View
1
Download
0
Category
Preview:
Citation preview
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
IB Questionbank Mathematics Higher Level 3rd edition 2
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
IB Questionbank Mathematics Higher Level 3rd edition 3
(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
IB Questionbank Mathematics Higher Level 3rd edition 4
(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
IB Questionbank Mathematics Higher Level 3rd edition 5
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
IB Questionbank Mathematics Higher Level 3rd edition 6
(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
IB Questionbank Mathematics Higher Level 3rd edition 7
(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
IB Questionbank Mathematics Higher Level 3rd edition 8
(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
IB Questionbank Mathematics Higher Level 3rd edition 9
(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
IB Questionbank Mathematics Higher Level 3rd edition 10
(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
IB Questionbank Mathematics Higher Level 3rd edition 11
(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
IB Questionbank Mathematics Higher Level 3rd edition 12
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
IB Questionbank Mathematics Higher Level 3rd edition 13
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
IB Questionbank Mathematics Higher Level 3rd edition 14
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
IB Questionbank Mathematics Higher Level 3rd edition 15
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
IB Questionbank Mathematics Higher Level 3rd edition 16
(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
IB Questionbank Mathematics Higher Level 3rd edition 17
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
IB Questionbank Mathematics Higher Level 3rd edition 18
(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
IB Questionbank Mathematics Higher Level 3rd edition 19
(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
IB Questionbank Mathematics Higher Level 3rd edition 20
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
IB Questionbank Mathematics Higher Level 3rd edition 21
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
IB Questionbank Mathematics Higher Level 3rd edition 22
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
IB Questionbank Mathematics Higher Level 3rd edition 23
(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
IB Questionbank Mathematics Higher Level 3rd edition 24
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
IB Questionbank Mathematics Higher Level 3rd edition 25
(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
IB Questionbank Mathematics Higher Level 3rd edition 26
(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
IB Questionbank Mathematics Higher Level 3rd edition 27
(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
IB Questionbank Mathematics Higher Level 3rd edition 28
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
IB Questionbank Mathematics Higher Level 3rd edition 29
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
IB Questionbank Mathematics Higher Level 3rd edition 30
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
IB Questionbank Mathematics Higher Level 3rd edition 31
(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
IB Questionbank Mathematics Higher Level 3rd edition 32
(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
IB Questionbank Mathematics Higher Level 3rd edition 33
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
IB Questionbank Mathematics Higher Level 3rd edition 34
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
IB Questionbank Mathematics Higher Level 3rd edition 35
(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
IB Questionbank Mathematics Higher Level 3rd edition 36
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
IB Questionbank Mathematics Higher Level 3rd edition 37
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
IB Questionbank Mathematics Higher Level 3rd edition 38
(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
IB Questionbank Mathematics Higher Level 3rd edition 39
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
IB Questionbank Mathematics Higher Level 3rd edition 40
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
IB Questionbank Mathematics Higher Level 3rd edition 41
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
IB Questionbank Mathematics Higher Level 3rd edition 42
(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
IB Questionbank Mathematics Higher Level 3rd edition 43
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
IB Questionbank Mathematics Higher Level 3rd edition 44
(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
IB Questionbank Mathematics Higher Level 3rd edition 45
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
IB Questionbank Mathematics Higher Level 3rd edition 46
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
IB Questionbank Mathematics Higher Level 3rd edition 47
(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
IB Questionbank Mathematics Higher Level 3rd edition 48
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
IB Questionbank Mathematics Higher Level 3rd edition 49
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
IB Questionbank Mathematics Higher Level 3rd edition 50
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
IB Questionbank Mathematics Higher Level 3rd edition 51
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
IB Questionbank Mathematics Higher Level 3rd edition 52
[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
IB Questionbank Mathematics Higher Level 3rd edition 53
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
IB Questionbank Mathematics Higher Level 3rd edition 54
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
IB Questionbank Mathematics Higher Level 3rd edition 55
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
IB Questionbank Mathematics Higher Level 3rd edition 56
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
IB Questionbank Mathematics Higher Level 3rd edition 57
(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
IB Questionbank Mathematics Higher Level 3rd edition 58
(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
IB Questionbank Mathematics Higher Level 3rd edition 59
(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
IB Questionbank Mathematics Higher Level 3rd edition 60
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
IB Questionbank Mathematics Higher Level 3rd edition 61
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
IB Questionbank Mathematics Higher Level 3rd edition 62
(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
IB Questionbank Mathematics Higher Level 3rd edition 63
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
IB Questionbank Mathematics Higher Level 3rd edition 64
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
IB Questionbank Mathematics Higher Level 3rd edition 65
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
IB Questionbank Mathematics Higher Level 3rd edition 66
(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
IB Questionbank Mathematics Higher Level 3rd edition 67
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
Correctly labelled graph of f (x) for x < 0 denoting the maximum f (x) R1
(e.g. f (0.6) = 0.857 and f (0.4) = 1.05 stated) A1 N2
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
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
IB Questionbank Mathematics Higher Level 3rd edition 68
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
Correctly labelled graph of f (x) for x < 0 denoting the maximum f (x) R1
(e.g. f (0.6) = 0.857 and f (0.4) = 1.05 stated) A1 N2
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
IB Questionbank Mathematics Higher Level 3rd edition 69
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
IB Questionbank Mathematics Higher Level 3rd edition 70
(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
IB Questionbank Mathematics Higher Level 3rd edition 71
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
IB Questionbank Mathematics Higher Level 3rd edition 72
(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
IB Questionbank Mathematics Higher Level 3rd edition 73
(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
IB Questionbank Mathematics Higher Level 3rd edition 74
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
IB Questionbank Mathematics Higher Level 3rd edition 75
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
IB Questionbank Mathematics Higher Level 3rd edition 76
(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
IB Questionbank Mathematics Higher Level 3rd edition 77
(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
IB Questionbank Mathematics Higher Level 3rd edition 78
(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
IB Questionbank Mathematics Higher Level 3rd edition 79
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
IB Questionbank Mathematics Higher Level 3rd edition 80
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
IB Questionbank Mathematics Higher Level 3rd edition 81
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
IB Questionbank Mathematics Higher Level 3rd edition 82
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
IB Questionbank Mathematics Higher Level 3rd edition 83
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
IB Questionbank Mathematics Higher Level 3rd edition 84
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
IB Questionbank Mathematics Higher Level 3rd edition 85
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
IB Questionbank Mathematics Higher Level 3rd edition 86
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
IB Questionbank Mathematics Higher Level 3rd edition 87
(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
IB Questionbank Mathematics Higher Level 3rd edition 88
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
IB Questionbank Mathematics Higher Level 3rd edition 89
(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
IB Questionbank Mathematics Higher Level 3rd edition 90
(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
IB Questionbank Mathematics Higher Level 3rd edition 91
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]
Recommended