CHAPTER 5 REVIEW SOLUTIONS – SL
1.
(a) f′(x) = –sin 2x × 2 (= –2 sin 2x)
Note: Award A1 for 2, A1 for sin 2x.
1 
3 
(b) g′(x) = 3 ×


3x  5  3x  5 
1
Note: Award A1 for 3, A1 for
.
3x  5
(c) evidence of using product rule
 3 
h′(x) = (cos 2x) 
 + ln(3x – 5)(–2 sin 2x)
 3x  5 
A1A1
N2
A1A1
N2
(M1)
A1
N2
[6]
2.
evidence of choosing the product rule
f′(x) = ex × (–sin x) + cos x × ex (= ex cos x – ex sin x)
substituting π
e.g. f′(π) = eπ cos π – eπ sin π, eπ(–1 – 0), –eπ
taking negative reciprocal
1
e.g. 
f ( π)
1
gradient is π
e
(M1)
A1A1
(M1)
(M1)
A1
N3
[6]
3.
(a)
(i)
–3e–3x
π

cos x  
3

evidence of choosing product rule
e.g. uv′ + vu′
correct expression
π
π


e.g. –3e–3x sin  x    e 3 x cos x  
3
3


π
complete correct substitution of x =
3
(ii)
(b)
e.g.  3e
3
π
3
π
h′   = e–π
3
A1
N1
A1
N1
(M1)
A1
(A1)
π
 π π  3
π π
sin     e 3 cos  
3 3
3 3
A1
N3
[6]
IB Questionbank Maths SL
1
4.
(a)
(b)
(c)
dy
 3 cos 3x
dx
dy
x
accept x sec2 x + tan x

 tan x
2
dx cos x
METHOD 1
Evidence of using the quotient rule
1
x   ln x
dy
x

dx
x2
dy 1  ln x

dx
x2
METHOD 2
y = x1 In x
Evidence of using the product rule
dy
1
 x 1   ln x 1 x 2
dx
x
dy 1 ln x


dx x 2 x 2
 
A1
N1
A1A1
N2
(M1)
A1A1
N3
(M1)
A1A1
N3
[6]
5.
(a)
(b)
(c)
f (x) = 5e5x
g (x) = 2 cos 2x
h = fg + gf ′
= e5x (2 cos 2x) + sin 2x (5e5x)
A1A1
A1A1
(M1)
A1
N2
N2
N2
[6]
6.
METHOD 1 (quotient)
derivative of numerator is 6
derivative of denominator is –sin x
attempt to substitute into quotient rule
correct substitution
cos x 6  6 x  sin x 
e.g.
cos x 2
substituting x = 0
cos 06  6  0 sin 0
e.g
cos 02
h′(0) = 6
(A1)
(A1)
(M1)
A1
(A1)
A1
N2
[6]
METHOD 2 (product)
h(x) = 6x × (cos x)–1
derivative of 6x is 6
derivative of (cos x)–1 is (–(cos x)–2 (–sin x))
attempt to substitute into product rule
correct substitution
e.g. (6x) (–(cos x)–2 (–sin x)) + (6) (cos x)–1
substituting x = 0
e.g. (6× 0) (–(cos 0)–2 (–sin 0)) + (6) (cos 0)–1
h′(0) = 6
(A1)
(A1)
(M1)
A1
(A1)
A1
N2
[6]
IB Questionbank Maths SL
2
7.
(a)
(b)
evidence of choosing the product rule
e.g. uv′ + vu′
correct derivatives cos x, 2
g′(x) = 2 x cos x + 2 sin x
attempt to substitute into gradient function
e.g. g′(π)
correct substitution
e.g. 2π cos π + 2 sin π
gradient = –2π
(M1)
(A1)(A1)
A1
(M1)
(A1)
A1
N4
N2
[7]
8.
(a)
evidence of choosing the product rule
e.g. x × (– sin x) + 1 × cos x
f′(x) = cos x – x sin x
(M1)
A1A1
N3
A1A1A1A1
N4
(b)
Note: Award A1 for correct domain, 0 ≤ x ≤ 6 with
endpoints in circles,
A1 for approximately correct shape,
A1 for local minimum in circle,
A1 for local maximum in circle.
[7]
IB Questionbank Maths SL
3
9.
(a)
(b)
d
d
sin x  cos x, cos x   sin x (seen anywhere)
dx
dx
evidence of using the quotient rule
correct substitution
sin x( sin x)  cos x(cos x)  sin 2 x  cos 2 x
,
e.g.
sin 2 x
sin 2 x
2
2
 (sin x  cos x)
f′(x) =
sin 2 x
1
f′(x) =
sin 2 x
METHOD 1
appropriate approach
e.g. f′(x) = –(sin x)–2
 2 cos x 
f″(x) = 2(sin–3 x)(cos x)  

 sin 3 x 
Note: Award A1 for 2 sin–3 x, A1 for cos x.
METHOD 2
derivative of sin2 x = 2 sin x cos x (seen anywhere)
evidence of choosing quotient rule
sin 2 x  0  (1)2 sin x cos x
e.g. u = –1,v = sin2 x, f″(x) =
(sin 2 x) 2
(A1)(A1)
M1
A1
A1
AG
(M1)
A1A1
(d)
N3
A1
(M1)
2 sin x cos x  2 cos x 


(sin 2 x) 2  sin 3 x 
π
evidence of substituting
2
π
2 cos
1
2
e.g.
,
π
π
sin 2
sin 3
2
2
p = –1, q = 0
second derivative is zero, second derivative changes sign
f″(x) =
(c)
N0
A1
N3
M1
A1A1 N1N1
R1R1 N2
[13]
10.
(a)
METHOD 1
evidence of recognizing the amplitude is the radius
e.g. amplitude is half the diameter
8
a
2
a=4
METHOD 2
evidence of recognizing the maximum height
e.g. h = 6, a sin bt + 2 = 6
correct reasoning
e.g. a sin bt = 4 and sin bt has amplitude of 1
a=4
IB Questionbank Maths SL
(M1)
A1
AG
N0
2
N0
2
(M1)
A1
AG
4
(b)
(c)
(d)
METHOD 1
period = 30
2
b
30

b
15
METHOD 2
correct equation
e.g. 2 = 4 sin 30b + 2, sin 30b = 0
30b = 2π

b
15
recognizing h′(t) = –0.5 (seen anywhere)
attempting to solve
e.g. sketch of h′, finding h′
correct work involving h′
4π
π 
cos t 
e.g. sketch of h′ showing intersection, –0.5 =
15
 15 
t = 10.6, t = 19.4
METHOD 1
valid reasoning for their conclusion (seen anywhere)
e.g. h(t) < 0 so underwater; h(t) > 0 so not underwater
evidence of substituting into h
19.4π
2
e.g. h(19.4), 4 sin
15
correct calculation
e.g. h(19.4) = –1.19
correct statement
e.g. the bucket is underwater, yes
METHOD 2
valid reasoning for their conclusion (seen anywhere)
e.g. h(t) < 0 so underwater; h(t) > 0 so not underwater
evidence of valid approach
e.g. solving h(t) = 0, graph showing region below x-axis
correct roots
e.g. 17.5, 27.5
correct statement
e.g. the bucket is underwater, yes
(A1)
A1
AG
N0
2
N0
2
N3
6
N0
4
N0
4
(A1)
A1
AG
R1
(M1)
A2
A1A1
R1
(M1)
A1
A1
R1
(M1)
A1
A1
[14]
IB Questionbank Maths SL
5
11.
(a)
(b)
(c)
y = e2x cos x
dy
= e2x (–sin x) + cos x (2e2x)
dx
= e2x (2 cos x – sin x)
d2 y
= 2e2x (2 cos x – sin x) + e2x (–2 sin x – cos x)
2
dx
= e2x (4 cos x – 2 sin x – 2 sin x – cos x)
= e2x (3 cos x – 4 sin x)
d2 y
(i)
At P,
=0
dx 2
 3 cos x = 4 sin x
3
 tan x =
4
3
At P, x = a, ie tan a =
4
(ii) The gradient at any point e2x (2 cos x – sin x)
Therefore, the gradient at P = e2a (2 cos a – sin a)
3
4
3
When tan a = , cos a = , sin a =
4
5
5
(by drawing a right triangle, or by calculator)
8 3
Therefore, the gradient at P = e2a   
5 5
2a
=e
(A1)(M1)
(AG)
2
(A1)(A1)
(A1)
(A1)
4
(R1)
(M1)
(A1)
(M1)
(A1)(A1)
(A1)
(A1)
8
[14]
12.
(a)
(b)
evidence of finding height, h
h
e.g. sin θ = , 2 sin θ
2
evidence of finding base of triangle, b
b
e.g. cos θ = , 2 cos θ
2
attempt to substitute valid values into a formula for the area
of the window
e.g. two triangles plus rectangle, trapezium area formula
correct expression (must be in terms of θ)
1
1

e.g. 2  2 cos  2 sin    2  2 sin  , 2 sin  2  2  4 cos 
2
2

attempt to replace 2sinθ cosθ by sin 2θ
e.g. 4 sin θ + 2(2 sin θ cos θ)
y = 4 sin θ + 2 sin 2θ
correct equation
e.g. y = 5, 4 sin θ + 2 sin 2θ = 5
evidence of attempt to solve
e.g. a sketch, 4 sin θ + 2 sin θ – 5 = 0
θ = 0.856 (49.0º), θ = 1.25 (71.4º)
IB Questionbank Maths SL
(A1)
(A1)
(M1)
A1
M1
AG
A1
N0
5
N3
4
(M1)
A1A1
6
(c)
recognition that lower area value occurs at θ =

2


 


e.g. 4 sin    2 sin  2   , draw square
2
2
 


A=4
recognition that maximum value of y is needed
A = 5.19615…
4 < A < 5.20 (accept 4 < A < 5.19)
finding value of area at θ =

2
(M1)
(M1)
(A1)
(M1)
(A1)
A2
N5
7
[16]
IB Questionbank Maths SL
7