly stated the amplitude but then had difficulty finding the correct period. Few students faced problems in sketching the graph of the given function, even if they had found the wrong period, thus indicating a lack of understanding of the term ‘period’ in part a(ii). Most sketches were good although care should be taken to observe the given domain and to draw a neat curve.

23b. [4 marks]

Markscheme

     A1A1A1A1     N4

[4 marks]

Examiners report

Almost all candidates correctly stated the amplitude but then had difficulty finding the correct period. Few students faced problems in sketching the graph of the given function, even if they had found the wrong period, thus indicating a lack of understanding of the term ‘period’ in part a(ii). Most sketches were good although care should be taken to observe the given domain and to draw a neat curve.

24a. [2 marks]

Markscheme

recognition that the -coordinate of the vertex is  (seen anywhere)     (M1)

egaxis of symmetry is , sketch, 

correct working to find the zeroes     A1

eg

and      AG     N0

[2 marks]

Examiners report

As a ‘show that’ question, part a) required a candidate to independently find the answers. Again, too many candidates used the given answers (of 3 and ) to show that the two zeros were 3 and (a circular argument). Those who were able to recognize that the -coordinate of the vertex is tended to then use the given answers and work backwards thus scoring no further marks in part a).

24b. [4 marks]

Markscheme

METHOD 1 (using factors)

attempt to write factors     (M1)

eg

correct factors     A1

eg

   A1A1     N3

METHOD 2 (using derivative or vertex)

valid approach to find     (M1)

eg

   A1

correct substitution     A1

eg

   A1

   N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations     (M1)

eg

one correct value

eg     A1

correct substitution     A1

eg

second correct value     A1

eg

   N3

[4 marks]

Examiners report

Answers to part b) were more successful with a good variety of methods used and correct solutions seen.

25a. [2 marks]

Markscheme

   A1A1     N2

[2 marks]

Examiners report

Nearly all candidates performed well on this question, earning full marks on all three question parts.

25b. [2 marks]

Markscheme

   A1A1     N2

[2 marks]

Examiners report

Nearly all candidates performed well on this question, earning full marks on all three question parts. In part (b), there were some candidates who factored the quadratic expression correctly, but went on to give negative values for and .

25c. [2 marks]

Markscheme

attempt to substitute into their     (M1)

eg

   A1     N2

[2 marks]

Examiners report

Nearly all candidates performed well on this question, earning full marks on all three question parts.

26a. [5 marks]

Markscheme

METHOD 1

recognizing      (M1)

recognizing displacement of P in first 5 seconds (seen anywhere)     A1

(accept missing )

eg

valid approach to find total displacement     (M1)

eg

0.284086

0.284 (m)     A2     N3

METHOD 2

recognizing      (M1)

correct integration     A1

eg (do not penalize missing “”)

attempt to find     (M1)

eg

attempt to substitute into their expression with     (M1)

eg

0.284086

0.284 (m)     A1     N3

[5 marks]

Examiners report

This question was not well done throughout. Analytical approaches were almost always unsuccessful as a result of poor integration and differentiation skills and many of the errors were a result of having the GDC in degree mode. In (a), most candidates recognized the need to integrate to find the displacement, although a significant number differentiated . Of those that integrated, many assumed incorrectly that the initial displacement was the value of the constant of integration. Some candidates integrated and obtained no marks for an invalid approach. In the case where a correct definite integral was given, it was disappointing to see many candidates try to evaluate it analytically rather than using their GDC.

26b. [2 marks]

Markscheme

recognizing that at rest,      (M1)

     A1     N2

[2 marks]

Examiners report

This question was not well done throughout. Analytical approaches were almost always unsuccessful as a result of poor integration and differentiation skills and many of the errors were a result of having the GDC in degree mode.  In part (b), many candidates did not read the question carefully and gave the two occasions, in the given domain, where the particle was at rest.

26c. [2 marks]

Markscheme

recognizing when change of direction occurs     (M1)

eg crosses axis

2 (times)     A1     N2

[2 marks]

Examiners report

This question was not well done throughout. Analytical approaches were almost always unsuccessful as a result of poor integration and differentiation skills and many of the errors were a result of having the GDC in degree mode. In part (c), many candidates did not appreciate that velocity is a vector and that the particle would change direction when its velocity changes sign. Consequently, many candidates gave the incorrect answer of four changes in directions, rather than the correct two direction changes.

26d. [2 marks]

Markscheme

acceleration is (seen anywhere)     (M1)

eg

0.743631

     A1     N2

[2 marks]

Examiners report

This question was not well done throughout. Analytical approaches were almost always unsuccessful as a result of poor integration and differentiation skills and many of the errors were a result of having the GDC in degree mode. Part (d), was done very poorly, with candidates struggling to differentiate sine and cosine correctly and to evaluate their derivative. As with question 3, many candidates worked with the incorrect angle setting on their calculator.

26e. [3 marks]

Markscheme

valid approach involving max or min of     (M1)

eg, graph

one correct co-ordinate for min     (A1)

eg

     A1     N2

[3 marks]

Examiners report

This question was not well done throughout. Analytical approaches were almost always unsuccessful as a result of poor integration and differentiation skills and many of the errors were a result of having the GDC in degree mode. Few candidates attempted part (e). Of those that did, many attempted to find the largest local maximum of the graph rather than least local minimum as they did not recognise speed as .

27a. [2 marks]

Markscheme

(correct equation only)     A2     N2

[2 marks]

Examiners report

Part (a) was in general well answered. Many candidates lost the marks for writing 2 or instead of .

27b. [2 marks]

Markscheme

valid approach     (M1)

eg

   A1     N2

[2 marks]

Examiners report

In part (b) some candidates got confused and found instead of . When calculating the derivative, two types of approaches were seen. Most of the ones who rewrote the function as , applied the chain rule correctly. Those who tried to apply the quotient rule made various mistakes: incorrect derivative of a constant, incorrect multiplication by zero, wrong subtraction order in the numerator, omitted the negative sign in the answer.

27c. [2 marks]

Markscheme

correct equation for the asymptote of

eg     (A1)

     A1     N2

[2 marks]

Examiners report

In (c), most candidates were coherent and obtained the same value as the one written in part (a).

27d. [4 marks]

Markscheme

correct derivative of g (seen anywhere)     (A2)

eg

correct equation     (A1)

eg

7.38905

     A1     N2

[4 marks]

Examiners report

In part (d) many candidates did not manage to differentiate the function g correctly. Of those who could, the equation was generally well solved algebraically.

27e. [4 marks]

Markscheme

attempt to equate their derivatives     (M1)

eg

valid attempt to solve their equation     (M1)

egcorrect value outside the domain of such as 0.522 or 4.51,

correct solution (may be seen in sketch)     (A1)

eg

gradient is      A1     N3

[4 marks]

Examiners report

For part (e), not many candidates wrote a correct equation with their derivatives. There was mixed performance for this question, as those who knew they needed to use their GDC managed to obtain an answer, while many got tangled in unsuccessful attempts to solve the equation algebraically. Many candidates tried to solve quite complex equations ‘manually’ instead of trying to graph the expressions on their calculators and finding the value of at the point of intersection. Of those students who tried to solve graphically only a small percentage actually sketched the two curves that they were considering. This sketch is particularly useful to examiners to see how the student is thinking, or what steps s/he is taking to solve the equations.

Only a few realized that the question asked for the gradient, which was represented by the -coordinate of the point of intersection, rather than the -coordinate.

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