MATH HL2 EXAM PREP CORE TOPICS FUNCTIONS & …

MATH HL2 EXAM PREP – CORE TOPICS – FUNCTIONS & EQUATIONS (SOLUTIONS)

IB Questionbank Mathematics Higher Level 3rd edition 1

1. (a) (i) (g ○ f)(x) = 2

3,

32

1

x

x (or equivalent) A1

(ii) (f ○ g)(x) = x

2 + 3, x ≠ 0 (or equivalent) A1

(b) EITHER

f(x) = (g–1

○ f ○ g)(x) (g ○ f)(x) = (f ○ g)(x) (M1)

32

32

1

xx A1

OR

(g–1

○ f ○ g)(x) =

32

1

x

A1

2x + 3 =

32

1

x

M1

THEN

6x2 + 12x + 6 = 0 (or equivalent) A1

x = –1, y = 1 (coordinates are (–1, 1)) A1 [6]

2. (a) f (x – a) ≠ b (M1)

x ≠ 0 and x ≠ 2a (or equivalent) A1



(b) vertical asymptotes x = 0, x = 2a A1

horizontal asymptote y = 0 A1

Note: Equations must be seen to award these marks.

maximum

ba

1, A1A1

Note: Award A1 for correct x-coordinate and A1 for correct y-coordinate.

one branch correct shape A1

other 2 branches correct shape A1

[8]

3. (a) METHOD 1

f′(x) = q – 2x = 0 M1

f′(3) = q – 6 = 0

q = 6 A1

f(3) = p + 18 – 9 = 5 M1

p = –4 A1

METHOD 2

f(x) = –(x – 3)2 + 5 M1A1

= –x2 + 6x – 4

q = 6, p = –4 A1A1



(b) g(x) = –4 + 6(x – 3) – (x – 3)2 (= –31 + 12x – x

2) M1A1

Note: Accept any alternative form that is correct.

Award M1A0 for a substitution of (x + 3). [6]

4. (a)

A3

Note: Award A1 for each correct branch with position of asymptotes

clearly indicated. If x = 2 is not indicated, only penalise once.

(b)

A3

Note: Award A1 for behaviour at x = 0, A1 for intercept at x = 2,

A1 for behaviour for large │x│. [6]



5. x = 2ey

ye

1 M1

Note: The M1 is for switching the variables and may be awarded at any

stage in the process and is awarded independently. Further marks

do not rely on this mark being gained.

xey = 2e

2y – 1

2e2y

– xey – 1 = 0 A1

4

8e

2

xxy M1A1

y = ln

4

82xx

therefore h–1

(x) = ln

4

82xx A1

since ln is undefined for the second solution R1

Note: Accept y = ln

4

82xx.

Note: The R1 may be gained by an appropriate comment earlier. [6]

6. (a) (i)

A2

Note: Award A1 for correct sin x, A1 for correct sin 2x.

Note: Award A1A0 for two correct shapes with 2

π and/or 1 missing.

Note: Condone graph outside the domain.

(ii) sin 2x = sin x, 0 ≤ x ≤ 2

π

2 sin x cos x – sin x = 0 M1



sin x (2 cos x – 1) = 0

x = 0, 3

π A1A1 N1N1

(iii) area = 3

π

0d )sin(sin2 xxx M1

Note: Award M1 for an integral that contains limits, not necessarily

correct, with sin x and sin 2x subtracted in either order.

= 3

π

0

cos2cos2

1

xx A1

=

0cos0cos

2

1

3

πcos

3

π2cos

2

1 (M1)

= 2

1

4

3

= 4

1 A1

(b)

6

π

0 2

21

0 sin44

sin4d

4

x

x

x × 8 sin θ cos θ dθ M1A1A1

Note: Award M1 for substitution and reasonable attempt at finding

expression for dx in terms of dθ, first A1 for correct limits,

second A1 for correct substitution for dx.

dsin86

π

0

2

A1

d2cos446

π

0 M1

= 6

π

02sin24 A1

= 03

πsin2

3

π2

(M1)

= 33

π2 A1



(c) (i)

M1

from the diagram above

the shaded area =

ba

yyfabxxf0

1

0d)(d)( R1

= ab –

b

xxf0

1 d)( AG

(ii) f(x) = arcsin 4

x f–1

(x) = 4 sin x A1

6

π

0

2

0dsin4

3

πd

4arcsin xxx

x M1A1A1

Note: Award A1 for the limit 6

π seen anywhere, A1 for all else correct.

= 60cos4

3

π

x A1

= 3243

π A1

Note: Award no marks for methods using integration by parts. [25]

7. (a) (i) median = 104 grams A1

Note: Accept 105.

(ii) 30th

percentile = 90 grams A1



(b) 80 – 49 (M1)

= 31 A1

Note: Accept answers 30 to 32. [4]

8. (a) f(a) = 4a3 + 2a

2 – 7a = –10 M1

4a3 + 2a

2 – 7a + 10 = 0

(a + 2) (4a2 – 6a + 5) = 0 or sketch or GDC (M1)

a = –2 A1

(b) substituting a = –2 into f(x)

f(x) = 4x3 – 4x + 14 = 0 A1

EITHER

graph showing unique solution which is indicated (must include

max and min) R1

OR

convincing argument that only one of the solutions is

real (–1.74, 0.868±1.12i) R1 [5]



9.

M1A1A1A1A1A1A1

Note: Award A1 for both vertical asymptotes correct,

M1 for recognizing that there are two turning points near the origin,

A1 for both turning points near the origin correct, (only this

A mark is dependent on the M mark)

A1 for the other pair of turning points correct,

A1 for correct positioning of the oblique asymptote,

A1 for correct equation of the oblique asymptote,

A1 for correct asymptotic behaviour in all sections. [7]

10. using the factor theorem or long division M1

–A + B – 1 + 6 = 0 A – B = 5 A1

8A + 4B + 2 + 6 = 0 2A + B = –2 A1

3A = 3 A = 1 A1

B = –4 A1 N3

Note: Award M1A0A0A1A1 for using (x – 3) as the third factor, without

justification that the leading coefficient is 1. [5]



11. 22x–2

= 2x + 8 (M1)

4

12

2x = 2

x + 8 (A1)

22x

– 4 × 2x – 32 = 0 A1

(2x – 8)(2

x + 4) = 0 (M1)

2x = 8 x = 3 A1

Notes: Do not award final A1 if more than 1 solution is given. [5]

12. (a) an attempt to use either asymptotes or intercepts (M1)

a = –2, b = 1, c = 2

1 A1A1A1

(b)

A4

Note: Award A1 for both asymptotes,

A1 for both intercepts,

A1, A1 for the shape of each branch, ignoring shape at (x = –2). [8]

13. (a) attempt at completing the square (M1)

3x2 – 6x + 5 = 3(x

2 – 2x) + 5 = 3 (x – 1)

2 – 1 + 5 (A1)

= 3(x – 1)2 + 2 A1

(a = 3, b = –1, c = 2)



(b) definition of suitable basic transformations:

T1 = stretch in y direction scale factor 3 A1

T2 = translation

0

1 A1

T3 = translation

2

0 A1

[6]

14. (a) Note: Interchange of variables may take place at any stage.

for the inverse, solve for x in

y = 1

32

x

x

y(x – 1) = 2x – 3 M1

yx – 2x = y – 3

x(y – 2) = y – 3 (A1)

x = 2

3

y

y

2

3)(1

x

xxf (x ≠ 2) A1

Note: Do not award final A1 unless written in the form f–1

(x) = ...

(b) ±f–1

(x) = 1 + f–1

(x) leads to

2

32

x

x = –1 (M1)A1

x = 3

8 A1

[6]

15. (a) (A + B)2 = A

2 + AB + BA + B

2 A2

(b) (A – kI)3 = A

3 – 3kA

2 + 3k

2A – k

3I A2

(c) CA = B C = BA–1

A2

Note: Award A1 in parts (a) to (c) if error is correctly identified, but not corrected. [6]

16. (a) x ≥ 0 and x ≠ 16 A1A1

(b)



graph not to scale (M1)

finding crossing points

e.g. 4 – x2 = 4 – x

x = 0 or x = 1 (A1)

0 ≤ x ≤ 1 or x > 16 A1A1

Note: Award M1A1A1A0 for solving the inequality only for the case x < 16 [6]

17. (a) rewrite the equation as (4x – 1)ln 2 = (x + 5)ln 8 + (1 – 2x) log216 (M1)

(4x – 1)ln 2 = (3x + 15)ln 2 + 4 – 8x (M1)(A1)

x = 2ln8

2ln164

A1

(b) x = a2 (M1)

a = 1.318 A1

Note: Treat 1.32 as an AP.

Award A0 for ±. [6]



18. EITHER

│x – 1│ > │2x – 1│ (x – 1)2 > (2x – 1)

2 M1

x2 – 2x + 1 > 4x

2 – 4x + 1

3x2 – 2x < 0 A1

0 < x < 3

2 A1A1 N2

Note: Award A1A0 for incorrect inequality signs.

OR

│x – 1│ > │2x – 1│

x – 1 = 2x – 1 x – 1 = 1 – 2x M1A1

–x = 0 3x = 2

x = 0 x = 3

2

Note: Award M1 for any attempt to find a critical value. If graphical

methods are used, award M1 for correct graphs, A1 for correct

values of x.

0 < x < 3

2 A1A1 N2

Note: Award A1A0 for incorrect inequality signs [4]

19. (a) 4

π – arccos x ≥ 0

arccos x ≤ 4

π (M1)

x ≥

2

1accept

2

2x (A1)

since –1 ≤ x ≤ 1 (M1)

1

2

1accept 1

2

2xx A1

Note: Penalize the use of < instead of ≤ only once.



(b) y =

2

4

πcosarccos

4

πyxx M1A1

f–1

: x → cos

2

4

πx A1

0 ≤ x ≤ 4

π A1

[8]

20. EITHER

5ln2ln3

1lnln

5lnln

1ln

23yx

yx

yx

y

x

M1A1

solve simultaneously

5

2ln

5

7ln

y

x

M1

x = 5

7

e (= 4.06) and y = 5

2

e (= 1.49) A1A1

OR

lny

x = 1

x = ey A1

ln x3 + ln y

2 = 5

ln x3y

2 = 5

x3y

2 = e

5 M1

e3y

5 = e

5

y5 = e

2 M1

y = 5

7

5

2

e,e x A1A1

[5]



21.

correct concavities A1A1

Note: Award A1 for concavity of each branch of the curve.

correct x-intercept of g

f (which is EXACTLY the x-intercept of f) A1

correct vertical asymptotes of g

f (which ONLY occur when x equals the

x-intercepts of g) A1A1 [5]

22. g(x) = 0

log5|2log3x| = 0 (M1)

|2log3x| = 1 A1

log3x = ±2

1 (A1)

x = 2

1

3

A1

so the product of the zeros of g is 2

1

2

1

33

= 1 A1 N0 [5]



23. (a)

A1A1

Note: Award A1 for the correct x-intercept,

A1 for completely correct graph.

(b) METHOD 1

the area under the graph of y = 2

ax for –a ≤ x ≤ a, can be divided

into ten congruent triangles; M1A1

the area of eight of these triangles is given by 0

d2a

xa

x

and the areas of the other two by a

xa

x0

d2

M1A1

so, xa

xxa

xa

ad

24d

2 0

0

k = 4 A1 N0



METHOD 2

use area of trapezium to calculate M1

20

22

3

2

1d

2a

aaax

ax

a

A1

and area of two triangles to obtain M1

422

12d

2

22

0

aax

ax

a

A1

so, k = 4 A1 N0

METHOD 3

use integration to find the area under the curve

xa

xxa

xaa

d2

d2

00

M1

= 222

02

2222a

aax

ax

a

A1

and

xa

xxa

xxa

xa

a

aa

d2

d2

d2

2

2

00 M1

= 44822482222

2222222

2

22

0

2 aaaaaaax

axx

axa

a

a

A1

so, k = 4 A1 N0 [7]

24. (a) METHOD 1

V = a3 –

3

1

a A1

x3 =

31

aa M1

= a3 – 3a +

3

13

aa

= a3 –

aa

a

13

13

(or equivalent) (A1)

xxa

a 31 3

3

3

V = x3 + 3x A1 N0



METHOD 2

V = a3 –

3

1

a A1

attempt to use difference of cubes formula, x3 – y

3 = (x – y) (x

2 + xy + y

2) M1

V =

2

2 11

1

aa

aa

=

3

112

aa

aa (A1)

= x(x2 + 3) or x

3 + 3x A1 N0

METHOD 3

diagram showing that the solid can be decomposed M1

into three congruent x × a × a

1 cuboids with volume x A1

and a cube with edge x with volume x3 A1

so, V = x3 + 3x A1 N0



(b) Note: Do not accept any method where candidate substitutes

the given value of a into x = a – a

1.

METHOD 1

V = 4x x3 + 3x = 4x x

3 – x = 0

x(x – 1)(x + 1) = 0 M1

x = 1 as x > 0 A1

so, a – 11

a a

2 – a – 1 = 0

2

51 a M1A1

as a > 1, a = 2

51 AG N0

METHOD 2

a3 –

aa

a

14

13

= a6 – 4a

4 + 4a

2 – 1 = 0

(a2 – 1)(a

4 – 3a

2 + 1) = 0 M1A1

as a > 1

2

222

2

51

2

53,1

aaa M1A1

2

51 a AG N0

[8]

25. (a) f(1) = 1 – arctan1 = 1 4

π A1

3

π3)3arctan(3)3( f A1

(b) f(–x) = – x – arctan(–x) M1

= –x + arctan x A1

= –(x – arctan x)

= –f(x) AG N0



(c) as –2

πarctan

2

π x , for any x A1

2

πarctan

2

π x , for any x

then by adding x (or equivalent) R1

we have x – 2

πarctan

2

π xxx AG N0

(d) f′(x) = 1 2

2

2 1or

1

1

x

x

x A1A1

f″(x) = 2222

32

)1(

2or

)1(

2)1(2

x

x

x

xxx

M1A1

f′(0) = f″(0) = 0 A1A1

EITHER

as f ′(x) ≥ 0 for all values of x

((0,0) is not an extreme of the graph of f (or equivalent)) R1

OR

as f″(x) > 0 for positive values of x and f″(x) < 0 for

negative values of x R1

THEN

(0, 0) is a point of inflexion of the graph of f (with zero gradient) A1 N2



(e)

A1A1A1

Note: Award A1 for both asymptotes.

A1 for correct shape (concavities) x < 0.

A1 for correct shape (concavities) x > 0.

(f) (see sketch above)

as f is increasing (and therefore one-to-one) and its range is ,

f–1

is defined for all x R1

use the result that the graph of y = f–1

(x) is the reflection

in the line y = x of the graph of y = f(x) to draw the graph of f–1

(M1)A1 [20]

26. q(–1) = k + 9 M1A1

q(–2) = 4k + 9 A1

k + 9 = 7(4k + 9) M1

k = –2 A1

Notes: The first M1 is for one substitution and the consequent equations.

Accept expressions for q(–1) and q(–2) that are not simplified. [5]



27. (a)

A1

Note: Award A1 for correct concavity, passing through (0, 0) and increasing.

Scales need not be there.

(b) a statement involving the application of the Horizontal Line Test or equivalent A1

(c) y = xk

for either x = 2

2

or k

yxyk A1

f–1

(x) = 2

2

k

x A1

dom(f–1

(x)) = [0, ∞[ A1

(d) xkk

x

2

2

or equivalent method M1

k = x

k = 2 A1

(e) (i) A = b

axyy d)( 21 (M1)

A = xxx d4

12

4

0

22

1

A1

=

4

0

32

3

12

1

3

4

xx A1

= 3

16 A1



(ii) attempt to find either f′(x) or (f–1

)′(x) M1

f′(x) =

2)()(,

1 1 xxf

x A1A1

2

1 c

c M1

c = 3

2

2 A1 [16]

28.

f′(x) = 2)1(

2

x

M1A1

Note: Alternatively, award M1A1 for correct sketch of the derivative.

find at least one point of intersection of graphs (M1)

y = f(x) and y = f′(x) for x = 3 or 1.73 (A1)

y = f(x) and y = g (x) for x = 0 (A1)

forming inequality 0 ≤ x ≤ 3 (or 0 ≤ x ≤ 1.73) A1A1 N4

Note: Award A1 for correct limits and A1 for correct inequalities. [7]



29. (a) EITHER

translation of 2

1 parallel to the x-axis

stretch of a scale factor of 2

1 parallel to the x-axis A1A1

OR

stretch of a scale factor of 2

1 parallel to the x-axis

translation of –1 parallel to the x-axis A1A1

Note: Accept clear alternative terminologies for either transformation.

(b) EITHER

1.16 < x < 5.71 6.75 < x ≤ 10 A1A1A1A1

OR

]1.16, 5.71[ ]6.75,10] A1A1A1A1

Note: Award A1 for 1 intersection value, A1 for the other 2,

A1A1 for the intervals. [6]

30. (a) x

y

d

d = 24x

2 + 2bx + c (A1)

24x2 + 2bx + c = 0 (M1)

∆ = (2b)2 – 96(c) (A1)

4b2 – 96c > 0 A1

b2 > 24c AG

(b) 1 + cb2

1

4

1 + d = –12

6 + b + c = 0

–27 + cb2

3

4

9 + d = 20

54 – 3b + c = 0 A1A1A1

Note: Award A1 for each correct equation, up to 3, not necessarily simplified.

b = 12, c = –18, d = –7 A1 [8]



31. (a) f(1) = 3 – a + b (A1)

f(–1) = –3 + a + b (A1)

3 – a + b = – 3 + a + b M1

2a = 6

a = 3 A1 N4

(b) b is any real number A1 [5]

32. (a) –l ≤ ln x ≤ 1 (M1)

e

1 ≤ x ≤ e A1A1

(b) y = arcsin (ln x) ln x = sin y (M1)

ln y = sin x y = esinx

(M1)

f–1

(x) = esinx

A1 [6]

33. (a) EITHER

graph of the cubic is shifted horizontally one unit to the right (M1)

x = –0.796 A1

OR

(x – 1) = –1.796 (M1)

x = –0.796 A1

(b) EITHER

stretch factor of 0.5 in the x-direction (M1)

2x = –1.796 (M1)

Note: At least one of the above lines must be seen to award the M marks.

x = –0.898 A1

OR

8x3 – 2x + 4 = (2x)

3 – 2x + 4 = 0 (M1)

2x = –1.796 (M1)

Note: At least one of the above lines must be seen to award the M marks.

x = –0.898 A1 [5]

34. (a)



A3

Note: Award A1 for each graph

A1 for the point of tangency.

point on curve and line is (a, ln a) (M1)

y = ln (x)

ax

y

xx

y 1

d

d1

d

d (when x = a) (M1)A1

EITHER

gradient of line, m, through (0, 0) and (a, ln a) is a

aln (M1)A1

e

1e1ln

1ln maa

aa

a M1A1

OR

y – ln a = a

1(x – a) (M1)A1

passes through 0 if

ln a – 1 = 0 M1

a = e e

1 m A1

THEN

xye

1 A1

(b) the graph of ln x never goes above the graph of y = xe

1, hence ln x ≤

e

x R1

(c) lnx xxxxx

elnlnee

M1A1

exponentiate both sides of lnxe ≤ x x

e ≤ e

x R1AG



(d) equality holds when x = e R1

letting x = π πe < e

π A1 N0

[17]

35. from GDC, sketch a relevant graph A1

maximum: y = 3 or (–1, 3) A1

minimum: y = 1.81 or (0.333, 1.81)

27

49,

3

1or

27

49or y A1

hence, 1.81 < k < 3 A1A1 N3

Note: Award A1 for 1.81 ≤ k ≤ 3. [5]

36. (a) a = 2.24 )5( A1

(b) (i)

A2

Note: Award A1 for end point

A1 for its asymptote.

(ii) sketch of g–1

(see above) A2

Note: Award A1 for end point

A1 for its asymptote.



(c) y = 25

3

x

xyx

2 – 3x + 5y = 0 M1

y

yx

2

2093 2 A1

x

xxg

2

2093)(

21

A1

[8]

37. (a) h (x) = g ◦ f (x) = 0,3e

12

xx

(M1)A1

(b) 0 < x 4

1 A1A1

Note: Award A1 for limits and A1 for correct inequality signs.

(c) 3e

12

xy

13e2

yy x M1

y

yx 31e

2 A1

x2 = ln

y

y31 M1

x = y

y31ln

h1

(x) =

3

1ln

31ln

xx

x A1

[8]



38. METHOD 1

As (x + 1) is a factor of P(x), then P(1) = 0 (M1)

a b + 1 = 0 (or equivalent) A1

As (x 2) is a factor of P(x), then P(2) = 0 (M1)

4a + 2b + 10 = 0 (or equivalent) A1

Attempting to solve for a and b M1

a = 2 and b = 1 A1 N1

METHOD 2

By inspection third factor must be x 1. (M1)A1

(x + 1)(x 2)(x 1) = x3 2x

2 x + 2 (M1)A1

Equating coefficients a = 2, b = 1 (M1)A1 N1

METHOD 3

Considering

22 xx

xP or equivalent (M1)

2

2231

2 22

xx

axbaax

xx

xP A1A1

Recognizing that (a + b + 3) x + 2 (a + 2) = 0 (M1)

Attempting to solve for a and b M1

a = 2 and b = 1 A1 N1 [6]

39. (a)

2

4

xgxh (M1)

=

x

x

x 2

21

2

4 A1

(b) METHOD 1

12

4

yx (interchanging x and y) M1

Attempting to solve for y M1

(y + 2)(x + 1) = 4

1

42

xy (A1)

121

41

xx

xh A1 N1

METHOD 2



y

yx

2

2 (interchanging x and y) M1

Attempting to solve for y M1

xy + y = 2 2x (y(x + 1) = 2(1 x)) (A1)

11

121

x

x

xxh A1 N1

Note: In either METHOD 1 or METHOD 2 rearranging first

and interchanging afterwards is equally acceptable. [6]

40. (a)

A1A1A1A1

Note: Award A1 for y-intercept

A1A1 for x-intercepts

A1 for shape.

(b) correct line A1

5 solutions A1 [6]

41. (a) y = arccos (1.2 cos x) A1

y = arcsin (1.4 sin x) A1

(b) The solutions are

x = 1.26, y = 0.464 A1A1



x = 0.464, y = 1.26 A1A1 [6]

42.

A1A1A1A1A1

Notes: Award A1 for vertical asymptotes at x = 1, x = 2 and x = 5.

A1 for x 2,

01

xf

A1 for x 8,

11

xf

A1 for local maximum at

2

1,0 (branch containing

local max. must be present)

A1 for local minimum at (3, 1) (branch containing local

min. must be present)

In each branch, correct asymptotic behaviour must be

displayed to obtain the A1.

Disregard any stated horizontal asymptotes such as y = 0 or y = 1. [5]



43. Attempting to solve |0.1x2 2x + 3| = log10 x numerically or graphically. (M1)

x = 1.52, 1.79 (A1)(A1)

x = 17.6, 19.1 (A1)

(1.52 < x < 1.79) (17.6 < x < 19.1) A1A1 N2 [6]

44. f(2) = 16 + 24 + 4p – 4 + q = 15 M1

4p + q = –21 A1

f(–3) = 81 – 81 + 9p + 6 + q = 0 M1

9p + q = –6 A1

p = 3 and q = –33 A1A1 N0 [6]

45. ln(x2 – 1) – ln(x + 1)

2 + lnx(x + 1) (A1)

= 2

2

)1(

)1)(1(ln

x

xxx (M1)A1

= 2)1(

)1)(1)(1(ln

x

xxxx (A1)

= ln x(x – 1) (= ln(x2 – x)) A1

[5]



46. (a)

A1

Note: Award A1 for shape.

x intercepts 0.354, 1.36, 2.59, 2.95 A2

Note: Award A1 for three correct, A0 otherwise.

maximum = (1.57, 0.352) =

352.0,

2

π A1

minimum = (1, –0.640) and (2.77, –0.0129) A1

(b) 0 < x < 0.354, 1.36 < x < 2.59, 2.95 < x < 4 A2

Note: Award A1 if two correct regions given. [7]



47.

A1A1

Note: Award A1 for each graph.

2x = 1 – x 3

1 x M1A1

3

1 x A1

[5]

48. Attempting to find f (2) = 8 + 12 + 2a + b (M1)

= 2a + b + 20 A1

Attempting to find f (–1) = –1 + 3 – a + b (M1)

= 2 – a + b A1

Equating 2a + 20 = 2 – a A1

a = –6 A1 N2 [6]

49. (a) f : x ex f

–1 : x ln x

f–1

(3) = ln 3 A1

g : x x + 2 g–1

: x x – 2

g–1

(3) = 1 A1

f–1

(3) × g–1

(3) = ln3 A1 N1



(b) EITHER

f ○ g(x) = f(x + 2) = ex+2

A1

ex+2

= 3 x + 2 = ln3 M1

x = ln 3 – 2 A1 N0

OR

f ○ g(x) = ex+2

f ○ g–1

(x) = ln (x) – 2 A1

f ○ g–1

(3) = ln (3) – 2 M1

x = ln 3 – 2 A1 N0 [6]

50. METHOD 1

Graph of f (x) – g(x) M1

A1A1A1

Note: Award A1 for each branch.

x < – 1 or 4 < x ≤ 14 A1A1 N3

Note: Each value and inequality sign must be correct.



METHOD 2

04

2

1

4

x

x

x

x M1

0)4)(1(

216 22

xx

xxx

0)4)(1(

14

xx

x A1

Critical value of x = 14 A1

Other critical values x = –1 and x = 4 A1

x < –1 or 4 < x ≤ 14 A1A1 N3

Note: Each value and inequality sign must be correct. [6]

51. (a)

A3

Note: Award A1 for each correct shape,

A1 for correct relative position.

(b) e–x

sin (4x) = 0 (M1)

sin (4x) = 0 A1

4x = 0, π, 2π, 3π, 4π, 5π A1

x = 0, 4

π5,

4

π4,

4

π3,

4

π2,

4

π AG



(c) e–x

= e–x

sin (4x) or reference to graph

sin 4x = 1 M1

4x = 2

π9,

2

π5,

4

π A1

x = 8

π9,

8

π5,

8

π A1 N3

(d) (i) y = e–x

sin 4x

x

y

d

d = –e

–x sin 4x + 4e

–x cos 4x M1A1

y = e–x

x

y

d

d = –e

–x A1

verifying equality of gradients at one point R1

verifying at the other two R1

(ii) since x

y

d

d ≠ 0 at these points they cannot be local maxima R1

(e) (i) maximum when y′ = 4e–x

cos 4x – e–x

sin 4x = 0 M1

x = ...,4

π2)4arctan(,

4

π)4arctan(,

4

)4arctan(

maxima occur at

x = 4

π4)4arctan(,

4

π2)4arctan(,

4

)4arctan( A1

so y1 = ))4(arctan(

4

1

e

sin(arctan (4)) (= 0.696) A1

y2 = )π2)4(arctan(

4

1

e

sin(arctan (4) + 2π) A1

145.0))4(sin(arctane)π2)4(arctan(

4

1

y3 = )π4)4(arctan(

4

1

e

sin(arctan (4) + 4π) A1

0301.0))4(sin(arctane)π4)4(arctan(

4

1

N3



(ii) for finding and comparing 1

2

2

3 and y

y

y

y M1

r = 2

π

e

A1

Note: Exact values must be used to gain the M1 and the A1. [22]

52. (a) METHOD 1

using GDC

a = 1, b = 5, c = 3 A1A2A1

METHOD 2

x = x + 2 cos x cos x = 0 M1

...2

π3,

2

π x

a = 1, c = 3 A1

1 – 2 sin x = 0 M1

6

5πor

6

π

2

1sin xx

b = 5 A1

Note: Final M1A1 is independent of previous work.

(b) 36

π5

6

π5

f (or 0.886) (M1)

f(2π) = 2π + 2 (or 8.28) (M1)

the range is

2π2,3

6

π5 (or [0.886, 8.28]) A1

(c) f′(x) = 1 – 2 sin x (M1)

f′

2

π3 = 3 A1

gradient of normal = 3

1 (M1)

equation of the normal is y

2

π3

3

1

2

π3x (M1)

y = 3

1 x + 2π (or equivalent decimal values) A1 N4



(d) (i) V = π 2

π3

2

π

22 d))cos2(( xxxx (or equivalent) A1A1

Note: Award A1 for limits and A1 for π and integrand.

(ii) V = xxxx d))cos2((π 2

π3

2

π

22

= xxxx d)cos4cos4(π 2

π3

2

π

2

using integration by parts M1

and the identity 4cos2 x = 2cos 2x + 2, M1

V = –π 2

π3

2

π)]22(sin)cos4sin4[( xxxxx A1A1

Note: Award A1 for 4x sin x + 4 cos x and A1 for sin 2x + 2x.

=

ππsin

2

πcos4

2

πsinπ2π3π3sin

2

π3cos4

2

π3sinπ6π

A1

= –π(–6π + 3π – 2π – π)

= 6π2 AG N0

Note: Do not accept numerical answers. [19]

53. (a) g ○ f(x) = xe1

1

A1

1 < 1 + ex < ∞ (M1)

range g ○ f is ]0, 1[ A1 N3



(b) Note: Interchange of variables and rearranging can be done in either order.

attempt at solving y = xe1

1

M1

rearranging

ex =

y

y1 M1

(g ○ f)–1

(x) = ln

x

x1 A1

Note: The A1 is for RHS.

domain is ]0, 1[ A1

Note: Final A1 is independent of the M marks.

(c) (i) y = f ○ g ○ h = 1 + ecos x

M1A1

x

y

d

d = –sin xe

cos x M1A1

= (1 – y)sin x AG

Note: Second M1A1 could also be obtained by solving the differential

equation.

(ii) EITHER

rearranging

y sin x = sin x – x

y

d

d A1

xx

yxxxxy d

d

ddsindsin M1

= –cos x – y(+ c) A1

= –cos x – ecos x

(+ d) A1

OR

xxxxy x dsin)e1(dsin cos A1

= xxxx xdesindsin cos

Note: Either the first or second line gains the A1.

= –cos x – ecos x

(+ d) A1M1A1



(iii) use of definition of y and the differential equation or

GDC to identify first minimum at x = π (3.14...) (M1)A1

EITHER

the required integral is

xxy

ydπ

max

min

2

M1A1

Note: ymax = 1 + e and ymin = 1 + e–1

but these do not need to be specified.

...32.4πdesinπ cos0

π

2 xxx x = 13.6 (M1)A1

OR

the required integral is

e1

e1

2

1dπ yx M1A1

=

e1

e1

2

1)1ln(arccosπ y dy = π × 4.32... = 13.6 M1A1

Note: 1 + e = 3.7182... and 1 + e–1

= 1.3678... [21]

54. (a) the expression is

!2)!22(

)!2(

!3)!3(

!

n

n

n

n (A1)

2

)12(2

6

)2)(1(

nnnnn M1A1

=

6

)815

6

)815( 32 nnnnnn A1

(b) the inequality is

6

815 23 nnn > 32n

attempt to solve cubic inequality or equation (M1)

n3 – 15n

2 – 184n > 0 n(n – 23)(n + 8) > 0

n > 23 (n ≥ 24) A1 [6]



55. g(x) = 0 or 3 (M1)(A1)

x = –1 or 4 or 1 or 2 A1A1

Notes: Award A1A1 for all four correct values,

A1A0 for two or three correct values,

A0A0 for less than two correct values.

Award M1 and corresponding A marks for correct attempt to

find expressions for f and g. [4]

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