IB Math – Standard Level – Calculus Practice Problems - Markscheme Alei - Desert Academy \\.psf\Home\Documents\Desert 2011-12\SL 2011-12\7Calculus\LP_SL2Calculus.doc on 01/08/2013 at 10:51 AM 1 of 24 Diagrams for Calc Practice problems (they don’t print from Mac when text wrapping is on). 10 4 2 –2 –4 –6 –8 –10 –12 –1 1 2 3 4 5 6 x y P(3, 2) 13 x y 0 17 f g p x y 16 12 8 4 0.5 1 1.5 18 x y 16 12 8 4 0.5 1 1.5 5 a 19 25 4 3 2 1 1 1 3 y = 1+ 1 x – 0 1 2 3 4 27 A C B y x O 34 x y 4 3π π
24
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IB Math – Standard Level – Calculus Practice Problems - Markscheme Alei - Desert Academy
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Diagrams for Calc Practice problems (they don’t print from Mac when text wrapping is on).
10
4
2
–2
–4
–6
–8
–10
–12
–1 1 2 3 4 5 6x
y
P(3, 2)
13x
y
0
17
f
g
p
x
y
16
12
8
4
0.5 1 1.5 18x
y
16
12
8
4
0.5 1 1.5
5
a
19 25
4
3
2
11 1
3
y = 1+1x–
0 1 2 3 4
27
A
C
B
y
x
O
34x
y
43π π
IB Math – Standard Level – Calculus Practice Problems - Markscheme Alei - Desert Academy
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35
–1 0 1 2
A
B
C D
y
x
37
6
4
2
1 2 3–2 –1
D
C(a, b)
y
x
39
y
4
4
2
20
–2
–2–4 x
40 a
5
4
3
2
1
–1
–2
–3
–4
–5
–5 –4 –3 –2 –1 21 3 4 50 x
y
40b
5
4
3
2
1
–1
–2
–3
–4
–5
–5 –4 –3 –2 –1 21 3 4 50 x
y
A
43
R
1
2
–1
–1
112
12
–x
y
44
–5 –4 –3 –2 –1 1 2 3 4 5 x
y
S
( , 0)b( , 0)a
R
4
3
2
1
–1
–2
–3
–4
–5
–6
–7 50
IB Math – Standard Level – Calculus Practice Problems - Markscheme Alei - Desert Academy
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SL Calculus Practice Problems - MarkScheme 1. y = sin (2x – 1)
x
y
d
d = 2 cos (2x – 1) (A1)(A1)
At
0,2
1, the gradient of the tangent = 2 cos 0 (A1)
= 2 (A1) (C4)
2. (a) (i) a = –3 (A1)
(ii) b = 5 (A1) 2
(b) (i) f ′(x) = –3x2
+ 4x + 15
(A2)
(ii) –3x2
+ 4x + 15 = 0
–(3x + 5)(x – 3) = 0 (M1)
x = –3
5 or x = 3 (A1)(A1)
OR
x = –3
5 or x = 3 (G3)
(iii) x = 3 ⇒ f (3) = –33
+ 2(32) + 15(3) (M1)
= –27 + 18 + 45 =36 (A1)
OR
f (3) = 36 (G2) 7
(c) (i) f ′(x) = 15 at x = 0 (M1)
Line through (0, 0) of gradient 15
⇒ y = 15x (A1)
OR
y = 15x (G2)
(ii) –x3
+ 2x2
+ 15x = 15x (M1)
⇒ –x3
+ 2x2
= 0
⇒ –x2
(x – 2) = 0
⇒ x = 2 (A1)
OR
x = 2 (G2) 4
(d) Area =115 (3 sf) (G2)
OR
Area = ∫
++−=++−
6
0
5
0
23423
215
32
4d)152(
xxxxxxx (M1)
= 12
1375 = 115 (3 sf) (A1) 2
[15]
3. (a) (i) v(0) = 50 – 50e0 = 0 (A1)
(ii) v(10) = 50 – 50e–2
= 43.2 (A1) 2
(b) (i) a = t
v
d
d = –50(–0.2e
–0.2t) (M1)
= 10e–0.2t
(A1)
(ii) a(0) = 10e0 = 10 (A1) 3
(c) (i) t → ∞ ⇒ v → 50 (A1)
(ii) t → ∞ ⇒ a → 0 (A1)
(iii) when a = 0, v is constant at 50 (R1) 3
(d) (i) y = ∫vdt (M1)
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= 50t –2.0
e –0.2
−
t
+ k (A1)
= 50t + 250e–0.2t
+ k (AG)
(ii) 0 = 50(0) + 250e0
+ k = 250 + k (M1)
⇒ k = –250 (A1)
(iii) Solve 250 = 50t + 250e–0.2t
– 250 (M1)
⇒ 50t + 250e–0.2t
– 500 = 0
⇒ t + 5e–0.2t
– 10 = 0
⇒ t = 9.207 s (G2) 7
[15]
4. METHOD 1
0 x
y
Using gdc coordinates of maximum are
(0.667, 26.9) (G3)(G3)(C6)
METHOD 2
At maximum x
y
d
d = 3x
2 – 20x + 12 = 0 = (3x – 2) (x – 6) (M1)(A1)(M1)
=> x = 3
2 must be where maximum occurs (A1)
x = 3
2 => y =
3
3
2
– 10
2
3
2
+ 12
3
2 + 23 =
27
725 (= 26.9, 3 sf) (M1)(A1)
Maximum at
27
725,
3
2 (C4)(C2)
[6]
5. (a) t
s
d
d= 30 – at => s = 30t – a
2
2t + C (A1)(A1)(A1)
Note: Award (A1) for 30t, (A1) for a2
2t, (A1) for C.
t = 0 => s = 30(0) – a( )2
02
+ C = 0 + C => C = 0 (M1)
=> s = 30t – 2
1at
2 (A1) 5
(b) (i) vel = 30 – 5(0) = 30 m s–1
(A1)
(ii) Train will stop when 0 = 30 – 5t => t = 6 (M1)
Distance travelled = 30t – 2
1at
2
= 30(6) – 2
1(5) (6
2) (M1)
= 90m (A1)
90 < 200 => train stops before station. (R1)(AG) 5
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(c) (i) 0 = 30 – at => t = a
30 (A1)
(ii) 30
a
30–
2
1(a)
230
a = 200 (M1)(M1)
Note: Award (M1) for substituting a
30, (Ml) for setting equal to 200.
=> aaa
450450–
900= = 200 (A1)
=> a = 4
9
200
450= = 2.25 m s
–2 (A1) 5
Note: Do not penalize lack of units in answers.
[15]
6. (i) At x = a, h (x) = a 5
1
h′ (x) = 5
1x 5
4–
=> h′ (a) =
5
4
5
1
a
= gradient of tangent (A1)
=> y – a 5
1
=
5
4
5
1
a
(x – a) =
5
4
5
1
a
x – 5
1a 5
1
(M1)
=> y =
5
4
5
1
a
x + 5
4a 5
1
(A1)
(ii) tangent intersects x-axis => y = 0
=>
5
4
5
1
a
x = –5
4a 5
1
(M1)
=> x = 5a 5
4
5
1
5
4– a = –4a (M1)(AG) 5
[5]
7. (a) (i) When t = 0, v = 50 + 50e0
(A1)
= 100 m s–1
(A1)
(ii) When t = 4, v = 50 + 50e–2
(A1)
= 56.8 m s–1
(A1) 4
(b) v = t
s
d
d
⇒ s = ∫ tv d
( ) tt∫ +4
0
0.5- de5050 (A1)(A1)(A1) 3
Note: Award (A1) for each limit in the correct position and (A1) for the function.
(c) Distance travelled in 4 seconds = ( ) tt∫ +4
0
0.5- de5050
= [50t – 100e–0.5t
]4
0 (A1)
= (200 – 100e–2
) – (0 – 100e0)
= 286 m (3 sf) (A1)
Note: Award first (A1) for [50t – 100e–0.5t
], ie
limits not required.
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OR
Distance travelled in 4 seconds = 286 m (3 sf) (G2) 2
(d)
100
50
velocity
(t = 4)
121086420 t
v
time Notes: Award (A1) for the exponential part, (A1) for the straight line through (11, 0),
Award (A1) for indication of time on x-axis and velocity on
y-axis,
(A1) for scale on x-axis and y-axis.
Award (A1) for marking the point where t = 4.
5
(e) Constant rate = 7
8.56 (M1)
= 8.11 m s–2
(A1) 2
Note: Award (M1)(A0) for –8.11.
(f) distance = 2
1(7)(56.8) (M1)
= 199 m (A1) 2
Note: Do not award ft in parts (e) and (f) if candidate has not used a straight line for t = 4 to t
= 11 or if they continue the exponential beyond t = 4.
[18]
8. y
x
(A2)(A1)(A1)(A2) (C6)
Note: Award A2 for correct shape (approximately parabolic), A1 A1 for intercepts at 0 and 4, A2
for minimum between x = 1.5 and x = 2.5.
[6]
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9. (a) (i) f ′(x) = –2e–2x
(A1)
(ii) f ′(x) is always negative (R1) 2
(b) (i) y = 1 + 2
1–2–
e×
(= 1 + e) (A1)
(ii) f ′ 2
1–2–
e22
1 ×−=
− (= –2e) (A1) 2
Note: In part (b) the answers do not need to be simplified.
(c) y – (1 + e) = –2e
+
2
1x (M1)
y = –2ex + 1 ( y = –5.44 x + 1) (A1)(A1) 3
(d) (i) (ii) (iii)
P
(A1)(A1)(A1)
Notes: Award (A1) for each correct answer. Do not allow (ft) on an incorrect answer to part (i).
The correct final diagram is shown below. Do not penalize if the horizontal asymptote is missing.
Axes do not need to be labelled.
(i)(ii)(iii)
–1 1 2x
y
8
6
4
2
1
P
12
–
(iv) Area = ∫−− +−−+
0
2
12 d)]1e2()e1[( xxx
(or equivalent) (M1)(M1)
Notes: Award (M1) for the limits, (M1) for the function.
Accept difference of integrals as well as integral of difference. Area below line may be calculated
geometrically.
Area = ∫−− +
0
2
12 d)e2e[( xxx
=
0
2
1
22 ee2
1
−
−
+− xx
(A1)
= 0.1795 …= 0.180 (3 sf) (A1)
OR
Area = 0.180 (G2) 7
[14]
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10. (a) x = 1 (A1)1
(b) (i) f (–1000) = 2.01 (A1)
(ii) y = 2 (A1) 2
(c) f ′(x) = 4
22
)1(
)20132)(1(2)134()1(
−+−−−−−
x
xxxxx (A1)(A1)
= 3
22
)1(
)40264()13174(
−+−−+−
x
xxxx (A1)
= 3)1(
279
−−
x
x (AG) 3
Notes: Award (M1) for the correct use of the quotient rule, the first (A1) for the placement of the
correct expressions into the quotient rule.
Award the second (A1) for doing sufficient simplification to make the given answer reasonably
obvious.
(d) f ′(3) = 0 ⇒ stationary (or turning) point (R1)
f ″(3) = 16
18 > 0 ⇒ minimum (R1) 2
(e) Point of inflexion ⇒ f ″(x) = 0 ⇒ x = 4 (A1)
x = 4 ⇒ y = 0 ⇒ Point of inflexion = (4, 0) (A1)
OR
Point of inflexion = (4, 0) (G2) 2
[10]
11. (a) d = ∫ −+4
0
– )e554( tt dt (M1)(A1)(A1)(C3)
Note: Award (M1) for ∫, (A1) for both limits, (A1) for 4t + 5 – 5e–t
(b) d = 40
2 ]e552[ -ttt ++ (A1)(A1)
Note: Award (A1) for 2t2
+ 5t, (A1) for 5e–t
.
= (32 + 20 + 5e–4
) – (5)
= 47 + 5e–4
(47.1, 3sf ) (A1) (C3)
[6]
12. (a) Velocity is d
d
s
t. (M1)
d10
d
st
t= − (A1)
10 (m s–1
) (A1) (C3)
(b) The velocity is zero when d
0d
s
t= (M1)
10 0t− =
10t = (secs) (A1)(C2)
(c) s = 50 (metres) (A1) (C1)
Note: Do not penalize absence of units.
[6]
13. (a) 3h = (A1)
2k = (A1) 2
(b) ( )f x2( 3) 2x= − − +
2 6 9 2x x= − + − + (must be a correct expression) (A1)
2 6 7x x= − + − (AG)1
(c) ( ) 2 6f x x′ = − + (A2) 2
(d) (i) tangent gradient 2= − (A1)
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gradient of L 1
2= (A1) (N2) 2
(ii) EITHER
equation of L is 1
2y x c= + (M1)
1c = − . (A1)
11
2y x= −
OR
11 ( 4)
2y x− = − (A2) (N2) 2
(iii) EITHER
2 16 7 1
2x x x− + − = − (M1)
22 11 12 0x x− + = (may be implied) (A1)
(2 3)( 4) 0x x− − = (may be implied) (A1)
1.5x = (A1) (N3) 4
OR
2 16 7 1
2x x x− + − = − (or a sketch) (M1)
1.5x = (A3) (N3) 8
[13]
14. (a) 4x = (A1)
g″ changes sign at 4x = or concavity changes (R1) 2
(b) 2x = (A1)
EITHER
g′ goes from negative to positive (R1)
OR
g′ (2) = 0 and g″ (2) is positive (R1) 2
(c)
1 2 3 4 5 6 7 8
P
M
(A2)(A1)(A1) 4
Note: Award (A2) for a suitable cubic curve through (4, 0), (A1) for M at x = 2, (A1) for P at (4,
0).
[8]
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15. (a) tva
dd= (M1)
= –10 A1 3
(b) s = ∫vdt (M1)
= 50t – 5t2
+ c A1
40 = 50(0) – 5(0) + c ⇒ c = 40 A1
s = 50t – 5t2
+ 40 A1 3
Note: Award (M1) and the first (A1) in part (b) if c is missing, but do not award the final 2
marks.
[6]
16. (a) (i) f ′(x) = –x + 2 A1
(ii) f ′(0) = 2 A12
(b) Gradient of tangent at y-intercept = f ′(0) = 2
⇒ gradient of normal = 21
(= –0.5) A1
Finding y-intercept is 2.5 A1
Therefore, equation of the normal is
y – 2.5 = ~(x – 0) (y – 2.5 = –0.5x) M1
(y = –0.5x + 2.5 (AG) 3
(c) (i) EITHER
solving –0.5x2
+ 2x + 2.5 = –0.5x + 2.5 (M1)A1
⇒ x = 0 or x = 5 A1 2
OR
f x( )
y
x
g x( )
M1
Curves intersect at x = 0, x = 5 (A1)
So solutions to f (x) = g (x) are x = 0, x = 5 A12
OR
⇒ 0.5x2
– 2.5x = 0 (A1)
⇒ – 0.5x(x – 5) = 0 M1
⇒ x = 0 or x = 5 A12
(ii) Curve and normal intersect when x = 0 or x = 5 (M2)
Other point is when x = 5
⇒ y = –0.5(5) + 2.5 = 0 (so other point (5, 0) A1 2
(d) (i) Area = ∫ ∫
××−++−−
5
0
5
0
2 5.2521d)5.225.0(or d))()(( xxxxxgxf
A1A1A1 3
Note: Award (A1) for the integral, (A1) for both correct
limits on the integral, and (A1) for the difference.
(ii) Area = Area under curve – area under line (A = A1 – A2) (M1)
(A1) = 425,
350
2 =A
Area = 12125
425
350 =− (or 10.4 (3sf) A12
[16]
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