1.

evidence of finding intersection points (M1)

e.g. f (x) = g (x), cos x2 = ex, sketch showing intersection
x = –1.11, x = 0 (may be seen as limits in the integral) A1A1
evidence of approach involving integration and subtraction (in any order)(M1)
0
e.g. 1.11  
cos x 2  e x , (cos x 2  e x ) dx, g  f

area = 0.282 A2 N3
[6]

2. (a) METHOD 1
evidence of recognizing the amplitude is the radius (M1)
e.g. amplitude is half the diameter
8
a A1
2
a=4 AG N0 2

METHOD 2
evidence of recognizing the maximum height (M1)
e.g. h = 6, a sin bt + 2 = 6
correct reasoning
e.g. a sin bt = 4 and sin bt has amplitude of 1 A1
a=4 AG N0 2

(b) METHOD 1
period = 30 (A1)
2
b A1
30


b AG N0 2
15

METHOD 2
correct equation (A1)
e.g. 2 = 4 sin 30b + 2, sin 30b = 0
30b = 2π A1

b AG N0 2
15
 (c) recognizing h′(t) = –0.5 (seen anywhere) R1
attempting to solve (M1)
e.g. sketch of h′, finding h′
correct work involving h′ A2
4π π 
e.g. sketch of h′ showing intersection, –0.5 = cos t 
15  15 

t = 10.6, t = 19.4 A1A1 N3 6

(d) METHOD 1
valid reasoning for their conclusion (seen anywhere) R1
e.g. h(t) < 0 so underwater; h(t) > 0 so not underwater
evidence of substituting into h (M1)
19.4 π
e.g. h(19.4), 4 sin 2
15
correct calculation A1
e.g. h(19.4) = –1.19
correct statement A1 N0 4
e.g. the bucket is underwater, yes

METHOD 2
valid reasoning for their conclusion (seen anywhere) R1
e.g. h(t) < 0 so underwater; h(t) > 0 so not underwater
evidence of valid approach (M1)
e.g. solving h(t) = 0, graph showing region below x-axis
correct roots A1
e.g. 17.5, 27.5
correct statement A1 N0 4
e.g. the bucket is underwater, yes
[14]

3. (a) B, D A1A1 N2 2

2
(b) (i) f′(x) =  2 xe  x A1A1 N2
2
Note: Award A1 for e  x and A1 for –2x.
 (ii) finding the derivative of –2x, i.e. –2 (A1)
evidence of choosing the product rule (M1)
2 2
e.g.  2e  x  2 x  2 xe  x
2 2
 2e  x  4 x 2 e  x A1
2
f ′′(x) = (4x2 – 2) e  x AG N0 5

(c) valid reasoning R1

e.g. f ′′(x) = 0
attempting to solve the equation (M1)
e.g. (4x2 – 2) = 0, sketch of f ′′(x)

 1   1 
p = 0.707    , q  0.707     A1A1 N3 4
 2  2

(d) evidence of using second derivative to test values on either side of POI M1
e.g. finding values, reference to graph of f′′, sign table
correct working A1A1
e.g. finding any two correct values either side of POI,
checking sign of f ′′ on either side of POI
reference to sign change of f ′′(x) R1 N0 4
[15]

4. METHOD 1
evidence of antidifferentiation (M1)
e.g. ∫(10e2x – 5)dx
y = 5e2x – 5x + C A2A1
2x
Note: Award A2 for 5e , A1 for –5x. If “C” is omitted, award
no further marks.
substituting (0, 8) (M1)
e.g. 8 = 5 + C
C = 3 (y = 5e2x – 5x + 3) (A1)
substituting x = 1 (M1)
y = 34.9 (5e2 – 2) A1 N4 8
 METHOD 2
evidence of definite integral function expression (M2)

 f t dt  f x   f a ,  10e 
x x
' 2x
e.g. 5
a 0

initial condition in definite integral function expression (A2)

 10e   10e 
x x
2t 2x
e.g.  5 dt  y  8,  5 dx  8
0 0

correct definite integral expression for y when x =1 (A2)

 10e 
1
2x
e.g.  5 dx  8
0

y = 34.9 (5e2 – 2) A2 N4 8

5. (a) evidence of finding height, h (A1)

h
e.g. sin θ = , 2 sin θ
2
evidence of finding base of triangle, b (A1)
b
e.g. cos θ = , 2 cos θ
2
attempt to substitute valid values into a formula for the area
of the window (M1)
e.g. two triangles plus rectangle, trapezium area formula
correct expression (must be in terms of θ) A1
1  1
e.g. 2  2 cos   2 sin    2  2 sin  , 2 sin  2  2  4 cos  
2  2

attempt to replace 2sinθ cosθ by sin 2θ M1

e.g. 4 sin θ + 2(2 sin θ cos θ)
y = 4 sin θ + 2 sin 2θ AG N0 5

(b) correct equation A1

e.g. y = 5, 4 sin θ + 2 sin 2θ = 5
evidence of attempt to solve (M1)
e.g. a sketch, 4 sin θ + 2 sin θ – 5 = 0
θ = 0.856 (49.0º), θ = 1.25 (71.4º) A1A1 N3 4
 
(c) recognition that lower area value occurs at θ = (M1)
2


finding value of area at θ = (M1)
2

  
e.g. 4 sin    2 sin  2   , draw square
2  2

A=4 (A1)
recognition that maximum value of y is needed (M1)
A = 5.19615… (A1)
4 < A < 5.20 (accept 4 < A < 5.19) A2 N5 7
[16]

6. (a) evidence of choosing the product rule (M1)

e.g. x × (– sin x) + 1 × cos x
f′(x) = cos x – x sin x A1A1 N3
(b)

A1A1A1A1 N4
Note: Award A1 for correct domain, 0 ≤ x ≤ 6 with
endpoints in circles,
A1 for approximately correct shape,
A1 for local minimum in circle,
A1 for local maximum in circle.
[7]
7. evidence of integrating the acceleration function (M1)
1 

e.g.   3 sin 2t dt
t 
3
correct expression ln t – cos 2t + c A1A1
2
evidence of substituting (1, 0) (M1)
3
e.g. 0 = ln 1 – cos 2 + c
2
 3 3 
c = –0.624   cos 2  ln 1 or cos 2  (A1)
 2 2 
3  3 3 3 3 
v = ln t – cos 2t  0.624   ln t  cos 2t  cos 2 or lnt  cos 2t  cos 2  ln 1 (A1)
2  2 2 2 2 
v(5) = 2.24 (accept the exact answer ln 5 – 1.5 cos 10 + 1.5 cos 2) A1 N3
[7]

8. (a) substituting (0, 13) into function M1

e.g. 13 = Ae0 + 3
13 = A + 3 A1
A = 10 AG N0

(b) substituting into f(15) = 3.49 A1

e.g. 3.49 = 10e15k + 3, 0.049 = e15k
evidence of solving equation (M1)
e.g. sketch, using ln

 ln0.049 
k = –0.201  accept  A1 N2
 15 

(c) (i) f(x) = 10e–0.201x + 3

f′(x) = 10e–0.201x × –0.201 (= –2.01e–0.201x) A1A1A1 N3
Note: Award A1 for 10e–0.201x, A1 for × –0.201,
A1 for the derivative of 3 is zero.

(ii) valid reason with reference to derivative R1 N1

e.g. f′(x) < 0, derivative always negative

(iii) y=3 A1 N1
 (d) finding limits 3.8953…, 8.6940… (seen anywhere) A1A1
evidence of integrating and subtracting functions (M1)
correct expression A1
8.69 8.69
e.g. 3.90
g ( x)  f ( x)dx, 
3.90
[( x 2  12 x  24)  (10e 0.201x  3)]dx

area = 19.5 A2 N4
[16]

9. (a) 2.31 A1 N1

(b) (i) 1.02 A1 N1

(ii) 2.59 A1 N1

q
(c) p
f ( x )dx = 9.96 A1 N1

split into two regions, make the area below the x-axis positive R1R1 N2
[6]

10. (a) n = 800e0 (A1)

n = 800 A1 N2

(b) evidence of using the derivative (M1)

n′(15) = 731 A1 N2

(c) METHOD 1
setting up inequality (accept equation or reverse inequality) A1
e.g. n′(t) > 10 000
evidence of appropriate approach M1
e.g. sketch, finding derivative
k = 35.1226... (A1)
least value of k is 36 A1 N2

METHOD 2
n′(35) = 9842, and n′(36) = 11208 A2
least value of k is 36 A2 N2
[8]
11. (a) (i) –1.15, 1.15 A1A1 N2

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

e.g. x = –1.15, x = 1.15
k = –1.13, k = 1.13 A1A1 N3

(b) evidence of choosing the product rule (M1)

e.g. uv′ + vu′
derivative of x3 is 3x2 (A1)
 2x
derivative of ln (4 – x2) is (A1)
4  x2
correct substitution A1
 2x
e.g. x 3   ln(4  x 2 )  3 x 2
4  x2
 2x 4
g′(x) =  3 x 2 ln(4  x 2 ) AG N0
4  x2

(c)

A1A1 N2

(d) w = 2.69, w < 0 A1A2 N2

[14]
12. (a)

A1A1A1 N3
Note: Award A1 for approximately correct shape, A1 for right
endpoint at (25, 0) and A1 for maximum point in circle.

(b) (i) recognizing that d is the area under the curve (M1)
e.g.  v(t )
correct expression in terms of t, with correct limits A2 N3
9 9
e.g. d = 0 
(15 t  3t )dt , d  vdt
0

(ii) d = 148.5 (m) (accept 149 to 3 sf) A1 N1

[7]

13. (a) valid approach R1

e.g. f″(x) = 0, the max and min of f′ gives the points of inflexion on f
–0.114, 0.364 (accept (–0.114, 0.811) and (0.364, 2.13)) A1A1 N1N1
 (b) METHOD 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
g″(x) = –144 R1 N1
therefore no points of inflexion as g″(x) ≠ 0 R1 N1
[5]

14. (a) evidence of valid approach (M1)

e.g. f(x) = 0, graph
a = –1.73, b = 1.73 (a   3 , b  3 ) A1A1 N3

(b) attempt to find max (M1)

e.g. setting f′(x) = 0, graph
c = 1.15 (accept (1.15, 1.13)) A1 N2

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

  f ( x) dx, π  x ln(4  x ) , π 

c 2 2 1.149...
2
e.g. V = π y 2 dx
0 0

V = 2.16 A2 N2

(d) valid approach recognizing 2 regions (M1)

e.g. finding 2 areas
correct working (A1)
1.73... 1.149... 0 1.149...
e.g. 
0
f ( x)dx  
0
f ( x)dx;  
1.73...
f ( x)dx  0
f ( x)dx

area = 2.07 (accept 2.06) A2 N3

[12]

15. (a) attempt to expand (M1)

(x + h)3 = x3 + 3x2h + 3xh2 + h3 A1 N2
 (b) evidence of substituting x + h (M1)
correct substitution A1
( x  h) 3  4( x  h)  1  ( x 3  4 x  1)
e.g. f′(x) = lim
h0 h
simplifying A1
( x 3  3x 2 h  3xh 2  h 2  4 x  4h  1  x 3  4 x  1)
e.g.
h
factoring out h A1
h(3x 2  3 xh  h 2  4)
e.g.
h
f′(x) = 3x2 – 4 AG N0

(c) f′(1) = –1 (A1)

setting up an appropriate equation M1
e.g. 3x2 – 4 = –1
at Q, x = –1, y = 4 (Q is (–1, 4)) A1A1 N3

(d) recognizing that f is decreasing when f′(x) < 0 R1

correct values for p and q (but do not accept p = 1.15, q = –1.15) A1A1 N1N1
2
e.g. p = –1.15, q = 1.15;  ; an interval such as –1.15 ≤ x ≤ 1.15
3

(e) f′(x) ≥ –4, y ≥ –4, [–4, ∞[ A2 N2

[15]

16. (a) gradient is 0.6 A2 N2

(b) at R, y = 0 (seen anywhere) A1

at x = 2, y = ln 5 (= 1.609...) (A1)
gradient of normal = –1.6666... (A1)
evidence of finding correct equation of normal A1
5
e.g. y – ln 5 =  (x – 2), y = – 1.67x + c
3
x = 2.97 (accept 2.96) A1
coordinates of R are (2.97, 0) N3
[7]
17. (a) finding the limits x = 0, x = 5 (A1)
integral expression A1
5
e.g.  f ( x)dx
0

area = 52.1 A1 N2

 πy
2
(b) evidence of using formula v = dx (M1)
correct expression A1
5
e.g. volume = π 
0
x 2 ( x  5) 4 dx
volume = 2340 A2 N2

a
(c) area is  x(a  x)dx
0
A1
a
 ax 2 x 3 
=    A1A1
 2 3 0
substituting limits (M1)
a3 a3
e.g. 
2 3
setting expression equal to area of R (M1)
correct equation A1
a2 a3
e.g.  = 52.1, a3 = 6 × 52.1,
2 3
a = 6.79 A1 N3
[14]

18. (a)

A1A1A1 N3
Note: Award A1 for approximately sinusoidal shape,
A1 for end points approximately correct, (–2π, 4),
(2π, 4) A1 for approximately correct position of graph,
(y-intercept (0, 4) maximum to right of y-axis).
(b) (i) 5 A1 N1

(ii) 2π (6.28) A1 N1

(iii) –0.927 A1 N1

(c) f(x) = 5 sin (x + 0.927) (accept p = 5, q = 1, r = 0.927) A1A1A1 N3

(d) evidence of correct approach (M1)

e.g. max/min, sketch of f′(x) indicating roots

one 3 s.f. value which rounds to one of –5.6, –2.5, 0.64, 3.8 A1 N2

(e) k = –5, k = 5 A1A1 N2

(f) METHOD 1
graphical approach (but must involve derivative functions) M1
e.g.

each curve A1A1

x = 0.511 A2 N2
 METHOD 2
1
g′(x) = A1
x 1
f′(x) = 3 cos x – 4 sin x (5 cos(x + 0.927)) A1
evidence of attempt to solve g′(x) = f′(x) M1
x = 0.511 A2 N2
[18]

19. (a) f′(x) = –sin 2x × 2 (= –2 sin 2x) A1A1 N2

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

1  3 
(b) g′(x) = 3 ×   A1A1 N2
3x  5  3x  5 

1
Note: Award A1 for 3, A1 for .
3x  5

(c) evidence of using product rule (M1)

 3 
h′(x) = (cos 2x)   + ln(3x – 5)(–2 sin 2x) A1 N2
 3x  5 
[6]

20. substituting x = 1, y = 3 into f(x) (M1)

3=p+q A1
finding derivative (M1)
f′(x) = 2px + q A1
correct substitution, 2p + q = 8 A1
p = 5, q = –2 A1A1 N2N2
[7]
21. METHOD 1
correct expression for second side, using area = 525 (A1)
525
e.g. let AB = x, AD =
x
attempt to set up cost function using $3 for three sides and $11 for one side (M1)
e.g. 3(AD + BC + CD) + 11AB
correct expression for cost A2
525 525 525 525 3150
e.g. 3  3  11x  3x, 3  3  11AB  3AB,  14 x
x x AB AB x
EITHER
sketch of cost function (M1)
identifying minimum point (A1)
e.g. marking point on graph, x = 15
minimum cost is 420 (dollars) A1 N4
OR
correct derivative (may be seen in equation below) (A1)
 1575  1575
e.g.C′(x) =   14
x2 x2
setting their derivative equal to 0 (seen anywhere) (M1)
 3150
e.g.  14  0
x2
minimum cost is 420 (dollars) A1 N4
METHOD 2
correct expression for second side, using area = 525 (A1)
525
e.g. let AD = x, AB =
x
attempt to set up cost function using $3 for three sides and $11 for one side (M1)
e.g. 3(AD + BC + CD) + 11AB
correct expression for cost A2
 525  525  525  525 7350
e.g. 3 x  x    11, 3 AD  AD    11, 6 x 
 x  x  AD  AD x

EITHER
sketch of cost function (M1)
identifying minimum point (A1)
e.g. marking point on graph, x = 35
minimum cost is 420 (dollars) A1 N4
OR
correct derivative (may be seen in equation below) (A1)
7350
e.g. C′(x) = 6 –
x2
setting their derivative equal to 0 (seen anywhere) (M1)
7350
e.g. 6 – 0
x2
minimum cost is 420 (dollars) A1 N4
[7]
22. (a)

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 for two intersection points in approximately
correct position.

(b) (i) (2,0) (accept x = 2) A1 N1

(ii) period = 8 A2 N2

(iii) amplitude = 5 A1 N1

(c) (i) (2, 0), (8, 0) (accept x = 2, x = 8) A1A1 N1N1

(ii) x = 5 (must be an equation) A1 N1

(d) METHOD 1
intersect when x = 2 and x = 6.79 (may be seen as limits of integration) A1A1
evidence of approach (M1)
6.79   π 
e.g.  g  f ,  f ( x ) dx   g ( x )dx , 
2
 ( 0.5 x 2  5 x  8   5 cos x  
  4 

area = 27.6 A2 N3
 METHOD 2
intersect when x = 2 and x = 6.79 (seen anywhere) A1A1
evidence of approach using a sketch of g and f, or g – f. (M1)

e.g. area A + B – C, 12.7324 + 16.0938 – 1.18129...

area = 27.6 A2 N3
[15]

23. (a)
y

p q r x

A1A1 N2
Note: Award A1 for ne.g.ative gradient throughout,
A1 for x-intercept of q. It need not be linear.
 (b)

x-coordinate
(i) Maximum point on f r A1 N1
(ii) Inflexion point on f q A1 N1

(c) METHOD 1
Second derivative is zero, second derivative changes sign. R1R1 N2
METHOD 2
There is a maximum on the graph of the first derivative. R2 N2
[6]

24. (a) (i) intersection points x =3.77, x = 8.30 (may be seen as

the limits) (A1)(A1)
approach involving subtraction and integrals (M1)
fully correct expression A2
8.30
e.g.   4 cos 0.5 x   2  ln 3x  2  1 dx ,
3.77

8.30 8.30
 g  x  dx   f  x  dx N5
3.77 3.77

(ii) A = 6.46 A1 N1

3
(b) (i) f  x   A1A1 N2
3x  2
Note: Award A1 for numerator (3), A1 for
denominator (3x  2), but penalize
1 mark for additional terms.
(ii) g(x) = 2 sin (0.5x) A1A1 N2
Note: Award A1 for 2, A1 for sin (0.5x), but
penalize 1 mark for additional terms.

(c) evidence of using derivatives for gradients (M1)

correct approach (A1)
e.g. f(x) = g(x), points of intersection
x = 1.43, x = 6.10 A1A1 N2N2
[14]
25. (a) intercepts when f (x) = 0 (M1)
(1.54, 0) (4.13, 0) (accept x = 1.54 x = 4.13) A1A1 N3

(b)
y
3
2
1

–2 –1 0 1 2 3 4 5 6 x
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10

A1A1A1 N3
Note: Award A1 for passing through
approximately (0,  4), A1 for correct
shape, A1 for a range of approximately
9 to 2.3.

(c) gradient is 2 A1 N1
[7]

26. (a) evidence of using the product rule M1

f (x) = ex(1  x2) + ex(2x) A1A1
Note: Award A1 for ex(1 x2), A1 for ex(2x).
f (x) = ex(1  2x x2) AG N0

(b) y=0 A1 N1

(c) at the local maximum or minimum point

f (x) = 0 (ex(1  2x  x2) = 0) (M1)
Þ 1  2x  x2 = 0 (M1)
r = 2.41 s = 0.414 A1A1 N2N2
 (d) f(0) = 1 A1
gradient of the normal = 1 A1
evidence of substituting into an equation for a straight line (M1)
correct substitution A1
e.g. y  1 = 1(x  0), y  1 = x, y = x + 1
x+y=1 AG N0

(e) (i) intersection points at x = 0 and x = 1 (may be seen as the limits) (A1)
approach involving subtraction and integrals (M1)
fully correct expression A2 N4

 e 1  x  1  x dx ,  f x  dx   1  x  dx
1 1 1
x 2
e.g.
0 0 0

(ii) area R = 0.5 A1 N1

[17]

27. (a)

A1A2 N3
Notes: Award A1 for correct domain, 0 ≤ x ≤ 3.
Award A2 for approximately correct shape, with
local maximum in circle 1 and right endpoint
in circle 2.

(b) a = 2.31 A1 N1
   f ( x) dx
2
(c) evidence of using V = π (M1)
fully correct integral expression A2
2.31 2.31
e.g. V = π 
0
[ x cos( x  sin x)] 2 dx, V  π 
0
[ f ( x)] 2 dx
V = 5.90 A1 N2
[8]

28. (a) correctly finding the derivative of e2x, i.e. 2e2x A1

correctly finding the derivative of cos x, i.e. –sin x A1
evidence of using the product rule, seen anywhere M1
e.g. f′(x) = 2e2x cos x – e2x sin x
f′(x) = e2x(2 cos x – sin x) AG N0

(b) evidence of finding f(0) = 1, seen anywhere A1

attempt to find the gradient of f (M1)
e.g. substituting x = 0 into f′(x)
value of the gradient of f A1
e.g. f′(0) = 2, equation of tangent is y = 2x + 1
1
gradient of normal =  (A1)
2
1  1 
y–1=  x  y   x  1 A1 N3
2  2 

(c) (i) evidence of equating correct functions M1

1
e.g. e2x cos x =  x  1 , sketch showing intersection of graphs
2
x = 1.56 A1 N1

(ii) evidence of approach involving subtraction of integrals/areas (M1)

e.g.  [ f ( x)  g ( x)]dx,  f ( x)dx – area under trapezium
fully correct integral expression A2
1.56   1  1.56
e.g.
0 

e 2 x cos x    x  1dx,
 2  0
e 2 x cos xdx  0.951...

area = 3.28 A1 N2
[14]
29. (a) Using the chain rule (M1)
f′(x) = (2 cos(5x – 3))5 (= 10 cos(5x – 3)) A1
f′′(x) = –(10 sin(5x – 3))5
= –50sin(5x – 3) A1A1 N2
Note: Award A1 for sin (5x – 3), A1 for –50.

2
(b)  f ( x)dx   5 cos(5x  3)  c A1A1 N2

2
Note: Award A1 for cos(5x – 3), A1 for  .
5
[6]

30. (a) Curve intersects y-axis when x = 0 (A1)

Gradient of tangent at y-intercept = 2 A1
1
Þ gradient of N =  (= –0.5) A1
2
Finding y-intercept, 2.5 A1
Therefore, equation of N is y = –0.5x + 2.5 AG N0

(b) N intersects curve when –0.5x2 + 2x + 2.5 = –0.5x + 2.5 A1

Solving equation (M1)
e.g. sketch, factorising
Þ x = 0 or x = 5 A1

Other point when x = 5 (R1)

x = 5 Þ y = 0 (so other point (5, 0)) A1 N2

(c)

Using appropriate method, with subtraction/correct expression, correct limitsM1A1

5 5 5
e.g. 
0 
f ( x)dx  g ( x)dx,
0 
0
(0.5 x 2  2.5 x)dx
Area = 10.4 A2 N2
[13]
31. (a) (i) p = 1, q = 5 (or p = 5, q = 1) A1A1 N2
(ii) x=3 (must be an equation) A1 N1

(b) y = (x  1)(x  5)
= x2  6x + 5 (A1)
= (x  3)2  4 (accept h = 3, k = 4) A1A1 N3

dy
(c)  2 x  3  2 x  6 A1A1 N2
dx

dy
(d) When x = 0,  6 (A1)
dx

y  5 = 6(x  0) (y = 6x + 5 or equivalent) A1 N2

[10]

32. (a)  (3.14) (accept (, 0), (3.14, 0)) A1 N1

(b) (i) For using the product rule (M1)

f (x) = ex cos x + ex sin x = ex(cos x + sin x) A1A1 N3
(ii) At B, f (x) = 0 A1 N1

(c) f ²(x) = ex cos x  ex sin x + ex sin x + ex cos x A1A1

= 2ex cos x AG N0

(d) (i) At A, f ²(x) = 0 A1 N1

(ii) Evidence of setting up their equation (may be seen in part
(d)(i)) A1
eg 2ex cos x = 0, cos x = 0


x 1.57, y  e 2  4.81 A1A1
2

  
Coordinates are  , e 2  1.57, 4.81 N2
2 
 
  
e  f  x  dx
x
(e) (i) sin x dx or A2 N2
0 0

(ii) Area = 12.1 A2 N2

[15]

33. (a)

A1A1A1 N3
Notes: Award A1 for both asymptotes shown.
The asymptotes need not be labelled.
Award A1 for the left branch in
approximately correct position,
A1 for the right branch in
approximately correct position.

5
(b) (i) y = 3, x = (must be equations) A1A1 N2
2

14 7  14  
(ii) x=  or 2.33 , also accept  , 0   A1 N1
6 3  6 

14   14  
(iii) y=  y  2.8  accept  0 ,  or 0 , 2.8 A1 N1
6   5 

 6 1 
(c) (i)   9  2 x  5  2 x  5 2
 dx  9 x 


1
3 ln 2 x  5  C A1A1A1
22 x  5
A1A1 N5
 b
(ii) Evidence of using V =  a
y 2 dx (M1)

Correct expression A1
2
 1  a 
9  6  1
a
eg  3
 3 

 dx , 
2 x  5   3 
 2 x  5 2 x  52
 dx ,


a
 1 
9 x  3 ln 2 x  5  
 22 x  5  3

 1   1
Substituting  9a  3 ln 2a  5     27  3ln 1   A1
 22a  5   2

Setting up an equation (M1)

1 1  28 
9a   27   3 ln 2a  5  3 ln 1    3 ln 3 
22a  5 2  3 

Solving gives a = 4 A1 N2
[17]

34. (a) (i) p=2 A1 N1

(ii) q=1 A1 N1

(b) (i) f (x) = 0 (M1)

3x
2 =0 (2x2  3x  2 = 0) A1
x 2 1

1
x=  x=2
2

 1 
  , 0 A1 N2
 2 
b
(ii) Using V =  a
y2dx (limits not required) (M1)

2
0  3x 
V= ∫ 1   2  2  dx A2
2  x 1 

V = 2.52 A1 N2
(c) (i) Evidence of appropriate method M1
eg Product or quotient rule
Correct derivatives of 3x and x2  1 A1A1
Correct substitution A1

 3 ( x 2  1)  ( 3 x ) (2 x )
eg
( x 2  1) 2

 3x 2  3  6 x 2
f ′ (x) = A1
( x 2  1) 2

3x 2  3 3( x 2  1)
f ′ (x) = = AG N0
( x 2  1) 2 ( x 2  1) 2

(ii) METHOD 1
Evidence of using f ′(x) = 0 at max/min (M1)
3 (x2 + 1) = 0 (3x2 + 3 = 0) A1
no (real) solution R1
Therefore, no maximum or minimum. AG N0
METHOD 2
Evidence of using f ′(x) = 0 at max/min (M1)
Sketch of f ′(x) with good asymptotic behaviour A1
Never crosses the x-axis R1
Therefore, no maximum or minimum. AG N0
METHOD 3
Evidence of using f ′ (x) = 0 at max/min (M1)
Evidence of considering the sign of f ′ (x) A1
f ′ (x) is an increasing function (f ′ (x) > 0, always) R1
Therefore, no maximum or minimum. AG N0
(d) For using integral (M1)
a  a a 3x 2  3 
Area =  0
g ( x ) dx  or

 0
f  ( x) dx or  0 ( x 2  1) 2
d x 


A1

a a
Recognizing that  0
g ( x ) dx  f ( x )
0
A2

Setting up equation (seen anywhere) (M1)

Correct equation A1

3x 2  3  3a 
dx = 2, 2  2   2  0 = 2, 2a2 + 3a  2 = 0
a
eg  0 2
( x  1) 2
 a  1

1
a= a=2
2

1
a= A1 N2
2
[24]
35. (a)
y

20

10

x
–2 –1 1 2

–10

–20

A1A1A1 N3
Note: Award A1 for the left branch asymptotic
to the x-axis and crossing the y-axis,
A1 for the right branch approximately
the correct shape,
A1 for a vertical asymptote at
1
approximately x = .
2

1
(b) (i) x (must be an equation) A1 N1
2
2
(ii)  0
f ( x ) dx A1 N1

(iii) Valid reason R1 N1

eg reference to area undefined or discontinuity
Note: GDC reason not acceptable.
 1. 5
(c) (i) V=  f  x  2 dx A2 N2
1

(ii) V = 105 (accept 33.3 ) A2 N2

(d) f (x) = 2e2x  1  10(2x  1)2 A1A1A1A1 N4

(e) (i) x = 1.11 (accept (1.11, 7.49)) A1 N1

(ii) p = 0, q = 7.49 (accept 0 £ k < 7.49) A1A1 N2
[17]

36. Note: Accept exact answers given in terms of .

(a) Evidence of using l = r (M1)
arc AB = 7.85 (m) A1 N2

1
(b) Evidence of using A  r 2 θ (M1)
2

Area of sector AOB = 58.9 (m2) A1 N2

(c) METHOD 1

π6
2 π
3


angle = 30 (A1)
6


attempt to find 15 sin M1
6


height = 15 + 15 sin
6
= 22.5 (m) A1 N2
METHOD 2

π
3


angle = 60 (A1)
3


attempt to find 15 cos M1
3


height = 15 + 15 cos
3
= 22.5 (m) A1 N2
     
(d) (i) h  15 15 cos    (M1)
4 2 4
= 25.6 (m) A1 N2

 
(ii) h(0) = 15  15 cos  0   (M1)
 4

= 4.39(m) A1 N2
(iii) METHOD 1
Highest point when h = 30 R1

 
30 = 15  15 cos  2t   M1
 4

 
cos  2t   = 1 (A1)
 4

 3 
t = 1.18  accept  A1 N2
 8 

METHOD 2
h
30

2π t

Sketch of graph of h M2
Correct maximum indicated (A1)
t = 1.18 A1 N2
METHOD 3
Evidence of setting h(t) = 0 M1

 
sin  2t    0 (A1)
 4

Justification of maximum R1
eg reasoning from diagram, first derivative test, second
derivative test

 3 
t = 1.18  accept  A1 N2
 8 
  
(e) h(t) = 30 sin  2t   (may be seen in part (d)) A1A1 N2
 4

(f) (i)
h(t)
30

π π t
2

–30

A1A1A1 N3
Notes: Award A1 for range 30 to 30, A1
for two zeros.
Award A1 for approximate correct
sinusoidal shape.
(ii) METHOD 1
Maximum on graph of h (M1)
t = 0.393 A1 N2
METHOD 2
Minimum on graph of h (M1)
t = 1.96 A1 N2
METHOD 3
Solving h(t) = 0 (M1)
One or both correct answers A1
t = 0.393, t = 1.96 N2
[22]
 3
37. (a) (i) f (x) =  x 1 A1A1 N2
2
(ii) For using the derivative to find the gradient of the tangent (M1)
f (2) =  2 (A1)

1
Using negative reciprocal to find the gradient of the normal   M1
2

1  1 
y  3  ( x  2)  or y  x  2  A1 N3
2  2 

3 2 1
(iii) Equating  x  x  4  x  2 (or sketch of graph) M1
4 2

3x2  2x  8 = 0 (A1)
(3x + 4)(x  2) = 0
4  4 4 4
x=    1.33 (accept   ,  or x   , x  2) A1 N2
3  3 3  3

(b) (i) Any completely correct expression (accept absence of dx) A2

2
 3 2
2   1 3 1 2 
eg   x  x  4  dx ,  x  x  4 x 
1  4   4 2  1
N2

45
(ii) Area = 11.25 (accept 11.3) A1 N1
4
(iii) Attempting to use the formula for the volume (M1)
2
2  3  2  3 2 
eg     x 2  x  4  dx , 
1  4 
   x  x  4  dx
1  4 
A2 N3

k
k  1 1 
(c) 1
f ( x) dx   x 3  x 2  4 x 
 4 2 1
A1A1A1

1 3 1
Note: Award A1 for  x , A1 for x 2 , A1 for 4x.
4 2

 1 1   1 1 
Substituting   k 3  k 2  4k       4  (M1)(A1)
 4 2   4 2 

1 3 1 2
=  k  k  4k  4.25 A1 N3
4 2
[21]
38. (a) METHOD 1
Attempting to interchange x and y (M1)
Correct expression x = 3y  5 (A1)

1 x5
f ( x)  A1 N3
3
METHOD 2
Attempting to solve for x in terms of y (M1)
y5
Correct expression x  (A1)
3

1 x5
f ( x)  A1 N3
3

(b) For correct composition (g1◦ f) (x) = (3x  5) + 2 (A1)

(g1◦ f) (x) = 3x  3 A1 N2

x3
(c)  3 x  3 x  3  9 x  9  (A1)
3

12
x A1 N2
8
(d) (i)

y=3

x=2

A1A1A1 N3
Note: Award A1 for approximately correct x
and y intervals, A1 for two branches of
correct shape, A1 for both asymptotes.
(ii) (Vertical asymptote) x = 2, (Horizontal asymptote) y = 3 A1A1 N2
(Must be equations)

(e) (i) 3x + ln (x  2) + C (3x + ln x  2 + C) A1A1 N2

(ii) 3 x  ln x  253 (M1)

= (15 + ln 3)  (9 + ln1) A1
= 6 + ln 3 A1 N2

(f) Correct shading (see graph). A1 N1

[18]
39. (a)
y

Q
1
P
R

x
1 2 3

A1A1A1 N3
Note: Award A1 for the shape of the curve,
A1 for correct domain,
A1 for labelling both points P and
Q in approximately correct positions.

(b) (i) Correctly finding derivative of 2x + 1 ie 2 (A1)

Correctly finding derivative of ex ie ex (A1)
Evidence of using the product rule (M1)
f  (x) = 2ex + (2x + 1)(ex) A1
= (1  2x)ex AG N0
(ii) At Q, f (x) = 0 (M1)

x = 0.5, y = 2e0.5 A1A1

Q is (0.5, 2e0.5) N3

(c) 1 £ k < 2e0.5 A2 N2

(d) Using f ² (x) = 0 at the point of inflexion M1

ex (3 + 2x) = 0

This equation has only one root. R1
So f has only one point of inflexion. AG N0
 (e) At R, y = 7e3 (= 0.34850 ...) (A1)
7 e 3  1
Gradient of (PR) is   0.2172 (A1)
3

 7 e 3  1 
Equation of (PR) is g (x) =   x  1  0.2172 x  1 A1

 3 
Evidence of appropriate method, involving subtraction of integrals
or areas M2
Correct limits/endpoints A1
3
eg   f x   g x  dx, area under curve  area under PR
0

3  3 
 2 x  1 e  x   7e  1 x  1  dx
Shaded area is  0

 3




= 0.529 A1 N4
[21]

40. (a) (i) f (x) = –x + 2 A1

(ii) f (0) = 2 A1 2
(b) Gradient of tangent at y-intercept = f (0) = 2
Þ gradient of normal = 1 (= –0.5) A1
2
Finding y-intercept is 2.5 A1
Therefore, equation of the normal is
y – 2.5 = ~(x – 0) (y – 2.5 = –0.5x) M1
(y = –0.5x + 2.5 (AG) 3
(c) (i) EITHER
solving –0.5x2 + 2x + 2.5 = –0.5x + 2.5 (M1)A1
Þ x = 0 or x = 5 A1 2
OR
y

f(x)
g(x)

M1
Curves intersect at x = 0, x = 5 (A1)
So solutions to f (x) = g (x) are x = 0, x = 5 A1 2

OR
Þ 0.5x2 – 2.5x = 0 (A1)
Þ – 0.5x(x – 5) = 0 M1
Þ x = 0 or x = 5 A1 2

(ii) Curve and normal intersect when x = 0 or x = 5 (M2)

Other point is when x = 5
Þ y = –0.5(5) + 2.5 = 0 (so other point (5, 0) A1 2

( f ( x)  g ( x))dx or (0.5 x 2  2 x  2.5)dx  1  5  2.5 

5 5
(d) (i) Area =  0   0 2 
A1A1A1 3
Note: Award (A1) for the integral, (A1) for both correct
limits on the integral, and (A1) for the difference.
(ii) Area = Area under curve – area under line (A = A1 – A2) (M1)
(A1) = 50 , A2  25
3 4
Area = 50  25  125 (or 10.4 (3sf) A1 2
3 4 12
[16]
41. (a) (i) p = (10x + 2) – (1 + e2x) A2 2
2x
Note: Award (A1) for (l + e ) – (10x + 2).
dp
(ii) = 10 – 2e2x A1A1
dx

dp
= 0 (10 – 2e2x = 0) M1
dx

1n 5
x= (= 0.805) A1 4
2

(b) (i) METHOD 1

x = 1 + e2x M1
1n(x – 1) = 2y A1
1n ( x  1)  1n ( x  1) 
f –1(x) =  Allow y   A1 3
2  2 
METHOD 2
y – 1 = e2x A1
ln( y  1)
=x M1
2
1n ( x  1)  1n ( x  1) 
f –1(x) =  Allow y   A1 3
2  2 

1n (5  1)  1
  1n 2 
2
(ii) a= M1
2  2 

= 1 × 21n2 A1
2
= 1n 2 AG 2

b
(c) Using V =  a
πy 2 d x (M1)

π (1  e 2 x ) 2 dx  or π (1  e 2 x ) 2 dx 
ln 2 0.805
Volume =  0   0 
A2 3

[14]
42. (a) x=1 (A1) 1
(b) Using quotient rule (M1)
2
( x  1) (1)  ( x  2)[2( x  1)]
Substituting correctly g(x) = A1
( x  1) 4

( x  1)  (2 x  4)
= (A1)
( x  1) 3

= 3  x (Accept a = 3, n = 3) A1 4
( x  1) 3

(c) Recognizing at point of inflexion g²(x) = 0 M1

x=4 A1

Finding corresponding y-value = 2 = 0.222 ie P  4, 2  A1 3

9  9
[8]

43. (a) (i) p  2 q  4 (or p  4, q  2 ) (A1)(A1) (N1)(N1)

(ii) y  a ( x  2)( x  4)
8  a (6  2)(6  4) (M1)
8  16a

1
a (A1) (N1)
2

1
(iii) y  ( x  2)( x  4)
2

1
y  ( x 2  2 x  8)
2

1 2
y x x4 (A1) (N1) 5
2

dy
(b) (i)  x 1 (A1) (N1)
dx
(ii) x 1  7 (M1)
x  8, y  20  P is (8, 20)  (A1)(A1) (N2) 4
 (c) (i) when x = 4, gradient of tangent is 4 – 1 = 3 (may be implied) (A1)
1
gradient of normal is  (A1)
3

1  1 4
y  0   ( x  4) y  x  (A1) (N3)
3  3 3

1 2 1 4
(ii) x  x  4   x  (or sketch/graph) (M1)
2 3 3

1 2 2 16
x  x 0
2 3 3

3 x 2  4 x  32  0 (may be implied) (A1)

(3x  8)( x  4)  0

8
x   or x  4
3

8
x   (2.67) (A1) (N2) 6
3
[15]

1
44. (a) x or 5 x  1  0 (A1) (N1) 1
5

(5 x  1)(6 x)  (3 x 2 ) (5)
(b) f ( x)  (M1)(A1)
(5 x  1)2

30 x 2  6 x  15 x 2
 (may be implied) (A1)
(5 x  1) 2

15 x 2  6 x
 (accept a = 15, b = –6) (A1) (N2) 4
(5 x  1)2
[5]
45. (a) x=1 (A1)
EITHER
The gradient of g ( x) goes from positive to negative (R1)

OR
g ( x) goes from increasing to decreasing (R1)

OR
when x  1, g ( x ) is negative (R1) 2

(b) 3  x  2 and 1  x  3 (A1)

g ( x ) is negative (R1) 2

1
(c) x (A1)
2
EITHER
g ( x ) changes from positive to negative (R1)

OR
concavity changes (R1) 2

(d)

–3 –2 –1 1 2 3

(A3) 3
[9]
 4 3
46. (a) s = 25t  t c (M1)(A1)(A1)
3
Note: Award no further marks if “c” is
missing.
Substituting s = 10 and t = 3 (M1)
4 3
10 = 25  3  (3)  c
3

10 = 75  36 + c
c =  29 (A1)
4
s = 25t  t 3  29 (A1) (N3)
3

(b) METHOD 1
ds
s is a maximum when v =  0 (may be implied) (M1)
dt

25  4t2 = 0 (A1)
25
t2 =
4

5
t= (A1) (N2)
2
METHOD 2
2
Using maximum of s ( 12 , may be implied) (M1)
3

4 3 2
25t  t  29 12 (A1)
3 3
t = 2.5 (A1) (N2)

4 3
(c) 25t  t  29 > 0 (accept equation) (M1)
3
m = 1.27, n = 3.55 (A1)(A1) (N3)
[12]
47.
Note: There are many approaches possible.
However, there must be some evidence
of their method.
k
Area =  0
sin 2 xdx (must be seen somewhere) (A1)

Using area = 0.85 (must be seen somewhere) (M1)

EITHER
k
1 
Integrating  cos 2 x 
2 0
 1 1 
 cos 2k  cos 0  (A1)
 2 2 

1
Simplifying cos 2k  0.5 (A1)
2

1
Equation cos 2k  0.5 = 0.85 (cos 2k =  0.7)
2
OR
Evidence of using trial and error on a GDC (M1)(A1)


Eg  0
2 sin 2 xdx = 0.5 ,
2
too small etc

OR
k
Using GDC and solver, starting with  0
sin 2 xdx  0.85 = 0 (M1)(A1)

THEN
k = 1.17 (A2) (N3)
[6]
48. (a)
y

(A1)(A1) 2
Note: Award (A1) for a second branch in approximately the
correct position, and (A1) for the second branch having
positive x and y intercepts. Asymptotes need not be drawn.

1 1  1
(b) (i) x-intercept =  Accept  ,0 , x   (A1)
2 2  2 
y-intercept = 1 (Accept (0, 1), y = 1) (A1)

(ii) horizontal asymptote y = 2 (A1)

vertical asymptote x = 1 (A1) 4

 1 
(c) (i) f (x) = 0 – (x – 1)–2   2

 (A2)
 ( x  1) 
(ii) no maximum / minimum points.
1
since ¹0 (R1) 3
( x  1) 2

(d) (i) 2x + ln (x – 1) + c (accept lnïx – 1ï) (A1)(A1)(A1)

 4 1  4
dx, 2 x  ln ( x  1) 2 
4
(ii) A=  2


f ( x )dx Accept  2 
2 x 1 
(M1)(A1)

Notes: Award (A1) for both correct limits.

Award (M0)(A0) for an incorrect function.
 (iii) A = 2 x  ln ( x  1)42
= (8 + ln 3) – (4 + ln 1) (M1)
= 4 + ln 3(= 5.10, to 3 sf) (A1) (N2) 7
[16]

49. (a) h3 (A1)

k2 (A1) 2

(b) f ( x)   ( x  3)2  2

  x 2  6 x  9  2 (must be a correct expression) (A1)

  x2  6 x  7 (AG) 1

(c) f ( x )  2 x  6 (A2) 2

(d) (i) tangent gradient  2 (A1)

1
gradient of L  (A1) (N2) 2
2

(ii) EITHER
1
equation of L is y  xc (M1)
2
c  1 . (A1)

1
y x 1
2
OR
1
y  1  ( x  4) (A2) (N2) 2
2
 (iii) EITHER
1
 x2  6x  7  x 1 (M1)
2

2 x 2  11x  12  0 (may be implied) (A1)

(2 x  3)( x  4)  0 (may be implied) (A1)

x  1.5 (A1) (N3) 4

OR
1
 x2  6x  7  x  1 (or a sketch) (M1)
2
x  1.5 (A3) (N3) 8
[13]

50. (a) (i) f ( x )   6sin 2 x (A1)(A1)

(ii) EITHER
f ( x )  12sin x cos x  0
Þ sin x  0 or cos x  0 (M1)

OR
sin 2 x  0 ,
for 0 £ 2 x £ 2  (M1)
THEN
π
x  0, , π (A1)(A1)(A1) (N4) 6
2

(b) (i) translation (A1)

in the y-direction of –1 (A1)
(ii) 1.11 (1.10 from TRACE is subject to AP) (A2) 4
[10]

51. (a) (i) a 1   accept (1   , 0)  (A1)

(ii) b  1   accept (1   , 0)  (A1) 2

 1 2
(b) (i)   2.14
h ( x)dx   h ( x )dx
1
(M1)(A1)(A1)

OR
1 2
  2.14
h ( x ) dx   1
h ( x ) dx (M1)(A1)(A1)

OR
1 1
  2.14
h ( x)dx   h ( x)dx
2
(M1)(A1)(A1)

(ii) 5.141...  ( 0.1585...)

= 5.30 (A2) 5

(c) (i) y = 0.973 (A1)

(ii)  0.240  k  0.973 (A3) 4
[11]

52. (a) y0 (A1) 1

2 x
(b) f ( x)  (A1)(A1)(A1) 3
(1  x 2 ) 2

6x2  2
(c)  0 (or sketch of f ( x ) showing the maximum) (M1)
(1  x 2 )3

6x2  2  0 (A1)

1
x (A1)
3

1
x ( 0.577) (A1) (N4) 4
3

0.5 1  0.5 1 0 1 
(d)   0.5 1 x 2
dx  = 2 
 0 1 x 2
dx = 2 
 0.5 1 x 2
dx 

(A1)(A1) 2

[10]
53. (a) x4 (A1)

g² changes sign at x  4 or concavity changes (R1) 2

(b) x2 (A1)

EITHER
g goes from negative to positive (R1)
OR
g (2) = 0 and g² (2) is positive (R1) 2

(c)

1 2 3 4 5 6 7 8
P

(A2)(A1)(A1) 4
Note: Award (A2) for a suitable cubic curve through (4, 0),
(A1) for M at x = 2, (A1) for P at (4, 0).
[8]

54. (a) (i) When t = 0, v = 50 + 50e0 (A1)

= 100 m s–1 (A1)

(ii) When t = 4, v = 50 + 50e–2 (A1)

= 56.8 m s–1 (A1) 4
 ds
(b) v= Þ s =  v dt
dt

 50  50e dt

4
- 0.5 t
(A1)(A1)(A1) 3
0

Note: Award (A1) for each limit in the correct position and
(A1) for the function.

 50  50e dt

4
- 0.5 t
(c) Distance travelled in 4 seconds =
0

= [50t – 100e–0.5t] 04 (A1)

= (200 – 100e–2) – (0 – 100e0)
= 286 m (3 sf) (A1)
–0.5t
Note: Award first (A1) for [50t – 100e ], ie
limits not required.
OR
Distance travelled in 4 seconds = 286 m (3 sf) (G2) 2

(d)

100

velocity
(t = 4)
50

0 2 4 6 8 10 12 t
time
Notes: Award (A1) for the exponential part, (A1) for the
straight line through (11, 0),
Award (A1) for indication of time on x-axis and velocity on
y-axis,
(A1) for scale on x-axis and y-axis.
Award (A1) for marking the point where t = 4.
5
 56.8
(e) Constant rate = (M1)
7
= 8.11 m s–2 (A1) 2
Note: Award (M1)(A0) for –8.11.

1
(f) distance = (7)(56.8) (M1)
2
= 199 m (A1) 2
Note: Do not award ft in parts (e) and (f) if candidate has not
used a straight line for t = 4 to t = 11 or if they continue the
exponential beyond t = 4.
[18]

 π 1  π 1
55. (a) (i) cos  –  , sin  –   – (A1)
 4 2  4 2
 π  π
therefore cos  –   sin  –  = 0 (AG)
 4  4

(ii) cos x + sin x = 0 Þ 1 + tan x = 0

Þ tan x = –l (M1)
3π
x= (A1)
4
Note: Award (A0) for 2.36.

OR
3π
x= (G2) 3
4

(b) y = ex(cos x + sin x)

dy
= ex(cos x + sin x) + ex(–sin x + cos x) (M1)(A1)(A1) 3
dx
= 2ex cos x
 dy
(c) = 0 for a turning point Þ 2ex cos x = 0 (M1)
dx
Þ cos x = 0 (A1)
π π
Þx= Þa= (A1)
2 2
π π
π π
y = e 2 (cos + sin ) = e 2
2 2
π
2
b=e (A1) 4
Note: Award (M1)(A1)(A0)(A0) for a = 1.57, b = 4.81.

d2 y
(d) At D, =0 (M1)
dx 2
2ex cos x – 2exsin x = 0 (A1)
2ex (cos x – sin x) = 0
Þ cos x – sin x = 0 (A1)
π
Þx= (A1)
4
π
π π
Þ y = e 4 (cos + sin ) (A1)
4 4
π
= 2e 4
(AG) 5

3
(e) Required area = 0
4 e x (cos x + sin x)dx (M1)

= 7.46 sq units (G1)

OR
Αrea = 7.46 sq units (G2) 2
Note: Award (M1)(G0) for the answer 9.81 obtained if the
calculator is in degree mode.
[17]

56. (a) (i) f (x) = –2e–2x (A1)

(ii) f (x) is always negative (R1) 2
 1
– 2 –
(b) (i) y=1+ e 2 (= 1 + e) (A1)
1
 1 – 2 –
(ii) f    2e 2 (= –2e) (A1) 2
 2
Note: In part (b) the answers do not need to be simplified.

 1
(c) y – (1 + e) = –2e  x   (M1)
 2
y = –2ex + 1 ( y = –5.44 x + 1) (A1)(A1) 3

(d)

(i) (ii) (iii)

(A1)(A1)(A1)
Notes: Award (A1) for each correct answer. Do not allow (ft)
on an incorrect answer to part (i). The correct final diagram is
shown below. Do not penalize if the horizontal asymptote is
missing. Axes do not need to be labelled.

(i)(ii)(iii)
y
8

P 4

2
1

x
–1 – 12 1 2
 0
(iv) Area =  
1 [(1 
2
e  2 x )  (2ex  1)]dx (or equivalent) (M1)(M1)

Notes: Award (M1) for the limits, (M1) for the function.
Accept difference of integrals as well as integral of difference.
Area below line may be calculated geometrically.
0

2 x
Area = 1 [(e  2ex)dx

2
0
 1 
=  e  2 x  ex 2  (A1)
 2  1
2

= 0.1795 …= 0.180 (3 sf) (A1)

OR
Area = 0.180 (G2) 7
[14]

57. Note: Do not penalize missing units in this question.

(a) (i) At release(P), t = 0 (M1)
s = 48 + 10 cos 0
= 58 cm below ceiling (A1)

(ii) 58 = 48 +10 cos 2πt (M1)

cos 2πt = 1 (A1)
t = 1sec (A1)
OR
t = 1sec (G3) 5

ds
(b) (i) = –20π sin 2πt (A1)(A1)
dt
Note: Award (A1) for –20π, and (A1) for sin 2t.
ds
(ii) v= = –20π sin 2πt = 0 (M1)
dt
sin 2 πt = 0
1
t = 0, ... (at least 2 values) (A1)
2
s = 48 + 10 cos 0 or s = 48 +10 cos π (M1)
= 58 cm (at P) = 38 cm (20 cm above P) (A1)(A1) 7
Note: Accept these answers without working for full marks.
May be deduced from recognizing that amplitude is 10.
 (c) 48 +10 cos 2πt = 60 + 15 cos 4πt (M1)
t = 0.162 secs (A1)
OR
t = 0.162 secs (G2) 2

(d) 12 times (G2) 2

Note: If either of the correct answers to parts (c) and (d) are
missing and suitable graphs have been sketched, award (G2)
for sketch of suitable graph(s); (A1) for t = 0.162; (A1) for 12.
[16]

58. (a) x=1 (A1) 1

(b) (i) f (–1000) = 2.01 (A1)
(ii) y=2 (A1) 2

( x  1) 2 ( 4 x  13)  2( x  1)(2 x 2  13 x  20)

(c) f (x) = (A1)(A1)
( x  1) 4
( 4 x 2  17 x  13)  ( 4 x 2  26 x  40)
= (A1)
( x  1) 3
9 x  27
= (AG) 3
( x  1) 3
Notes: Award (M1) for the correct use of the quotient rule, the
first (A1) for the placement of the correct expressions into the
quotient rule.
Award the second (A1) for doing sufficient simplification to
make the given answer reasonably obvious.

(d) f ′(3) = 0 Þ stationary (or turning) point (R1)

18
f ²(3) = > 0 Þ minimum (R1) 2
16

(e) Point of inflexion Þ f ²(x) = 0 Þ x = 4 (A1)

x=4Þy=0 Þ Point of inflexion = (4, 0) (A1)
OR
Point of inflexion = (4, 0) (G2) 2
[10]
 1
1
-kx  1 -kx 
59. (a) 0 e dx   k e  0 (A1)

1 –k 0
=– (e – e ) (A1)
k
1
= – (e–k – 1) (A1)
k
1
= – (1 – e–k) (AG) 3
k

(b) k = 0.5
(i)

(0,1)
1

x
–1 0 1 2 3

(A2)
Note: Award (A1) for shape, and (A1) for the point (0,1).

(ii) Shading (see graph) (A1)

1
-kx dx for k = 0.5
(iii) Area = e
0
(M1)

1
= (1 – e0.5)
0 .5
= 0.787 (3 sf) (A1)
OR
Area = 0.787 (3 sf) (G2) 5

dy
(c) (i) = –ke–kx (A1)
dx

(ii) x=1 y = 0.8 Þ 0.8 = e –k (A1)

ln 0.8 = –k
k = 0.223 (A1)
 dy
(iii) At x = 1 = –0.223e–0.223 (M1)
dx
= –0.179 (accept –0.178) (A1)
OR
dy
= –0.178 or – 0.179 (G2) 5
dx
[13]

60. (a) (i) Vertical asymptote x = –l (A1)

(ii) Horizontal asymptote y = 0 (A1)
(iii)

4
y
2

–3 –2 –1 0 1 x 2 3

–2

(A1)(A1)
Note: Award (A1) for each branch.
4
 – 6x 2
(b) (i) f ' (x) =
1  x 
3 2

f '' (x) =
1  x  – 12 x   6 x (2)(1  x ) 3x 
3 2 2 3 1 2
(M1)
1  x  3 4

=
1  x – 12x   36 x
3 4
(A1)
1  x  3 3

– 12 – 12 x 4  36 x 4
= (A1)
1  x  3 3

12 x 2 x – 13
= (AG)
1  x  3 3

(ii) Point of inflexion => f " (x) = 0 (M1)

1
=> x = 0 or x = 3
2
x = 0 or x = 0.794 (3 sf) (A1)(A1)
OR
x = 0, x = 0.794 (G1)(G2) 6

3 b–a 2
(c) (i) Approximate value of  1
f ( x ) dx , h =
n

5
(A1)

1
= [1 + 1.068377 + ... + 0.215332 + 0.071429] (A1)
5
1
= (3.284025)
5
= 0.656805 (A1)
 3
(ii) 1
f ( x ) dx = 0.637599

2
y
1

–1 0 1 x 2 3

–1
(A1)
Between 1 and 3, the graph is 'concave up', so that the straight lines
forming the trapezia are all above the graph. (R1) 5
[15]

61. (a) At A, x = 0 => y = sin (e0) = sin (1) (M1)

=> coordinates of A = (0,0.841) (A1)
OR
A(0, 0.841) (G2) 2

(b) sin (ex) = 0 => ex =  (M1)

=> x = ln  (or k = π) (A1)
OR
x = ln  (or k = π) (A2) 2

(c) (i) Maximum value of sin function = 1 (A1)

dy
(ii) = ex cos (ex) (A1)(A1)
dx
Note: Award (A1) for cos (ex) and (A1) for ex.

dy
(iii) = 0 at a maximum (R1)
dx
ex cos (ex) = 0
=> ex = 0 (impossible) or cos (ex) = 0 (M1)
π π
=> ex = => x = ln (A1)(AG) 6
2 2
 ln 
(d) (i) Area = 0
sin (e x ) dx (A1)(A1)(A1)

Note: Award (A1) for 0, (A1) for ln π, (A1) for sin (ex).
(ii) Integral = 0.90585 = 0.906 (3 sf) (G2) 5

(e)

y = x3

(M1)
At P, x = 0.87656 = 0.877 (3 sf) (G2) 3
[18]

ds t2
62. (a) = 30 – at => s = 30t – a +C (A1)(A1)(A1)
dt 2
t2
Note: Award (A1) for 30t, (A1) for a , (A1) for C.
2

t = 0 => s = 30(0) – a
0  + C = 0 + C => C = 0
2
(M1)
2
1 2
=> s = 30t – at (A1) 5
2

(b) (i) vel = 30 – 5(0) = 30 m s–1 (A1)

(ii) Train will stop when 0 = 30 – 5t => t = 6 (M1)
1 2
Distance travelled = 30t – at
2
1
= 30(6) – (5) (62) (M1)
2
= 90m (A1)
90 < 200 => train stops before station. (R1)(AG) 5
 30
(c) (i) 0 = 30 – at => t = (A1)
a
2
 30  1  30 
(ii) 30   – (a)   = 200 (M1)(M1)
 a  2  a 
30
Note: Award (M1) for substituting , (Ml) for setting equal
a
to 200.
900 450 450
=> –  = 200 (A1)
a a a
450 9
=> a =  = 2.25 m s –2 (A1) 5
200 4
Note: Do not penalize lack of units in answers.
[15]

1
1
  2 x  +(2)(1 + x2) 2
1
dy –
63. (a) = (2x)  1  x 2 2 (M1)(M1)
dx 2 
Note: Award (M1) for correct use of product rule,
(M1) for correct use of chain rule.

dy 2x2
= +2 1  x 2 (A1) 3
dx 1 x 2

du
(b) u = 1 + x2 => = 2x (or du = 2xdx) (M1)
dx
1 1
 du 
 2 x 1  x dx   u 2  dx  dx   u 2 du
2
=> (M1)

3 3
2 2
= u 2 + C = (1 + x2) 2 + C (A1)(AG) 3
3 3
Note: Accept proof by differentiation.
 k
2
  
k 3
(c) R= 0
2 x 1  x 2 dx   1  x 2
3
2
 = 1
0
(M1)

3

1  k 2 2  1 = 1 => (1 + k2) 2 =
3
2 2 5
=> (M1)
3 3 2
=> k = 0.9176 = 0.918 (A1)
OR
k = 0.918 (G3) 3
[9]

1
64. (i) At x = a, h (x) = a 5
4
1 – 1
h (x) = x 5 => h (a) = 4
= gradient of tangent (A1)
5
5a 5

1 1
1 1 1 5
=> y – a 5 = 4
(x – a) = 4
x– a (M1)
5
5a 5
5a 5

1
1 4 5
=> y = 4
x+ a (A1)
5
5a 5

(ii) tangent intersects x-axis => y = 0

1
1 4 5
=> 4
x=– a (M1)
5
5a 5

4
 4 15 
=> x = 5a 5  – 
 5 a  = –4a (M1)(AG) 5
 
[5]
  1 
x  2   (ln 2 x  1)
2 x
65. (a) (i) f (x) =   (M1)(M1)
x2
Note: Award (M1) for the correct use of the quotient rule and
(M1) for correct substitution.
1  ln 2 x
= (AG)
x2

(ii) f (x) = 0 for max/min. (R1)

1  ln 2 x e
2
= 0 only at 1 point, when x = (R1)
x 2
Note: Award no marks if the reason given is of the sort “by
looking at the graph”.

(iii) Maximum point when f (x) = 0.

e
f (x) = 0 for x = (= 1.36) (A1)
2
e 2
y=f   = (= 0.736) (A1) 6
2 e
Note: Award (A1) per correct coordinate if the answer is found
using the GDC, regardless of method. If one or both
coordinates are wrong, you may award up to 1 mark for
method.

1
  2  x 2  (1  ln 2 x ) 2 x
(b) f ²(x) = 2x (M1)(M1)
x4
2 ln 2 x  3
= (AG)
x3
Inflexion point Þ f ²(x) = 0 (M1)
Þ 2ln 2x = 3 (M1)
e 1.5
x= (= 2.24) (A1) 6
2
(c) (i) The trapezium rule would underestimate the area of S. (A1)
y

x
Shaded area not included when using the trapezium rule
(or similar reasonable explanation). (R2) 3

1 1
(ii) u = ln 2x; du =  2dx  dx (M1)
2x x
ln 2
 x 
dx  udu (M1)

u2
= +C (A1)
2
(ln 2 x ) 2
= +C (A1) 4
2

e
ln 2 x
(iii) Area of S =  2
0.5 x
dx (M1)(A1)

Note: Award (M1) for the integral expression,

and (A1) for the limits. (M1)
2
  e 
 ln 2  
 2  (ln (2  0.5)) 2
=   (M1)
2 2
1 1
= 0 (A1) 4
2 2
Note: Award only (A1)(M0)(M0)(A1) if the area (to 3 sf or
exactly) is found on the GDC.

(d) (i) If x1 = 1, then x2 = –1.26 (M1)

f (x2) = f (–1.26) does not exist, so x3 cannot be calculated. (R2) 3
 f (0.4)
(ii) x2 = 0.4 – = 0.47297 (A1)
f (0.4)
Absolute error = ½0.5 – 0.47297½ = 0.02703 (A1)
f (x2 )
x3 = x2 – = 0.49787 (A1)
f ( x 2 )
Absolute error = ½0.5 – 0.49787½ = 0.00213 (A1) 4
which is less than 0.01.
Notes: Absolute errors need not be explicitly given.
Award (A3) if further terms are listed, without stating that they
are unnecessary.
[30]

66. (a) (i) a = –3 (A1)

(ii) b=5 (A1) 2

(b) (i) f (x) = –3x2 + 4x + 15 (A2)

(ii) –3x2 + 4x + 15 = 0
–(3x + 5)(x – 3) = 0 (M1)
5
x = – or x = 3 (A1)(A1)
3
OR
5
x=– or x = 3 (G3)
3

(iii) x = 3 Þ f (3) = –33 + 2(32) + 15(3) (M1)

= –27 + 18 + 45 =36 (A1)
OR
f (3) = 36 (G2) 7

(c) (i) f (x) = 15 at x = 0 (M1)

Line through (0, 0) of gradient 15
Þ y = 15x (A1)
OR
y = 15x (G2)
 (ii) –x3 + 2x2 + 15x = 15x (M1)
Þ –x3 + 2x2 = 0
Þ –x2 (x – 2) = 0
Þx=2 (A1)
OR
x=2 (G2) 4

(d) Area =115 (3 sf) (G2)

OR
5
6  x4 x3 x2 

3 2
Area = ( x  2 x  15 x)dx    2  15  (M1)
0
 4 3 2 0
1375
= = 115 (3 sf) (A1) 2
12
[15]

67. (a) (i) v(0) = 50 – 50e0 = 0 (A1)

(ii) v(10) = 50 – 50e–2 = 43.2 (A1) 2

dv
(b) (i) a= = –50(–0.2e–0.2t) (M1)
dt
= 10e–0.2t (A1)

(ii) a(0) = 10e0 = 10 (A1) 3

(c) (i) t  ¥ Þ v  50 (A1)

(ii) t¥Þa0 (A1)
(iii) when a = 0, v is constant at 50 (R1) 3

(d) (i) y = vdt (M1)

–0.2t
e
= 50t – +k (A1)
 0.2
= 50t + 250e–0.2t + k (AG)

(ii) 0 = 50(0) + 250e0 + k = 250 + k (M1)

Þ k = –250 (A1)
 (iii) Solve 250 = 50t + 250e–0.2t – 250 (M1)
Þ 50t + 250e–0.2t – 500 = 0
Þ t + 5e–0.2t – 10 = 0
Þ t = 9.207 s (G2) 7
[15]

5
68. (a) (i) x=– (A1)
2

3
(ii) y= (A1) 2
2

(b) By quotient rule (M1)

dy (2 x  5)(3)  (3 x  2)(2)
 (A1)
dx (2 x  5) 2
19
= (A1) 3
( 2 x  5) 2

(c) There are no points of inflexion. (A1) 1

[6]
69. (a)
y

4
{0.5<
(A1)
x<1
3.5<y<4
MAXIMUM
POINT

integers (A1)
{
1 on axis

x
1 2 3 4 5
LEFT RIGHT
INTERCEPT 3<x<3.5 (A1) 3.5<x<4 (A1) INTERCEPT
–1
(A1) {3.2< x<3.6
–0.2<y <0
MINIMUM
POINT
5

(b)  is a solution if and only if  +  cos  = 0. (M1)

Now  +  cos  =  + (–1) (A1)
=0 (A1) 3

(c) By using appropriate calculator functions x = 3.696 722 9... (M1)

Þ x = 3.69672 (6sf) (A1) 2

(d) See graph: (A1)

π
 (π  x cos x)dx
0
(A1) 2
 π
(e) EITHER  (π  x cos x)dx = 7.86960 (6 sf)
0
(A3) 3

Note: This answer assumes appropriate use of a calculator eg

 fnInt (Y1 , X , 0, π)  7.869604401
‘fnInt’: 
with Y1  π  x cos x
π
 (π  x cos x)dx  [πx  x sin x  cos x]
π
OR 0
0

= ( – 0) + ( sin  – 0 × sin 0) + (cos  – cos 0) (A1)

= 2 + 0 + –2 = 7.86960 (6 sf) (A1) 3
[15]

70. (a) When t = 0, (M1)

h = 2 + 20 × 0 – 5 × 02 = 2 h=2 (A1) 2

(b) When t = 1, (M1)

h = 2 + 20 × 1 – 5 × 12 (A1)
= 17 (AG) 2

(c) (i) h = 17 Þ 17 = 2 + 20t – 5t2 (M1)

(ii) 5t2 – 20t + 15 = 0 (M1)
Û 5(t2 – 4t + 3) = 0
Û (t – 3)(t – 1) = 0 (M1)
Note: Award (M1) for factorizing or using the formula
Û t = 3 or 1 (A1) 4
Note: Award (A1) for t = 3

(d) (i) h = 2 + 20t – 5t2

dh
Þ = 0 + 20 – 10t
dt
= 20 – 10t (A1)(A1)

(ii) t=0 (M0)

dh
Þ = 20 – 10 × 0 = 20 (A1)
dt

dh
(iii) =0 (M1)
dt
Û 20 – 10t = 0 Û t = 2 (A1)
 (iv) t=2 (M1)
Þ h = 2 + 20 × 2 – 5 × 22 = 22 Þ h = 22 (A1) 7
[15]

71. (a) (i) Note: Range of f = {y : 0 £ y £ 2} (graphic display calculator)

So let a = 0 – Î, b = 2 + Î, with 0 < Î < 2
For example, a = –1 b = 3 etc. (A1)(A1)

2x
(ii) As x  ¥,  0, f (x)  1; y = 1 (A1) 3
1 x2

d  2x 
(b) f (x) = 1  
dx  1  x 2 
 (1  x 2 )  (2) – (2 x)(2 x) 
= 0 –  
 (A1)(A1)(A1)
 (1  x 2 ) 2 
4 x 2  2(1  x 2 )
= (A1)
(1  x 2 ) 2
2x 2  2
= (AG) 4
(1  x 2 ) 2

f ( x)  0 Û 2 x 2  2  0
(c)  (M1)
Û x  1 
From graphic display calculator inspection, or f (x) on each side
of –1, max when x = –1 (M1)
2
f (–1) = 1 – =1+1=2
11
(–1, 2) (A1) 3
  2x 
(d) (i)  f ( x)dx   1  1  x 2
dx

1
=x–  u du (A1)(M1)

1
Note: Award (A1) for x, and (M1) for  u du.
= x – ln u + C (M1)
Notes: Award (M1) for ln u or award (A2) by inspection.
= x – ln(1 + x2) + C (A1) 4
2
Note: Award (A1) for ln(1 + x ).

1
(ii) Area =  0
f ( x )dx (A1)

Note: Award (A1) for upper and lower limits.

= [ x  ln(1  x 2 )]10 (M1)

= (1 – 0) – (ln 2 – ln 1) (A1)
= 1 – ln 2 (A1) 4
Note: Award (A0) for 0.307
[18]

72. (a) (i) t = 0 s = 800

t = 5 s = 800 + 500 – 100 = 1200 (M1)
distance in first 5 seconds = 1200 – 800
= 400 m (A1) 2

ds
(ii) v= = 100 – 8t (A1)
dt
At t = 5, velocity = 100 – 40 (M1)
= 60 m s–1 (A1) 3

(iii) Velocity = 36 m s–1 Þ 100 – 8t = 36 (M1)

t = 8 seconds after touchdown. (A1) 2

(iv) When t = 8, s = 800 + 100(8) – 4(8)2 (M1)

= 800 + 800 – 256 (A1)
= 1344 m (A1) 3
 (b) If it touches down at P, it has 2000 – 1344 = 656 m to stop. (M1)
To come to rest, 100 – 8t = 0 Þ t = 12.5 s (M1)
Distance covered in 12.5 s = 100(12.5) – 4(12.5)2 (M1)
= 1250 – 625
= 625 (A1)
Since 625 < 656, it can stop safely. (R1) 5
[15]

73. (a) y =  sin x – x

y
3

2 (1.25, 1.73)

1
(2.3, 0)

–3 –2 –1 x
1 2 3
(–2.3, 0)
–1

(–1.25, –1.73) –2

–3
(A5) 5
Notes: Award (A1) for appropriate scales marked on the axes.
Award (A1) for the x-intercepts at (2.3, 0).
Award (A1) for the maximum and minimum points at (1.25,
1.73).
Award (A1) for the end points at (3, 2.55).
Award (A1) for a smooth curve.
Allow some flexibility, especially in the middle three marks
here.

(b) x = 2.31 (A1) 1

 x2
(c)  ( π sin x  x )dx   π cos x 
2
C (A1)(A1)

Note: Do not penalize for the absence of C.

1
Required area =  ( π sin x  x)dx
0
(M1)

= 0.944 (G1)
OR area = 0.944 (G2) 4
[10]

74. (a) y = e2x cos x

dy
= e2x (–sin x) + cos x (2e2x) (A1)(M1)
dx
= e2x (2 cos x – sin x) (AG) 2

d2 y
(b) 2
= 2e2x (2 cos x – sin x) + e2x (–2 sin x – cos x) (A1)(A1)
dx
= e2x (4 cos x – 2 sin x – 2 sin x – cos x) (A1)
= e2x (3 cos x – 4 sin x) (A1) 4

d2 y
(c) (i) At P, =0 (R1)
dx 2
Þ 3 cos x = 4 sin x (M1)
3
Þ tan x =
4
3
At P, x = a, ie tan a = (A1)
4

(ii) The gradient at any point e2x (2 cos x – sin x) (M1)

Therefore, the gradient at P = e2a (2 cos a – sin a)
3 4 3
When tan a = , cos a = , sin a = (A1)(A1)
4 5 5
(by drawing a right triangle, or by calculator)
8 3
Therefore, the gradient at P = e2a    (A1)
5 5
= e2a (A1) 8
[14]
75. (a) t = 2 ⇒ h = 50 – 5(22) = 50 – 20
= 30 (A1)
OR
h = 90 – 40(2) + 5(22)
= 30 (A1) 1

(b)

50
h
40
30
20
10

1 2 3 4 5 t

(A4) 4
Note: Award (A1) for marked scales on each axis, (A1) for each
section of the curve.

dh d
(c) (i)  (50 – 5t2)
dt dt
= 0 – 10t = – 10t (A1)

dh d
(ii)  (90 – 40t + 5t2)
dt dt
= 0 – 40 + 10t = –40 + 10t (A1) 2

dh dh
(d) When t = 2 (i) = –10(2) or = –40 + 10 × 2 (M1)
dt dt
= –20 = –20 (A1) 2

dh
(e) = 0 Þ –10t = 0(0 ≤ t ≤ 2) or –40 + 10t = 0(2 ≤ t ≤ 5) (M1)
dt
t=0 or t=4 (A1)(A1) 3
 (f) When t = 4 (M1)
h = 90 – 40(4) + 5(42) (M1)
= 90 – 160 + 80
= 10 (A1) 3
[15]

76. (a)(i) & (c)(i)

y

(1.1, 0.55)

(1.51, 0)
0 x
1 2

–1

(2, –1.66)

–2

(A3)
Notes: The sketch does not need to be on graph paper. It
should have the correct shape, and the points (0, 0), (1.1, 0.55),
(1.57, 0) and (2, –1.66) should be indicated in some way.
Award (A1) for the correct shape.
Award (A2) for 3 or 4 correctly indicated points, (A1) for 1 or 2
points.

(ii) Approximate positions are

positive x-intercept (1.57, 0) (A1)
maximum point (1.1, 0.55) (A1)
end points (0, 0) and (2, –1.66) (A1)(A1) 7
 (b) x2 cos x = 0 x ≠ 0 ⇒ cos x = 0 (M1)
π
Þx= (A1) 2
2
Note: Award (A2) if answer correct.

(c) (i) see graph (A1)


(ii)  0
2 x 2 cos x dx (A2) 3

Note: Award (A1) for limits, (A1) for rest of integral correct
(do not penalize missing dx).

(d) Integral = 0.467 (G3)

OR


Integral = x 2 sin x  2 x cos x  2 sin x 0 π/2
(M1)
π2 π 
=  (1)  2 (0)  2(1) – [0 + 0 – 0] (M1)
 4 2 
π
= – 2 (exact) or 0.467 (3 sf) (A1) 3
2
[15]

77. (a) From graph, period = 2π (A1) 1

(b) Range = {y |–0.4 < y < 0.4} (A1) 1

d
(c) (i) f (x) = {cos x (sin x)2}
dx
= cos x (2 sin x cos x) – sin x (sin x)2 or –3 sin3 x + 2 sin x(M1)(A1)(A1)
Note: Award (M1) for using the product rule and (A1) for each
part.
 (ii) f (x) = 0 (M1)
Þ sin x{2 cos x – sin2 x} = 0 or sin x{3 cos x – 1} = 0 (A1)
Þ 3 cos2 x – 1 = 0
1
Þ cos x = ±   (A1)
 3
1
At A, f (x) > 0, hence cos x =   (R1)(AG)
 3

 2

 1    1   
(iii) f (x) =   1 –    (M1)
 3    3   
 
2 1 2
=   3 (A1) 9
3 3 9

π
(d) x= (A1) 1
2

1
 (cos x)(sin x)
2
(e) (i) dx  sin 3 x  c (M1)(A1)
3

1 ï ï
3
π/2 π
 (cosx)(sin x) dx   sin   (sin 0) 3 
2
(ii) Area = (M1)
0 3 ï 2 ï
1
= (A1) 4
3

(f) At C f ²(x) = 0 (M1)

Û 9 cos3 x – 7 cos x = 0
Û cos x(9 cos2 x – 7) = 0 (M1)
π 7
Þx= (reject) or x = arccos = 0.491 (3 sf) (A1)(A1) 4
2 3
[20]
 2x  1
78. (a) (i) f (x) =
x3
7
=2+ by division or otherwise (M1)
x3
Therefore as x  ¥ f (x)  2 (A1)
Þ y = 2 is an asymptote (AG)
2x  1
OR lim =2 (M1)(A1)
x ¥ x  3

Þ y = 2 is an asymptote (AG)
OR make x the subject
yx – 3y = 2x + 1
x(y – 2) = 1 + 3y (M1)
1  3y
x= (A1)
y2
Þ y = 2 is an asymptote (AG)
Note: Accept inexact methods based on the ratio of the
coefficients of x.

(ii) Asymptote at x = 3 (A1)

(iii) P(3, 2) (A1) 4

1  1 
(b) f (x) = 0 Þ x = –   , 0 (M1)(A1)
2 2 
1  1
x = 0 Þ f (x) = –  0,   (M1)(A1) 4
3  3
Note: These do not have to be in coordinate form.
(c)
y

3 x

(A4) 4
Note: Asymptotes (A1)
Intercepts (A1)
“Shape” (A2).

( x  3)(2)  (2 x  1)
(d) f (x) = (M1)
( x  3) 2
7
= (A1)
( x  3) 2
= Slope at any point
Therefore slope when x = 4 is –7 (A1)
And f (4) = 9 ie S(4, 9) (A1)
Þ Equation of tangent: y – 9 = –7(x – 4) (M1)
7x + y – 37 = 0 (A1) 6

7
(e) at T, = –7 (M1)
( x  3) 2
Þ (x – 3)2 = 1 (A1)
x – 3 = ±l (A1)
x  4 or 2  S ( 4, 9)
 (A1)(A1) 5
y  9 or – 5 T ( 2,  5)

 4  2 9 5
(f) Midpoint [ST] =  , 
 2 2 
= (3, 2)
= point P (A1) 1
[24]
79. (a) f ²(x) = 2x – 2
Þf (x) = x2 – 2x + c (M1)(M1)
= 0 when x = 3
Þ 0 =9–6+c
c = –3 (A1)
f (x) = x2 – 2x – 3 (AG)
x3
f (x) = – x2 – 3x + d (M1)
3
When x = 3, f (x) = –7
Þ –7 = 9 – 9 – 9 + d (M1)
Þ d =2 (A1) 6
x3
Þ f (x) = – x2 – 3x + 2
3

(b) f (0) = 2 (A1)

1
f (–1) = – –1+3+2
3
2
=3 (A1)
3
f (–1) = 1 + 2 – 3
=0 (A1) 3

 2
(c) f (–1) = 0 Þ   1, 3  is a stationary point
 3
y

–1, 3 23
2
x

(3, –7)

(A4) 4
Note: Award (A1) for maximum, (A1) for (0, 2)
(A1) for (3, –7), (A1) for cubic.
[13]

80. (a) y = ex/2 at x = 0 y = e0 = 1 P(0, 1) (A1)(A1) 2

 ln 2
(b) V=  0
(e x / 2 ) 2 dx (A4) 4

Notes: Award (A1) for 

(A1) for each limit
(A1) for (ex/2)2.

ln 2
(c) V=  0
e x dx (A1)

=  [e x ] ln0 2 (A1)
ln2 0
= [e – e ] (A1)
= [2 – 1] =  (A1)(A1)
= (AG) 5
[11]

81. (a)
y
y 2 = 9x
P

Q
x

y2 = 9x
62 = 9(4) (M1)
36 = 36 (A1) 2
Þ (4, 6) on parabola
(b) (i) y=3 x
dy 3
 (M1)
dx 2 x
= Slope at any point
3
Therefore at (4, 6), slope of tangent = (A1)
4
4
Þ Slope of normal = – (A1)
3
4
Therefore equation of normal is y – 6 = – (x – 4) (M1)
3
3y – 18 = –4x + 16
4x + 3y – 34 = 0 (A1) 5
Notes: Candidates may differentiate implicitly to obtain
dy 9
 .
dx 2 y
Answer must be given in the form ax + by + c = 0.

(ii) Coordinates of Q:
y = 0, 4x = 34
17
x= (A1)
2
 17 
Q  , 0 (A1) 2
 2 

2
9  2
(c) SP =   4   (0  6) (M1)
4 
49
=  36
16
25
= (A1)
4
17 9
SQ =  (M1)
2 4
34 9
= 
4 4
25
= (A1) 4
4
 (d) SP  = SQ  Þ SPˆ Q  SQˆ P (M1)
But SQˆ P  MPˆ Q (alternate angles) (A1)
Þ MPˆ Q  SPˆ Q (A1) 3
[16]

82. (a) f (1) = 3 f (5) = 3 (A1)(A1) 2

(b) EITHER distance between successive maxima = period (M1)

=5–1 (A1)
=4 (AG)
2π
OR Period of sin kx = ; (M1)
k
2π
so period = (A1)
π
2
=4 (AG) 2

π  3π 
(c) EITHER A sin   + B = 3 and A sin   + B = –1 (M1) (M1)
2
   2 
Û A + B = 3, – A + B = –1 (A1)(A1)
Û A = 2, B = 1 (AG)(A1)
OR Amplitude = A (M1)
3  (1) 4
A=  (M1)
2 2
A=2 (AG)
Midpoint value = B (M1)
3  (1) 2
B=  (M1)
2 2
B=1 (A1) 5
Note: As the values of A = 2 and B = 1 are likely to be quite
obvious to a bright student, do not insist on too detailed a
proof.
 π 
(d) f (x) = 2 sin  x + 1
2 
π π 
f (x) =  2 cos  x  + 0 (M1)(A2)
2 2 
π
Note: Award (M1) for the chain rule, (A1) for   , (A1) for
2
π 
2 cos  x  .
2 

π 
=  cos  x  (A1) 4
2 
Notes: Since the result is given, make sure that reasoning is
valid. In particular, the final (A1) is for simplifying the result of
the chain rule calculation. If the preceding steps are not valid,
this final mark should not be given. Beware of “fudged”
results.

π 
(e) (i) y = k – x is a tangent Þ – =  cos  x  (M1)
2 
π 
Þ –1 = cos  x  (A1)
2 
π
Þ x =  or 3 or ...
2
Þ x = 2 or 6 ... (A1)
Since 0 £ x £ 5, we take x = 2, so the point is (2, 1) (A1)

(ii) Tangent line is: y = –(x – 2) + 1 (M1)

y = (2 + 1) – x
k = 2 + 1 (A1) 6

π 
(f) f (x) = 2 Þ 2 sin  x + 1 = 2 (A1)
2 
π  1
Þ sin  x   (A1)
2  2
π π 5π 13π
Þ x  or or
2 6 6 6
1 5 13
x= or or (A1)(A1)(A1) 5
3 3 3
[24]
 dy 1
83. (a) y = ln x Þ  (A1)
dx x
dy 1
when x = e, 
dx e
1
tangent line: y =   (x – e) + 1 (M1)
e
1 x
y= (x) – 1 + 1 = (A1)
e e
0
x=0Þy= =0 (M1)
e
(0, 0) is on line (AG) 4

d 1
(b) (x ln x – x) = (1) × ln x + x ×   – 1 = ln x (M1)(A1)(AG) 2
dx  x
Note: Award (M1) for applying the product rule, and (A1) for
1
(1) × ln x + x ×   .
 x

(c) Area = area of triangle – area under curve (M1)

1  e

2  1 
=   e  1  ln xdx (A1)

e
=  [ x ln x  x]1e (A1)
2
e
= – {(e ln e – 1 ln1) – (e – 1)} (A1)
2
e
= – {e – 0 – e + 1}
2
1
= e – 1. (AG) 4
2
[10]
84. (a) y = x(x – 4)2
(i) y = 0 Û x = 0 or x = 4 (A1)
dy
(ii) = 1(x – 4)2 + x × 2(x – 4) = (x – 4)(x – 4 + 2x)
dx
= (x – 4)(3x – 4) (A1)
dy 4
= 0 Þ x = 4 or x = (A1)
dx 3
dy 
x 1Þ  (3)(1)  3 > 0 ï
dx ï 4
 Þ is a maximum (R1)
dy 3
x2Þ  (2)(2)  4  0ï
dx ï

Note: A second derivative test may be used.
2 2
4 4 4  4  8 4 64 256
x= Þ y =    4       
3 3 3  3  3  3 9 27
 4 256 
 ,  (A1)
 3 27 
 4 256 
Note: Proving that  ,  is a maximum is not necessary to
 3 27 
receive full credit of [4 marks] for this part.

d2 y d d
(iii) 2
 ( x  4)(3x  4)   (3x2 – 16x + 16) = 6x – 16 (A1)
dx dx dx
d2 y
= 0 Û 6x – 16 = 0 (M1)
dx 2
8
Ûx= (A1)
3
2 2
8 88  8 4 8 16 128
x= Þ y =   4      
3 33  3 3  3 9 27

 8 128 
 ,  (A1) 9
 3 27 
Note: GDC use is likely to give the answer (1.33, 9.48). If this
answer is given with no explanation, award (A2), If the answer
is given with the explanation “used GDC” or equivalent,
award full credit.
(b)
y
max pt.
10

pt. of inflexion

0
0 1 2 3 4 x
x–intercepts (A3) 3
Note: Award (A1) for intercepts, (A1) for maximum and (A1)
for point of inflexion.

(c) (i) See diagram above (A1)

(ii) 0 < y < 10 for 0 £ x £ 4 (R1)
4 4 4 4
So  0dx  
0 0
ydx   0
10dx Þ 0   0
ydx  40 (R1) 3

[15]