CALCULUS CORE - Shelby Bryant

CALCULUS CORE

IB Questionbank Mathematics Higher Level 3rd edition 1

1. METHOD 1

area = 3

0darctan xx A1

attempting to integrate by parts M1

= xx

xxx d1

1]arctan[

2

3

0

30

A1A1

=

3

0

230 )1ln(

2

1]arctan[

xxx A1

Note: Award A1 even if limits are absent.

= 4ln2

1

3

π A1

2ln

3

3π

METHOD 2

area = 3

π

0dtan

3

3πyy M1A1A1

= 3

π

0]cos[ln3

3πy M1A1

=

2ln

3

3π

2

1ln

3

3π A1

[6]

CALCULUS CORE


2. Attempt at implicit differentiation M1

e(x+y)

y

x

yxxy

x

y

d

d)sin(

d

d1 A1A1

let x = 0, y = 0 M1

e0

x

y

d

d1 = 0

x

y

d

d = –1 A1

let x = π2,π2 y

e0

y

x

yx

x

y

d

dπ)2sin(

d

d1 = 0

so x

y

d

d = –1 A1

since both points lie on the line y = –x this is a common tangent R1

Note: y = –x must be seen for the final R1. It is not sufficient to note

that the gradients are equal. [7]

3. (a) (i) f′(x) = 2

ln1

x

xx

x

M1A1

= 2

ln1

x

x

so f′(x) = 0 when ln x = 1, i.e. x = e A1

(ii) f′(x) > 0 when x < e and f′(x) < 0 when x > e R1

hence local maximum AG

Note: Accept argument using correct second derivative.

(iii) y ≤ e

1 A1

CALCULUS CORE


(b) f′″(x) = 4

2 2)ln1(1

x

xxx

x

M1

= 4

ln22

x

xxxx

= 3

ln23

x

x A1

Note: May be seen in part (a).

f″(x) = 0

–3 + 2 ln x = 0 M1

x = 2

3

e

since f″(x) < 0 when x < 2

3

e and f″(x) > 0 when x > 2

3

e R1

then point of inflexion

2

3

2

3

2e

3,e A1

(c)

A1A1A1

Note: Award A1 for the maximum and intercept, A1 for a vertical

asymptote and A1 for shape (including turning concave up).

CALCULUS CORE


(d) (i)

A1A1

Note: Award A1 for each correct branch.

(ii) all real values A1

(iii)

(M1)(A1)

Note: Award (M1)(A1) for sketching the graph of h, ignoring

any graph of g.

–e2 < x < –1 (accept x < –1) A1

[19]

CALCULUS CORE


4. (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]

5. (a) 3

2

12

d

dxx

x

y A1

2

2

12 xx = 0

x = 0, ±2

8

25,2,

8

25,2,

8

90,at 0

d

d

x

y A1A1A1

Note: Award A2 for all three x-values correct with errors/omissions in y-values.

(b) at x = 1, gradient of tangent = 2

3 (A1)

Note: In the following, allow FT on incorrect gradient.

equation of tangent is y – 2 = 2

3(x – 1)

2

1

2

3xy (A1)

meets x-axis when y = 0, –2 = 2

3(x – 1) (M1)

x = 3

1

coordinates of T are

0,

3

1 A1

CALCULUS CORE


(c) gradient of normal = –3

2 (A1)

equation of normal is y – 2 =

3

8

3

2)1(

3

2xyx (M1)

at x = 0, y = 3

8 A1

Note: In the following, allow FT on incorrect coordinates of T and N.

lengths of PN = 9

52PT ,

9

13 A1A1

area of triangle PTN = 9

52

9

13

2

1 M1

= 9

13 (or equivalent e.g.

18

676) A1

[15]

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.

CALCULUS CORE


(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

CALCULUS CORE


(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) f ′(x) = 3x2 – 6x – 9 (= 0) (M1)

(x + 1)(x – 3) = 0

x = –1; x = 3

(max)(–1, 15); (min)(3, –17) A1A1

Note: The coordinates need not be explicitly stated but the values

need to be seen.

y = –8x + 7 A1 N2

CALCULUS CORE


(b) f″(x) = 6x – 6 = 0 inflexion (1, –1) A1

which lies on y = – 8x + 7 R1AG [6]

8. (a) equation of line in graph a = t60

25 + 15 A1

15

12

5ta

(b) 1512

5

d

d t

t

v (M1)

v = 2

24

5t + 15t + c (A1)

when t = 0, v = 125 m s–1

v = 2

24

5t + 15t + 125 A1

from graph or by finding time when a = 0

maximum = 395 m s–1

A1

CALCULUS CORE


(c) EITHER

graph drawn and intersection with v = 295 m s

–1 (M1)(A1)

t = 57.91 – 14.09 = 43.8 A1

OR

295 = 2

24

5t + 15t + 125 t = 57.91...; 14.09... (M1)(A1)

t = 57.91… – 14.09… = 43.8 (8 30 ) A1

[8]

9. (a) volume = h

yx0

2dπ (M1)

h

yy0

dπ M1

= 2

π

2π

2

0

2hy

h

A1

CALCULUS CORE


(b) t

V

d

d = –3 × surface area A1

surface area = πx2 (M1)

= πh A1

since V = π

2

2

π 2 Vh

h M1A1

π

2π3

d

d V

t

V A1

Vt

Vπ23

d

d AG

Note: Assuming that t

h

d

d= –3 without justification gains no marks.

[6]

(c) V0 = 5000π (= 15700 cm3) A1

Vt

Vπ23

d

d

attempting to separate variables M1

EITHER

tV

Vdπ23

d A1

ctV π232 A1

c = π50002 A1

V = 0 M1

3133

2π

5000π

3

2 t hours A1

OR

T

tV

V

0

0

π5000dπ23

d M1A1A1

Note: Award M1 for attempt to use definite integrals, A1 for correct

limits and A1 for correct integrands.

TV π2320

π5000 A1

T = 3133

2π

5000π

3

2 hours A1

[16]

CALCULUS CORE


10. (a) a = 10e–0.2t

(M1)(A1)

at t = 10, a = 1.35 (m s–2

) (accept 10e–2

) A1

(b) METHOD 1

d =

10

0

2.0 d)e1(50 tt (M1)

= 283.83... A1

so distance above ground = 1720 (m) (3 s.f.) (accept 1716 (m)) A1

METHOD 2

s = ∫50(1 – e–0.2t

)dt = 50t + 250e–0.2t

(+ c) M1

Taking s = 0 when t = 0 gives c = –250 M1

So when t = 10, s = 283.3...

so distance above ground = 1720 (m) (3 s.f.) (accept 1716 (m)) A1 [6]

11. let x = distance from observer to rocket

let h = the height of the rocket above the ground

METHOD 1

t

h

d

d = 300 when h = 800 A1

x = 2

1

22 )360000(360000 hh M1

360000d

d

2

h

h

h

x A1

when h = 800

t

h

h

x

t

x

d

d

d

d

d

d M1

= 360000

300

2 h

h A1

= 240 (m s–1

) A1

CALCULUS CORE


METHOD 2

h2 + 600

2 = x

2 M1

2h = 2xh

x

d

d A1

x

h

h

x

d

d

=

5

4

1000

800 A1

t

h

d

d = 300 A1

t

h

h

x

t

x

d

d

d

d

d

d M1

= 5

4 × 300

= 240 (m s–1

) A1

METHOD 3

x2 = 600

2 + h

2 M1

2xt

hh

t

x

d

d2

d

d A1A1

when h = 800, x = 1000

t

h

t

x

d

d

1000

800

d

d M1A1

= 240 m s–1

A1

METHOD 4

Distance between the observer and the rocket = 2

1

22 )800600( = 1000 M1A1

Component of the velocity in the line of sight = sin θ × 300

(where θ = angle of elevation) M1A1

sin θ = 1000

800 A1

component = 240 (m s–1

) A1 [6]

CALCULUS CORE


12. 2

1

2

1

2

1

ayx

0d

d

2

1

2

12

1

2

1

x

yyx M1

x

y

y

x

x

y

2

1

2

1

d

d A1

Note: Accept

2

1

2

1

1d

d

x

a

x

y from making y the subject of the equation,

and all correct subsequent working

therefore the gradient at the point P is given by

p

q

x

y

d

d A1

equation of tangent is y – q = )( pxp

q M1

(y = pqqxp

q )

x-intercept: y = 0, n = ppqpq

pq A1

y-intercept: x = 0, m = qpq A1

n + m = ppqppq M1

= qppq 2

= 2)( qp A1

= a AG [8]

13. (a) METHOD 1

∫e2x sin x dx = –cos xe

2x + ∫2e

2x cos x dx M1A1A1

= –cos xe2x

+ 2e2x

sin x – ∫4e2x sin x dx A1A1

5∫e2x

sin x dx = –cos xe2x

+ 2e2x

sin x M1

∫e2x

sin x dx = 5

1e

2x(2 sin x – cos x) + C AG

CALCULUS CORE


METHOD 2

∫sin xe2x

dx = xxx xx

d2

ecos

2

esin 22

M1A1A1

= xxxx xxx

d4

esin

4

ecos

2

esin 222

A1A1

4

ecos

2

sinedsine

4

5 222

xxx xx

xx M1

Cxxxx xx )cossin2(e5

1dsine 22 AG

(b) xxy

y x dsine1

d 2

2

M1A1

arcsin y = 5

1e

2x (2 sin x – cos x)(+ C) A1

when x = 0, y = 0 5

1 C M1

y = sin

5

1)cossin2(e

5

1 2 xxx A1

(c) (i)

A1

P is (1.16, 0) A1

Note: Award A1 for 1.16 seen anywhere, A1 for complete sketch.

Note: Allow FT on their answer from (b)

CALCULUS CORE


(ii) V = xy dπ....162.1

0

2

M1A1

= 1.05 A2

Note: Allow FT on their answers from (b) and (c)(i). [25]

14. xx x ee x = 0 or 1 (A1)

attempt to find xy d2

M1

V1 = xxdeπ1

0

2

=

1

0

22e2

1π

x

= 2

πe2

A1

V2 = π xx xde1

0

2

= π

1

0

2

1

0

2x de2

1e

2

1xx x M1A1

Note: Award M1 for attempt to integrate by parts.

=

1

0

22

e4

1π

2

πe

x

finding difference of volumes M1

volume = V1 – V2

=

1

0

2e4

1π

x

= 4

1π(e

2 – 1) A1

[7]

CALCULUS CORE


15. (a) u = xx

ux

d1

d1

2 M1

2

dd

u

ux A1

2

1

1 21 2

d

11

1d

1

1

u

u

u

xx

A1M1A1

Note: Award A1 for correct integrand and M1A1 for correct limits.

= uu

d1

11

1 2

(upon interchanging the two limits) AG

(b) 111 arctanarctan

ux A1

arctan α

1arctan

4

π

4

π A1

arctan α or + arctan2

π1

AG

[7]

16. (a) x2 + 5x + 4 = 0 x = –l or x = –4 (M1)

so vertical asymptotes are x = –1 and x = –4 A1

as x → ∞ then y → 1 so horizontal asymptote is y = 1 (M1)A1

(b) x2 – 5x + 4 = 0 x = 1 or x = 4 A1

x = 0 y = 1

so intercepts are (1, 0), (4, 0) and (0, 1)

(c) (i) f′(x) = 22

22

)45(

)52)(45()52)(45(

xx

xxxxxx M1A1A1

=

2222

2

)45(

)2)(2(10

)45(

4010

xx

xx

xx

x A1

f′(x) = 0 x = ±2 M1

so the points under consideration are (–2, –9) and

9

1,2 A1A1

looking at the sign either side of the points (or attempt

to find f″(x)) M1

e.g. if x = –2– then (x – 2)(x + 2) > 0 and if x = –2

+ then

(x – 2)(x + 2) < 0, therefore (–2, –9) is a maximum A1

(ii) e.g. if x = 2– then (x – 2)(x + 2) < 0 and if x = 2

+ then

CALCULUS CORE


(x – 2)(x + 2) > 0, therefore

9

1,2 is a minimum A1

Note: Candidates may find the minimum first.

(d)

A3

Note: Award A1 for each branch consistent with and including the

features found in previous parts.

(e) one A1 [20]

17. (a)

22

d

d;5.122 x

x

uxxu

x

u

u

f

x

f

d

d

d

d

d

d = e

x (2x – 2) (M1)

= 2(x – 1)5.122

e xx A1

CALCULUS CORE


(b) 2

5.125.12

)1(

e1e)1(2)1(

d

d22

x

xx

x

y xxxx

M1A1

= 5.12

2

22

e)1(

142

xx

x

xx (A1)

minimum occurs when 0d

d

x

y (M1)

x = 1

4

84accept

2

1x A1

a = 1 +

4

84accept

2

1a R1

[8]

18. EITHER

differentiating implicitly:

1 × e–y

– xe–y

x

y

x

y y

d

de

d

d = 1 M1A1

at the point (c, ln c)

1d

d

d

d11

x

yc

x

y

cc

c M1

cx

y 1

d

d (c ≠ 1) (A1)

OR

reasonable attempt to make expression explicit (M1)

xe–y

+ ey = 1 + x

x + e2y

= ey(1 + x)

e2y

– ey(1 + x) + x = 0

(ey – 1)(e

y – x) = 0 (A1)

ey = 1, e

y = x

y = 0, y = ln x A1

Note: Do not penalize if y = 0 not stated.

2

1

d

d

x

y

gradient of tangent = c

1 A1

Note: If candidate starts with y = ln x with no justification,

award (M0)(A0)A1A1.

CALCULUS CORE


THEN

the equation of the normal is

y – ln c = –c(x – c) M1

x = 0, y = c2 + 1

c2 + 1 – ln c = c

2 (A1)

ln c = 1

c = e A1 [7]

19. EITHER

attempt at integration by substitution (M1)

using u = t + 1, du = dt, the integral becomes

2

1dln)1( uuu A1

then using integration by parts M1

uu

uu

uuu

uuu d1

2ln

2dln)1(

2

1

22

1

22

1

A1

=

2

1

2

4

u

u (A1)

= 4

1 (accept 0.25) A1

OR

attempt to integrate by parts (M1)

correct choice of variables to integrate and differentiate M1

tt

tt

tttt d

1

1

2)1ln(

2d)1ln(

1

0

21

0

21

0

A1

tt

ttt

d1

11

2

1)1ln(

2

1

0

1

0

2

A1

1

0

21

0

2

)1ln(22

1)1ln(

2

tt

tt

t (A1)

= 4

1 (accept 0.25) A1

[6]

CALCULUS CORE


20. (a) the differential equation is separable and can be written as (M1)

xxyy dcosd 22

(or equivalent) A1

= xxd

2

2cos1

A1

)(2sin4

1

2

11Cxx

y A1A1

when x = 0, y = 1 M1

C = 1

y =

12sin4

1

2

1

1

xx

A1

(b) (i) recognizing use of (1 + tan x)2 (M1)

(1 + tan x)2 = 1 + 2 tan x + tan

2 x ≥ 1 + tan

2 x = sec

2 x A1

(since all terms are positive)

(1 + tan x)2 ≥ sec

2 x

sec2 x = 1 + tan

2 x ≥ 1 A1

(1 + tan x)2 ≥ sec

2 x ≥ 1

since all terms are positive, taking square root gives R1

1 ≤ sec x ≤ 1 + tan x AG

(ii) 4

π

0

4

π

0

4

π

0dtan 1d secd xxxxx M1

4

π

04

π

0

4

π

0 coslnd sec xxxxx M1A1

2

1ln

4

πd sec

4

π4

π

0 xx A1

2ln2

1

4

πd sec

4

π4

π

0 xx AG

[15]

CALCULUS CORE


21. 32

3x = 0 when x = 2 (A1)

the equation of the parabola is y = p(x – 2)2 – 3 (M1)

through (0, 3) 3 = 4p – 3 2

3 p (M1)

the equation of the parabola is y =

36

2

33)2(

2

3 22 xxx A1

area =

36

2

3

2

332 2

2

0xxx dx M1M1A1

Note: Award M1 for recognizing symmetry to obtain 2

02 ,

M1 for the difference,

A1 for getting all parts correct.

= 2

0

2 d)93( xxx A1

[8]

22. (a) BAO = π – θ (allied) A1

recognizing OAB as an isosceles triangle M1

so OBA = π – θ A1

COB = π – θ (alternate) AG

Note: This can be done in many ways, including a clear diagram.

(b) area of trapezium is T = area∆BOC + area∆AOB (M1)

= )π2sin(2

1)πsin(

2

1 22 rr M1A1

= 2sin2

1sin

2

1 22 rr AG

(c) (i)

2coscos2

1

d

d 22 rrT

M1A1

for maximum area 2coscos2

1 22 rr = 0 M1

cos θ = 2 cos 2θ AG

CALCULUS CORE


(ii) θmax = 2.205... (A1)

maxmax 2sin2

1sin

2

1 = 0.880 A1

[11]

23. (a)

200

640032

200d

d 22 vv

t

v (M1)

vv

tVT

d80

200d

40 220

M1A1A1

T = 200 vvV

d80

140

22

AG

(b) (i) a = t

s

s

v

t

v

d

d

d

d

d

d R1

= s

vv

d

d AG

(ii) 200

80

d

d 22

v

s

vv (M1)

vv

vs

VS

d80

200d

40 220

M1A1A1

vv

vs

V

S

d80

200d

40

220

M1

S = vv

v

Vd

80200

40

22

A1

(c) letting V = 0 (M1)

distance = vv

vd

80200

40

0 22

= 22.3 metres A1

time = vv

d80

1200

40

0 22

= 1.16 seconds A1

[14]

CALCULUS CORE


24. (a) the distance of the spot from P is x = 500 tan θ A1

the speed of the spot is

tt

x

d

dsec500

d

d 2 (= 4000 π sec

2 θ) M1A1

when x = 2000, sec2 θ = 17 (θ = 1.32581...)

π8

d

d

t

t

x

d

d = 500 × 17 × 8π M1A1

speed is 214000 (metres per minute) AG

Note: If their displayed answer does not round to 214 000, they lose the final A1.

(b)

2

22

2

2

d

dtan2sec500

d

dtansecπ8000

d

d

ttt

x

or M1A1

0

d

d since

2

2

t

= 43000000 (= 4.30 × 107) (metres per minute

2) A1

[8]

25. (a) solving to obtain one root: 1, –2 or –5 A1

obtain other roots A1

(b) D = x [–5, –2] [1, ∞) (or equivalent) M1A1

Note: M1 is for 1 finite and 1 infinite interval.

(c) coordinates of local maximum –3.73 (– 2 – 3 ), 3.22( 36 ) A1A1

(d) use GDC to obtain one root: 1.41, –3.18 or –4.23 A1

obtain other roots A1

CALCULUS CORE


(e)

A1A1A1

Note: Award A1 for shape, A1 for max and for min clearly in correct

places, A1 for all intercepts.

Award A1A0A0 if only the complete top half is shown.

(f) required area is twice that of y = f(x) between –5 and –2 M1A1

answer 14.9 A1 N3

Note: Award M1A0A0 for

2

5d)( xxf = 7.47... or N1 for 7.47.

[14]

26. (a) g ○ f(x) = xe1

1

A1

1 < 1 + ex < ∞ (M1)

range g ○ f is ]0, 1[ A1 N3

CALCULUS CORE


(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

CALCULUS CORE


(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]

CALCULUS CORE


27. (1 + x3) x

x

x

y

yyx

y

yd

1

2

tan

dtan2

d

d3

22

M1

xx

xy

y

yd

1

3

3

2d

sin

cos3

2

(A1)(A1)

Cxy 31ln3

2sinln A1A1

Notes: Do not penalize omission of modulus signs.

Do not penalize omission of constant at this stage.

EITHER

01ln3

2

2

πsinln CC M1

OR

|sin y| = 3

2

31 xA , A = eC

1012

πsin 3

2

3 AA M1

THEN

y = arcsin

3

2

3 )1( x A1

Note: Award M0A0 if constant omitted earlier. [7]

28. (a) (i) xex = 0 x = 0 A1

so, they intersect only once at (0, 0)

(ii) y′ = ex + xe

x = (1 + x)e

x M1A1

y′(0) = 1 A1

θ = arctan1 = 4

π (θ = 45°) A1

(b) when k = 1, y = x

xex = x x(e

x – 1) = 0 M1

x = 0 A1

y′(0) = 1, which equals the gradient of the line y = x R1

so, the line is tangent to the curve at origin AG

Note: Award full credit to candidates who note that the equation x(ex – 1) = 0

has a double root x = 0 so y = x is a tangent.

(c) (i) xex = kx x(e

x – k) = 0 M1

x = 0 or x = ln k A1

CALCULUS CORE


k > 0 and k ≠ 1 A1

(ii) (0, 0) and (ln k, k ln k) A1A1

(iii) A = k

x xxkxln

0de M1A1

Note: Do not penalize the omission of absolute value.

(iv) attempt at integration by parts to find ∫xex dx M1

∫xexdx = xe

x – ∫e

xdx = e

x(x – 1) A1

as 0 < k < 1 ln k < 0 R1

A =

0

ln

20

lne)1(

2de

k

x

k

x xxk

xxkx

A1

=

kkk

k)1(ln)(ln

21 2

A1

= )2ln2)((ln2

1 2 kkk

= )1)1((ln2

1 2 kk

M1A1

since 2

k((ln k – 1)

2 + 1) > 0 R1

A < 1 AG [23]

29. x3y

3 – xy = 0

3x2y

3 + 3x

3y

2y′ – y – xy′ = 0 M1A1A1

Note: Award A1 for correctly differentiating each term.

x = 1, y = 1 3 + 3y′ – 1 – y′ = 0

2y′ = –2

y′ = –1 (M1)A1

gradient of normal = 1 (A1)

equation of the normal y – 1 = x – 1 A1 N2

y = x

Note: Award A2R5 for correct answer and correct justification. [7]

CALCULUS CORE


30. EITHER

y = 2)1(

1

1

1

xy

x

M1A1

solve simultaneously M1

mx

mxmx

2)1(

1and )(

1

1

22 )1(

1

)1(

1

1

1

xx

xx A1

Note: Accept equivalent forms.

(1 – x)3 – x(1 – x)

2 + 1 = 0, x ≠ 1

x = 1.65729... 65729.11

1

y = –1.521379...

tangency point (1.66, –1.52) A1A1

m = (–1.52137...)2 = 2.31 A1

OR

(1 – x)y = 1

m (1 – x) (x – m) = 1 M1

m (x – x2 – m + mx) = 1

mx2 – x(m + m

2) + (m

2 + 1) = 0 A1

b2 – 4ac = 0 (M1)

(m + m2)2 – 4m(m

2 + 1) = 0

m = 2.31 A1

substituting m = 2.31... into mx2 – x(m + m

2) + (m

2 + 1) = 0 (M1)

x = 1.66 A1

y = 65729.11

1

= –1.52 A1

tangency point (1.66, –1.52) [7]

31. (a) f′(x) = 2)e(

)e(e)e(e

ba

baababx

xxxx

M1A1

= 2

2222

)e(

eeee

ba

abababx

xxxx

A1

= 2

22

)e(

e)(

ba

abx

x

AG

CALCULUS CORE


(b) EITHER

f′(x) = 0 (b2 – a

2)e

x = 0 b = ±a or e

x = 0 A1

which is impossible as 0 < b < a and ex > 0 for all x R1

OR

f′(x) < 0 for all x since 0 < b < a and ex > 0 for all x A1R1

OR

f′(x) cannot be equal to zero because ex is never equal to zero A1R1

(c) EITHER

f″(x) = 4

22222

)e(

e))(e(e2)e(e)(

ba

abbaabaabx

xxxxx

M1A1A1

Note: Award A1 for each term in the numerator.

= 3

22

)e(

)e2e(e)(

ba

abaabx

xxx

= 3

22

)e(

e)e()(

ba

ababx

xx

OR

f′(x) = (b2 – a

2)e

x(ae

x + b)

–2

f″(x) = (b2 – a

2)e

x(ae

x + b)

–2 + (b

2 – a

2)e

x(–2ae

x)(ae

x + b)

–3 M1A1A1

Note: Award A1 for each term.

= (b2 – a

2)e

x(ae

x + b)

–3((ae

x + b) – 2ae

x)

= (b2 – a

2)e

x(ae

x + b)

–3(b – ae

x)

THEN

f″(x) = 0 b – aex = 0

a

bx ln M1A1

ab

ba

a

bf

2ln

22

A1

coordinates are

ab

ba

a

b

2,ln

22

CALCULUS CORE


(d) b

ay

b

axf

x

)(lim horizontal asymptote A1

a

by

a

bxf

x

)(lim horizontal asymptote A1

0 < b < a aex + b > 0 for all x (accept ae

x + b ≠ 0)

so no vertical asymptotes R1

Note: Statement on vertical asymptote must be seen for R1.

(e) y = 1e4

e4

x

x

y = 2

7ln

2

1 x (or 1.25 to 3 sf) (M1)(A1)

V =

2

7ln

0 4

1

1e4

e4π

x

x

dx (M1)A1

= 1.09 (to 3 s.f.) A1 N4 [19]

32. (a)

A1A1

Note: Award A1 for the correct x-intercept,

A1 for completely correct graph.

(b) METHOD 1

CALCULUS CORE


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]

CALCULUS CORE


33. (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

CALCULUS CORE


(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]

34. (a) from f(x + y) = f(x)f(y)

for x = y = 0 M1

we have f(0 + 0) = f(0)f(0) f(0) = (f(0))2 A1

as f(0) ≠ 0, this implies that f(0) = 1 R1AG N0

(b) METHOD 1

from f(x + y) = f(x)f(y)

for y = –x, we have f(x – x) = f(x)f(–x) f(0) = f(x)f(–x) M1A1

as f(0) ≠ 0 this implies that f(x) ≠ 0 R1AG N0

CALCULUS CORE


METHOD 2

suppose that, for a value of x, f(x) = 0 M1

from f(x + y) = f(x)f(y)

for y = –x, we have f(x – x) = f(x)f(–x) f(0) = f(x)f(–x) A1

substituting f(x) by 0 gives f(0) = 0 which contradicts part (a) R1

therefore f(x) ≠ 0 for all x. AG N0

(c) by the definition of derivative

f′(x) =

h

xfhxf

h

)()(lim

0 (M1)

=

h

fxfhfxf

h

)0()()()(lim

0 A1(A1)

= )()0()(

lim0

xfh

fhf

h

A1

= f′(0)f(x) (= kf(x)) AG N0

(d)

xkxxf

xfdd

)(

)(ln f(x) = kx + C M1A1

ln f(0) = C C = 0 A1

f(x) = ekx

A1 N1

Note: Award M1A0A0A0 if no arbitrary constant C. [14]

35. (a) )3)(1(

)1(2)3(3

3

2

1

3

xx

xx

xx M1

= 34

22932

xx

xx A1

= 34

1152

xx

x AG

(b) xxx

xxx

xd

3

2

1

3d

34

115 2

0

2

0 2

M1

= 203)] ln( 2 1) ln( [3 xx A1

= 3 ln 3 + 2 ln 5 – 3 ln 1 – 2 ln 3 (= 3 ln 3 + 2 ln 5 – 2 ln 3) A1

= ln 3 + 2 ln 5

= ln 75 (k = 75) A1 [6]

36. (a) 8x + 2yx

y

d

d = 0 M1A1

CALCULUS CORE


Note: Award M1A0 for 8x + 2yx

y

d

d = 4

y

x

x

y 4

d

d A1

(b) –4 A1

(c) V = xy dπ 2

or equivalent M1

V = 1

0

2 d)4(4π xx A1

=

1

0

3

3

44π

xx A1

= 3

π8 A1

Note: If it is correct except for the omission of π, award 2 marks. [8]

37. (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

CALCULUS CORE


(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]

CALCULUS CORE


38.

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]

39. (a)

d

cos1

)cos(1d

cos1

sin

= ln (1 – cos θ) + C (M1)A1A1

Note: Award A1 for ln (1 – cos θ) and A1 for C.

(b) 2

1cos1ln(

2

1d

cos1

sin

2

π

2

π

aa

M1

1 – cos a = )e1arccos(e 2

1

a or 2.28 A1 N2

[5]

CALCULUS CORE


40. (a) rearrange yxxx

yx yy y

ydeedcosobtain to0

d

de

e

cos e2e2

(M1)

as 12 )2sin(

4

1

2

1d

2

)2cos(1dcos Cxxx

xxx

M1A1

and 2ee edee Cy

yyy A1

Note: The above two integrations are independent and should not be

penalized for missing Cs.

a general solution of Cxxx

yx yy

y ee

2

e)2sin(4

1

2

1 is 0

d

de

e

cos A1

given that y = 0 when x = π, C = e2

πe)π2sin(

4

1

2

π 0e (or – 1.15)(M1)

so, the required solution is defined by the equation

2

πe)2sin(

4

1

2

1lnln e

2

πe)2sin(

4

1

2

1 e xxyxxy

or A1 N0

(or equivalent)

(b) for x =

4

πelnln,

2

πy (or –0.417) A1

[8]

CALCULUS CORE


41. (a) t

y

y

m

t

m

d

d

d

d

d

d (M1)

1000arcsinarcsinsec 2 r

r

y

r

y

= 1000

1

1

arcsincos

1

22

r

r

y

r

r

y

(or equivalent) A1A1A1

= 1000

1

2

22

22r

r

yr

yr

(A1)

= 3223

3

)(10 yr

r

(or equivalent) A1

=

3

2210

yr

r AG N0

(b) t

m

d

d represents the rate of change of the gradient of the line OP A1

[7]

42. (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.

CALCULUS CORE


(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]

43. (a) f′(x) =

222 1

21

1

2

1

1

x

x

x

x

x M1A1A1

CALCULUS CORE


Note: Award A1 for first term,

M1A1 for second term (M1 for attempting chain rule).

(b) f′(x) = 0 (M1)

x = 0.5, y = 2.26 or 36

π (accept (0.500, 2.26)) A1A1 N3

[6]

44. vt

v

2

1

d

d A1

tv

vd

2

1d (A1)

ln v = ct 2

1 (A1)

v =

tct

A 2

1

2

1

ee (A1)

t = 0, v = 40, so A = 40 M1

v = t

2

1

e40

(or equivalent) A1 [6]

45. (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

CALCULUS CORE


(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]

46. xx d4 2

x = 2 sin θ

dx = 2 cos θ dθ A1

= dcos2sin44 2 M1A1

= dcos2cos2

= dcos4 2

now dcos 2

= d2

12cos

2

1

M1A1

=

2

1

4

2sin A1

so original integral

= sin 2θ + 2θ

= 2 sin θ cos θ + 2θ

=

2arcsin2

2

4

22

2 xxx

= Cxxx

2arcsin2

2

4 2

A1A1

Note: Do not penalise omission of C.

2,

2

1BA

[8]

CALCULUS CORE


47. (a) let QPH = θ

tan θ = 40

h

t

h

t d

d

40

1

d

dsec 2

M1

2sec4

1

d

d

t (A1)

=

0.6435or

4

5sec

254

16 A1

= 0.16 radians per second AG

(b) x 2 = h

2 + 1600, where PH = x

2xt

hh

t

x

d

d2

d

d M1

10d

d

x

h

t

x A1

= 1600

10

2 h

h (A1)

h = 30, t

x

d

d = 6 m s

–1 A1

Note: Accept solutions that begin x = 40 sec θ or use h = 10t. [7]

CALCULUS CORE


48. (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

CALCULUS CORE


(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]

CALCULUS CORE


49. (a) EITHER

let u = tan x; du = sec2 x dx (M1)

consideration of change of limits (M1)

u

u

xx

xd

1d

tan

sec3

π

4

π

3

13

π

4

π 3

2

(A1)

Note: Do not penalize lack of limits.

=

3

1

3

2

2

3

u

A1

=

2

333

2

3

2

33 33

2

A1A1 N0

OR

3

π

4

π

3

2

3

π

4

π 3

2

2

)(tan3d

tan

sec

x

xx

x M2A2

=

2

333

2

3

2

33 33

2

A1A1 N0

(b) xxxxx d)1(sectandtan 23

M1

= xxxx d)tansec(tan 2

= Cxx seclntan2

1 2 A1A1

Note: Do not penalize the absence of absolute value or C. [9]

CALCULUS CORE


50. (a) t

y

d

d = ky cos (kt)

y

yd = k cos(kt)dt (M1)

tktky

yd)cos(

d M1

ln y = sin(kt) + c A1

y = Aesin(kt)

t = 0 y0 = A (M1)

y = y0esin kt

A1

(b) – l ≤ sin kt ≤ 1 (M1)

y0e–1

≤ y ≤ y0e1

so the ratio is e

1 : e or 1 : e

2 A1

[7]

51.

A5

Note: Award A1 for origin

A1 for shape

A1 for maximum

A1 for each point of inflexion. [5]

CALCULUS CORE


52. y = ex x = ln y

volume = yy d)(lnπ5

1

2

(M1)A1

using integration by parts (M1)

5

1

5

1

25

1

2 dln2)(lnπd)ln(π yyyyyy A1A1

= 512 2ln2)(lnπ yyyyy A1A1

Note: Award A1 marks if π is present in at least one of the above lines.

xy d)(lnπ5

1

2

= π 5(ln 5)2 – 10 ln 5 + 8 A1

[8]

CALCULUS CORE


53. (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

CALCULUS CORE


[17]

54. (a) (i) x – xa = 0 M1

xx – a = 0 (A1)

x = 0, x = a2 A1 N2

(ii) f′(x) = 1 – x

a

2 A1

f is decreasing when f′ < 0 (M1)

40

2

20

21

4ax

x

ax

x

a

A1

(iii) f is increasing when f′ > 0

40

2

20

21

4ax

x

ax

x

a

A1

Note: Award the M1 mark for either (ii) or (iii).

(iv) minimum occurs at x = 4

4a

minimum value is y = 4

2a (M1)A1

hence y ≥ 4

2a A1

(b) concave up for all values of x R1 [11]

CALCULUS CORE


55. (a)

td

d = 3 (A1)

y = 10 sin θ A1

d

dy = 10 cos θ M1

t

y

t

y

d

d

d

d

d

d

= 30 cos θ M1

at y = 6, cos θ = 10

8 (M1)(A1)

t

y

d

d = 24 (metres per minute) (accept 24.0) A1

(b) α = 4

π

2

M1A1

tt d

d

2

1

d

d = 1.5 A1

[10]

56. (a) 3

1

21

x

xf A1

1 8202

3

1

3

1 xx

x

A1

(b) 3

4

3

2

x

xf A1

f (8) > 0 at x = 8, f (x) has a minimum. M1A1 [5]

57. 2 + x x2 = 2 3x + x

2 M1

2x2 4x = 0

CALCULUS CORE


2x(x 2) = 0

x = 0, x = 2 A1A1

Notes: Accept graphical solution.

Award M1 for correct graph and A1A1 for correctly labelled roots.

xxxxx d322A2

0

22

(M1)

= equivalentor d242

0

2 xxx A1

=

2

0

32

3

22

xx A1

=

3

22

3

8 A1

[7]

CALCULUS CORE


58. METHOD 1

V = xx

xd

lnπ

2e

1

M1

Integrating by parts:

2

2 1

d

d,ln

xx

vxu (M1)

xv

x

x

x

u 1,

ln2

d

d

V =

x

x

x

x

xd

ln2

ln2

2

A1

u = ln x, 2

1

d

d

xx

v (M1)

xv

xx

u 1,

1

d

d

xx

xx

xx

xx

x

x 1lnd

1lnd

ln22

A1

V =

e

1

21ln

2ln

xx

x

x

x

= 2 e

5 A1

CALCULUS CORE


METHOD 2

V = xx

xd

lnπ

2e

1

M1

Let ln x = u x = eu, u

x

xd

d (M1)

uuuuuu

ux

x

x uuu

ude2eded

ed

ln 2222

A1

= uuuuuu uuuuu e2e2edee2e 22

= 22e 2 uuu A1

When x = e, u = 1. When x = 1, u = 0.

1

0

2 22eπVolume uuu M1

=

e

π5π22e5π 1

A1

[6]

59. (a) f (x) = (1 + 2x) e2x

A1

f (x) = 0 M1

(1 + 2x)e2x

= 0 x = 2

1 A1

f (x) = (22x + 2 2

2 1)e

2x = (4x + 4)e

2x A1

f e

2

2

1

A1

e

2 > 0 at x =

2

1, f (x) has a minimum. R1

P

e2

1,

2

1 A1

(b) f(x) = 0 4x + 4 = 0 x = 1 M1A1

Using the 2nd

derivative f e

2

2

1

and f (2) = ,

e

44

M1A1

the sign change indicates a point of inflexion. R1

CALCULUS CORE


(c) (i) f (x) is concave up for x > 1. A1

(ii) f (x) is concave down for x < 1. A1

(d)

A1A1A1A1

Note: Award A1 for P and Q, with Q above P,

A1 for asymptote at y = 0,

A1 for (0, 0),

A1 for shape.

CALCULUS CORE


(e) Show true for n = 1 (M1)

f (x) = e2x

+ 2xe2x

A1

= e2x

(1 + 2x) = (2x + 20) e

2x

Assume true for n = k, ie f (k)

x = (2k x + k 2

k 1) e

2x, k 1 M1A1

Consider n = k + 1, ie an attempt to find .)(d

dxf

x

k M1

f (k + 1)

(x) = 2k e

2x + 2e

2x (2

k x + k 2

k 1) A1

= (2k + 2 (2

k x + k 2

k 1)) e

2x

= (2 2k x + 2

k + k 2 2

k 1) e

2x

= (2k + 1

x + 2k + k 2

k) e

2x A1

= (2k + 1

x + (k + 1) 2k) e

2x A1

P(n) is true for k P(n) is true for k + 1, and since true

for n = 1, result proved by mathematical induction n+ R1

Note: Only award R1 if a reasonable attempt is made to

prove the (k + 1)th

step. [27]

60. (a) crt

V

d

d A1

3

3

4rV

t

rr

t

V

d

d4

d

d 2 M1A1

crt

rr

d

d4 2

M1

r

c

t

r

4d

d A1

= r

k AG

CALCULUS CORE


(b) t

r

d

d =

r

k

tkrr dd M1

dktr

2

2

A1

An attempt to substitute either t = 0, r = 8 or t = 30, r = 12 M1

When t = 0, r = 8

d = 32 A1

322

2

ktr

When t = 30, r = 12

32302

122

k

3

4 k A1

323

4

2

2

tr

When t = 15, 32153

4

2

2

r

M1

r2 = 104 A1

r 10 cm A1

Note: Award M0 to incorrect methods using proportionality

which give solution r =10 cm. [13]

61. (a) Attempting implicit differentiation M1

0d

d2

d

d2

x

yy

x

yxyx A1

EITHER

Substituting x = 1, y = k 0d

d2

d

d2eg

x

yk

x

yk M1

Attempting to make x

y

d

d the subject M1

OR

CALCULUS CORE


Attempting to make x

y

d

d the subject eg

x

y

d

d =

yx

yx

2

2

M1

Substituting x = 1, y = k into x

y

d

d M1

THEN

12

2

d

d

k

k

x

y A1 N1

(b) Solving x

y

d

d = 0 for k gives k = 2 A1

[6]

62. Using integration by parts (M1)

xvxx

v

x

uxu 2cos

2

1and2sin

d

d,1

d

d, (A1)

xxxx d2cos2

12cos

2

16

π

0

6

π

0

A1

= 6

π

0

6

π

0

2sin4

12cos

2

1

xxx A1

Note: Award the A1A1 above if the limits are not included.

24

π2cos

2

1 6

π

0

xx A1

8

32sin

4

1 6

π

0

x A1

6

π

0 24

π

8

3d2sin xxx AG N0

Note: Allow FT on the last two A1 marks if the expressions are the

negative of the correct ones. [6]

63. xd

d(arctan (x 1)) =

211

1

x (or equivalent) A1

mN = 2 and so mT = 2

1 (R1)

CALCULUS CORE


Attempting to solve 211

1

x =

2

1 (or equivalent) for x M1

x = 2 (as x > 0) A1

Substituting x = 2 and 4

y to find c M1

c = 4 + 4

A1 N1

[6]

64. (a) AQ = 42 x (km) (A1)

QY = (2 x) (km) (A1)

QY5AQ55 T (M1)

= xx 25455 2 (mins) A1

(b) Attempting to use the chain rule on 455 2 x (M1)

xd

d xxx 242

155455 2

122

A1

4

55

2x

x

525d

d x

x A1

54

55

d

d

2

x

x

x

T AG N0

CALCULUS CORE


(c) (i) 45 2 xx or equivalent A1

Squaring both sides and rearranging to

obtain 5x2 = x

2 + 4 M1

x = 1 A1 N1

Note: Do not award the final A1 for stating a negative solution

in final answer.

(ii) 1254155 T M1

= 30 (mins) A1 N1

Note: Allow FT on incorrect x value.

(iii) METHOD 1

Attempting to use the quotient rule M1

1d

d42

x

u,xv,xu and 2

12 4

d

d xx

x

v (A1)

4

242

14

55d

d2

22

122

2

2

x

xxx

x

T A1

Attempt to simplify (M1)

22

2

32

4

4

55xx

x

or equivalent A1

23

2 4

520

x

AG

When x = 1,

2

32 4

520

x

> 0 and hence T = 30

is a minimum R1 N0

Note: Allow FT on incorrect x value, 0 x 2.

CALCULUS CORE


METHOD 2

Attempting to use the product rule M1

1d

d,4, 2

x

uxvxu and 2

12 4

d

d xx

x

v (A1)

xxx

xx

T24

2

55455

d

d2

322

12

2

2

A1

2

32

2

2

12 4

55

4

55

x

x

x

Attempt to simplify (M1)

2

32

22

2

32

22

4

455

4

55455

x

xx

x

xx A1

23

2 4

520

x

AG

When x = 1,

2

32 4

520

x

> 0 and hence T = 30 is a

minimum R1 N0

Note: Allow FT on incorrect x value, 0 x 2. [18]

CALCULUS CORE


65. METHOD 1

x

yy

x

yyxyx

d

dπsinπ

d

d23 322 A1A1A1

At (1, 1), 3 2 0d

d

x

y M1A1

2

3

d

d

x

y A1

METHOD 2

x

yy

x

yyxyx

d

dπsinπ

d

d23 322 A1A1A1

yxy

yx

x

y3

22

2πsinπ

3

d

d

A1

At (1, 1),

2

3

112πsinπ

113

d

d3

22

x

y M1A1

[6]

66. Let u = ln y du = yy

d1

A1(A1)

uuy

y

ydtand

lntan A1

= cuuu

u |cos|lnd

cos

sin A1

EITHER

cyy

y

y |lncos|lnd

lntan A1A1

OR

cyy

y

y |lnsec|lnd

lntan A1A1

[6]

CALCULUS CORE


67. (a) (i) 123 22 kxxkxf k A1

kxkxf k 26 2 A1

(ii) Setting f (x) = 0 M1

6k2x 2k = 0 x =

k3

1 A1

kkk

kk

kf

3

1

3

1

3

1

3

123

2 M1

= k27

7 A1

Hence, Pk is

kk 27

7,

3

1

(b) Equation of the straight line is xy9

7 A1

As this equation is independent of k, all Pk lie on this straight line R1

(c) Gradient of tangent at Pk:

3

21

3

12

3

13

3

12

2

kk

kk

kfPf k M1A1

As the gradient is independent of k, the tangents are parallel. R1

kcc

kk 27

1

3

1

3

2

27

7 (A1)

The equation is y = k

x27

1

3

2 A1

[13]

68. (a) Either solving e x

x + 1 = 0 for x, stating e x

x + 1 = 0,

stating P(x, 0) or using an appropriate sketch graph. M1

x = 1.28 A1 N1

Note: Accept P(1.28, 0).

CALCULUS CORE


(b) Area = ...278.1

0d1e xxx M1A1

= 1.18 A1 N1

Note: Award M1A0A1 if the dx is absent. [5]

69. METHOD 1

EITHER

Using the graph of y = f (x) (M1)

A1

The maximum of f (x) occurs at x = 0.5. A1

OR

Using the graph of y = f ″(x). (M1)

A1

The zero of f (x) occurs at x = 0.5. A1

THEN

Note: Do not award this A1 for stating x = 0.5 as the final answer for x.

f (0.5) = 0.607 (= e0.5

) A2

Note: Do not award this A1 for also stating (0.5, 0.607) as a coordinate.

CALCULUS CORE


EITHER

Correctly labelled graph of f (x) for x < 0 denoting the maximum f (x) R1

(e.g. f (0.6) = 1.17 and f (0.4) = 1.16 stated) A1 N2

OR



OR

f (0.5) 1.21. f (x) < 1.21 just to the left of 2

1x

and f (x) < 1.21 just to the right of 2

1x R1

(e.g. f (0.6) = 1.17 and f (0.4) =1.16 stated) A1 N2

OR

f (x) > 0 just to the left of 2

1x and f (x) < 0 just to the right

of 2

1x R1


METHOD 2

f (x) = 4x22e x A1

f ″(x) = –422e x + 16x

2 22e x 222 e416 xx A1

Attempting to solve f ″(x) = 0 (M1)

2

1x A1

Note: Do not award this A1 for stating 2

1x as the final answer for x.

607.0e

1

2

1

f A1

Note: Do not award this A1 for also stating

e

1,

2

1 as a coordinate.

CALCULUS CORE


EITHER

Correctly labelled graph of f ′(x) for x < 0 denoting the maximum f (x) R1


OR



OR

f (0.5) 1.21. f (x) < 1.21 just to the left of 2

1x

and f (x) < 1.21 just to the right of 2

1x R1

(e.g. f (0.6) = 1.17 and f (0.4) =1.16 stated) A1 N2

OR

f (x) > 0 just to the left of 2

1x and f (x) < 0 just to the right

of 2

1x R1

(e.g. f (0.6) = 0.857 and f (0.4) = 1.05 stated) A1 N2 [7]

CALCULUS CORE


70. (a) (i) EITHER

Attempting to separate the variables (M1)

50

d

1

d2

t

vv

v

(A1)

OR

Inverting to obtain v

t

d

d (M1)

21

50

d

d

vvv

t

(A1)

THEN

10

5 2

5

10 2d

1

150d

1

150 v

vvv

vvt A1 N3

(ii)

sec

101

104ln25sec732.0t A2 N2

(b) (i) x

vv

t

v

d

d

d

d (M1)

Must see division by v (v > 0) A1

50

1

d

d2v

x

v AG N0

(ii) Either attempting to separate variables or inverting to obtain

v

x

d

d (M1)

xv

vd

50

1

1

d2

(or equivalent) A1

Attempting to integrate both sides M1

arctan v = Cx

50 A1A1

Note: Award A1 for a correct LHS and A1 for a correct

RHS that must include C.

When x = 0, v = 10 and so C = arctan10 M1

x = 50(arctan10 arctan v) A1 N1

CALCULUS CORE


(iii) Attempting to make arctan v the subject. M1

arctan v = arctan10 50

x A1

v = tan

5010arctan

x M1A1

Using tan (A B) formula to obtain the desired form. M1

50tan101

50tan10

x

x

v

AG N0

[19]

71. (a) xxxx

x

e

1lim

elim

M1A1

= 0 AG

(b) Using integration by parts M1

axax

ax xxxx

00

0deede A1A1

axaa 0ee A1

= 1 aea

ea

A1

(c) Since ea

and aea

are both convergent (to zero), the integral is

convergent. R1

Its value is 1. A1 [9]

CALCULUS CORE


72. Recognition of integration by parts M1

xx

xx

xxxx d

1

3ln

3dln

332

A1A1

= xx

xx

d3

ln3

23

=

9ln

3

33 xx

x A1

9

1e2

9

10

9

e

3

edln

333e

1

2 xxx A1

[5]

73. 5y2 + 10xy

x

y

d

d – 4x = 0 A1A1A1

Note: Award A1A1 for correct differentiation of 5xy2.

A1 for correct differentiation of –2x2 and 18.

At the point (1, 2), 20 + 20x

y

d

d – 4 = 0

5

4

d

d

x

y (A1)

Gradient of normal = 4

5 A1

Equation of normal y – 2 = 4

5(x – 1) M1

y = 4

8

4

5

4

5x

y = 4

3

4

5x (4y = 5x + 3) A1

[7]

74. (a) y = x

x

cos

sin

x

xx

x

y2

22

cos

sincos

d

d M1A1

= x2cos

1 A1

= sec2x AG

CALCULUS CORE


(b) y = arctan x

x = tan y (M1)

y

x

d

d = sec

2 y A1

EITHER

y

x

d

d = 1 + tan

2 y (A1)

= 1 + x2 A1

21

1

d

d

xx

y

AG

OR

yx

y2sec

1

d

d = cos

2 y (A1)

A1

= 2

2

2 1

1

1

1

xx

AG

[7]

75. (a) Area of hexagon = 6 × 2

1 × x × x × sin 60° M1

= 2

33 2x AG

CALCULUS CORE


(b) Let the height of the box be h

Volume = 2

33 2hx = 90 M1

Hence h = 23

60

x A1

Surface area, A = 3 3 x2 + 6hx M1

= 12

3

36033 xx A1

2

3

36036

d

d xxx

A A1

0

d

d

x

A

3

36036 3 x M1

x3 = 20

x = 3 20 AG

3

72036

d

d 3

2

2

x

x

A

which is positive when x = 3 20 , and hence gives a minimum value. R1

[8]

CALCULUS CORE


76. y =

)e1(

3

1ln 2x

EITHER

)e1(3

1

e3

2

d

d

2

2

x

x

x

y

M1A1

x

x

x

y2

2

e1

e2

d

d

A1

ey =

3

1(1 + e

–2x) M1

Now e–2x

= 3ey – 1 A1

13e1

)1e3(2

d

d

y

y

x

y A1

= )1e3(e3

2 y

y

= )e3(3

2 y A1

= )3e(3

2 y

AG

OR

ey =

3

1(1 + e

–2x) M1A1

xy

x

y 2e3

2

d

de M1A1

Now e–2x

= 3ey – 1 (A1)

)1e3(3

2

d

de yy

x

y

)1e3(e3

2

d

d yy

x

y (A1)

= 3

2(–3 + e

–y) (A1)

= 3

2(e

–y – 3) AG

Note: Only two of the three (A1) marks may be implied. [7]

CALCULUS CORE


77. Let the number of mosquitoes be y.

kyt

y

d

d M1

tkyy

dd1

M1

ln y = –kt + c A1

y = e–kt+c

y = Ae–kt

when t = 0, y = 500 000 A = 500 000 A1

y = 500 000e–kt

when t = 5, y = 400 000

400 000 = 500 000e–5k

M1

5

4 = e

–kt

–5k = 5

4ln

k = 5

4ln

5

1 (= 0.0446) A1

250 000 = 500 000e–kt

M1

2

1 = e

–kt

kt2

1ln

t = 2

1ln

5

4ln

5 = 15.5 years A1

[8]

78. (a) For 29 xx , – 3 ≤ x ≤ 3 and for 2arcsin

3

x, – 3 ≤ x ≤ 3 A1

D is – 3 ≤ x ≤ 3 A1

(b) V = xx

xx d3

arcsin29π

28.2

0

2

M1A1

= 181 A1

CALCULUS CORE


(c)

91

3

2

)9(

)9(d

d

22

1

2

2

2

1

2

xx

xx

x

y

M1A1

=

2

1

22

1

2

2

2

1

2

)9(

2

)9(

)9(

xx

xx

A1

=

2

1

2

22

)9(

29

x

xx

A1

= 2

2

9

211

x

x

A1

(d)

p

p

p

p

xxxx

x

x

3

arcsin29d9

211 2

2

2

M1

= 3

arcsin293

arcsin29 22 ppp

ppp A1

=

3arcsin492 2 p

pp AG

(e) 11 – 2p2 = 0 M1

p = 2.35

2

11 A1

Note: Award A0 for p = ±2.35.

(f) (i) f″(x) = 2

2

1

222

1

2

9

)9)(211()4()9(

x

xxxxx

M1A1

=

2

3

2

22

)9(

)211()9(4

x

xxxx

A1

=

2

3

2

33

)9(

211436

x

xxxx

A1

=

2

3

2

2

)9(

)252(

x

xx

AG

CALCULUS CORE


(ii) EITHER

When 0 < x < 3, f″(x) < 0. When – 3 < x < 0, f″(x) > 0. A1

OR

f″(0) = 0 A1

THEN

Hence f″(x) changes sign through x = 0, giving a point of inflexion. R1

EITHER

x = ±2

25 is outside the domain of f. R1

OR

x = ±2

25 is not a root of f″(x) = 0. R1

[21]

79. (a) this separable equation has general solution

∫sec2 y dy = ∫cos x dx (M1)(A1)

tan y = sin x + c A1

the condition gives

tan4

π = sin π + c c = 1 M1

the solution is tan y = 1 + sin x A1

y = arctan (1 + sin x) AG

CALCULUS CORE


(b) the limit cannot exist unless a = arctan

2

πsin1 = arctan 2 R1A1

in that case the limit can be evaluated using l’Hopital’s rule (twice)

limit is

2

π2

lim

2

π2

))sin1(arctan(lim

2

π

2

π

x

y

x

x

xx

M1A1

where y is the solution of the differential equation

the numerator has zero limit (from the factor cos x in the differential equation) R1

so required limit is

2lim

2

π

y

x

M1A1

finally,

y″ = –sin x cos2 y – 2 cos x cos y sin y × y′(x) M1A1

since 5

1

2

πcos

y A1

y″ = 2

πat

5

1 x A1

the required limit is 10

1 A1

[17]

80. (a) Using the chain rule f″(x) = 52

π5cos2

x (M1)

= 10 cos

2

π5x A1 N2

(b) f(x) = xxf d)(

= cx

2

π5cos

5

2 A1

Substituting to find c,

2

π

2

π5cos

5

2

2

πf + c = 1 M1

c = 1 + 5

2cos 2π = 1 +

5

7

5

2 (A1)

f(x) = 5

7

2

π5cos

5

2

x A1 N2

[6]

CALCULUS CORE


81. Attempting to differentiate implicitly (M1)

3x2y + 2xy

2 = 2 0

d

d42

d

d36 22

y

yxyy

x

yxxy A1

Substituting x = 1 and y = –2 (M1)

–12 + 3 0d

d88

d

d

x

y

x

y A1

5

4

d

d4

d

d5

x

y

x

y A1

Gradient of normal is 4

5 A1 N3

[6]

82. 1d

d,1

d

d 22 yx

yxy

x

yx

Separating variables (M1)

x

x

y

y d

1

d2

A1

arctan y = ln x + c A1A1

y = 0, x = 2 arctan 0 = ln 2 + c

–ln 2 = c (A1)

arctan y = ln x – ln 2 = ln2

x

y =

2lntan

x A1 N3

[6]

CALCULUS CORE


83. (a)

A1A1A1

Note: Award A1 for correct shape, A1 for points of intersection

and A1 for symmetry.

(b) A = 1

0

2 d)(2 xxx M1

=

1

0

32

322

xx A1

=

3

1

2

12 (A1)

= 3

1 square units A1

[7]

CALCULUS CORE


84. (a) a = 1

22 s

s

a = s

vv

d

d M1

1

2

d

d2

s

s

s

vv

ss

svv d

1

2d

2

M1

2

2v = ln│s

2 + 1│ + k A1A1

Note: Do not penalize if k is missing.

When s = 1, v = 2

2 = ln 2 + k M1

k = 2 – ln 2 A1

2

2

1ln2ln21ln

2

22

2 ss

v A1

(b) EITHER

22

26ln

2

2

v

M1

v2 = 2 ln│13│+ 4

413ln2 v A1

OR

2

2v = ln│26│+ 2 – ln 2 M1

v2 = 2 ln│26│+ 4 – 2 ln 2

v = 2ln2426ln2 A1

[9]

CALCULUS CORE


85. xey = x

2 + y

2

x

yyx

x

yx yy

d

d22

d

dee M1A1A1A1A1

(1, 0) fits x

y

d

d1 = 2 + 0

x

y

d

d = 1 A1

Equation of tangent is y = x + c

(1, 0) fits c = –1

y = x – 1 A1 [7]

86. f′(x) = 2

))2(ln(2

x

x M1A1

f″(x) = 2)2(

1)2ln(22

1)2(

x

xx

x

M1A1

= 2)2(

)2ln(22

x

x A1

f″(x) = 0 for point of inflexion (M1)

2 – 2 ln(x – 2) = 0

ln (x – 2) = 1 A1

x – 2 = e

x = e + 2 A1

f(x) = (ln(e + 2 – 2))2 = (ln e)

2 = 1 A1

( coordinates are (e + 2, 1)) [9]

87. EITHER

xx

xd

)(lne

1

3

y = (ln x)4 M2

A1

e14e

1

3

)(ln4

1d

)(lnxx

x

x A1

= 4

1]01[

4

1 A1

CALCULUS CORE


OR

Let u = ln x M1

xx

u 1

d

d A1

When x = 1, u = 0 and when x = e, u = 1 A1

1

0

3duu A1

4

1

4

11

0

4

u A1

[5]

88. x = 2 sin θ

x2 = 4 sin

2 θ

4 – x2 = 4 – 4 sin

2 θ

= 4(1 – sin2 θ)

= 4 cos2 θ

cos24 2 x A1

d

dx = 2 cos θ M1

When x = 1, 2 sin θ = 1

sin θ = 2

1

6

π A1

When x = 3sin2,3

sin θ = 2

3

3

π A1

Let I = xx d43

1

2

3

π

6

πdcos2cos2 I

dcos4 3

π

6

π

2

I A1

Now cos2 θ =

2

1(cos 2θ + 1)

CALCULUS CORE


3

π

6

πd12cos2 I M1A1

3

π

6

π2sin

2

12

I M1A1

6

π

3

πsin

2

12

3

π

3

π2sin

2

12I (M1)

3

π

3

π

2

3

3

π2

2

3I A1

[11]

89. (a) OP = 222 )5( aa M1

= 2510 242 aaa A1

= 259 24 aa AG

(b) EITHER

Let s = 259 24 aa

s2 = a

4 – 9a

2 + 25

a

s

d

d 2

= 4a3 – 18a = 0 M1A1

a

s

d

d 2

= 0 for minimum (M1)

2a(2a2 – 9) = 0

a2 =

2

9

2

23

2

3a A1A1

CALCULUS CORE


OR

s = 2

1

24 )259( aa

)184()259(2

1

d

d 32

1

24 aaaaa

s

M1A1

0d

d

a

s for a minimum (M1)

4a3 – 18a = 0

2a(2a2 – 9) = 0

2

92 a

2

23

2

3a A1A1

[7]

90. EITHER

xxxx

xd(cos)sind

cos

sin4

π

0

2

1

4

π

0

(M1)

=

4

π

0

2

1

2

1

cos

x

(M1)A1A1

= 4

π

0cos2 x

= 0cos24

πcos2 A1A1

= 2 – 4

3

2 A1

CALCULUS CORE


OR

Let u = cos x (M1)

xx

usin

d

d (M1)

when x = 2

1,

4

πu A1

when x = 0, u = 1 A1

2

1

1

2

1

2

1

1

2

14

π

0dd

1d

cos

sinuuu

u

xx

x (M1)

= 2

1

1

2

1

2

u A1

=

4

3

4

1222

2

2 A1

[7]

91. Substituting u = x + 2 u – 2 = x, du = dx (M1)

uu

ux

x

xd

)2(d

)2( 2

3

2

A1

= uu

uuud

81262

23

A1

= uuuu

uuu d8d12

d)6(d 2

A1

= 2

u – 6u + 12 ln │u│– 8u

–1 + c A1

= cx

xxx

2

82ln12)2(6

2

)2( 2

A1 N0

[6]

92. (a) (i) 18(x – 1) = 0 x = 1 A1

(ii) vertical asymptote: x = 0 A1

horizontal asymptote: y = 0 A1

CALCULUS CORE


(iii) 18(2 – x) = 0 x = 2 M1A1

f″(2) = 2

9

2

)32(363

< 0 hence it is a maximum point R1

When x = 2, f(x) = 2

9 A1

2

92,at maximum a has )(xf

(iv) f(x) is concave up when f″(x) > 0 M1

36(x – 3) > 0 x > 3 A1

(b)

A1A1A1A1A1

Note: Award A1 for shape, A1 for maximum, A1 for

x-intercept, A1 for horizontal asymptote and A1 for

vertical asymptote. [14]

CALCULUS CORE


93. (a) (i) Attempting to use quotient rule f′(x) = 2

1ln1

x

xx

x

(M1)

f′(x) = 2

ln1

x

x A1

f″(x) = 4

2 2)ln1(1

x

xxx

x

(M1)

f″(x) = 3

3ln2

x

x A1

Stationary point where f′(x) = 0 M1

i.e. ln x = 1 , (so x = e) A1

f″(e) < 0 so maximum. R1AG N0

(ii) Exact coordinates x = e, y = e

1 A1A1 N2

(iii) Solving f″(0) = 0 M1

ln x = 2

3 (A1)

x = 2

3

e A1 N2

CALCULUS CORE


(b) Area = xx

xd

ln5

1 A1

EITHER

Finding the integral by substitution/inspection

u = ln x, du = x

1dx (M1)

2

)(ln

2d

22 xuuu M1A1

Area = 22

5

1

2

)1(ln)5(ln2

1

2

)(ln

x A1

Area = 2)5(ln2

1 A1 N2

OR

Finding the integral I by parts (M1)

u = ln x, dv = x

ux

1d

1 , v = ln x

I = uv – Ixxx

xxvu 22 )(lnd

1ln)(lnd M1

2

)(ln)(ln2

22 x

IxI A1

Area = 22

5

1

2

)1(ln)5(ln2

1

2

)(ln

x A1

Area = 2)5(ln

2

1 A1 N2

[18]

94. (a) ln e2–2x

= ln2e–x

M1

2 – 2x = ln(2e–x

) (A1)

= ln 2 – x (A1)

x = 2 – ln 2 A1

2

eln2lneln

22x

CALCULUS CORE


(b) xx

x

y e2e2d

d 22 M1A1

x

y

d

d = 0 for a minimum point (M1)

–2e2–2x

+ 2e–x

= 0

e2–2x

= e–x

(A1)

2 – 2x = –x (A1)

x = 2 A1

y = e–2

– 2e–2

= –e–2

A1

( minimum point is (2, –e–2

))

(c)

A1A1A1

(d) 2 distinct roots provided –e–2

< k < 0 A1A1 [16]

CALCULUS CORE


95. (a) f′(x) =

xx

xx xxxx

e

2

e

ee2 2

2

2

M1A1

For a maximum f′(x) = 0 (M1)

2x – x2 = 0

giving x = 0 or 2 A1A1

f″(x) =

xx

xxxxxxx

e

24

e

)2(ee)22( 2

2

2

M1A1

f″(0) = 2 > 0 minimum R1

f″(2) = 0e

22

maximum R1

Maximum value = 2e

4 A1

(b) For a point of inflexion,

f″(x) = 0e

242

x

xx M1

giving x = 2

8164 (A1)

= 2 ± 2 A1

(c) xxxxx xxx de2ede1

0

1

02

1

0

2

M1A1

= –e– 1

– xx xx de2e21

0

1

0 A1M1A1

= –e–1

– 2e–1

– 10e2 x A1A1

= –3e–1

– 2e–1

+ 2 (= 2 – 5e–1

) A1 [21]

96. 4 – x2 ≥ 0 for 0 ≤ x ≤ 2 A1

and 4 – x2 ≤ 0 for 2 ≤ x ≤ 4 A1

I = 4

2

22

0

2 d)4(d)4( xxxx M1A1

=

4

2

32

0

3

433

4

x

xxx A1A1

= )16(83

816

3

64

3

88 A1

[7]

CALCULUS CORE


97. exy

+ ln (y2) + e

y = 1 + e

exy

x

y

x

y

yx

yxy y

d

de

d

d2

d

d

= 0, at (0, 1) A1A1A1A1A1

1(1 + 0) + 2x

y

x

y

d

de

d

d = 0

1 + 2x

y

x

y

d

de

d

d = 0

e2

1

d

d

x

y (= –0.212) M1A1 N2

[7]

Documents

CALCULUS CORE - Shelby Bryant