Download Year 12 2/3 Unit Math Homework

Yimin Math Centre
2/3 Unit Math Homework for Year 12 (Answers)
Grade:
Date:
Score:
nt
re
Student Name:
Table of contents
Ce
Integration 2
1
Rules of Differentiation and Integration . . . . . . . . . . . . . . . . . . . . . . . . . .
1
4.2
Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
4.3
Miscellaneous Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4.4
Practical Exam Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
at
h
4.1
Yi
m
in
M
4
This edition was printed on December 11, 2013.
Camera ready copy was prepared with the LATEX2e typesetting system.
Copyright © 2000 - 2013 Yimin Math Centre (www.yiminmathcentre.com)
2/3 Unit Math Homework for Year 12
Year 12 Topic 4 Worked Answers
Rules of Differentiation and Integration
f (x)
f 0 (x)
f (x)
kx
k
k
xn
nxn−1
xn
axn + bx + c
anxn−1 + b
axn
f (h(x))
f 0 (h(x)) .h0 (x)
(ax + b)n
g(x)
h(x)
g 0 (x).h(x)−g(x).h0 (x)
[h(x)]2
0
0
y 0 = u v−uv
2
v
x
y=
u
v
ex
e
xn+1
n+1 + c
axn+1
n+1 + c
(ax+b)n+1
a(n+1) + c
1
n+1
+
n+1 (f (x))
c
nt
re
y 0 = uv 0 + u0 v
f (x).dx
kx + c
g 0 (x).h(x) + g(x).h0 (x) f 0 (x).(f (x))n
y = uv
R
Ce
g(x).h(x)
ex
= ex + c
ekx
1 kx
ke
1
x
k
x
f 0 (x)
f (x)
ln x + c
ln f (x) + c
kekx
ln x
ln f (x)
1
x
k
x
f 0 (x)
f (x)
sin x
cos x
sin x
− cos x + c
sin kx
k cos kx
sin kx
− k1 cos kx + c
− sin x
cos x
sin x + c
tan x
− ln(cos x) + c
cot x
ln(sin x) + c
sec x
ln(sec x + tan x) + c
cos x
at
M
ln kx
h
ekx
in
4.1
Integration 2
Yi
m
4
Page 1 of 14
cos kx
−k sin kx
cos kx
tan x
sec2 x
sec2 x
tan kx
k sec2 kx
sec2 kx
√ 1
a2 −x2
− √a21−x2
a
a2 +x2
√ 1
a2 −x2
sin−1
x
a
−1 x
cos a
tan−1 xa
1
a2 +x2
1
k
1
k
+c
ln x + c
sin kx + c
tan x + c
1
k
tan kx + c
sin−1 xa c + c
1
a
tan−1 xa + c
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Year 12 Topic 4 Worked Answers
4.2
Page 2 of 14
Definite Integral
Definition:
Z
b
f 0 (x) dx = [f (x)]ba = f (b) − f (a)
a
Example 4.2.1
R9
1. Find 1 (2 +
√
x) dx
Solution:
Z
9
(2 +
√
Z
9
1
(2 + x 2 ) dx
1
"
#9
1
x( 2 +1)
= 2x + 1
( 2 + 1)
1
9
2 3
= 2x + x 2
3
1
2 3
2
= 2(9) + (9) 2 − 2(1) + (1)
3
3
1
2
= [18 + 18] − [2 ] = 33 .
3
3
x) dx =
R3
1
(x2 + 2e2x + 1) dx
Solution:
Z
3
2
M
2. Find
at
h
Ce
nt
re
1
(x + 2e
2x
Yi
m
in
1
3. Find
R2
1
1 x2
Solution:
3
1 3 1
2x
+ 1) dx =
x + × 2e + x
3
2
1
1 3
1 3
6
2
(3) + e + 3 − (1) + e + 1
=
3
3
1
= [9 + 403.43 + 3] − [ + 7.39 + 1]
3
= 406.71
dx
Z
1
2
1
dx =
x2
Z
2
x−2 dx
1
#2
x(−2+1)
=
(−2 + 1)
1
1 2
= −
x 1
1
1
1
= (− ) − (− ) = .
2
1
2
"
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Year 12 Topic 4 Worked Answers
Page 3 of 14
Exercise 4.2.1
R2
e2x dx
Solution:
2.
R1
2
−1 x+2
R2
1
2
e2x dx = 21 e2x 1
= ( 21 e4 ) − ( 21 e2 )
= 29.30 − 3.69
= 23.61
dx
R1
2
−1 x+2
R1 1
dx = 2 −1 x+2
dx
= 2 [ln (x + 2)]1−1
= 2(ln 3 − ln 1)
= 2 ln 3 or = 2.197
0
(e3x − 1) dx
4.
R π3
π
6
R1
0
1
(e3x − 1) dx = 31 e3x − x 0
= 13 e3 − 1 − 31 e0 − 0
= 13 e3 − 43
= 31 (e3 − 4) or = 5.36
Yi
m
Solution:
M
R1
in
3.
at
h
Solution:
nt
re
1
Ce
1.
sec2 x dx
Solution:
R π3
π
6
2
π3
sec x dx = tan x
=
=
=
π
6
tan π − tan π6
√ 3 √3
3− 3
√
2 3
or 1.154
3
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Year 12 Topic 4 Worked Answers
4.3
Page 4 of 14
Miscellaneous Exercise
Exercise 4.3.1
R3
1. −1 (2x2 − 8x + 8) dx
=
=
=
=
(−x3 )dx
h 4 i− 12
x
3
(−x
)dx
=
−
4
−1
1 4
−1
1
4
= − 4 (− 2 ) − (−1)
1
− 1)
= − 41 ( 16
1
= − 4 × (− 15
) = 15
16
64
R −1
Solution:
0
1
at
R1
2
1
(x 5 − x 3 ) dx
R1
M
3.
1
h
1
Yi
m
R 12
1
4
( x12 −
−
i1
1
x 3 +1
1
+1 0
3
0
= ( 56 − 43 ) − (0)
18
2
− 24
= 24
=
= 20
24
4.
1
x 5 +1
1
+1
5
in
(x 5 − x 3 ) dx =
0
h 6
i1
5 5
3 34
= 6x −4x
Solution:
Ce
2
−1
+ 8(−1)
h
R −1
h
2
nt
re
Solution:
2.
i3
2 3
8x2
(2x − 8x + 8) dx = 3 x − 2 + 8x
−1
−1
2
2 3
2
3
2
(3)
−
4(3)
+
8(3)
−
(−1)
−
4(−1)
3
3
( 32 × 27 − 4 × 9 + 24) − (− 32 − 4 − 8)
(18 − 36 + 24) − (−12 23 )
6 + 12 32 = 18 23
R3
1
12
1
x3 ) dx
Solution:
R 21
=
1
4
h
( x12 −
x−2+1
−2+1
1
) dx
x3
−
=
i1
−3+1 2
x
−3+1
R 21
1
4
(x−2 − x−3 ) dx
1
14
−2 2
= − x−1 + 21 x 1
4 = −( 21 )−1 + 12 ( 12 )−2 − −( 14 )−1 + 12 ( 14 )−2
= (−2 + 2) − (−4 + 8) = 0 − 4 = −4
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Year 12 Topic 4 Worked Answers
Page 5 of 14
Exercise 4.3.2
1. Use the substitution u = 2x2 + 1, or otherwise. find
Solution:
Z
R
5
x(2x2 + 1) 4 dx.
Let u = 2x2 + 1 ⇒ du = 4x dx.
Z
5
5
1
2
x(2x + 1) 4 dx =
u 4 du
4
1 4 9
= × u4 + c
4 9
9
1
= (2x2 + 1) 4 + c.
9
nt
re
2. Differentia e3x (cos x − 3 sin x).
Ce
Solution: d e3x (cos x − 3 sin x) = 3e3x (cos x − 3 sin x) + e3x (− sin x − 3 cos x)
dx
= e3x (3 cos x − 9 sin x − sin x − 3 cos c)
R
Z
Rπ
4
1 3x
e (cos x − 3 sin x) + c
10
cos x sin2 x dx.
Yi
m
0
e3x sin x dx = −
in
Solution:
4. Evaluate
e3x sin x dx.
M
3. Hence, or otherwise find
at
h
= −10e3x sin x.
Solution:
Z
π
4
Let u = sin x, then du = cos dx.
When x = 0, ⇒ u = 0; when x =
2
Z
cos x sin x dx =
0
1
√
2
π
1
⇒ u= √ .
4
2
u2 du
0
√1
u3 2
=
3 0
( √12 )3
=
√3
2
=
.
12
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Year 12 Topic 4 Worked Answers
Page 6 of 14
Multiplication of Polynomials:
• Expansion of perfect square: (a ± b)2 = a2 ± 2ab + b2
• Difference of perfect square: (a − b)(a + b) = a2 − b2
• Expansion of cube: (a ± b)3 = a3 ± 3a2 b + 3ab2 ± b3
• Sum and difference of two cubes: a3 ± b3 = (a ± b)(a2 ∓ ab + b2 )
Exercise 4.3.3 Factorising the following:
nt
re
1. (2a − b)3 + (a − 2b)3
Solution:
(2a − b)3 + (a − 2b)3 = [(2a − b) + (a − 2b)][(2a − b)2 − (2a − b)(a − 2b) + (a − 2b)2 ]
Ce
= (3a − 3b)[4a2 − 4ab + b2 − (2a2 − 4ab − ab + 2b2 ) + a2 − 4ab + 4b2 ]
= 3(a − b)(3a2 − 3ab + 3b2 )
at
h
= 9(a − b)(a2 − ab + b2 ).
(3x + y)3 + 8y 3 = 27x3 + 27x2 y + 9xy 2 + y 3 + 8y 3
Yi
m
in
Solution:
M
2. (3x + y)3 + 8y 3
= 27x3 + 27x2 y + 9xy 2 + 9y 3
= 27x2 (x + y) + 9y 2 (x + y)
= (x + y)(27x2 + 9y 2 )
= 9(x + y)(3x2 + y 2 )
3. x8 + x2
Solution:
x8 + x2 = x2 (x6 + 1)
= x2 [(x2 )3 + 1]
= x2 [(x2 + 1)(x2 − x2 + 1)]
= x2 (x2 + 1)[x4 + 2x2 + 1) − 3x2 ]
√
= x2 (x2 + 1)[(x2 + 1)2 − ( 3x)2 ]
√
√
= x2 (x2 + 1)(x2 + 1 + 3x)(x2 + 1 − 3x).
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Year 12 Topic 4 Worked Answers
Page 7 of 14
Exercise 4.3.4 Factorising the following:
1. 3x2 + 10x − 8
Solution:
Factors of -8 are -2 and 4 or 2 and -4 or 1 nad -8 or -1 and 8
3x × 4 + 1x × (−2) = 10x
So 3x2 + 10x − 8 = (3x − 2)(x + 4).
2. 3x2 − 7x − 6
Solution:
nt
re
3x × (−3) + x × 2 = −7x
So 3x2 − 7x − 6 = (3x + 2)(x − 3).
Solution:
Let (3x + 1) be a
Ce
3. (3x + 1)2 + 3(3x + 1) − 18
h
(3x + 1)2 + 3(3x + 1) − 18 = a2 + 3a − 18
Yi
m
4. x4 + 5x2 + 9
Solution:
= (3x + 1 + 6)(3x + 1 − 3)
= (3x + 7)(3x − 2)
in
M
at
= (a + 6)(a − 3)
x4 + 5x2 + 9 = x4 + 6x2 + 9 + 5x2 − 6x2
= x4 + 6x2 + 9 − x2
= (x2 + 3)2 − x2
= (x2 − x + 3)(x2 + x + 3).
5. x2 − 4x − 8
Solution:
x2 − 4x − 8 = x2 − 4x + 4 − 8 − 4
= (x − 2)2 − 12
√
√
= (x − 2 − 2 3)(x − 2 + 2 3).
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Year 12 Topic 4 Worked Answers
Page 8 of 14
Exercise 4.3.5
1. Use the substitution u = cos θ to evaluate
Solution:
Rπ
0
2 sin θ cos2 θ dθ.
1
Let u = cos θ ⇒ du = − sin θ dθ, ⇒ d θ =
du.
−
sin
θ

θ = 0, ⇒ u = 1
Limits:
θ = π, ⇒ u = −1
Z 1
Z π
2
u2 du
2 sin θ cos θ dθ = 2
I=
0
=2
−1
1
u3
3
−1
1
1 1
=1 .
=2
+
3 3
3
x, to evaluate:
R4
1√
1 x+ x
dx.
√
1 1
x, ⇒ then du = x 2 dx
2 
x = 4, ⇒ u = 2
∴ dx = 2u du, When
x = 1, ⇒ u = 1.
Z 4
Z 2
1
2u
√ dx =
I=
du
2
1 x+ x
1 u +u
Z 2
2
=
du
1 1+u
Let u =
M
at
h
Solution:
√
Ce
2. Use the substitution u =
nt
re
in
= 2[ln(1 + u)]21
Yi
m
3
= 2(ln 3 − ln 2) = 2 ln .
2
3. Evaluate
R0
−1
Solution:
√
x 1 + x dx using the substitution u = 1 + x.

x = 0, ⇒ u = 1
Let u = 1 + x, du = dx,
x = −1, ⇒ u = 0
Z 0
Z 1
√
√
x 1 + x dx =
(u − 1) u du
−1
0
Z
1
3
1
u 2 − u 2 du
0
2 5 2 3 1
=
u2 − u2
5
3
0
2 2
4
=
−
− (0 − 0) = − .
5 3
15
=
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Year 12 Topic 4 Worked Answers
4.4
Page 9 of 14
Practical Exam Questions
Exercise 4.4.1
Rπ
sin 2x dx
dx
Solution:
3.
R ln 5
0
R e2
3
x
1
2
dx = 3 [loge x]e1
= 3(loge e2 − loge 1)
= 3(2loge e − 0)
=3×2=6
e−x dx
R ln 5
0
−x
e
4.
R3
0
−x
ln 5
dx = −e
Yi
m
Solution:
Ce
3
x
1
0
π
sin 2x dx = − 12 cos 2x 02
= − 12 cos 2( π2 ) − cos 2(0)
= − 12 [cos π − cos 0]
= − 21 (−1 − 1)
= − 21 × (−2) = 1
h
R e2
2
at
2.
Rπ
nt
re
Solution:
M
2
0
in
1.
0
= −e−ln 5 − (−e0 )
= − eln1 5 − (−1)
= − 15 + 1
= 54
e3x dx
Solution:
R2
0
3
e3x dx = 31 e3x 0
= 13 e9 − 31 e0
= 13 (e9 − 1)
= 2701.03 − 13
= 2700.70
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Year 12 Topic 4 Worked Answers
Page 10 of 14
Exercise 4.4.2
R e3 2
dx.
1. e−3 3x
Solution:
Z
e3
e−3
0
e2x dx
Solution:
R1
0
e2x dx =
=
=
3.
R3
1
1
x+1
i1
e2x
2
0
1 2
(e
−
2
1 2
(e −
2
h
e0 )
1) = 3.19
nt
re
R1
dx
Solution:
R3
1
1
x+1
3
dx = loge (x + 1)
Ce
2.
2
2
3
dx = [ln x]ee−3 = 4.
3x
3
4
0
cos 2x dx
Rπ
in
Rπ
π
cos 2x dx = 12 sin 2x 04
= 12 sin 2 π4 − sin 2 (0)
= 21 [sin π2 − sin 0]
= 21
Yi
m
4.
M
at
h
1
= loge 4 − loge 2
= loge 42
= loge 2 = 0.693
Solution:
5.
R π2
π
3
4
0
cos x dx
Solution:
R π2
π
3
π2
cos x dx = sin x
=
=
=
π
3
sin π2 − sin π3
√
1 − 23
√
2− 3
. or =
2
0.13397
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Year 12 Topic 4 Worked Answers
Page 11 of 14
Exercise 4.4.3 Evaluate the following:
0
√ dx
.
16−x2
Z
Solution:
Standard formula:
Z
2
0
2.
Rπ
6
0
1
x
= sin−1 ( )
2
a
−x
h
x i2
dx
√
= sin−1
4 0
16 − x2
1
= sin−1
− sin−1 0
2
π
= −0
6
π
= .
6
√
a2
sec 2x tan 2x dx.
Z
Solution:
1
sec ax.
a
π6
1
sec 2x tan 2x dx =
sec 2x
2
0
π
1
sec − sec 0
=
2 3
1
1
1
=
−
2 cos π3
cos 0
1
= (2 − 1)
2
1
= .
2
sec ax tan ax dx =
Standard formula:
π
6
Z
3.
Rπ
0
Yi
m
in
M
at
h
0
nt
re
R2
Ce
1.
cos2 3x dx.
Solution:
1
cos 2A = 2 cos2 A − 1 ⇒ cos2 A = (1 + cos2 A)
2
1
∴ cos2 3x = (1 + cos 6x)
Z π
Z2 π
1
∴
cos2 3x dx =
(1 + cos 6x) dx
0
0 2
π
1
1
=
x + sin 6x
2
6
0
1
1
1
=
(π + sin 6π) − (0 + sin 0)
2
6
6
π
= .
(Noting sin 6π = sin 2π = sin 0 = 0).
2
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Year 12 Topic 4 Worked Answers
Page 12 of 14
Exercise 4.4.4
1.
R
ex
1+ex
dx.
Solution:
Z
Let u = 1 + ex ⇒ du = ex dx
ex dx
du
=
x
1+e
u
= ln u + c
= ln(1 + ex ) + c.
R0
−1
√
x 1 + x dx.
nt
re
2. Use the substitution u = 1 + x to evaluate 15
Solution:
Let u = 1 + x, ⇒ then x = u − 1 ⇒ dx = du,
Yi
m
in
M
at
h
Ce
When x = −1, u = 1 + (−1) = 0; when x = 0, u = 1 + 0 = 1.
Z 0
Z 1
√
√
x 1 + x dx = 15
(u − 1) u du
15
−1
0
Z 1
1
(u − 1)u 2 du
= 15
Z0 1
3
1
(u 2 − u 2 ) du
= 15
0
1
2 5 2 3
= 15 u 2 − u 2
5
3
0
2
2
= 15 ( × 1 − × 1) − 0
5
3
= −4
3. Use the substitution u = 1 − x2 to evaluate
Solution:
R3
2x
2 (1−x2 )
dx.
Let u = 1 − x2 ⇒ du = −2x dx
When x = 2 u = 1 − 4 = −4; when x = 4 u = 1 − 9 = −8.
Z 3
Z −8
2x dx
−1
=
du
2
2
2 (1 − x )
−3 u
Z −3
=
u−2 du
−8
−1 −3
u
−1 −8
1
5
1
=−
−
= .
−3 −8
24
=
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Year 12 Topic 4 Worked Answers
Page 13 of 14
Exercise 4.4.5
R1
√ 1
2−x2
0
dx
Z
Solution:
1
1
−1 x
√
dx = sin √
2 − x2
2 0
1
= sin−1 √ − sin−1 0
2
π
=
4
2. Use the substitution u = x2 + 1 to evaluate
Solution:
R2
x
0 (x2 +1)3
dx
du
= x dx.
2
When x = 0, ⇒ u = 02 + 1 = 1; when x = 2, ⇒ u = 22 + 1 = 5
Z 5 du
x dx
2
=
2
3
3
(x + 1)
1 u
Z 5
1
=
u−3 du
2 1
5
1 u−2
=
2 −2 1
1 1
=−
−1
4 25
6
= .
25
Z
in
M
at
h
0
2
Ce
Let u = x2 = 1, thendu = 2dx ⇒
nt
re
1. Evaluate
Yi
m
3. Use the substitution u = x − 3 to evaluate
Solution:
R4 √
x x − 3 dx.
3
Let u = x − 3, ⇒ du = dx,
when x = 3, ⇒ u = 3 − 3 = 0; when x = 4 ⇒ u = 4 − 3 = 1.
Z 1
4 √
1
x x − 3 dx =
(u + 3)u 2 du
3
Z0 1 3
1
=
u 2 + 3u 2 du
0
1
2 5
2 3
2
2
= u +3× u
5
3
0 2
2
=
×1+2×1 −
×0+2×0
5
5
2
=2 .
5
Z
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Year 12 Topic 4 Worked Answers
Page 14 of 14
Exercise 4.4.6 Evaluate the following:
R π sin x
dx.
1. 03 2−cos
x
Solution:
Let f (x) = 2 − cos x, ⇒ f 0 (x) = sin x,
Z π
π
3
sin x
∴
dx = [ln(2 − cos x]03
0 2 − cos x
= ln(1.5).
Solution: f (−x) = (−x)3 cos(−x) = −x3 cos x = −f (x).
∴ the function x3 cos x is an odd function.
R
π
− 2
π
2
x3 cos x dx.
nt
re
2. State whether the function x3 cos x is odd, even or neither.
Ce
Since the function is odd, whatever happens in the first quadrant must happen
in reverse in the the 3rd quadrant.
Hence the value of the ntegral in the first quadrant will be equal in value
at
h
but opppsite in sign in the third.
Z 2
Thus
x3 cos x dx = 0.
Rπ
6
0
sin 2x cos 2x dx.
π
6
in
3.
M
−2
Z
Solution:
Z
1
=
2
Z
Yi
m
0
1
sin 2x cos 2x dx =
2
4. Show that
1
x
−
1
x+3
=
π
6
2 sin 2x cos 2x dx
0
π
6
sin 4x dx
0
π
6
1
1
=
− cos 4x
2
4
0
3
= .
16
3
.
x(x+3)
Solution:
Z
∴
1
3
Hence, evaluate
R3
2
1 x2 +3x
dx.
1
1
x+3
x
3
−
=
−
=
.
x x+3
x(x + 3) x(x + 3)
x(x + 3)
Z 3
3
1
1
dx =
−
dx
x2 + 3x
x x+3
1
= [ln x − ln(x + 3)]31 = ln 2.
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