Page 1 of 6
Chapter 1 Reasoning and Logic
Try these 1.1
(a)
(b)
(c)
Today is not Sunday.
The music is not loud.
Lorraine does not teach physics.
Try these 1.2
(a)
(i)
(ii)
(b)
(i)
(ii)
Let p be ‘the measuring cylinder is full’
q ‘the beaker is empty’
p∨q
b ‘breakfast is at the house’
c ‘the coffee is hot’
b∧c
‘The air-condition unit is not broken’
‘December 25th is not Christmas’
Try these 1.3
(a)
(b)
(c)
Paris is in France is true
Rome is in Italy is also true
∴ Paris is in France if and only if Rome is in Italy is true
8 – 4 = 12 is false
30 + 24 = 64 is false
∴ 8 – 4 ↔ 30 + 24 = 64 is true
4 1
= is false
6 2
4 1
∴ Port of Spain is the capital of Trinidad and Tobago if and only if = is false
6 2
Port of Spain is the capital of Trinidad and Tobago is true
Exercise 1A
1
(a)
(b)
(c)
(d)
(e)
Yes
No
Yes
No
Yes
2
(a)
Let m: It is May
c: CAPE examination begins
m∧c
S: takes Spanish
A: takes additional mathematics
s∨a
l: the music is loud
r: ryan is speaking
l∧r
(b)
(c)
Unit 1 Answers: Chapter 1
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Page 2 of 6
(d)
3
4
5
6
7
8
m: drive to Montego Bay
b: go by bus
p: go by plane
∼ m ∧ (b ∨ p)
Conjunction
(a) Alvin is tall or Sintra is not short
(b) Alvin is not tall or Sintra is not short
(c) Alvin is tall and Sintra is not short
(d) Alvin is not tall and Sintra is not short
(a) 4 is not a complete square
(b) the iPod is not white
(c) Robin likes to work overtime
(d)
7 is rational.
(a) ∼ (s ∨ p) = ∼ s ∧ ∼ p
Statistics is easy and probability is difficult
(b) ∼ (s ∧ ∼ p) = ∼ s ∨ p
Statistics or probability is easy
(c) ∼ (∼s∧ p) = s ∨ ∼ p : Statistics or probability is difficult
(a) ∼ (p ∨ q)
= ∼ (p ∨ q)
=~p∧∼q
Jamaica is not beautiful and Watson does not like jerk chicken
(b) ∼ (p ∧ ∼ q)
=∼p∨q
Jamaica is not beautiful or Watson likes jerk chicken
(c) ∼ (∼ p ∧ ∼ q)
=p∨q
Jamaica is beautiful or Watson likes jerk chicken
(a) x ∧ y
(b) ∼ (x ∨ y)
(c) ∼ (x ∧ y)
9
p
q
p∨q
T
T
T
T
F
T
F
T
T
F
F
F
Only if both ‘p’ and ‘q’ are false, the disjunction p ∨ q is false
10
r
c
r∧c
T
T
T
T
F
F
Unit 1 Answers: Chapter 1
© Macmillan Publishers Limited 2013
Page 3 of 6
F
T
F
F
F
F
The conjunction r ∧ c is true only if both r is true and c is true
Exercise 1B
1
2
3
4
5
6
7
The music is not good if I don’t dance
(a) I will go running and the sun is shining
(b) The sun is not shining
(c) Either the sun is shining or I am going running
(a) If the teacher will not write then the board will not be clean
(b) Chris will not pass calculus if he does not study
p
q
q∨p
∼ (q ∨ p)
p → ∼ (q ∨ p)
T
T
T
F
F
T
F
T
T
T
F
T
T
T
T
F
F
F
T
T
(p ∨ ∼ q) ∧ (∼ p ∧ q)
= (p ∨ ∼ q) ∧ ∼ (p ∨ ∼ q)
=F
Let A = (p ∨ ∼ q)
A∧∼A=F
∴ (p ∨ ∼ q) ∧ (∼ p ∧ q) is a contradiction since the statement is always false
(∼ (x ∨ y)) ∨ ((∼ x) ∧ y)
= (∼ x ∧ ∼ y) ∨ (∼ x ∧ y)
= ∼ x ∧ (∼ y ∨ y)
=∼x∧T
=~x
Contrapositive: The sun is not shining whenever the Trinidad and Tobago cricket team loses
Converse: The sun is shining whenever the Trinidad and Tobago cricket team wins
Inverse: The Trinidad and Tobago cricket team loses whenever the sun is not shining
8
9
p
q
∼q
(p ∨ ∼ q)
p∧q
T
T
F
F
(a)
p
T
T
F
F
T
F
T
F
F
T
F
T
T
T
F
T
T
F
F
F
q
T
F
T
F
∼p
F
F
T
T
q
T
∼p
F
(b)
p
T
Unit 1 Answers: Chapter 1
∼q
F
T
F
T
∼q
F
(p ∨ ∼ q)
⇒ (p ∧ q)
T
F
T
F
(∼ p ∨ ∼ q)
F
T
T
T
(∼ p ∧ ∼ q)
F
∼ (∼ p ∨ ∼ q)
T
F
F
F
∼ (∼ p ∧ ∼ q)
T
© Macmillan Publishers Limited 2013
Page 4 of 6
T
F
F
F
T
F
F
T
T
T
F
T
F
F
T
T
T
F
(c)
p
q
p∨q
p∧q
∼ (p ∧ q)
T
F
T
F
T
T
T
F
T
F
F
F
F
T
T
T
(p ∨ q) ∧
∼ (p ∧ q)
F
F
F
F
T
T
F
F
(d)
10
11
12
13
14
p
q
r
(q ∨ r)
p ∧ (q ∨ r)
T
T
T
T
T
T
T
F
T
T
T
F
T
T
T
T
F
F
F
F
F
T
T
T
F
F
T
F
T
F
F
F
T
T
F
F
F
F
F
F
s ‘access the staff room’
m ‘member of staff’
f ‘first year student’
s ∧ (m ∨ ∼ f)
w ‘watch the movie’
o ‘under 13 years old’
a ‘accompanied by an adult’
∼ w ∧ (o ∧ ∼ a )
(a) I did not buy peanut butter this week
(b) I either bought peanut butter this week or made peanut punch on Saturday
(c) I bought peanut butter this week and made peanut punch on Saturday
(d) We can simplify the expression:
∼ p ∨ (p ∧ q) = (∼ p ∨ p) ∧ (∼ p ∨ q)
= T ∧ (∼ p ∨ q)
=∼p∨q
I did not buy peanut butter this week or I made peanut punch on Saturday
(a) ∼ p
(b) p ∧ ∼ q
(c) p → q
(d) q → p
(a)
p
T
T
F
F
q
T
F
T
F
(b)
p
q
Unit 1 Answers: Chapter 1
∼q
F
T
F
T
(p ∨ q)
(p ∨ ∼ q)
T
T
F
T
(p ∧ q)
(p ∨ ∼ q) → q
T
F
T
F
(p ∨ q) →
(p ∧ q)
© Macmillan Publishers Limited 2013
Page 5 of 6
T
T
F
F
15
T
F
T
F
17
(p → q)
T
F
F
F
T
F
F
T
(p → q) ∧
p↔q
(q → p)
T
T
T
T
T
T
T
F
F
T
F
F
F
T
T
F
F
F
F
F
T
T
T
T
For the same combinations of p and q, the output for (p → q) ∧ (q → p) and p ↔ q are the
same therefore the statements are logically equivalent
∼ (p ∨ (∼ p ∧ q))
= ∼ ((p ∨ ∼ p) ∧ (p ∨ q))
= ∼ (T ∧ (p ∨ q))
= ∼ (p ∨ q)
=∼p∧∼q
(a) Either the mango is sweet or yellow
(b) The mango is not yellow
(c) The mango is both sweet and yellow
p
16
T
T
T
F
q
q→p
18
∼q
F
T
F
T
p
q
T
T
T
F
F
T
F
F
This is not a contradiction
19
p
q
∼p
∼q
(p ∨ ∼ q)
(∼ p ∧ q)
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
T
T
F
T
F
F
T
F
p
q
∼p
∼q
∼ (p ∨ q)
(∼ p ∧ ∼ q)
F
F
T
T
F
T
F
T
F
F
F
T
F
F
F
T
20
T
T
T
F
F
T
F
F
The statement is always true
21
p
T
(p ↔ ∼ q)
F
T
T
F
q
T
(p ∧ q)
T
Unit 1 Answers: Chapter 1
(p ∨ ∼ q)
→
(∼p ∧ q)
F
F
T
F
∼ (p ∨ q)
↔
(∼ p ∧ ∼ q)
T
T
T
T
p∧q→p
T
© Macmillan Publishers Limited 2013
Page 6 of 6
F
T
F
T
T
F
F
T
F
F
F
T
Since the statement is always true, the statement is a tautology
22
p
T
T
F
F
23
24
25
q
p∨q
∼q
(p ∨ q)
∧∼q
∼ (p ∧ q)
(p ∧ ∼ q)
T
T
F
F
F
F
F
T
T
T
T
T
T
T
F
F
T
F
F
F
T
F
T
F
(p ∨ q) ∧ ∼ q ≡ (p ∨ q) ∧ ∼ (p ∧ q)
(p ∨ q) ∧ ∼ (p ∧ q) ≡ ∼ (p ∧ q) ∧ (p ∧ ∼ q)
But (p ∨ q) ∧ ∼ q ≡ ∼ (p ∧ q) ∧ (p ∧ ∼ q)
(a) Deepak works out everyday and has muscles
Deepak doesn’t work out everyday and does not have muscles
(b) (i)
q→p
∼ (p ∧ q)
∧
(p ∧ ∼ q)
F
T
F
F
(p ∨ q)
∧
∼ (p ∧ q)
F
T
T
F
(ii)
p→q
(iii) q → p
(a) p → q
(b) p ∧ q
(c) p
q
p → qp ∧ q
T
F
F
F
∼ ((p ∧ q) ∧ (p ∨ ∼ q))
= ∼ ((p ∧ q) ∧ p) ∨ ((p ∧ q) ∧ ∼ q)
= ∼ ((p ∧ q) ∨ F)
= ∼ (p ∧ q)
Unit 1 Answers: Chapter 1
© Macmillan Publishers Limited 2013
Page 1 of 7
Chapter 2 The Real Number System
Try these 2.1
Let a ∈ ℤ, b ∈ ℤ
Since a and b are integers, when we multiply two integers we get an integer
∴a×b∈ℤ
Hence the set of integers is closed with respect to multiplication.
a c
(b) Let , ∈ ℚ
b d
a c
Now ÷
b d
a d
=
×
b c
ad
=
bc
Since a, b, c, d ∈ ℤ ⇒ ad ∈ ℤ and bc ∈ ℤ
ad
∴
∈ℚ
bc
Hence ℚ is closed with respect to division.
(c) Let a, b ∈ 
(a)
a + b ∈
∴  is closed with respect to addition.
Try these 2.2
(a)
Let a, b, c ∈ ℝ.
Now a × (b + c) and a × b + a × c give the same value.
∴ multiplication distributes over addition.
2 + (3 × 4) = 2 + 12 = 14.
(2 + 3) × (2 + 4) = 5 × 6 = 30.
14 ≠ 30, ∴ addition does not distribute over multiplication.
Try these 2.3
Let a, e ∈ ℕ
a
=a
e
a
⇒ e = =1
a
∴ the identity for division of Natural numbers is 1
(b)
Let a, e ∈ ℝ
Now a ∗ e = a + 2e + 4
If e is the identity then a ∗ e =
a
⇒ a + 2e + 4 = a
2e + 4 = 0
2e = −4
Unit 1 Answers: Chapter 2
(a)
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Page 2 of 7
⇒ e = −2
∴ There is an identity element which is −2
Try these 2.4
Let a, b ∈ ℤ
If b is the inverse of a with respect to multiplication then ab = 1
1
⇒ a=
b
1
∈ ℤ only if b = 1
b
1
= 1
1
Since the set is the set of integers the only element that has an inverse with respect to
multiplication is the identity 1.
⇒ a=
Exercise 2
1
Since 2 is a prime number and 2 is not odd
⇒ the statement is not true.
2
RTP if x = 4n then x = (a)2 − (b)2
Proof:
Since x is divisible by 4
⇒ x = 4n = n2 + 2n + 1 − n2 + 2n − 1
= (n + 1)2 − (n − 1)2
= a2 − b2 where a = n + 1, b = n − 1
Hence if x is an integer divisible by 4, then x is the difference of two squares
3
6 is an even number but 2 × 3 = 6
2 is even and 3 is odd
The statement is false
4
R.T.P if x, y ∈ ℝ, x2 + y2 ≥ 2xy
Proof:
Since x, y are real
(x − y)2 ≥ 0
⇒ x2 − 2xy + y2 ≥ 0
⇒ x2 + y2 ≥ 2xy
5
The statement in false
1
1 1
Let x = ,
=
4
4 2
1 1
Since >
2 4
1 1
⇒
>
4 4
6
Let a, b ∈ ℝ. ab = 0 ⇔ a = 0 or b = 0
Proof:
Suppose that ab = 0. Then either a = 0 or a ≠ 0
If a = 0 ⇒ a = 0 or b = 0
if a ≠ 0, then a−1 exists
Unit 1 Answers: Chapter 2
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Page 3 of 7
7
8
∴ ab = 0
⇒ a−1 (ab) = a−1 0
⇒ (a−1a) b = 0
⇒ 1b = 0
b=0
⇒ b = 0, since any number multiplied by 0 gives 0
Hence if ab = 0 ⇒ a = 0 or b = 0
Now if a = 0 ⇒ ab = 0b
⇒ab = 0
If b = 0, ⇒ ab = a0
⇒ ab = 0
Hence if a = 0 or b = 0 then ab = 0
∴ ab = 0 ⇔ a = 0 or b = 0
1
(a + b)2
Let a = 2, b = 4
1
1
=
2
(2 + 4)
36
1
1
1
1
+ 2 = 2 + 2
2
a
b
2
4
1 1
=
+
4 16
4 +1
=
16
5
=
16
1
5
Since
≠
36 16
1
1
1
⇒
≠ 2 + 2 when a = 2, b = 4
2
(a + b)
a
b
1
1
1
Hence
is not equivalent to 2 = 2
2
(a + b)
a
b
a∗b=a + b+5
(a) Let a, b ∈ ℝ
Since a and b are real numbers
a + b is also real
a+b+5 is real
⇒a+b+5∈ℝ
∴ a ∗ b∈ .
Hence ℝ is closed wrt *
(b) Let e be the identity:
Now a ∗ e =
a
⇒a+e+5=a
e+5=0
e =−5
∴ the identity is −5
For the inverse of a
a ∗ a −1 =
e
Unit 1 Answers: Chapter 2
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Page 4 of 7
∴ a + a −1 + 5 =− 5
a −1 = −10 − a
the inverse of a is − 10 − a
9
a ∗ b= 3(a + b)
a  b = 2ab
Let a, b, c ∈ R
(a ∗ b) ∗ c= 3(a + b) ∗ c
= 3[3a + 3b + c]
= 9a + 9b + 3c
a ∗ (b ∗ c) =a ∗ 3(b + c)
=∗
a (3b + 3c)
= 3[a + 3b + 3c]
= 3a + 9b + 9c
Since (a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
⇒ * is not associative
Now (a  b)  c = 2ab  c
= 2(2ab)(c)
= 4 abc
and a  (b  c) = a  (2bc)
= 2 a (2 bc)
= 4 abc
∴ (a  b)  c = a  (b  c)
⇒  is associative
10
2n + 3
When n = 5, 2n + 3
= 25 + 3
= 32 + 3
= 35
Since 35 is not prime, the statement is false
11
(a) Since a and b are real
a − b is real
and | a − b | is real and positive
∴ For every a, b ∈ A, | a − b| ∈ A
⇒ A is closed with respect to ∗
(b) Let e be the identity
a∗e=a
|a−e|=a
⇒ e = 0 since a ≥ 0
The identity is 0
(c) If a has an inverse then a ∗ a−1 = e
⇒ |a − a−1| = |0| = 0, a − a−1 = 0, a−1 = a
Hence the inverse of a is a
∴ the elements are self inverses
(d) (1 ∗ 2) ∗ 3 = ||1 − 2| − 3| = |1 – 3| = 2
1 ∗ (2 ∗ 3) = |1 − |2 – 3|| = |1 – 1| = 0
so by counterexample ∗ is not associative
12
a ∆ b = aln b
(a) R.T.S. (a ∆ b) ∆c = a ∆(b ∆ c) ∀ a, b, c ∈ ℝ+
Solution:
(a ∆ b) ∆ c = (aln b) ∆ c
= (aln b)ln c = aln b × ln c
Unit 1 Answers: Chapter 2
© Macmillan Publishers Limited 2013
Page 5 of 7
Also a ∆ (b ∆ c) =a ∆(b ln c ) =a ln b
∴(a ∆ b) ∆ c = a ∆ (b ∆ c)
(b) Let a, b, ∈ ℝ+
Now a ∆ b = a
∴ a ln b =
a
ln c
=a ln c ln b
ln a ln b = ln a
13
ln b ln a = ln a
ln b = 1
b=e
∴ the identity is e
(c) Let a, b, c ∈ ℝ+
a ∆ (b × c) = a ∆ b c = aln bc
and (a ∆ b) × (a ∆ c) = aln b × aln c = aln b + ln c=aln bc
Since a ∆ (b × c) = (a ∆ b) × (a ∆ c)
⇒ ∆ distributes over multiplication
(a) Let x, y, x1, y1, ∈ ℝ
(x + y 3) + (x1 + y1 3)
=(x + x1 ) + y 3 + y1 3
= (x + x1 ) + (y + y1 ) 3
Since x, x1, y, y1, are integers
⇒ x + x1 and y + y1 are integers
∴ (x + x1) + (y + y1) 3 ∈ X
Hence X is closed under addition
(x + y 3 ) (x1 + y1 3 )
= xx1 + xy1
3 + x1 y
3 + 3yy1
= (xx1 + 3yy1) + [xy1 + x1 y] 3
Since x, x1, y, y1, are integers
⇒ xx1, yy1, xy1 and x1y are integers
∴ (xx1 + 3yy1) + [xy1 + x1 y] 3 ∈ X
Hence X is closed under multiplication
(b) Let e1, e2 ∈ ℝ
(x + y 3 ) + (e1 + e2 3 ) = x + y 3
⇒ e1 + e2 3 = 0 + 0 3
⇒ e1 = 0, e2 = 0
∴ the identity with respect to addition is 0 + 0
(c) (x + y
3 ) (e1 + e2
3)=x+y
3
3
e1 + e2 3 = 1
e1 = 1, e2 = 0
∴ identity with respect to multiplication is 1
(d) For a ∈ X, inverse of a =
1
a
For a = 0 + 0 3 = 0 ∈ X,
inverse of a =
1
∉X
0
∴ not every element of X has an inverse with respect to multiplication
Unit 1 Answers: Chapter 2
© Macmillan Publishers Limited 2013
Page 6 of 7
14)
x ∆ y=
x 2 + y2
(a) x, y, z ∈ ℝ
(x ∆ y)∆ z=
x2 + y2 ∆ z
=
( x +y ) +z
=
x2 + y2 + z2
2
2
2
2
x ∆(y ∆ z)= x ∆ y 2 + z 2
=
=
x2 +
(
y2 + z2
)
2
x 2 + y2 + z2
∴ (x ∆ y) ∆ z = x ∆ (y ∆ z)
⇒ ∆ is associative
(b) Let e be the identity:
x∆e=x
x2 + e2 =
x
2
e = 0, e = 0
∴ the identity is 0
(c) x ∆ (y ∆ z) = x ∆
=
y2 + z2
x2 + y2 + z2
(x ∆ y) ∆ (x ∆=
z)
15
=
x 2 + y2 + x 2 + z2
=
2x 2 + y 2 + z 2
x 2 + y2 ∆ x2 + z2
⇒ x ∆ (y ∆ z)≠ (x ∆ y) ∆ (x ∆ z)
∆ does not distribute over ∆
a ∗ b = a + b − ab, a, b ∈ R
(a) a ∗ e =
a
a ∗ e = a + e − ea
a + e − ea = a
e(1 − a)= 0
e=0
identity is 0
(b) a ∗ a −1 =
e
−1
a + a − aa−1 = 0
a = aa−1 − a−1
= (a − 1)a−1
a
a −1 =
a −1
a
∴ the inverse of a is
,a≠1
a −1
(c) a ∗ b = a + b − ab
b ∗ a = b + a − ba = a + b − ab
Unit 1 Answers: Chapter 2
© Macmillan Publishers Limited 2013
Page 7 of 7
(d)
16
(a)
(b)
(c)
17
(a)
(b)
Since a ∗ b = b ∗ a
∗ is commutative
a ∗ (a ∗ 2) =
10
⇒ a ∗ (a + 2 − 2a) =
10
a*(2−a)=10
a + 2 − a − a(2 − a) = 10
2−2a+a2=10
a2 − 2a − 8 = 0
(a − 4) (a + 2) = 0
a = 4, −2
Since all elements in the table belong to S
⇒ S in closed with respect to ∆
identity is a
Element
inverse
a
a
b
c
c
b
d
d
Identity is q
Element
Inverse
p
r
q
q
r
p
s
s
Unit 1 Answers: Chapter 2
© Macmillan Publishers Limited 2013
Page 1 of 20
Chapter 3 Principle of Mathematical Induction
Try these 3.1
12
(a)
∑r =
2
r =1
12(12 + 1) (2(12) + 1)
6
(12)(13)(25)
6
= 650
=
30
r2
∑=
(b)
30
9
∑ r2 − ∑ r2
=
r 10=
r 1=
r 1
30 (31) (61) 9(9 + 1)(18 + 1)
−
6
6
= 9455 − 285 = 9170
=
Try these 3.2
20
(a)
∑ r(r + 3)
r =1
20
20
∑ r + 3∑ r
=
2
r 1=
r 1
=
20 (21) (41)
 20 (21) 
+3
 = 3500
6
 2 
=
25
25
9
1) ∑ 2r(r + 1) − ∑ 2r(r + 1)
∑ 2r(r +=
(b)
=r 10=r 1 =r 1
25
25
9
9
= 2∑ r 2 + 2∑ r − 2∑ r 2 − 2∑ r
=r 1 =r =
1
r 1 =r 1
2(25)(26)(51) 2(25)(26) 2(9)(10)(19) 2(9)(10)
+
−
−
6
2
6
2
= 11 040
=
n
(c)
∑ r(r 2 + 2r)=
=r 1
n
n
∑ r 3 + 2∑ r 2
=r 1 =r 1
n 2 (n + 1) 2 2n(n + 1)(2n + 1)
=
+
4
6
n(n + 1)
=
[3n(n + 1) + 4(2n + 1)]
12
n(n + 1)
1
=
[3n 2 +=
11n + 4]
n(n + 1)(3n 2 + 11n + 4)
12
12
Exercise 3A
1
1 1 1 1 1
, , , ,
,
3 9 27 81 343
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 2 of 20
1
1
1
=
, u2 =
, u3
...
1
2
3
3
33
1
un = n
3
16, 13, 10, 7, 4, …
Sequence decreasing by 3
un = an + b
un = −3n + b
u1 = 16 ⇒ 16 = −3 + b
b = 19
un = −3n + 19
1
1
1
1
,
,
,
,
2 × 5 3 × 7 4 × 9 5 × 11
2, 3, 4, ….
n+1
5, 7, 9, …
2n + 3
1
∴un =
(n + 1) (2n + 3)
8 + 16 + 32 + 64 + 128 + 256 + 512
u = 2r+2
=
u1
2
3
4
7
∑2
r+2
r =1
5
9 + 12 + 15 + … + 30
ur = 3r + 6
8
∑ (3r + 6)
r=1
6
4 × 5 + 5 × 6 + 6 × 7 + … + 10 × 11
ur = (r + 3) (r + 4)
7
∑ (r + 3) (r + 4)
r =1
n
7
∑ (6r − 5)
r =1
un = 6n − 5
n
8
∑ (4r − 3)
2
r =1
un = 4n2 − 3
2n
9
∑ (r + r )
3
2
r =1
u n = n3 + n2
4n
10
∑ (6r + 2)
3
r =1
un = 6n3+ 2
n+2
11
∑3
2r − 1
r =1
12
un = 32n−1
u16 = 7(16) + 3 = 115
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 3 of 20
13
14
u8 = 3(9)2 − 1 = 242
1
1
r = 14, u10 =
=
4(14) − 2 54
25
15
25
25
∑ (r − 2)= ∑ r − ∑ 2
=r 1 =
r=1
r 1
25(26)
− 25(2)
2
= 275
=
30
16
30
30
+ 3) 6∑ r + ∑ 3
∑ (6r=
r=1
r 1
=r 1 =
= 6
(30) (31)
+ 30 (3)
2
= 2880
50
17
∑ r(r + 2)
r=1
50
50
∑ r + 2∑ r
=
2
r =1
r=1
50 (51) (101) 2 (50) (51)
+
6
2
= 45 475
=
10
18
∑ r (r + 4)
2
r=1
10
10
= ∑ r 3 + 4∑ r 2
r=1
r=1
2
(10) (11)2 4(10) (11) (21)
+
4
6
= 3025 + 1540
= 4565
=
45
19
∑ 6r (r +1)
r =1
45
45
r=1
r=1
= 6∑ r 2 + 6∑ r
6 (45) (46) (91) 6(45) (46)
=
+
6
2
= 194 580
12
20
12
4
∑ (r + 4)= ∑ (r + 4) − ∑ (r + 4)
=r 5 =r 1 =r 1
12
12
4
4
= ∑r+ ∑4− ∑r − ∑4
=
r 1=
r 1=
r 1=
r 1
12 (13)
(4) (5)
+ (4) (12) −
− (4) (4)
2
2
= 78 + 48 – 10 – 16
= 100
=
25
21
∑ (r − 3)
2
r =10
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 4 of 20
25
9
∑ (r 2 − 3) − ∑ (r 2 − 3)
=
=r 1 =r 1
25
25
9
9
= ∑ r2 − ∑ 3 − ∑ r2 + ∑ 3
=r 1 =r =
1
r 1 =r 1
(25) (26) (51)
9 (10) (19)
− (25) (3) −
+ (9) (3)
6
6
= 5192
=
30
∑ r(3r − 2)
22
r = 15
30
30
14
14
= 3∑ r 2 − 2∑ r − 3∑ r 2 + 2∑ r
=r 1 =r 1 =r 1 =r 1
3 (30) (31) (61) 2 (30) (31) 3 (14) (15) (29) 2 (14) (15)
−
−
+
6
2
6
2
= 24 600
=
40
23
∑ (2r +1) (5r + 2)
r=9
40
8
∑ (10r + 9r + 2) − ∑ (10r + 9r + 2).
=
2
2
=r 1 =r 1
40
40
40
8
=r 1 =r 1
=r 1
8
8
= 10∑ r 2 + 9∑ r + ∑ 2 − 10∑ r 2 − 9∑ r − ∑ 2
=r 1
=r 1 =r 1
10 (40) (41) (81) 9 (40) (41)
10 (8) (9) (17) 9 (8) (9)
+
+ (40) (2) −
−
− (8) (2)
6
2
6
2
= 226 480
=
n
24
∑ (r + 4)
r =1
n
n
∑r + ∑4
=
=r 1 =r 1
=
n(n +1)
+ 4n
2
n
= (n + 1 + 8)
2
1
= n (n + 9)
2
n
25
∑ 3r(r +1)
r =1
n
= ∑ (3r 2 + 3r)
r =1
n
n
= 3∑ r 2 + 3∑ r
=r 1 =r 1
3n(n + 1)(2n + 1) 3n(n + 1)
+
6
2
n(n + 1)
=
[2n + 1 + 3]
2
=
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 5 of 20
n(n +1)
(2n + 4)
2
= n (n + 1) (n + 2)
=
n
26
∑ 4r(r − 1)
r =1
n
n
= 4∑ r 2 − 4∑ r
=r 1 =r 1
4n(n + 1)(2n + 1) 4n(n + 1)
−
6
2
n(n +1)
=
[2(2n + 1) − 6]
3
n(n +1)
=
(4n − 4)
3
4n(n + 1) (n − 1)
=
3
=
n
27
∑ r (r + 3)
2
r=1
n
n
∑ r + 3∑ r
=
3
r=1
2
r=1
n 2 (n + 1) 2 3n(n + 1) (2n + 1)
+
4
6
n(n + 1)
=
[n(n + 1) + 2(2n +1)]
4
n(n + 1) 2
=
[n + 5n + 2]
4
=
2n
∑ 2r(r − 1)
28
r=n+1
2n
n
∑ (2r − 2r) − ∑ (2r − 2r)
=
2
2
r =1
r =1
2n
2n
n
n
r=1
r=1
r=1
r=1
= 2∑ r 2 − 2∑ r − 2∑ r 2 + 2∑ r
2(2n) (2n + 1) (4n +1) 2(2n)(2n + 1) 2n(n + 1)(2n +1) 2n(n + 1)
=
−
−
+
6
2
6
2
1
2

= n  (2n + 1)(4n +1) − 2(2n + 1) − (n + 1)(2n + 1) + (n + 1)
3
3

n
=  2(8n 2 + 6n +1) − (12n + 6) − (2n 2 + 3n + 1) + 3n + 3
3
n
2n(7n 2 − 1)
=
(14n 2 −=
2)
3
3
2n
29
∑ r(r + 4)
r=n+1
=
2n
n
r=1
r=1
∑ (r 2 + 4r) − ∑ (r 2 + 4r)
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 6 of 20
2n
2n
n
n
r=1
r=1
r=1
r=1
= ∑ r2 + 4 ∑ r − ∑ r2 − 4 ∑ r
2n (2n + 1) (4n + 1) 4 (2n) (2n +1) n(n + 1)(2n + 1) 4n(n +1)
=
+
−
−
6
2
6
2
n
=
[2(8n 2 + 6n + 1) + 24 (2n +1) − (2n 2 + 3n + 1) − 12 (n + 1)]
6
n
=
[14n 2 + 45n + 13]
6
2n
∑ (r + 1) (r − 1)
30
r=n+1
2n
n
∑ (r 2 − 1) − ∑ (r 2 − 1)
=
r=1
=
r=1
2n
2n
n
n
∑ r2 − ∑1 − ∑ r2 + ∑1
=r 1 =r =
1
r 1 =r 1
=
=
=
=
=
2n(2n + 1) (4n + 1)
n(n + 1) (2n + 1)
− 2n −
+n
6
6
2n(8n 2 + 6n + 1) n(2n 2 + 3n + 1)
−
−n
6
6
n
[16n 2 + 12n + 2 − 2n 2 − 3n − 1 − 6]
6
n
(14n 2 + 9n − 5)
6
n
(14n − 5)(n + 1)
6
Exercise 3B
n
1
RTP ∑ (3r −=
2)
r=1
1
n (3n − 1)
2
1
LHS = ∑ (3r − 2)= 3(1) − 2= 1 , RHS =
Proof: When n = 1,
r =1
∴
1
1
(1) (3(1) − 1) = × 2 = 1
2
2
LHS = RHS
n
1
2)
n(3n − 1)
∑ (3r −=
2
Hence when n=1,
r =1
Assume that the statement is true for n = k
k
1
i.e.
(3r −=
2)
k (3k − 1)
∑
2
r =1
RTP the statement true for n = k + 1
k +1
1
i.e. ∑ (3r − 2)=
(k + 1) (3(k + 1) − 1)
2
r =1
k +1
Proof:
k
=
∑ (3r − 2)
∑ (3r − 2) + 3(k + 1) − 2
r=1
r=1
1
=
k (3k − 1) + (3k + 1)
2
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 7 of 20
1
[3k 2 − k + 6k + 2]
2
1
= (3k 2 + 5k + 2)
2
1
= (k +1) (3k + 2)
2
1
=
(k + 1) (3(k + 1) − 1)
2
k +1
1
Hence ∑ (3r − 2)=
(k + 1) (3(k + 1) − 1)
2
r=1
n
1
∴ by PMI
(3r −=
2)
n (3n − 1)
∑
2
r=1
=
n
2
RTP ∑ (4r − 3)
= n (2n − 1)
r=1
Proof:
1
When n = 1, ∑ (4r − 3) = 4(1) − 3 = 4 − 3 = 1
r =1
RHS = 1(2(1)−1) = 2−1=1
⇒ LHS = RHS
n
Hence when n = 1, ∑ (4r − 3)= n(2n − 1)
r =1
k
Assume true for n = k i.e. ∑ (4r − 3) = k (2k − 1)
r=1
k +1
RTP true for n = k + 1 i.e. ∑ (4r − 3) = (k + 1) (2(k + 1) − 1)
r =1
k +1
k
∑ (4r − 3)= ∑ (4r − 3) + 4(k +1) − 3
Proof:
=r 1 =r 1
= k (2k − 1) + 4k + 1
= 2k2 – k + 4k + 1
= 2k2 + 3k + 1
= (2k + 1) (k + 1)
= (k + 1) (2(k + 1) − 1)
k +1
∑ (4r − 3) = (k + 1)(2(k +1) − 1)
∴
r=1
n
Hence by PMI
∑ (4r − 3)= n(2n − 1)
r =1
n
3
RTP ∑ (2r − 1) (2r) =
r=1
1
n(n + 1)(4n − 1)
3
Proof:
1
When n = 1, LHS = ∑ (2r − 1) (2r) =(2(1) − 1) (2(1)) =2
r=1
1
1
RHS = (1) (1 + 1)(4(1) − 1) =
×2 × 3 = 2
3
3
∴
LHS = RHS
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 8 of 20
k
1
(2r)
k(k + 1) (4k − 1)
∑ (2r − 1)=
3
Assume true for n = k, i.e.
r=1
k +1
1
(k +1) (k + 1 + 1) (4(k + 1) − 1)
3
RTP true for n = k + 1, i.e. =
∑ (2r − 1) (2r)
r=1
Proof:
k +1
k
r=1
r=1
∑ (2r − 1) (2r) = ∑ (2r − 1) (2r) + (2(k + 1) − 1) (2(k + 1))
1
=
k(k +1) (4k − 1) + (2k +1) (2k + 2)
3
1
=
(k +1) [k(4k − 1) + 6(2k + 1)]
3
1
=
(k +1) [4k 2 − k + 12k + 6]
3
1
=
(k +1) (4k 2 + 11k + 6)
3
1
=
(k +1) (k + 2) (4k + 3)
3
1
=
(k +1) (k + 1 + 1) (4(k + 1) − 1)
3
k +1
1
Hence
=
(2r − 1) (2r)
(k + 1) (k +1+1) (4(k + 1) − 1)
∑
3
r=1
n
1
Hence by PMI ∑ (2r − 1)
=
(2r)
n(n + 1) (4n − 1)
3
r=1
n
n(n +1) (n + 2) (3n +1)
4
RTP ∑ (r 2 + r 3 ) =
12
r=1
Proof:
1
When n = 1, LHS = ∑ (r 2 + r 3 ) = 12 +13 = 2
r=1
(1) (1 + 1) (1 + 2) (3(1) + 1) 2 × 3 × 4
=2
=
12
12
∴
LHS = RHS
n
n(n +1) (n + 2) (3n +1)
When n = 1, ∑ (r 2 + r 3 ) =
12
r=1
k
k(k +1) (k + 2) (3k +1)
Assume true for n = k i.e. ∑ (r 2 + r 3 ) =
12
r=1
k +1
(k + 1) (k + 1 + 1) (k + 1 + 2) (3(k + 1) + 1)
RTP true for n = k+1 i.e. ∑ (r 2 + r 3 ) =
12
r=1
RHS =
k +1
2
Proof: ∑ (r=
+ r3 )
r=1
k
∑ (r + r ) + (k + 1) + (k + 1)
2
3
2
3
r=1
k(k +1) (k + 2) (3k +1)
=
+ (k +1) 2 + (k +1)3
12
k +1
=
[k(k + 2) (3k +1) + 12(k +1) + 12(k +1) 2 ]
12
1
=
(k + 1) [3k 3 + 7k 2 + 2k + 12k + 12 + 12k 2 + 24k + 12]
12
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 9 of 20
1
(k + 1) (3k 3 + 19k 2 + 38k + 24)
12
1
=
(k + 1) (k + 2) (3k 2 + 13k +12)
12
1
=
(k +1) (k + 2) (k + 3) (3k + 4)
12
1
=
(k +1) ((k +1) + 1) ((k +1) + 2) (3(k +1) + 1)
12
n
n(n +1) (n + 2)(3n +1)
Hence by PMI ∑ (r 2 + r 3 ) =
12
r =1
=
n
5
RTP ∑ r 3 =
r=1
n 2 (n +1)2
4
1
3
3
Proof: when n = 1, LHS = ∑ r=
1=
1
r=1
(1) (1 + 1)
4
= = 1
4
4
∴
LHS = RHS
n
n 2 (n +1) 2
When n = 1, ∑ r 3 =
4
r=1
2
RHS=
2
k
Assume true for n = k i.e. ∑ r 3 =
r=1
k+1
RTP true for n = k + 1 i.e. ∑ r 3 =
r=1
k+1
Proof:
k 2 (k +1) 2
4
(k +1)2 ((k +1) + 1)2
4
k
∑ r = ∑ r + (k + 1)
3
r=1
3
3
r=1
k (k + 1)
+ (k + 1)3
4
(k + 1)2 2
=
[k + 4(k + 1)]
4
1
= (k + 1)2 (k 2 + 4k + 4)
4
1
= (k + 1) 2 (k + 2) 2
4
2
=
∴
2
n
∑ r3 =
by PMI
r =1
n 2 (n + 1)2
4
n
6
RTP
1
n
=
∑
n +1
r=1 r(r +1)
1
1
1
1
Proof: When n=1, LHS = ∑ = =
1(1+1) 2
r = 1 r(r +1)
RHS =
∴
1
1
=
1+1 2
LHS = RHS
n
1
n
=
n +1
r=1 r(r +1)
Hence when n=1, ∑
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 10 of 20
k
1
k
=
k +1
r = 1 r(r +1)
Assume true for n = k i.e. ∑
k +1
1
k +1
=
(k + 1) +1
r=1 r(r +1)
k+1
k
1
1
1
Proof: ∑
= ∑
+
(k +1) (k +1+1)
r=1 r(r +1)
r=1 r(r +1)
k
1
=
+
k +1 (k +1)(k + 2)
1 
1 
=
k+

k +1 
k + 2 
RTP true for n = k + 1 i.e. ∑
=
1  k (k + 2) + 1 


k +1  k + 2 
=
1  k 2 + 2k +1 


k +1  k + 2 
1 (k +1) 2
k +1 k + 2
k +1
k +1
= =
k + 2 (k +1) +1
=
k
Hence by PMI
1
n
∑ r(r +1) = n +1
r=1
(−1) n +1 (n)(n +1)
r+1 2
(
−
1)
r
=
∑
2
r=1
n
7
RTP
1
Proof: when n = 1, LHS = ∑ (−1) r +1 r 2 =
(−1) 2 (1) 2 =
1
r=1
(−1) (1) (2)
= 1
2
LHS = RHS
2
RHS
=
n
(−1) n +1 n(n +1)
Hence when n = 1, ∑ (−1) r+1 r 2 =
2
r=1
k
( −1) k +1 (k)(k +1)
Assume true for n = k, i.e. ∑ ( −1) r+1 r 2 =
2
r=1
k+2
k +1
( −1) (k +1)(k +1+1)
RTP true for n = k + 1 i.e. ∑ ( −1) r+1 r 2 =
2
r=1
k +1
Proof:
k
∑ (−1) r =∑ (−1) r + (−1)
r+1 2
r=1
( −1)
r+1 2
k+2
(k +1) 2
r=1
k+1
(k) (k +1)
+ ( −1)k+2 (k +1)2
2
(−1)k +1 (k +1)
=
[k + (−1)1 2(k + 1)]
2
(−1)k +1 (k +1)
=
[− k − 2]
2
=
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 11 of 20
( −1)k+1 (k +1) ( −1)1 (k + 2)
2
k+2
( −1) (k + 1) (k + 2)
=
2
=
(−1) n +1 (n)(n +1)
r +1 2
(
−
1)
r
=
∑
2
r =1
n
Hence by PMI
n
8
1
n(n + 3)
=
4(n +1)(n + 2)
r =1 r(r +1)(r + 2)
1
1
1
1
1
Proof: when n = 1, LHS
= ∑
=
= =
1(1 + 1) (1 + 2) 2 × 3 6
r =1 r(r +1)(r + 2)
(1) (1 + 3)
4
1
RHS
=
= =
4(1 + 1) (1 + 2) 4 × 2 × 3 6
∴
LHS = RHS
n
1
n(n + 3)
When n = 1, ∑
=
4(n + 1)(n + 2)
r = 1 r(r + 1) (r + 2)
RTP ∑
k
1
k(k + 3)
=
4(k + 1) (k + 2)
r = 1 r(r + 1) (r + 2)
Assume true for n = k i.e. ∑
1
(k +1) ((k +1) + 3)
=
4((k+1) + 1) ((k + 1) + 2)
r=1 r(r +1) (r +2)
k +1
RTP true for n = k + 1, i.e. ∑
k +1
k
1
1
1
=
+
∑
∑
(k + 1) (k + 2) (k + 3)
r = 1 r(r +1) (r + 2)
r = 1 r(r + 1) (r + 2)
k(k + 3)
1
=
+
4(k +1)(k + 2) (k +1) (k + 2) (k + 3)
Proof:
1
4 

k(k + 3) +

4(k +1) (k + 2) 
k + 3 
1
 k(k + 3) (k + 3) + 4 
=

4(k +1) (k + 2) 
k+3

=
=
 k 3 + 6k 2 + 9k + 4 
1


4(k +1) (k + 2) 
k+3

=
1
(k +1) (k 2 + 5k + 4)
4(k +1) (k + 2)
k+3
=
(k +1) (k + 4) (k +1)
4 (k +1) (k + 2) (k + 3)
(k +1) (k + 4)
(k +1) ((k +1 ) + 3)
=
4(k + 2) (k + 3) 4((k +1) + 1) ((k +1) + 2)
1
n(n + 3)
Hence by PMI ∑
=
4(n +1) (n + 2)
r =1 r(r +1) (r + 2)
n
1
n
RTP ∑
=
6n + 4
r=1 (3r − 1) (3r + 2)
=
9
Proof: when=
n = 1, LHS
1
1
1
1
=
∑ (3r=
− 1) (3r + 2) (3 − 1) (3 + 2) 10
r=1
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 12 of 20
1
1
=
6(1) + 4 10
⇒ LHS = RHS
RHS
=
n
When n = 1,
1
n
∑ (3r − 1)(3r + 2) = 6n + 4
r =1
k
1
k
1
k +1
∑ (3r − 1) (3r + 2) = 6k + 4
Assume true for n = k i.e.
r=1
k +1
∑ (3r − 1) (3r + 2) = 6(k +1) + 4
RTP true for n = k + 1 i.e.
r =1
k +1
Proof:
k
1
1
1
= ∑
+
∑
(3r − 1) (3r + 2)
(3r − 1) (3r + 2) (3(k +1) − 1) (3(k +1) + 2)
r=1
r=1
k
1
=
+
6k + 4 (3k + 2) (3k + 5)
k
1
=
+
2(3k + 2) (3k + 2) (3k + 5)
1
2 

=
k+

2(3k + 2) 
3k + 5 
=
 k(3k + 5) + 2 
1
2(3k + 2)  3k + 5 
=
 3k 2 + 5k + 2 
1


2(3k + 2)  3k + 5 
=
=
(3k + 2) (k +1)
2(3k + 5) (3k + 2)
k +1
k +1
=
6k + 10 6(k + 1) + 4
n
Hence by PMI
1
n
∑ (3r − 1) (3r + 2) = 6n + 4
r=1
10
11
RTP 34n – 1 = 16 A, A ∈ ℤ, n ≥ 1
Proof: when n = 1, LHS = 34(1) – 1 = 34 – 1 = 81 − 1 = 80
= 16(5)
∴ when n = 1, 34n – 1 is divisible by 16
Assume true for n = k, i.e. 34k −1 = 16A
RTP true for n = k + 1, i.e. 34(k+1) − 1 = 16B
Proof: 34k + 4 – 1
= 34k + 4 + 16A – 34k
= 34k × 34 − 34k + 16A
= 34k (34 − 1) + 16A
=34k (80) + 16A
= 16 [5(34k) + A]
= 16B, B = 5(34k) + A ∈ ℤ
Hence by PMI 34n – 1 is divisible by 16
RTP n4 + 3n2 = 4A, A ∈ ℤ, n ≥ 1
Proof: when n = 1, LHS = 14 + 3(1)2
=1+3=4
= 4 (1)
Unit 1 Answers: Chapter 3
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Page 13 of 20
∴ when n = 1, n4 + 3n2 is divisible by 4
Assume true for n = k, i.e. k4 + 3k2 = 4A
RTP true for n = k + 1, i.e. (k + 1)4 + 3 (k+1)2 = 4B
Proof:
(k+1)4 + 3(k+1)2
= k4 + 4k3 + 6k2 + 4k + 1 + 3k2 + 6k + 3
= (k4 + 3k2) + 4k3 + 4k + 4 + 6k2 + 6k
= 4A + 4k3 + 4k + 4 + 6k (k + 1)
= 4A + 4k3 + 4k + 4 + 6 (2c)
Since k(k + 1) is the product of two consecutive, integers
then k(k + 1) is divisible by 2. i.e. k(k + 1) = 2c
= 4 [A + k3 + k + 1 + 3c]
= 4B
Hence by PMI n4 + 3n2 is divisible by 4
Review exercise 3
1
6 × 7 + 8 × 10 + 10 × 13 + …
(a) un = (2n + 4) (3n + 4)
n
n
∑ u = ∑ (2r + 4) (3r + 4)
(b)
r
r=1
=
r=1
n
∑ (6r + 20r + 16)
2
r=1
n
n
n
r=1
r=1
r=1
= 6∑ r 2 + 20∑ r + ∑ 16
6n(n + 1)(2n + 1) 20n(n +1)
+
+ 16n
6
2
= n (n + 1) (2n + 1) + 10n (n + 1) + 16n
= n [2n2 + 3n + 1 + 10n + 10 + 16]
= n (2n2 + 13n + 27)
=
n
2
(a) ∑ r(3r − 2)
r=1
n
n
r=1
r=1
= 3∑ r 2 − 2∑ r
3n(n +1)(2n +1) 2n(n +1)
=
−
6
2
n(n + 1)
=
[2n + 1 − 2]
2
n(n + 1) (2n − 1)
=
2
20
20 (21) (39)
(b) (i) ∑ r(3r
=
− 2)
= 8190
2
r =1
100
(ii)
100
20
∑ r(3r − 2) = ∑ r(3r − 2) − ∑ r(3r − 2)
=
r 1=
r =1
r 1
(100) (101) (199)
− 8190
2
= 996 760
=
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 14 of 20
n
3
RTP
∑ 2r(r − 5) =
r=1
2n(n +1) (n − 7)
3
Proof:
When n=1,=
LHS
1
∑ 2r(r − 5)
r =1
= 2(1 − 5) = − 8
2(1) (1 + 1)(1 − 7) 2 × 2 × ( −6)
RHS =
=
= −8
3
3
∴ when n=1, LHS = RHS
n
2n(n + 1)(n − 7)
⇒ ∑ 2r(r − 5) =
when n = 1
3
r=1
k
2k(k + 1) (k − 7)
Assume true for n = k, i.e. ∑ 2r(r − 5) =
3
r=1
k +1
RTP true for n = k + 1, i.e. ∑ 2r(r − 5) =
r =1
2(k + 1) (k + 1 + 1) (k + 1 − 7)
3
Proof:
k +1
k
r =1
r=1
5) ∑ 2r(r − 5) + 2(k+1) (k + 1 − 5)
∑ 2r(r −=
2k(k + 1) (k − 7)
+ 2(k + 1) (k − 4)
3
2(k + 1)
=
[k(k − 7) + 3(k − 4)]
3
2(k + 1) 2
=
[k − 4k − 12]
3
2(k + 1)
=
(k + 2) (k − 6)
3
2(k+1) (k+1+1) (k+1 − 7)
=
3
n
2n(n + 1)(n − 7)
Hence by PMI ∑ 2r(r − 5) =
3
r =1
2n–1
+1
4
an = 3
an+1 = 32(n+1) –1 + 1 = 32n+1 + 1
an+1 − an = 32n+1 + 1 − 32n–1 – 1
= 32n+1 − 32n–1
= 32n–1 [32 − 1]
= 8 (32n–1)
RTP an = 32n–1 + 1 = 4A , A ∈ Z for all n ≥ 1
Proof: when n=1, 32-1 + 1 = 3 + 1 = 4(1)
∴ when n=1, an is divisible by 4
Assume true for n = k i.e. ak = 4A
RTP true for n = k + 1 i.e. ak+1 = 4B,
Proof: From above
an+1 − an = 8(32n−1)
⇒ ak+1 – ak = 8(32k−1)
⇒ ak+1 – 4A = 8(32k−1)
ak+1 = 4A + 8(32k−1)
= 4 [A + 2(32k−1)]
=
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
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= 4B
∴ ak+1 is divisible by 4
Hence by PMI an is divisible by 4
n
5
∑ 2r(r − 1)
2
r =1
n
n
r=1
2
r=1
2
= 2∑ r 3 − 2∑ r
2n (n + 1)
2n(n + 1)
−
4
2
n(n + 1)
=
[n(n + 1) − 2]
2
n(n + 1) 2
=
[n + n − 2]
2
n(n + 1)(n + 2)(n − 1)
=
2
Proof by induction:
n
n(n + 1)(n + 2)(n − 1)
RTP ∑ 2r(r 2 − 1) =
2
r=1
Proof: when n=1, LHS = 2(12 − 1) = 0
(1) (2) (3) (0)
RHS
=
= 0
2
∴ LHS = RHS
n
n(n + 1)(n − 1)(n + 2)
Hence when n = 1, ∑ 2r(r 2 − 1) =
2
r=1
k
k (k + 1) (k − 1) (k + 2)
Assume true for n = k i.e. ∑ 2r(r 2 − 1) =
2
r=1
k +1
(k + 1) (k + 1 + 1) (k + 1 − 1) (k + 1 + 2)
RTP true for n = k + 1 i.e. ∑ 2r(r 2 − 1) =
2
r=1
=
k +1
∑ 2r(r − 1)
2
Proof:
r=1
k
=
∑ 2r(r − 1) + 2(k + 1) ((k +1)2 − 1)
2
r=1
k(k + 1) (k − 1) (k + 2)
+ 2(k + 1) (k 2 + 2k)
2
k(k +1)
=
[(k − 1) (k + 2) + 4(k + 2)]
2
k(k + 1) (k + 2)
=
(k − 1 + 4)
2
k(k + 1) (k + 2)(k + 3) (k + 1) (k + 1 + 1) (k + 1 − 1) (k + 1 + 2)
=
2
2
k
n(n + 1) (n − 1) (n + 2)
Hence by PMI
2r(r 2 − 1) =
∑
2
r=1
6
an = 52n+1 + 1
an + 1 = 52(n + 1) + 1 + 1
= 52n+3 + 1
an + 1 − an = 52n+3 + 1 − 52n+1− 1
=
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 16 of 20
= 52n + 3 − 52n + 1
= 52n+1(52 − 1)
= (24) (52n + 1)
RTP: an = 6A, ∀n ≥ 0
Proof: when n = 0, a0 = 51 + 1 = 6 = 6 (1)
Hence when n = 0, an is divisible by 6
Assume true for n = k i.e. ak = 6A
RTP true for n = k + 1, i.e. ak+1 = 6B
Proof: ak+1 − ak = 24(52k+1), from above
⇒ ak+1 – 6A = 6(4) (52k+1)
ak+1 = 6A + 6(4) 52k+1
= 6 [A + 4(52k+1)]
= 6B
Hence by PMI an is divisible by 6. ∀n ≥ 0
n
7
∑ (6r + 2)
3
r=1
n
n
r=1
r=1
= 6∑ r 3 + ∑ 2
6n 2 (n + 1) 2
=
+ 2n
4
n
=
[3n(n +1) 2 + 4]
2
n
=
(3n 3 + 6n 2 + 3n + 4)
2
n
n
RTP ∑ (6r 3 + 2) = (3n 3 + 6n 2 + 3n + 4)
2
r=1
Proof: n = 1, LHS = 6(1)3 + 2 = 8
1
16
RHS =
(3 + 6 + 3 + 4) =
= 8
2
2
∴ LHS = RHS
n
n
Hence when n = 1, ∑ (6r 3 + 2) = (3n 3 + 6n 2 + 3n + 4)
3
r=1
k
k
Assume true for n = k, i.e. ∑ (6r 3 + 2)=
(3k 3 + 6k 2 + 3k + 4)
2
r=1
k+1
k +1
RTP true for n = k + 1 i.e. ∑ (6r 3 =
+ 2)
(3(k +1)3 + 6(k +1)2 + 3(k +1) + 4)
2
r=1
Proof:
k+1
k
2) ∑ (6r + 2) + 6(k +1) + 2
∑ (6r +=
3
r=1
3
3
r=1
k
(3k 3 + 6k 2 + 3k + 4) + 6(k +1)3 + 2
2
1
= [3k 4 + 6k 3 + 3k 2 + 4k +12(k 3 + 3k 2 + 3k +1) + 4]
2
1
= [3k 4 +18k 3 + 39k 2 + 40k +16]
2
1
=
(k +1) (3k 3 + 15k 2 + 24k +16)
2
=
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 17 of 20
1
(k +1)[3(k +1)3 + 6(k +1)2 + 3(k +1) + 4]
2
n
n
Hence by PMI ∑ (6r 3 + 2) = (3n 3 + 6n 2 + 3n + 4)
2
r=1
n
1
RTP ∑ (r + 4) = n(n + 9)
2
r=1
Proof:
when n = 1, L.H.S = 1 + 4 = 5
1
10
R.H.S = (1) (1 + 9) =
= 5
2
2
∴ LHS = RHS
n
1
when n = 1, ∑ (r + 4) = n(n + 9)
2
r=1
k
1
Assume true for n = k, i.e. ∑ (r + 4) = k(k+9)
2
r =1
k +1
1
RTP true for n = k+1, i.e. ∑ (r + 4) = (k+1) (k + 1 + 9)
2
r=1
=
8
Proof:
k+1
+ 4)
∑ (r=
k
∑ (r + 4) + (k+1+4)
r=1
r=1
1
= k(k + 9) + (k + 5)
2
1 2
=
[k + 9k + 2k + 10]
2
1
= [k 2 + 11k + 10]
2
1
= (k+1) (k+10)
2
1
= (k + 1) (k + 1 + 9)
2
9
n
1
Hence by PMI ∑ (r + 4) =n(n + 9)
2
r=1
n
4n (n +1) (n −1)
RTP ∑ 4r (r −1) =
3
r=1
Proof:
When n = 1, LHS = 4(1)(1 − 1) = 0
4(1) (1 +1) (1 −1) 4 × 2 × 0
RHS
=
= = 0
3
3
∴ LHS = RHS
n
4n (n +1) (n −1)
∴ ∑ 4r (r −1) =
3
r=1
k
4k (k +1) (k −1)
Assume true for n = k i.e. ∑ 4r (r −1) =
3
r=1
k +1
4(k +1) (k +1 + 1) (k + 1 − 1)
RTP true for n = k + 1
i.e. ∑ 4r (r −1) =
3
r =1
Proof:
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 18 of 20
k +1
− 1)
∑ 4r (r=
r =1
k
∑ 4r(r − 1) + 4(k + 1)(k + 1 − 1)
r=1
4 k (k + 1) (k − 1)
=
+ 4(k + 1) (k)
3
4
=
(k + 1) k [k − 1 + 3]
3
4
=(k + 1) (k) (k + 2)
3
4
=
(k + 1) (k + 1 + 1) (k + 1 − 1)
3
n
4
Hence by PMI ∑ 4r(r − =
1)
n(n + 1) (n − 1)
3
r =1
n
1
n
10 RTP ∑
=
(n + 1)
r=1 r(r + 1)
Proof:
1
1
when n = 1,=
LHS
=
1(1 + 1) 2
1
1
RHS
= =
1+1 2
L.H.S = RHS
n
1
n
∴ when n = 1, ∑
=
(n + 1)
r=1 r(r + 1)
k
1
k
=
(k + 1)
r=1 r(r + 1)
Assume true for n = k, i.e. ∑
1
k +1
=
k +1+1
r =1 r(r + 1)
k+1
k
1
1
1
Proof: ∑
= ∑
+
(k +1)(k + 2)
r=1 r(r +1)
r=1 r(r +1)
k
1
=
+
(k +1) (k +1) (k + 2)
k(k + 2) + 1
=
(k +1) (k + 2)
k +1
RTP true for n = k + 1, i.e. ∑
=
=
k 2 + 2k +1
(k +1) (k + 2)
(k +1) 2
k +1
=
(k +1) (k + 2) k + 2
n
Hence by PMI
1
n
∑ r(r +1) = n +1
r=1
n
11
RTP ∑ 3(2r −1 ) = 3(2n − 1)
r=1
Proof:
when n = 1, LHS = 3(21−1) = 3
RHS = 3(21 − 1) = 3
LHS = RHS
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 19 of 20
n
when n = 1, ∑ 3(2r −1 ) = 3(2n − 1)
r=1
k
Assume true for n = k, i.e. ∑ 3(2 r −1 ) = 3(2 k − 1)
r=1
k+1
RTP true for n = k + 1, i.e. ∑ 3(2 r −1 ) = 3(2 k+1 − 1)
r=1
k+1
k
r=1
r=1
Proof: ∑ 3(2 r −1 ) = ∑ 3(2 r −1 ) + 3(2 k+1−1 )
= 3 (2k − 1) + 3 (2k)
= 3[2k – 1 + 2k]
= 3[2 × 2k − 1]
= 3 (2k+1 − 1)
n
by PMI ∑ 3(2r − 1 ) = 3(2n − 1)
r=1
12
13
RTP
n (n2 + 5) = 6A, n ∈ ℤ+
Proof:
when n = 1,
n (n2 + 5)
2
= 1(1 + 5) = 6
= 6(1)
Hence when n = 1, n (n2 + 5) is divisible by 6
Assume true for n = k, i.e. k (k2 + 5) = 6A
RTP true for n = k +1, i.e. (k+1) ((k + 1)2 + 5) = 6B
Proof:
(k + 1) ((k + 1)2 + 5)
= k ((k + 1)2 + 5) + (k + 1)2 + 5
= k (k2 + 2k +6) + (k2 + 2k + 6)
= k (k2 + 5) + k (2k + 1) + k2 + 2k + 6
= k (k2 + 5) + 3k2 + 3k + 6
= k (k2 + 5) + 3k (k + 1) + 6
Since k (k + 1) is the product of two consecutive integers,
k (k + 1) is an even number and hence divisible by 2
∴ 3k (k + 1) is divisible by 6
= 6A + 6C + 6
= 6 (A + C + 1)
= 6B
Hence by PMI n (n2 + 5) is divisible by 6 for all positive integers n
RTP n5 – n = 5A
Proof:
When n = 1, 15 – 1 = 0 which is divisible by 5
Hence when n = 1, n5 – n = 5A
Assume true for n = k
i.e.
k5 – k = 5A
RTP true for n = k + 1, i.e.
(k + 1)5 – (k + 1) = 5B
Proof:
(k + 1)5 – (k + 1)
= k5 + 5k4 + 10 k3 + 10 k2 + 5k + 1 – k − 1
= (k5 – k) + 5 k4 + 10 k3 + 10 k2 + 5 k
= 5 A + 5 (k4 + 2 k3 + 2 k2 + k)
= 5 [A + k4 + 2 k3 + 2 k2 + k]
=5B
Hence by PMI n5 – n is divisible by 5 for any positive integers n
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 20 of 20
k
14
1
1
1
=
−
4 k+4
n =1 (n + 4) (n + 3)
Proof:
1
1
1
k = 1, LHS =
= =
(1 + 4) (1 + 3) 5 × 4 20
1 1 5−4 1
RHS = − =
=
4 5
20
20
LHS = RHS
k
1
1
1
when k = 1, ∑
=
−
4 k+4
n =1 (n + 4) (n + 3)
RTP ∑
r
1
1
1
=
−
4 r+4
n=1 (n + 4) (n + 3)
r+1
1
1
1
RTP true for k = r + 1, i.e. ∑
=
−
4 (r + 1) + 4
n =1 (n + 4) (n + 3)
Assume true for k = r, i.e. ∑
r+1
Proof:
1
r
1
1
∑=
∑ (n + 4) (n + 3) + (r + 5) (r + 4)
(n + 4) (n + 3)
n 1=
n 1
=
1
1
1
−
+
4 r + 4 (r + 5) (r + 4)
1
(r + 5) − 1
1
r+4
1
1
= −
=
−
=
−
4 (r + 5) (r + 4) 4 (r + 4) (r + 5) 4 (r + 1) + 4
=
Unit 1 Answers: Chapter 3
© Macmillan Publishers Limited 2013
Page 1 of 22
Chapter 4 Polynomials
Try these 4.1
(a)
(b)
(x – 2) (x + 1) (x + 3)
= (x2 – x – 2) (x + 3)
= x3 + 3x2 – x2 – 3x – 2x – 6
= x3 + 2x2 – 5x – 6
⇒ x3 + 2x2 – 5x – 6 = Ax3 + Bx2 + Cx + D
⇒ A = 1, B = 2, C = – 5, D = – 6
2x2 – 3x + 1 = A (x + 1)2 + Bx + C
= A (x2 + 2x + 1) + Bx + C
= Ax2 + 2Ax + Bx + A + C
Equating coefficients
⇒A=2
2A + B = – 3
4+B=–3
B=–7
A+C=1
2+C=1
C = –1
∴ A = 2, B = – 7, C = – 1
Exercise 4A
1
2
x2 + x + b ≡ (x + b) (x – 2) + a
x2 + x + b = x2 – 2x + bx – 2 b + a
Equating coefficients of x
⇒1=–2+b⇒b=3
Equating constants:
b = – 2b + a
3=–6+a
a=9
∴ a = 9, b = 3
4x2 + 6x + 1 = p (x + q) 2 + r
4x2 + 6x + 1 = p [x2 + 2qx + q2] + r
Coeff of x2 ⇒ 4 = p
Coeff of x ⇒ 6 = 2pq, 6 = 2 (4) q
6 3
q= =
8 4
Constants ⇒ 1 = pq2 + r
2
⇒
=
1 4 3 +r
4
9
5
r =−
1
=
−
4
4
3
−5
Hence=
p 4,=
q =
, r
4
4
8x3 + 27x2 + 49x + 15 = (ax + 3) (x2 + bx + c)
= ax3 + abx2 + acx + 3x2 + 3bx + 3c
( )
3
Unit 1 Answers: Chapter 4
(8x + 3) (x2 + 3x + 5)
© Macmillan Publishers Limited 2013
Page 2 of 22
= ax3 + x2 (ab + 3) + x (ac + 3b) + 3c
4
5
6
7
8
9
10
8x3 + 24x2 + 40 x + 3x2 + 9x + 15
= 8x2 + 27x2 + 49x + 15
Equating coeff of x3: a = 8
Equating coeff of x2: ab + 3 = 27 ⇒ 8b + 3 = 27 ⇒ b = 3
Coeff of x: ac + 3b = 49 ⇒ 8c + 9 = 49 ⇒ c = 5
Hence a = 8, b = 3, c = 5
x3 + px2 – 7x + 6 = (x – 1) (x – 2) (qx + r)
Coeff of x3 ⇒ q = 1
Constants ⇒ 6 = (– 1) (– 2) r ⇒ r = 3
Coeff of x2 ⇒ p = − 3q + r ⇒ p = −3 + 3 = 0
Hence p = 0, q = 1, r = 2
2x3 + 7x2 – 7x – 30 = (x – 2) (ax2 + bx+ c)
= ax3 + bx2 + cx – 2ax2 – 2bx – 2c
= ax3 + x2 (b – 2a) + x (c – 2b)– 2 c
Equating coeff of x3 ⇒ a = 2
Equating coeff of x2 ⇒ b – 2a = 7 ⇒ b = 11
Equating coeff of x ⇒ c – 2b = − 7 ⇒ c = 15
Hence a = 2, b = 11, c = 15
x3 – 3x2 + 4x + 2 ≡ (x – 1) (x2 – 2x + a) + b
= x3 – 2x2 + ax – x2 + 2x – a + b
= x3 – 3x2 + x (a + 2) + b – a
Equating coeff of x ⇒ 4 = a + 2
a=2
Equating constants ⇒ 2 = b – a
b=4
Hence a = 2 , b = 4
4x3 + 3x2 + 5x+ 2 = (x + 2) (ax2 + bx + c)
Equating coeff of x3: a = 4
Equating constants: 2 = 2c ⇒ c = 1
Coeff of x2: 3 = 2a + b ⇒ 3 = 8 + b
b=−5
Hence a = 4, b = −5, c = 1
2x3 + Ax2 – 8x – 20 = (x2 − 4) (Bx + C)
= Bx3 + Cx2 – 4Bx – 4 C
Coeff of x3 ⇒ B = 2
Coeff of x2 ⇒ C = A
Coeff of x ⇒ − 8 = − 4B ⇒ B = 2
Constants ⇒ − 20 = − 4C, C = 5
Hence A = 5, B = 2, C = 5
ax3 + bx2 + cx + d = (x + 2) (x + 3) (x + 4)
= (x2 + 5x + 6) (x + 4)
= x3 + 4x2 + 5x2 + 20 x + 6x + 24
= x3 + 9x2 + 26x + 24
∴ a = 1, b = 9, c = 26, d = 24
ax3 + bx2 + cx + d = (4x + 1) (2x – 1) (3x + 2)
= (8x2 – 2x – 1) (3x + 2)
= 24x3 + 16x2 – 6x2 – 4x – 3x – 2
= 24x3 + 10x2 – 7x – 2
∴ a = 24, b = 10, c = − 7, d = − 2
11
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 3 of 22
4 x2 + 5 x + 12
x − 2 4x 3 − 3x 2 + 2x + 1
−( 4x 3 − 8x 2 )
5x 2 + 2x + 1
5x 2 − 10x
12x + 1
12x − 24
25
∴ Quotient is 4x + 5x + 12
Remainder is 25.
2
12
5 x2 + 4 x + 10
x − 2 5x 3 − 6x 2 + 2x + 1
−( 5x 3 − 10x 2 )
4x 2 + 2x + 1
4x 2 − 8x
10x + 1
10x − 20
21
5x − 6x + 2x + 1
21
= 5x 2 + 4x + 10 +
x−2
x−2
∴ A = 5, B = 4, C = 10
∴
13
3
2
x3 − 2 x2 − 2x + 3
x + 1 x − 2x 4 − x 3 + x 2 + x + 1
2
5
−( x 5 + x 3 )
− 2x 4 − 2x 3 + x 2 + x + 1
− 2x 4 − 2x 2
− 2x 3 + 3x 2 + x + 1
− 2x 3 − 2x
3x 2 + 3x + 1
3x 2 + 3
3x − 2
∴ Quotient is x – 2x – 2x + 3
Remainder is 3x – 2
2
3
(a)
+
x +1 x − 2
2 (x − 2) + 3 (x + 1)
=
(x + 1) (x − 2)
2x − 4 + 3x + 3
=
(x + 1) (x − 2)
3
14
Unit 1 Answers: Chapter 4
2
© Macmillan Publishers Limited 2013
Page 4 of 22
5x − 1
(x + 1) (x − 2)
x + 1 2x + 1
−
x + 3 2x − 4
=
(b)
=
(x + 1) (2x − 4) − (2x + 1) (x + 3)
(x + 3) (2x − 4)
2x 2 − 2 x − 4 − 2 x 2 − 7 x − 3
(x + 3) (2x − 4)
− 9x − 7
=
(x + 3) (2x − 4)
x
x −1
−
2
x + 2x + 1 x + 2
=
(c)
=
x (x + 2) − (x − 1) (x 2 + 2x + 1)
(x 2 + 2x + 1) (x + 2)
=
x 2 + 2x − x 3 − 2x 2 − x + x 2 + 2x + 1
(x 2 + 2x + 1) (x + 2)
− x 3 + 3x + 1
(x 2 + 2x + 1) (x + 2)
3x + 4
x
x+2
−
+
x − 1 x + 1 2x + 1
(3x + 4) (x + 1) (2x + 1) − x (x − 1) (2x + 1) + (x + 2) (x − 1) (x + 1)
=
(x − 1) (x + 1) (2x + 1)
=
(d)
=
6x 3 + 17x 2 + 15x + 4 − 2x 3 + x 2 + x + x 3 + 2x 2 − x − 2
(x − 1) (x + 1) (2x + 1)
=
5x 3 + 20x 2 + 15x + 2
(x + 1) (x − 1) (2x + 1)
x2
x2
−
2− x 3−x
(e)
=
=
x 2 (3 − x) − x 2 (2 − x)
(2 − x) (3 − x)
3x 2 − x 3 − 2x 2 + x 3
(2 − x) (3 − x)
=
x2
(2 − x) (3 − x)
Try these 4.2
(a)
Let f (x) = 6x3 – 3x2 + x – 2
(i) f (2) = 6(2)3 – 3(2)2 + 2 – 2
= 48 – 12 + 2 – 2
= 36
(ii) f (− 1) = 6(− 1)3 – 3(− 1)2 + (− 1) – 2
=−6–3–1–2
= − 12
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 5 of 22
3
(b)
(c)
2
1
 1
 1
 1
(iii) f =
6  −3  + − 2

 2
 2
 2
2
3 3 1
3
= − + − 2 =−
4 4 2
2
f(x) = x4 + ax2 – 2x + 1
f(1) = 4
⇒1+a–2+1=4
a=4
f(x) = x3 – 4x2 + ax + b
1
1
1
f   =1⇒ − 1 + a + b =1
8
2
2
1
15
…(1)
∴ a+b=
2
8
f(1) = 2 ⇒ 1 – 4 + a + b = 2
a + b = 5 …(2)
1
25
(2) – (1) ⇒ a =
2
8
25
1
=
a = 6
4
4
1
6 +b=5
4
−5
b=
4
1
1
∴a = 6 , b = − 1
4
4
Exercise 4B
1
(a)
(b)
(c)
f(x) = ax4 + 3x2 – 2x + 1
f(1) = a + 3 – 2 + 1
= a + 2.
f(x) = 3x3 + 6x2 – 7x + 2
f(−1) = 3 (− 1)3 + 6 (− 1)2 – 7 (− 1) + 2
= 12
f(x) = x5 + 6x2 – x + 1
5
2
 1  1
 1
 1
f  −  =−
+ 6 −  − −  +1


 2  2
 2
 2
95
=
32
(d) f(x) = (4x + 2) (3x2 + x + 2) + 7
f (2) = (10) (12 + 2 + 2) + 7
= 167
(e) f (x) = x7 + 6x2 + 2
f (− 2) = (− 2)7 + 6 (− 2)2 + 2
= − 102
(f) f(x) = 4x3 – 3x2 + 5
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 6 of 22
 3
 − 3
 − 3
f  −=
4
−3
+5


 2
 2 
 2 
− 61
=
4
(g) f(x) = 3x4 – 4x3 + x2 + 1
f (3) = 3(3)4 – 4(3)3 + (3)2 + 1
= 145
f(2) = a
⇒ 22 – a (2) + 2 = a
6 = 3a
a=2
f(x) = 5x2 – 4x + b
 1
f−  =
2
 2
3
2
3
2
2
4
5
 1
 1
⇒ 5−  − 4 −  + b =
2
 2
 2
−5
b=
4
f(x) = 3x3 + ax2 + bx + 1
f(1) = 2
⇒3+a+b+1=2
a+b=−2
f(2) = 13
⇒3(2)3 + a (2)2 + b(2) + 1 = 13
⇒ 4a + 2b = −12
2a + b = −6
[2] – [1] ⇒ a = − 4
b=2
∴ a = − 4, b = 2
f(x) = x3 + px2 + qx + 2
f(−1) = − 3
⇒−1+p−q+2=−3
p – q = −4
f(2) = 54
(2)3 + p(2)2 + q (2) + 2 = 54
4p + 2q = 44
2p + q = 22
[1] + [2] ⇒ 3p = 18, p = 6
q = 10
∴ f(x) = x3 + 6x2 + 10x + 2
3
6
[1]
[2]
[1]
[2]
2
− 13
 − 1  − 1
 − 1
 − 1
+ 6   + 10   =
+2
f =



 2   2 
 2 
 2 
8
3
2
f(x) = 2x – 3x – 4x + 1
f(a) = f(− a)
2a3 – 3a2 – 4a + 1 = − 2a3 – 3a2 + 4a + 1
⇒ 4a3 – 8a = 0
⇒ 4a (a2 − 2) = 0
a = 0, a2 = 2 ⇒ a = 2 , − 2
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 7 of 22
7
8
9
10
Hence a = 0, 2 , − 2
f(x) = 2x3 – x2 – 2x – 1
f(2) = 2 f(2a)
⇒ 2(2)3 – (2)2 – 2 (2) – 1 = 2 [2(2a)3 – (2a)2 – 2 (2a) – 1]
⇒ 16 – 4 – 4 – 1 = 32 a3 – 8a2 – 8a – 2
32 a3 – 8a2 – 8a – 9 = 0
f (x) = 2x3 – 5x2 – 4x + b
f (− 2) = 2 f(1)
2 (− 2)3 – 5 (− 2)2 – 4 (− 2) + b = 2 [2 – 5 – 4 + b]
⇒ − 16 – 20 + 8 + b = − 14 + 2b
b = − 14
f (x) = x3 + (λ + 5) x + λ
f(1) + f (− 2) = 0
f (1) = 1 + λ + 5 + λ = 2 λ + 6
f (− 2) = − 8 – 2 (λ + 5) + λ = − λ − 18
2λ + 6 − λ − 18 = 0
λ − 12 = 0
λ = 12
f(x) = 3x3 + kx2 + 15
f(3) = 3 (3)3 + k(3)2 + 15
= 9k + 96
3
2
1
1
1
f   =3   + k   + 15
3
3
 
 
3
1
136
=
k+
9
9
1  1
f(3) = f  
3  3
1
136
9K + 96=
k+
27
27
1
136
9K −
k=
− 96
27
27
242
− 2456
k=
27
27
− 1228
k=
121
11
3x + (p − 6)
x 2 + 2x + 3 3x 3 + px 2 + qx + 2
− 3x 3 + 6x 2 + 9x
(p − 6) x 2 + (q − 9) x + 2
(p − 6) x 2 + 2 (p − 6) x + 3 (p − 6)
[(q − 9) − 2 (p − 6)] x + 2 − 3(p − 6)
∴ (q – 9 – 2p + 12) x + (20 – 3p) = x + 5
∴ 20 – 3p = 5 ⇒ p = 5
q – 2p + 3 = 1
q – 10 + 3 = 1
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 8 of 22
12
q=8
p= 5, q = 8
f (x) = 8x3 + px2 + qx + 2
1 7
f  =
2 2
3
13
2
7
 1
 1
 1
⇒8  + P  + q  + 2=
 2
 2
 2
2
1
1
1
⇒ p+ q=
4
2
2
f (−1) = −1
⇒ 8(−1)3 + p(−1)2 + q(−1) + 2 = − 1
p–q=5
1
p+q=1
2
3
[1] +[2] ⇒ p= 6
2
p=4
q = −1
∴ p = 4, q = − 1
f (x) = 6x5 + 4x3 − ax + 2
f(− 1) = 15
6(−1)5 + 4(−1)3 – a (− 1) + 2 = 15
⇒– 6 – 4 + a + 2 = 15
a = 23
∴ f (x) = 6x5 + 4x3 – 23x + 2
f (2) = 6 (2)5 + 4 (2)3 – 23 (2) + 2
= 180
Remainder is 180
[1]
[2]
Try these 4.3
(a)
Let f (x) = 3x3 – x2 – 3x + 1
(i)
f (1) = 3 – 1 – 3 + 1 = 0
Since f (1) = 0 ⇒ x – 1 is a factor of f (x)
3
(ii)
3
(iii)
2
 1
 1  1
 1
f  −  =3  −  −  −  − 3  −  + 1
 2
 2  2
 2
−3 1 3
=
− + +1
8
4 2
15
=
8
 1
Since f  −  ≠ 0 ⇒ 2x + 1 is not a factor of f (x)
 2
2
 1
 1  1
 1
f   = 3  −  −3  +1
 3
 3  3
 3
1 1
= − −1+1
9 9
=0
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 9 of 22
(b)
 1
Since f   = 0 ⇒ 3x – 1 is a factor of f (x)
 3
3
Let f (x) = 4x + px2 – qx – 6
3
2
 1
 1
 1
 1
f −  = 4 −  + p −  − q −  − 6
 4
 4
 4
 4
1
1
1
= −
+
p+ q–6
16 16
4
1
1
1
p+ q–6
=0
∴
16
4
16
1
1
97
p+ q=
16
4
16
p + 4q = 97
[1]
f (1) = − 20
⇒ 4 – p + q – 6 = − 20
p–q=−8
[2]
[1] – [2] ⇒ 5q = 115
115
q=
= 23
5
p + 92 = 97
p=5
Hence p = 5, q = 23
Try these 4.4
(a)
Let f (x) = 2x3 + 5x2 – 4x – 3
f(1) = 2(1)3 + 5(1)2 – 4(1) − 3
=2+5–4–3=0
Since f(1) = 0 ⇒ x – 1 is a factor of f(x)
2 x2 + 7x + 3
3
x − 1 2 x + 5x2 − 4x − 3
2 x3 − 2 x 2
7 x2 − 4 x − 3
7 x2 − 7 x
3x − 3
3x − 3
0
∴ 2x + 5x – 4x – 3 = (x − 1) (2x2 + 7x + 3)
= (x − 1) (2x + 1) (x + 3)
⇒ (x − 1) (2x + 1) (x + 3) = 0
∴ x – 1 = 0, 2x + 1 = 0, x + 3 = 0
1
Hence x = 1, − , − 3
2
f(x) = x3 – 2x2 – 5x + 6
f(1) = 1 – 2 – 5 + 6 = 0
∴ x – 1 is a factor of f (x)
3
(b)
2
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 10 of 22
x2 − x − 6
x − 1 x − 2 x 2 − 5x + 6
3
x3 − x2
− x 2 − 5x + 6
− x2 + x
− 6x + 6
− 6x + 6
0
∴ f(x) = (x − 1) (x2 – x − 6) = (x − 1) (x − 3) (x + 2)
f(x) = 0 ⇒ x – 1 = 0, x – 3 = 0, x + 2 = 0
x = 1, 3, − 2
∴ factors of f (x) are (x − 1), (x − 3), (x + 2)
Roots of f (x) = 0 are 1, 3, − 2
Exercise 4c
1
f(x) = x3 + 2x2 – x – 2.
(a)
By trial and error:
f(1) = 1 + 2 – 1 – 2 = 0
∴ x – 1 is a factor of f (x)
x 2 + 3x + 2
3
x − 1 x + 2x 2 − x − 2
x3 − x 2
3x 2 − x − 2
3x 2 − 3x
2x − 2
2x − 2
(b)
0
∴ f(x) = (x − 1) (x2 + 3x + 2)
= (x − 1) (x + 1) (x + 2)
f(x) = x3 + 6 x2 + 11 x + 6
f(− 2) = (− 2)3 + 6 (− 2)2 + 11 (− 2) + 6
= −8 + 24 – 22 + 6
=0
∴ x + 2 is a factor of f (x)
x2 + 4 x + 3
x + 2 x 3 + 6 x 2 + 11x + 6
x3 + 2 x2
4 x 2 + 11x + 6
4x 2 + 8x
3x + 6
3x + 6
0
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 11 of 22
(c)
∴ f (x) = (x + 2) (x2 + 4x + 3)
= (x + 2) (x + 1) (x + 3)
f (x) = x3 – 7x + 6
f (1) = 1 – 7 + 6 = 0
∴ x – 1 is factor of f (x)
x2 + x − 6
x − 1 x 3 − 7x + 6
x3 − x 2
x 2 − 7x + 6
x2 − x
− 6x + 6
−6x + 6
(d)
0
∴ f (x) = (x − 1) (x2 + x − 6)
= (x − 1) (x + 3) (x − 2)
f (x) = x3 – 4 x2 + x + 6
f (2) = 23 – 4 (2)2 + 2 + 6
= 8 – 16 + 2 + 6
=0
⇒ x – 2 is a factor of f(x)
x2 − 2 x − 3
3
x − 2 x − 4x2 + x + 6
x3 − 2 x 2
− 2x2 + x + 6
− 2x2 + 4x
− 3x + 6
− 3x + 6
0
f (x) = (x − 2) (x – 2x − 3)
= (x − 2) (x − 3) (x + 1)
f (x) = x3 – 7x − 6
f (−1) = – 1 + 7 – 6 = 0
x + 1 is a factor of f (x)
x2 − x − 6
x + 1 x3 − 7 x − 6
2
(e)
x3 + x 2
− x2 − 7x − 6
− x2 − x
− 6x − 6
−6 x − 6
(f)
0
∴f (x) = (x + 1) (x2 – x − 6)
= (x + 1) (x − 3) (x + 2)
f(x) = 6x3 + 31 x2 + 3x −10
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 12 of 22
f(−5) = 6(−5)3 + 31 (−5)2 + 3 (−5) – 10
= –750 + 775 – 15 – 10
=0
6 x2 + x − 2
x + 5 6 x 3 + 31x 2 + 3 x − 10
6 x 3 + 30 x 2
x 2 + 3 x − 10
x 2 + 5x
− 2 x − 10
− 2 x − 10
0
∴ f (x) = (x + 5) (6x + x − 2)
= (x + 5) (3x + 2) (2x − 1)
3x3 + x2 – 20x + 12 = 0
f(x) = 3x3 + x2 – 20x + 12
f(2) = 3(2)3 + (2)2 – 20(2) + 12
= 24 + 4 – 40 + 12 = 0
∴ x – 2 is a factor of f(x)
Now 3x3 + x2 – 20 x + 12 = (x − 2) (a x2 + bx + c)
Equating coeff:
x3 ⇒ 3 = a
x2 ⇒ 1 = −2 a + b ⇒ b = 7
Constants ⇒ 12 = − 2 c ⇒ c = − 6
∴ 3 x3 + x2 – 20 x + 12 = (x − 2) (3x2 + 7x − 6)
= (x − 2) (3 x − 2) (x + 3)
∴ (x − 2) (3x − 2) (x + 3) = 0
⇒ x – 2 = 0, 3x – 2 = 0, x + 3 = 0
2
Hence x = 2, x = , x = − 3
3
2x3 + 13 x2 + 17x – 12 = 0
f (x) = 2x3 + 13x2 + 17x – 12
f (− 4) = 2 (− 4)3 + 13 (− 4)2 + 17 (− 4) – 12
= − 128 + 208 – 68 – 12
=0
∴ x + 4 is a factor of f (x)
2x3 + 13x2 + 17x – 12 = (x + 4) (ax2 + bx + c)
a = 2, 4c = − 12 ⇒ c = − 3
13 = 4a + b ⇒ b = 5
∴ f (x) = (x + 4) (2x2 + 5x - 3)
= (x + 4) (2x − 1) (x + 3)
(x + 4) (2x − 1) (x + 3) = 0
⇒ x + 4 = 0, 2x – 1 = 0, x + 3 = 0
1
x = − 4, , − 3
2
2x3 – 11x2 + 3x + 36 = 0
f(x) = 2x3 – 11x2 + 3x + 36
f(4) = 2(4)3 – 11(4)2 + 3(4) + 36
= 128 – 176 + 12 + 36
=0
2
2
(a)
(b)
(c)
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 13 of 22
(d)
3
(a)
(b)
(c)
(d)
x – 4 is a factor of f (x)
Now 2x3 – 11x2 + 3x + 36 = (x – 4) (ax2 + bx + c)
x3: a = 2
Constants: 36 = − 4 c ⇒ c = − 9
x2: − 11 = − 4a + b ⇒ b = − 3
2x3 – 11x2 + 3x + 36 = (x – 4) (2x2 – 3x – 9)
= (x – 4) (2x + 3) (x – 3)
(x – 4) (2x + 3) (x – 3) = 0
⇒ x – 4 = 0, 2x + 3 = 0, x – 3 = 0
−3
x = 4, x =
,x=3
2
f (x) = 3x3 + 10x2 + 9x + 2
f (− 1) = 3 (− 1)3 + 10 (− 1)2 + 9 (− 1) + 2
= − 3 + 10 – 9 + 2
= 0.
x + 1 is a factor of f (x)
3x3 + 10x2 + 9x + 2 = (x + 1) (ax2 + bx + c)
a = 3, c = 2
10 = a + b ⇒ b = 7
∴ 3x3 + 10x2 + 9x + 2 = (x + 1) (3x2 + 7x + 2)
= (x + 1) (3x + 1) (x + 2)
(x + 1) (3x + 1) (x + 2) = 0
x + 1 = 0, 3x + 1 = 0, x + 2 = 0
1
x = − 1, x = − , x = − 2
3
f (x) = x3 + 2x2 + 2x + 1
f (− 1) = (− 1)3 + 2 (− 1)2 + 2 (− 1) + 1
=–1+2–2+1=0
True
f (x) = 2x4 + 3x2 – x + 2
f(1) = 2 + 3 – 1 + 2 = 6
Since f (1) ≠ 0
False
f(x) = x5 – 4x4 + 3x3 – 2x2 + 4
f(2) = (2)5 – 4(2)4 + 3(2)3 – 2 (2)2 + 4
= 32 – 64 + 24 – 8 + 4
= − 12
False
f(x) = 2x3 + 9x2 + 10x + 3
3
4
5
2
 1
 1
 1
 1
f  −  = 2  −  + 9  −  + 10  −  + 3 = 0
 2
 2
 2
 2
True
f (x) = x3 – 3x2 + kx + 2
f(1) = 1 – 3 + k + 2 = 0
⇒ k = 0.
f (x) = x3 – 12x + 16
f(2) = (2)3 – 12 (2) + 16
= 0.
∴ x – 2 is a factor of f(x)
x3 – 12x + 16 = (x − 2)(ax2 + bx + c)
a = 1, − 2c = 16
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 14 of 22
6
7
8
c=−8
0 = − 2a + b, b = 2a = 2
∴ x3 – 12x + 16 = (x − 2) (x2 + 2x – 8)
= (x − 2) (x + 4) (x− 2)
factors are (x − 2) (x + 4) (x − 2)
f(x) = 4x3 – 3x2 + 5x + k
f (− 1) = 0
⇒ 4 (− 1)3 – 3 (− 1)2 + 5 (− 1) + k = 0
⇒−4–3–5+k=0
k = 12
f(x) = x3 + 3x2 – 6x – 8
f (2) = (2)3 + 3(2)2 – 6(2) – 8
= 8 + 12 – 12 − 8
=0
x3 + 3x2 – 6x – 8 = (x − 2) (x2 + 5x + 4)
= (x − 2) (x + 1) (x + 4)
(x − 2) (x + 1) (x + 4) = 0
x = 2, − 1, − 4
f(x) = 3x3 – kx2 + 5x + 2
 1
f  =0
 2
3
9
10
2
1
1
1
⇒3  − k   + 5  + 2 =
0
2
2
2
3 1
5
⇒ − k+ +2=
0
8 4
2
1
39
k=
4
8
39
k=
2
x3 + 5x2 + 3x – 1 = 0
f(x) = x3 + 5x2 + 3x – 1
f(− 1) = − 1 + 5 – 3 – 1 = 0
x3 + 5x2 + 3x – 1 = (x + 1) (x2 + 4x − 1)
(x + 1) (x2 + 4x − 1) = 0
x = − 1, x2 + 4x – 1 = 0
− 4 ± 20
x=
2
−4±2 5
=
2
=− 2 ± 5
(a)
f(x) = 4x3 + ax2 + bx + 3
f (− 3) = 4 (− 3)3 + a (− 3)2 + b (− 3) + 3 = 0
− 108 + 9a – 3b + 3 = 0
9a – 3b = 105
3a – b = 35
f (2) = 4 (2)3 + a(2)2 + b(2) + 3
= 32 + 4a + 2b + 3
= 4a + 2b + 35
f (2) = 165
Unit 1 Answers: Chapter 4
[1]
© Macmillan Publishers Limited 2013
Page 15 of 22
11
12
4a + 2b + 35 = 165
4a + 2b = 130
2a + b = 65
[2]
[1] + [2] ⇒ 5a = 100, a = 20
40 + b = 65, b = 25
∴ a = 20, b = 25
(b)
f (x) = 4x3 + 20x2 + 25x + 3
4x3 + 20x2 + 25x + 3 = (x + 3) (ax2 + bx + c)
= (x + 3) (4x2 + 8x + 1)
(a)
f (x) = x4 + x3 + ax2 + bx + 10
Now x2 – 3x + 2 = (x − 1) (x − 2)
∴ x – 1 and x – 2 are factor of f (x)
a + b = − 12
[1]
f (2) = 0 ⇒ 24 + 23 + a (2)2 + b(2) + 10 = 0
4a + 2b = − 34
2a + b = − 17
[2]
[2] – [1] ⇒ a = − 5
− 5 + b = − 12
b=−7
∴ a = − 5, b = − 7
(b)
f (x) = x4 + x3 – 5x2 – 7x + 10
x4 + x3 – 5x2 – 7x + 10 = (x2 – 3x + 2) (x2 + 4x + 5)
Other quadratic factor is x2 + 4x + 5
f(x) = 36x3 + ax2 + bx – 2
3
13
2
 1
 1
 1
 1
f  −  =0 ⇒ 36  −  + a  −  + b  −  − 2 =0
6
6
6






 6
1
1
13
⇒
a− b=
36
6
6
3
2
 2
 2
 2
 2
f   = 0 ⇒ 36   + a   + b   − 2 = 0
 3
 3
 3
 3
4
2
− 26
a+ b=
9
3
3
1
[1]
a−b=
13
6
2
[2]
a + b =− 13
3
5
[1] + [2] ⇒ a = 0
6
a=0
a = 0, b = − 13
f (x) = 36 x3 – 13x – 2
36x3 – 13x – 2 = (6x + 1) (3x − 2) (2x + 1)
f (x) = 4x3 + ax2 + bx + 3
f (− 3) = 0
⇒ 4 (− 3)3 + a (− 3)2 + b (− 3) + 3 = 0
⇒ 9a – 3b = 105
3a – b = 35
[1]
f (1) = − 12
⇒ 4 + a + b + 3 = − 12
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 16 of 22
14
a + b = − 19
[1] + [2] ⇒ 4a = 16
a=4
4 + b = − 19
b = − 23
a = 4, b = − 23
g(x) = 4x4 + px3 – 21x2 + qx + 27
g(3) = 0
⇒4 (3)4 + p (3)3 – 21 (3)2 + q (3) + 27 = 0
27p + 3q = − 162
9p + q = − 54
g (− 1) = 0
⇒ 4 – p – 21 – q + 27 = 0
p + q = 10
[1] − [2] ⇒ 8p = − 64
p=−8
− 8 + q = 10
q = 18
Hence p = − 8, q = 18
4x4 – 8x3 – 21x2 + 18x + 27 = (x − 3) (x + 1) (4x2 – 9)
= (x − 3) (x + 1) (2x − 3) (2x + 3)
a = 4, b = − 9
[2]
[1]
[2]
Try these 4.5
(a)
(b)
(c)
27x3 – 64 = (3x)3 – 43 = (3x – 4) ((3x)2 + (3x) (4) + 42)
= (3x − 4) (9x2 + 12x + 16)
81x4 – 16 = (3x)4 – 24 = ((3x)2 – (2)2) ((3x)2 + (2)2)
= (3x − 2) (3x + 2) (9x2 + 4)
x6 – y6 = (x3 – y3) (x3 + y3) = (x − y) (x2 + xy + y2) (x + y) (x2 – xy + y2)
= (x − y) (x + y) (x2 + xy + y2) (x2 – xy + y2)
Review exercise 4
1
2
3
x6 – 64 = (x3)2 – (23)2
= (x3 − 8) (x3 + 8)
= (x − 2) (x2 + 2x + 4) (x + 2) (x2 – 2x + 4)
32x5 − 243
= (2x)5 − 35
= (2x − 3) ((2x)4 + (2x)3 (3) + (2x)2 (3)2 + (2x) (3)3 + 34)
= (2x − 3) (16x4 + 24x3 + 36x2 + 54x + 81)
(a)
y = 6x3 + x2 – 5x – 2
if x = 1, y = 6 + 1 – 5 – 2 = 0
∴ x – 1 is a factor
∴ y = (x − 1) (6x2 + 7x + 2)
= (x − 1) (3x + 1) (x + 2)
y = (x − 1) (3x + 1) (x + 2)
when x = 0, y = − 2, (0, − 2)
(b)
y = 3x3 – 2x2 – 7x – 2
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 17 of 22
(c)
4
(a)
(b)
5
(a)
if x = − 1, y = − 3 – 2 + 7 – 2 = 0
∴ x + 1 is a factor
3x3 – 2x2 – 7x – 2 = (x + 1) (3x2 – 5x – 2)
= (x + 1) (3x + 1) (x – 2)
y = (x + 1) (3x + 1) (x – 2).
when x = 0, y = – 2
1
when y = 0, x = 2, − 1, −
3
y = 4x3 + 9x2 – 10x – 3
if x = 1, y = 4 + 9 – 10 – 3 = 0
∴ x − 1 is a factor
4x3 + 9x2 – 10x – 3 = (x – 1) (4x2 + 13x + 3)
= (x − 1) (4x + 1) (x + 3)
y = (x − 1) (4x + 1) (x + 3)
when x = 0, y = − 3
1
when y = 0, x = 1, − , − 3
4
f(x) = x3 – 2x2 – 4x + 8 = 0
f(2) = 8 – 8 – 8 + 8 = 0
∴ x – 2 is a factor of f (x)
x3 – 2x2 – 4x + 8 = (x – 2) (x2 – 4)
= (x − 2) (x − 2) (x + 2)
∴ x – 2 = 0, x + 2 = 0
x = 2, − 2
f(x) = 2x3 – 7x2 – 10x + 24 = 0
f(4) = 2(4)3 – 7(4)2 – 10 (4) + 24
=0
∴ x – 4 is a factor of f (x)
∴ 2x3 – 7x2 – 10x + 24 = (x − 4) (2x2 + x − 6)
= (x − 4) (2x − 3) (x + 2)
3
x = 4, x = , x = − 2
2
f(x) = 7x3 – 5x2 + 2x + 1
3
(b)
(c)
2
59
 2
 2
 2
 2
+1
f =
7   − 5   + 2  =

 3
 3
 3
 3
27
f(x) = 6x3 + 7x + 1
f(– 3) = 6 (– 3)3 + 7 (− 3) + 1 = − 182
f(x) = x4 + 2
4
6
657
 5  − 5
f  −=
=
+2



 2  2 
16
3
2
f (x) = x – ax + 2ax − 8
f (2) = (2)3 – a (2)2 + 2a (2) – 8
= 8 – 4a + 4a − 8
=0
∴ x – 2 is a factor of f(x)
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 18 of 22
x 2 + ( −a + 2) x + 4
x − 2 x 3 − ax 2 + 2ax − 8
x 3 − 2x 2
(− a + 2) x 2 + 2ax − 8
(− a + 2) x 2 − 2 (− a + 2) x
4x − 8
4x − 8
0
f(x) = (x − 2) (x + (2 − a) x) + 4
x2 + (2 − a) x + 4 = 0
Since all roots are real b2 – 4ac ≥ 0
(2 − a)2 – 4 (4) ≥ 0
a2 – 4a – 12 ≥ 0
(a − 6) (a + 2) ≥ 0
a ≤ – 2, a ≥ 6
2
7
(a)
(b)
8
(a)
f (x) = (3x + 2) (x – 1) (x – 2)
= (3x2 – x – 2) (x – 2)
= 3x3 – 6x2 – x2 + 2x – 2x + 4
= 3x3 – 7x2 + 4
= Ax3 + Bx2 + Cx + D
A = 3, B = − 7, C = 0, D = 4
f (− 2) – 2b = 0
2b = f (− 2)
2b = (− 6 + 2) (− 2 – 1) (− 2 − 2)
2b = − 48
b = − 24
f (x) = x4 + 5x3 + 4x2 – 10x – 12
f ( 2) = ( 2)4 + 5 ( 2)3 + 4 ( 2)2 − 10 2 − 12
= 4 + 10 2 + 8 − 10 2 − 12 = 0
⇒ x − 2 is a factor of f (x)
f (− 2) =
(− 2)4 + 5 (− 2)3 + 4 (− 2)2 − 10 (− 2) − 12
= 4 − 10 2 + 8 + 10 2 − 12
=0
⇒ x + 2 is a factor of f (x).
(b)
A quadratic factor is (x − 2) (x + 2) =x 2 − 2
x4 + 5x3 + 4x2 – 10x – 12 = (x2 – 2) (x2 + 5x + 6)
The second quadratic factor is x2 + 5x + 6
f (x) = x4 + 5x3 + 4x2 – 10x – 12
f (– 2) = (− 2)4 + 5 (− 2)3 + 4 (− 2)2 – 10 (− 2) – 12 = 0
x + 2 is a factor of f (x)
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 19 of 22
x3 + 3x 2 − 2x − 6
x + 2 x 4 + 5x 3 + 4x 2 − 10x − 12
x 4 + 2x 3
3x 3 + 4x 2 − 10x − 12
3x 3 + 6x 2
− 2x 2 − 10x − 12
− 2x 2 − 4x
− 6x − 12
− 6x − 12
0
∴ f (x) = (x + 2) (x + 3x – 2x – 6)
= (x + 2) (x + 3) (x2 − 2)
= (x + 2) (x + 3) (x − 2 ) (x + 2 )
∴ x + 2 = 0, x + 3 = 0, x − 2 = 0, x + 2 = 0
x = − 2, − 3, 2 , − 2
q (x) = x4 – ax3 + bx2 + x + 6
q (2) = 0
⇒ (2)4 – a (2)3 + b (2)2 + 2 + 6 = 0
⇒ 16 – 8a + 4b + 8 = 0
− 8a + 4b = − 24
− 2a + b = − 6
[1]
q (3) = 0
(3)4 – a (3)3 + b (3)2 + 3 + 6 = 0
81 – 27a + 9b + 9 = 0
− 27a + 9b = − 90
− 3a + b = − 10
[2]
[1] – [2] ⇒ a = 4
b=2
Hence a = 4, b = 2
(2x + 1) (x − 2) (3x + 4)
= (2x2 – 3x − 2) (3x + 4)
= 6x3 + 8x2 – 9x2 – 12x – 6x – 8
= 6x3 – x2 – 18x – 8
= Ax3 + Bx2 + Cx + D
A = 6, B = − 1, C = − 18, D = − 8
3x4 + Bx3 + Cx2 + Dx + 2 = (3x2 + 2x + 1) (x2 – 4x + 2)
= 3x4 − 12x3 + 6x2 + 2x3 – 8x2 + 4x + x2 – 4x + 2
= 3x4 − 10x3 – x2 + 2
∴ B = − 10, C = − 1, D = 0
f (x) = ax3 – bx2 + 8x + 2
f (− 2) = − 110
⇒ − 8a – 4b – 16 + 2 = − 110
⇒ 8a + 4b = 96
2a + b = 24
[1]
f (1) = a – b + 8 + 2
= a – b + 10
a – b + 10 = 13
3
9
10
11
12
Unit 1 Answers: Chapter 4
2
© Macmillan Publishers Limited 2013
Page 20 of 22
a–b=3
[1] + [2] ⇒ 3a = 27
a=9
b=6
∴ f (x) = 9x3 – 6x2 + 8x + 2
3
13
14
15
[2]
2
 − 2
 − 2
 − 2
 − 2
f
= 9
−6
+8
+2



 3 
 3 
 3 
 3 
8 8 16
=
− − −
+2
3 3
3
− 26
=
3
f (x) = 2x3 – x2 – 5x – 2.
f (− 1) = 2 (− 1)3 – (− 1)2 – 5 (− 1) – 2
=−2–1+5–2=0
⇒ x + 1 is a factor of f (x)
2x3 – x2 – 5x – 2 = (x + 1) (2x2 – 3x − 2)
= (x + 1) (2x + 1) (x – 2)
f (x) = 6x3 – 5x2 – 13x − 2
f (− 1) = − 6 – 5 + 13 – 2 = 0
x + 1 is a factor of f (x)
6x3 – 5x2 – 13x – 2 = (x + 1) (6x2 – 11x − 2)
= (x + 1) (6x + 1) (x − 2)
∴ x + 1 = 0, 6x + 1 = 0, x – 2 = 0
1
x = − 1, − , 2
6
f (x) = x4 + x3 – 11x2 – 27x - 36
f (− 3) = 81 – 27 – 99 + 81 – 36
= 0.
∴ x + 3 is a factor of f (x)
x3 − 2x 2 − 5x − 12
4
x + 3 x + x 3 − 11x 2 − 27x − 36
x 4 + 3x 3
− 2x 3 − 11x 2 − 27x − 36
− 2x 3 − 6x 2
− 5x 2 − 27x − 36
− 5x 2 − 15x
− 12x − 36
− 12x − 36
0
f (x) = (x + 3) (x – 2x – 5x – 12)
= (x + 3) (x – 4) (x2 + 2x + 3)
∴ x + 3 = 0, x – 4 = 0
x = − 3, x = 4
x2 + 2x + 3 = 0
− 2 ± 4 − 12
, so no more real solutions
x=
2
3
Unit 1 Answers: Chapter 4
2
© Macmillan Publishers Limited 2013
Page 21 of 22
16
(a)
(b)
h (t) = 0.16t3 – 2.3t2 + 9.3 t – 2.2
1971 to 1980, t = 6
∴ h (6) = 0.16 (6)3 – 2.3 (6)2 + 9.3 (6) – 2.2
= 5.36
(c)
17
(d)
(e)
(a)
Only for a section of the data. The model may need refining
The model predicts approximately 9 hurricanes, the model does not support 5
Let one side of the square be x cm
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 22 of 22
18
V = (24 – 2x) (16 – 2x) (x)
= 384x – 80x2 + 4x3
(b)
V = 512
4x3 – 80x2 + 384x – 512 = 0
if x = 4, 4(4)3 – 80 (4)2 + 384(4) – 512 = 256 – 1280 + 1536 – 512
=0
4x3 – 80x2 + 384x – 512 = (x – 4) (4x2 – 64x – 128)
= 4 (x − 4) (x2 – 16x − 32)
Ht. x = 4, x2 – 16x – 32 = 0
16 ± 384 16 ± 19.6
=
x =
2
2
= 17.8, - 1.8
So the only possible size for the corner cut out is x = 4
(a)
f (x) = 2x3 – 7x2 – 10x + 24
= (x − 4) (2x2 + x – 6)
= (x – 4) (2x – 3) (x + 2)
3
(b)
x = 4, , − 2
2
Unit 1 Answers: Chapter 4
© Macmillan Publishers Limited 2013
Page 1 of 32
Chapter 5 Indices, Surds and Logarithms
Exercise 5A
1
1
3 3
(125)
=3 (5=
) 5
(b)
4 4
=
814 (3=
) 3
(c)
4 4
=
16 4 (2=
) 2
(d)
(e)
(f)
(g)
(h)
2
1
(a)
1
1
1
1
2
6× 2
4
3
3
(46 )=
4 =
4=
256
1
−1
−1
−1
8=3 (23 )=3 2=
2
−2
1
−1 −2
2
  = (4 ) = 4= 16
4
3
3
3
2
(121)=
(112 =
) 2 11
=
1331
1
1
1
−
−
−1
4
4
81=
(34 ) =
3=
3
3
3+1
1
(a)
a 4 × a =4 a 4 4= a
(b)
3
3
4
a=
a 4 a=
a 12
(c)
4
4
y 4 ÷ y=
y 4 =
y2
(d)
y2 y4
2+4−3
=
y=
y3
y3
−2
−2 + 3
3
5
−1
−3
−1 −5
1
3
1
1
1
1
64 3 × 216 3 (43 ) 3 × (63 ) 3
(a)
=
8
8
4×6
= = 3
8
6
4 × 54 4 6 × 54
(b)
4 6 − 7 × 54 − 7
= 7 =
207
4 × 57
= 4 −1 × 5−3
1
1
1
=×
=
4 125 500
1
(c)
1
32 6 × 16 12
1
1
86 × 43
=
1
1
1
1
(2 5 ) 6 × (2 4 )12
(23 ) 6 × (22 ) 3
5+1−1−2
4
(a)
= 26 3 2 3
20 = 1
82x × 25x ÷ 46x
= (23 ) 2x × 25x ÷ (22 )6x
6x + 5x −12x
= 2=
2− x
(b)
273x × 93x ÷ 815x
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 2 of 32
= (33 )3x × (32 )3x ÷ (34 )5x
9x + 6x − 20x
= 3=
3−5x
1x
2x
2x
10 3 × 20 3 ÷ 2 3
(c)
1x
2x
2x
2x
= 10 3 × 2 3 × 10 3 ÷ 2 3
1x + 2x
3
3
= 10
=
10 x
5x
3x
(16 4 ÷ 8 3 ) × 4x + 1
(d)
5x
3x
= (2 4 ) 4 ÷ (23 ) 3 × (2 2 )x + 1
2
2
= 23x − 5x + 2x +=
2=
4
3
5
(1 − x) 2
(a)
(1 − x)
3−1
=
(1 − x) 2 2 =
1− x
1
2
1
3
(1 + x) 3 + (1 + x) 4
(b)
(1 + x)
3−1
=1 + (1 + x) 4 3
1
3
5
=1 + (1 + x)12
310 + 39
=
311 + 312
6
x
7
39 (3 + 1)
=
311 (3 + 1)
1
1
=
2
3
3
1
1
2
+ 2(1 + x) 2
(1 + x)
x + 2(1 + x) 2 + 3x
= =
1
1
(1 + x) 2
(1 + x) 2
8
x
1
(x + 1) 3 +
2
3(1 + x) 3
3(x + 1) + x
4x + 3
= =
2
2
3(1 + x) 3
3(1 + x) 3
9
4x + 3
11
x −5
2 x − 5 5 4x + 3
5(4x + 3) + 2(x − 5)
=
10 x − 5 4x + 3
=
10
+
22x + 5
10 x − 5 4x + 3
3
1
3
(x + 1) 2 + x (x + 1) 2
2
1 
3 
= (x + 1) 2  x + 1 + x 
2 

1 5

=
(x + 1) 2  x + 1
2


4
1
3(x 2 + 4) 3 + 8x 2 (x 2 + 4) 3
1
= (x 2 + 4) 3 [3(x 2 + 4) + 8x 2 ]
1
=
(x 2 + 4) 3 (11x 2 + 12)
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 3 of 32
4
1
2x(3x + 4) 3 + 4x 2 (3x + 4) 3
12
1
= 2x(3x + 4) 3 (3x + 4 + 2x)
1
=2x(3x + 4) 3 (5x + 4)
Try these 5.1
75 + 48 − 2 675
(a)
= 5 3 + 4 3 − 2(15 3)
= 9 3 − 30 3
= − 21 3
(b)
4 288 − 3 882
= 48 2 − 63 2
= − 15 2
(c)
5 80 + 3 20 − 125
= 20 5 + 6 5 − 5 5
= 21 5
Try these 5.2
(a)
2
2
1+ 3
=
×
1− 3 1− 3 1+ 3
=
(b)
2(1 + 3)
1− 3
2(1 + 3)
=
=− 1 − 3
−2
4
4
2− 7
=
×
2+ 7 2+ 7 2− 7
4(2 − 7)
4
=
=
− (2 − 7)
4−7
3
(c)
1
=
2+ 5
1
2+ 5
×
2− 5
2− 5
2− 5
1
=
=
− ( 2 − 5)
2−5
3
Try these 5.3
(a)
1
1
2− x 2− x
=
×
=
4−x
2+ x 2+ x 2− x
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 4 of 32
(b)
(c)
4 + x +1
3
3
=
×
4 − x +1 4 − x +1 4 + x +1
=
3(4 + x + 1)
16 − (x + 1)
=
3(4 + x + 1)
15 − x
2 + x +1 2 + x +1 3 + x
=
×
3− x
3− x
3+ x
=
(2 + x + 1)(3 + x )
9−x
Exercise 5B
1
2
(a)
1083 = 19 3
(b)
1445 = 17 5
(c)
1058 = 23 2
(a)
147 − 5 192 + 108
=
7 3 − 40 3 + 6 3 =
− 27 3
2 847 + 4 576 − 4 1008
(b)
= 22 7 + 96 − 48 7
= 96 − 26 7
3
2− 5
2− 5 2−3 5
=
×
2+3 5 2+3 5 2−3 5
(a)
=
4 − 8 5 + 15
4 − 45
19 8 5
−
−41 −41
−19 8
=
+
5
41 41
1
1
5+4
(b)
=
×
5−4
5 −4
5+4
=
5+4
1
4
5−
= =
−
5 − 16
11
11
(c)
4
4
2 + 7 4(2 + 7) − 4
=
×
=
= (2 + 7)
4−7
3
2− 7 2− 7 2+ 7
(d)
12
12
4 + 11
=
×
4 − 11 4 − 11 4 + 11
=
12(4 + 11) 12
=
(4 + 11)
16 − 11
5
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 5 of 32
3 −1
3 −1 2 3 − 3
=
×
2 3 +3 2 3 +3 2 3 −3
(e)
=
2(3) − 3 3 − 2 3 + 3
12 − 9
9−5 3
3
5 3
= 3−
3
1
1
2 3−5
=
×
2 3+5 2 3+5 2 3−5
=
(f)
=
=
4
(a)
2 3 −5
12 − 25
2 3 −5
−13
1
1
+
3+ 7
3− 7
=
3− 7+ 3+ 7
( 3 + 7) ( 3 − 7)
=
2 3 2 3
1
=
= −
3
3 − 7 −4
2
1+ 5 1− 5
−
5+2
5−2
(b)
=
(1 + 5) ( 5 − 2) − (1 − 5) ( 5 + 2)
( 5 + 2) ( 5 − 2)
5 −2+5− 2 5 − 5 −2+5+ 2 5
5−4
=
=6
1
(c)
+
1
(1 − 3)
(1 + 3) 2
1
1
=
+
4−2 3 4+2 3
=
2
4+2 3+4−2 3
8
= = 2
(4 − 2 3) (4 + 2 3) 16 − 12
1
2+ 2
5
=
1
2− 2
×
2+ 2 2− 2
2− 2
2
2
2
=
−
2
2
=
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 6 of 32
6
= 1−
2
2
(a)
x
=
x−2
=
=
=
x−2
x( x − 2)
x−2
2− x
4−x
x(5 + 4 x )
25 − 16x
1
=
x+2 −3
(d)
=
x+2 +3
(x + 2) − 9
=
x+2 +3
x−7
x +1
=
x +1 − 4
(e)
x+2 +3
1
×
x+2 −3
x +1
×
x +1 − 4
x+2 +3
x +1 + 4
x +1 + 4
(x + 1) ( x + 1 + 4) (x + 1) ( x + 1 + 4)
=
x + 1 − 16
x − 15
6
2x − 3 + 4
=
(f)
2x − 3 − 4
6
×
2x − 3 + 4
2x − 3 − 4
=
=
6( 2x − 3 − 4)
2x − 3 − 16
=
6( 2x − 3 − 4)
2x − 19
x − 13
=
x+3+4
=
x−2
x
x
5+ 4 x
=
×
5−4 x 5−4 x 5+ 4 x
(c)
=
x−2
×
1
1
2− x
=
×
2+ x 2+ x 2− x
(b)
7
x
x+3−4
x − 13
×
x+3+4
x+3−4
(x − 13) ( x + 3 − 4)
x + 3 − 16
(x − 13) ( x + 3 − 4)
x − 13
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 7 of 32
=
x+3−4
+6 +3
( 2x + 6 − 3) × 2x
2x + 6 + 3
2x + 6 −=
3
8
=
2x + 6 − 9
2x + 6 + 3
=
2x − 3
2x + 6 + 3
2x − 17
=
2x − 1 + 4
9
2x − 1 − 4
2x − 17
×
2x − 1 + 4
2x − 1 − 4
=
(2x − 17) ( 2x − 1 − 4)
2x − 1 − 16
=
(2x − 17) ( 2x − 1 − 4)
=
2x − 17
2x − 1 − 4
1 + tan 30°
1 − tan 30°
10
1
1+
3
= =
1
1−
3
=
=
=
3 +1
3
3 −1
3
3 +1
3 −1
3 +1
3 +1
×
3 −1
3 +1
3+ 2 3 +1
3 −1
4+2 3
2
= 2+ 3
=
Try these 5.4
(a)
y = ex−3
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 8 of 32
(b)
4000
1 + 3999 e − 0.2t
t = 0,
x=1
t→ ∞,
x→4000
x=
Exercise 5C
1
(a)
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 9 of 32
(b)
(c)
2
x = 3 + 4e−2t
t = 0, x = 3 + 4 = 7
t → ∞, e−2t → 0 ⇒ x = 3 + 4(0) = 3
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 10 of 32
3
θ = et
Reflect the graph of et in the y-axis to obtain e−t
e−t is then stretched along the y-axis by factor 2 to obtain 2 e−t
2 e−t is shifted up the y-axis by factor 5 to obtain the graph of 5 + 2 e−t
4
y = 2x+1
9e-x = 2x + 1
9 = 2xex + ex
9 – 2xex – ex = 0
The equation has one real root
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 11 of 32
5
x=
5
4 + 10e − t
(a)
t ==
0, x
(b)
5
5
=
4 + 10 14
5
t → ∞, e−t → 0, x =
4
6
(a)
(b)
(c)
400
Population becomes infinite
7
(a)
(b)
(c)
θ = 64.62°C
62.08°C
8
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 12 of 32
9
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 13 of 32
10
Try these 5.5
(a)
(b)
32x = 243x + 1
32x = (35)x + 1
32x = 35x + 5
⇒2x = 5x + 5
−3x = 5
−5
x=
3
1
e x + 2 = 3x − 1
e
x+2
e = e −3x + 1
⇒ x + 2 = −3x + 1
4x = −1
−1
x=
4
Try these 5.6
(a)
x = a 2 ⇒ 2 = log a x
(b)
16 = 42 ⇒ 2 = log 4 16
(c)
10 = a x ⇒ x = log a 10
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 14 of 32
Try these 5.7
(a)
(b)
log3 y = 6 ⇒ y = 36
=
x log 3 10 ⇒ 10
= 3x
(c)
log x 6 = 4 ⇒ 6 = x 4
Exercise 5D
1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
102 = 100
⇒ log10100 = 2
x° = 1
⇒ log x 1 =
0
5° = 1
⇒ log5 1 = 0
a1 = a
⇒ loga a = 1
x3 = 8
logx 8 = 3
42 = 16
⇒ log4 16 = 2
53 = 125
⇒ log5 125 = 3
y2 = 9
⇒ logy 9 = 2
1
x2 = 4
⇒ log x 4 =
2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
1
2
logx 4 = 2
⇒ x2 = 4
x=2
logx 625 = 4
⇒ x4 = 625
⇒x=5
log2 64 = x
⇒ x2 = 64
⇒x = 8.
log4 x = 2
⇒ x = 42 = 16
log9 x = 0
x = 9° = 1
logx 8 = 3
⇒ x3 = 8
⇒x=2
logx 9 = 2
⇒ x2 = 9
⇒x=3
logx 216 = 3
⇒ x3 = 216
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 15 of 32
(b)
⇒x=6
log8 x = 2
⇒ x = 82 = 64
1
− 1 log8 8 =
−1
log8 =
log8 8−1 =
8
6
log=
log
=
6 log
=
6
3 729
33
33
(c)
2
log
=
log
=
2log
=
2
8 64
88
88
(i)
3
(a)
(d)
(e)
(f)
(g)
(h)
(i)
4
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
5
(a)
(b)
(c)
(d)
(e)
5
log
=
log
=
5=
log 2 2 5
2 32
22
1
=
− 3log 5 5 =
−3
log 5
log 5 5−3 =
125
=
log x x10 10
=
log x x 10
1
=
− 3 log 7 7 =
−3
log 7
log 7 7 −3 =
343
2
log8 =
4096 log
=
log
=
2=
log8 8 2
8 64
88
4
log 6=
1296 log
=
4=
log 6 6 4
66
log10 4
=
log 3 4 =
1.262
log10 3
log10 7
= 1.086
log10 6
log10 2
=
log 5 2 =
0.431
log10 5
=
log6 7
log10 6
= 0.815
log10 9
log10 18
=
log 5 18 =
1.796
log10 5
log10 17
=
log 2 17 =
4.087
log10 2
log10 4
=
log 9 4 =
0.631
log10 9
log10 6
=
log 5 6 =
1.113
log10 5
log10 29
=
log 4 29 =
2.429
log10 4
ln 12
log=
= 1.544
5 12
ln 5
ln 22
log=
= 2.814
3 22
ln 3
ln 18
log=
= 1.796
5 18
ln 5
=
log 9 6
ln 17
= 2.044
ln 4
ln 32
log=
= 3.155
3 32
ln 3
log=
4 17
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 16 of 32
(f)
(g)
6
ln 41
= 2.307
ln 5
ln 62
= 2.977
log=
4 62
ln 4
log=
5 41
ln 28
= 4.807
ln 2
(h)
log=
2 28
(a)
2 x = 8 ⇒ 2 x = 23 ⇒ x = 3.
(b)
3x = 81 ⇒ 3x = 34 ⇒ x = 4
4x =
3x + 1 ⇒ log 4 x =
log 3x + 1
⇒ x log 4 =
(x + 1) log 3
⇒ x log 4 = x log 3 + log 3
⇒ x log 4 − x log 3 =
log 3
x(log 4 − log 3) =
log 3
log 3
x=
log 4 − log 3
= 3.819
(d) 2x+2 = 4x−3
2x+2 = (22)x−3
2x+2 = 22x−6
⇒ x + 2 = 2x − 6
8=x
(e) 32x = 4x+1
log 32x = log 4 x + 1
(2x) log 3 = (x + 1) log 4
(2 log 3) x − x log 4 = log 4
x [2 log 3 − log 4] = log 4
log 4
=
x = 1.710
2 log 3 − log 4
2x
7
(a) 2 − 5(2x) + 4 = 0
⇒ (2x)2 − 5(2x) + 4 = 0
Let y = 2x
⇒ y2 − 5y + 4 = 0
⇒ (y − 1) (y − 4) = 0
y = 1, 4.
2x = 1, 2x = 4
2x = 2°, 2x = 22
⇒ x = 0, x = 2
(b) 32x − 4 (3x + 1) + 27 = 0
⇒ (3x)2 − 4(3x) (3) + 27 = 0
Let y = 3x
⇒ y2 − 12y + 27 = 0
⇒ (y − 3) (y − 9) = 0
y = 3, y = 9
⇒ 3x = 3, 3x = 9
x = 1, 3x = 32
x=2
(c) e2x − 6ex + 9 = 0
(c)
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 17 of 32
8
(ex)2 − 6ex + 9 = 0
y = ex
y2 − 6y + 9 = 0
(y − 3) (y − 3) = 0
y=3
⇒ ex = 3
x = ln 3 = 1.099
(a) 8x = 64
⇒ 8x = 82
⇒x=2
(b) 9x = 27
(32)x = 33
32x = 33
2x = 3
3
x=
2
(c) 4x = 128
(22)x = 27
22x = 27
2x = 7
7
x=
2
1
(d)
2x 2x +1 =
2
⇒ 2x+x+1 = 2−1
⇒ 22x+1 = 2−1
⇒ 2x + 1 = −1
2x = −2
x = −1
(e) 42x − 5(22x−1) + 1 = 0
5
⇒ (22x ) 2 − (22x ) + 1 =
0
2
Let y = 22x
5
⇒ y2 − y + 1 =
0
2
2y 2 − 5y + 2 =
0
(2y − 1) (y − 2) = 0
1
y= ,2
2
1 2x
=
22x =
,2
2
2
22x = 2−1 , 22x = 21
2x = −1, 2x = 1
1
−1
, x=
x=
2
2
1 2x
4 x
(f)
(3 ) − (3 ) + 1 =
0
27
9
(3x)2 − 12(3x) + 27 = 0
Let y = 3x
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 18 of 32
y2 − 12y + 27 = 0
(y − 3) (y − 9) = 0
y = 3, 9
3x = 3, 3x = 9 = 32
x = 1, x = 2
(g)
16 2 − 3(2 x + 1 ) + 8 =
0
x
x
(24 ) 2 − 3(2x )(2) + 8 =
0
2 2x − 6(2 x ) + 8 =
0
9
(a)
(2 x ) 2 − 6(2 x ) + 8 =
0
Let y = 2x
y2 − 6y + 8 = 0
(y − 2) (y − 4) = 0
y = 2, y = 4
2x = 2, 2x = 4 =22
x = 1, x = 2
(1 + i) −9 = 0.95
−1
(1 + i) =
(0.95) 9
−1
=
i (0.95) 9 − 1
= 0.00572
(b) (1 + i)4 = (1.01)9
9
1 + i =(1.01) 4
9
10
=
i (1.01) 4 −=
1 0.02264
3 2
(a) ln(x y )
= lnx3 + lny2
= 3ln x + 2ln y
(b)
1
ln xy = ln(xy) 2
1
ln (xy)
2
1
1
=
ln x + ln y
2
2
4
x 
(c)
ln  =
ln x 4 − ln y

 y
= 4 ln x − ln y
 xy3 
 y3 
(d)
ln  6  = ln  5 
 x 
x 
3
5
= In y − ln x
= 3ln y − 5 ln x
(a) ln 14 − ln 21 + ln 8
2
14
×8
= ln
213
=
11
(b)
 16 
= ln  
 3
1
4 ln 2 + ln 8
3
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 19 of 32
1
= ln 2 4 + ln 8 3
1
= ln (24 × (23 ) 3 )
= ln (24 × 2)= ln 25= 5 ln 2
(c)
 4
 8
2 ln   − ln  
 9
 27 
2
 4
 8
= ln   − ln  
 9
 27 
 16 27 
ln  × 
 81 8 
(d)
2
= ln  
3
3ln 4 + 4 ln 2 − 4 ln 6
= ln 43 + ln 24 − ln 64
 43 × 2 4 
= ln 

4
 6 
 64 
= ln  
 81 
4 ln 5 − ln 25 + 2 ln 2
= ln 54 − ln 25 + ln 22
 625

= ln 
× 4
 25

= ln 100
y = logb x
2
(a)
log
=
2log
=
2y
b x
b x
(e)
12
(b)
(c)
=
log b x 4 4=
log b x 4 y
2
2
2
=
=
=
log
2log
x b
x b
log b x y
2
6 3
(d)
log x (b
=
x)3 log x (b
=
x ) log x b6 + log x x 3
= 6 log x b + 3 log x x
6
=
+3
log b x
6
=
+3
y
(e)
13
 b6 
log x  4 
x 
= log x b 6 − log x x 4
= 6 log x b − 4 log x x
6
=
−4
log b x
6
=
−4
y
(a)
log 4 x + log x 16 =
3
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 20 of 32
log 4 x + log x 42 =
3
log 4 x + 2log x 4 =
3
log 4 x +
(b)
2
=
3
log 4 x
(log 4 x) 2 + 2 =
3 log 4 x
Let y = log 4 x ⇒ y2 + 2 = 3y
y2 − 3y + 2 = 0
(y − 1) (y − 2) = 0
y = 1, 2
∴ log4 x = 1, log4 x = 2
x = 41, x = 42
x = 4, x = 16
3 log 6 x + 2 log x 6 =
5
2
⇒ 3 log 6 x +
=
5
log 6 x
Let y = log6 x
2
⇒ 3y + =
5
y
⇒ 3y 2 + 2 =
5y
⇒ 3y 2 − 5y + 2 =
0
(3y + 1) (y − 2) =
0
1
y=
− ,y=
2
3
1
− , log 6 x =
log 6 x =
2
3
−1
(c)
2
=
x 6 3, =
x 6=
36
log 2 x = 4 log x 2
4
log 2 x =
log 2 x
⇒ (log 2 x) 2 =
4
log 2 x = 2, log 2 x = − 2
2
=
x 2=
, x 2 −2
1
=
x 4,=
x
4
(d)
log 4 (2x) + log 4 (x + 1) =
1
⇒ log 4 2x (x + 1) =
log 4 4
⇒ 2x (x + 1) =
4
2x 2 + 2x − 4 =
0
(e)
x2 + x − 2 =
0
(x + 2) (x − 1) =
0
x = −2, 1
x cannot be negative for log x to exist, so
x=1
log 9 x = log 3 (3x)
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 21 of 32
log 3 x
=
log 3 3x
log 3 9
log 3 x
⇒
=log 3 3 + log 3 x
log 3 32
1
⇒ log3 x =
1 + log3 x.
2
1
=
− log3 x =
1
2
log 3 x = − 2
1
−2
=
x 3=
9
(f)
log 2 x + log x 2 =
2
1
= log 2 x +
=2
log 2 x
1
Let=
y log 2 x ⇒ y += 2
y
2
y − 2y + 1 = 0
(y − 1)2 = 0
y=1
∴ log2 x = 1
x = 21 = 2
(a) 0.6x = 9.7
x log 0.6 = log 9.7
log 9.7
x=
log 0.6
= −4.448
(b) (1.5)x+2 = 9.6
log(1.5)x+2 = log (9.6)
(x + 2) log 1.5 = log 9.6
log 9.6
x= 2=
log 1.5
log 9.6
x =− 2 +
=3.578
log 1.5
1
(c)
= 0.9
0.8 x − 2
⇒ 0.8−x+2 = 0.9
⇒ log 0.8−x+2 = log 0.9
⇒ (−x + 2) log 0.8 = log 0.9
log 0.9
−x + 2 =
log 0.8
log 0.9
=
−x
−2
log 0.8
x = 1.528
⇒
14
(d)
 1 


0.24 
2+ x
= 1.452 x +1
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 22 of 32
 1 
⇒ log 
 0.24 
(e)
15
(a)
2+ x
=
log(1.452 x +1 )
 1 
⇒ (2 + x) log 
= (x + 1) log (1.452)
 0.24 
⇒ (2 + x ) (0.6198) = ( x + 1) (0.1620)
1.2396 + 0.6198x = 0.1620 x + 0.1620
0.6198x − 0.1620x = 0.1620 − 1.2396
0.4578x = −1.0776
x = −2.354
2.79x−1 = 3.377x
log 2.79x−1 = log 3.377x
(x − 1) log 2.79 = x log 3.377
0.4456 x − 0.4456 = 0.5285 x
−0.4456 = 0.0829 x
x = −5.375
22x + 2 x + 2 − 32 =
0
(2 x )2 + 2 x × 2 2 − 32 =
0
Let y = 2x
y2+ 4y −32 = 0
(y + 8) (y − 4) = 0
y = 4, −8
2x = 4,
2x = −8
⇓
no solutions
2 x = 22
x=2
(b) 22x − 9(2x−1) + 2 = 0
9
(2 x ) 2 − 2 x + 2 =
0
2
Let y = 2x
9
y2 − y + 2 =
0
2
2y 2 − 9y + 4 =
0
(2y − 1) (y − 4) =
0
1
=
y =
,y 4
2
1
=
2x =
, 2x 4
2
−1
=
2 x 2=
, 2 x 22
x = −1, x = 2
(c) 9x = 3x+2 − 8
(32)x = 3x 32 − 8
(3x)2 − 9(3x) + 8 = 0
Let y = 3x
y2 − 9y + 8 = 0
(y − 1) (y − 8) =
0
y = 1, 8
3x = 1 , 3x = 8
3x = 3°, x log 3 = log 8
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 23 of 32
log 8
= 1.893
log 3
(d) 25y − 7(5y) = 8
(5y)2 − 7(5y) − 8 = 0
Let x = 5y
x2 – 7x – 8 = 0
(x − 8) (x + 1) = 0
x = 8, −1
5y = 8,
5y = −1
⇓
no solutions
log 5y = log 8
y log 5 = log 8
log 8
=
y = 1.292
log 5
kt
P = P0 e
(a) t = 7, P = 2P0
2 P 0 = P 0 e7k
ln2 = 7k
1
k = ln 2
7
(b) =
P 3 P0 ⇒ 3 P0 = P0 e kt
x = 0,
=
x
16
17
ln 3 = kt
1
ln 3 = ln 2 t
7
7 ln 3
=
t = 11.1 hrs
ln 2
0.8
P=
1 + 21.7 e −0.16 t
0.8
(a) t=
= 0, P = 0.035
1 + 21.7
0.8
(b) t → ∞, e−0.16 t → 0 P →
=
0.8
1
Pmax = 0.8
(c)
(d)
40
0.8
(0.8) =
100
1 + 21.7 e −0.16 t
5
1 + 21.7 e −0.16 t =
2
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 24 of 32
18
19
21.7 e-0.16t = 1.5
1.5
e −0.16 t =
21.7
1.5
− 0.16t =
ln
=
−2.672
21.7
t = 16.7 yrs
M = 800 (1.04)t
(a) t = 6, M = 800(1.04)6 = 1012.26
M
(b)
= (1.04) t
800
 M
ln 
= ln 1.04 t
 800 
= t ln 1.04
 M 
ln 

 800  25.5 ln  M 
=
t =


ln 1.04
 800 
i = 15(1 − e−250 t)
(a)
i
= 1 − e −250 t
15
1
e −250 t= 1 − i
15
1 

−250 t = ln 1 − i 
 15 
1
1 

t=
−
ln 1 − i 
250  15 
1
2

(c) i = 2, t =
−
ln 1 − 

250
15 
1
 13 
= −
ln  
250  15 
= 0.000572
4000
x=
1 + 3999 e −0.2 t
As t → ∞, e−0.2 t → 0
xmax = 4000
40
4000
( 4000 ) =
1 + 3999 e −0.2 t
100
5
1 + 3999 e −0.2 t =
2
(b)
20
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 25 of 32
3999 e −0.2 t =
e −0.2 t =
3
2
3
2(3999)


3
−0.2 t =
ln 
−7.888
=
 2(3999) 
t = 39.4 days
−1 t
θ= 10 + 90 e 8
21
(a)
22
50
= 10 + 90 e 8
40
−1 t
=e 8
90
 4
−8 ln   =
t
 9
t = 6.5
2x × 42y = 1
⇒ 2x × (22)2y = 1
⇒ 2x+4y = 2°
⇒ x+4y = 0
1
55x × 25y =
25
55x × (52 ) y =
5−2
(b)
23
55x + 2 y = 5−2
⇒ 5x + 2y = −2
From [1], x = −4y
−20y + 2y = −2
−18y = −2
1
y=
9
−4
x=
9
x
9
=3
3y
32x
=3
3y
32x − y = 3
⇒ 2x − y = 1
lg (2x + 2y) = 1
2x + 2y = 10
x+y=5
Unit 1 Answers: Chapter 5
−1 t
[1]
[1]
[2]
© Macmillan Publishers Limited 2013
Page 26 of 32
[1] + [2] ⇒
3x = 6
x=2
y=3
Review Exercise 5
1
(a)
(b)
5x+3 = 1
ln 5x+3 = ln 1
(x + 3) ln 5 = 0
x ln 5 = −3 ln 5
x = −3
OR:
5x+3 = 50
x+3=0
x = −3
2 x −1 = 4 x + 2
2 x −1 = (22 ) x + 2
(c)
2 x −1 = 22x + 4
⇒ x − 1 = 2x + 4
x = −5
32x 3x −1 = 27
32x + x −1 = 33
2
(a)
33x −1 = 33
3x – 1 = 3
4
x=
3
100 (1 + i) −25 =
35
35
(1 + i) −25 =
100
−1
(b)
 35  25
1 + i =
 100 
i = 0.043
800 (1 − d)11 =
500
500
(1 − d)11 =
800
1
 5  11
(1 − d ) =
 
8
1
3
4
 5 11
d= 1 −  
8
= 0.0418
RTP: (logab) (logbc) = logac
Proof:
log a c
(logab) (log
=
log a b ×
bc)
log a b
= logac
(a) ln(3x) − 2ln y + 3ln x2
= ln(3x) − lny2 + ln(x2)3
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 27 of 32
 3x 
= ln  2  x 6
y 
 3x 7 
= ln  2 
 y 
(b)
1
ln (9x 2 ) + 2ln x − 3ln 2x
2
1
= ln (9x 2 ) 2 + ln x 2 − ln (2x)3
= ln 3x + ln x 2 − ln 8x 3
3x × x 2
3
= ln=
ln  
3
8x
8
2
(c)
ln (x − 1) − ln (x − 1) + 1n (x 2 + 3x + 2)
 x −1
= ln  2  (x 2 + 3x + 2)
 x − 1


x −1
( x + 1) ( x + 2) 
= ln 
 ( x − 1) ( x + 1)



= ln (x + 2)
5
log
=
log 9 (x + 6)
3x
Changing log9 (x + 6) to base 3
log (x + 6)
⇒ log 9 (x + 6) = 3
log 3 9
log 3 (x + 6) log 3 (x + 6) 1
= =
=
log 3 (x + 6)
log 3 32
2 log 3 3
2
1
∴ log 3 x=
log 3 (x + 6)
2
1
⇒ log3 x = log3 (x + 6) 2
1
⇒ x = (x + 6) 2
⇒ x 2 =x + 6
x2 − x − 6 =
0
(x − 3) (x + 2) = 0
x = 3, −2
x=3
6
=
log 2 x log 4 (8x − 16)
log 2 (8x − 16) log 2 (8x − 16) log 2 (8x − 16)
=
log 2 x =
=
log 2 4
log 2 22
2log 2 2
1
⇒ log 2 =
x
log (8x − 16)
2
⇒ 2log 2 x = log (8x − 16)
2
log
=
log (8x − 16)
2 x
2
x = 8x − 16
x2 − 8x + 16 = 0
(x − 4)2 = 0
x=4
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 28 of 32
7
8
(a)
32x +1 + 9 =
28 (3x )
⇒ 3 × (3x ) 2 + 9 − 28 (3x ) =
0
x
Let y = 3
3y2 − 28y + 9 = 0
(3y − 1) (y − 9) = 0
1
=
y =
,y 9
3
1
=
3x =
, 3x 9
3
−1
=
3x 3=
, 3x 32
x = −1, x = 2
x
(b)
64
=
8x +1 − 16
(82)x = 8(8x )− 16
(8x)2 = 8 (8x) − 16
Let y = 8x
y2 = 8y − 16
y2 − 8y + 16 = 0
(y − 4)2 = 0
y=4
8x = 4
(23)x = 22
3x = 2
2
x=
3
x = 400 e0.05 t
(a) t = 0, x = 400
(b) t = 5, x = 400 e0.05(5) = 513.61
(c) x = 450
⇒ 450 = 400 e0.05 t
450
e0.05 t =
400
 450 
0.05 t = ln 
 400 
1
 450 
=
ln 
 2.36
0.05  400 
(d) x = 800
⇒
800 = 400 e0.05t
0.05t
2=e
1n 2 = 0.05t
1
=
t =
ln 2 13.86
0.05
x = x0 ekt
(a) t = 4, x = 3x0
3x0 = x0 e4k
3 = e4k
ln 3 = 4k
1
k = ln 3
4
=
t
9
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 29 of 32
1
ln 3
4
(b)
=
x 2x
=
0, k
1 t ln 3
⇒ 2x 0 =
x 0e 4
1 t ln 3
2 = e4
1
t ln 3
4
4ln 2
=
t = 2.52 days
ln 3
N = 450 e0.03t
(a) t = 10, N = 450 e0.03(10) = 607.4
N ≈ 607
(b) When t = 0, N = 450
N = 900 ⇒ 900 = 450 e0.03 t
2 = e0.03t
0.03t = ln2
1
=
t =
ln 2 23.1 days
0.03
(c) N = 1000 ⇒
1000 = 450 e0.03t
1000
= e0.03t
450
 1000 
ln 
= 0.03t
 450 
ln 2 =
10
(a)
1
 1000 
=
ln 
 26.6 days
0.03  450 
0.6
P=
1 + 60 e −0.25t
0.6
when=
t 0,=
P
= 0.00984
1 + 60
(b)
When the denominator is minimum t → ∞ ⇒ e−0.25t → 0 ∴ P →
=
t
11
0.6
=
0.6
1+ 0
Max P is 0.6
0.6
0.6
⇒ 1 + 60 e −0.25 t=
−0.25 t
1 + 60 e
0.3
1
60 e−0.25t = 1 ⇒ e−0.25t =
60
1
−0.25t = ln , t = 16.4
60
Year: 2025
n = λ e5t
(a) t = 0, n = λ
(b) 2λ = λe5t
e5t = 2
5t = ln2
1
=
t =
ln 2 0.14
5
(a) x = 7000(1.06)n
n = 8, x = 7000(1.06)8 = 11156.94
(c)
12
13
P= 0.3 ⇒ 0.3=
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 30 of 32
(b)
14
15
18000 = 7000(1.06)n
18
= 1.06n
7
 18 
ln   = n ln(1.06)
 7
 18 
ln  
 7
=
n = 16.2
ln(1.06)
Year: 2027
(a) θ = 70 e−0.02t
t = 12, θ = 70 e−0.02(12) = 55°
(b) θ = 40
⇒
40 = 70 e−0.02t
4
= e −0.02t
7
4
ln   = ln e −0.02t = − 0.02t
7
 4
ln  
 7
=t
−0.02
t = 27.98 mins
(a) =
Q Q1 (1 − e − xt )
Q
= 1 − e − xt
Q1

Q
e − xt= 1 − 
 Q1 

Q
− xt= ln 1 −

 Q1 
1 
Q
t=
− ln 1 −

x  Q1 
(b)
(c)
1
Q 
1
1 
Q=
Q1 , t =
− ln 1 − 2 1 
2
x 
Q1 
1 
1
=
− ln 1 − 
x 
2
1 1
= − ln  
x 2
1
= ln 2
x
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 31 of 32
3p 9q = 2187
3p+2q = 37
p + 2q = 7
ln(4p − q) = ln2 + ln5 = ln(10)
4p − q = 10
8p − 2q = 20
[1] + [2] ⇒ 9p = 27, p = 3
3 + 2q = 7 ⇒ q = 2
p = 3, q = 2
17
log4 (2x − 3y) = 2
2x − 3y = 42
2x − 3y = 16
log3x − log3 (2y + 1) = 1
 x 
log 3 
=1
 2y + 1
x
=3
2y + 1
x = 6y + 3
x −6y = 3
2x − 12y = 6
[1] – [2] ⇒ 9y = 10
10
y=
9
20
29
 10 
x = 6   + 3=
+ 3=
 9
3
3
29
10
=
x =
, y
3
9
1+ 7 1+ 7 2 + 7
18
=
×
2− 7 2− 7 2+ 7
16
=
[1]
[2]
[1]
[2]
2+ 7 +2 7 +7
4−7
9+3 7
−3
=− 3 − 7
=
2− 7
1
1
−3 + 7
Now = =
×
1 + 7 − 3 − 7 −3 − 7 −3 + 7
=
−3 + 7 −3 + 7
=
9−7
2
3 1
=
− +
7
2 2
1+ 7 2 − 7
3 1
∴
+
=− 3 − 7 − +
7
2
2
2 − 7 1+ 7
−9 1
=
−
7
2 2
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 32 of 32
Hence
19
1+ 7 2 − 7
9 1
+
=− −
7
2 2
2 − 7 1+ 7
(1 − x 2 ) (2 + x)
1− x
(1 − x) (1 + x) (2 + x) 1 + x
=
×
since (1 − x) 2 =(1 − x) (1 + x)
1− x
1+ x
=
(1 − x) (1 + x) (2 + x) (1 + x )
1− x
=+
(1 x) (2 + x) (1 + x )
Unit 1 Answers: Chapter 5
© Macmillan Publishers Limited 2013
Page 1 of 34
Chapter 6 Functions
Try these 6.1
(a)
(b)
(c)
This relation is a function.
This relation is not a function since e does not map onto any values in the range.
1 → 7 and 1 → 8, Since 1 maps onto two different values, this relation is not a function.
Try these 6.2
(a)
(b)
(c)
x – 4 must be positive in order to find a real solution for √(x – 4)
∴ x – 4 ≥ 0, x ≥ 4
Hence, domain is x: x ∈ ℝ, x ≥ 4
Range is y ∈ ℝ
x + 1 > 0, x > -1
Hence, domain is x: x ∈ ℝ, x > -1
Range is y ∈ ℝ
Domain is x ∈ ℝ
As x → ∞, ex → ∞, y → ∞
As x → −∞, ex → 0, y → 1
Range is y: y ∈ ℝ, y > 1
Exercise 6A
1
2
3
This is not a function since 7 → 8 and 7 → 9
i.e. the value in the domain maps onto two different values in the range.
(b) This relation is a function and the mapping is:
–1→2
1→2
–3→8
3→8
(c) Not a function since c maps onto two different values.
(d) This relation is a function and the mapping is:
1→2
2→4
3→6
4→8
(a) This is a function {(a,1), (b,1), (c,2), (d,3)}
(b) This is not a function since t maps onto two different values.
(c) This is a function since each value in the domain maps onto one value in the range.
{(2, a), (4, b), (6, c), (8, d), (10, c), (12, b)}
g(x) = 4x–5
(a) x = 0,
g(x) = –5
x = 1,
g(x) = 4 – 5 = –1
x = 2,
g(x) = 8 – 5 = 3
(a)
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 2 of 34
4
5
x = 3,
g(x) = 12 – 5 = 7
x = 4,
g(x) = 16 – 5 = 11
Range of g is {–5, –1, 3, 7, 11}
(b) {(0, – 5), (1, –1), (2, 3), (3, 7) (4, 11)}
f (x) =
x+2
if x < 2
x–1
if x > 8
2
(x + 1)
if 2 ≤ x ≤ 8
(a) f (–4) = – 4 + 2 = –2
(b) f (9) = 9 – 1 = 8
(c) f (2) = (2 + 1)2 = 9
(d) f (8) = (8 + 1)2 = 81
f(x) = 5x – 2,
x∈ℝ
(a)
6
7
Any line y = c drawn parallel to the x-axis will cut the graph only once.
⇒ f (x) is one-to-one.
(b)
Any line drawn parallel to the x-axis cuts the graph at least once (exactly once in this case):
for every y there is a corresponding x mapping onto it. Hence f(x) is an onto function.
1
f (x) =
x−4
1
f (a) =
a−4
1
f (b) =
b−4
f (a) = f (b)
1
1
=
⇒
a−4
b−4
⇒a–4=b–4
a= b
Since f (a) = f (b) ⇒ a = b
f (x) is one-to-one or injective
g: ℝ → ℝ
Let us draw the graph of g (x) and use the graph to answer both (a) and (b)
g (x) = x + 2, x ≥ 0
g (x) = x, x < 0
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 3 of 34
8
9
(a) g (x) is injective since every x maps onto one and only one y
(b) g (x) is not surjective since there are no x-values mapping onto the y-values from 0 to 2
g(x) = (x – 2)2
Any line drawn parallel to the x-axis will cut the graph at least once for y ∈ ℝ +
Hence every y has a corresponding x mapping onto. Therefore g (x) is surjective.
(a) f(x) = x2 + 2, x ≥ 0
x + 2, x < 0
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 4 of 34
(b)
Any line drawn parallel to the x-axis cuts the graph once.
Hence f(x) is one-to-one and onto
∴ f (x) is bijective
Let y = x2 + 2
To find the inverse, let
x = y2 + 2
y2 = x – 2
y= ± x−2
Since the range of f(x) ≥ 0,
f-1 (x) = x − 2 for x ≥ 2
Let y = x + 2
x=y+2
y=x–2
 x − 2, x ≥ 2
∴ f–1 (x) = 
 x - 2, x < 2
10
(a)
(b)
Since a line drawn parallel to the x-axis cuts the graph at most once ⇒ g(x) is one-to-one.
When y = 0, there is no x-value mapping onto it
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 5 of 34
11
12
Hence y is not surjective.
f (x) = 2x + 1, x ≥ 0
x – 1, x < 0
From the graph there are no x-values mapping onto y from –1 to 1 Hence f (x) is not surjective.
For f (x) to be bijective , f (x) must be both injective and surjective.
∴ f (x) is not bijective.
(a)
(b)
(c)
(d)
Any line drawn parallel to the x-axis for y > 4 will cut the graph twice.
Hence f (x) is not injective.
Any line drawn parallel to the x-axis for y < 4 will not cut the graph, hence f (x) is not
surjective.
g: x → x2 + 4,
x ≥ 0, f (x) ≥ 4
y = x2 + 4
x = y2 + 4
y2 = x – 4
y= x−4
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 6 of 34
Hence g–1(x) =
13
(a)
(b)
x−4,x≥4
α → 1 and α → 4, since the same x-value maps onto two different y-values, g is not a
function. Also µ has no value assigned to it.
Removing the mapping α → 4, and assigning µ → 3
we have f: {(α, 1), (β, 2), (δ, 3), (ε, 4), (θ, 2), (µ, 3)}
Try these 6.2
(a)
(b)
(c)
This relation is a function.
This relation is not a function since e does not map onto any values in the range.
1 → 7 and 1 → 8, Since 1 maps onto two different values, this relation is not a function.
Exercise 6B
1
2
3
4
f(x) = 4x – 2
g(x) = 6x +1
gf = g(4x – 2)
gf: x → 24x–11
= 6 (4x – 2) + 1
= 24x – 11
fg = f (6x + 1)
fg: x → 24x + 2
= 4 (6x + 1) –2
= 24x + 2
f(x) = 3x + 5
g(x) = 2x2 + x + 1
gf = g (3x + 5)
gf: x → 18x2 – 63x + 56
2
= 2(3x + 5) + 3x + 5 + 1
= 2 [9x2 + 30x +25] + 3x + 6
= 18x2 + 63x + 56
fg = f (2x2 + x + 1) fg : x → 6x2 + 3x + 8
= 3(2x2 + x + 1) + 5
= 6x2 + 3x + 8
f(x) = x + 4
2
g(x) =
x
2
gf = g (x + 4) =
x+4
2
2
fg = f   =
+4
x
x
2
gf: x →
,x≠–4
x+4
2
fg: x → + 4, x ≠ 0
x
f(x) = 1 + 5x
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 7 of 34
x +1
x -1
gf = g(1 + 5x)
1 + 5x + 1
=
1 + 5x -1
5x + 2
=
5x
 x +1
fg = f 
 x -1 
g(x) =
5
6
7
 x +1
=1+5 
 x -1 
x − 1 + 5x + 5
=
x -1
6x + 4
=
x -1
5x + 2
gf: x →
,x≠0
5x
6x+4
fg: x →
, x ≠1
x-1
g(x) = 2x – 1
g2 = g (2x–1) = 2(2x–1) – 1 = 4x – 3
g3 = gg2 = g (4x – 3) = 2(4x – 3) – 1 = 8x – 7
x
g(x) =
2x+1
x
 x 
2x + 1
g2 = g 
=

2x
 2x + 1
+1
2x + 1
x
2x + 1
=
2x + 2x + 1
2x + 1
x
1
1
=
,x≠− ,x≠−
4x + 1
2
4
x
x


x
x
−1
−1
−1
4x + 1
4x + 1
g3 = g 
, x≠ , x≠ , x≠
=
=
=

2x
2x
+
4x
+
1
4x
+
1
6x
+
1
2
4
6


+1
4x + 1
4x + 1
2x + 1
g(x) =
x +1
 2x + 1
g2 = g 
 x + 1 
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 8 of 34
 2x + 1
2
+1
 x + 1 
=
2x + 1
+1
x +1
4x + 2 + x + 1
x +1
=
2x + 1 + x + 1
x +1
5x + 3 x + 1
=
×
x + 1 3x + 2
2
5x + 3
=
, x ≠ − 1, x ≠ –
3
3x + 2
 5x + 3 
g3 = g 
 3x + 2 
8
 5x + 3 
2
+1
 3x + 2 
=
5x + 3
+1
3x + 2
10x + 6 + 3x + 2
3x + 2
=
5x + 3 + 3x + 2
3x + 2
13x + 8
2
5
=
, x ≠ − 1, x ≠ − , x ≠ − .
8x + 5
3
8
3
g(x) =
4x − 2
 3 
g2 = g 
 4x − 2 
3
=
12
−2
4x − 2
3
12 − 8x + 4
=
4x − 2
3(4x − 2)
=
16 − 8x
6(2x − 1)
=
8(2 − x)
3(2 x − 1)
=
4(2 − x)
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 9 of 34
9
10
 3(2x − 1) 
g3 = g 
 4(2 − x) 
3
=
6x − 3
−2
2−x
3(2 − x)
=
6x − 3 − 4 + 2x
3(2 − x)
=
8x − 7
g(x) = x2 + 2x + 3
h(x) = x − 2
gh = g (x − 2)
= (x − 2)2 + 2(x − 2) + 3
= x2 − 4x + 4 + 2x − 4 + 3
= x2 − 2x + 3
hg = h (x2 + 2x + 3)
= x2 + 2x + 3 − 2
= x2 + 2x + 1
Now gh = hg
⇒ x 2 − 2x + 3 = x 2 + 2x + 1
2 = 4x
1
x=
2
f(x) = 3x − 4
x
g(x) =
x+2
 x 
3x
fg = f 
−4
 =
x
+2
x
+
2


3x − 4x − 8
=
x+2
−x − 8
=
x+2
gf = g (3x − 4)
3x − 4
3x − 4
=
=
3x − 4 + 2 3x − 2
fg = gf
− x − 8 3x − 4
⇒
=
x+2
3x − 2
⇒ (−x − 8) (3x − 2) = (3x − 4) (x + 2)
⇒ − 3x 2 + 2x − 24x + 16 = 3x 2 + 6x – 4x – 8
6x2 + 24x − 24 = 0
x2 + 4x − 4 = 0
(x–2)2 = 0, x = 2
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 10 of 34
11
(a)
y = 4x − 3
x = 4y – 3
x+3
=y
4
∴ f–1: x →
(b)
(c)
12
(a)
x+3
4
5
x−2
5
x=
y−2
xy − 2x = 5
5 + 2x
y=
x
5 + 2x
,x≠0
∴ f–1: x →
x
3x − 1
y=
x+2
3y − 1
x=
y+2
⇒ xy + 2x = 3y − 1
⇒ xy − 3y = − 1 − 2x
y(x − 3) = − 1 − 2x
− 1 − 2x
y=
x−3
2x + 1
=
3−x
2x + 1
∴ f −1: x →
,x≠3
3−x
4x2 + 12x + 3 = a(x + b)2 + c
= ax2 + 2abx + ab2 + c.
Equating coefficients:
⇒a=4
12
3
2ab = 12 ⇒ 8b = 12, b =
=
8
2
2
 3
ab2 + c = 3 ⇒ 4   + c = 3, c = − 6
 2
Hence
y=
2
3

4x2 + 12x + 3 = 4  x +  − 6

2
The range of the function is y ≥ − 6
(b)
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 11 of 34
13
(a)
(b)
A line y = c drawn parallel to the x-axis cuts the graph more than once hence the function is
not one-to-one and therefore has no inverse.
4x − 1
f(x) =
3x + 2
4x − 1
Let y =
3x + 2
4y − 1
x=
3y + 2
x (3y + 2) = 4y − 1
3xy + 2x = 4y − 1
3xy – 4y = − 2x − 1
y(3x − 4) = − 2x − 1
− 2x − 1
y=
3x − 4
1 + 2x
=
4 − 3x
1 + 2x
4
∴ f–1 (x) =
,x≠
4 − 3x
3
1+ 2
f–1 (1) =
=3
4−3
1− 2
1
f–1 (−1) =
= −
4+3
7
f–1 (x) = x
1 + 2x
=x
⇒
4 − 3x
⇒1 + 2x = x(4 − 3x)
⇒ 1 + 2x = 4x − 3x2
⇒ 3x2 − 2x + 1 = 0
2 ± 4 − 12
x=
6
2± −8
=
6
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 12 of 34
Since
14
− 8 is not real
⇒ there is no real x for which
f–1 (x) = x
4x
, g(x) = x + 4
f (x) =
x +1
(a) k = −1
(b) fg = f(x + 4)
4(x + 4)
=
x + 4 +1
4x + 16
=
x+5
fg is not defined when x = −5
4x
(c) Let y =
x +1
4y
x=
y +1
xy + x = 4y
xy − 4y = − x
y(x − 4) = − x
−x
x
y=
=
x−4 4−x
x
f–1 (x) =
,x≠ 4
4−x
a
(d) f–1 (a) =
, g( −1) =
3
4−a
a
= 3 ⇒ a = 12 − 3a, 4a = 12, a = 3
4−a
Try these 6.3
(a)
y=
x +1
x−2
1
x − 2 x +1
x−2
3
∴y=1+
f(x) =
3
x −2
1
x
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 13 of 34
1
x−2
3
3 f(x − 2) =
x−2
f(x − 2) =
1 + 3 f(x − 2) = 1 +
3
x−2
1
is shifted to the right by 2 units and stretched along the y-axis by factor 3. This graph is then
x
moved upwards by 1 unit.
(b)
y=
2x +1
x +1
2
x + 1 2x + 1
2x + 2
−1
∴y=2−
1
x +1
1
1
1
⇒ f (x + 1)
⇒ – f (x + 1) = −
x +1
x +1
x
1
⇒ 2 − f(x + 1) = 2 −
x +1
1
∴ is shifted to the left by 1 unit, then reflected in the x-axis and moved upwards by 2 units.
x
f(x) =
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 14 of 34
Exercise 6c
1
2
y = 2 f(x)
y = f(x) − 4
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 15 of 34
3
4
2x2 – 4x + 6
= 2(x2 − 2x) + 6
= 2 (x – 1)2 + 6 – 2
= 2(x − 1)2 + 4
y = 2(x − 1)2 + 4
Starting with y = x2, shift to the right by 1 unit, then stretch along the y-axis by factor 2 and up the
y-axis by 4 units.
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 16 of 34
5
1
x + 2 x +1
x+2
−1
∴y=1−
f(x) =
1
x+2
1
x
f(x + 2) =
1
x+2
− f(x + 2) = −
1
x+2
1 − f(x + 2) = 1 −
1
x+2
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 17 of 34
Transformations are:
6
shift
1
to the left by 2 units followed by a reflection in the x-axis and a
x
translation upwards by 1 unit.
(a)
(b)
7
(a)
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 18 of 34
(b)
(c)
8
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 19 of 34
9
3
10
x + 2 3x + 1
3x + 6
−5
5
y=3−
x+2
1
Shift
to the left by 2 units, reflect in the x-axis and stretch by factor 5 units along the y-axis
x
finally move upwards by 3 units.
2x 2 − 8x + 9
y= 2
x − 4x + 4
2
x 2 − 4x + 4 2x 2 − 8x + 9
2x 2 − 8x + 8
1
1
x − 4x + 4
1
= 2+
( x − 2) 2
1
Shift to the right by 2 units move upwards by 2 units
x
y=2+
2
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 20 of 34
11
(a)
y=
4x − 2
x−4
4
x − 4 4x − 2
4x − 16
14
14
y=4+
x−4
(b)
y=
4x 2 + 4x + 2
4x 2 + 4x +1
1
4x 2 + 4x + 1 4x 2 + 4x + 2
4x 2 + 4x + 1
1
1
y=1+
4x 2 + 4 x + 1
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 21 of 34
= 1+
1
1
= 1+
2
2
(2x + 1)
1

4 x + 

2
Review Exercise 6
1
2
3
(a)
(b)
(c)
(d)
(e)
(f)
(a)
(b)
(c)
Function
Not a function, since a maps onto two different values and d has no correspondence
Function
Function
Not a function since β has no corresponding mapping.
Function
Function
Function
Not a function since 6 → 7 and 6 → 8. Every value in the domain must map onto one value
in the range.
(d) Not a function since c → d and c → e
f(x) = 4x − 3
f(4) = 16 –3 = 13
f(−3) = – 12 – 3 = – 15
 1
f  = 2 – 3 = – 1
 2
1
7
 1
f−  = – – 3 = –
2
8
2


∴ Images are 13, – 15, – 1, –
Unit 1 Answers: Chapter 6
7
2
© Macmillan Publishers Limited 2013
Page 22 of 34
4
f(x) = 4 –
3 13
=
4
4
3
5
(b) f(2) = 4 – =
2
2
3
(c) f (0) = 4 – (0) = 4
4
3
(d) f (4) = 4 – (4) = 1
4
f (x) = 3x – 7
g(x) = 4x + 2
(a) f(0) + g(2) = 3(0) – 7 + 4(2) + 2
=3
(b) 2 f(3) + g(1)
= 2[3(3) − 7] + 4(1) + 2
= 4 + 4 + 2 = 10
(c) 2 f(1) − 3 g (2)
= 2 [3 − 7] − 3[8 + 2]
= − 8 − 30
= − 38
(d) 4 f(–1) + 3g (− 2)
= 4 [−3 − 7] + 3[− 8 + 2]
= − 40 − 18
= − 58
1
3
(a) f(x) =
x+
2
4
5
2
g(x) = x +
6
3
f(x) = g(x)
1
3
5
2
x+ =
x+
2
4
6
3
3
2
2
− =
x
4
3
6
1
2
=
x
12
6
1
x=
4
1
(b) f(x) = x
4
1
3
1
= x
⇒ x+
2
4
4
1
3
x=−
4
4
x=−3
(a)
5
6
3
x
4
f(1) = 4 –
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 23 of 34
(c)
(d)
7
(a)
(b)
f(2x) = 3g (x)
1
3
2
5
(2x) + = 3  x + 
3
2
4
6
3
5
⇒x + =
x+2
4
2
3
3
−2= x
4
2
−5 3
= x
4
2
5
x= −
6
1 
1 
f  x = g  x
4 
2 
1 1 
3
51  2
=  x +
 x +
6 4  3
2 2
4
1
3 5
2
x+ =
x+
4
4 24
3
1
5
2 3
x−
x= −
4
24
3 4
1
1
x=−
24
12
x=−2
f(x) = 4x + 1, − 2 ≤ x ≤ 2
x = 2, f(x) = 8 + 1 = 9
x = − 2, f(x) = − 8 + 1 = −7
Range : −7 ≤ y ≤ 9
f (x) = 2 − 3x, − 1 ≤ x ≤ 1
x = 1, f (x) = 2 − 3 = − 1
x = −1, f (x) = 2 − 3(− 1) = 5
Range : −1 ≤ y ≤ 5
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 24 of 34
(-1,5)
2
8
(c)
f(x) = 2 + x, x ≥ 1
Range : y ≥ 3
(d)
f(x) = − 3x + 6, x ≤ 0
x = 0, f(x) = 6
Range : y ≥ 6
(a)
f(x) = x2 − 4
Domain: x ∈ ℝ
Range:
y≥−4
f (x) = x2 + 2x + 3
= (x + 1)2 + 2
Min pt. at (−1, 2)
Domain: x ∈ ℝ
Range:
y≥2
2
f(x) = − 4x + x + 1
(b)
(c)
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 25 of 34
1 

= − 4  x2 − x  + 1

4 
2
17
1

= − 4x−  +
8
16

 1 17 
Max pt. at  , 
 8 16 
Domain: x ∈ ℝ
y≤
Range:
(d)
17
16
f (x) = − (x2 + 3x) + 5
2
9
3

= –x +  + 5 +
4
2

2
9
(a)
29
3

= −x +  +
4
2

 3 29 
Max pt at  − , 
 2 4
Domain: x ∈ ℝ
29
Range:
y≥
4
Minimum value = 1
When x = 2
(b)
10
(a)
2
x
Rearrange to make x the subject
y=1−
1–y=
(b)
2
2
,x=
x
1− y
Domain: x ∈ ℝ, x ≠ 0
Range: y ∈ ℝ, y ≠ 1
4x + 2
y=
x−3
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 26 of 34
y(x – 3) = 4x + 2
xy – 4x = 2 + 3y
x=
11
12
13
2 + 3y
y−4
Domain: x ∈ ℝ, x ≠ 3
Range: y ∈ ℝ, y ≠ 4
(c) y = x − 4
y2 = x – 4
x = y2 + 4
Domain: x ∈ ℝ, x > 4
Range: y ∈ ℝ, y ≥ 0
f(x) = 6x + 2
g(x) = 7x − 1
fg = f (7x − 1)
= 6(7x − 1) + 2
= 42x − 4
gf= g (6x + 2)
= 7 (6x + 2) − 1
= 42x + 13
fg(0) = 42(0) − 4 = − 4
fg(−2) = 42(− 2) − 4 = −84 − 4 = − 88
gf(0) = 42(0) + 13 = 13
gf (– 2) = 42(− 2) + 13 = −84 + 13 = –71
g(x) = x + 3
gh = x2 + 3x – 2
h = g–1 (x2 + 3x – 2)
Let y = x + 3
x=y+3
y=x–3
g–1 (x) = x – 3
∴ h(x) = x2 + 3x – 2 – 3
= x2 + 3x – 5
x +1
fg =
x+2
 x + 1
g = f–1 
 x + 2 
f(x) = 4x – 2
y = 4x – 2
x = 4y – 2
x+2
y=
4
x+2
f– 1 (x) =
4
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 27 of 34
14
x +1
+2
x+2
g=
4
x + 1 + 2x + 4
=
4(x + 2)
3x + 5
=
4(x + 2)
3
f(x) =
,x≠−1
x +1
g(x) = 4x − 2
(a) (i)
fg = f (4x − 2)
3
=
4x − 2 + 1
3
1
=
,x≠ ,x≠−1
4x − 1
4
(ii)
15
 3 
gf = g 
 x + 1
 3 
=4 
−2
 x + 1
12
=
−2
x +1
12 − 2x − 2
=
x +1
10 − 2x
=
, x ≠ −1
x +1
(b) fg = gf
3
10 − 2x
⇒
=
4x − 1
x +1
⇒ 3(x + 1) = (10 – 2x) (4x –1)
⇒ 3x + 3 = 40x – 10 – 8x2 + 2x
⇒ 8x2 – 39x + 13 = 0
b2 – 4a c = (– 39)2 – 4(8) (13)
= 1105
Since b2 – 4ac > 0 ⇒ there are two real and distinct solutions to fg = gf
6
,x≠k
f(x) =
x−3
g(x) = 5x – 3
(a) k = 3
 6 
(b) (i)
gf = g 
 x − 3 
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 28 of 34
30
−3
x−3
30
gf: x →
− 3 , x ∈ ℝ, x ≠ 3
x−3
6
(ii) y =
x−3
6
x=
y−3
xy − 3x = 6
xy = 6 + 3x
6 + 3x
y=
x
6 + 3x
,x≠0
f −1 (x) : x →
x
y = 5x − 3
x = 5y − 3
x+3
y=
5
x+3
g −1 (x) =
5
7
−1
g (4) =
5
=
(c)
7
4
fg −1 (4) = f  
=
6
7
−3
5
15
= −
4
16
4x + 1
3x − 2
4y + 1
x=
3y − 2
3xy − 2x = 4y + 1
3xy − 4y = 2x + 1
y (3x − 4) = 2x + 1
2x + 1
y=
3x − 4
2x + 1
4
f−1 : x →
,x≠
3x − 4
3
2(2) + 1 5
(a) f − 1 (2) =
=
3(2) − 4 2
y=
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 29 of 34
(b)
1
4  + 1
3
2
=
= −6
1
1
3  − 2 −
2
2
− 24 + 1 23
 1
ff   = f(− 6) =
=
 2
−18 − 2 20
ff − 1 (4) = 4
2(4) + 1 9
or f − 1 (4) =
=
3(4) − 4 8
 1
f  =
 2
(c)
 9
4   +1
 8
 9
ff − 1 (4) = f   =
 8
 9
3  −2
 8
11
= 2
11
8
=4
2
17
f(x) =
x −1
g(x) = λx2 − 1
2
(a) f(3) =
=1
3 −1
g(1) = λ − 1
1
gf (3) =
5
1
⇒λ−1=
5
1
λ=1
5
(b) f2 (x) = ff (x)
 2 
= f
 x − 1
2
=
2
−1
x −1
2
2 − x +1
=
x −1
2x − 2
=
,x≠3
3−x
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 30 of 34
18
a = 2, b = − 2
c=−1,d=3
f(x) = 3x − 2, x ∈ ℝ
2
g(x) =
, x ∈ ℝ, x ≠ 2
x−2
(a) y = 3x − 2
x = 3y − 2
x+2
y=
3
x+2
f − 1 (x) =
3
2
y=
x−2
2
x=
y−2
xy − 2x = 2
2 + 2x
y=
x
2 + 2x
g − 1 (x) =
,x≠0
x
(b) fg (x) = x
 2 
f
=x
 x − 2 
6
⇒
−2=
x
x−2
⇒ 6 − 2(x − 2) = x (x − 2)
⇒ 6 − 2x + 4 = x2 − 2x
x2 = 10
x = ± 10
(c)
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 31 of 34
19
(a)
(b)
(c)
(d)
20
(a)
k=−3
fg = f(2x + 1)
3
=
2x + 1 + 3
3
=
2x + 4
3
fg : x →
, x ∈ ℝ, x ≠ − 2
2x + 4
fg is not defined at x = − 2
3
y=
x+3
3
x=
y+3
xy + 3x = 3
3 − 3x
y=
x
3 − 3x
−1
f (x) =
, x≠0
x
f − 1 (a) = g(4)
3 − 3a
= 2(4) + 1
a
3 − 3a = 9a
12a = 3
1
a=
4
4x
x+λ
f(x) =
, g(x) =
x −1
x
4x
y=
x −1
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 32 of 34
4y
y −1
xy − x = 4y
xy − 4y = x
x
y=
x−4
x=
x
x−4
x
f − 1: x →
, x ∈ ℝ, x ≠ 4.
x−4
5
f − 1 (5) =
= 5.
5−4
g(5) = 5.
5+ λ
=5
5
λ = 20.
3x2 + 12x + 5 = a (x + b)2 + c
= ax2 + 2abx + ab2 + c
Coefficient of x2 : a = 3
Coefficient of x: 2ab = 12
b=2
Constants: ab2 + c = 5
12 + c = 5
c=−7
3x2 + 12x + 5 = 3 (x + 2)2 − 7
(i)
y≥−7
(ii) x = 0, f(0) = 3(2)2 − 7 = 5
f − 1 (x) =
(b)
21
(a)
(b)
The line y = c, c > -7, cuts the graph more than once ⇒ f(x) is not one-to-one.
g(x) = 3 (x + 2)2 − 7, x ≥ −2
(i)
k=−2
(ii) y = 3(x + 2)2 − 7
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 33 of 34
x = 3(y + 2)2 − 7
x+7
(y + 2)2 =
3
x+7
y+2=±
3
y=−2±
x+7
3
g − 1 (x) = − 2 +
22
23
x+7
3
f(1) = 4(1) − 1 = 3
f(3·5) = 2(3·5) + 5 = 12
f(5) = 52 + 3 = 28
  1 
 1
(d) ff   = f  4   − 1 = f(1) = 4 − 1 = 3
 2
  2 
2
For x < 0 ; Gradient =
=+1
+2
Equation: y − 2 = 1(x − 0)
y=x+2
For x = 1, y = 2
4−0 4
For x > 1 , Gradient =
=
4 −1 3
4
y − 4 = (x − 4)
3
4
16
x−
+4
y=
3
3
4
4
y=
x−
3
3

x + 2, x < 0

∴ f(x) =  2, x = 1
4
4
 x − , x >1
3
3
(a)
(b)
(c)
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 34 of 34
24
Unit 1 Answers: Chapter 6
© Macmillan Publishers Limited 2013
Page 1 of 9
Chapter 7 Cubic Polynomials
Review Exercise 7
1
(a)
2x2 = 3x – 5
2x2 – 3x + 5 = 0
3
α+β=
2
5
αβ =
2
α
β
+
2β + 1 2α + 1
α (2α + 1) + β (2β + 1)
=
(2β + 1) (2α + 1)
=
2α 2 + α + 2β2 + β
4αβ + 2β + 2α + 1
=
2 (α 2 + β2 ) + α + β
4αβ + 2 (α + β) + 1
=
2 [(α + β)2 − 2αβ] + (α + β)
4αβ + 2 (α + β) + 1
  3 2
 5  3
2   − 2   +
 2 2
 2
=
 5
 3
4  + 2  +1
 2
 2
− 11 3
+
4 −2
2 =
2 −=
=
10 + 3 + 1 14
7
α
β
−2
(b) Sum of the roots =
+
=
2β + 1 2α + 1 7
2
 α  β 
Product of roots = 
 2β + 1  2α + 1
5
αβ
5
=
= 2=
(2β + 1) (2α + 1) 14 28
The equation is:
2
5
x2 + x +
=
0
7
28
28x2 + 8x + 5 = 0
2
2x – 5x + 7 = 0
5
α+β=
2
7
αβ =
2
Unit 1 Answers: Chapter 7
© Macmillan Publishers Limited 2013
Page 2 of 9
Sum of the roots = α +
= (α + β) +
1
1
+β+
α
β
β+α
αβ
5
5 2
=
+
2 7
2
5 5 45
= + =
2 7 14
1 
1

Product of the roots =  α +   β + 


α 
β
=
αβ +
α β
1
+ +
β α αβ
= αβ +
α 2 + β2
1
+
αβ
αβ
= αβ +
(α + β)2 − 2αβ 1
+
αβ
αβ
2
 5
 7
−2 
 2 1
7  2 
=
+
+
7
7
2
2
2
7  3 2
= + −  +
2  14  7
25
=
7
The equation with roots α +
1
1
, β + is
α
β
45
25
x+
=
0
14
7
14x2 – 45x + 50 = 0
(a) 3x2 = – (2x – 1)
⇒ 3x2 + 2x – 1 = 0
−2
α+β=
3
1
αβ = −
3
x2 −
3
4
−2
α 2 + β2 + 2αβ = (α + β) 2 = 
 =
 3  9
4
4
2
2
α −β +α −β
2
(b)
= (α 2 − β2 ) (α 2 + β2 ) + α 2 − β2
= (α 2 − β2 ) (α 2 + β2 + 1)
= (α − β) (α + β) ((α + β)2 − 2αβ + 1)
Unit 1 Answers: Chapter 7
© Macmillan Publishers Limited 2013
Page 3 of 9
2
 − 2  − 2
 1 
= 
− 2  −  + 1 (α − β)




 3   3 
 3 
=
− 38
(α − β)
27
−2
 1  4 4 16
(α − β) = (α + β) − 4αβ = 
 − 4 −  = + =
 3 
 3 9 3 9
16 4
α −=
β
=
, α >β
9 3
− 38 4 − 152
∴ α 4 − β4 + α 2 −=
β2
×=
27 3
81
x3 – 10x + 6 = 0
α+β+γ=0
α β + α γ + β γ = – 10
αβγ=–6
2
2
4
(a)
(b)
(c)
2
∑ α= ( ∑ α ) − 2 ∑ αβ
2
2
= (0)2 – 2 (– 10)
= 20.
x = α ⇒ α3 – 10α + 6 = 0
x = β ⇒ β3 – 10β + 6 = 0
x = γ ⇒ γ3 – 10γ + 6 = 0
Add] ⇒ ∑ α 3 − 10 ∑ α + 18 =0
∑ α − 10 (0) + 18 =0
∴ ∑ α =− 18.
3
3
5
(a)
2x3 – x2 – 10x – 6 = 0
1
∑α = α + β + γ = 2
∑ αβ = − 5
− (− 6)
= 3
2
1
∑ α =2
=
αβγ
(b)
∑ α= ( ∑ α ) − 2 ∑ αβ
2
2
2
(c)
 1
=   − 2 (− 5)
 2
1
= 10
4
x = α, 2α 3 − α 2 − 10α − 6 = 0 

x = β, 2β3 − β2 − 10β − 6 = 0  add
x = γ , 2 γ 3 − γ 2 − 10 γ − 6 = 0 
⇒ 2 ∑ α 3 − ∑ α 2 − 10 ∑ α − 18 =0
 1
 1
∴ 2 ∑ α 3 −  10  − 10   − 18 =0
 4
 2
Unit 1 Answers: Chapter 7
© Macmillan Publishers Limited 2013
Page 4 of 9
1
1
α
18 + 5 + 10 
∑=
2 
4
3
133
8
(a) x3 + 4x + 1 = 0
Since the roots are p, q and r
p+q+r=0
pq + pr + qr = 4
pqr = – 1
p2 + q2 + r2 = (p + q + r)2 – 2 (pq + pr + qr)
= (0)2 – 2 (4)
=–8
(b) p3 + 4p + 1 = 0
q3 + 4q + 1= 0
r3 + 4r + 1= 0
p3 + q3 + r3 + 4 (p + q + r) + 3 = 0
p3 + q3 + r3 + 4 (0) + 3 = 0
∴ p3 + q3 + r3 = – 3
2x3 – 4x2 + 6x – 1 = 0
− (− 4)
α +=
β+γ
= 2
2
6
αβ + αγ + βγ= = 3
2
1
αβγ =
2
1 1 1
Sum of the roots = + +
α β γ
βγ + αγ + αβ
=
αβγ
3
= = 6
1
2
Sum of product of pairs
 1   1  1   1  1  1
=    +     +    
 α   β  α   γ   β  γ 
=
6
7
=
γ +β+α 2
= = 4
1
αβγ
2
 1   1  1
Product of the roots =      
 α   β  γ 
1
αβγ
1
= = 2
1
2
=
The equation with roots
Unit 1 Answers: Chapter 7
1 1 1
, , is
α β γ
© Macmillan Publishers Limited 2013
Page 5 of 9
x3 – 6x2 + 4x – 2 = 0
OR:
1
1
Let y = ⇒ x =
x
y
1
Substituting x = into the equation
y
3
8
9
2
1
1
1
2   − 4   + 6   −1=
0
y
y
y
× y3 ⇒ 2 – 4y + 6y2 – y3 = 0
∴ y3 – 6y2 + 4y – 2 = 0
1 1 1
Since x = α, β, γ ⇒ y = , ,
α β γ
1 1 1
The equation with roots , , is
α β γ
y3 – 6y2 + 4y – 2 = 0
The roots of the equation are – 2, – 3 and 4
∴ (x + 2) (x + 3) (x – 4) = 0
⇒ (x2 + 5x + 6) (x – 4) = 0
⇒ x3 – 4x2 + 5x2– 20x + 6x – 24 = 0
⇒ x3 + x2 – 14x – 24 = 0
The equation with roots – 2, – 3 and 4 is
x3 + x2 – 14x – 24 = 0
Comparing with x3 + αx2 + βx+ γ = 0
⇒ α = 1, β = – 14, γ = – 24
OR:
Sum of the roots = (– 2) + (– 3) + 4
=–1
Sum of the product of pairs = (– 2) (– 3) + (– 2) (4) + (– 3) (4)
= 6 – 8 – 12 = – 14
Product of the roots = (– 2) (– 3) (4) = 24.
The equation is
x3 + x2 – 14x – 24 = 0
∴ α = 1, β = – 14, γ = – 24
(a) x3 + 6x2 + 10x + 14 = 0
α + β + γ = – 6, αβ + αγ + βγ = 10, αβγ = – 14
Let y = x 2 ⇒ x = y
Substituting into the equation:
( y)3 + 6 ( y) 2 + 10 y + 14 =
0
y y + 6y + 10 y + 14 =
0
(b)
y (y + 10) =
− 14 − 6 y
Squaring: y (y + 10)2 = (– 14 – 6 y)2
y (y2 + 20y + 100) = 196 + 168y + 36y2
y3 – 16y2 – 68y – 196 = 0
Hence the equation with roots α2, β2 and γ2 is
y3 – 16y2 – 68y – 196 = 0
Let y = x + 3
⇒x=y–3
Substituting into the equation:
Unit 1 Answers: Chapter 7
© Macmillan Publishers Limited 2013
Page 6 of 9
(y – 3)3 + 6 (y – 3)2 + 10 (y – 3) + 14 = 0
⇒ y3 – 9y2 + 27y – 27 + 6(y2 – 6y + 9) + 10y – 30 + 14 = 0
⇒ y3 – 3y2 + y + 11 = 0
The equation with roots α + 3, β + 3, γ + 3 is
y3 – 3y2 + y + 11 = 0
OR:
Sum of the roots = α + 3 + β + 3 + γ + 3
= (α + β + γ) + 9
=–6+9=3
Sum of the product of pairs = (α + 3) (β + 3) + (α + 3) ( γ + 3) + (β + 3) ( γ + 3)
= αβ + 3α + 3β + 9 + αγ + 3α + 3γ + 9 + βγ + 3β + 3γ + 9
= (αβ + αγ + βγ ) + 6 (α + β + γ ) + 27
= 10 + 6 (– 6) + 27
=1
10
Product of the roots
= (α + 3) (β + 3) (γ + 3)
= (αβ + 3α + 3β + 9) (γ + 3)
=αβγ + 3αβ + 3αγ + 9α + 3βγ + 9β + 9γ + 27
= αβγ + 3 (αβ + αγ + βγ ) + 9 (α + β + γ ) + 27
= – 14 + 3 (10) + 9 (– 6) + 27
= – 11
The equation with roots α + 3, β + 3, γ + 3 is
x3 – 3x2 + x + 11 = 0
3x3 – 4x2 + 8x – 7 = 0
1
1
Let y = ⇒ x =
x
y
3
2
 1
 1
 1
∴ 3  − 4   + 8   − 7 =
0
 y
 y
 y
⇒ 3 – 4y + 8y2 – 7y3 = 0
1 1 1
The equation with roots , , is
α β γ
7y3 – 8y2 + 4y – 3 = 0
OR:
3x3 – 4x2 + 8x – 7 = 0
4
∑ α =3
8
∑ αβ = 3
7
∑ αβγ = 3
8
1 1 1 αβ + βγ + γα 3 8
= =
Sum of the roots = + + =
7 7
α β γ
αβγ
3
 1   1  1   1  1  1
Sum of the product of pairs =     +     +    
 α   β  α   γ   β  γ 
Unit 1 Answers: Chapter 7
© Macmillan Publishers Limited 2013
Page 7 of 9
11
4
α+β+γ 3 4
=
= =
7 7
αβγ
3
 1   1  1
Product of roots =      
 α   β  γ 
1
1 3
=
= =
7
αβγ
7
3
1 1 1
The equation with roots , , is
α β γ
8
4
3
x3 − x2 + x − =
0
7
7
7
7x3 – 8x2 + 4x – 3 = 0
(a) 2x3 – x2 – 10x − 6 = 0
1
α+β+γ =
2
(b)
α 2 + β2 + =
γ2
( ∑ α ) − 2 ∑ αβ
2
2
 1
=   − 2 (− 5)
 2
1
= 10
4
(c)
9β2 + 9 γ 2 = 32α 2
32 2
2
β 2 + γ=
α
9
Substitute into α 2 + β2 + γ 2 =10
1
4
32 2 41
α =
9
4
41 2 41
9
−3
α =
⇒ α2 =
, α=
,α<0
9
4
4
2
1
1 3
α + β + γ=
⇒ β + γ=
+ = 2
2
2 2
γ= 2 − β
⇒ α2 +
 3
αβγ = 3 ⇒  −  β (2 − β) = 3
 2
2
2β − β = − 2
β 2 − 2β − 2 = 0
2± 4+8
2
2 ± 12
=
2
= 1± 3
β=
β > 0 ∴ β= 1 + 3 , γ= 1 − 3
Unit 1 Answers: Chapter 7
© Macmillan Publishers Limited 2013
Page 8 of 9
−3
, β = 1 + 3, γ = 1 − 3
2
x3 + ax2 + bx + c = 0
α+β+ γ = −a
αβ + βγ + γα = b
αβγ = − c
α + β + γ = 6 ⇒ − a= 6
a=–6
∑ α 2 =14
α=
12
∑ α= ( ∑ α ) − 2 ∑ αβ
2
2
= (−6)2 – 2b
36 – 2b = 14
2b = 22
b = 11
∑ α3 + a ∑ α 2 + b ∑ α + 3c =0
13
36 + (– 6) (14) + (11) (+ 6) + 3c = 0
36 – 84 + 66 + 3c = 0
c=6
∴ a = – 6, b = 11, c = 6
x3 + ax2 + bx + c = 0
∑ α = − a, ∑ αβ = b, αβγ = − c
Now ∑ α = 0 ⇒ a = 0
∑ α= 14 ⇒ ( ∑ α ) − 2 ∑ αβ= 14
2
2
– 2b = 14
b=–7
∑ α 3 + a ∑ α 2 + b ∑ α + 3c =0
14
18 + 0 – 7 (0) + 3c = 0
c=–6
a = 0, b = – 7, c = – 6
x3 – 6x + 3 = 0
x +1
Let y =
x
⇒ xy = x + 1
⇒ xy – x = 1
x (y – 1) = 1
1
x=
y −1
Substituting into the equation:
3
 1 
 1 
0
 y − 1 − 6  y − 1 + 3 =
× (y – 1)3 ⇒ 1 – 6 (y – 1)2 + 3 (y – 1)3 = 0
1 – 6 [y2 – 2y + 1] + 3 [y3 – 3y2 + 3y – 1] = 0
∴ 3y3 – 15y2 + 21y – 8 = 0
α +1 β +1 γ +1
is
∴ The equation with roots
,
,
α
β
γ
Unit 1 Answers: Chapter 7
© Macmillan Publishers Limited 2013
Page 9 of 9
15
3x3 – 15x2 + 21x – 8 = 0
(a) x3 – 2x2 + 4x + 5 = 0
α + β + γ =2
αβ + βγ + γα = 4
αβγ = − 5
Sum of the roots = 2α + 2β + 2 γ
= 2 (α + β + γ )
= 2 (2) = 4
Sum of the product of pairs = (2α) (2β) + (2α) (2 γ) + (2β) (2 γ)
= 4 [αβ + αγ + βγ ]
= 4 (4) = 16
Product of the roots = (2α ) (2β) (2 γ )
= 8 αβγ = 8 (− 5) = − 40
The equation is
x3 – 4x2 + 16x + 40 = 0
OR:
3
(b)
2
y
 y
 y
 y
y = 2x ⇒ x =
∴   −2  + 4  +5= 0




 2
2
2
2
3
2
× 8 ⇒ y – 4y + 16y + 40 = 0
1
1
=
y
⇒x=
x
y
3
2
1
1
1
∴  − 2   + 4   + 5 =
0
y
y
y
× y3 ⇒ 1 – 2y + 4y2 + 5y3 = 0
The equation with roots
1 1 1
, , is
α β γ
5x3 + 4x2 – 2x + 1 = 0
Unit 1 Answers: Chapter 7
© Macmillan Publishers Limited 2013
Page 1 of 18
Chapter 8 Inequalities and the Modulus Function
Try these 8.1
(a)
(b)
3x < x2 − 4
⇒ x2 − 3x − 4 > 0
(x − 4)(x + 1) > 0
{x: x < − 1} ∪ {x: x > 4}
6x2 − 11x − 7 ≥ 0
(2x + 1) (3x − 7) ≥ 0
1 
7

x : x ≤ −  ∪ x : x ≥ 
2 
3

Try these 8.2
(a)
3x + 2
<0
4x − 1
× (4x − 1)2 ⇒ (3x + 2) (4x − 1) < 0
1
4
(b)
1
 −2
∴ x :
<x< 
3
4

3x − 4
<5
x−2
× (x − 2)2 ⇒ (3x − 4) (x − 2) < 5 (x − 2)2
⇒ (3x − 4) (x − 2) − 5 (x − 2)2 < 0
⇒ (x − 2) [3x − 4 − 5x + 10] < 0
⇒ (x − 2) (−2x + 6) < 0
{x: x < 2} ∪ {x: x > 3}
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 2 of 18
(c)
x +1
>3
x+2
× (x + 2)2 ⇒ (x + 1) (x + 2) > 3(x + 2)2
⇒ (x + 1) (x + 2) − 3(x + 2)2 > 0
⇒ (x + 2) [x + 1 − 3 (x + 2)] > 0
⇒ (x + 2) (− 2x – 5) > 0
 −5

⇒ x :
< x < − 2
 2

-5
2
-2
Exercise 8A
1
2
x2 + 8x + 15 < 0
⇒ (x + 3) (x + 5) < 0
⇒ {x: −5 < x < −3}
x2 + 3x − 4 < 0
⇒ (x + 4) (x − 1) < 0
{x: −4 < x < 1}
3
x2 − x < 6
⇒ x2 − x − 6 < 0
∴ (x − 3) (x + 2) < 0
⇒ {x: −2 < x < 3}
4
3x2 + 4x < − 3x − 2
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 3 of 18
⇒ 3x2 + 7x + 2 < 0
⇒ (3x + 1) (x + 2) < 0
1

∴ x : − 2 < x < − 
3

5
6x2 + 7x + 2 < 0
(3x + 2) (2x + 1) < 0
−1 
 −2
<x< 
x :
3
2

6
5x2 + 6x + 1 < 0
(5x + 1) (x + 1) < 0
1

x : − 1 < x < − 
5

7
x2 − 2 > 0
(x − 2 ) (x +
2)>0
{x : x < − 2 } ∪ {x : x > 2 }
8
kx2 + 2kx + 2x + 7 = 0
kx2 + x (2k + 2) + 7 = 0
For real roots, b2 − 4ac ≥ 0
∴ (2k + 2)2 − 4 (k) (7) ≥ 0
⇒ 4k2 + 8k + 4 − 28k ≥ 0
4k2 − 20k + 4 ≥ 0
k2 − 5k + 1 ≥ 0
5 ± 21
k=
2
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 4 of 18
(
)

5 + 21  
5 − 21 
k −
 k −
≥0
2 
2

 

5 − 21  
5 + 21 
k : k ≤
 ∪ k : k ≥

2  
2 

9
10
11
12
2x2 + 5x + k + 2 = 0
For real and distinct roots b2 − 4ac > 0
⇒ (5)2 − 4(2) (k + 2) > 0
⇒ 25 − 8k − 16 > 0
8k < 9
9
k<
8
(p + 2) x2 − 4x + 3 = 0
For real and distinct roots b2 − 4ac > 0
⇒ (− 4)2 − 4(p + 2) (3) > 0
16 − 12p − 24 > 0
12p < − 8
−2
p < − 8/12 =
3
4x2 − 2kx + 3 = 0
For real roots b2 − 4ac ≥ 0
⇒ (−2k)2 − 4(4)(3) ≥ 0
4k2 − 48 ≥ 0
k2 − 12 ≥ 0
k2 ≥ 12
The smallest possible positive integer value for k is 4
x+4
(a)
>2
x+5
x+4
⇒
−2>0
x +5
x + 4 − 2 (x + 5)
⇒
>0
x+5
−x − 6
⇒
>0
x +5
Critical values: −6, −5
−x − 6
x+5
−x − 6
x +5
(b)
+ve
x < −6
−6 < x < −5
−ve
x > −5
−ve
∴ {x: − 6 < x < − 5}
2x − 1
>1
3x + 1
2x − 1
⇒
−1> 0
3x + 1
Unit 1 Answers: Chapter 8
−ve
−ve
+ve
−ve
+ve
−ve
© Macmillan Publishers Limited 2013
Page 5 of 18
2x − 1 − 3x − 1
>0
3x + 1
−x−2
⇒
>0
3x + 1
⇒
Critical values: −2, −1/3
−x−2
x < −2
−2 < x < −
x>−
(c)
1
3
1
3
−ve
−ve
−ve
+ve
−ve
2x−3
x+1
−ve
−ve
−ve
+ve
2x − 3
x +1
+ve
−ve
+ve
+ve
+ve
1

∴ x, − 2 < x < − 
3

7x + 2
>5
x +1
7x + 2
−5>0
x +1
7x + 2 − 5x − 5
>0
x +1
2x − 3
>0
x +1
Critical values:
x < −1
3
−1 < x <
2
x>
(d)
+ve
−ve
−x − 2
3x + 1
−ve
+ve
3x+1
3
2
3
, −1
2
∴ {x: x < −1} ∪ {x: x > 2}
3x − 1
<1
x−2
3x − 1
⇒
−1< 0
x−2
3x − 1 − x + 2
⇒
<0
x−2
2x + 1
⇒
<0
x−2
1
Critical values: 2, −
2
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 6 of 18
2x+1
x−2
1
2
−ve
−ve
2x + 1
x−2
+ve
1
− <x<2
2
x>2
+ve
−ve
−ve
+ve
+ve
+ve
x<−
13
1


∴ x : − < x < 2
2


x +1
x
>
x −2 x+3
x +1
x
⇒
−
>0
x −2 x+3
(x + 1) (x + 3) − x (x − 2)
⇒
>0
(x − 2) (x + 3)
x 2 + 4x + 3 − x 2 + 2x
>0
(x − 2) (x + 3)
6x + 3
⇒
> 0.
(x − 2) (x + 3)
1
Critical values are : −3, − , 2
2
6x+3
⇒
x−2
x+3
6x + 3
(x − 2) (x + 3)
−ve
−ve
−ve
−ve
1
−3 < x < −
2
+ve
−ve
1
− <x<2
2
x>2
+ve
+ve
1

∴
 x : − 3 < x < −  ∪ {x : x > 2}
2

2x + 1
1
<
x −3 x +2
2x + 1
1
⇒
−
<0
x−3 x+2
(2x + 1) (x + 2) − (x − 3)
⇒
<0
(x − 3) (x + 2)
−ve
+ve
−ve
+ve
+ve
−ve
+ve
+ve
x < −3
14
⇒
2x 2 + 5x + 2 − x + 3
<0
(x − 3) (x + 2)
2x 2 + 4x + 5
<0
(x − 3) (x + 2)
Critical values −2, 3
⇒
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 7 of 18
2x2+4x+5
+ve
x < −2
+ve
−2 < x < 3
x>3
+ve
∴
{x: − 2 < x < 3}
x−3
x+2
2x 2 + 4x + 5
(x − 3) (x + 2)
−ve
−ve
+ve
−ve
+ve
+ve
+ve
−ve
+ve
Try these 8.3
(a)
|x + 1| = 3
⇒ x + 1 = 3, x + 1 = − 3
x = 2, x = − 4
Hence x = 2, − 4
(b)
|4x − 3| = 7
⇒ 4x − 3 = 7, 4x − 3 = −7
4x = 10, 4x = − 4
10
5
x=
= ,x=−1
4
2
5
Hence x = , − 1
2
(c)
|2x + 5| = |4x − 7|
⇒ |2x + 5|2 = |4x − 7|2
⇒ (2x + 5)2 = (4x − 7)2
⇒ 4x2 + 20x + 25 = 16x2 − 56x + 49
12x2 − 76x + 24 = 0
3x2 − 19x + 6 = 0
(3x − 1) (x − 6) = 0
1
x= ,6
3
Exercise 8B
1
(a)
(b)
(c)
|2x + 3| = 7
2x + 3 = 7, 2x + 3 = − 7
2x = 4, 2x = − 10
x = 2, x = −5
|5x − 1| = 8
⇒ 5x − 1 = 8, 5x − 1 = −8
5x = 9, 5x = − 7
9
−7
x= ,x=
5
5
|4x + 3| = 1
⇒ 4x + 3 = 1, 4x + 3 = − 1
4x = −2, 4x = −4
−1
x=
, x = −1
2
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 8 of 18
(d)
2
(a)
(b)
(c)
3
(a)
(b)
|1 − 2x| = 6
⇒ 1 − 2x = 6, 1 − 2x = −6
2x = − 5, 2x = 7
−5
7
x=
,x=
2
2
|3x + 1| = |2x − 4|
squaring both sides ⇒ |3x + 1|2 = |2x − 4|2
(3x + 1)2 = (2x − 4)2
9x2 + 6x + 1 = 4x2 − 16x + 16
5x2 + 22x − 15 = 0
(5x − 3) (x + 5) = 0
3
x= ,−5
5
|x − 1| = |x + 2|
squaring both sides ⇒ |x − 1|2 = |x + 2|2
(x − 1)2 = (x + 2)2
x2 − 2x + 1 = x2 + 4x + 4
6x = −3
−3
−1
=
x=
6
2
|7x + 1| = |5x + 3|
squaring both sides ⇒ |7x + 1|2 = |5x + 3|2
(7x + 1)2 = (5x + 3)2
49x2 + 14x + 1 = 25x2 + 30x + 9
⇒ 24x2 − 16x − 8 = 0
3x2 − 2x − 1 = 0
(3x + 1) (x −1) = 0
−1
x=
,1
3
OR
using |x| = x
|7x + 1| = |5x + 3|
⇒ 7x + 1 = 5x + 3
2x = 2
x=1
using |x| = -x
7x + 1 = − (5x + 3)
7x + 1 = − 5x − 3
12x = − 4
−4
−1
x=
=
3
12
|x| = 2 − |x|
⇒ 2|x| = 2
|x| = 1
⇒ x = 1 or −1
2|x| = 3 + 2x − x2
|x| = x ⇒ 2x = 3 + 2x − x2
x2 = 3 ⇒ x = 3 , − 3 .
|x| = −x ⇒ − 2x = 3 + 2x − x2
x2 − 4x − 3 = 0
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 9 of 18
4 ± 28
4±2 7
=2± 7
=
2
2
|x2 − 1| − 1 = 3x − 2
|x| = x ⇒ x2 − 1 − 1 = 3x − 2
x2 − 3x = 0
x (x − 3) = 0
x = 0, 3
|x| = -x ⇒ − (x2 − 1) − 1 = 3x − 2
− x2 + 1 − 1 = 3x − 2
x2 + 3x − 2 = 0
x=
(c)
x=
4
5
6
7
− 3 ± 17
2
2x + 1
=2
3x − 4
⇒ |2x + 1| = 2|3x − 4|
⇒ |2x + 1|2 = 4|3x − 4|2
⇒ (2x + 1)2 = 4(3x − 4)2
⇒ 4x2 + 4x + 1 = 4(9x2 − 24x + 16)
⇒ 4x2 + 4x + 1 = 36x2 − 96x + 64
32x2 − 100 x + 63 = 0
(4x − 9) (8x − 7) = 0
9 7
x= ,
4 8
|4x − 1|2 − 6 |4x − 1| + 5 = 0
Let y = |4x − 1|
⇒ y2 − 6y + 5 = 0
(y − 1) (y − 5) = 0
y = 1, 5
|4x − 1| = 1, |4x − 1| = 5
4x − 1 = 1, 4x − 1 = − 1, 4x − 1 = 5, 4x − 1 = − 5
1
3
x = , x = 0, x = , x = − 1
2
2
|3x + 2|2 − 9|3x + 2| + 20 = 0
y = |3x + 2|
y2 − 9y + 20 = 0
(y − 4) (y − 5) = 0
y = 4, 5
|3x + 2| = 4, |3x + 2| = 5
3x + 2 = 4, 3x + 2 = − 4, 3x + 2 = 5, 3x + 2 = − 5
2
−7
x = , x = − 2, x = 1, x =
3
3
(a) 2x2 − 5 |x| + 2 = 0
Since |x|2 = x2
⇒ 2|x|2 − 5|x| + 2 = 0
y = |x| ⇒ 2y2 − 5y + 2 = 0
(2y − 1) (y − 2) = 0
1
y= ,2
2
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 10 of 18
1
, |x| = 2
2
1 −1
, 2, − 2
⇒x= ,
2 2
(b) 3x2 − 19|x| + 20 = 0
Since |x|2 = x2
⇒ 3|x|2 − 19 |x| + 20 = 0
y =|x|
⇒ 3y2 − 19y + 20 = 0
⇒ (3y − 4) (y − 5) = 0
4
y= ,5
3
4
=
|x| =
, |x| 5
3
4 −4
, 5, − 5
∴x= ,
3 3
OR:
|x| = x
⇒ 3x2 − 19x + 20 = 0
(3x − 4) (x − 5) = 0
4
x= ,5
3
|x| = − x
⇒ 3x2 + 19x + 20 = 0
(3x + 4) (x + 5) = 0
−4
=
x
,−5
3
4 −4
Hence x = ,
, 5, − 5
3 3
|x| =
8
(a)
(b)
(c)
(d)
|4x − 1| < 3
⇒ − 3 < 4x – 1 < 3
− 2 < 4x < 4
−1
< x <1
2
|2 x + 4| < 5
⇒ − 5 < 2x + 4 < 5
− 9 < 2x < 1
−9
1
<x<
2
2
|3x − 1| > 6
⇒ 3x − 1 > 6, 3x − 1 < − 6
3x > 7, 3x < − 5
7
−5
x> x<
3
3
|5x + 2| > 9
⇒ 5x + 2 > 9, 5x + 2 < − 9
5x > 7, 5x < − 11
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 11 of 18
7
− 11
x<
5
5
We can also square both sides and solve
|3x + 1| < |2x − 5|
⇒ (3x + 1)2 < (2x − 5)2
⇒ 9x2 + 6x + 1 < 4x2 − 20x + 25
⇒ 5x2 + 26x − 24 < 0
⇒ (5x − 4) (x + 6) < 0
x>
9
(a)
4

∴ x : − 6 < x < 
5

(b)
|7x + 1| < |3x + 5|
⇒ (7x + 1)2 < (3x + 5)2
⇒ 49x2 + 14x + 1 < 9x2 + 30x + 25
⇒ 40x2 − 16x − 24 < 0
5x2 − 2x − 3 < 0
(5x + 3) (x − 1) < 0
3


 x : − < x < 1
5


(c)
4|x+2|<|x−1|
⇒ 16 (x + 2)2 < (x − 1)2
⇒ 16(x2 + 4x + 4) < x2 − 2x + 1
⇒ 15x2 + 66x + 63 < 0
5x2 + 22x + 21 < 0
(5x + 7) (x + 3) < 0
7

∴ x : − 3 < x < − 
3

Review Exercise 8
1
2x + 1 > 4x − 5
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 12 of 18
2
2x − 4x > −5 −1
−2x > −6
x<3
x 2 − 2 3x > 6
x 2 − 2 3x − 6 > 0
2 3 ± 12 + 24
2
2 3±6
=
2
= 3±3
x=
(x − ( 3 + 3) (x − ( 3 − 3)) > 0
x < 3 − 3,
3
x> 3+3
(p + 3) x − 2px + p + 2 =
0
2
b 2 − 4ac > 0
( −2p) 2 − 4 (p + 3) (p + 2) > 0
4
4p 2 − 4p 2 − 20p − 24 > 0
−20p > 24
24
5
p<−
=−
20
4
2
4x − 4λ x = 5λ − 12x − 15
⇒ 4x 2 − 4λx + 12 x − 5λ + 15 = 0
⇒ 4 x 2 + x ( −4λ + 12) − 5λ + 15 =0
a = 4, b = −4λ + 12, c = −5λ + 15
b2 − 4ac < 0
(−4λ + 12)2 − 4(4) (−5 λ + 15) < 0
16λ2 − 96λ + 144 + 80λ − 240 < 0
16λ2 – 16λ – 96 < 0
λ2 − λ − 6 < 0
(λ − 3) (λ + 2) < 0
−2 < λ < 3
5
(2 − 3θ) x2 = (θ − 4) x − 2
(2 − 3θ) x2 − (θ − 4) x + 2 = 0
b2 − 4ac < 0
⇒ [−(θ − 4)]2 − 4(2 − 3θ) (2) < 0
θ2 − 8θ + 16 − 16 + 24θ < 0
θ2 + 16θ < 0
θ(θ + 16) < 0
−16 < θ < 0
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 13 of 18
6
7
(λ + 1) y2 + (2λ + 3) y + λ + 2 = 0
For real roots b2 − 4ac ≥ 0
(2λ + 3)2 − 4 (λ + 1) (λ + 2) ≥ 0
4λ2 + 12λ + 9 − 4λ2 − 12λ − 8 ≥ 0
⇒1≥0
⇒ ∀ λ ∈ R b2 − 4ac ≥ 0
(a) x2 + 11x + 30 > 0
(x + 5) (x + 6) > 0
x < −6, x > −5
x2 + 11x + 30 < 0
(x +5) (x + 6) < 0
−6 < x < −5
4x2 − 3x + 2 > 3x2
x2 − 3x + 2 > 0
(x − 1) (x − 2) > 0
x < 1, x > 2
(b)
8
9
10
| 3x + 2 |2 − 9| 3x + 2 | + 20 = 0
y = | 3x + 2 |
y2 − 9y + 20 = 0
(y − 4) (y − 5) = 0
y = 4, 5
| 3x + 2 | = 4,
| 3x + 2 | = 5
3x + 2 = 4, 3x + 2 = −4, 3x + 2 = 5, 3x + 2 = −5
2
−7
x = −2,
x = 1, x =
x= ,
3
3
2x + 1
=1
3x − 2
| 2x + 1 | = | 3x − 2 |
(2x + 1)2 = (3x − 2)2
4x2 + 4x + 1 = 9x2 − 12x + 4
5x2 − 16x + 3 = 0
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 14 of 18
11
12
(5x − 1) (x − 3) = 0
1
x = ,3
5
|2−x|=2|x+2|
(2 − x)2 = 4(x + 2)2
4 − 4x + x2 = 4[x2 + 4x + 4]
4 − 4x + x2 = 4x2 + 16x + 16
3x2 + 20x + 12 = 0
(3x + 2) (x + 6) = 0
−2
=
x
,−6
3
(a) x2 + 5x + 6 < 0
(x + 2) (x + 3) < 0
−3 < x < −2
(b)
x2 + 2x − 8 > 0
(x + 4) (x − 2) > 0
x < − 4, x > 2
(c)
x2 < x + 20
x2 − x − 20 < 0
(x − 5) (x + 4) < 0
−4 < x < 5
(d)
(x + 3) (x − 2) > 2(x + 3)
x2 + x − 6 > 2x + 6
x2 − x − 12 > 0
(x − 4) (x + 3) > 0
x < −3, x > 4
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 15 of 18
13
14
15
y = 3x + p
x2 + y2 = 64
⇒ x2 + (3x + p)2 = 64
⇒ x2 + 9x2 + 6px + p2 = 64
⇒ 10x2 + 6px + p2 − 64 = 0
For no real values of x, b2 − 4ac < 0
(6p)2 − 4(10) (p2 − 64) < 0
36p2 − 40p2 + 2560 < 0
4p2 − 2560 > 0
p2 − 640 > 0
(p − 640) (p + 640) > 0
{p : p < − 640} ∪ {p : p > 640}
y = 3 + px
x2 + 2xy + 1 = 0
⇒ x2 + 2x(3 + px) + 1 = 0
x2 + 6x + 2px2 + 1 = 0
x2(1 + 2p) + 6x + 1 = 0
For no real roots b2 − 4ac < 0
(6)2 − 4(1 + 2p) (1) < 0
36 − 4 − 8p < 0
32 < 8p
4<p
2x + 1
<2
3x − 4
2x + 1
−2<0
3x − 4
2x + 1 − 6x + 8
<0
3x − 4
−4x + 9
<0
3x − 4
4 9
,
3 4
−4x + 9
3x − 4
4
3
+ve
−ve
−4x + 9
3x − 4
+ve
4
9
<x<
3
4
9
x>
4
+ve
+ve
+ve
−ve
+ve
−ve
Critical values are
x<
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 16 of 18
16
4
9
x< ,x>
3
4
f(x) > 1
2−x
⇒
>1
x+3
2−x
⇒
−1> 0
x+3
2−x −x −3
⇒
>0
x+3
−2x − 1
⇒
>0
x+3
2x + 1
<0
x+3
−1
, −3
2
2x+1
x+3
The critical values are
x < −3
−3 < x <
x>
−1
2
−1
2
2x + 1
x+3
−ve
−ve
−ve
+ve
+ve
−ve
+ve
+ve
+ve
1
2
17
(a) 2x2 − 5| x | + 2 = 0
2| x |2 − 5| x | + 2 = 0
(2 | x | − 1) (| x | − 2) = 0
1
=
|x| =
,
|x| 2
2
1 −1
x=
, , x= 2, − 2
2 2
(b) 3x2 − 19 | x | + 20 = 0
3| x |2 − 19| x | + 20 = 0
(3 | x | − 4) (| x | − 5) = 0
4
=
|x| =
, |x| 5
3
4 −4
=
x
, , 5, − 5
3 3
4x + 2
18
+ 2>0
x −1
4x + 2 + 2x − 2
>0
x −1
6x
>0
x −1
The critical values are 0, 1
∴− 3 < x < −
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 17 of 18
6x
19
20
21
x−1
6x
x −1
+ve
−ve
+ve
x<0
−ve
−ve
0<x<1
+ve
−ve
x>1
+ve
+ve
{x : x < 0} ∪ {x : x > 1}
p−x
y=
2
−8
=
y
−x
x
p x −8
− =
−x
2 2 x
× 2x ⇒ px − x 2 =
− 16 − 2x 2
x2 + px + 16 = 0
b2 − 4ac ≥ 0
p2 − 4(1) (16) ≥ 0
p2 − 64 ≥ 0
(p − 8) (p + 8) ≥ 0
p ≤ −8, p ≥ 8
s = −16t2 + 80t + 6
s > 6 ⇒ −16t2 + 80t + 6 > 6
⇒ −16t2 + 80t > 0
⇒ −16t(t − 5) > 0
⇒0<t<5
The ball will be above 6 metres for anywhere between 0 and 5 seconds.
P = 8x − 0.02x2
P>4
⇒ 8x − 0.02x2 > 4
⇒ 0.02x2 − 8x + 4 < 0
⇒ x2 − 400x + 200 < 0
When x2 − 400x + 200 = 0
400 ± (−400) 2 − 4(200)
2
= 399.5, 0.5
⇒ (x − 341.42) (x − 58.58) < 0
∴ 58.58 < x < 341.42
x=
22
When x = 58, p = 8(58) − 0.02(58)2 = 39.72
x = 59, p = 8(59) − 0.02(59)2 = 402.38
x = 341, p = 8(341) − 0.02(341)2 = 402.38
x = 342, p = 8(342) − 0.02(342)2 = 396.72
∴ 59 ≤ x 341
y = −4x2 + 4 000x
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 18 of 18
(a)
(b)
y = 0 ⇒ −4x2 + 4 000x = 0
−4x(x − 1 000) = 0
x = 0, 1 000
∴ Revenue is $0 when x = $0 and x = $1 000.00
When y > 800 000
−4x2 + 4 000x > 800 000
⇒ −4x2 + 4 000x − 800 000 > 0
⇒ x2 − 1 000 x + 200 000 < 0
Now x2 − 1 000 x + 200 000 = 0
1 000 ± (− 1000) 2 − 4 (200000)
2
1000 ± 447.21
x=
2
= 723.61, 276.40
∴ (x − 723.61) (x − 276.40) < 0
∴276.40 < x < 723.61
The range of prices is
$276.40 < x < $723.61
⇒x =
Unit 1 Answers: Chapter 8
© Macmillan Publishers Limited 2013
Page 1 of 44
Chapter 9 Trigonometry
Try these 9.1
(a)
sin 4x = 0.28
4x = sin – 1 (0. 28)
4x = 180°n + (–1)n (16.3°), n ∈ ℤ
1
=
x
[180n + ( − 1) n (16.3°)], n ∈ 
4
(b) cos (x + 30°) = 0.6
x + 30 = cos–1 (0.6)
x + 30 = 360n ± 53.13°
x = 360n + 23.13°
 n∈
x = 360n − 83.13° 
(c) tan (2x + 45°) = 0.7
2x + 45° = tan – 1 (0.7)
2x + 45 = 180°n + 35°
2x = 180n – 10
x = 90°n – 5° , n ∈ 
Hence x = 90n – 5°, n ∈ 
Exercise 9A
1
sin 2θ =
−1
2
−π
2θ = nπ + (− 1)n 
, n ∈
 6 
π
n= 2p ⇒ 2θ= 2pπ −
6
π
θ = pπ −
12
π
6
π
13π
 2p + 1 
θ= 
= pπ +
π+
12
12
 2 
π 
Hence θ = pπ −
12  p 
 ∈
13π 
pπ +
12 
cos 3θ = 0
π
⇒ 3θ= 2nπ ± , n ∈Ζ
2
2
π
=
θ
nπ ± , n ∈Ζ
3
6
n= 2p + 1 ⇒ 2θ
= (2p + 1) π +
2
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 2 of 44
3
π

tan  2θ +  =1

3
π
π
= nπ +
3
4
π π
2θ = nπ + −
4 3
1
π
=
θ
nπ − 

2
12 
1
π
=
θ
nπ − , n ∈ 
2
24
π

 1
cos  2θ −  =
4 2

π
π
2θ − = 2nπ ±
4
3
π π
2θ= 2nπ ± +
3 4
7π
π
2θ= 2nπ +
, 2nπ −
12
12
7π 
θ = nπ +
24  n 
 ∈
π 
θ = nπ −
24 
2θ +
4
5
π 1

sin  3θ −  =
3
2

π
π
3θ − = nπ + (− 1)n , n ∈ 
3
4
π
π
n= 2p] ⇒ 3θ − = 2pπ +
4
3
π π
3θ= 2pπ + +
4 3
1
7π 
=
θ
2pπ +
,p∈

3
12 
π
π
n = 2p + 1] ⇒ 3θ − = (2p + 1) π −
3
4
π π
3=
θ (2p + 1) π − +
4 3
π π
3θ= 2pπ + π − +
4 3
1
13 
=
θ
2pπ +
π , p∈
3 
12 
7π  
1
2pπ +
=
∴θ

12 
3
 p∈
1
13π  
2pπ +
3 
12  
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 3 of 44
6
7
π
1
tan  θ −  = − 1
2
2
1
π
π
θ − = nπ −
2
2
4
1
π π
θ = nπ − +
2
4 2
π

=
θ 2  nπ +  , n ∈ 
4

sin 3x = 3 cos 3x
sin 3x
⇒
=
3
cos 3x
tan 3x = 3
3x = nπ + 1.25
1
=
x
nπ + 0.416, n ∈ 
3
Try these 9.2
(a)
RTP:
secθ − cosθ
sin 2 θ
=
secθ + cosθ 1 + cos 2 θ
Proof:
1
− cos θ
secθ − cosθ cos θ
=
1
secθ + cosθ
+ cos θ
cos θ
1 − cos2 θ
cos θ
=
1 + cos2 θ
cos θ
(b)
(c)
=
1 − cos2 θ
1 + cos2 θ
=
sin 2 θ
1 + cos 2 θ
1 − cos 2 θ
= sin θ
sin θ
Proof:
1 − cos 2 θ sin 2 θ
=
sin θ
sin θ
= sin θ
sec θ + tan θ sin θ
RTP:
=
cot θ + cosθ cos 2 θ
Proof:
sec θ + tan θ
cot θ + cos θ
RTP:
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 4 of 44
1
sin θ
+
cos θ cos θ
=
cos θ
+ cos θ
sin θ
1 + sin θ
cos θ
=
cos θ + cos θ sin θ
sin θ
(1 + sin θ) sin θ
=
(cos θ + cos θ sin θ) cos θ
=
=
(1 + sin θ) sin θ
cos2 θ (1 + sin θ)
sin θ
cos 2 θ
Try these 9.3
(a)
(b)
3 sin2θ = 1 + cos θ
3 (1 – cos2θ) = 1 + cos θ
3 cos2θ + cos θ – 2 = 0
(3 cos θ – 2) (cos θ + 1) = 0
2
cos θ = , cos θ = − 1
3
θ = 48.2°, 311.8°, 180°
(ii) 4 cosec2θ – 4 cot θ – 7 = 0
4 (1 + cot2θ) – 4 cotθ – 7 = 0
4 cot2θ – 4 cot θ – 3 = 0
(2 cot θ + 1) (2 cot θ – 3) = 0
1
3
cot θ = − , cot θ =
2
2
2
tan θ = − 2, tan θ =
3
θ = 116.6°, 296.6°, 33.7°, 213.7°
20 sec2θ – 3 tan θ – 22 = 0
20 (1 + tan2θ) – 3 tan θ – 22 = 0
20 tan2θ – 3 tan θ – 2 = 0
(4 tan θ + 1) (5 tan θ – 2) = 0
1
2
tan θ = − , tan θ =
4
5
1
tan θ = − ⇒ θ = 180n − 14°, n ∈ 
4
2
tan θ=
⇒ θ= 180n + 21.8, n ∈ 
5
(i)
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 5 of 44
Exercise 9B
(sin θ + cos θ)2 – 1
= sin2θ + 2 sin θ cos θ + cos2θ – 1
= 1 + 2 sin θ cos θ – 1, since sin2θ + cos2θ = 1
= 2 sin θ cos θ
2
sin x (sin x – cot x cosec x)
cos x
1
=
×
sin 2 x − sin x
sin x sin x
cos x
= sin 2 x −
sin x
2
= sin x – cot x
3
sin4θ – cos4θ
= (sin2θ – cos2θ) (sin2θ + cos2θ)
= sin2θ – cos2θ
4
sin2θ (cot2θ + cosec2θ)
 cos2 θ
1 
=
sin 2 θ  2 +
2 
 sin θ sin θ 
1
5
6
7
 cos2 θ + 1 
= sin 2 θ 

2
 sin θ 
= 1 + cos2θ
sin 2 θ − 1 − (1 − sin 2 θ)
=
cos2 θ
cos2 θ
− (cos2 θ)
=
= −1
cos2 θ
sec 2 θ − tan 2 θ sec 2 θ − (sec 2 θ − 1)
=
sin θ
sin θ
1
=
sin θ
= cosec θ
sin 2 θ
RTP: 1 −
=
− cos θ
1 − cos θ
Proof:
sin 2 θ
1−
1 − cos θ
1 − cos 2 θ
1 − cos θ
(1 − cos θ) (1 + cos θ)
= 1−
1 − cos θ
= 1 – (1 + cos θ)
= – cos θ
cos 2 θ
RTP:
− 1= sin θ
1 − sin θ
Proof:
= 1−
8
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 6 of 44
cos 2 θ
1 − sin 2 θ
=
−1
−1
1 − sin θ
1 − sin θ
(1 − sin θ)(1 + sin θ)
=
−1
1 − sin θ
= 1 + sin θ – 1
= sin θ
cosec θ cos θ
9
RTP:
−
=
tan θ
cos θ
sin θ
Proof:
cosec θ cos θ
−
cos θ
sin θ
1
cos θ
−
sin θ cos θ sin θ
=
=
1 − cos 2 θ
sin θ cos θ
=
sin 2 θ
sin θ
=
sin θ cos θ cos θ
= tan θ
1 − cos 2 θ
10
RTP:
=
− cos 2 θ
1 − sec 2 θ
Proof:
1 − cos2 θ
sin 2 θ
=
1 − sec 2 θ − tan 2 θ
cos 2 θ
= sin 2 θ × −
sin 2 θ
= – cos2θ
11
RTP: sec4 x – sec2 x = tan4 x + tan2 x
Proof:
sec4x – sec2x
= sec2x (sec2x – 1)
= (1 + tan2 x) (tan2 x), since sec2 x – 1 = tan 2 x
= tan2 x + tan4 x
cos x
sin x
12
RTP:
+
=
sin x + cos x
1 − tan x 1 − cot x
Proof:
cos x
sin x
+
1 − tan x 1 − cot x
cos x
sin x
=
+
sin x
cos x
1−
1−
cos x
sin x
cos x
sin x
=
+
cos x − sin x sin x − cos x
cos x
sin x
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 7 of 44
cos2 x
sin 2 x
+
cos x − sin x sin x − cos x
=
=
=
13
cos2 x − sin 2 x
cos x − sin x
(cos x − sin x) (cos x + sin x)
cos x − sin x
= cos x + sin x
1 − cos x
RTP: = (cosec x − cot x) 2
1 + cos x
Proof:
Now (cosec x – cot x)2
 1
cos x 
= 
−
 sin x sin x 
 1 − cos x 
=
 sin x 
2
2
(1 − cos x)2
sin 2 x
(1 − cos x)2
=
1 − cos2 x
=
=
(1 − cos x) (1 − cos x)
(1 − cos x) (1 + cos x)
1 − cos x
1 + cos x
1
1
RTP:
+
=
2 sec 2 x
sin x + 1 1 − sin x
Proof:
1
1
1 − sin x + 1 + sin x
+
=
sin x + 1 1 − sin x (1 + sin x) (1 − sin x)
2
=
1 − sin 2 x
2
=
cos2 x
= 2 sec2 x
RTP: sin4θ – sin2θ = cos4θ – cos2θ
Proof:
sin4θ – sin2θ
= sin2θ (sin2θ – 1)
= (1 – cos2θ) (– cos2θ) = – cos2θ + cos4θ
cot 2 x − 1
RTP:
= − cot 2 x
tan 2 x − 1
Proof:
=
14
15
16
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 8 of 44
cos2 x
−1
cot 2 x − 1 sin 2 x
=
tan 2 x − 1 sin 2 x
−1
cos2 x
=
cos2 x − sin 2 x
sin 2 x
×
cos2 x
sin 2 x − cos2 x
−1
cos x
sin 2 x
= – cot2 x
4 sec x – tan x = 6 cos x
Converting to sin x and cos x:
4
sin x
−
=
6 cos x
cos x cos x
⇒ 4 – sin x = 6 cos2x
⇒ 4 – sin x = 6 [1 – sin2x]
4 – sin x = 6 – 6 sin2x
6 sin2 x – sin x – 2 = 0
Let y = sin x
6y2 – y – 2 = 0
(3y – 2) (2y + 1) = 0
2 1
=
y
,−
3 2
2
1
sin x = , sin x = −
3
2
x = 41.8°, 138.2°, x = 210°, 330°
∴ x = 41.8°, 138.2°, 210°, 330°
3 tan2 x – sec x – 1 = 0
Replacing tan2 x = sec2 x – 1
⇒ 3 [sec2 x – 1] – sec x – 1 = 0
3 sec2 x – sec x – 4 = 0
y = sec x
3y2 – y – 4 = 0
(3y – 4) (y + 1) = 0
4
=
y
, −1
3
4
Now sec x = , sec x = − 1
3
3
⇒ cos x =cos x =
−1
4
x = 41.4°, 318.6°, x = 180°
∴ x = 41.4°, 180°, 318.6°
2 cot2 x + cosec x = 1
Replacing cot2 x = cosec2 x – 1
⇒ 2 [cosec2 x – 1] + cosec x = 1
⇒ 2 cosec2 x + cosec x – 3 = 0
y = cosec x
2y2 + y – 3 = 0
⇒ (2y + 3) (y – 1) = 0
= −
17
18
19
2
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 9 of 44
3
y=
− ,y=
1
2
−3
=
cosec x =
, cosec x 1
2
−2
=
sin x =
, sin x 1
3
x = 221.8°, 318.2°, x = 90°
x = 90°, 221.8°, 318.2°
20
3 cos2 x = 4 sin x – 1
⇒ 3 (1 – sin2 x) = 4 sin x – 1
∴ 3 sin2 x + 4 sin x – 4 = 0
y = sin x
3y2 + 4y – 4 = 0
(3y – 2) (y + 2) = 0
2
=
y
,−2
3
2
∴ sin x =
, sin x =
− 2 (invalid)
3
x = 41.8°, 138.2°
21
2 cot x = 3 sin x
2 cos x
⇒
=
3 sin x
sin x
⇒ 2 cos x = 3 sin2 x
2 cos x = 3 (1 – cos2 x)
3 cos2 x + 2 cos x – 3 = 0
− 2 ± 40
6
cos x = – 1.387, 0.72076
cox x = – 1.387 (invalid)
cos x = 0.72076
⇒ x = 43.9°, 316.1°
sin2 x = 3 cos2 x + 4 sin x
cos2 x = 1 – sin2 x
∴ sin2 x = 3 (1 – sin2 x) + 4 sin x
⇒ sin2 x = 3 – 3 sin2 x + 4 sin x
4 sin2 x – 4 sin x – 3 = 0
(2 sin x + 1) (2 sin x – 3) = 0
1
3
sin x =
− , sin x =
(invalid)
2
2
x = 210°, 330°
2 cos x = tan x
sin x
2 cos x =
cos x
⇒ 2 cos2 x = sin x
⇒ 2 [1 – sin2 x] = sin x
2 sin2 x + sin x – 2 = 0
cos x =
22
23
sin x =
− 1 ± 17
4
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 10 of 44
24
25
26
− 1 ± 17
∴ sin x =
4
x = 51.3°, 128.7°
2 cot x = 1 + tan x
2 cos x
sin x
= 1+
sin x
cos x
2
⇒
=
1 + tan x
tan x
2 = tan x + tan2 x
∴ tan2 x + tan x – 2 = 0
(tan x – 1) (tan x + 2) = 0
tan x = 1, tan x = – 2
x = 45°, 225°, x = 116.6°, 296.6°
2 + 3 sin z = 2 cos2 z
2 + 3 sin z = 2 (1 – sin2 z)
2 sin 2 z + 3 sin z = 0
sin z (2 sin z + 3) = 0
−3
=
sin z 0,=
sin z
(invalid)
2
z = nπ + (– 1)n (0)
z = nπ, n ∈ 
2 cot2 x + cosec x = 4
⇒ 2 (cosec2 x – 1) + cosec x – 4 = 0
⇒ 2 cosec2 x + cosec x – 6 = 0
⇒ (2 cosec x – 3) (cosec x + 2) = 0
3
cosec x = , cosec x = − 2
2
2
1
∴ sin x =
, sin x =
−
3
2
n
x = nπ + (– 1) (0.730) n ∈ 
−π
x = nπ + (− 1) n 
, n ∈ 
 6 
27
2 sec x + 3 cos x = 7
2
+ 3 cos x =
7
cos x
2 + 3 cos2 x = 7 cos x
3 cos2 x – 7 cos x + 2 = 0
(3 cos x – 1) (cos x – 2) = 0
1
=
cos x =
, cos x 2 (invalid)
3
x = 2nπ ± 1.23, n ∈ 
(28) 5 cos x = 6 sin2 x
5 cos x = 6 (1 – cos2 x)
6 cos2 x + 5 cos x – 6 = 0
(3 cos x – 2) (2 cos x + 3) = 0
2
−3
=
cos x =
, cos x
(invalid)
3
2
x = 2nπ ± (0.841) n ∈ 
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 11 of 44
Try these 9.4
(a)
 7π 
 π π
cos=
  cos  + 
12
3 4
 π
 π
 π
 π
= cos   cos   − sin   sin  
 3
 4
 3
 4
 1  2   3   2 
=  
−
 2   2   2   2 
2
6
−
4
4
2
=
(1 − 3)
4
 5π 
 π π
(b)
sin =
  sin  + 
12
4 6
 π
 π
 π
 π
= sin   cos   + cos   sin  
 4
 6
 4
 6
=
 2  3  2 1
= 
 
 + 
  
 2  2   2 2
2
( 3 + 1)
4
 7π 
 π π
(c)
sin =
  sin  + 
12
3 4
π
π
π
π
= sin cos + cos sin
3
4
3
4
 3   2   1  2 
= 

 +  

 2   2   2  2 
=
=
2
( 3 + 1)
4
Exercise 9C
1
sin 75 = sin (30 + 45)
= sin 30 cos 45 + cos 30 sin 45
1
2
3
2
= ×
+
×
2
2
2
2
2
6
=
+
4
4
2
=
( 3 + 1)
4
sin (A − B) 5
2
=
sin (A + B) 13
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 12 of 44
3
sin A cos B − cos A sin B 5
=
sin A cos B + cos A sin B 13
13 sin A cos B – 13 cos A sin B = 5 sin A cos B + 5 cos A sin B
18 cos A sin B = 8 sin A cos B
sin B
sin A
18
=8
cos B
cos A
9 tan B = 4 tan A
5
tan α =
12
cos β = −
(a)
3
5
sin (α + β=
) sin α cos β + cos α sin β
 − 5   − 3   − 12   4 
= 
+
 13   5   13   5 
=
15 48 − 33
−
=
65 65 65
5 −4
−

tan α − tan β
63
12  3 
(b) =
=
tan (α − β) =
1 + tan α tan β
 5   − 4  16
1+   

 12   3 
(c)
cos (α − β=
) cos α cos β + sin α sin β
4
 − 12   − 3   − 5   4  16
= 

+
  =
 13   5   13   5  65
sin (θ + 30) + 3 cos (θ + 30)
= sin θ cos 30 + cos θ sin 30 + 3 cos θ cos 30 − 3 sin θ sin 30
=
 3

3
1
3
sin θ + cos θ + 3 
cos θ −
sin θ
2
2
2
 2

1
3
cos θ + cos θ
2
2
= 2 cosθ.
sin (θ + 30) = 2 cos (θ + 60)
=
5
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 13 of 44
6
sin θ cos 30 + cos θ sin 30 = 2 cos θ cos 60 – 2sin θ sin 60
3
1
1
3
sin θ + =
cos θ
(2) cos θ − 2
sin θ
2
2
2
2
3
1
sin θ + 3 sin=
θ cos θ − cos θ
2
2
3 3
1
sin
=
θ
cos θ
2
2
3 3 sin=
θ cos θ
sin (θ +
=
α) k sin (θ − α)
sin θ cos α + cos θ sin=
α k sin θ cos α − k cos θ sin α
cos θ sin α + k cos θ sin=
α k sin θ cos α − sin θ cos α
cos θ sin α (k + 1) = sin θ cos α (k − 1)
sin α  k + 1  sin θ

=
cos α  k − 1  cos θ
 k +1
tan θ 
=
 tan α
 k −1
− 12
7
cos α =
13
(a)
(b)
(c)
+5
13
−5
tan α =
12
cos (α + 30)
= cos α cos 30 − sin α sin 30
sin α =
 − 12   3   5   1 
= 
−
 13   2   13   2 
− 12 3 5 − 1
− =
(5 + 12 3)
26
26 26
12
sin α =
13
=
8
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 14 of 44
∴ cos α =
sin β =
−5
13
4
5
−3
5
both α and β are in the second quadrant
(a) sin (α + β) = sin α cos β + cos α sin β
 12   − 3   − 5   4 
=  
+
 13   5   13   5 
− 36 − 20 − 56
= =
65 65
65
(b) cos (α + β) = cos α cos β – sin α sin β
 − 5   − 3   12   4 
= 
−
 13   5   13   5 
15 48 − 33
=
−
=
65 65
65
− 56
sin (α + β)
56
65
(c)
tan (α +=
β)
= =
cos (α + β) − 33 33
65
1
9
cos α =
4
cos β =
Since α is in the 4th quadrant sin α is negative
− 15
(a)
sin α =
4
π
π
π

(b)
sin  α − =
 sin α cos − cos α sin

6
6
6
 − 15   3   1   1 
= 

 −   
 4   2   4  2
=
−3 5 1
−
8
8
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 15 of 44
(c)
π
π
π

cos  α + =
 cos α cos − sin α sin

3
3
3
 1   1   − 15   3 
=     −
 4   2   4   2 
1 3 5
+
8
8
10
(a) tan (2π − θ)
tan 2π − tan θ 0 − tan θ
= =
1 + tan 2π tan θ
1+ 0
= −tan θ
 3π

(b)
sin 
+ θ
 2

=
 3π 
 3π 
= sin   cos θ + cos   sin θ
 2
 2
= (−1) cosθ + (0) sinθ
= −cosθ
11
tan A = y + 1
tan B = y − 1
tan A − tan B
tan(A − B) =
1 + tan A tan B
y + 1 − (y − 1)
=
1 + (y + 1) (y − 1)
2
=
1 + y2 − 1
2
= 2
y
2
2 cot(A − B) =
tan(A − B)
2
=
2
y2
= y2
1 + tan θ
12
(a)
1 − tan θ
π
+ tan θ
4
=
π
1 − tan tan θ
4
π

= tan  + θ
4

1
1
(b)
cos θ +
sin θ
2
2
π
π
= cos cos θ + sin sin θ
4
4
tan
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 16 of 44
13
14
π

= cos  − θ
4

cot (θ − α) = 4
1
⇒ tan(θ − α ) =
4
tan θ − tan α 1
1
⇒
=
cot α=
⇒ tan α= 2
1 − tan θ tan α 4
2
tan θ − 2 1
⇒
=
1 − 2 tan θ 4
⇒ 4 tanθ − 8 = 1 − 2 tanθ
6 tanθ = 9
9 3
tan θ= =
6 2
2
⇒ cot θ =
3
cos(α − β) − cos(α + β)
sin(α + β) + sin(α − β)
=
cos α cos β + sin α sin β − cos α cos β + sin α sin β
sin α cos β + cos α sin β + sin α cos β − cos α sin β
2 sin α sin β
=
=
2 sin α cos β
sin β
cos β
= tan β
15
tan (α + β) = b
1
tan β =
2
tan (α + β) = b
⇒
tan α +
⇒
tan α + tan β
=
b
1 − tan α tan β
1
2 =
b
1
tan α
2
2 tan α + 1
=b
2 − tan α
⇒ 2 tan α + 1 = 2b − b tan α
⇒ 2 tan α + b tan α = 2b − 1
tan α(2 + b) = 2b − 1
2b − 1
tan α =
b+2
1−
=
P VI cos φ sin 2 (ωt ) − VI sin φ sin ωt cos ωt
= VI sin ωt [cos φ sin ωt − sin φ cos ωt ]
= VI sin ωt sin(ωt − φ)
16
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 17 of 44
Try these 9.5
(a)
3sin θ − cos θ = r sin (θ − α)
= r sin θ cos α − r cos θ sin α
⇒ r cos α = 3
r sin α = 1
1
[2] ÷ [1] ⇒ tan =
α
, =
α 18.4°
3
[1]2 + [2]2⇒ r2 [cos2 α + sin2 α] = 32 + 12
r2 = 10
r = 10
∴ 3 sin θ =
− cos θ
[1]
[2]
10 sin (θ − 18.4°)
10 sin(θ − 18.4°) =2
2
sin(θ − 18.4) =
10
 2 
θ − 18.4 =sin −1 
 10 
θ − 18.4
= 39.2°,140.8°
=
θ 57.6°,159.2°
(b)
(c)
3 cos 2θ − sin
=
2θ r cos(2θ + α)
= r cos 2θ cos α − r sin 2θ sin α
⇒ r cos α = 3
r sin α = 1
1
[2] ÷ [1] ⇒ tan α=
⇒ α= 30°
3
[1]2 + [2]2] ⇒ r2 = 3 + 1 ⇒ r = 2
∴ 3 cos 2θ − sin
=
2θ 2 cos (2θ + 30°)
2 cos(2θ + 30°) = −1
−1
cos(2θ + 30°) =
2
2θ + 30° = 360n ± 120°
2θ = 360n + 90°
2θ = 360n − 150
⇒
=
θ 180n + 45° 
n ∈
180n − 75°
2 sin x − cos x = r sin (x − α)
= r sinx cos α − r cos x sin α
r cos α = 2
r sin α = 1
1
tan α=
⇒ α= 26.6°
2
r 2 = 22 + 12 ⇒ r = 5
2 sin x − cos
=
x
[1]
[2]
5 sin (x − 26.6°)
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 18 of 44
max f (x) = 5
when sin(x − 26.6) =1, x − 26.6 =90
x = 116.6˚
min f (x) = − 5
when sin(x − 26.6) =
−1
x − 26.6 = 270
x = 296.6˚
Try these 9.6
(a)
(b)
(c)
cos 2θ − cos θ = 0
2 cos2 θ − cos θ − 1 = 0
(2 cos θ + 1) (cos θ − 1) = 0
1
cos θ = − , cos θ = 1
2
θ = 120°, 240°, 0°, 360°
Hence θ = 0°, 120°, 240°, 360°
cosθ = sin 2θ
cosθ − 2 sin θ cos θ = 0
cosθ (1 – 2 sinθ) = 0
1
cosθ = 0, sin θ =
2
θ = 0°, 180°, θ = 30°, 150°
Hence θ = 0°, 30°, 150°, 180°
cos 2θ − 2cos θ = 3
2 cos2θ − 1 − 2cos θ − 3 = 0
2 cos2θ − 2cos θ − 4 = 0
cos2θ − cos θ − 2 = 0
(cos θ + 1) (cos θ − 2) = 0
cos θ = −1, cos θ = 2.
θ = 180°
cos θ = 2 has no solutions
Hence θ = 180°
Exercise 9D
1
RTP:
sin 2x + cos x
= cot x
2 − 2 cos 2 x + sin x
Proof:
sin 2x + cos x
2 − 2cos 2 x + sin x
2 sin x cos x + cos x
=
2 sin 2 x + sin x
=
cos x [2 sin x + 1]
sin x [2 sin x + 1]
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 19 of 44
cos x
sin x
= cot x
1 − cos 2x
RTP:
= tan 2 x
1 + cos 2x
Proof:
1 − cos 2x
1 + cos 2x
=
2
2 sin 2 x
2 cos 2 x
= tan2x
RTP: tan x − cot x = −2 cot 2x
Proof:
sin x cos x
tan x − cot
x
=
−
cos x sin x
sin 2 x − cos 2 x −[cos 2 x − sin 2 x] −2 cos 2x
=
=
=
= − 2 cot 2x
1
cos x sin x
sin
2x
sin 2x
2
cos 2x
RTP:
= cos x + sin x
cos x − sin x
Proof:
cos 2x
cos 2 x − sin 2 x
=
cos x − sin x
cos x − sin x
=
3
4
=
5
(cos x − sin x) (cos x + sin x)
cos x − sin x
= cos x + sin x
1 − cos 2A + sin A
RTP:
= tan A
sin 2A + cos A
Proof:
1 − cos 2A + sin A
sin 2A + cos A
=
=
2 sin 2 A + sin A
2 sin A cos A + cos A
sin A (2 sin A + 1)
cos A (2 sin A + 1)
sin A
cos A
= tan A
1 − cos 4θ
RTP:
= tan 2θ
sin 4θ
Proof:
=
6
1 − cos 4θ
2 sin 2 2θ
=
sin 4θ
2 sin 2θ cos2θ
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 20 of 44
sin 2θ
cos 2θ
= tan 2θ
=
tan 2x =
7
(a)
(b)
1
4
4
4 17
=
17
17
cos 2x = 1 − 2 sin2x
2 sin2 x = 1 − cos 2x
1
4 
2
sin=
x
1−


2
17 
cos=
2x
sin=
x
1
2
−
2
17
1 2 17
−
2
17
sin(x + y) − sin(x − y)
RTP:
= tan y
cos(x + y) + cos(x − y)
Proof:
sin(x + y) − sin(x − y) sin x cos y + cos x sin y − sin x cos y + cos x sin y
=
cos(x +y) + cos(x - y) cos x cos y − sin x sin y + cos x cos y + sin x sin y
=
8
2 cos x sin y sin y
= = tan y
2 cos x cos y cos y
sin(θ − α ) 4
=
sin(θ + α ) 5
5 sin(θ − α) = 4 sin (θ + α)
⇒ 5 sin θ cos α − 5 cos θ sin=
α 4 sin θ cos α + 4 cos θ sin α
5 sin θ cos α − 4 sin θ cos
=
α 5 cos θ sin α + 4 cos θ sin α
sin θ cos α = 9 cos θ sin α
sin θ
sin α
=9
cos θ
cos α
tan θ = 9 tan α
1
 1
tan
=
α
, tan
=
θ (9) =
  3
3
3
2 tan θ
tan 2θ =
1 − tan 2 θ
=
9
=
2(3)
6 −3
=
=
2
1 − (3)
−8 4
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 21 of 44
10
5
tan α =
12
5
sin α =
13
12
cos α =
13
cos 2α = 2cos2α − 1
2
119
 12 
= 2  =
−1
169
 13 
2
cos 4α = 2 cos (2α) − 1
−239
 119 
= 2 =
 −1
28 561
 169 
11
cos θ = p
2
(a)
sin 2θ = 2 sin θ cos θ
= 2p 1 − p 2
2
 1 − p2  1 − p2

(b)=
tan θ  =
 p 
p2


(c) sin 4θ = 2 sin 2θ cos 2θ
= 2 sin 2θ [2 cos2 θ − 1]
2
=4p 1 − p 2 [2p 2 − 1]
12
tan 2α = 1
2 tan α
=1
1 − tan 2 α
2 tan α = 1 − tan2 α
tan2 α + 2 tan α − 1 = 0
−2 ± 8
tan α =
2
−2 ± 2 2
=
2
tan α = − 1 + 2, − 1 − 2
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 22 of 44
13
14
Since 0° < α < 90° ⇒ tan =
α
2 tan α
tan 2α =
1 − tan 2 α
1
α =67
2
2 −1
1
2 tan 67
1


2
⇒ tan 2  67  =
1
 2
2
1 − tan 67
2
1
2 tan 67
2
tan135 =
1
2
1 − tan 67
2
1
2 tan 67
2
−1 =
1
2
1 − tan 67
2
1
1
⇒ − 1 + tan 2 67 =2 tan 67
2
2
1
1
⇒ tan 2 67 − 2 tan 67 − 1 =
0
2
2
1 2± 8
tan 67 =
2
2
2±2 2
=
2
= 1± 2
1°
∴ tan 67
=
1 + 2, since the angle is acute
2
3
(a)
tan θ =
4
tan(θ + β) = −2
tan θ + tan β
= −2
1 − tan θ tan β
3
+ tan β
4
= −2
3
1 − tan β
4
3
3
+ tan β = − 2 + tan β
4
2
3
3
=
+2
tan β − tan β
4
2
11 1
=
tan β
4 2
11
11
tan β=
× 2=
4
2
(b)
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 23 of 44
sin θ =
3
5
sin β =
=
11
125
11
5 5
11 5
25
cos(A − B) 5
=
cos(A + B) 2
2 cos(A − B) = 5 cos(A + B)
2 cos A cos B + 2 sin A sin B = 5 cos A cos B − 5 sin A sin B
2 sin A sin B + 5 sin A sin B = 5 cos A cos B − 2 cos A cos B
⇒ 7 sin A sin B = 3 cos A cos B
7 sin A 3 cos B
=
cos A
sin B
7 tan A = 3 cot B
tan B = 3
1
cot B =
3
3 1 1
tan A = × =
7 3 7
tan (A + B)
tan A + tan B
=
1 − tan A tan B
=
15
(a)
1
+3
7
=
3
1−
7
22
= 7
4
7
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 24 of 44
22 11
=
4
2
=
(b)
1
50
=
50
50
cos 2A = 1 − 2sin2A
sin
=
A
(c)
2
16
 50 
= 1− 2 

 50 
2
= 1−
50
24
=
25
Since α, β and θ are the angles of a triangle:
α + β + θ = 180°
θ = 180 − (α + β)
tan θ = tan (180 − (α + β))
tan180 − tan(α + β)
=
1 − tan 180 tan(α + β)
= −tan(α + β)
 tan α + tan β 
= −

1 − tan α tan β 
=
tan α + tan β
tan α tan β − 1
17
(a)
cos 2θ = 1 − 2 sin2 θ
 1
= 1− 2  
 4
= 1−
=
2
2
16
7
8
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 25 of 44
cos 4θ = 2 cos2(2θ) − 1
2
 7
= 2   −1
 8
49
=
−1
32
17
=
32
3 cos x + 2 sin x = r cos(x − α)
3 cos x + 2 sin x = r [cos x cos α + sin x sin α]
Comparing coefficients of cos x and sin x
⇒ r cos α = 3
r sin α = 2
2
[2] ÷ [1] ⇒ tan α =
3
α = 33.7°
[1]2 + [2]2⇒ r2 = 32 + 22
r = 13
(b)
18
∴ 3 cos x + 2 sin
=
x
(a)
(b)
19
(a)
13 cos(x − 33.7°)
Max values = 13
When cos (x − 33.7°) = 1
⇒ x − 33.7° = 0
x = 33.7°
3 cos x + 2 sin x = 2
⇒ 13 cos(x − 33.7°) =2
2
cos(x − 33.7°) =
13
 2 
x − 33.7° =cos −1 
 13 
x − 33.7° = 360° n ± 56.3°
x = 360° n + 90° 
 n∈ 
x = 360° n − 22.6°
2 sin x + 4 cos x = r sin (x + α)
2 sin x + 4 cos x = r sin x cos α + r cos x sin α
Equating coefficients of sin x and cos x
⇒ r cos α = 2
[1]
r sin α = 4
[2]
4
[2] ÷ [1] ⇒ tan α= = 2 ⇒ α= 63.4°
2
[1]2 + [2]2 ⇒ r2 = 22 + 42
r = 20
∴ 2 sin x + 4 cos
=
x
(b)
[1]
[2]
20 sin (x + 63.4°)


2
Max 

 2 sin x + 4 cos x 


2
= Max 

 20 sin(x + 63.4°) 
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 26 of 44
2 20
20
4 5)
5
= =
20
5
4 cos x − 3 sin x = r cos (x + α)
4 cos x − 3 sin x = r cos x cos α − r sin x sin α
(a) Equating coefficients of cos x and sin x
⇒ r cos α = 4
r sin α = 3
3
[2] ÷ [1] ⇒ tan α=
⇒ α= 36.7°
4
[1]2 + [2]2 ⇒ r2 = 32 + 42
=
r =
25 5
∴ 4 cos x − 3 sin x = 5 cos (x + 36.7°)
(b) 5 cos (x + 36.9°) = 2
2
cos (x + 36.9°) =
5
x + 36.9° = 66.4°, 293.6°
x = 29.5°, 256.7°
(c) max (4 − 5 cos (x + 36.9°))
=4+5
=9
=
20
[1]
[2]
cos(x + 36.9) = 1
x + 36.9 = 0°, 360
x = −36.9, 323.1
Try these 9.7
(a)
sin (C + D) = sin C cos D + cos C sin D
[1]
sin (C − D) = sin C cos D − cos C sin D
[2]
[1] − [2] ⇒ sin(C + D) − sin(C − D) = 2 cos C sin D
A +B
A−B
Let C =
,D=
2
2
 A + B  A - B 
 A + B A − B
 A + B
A − B
⇒ sin 
−
+
 − sin 
  = 2 cos 
 sin 

2
2
2
2
2






 2 

 A + B
 A − B
⇒ sin A − sin B =
2 cos 
 sin 

 2 
 2 
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 27 of 44
cos(C + D) = cos C cos D − sin D sin C
[1]
cos(C − D) = cos C cos D + sin C sin D
[2]
[1] + [2] ⇒ cos(C + D) + cos (C − D) = 2 cos C cos D
A+B
A−B
=
Let C =
, D
2
2
 A + B  A − B 
 A + B A − B
 A + B
 A − B
2 cos 
cos 
⇒ cos 
+
+ cos 
−
=




 2
 2 
 2 
 2 
 2
2 
(b)
 A + B
 A − B
⇒ cos A + cos B =
2 cos 
cos 

 2 
 2 
(c) cos(C + D) = cos C cos D − sin C sin D
[1]
cos(C − D) = cos C cos D + sin C sin D
[2]
[1] − [2] ⇒ cos (C + D) − cos(C − D) = −2 sin C sin D
A+B
A−B
=
Let C =
,D
2
2
 A + B  A − B 
 A + B A − B
 A + B
 A − B
sin 
⇒ cos 
+
− cos 
−
=
− 2 sin 




 2
 2 
 2 
 2 
 2
2 
 A + B
 A − B
⇒ cos A − cos B =
− 2 sin 
sin 

 2 
 2 
Exercise 9E
1
2
sin 4 x − sin x
 4x + x 
 4x − x 
= 2 cos 
sin 

 2 
 2 
5 
3 
= 2 cos  x  sin  x 
2 
2 
cos 3x + cos 2x
 3x + 2x 
 3x − 2x 
= 2 cos 
cos 




2 
2 
5 
1 
= 2 cos  x  cos  x 
2 
2 
3
4
5
6
 5A + A 
 5A − A 
cos5A − cos A =
− 2sin 
sin 

 2 
 2 
= −2 sin 3A sin 2A
 4A + 4B 
 4A − 4B 
sin 4A + sin 4B =
2 sin 
 cos 


2
2
= 2 sin 2(A + B) cos 2(A − B)
 6A + 4A 
 6A − 4A 
cos 6A − cos 4A =
− 2 sin 
sin 





2
2
= − 2 sin 5A sin A
 2A + 8A 
 2A − 8A 
cos 2A − cos8A =
− 2 sin 
sin 





2
2
= − 2 sin 5A sin (−3A)
= 2 sin 5A sin 3A
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 28 of 44
7
8
9
10
11
 7x + 3x 
 7x − 3x 
sin 7x + sin 3x =
2 sin 
 cos 


2
2 
= 2 sin 5x cos 2x
 5x + 3x 
 5x − 3x 
cos5x + cos3x =
2 cos 
 cos 


2
2 
= 2 cos 4x cos x
 6x + 2x 
 6x − 2x 
sin 6x − sin 2x =
2 cos 
sin 




2 
2 
= 2 cos 4x sin 2x
 7x + 5x 
 7x − 5x 
sin 7x + sin 5x =
2 sin 
cos 




2 
2 
= 2 sin 6x cos x
5π
π
(a)
cos
+ cos
12
12
5
π
π

 5π π 
+
−




= 2 cos  12 12  cos  12 12 
2
2




 π
 π
= 2 cos   cos  
 4
 6
 2  3
= 2
 
 2  2 
6
2
5π
π
cos
− cos
12
12
π
π
= − 2 sin   sin  
4
6
 2   1
= −2
 
 2   2
=
(b)
− 2
2
 5π 
 π
sin   − sin  
 12 
 12 
=
(c)
 π
 π
= 2 cos   sin  
 4
 6
=2
2  1
 
2  2
2
2
cos α + cos β
sin α + sin β
=
12
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 29 of 44
=
 α + β
 α − β
cos 
2 cos 

 2 
 2 
 α + β
 α − β
2 sin 
cos 
 2 
 2 
 α + β
cos 
 2 
 α + β
= = cot 


2 
 α + β
sin 
 2 
13
sin 40 + cos 70
= cos 50 + cos 70
50 + 70
50 − 70
= 2 cos
cos
2
2
= 2 cos 60 cos (−10)
= 2 cos 60 cos 10
14
cos 5x + cos x = 0
⇒ 2 cos 3x cos 2x = 0
∴ cos 3x = 0, cos 2x = 0
π
π
3x= 2nπ ± OR 2x= 2nπ ±
2
2
2π
π
x=π±
n or x = nπ ± , n ∈ 
3
6
4
15
sin 6x + sin 2x = 0
2 sin 4x cos 2x = 0
sin 4x = 0, cos 2x = 0
π
4x = nπ, 2x= 2nπ ±
2
nπ
π
x=
or
π ±x =, nn
∈
4
4
16
cos 6x − cos 4x = 0
⇒ −2 sin 5x sin x = 0
sin 5x = 0, sin x = 0
5x = nπ + (−1)n or x = nπ + (−1)n (0)
nπ
x=
or
π, xn = n
∈
5
17
sin 3x = sin x
⇒ sin 3x − sin x = 0
2 cos 2x sin x = 0
cos 2x = 0, sin x = 0
π
2x = 2nπ ± , x = nπ
2
π
∴ x = nπ + 
4  n ∈

x = nπ
18
cos 6x + cos 2x = cos 4x
⇒ 2 cos 4x cos 2x = cos 4x
⇒ 2 cos 4x cos 2x − cos 4x = 0
⇒ cos 4x (2 cos 2x − 1) = 0
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 30 of 44
19
20
cos 4x = 0, 2 cos 2x − 1 = 0
1
cos 2x =
2
π
π
4x = 2nπ ± , 2x = 2nπ ±
2
3
2nπ π
2nπ π
x=
± , x=
±
4
8
2
6
nπ π 
x=
+
2 8 
 n ∈
π
nπ ±
6 
sin 7x + sin x = sin 4x
⇒ sin 7x + sin x − sin 4x = 0
⇒ 2 sin 4x cos 3x − sin 4x = 0
sin 4x [2 cos 3x − 1] = 0
1
sin 4x = 0, cos3 x =
2
π
4x = nπ,
3x = 2nπ ±
3
nπ
2
π
x=
,π ± x,=n n
∈
4
3
9
cos 5x − sin 3x − cos x = 0
⇒ cos 5x − cos x − sin 3x = 0
⇒ −2 sin 3x sin 2x − sin 3x = 0
⇒ −sin 3x [2sin 2x + 1] = 0
1
sin 3x = 0, sin 2x = −
2
 −π 
3x = nπ, 2x = nπ + ( −1) n  
 6 
nπ
nπ
 −π 
, x=
+ ( − 1) n 
 ,n ∈
3
2
 12 
sin 3x + sin 4x + sin 5x = 0
sin 5x + sin 3x + sin 4x = 0
2 sin 4x cos x + sin 4x = 0
sin 4x (2 cos x + 1) = 0
1
sin 4x = 0, cos x = −
2
2π
4x = πn, x = 2n π ±
3
nπ
2π
x=
orπ x± = 2n, n
∈
4
3
sin x + 2 sin 2x + sin 3x = 0
sin 3x + sin x + 2 sin 2x = 0
2 sin 2x cos x + 2 sin 2x = 0
2 sin 2x (cos x + 1) = 0
sin 2x = 0, cos x + 1 = 0
2x = nπ,
cos x = −1
x = 2nπ ± π
x=
21
22
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 31 of 44
nπ
orπ x± =π,2nn
∈
2
cos 3x + cos x + 2 cos 2x = 0
2 cos 2x cos x + 2 cos 2x = 0
2 cos 2x (cos x + 1) = 0
cos 2x = 0, cos x = −1
π
2x= 2nπ ± , x= 2nπ ± π
2
π
x = nπ ± ,π x± =π,2nn
∈
4
sin 4θ + sin θ
cos 4θ + cos θ
x=
23
24
=
 5θ 
 3θ 
2 sin   cos  
 2
 2
 5θ 
 3θ 
2 cos   cos  
 2
 2
 5θ 
= tan  
 2
25
sin 6θ − sin 2θ
cos 6θ + cos 2θ
=
26
2 cos 4θ cos 2θ
= tan 2θ
sin 8θ + sin 4θ
cos8θ − cos 4θ
=
27
2 cos 4θ sin 2θ
2 sin 6θ cos 2θ
−2 sin 6θ sin 2θ
= − cot 2θ
cos 7θ + cos θ
sin 7θ + sin θ
2 cos 4θ cos3θ
2 sin 4θ cos3θ
= cot 4θ
sin x + 2 sin 3x + sin 5x
sin 3x + 2 sin 5x + sin 7x
sin 5x + sin x + 2sin 3x
=
sin 7x + sin 3x + 2 sin 5x
2 sin 3x cos 2x + 2 sin 3x
=
2 sin 5x cos 2x + 2 sin 5x
=
28
=
2 sin 3x( cos 2x + 1)
2 sin 5x (cos 2x + 1)
Unit 1 Answers: Chapter 9
=
sin 3x
sin 5x
© Macmillan Publishers Limited 2013
Page 32 of 44
29
sin x + sin 2x
=
cos x − cos 2x
 3x 
 x
2 sin   cos  
 2
 2
 −x 
 3x 
−2 sin   sin  
 2 
 2
 x
cos  
 2
=
 x
sin  
 2
 x
= cot  
 2
 5x 
x
2 sin   cos  
sin 3x + sin 2x
 2 
 2  tan  5x  cot  x 
30 = =
 
 
5x
sin 3x − sin 2x
 
x
 2 
2
2 cos   sin  
 2 
2
cos3x + cos x 2 cos 2x cos x
31
=
sin 3x + sin x
2 sin 2x cos x
= cot 2x
sin 7θ + sin θ
2 sin 4θ cos3θ
32
=
cos 7θ + cos θ 2 cos 4θ cos3θ
= tan 4θ
cos θ + 2 cos2θ + cos3θ
33
cos θ − 2 cos2θ + cos3θ
cos3θ + cos θ + 2 cos 2θ
=
cos3θ + cos θ − 2 cos 2θ
2 cos 2θ cos θ + 2 cos 2θ
=
2 cos 2θ cos θ − 2 cos 2θ
=
=
2 cos 2θ (cos θ + 1)
2 cos 2θ (cos θ − 1)
cos θ + 1
cos θ − 1
θ
2 = − cot 2  θ 
=
 
θ
2
−2 sin 2
2
sin 4θ + sin 6θ + sin 5θ
cos 4θ + cos 6θ + cos5θ
sin 6θ + sin 4θ + sin 5θ
=
cos 6θ + cos 4θ + cos5θ
2 sin 5θ cos θ + sin 5θ
=
2 cos5θ cos θ + cos5θ
2 cos 2
34
=
sin 5θ [2 cos θ + 1]
cos5θ [2 cos θ + 1]
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 33 of 44
sin 5θ
cos5θ
= tan 5θ
sin 6θ + sin 7θ + sin θ + sin 2θ
cos 2θ + cos θ + cos 6θ + cos 7θ
sin 7θ + sin 2θ + sin 6θ + sin θ
=
cos 7θ + cos 2θ + cos 6θ + cos θ
=
35
 9θ 
 5θ 
 7θ 
 5θ 
2 sin   cos   + 2 sin   cos  
 2
 2
 2
 2
=
 9θ 
 5θ 
 7θ 
 5θ 
2 cos   cos   + 2 cos   cos  
 2
 2
 2
 2
=
36
 5θ    9θ 
 7θ  
2 cos   sin   + sin   
 2   2 
 2 
 9θ 
 7θ  
 5θ  
2 cos   cos   + cos   
 2  
 2 
 2 
 9θ 
 7θ 
sin   + sin  
 2 
 2 
=
 9θ 
 7θ 
cos   + cos  
 2 
 2 
θ
2 sin 4θ cos  
2
=
θ
2 cos 4θ cos  
2
sin 4θ
=
cos 4θ
= tan 4θ
sin θ + sin 3θ + cos5θ + cos 7θ
sin 4θ + cos8θ + cos 4θ
2 sin 2θ cos θ + 2cos 6θ cos θ
=
2 sin 2θ cos 2θ + 2cos 6θ cos 2θ
=
=
2 cos θ [sin 2θ + cos 6θ]
2 cos 2θ [sin 2θ + cos 6θ]
cos θ
2 cos 2 θ − 1
cos θ
sec θ
cos 2 θ
= =
1
−
2
sec 2 θ
2−
cos 2 θ
cos θ − 2 cos3θ + cos 7θ
37
cos θ + 2cos3θ + cos 7θ
2 cos 4θ cos3θ − 2 cos3θ
=
2 cos 4θ cos3θ + 2 cos3θ
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 34 of 44
=
=
2 cos3θ [cos 4θ − 1]
2 cos3θ [cos 4θ + 1]
cos 4θ − 1
cos 4θ + 1
−2 sin 2 2θ
2 cos 2 2θ
= −tan2(2θ)
cos5θ + 2 cos 7θ + cos9θ
cos5θ − 2 cos 7θ + cos9θ
2 cos 7θ cos 2θ + 2 cos 7θ
=
2 cos 7θ cos 2θ − 2 cos 7θ
=
38
=
39
40
41
42
43
44
45
46
2 cos 7θ [cos 2θ + 1]
cos 2θ + 1 2 cos 2 θ
=
=
=
− cot 2 θ
2 cos 7θ [cos 2θ − 1] cos 2θ − 1 −2 sin 2 θ
2 sin 6θ cos θ = sin 7θ + sin 5θ
−2 sin 8θ cos 4θ = − [2 sin 8θ cos 4θ]
= −[sin 12θ + sin 4θ]
2 cos 6θ cos 2θ = cos 8θ + cos 4θ
2 sin 7θ sin θ = −[−2 sin 7θ sin θ]
= −[cos 8θ − cos 6θ]
= cos 6θ − cos 8θ
2 cos 7θ cos 3θ = cos 10θ + cos 4θ
−2 sin 7θ cos 3θ = −[2 sin 7θ cos 3θ]
= −[sin 10θ + sin 4θ]
= −sin 10θ − sin 4θ
RTP: 2 cos x (sin 3x − sin x) = sin 4x
Proof:
2 cos x (sin 3x − sin x)
= 2 cos x [2 cos 2x sin x]
= 2 cos 2x [2 sin x cos x]
= 2 cos 2x sin 2x
= sin 4x
 5x + x 
 5x − x 
sin 5x + sin x =
2 sin 
 cos 

 2 
 2 
= 2 sin 3x cos 2x.
sin 5x + sin x + cos 2x = 0
⇒ 2 sin 3x cos 2x + cos 2x = 0
cos 2x [2 sin 3x + 1] = 0
1
cos 2x = 0, sin 3x = −
2
π
 −π 
2x = 2nπ ± , 3x = nπ + ( −1) n  
 6 
2
47
π
4


 n ∈
nπ
n  −π 
x=
+ ( − 1) 

3
 18  
(a) RTP: sin 2P + sin 2Q + sin 2R = 4 sin P sin Q sin R where P + Q +R = 180°
x = nπ ±
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 35 of 44
Proof:
sin 2P + sin 2Q + sin 2R
= sin 2P + 2 sin (Q + R) cos (Q − R)
= sin 2P + 2 sin (180 − P) cos (Q − R),
= 2 sin P cos P + 2 sin P cos (Q − R),
= 2 sin P [cos P + cos (Q − R)]

P+Q−R
 P − Q + R
cos 
= 2 sin P  2 cos
 

2
2


= 2 sin P [2 cos (90 − R) cos (90 − Q)]
= 4 sin P sin Q sin R
(b)
48
(a)
(b)
49
(a)
(b)
(c)
[Q + R = 180 − P]
[sin (180 − P) = sin P]
[P + Q − R = 180 − 2R
P+Q−R
= 90 − R
2
P + R − Q = 180 − 2Q
P+R −Q
= 90 − Q ]
2
[cos(90 − R) = sin R
cos (90 − Q) = sin Q]
sin 2P + sin 2Q − sin 2R
= 2 sin P cos P + 2 cos (Q + R) sin (Q − R)
= 2 sin P cos P + 2 cos(180 − P) sin (Q − R)
= 2 sin P cos P − 2 cos P sin (Q − R).
= 2 cos P [sin P − sin (Q − R)]
P+Q−R
P + R − Q

= 2 cos P  2 cos
sin

2
2

= 2 cos P [2 cos(90 − R) sin (90 − Q)]
= 4 cos P cos Q sin R
Proof:
P + Q + R = 180°
sin(Q + R) = sin(180 − P)
= sin P
cos(Q + R) = cos(180 − P)
= −cos P
α + β + γ = 180°.
sin β cos γ + cos β sin γ
= sin (β + γ)
= sin (180 − α)
= sin α
cos γ + cos β cos α
= cos (180 − (β + α)) + cos β cos α
= − cos(β + α) + cos β cos α
=
− cos β cos α + sin β sin α + cos β cos α
= sin β sin α
sin α − cos β sin γ
= sin(180α)
− cosβ
− sin γ
= sin β cos γ + cos β sin γ − cos β sin γ
50
= sin β cos γ
1 + cos 2θ + cos 4θ + cos 6θ
= 1 + cos 2θ + 2 cos 5θ cos θ
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 36 of 44
51
= 2 cos2θ + 2 cos 5θ cos θ
= 2 cos θ [cos θ + cos 5θ]
= 2 cos θ [2 cos 3θ cos 2θ]
= 4 cos θ cos 2θ cos 3θ
Now 1 + cos 2θ + cos 4θ + cos 6θ = 0
⇒ 4 cos θ cos 2θ cos 3θ = 0.
⇒ cos θ = 0, cos 2θ = 0, cos 3θ = 0
π
π
π
θ = 2nπ ± , 2θ = 2nπ ± , 3θ = 2nπ ±
2
2
2
π
θ = 2nπ ± 
2

π 
θ = nπ ±
 n ∈
4 
2nπ π 
=
θ
± 
3
6
1 − cos 2θ + cos 4θ − cos 6θ
= 2 sin2θ + [−2 sin(−θ) sin 5θ]
= 2 sin2 θ + 2 sin θ sin 5θ
= 2 sin θ [sin θ + sin 5θ]
= 2 sin θ [2 cos 2θ sin 3θ]
= 4 sin θ cos 2θ sin 3θ
Now 1 − cos 2θ + cos 4θ − cos 6θ = 0
⇒ 4 sin θ cos 2θ sin 3θ = 0
⇒ sin θ = 0, cos 2θ = 0, sin 3θ = 0
π
θ = nπ, 2θ = 2nπ ± , 3θ = nπ
2


θ = nπ

π
nπ ±  n ∈ 
4
nπ


3
Review Exercise 9
1
(a)
RTP:
cos θ
= sec θ + tan θ
1 − sin θ
Proof:
cos θ
cos θ
1 + sin θ
=
×
1 − sin θ 1 − sin θ 1 + sin θ
cos θ(1 + sin θ)
=
1 − sin 2 θ
cos θ(1 + sin θ)
=
cos 2 θ
1 + sin θ
=
cos θ
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 37 of 44
1
sin θ
+
cos θ cos θ
= sec θ + tan θ
1
1
sec x + tan x
(b)
=
×
sec x − tan x sec x − tan x sec x + tan x
sec x + tan x
=
sec 2 x − tan 2 x
= sec x + tan x
[since sec2x − tan2x = 1]
2
(a) 3 cos x = 1 + sin x
⇒ 3(1 − sin2x) = 1 + sin x
⇒ 3 − 3 sin2x = 1 + sin x
⇒ 3 sin2x + sin x − 2 = 0.
y = sin x
3y2 + y − 2 = 0
(3y − 2) (y + 1) = 0
2
=
y
, −1
3
2
sin x = , sin x = − 1
3
n

x = nπ + ( −1) (0.730)

n ∈
n  −π  
x = nπ + ( −1)  , 
 2  
2
(b) 3 cos x = 2 sin x
3 cos x = 2(1 − cos2x)
2 cos2 x + 3 cos x − 2 = 0
y = cos x
2y2 + 3y − 2 = 0
(2y − 1) (y + 2) = 0
1
=
y
,−2
2
1
cos x = , cos x =− 2 ⇒ No solutions
2
π
x= 2nπ ±
3
9 + 9 cos6α
9(1 + cos6α )
=
2
2
=
2
3
=
4
9 cos 2 (3α )
= 3 cos 3α
cos3α sin 3α
+
cos α
sin α
cos3α sin α + sin 3α cos α
=
cos α sin α
sin(α + 3α )
=
1
sin 2α
2
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 38 of 44
sin 4α
1
sin 2α
2
2 sin 2α cos2α
=
1
sin 2α
2
= 4 cos 2α
RTP: sin (α + β) sin(α − β) = sin2 α − sin2 β
Proof:
sin (α + β) sin (α − β)
= (sin α cos β + cos α sin β) (sin α cos β − cos α sin β)
= sin2α cos2β − cos2α sin2β
= sin 2 α (1 − sin 2 β) − sin 2 β(1 − sin 2 α )
=
5
= sin 2 α − sin 2 α sin 2 β − sin 2 β + sin 2 β sin 2 α
6
= sin 2 α − sin 2 β
5
sin θ =
13
12
13
5
tan θ =
12
cos θ =
−3
5
−4
cos α =
5
3
tan α =
4
(a) sin (θ + α)
= sin θ cos α + cos θ sin α
 5   −4   12   −3 
=     +   
 13   5   13   5 
sin α =
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 39 of 44
−20 36
−
65 65
−56
=
65
(b) cos(θ + α) = cosθ cosα − sin θ sin α
 12   −4   5   3 
=     −   − 
 13   5   13   5 
−48 15 −33
=
+
=
65 65 65
(a) RTP:
cos θ − 2cos3θ + cos 7θ
=
− tan 2 (2θ)
cos θ + 2cos3θ + cos 7θ
Proof:
cos θ − 2cos3θ + cos 7θ
cos θ + 2cos3θ + cos 7θ
2 cos 4θ cos3θ − 2 cos3θ
=
2 cos 4θ cos3θ + 2 cos3θ
=
7
=
2cos3θ (cos 4θ − 1)
2 cos3θ (cos 4θ + 1)
=
cos 4θ − 1
cos 4θ + 1
=
=
1 − 2sin 2 2θ − 1
2 cos 2 2θ − 1 + 1
− 2 sin 2 2θ
2 cos 2 2θ
− sin 2 2θ
cos 2 2θ
= − tan2(2θ)
RTP:
cos5θ + 2 cos 7θ + cos9θ
=
− cot 2 θ
cos5θ − 2 cos 7θ + cos9θ
Proof:
cos5θ + 2 cos 7θ + cos9θ
cos5θ − 2 cos 7θ + cos9θ
2 cos 7θ cos 2θ + 2 cos 7θ
=
2 cos 7θ cos 2θ − 2 cos 7θ
=
(b)
=
=
2 cos 7θ [cos 2θ + 1]
2 cos 7θ [cos 2θ − 1]
2 cos 2 θ − 1 + 1
1 − 2 sin 2 θ − 1
2 cos 2 θ
− 2 sin 2 θ
= −cot2θ
=
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 40 of 44
cos 15° = cos(60° − 45°)
= cos 60 cos 45 − sin 60 sin 45
2
2 3
=
−
4
2 2
2
=
[1 − 3]
4
sin(θ − α ) sin θ cos α − cos θ sin α
9
=
sin θ sin α
sin θ sin α
8
sin θ cos α cos θ sin α
−
sin θ sin α sin θ sin α
cos α cos θ
=
−
sin α sin θ
= cot α − cot θ
π

π

10
sin  + x  cos  + x 
4

4

=
1
π

sin  + 2x 
2

2
1
= cos 2x
2
Since sin (90 + α=
) cos α
f(θ) = 3 sin θ + 4 cos θ
3 sin θ + 4 cos θ = r sin (θ + α)
= r sin θ cos α + r cos θ sin α
Comparing coefficients of sin θ and cos θ
⇒ r cos α = 3
[1]
r sin α = 4
[2]
r sin α 4
4
4


[2] ÷ [1] ⇒
=
⇒ tan α=
, α= tan −1  = 53.1°
r cos α 3
3
3
=
11
[1]2 + [2]2 ⇒ r 2 sin 2 α + r 2 cos 2 α= 42 + 32
r2 = 25
r=5
∴ f(θ) = 5 sin (θ + 53.1°)
max f(θ) = 5


1
1
1
min 
=
 =
10 + f (θ)  10 + 5 15
12=
f (x)
(a)
(b)
1
π

sin  4x + 

2
2
−1
1
Range : ≤ f (x) ≤
2
2
π
Period :
2
(c)
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 41 of 44
13
(a)
2 cos(2x) + 3 sin 2x = r cos (2x − θ)
= r cos 2x cos θ + r sin 2x sin θ
Equating coefficients of cos 2x and sin 2x
⇒ r cos θ = 2
r sin θ = 3
r sinθ 3
[2] ÷ [1] ⇒
=
r cosθ 2
tan θ=
[1]
[2]
3
 3
⇒ θ= tan −1  = 56.3°
 2
2
[1]2 + [2]2 ⇒ r 2 cos 2 θ + r 2 sin 2 θ= 22 + 32
r2 = 13
r = 13
∴ 3 cos 2x + 3 sin 2x = 13 cos (2x − 56.3°)
(b)
14
15
16
17
13 cos(2 x − 56.3°) =2
2
cos(2x − 56.3°) =
13
 2 
2x − 56.3 =
cos −1 
 13 
⇒ 2x − 56.3 = 360n ± 56.3°
2x = 360n + 112.6°, 360°n
x = 180°n + 56.3°, 180°n, n ∈ ℤ
(c)
Maximum value = 13
cos(2x − 56.3) = 1
2x − 56.3 = 0
x = 28.2°
2 sin 6θ cos θ = sin 7θ + sin 5θ
−2 sin 8θ cos 4θ = −[sin 12θ + sin 4θ]
= −sin 12θ − sin 4θ
2 cos 6θ cos 2θ = cos 8θ + cos 4θ
(a) 2 tan x− 1 = 3 cot x
3
2 tan x − 1 =
tan x
⇒ 2 tan2 x − tan x − 3 = 0
(2 tan x −3) (tan x + 1) = 0
3
tan x = , tan x = − 1
2
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 42 of 44
x = nπ + 0.983 

 n ∈
π
x = nπ −

4
(b) 6 sec2 z = tan z + 8
6 (1 + tan2 z) = tan z + 8
6 tan2 z − tan z − 2 = 0
(2 tan z + 1) (3 tan z − 2) = 0
−1
2
=
tan z =
tan z
2
3
z = nπ −0.464 
 n ∈
z = nπ + 0.588 
18
(a)
RTP:
sin x
sin x
+
=
− 2 cos x cot x
1 − sec x 1 + sec x
Proof:
sin x
sin x
+
1 − sec x 1 + sec x
sin x(1 + sec x) + sin x(1 − sec x)
=
(1 − sec x) (1 + sec x)
=
=
sin x + sin x sec x + sin x − sin x sec x
1 − sec 2 x
2 sin x
1 − 1 − tan 2 x
2 sin x
−2 cos 2 x
=
− sin 2 x
sin x
2
cos x
= −2 cot x cos x
1 − sin x
cos x
RTP:
=
cos x
1 + sin x
Proof:
1 − sin x 1 − sin x 1 + sin x
=
×
cos x
cos x
1 + sin x
=
(b)
=
1 − sin 2 x
cos x(1 + sin x)
cos 2 x
cos x (1 + sin x)
cos x
=
1 + sin x
OR:
x
x
x
x
cos2 + sin 2 − 2 sin cos
1 − sin x
2
2
2
2
=
x
cos x
 x
cos2 − sin 2  
 2
2
=
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 43 of 44
2
x
x

 cos − sin 
2
2
=

x
x
x
x
 cos 2 − sin 2   cos 2 + sin 2 
x
x
− sin
2
2
=
x
x
cos + sin
2
2
x
x
x
x
cos − sin
cos + sin
2
2 ×
2
2
=
x
x
x
x
cos + sin
cos + sin
2
2
2
2
x
x
cos2 − sin 2
2
2
=
x
x
x
x
cos2 + 2sin cos + sin 2
2
2
2
2
cos x
=
1 + sin x
(c) cot θ + tan θ
cos θ sin θ
=
+
sin θ cos θ
cos 2 θ + sin 2 θ
=
sin θ cos θ
1
1
=
×
sin θ cos θ
= sec θ cosec θ
19
(a) cos 5x − sin 3x − cos x = 0
⇒ cos 5x − cos x − sin 3x = 0
⇒ −2sin 3x sin 2x − sin 3x = 0
⇒ −sin 3x (2 sin 2x + 1) = 0
−1
=
sin 3x 0,=
sin 2x
2
nπ
3x = nπ ⇒ x =
3
 −π 
2x = nπ + ( −1) n  
 6 
cos

nπ
n  −π 
 x = 2 + ( − 1)  12  

 n ∈


nπ
x =

3


(b) sin 3x + sin 4 x + sin 5x = 0
sin 4x + sin 3x + sin 5x = 0
sin 4x + 2 sin 4x cos x = 0
sin 4x (1 + 2 cos x) = 0
−1
=
sin 4x 0,=
cos x
2
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 44 of 44
4x = nπ,
(a)
2π
3


 n ∈
π
x = 2nπ ±
3 
sin 7θ − sin θ
cos 7θ + cos θ
x=
20
x = 2nπ ±
=
nπ
4
2 cos 4θ sin 3θ
2 cos 4θ cos3θ
sin 3θ
cos3θ
= tan 3θ
cos θ + 2 cos 2θ + cos3θ
cos θ − 2 cos 2θ + cos3θ
2 cos 2θ cos θ + 2 cos 2θ
=
2 cos 2θ cos θ − 2 cos 2θ
=
(b)
=
2 cos 2θ (cos θ + 1)
2 cos 2θ (cos θ − 1)
 θ
2 cos2  
 2
=
 θ
− 2 sin 2  
 2
 θ
= − cot 2  
 2
Unit 1 Answers: Chapter 9
© Macmillan Publishers Limited 2013
Page 1 of 20
Chapter 10 Coordinate Geometry
Try these 10.1
(a)
(ii)
(i)
P (4, 5)
Q (5, 4)
4−5
Gradient of PQ =
= −1
5−4
Gradient of perpendicular bisector = 1
 4 + 5 5 + 4
Mid point of PQ = 
,

 2
2 
9 9
= , 
2 2
Equation of the perpendicular bisector is
9
9

y − = 1x − 

2
2
y=x
Q (5, 4)
R (6, 1)
1− 4
Gradient of QR =
= −3
6−5
1
Gradient of perpendicular =
3
5
+
6
4
+ 1

Mid-point of QR = 
,

2 
 2
 11 5 
= , 
 2 2
Equation of the perpendicular bisector is:
5 1
11
y− =
 x − 
2 3
2
1
11 5
x− +
3
6 2
1
2
=
y
x+
3
3
(iii) Centre of the circle is found by solving the equation of the perpendicular bisectors
simultaneously:
y=x
1
2
=
y
x+
3
3
1
2
=
x
x+
3
3
2
2
x = ⇒ x = 1, y = 1
3
3
y=
(b)
Centre (1, 1), radius : (4 − 1) 2 + (5 − 1) 2 =
Equation is (x – 1)2 + (y – 1)2 = 52
A (1, 3)
B (– 2, 6) C (4, 2)
Let the equation of the circle be
Unit 1 Answers: Chapter 10
9 + 16 = 5
© Macmillan Publishers Limited 2013
Page 2 of 20
(x – a)2 + (y – b)2 = r2
when x =1, y = 3 ⇒ (1 – a)2 + (3 – b)2 = r2
x = – 2, y = 6 ⇒ (– 2 – a)2 + (6 – b)2 = r2
x = 4, y = 2 ⇒ (4 – a)2 + (2 – b)2 = r2
1 – 2a + a2 + 9 – 6b + b2 = r2
[1]
2
2
2
4 + 4a + a + 36 – 12b + b = r
[2]
16 – 8a + a2 + 4 – 4b + b2 = r2
[3]
[2] – [1] ⇒ 3 + 6a + 27 – 6b = 0
6a – 6b = –30
a – b = –5
[4]
[3] – [1] ⇒ 15 – 6a – 5 + 2b = 0
–6a + 2b = –10
–3a + b = –5
[5]
[4] + [5] ⇒ –2a = –10
a=5
a – b = –5 ⇒ 5 – b = –5, b = 10
when a = 5, b = 10 ⇒ 1 – 10 + 25 + 9 – 60 + 100 = r2
r2 = 65
∴ Equation of the circle is
(x – 5)2 + (y – 10)2 = 65
Try these 10.2
(a)
(b)
x2 + y2 = 20
[1]
(x – 6)2 + (y – 3)2 = 5 ⇒ x2 – 12x + 36 + y2 – 6y + 9 = 5
x2 + y2 – 12x – 6y = – 40
[2]
[1] – [2] ⇒ 12x + 6y = 60
2x + y = 10
y = 10 – 2x
∴ x2 + (10 – 2x)2 = 20
⇒ x2 + 100 – 40x + 4x2 – 20 = 0
5x2 – 40x + 80 = 0
x2 – 8x + 16 = 0
(x – 4)2 = 0
x=4
when x = 4, y = 10 – 2 (4) = 2
Point of intersection is (4,2)
C (– 2, 3) P (– 5, – 1)
−1− 3
−4 4
(i)
Gradient of CP
=
= =
− 5 − ( − 2) − 3 3
Equation of CP:
4
4
17
y − 3=
(x + 2) ⇒ y=
x+
3
3
3
(ii)
(iii)
(iv)
r = (− 5 − (− 2)) 2 + (− 1 − 3) 2 = 9 + 16 = 25 = 5
Equation of the circle is (x + 2)2 + (y – 3)2 = 52
−3
Gradient of the tangent =
4
Equation of the tangent:
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 3 of 20
−3
(x + 5)
4
−3
19
=
y
x−
4
4
(v) (x + 2)2 + (y – 3)2 = 52
x2 + y2 + 4x – 6y + 4 + 9 = 25
x2 + y2 + 4x – 6y = 12
[1]
x2 + y2 + 6x – 7y = 10
[2]
[1] – [2] ⇒ – 2x + y = 2
y = 2 + 2x
substitute into [1]
⇒ x2 + (2 + 2x)2 + 4x – 6 (2 + 2x) – 12 = 0
x 2 + 4 + 8x + 4x 2 + 4x − 12 − 12x − 12 =
0
2
5x – 20 = 0
x2 = 4 ⇒ x = ± 2
when x = 2, y = 2 + 2 (2) = 6
x = – 2, y = 2 + 2 (– 2) = – 2
∴ Pt. of intersection (2, 6) and (– 2, – 2)
y=
+1
Exercise 10 A
1
2
3
4
5
6
7
(x – 1)2 + (y – 1)2 = 42 = 16
⇒ x2 – 2x + 1 + y2 – 2y + 1 = 16
x2 + y2 – 2x – 2y – 14 = 0
(x + 2)2 + (y – 3)2 = 25
(x – 4)2 + (y – 2)2 = 49
x2 + (y – 2)2 = 1
(x + 1)2 + (y – 1)2 = 4
(x – 3)2 + y2 = 2
Centre (– 1, 2)
Point (4, 1)
radius=
(4 − ( − 1))2 + (1 − 2)2
= 26
Equation: (x + 1)2 + (y – 2)2 = 26
8
radius
=
(2 − ( − 3))2 + (2 − 1)2
= 26
Equation: (x + 3)2 + (y – 1)2 = 26
9
radius =
(4 − 1)2 + (6 − 1)2
10
= 9 + 25 = 34
Equation: (x – 1)2 + (y – 1)2 = 34
A (2, 4) B(– 1, 6)
 2 − 1 4 + 6
Centre of the circle in the mid-point of AB: 
,

 2
2 
1 
=  , 5
2 
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 4 of 20
Radius of the circle =
1
length of AB
2
1
( − 1 − 2)2 + (6 − 4)2
2
1
=
9+4
2
1
=
13
2
2
2
1
2

1

Equation is  x −  + ( y − 5 ) =
13


2

2

13
=
4
2
1

4  x −  + 4 (y − 5) 2 =
13
2

11
(a) 6x2 + 6y2 – 4x – 5y – 2 = 0
2
5
1
⇒ x 2 − x + y2 − y =
3
6
3
2
2
1
1 
5
25 1

⇒ x −  − + y −  −
=




3
9
12
144 3
=
2
2
1
5
1 25 1 89


+ =
 x −  +  y −  = +
3
12
9 144 3 144
(b)
(c)
89
1 5 
Centre  ,  , radius =
 3 12 
12
x2 + y2 + 6x + 8y – 1 = 0
x2 + 6x + y2 + 8y = 1
(x + 3)2 – 9 + (y + 4)2 – 16 = 1
(x + 3)2 + (y + 4)2 = 26
Centre (– 3, – 4) radius 26
3x2 + 3y2 – 4x + 8y – 2 = 0
4
8
2
x 2 − x + y2 + y =
3
3
3
2
2
2
4 
4
16 2

x
−
−
+
y
+
=



 −
3
9
3
9 3
2
12
2
2 
4
4 16 6

+
x −  + y +  = +
3 
3 9 9 9

26
=
9
26
 2 − 4
Centre  ,
radius

3 3 
3
2
2
x + y – 2x + 4y = 0
(x – 1)2 – 1 + (y + 2)2 – 4 = 0
(x – 1)2 + (y + 2)2 = 4
Centre (1, – 2)
Point (2, 0)
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 5 of 20
−2−0
= +2
1− 2
1
Gradient of the tangent = −
2
Equation of the tangent at (2, 0) is
1
y=
− (x − 2)
2
2y = – x + 2
Equation: 2y + x = 2
x2 + y2 – 6x + 4y + 3 = 0
(x – 3)2 – 9 + (y + 2)2 – 4 + 3 = 0
(x – 3)2 + (y + 2)2 = 10
Centre (3, – 2)
Point (0, – 3)
− 2 − (− 3) 1
Gradient of the normal =
=
3−0
3
Gradient of the tangent = – 3
Equation of the tangent at (0, – 3) is
y + 3 = – 3x
Equation: y + 3x + 3 = 0
2x2 + 2y2 – x + 4y – 15 = 0
1
15
x 2 + y 2 − x + 2y =
2
2
2
1
1
15

2
 x −  − + (y + 1) − 1 =
4  16
2

Gradient of the normal =
13
14
2
1
15
1

2
+1+
 x −  + (y + 1) =
4
2
16
137
=
16
1

Centre  , − 1
4

Point (3, 0)
−1− 0 4
Gradient of the normal
= =
1
− 3 11
4
− 11
Gradient of the tangent =
4
Equation of the tangent at (3, 0) is
− 11
=
y
(x − 3)
4
4y + 11x = 33
15
x2 + 6x + y2 = 16
(x + 3)2 – 9 + y2 = 16
(x + 3)2 + y2 = 25
Centre (–3, 0)
Point. (1, – 3)
0 − (− 3) − 3
Gradient of the=
normal =
4
− 3 −1
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 6 of 20
16
Equation of the normal at (1, – 3) is
−3
y=
+3
(x − 1)
4
4y + 12 = – 3x + 3
4y + 3x + 9 = 0
3x2 + 3y2 – 6x + 12y = 0
x2 – 2x + y2 + 4y = 0
(x – 1)2 – 1 + (y + 2)2 – 4 = 0
(x – 1)2 + (y + 2)2 = 5
Centre (1, –2)
Point (0, – 4)
− 2 − ( − 4)
Gradient of the normal =
1− 0
=2
Equation of the normal at (0, – 4) is
y + 4 = 2x
y – 2x + 4 = 0
Exercise 10 B
1
x = t2 + t + 1
y = t2 – 2
⇒=
t
y+2
[1]
[2]
substitute in [1]: x = (y + 2) + ( y + 2) + 1
⇒ x − y − 3=
y+2
⇒ [x – (y + 3)] ≡ y + 2
x2 + (y + 3)2 – 2x (y + 3) = y + 2
x2 + y2 + 6y + 9 – 2xy – 6x – y – 2 = 0
Cartesian equation: x2 + y2 – 2xy – 6x + 5y + 7 = 0
1
[1]
x=t+
t
1
[2]
y=t−
t
t2 + 1
[1]: x =
t
2
t −1
[2]: y =
t
x+y
[1] + [2] ⇒ x + y = 2t ⇒ t =
2
2
 x + y

 +1
2 
in [1]: x =
 x + y


2 
2
2
 (x + y)2 + 4   2 
x=
 ×  x + y 
4

Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 7 of 20
 (x + y)2 + 4 
x=
 2 (x + y) 
3
4
⇒ 2 (x2 + xy) = x2 + y2 + 2xy + 4
y2 – x2 + 4 = 0
Cartesian equation: x2 = y2 + 4
x+2
x = 4t2 – 2 ⇒ t 2 =
4
y
−
5
y = 3t2 + 5 ⇒ t 2 =
3
[1] = [2]:
x +2 y−5
So
=
4
3
⇒ 3 (x + 2) = 4 (y – 5)
3x + 6 = 4y – 20
Cartesian equation: 4y = 3x + 26
t+2
x=
4
4t 2 − 3
y=
t
[1] ⇒ =
t 4x − 2
[1]
[2]
[1]
[2]
4 (4x − 2) 2 − 3 64x 2 − 32x + 16 − 3
=
4x − 2
4x − 2
2
⇒ 4xy – 2y = 64x – 32x + 13
⇒ 64x2 – 32x + 2y – 4xy + 13 = 0
4x2 – 16x + 16 + 9y2 + 18y + 9 = 36
4x2 + 9y2 – 16x + 18y – 2 = 0
Cartesian Equation
y = 3 cos t
[1]
2
x = 4 sin t
[2]
2
y
[1]2: y2 = 9 cos2 t ⇒ cos 2 t =
9
x
[2] ⇒ sin2 t =
4
y2 x
adding: (sin2 t + cos2 t) =
+ =
1
9 4
⇒ 4y2 + 9x = 36
Cartesian Equation of curve: 4y2 + 9x = 36
x = 3 sin t + 2
[1]
y = 2 cos t – 1
[2]
x−2
[1]:
= sin t
3
y +1
[2]:
= cos t
2
So [1]2 + [2]2:
[1] in [2]: y
=
5
6
 x − 2   y + 1
sin2 t + cos2 t = 
1
 +
 =
 3   2 
(x − 2) 2 (y + 1) 2
⇒
+
=
1
9
4
2
Unit 1 Answers: Chapter 10
2
© Macmillan Publishers Limited 2013
Page 8 of 20
7
8
⇒ 4(x – 2)2 + 9(y + 1)2 = 36 Cartesian Equation
x = 5 sin t
[1]
y = tan t – 1
[2]
1
5
5
[1]:
=⇔ cosec t =
sin t x
x
1
[2]: tan t = y + 1 ⇒ cot t =
y +1
25
[1]2: cosec 2 t = 2
x
1
[2]2: cot 2 t =
(y + 1) 2
Since cot2 t + 1 = cosec2 t ∀ t:
1
25
1+
=
2
(y + 1)
x2
x2 (y + 1)2 + x2 = 25 (y + 1)2
x2 y2 + 2x2 y + x2 + x2 = 25y2 + 50y + 25
So 2x2 – 25y2 – 50y + 2x2 y+ x2 y2 – 25 = 0 Cartesian Equation
x = 4 sec t – 1
[1]
y = 3 tan t + 7
[2]
x +1
[1]: sec t =
4
y−7
[2]: tan t =
3
Since tan2 t + 1 = sec2 t:
2
 y − 7
 x + 1

 + 1 =


3
4 
2
(y − 7)2 9 (x + 1)2
+ =
9
9
16
16 (y – 7)2 + 144 = 9 (x + 1)2
9 (x + 1)2 – 16 (y – 7)2 = 144
9 (x2 + 2x + 1) – 16 (y2 – 14y + 49) = 144
9x2 – 16y2 + 18x + 224 y – 919 = 0
Cartesian Equation
x
6
[1]
x = 6 cosec θ ⇒ cosec θ=
⇒ sin θ=
6
x
y
[2]
y = 2 cos θ ⇒ cos θ =
2
36
so sin 2 θ = 2
[3]
x
y2
[4]
cos2 θ =
4
[3] + [4]: sin2θ + cos2θ = 1 ⇒
36 y 2
+
=
1
x2
4
144 + x2 y2 = 4x2
4x2 – x2 y2 = 144 Cartesian Equation
x−2
[1]
x = 3 cos θ + 2 ⇒ cos θ =
3
⇒
9
10
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 9 of 20
y = 5 sin θ + 2 ⇒ sin θ =
y−2
5
[2]
2
11
12
13
 x −2  y−2
[1]2 + [2]2: sin2θ + cos2θ = 
 +
 =1
 3   5 
(x − 2) 2 (y − 2) 2
⇒
+
=
1
9
25
Equation of an ellipse: centre (2, 2)
Minor & Major axis lengths: 6 units, and 10 units
x2 y2
+
=
1
9 25
(a) x = 0, y2 = 25, y = ± 5
y = 0, x2 = 9, x = ± 3
Intercepts (0, 5) (0, – 5) (3, 0) (– 3, 0)
(b) c2 = a2 – b2
= 25 – 9 = 16
c=±4
Coordinates of the foci: (0, 4), (0, – 4)
(c) Length of the major axis = 10 units
Length of the minor axis = 6 units
(d)
2
x2 + 4x + 4y2 – 8y + 4 = 0
(x + 2)2 – 4 + 4 (y2 – 2y) + 4 = 0
(x + 2)2 + 4 (y – 1)2 – 4 = 0
(x + 2)2 + 4 (y – 1)2 = 4
(x + 2) 2
+ (y − 1) 2 =
1
4
Centre (– 2, 1)
(b) 9x2 – 18x + 4y2 + 16y – 11 = 0
9 (x2 – 2x) + 4 (y2 + 4y) – 11 = 0
9 (x – 1)2 – 9 + 4 (y + 2)2 – 16 – 11 = 0
9 (x – 1)2 + 4 (y + 2)2 = 36
9 (x − 1)2 4 (y + 2)2
+
=
1
36
36
(x − 1)2 (y + 2)2
+
=
1
4
9
Centre (1, – 2)
(x − 1)2 (y + 3)2
+
=
1
4
9
(a)
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 10 of 20
(a)
(b)
(c)
Centre (1, – 3)
c2 = 32 – 22 = 5
major axis parallel to the y – axis
c=± 5
Foci (1, –3 + 5 ), (1, – 3 – 5 )
Parametric equations are
x = 1 + 2 cos t, y = – 3 + 3 sin t
(d)
(a)
(y + 3) 2 3
=
9
4
27
(y + 3) 2 =
4
3 3
y+3=
2
3 3
y =− 3 ±
2
2
x=t ⇒t= x
(b)
y = 4t ⇒ y = 4 x
Cartesian equation
y2 = 16x
x = 6t2, y = 12t
x = 0,
14
t=
x
6
x
6
144
y2 =
x
6
y2 = 24x
Cartesian equation
(c) x = t – 1 ⇒ t = x + 1
y = t2 + 1
⇒ y = (x + 1)2 + 1 Cartesian equation
y2 = 16x ⇒ y2 = 4ax, a = 4
(1, 4)
at2 = 1
4t2 = 1
1
t=
2
y = 12
15
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 11 of 20
1 1
=2
=
t 1
2
1
Gradient of the normal = −
2
Equation of the tangent at (1, 4) is
y – 4 = 2 (x – 1)
y = 2x + 2
Equation of the normal at (1, 4) is
1
y − 4 = − (x − 1)
2
2y – 8 = – x + 1
2y + x = 9
Gradient of the tangent=
Review Exercise 10
1
3x + 4y = 25
x2 + y2 = 25
25 − 4y
[1]: x =
3
[1]
[2]
 25 − 4y 
2
Substituting [1] in [2]: 
25
 +y =
3


⇒ (25)2 + 16y2 – 200y + 9y2 = 225
25y2 – 200y + 400 = 0
y2 – 8y + 16 = 0
Discriminant: (– 8)2 – 4 [1] (16) = 0, so only one root
(y – 4)2 = 0 ⇒ y = 4
25 − 16 9
in [1]: x=
= = 3
3
3
So point of intersection of line & circle is (3, 4)
As the line touches the circle at only one point, it must be a tangent to the circle with (3, 4) the
point of tangency.
x2 + y2 – 2x + 4y – 1 = 0
x2 – 2x + y2 + 4y – 1 = 0
(x – 1)2 – 1 + (y + 2)2 – 4 – 1 = 0
(x – 1)2 + (y + 2)2 = 6
Centre of circle is (1, – 2)
y − y1
x − x1
In
=
y 2 − y1 x 2 − x1
y − (− 2) x − 1
⇒
=
1 − (− 2) 3 − 1
y + 2 x −1
⇒
=
⇒ 2y + 4 = 3x − 3
3
2
So 3x – 2y = 7
Equation of diameter
The equation of the circle is:
(x – 4)2 + (y + 1)2 = 2
[1]
x+y=1
[2]
[2]: y = 1 – x
Substituting in [1]: (x – 4)2 + (2 – x)2 = 2
2
2
3
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 12 of 20
⇒ 2x2 – 12x + 18 = 0
x2 – 6x + 9 = 0, and discriminant = 0, so only one root
(x – 3)2 = 0 ⇒ x = 3, y = – 2
So the line [2] touches [1] at exactly one point, (3, – 2) the point of tangency
4
(5, 1) (4, 6) (2, – 2)
Let the circle have equation:
(x – a)2 + (y – b)2 = r2
⇒ (5 – a)2 + (1 – b)2 = r2
[1]
(4 – a)2 + (6 – b)2 = r2
[2]
(2 – a)2 + (2 + b)2 = r2
[3]
[1] – [2]: (5 – a)2 – (4 – a)2 + (1 – b)2 − (6 − b)2 = 0
⇒ – 2a + 10b – 26 = 0
– a + 5b – 13 = 0
⇒ a = 5b – 13
[4]
2
2
2
2
[2] – [3] ⇒ (4 – a) – (2 – a) + (6 – b) – (2 + b) = 0
12 – 8a + 4a + 32 – 12b – 4b = 0
– 4a – 16b + 44 = 0
– a – 4b + 11 = 0
⇒ a = 11 – 4b
[5]
So [4] = [5] ⇒ 11 – 4b = 5b – 13
⇒ 24 = 9b
8
⇒b=
3
 8  33 32 1
in [5]: a =11 − 4   = −
=
3 3
3 3
1
8
so=
a =
,b
in [3]:
3
3
2
2
2
2
1 
8   5   14 

2
r =2 −  + 2 +  =  +  
3 
3  3  3 

25 + 196 221
= =
9
9
∴ Equation of circle is:
2
2
1 
8
221

x −  + y −  =
3 
3
9

OR:
2
2
1
8


9x −  + 9y−  =
221
3
3


5
221
 1 8
Centre  ,  radius
units
 3 3
3
Let the equation of the circle be:
(x – a)2 + (y – b)2 = 13
so (1, 1) in [1]:
(1 – a)2 + (1 − b)2 = 13
2
1
2
Gradient of tangent :=
y
x+
is
3
3
3
−3
so gradient of radius is
2
Unit 1 Answers: Chapter 10
[1]
[2]
© Macmillan Publishers Limited 2013
Page 13 of 20
−3 1− b
=
2
1−a
⇒ – 3 (1 – a) = 2 (1 – b)
– 3 + 3a = 2 – 2b
∴ 3a = 5 – 2b
5 − 2b
a=
3
∴
[3]
 3 − 5 + 2b 
2
[3] in [2]: 
13
 + (1 + b − 2b) =
3


(2b – 2)2 + 9 + 9b2 – 18b = 117
13b2 – 26b – 104 = 0
b2 – 2b – 8 = 0
(b – 4) (b + 2) = 0
∴ b = – 2, b = 4
b=–2⇒a=3
b=4⇒a=–1
∴ possible Circles are:
(x – 3)2 + (y + 2)2 = 13 and
(x + 1)2 + (y – 4)2 = 13
x = 4 cos t – 3
[1]
y = 4 sin t + 4
[2]
x+3
[1]: cos t =
4
y−4
[2] sin t =
4
(y − 4) 2 (x + 3) 2
sin 2 t + cos 2 t=
+
= 1
16
16
(x + 3)2 + (y – 4)2 = 16
Equation of a circle, centre (– 3, 4) radius = 4 units
C1: x2 + y2 – 6x + 8y = 5
C1: x2 + y2 – 2x (3) – 2y (– 4) = 5
∴ C1 has centre (3, – 4) and
Radius = (9 + 16 + 5)1/2 = 30
C2: 4x2 + 4y2 – 12x + 16y – 12 = 0
⇒ x2 + y2 – 3x + 4y – 3 = 0
3
⇒ x 2 + y 2 − 2x   − 2y ( − 2) − 3 =
0
2
3

∴ centre of C2 is  , − 2 and
2

2
6
7
1/ 2
1/ 2
37
9

 37 
radius =  + 4 + 3  =   =
2
4

 4 
So length of line joining centres of C1 to C2
3

(− 4 − (− 2)) 2 +  3 − 
2

=
=
4+
9
=
4
2
25
4
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 14 of 20
5
2
C1 has centre (3, – 4) and radius 30 units , distance between centres is 2.5 units
Let the circle be represented by C1 & C2:
C1: x2 + y2 + 6x – 2y – 54 = 0
[1]
C2: x2 + y2 – 22x – 8y – 12 = 0
[2]
C1: x2 + y2 – 2x (– 3) – 2y (1) – 54 = 0
C1 has centre (– 3, 1) & radius r1 where r12 = 9 + 1 + 54 = 64 ⇒ r1 = 8 units
C2: x2 + y2 – 2x (11) – 2y (4) – 12 = 0
C2 has centre (11, 4) & radius r2 where r22 = 121 + 16 + 12 = 149 ⇒ r2 = 149 units
Distance between centres of C1 & C2, d:
d = [(4 – 1)2 + (11 – (– 3))2]1/2
=
(9 + 196)1/ 2 =
205 units
[1] – [2]: 28x + 6y – 42 = 0
14x + 3y – 21 = 0
∴ 3y = 21 – 14x
14
y= 7 − x
3
2
14 

 14 
Substituting in [2]: x 2 +  7 − x  − 22x − 8 7 − x  – 12 =
0
3 
3 


196 2
196
112
x2 +
x + 49 −
x − 22x − 56 +
x − 12 =
0
9
3
3
9x2 + 196 x2 + 441 − 588 x – 198 x – 504 + 336 x – 108 = 0
∴ 205x2 – 450 x – 171 = 0
(Discriminant > 0 ⇒ roots are real & distinct)
=
8
x=
450 ± (450) 2 − 4 (205) (− 171)
2 (205)
450 ± (342720)1/ 2
410
x 2.525, − 0.3303
=
So y = − 4.783,
8.5414
So points of intersection are:
(2.53, − 4.78), ( − 0.33, 8.54)
Therefore there exist two point of intersection so C1 & C2 intersect.
OR:
Since: d < r1 + r2
the circles C1 & C2 intersect at two points.
9
x = cos t – 1
[1]
y = cos 2t
[2]
(a) When t = 0:
x = 0,
y=1
π
−1
−1
& When
=
t =
:x
=
,y
3
2
2
 − 1 − 1
Distance between (0, 1) and  ,  : d
 2 2 
=
   − 1  2  1  2 
d =−
 1    +   
 2 
  2  

Unit 1 Answers: Chapter 10
1/ 2
© Macmillan Publishers Limited 2013
Page 15 of 20
1/ 2
10
9 1
=
 +  = units
2
4 4
= 1.58 units
(b) [1]: cos t = x + 1
[2]: cos 2t = 2 cos2t – 1 = y
[1] in [2] ⇒ 2 (x + 1)2 – 1 = y
∴ y = 2x2 + 4x + 2 – 1
so y = 2x2 + 4x + 1
Cartesian equation
10
(0, 6) (8, – 8)
Centre = midpoint of (0, 6) & (8, – 8)
= (4, – 1)
1
Radius = [Diameter Length]
2
1
=
[(6 − ( − 8))2 + (0 − 8)2 ]1/ 2
2
1
=
(196 + 64)1/ 2
2
260
= = 65 units
2
∴ Equation of circle:
(x – 4)2 + (y + 1)2 = 65
11
(– 3, 2) (5, – 6)
Centre of circle: (1, –2)
1
Radius
=
[(2 − ( − 6))2 + ( − 3 − 5)2 ]1/ 2
2
1
128 4 8
= [64 + 64]1/ 2 =
=
2
2
2
= 2 8 units
Equation of circle:
(x – 1)2 + (y + 2)2 = 32
12
x2 + y2 – 2x – 4y + 3 = 0
[1]
2x2 + 2y2 + 4x + 6y + 9 = 0
[2]
[1]: x2 + y2 – 2x (1) – 2y (2) + 3 = 0
∴ Centre: (1, 2)
9
=0
[2]: x2 + y2 + 2x + 3y +
2
 − 3 9
x 2 + y 2 − 2x ( − 1) − 2y 
0
+ =
 2  2
− 3

∴ Centre:  − 1,

2 

y − y1
x − x1
In
=
y 2 − y1 x 2 − x l
y−2
x −1
y − 2 x −1
=
⇔
=
−3
−7 −2
1
1
−
−

−2
 2 
2


y−
=
2
−7
 − 1
(x − 1)  
 2 
2
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 16 of 20
7
(x − 1)
4
7
7
7
1
⇒ y=
x − + 2=
x+
4
4
4
4
4y = 7x + 1
Equation of the line containing circles’ centres
y – 7x = 2
[1]
x2 + y2 + 8x + 2y – 8 = 0
[2]
[1]: y = 2 + 7x
Substitute in [2]:
x2 + (7x + 2)2 + 8x + 2 (2 + 7x) – 8 = 0
x2 + 49x2 + 4 + 28x + 8x + 4 + 14x – 8 = 0
50x2 + 50x = 0
x (x + 1) = 0 Discriminant > 0 ⇒ intersect at two points
⇒ x = 0, – 1
y = 2, – 5
So the line intersects the circle at two points (0, 2), (– 1, – 5)
x2 + y2 = 4
[1]
x2 + y2 – 4x – 4y + 4 = 0
[2]
[1] – [2]: 4x + 4y – 4 = 4
⇒x+y=2
⇒y=2–x
Substituting in [1]:
x2 + (2 – x)2 = 4
2x2 – 4x = 0 ⇒ x2 – 2x = 0
∴ x (x – 2) = 0 so x = 0, 2 and y = 2, 0
points of intersection are:
(0, 2) & (2, 0)
x2 + y2 – 2x – 4y = 0
[1]
x2 + y2 – 2y – 2 = 0
[2]
[1] – [2]: – 2x – 2y + 2 = 0
x+y=1
⇒y=1–x
substitute into [2]:
x2 + (1 – x)2 – 2 (1 – x) – 2 = 0
x2 + 1 – 2x + x2 – 2 + 2x – 2 = 0
2x2 – 3 = 0
3
6
x=
±
=
±
2
2
6
6
y=
1−
,1 +
2
2
Points of intersection are:
 6
− 6
6
6
,1 −
,1 +

 and 

2 
2 
 2
 2
The two circles are C1, C2:
C1: x2 + y2 = 1
[1]
2
2
C2: x + (y – 3) = 4
[2]
[2]: x2 + y2 – 6y + 9 = 4
Substituting [1] in [2]: 1 – 6y + 9 = 4
⇒ 6y = 6 ⇒ y = 1
into [1]: x2 + (1)2 = 1 ⇒ x2 = 0
⇒x=0
Point of intersection is (0, 1)
C1: x2 + y2 = 1
y−=
2
13
14
15
16
17
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 17 of 20
C2: (x – 3)2 + (y – 1)2 = 1
Let the equations C1, C2 represent the two circles with radii r1 & r2 respectively
Distance between centres: d
d = [(3 – 0)2 + (1 – 0)2]1/2 = 10
18
Since d > r1 + r2 i.e. 10 > (1) + (1)
C1 and C2 cannot intersect
OR:
C2: x2 + y2 – 6x – 2y + 9 + 1 = 1
C1 in C2: 1 – 6x – 2y + 10 = 1
6x + 2y = 10 ⇔ 3x + y = 5
So y = 5 – 3x, into C1:
x2 + 25 + 9x2 – 30x = 1
⇒ 10x2 – 30x + 24 = 0
Discriminant = (– 30)2 – 4 (10) (24)
= – 60 < 0
⇒ No roots so C1 and C2 do not intersect
y − y1
x − x1
(a) In
=
y 2 − y1 x 2 − x1
y − ( − 1)
x − ( − 5)
=
3 − ( − 1) − 2 − ( − 5)
y +1 x + 5
⇒
=
4
3
3y + 3 = 4x + 20
3y = 4x + 17
Equation of the line joining the two points
(b) Radius = [(3 – (– 1))2 + (– 2 – (– 5))2]1/2
= (16 + 9)1/2
= 5 units
(c) Equation of circle: (x + 2)2 + (y – 3)2 = 25
4
(d) Gradient of radius =
3
−3
Gradient of tangent =
4
In y – y1 = m (x – x1)
−3
y – (–=
1)
(x − ( − 5))
4
−3
y+1 =
(x + 5)
4
4y + 4 = – 3x + (– 15)
4y + 3x = – 19
Equation of tangent.
(e) (x + 2)2 + (y – 3)2 = 25
[1]
x2 + y2 + 6x – 7y – 10 = 0
[2]
2
2
[1] x + y + 4x – 6y + 4 + 9 = 25
x2 + y2 + 4x – 6y – 12 = 0
[3]
[2] – [3]: 2x – y + 2 = 0
⇒ y = 2x + 2
[4]
[4] in [2]: x2 + 4x2 + 4 + 8x + 6x – 14x – 14 – 10 = 0
5x2 – 20 = 0 ⇒ 5 (x2 – 4) = 0
⇒ x2 – 4 = 0
x = – 2, 2
y = – 2, 6
So points of intersection are:
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 18 of 20
19
(–2, – 2) and (2, 6)
3
y=5+ x
4
2
y = 15x
20
3 

⇒ 5 + x  =
15x
4 

15
9 2
⇒ 25 + x +
x =
15x
2
16
15
9 2
⇒ 25 − x +
x =
0
2
16
9x2 – 120 x + 400 = 0
b2 – 4ac = (– 120)2 – 4 (9) (400)
=0
Since b2 – 4ac = 0 ⇒ the line is a tangent to the curve
4x2 + 9y2 = 36
(a) (i)
x = 0, y2 = 4, y = ± 2
y = 0, x2 = 9, x = ± 3
Intercepts (0, 2), (0, – 2), (3, 0), (– 3, 0)
(ii) 4x2 + 9y2 = 36
4x 2 9y 2
⇒
+
=
1
36
36
x 2 y2
⇒
+
=
1
9
4
c2 = a2 – b2
c2 = 9 – 4 = 5 ⇒ c = ± 5
2
(iii)
21
Coordinate of foci: ( 5, 0) ( − 5, 0)
Length of the major axis = 6 units
Length of the minor axis = 4 units
(b) x = 3 cos θ , y = 2 sin θ
x+y=5⇒y=5–x
x2 y2
+
=
1
16 9
x 2 (5 − x)2
⇒
+
=
1
16
9
x 2 25 − 10x + x 2
⇒
+
=
1
16
9
⇒ 9x2 + 400 – 160x + 16x2 = 144
∴ 25x2 – 160x + 256 = 0
x=
160 ± ( − 160) 2 − 4(25) (256)
50
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 19 of 20
160 16
=
50
5
16
16 9
When x = , y =5 − =
5
5 5
=
22
23
 16 9 
Coordinate of the point of intersection is  , 
 5 5
(ii) x = 4 cos θ , y = 3 sin θ
y=x+1
x2 y2
−
=
1
3
2
x 2 (x + 1)2
Substituting
−
=
1
3
2
2x2 – 3x2 – 6x – 3 = 6
x2 + 6x + 9 = 0
b2 – 4ac = 62 – 4 (9) (1)
= 36 – 36 = 0
Since b2 − 4ac = 0 ⇒ the line is a tangent to the curve
4x2 + 32x + 9y2 + 36y + 64 = 0
4 (x2 + 8x) + 9 (y2 + 4y) + 64 = 0
4 (x + 4)2 – 4 (16) + 9 (y + 2)2 – 9 (4) + 64 = 0
4 (x + 4)2 + 9 (y + 2)2 = 36
(x + 4)2 (y + 2)2
+
=
1
9
4
(x + 4)2 (y + 2)2
1
+
=
32
22
centre (– 4, – 2)
y=x+2
(x + 4)2 (y + 2)2
(x + 4)2 (x + 4)2
1
1
+
=
⇒
+
=
4
9
4
32
4 (x + 4)2 + 9 (x + 4)2 = 36
13 (x + 4)2 = 36
36
x+4=±
13
36
6 13
x =− 4 ±
=− 4 ±
13
13
6 13
6 13
6 13
x =− 4 +
, y =− 4 +
+ 2 =− 2 +
13
13
13
6 13
6 13
6 13
x =− 4 −
, y =− 4 −
+ 2 =− 2 −
13
13
13
points of intersection are

6 13
6 13  
6 13
6 13 
, −2 +
, −2 −
 −4 +
 ,  −4 −

13
13  
13
13 

24
x = – 4 + 3 cos θ , y = – 2 + 2 sin θ
y2 = 4x, (t2, 2t)
1
Gradient of the tangent =
t
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 20 of 20
Gradient of the normal = – t
Equation of the normal at (t2, 2t) is
y – 2t = – t (x – t2)
y = – tx + t3 + 2t
At P, y = – 2x + 8 + 4
y = – 2x + 12
1
1
At Q, y =
− x + +1
2
8
1
9
y=
− x+
2
8
1
9
– 2x + 12 = − x +
2
8
−3
9
x=
− 12
2
8
−3
87
x= −
2
8
29
 29 
x=
,y=
− 2   + 12
 4
4
− 29 24 − 5
=
+ =
2
2
2
 29 − 5 
C ,

 4 2 
Unit 1 Answers: Chapter 10
© Macmillan Publishers Limited 2013
Page 1 of 25
Chapter 11 Vectors in Three Dimensions (ℝ3)
Exercise 11A
1
(a)
 1  −1
 2 .  2  =− 1 + 4 + 12 =15
   
 3  4 
(b)
 0  2
 1 .  1 = 0 + 1 + 20 = 21
   
 4  5
(c)
 4   −1
 2  .  2  =− 4 + 4 − 3 =− 3
   
 −3  1 
(a)
 4  3 
   
 2  .  4  = 12 + 8 − 8 = 12
 2   −4 
   
→
OA = 42 + 22 + 12 = 21
(b)
→
OB=
2
32 + 22 + (−2)=
(c)
→
OC=
2
42 + (−1) 2 + (−2)=
(d)
2
3
(a)
(b)
(c)
(d)
4
(a)
17
21
 1   3
   
 2 .  2  = 3 + 4 − 2 = 5
 −1  2 
   
Not Perpendicular
 1  0 
 4 .  1  = 0 + 4 − 4 = 0
   
 2  −2
Perpendicular
 2   −2
 3  .  2  =− 4 + 6 − 2 =0
   
 −2  1 
Perpendicular
 4  0 
 2 .  1  =0 + 2 − 5 =− 3
   
 5  −1
Not perpendicular
→ → →
AB
= OB − OA
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 2 of 25
 −2   2   −4 
     
=
 1  −  4 =
 −3 
 3  1  2 
     
→
AB =
(−4) 2 + (−3) 2 + 22 =
29
→
(b)
(c)
A unit vector in the direction of AB is
 −4 29 


 −3 29 


 2 29 
→
→ →
AC
= OC − OA
 −1  2   −3 
     
= −1 −  4  = −5 
 1  1  0 
     
→
AC = (−3) 2 + (−5) 2 = 34
→
A unit vector in the direction of AB is
 −3 34 


 −5 34 


 0 


→ → →
BC
= OC − OB
 −1  −2
= −1 −  1 
   
 1  3
1
 
=  −2 
 −2 
 
→
BC=
=
(1) 2 + (−2) 2 + (−2) 2
=
9 3
→
A unit vector in the direction of AB is
 1/ 3 


 −2 / 3 
 −2 / 3 


5
(a)
Let θ be the angle between a and b
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 3 of 25
 2   3
 4  .  2
   
 −1  2
cos θ =
 2   3
 4   2
   
 −1  2
6+4−2
21 17
8
=
21 17
⇒ θ = 65°
a.c = 0
 1  2

  
⇒  p .  4  =
0
 2p − 1  −1

  
=
(b)
⇒ 2 + 4p − 2p + 1 = 0
2p = −3
−3
p=
2
6
 2   3   −1
→    
=
BA  1  −  2  =  −1
 −4   1   −5
   
 1   3   −2 
→      
=
BC  1  −  2  =  −1 
 −1  1   −2 
     
→ →
BA . BC
cos θ = → →
BA BC
 −1  −2
 −1 .  −1
   
 −5  −2
=
 −1  −2
 −1  −1
   
 −5  −2
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 4 of 25
2 + 1 + 10
27 9
13
=
9 3
⇒ θ = 33.5°
6
→   →
7 =
OA =
2  , OB
1
 
=
→
→
 3
 
1
 2
 
AC = 2 AB
→ →
→ → 
⇒ OC − OA= 2 OB − OA 


→
→
→ →
OC = 2 OB − 2 OA + OA
→ →
= 2 OB − OA
 3  6
= 2  1 −  2
   
 2  1
 6  6
=  2 −  2
   
 4  1
 0
=  0
 
 3
→
OC
= =
9 3
8
 0 / 3  0 
→ 
  
∴ A unit vector in the direction of OC is  0 / 3  =
0
 3 / 3 1

  
→ → →
(a)
PQ
= OQ − OP
 3  2
=  2 −  3
   
 4  2
 1
=  −1
 
 2
→
PQ
= (1) 2 + (−1) 2 + =
22
6
1 6 


A unit vector in the direction of PQ is  − 1 6 


2 6
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 5 of 25
(b)
→ →
OP . OR = 0
 3  −2
⇒  2 .  1  =
0
   
 4  a 
⇒ −6 + 2 + 4a = 0
4a = 4
a=1
→ → →
(c)
PS
= OS − OP
 −1  2
=  3  −  3
   
 b   2
 −3 
= 0 


 b − 2
→
PS = 5
⇒ (−3) 2 + (0) 2 + (b − 2) 2 =
5
⇒ 9 + 0 + (b − 2) 2 =
25
9
(a)
(b − 2) 2 =
16
b − 2 =± 4
b = 6, −2
 −2 
→  
OA =  4 
 2
 
 3
→  
OB =  1 
1
 
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 6 of 25
 −2  3
 4  .  1
   
 2   1
cos θ =
 −2  3
 4   1
   
 2   1
−6 + 4 + 2
24 11
=0
⇒ θ = 90°
→ → →
AB
= OB − OA
=
(b)
 3   −2 
   
= 1 −  4 
p  2 
   
 5 


=  −3 
 p − 2


→
AB = 8
⇒ (5) 2 + (−3) 2 + (p − 2) 2 =8
⇒ 34 + (p − 2) 2 =
64
(p − 2) 2 =
30
p − 2 =± 30
p= 2 ± 30
10
(a)
 3
→   →
OP =  2  , OQ =
5
 
 5
 
 −4 
 5
 
 3  5 
 2 .  −4
   
 5  5 
cos θ =
 3  5 
 2  −4
   
 5  5 
=
15 − 8 + 25
38 66
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 7 of 25
32
38 66
θ = 50.3°
=
(b)
 5
→
 
Since A is on OQ ⇒ OA =λ  −4 
 5
 
 5λ 


=  −4λ 
 5λ 


→ → →
PA
= OA − OP
 5λ   3  5λ − 3 
=  −4λ  −  2 =  −4λ − 2

   

 5λ   5  5λ − 5 
→ →
PA ⋅ OQ =
0
 5λ − 3   5 

  
⇒  −4λ − 2  .  −4  =0
 5λ − 5   5 

  
⇒ 25λ + 16λ + 25λ − 15 + 8 − 25 =0
66λ =32
16
λ=
33
Exercise 11B
1
1
 2
 
 
(a) =
r  2 + λ  1 , λ ∈ 
 4
 3
 
 
1
 3
 
 
(b)
r =  −1 + λ  2  , λ ∈ 
4
 −4 
 
 
(c)
2
(a)
 4
 −3 
 
 
r =+
 1  t  0 , t ∈ 
 4
2
 
 
Using r = a + λ (b − a), λ ∈ 
 4   2  
 2
    
 
=
r  1  + λ  1  −  1   , λ ∈ 
1
 3   1  
 
 2
 2
=
r  1 + λ  0
 
 
 1
 2
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 8 of 25
(b)
 4  −1 
 −1




r = 0 + t  2 −  0  
 
   
 4  2  
 2
 −1  5 
   
r=
 0  + t  2 ,t ∈
 2   2
   
(c)
3
 4
 2
 
 
r =  4  + s  −2  , s ∈ 
5
 2
 
 
Using r. n = a. n.
 1  2   1
(a)
r.  4 =  4  .  4
     
 1  −2  1
(b)
(c)
4
  6  4 
 4


r =4 + s   2 −  4 
   
 
  7  5 
 5
1
 
r.  4  = 2 + 16 − 2 = 16
1
 
 1
r.  4 = 16
 
 1
⇒ x+ 4y + z = 16
 4  2  4
r.  2 =  2 .  2
     
 2  5  2
 4
 2


⇒ r. 2 =
8 + 4 + 10 =
22 ⇒ r.  1 =
11
 
 
 2
 1
4x + 2y + 2z = 22
2x + y + z = 11
 −1  6  −1
r.  0  =  7 .  0 
     
 −4  3  −4
 −1
 1


r. 0 =− 6 − 12 =− 18 ⇒ r.  0 =
18
 
 
 −4
 4
−x − 4z = −18
x + 4z = 18
vector equation of l :
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 9 of 25
 2
 −1
 
 
=
r  1  + λ  2 , λ ∈ 
 3
1
 
 
 x   2   −λ 
     
 y  =  1  +  2λ 
 z   3  λ 
     
 x
 y =
 
 z
 2−λ
 1 + 2λ 


 3+ λ
⇒ x= 2 − λ 

y = 1 + 2λ  λ ∈  parametric equations
z = 3 + λ 
λ=2−x
y −1
λ=
2
λ= z − 3
∴2 − x =
5
6
y −1
= z − 3 → cartesian equation
2
x−2
⇒ 4λ= x − 2
4
x = 2 + 4λ
y−3
λ=
⇒ 5λ = y − 3
5
y = 5λ + 3
2−z
λ=
⇒ 3λ = 2 − z
3
z = 2 − 3λ
∴ x = 2 + 4λ
y = 3 + 5λ
z = 2 − 3λ
 x   2
 4
 y=
  3 + λ  5 
   
 
 z   2
 −3
λ=
 2
4
 
 
∴
=
r  3 + λ  5 , λ ∈ 
 2
 −3 
 
 
x=2+λ
y = 3 + 4λ
z = 2 + 2λ
 x   2
1
   
 
⇒  y=
  3 + λ  4
 z   2
 2
   
 
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 10 of 25
7
 2
1
 
 
⇒ r =  3λ +4 , λ  ∈ 
 2
 2
 
 
2
 4
 
 
(a)
r = −3  + t  5  , t ∈ 
4
 3
 
 
(b)
 x   2 + 4t 
 y =  −3 + 5t 
  

 z   4 + 3t 
⇒ x = 2 + 4t 

y = − 3 + 5t  t ∈  , parametric equations
z = 4 + 3t 
x−2
4
y+3
y =− 3 + 5t ⇒ t =
5
z−4
z = 4 + 3t ⇒ t =
3
x−2 y+3 z−4
∴
=
=
is the cartesian equation
4
5
3
A vector parallel to li is: −i −2j −k
A point on l1 is : 2i – 5j – k
A vector parallel to l2 is − 3i + 5j − k
A point on l2 is 4i + j + 2k
x + 2 y −1 z + 2
= =
3
4
−1
x − (−2) y − 1 z − (−2)
⇒
= =
3
4
−1
 −2 
3
 
 
A point on l3 is  1  and a vector parallel to l3 is  4 
 −2 
 −1
 
 
(c)
8
x = 2 + 4t ⇒ t =
6
12 
→   →  
9=
OC =
4  , OD  p 
5
q
 
 
6
 
 1
(a) =
r  4 + λ  −1
 
 
 5
 0
 12  6 
 6


Equation of CD =
: r  4 + λ   p  −  4 
 
   
 q   5 
 5
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 11 of 25
 6
 6 


=
r
4 + λ  p − 4
 


 5
 q − 5


1


 6
 p − 4

∴=
r  4 + 6λ 
 
 6 
 5
 q − 5


6 
Comparing the direction vectors:
p−4
=− 1 ⇒ p =− 6 + 4 =− 2
6
q−5
=0⇒q =5
6
 6
 1


(b) =
r
4 + λ  −1
 
 
 5
 0
 6 + λ
=  4 − λ


 5 
Since A is on the line
6 + λ
→ 

OA=  4 − λ 
 5 


→
→
Since OA is perpendicular to CD
1
→  
OA .  −1 =
0
0
 
 6 + λ  1 
⇒  4 − λ  .  −1 =0

  
 5   0
⇒6+λ−4+λ=0
2λ + 2 = 0
λ = −1
 6 − 1   5
→ 
  
∴ OA=  4 − (−1) =  5 
 5   5

  
10
1
 2
 
 
l: r =  −1 + λ  1  , λ ∈ 
2
 3
 
 
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 12 of 25
 1 + 2λ 
→ 

OP =  −1 + λ 
 2 + 3λ 


→
Since OP is perpendicular to l
 2
→  
0
⇒ OP .  1  =
 3
 
 1 + 2 λ   2
⇒  −1 + λ  .  1 =0

  
 2 + 3λ   3
⇒ 2 − 1 + 6 + 4λ + λ + 9λ =0
14λ = − 7
−1
λ=
2

 1
1 + 2 −  
 2

 0 
→ 
1  

∴ OP =  −1 −
 =  −3 / 2 
2

  1/ 2 


 1  
+
−
2
3




 2 

Since Q is on l:
 1 + 2λ 
→ 

OQ =  −1 + λ 
 2 + 3λ 


→
OQ = 4
⇒ (1 + 2λ) 2 + ( −1 + λ) 2 + (2 + 3λ) 2 =
4
⇒ (1 + 2λ ) 2 + ( −1 + λ ) 2 + (2 + 3λ ) 2 =
42
⇒ 1 + 4λ + 4λ 2 + 1 − 2λ + λ 2 + 4 + 12λ + 9λ 2 =16
⇒ 14λ 2 + 14λ − 10 =0
7 λ 2 + 7 λ − 5 =0
−7 ± 49 − (7) ( −5)4
14
−7 ± 189
=
14
−7 ± 3 21
=
14
0
 −2   −2s 
 
  

r =+
1 s  2  =
1 + 2s 
0
 3   3s 
 
  

λ=
11
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 13 of 25
 -9 
 5   -9 + 5t 
 
  

r =  36  + t  2  =  36 + 2t 
1
 5   1 + 5t 
 
  

 -2s   -9 + 5t 

 

∴ 1 + 2s  =  36 + 2t 
 3s   1 + 5t 

 

⇒ −2s = −9 + 5t
[1]
1 + 2s = 36 + 2t
[2]
3s = 1 + 5t
[3]
[3] − [1] ⇒ 5s = 10
s=2
Subst. into [1] ⇒ −4 = −9 + 5t
t=1
Subst. s = 2, t = 1 into [3]⇒ 3(2) = 1 + 5(1)
6=6
Since all three equations are satisfied by s = 2 and t = 1 ⇒ the lines intersect.
 -2s 


Point of intersection: Substituting s = 2 into 1 + 2s 
 3s 


12
 −2(2)   −4 

  
We get 1 + 2(2)  =
 5
 3(2)   6 

  
Using r . n = a. n, we have
 0  1   0
r .  2 =  2  .  2
     
 3  −1  3
 0
⇒ r.  2 =
4−3
 
 3
0
 
r.  2  = 1
 3
 
⇒ 2y + 3z = 1
 2
 4
→   →  
13=
OA =
3  , OB  2 
 −2 
 2
 
 
Equation of the plane:
 2   4  2 
     
r .  3  =  2 .  3 
 −2   2   −2 
     
 2
 
⇒ r . 3  =8 + 6 − 4
 −2 
 
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 14 of 25
 2
 
r .  3  = 10
 −2 
 
 2
 
Now  3 =
 −2 
 
14
2
22 + 32 + (−2)=
17
 2 17 

 10
∴ r .  3 17  =
17


 −2 17 


Which is of the form r.nˆ = d
∴ The distance from the origin to the plane:
10 10 17
=
17
17
→
→
→
OP =
2i + j + k, OQ =
− 32i + 4j + 2k, OR =
2i + j + 4k
(a) Equation of PQ:
→
 → →
r=
OP + t OQ− OP  , t ∈ 


 −32  2 
 2


∴=
r  1 + t   4  −  1 
 

  
 2   1 
 1
 2
 −34 
 


r =  1  + t  3 , t ∈ 
1
 1 
 


(b) Since PQ is perpendicular to the plane,
 −34 


 3  is a vector perpendicular to the plane
 1 


Using r . n = a . n
 −34  2  −34
 1 .  3 
⇒ r. 3  =

   

 1   4  1 
 −34 


⇒ r .  3  =− 68 + 3 + 4
 1 


 −34
r .  3  = − 61


 1 
 −34 


The equation of the plane is r .  3  = − 61
 1 


Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 15 of 25
15
 2
 
A vector parallel to l1 is  3 
 4
 
3
 
A vector parallel to l2 is  −1
4
 
Let θ be the angle between l1 and l2:
 2  3 
   
 3  .  −1
 4  4 
cos θ =    
 2  3 
   
 3   −1
 4  4 
   
6 − 3 + 16
19
=
29 26
29 26
θ = 46.2°, No
 -1
1
 
 
r =  1  + t  2
0
 2
 
 
=
16
(a)
 p   -1 + t 
  

 q  =  1+2t  ⇒ 3 = 2t
 3   2t 
  

t=
3
2
3 1
=
2 2
3
q = 1 + 2  = 4
2
∴ p = −1 +
(b)
(c)
 −1 + 2   1 
→ 
  
t=
2, OB =
1 + 2(2)  =
5
 2(2)   4 

  
 -1 + t 
→ 

OC = 1 + 2t 
 2t 


→
Since OC is perpendicular to l
1
→  
⇒ OC .  2  =
0
 2
 
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 16 of 25
 -1 + t   1 

  
⇒ 1 + 2t  .  2  =
0
 2t   2 

  
⇒ −1 + t + 2 + 4t + 4t = 0
9t + 1 = 0
1
t= −
9

1 
 −1 −

9  −10 / 9



→ 
 1  

∴ OC = 1 + 2  −   =  7 / 9 
9

 

−2 / 9 
  1  
 2 −  
  9 
Review Exercise 11
1
 −2   2   −4 
→ → →      
BA =OB − OA = −1  −  4  = −5 
 −1   5   −6 
     
1  −2   3 
→ → →      
BC= OC − OB= 1 −  −1 =  2 
1  −1   2 
     
Let θ be angle ABC:
 −4  3
 −5 .  2
   
 −6  2
−12 − 10 − 12
=
= =
cos θ
77 17
 −4  3
 −5  2
   
 −6  2
2
−34
77 17
θ = 160°
 2
r .  1 = 6
 
 2
 2
 1 =
 
 2
22 + 12 + 22 =
Unit 1 Answers: Chapter 11
9= 3
© Macmillan Publishers Limited 2013
Page 17 of 25
3
 2 / 3

 6
r .  1 / 3 = = 2
 2 / 3 3


The equation is of the form r . nˆ = d
∴d =
2
The distance from the origin to the plane is 2 units
 1
 −1


=
r
0 +λ 2
 
 
 1
 3
 −4 
1
 
 
=
r  2 +µ 3 
 −1 
 −2 
 
 
 1 − λ   −4 + µ 

 

 2λ =  2 + 3µ 
1 + 3λ   −1 − 2µ 

 

4
⇒ 1 − λ = −4 + µ
[1]
2λ = 2 + 3µ
[2]
1 + 3λ = −1 − 2µ
[3]
[1] × 2 ⇒ 2 − 2λ = −8 + 2µ
[4]
[2] + [4] ⇒ 2 = −6 + 5µ
8
µ=
5
24
2λ = 2 +
5
17
λ=
5
17
8
51
16
Substitute λ =
and µ = int o [3] ⇒ 1 + = − 1 −
which in inconsistent
5
5
5
5
Since the lines are not parallel and do not intersect, the lines are skew
x − 2 y − 4 z −1
l1 : = =
3
2
2
 2
 3
 
 
⇒=
r  4 + λ  2, λ ∈ 
1
 2
 
 
0
 −1
 
 
r =  2 + t  4 
 3
1
 
 
 2 + 3λ   − t 
Equating :  4 + 2λ  =  2 + 4t 

 

 1 + 2λ   3 + t 
⇒ 2+ 3λ = − t
4 + 2λ = 2 + 4t
1 + 2λ = 3 + t
Solving [1] and [3] simultaneously :
Unit 1 Answers: Chapter 11
[1]
[2]
[3]
© Macmillan Publishers Limited 2013
Page 18 of 25
5
[1] + [3] ⇒ 3 + 5λ = 3
λ=0
Substitute into [1] ⇒ 2 = −t
t = −2.
When λ = 0, t = −2, substitute into [2]
⇒ 4 = 4(−2) + 2
4 = −6
∴ l1 does not intersect l2
Since the lines are not parallel and do not intersect, the lines are skew
x + 2y + z = 4
2x − y − z = 1
1
2
 
 
The normal to the planes are  2  and  −1
1
 −1
 
 
=
cos θ
6
1  2 
   
 2  .  −1
   −1
2 − 2 − 1 −1
1 =
 
=
6
6 6
1  2 
   
 2   −1
 1   −1
   
θ = 99.6°
2
→  
OA =  1 
 −1
 
1
→  
OB =  −2 
 3
 
→ → →
AC
= OC − OA
→ → →
CB
= OB − OC
→ →
→ →
∴  OC− OA  (1 − =
p) p  OB − OC 




→
→
→
→


(1 − p) OC − (1 − p)  OA=
 p OB − p OC


→
→
(1 − p) OC+ p OC= p(i − 2 j + 3k) + (1 − p) (2i + j − k)
→
OC = pi + 2i - 2pi - 2pj + j - pj + 3pk - k + pk
= (2 − p) i + (1 − 3p)j + (4p − 1)k
→ → →
AB = OB − OA = (i - 2j + 3k) - (2i + j - k)
= −i −3j + 4k
→ →
OC . AB = 0
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 19 of 25
 2 − p   −1
⇒  1 − 3p  .  −3 =
0

  
 4p − 1  4 
⇒ −2 + p −3 + 9p + 16p − 4 = 0
26p = 9
9
p=
26
→ 
9  
27 
 18 
∴ OC = 2 −  i + 1 −
 j +  − 1 k
26  
26 

 13 
43
1
5
=
i−
j+ k
26
26
13
cos θ
=
 2  1 
  
 1   −2 
 −1  3 
  
=
2 1
   
 1   −2 
 −1  3 
   
cos θ
=
 2−p  1 

  
 1 − 3p  .  −2 

  
 4p − 1  3 
=
 2−p   1 

  
 1 − 3p   −2 
 4p − 1  3 

  
−3
6 14
−3
6 14
2 − p − 2 + 6p + 12 p − 3
(2 − p) 2 + (1 − 3p) 2 + (4p − 1) 2 14
=
-3
16 14
−3
(2 − p) 2 + (1 − 3p) 2 + (4p − 1) 2
16
3
(2 − p) 2 + (1 − 3p) 2 + (4p − 1) 2 
⇒ (17p − =
3) 2
16 
⇒ 17p
=
−3
289p2 – 102p + 9 =
3
(4 – 4p + p2 + 1 – 6p + 9p2 + 16 p2 – 8p + 1)
16
4624p2 – 1632p + 144 = 78p2 – 54p + 18
4546p2 – 1578p + 126 = 0
2273p2 – 789p + 63 = 0
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 20 of 25
789 ± 7892 − 4(2273)(63)
p=
2(2273)
=
789 ± 49725
4546
= 0.223, 0.125
7
8
 4 + t   −3 + 2s 

 

 9 + 6t  = −15 + 8s 
12 + 5t   −19 + 8s 

 

⇒ 4 + t = −3 + 2s
9 + 6t = −15 + 8s
12 + 5t = −19 + 8s
Solving [2] and [3] :
[2] − [3] ⇒ − 3 + t = 4
t=7
∴ 9 + 6(7) = −15 + 8s
s = 8.25
Substituting into [1] ⇒ 4 + 7 = −3 + 2(8.25)
11 = 13.5
⇒ the lines do not intersect
1
0
 
 
=
r  2  + λ 1
 −1
 4
 
 
[1]
[2]
[3]
3
→  
OA =  2 
 −1
 
 1 
→ 

ON
=  2+λ 
 −1 + 4λ 


 1  3
→ → → 
  
AN = ON − OA =  2 + λ  −  2  =
 −1 + 4λ   −1

  
 −2 
 
λ
 4λ 
 
0
→  
AN .  1  = 0
 4
 
 −2   0 
   
⇒  λ  .1 =
0
 4λ   4 
   
⇒ λ + 16λ = 0
λ=0
1
→  
ON =  2 
 −1
 
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 21 of 25
 −2 
→  
AN =  0 
 0
 
→
⇒ AN = (−2) 2 =2
9
 1
 2
l1 : r =  −1 + λ  1 
 
 
 0
 −2
1
 −3 
 
 
l 2=
: r  2 + µ  0 
 2
4
 
 
1
2
+
λ

  1 − 3µ 

 

⇒  −1 + λ  = 2 
 −2λ   2 + 4µ 

 

⇒ 1 + 2λ = 1 − 3µ
[1]
−1 + λ = 2
[2]
−2 λ = 2 + 4µ
[3]
From [2] λ = 3
Substitute into [3] ⇒ −6 = 2 + 4µ
µ = −2
Substitute λ = 3, µ = −2 into [1] ⇒ 1 + 6 = 1 + 6
⇒ All three equations are satisfied by λ = 3, µ = −2
∴ l1 and l2 intersect and hence l1 and l2 are not skew
1
 −2 
→   →  
10 =
OA =
3  , OB  1 
1
1
 
 
→ →
→
= OA + λ AB
OC
 1   −2 
→ →    
(a)
OA. OB = 3  .  1  =− 2 + 3 + 1 =2
1  1 
   
→ →
OA . OB
2
(b)
cos=
θ →
θ 75.7°
→= 11 6 ⇒=
OA OB
(c)
1
 −3   1 − 3λ 
→  
  

OC=  3  + λ  −2 =  3 − 2λ 
1
 0  1 
 
  

 1 − 3λ   −3 
→ → 
  
OC . AB =  3 − 2λ  .  −2  = 0
 1   0

  
⇒ − 3 + 9 λ − 6 + 4 λ =0
13λ =9
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 22 of 25
(d)
9
λ=
13
→
OC = (1 − 3λ ) 2 + (3 − 2λ ) 2 + 12
 1 − 3λ   1   −3λ 
→ → →
   

AC = OC − OA  3 − 2λ  −  3  =  −2λ 
 1  1  0 

   

→
AC = 9λ 2 + 4λ 2 = 13λ 2
→
→
∴ OC =
AC
⇒ 1 − 6λ + 9λ 2 + 9 − 12λ + 4λ 2 +=
1
11
13λ 2
⇒ 13λ 2 − 18λ + 11= 13λ 2
18λ = 11
11
λ=
18
5
 −4 
→   →  
OA =
 −1  , OB =
 4
 −3 
 −1 
 
 
(a)
(b)
=
θ
cos
 5   −4
 −1 .  4 
   
 −3  −1
=
 5   −4
 −1  4 
   
 −3  −1
−21
⇒
=
θ 128.2°
35 33
Eq. of BC is
  5   −4  
 −4 
   
 
r  4  + λ   −2  −  4  
=
 
   
 −1 
  11   −1  
 −4 
9
 
 
=  4  + λ  −6  , λ ∈ 
 −1 
 12 
 
 
 −1  2   −1
     
(c)
r . 1  =  1 . 1 
 2   −4   2 
     
 −1
 
r . 1  = −9
2
 
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 23 of 25
(d)
12
cos=
α
 −1  9 
   
 1  .  −6 
 2   12 
   
 −1  9 
   
 1   −6 
 2   12 
   
9
6 261
⇒=
α 76.9°
Angle between the line and the plane = 180 − (90 + 76.9) = 13.1°
10
  s
   
 t = 5 
 − t   5s 
   
∴ 10 = s
t=5
−t = 5s
∴ −5 = 5(10) Inconsistent
Since the lines are not parallel and do not intersect they are skew
3
5
→   →  
13
OA =
4  , OB  7 
=
 −1
6
 
 
(a)
 5  3   2
     
Direction of AB =  7  −  4  =
 3
 6   −1  7 
     
Equation of the plane:
 2  3   2
r .  3 =  4  .  3
     
 7  −1  7
 2
 
11
⇒ r .  3 =
7
 
3
6
→   →  
(b)
=
OA =
4  , OC  23 
 −1
8
 
 
Equation of plane :
 6   3  
3
    
 
=
r  4  + µ  23  −  4  
 −1
 8   −1 
 
3
3
 
 
=
r  4  + µ 19 
 −1
9
 
 
∴ x = 3 +μ3 

y = 4 + 19μ  µ ∈ 
z = -1 + 9µ 
(c) Substitute the line into the plane :
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 24 of 25
2(3 + 3µ) + 3(4 + 19µ) + 7(−1 + 9µ) = 11
⇒ 11 + 126µ =11
µ =0
∴ The point of intersection is
x=3
y=4
z = −1
3
 
4
 −1
 
14
π : 2x − 3y + z =
6
 2
 1


=
l:r
1 + t  2
 
 
 −1
 2
(a)
(b)
(c)
15
(a)
x=2+t 

y = 1 + 2t  t ∈ 
z = -1 + 2t 
2(2 + t) – 3(1 + 2t) + (−1 + 2t) = 6
⇒ 2t − 6t + 2t = 6
−2t = 6
t = −3
 2−3   1 

  
Point of intersection is  1 − 6  =
 −5 
 −1 − 6   −7 

  
=
θ
cos
1  2 
   
 2  .  −3 
 2  1 
   =

1  2 
   
 2   −3 
 2  1 
   
−2
9 14
=
⇒ θ 100.26°
Acute angle is 180 − 100.26 = 79.7°
2
→  
OA =  −1
1
 
 3
 −3 
 
 
=
r  4 + s  1 
1
2
 
 
 3 − 3s 
→ 

OB
=  4+s 
 1 + 2s 


Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 25 of 25
 3 − 3s   2 
→ → → 
  
AB= OB − OA=  4 + s  −  −1=
 1 + 2s   1 

  
Since AB is perpendicular to l:
 −3 
→  
AB .  1  = 0
2
 
(b)
1 − 3s 


5+s 
 2s 


1 − 3s   −3 

  
⇒  5 + s  .  1  =− 3 + 9s + 5 + s + 4s =0
 2s   2 

  
1
14s =
−2,s=
−
7
3

3 + 7 
  24 / 7 
→ 
1 

∴ OB = 4 −
= 27 / 7 

7  

  5 / 7 
2
1− 


7

3

1 + 7 
  10 / 7 
→ 
1 

AB = 5 −
= 34 / 7 

7  

  −2 / 7 
2
−




 7 
→
=
AB
2
2
2
 10   34   2 
  +   +  −=
 5.1
 7   7   7
Unit 1 Answers: Chapter 11
© Macmillan Publishers Limited 2013
Page 1 of 14
Chapter 12 Limits and Continuity
Try these 12.1
(a)
lim (4x 2 + 6x + 1) = 4(2) 2 + 6(2) + 1 = 16 + 12 + 1 = 29
(b)
 x2 + 1  1 + 1
2
= −
lim 
=
x→1
3
 x − 4  1− 4
x→ 2
(d)
t +1
0 +1 1
lim = =
t →0
t+9
0+9 3
lim (3t + 1)3= (3(4) + 1)3= 2197
(e)
lim (t + 1) 2 (4t − 2) = (4 + 1) 2 (4(4) − 2) = 350
(f)
lim (3x − 4) (2x 2 + 7) =
(3(0) − 4) (2(0) 2 + 7) =
(−4) (7) =
−28
(c)
t →4
t →4
x →0
lim (4x 3 − 3x 2 + 5) =4(−2)3 − 3(−2) 2 + 5 =−39
(g)
x →−2
(h)
−1 + 1
 x+1 
lim =
= 0

x→−1
 2x + 3  2(−1) + 3
(i)
lim
x 2 + 2x + 1
=
x+2
1
=
2
(j)
lim
(x + 1)3
=
4x − 2
(2 + 1)3
=
4(2) − 2
x→ 0
x→ 2
2
2
27
=
6
3
3
=
2
2 2
Try these 12.2
(a)
x3 − 1
x →1 x − 1
(x − 1) (x 2 + x + 1)
, factoring x 3 − 1
= lim
x →1
(x − 1)
lim
= lim (x 2 + x + 1)
x →1
= (1) 2 + 1 + 1
=3
x3 + 1
(b)
lim
x→−1 x + 1
(x + 1) (x 2 − x + 1)
= lim
x →−1
(x + 1)
=
(c)
lim (x 2 − x + 1)
x →−1
= ( −1) 2 − ( −1) + 1
=3
 x2 + x 
lim 

x →0
 x 
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 2 of 14
= lim (x + 1)
x →0
= 0 +1
=1
Try these 12.3
5− x
25 − x
5− x 5+ x
= lim
×
x → 25
25 − x
5+ x
(25 − x)
= lim
x→ 25
(25 − x) (5 − x )
(a)
lim
x→ 25
1
5+ x
1
1
= =
5 + 25 10
x − 19
(b)
lim
x →19 5 −
x+6
= lim
x→ 25
x − 19
5+ x +6
×
x →19 5 −
x +6 5+ x +6
(x − 19)(5 + x + 6)
= lim
x →19
25 − (x + 6)
= lim
(x − 19)(5 + x + 6)
19 − x
= lim − (5 + x + 6)
= lim
x →19
x →19
=
− (5 + 25)
= – 10
x −5
(c)
lim
x→5
x −5
x −5
x −5
= lim
×
x →5
x −5
x −5
= lim
(x − 5) x − 5
(x − 5)
x →5
= lim x − 5
x →5
=
5−5 = 0
Try these 12.4
(a)
lim
x→∞
x 2 + 6x + 9
2x 2 + 7x + 8
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 3 of 14
6 9
+
x x2
= lim
x →∞
7 8
2+ + 2
x x
1
=
2
x 3 + 7x 2 + 2
lim
x→∞
3x 3 + 6x + 1
7 2
1+ + 3
x x
= lim
x →∞
6
1
3+ 2 + 3
x
x
1
=
3
1+
(b)
Exercise 12A
1
lim (x 2 + 4x + 5) =2 2 + 4(2) + 5
2
= 17
x
ο
lim 3=
3=
1
x→ 2
x→ 0
3
4
5
6
3x + 2 3(0) + 2
=
x→ 0 4x − 1
4(0) − 1
= −2
3x 2 + 5x + 2 3(2)2 + 5(2) + 2
lim
=
x→ 2
2−4
x−4
24
=
= −12
−2
4x 2 − 3x + 2 4( −1)2 − 3( −1) + 2
=
lim 2
x→−1 x + x + 2
( −1)2 + ( −1) + 2
9
=
2
2
lim (4x + 3) =(4 + 3) 2 =49
lim
x→1
7
8
(x − 5) (x + 2)
= lim (x + 2)
x→5
x →5
x −5
=5+ 2 =7
(x − 3) (2x + 1)
lim
= lim (2x + 1)
x→3
x →3
(x − 3)
lim
= 2(3) +=
1 7
 3x 3 − 4x 2 
x 2 (3x − 4)
lim 
=
lim

x →0
x2
x2

 x →0
= lim (3x − 4)
9
x →0
= 3(0) − 4
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 4 of 14
= −4
10
( x + 3) ( x + 2)
x 2 + 5x + 6
= lim
x →−2
x →−2
x+2
x+2
= lim (x + 3)
lim
x →−2
= –2 + 3 = 1
x 2 + 9x + 20
x →−5
x+5
= lim (x + 4)
11
lim
= lim
(x + 4) (x + 5)
x →−5
x+5
x→−5
=–5 + 4 =−1
(x − 2) (x + 2) (x + 4)
x 4 − 16
= lim
lim
x →−2
x →−2 x + 2
x+2
2
12
= lim (x − 2)(x 2 + 4)
x →−2
13
= ( −2 − 2) (( −2) 2 + 4)
= −32
x + 11 − 4
x + 11 + 4
x + 11 − 4
= lim
×
lim
x →5
x →5
x −5
x −5
x + 11 + 4
x + 11 − 16
= lim
x →5 (x − 5)( x + 11 + 4)
x −5
= lim
x − 5 ( x + 11 + 4)
1
= lim
x →5
x + 11 + 4
1
1
1
=
= =
16 + 4 4 + 4 8
x →5
14
x+2
x+6+2
x+2
= lim
×
lim
x
→−
2
x+6 −2
x+6+2
x+6 −2
(x + 2)[ x + 6 + 2]
= lim
x →−2
x+6−4
(x + 2) ( x + 6 + 2)
= lim
x →−2
x+2
x →−2
= lim ( x + 6 + 2)
x →−2
=
15
−2 + 6 + 2
= 4 +2=2+2= 4
x − 16
 x − 16 
×
lim=
 xlim
→16
x →16 
x−4
 x −4
= lim
x +4
x +4
(x − 16) ( x + 4)
x →16
x − 16
= lim ( x + 4)=
x →16
16 + 4
= 4+ 4 =8
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 5 of 14
16
17
18
19
1
3+
3x + 1
x =2
= lim
lim
x →∞
x →∞ 2x − 1
1 2
2−
x
5
6−
6x − 5
x= 6= 3
= lim
lim
x →∞
x →∞ 4x − 1
1 4 2
4−
x
3 1
1+ + 2
x 2 + 3x + 1
x x =1
= lim
lim
x →∞
x →∞ x 2 − 2x − 1
2 1
1− − 2
x x
6 1
1− 2 + 3
x 3 − 6x + 1
x x
= lim
lim
x →∞
x →∞
2 1
2x 2 − 1
−
x x3
1
=
→∞
0
Try these 12.5
(a)
(b)
(c)
sin 8θ
8sin 8θ
= lim
θ→ 0
θ
8θ
sin 8θ
= 8lim
θ→ 0
8θ
= 8(1)
= 8
lim
θ→ 0
 tan 7θ 
 sin 7θ 
lim 
 = lim


θ→ 0
θ→
0
 θ 
 θ cos 7θ 
1
 7sin 7θ 
= lim 
 lim
θ→ 0 
θ→
0
7θ
cos7θ
= (7)(1)
=7
 sin 6θ 
sin 6θ


lim
= lim  θ 
θ→ 0 sin 5θ
θ→ 0
sin 5θ


 θ 
 6sin 6θ 
lim 

θ→ 0 
6θ 
=
 5sin 5θ 
lim 

θ→ 0 
5θ 
6(1)
=
5(1)
6
=
5
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 6 of 14
Exercise 12B
1
2
4 + 5 −9
4x 3 + 5
= =
x→1 x − 3
1− 3
2
lim (4x +=
2)5 (4(0) + 2)5
lim
x→ 0
5
= 2=
32
3
lim 4x=
+2
4
= 10
lim (x 2 − 2x + 1)6 = ((1)2 − 2(1) + 1)6
5
=0
5. lim (4x + 3)3 (2x + 1) 2 = (4( −1) + 3)3 (2( −1) + 1)2
6
7
8
x→ 2
4(2) + 2
x→1
x →−1
=
( −1)3 ( −1)2
= −1
(6x − 2)3 (6(0) − 2)3
=
lim
x→ 0 (3x + 1) 2
(3(0) + 1)2
= −8
x 2 − 3 (−3) 2 − 3
=
lim
x→−3 2x + 3
2(−3) + 3
6
=
= −2
−3
(2x + 1) (x + 2)
2x 2 + 5x + 2
= lim
lim
1
1
2x + 1
2x + 1
x→−
x →−
2
2
= lim (x + 2)
x →−
1
2
1
3
=− + 2 =
2
2
5
(x + 2) (x 4 − 2x 3 + 4x 2 − 8x + 16)
x + 32
9
= lim
lim
x →−2
x→−2 x + 2
x+2
4
3
2
= lim (x − 2x + 4x − 8x + 16)
x →−2
=
(−2) 4 − 2(−2)3 + 4(−2) 2 − 8(−2) + 16
= 16 + 16 + 16 + 16 + 16 = 80
10
(x − 5) (x + 7)
x 2 + 2x − 35
= lim
x →5 (x − 5) (x − 5)
x→5 x 2 − 10x + 25
lim
x+7
x −5
5 + 7 12
=
=
= ∞
5−5 0
(x − 1) (x + 2)
x2 + x − 2
= lim
lim
2
x
→−
1
x→−1
x −1
(x − 1) (x + 1)
= lim
x →5
11
x+2
x →−1 x + 1
= lim
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 7 of 14
−1 + 2 3
= = ∞
−1 + 1 0
(x − 1) (2x 2 + 3x + 1)
2x 3 + x 2 − 2x − 1
=
lim
lim
x →1
x→1
3x 2 − 2x − 1
(3x + 1) (x − 1)
=
12
2x 2 + 3x + 1
x →1
3x + 1
2 + 3 +1 6 3
=
= =
3 +1
4 2
3
(3x − 1) (9x 2 + 3x + 1)
27x − 1
= lim
lim
1 3x − 1
x →1/ 3
3x − 1
x→
= lim
13
3
= lim (9x 2 + 3x + 1)
x→
1
3
2
 1
 1
= 9   + 3  + 1
 3
 3
= 1+1+1 = 3
(x − 3) (x + 3x + 9x + 27x + 81)
x 5 − 243
14
= lim
lim
x →3
x →3
x −3
x −3
4
3
2
= lim (x + 3x + 9x + 27x + 81)
4
3
2
x →3
=(3)4 + 3(3)3 + 9(3)2 + 27(3) + 81
= 405
15
9−x
x +3
9−x
= lim
×
lim
x →9
x +3
x − 3 x →9 x − 3
= lim
x →9
− (x − 9) ( x + 3)
x −9
=lim − ( x + 3)
x →9
=
−( 9 + 3) =
−6
16
5 − x −5
5 + x −5
5 − x −5
= lim
×
lim
x →10
x→10
10 − x
10 − x
5 + x −5
5 − (x − 5)
= lim
x → 0 (10 − x)( 5 +
x − 5)
= lim
x →10
= lim
10 − x
(10 − x) 5 + x − 5
1
5 + x −5
1
1
5
=
= =
5 + 5 2 5 10
2− x 2+ x
2− x
17
= lim
×
lim
x→4 4 − x
x→ 4 4 − x
2+ x
x →10
= lim
x→4
4−x
(4 − x) (2 + x )
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 8 of 14
1
x → 4 (2 +
x)
1
1
1
=
= =
2+ 4 2+2 4
= lim
18
lim
x →1
= lim
x →1
= lim
x →1
= lim
x →1
( 3+ x − 5 − x) ( 3+ x + 5 − x)
3+ x − 5− x
= lim
2
→
x
1
x −1
(x 2 − 1) ( 3 + x + 5 − x )
(3 + x) − (5 − x)
(x 2 − 1)( 3 + x + 5 − x )
−2 + 2x
(x − 1) (x + 1) ( 3 + x + 5 − x )
2 (x − 1)
(x − 1) (x + 1)( 3 + x + x − 5)
2
(x + 1)( 3 + x + 5 − x )
2
2
1
=
= =
2( 4 + 4) 2(4) 4
= lim
x →1
19
lim
x →0
x2
x 2 + 12 − 12
= lim
x →0
x2
x 2 + 12 − 12
×
x 2 + 12 + 12
x 2 + 12 + 12
x 2  x 2 + 12 + 12 


= lim
x →0
x 2 + 12 − 12
= lim ( x 2 + 12 + 12)
x →0
=
12 + 12= 2 4 × 3= 4 3
5sin 5θ
sin 5θ
20
= lim
lim
θ→ 0
θ→0
5θ
θ
sin 5θ
= 5 lim
θ→0
5θ
= 5(1)
= 5
3sin 3θ
sin 3θ
21
= lim
lim
θ→
0
θ→0
5(3θ)
5θ
3
sin 3θ
= lim
5 θ→ 0 3θ
3
3
= =
(1)
5
5
sin 2θ
2 sin 2θ
22
lim
= lim
θ→0
θ→
0
3θ
3 2θ
2
sin 2θ
= lim
3 θ→0 2θ
2
2
= =
(1)
3
3
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 9 of 14
23
  sin 3θ  
  θ  
sin 3θ

lim
= lim 
θ→ 0 sin 5θ
θ→0  sin 5θ 


θ


 sin 3θ 
lim 

θ→0 
θ 
=
 sin 5θ 
lim 

θ→0 
θ 
 sin 3θ 
lim 3 

θ→ 0
 3θ 
=
 sin 5θ 
lim 5 

θ→ 0
 5θ 
3
=
5
24
 8sin8θ 


sin
8
θ
8θ 


lim
=
lim 
 θ→0 4sin 4θ
θ→ 0 sin 4θ






4θ 
 sin8θ 
8 lim 

θ→0  8θ 
=
4 lim  sin 4θ 
θ→0 
4θ 
8
= 2
4
cos 2θ −1
25
lim
=0
θ→0
2θ
(cos 2θ −1)
 1 − cos 2θ 
26
=0
lim 
=
−
 lim
θ→0
θ→0
2θ
2θ


tan2θ
 sin 2θ  1 
27
lim
= lim 
 
θ→0
θ→0
2θ
 cos2θ  2θ 
 sin 2θ 
 1 
= lim 
 × lim


θ→ 0  2θ 
θ→ 0  cos2θ 
= (1)
=
(1) 1
=
28
 sin 6θ   1 
 tan 6θ 
lim 

 
 = lim
θ→0
θ→
0
cos 6θ   4θ 
 4θ 
 sin 6θ 
 1 
= lim 
 lim 

θ→0  4θ  θ→0  cos 6θ 
6sin 6θ
1
= lim
lim
θ→ 0 4(6θ)
θ→ 0 cos 6θ
 6
=   (1)(1)
 4
=
3
2
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 10 of 14
29
 x 
x


lim
= lim  x 
x→ 0 sin 3x
x → 0 sin 3x


x 
 x
lim  
x →0  x 
=
 sin 3 x 
lim 3 

x →0
 3x 
1
=
3
30
 x 
x


lim
= lim  x 
x→ 0 tan x
x →0
tan x


x 
x 
lim  
x →0
x 
=
 tanx 
lim 

x →0
 x 
lim1
= x →0
 sin x 
lim 

x → 0  x cos x 
1
1
 sin x 
lim 
 lim
x →0
x
→
0
cos x
 x 
1
= = 1
(1)(1)
=
31
32
33
3
1
2

4+ 2 + 3 − 4
 4x 4 + 3x 2 + x − 2 


x
x
x
lim 
 = xlim
4


x →∞
→∞
1
x −1


1− 4


x
4
= = 4
1
 3 
 2 
 3x 2 
lim  4
= lim  x 

x →∞ x − 1

 x →∞  1 − 1 
x4 

0
= = 0
1
10
(a)
lim3f (x) =
10 ⇒ lim f (x) =
x →2
x →2
3
10
34
lim f (x) + 2x 2  =
lim f (x) + 2lim x 2 = + 2(2) 2 =
x →2
x →2
x →2
3
3
(b)
x 2 + 7x + 12 =
0
(x + 3)(x + 4) =
0
x =−3, −4
the function is not continuous when x = –3, –4
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 11 of 14
34
(a)
=
(b)
x − 6 − 1 ( x − 6 − 1) ( x − 6 + 1)
=
7−x
(7 − x) ( x − 6 + 1)
x − 6 −1
(7 − x)( x − 6 + 1)
=
x−7
− (x − 7) ( x − 6 + 1)
=
−1
( x − 6 + 1)
 ( x − 6 − 1) 
1
lim 
 = − lim
x →7
x
→
7
7−x
x − 6 +1


1
= −
1 +1
−1
=
2
 sin 2 θ   1 
tan 2 θ
= lim 
lim

2 
θ→0
θ→0
θ
 θ   cos θ 
 sin θ 
 1 
= lim (sin θ) lim 
 lim 

θ→0
θ→0  θ  θ→0  cos 2 θ 
= (0)
=
(1) (1) 0
 5sin x + cos x − 1 
lim 

x →0
4x


5  sin x 
 cos x − 1
= lim 
 + lim


x →0 4 
x
→
0
x
4x 
35
5
 sin x  1
 cos x − 1
lim 
 + lim


x
0
x
0
→
→

4
x
4
x 
5
=
4
sin 2 x + sin x (cos x − 1)
36
lim
x →0
x2
 sin 2 x  sin x   cos x − 1 
= lim  2 + 

x →0
 x  
x  
 x
=
 sin x 
 sin x 
 sin x 
 cos x − 1
lim 
 lim

 + lim

 lim


x →0 
x
→
0
x
→
0
x
→
0
x
x
x
x 
= (1) (1) + (1) (0)
=1
Review exercise 12
4x + x + x − 2
= lim
x →∞
4x 3 + 5
5
1
lim
x →∞
2
Unit 1 Answers: Chapter 12
4+
1
1
2
+ 4− 5
3
x
x
x
4
5
+
x 2 x5
© Macmillan Publishers Limited 2013
Page 12 of 14
4
→∞
0
=
2
7x 3 − 2x 2 + x
= lim
lim
x →∞
x →∞
3x 2 − 1
2 1
+
x x3
3 1
−
x x3
7−
7
→∞
0
4x 2 − 3x + 2 4 + 3 + 2 9
lim 2 = =
x →−1 x + x + 2
1−1+ 2 2
3
2
(x − 2) (x 2 + 3x + 7)
 x + x + x − 14 
(a)
lim 
 = lim
3
2
x →2
 x − x − 6  x → 2 (x − 2) (x + 2x + 3)
=
3
4
 x 2 + 3x + 7 
= lim  2
x → 2  x + 2x + 3 

4 + 6 + 7 17
= =
4 + 4 + 3 11
(4x + 5) (3x + 2)
12x 2 + 23x + 10
(b)
=
lim
lim
−5
−5 4x 2 + 13x + 10
(4x + 5) (x + 2)
x→
x→
4
4
 −5 
3   + 2 −7
−7
 4  = =
4
3
−5
3
+2
4
4
4
 sin 2θ 
lim 2
sin 2θ θ→0  2θ 
5
=
lim
θ→0 sinθ
 sin θ 
lim 

θ→0
 θ 
2(1)
= = 2
1
cos 6θ −1
cos 6θ −1
6
lim
= 3 lim
θ→ 0
θ→0
2θ
6θ
= 3(0)
= 0
 3x + 2 
= lim 
=

−5
x→
 x+2 
sin 3x sin 5x
 sin 3x  sin 5x 
= lim 


2
x
→
0
7x
 7x  x 
1
 3sin 3x 
 5sin 5x 
= lim 
 lim


x
→
0
x
→
0

7
3x
5x 
1
15
= =
(3) (1) (5) (1)
7
7
3 + 15
18 −3 2
3 − 5x
8
lim
=
= =
x →−3
−2
−2
2
x +1
2
(2x + 1) (x + 2)
2x + 5x + 2
9
(a)
= lim
lim
2
−1 2x + 9x + 4
1
x→
x →− (2x + 1) (x + 4)
7
lim
x→ 0
2
Unit 1 Answers: Chapter 12
2
© Macmillan Publishers Limited 2013
Page 13 of 14
10
11
1
3
2−
(x + 2)
3
2 =
2
= lim
= =
−1 (x + 4)
1 7/2 7
x→
4−
2
2
(b)
x 2 + 3x + 2 =
0
(x + 1) (x + 2) =
0
x =− 1, − 2
f(x) is continuous everywhere except at x = –2, –3
sin 5θ
5sin 5θ
(a)
lim = lim
= 5(1)
= 5
θ→0
θ→0
θ
5θ
(b)
2x − 3 − 5 =
0
2x − 3 =
5
2x–3 = 5, 2x–3 = –5
x = 4, x= –1
The function is continuous everywhere except at x = 4, x = −1
 sin 5θ 
 tan 5θ 
lim 


 = lim
θ→0
θ→0
 θ 
 θcos5θ 
1
 sin 5θ 
= lim5 
 lim
θ→0
θ→0 cos5θ
5θ


= 5(1)
=
(1) 5
sin 2x
− 4x
tan 2x − 4x
12
lim
= lim cos 2x
x→ 0 sin 3x − 7x
x → 0 sin 3x − 7x
sin 2x
4x
−
x
cos
2x
x
= lim
x → 0 sin 3x
7x
−
x
x
 2sin 2x 
 1 
lim 
 lim

 − lim 4
x →0 
x
→
0
2x
cos 2x  x → 0
=
 3sin 3x 
lim 
 − lim 7
x →0 
3x  x → 0
=
13
(2)(1) − 4 −2 1
= =
3(1) − 7
−4 2
(a)
4x 2 − 11x − 3 =
0
(4x + 1)(x – 3) = 0
1
x = − ,3
4
1
f(x) is continuous everywhere except at x = − , 3
4
sin 4x
 sin 4x 
(b)
= lim 4 
=
= 4
lim
 4(1)
x →0
x
→
0
x
 4x 
 sin 4x  lim  4sin 4x 


sin 4x
 x  x → 0  4x 
=
=
lim
lim
x → 0 sin 5x
x → 0  sin 5x 
 5sin 5x 

 lim 
x
→
0
 5x 
x
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 14 of 14
4
5
x + 4 −3
x + 4 −3
x+4 +3
=
×
x −5
x −5
x+4 +3
(x + 4) − 9
=
(x − 5)( x + 4 + 3)
=
14
(a)
=
x −5
(x − 5) ( x + 4 + 3)
1
x+4 +3
x+4 −3
1
lim
= lim
x →5
x
→
5
x −5
x+4 +3
1
1
= =
9 +3 6
1 − cos 2x
(b)
lim
x →0
x2
2sin 2 x
 sin x  sin x 
= lim
=
lim 2 


2
→
x →0
x
0
x
 x  x 
 sin x 
 sin x 
= 2 lim 
 lim


x →0 
x
→
0
x
x 
= 2(1) (1) = 2
(c) |3x – 1| – 8 = 0
|3x – 1| = 8
3x – 1 = 8, 3x – 1 = –8
x = 3, 3x = −7
x = –7/3
=
the function is continuous everywhere except at x = 3, x =
15
−7
3
9x 2 − 3x − 2
3x 2 + 13x + 4
(3x + 1)(3x − 2)
=
(3x + 1)(x + 4)
f(x) =
∴ Discontinuous at x =
−1
, x = −4
3
−1
, point discontinuity which is removable
3
x = 4, infinite discontinuity which is non-removable
At x =
Unit 1 Answers: Chapter 12
© Macmillan Publishers Limited 2013
Page 1 of 39
Chapter 13 Differentiation I
Try these 13.1
(a)
f (x) =
1
x
1
f (x + h) =
x+h
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
1
1
−
x+h x
= lim
h→0
h
x − (x + h)
(x + h) (x)
= lim
h→0
h
1
−h
= lim
×
h→0 h
(x + h) (x)
−1
= lim
h → 0 x(x + h)
1
= −
x(x)
1
= − 2
x
1
(b)
f (x) = 2
x
1
f (x + h) =
(x + h) 2
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
1
1
−
(x + h) 2 x 2
= lim
h→0
h
2
1  x − (x + h) 2 
= lim  2
2 
h→0 h
 x (x + h) 
1  x 2 − x 2 − 2xh − h 2 


h→0 h
x 2 (x + h) 2


= lim
−2x − h 2
h → 0 x 2 (x + h) 2
−2x
−2x
= =
2
2
x (x + 0)
x4
= lim
=
−2
x3
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 2 of 39
Try these 13.2
(a)
f(x) = x10
f ′ (x) = 10x 9
(b)
f (x) = x 4
3 −1
f ′ (x) = x 4
4
1
= =
f (x)
x −4
x4
(c)
(d)
3
−4
f ′ (x) =
− 4x −5 =
x5
1
= =
f (x)
x −8
8
x
−8
f ′(x) =
− 8 x −9 =
x9
Try these 13.3
(a)
(b)
(c)
f (x) = 10x 5
f ′ (x) = 50 x 4
1
−1
= = x 2
f (x)
x
1 −3
1
f ′(x) =
− x 2=
−
.
2
2x x
12
f (x)
= =
12x −3
3
x
f ′(x) = − 36x −4
−36
= 4
x
Try these 13.4
f(x) = g(x) – m(x)
f ′(x) = lim
[g(x + h) − m(x + h)] − [g(x) − m(x)]
h →0
= lim
h
[g(x + h) − g(x)] − [ m(x + h) − m(x)]
h →0
lim
[g(x + h) − g(x)]
h →0
h
= g′(x) − m′(x)
Unit 1 Answers: Chapter 13
h
− lim
h →0
[ m(x + h) − m(x)]
h
© Macmillan Publishers Limited 2013
Page 3 of 39
Try these 13.5
x3 − x 2
x5
1
1
y = 2 − 3 = x −2 − x −3
x
x
dy
=
− 2x −3 + 3x −4
dx
−2 3
= 3 + 4
x
x
4x + 1
(b)
y=
x
1
4x
1
−1
=
+
= 4x 2 + x 2
x
x
dy
 1 1 −1  1 −1 −1
= 4  x 2 +  −  x 2
 2
 2
dx
1 −3
−1
= 2x 2 − x 2
2
2
1
=
−
x 2x x
(c)
y =−
(x 2) (3x + 4)
(a)
y=
(d)
y = 3x 2 − 2x − 8
dy
= 6x − 2
dx
y =+
(x 5) (5x − 1)
y =5x 2 + 24x − 5
dy
= 10x + 24
dx
Try these 13.6
(a)
f(x) = 2x − 4
f(x + h) = 2(x + h) − 4
= 2x + 2h − 4
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
2x + 2h − 4 − (2x − 4)
= lim
h→0
h
 2h 
= lim  
h→0  h 
= lim
=
2 2
h→0
∴ f ′ (x) =
2
(b)
f (x) = 6x 2 + 2x + 1
f(x + h) = 6(x + h)2 + 2(x + h) + 1
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 4 of 39
= 6(x 2 + 2xh + h 2 ) + 2x + 2h + 1
= 6x 2 + 12xh + 6h 2 + 2x + 2h + 1
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
6x 2 + 12xh + 6h 2 + 2x + 2h + 1 − (6x 2 + 2x + 1)
= lim
h→0
h
2
12xh + 6h + 2h
= lim
h→0
h
= lim (12x + 6h + 2)
h→0
= 12x + 2
∴ f ′(x) = 12x + 2
1
1
(c)
=
f(x + h)
=
, f(x)
2(x + h) + 1
2x + 1
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
1
1
−
2x + 2h + 1 2x + 1
= lim
h→0
h

1 2x + 1 − 2x − 2h − 1 
= lim 
h → 0 h (2x + 2h + 1) (2x + 1) 


= lim
1
−2h

h → 0 h  (2x + 2h + 1) (2x + 1) 


−2
= lim
h → 0 (2x + 2h + 1)(2x + 1)
=
−2
(2x + 1) (2x + 1)
=
−2
(2x + 1) 2
−2
∴ f ′(x) = 2
(2x + 1)
Exercise 13A
1
f(x) = 4x − 7
f(x + h) = 4(x + h) − 7 = 4x + 4h − 7
By definition
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
(4x + 4h − 7) − (4x − 7)
= lim
h→0
h
4h
 
= lim  
h→0
 h 
4 4
= lim
=
(a)
h→0
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 5 of 39
(b)
f(x) = 3x + 9
f(x + h) = 3(x + h) + 9 = 3x + 3h + 9
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
3x + 3h + 9 − (3x + 9)
= lim
h→ 0
h
3h
= lim
h→0 h
= lim
=
3 3
h→0
f(x) = x2 + 2x + 5
f(x + h) = (x + h)2 + 2(x + h) + 5
= x2 + 2xh + h2 + 2x + 2h + 5
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
2
( x + 2xh + h 2 + 2x + 2h + 5 ) − ( x 2 + 2x + 5 )
= lim
h→0
h
2
2xh + h + 2h
= lim
h→0
h
= lim (2x + h + 2)
(c)
h→0
(d)
= 2x + 2 + 0
= 2x + 2
f (x) = 3x 2 − 4x + 1
f (x + h) = 3(x + h) 2 − 4(x + h) + 1
= 3[x 2 + 2xh + h 2 ] − 4x − 4h + 1
= 3x 2 + 6xh + 3h 2 − 4x − 4h + 1
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
2
3x + 6xh + 3h 2 − 4x − 4h + 1 − ( 3x 2 − 4x + 1 )
= lim
h→0
h
6xh + 3h 2 − 4h
= lim
h→0
h
= lim (6x + 3h − 4)
h→0
(e)
= 6x + 0 − 4
= 6x − 4
f(x) = 5x2 +2
f(x + h) = 5(x + h)2 + 2
= 5x2 + 10xh + 5h2 + 2
 f (x + h) − f (x) 
f ′ (x) = lim 

h→0 
h
5x 2 + 10xh + 5h 2 + 2 − 5x 2 − 2
h→0
h
2
10xh + 5h
= lim
h→0
h
= lim (10x + 5h)
= lim
h→0
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 6 of 39
(f)
= 10x +5(0)
= 10x
f (x) = x 3 + 3x + 1
f (x + h) = (x + h)3 + 3(x + h) + 1
= x 3 + 3x 2 h + 3xh 2 + h 3 + 3x + 3h + 1
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
3
2
x + 3x h + 3xh 2 + h 3 + 3x + 3h + 1 − x 3 − 3x − 1
= lim
h→0
h
2
2
3
3x h + 3xh + h + 3h
= lim
h→0
h
= lim (3x 2 + 3xh + h 2 + 3)
h→0
(g)
= 3x 2 + 3x(0) + (0) 2 + 3
= 3x2 + 3
1
f (x) =
(x + 2)3
1
(x + h + 2)3
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
1
1
−
(x + h + 2)3 (x + 2)3
= lim
h→0
h
3
(x + 2) − (x + 2 + h)3
= lim
h → 0 h(x + 2)3 (x + 2 + h)3
f(x + h) =
(x + 2)3 − (x + 2)3 − 3(x + 2) 2 h − 3(x + 2)h 2 − h 3
h→0
h(x + 2)3 (x + 2 + h)3
= lim
h [ −3(x + 2) 2 − 3(x + 2)h − h 2 ]
h→0
h (x + 2)3 (x + 2 + h)3
= lim
−3(x + 2) 2 − 3(x + 2)h − h 2
h→0
(x + 2)3 (x + 2 + h)3
= lim
=
2
(a)
(b)
(c)
(d)
(e)
−3(x + 2) 2 − 3(x + 2)(0) − 02 −3(x + 2) 2
−3
=
=
3
3
6
(x + 2) (x + 2 + 0)
(x + 2)
(x + 2) 4
d
[4x 3 + 5x − 6]= 12x 2 + 5
dx
d 5
[x − 3x 2 + 2] = 5x 4 − 6x
dx
d 5
[x + 7x 3 + 2x + 4] = 5x 4 + 21x 2 + 2
dx
d 
3 d
3
4x +  =
[4x + 3x −1 ] =
4− 2

dx 
x  dx
x
d  2 7 d
21
6x − 3  = [6x 2 − 7x −3 ] =12x + 4

dx 
x  dx
x
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 7 of 39
(f)
(g)
(h)
(i)
(j)
(k)
(l)
3
(a)
(b)
(c)
(d)
d 4
3 1 d
−12 6 1
+ 2 −  =[4x −3 + 3x −2 − x −1 ] =
− 12x −4 − 6x −3 + x −2 =
− +
3

dx  x
x
x  dx
x 4 x3 x 2
1
d 
d 
3 1
3
4x + 3 x − 2  =
4x + 3x 2 − 2  = 4 + x − 2 = 4 +


dx
dx 
2
2 x
d 
1
1 −3
1
 d 
−1
− 4 =
6x +
6x + x 2 − 4  = 6 − x 2 = 6 −


dx 
2
x
2( x )3
 dx 
3
3
1
d  52
9 1 1 -1
2x − 3x 2 + x 2  =
5x 2 − x 2 + x 2

dx 
2
2
3
1
1
1
d 
 = d 6x 2 + 5x 2  = 9x 2 + 5 x - 2 = 9 x + 5
6x
x
+
5
x
 dx 

dx 
2
2 x
5
3
3
1
d  2
3  d  2
3
6x x −
=
6x − 3x 2  = 15x 2 + x - 2




dx 
2
x  dx
d 
3 d
−3
 4 + 3x −1  = 2
4+ =

dx 
x  dx
x
2
x + 3x
y=
x
=x+3
dy
=1
dx
4x 3 − 3x 2 + 2
y=
x4
4x 3 3x 2
2
= 4 − 4 + 4
x
x
x
−1
−2
= 4x − 3x + 2x −4
dy
= − 4x −2 + 6x −3 − 8x −5
dx
−4 6
8
= 2 + 3− 5
x
x
x
3
x + 6x
y=
x2
6
=x+
x
dy
6
=1− 2
dx
x
6x + 1
y=
x
= 6x 2 + x − 2
1
dy
1 3
= 3x − 2 − x − 2
dx
2
3x x + 2
(e)
y=
x
1
1
= 3x + 2x − 2
3
dy
= 3 − x− 2
dx
1
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 8 of 39
(c)
7x 2 − 4x + 5
2x
7
5
=
x − 2 + x −1
2
2
dy 7 5 −2
=
− x
dx 2 2
7
5
=
− 2
2 2x
y = (x + 2) (7x − 1)
= 7x2 + 13x − 2
dy
= 14x + 13
dx
y = (3x + 4) (2x + 7)
= 6x2 + 29x + 28
dy
= 12x + 29
dx
y = x − 1 (x + 2)
(d)
= x 2 + 2x 2 − x − 2
dy 3 12
−1
= x + x 2 −1
dx 2
y=
( 6x + 2 ) x + 3
(f)
4
(a)
(b)
y=
(
)
3
1
(
3
2
)
1
2
= 6x + 18x + 2x + 6
1
dy
−1
= 9x 2 + 18 + x 2
dx
1
= 9 x + 18 +
x
(4x − 1) (x + 3)
(e)
y=
x
2
4x + 12x − x − 3
=
x
= 4x + 11 − 3x −1
dy
3
= 4 + 3x −2 = 4 + 2
dx
x
1
(4x − 2)x 2
(f)
y=
x
1
−1
= 4x 2 − 2x 2
dy
−1
−3
= 2x 2 + x 2
dx
(g) =
y
(3 x + 4x ) x
3
2
3
5
= 3x 2 + 4x 2
3
dy 9 12
= x + 10x 2
dx 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 9 of 39
(h)
3x − 5
x
y=
−1
1
5
(a)
(b)
(c)
=
y 3x 2 − 5x 2
dy 3 − 12 5 − 32
= x + x
dx 2
2
y = 3x3 − 4x2 + 2
dy
= 9x 2 − 8x
dx
dy
x = 1,
= 9 −8 =1
dx
y = 4x + 3x2
dy
= 4 + 6x
dx
dy
x = 1,
= 10
dx
y =(2x + 1) (x − 2)
= 2x 2 − 3x − 2
dy
= 4x − 3
dx
dy
x = 2,
=8−3=5
dx
(d)=
y
x (x + 4)
3
6
7
1
= x 2 + 4x 2
dy 3 12
−1
=
x + 2x 2
dx 2
dy 3
2
1
x = 16,
=
(4) + = 6
dx 2
4
2
y = 4x2 + x − 2
dy
= 8x + 1
dx
dy
= 17
dx
8x + 1 = 17
x=2
y = 4(2)2 + 2 − 2 = 16
(2, 16)
y = 4x3 − 3x2 + 2x + 1
dy
= 12x 2 − 6x + 2
dx
dy
= 8 ⇒ 12x 2 − 6x + 2 = 8
dx
12x2 − 6x − 6 = 0
2x2 − x − 1 = 0
(2x + 1) (x − 1) = 0
1
x = − ,1
2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 10 of 39
3
8
9
10
2
1
 1
 1
 1
x=
− , y=
4  −  − 3 −  + 2  −  + 1
 2
 2
 2
2
1 3
=− − − 1 + 1
2 4
1
= −1
4
x = 1, y = 4 − 3 + 2 + 1 = 4
 1 5
 − , −  and (1, 4)
 2 4
y = ax3 + bx
dy
= 3ax 2 + b
dx
When x = 1, y = 2
⇒2=a+b
[1]
dy
[2]
x = 1,
=
4⇒3a + b =
4
dx
[2] – [1] ⇒ 2a = 2 , a = 1, b = 1
a
y = 2 + bx
x
a
[1]
x = 2, y =
− 13 ⇒ + 2b =
− 13
4
dy
= − 2ax −3 + b
dx
dy
1
[2]
= −8⇒ −8 = − a + b
x = 2,
dx
4
[1] – [2] ⇒ 3b = −21,
b = −7
1
−8 =− a − 7
4
1
−1 =− a
4
a=4
a = 4, b = −7
p
y = + qx
x
1
[1]
x = 3, y = 13 ⇒ 13 = p + 3q
3
11
Gradient of the line =
3
dy 11
=
dx 3
dy
= − px −2 + q
dx
11
1
=
− p+q
3
9
1
[2]
11 = − p + 3q
3
[1] + [2] ⇒ 24 = 6q ⇒ q = 4
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 11 of 39
1
11 =
− p + 12
3
p=3
p = 3, q = 4
11
(a)
(b)
1
x
dy
1
=4− 2
dx
x
dy
x = 1,
= 4 −1= 3
dx
dy
1
=0⇒ 4 − 2 =0
dx
x
1
=4
x2
1
x2 =
4
1
x= ±
2
1
1 1
x = , y = 4  + = 2 + 2 = 4
2
 2  12
1
1
 1
x =− , y =4  −  +
=− 2 − 2 =− 4
2
 2  − 12
1  1

 , 4  − , − 4
2  2

y = 4x +
Try these 13.7
y = (6x + 2)7
dy
6
= 7(6) (6x + 2)=
42 (6x + 2)6
dx
3
(ii)
y= (3 − 2x) 4
dy 3
3
−1
−1
=
− (3 − 2x) 4
( −2) (3 − 2x) 4 =
dx 4
2
− 12
(b) =
y 4(3x − 2)
dy
1
−3
=
− (4)(3) (3x − 2) 2
dx
2
−3
=
− 6(3x − 2) 2
dy
−6 −3
−3
when x =
6,
=
− 6 (16) 2 =
=
dx
64 32
(a)
(i)
Try these 13.8
(a)
y = x2 (3x + 2)5
u = x2, v = (3x + 2)5
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 12 of 39
du
dv
= 2x,
= 15(3x + 2) 4
dx
dx
dy
= x 2 [15(3x + 2) 4 ]+ (3x + 2)5 (2x)
dx
= x(3x + 2) 4 [15x + 2(3x + 2)]
= x(3x + 2) 4 (21x + 4)
(b)
1
y=
(4x + 1) (6x − 2) 2
1
u =4x + 1, v =(6x − 2) 2
du
dv 1
−1
= 4, =
(6x − 2) 2 (6)
dx
dx 2
3
=
6x − 2
dy 3(4x + 1)
=
+ 4 6x − 2
dx
6x − 2
=
3(4x + 1) + 4(6x − 2)
6x − 2
36x − 5
6x − 2
y = (4x2 + 6x + 1)5 (6x + 2)
u = (4x2 + 6x + 1) 5, v = 6x + 2
du
dv
= 5(8x + 6) (4x 2 + 6x + 1) 4 , = 6
dx
dx
dy
= 6(4x 2 + 6x + 1)5 + (6x + 2) [5(8x + 6)] (4x 2 + 6x + 1) 4
dx
= (4x 2 + 6x + 1) 4 [6(4x 2 + 6x + 1) + 5(6x + 2) (8x + 6)]
=
(c)
= (4x 2 + 6x + 1) 4 [24x 2 + 36x + 6 + 240x 2 + 260x + 60]
= (4x 2 + 6x + 1) 4 (264x 2 + 296x + 66)
Try these 13.9
(a)
(i)
6x + 2
x2 + 1
u = 6x + 2, v = x2 + 1
du
dv
= 6,
= 2x
dx
dx
dy (x 2 + 1) (6) − (6x + 2) (2x)
=
dx
(x 2 + 1) 2
y=
=
6x 2 + 6 − 12x 2 − 4x
(x 2 + 1) 2
=
−6x 2 − 4x + 6
(x 2 + 1) 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 13 of 39
3x 2 − 2
x3 + 4
u = 3x2 − 2, v = x3 + 4
du
dv
= 6x,
= 3x 2
dx
dx
dy (x 3 + 4) (6x) − (3x 2 − 2)(3x 2 )
=
dx
(x 3 + 4) 2
(ii)
(b)
y=
y=
=
6x 4 + 24x − 9x 4 + 6x 2
(x 3 + 4) 2
=
−3x 4 + 6x 2 + 24x
(x 3 + 4) 2
x +1
1
(x − 2) 2
1
u = x + 1, =
v (x − 2) 2
du
dv 1
−1
= 1,
= (x − 2) 2
dx
dx 2
1
1
−1
(x − 2) 2 − (x + 1) (x − 2) 2
dy
2
=
1
2 2
dx
[(x − 2) ]
1
1
−1
9 2 − (12) (9) 2
dy
3−2 1
2
x 11,=
=
= =
dx
9
9
9
Try these 13.10
y = cos x
dy
cos(x + h) − cos x
= lim
h
→
0
dx
h
cos x cos h − sin x sin h − cos x
= lim
h →0
h
cos(cos h − 1) − sin x sin h
= lim
h →0
h
cos h − 1
sin h
= cos x lim
− sin x lim
h →0
h →0
h
h
= cos x (0) – sin x (1)
= – sin x
Try these 13.11
cos x
sin x
u = cos x, v = sin x
du
dv
= − sin x,
= cos x
dx
dx
(a) =
y cot
=
x
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 14 of 39
dy sin x( − sin x) − cos x(cosx)
=
dx
sin 2 x
−[sin 2 x + cos 2 x]
=
sin 2 x
1
= −
sin 2 x
= − cosec2 x
(b)
=
y cosec
=
x (sin x) −1
dy
= − (sin x) −2 (cos x)
dx
cos x
= − 2
sin x
cos x
1
=
−
×
sin x sin x
= − cosec x cot x
Try these 13.12
(a)
y = sin 4x
dy
= 4 cos 4x
dx
(b) y = cos(4x + π)
dy
= − 4sin(4x + π )
dx
π

(c)
y = x 2 cos  3x − 
2

π

2
=
u x=
, v cos  3x − 

2
du π dv


= 2x,
= − 3sin  3x − 
dx
dx
2

dyπ
π



= − 3x 2 sin  3x −  + 2x cos  3x − 
dx
2
2


(d) y = x cot x
dv
u = x,
= cot x
dx
du
= 1, v = − cosec 2 x
dx
dy
= − x cosec 2 x + cot x
dx
(e) y = x2 sec x
u = x2, v = sec x
du
dv
= 2x,
= sec x tan x
dx
dx
dy
= x 2sec x tan x + 2x sec x
dx
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 15 of 39
Exercise 13B
1
y = (6x + 1)5
dy
= 30(6x + 1) 4
dx
(b) y = (4 − 3x)6
dy
=
− 18 (4 − 3x)5
dx
1
(c) =
y (2x + 9) 2
dy 1
−1
=
(2)(2x + 9) 2
dx 2
1
=
2x + 9
(a)
(d) =
y
4
−1
= 4(5x + 1) 2
5x + 1
dy
 1
−3
=
4  −  (5)(5x + 1) 2
dx
 2
−3
=
− 10(5x + 1) 2
(e) y = (6x2 + 3x + 1)4
dy
= 4(12x + 3) (6x 2 + 3x + 1)3
dx
1
(f) =
y (6x 3 + 5x) 4
dy 1
−3
= (18x 2 + 5) (6x 3 + 5x) 4
dx 4
−1
(g) =
y 3(7x 2 − 5) 2
dy
 1
−3
= 3  −  (14x) (7x 2 − 5) 2
dx
 2
−3
=
− 21x(7x 2 − 5) 2
(h)
1
y = (4x 2 + 5)10
1
dy
-1
= 10(2x 2 ) (4x 2 + 5)9
dx
20(4 x + 5)9
=
x
1
(i) =
y (7x 2 + 3x 5 )7
1
dy
 7 −1

=
7  x 2 + 15x 4  (7x 2 + 3x 5 )6
dx
2

(j)
2
(a)
3
y = (x 3 − 5x + 2) 4
dy 3
−1
=
(3x 2 − 5) (x 3 − 5x + 2) 4
dx 4
θ = (3t + 1)(4t + 2)8
u = 3t + 1, v = (4t + 2)8
du
dv
= 3,
= 32 (4t + 2)7
dt
dt
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 16 of 39
dθ
= 32 (3t + 1) (4t + 2)7 + 3 (4t + 2)8
dt
= (4t + 2)7 [32 (3t + 1) + 3(4t + 2)]
= (4t + 2)7 [96t + 32 + 12t + 6]
= (4t + 2)7 (108t + 38)
(b)
θ = t 2 (7t + 1)3
u = t2, v = (7t + 1)3
du
dv
= 2t,
= 21(7t + 1) 2
dt
dt
dθ
= 21t 2 (7t + 1) 2 + 2t(7t + 1)3
dt
= t(7t + 1) 2 [21t + 2(7t + 1)]
= t(7t + 1) 2 (35t + 2)
(c)
1
θ = 4t 2t − 1 = 4t(2t − 1) 2
1
dθ
1
−1
= 4(2t − 1) 2 + (4t)   (2) (2t − 1) 2
dt
2
4t
= 4 2t − 1 +
2t − 1
4(2t − 1) + 4t
2t − 1
12t − 4
=
2t − 1
=
1
(d) =
θ 6t 2 (t 3 + 2t) 2
1
− 12
dθ
1
2
= 12t ( t 3 + 2t ) + ( 6t 2 )   ( 3t 2 + 2 )( t 3 + 2t )
dt
2
2
2
3t (3t + 2)
= 12t t 3 + 2t +
t 3 + 2t
=
=
(e)
12t(t 3 + 2t) + 3t 2 (3t 2 + 2)
t 3 + 2t
21t 4 + 30t 2
t 3 + 2t
=
3t 2 (7t 2 + 10)
t 3 + 2t
=
θ (5t 2 + 2)(7 − 3t) 4
dθ
= 10t (7 − 3t) 4 + ( − 12) (5t 2 + 2)(7 − 3t)3
dt
=(7 − 3t)3 [10t (7 − 3t) − 12(5t 2 + 2)]
= (7 − 3t)3 [ − 90t 2 + 70t − 24]
1
(f) =
θ t 3 (4t 3 + 3t 2 + 1) 4
1
dθ
1
−3
= 3t 2 (4t 3 + 3t 2 + 1) 4 + (t 3 )   (12t 2 + 6t)(4t 3 + 3t 2 + 1) 4
dt
4
1
3t 2 (4t 3 + 3t 2 + 1) 4 +
Unit 1 Answers: Chapter 13
3
−
1 4
t (12t + 6)(4t 3 + 3t 2 + 1) 4
4
© Macmillan Publishers Limited 2013
Page 17 of 39
(g)
θ = (7t 2 + 3t) (3t 2 + 1) − 2
1
1
1
dθ  7 − 12

 1
−1
−3
=  t + 3  (3t 2 + 1) 2 + (7t 2 + 3t)  −  (6t) (3t 2 + 1) 2
dt  2

 2
1
 7 -1

-1
-3
=  t 2 + 3  (3t 2 + 1) 2 − 3t (7t 2 + 3t)(3t 2 + 1) 2
2

(h)
3
(a)
1
3
=
θ t 2 (t + 5) 4
3
dθ 1 - 12
3 1
−1
= t (t + 5) 4 + t 2 (t + 5) 4
dt 2
4
3
1
3
−1
=
(t + 5) 4 +
t (t + 5) 4
4
2 t
4x + 2
y=
x −1
dy (x − 1)(4) − (4x + 2)
=
dx
(x − 1) 2
=
4x − 4 − 4x − 2
(x − 1) 2
−6
(x − 1) 2
3x − 5
y=
6x + 2
dy (6x + 2)(3) − (3x − 5)(6)
=
dx
(6x + 2) 2
18x + 6 − 18x + 30
=
(6x + 2) 2
=
(b)
=
(c)
36
(6x + 2) 2
y=
x2 + 1
2x + 5
dy (2x + 5) (2x) − (x 2 + 1) (2)
=
dx
(2x + 5) 2
=
4x 2 + 10x − 2x 2 − 2
(2x + 5) 2
2x 2 + 10x − 2
(2x + 5) 2
3x + 2
y=
4x − 1
=
(d)
1
−1
( 4x − 1) (3) − (3x + 2)   (4) (4x − 1) 2
dy
2
=
dx
( 4x − 1) 2
2(3x + 2)
3 4x − 1 −
4x − 1
=
4x − 1
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 18 of 39
(e)
=
3(4x − 1) − 6x − 4
(4x − 1) 4x − 1
=
6x − 7
(4x − 1) 4x − 1
y=
x 3 − 2x + 1
4x 2 + 5
dy (4x 2 + 5) (6x 2 − 2) − (x 3 − 2x + 1) (8x)
=
dx
(4x 2 + 5) 2
(f)
=
24x 4 − 8x 2 + 30x 2 − 10 − 8x 4 + 16x 2 − 8x
(4x 2 + 5) 2
=
16x 4 + 38x 2 − 8x − 10
(4x 2 + 5) 2
y=
x4 − 2
3x + 1
1
−1
3x + 1 (4x 3 ) − (x 4 − 2)   (3)(3x + 1) 2
dy
2
=
dx
( 3x + 1)2
3 4
(x − 2)
3
3
21 4
4x 3x + 1 − 2
4x 3 (3x + 1) − x 4 + 3
x + 4x 3 + 3
3x + 1
2
2
=
= =
3
3
3x + 1
(3x + 1) 2
(3x + 1) 2
x
(g)
y=
(x + 3)3
dy (x + 3)3 − x(3)(x + 3) 2
=
dx
(x + 3)6
(h)
=
(x + 3) 2 [x + 3 − 3x]
(x + 3)6
=
−2x + 3
(x + 3) 4
y=
x 4 + 2x
x2 − 7
dy (x 2 − 7) (4x 3 + 2) − (x 4 + 2x) (2x)
=
dx
(x 2 − 7)2
(i)
=
4x 5 + 2x 2 − 28x 3 − 14 − 2x 5 − 4x 2
(x 2 − 7) 2
=
2x 5 − 28x 3 − 2x 2 − 14
(x 2 − 7) 2
y=
(3x + 2) 2
(2x − 1)3
dy (2x − 1)3 (6)(3x + 2) − (3x + 2) 2 (6)(2x − 1) 2
=
dx
(2x − 1)6
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 19 of 39
6(2x − 1) 2 (3x + 2)[2x − 1 − (3x + 2)]
(2x − 1)6
6(3x + 2) ( − x − 3)
=
(2x − 1) 4
7x + 3
y=
5x − 1
=
(j)
dy (5x − 1) (7) − (7x + 3) (5)
=
dx
(5x − 1) 2
=
4
35x − 7 − 35 x − 15
−22
=
2
(5x − 1)
(5x − 1) 2
=
y (2x − 1)5
dy
= 10(2x − 1) 4
dx
dy
x = 1,
= 10
dx
(b) y = x2 (x + 1)3
dy
= 2x(x + 1)3 + 3x 2 (x + 1) 2
dx
dy
x = 0,
=0
dx
x2 + 1
(c)
y=
x+2
(a)
dy (x + 2) (2x) − (x 2 + 1) (1)
=
dx
(x + 2) 2
=
2x 2 + 4x − x 2 − 1
(x + 2) 2
x 2 + 4x − 1
(x + 2) 2
dy 4 + 8 − 1 11
x = 2,
=
=
dx
42
16
2
4
y = (x − 5x + 2)
dy
= 4(2x − 5) (x 2 − 5x + 2)3
dx
dy
x = 0,
= (4) ( − 5) (2)3 = − 160
dx
d
[sin 4x] = 4 cos 4x
dx
d
[sin 6x] = 6 cos 6x
dx
d
[cos 3x] = − 3 sin 3x
dx
d
[cos 7x] = − 7 sin 7x
dx
=
(d)
5
(a)
(b)
(c)
(d)
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 20 of 39
(e)
(f)
d
[cos 9x] = − 9 sin 9x
dx
dπ 
 π


sin  3x +   = 3 cos  3x + 

dx  
4 
4

dπ 
π 


sin  − 4x  = − 4 cos  − 4x 
dx
2
2




d π 3
3π



(h)
sin 
+ 4x  = 4 cos 
+ 4x 
dx
2
2




d
(i)
sin(8x =
+ 2) 8 cos(8x + 2)
dx
d
(j)
cos(3x − π) = − 3 sin (3x − π )
dx
d
(k)
cos (5 − 7x)
= 7 sin (5 − 7x)
dx
d
(l)
cos(2π − 9x) = 9 sin (2π − 9x)
dx
d
(a)
tan 2x = 2 sec 2 2x
dx
d
(b)
tan 5x = 5sec 2 (5x)
dx
d
(c)
tan (2x
=
+ π) 2 sec 2 (2x + π)
dx
dπ 
π 


(d)
tan  3x −  = 3 sec 2  3x − 
dx
2
2


d
(e)
sec(4x) = 4 sec(4x) tan(4x)
dx
d
(f)
sec(4x=
+ 3) 4 sec(4x + 3) tan (4x + 3)
dx
d
2
(g)
cot(6xπ)
−=
6−cosec (6x
π)
−
dt
d π 3 3
3
π 


(h)
cot  x −  = − cosec 2  x − 
dx
4
4
4
4
4
dπ
π 

 π



(i)
cosec  x −  = − cosec  x −  cot  x − 
dx
4
4
4



d
(j)
cosec(7x − 4) = − 7 cosec(7x − 4) cot(7x − 4)
dx
(a) y = sin x2
dy
= 2x cos x 2
dx
(b) =
θ sin(t 2 + 3)
dθ
= 2t cos(t 2 + 3)
dt
(c)=
θ cos(4x 2 + π)
dθ
= − 8x sin(4x 2 + π)
dx
(g)
6
7
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 21 of 39
(d)
=
y tan(7x 3 − 8)
dy
= 21x 2 sec 2 (7x 3 − 8)
dx
(e) =
v 8 tan(3x3 − 4x + 5)
dv
= 8(9x 2 − 4) sec 2 (3x 3 − 4x + 5)
dx
dy
(f)
y=
(7x + 5)10 ,
=
70(7x + 5)9
dx
dy
−12
(g)
y = (4x + 2) −3 ,
=
− 12(4x + 2) −4 =
dx
(4x + 2) 4
(h)
y= 8(7x 2 + 5x + 1) −6
dy
= − 48(14x + 5) (7x 2 + 5x + 1) −7
dx
1
(i)
1 2

=
x  t 3 − t −2 
4 

−1
dx 1  2 1 −3   3 1 −2  2
= 3t + t   t − t 
dt 2 
2 
4 
1 
1
3
=  t2 + 3 
4t  3 1
2
t − 2
4t
(j)
y = tan (5x + 1)6
dy
= 30 (5x + 1)5sec 2 (5x + 1)6
dx
1

(k)
=
y 5 cos  6x 2 + 

x
(l)
8
(a)
(b)
(c)
(d)
(e)
dy
1  
1

= − 5 12x − 2  sin  6x 2 + 
dx
x  
x

y = sec (x3 + 5)
dy
= 3x 2sec(x 3 + 5) tan(x 3 + 5)
dx
y = x sin x
dy
= sin x + x cos x
dx
y = x cos x
dy
= cos x - x sin x
dx
y = x2 tan x
dy
= 2x tan x + x 2sec 2 x
dx
y = x3 tan(3x + 2)
dy
= 3x 2 tan(3x + 2) + 3x 3sec 2 (3x + 2)
dx
y = (4x + 1) sin 4x
dy
= 4 sin 4x + 4(4x + 1)cos 4x
dx
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 22 of 39
(f)
(g)
(h)
(i)
(j)
(k)
9
(a)
y = x tan(x2)
dy
= tan(x 2 ) + 2x 2 sec 2 (x 2 )
dx
y = sin 2x tan 2x
dy
= 2 sin 2x sec 2 2x + 2 cos 2x tan 2x
dx
= 2 sin 2x sec 2 2x + 2 sin 2x = 2 sin 2x(sec 2 (2x) + 1)
y = (3x2 + 1) cos x
dy
= 6x cos x − (3x 2 + 1)sin x
dx
1

y =  x 3 − 2x  sec x
3

dy
1

= (x 2 − 2)sec x +  x 3 − 2x  sec x tan x
dx
3

4

y =  + 2x  cot 4x
x

dy
4

 −4

= − 4  + 2x  cosec 2 (4x) +  2 + 2  cot(4x)
dx
x

x

 6

y =  2 − 3x + 2  cosecx
x

dy
 6

 −12

= −  2 − 3x + 2  cosec x cot x +  3 − 3  cosec x
dx
x

 x

2
x
y=
x+2
dy (x + 2) (2x) − x 2
=
dx
(x + 2) 2
=
(b)
x 2 + 4x
(x + 2) 2
y=
x3 + 4
x − 6x + 1
2
dy (x 2 − 6x + 1) (3x 2 ) − (x 3 + 4) (2x − 6)
=
dx
(x 2 − 6x + 1) 2
(c)
=
3x 4 − 18x 3 + 3x 2 − 2x 4 + 6x 3 − 8x + 24
(x 2 − 6x + 1) 2
=
x 4 − 12x 3 + 3x 2 − 8x + 24
(x 2 − 6x + 1) 2
sin x
cos x + 2
dy (cos x + 2)cos x − sin x( − sin x)
=
dx
(cos x + 2) 2
y=
=
cos 2 x + 2cos x + sin 2 x
(cos x + 2) 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 23 of 39
1 + 2 cos x
(cos x + 2) 2
tan x + 1
y=
sec x + 3
=
(d)
dy (sec x + 3) (sec 2 x) − (tan x + 1) (sec x tan x)
=
dx
(sec x + 3) 2
sec3 x + 3 sec 2 x − sec x tan 2 x − sec x tan x
(sec x + 3) 2
x cos x
y= 2
x +5
=
(e)
dy (x 2 + 5) (cos x − x sin x) − (x cos x) (2x)
=
dx
(x 2 + 5) 2
=
x 2 cos x − x 3sin x + 5 cos x − 5x sin x − 2 x 2 cos x
(x 2 + 5) 2
− x 2 cos x − x 3 sin x + 5cos x − 5x sin x
(x 2 + 5) 2
7x
y=
x +1
1
−1
( x + 1) (7) − (7x) (x + 1) 2
dy
2
=
dx
( x + 1) 2
=
(f)
7x
2 x +1
=
x +1
7
7
7(x + 1) − x
x+7
2
2
= =
3
3
(x + 1) 2
(x + 1) 2
x + sin x
(g)
y=
x + cosx
7 x +1 −
dy (x + cos x) (1 + cos x) − (x + sin x) (1 − sin x)
=
dx
(x + cos x) 2
=
(h)
x + x cos x + cos x + cos 2 x − x + x sin x − sin x + sin 2 x
(x + cos x) 2
1 + x cos x + cos x + x sin x − sin x
=
(x + cos x) 2
cos(3x + 2)
y=
4x 3 + 2
dy (4x 3 + 2) ( − 3 sin(3x + 2)) − 12x 2 cos(3x + 2)
=
dx
(4x 3 + 2) 2
=
−12x 3sin(3x + 2) − 6 sin(3x + 2) − 12x 2 cos(3x + 2)
(4x 3 + 2) 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 24 of 39
(i)
y=
7x 2 + 2
x−4
dy (x − 4) (14x) − (7x 2 + 2)
=
dx
(x − 4) 2
(j)
=
14x 2 − 56x − 7x 2 − 2
(x − 4) 2
=
7x 2 − 56x − 2
(x − 4) 2
y=
x4
x2 − 1
dy (x 2 − 1) (4x 3 ) − x 4 (2x)
=
dx
(x 2 − 1) 2
=
4x 5 − 4x 3 − 2x 5
(x 2 − 1) 2
2x 5 − 4x 3 2x 3 (x 2 − 2)
=
(x 2 − 1) 2
(x 2 − 1) 2
(a) y = cos3x
dy
= − 3 cos 2 x sin x
dx
(b) y = sin3x
dy
= 3 sin 2 x cos x
dx
(c) y = tan4 x
dy
= 4 tan 3 x sec 2 x
dx
(d) y = x2 cos4 x
dy
= 2x cos 4 x − 4x 2 cos3 x sin x
dx
=2x cos3x [cos x − 2 x sin x]
(e) y = sec3(x + 2)
dy
= 3 sec 2 (x + 2) sec(x + 2) tan(x + 2)
dx
= 3 sec3 (x + 2) tan (x + 2)
(f)
y = tan2(3x + 2)
dy
=
6 tan(3x + 2) sec 2 (3x + 2)
dx
(g) y = (2x + 1) cot2 x
dy
= 2 cot 2 x − 2(2x + 1)cot x cosec 2 x
dx
(h)
y=
(4x 2 − x) sin 2 (x + 2)
dy
= (8x -1) sin 2 (x + 2) + 2(4x 2 − x) sin(x + 2) cos(x + 2)
dx
x
(i)
y=
cos ec 2 x + 1
=
10
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 25 of 39
dy (cosec 2 x + 1) + x (2 cosec x cosec x cot x)
=
dx
(cosec 2 x + 1) 2
cosec 2 x + 1 + 2x cosec 2 x cot x
(cosec 2 x + 1) 2
4
y = tan (3x)
dy
= 12 tan 3 (3x) sec 2 (3x)
dx
=
(j)
Try these 13.13
(a)
y = (2x + 1) cos x2
u = 2x + 1, v = cos x2
du
dv
= 2,
= − 2 x sin x 2
dx
dx
dy
= − 2x(2x + 1) sin x 2 + 2 cos x 2
dx
= (−4x 2 + 4x) sin x 2 + 2 cos x 2
d2 y
= (−8x − 2) sin x 2 + (−4x 2 − 2x)(2x)cos x 2 + 2(2x)( − sin x 2 )
2
dx
= – 2(6x + 1)sin x2 – 4x2(2x + 1)cos x2
x+2
(b)
y=
2x − 1
u = x + 2, v = 2x − 1
du
dv
= 1,= 2.
dx
dx
dy (2x − 1) (1) − (x + 2) (2)
=
dx
(2x − 1) 2
−5
= 2=
− 5(2x − 1) −2
(2x − 1)
d2 y
=−
( 5) ( −2) (2) (2x − 1) −3
dx 2
20
=
(2x − 1)3
x = 0,
d2 y
20
=
= − 20
2
dx
(−1)3
Exercise 13 C
1
=
y
( 4x + 7 )
1
2
dy 1
−1
=
(4) (4x + 7) 2
dx 2
−1
= 2 (4x + 7) 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 26 of 39
d2 y
1
−3
=
− (2) (4) (4x + 7) 2
2
dx
2
−4
−4
= =
3
2
(4x + 7)
(4x + 7)3
2
y = (2x + 3)10
dy
= 20(2x + 3)9
dx
d2 y
= 360(2x + 3)8
dx 2
1
3
=
= (5x − 3) −3
y
(5x − 3)3
dy
=
− 15(5x − 3) −4
(−3) (5x − 3) −4 (5) =
dx
d2 y
=
( −15) ( −4) (5) (5x − 3) −5
dx 2
300
=
(5x − 3)5
4
y = (x + 2) sin x
dy
= sin x + (x + 2) cos x
dx
d2 y
= cos x + cos x + (x + 2) ( − sin x)
dx 2
= 2 cos x − (x + 2) sin x
5
y = cos(x2)
dy
= − 2x sin(x 2 )
dx
d2 y
= − 2 sin(x 2 ) + ( − 2x) (2x) cos(x 2 )
dx 2
= −2 sin x2 − 4x2 cos(x2)
x2 + 2
6
y=
x +1
dy (x + 1) (2x) − (x 2 + 2)
=
dx
(x + 1) 2
=
2x 2 + 2x − x 2 − 2
(x + 1) 2
=
x 2 + 2x − 2
(x + 1) 2
d 2 y (x + 1) 2 (2x + 2) − (x 2 + 2x − 2) (2) (x + 1)
=
dx 2
(x + 1) 2
=
(x + 1) [2(x + 1) − 2(x 2 + 2x − 2)]
(x + 1) 4
−2x 2 − 2x + 6
(x + 1)3
y = sin(2x2 + 5x + 1)
=
7
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 27 of 39
8
dy
= (4x + 5) cos(2x 2 + 5x + 1)
dx
d2 y
= (4x + 5)[ − (4x + 5) sin(2x 2 + 5x + 1)] + 4 cos(2x 2 + 5x + 1)
2
dx
= − (4x + 5) 2 sin(2x 2 + 5x + 1) + 4 cos(2x 2 + 5x + 1)
y = x3 cos2 x
dy
= 3x 2 cos 2 x − 2x 3sin x cos x
dx
= 3x 2 cos 2 x − x 3sin 2x
d2 y
= 6x cos 2 x − 6x 2 cos x sin x − 3x 2sin 2x − 2x 3cos 2x
dx 2
= 6x cos 2 x − 3x 2 sin 2x − 3x 2 sin 2x − 2x 3cos 2x
9
10
= 6x cos 2 x − 6x 2sin 2x − 2x 3cos 2x
y = cos 3x + sin 4x
dy
= − 3 sin 3x + 4 cos 4x
dx
d2 y
= − 9 cos 3x − 16 sin 4x
dx 2
3x − 2
y=
x−4
dy (x − 4) (3) − (3x − 2)
=
dx
(x − 4) 2
3x − 12 − 3x + 2
=
(x − 4) 2
=
−10
(x − 4) 2
d2 y
20
=
2
dx
(x − 4)3
3 cos 2x
11
y=
x
dy −3
6
= 2 cos 2x − sin 2x
dx x
x
2
d y 6
6
6
12
= 3 cos 2x + 2 sin 2x + 2 sin 2x − cos 2x
2
dx
x
x
x
x
6
12
12
= 3 cos 2x + 2 sin 2x − cos 2x
x
x
x
d2 y
dy
+2
+ 4xy
2
dx
dx
6
12
6
12
= 2 cos 2x +
sin 2x − 12cos 2x − 2 cos 2x −
sin 2x + 12cos 2x
x
x
x
x
=0
x
12
1
=
y (x 2 + 1) 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 28 of 39
dy 1
−1
=
(2x) (x 2 +=
1) 2
dx 2
x
x2 + 1
d2y
 1
−1
−3
=(x 2 + 1) 2 + (x)  −  (2x) (x 2 + 1) 2
2
dx
 2
1
x2
−1
−3
= (x 2 + 1) 2 − x 2 (x 2 + 1) 2 =
− 2
x 2 + 1 (x + 1) x + 1
x2 + 1 − x2
(x 2 + 1) x + 1
1
= 2
(x + 1) x + 1
4
=
y 2x 2 +
x
dy
4
= 4x − 2
dx
x
2
d y
8
=4+ 3
2
dx
x
2
d
y
8
x 2 2 = 4x 2 +
dx
x
2
d y
x 2 2 = 2y
dx
x+2
y=
x −1
dy (x − 1) − (x + 2)
=
dx
(x − 1) 2
−3
=
(x − 1) 2
=
13
14
d2 y
6
=
2
dx
(x − 1)3
2
15
16
d2 y
 dy 
3 2 − 2   (x − 1)
 dx 
dx
18
9(x − 1)
18
18
=
−2
=
−
=0
3
4
3
(x − 1)
(x − 1)
(x − 1)
(x − 1)3
y = cos x + sin x
dy
=
− sin x + cos x
dx
d2 y
=
− cos x − sin x
dx 2
d2 y
= −y
dx 2
d2 y
+y=0
dx 2
y = x2 cos2 x
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 29 of 39
dy
= 2x cos 2 x − 2x 2 sin x cos x
dx
= 2x cos 2 x − x 2sin 2x
17
d2 y
=
2 cos 2 x − 4x sin x cos x − 2x 2 cos 2x − 2x sin 2x
dx 2
2
2
π d2 y
π π
x = ,=
2
=
 
2 dx 2
2
2
2
x +1
y= 2
x −1
dy (x 2 − 1) (2x) + (x 2 + 1) (2x)
=
dx
(x 2 − 1) 2
=
2x 3 − 2x + 2x 3 + 2x
(x 2 − 1) 2
=
4x 3
(x 2 − 1) 2
d 2 y (x 2 − 1) 2 (12x 2 ) − (4x 3 ) (2) (2x) (x 2 − 1)
=
dx 2
(x 2 − 1) 4
=
(x 2 − 1)[12x 2 (x 2 − 1) − 16x 4 ]
(x 2 − 1) 4
=
12x 4 − 12x 2 − 16x 4
(x 2 − 1)3
=
−4x 4 − 12x 2
(x 2 − 1)3
When x = 0,
18
d2 y 0
= = 0
dx 2 −1
2x
x+4
dy (x + 4) (2) − 2x
=
dx
(x + 4) 2
2x + 8 − 2x
8
=
=
= 8(x + 4) −2
2
(x + 4)
(x + 4) 2
y=
d2 y
−16
=
− 16(x + 4) −3 = 3
2
dx
(x + 4)
(x + 4)
d2 y
dy −16(x + 4)
16
+ 2=
+
2
3
dx
dx
(x + 4)
(x + 4) 2
−16
16
+
=0
2
(x + 4)
(x + 4) 2
1
=
= (x 2 −x) −1
y
2
x −x
dy
=
− (2x − 1) (x 2 − x) −2
dx
=
19
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 30 of 39
=
1 − 2x
(x 2 − x) 2
d 2 y (x 2 − x) 2 ( − 2) − (1 − 2x) (2) (2x − 1) (x 2 − x)
=
dx 2
(x 2 − x) 4
=
(x 2 − x) [(x 2 − x) ( − 2) + 2(2x − 1) 2 ]
(x 2 − x) 4
=
−2x 2 + 2x + 8x 2 − 8x + 2
(x 2 − x)3
=
6x 2 − 6x + 2
(x 2 − x)3
d 2 y 24 − 12 + 2
=
dx 2
8
14 7
= =
8 4
1
−1
20
y=
= (1 + x 2 ) 2
2
1+ x
x = 2,
dy
1
−3
−3
=
− (2x) (1 + x 2 ) 2 =
− x(1 + x 2 ) 2
dx
2
d2 y 3
−5
−3
=
x (2x) (1 + x 2 ) 2 + ( −1) (1 + x 2 ) 2
2
dx
2
−5
−3
2
= 3x (1 + x 2 ) 2 − (1 + x 2 ) 2
d2 y
dy
+ 3x
+y
2
dx
dx
1
−5
−3
−3
(1 + x 2 )[3x 2 (1 + x 2 ) 2 − (1 + x 2 ) 2 ] + 3x [− x(1 + x 2 ) 2 ] + (1 + x 2 ) − 2
=
(1 + x 2 )
−3
−1
−3
−1
= 3x 2 (1 + x 2 ) 2 − (1 + x 2 ) 2 − 3x 2 (1 + x 2 ) 2 + (1 + x 2 ) 2
=0
Review Exercise 13
1
f (x)= x +
1
x
1
x+h
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
1
1

x+h+
− x + 

x+h
x
= lim
h→0
h
1
1
−
h+
x+h x
= lim
h→0
h
f(x + h) = x + h +
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 31 of 39
x - (x + h)
h
(x + h) (x)
= lim + lim
h→0 h
h→0
h
−h


 (x + h) (x) 

= lim 1 + lim 
h→0
h→0 
h



1
= 1 − lim
h → 0 (x + h) (x)
1
= 1−
(x + 0)x
1
= 1− 2
x
2
f (x) = 2x 2 − 5x + 2
f(x + h) = 2(x + h) 2 − 5 (x + h) + 2
= 2[x 2 + 2xh + h 2 ] − 5x − 5h + 2
= 2x 2 + 4xh + 2h 2 − 5x − 5h + 2
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
2
( 2x + 4xh + 2h 2 − 5x − 5h + 2 ) − ( 2x 2 − 5x + 2 )
= lim
h→0
h
2
4xh + 2h − 5h
= lim
h→0
h
= lim (4x + 2h − 5) = 4x + 0 − 5
h→0
= 4x − 5
3
f (x) = x 3 − 2x + 1
f (x + h) = (x + h)3 − 2(x + h) + 1
= x 3 + 3x 2 h + 3xh 2 + h 3 − 2x − 2h + 1
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
3
2
x + 3x h + 3xh 2 + h 3 − 2x − 2h + 1 − ( x 3 − 2x + 1 )
= lim
h→0
h
3x 2 h + 3xh 2 + h 3 − 2h
= lim
h→0
h
2
= lim (3x + 3xh + h 2 − 2)
h→0
= 3x 2 + 3x(0) + 02 − 2
= 3x 2 − 2
4
f (x) =
1
2x + 3
f(x + h) =
1
2(x + h) + 3
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 32 of 39
f (x + h) − f (x)
h→0
h
1
1
−
2(x + h) + 3 2x + 3
= lim
h→0
h
2x + 3 − (2x + 2h + 3)
= lim
h → 0 h(2(x + h) + 3) (2x + 3)
−2h
= lim
h → 0 h(2(x + h) + 3) (2x + 3)
−2
= lim
h → 0 (2(x + h) + 3) (2x + 3)
−2
−2
=
(2(x + 0) + 3) (2x + 3) (2x + 3) 2
5
f(x) = sin 2x
f (x + =
h) sin 2(x + h)
= sin(2x + 2h)
sin(2x + 2h) − sin 2x
f ′ (x) = lim
h→0
h
sin 2x cos 2h + cos 2x sin 2h − sin 2x
= lim
h→0
h
 sin 2x cos 2h − sin 2x cos 2x sin 2h 
+
lim

h→0 

h
h
 cos 2h − 1
 sin 2h 
lim sin 2x 
cos 2x 
 + hlim
h→0
→
0

 h 
h
 2 cos 2h − 1
2  sin 2h 
sin 2x lim 
 + cos 2x hlim
h→0 
→0 
 2h 
2h
= (sin 2x) (2) (0) + (cos 2x) (2) (1)
= 2 cos 2x
OR
f (x + h) − f (x)
= sin(2x + 2h) − sin 2x
f ′ (x) = lim
 2x + 2h + 2x 
 2x + 2h − 2x 
= 2 cos 
 sin 


2
2
= 2 cos (2x + h) sin h.
2 cos(2x + h) sin h
∴ f ′ (x) =
lim
h→0
h
sin h
= 2 lim cos(2x + h) lim
h→0
h→0 h
= 2(cos 2x) (1)
= 2 cos 2x
6
f (x) = cos 2x
f (x + =
h) cos(2x + 2h)
f (x + h) − f (x)
f ′ (x) = lim
h→0
h
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 33 of 39
cos(2x + 2h) − cos 2x
h→0
h
cos 2x cos 2h − sin 2x sin 2h − cos 2x
= lim
h→0
h
 cos 2x cos 2 h − cos 2x sin 2x sin 2h 
lim
−

h→0 
h
h

sin 2h
 cos 2h − 1
lim 2 cos 2x 
− lim (2 sin 2x)

h→0
h
→
0
2h
2h


= (2 cos 2x) (0) − (2 sin 2x) (1)
= − 2sin 2x
5x − 2
7
y=
3x + 7
dy (3x + 7) (5) − (5x − 2) (3)
=
dx
(3x + 7) 2
15x + 35 − 15x + 6
=
(3x + 7) 2
41
=
= 41(3x + 7) −2
(3x + 7) 2
= lim
d2 y
=
(41) ( −2) (3) (3x + 7) −3
dx 2
−246
=
(3x + 7)3
8=
y A cos5x + B sin 5x
dy
=
− 5 A sin 5x + 5B cos5x
dx
d2 y
= − 25A cos 5x − 25 B sin 5x
dx 2
d2 y
=
− 25 [A cos5x + B sin 5x]
dx 2
d2 y
= − 25y
dx 2
d2 y
+ 25y = 0
dx 2
d 2
9
(a)
(x sin 5x) = 5x 2 cos 5x + 2x sin 5x
dx
d
(b)
=
[cos 4 (4x)] 4 cos3 (4x) ( − sin 4x) (4)
dx
= − 16 cos3 4x sin 4x
(c)
10
(a)
2
 1 − 2x   (1 + 2x) ( −2) − (1 − 2x) (2) 
d  1 − 2x  

 = 2



dx  1 + 2x  
(1 + 2x) 2
 1 + 2x  



 1 − 2x   −4  −8(1 − 2x)
= 2
=

2 
3
 1 + 2x   (1 + 2x)  (1 + 2x)
y=
(2x + 1) tan(4x + 5)
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 34 of 39
(b)
(c)
(d)
dy
= (2x + 1)[4 sec 2 (4x + 5)] + 2 tan(4x + 5)
dx
= 4(2x + 1) sec 2 (4x + 5) + 2 tan (4x + 5)
=
y sin(2x + 3) + x tan 4x
dy
= 2 cos(2x + 3) + tan 4x + 4x sec 2 (4x)
dx
=
y sin 3 (x 2 + 4x + 2)
dy
= 3 sin 2 (x 2 + 4x + 2) cos(x 2 + 4x + 2) [2x + 4]
dx
= (6x + 12) sin 2 (x 2 + 4x +2) cos(x 3 + 4x + 2)
1 + cos 2θ
x=
sin 2θ
dx sin 2θ ( −2 sin 2θ) − (1 + cos 2θ) (2 cos 2θ)
=
dθ
(sin 2θ) 2
−2 sin 2 (2θ) − 2 cos 2θ − 2 cos 2 2θ
=
sin 2 2θ
2
−2[sin (2θ) + cos 2 2θ] − 2 cos 2θ
=
sin 2 2θ
−2 − 2 cos 2θ −2 (1 + cos 2θ)
−4 cos 2 θ
1
=2
=2
=2
=
−
=
− cosec 2 θ
2
2
sin 2θ
sin 2θ
4 sin θ cos θ
sin θ
t = tan 2 θ sin 2 θ
dt
= tan 2 θ [2 sin θ cos θ] + sin 2 θ(2 tan θ sec 2 θ)
dθ
sin 2 θ
= 2 tan 2 θ sin θ cos θ + 2
tan θ
cos 2 θ
= 2 tan 2 θ sin θ cos θ + 2 tan 3 θ
(e)
= tan 2 θ(2 sin θ cos θ + 2 tan θ)
= tan 2 θ(sin 2θ + 2 tan θ)
(f)
r = sin 2 x + 2 cos3 x
dr
= 2 sin x cos x + 3(2) cos 2 x( − sin x)
dx
= 2 sin x cos x − 6 sin x cos 2 x
= 2 sin x cos x[1 − 3 cos x] = sin 2x (1 − 3 cos x)
11
=
y x 3 cos(2x 2 + π)
dy
= x 3 [4x( − sin(2x 2 + π))] + 3x 2 cos(2x 2 + π)
dx
= − 4x 4 sin(2x 2 + π) + 3x 2 cos(2x 2 + π)
dy
x = 0,
=0
dx
12 =
y
4x − 3
1
=
y (4x − 3) 2
dy 1
−1
=
(4) (4x − 3) 2
dx 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 35 of 39
−1
= 2 (4x − 3) 2
d2 y
1
−3
=
− (2) (4) (4x − 3) 2
2
dx
2
−4
−4
= =
3
y3
4x − 3
(
y=
13
)
x
x 2 + 32
1
1
−1
(x 2 + 32) 2 (1) −   (2x) (x 2 + 32) 2 (x)
dy
2
=
2
dx
x 2 + 32
(
x 2 + 32 −
)
x2
x 2 + 32
x 2 + 32
=
x 2 + 32 − x 2
=
=
=
(x 2 + 32) x 2 + 32
32
3
(x 2 + 32) 2
cos x
y=
1 − sin x
dy (1 − sin x) ( − sin x) − cos x( − cos x)
=
dx
(1 − sin x) 2
14
=
=
− sin x + sin 2 x + cos 2 x
(1 − sin x) 2
1 − sin x
1
=
2
(1 − sin x) 1 − sin x
dy
1
1
x=
π 4,
=
=
dx 1 − sin π 4
2
1−
2
2
=
2− 2
=
=
15
x 2 + 32
x 2 + 32
32
2
2+ 2
×
2− 2 2+ 2
2(2 + 2)
= 2+ 2
2
1 + cos x
y=
1 − sin x
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 36 of 39
dy (1 − sin x) ( − sin x) − (1 + cos x) ( − cos x)
=
dx
(1 − sin x) 2
− sin x + sin 2 x + cos x + cos 2 x
(1 − sin x) 2
1 − sin x + cos x
=
(1 − sin x) 2
=
π dy
When x = ,
=
3 dx
π
π
1 − sin   + cos  
3
3
2
π

1 − sin 
3

3 1
3
3
3− 3
1−
+
−
2
2 =
2 2
2
= =
2
3
7
−
4 3

3
1− 3 +
1 −

4
4
2 

=
2(3 − 3) 2(3 − 3) (7 + 4 3) 2(21 + 12 3 − 7 3 − 12)
=
=
49 − 48
7 − 4 3 (7 − 4 3) (7 + 4 3)
= 2(9 + 5 3)
x
16
y=
1 − 5x
dy (1 − 5x) (1) − x( −5)
(a)
=
dx
(1 − 5x) 2
1 − 5x + 5x
=
(1 − 5x) 2
1
(1 − 5x) 2
dy
1
Since
=
dx (1 − 5x) 2
=
(b)
x2
dy
x2
=
dx (1 − 5x) 2
2
dy  x 
⇒x
=
dx  1 − 5x 
dy
⇒ x2
= y2
dx
dy
= (1 − 5x) −2
dx
d2 y
=
− 2(1 − 5x) −3 ( −5)
dx 2
10
=
(1 − 5x)3
2
(c)
2
 x   1 
d2 y
10x 2
x
=
= 10 
 

2
3
dx
(1 − 5x)
 1 − 5x   1 − 5x 
2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 37 of 39
d2 y
y
= 10 y 2  
2
dx
x
2
3
d y 10y
x2 2 =
dx
x
x +1
y=
x −1
dy (x − 1) (1) − (x + 1) (1)
=
dx
(x − 1) 2
x2
17
(a)
x −1− x −1
(x − 1) 2
dy
−2
=
dx (x − 1) 2
dy
−2
Since
=
dx (x − 1) 2
=
(b)
⇒ (x + 1) 2
dy
(x + 1) 2
=
−2
dx
(x − 1) 2
2
(c)
 x + 1
= −2
 x − 1
=−2y2
dy
∴ (x + 1) 2
=
− 2y 2
dx
x +1
y=
x −1
dy
−2
=
− 2(x − 1) −2 = 2
dx
(x − 1)
d2 y
4
= 4(x − 1) −3 =
2
dx
(x − 1)3
(x + 1) 2

d2 y
dy 4(x + 1) 2
2(x + 1)   −2 
2(x
2y
1)
+
+
+
=
+ 2 x +
+ 1 
2
3
2
dx
dx (x − 1)
x −1

  (x − 1) 
=
4(x + 1) 2
4x
8(x + 1)
4
−
−
−
3
2
3
(x − 1)
(x − 1)
(x − 1)
(x − 1) 2
=
4(x + 1) 2 − 4x(x − 1) − 8(x + 1) − 4(x − 1)
(x − 1)3
4x 2 + 8x + 4 − 4x 2 + 4x − 8x − 8 − 4x + 4
(x − 1)3
0
= =
0
(x − 1)3
=
18
y=
x2
2 − 3x 2
dy (2 − 3x 2 ) (2x) − x 2 ( −6x)
=
dx
(2 − 3x 2 ) 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 38 of 39
=
4x − 6x 3 + 6x 3
(2 − 3x 2 ) 2
4x
(2 − 3x 2 ) 2
dy
4x
Since
=
dx (2 − 3x 2 ) 2
=
x3
dy
4x 4
=
dx (2 − 3x 2 ) 2
 x2 
dy
⇒x
=
4
2
dx
 2 − 3x 
2
3
dy
= 4y 2
dx
(a) y = cos3 x sin x
dy
=
cos3 x cos x + sin x [(3 cos 2 x) ( − sin x)]
dx
=
cos 4 x − 3 cos 2 x sin 2 x =
cos 2 x [cos 2 x − 3 sin 2 x]
x3
19
π dy 
π
2 π
2 π
=
,
 cos  − 3 cos sin
4 dx 
4
4
4
4
When
=
x
(b)
4
2
 2
 2  2
= 
− 3

 

 2 
 2   2 
2
4
 4
− 3 
 16 
16
1 3
=
−
4 4
1
= −
2
=
1
(a) =
θ sin(2t − π) 2
1
dθ 1
−1
=
(2t − π) 2 (2) cos(2t − π) 2
dt 2
1
=
cos 2t − π
2t − π
20
(
)
(b) =
θ t 4t 2 − 3t + 2
dθ
1

−1
= t  (4t 2 − 3t + 2) 2 (8t − 3)  + 4t 2 − 3t + 2
dt
2

1
t(8t − 3)
2
=
+ 4t 2 − 3t + 2
4t 2 − 3t + 2
1
t (8t − 3) + 4t 2 − 3t + 2
= 2
4t 2 − 3t + 2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 39 of 39
4t 2 −
=
(c)
3
9
t + 4t 2 − 3t + 2 8t 2 − t + 2
2
2
=
2
2
4t − 3t + 2
4t − 3t +2
1 + cos 2t
θ=
1 − sin 2t
dθ (1 − sin 2t) ( −2sin 2t) − (1 + cos 2t) ( −2cos 2t)
=
dt
(1 − sin 2t) 2
=
−2 sin2t + 2 sin 2 2t + 2 cos2t + 2 cos 2 2t
(1 − sin 2t) 2
2 (sin 2 2t + cos 2 2t) − 2 sin 2t + 2 cos2t
(1 − sin 2t) 2
2 − 2 sin 2t + 2 cos2t
=
(1 − sin 2t) 2
=
21
1
y= (1 + cos x) 2
dy 1
−1
=+
(1 cos x) 2 (− sin x)
dx 2
1
−1
=
− sin x(1 + cos x) 2
2
d2y 1
1
−3
−1
=
sin x (1 + cos x) 2 ( − sin x) − cos x(1 + cos x) 2
2
4
2
dx
3
1 2
1
−
−1
=
− sin x(1 + cos x) 2 − cos x(1 + cos x) 2
4
2
2
2
d y
 dy 
2y 2 + 2   + y 2
 dx 
dx
2
1
1
 1
 1
−3
−1 
−1 
= 2(1 + cosx)  − sin 2 x (1 + cosx) 2 − cosx(1 + cosx) 2  + 2  − sin x(1 + cos x) 2  + [(1 + cosx) 2 ]2
2
 4

 2

1
1
=
− sin 2 x(1 + cos x) −1 − cos x + sin 2 x(1 + cos x) −1 + 1 + cos x
2
2
=1
1
2
Unit 1 Answers: Chapter 13
© Macmillan Publishers Limited 2013
Page 1 of 42
Chapter 14 Applications of Differentiation
Try these 14.1
(a)
x −1
x +1
u = x – 1, v = x + 1
du
dv
= 1,= 1
dx
dx
dy (x + 1) − (x − 1)
=
dx
(x + 1) 2
y=
=
2
( x + 1) 2
2 − 1 1 dy
2
2
=
, =
=
2
2 + 1 3 dx (2 + 1)
9
Equation of the tangent is:
1 2
y− =
(x − 2)
3 9
2
4 1
y=
x− +
9
9 3
2
1
y= x−
9
9
The equation of the tangent at x = 2 is
2
1
y= x−
9
9
y = x2 sin x
u = x2, v = sin x
du
dv
= 2x,
= cos x
dx
dx
dy
= x 2 cos x + 2xsin x
dx
dy
when x = 0, y = 0,
=0
dx
Gradient of the normal →∞
∴ Equation of the normal at x = 0 is
x=0
=
x 2,=
y
(b)
Try these 14.2
(a)
(i)
(ii)
f (x) = x2 + 2x + 3
f ′(x)
= 2x + 2
f ′ (x) > 0
2x + 2 > 0
x > –1
∴ increasing for {x : x ≥ – 1}
f(x) = x3 – 2x2 + 5
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 2 of 42
f ′(x)
= 3x 2 − 4x
f ′ (x) > 0
3x2 – 4x > 0
x(3x – 4) > 0
(iii)
4

{x : x ≤ 0} ∪  x : x ≥ 
3

4
f(x) = x − x
f ′(x) = 4x 3 – 1
f ′ (x) > 0 ⇒ 4x3 – 1 > 0
1
=0.63
4
∴{x : x ≥ 0.63}
f(x) = 4x2 + 6x + 2
f ′(x)
= 8x + 6
f ′(x) < 0
8x + 6 < 0
−3
x<
4
−3 

 x: x ≤ 
4

x +1
f(x) =
x−2
(x − 2)(1) − (x + 1)(1)
f ′(x) =
(x − 2)2
−3
=
(x − 2)2
f ′(x) < 0 ∀ x since (x − 2) 2 > 0
x>
(b)
(i)
(ii)
3
Exercise 14A
1
2
y = 6x2 – 2x
dy
= 12x − 2
dx
dy
> 0 ⇒ 12x − 2 > 0
dx
1
x>
6
1

The function is increasing for  x : x ≥ 
6

4
2
x=t –t
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 3 of 42
dx
= 4t 3 − 2t
dt
= 2t (2t2 – 1)
dx
<0 ⇒ 2t(2t 2 − 1) < 0
dt
t (2t2 – 1) < 0
Critical values are 0, ±
t< −
1
2
1
, t (2t2 – 1) < 0
2
1
< t < 0, t (2t2 – 1) > 0
2
1
0<t<
, t (2t2 – 1) < 0
2
1
< t , t (2t2 – 1) > 0
2
−
1  
1 

The function is decreasing for  t : t ≤ −
 ∪ t : 0 ≤ t ≤

2 
2

3
1
2
x+
3
x
dy 1
2
−
=
dx
3 x2
x 2 − 6 (x − 6)(x + 6)
= =
>0
3x 2
3x 2
y=
√6
4
5
√6
x < –√6 , x > √6
Since 3x2 is always positive
∴y is increasing for {x: x ≤ –√6 } ∪ {x: x ≥ √6 }
s = 2 – 3t + t2
ds
=−3 + 2t
dt
ds
< 0 ⇒ −3 + 2t < 0
dt
3
t<
2
3

s is decreasing for  t : t ≤ 
2

y = 2x3 + 3x2 – 12x + 4
dy
= 6x 2 + 6x − 12
dx
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 4 of 42
dy
> 0 ⇒ 6x 2 + 6x − 12 > 0
dx
x2 + x – 2 > 0
(x + 2) (x – 1) > 0
6
7
8
9
x < –2, x > 1
y is decreasing for {x : x ≤ −2} ∪ {x : x ≥ 1}
y = 4x2 + 3x + 1
dy
= 8x + 3
dx
dy
x = 1,
= 8 + 3 = 11
dx
x = 1, y = 4 + 3 + 1 = 8
Equation of tangent y – 8 = 11 (x – 1)
y – 8 = 11x – 11
y = 11x – 3
The equation of the tangent is y = 11x – 3
4
y=
2x + 3
4
4  4
x=
= 2, y =
 2, 
2(2) + 3 7  7 
dy
−8
using the chain rule
=
dx (2x + 3) 2
dy − 8
x = 2,
=
dx 49
4 −8
y −=
(x − 2)
7 49
49y – 28 = –8x + 16
49y + 8x – 44 = 0
Equation of the tangent 49y + 8x – 44 = 0
4
y=
1 − 2x
4
x = 1, y =
= −4
1− 2
dy
8
=
dx (1 − 2x) 2
dy
x = 1,
=8
dx
y + 4 = 8(x – 1)
y = 8x – 12
4
y= 2, y=1
x
x2 = 4 ⇒ x = ± 2
(2, 1) (–2, 1)
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 5 of 42
dy −8
=
dx x 3
dy
x = 2,
= −1
dx
dy
x = –2,
=1
dx
At (2, 1), m = –1, y – 1 = –(x – 2)
y = –x + 3
(–2, 1) m = 1,
y–1=x+2
y=x+3
10
y = x cosx
dy
= cos x − x sin x
dx
π
π
π
dy
π π
π
π
x=
, y=
cos ==
0,
cos − sin =
−
2
2
2
dx
2 2
2
2
π
π
y – 0 = − x − 
2
2
2
−π
π
y
x+
=
2
4
11
y = sin (2x– π )
dy
= 2 cos(2x π)−
dx
x = π , y = sin(2 π – π ) = sin π = 0
dy
= 2cos(2π − π) = 2cos π = − 2
dx
1
Gradient of normal =
2
1
y – 0 = (x − π)
2
1
π
y= x−
2
2
12
y = x tanx
dy
= tan x + x sec 2 x
dx
π
π
π π dy
π π
π
=
x =
, y
tan
=
, = tan + sec 2  
4
4
4 4 dx
4 4
4
π 2+π
=1 + =
2
2
−2
Gradient of normal =
2+π
π
−2 
π
Equation of normal: y=
−
x − 
4 2+ π
4
π
π
(2 + π) y = −2x + + (2 + π)  
2
4
2
π
(2 + π )y + 2x = π +
4
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 6 of 42
13
π
2x + 1
, y=
2
sin 2 x
dy 2sin 2 x − (2x + 1)2 sin x cos x
=
dx
sin 4 x
π
π+1
dy 2 − 0
x= , y=
= π + 1,
=
=2
2
1
dx
1
x=
1
2
1
π
y – (π + 1) = − (x – )
2
2
1
5π
y= − x+
+1
2
4
Gradient of normal = −
x−2
2x + 1
dy (2x + 1) − (x − 2)(2)
=
dx
(2x + 1)2
5
=
(2x + 1)2
x−2
y = 1,
=1
2x + 1
2x + 1 = x – 2
x = –3
dy 5 1
= =
dx 25 5
Gradient of normal = –5, (–3, 1)
y – 1 = –5 (x + 3)
y = –5x – 14
15 =
y (3x − 2) −1/ 2
dy −3
=
(3x − 2)−3 2
dx 2
dy −3
x = 1, y = 1,
=
dx 2
2
Gradient of normal =
3
2
y – 1 = (x – 1)
3
2
1
y= x+
3
3
16
y = x2– 4x + 5
y + 3x = 4 ⇒ y = –3x + 4
dy
= −3
dx
dy
=2x − 4 =−3
dx
2x = 1
1
x=
2
14
y=
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 7 of 42
2
17
1
1
 1
 1
=
x
,=
y   − 4   +=
5 3
 2
 2
2
4
 1
 1,3  m = –3
4
Equation of the tangent is
13
y− =
−3(x − 1)
4
13
y =−3x + 3 +
4
25
y=
−3x +
4
4
y=
(2x − 1)2
dy
−16
=
dx (2x − 1)3
dy
1
x = 1,
=
−16, Gradient of normal =
dx
16
1
y–4=
(x – 1)
16
1
1
y=
x– +4
16
16
1
15
y=
x +3
16
16
1
63
y=0⇒
x= –
16
16
x = –63
A(–63, 0)
63
x = 0, y =
16
63


B  0, 
16


Length of AB =
 63 
(−63) 2 +  
 16 
2
= 63.12
18
3
1− x
dy
3
= 2−
, using the chain rule
dx
(1 − x)2
3
When x = 2, y = 2(2) –
=7
1− 2
dy
3
= 2−
= 2 − 3 = −1
dx
(1 − 2) 2
Gradient of the normal = 1
Equation of the normal:
y–7=x–2
y=x+5
y = 2x –
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 8 of 42
3
, if the normal meets the curve again:
1– x
3
x + 5 = 2x –
1− x
3
= x−5
1− x
3 = (x – 5)(1 – x)
3 = x – x2 – 5 + 5x
x2 – 6x + 8 = 0
(x – 2)(x – 4) = 0
x = 2, 4
3
When x = 4, y = 8 –
=9
1− 4
(4, 9)
1
(a) y = 3 + 4x – x 2
2
dy
= 4−x
dx
dy
x = 2,
=2
dx
1
Gradient of normal = −
2
Equation of normal:
1
y – 9 = − (x − 2)
2
1
y = − x + 10
2
1
1
(b)
– x + 10 =3 + 4x – x 2
2
2
1 2
1
x –4 x+7=
0
2
2
1 2 9
x – x+7=
0
2
2
x 2 − 9x + 14 =
0
(x – 2) (x – 7) = 0
x = 2, 7
−7
13
x = 7, y =
+ 10 =
2
2
13


A =  7,

2


dy
(c) x = 7,= 4=
– 7 –3
dx
13
y–
= –3(x – 7)
2
13
y = –3x + 21 +
2
Since y = 2x –
19
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 9 of 42
y = –3x +
55
2
y = 4x2cos(2x)
dy
= 8x cos2x − 8x 2 sin(2x)
dx
x = π , y = 4 π 2 cos(2 π ) = 4 π 2
dy
= 8π
dx
y − 4π2 = 8π( x − π)
y = 8πx – 4π2
20
Exercise 14B
1
2
y = x2 – 2x + 1
dy
= 2x − 2
dx
d2 y
=2
dx 2
dy
= 0 ⇒ 2x − 2 = 0
dx
x=1
x = 1, y = 1 – 2 + 1 = 0
d2 y
Minimum point since
>0
dx 2
∴ (1, 0) is a minimum point
y = x3 – 3x + 1
dy
= 3x 2 – 3
dx
d2 y
= 6x
dx 2
dy
=⇒
0 3x 2 =
3
dx
x2 = 1
x = 1, –1
When x = 1, y = 1 – 3 + 1 = –1,
∴ (1, –1) minimum point
x = –1, y = –1 + 3 + 1 = 3,
3
d2 y
= 6>0
dx 2
d2 y
=−6 < 0
dx 2
(–1, 3) maximum point
y = 2x3 – 3x2 – 12x + 1
dy
= 6x 2 − 6x − 12
dx
d2 y
= 12x − 6
dx 2
dy
=0 ⇒ 6x 2 − 6x − 12 =0
dx
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 10 of 42
4
x2 – x – 2 = 0
(x – 2)(x + 1) = 0
x = 2, –1
d2 y
=
0 ⇒ 12x – 6 =
0
dx 2
1
x=
2
1 dy
is –ve
x< ,
2 dx
1 dy
is –ve
x> ,
2 dx
1
 1
 1
1
Point of inflexion at x = , y = 2   – 3   – 12   + 1




8
4
2
2
1 3
= − − 6 +1
4 4
1
= −5
2
x = 2, y = 2(2)3 – 3(2)2 – 12(2) + 1 = 16 – 12 – 24 + 1 = –19
x = –1, y = 2(–1)3 – 3(–1)2 – 12(–1) + 1 = –2 – 3 + 12 + 1 = 8
d2 y
x = 2,
= 12(2) − 6 = 18 > 0 ⇒ minimum point
dx 2
d2 y
x = –1,
=−12 − 6 =−18 < 0 ⇒ maximum point
dx 2
(2, –19) minimum point
(–1, 8) maximum point
y = x3 – 6x2 + 9x – 2
dy
= 3x 2 − 12x + 9
dx
d2 y
= 6x − 12
dx 2
dy
= 0 ⇒ 3x 2 − 12x + 9 = 0
dx
x2 – 4x + 3 = 0
(x – 1) (x – 3) = 0
x = 1, 3
d2 y
=0 ⇒ 6x − 12 =0
dx 2
x=2
dy
<0
x < 2,
dx
dy
<0
x > 2,
dx
when x = 2, y = 8 – 24 + 18 – 2 = 0
(2, 0) point of inflexion
d2 y
x = 1, y = 1 – 6 + 9 – 2 = 2,
=6(1) – 12 =−6 < 0 maximum point
dx 2
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 11 of 42
x = 3, y = 27 – 54 + 27 – 2 = –2,
5
d2 y
= 6(3) – 12= 6 > 0 minimum point
dx 2
(1, 2) maximum point
(3, –2) minimum point
y = x4–2x2+3
dy
= 4x 3 − 4x
dx
d2 y
=
12x 2 − 4
dx 2
dy
=0 ⇒ 4x 3 − 4x =0
dx
4x (x2 – 1) = 0
4x (x – 1)(x + 1) = 0
x = 0, 1, –1
d2 y
When x = 0, y = 3,
=−4 < 0 maximum point
dx 2
d2 y
x=1, y = 1–2+3 = 2,
= 12 – 4= 8 > 0 minimum point
dx 2
d2 y
x=–1, y=1–2+3=2,
= 12 – 4= 8 > 0 minimum point
dx 2
∴ (0, 3) maximum point
(1, 2) minimum point
(–1, 2) minimum point
d2 y
= 0 ⇒ 12x2 – 4 = 0
dx 2
1
1
1
x2 =
,x=
,x= −
3
3
3
1 2
22
1
When x =
, y = − +3=
9 3
9
3
1 2
22
1
When x = −
, y = − +3=
9 3
9
3
 1 22 
 1 22 
,  and  −
,  are points of inflexion

3 9 
 3 9 

6
x
x +1
dy x 2 + 1 – x(2x)
=
dx
(x 2 + 1)2
y=
=
2
1 − x2
(x 2 + 1)2
d 2 y (x 2 + 1)2 ( −2x) – (1 − x 2 )(2)(x 2 + 1)(2x)
=
dx 2
(x 2 + 1)4
=
( −2x)(x 2 + 1) − 4x(1 − x 2 )
(x 2 + 1)3
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 12 of 42
−2x3 − 2x − 4x + 4x3 2x3 − 6x
=
(x 2 + 1)3
(x 2 + 1)3
dy
=0 ⇒ 1 − x 2 = 0, x =± 1
dx
1 d 2 y −4
When x = 1, y = , =
< 0 maximum point
2 dx 2
8
1 d2 y 4
x = –1, y = − ,
=
> 0 minimum point
2 dx 2 8
 1
1,  maximum point
 2
−1 

 −1,  minimum point
2 

2
d y
=0
dx 2
⇒ 2x (x2–3) = 0
x = 0, x = 3 , − 3
1
d2 y
d2 y
x= 3,y=
3 , x < 3, 2 < 0 x > 3, 2 > 0
4
dx
dx
1
( 3,
3 ) point of inflexion
4
1
d2 y
d2 y
x=– 3,y=–
3 , x < − 3, 2 < 0 x > − 3, 2 > 0
4
dx
dx
1


3  point of inflexion
 − 3, −
4


d2 y
d2 y
x = 0, y = 0, x < 0, 2 > 0 x > 0, 2 < 0
dx
dx
(0, 0) point of inflexion
x2 − 4
7
y= 2
x +4
2
x + 4 −8
=
x2 + 4
8
= 1− 2
x +4
dy
8(2x)
16x
= =
dx (x 2 + 4)2 (x 2 + 4)2
d 2 y (x 2 + 4) 2 (16) − 16x 2(2x)(x 2 + 4)
=
dx 2
(x 2 + 4) 4
=
16(x 2 + 4) − 64x 2 −48x 2 + 64
=
(x 2 + 4)3
(x 2 + 4)3
dy
=⇒
0 16x =
0,x =
0
dx
d 2 y 64
=
= 1 > 0 minimum point
dx 2 64
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 13 of 42
−4
= −1
4
∴ (0, –1) minimum point
d2 y
= 0 ⇒ −48x 2 + 64 = 0
dx 2
64 4
2
x=
=
48 3
2 3
x= ±
3
−8
2 3
−1
3
=
x
=
,y =
3
16 / 3 2
−2 3
−1
=
x =
, y
3
2
dy
Since
does not change sign:
dx
 −2 3 −1
2 3 1
x
, −  and 
,  are points of inflexion
 3

2
2
 3

3
8
Vol of cylinder = 20 cm
A = π r2 + 2 π rh
V= π r2h = 20
20
h= 2
πr
 20 
A = π r2 + 2 πr  2 
 πr 
40
=
πr 2 +
r
dA
40
= 2πr − 2
dr
r
2
d A
80
= 2π + 3
dr 2
r
dA
40
= 0 ⇒ 2πr − 2 = 0
dr
r
2 π r3 = 40
40
r3 =
2π
20
r=3
π
= 1.853
20
= 1.853
h=
π(1.853) 2
r = 1.853 cm, h = 1.853 cm
9
y=
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 14 of 42
p = 2x + 2y + π x.
15 = 2x + 2y + π x
15 − 2x − πx
y=
2
π 2
A
x + 2xy
=
2
π 2
15 − 2x − πx 
=
x + 2x 

2
2


π
= x 2 + 15x − 2x 2 − πx 2
2
dA
=πx + 15 − 4x − 2πx
dx
=15 – π x – 4x
dA
=0
dx
x(4+ π ) = 15
15
=
x = 2.10
4+π
Width = 2(2.10) = 4.20 cm
10
v = (16 – 2x)(10 – 2x)x
= (160 – 52x + 4x2)x
= 160x – 52x2 + 4x3
dv
=160 − 104x + 12x 2
dx
d2 v
=
−104 + 24x
dx 2
dv
=⇒
0 12x 2 − 104x + 160 =
0
dx
3x2 – 26x + 40 = 0
(3x – 20) (x – 2) = 0
x = 2,
20
3
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 15 of 42
d2 v
= –104 + 48 < 0 maximum
dx
20 d 2 v
= –104 + 160 > 0 minimum
x= ,
3 dx 2
Maximum volume when x = 2, V = 160(2) – 52(2)2 + 4(2)3
= 320 – 208 + 32
= 144 cm3
1
(a)
V= πr 2 h
3
x = 2,
11
r
(b)
h 8−r
=
16
8
8h = 16(8–r)
h = 2(8–r)
1
V = πr 2 (2)(8 − r)
3
2 2
= πr (8 − r)
3
16 2 2 3
V=
πr − πr
3
3
dv 32
= πr – 2πr 2
dr 3
dv
32
=
0⇒
π r – 2 π r2 = 0
dr
3
 16

r  − r  = 0 ⇒ r = 0, r = 16 / 3
 3

16
Since r ≠ 0 ⇒ r =
3
2
V=
12
3
3
16  16  2  16  1  16 
π   − π   = π   = 158.9cm3
3  3 3  3 3  3
(a)
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 16 of 42
4y + 2 π x = 120
2 π x = 120 – 4y
120 − 4 y
60 − 2 y
x=
=
2π
π
A = y2 + π x2
3600 − 240y + 4y
 60 − 2y 
2
A = π =
+ y2
 +y
π
π


2
2
4y + πy + 3600 − 240y
A=
π
2
y (4 + π) + 3600 − 240y
=
π
dA 2y(4 + π) − 240
(b)
=
= 0
dy
π
240
120
y=
=
2(4 + π) 4 + π
13
V= 3x × 2x × h = 6x2h
6x2h = 144
144 24
h=
=
6x 2 x 2
(a) A = 2(2x × 3x) + 2(2x × h) + 2(3x)(h)
= 12x2 + 4xh + 6xh
= 12x2 + 10xh
 24 
= 12x2 + 10x  2 
x 
240
= 12x2 +
x
dA
240
(b) (i)
= 24x – 2
dx
x
dA
240
=⇒
0 24x = 2
dx
x
x3 = 10
x = 3 10 = 2.15
2
(ii)
2
d2 A
480
=
24 + 3
2
dx
x
When x = 2.15,
Unit 1 Answers: Chapter 14
d2A
480
=+
24
> 0 minimum point
2
dx
2.153
© Macmillan Publishers Limited 2013
Page 17 of 42
x = 2.15, A = 12(2.15)2 +
240
2.15
= 167.1 cm2
2
14
V=xh
x2h = 64000
64000
h=
x2
A = 4xh + x2
64000 × 4
A=
+ x2
x
dA −64000 × 4
=
+ 2x
dx
x2
dA
=⇒
0 2x 3 =
64000 × 4
dx
64000 × 4
x3 =
2
64000 × 4
x= 3
= 50.40 cm
2
256000
when x = 50.40 cm, A =
+ (50.40) 2
(50.40)
2
= 7619.53 cm
64000
h=
= 25.2cm
(50.4) 2
Try these 14.3
(a)
(b)
x = t3 + 4t – 1 , y = t2 + 7t + 9
dx
dy
= 3t2 + 4,
= 2t + 7
dt
dt
dy 2t + 7
=
dx 3t 2 + 4
d  2t + 7  2(3t 2 + 4) − (2t + 7)6t
=
dt  3t 2 + 4 
(3t 2 + 4) 2
=
6t 2 + 4 − 12t 2 − 42t
(3t 2 + 4) 2
=
−6t 2 − 42t + 4
(3t 2 + 4) 2
d 2 y −6t 2 − 42t + 4
1
−6t 2 − 42t + 4
× 2
=
=
2
2
2
3t + 4
(3t + 4)
(3t 2 + 4)3
dx
x = tan t, y = 2 sin t + 1
dx
dy
= sec2 t,
= 2 cos t
dt
dt
dy 2cos t
= 2 cos3 t
=
dx sec 2 t
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 18 of 42
3
 2
π dy
π
2
t=
,
= 2 cos3 = 2 
 =

4 dx
4
2
 2 
d2 y
−6 cos 2 t sin t
=
= – 6 cos4t sin t
dx 2
sec 2 t
4
 2   2  −3 2
π d2 y
π
π
4
,
=
–
6
cos
sin
=
t=
6
−

 
 =
4 dx 2
4
4
4
 2   2 
Exercise 14C
1
(a)
x3y = 10
dx
= 0.5
dt
Where x = 5, y =
10
2
=
125
25
10
x3
dy −30
= 4
dx
x
dy dy dx −30
=
×
= 4 × 0.5
dt dx dt
x
−30
=4 × 0.5 =
− 0.024 unit per second
5
(b) y = 2, x3 = 5 ⇒ x = 3 5
dy
−30
=
× 0.5 =
−1.754 unit per second
dt ( 3 5) 4
1
1
1
=
− 2
2
50 x
y
dx
= 5cms −1
dt
dx
−2y −3
=
2x −3
dy
y=
2
dy
y3
= − 3
dx
x
When x = 10,
3
1
1
1
1
= −
=
2
50 100 100
y
y = 10
dy −1000
=
= −1
dx 1000
dy
∴ =−1 × 5 =−5cms −1
dt
dv
= 0.04 cm3s −1
dt
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 19 of 42
dA
when =
v 150π cm3
dt
4
v = πr 3
3
4 3
150 π=
πr
3
450 3
=r
4
450
r= 3
= 4.827 cm
4
dv dv dr
=
×
dt dr dt
0.05
dr
=
2
4π(4.827)
dt
= 0.0001708
A = 4 π r2
dA
= 8πr
dr
dA dA dr
=
×
dt
dr dt
4
5
6
= 8 π (4.827) × 0.0001708
= 0.02072 cm2s–1
v = x6(x2+ 4) = x8+ 4x6
dx
= 4cms −1
dt
dv
= 8x 7 + 24x 5
dx
dv
x = 1,
=8(1)7 + 24(1)5 =32
dx
dv dv dx
=
×
= 32 × 4 = 128cm3s −1
dt dx dt
dr
= 2cms −1
dt
A=
πr 2 =π
4
2
r = 4 ⇒ r = 2 cm
c = 2πr
dc
= 2π
dr
dc dc dr
=
× = 2π × 2 = 4π cms –1
dt dr dt
dx
= 0.05 cms−1
dt
v = 64 ⇒ x3 = 64, x = 4 cm
A = 6x2
dA
= 12x
dx
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 20 of 42
7
8
9
dA dA dx
=
×
= 12x × 0.05
dt
dx dt
= 12 × 4 × 0.05 = 2.4 cm2s–1
dc
= 4 cms –1
dt
(a) c = 2 π r
dc
= 2π
dt
dc dc dr
=
×
dt dr dt
dr
4 = 2π ×
dt
2 dr
=
π dt
dA
(b)
when r = 64 cm
dt
A = π r2
dA
= 2πr
dr
dA dA dr
=
×
dt dr dt
2
=
2π(64) ×
π
= 256 cm2s–1
dA
(a)
= 10 cm 2s −1
dt
A = 4 π r2
dA
= 8πr
dr
dA dA dr
=
×
dt
dr dt
dr
10 = 8 π r ×
dt
dr 10
=
dt 8πr
4
(b) v = πr 3
3
dv
= 4πr 2
dr
dv dv dr
=
×
dt dr dt
10
=
4π(42 ) ×
=
20 cm3s –1
8π(4)
4x + 2
y=
x +1
dy (x + 1)(4) − (4x + 2)
=
dx
(x + 1)2
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 21 of 42
4x + 4 − 4x − 2
(x + 1)2
2
=
(x + 1)2
y=5
4x + 2
=5
x +1
4x + 2 = 5x + 5
x = –3.
dy dy dx
=
×
dt dx dt
2
dx
0.3 =
×
2
dt
( −3 + 1)
dx
0.6 =
dt
dv
10
x = 4,
= 0.024 cm3s –1
dt
v = x3
dv
dv
2
2
= 3x=
=
, x 4,= 3(4)
48
dx
dx
dv dv dx
=
×
dt dx dt
dx
0.024 = 48
dt
dx 0.024
= = 0.0005 cms −1
dt
48
11
(a) pv = 600
600
p=
v
dp −600
= 2
dv
v
dp −600
(b) v = 20,
=
= −1.5
dv 202
12
(a)
=
Q
dθ π
= rad per sec
dt 3
S = rθ= 6θ
ds
=6
dθ
ds ds dθ
=
×
dt dθ dt
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 22 of 42
= 6 ( π / 3) = 2 π cm sec
1 2
rθ
2
dA 1 2
= r
dθ 2
dA dA dθ
=
×
dt
dθ dt
1 2 π
=
(6) ×
2
3
= 6 π cm2s–1
dv
(a)
= 16cm 3s −1
dt
v = 2x2 – 7x
v = 4 ⇒ 2x2 – 7x – 4 = 0
(2x + 1)(x – 4) = 0
1
x=– ,4
2
x=4
dv
(b)
= 4x − 7
dx
dv
x = 4,
=8−7 =1
dx
dv dv dx
=
×
dt dx dt
dx
16 =
dt
16 cms–1
(b)
13
A=
Try these 14.4
(a) y = 4 sin x,
period = 2π, amplitude = 4
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 23 of 42
(b) y = 3 cosx,
period = 2π, amplitude = 3
Try these 14.5
π
2π
π
π

, amplitude = 1, displacement = 12 =
(a) y = cos  3x −  ,period =
3
3 36
12 

π
1
2π
π

(b) y = 40 cos  2πx −  ,period =
= 1 , amplitude = 2, displacement = 8 =
2π
2π 16
8

π
π

(c) y = 2 sin  x −  ,period = 2π , amplitude = 2, displacement =
3
3

Try these 14.6
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 24 of 42
Try these 14.7
(a)
y=
2x + 3
4x − 1
3

2+

 2x + 3 
x = 2 = 1
Horizontal asymptote: lim 
 = lim
x →∞  4x − 1 
x →∞ 
1 4 2
 4 − 
x
1
2
Vertical asymptote: 4x – 1 = 0
1
x=
4
y=
∴ Horizontal Asymptote: y =
Vertical Asymptote: x =
(b)
(c)
y=
x +1
x−3
1
2
1
4
1

1+

 x + 1
x = 1
lim 
 = lim
x →∞  x − 3 
x →∞ 
3
 1 − 
x
Horizontal Asymptote y = 1
Vertical Asymptote x = 3
x 2 + 2x
y= 2
x − 7x + 12
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 25 of 42
 x 2 + 2x 
lim y = lim  2
x →∞
x →∞  x − 7x + 12 

2 

1+

x 
= lim 
x →∞
7 12 
 1 − + 2 
x x
=1
∴ y = 1 is a Horizontal asymptote
Vertical asymptote: x2 – 7x + 12 = 0
(x – 3)(x – 4) = 0
x = 3, 4
∴ Vertical asymptote: x = 3, x = 4
Exercise 14D
1
y = x2 + 2x + 1
dy
= 2x + 2
dx
d2 y
=2
dx 2
dy
= 0 ⇒ 2x + 2 = 0
dx
x = –1
when x = –1, y = 1 – 2 + 1 = 0.
∴ (–1, 0) is a min point
2
y = 12x – x3
dy
= 12 − 3x 2
dx
d2 y
= −6x
dx 2
dy
=0 ⇒ 12 − 3x 2 =0
dx
x2 = 4
x = ±2
d2 y
− ve Maximum point
dx 2
d2 y
x = –2, y = 12(–2) – (–2)3 = –16,
+ve min point
dx 2
when x = 2, y = 12(2) – (2)3 = 16,
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 26 of 42
3
y = x4 – 6x2
dy
= 4x 3 − 12x
dx
d2 y
=
12x 2 − 12
dx 2
dy
When
=⇒
0 4x 3 − 12x =
0
dx
4x(x2 – 3) = 0
x = 0, x = ± 3
d2 y
=12(0)2 − 12 =−12 < 0 ⇒ Max. pt at (0,0)
dx 2
4
2
2
d2 y
x = 3 , y = 3 − 6 3 = 9 – 18 = –9,
= 12 3 –12 = 24 > 0
2
dx
Minimum point at ( 3, –9)
When x = 0, y = 0,
( )
( )
(
x = − 3, y = − 3
)
4
(
( )
)
2
– 6 − 3 = –9,
(
)
2
d2 y
= 12 − 3 − 12 = 24 > 0
2
dx
Minimum point at ( − 3, −9)
4
y=
−1 2
x +x
40
 −1

x + 1 = 0 , x = 0, 40
 40

When y = 0, x 
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 27 of 42
dy −1
=
x +1
dx 20
dy
When
= 0 , x = 20, y = 10
dx
5
6
θ = –2t3 + 12t2 + 10
dθ
= −6t 2 + 24t
dt
d2θ
=
−12t + 24
dt 2
dθ
= 0 ⇒ −6t 2 + 24t = 0
dt
6t(–t + 4) = 0
t = 0, 4
d2θ
t = 0, θ = 10, 2 = 24 > 0 minimum point
dt
d2θ
t = 4, θ = –2(4)3 + 12(4)2 + 10 = 74. 2 –24 < 0 maximum point
dt
When t = 6, θ = –2(6)3+12(6)2+10=10
2x + 1
(a) y =
x−3
dy (x − 3)(2) − (2x + 1)
=
dx
(x − 3)2
−7
=
(x − 3)2
dy
= 0 ⇒ −7 = 0 ⇒ inconsistent
dx
⇒ y has no turning points
(b) Horizontal asymptotes when x – 3 = 0
x=3
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 28 of 42
2x + 1
=2
x →∞ x − 3
Vertical asymptotes when lim
(c)
7
y=2
Asymptotes x = 3, y = 2
1
when x = 0, y = –
3
1
y = 0, x = –
2
2
x +4
1
x = 0, y =
2
dy
−4x
=
dx (x 2 + 4)2
dy
=0⇒x=0
dx
dy
x < 0,
+ ve
dx
dy
x > 0,
− ve
dx
1
x = 0, y= : Maximum point
2
x → ±∞, y → 0
y=
2
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 29 of 42
8
y=x+
4
x
dy
4
= 1− 2
dx
x
2
d y 8
=
dx 2 x 3
dy
4
=0⇒1– 2 =0
dx
x
x2 = 4
x = ±2
4
d2 y 8
When x = 2, y = 2 + = 4,
=
> 0 Minimum point
2
dx 2 8
d2 y 8
x = –2, y = –2–2 = –4, =
< 0 Max point
dx 2 −8
(2, 4) minimum point
(–2, –4) maximum point
9
Asymptotes
y=x
x=0
2x + 1
y=
x−2
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 30 of 42
1
1
(0, − )
2
2
dy (x − 2)(2) − (2x + 1)
=
dx
(x − 2)2
−5
=
(x − 2)2
dy
= 0 ⇒ –5 = 0 Inconsistent, no turning points
dx
Asymptotes: x = 2
y=2
1
when x = 0, y = –
2
1
y = 0, x = –
2
x = 0, y = −
10
x 2 + 2x + 5
x+2
x
x+2
2
x + 2x + 5
2
x + 2x
5
5
∴y=x+
x+2
5
x = 0, y =
2
 5
 0, 
2
y=
dy
= 1 – 5(x + 2)2 Asymptotes
dx
d2 y
10
=
x = –2
2
dx
(x + 2)3
dy
5
y=x
= 0 ⇒1−
= 0
dx
(x + 2)2
(x + 2)2 = 5
x+2= ± 5
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 31 of 42
x = –2 ±
5
Review exercise 14
1
x
1 − x3
dy (1 − x 3 ) − x( −3x 2 )
=
dx
(1 − x 3 )2
1 − x 3 + 3x 3
=
(1 − x 3 ) 2
y=
=
1 + 2x 3
(1 − x 3 ) 2
2
2
=
3
−7
1− 2
3
dy 1 + 2(2) 17
= =
dx (1 − 23 )2 49
x = 2, y =
Gradient of normal = −
49
17
Equation of the normal:
2 17
(x – 2)
y+ =
7 49
17
34 2
y=
x+
−
49
49 7
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 32 of 42
17
20
x+
49
49
dv
(a)
= 40 cm3s –1
dt
v = 0.02 h3 + 0.4 h2 + 400 h
dv
= 0.06h 2 + 0.8h + 400
dh
dv
h = 10,
= 0.06(10)2 + 0.8(10) + 400 = 414
dh
dv dv dh
=
×
dt dh dt
dh
40 = 414
dt
dh 40
= = 0.097 cms –1
dt 414
dh
(b)
= 0.04
dt
dv dv dh
=
×
dt dh dt
40 = 0.04 [0.06 h2+ 0.8 h + 400]
1000 = 0.06 h2 + 0.8 h + 400
0.06 h2 + 0.8 h – 600 = 0
y=
2
−0.8 ± (0.8)2 + 4(0.06)(600)
2(0.06)
−0.8 ± 12.027
=
0.12
h = 93.6 cm
y = cosx + sinx
dy
=
− sin x + cos x
dx
y+x=3
y=–x+3
Gradient = –1
Gradient of tangent = 1
∴ –sinx + cosx = 1
cosx – sinx = Rcos (x + α)
= Rcosx cosα – Rsinx sinα
Rcosα = 1
Rsinα = 1
tanα = 1⇒ α = π /4
R= 2
2 cos (x + π /4) = 1
1
cos (x + π /4) =
2
π π 7π
x+ = ,
4 4 4
h=
3
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 33 of 42
6π
4
y = cos 0 + sin 0 = 1
∴ (0, 1) → coordinate
x +1
(a) y =
2x − 7
dy (2x − 7) − (x + 1)(2)
=
dx
(2x − 7)2
−9
=
(2x − 7)2
dy −9
x = 3,
=
= −9
dx 1
Equation of tangent y + 4 = –9(x – 3)
y = –9x + 27 – 4
y = –9x + 23
1
(b) Gradient of normal =
9
Equation of normal:
1
y + 4 = (x – 3)
9
1
1
y= x– –4
9
3
1
13
y= x–
9
3
23
(c) y = 0, x =
9
 23 
A  ,0
 9 
x = 0,
4
−13
 −13 
B  0,


3 
3
13
23 3
Area of ∆=
× = 5.54
9 2
dA
r = 4 m,
= 3 m 2s –1
dt
4
v = πr 3 , A =
4πr 2
3
dA dA dr
=
×
dt
dr dt
dA
= 8πr
dr
dr
3 = 8πr ×
dt
dr
3
3
=
= = 0.03 ms –1
dt 8π(4) 32π
x = 0, y =
5
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 34 of 42
6
7
dv
= 4πr 2
dr
dv dv dr
=
×
dt dr dt
3
= 4π(4) 2 ×
32π
= 6 m3s–1
Area of cross section = 20 h cm2
A = 20 h
dA
= 0.05
dt
dA
= 20
dh
dA dA dh
=
×
dt dh dt
dh
0.05 = 20
dt
dh 0.05
= = 0.0025 cms −1
dt
20
V = 20h × h = 20h2
dv
= 40h
dh
dv dv dh
=
×
dt dh dt
= 40h × 0.0025
= 40 (4) (0.0025)
2
= cm 3s −1
5
1
11
(a) y = x 4 − 2x 3 + x 2 − 6x + 1
4
2
dy
= x3 – 6x2 + 11x – 6
dx
d2 y
= 3x 2 − 12x + 11
dx 2
dy
For stationary points
=0
dx
x3 – 6x2 + 11x – 6 = 0
x3 – 6x2 + 11x – 6 = (x – 1)(x2 – 5x + 6)
= (x – 1)(x – 2)(x – 3)
=0
x = 1, 2, 3
1
11
−5 d 2 y
(b) x = 1, y = − 2 +
–6+1=
.,
=3 − 12 + 11 > 0 minimum point
4
2
4
dx 2
d2 y
x = 2, y = 4 – 16 + 22 – 12 + 1 = –1,
= 12 − 24 + 11 < 0 maximum point
dx 2
81
99
−5 d 2 y
x = 3, y =
,
− 54 +
− 18 + 1 =
= 27 − 36 + 11 > 0 Min pt.
4
2
4 dx 2
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 35 of 42
(c)
d2 y
= 0 ⇒ 3x2 – 12x + 11 = 0
2
dx
12 ± 144 − 132
x=
6
Inflexion points
12 ± 12
1
= 2±
3
6
3
= 2.58, 1.42
Points of inflexion at x = 2.58, y = –1.14
x = 1.42, y = –1.14
=
(d)
8
(a)
(b)
y=x3 + 3x2 – 24x + 10
dy
= 3x 2 + 6x − 24
dx
d2 y
= 6x + 6
dx 2
dy
= 0 ⇒ 3x 2 + 6x − 24 = 0
dx
x2 + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = –4, 2
d2 y
x = –4,
= – 24 + 6 < 0 maximum point
dx 2
d2 y
x = 2,
= 12 + 6 > 0 minimum point
dx 2
Point of inflexion
d2 y
= 0 , 6x + 6 = 0
dx 2
x = –1
x = –1, y = –1 + 3 + 24 + 10 = 36
Point of inflexion (–1, 36)
cos2x
y=
1 + sin 2x
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 36 of 42
dy (1 + sin 2x)( −2sin 2x) − (cos2x)(2 cos2x)
=
dx
(1 + sin 2x)2
−2sin 2x − 2[sin 2 2x + cos2 2x]
(1 + sin 2x)2
−2
=
1 + sin 2x
dy
−2
cos π
x = π/2,
=
= –2, y =
= –1
dx
1 + sin π
1 + sin π
1
Gradient of normal = , ( π / 2, − 1)
2
1
y + 1 = ( x − π / 2)
2
1
y+1=
x−π/4
2
1
y=0⇒1=
x−π/4
2
1
1+ π / 4 = x
2
x = 2 + π/2
p (2 + π / 2, 0 )
V = yx2
A = x2 + 4 xy
96 = x2 + 4 xy
96 − x 2
y=
4x
 96 − x 2 
V = x2 

 4x 
=
9
(a)
(b)
1
= 24x – x3
4
dV
3
= 24 – x 2
dx
4
dV
24 × 4
=
0 ⇒ x2 =
dx
3
x 2 = 32
x = 32
3
1
V = 24 32 –
32
4
= 24 32 – 8 32 = 16 32 = 64 2
y = 4x3 – 24x2 + 36x
dy
= 12x 2 − 48x + 36
dx
d2 y
= 24x − 48
dx 2
dy
=0 ⇒ 12x 2 − 48x + 36 =0
dx
x2 – 4x + 3 = 0
( )
10
(a)
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 37 of 42
(b)
(x –1) (x –3) = 0
x = 1, 3
x = 1, y = 4 – 24 + 36 = 16
x = 3, y = 4(3)3 – 24(3)2 + 36(3) = 0
(1, 16) (3, 0)
d2 y
=0
dx 2
24x – 48 = 0
x=2
y = 4(2)3–24(2)2 + 36(2) = 8
Point of inflexion (2,8)
(c)
Where x = 1,
d2 y
= 24 – 48 < 0 maximum point
dx 2
d2 y
24(3) – 48 > 0 minimum point
dx 2
A = 243 π
π r2+2 π rh = 243 π
243 − r 2
h=
2r
V = π r2h.
 243 − r 2 
2
= π r  2r 
x = 3,
11
(a)
π
r (243–r2)
2
π
= (243r – r 3 )
2
dv π
(b) =
(243 − 3r 2 )
dr 2
dv
=⇒
0 243 =
3r 2
dr
=
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 38 of 42
r2 = 81
r=9
r = 9, V =
π
(243(9) – 93)
2
= 729 π
d2 v π
= (–6r)
2
dr 2
d2 v
r = 9, 2 = –27 π < 0 maximum point
dr
12
dx
= 0.25
dt
(a)
(b)
13
(a)
4
4
=
(3)2
9
dy
−40
=
dx (5x − 2)3
dy −40
x = 1,
=
dx 27
dy dy dx −40
=
×
=
× 0.25
dt dx dt
27
−10
=
27
4
y = 9,
=9
(5x − 2)2
4
(5x – 2)2 =
9
5x – 2 = 2/3
8
5x =
3
x = 8/15
dy
−40
=
= −135
3
dx   8 

 5  15  – 2
x = 1, y =
dy
1
=
−135 ×
dt
4
−135
=
4
y = x3 + 3x2 + 3x +2
dy
= 3x 2 + 6x + 3
dx
3x2 + 6x + 3 = 0
x2 + 2x + 1 = 0
(x + 1)2 = 0
x = –1
x = –1, y = –1 + 3 – 3 + 2 = 1
(–1, 1)
d2 y
= 6x + 6 = 0
dx 2
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 39 of 42
x = –1
dy
>0
dx
dy
x > –1
>0
dx
x = –1 point of inflexion
x < –1,
(b)
14
36R
. R = 0, P = 0
R + 2R + 1
dP (R 2 + 2R + 1)(36) − 36R(2R + 2)
=
dR
(R 2 + 2R + 1) 2
P=
=
2
36R 2 + 72R + 36 – 72R 2 − 72R
(R 2 + 2R + 1)2
−36R 2 + 36
(R 2 + 2R + 1)2
dP
= 0 ⇒ 36 − 36R 2 = 0
dR
R2 = 1
R = ±1
36
When R = 1, P=
=9
1+ 2 +1
Maximum point at (1, 9)
=
15
y = cos32x + sin4 2x
x = π / 2 , y = cos3 π + sin4 π = –1
dy
=
−6cos 2 (2x) sin(2x) + 8sin 3 (2x) cos(2x)
dx
dy
= – 6 cos2 π sin π + 8 sin3 π cos π = 0
x =π/ 2,
dx
Equation of tangent:
y – (–1) = 0(x – π / 2 )
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 40 of 42
16
y+1=0
y = –1
9x
y= 2
x +9
x = 0, y = 0
4x
=
= 0
lim y lim
x →∞
x →∞ x 2 + 9
∴ y = 0 is an asymptote
dy (x 2 + 9)(9) − 9x(2x)
=
dx
(x 2 + 9)2
9x 2 + 81 − 18x 2
=
(x 2 + 9) 2
−9x 2 + 81
= 2
(x + 9)2
dy
=
0 ⇒ 9x 2 =
81
dx
x2 = 9
x = ±3
27
3
=
when x = 3, y =
18
2
−27
x = –3, y =
= −3 / 2
18
17
1 2
πr h
3
16 24
=
r
x
16x 2
=
r =
x
24 3
V=
2
1 2 
π x  x
3 3 
1
4
=
π × x3
3
9
4
=
πx 3
27
dv
= 0.05 cm3s–1
dt
V=
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 41 of 42
dv
=
dt
dv
=
dx
dv dx
×
dx dt
4 2
πx
9
dv 4
=
π(12)2
dx 9
When x = 12,
= 64 π
0.05 = 64 π ×
18
dx
dt
dx 0.05
= = 0.00025 cms −1
dt 64π
Surface area = 2400 cm2
⇒ x2 + 4xh = 2400
2400 − x 2
h=
4x
600 x
=
−
x
4
 2400 − x 2 
v = x2h = x 2 

 4x

1
= 600 x − x 3
4
dv
3
= 600 − x 2
dx
4
dv
3
=⇒
0 600 − x 2 =
0
dx
4
2400
x2=
= 800
3
x = 800
d 2 v −3
=
x
dx 2
2
d2 v
When x = 800,
= –ve ⇒ maximum
dx 2
3
1
Maximum volume = 600 ( 800 ) –
800 = 11313.7 cm3
4
V = 2000 cm3
πr 2 h =
2000
2000
h=
πr 2
(
19
)
A =πr 2 + 2πrh
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 42 of 42
 2000 
= πr 2 + 2 πr  2 
 πr 
4000
= πr 2 +
r
dA
4000
= 2 πr − 2
dr
r
dA
4000
= 0 ⇒ 2 πr=
dr
r2
4000
4000
3
⇒r=
= 8.603
r3 =
2π
2π
2000
h =
π(8.603)2
= 8.602
Unit 1 Answers: Chapter 14
© Macmillan Publishers Limited 2013
Page 1 of 32
Chapter 15 Integration
Exercise 15A
1
10
5
7
1
x dx
x +c
∫=
10
2
dx
x +c
∫ 5x =
7
3
dx
x
∫ 2 x=
3
4
dx
x +c
∫8 x =
7
5
dx ∫
+ dx
∫ x =
x
x
= ∫ x + x dx
9
6
1
1
1/2
6
8
3/2
+c
7
x +1
1
1
3
2
3
−2
−3
1
−1
−
+c
x 2x 2
2
2 ) dx
∫ ( 4x + 1) ( x + =
=
6
∫ 4x + x + 8x + 2 dx
3
2
1 3
x + 4x 2 + 2x + c
3
x −2
− 1/2
dx =−
x − 4x1/2 + c
∫ 1 2x dx =
x
= x4 +
7
∫
=
x−4 x +c
8
9
10
x 4 − 6x 2
1 2
−2
−4
∫ x 6 dx =∫ x − 6x dx =− x + x 3 + c
2
4 5 4 3
2
4
2
∫ ( 2x + 1) dx = ∫ 4x + 4x + 1 dx= 5 x + 3 x + x + c
( x + 1)2
x +2 x +1
dx
∫ x dx = ∫
x
∫x
=
1/2
+ 2 + x − 1/2 dx
2 3/2
x + 2x + 2x1/2 + c
3
2
=
x x + 2x + 2 x + c
3
2
2
11
∫ (x − 3 x ) dx =∫ x − 6x x + 9x dx.
=
=∫ x 2 − 6x 3/2 + 9x dx
1 3 12 5/2 9 2
x −
x + x +c
3
5
2
5
2
x + 6x
dx ∫ x − 2 + 6x − 5 dx
∫ x7 =
1 6
=
− − x− 4 + c
x 4
=
12
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 2 of 32
13
14
15
1
3
=
− − 4 +c
x 2x
3
x +4
−3
∫ x 3 dx= ∫1 + 4x dx
4
=
x − x −2 + c
2
2
=x − 2 + c
x
5
3
3
−2
∫ 4x − x 2 dx =∫ 4x − 5x dx
5
= x4 + + c
x
2
2
4
2
∫ (x + 2) dx = ∫ x + 4x + 4 dx
1 5 4 3
x + x + 4x + c
5
3
4 6
2 6
5
∫ 4x dx= 6 x + =c 3 x + c
7 4 3 −4
3
−5
∫ 7x − 3x dx =4 x + 4 x + c
7
3
= x4 + 4 + c
4
4x
3
∫ (4 + 2x) dx
=
16
17
18
2
2
3
=
∫16 + 3 (4) (2x) + 3 (4) (2x) + (2x) dx
2
3
=
∫16 + 96x + 48x + 8x dx
= 16x + 48 x 2 + 16x 3 + 2x 4 + c
19
∫ (x − 1) dx
= ∫ x − 3x + 3x − 1 dx
3
3
2
1 4
3
x − x3 + x 2 − x + c
4
2
2
 x + 1
∫  x 2  dx
=
20
x 2 + 2x + 1
dx
x4
1
2
1
= ∫ 2 + 3 + 4 dx
x
x
x
−2
−3
= ∫ x + 2x + x − 4 dx
=∫
1 1
1
− 2 − 3 +c
x x
3x
(1 – 3x + x2)2 = (1 – 3x + x2) (1 – 3x + x2)
=−
1 3x + x 2 − 3x + 9x 2 − 3x 3 + x 2 − 3x 3 + x 4
= 1 – 6x + 11x2 – 6x3 + x4
2 2
2
3
4
∫ (1 − 3x + x ) dx = ∫1 − 6x + 11x − 6x + x dx
=
−
21
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 3 of 32
=
x − 3x 2 +
22
4
11 3 3 4 1 5
x − x + x +c
3
2
5
∫ x + 6x + 10 x dx
= ∫ 4x + 6x + 10x dx
3
2
−2
3
1/2
4 3 4 20 3/2
+ x +
x +c
x 2
3
6
7
3
∫ 4x + 3x − x 3 dx
= ∫ 4x 7 + 3x 3 − 6x − 3 dx
=
−
23
=
24
1 8 3 4 3
x + x + 2 +c
2
4
x
x 2 + 4x
∫ x dx
=
∫x
3/2
+ 4x1/2 dx
2 5/2 8 3/2
x + x +c
5
3
3 2
−3
−2
∫ t 3 + t 2 dt = ∫ 3t + 2t dt
−3 −2 2 −1
=
t − t +c
2
1
−3 2
=
− +c
2t 2 t
dy
= 4x 3 − 8x + 2
dx
y = ∫ 4x 3 − 8x + 2 dx
=
25
26
y = x4 – 4x2 + 2x + c
x = 1, y = 0 ⇒ 0 = 1 – 4 + 2 + c
c=1
y = x4 – 4x2 + 2x + 1
dy
27
= 4x + 3
dx
=
y ∫ 4x + 3 dx
y = 2x2 + 3x + c
x = 2, y = 1 ⇒ 1 = 8 + 6 + c
c = – 13
y = 2x2 + 3x – 13
dy 4t − 3
28
= 3
dt
t
4 3
= 2 − 3
t
t
y = ∫ 4t −2 − 3t −3 dt
y=
3
−4
+ 2 +c
t
2t
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 4 of 32
3
−4
t = 2, y = 1 ⇒ 1 = +
+c
2
2 (2)2
3
1 =− 2 + + c
8
3 21
c=3− =
8 8
−4
3
21
y=
+ 2 +
t
2t
8
dy
1
2
29
= x −6+ x
dx
2
1
y= ∫ x 2 − 6 + x dx
2
1 3
1
y=
x − 6x + x 2 + c
3
4
1
1
x = 1, y = 1 ⇒ 1 = − 6 + + c
3
4
1 1
c =7 − −
3 4
77
=
12
1
1
77
y = x 3 + x 2 − 6x +
3
4
12
du
30
= 4t 2 − t
dt
=
u ∫ 4t 2 − t1/2 dt
4
2
u = t 3 − t 3/2 + c
3
3
4 2
u = 1, t = 1 ⇒ 1 = − + c
3 3
1
=c
3
4
2
1
u = t 3 − t 3/2 +
3
3
3
Try these 15.1
(a)
(i)
(ii)
(iii)
(b)
8
dx
∫ (4x + 2) =
1  (4x + 2)9 
1
+c
(4x + 2)9 + c

=
4
9
36

1
1 (3x + 5)3/ 2
2
+ c = (3x + 5)3/ 2 + c
3
32
9
1
1 (2x + 1)3/ 2
1
+ c = (2x + 1)3/ 2 + c
2
32
3
∫
3x + 5 dx = ∫ (3x + 5) 2 dx =
∫
2x + 1 dx = ∫ (2x + 1) 2 dx =
dy
= (3x − 1)6
dx
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 5 of 32
⇒ y = ∫ (3x − 1)6 dx
1 (3x − 1)7
+c
3
7
1
=
y
(3x − 1)7 + c
21
=
y
When x = 1, y = 1 ⇒=
1
27
+c
21
128
21
− 107
=
21
c= 1 −
Hence =
y
1
107
(3x − 1)7 −
21
21
Try these 15.2

π
π

1

π
4

π
π

(a)
∫ 4 sin  x + 2  dx = − 4 cos  x + 2  + c
(b)
∫ cos  2 − 3x dx = − 3 sin  2 − 3x + c
(c)
∫ 4 tan  3x + 2  dx = 3 l n sec  3x + 2  + c
(d)
∫ 3 cot (5x) dx = 5 ln sin 5x + c

π
3
Exercise 15 B
1
1
(a)
∫ sin 6x dx = − 6 cos 6x + c
(b)
∫ cos 3x dx = 3 sin 3x + c
(c)
cos 2x 2 sin 2x + c
∫ 4=
(d)
3x dx
tan 3x + c
∫ 2 sec=
3
(e)
∫ 2 cos 6x − sin 4x dx = 3 sin 6x + 4 cos 4x + c
(f)
dx ∫ cos 4x + sec 3x dx
∫ cos 4x + cos =
3x
1
2
2
1
1
1
2
2
=
1
1
sin 4x + tan 3x + c
4
3
1
1
(g)
− cos 4x + sin 5x + 3 sin x + c
∫ sin 4x + cos 5x + 3 cos x dx =
4
5
(h)
∫ x + 6 tan 2x dx = 3 x + 3ln | sec 2x| + c
2
Unit 1 Answers: Chapter 15
1
3
© Macmillan Publishers Limited 2013
Page 6 of 32
(i)
2
2
−3
3
− sec 2 ( 2x) dx
1
1
tan ( 2 x) + c
=
− 2 −
2x
2
4
2
(j)
+ 9)
tan (6x + 9) + c
∫ 4sec (6x=
6
2
=
tan (6x + 9) + c
3
1
(a) ∫ sin (3x − 1) dx =
− cos (3x − 1) + c
3
1
(b) ∫ cos=
sin (2x + 3) + c
(2x + 3) dx
2
1
π

π

(c)
− ln sec  − 4x  + c
∫ tan  3 − 4x  dx =
4
3

3π 


3π 
(d)
∫ sec  x − 4  dx= tan  x − 4  + c
(e)
=
dx
sin (6x + 9) + c
∫ 7 cos (6x + 9)
6
(f)
− 6) dx
cos (4x − 6) + c
∫ 6 sin (4x =
2
π) dx tan (8x − π) + c
∫ 8 sec (8x − =
(g)
3
1
dx ∫ x
∫ x − sec ( 2 x)=
2
7
−3
2
1
(h)
+ sin qx + c
∫ cos qx dx =
q
(i)
∫ sin (px + π) dx = − p cos (px + π) + c
(j)
− ln | sec (4 − rx) | + c
∫ tan (4 − rx) dx =
r
(a)
∫ (3t + 1) dt= 21 (3t + 1) + c
(b)
−   (1 − 4t) + c
∫ (1 − 4t) dt =
4  4
(c)
(d)
(e)
1
1
6
1
3
7
1  1
4
1
=
− (1 − 4t)4 + c
16
1 (2t + 7)−3
−1
−4
(2t
+
7)
=
=
+c
(2t + 7)− 3 + c
∫
2
−3
6
1
2
1/2
3/2
∫ 6t − 1 dt =∫ (6t − 1) dt =6 × 3 (6t − 1) + c
1
= (6t − 1)3 / 2 + c
9
4
dt ∫ 4 (2 − 3t) − 1/ 2 dt
∫ 2 − 3t=
=
4 (2 − 3t)1/ 2
+c
−3
12
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 7 of 32
−8
(2 − 3t)1/ 2 + c
3
4
dt ∫ 4 (7 − 6t)− 1/ 2 dt
∫ 7 − 6t=
=
(f)
(g)
4 (7 − 6t)1/ 2
=
−
+c
6
12
−4
=
(7 − 6t)1/ 2 + c
3
6
6
(3t − 1) −3 dt
dt
3
∫ 7 (3t − 1)=
7∫
6 1 (3t − 1) − 2
=
×
+c
−2
7 3
1
=
− (3t − 1) − 2 + c
7
2
(h) ∫
=
dt ∫ 2 (4t − 3) −5 dt
5
(4t − 3)
2
(4t − 3) −4
= 4
+c
−4
1
=
− (4t − 3)− 4 + c
8
−1
=
+c
8 (4t - 3) 4
(i)
− 4)8 dt
∫ 3(6t=
3 (6t − 4)9
+c
6
9
1
(6t − 4)9 + c
18
dy
1
=
dx (3x + 2) 4
=
4
=
y
∫ (3x + 2)
−4
dx
1 (3x + 2)−3
+c
3
−3
−1
y
=
+c
9 (3x + 2)3
=
y
x = 0, y = 2 ⇒
=
2
c= 2 +
−1
+c
9 (8)
1
72
145
72
∴ Equation of the curve is
1
145
y=
−
+
3
72
9 (3x + 2)
=
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 8 of 32
dx
π

5 = cos  2t − 

dt
4
π

=
x ∫ cos  2t −  dt

4
1
π

x=
+ sin  2t −  + c

2
4
π
x = 1, t =
4
1
π
=
1
sin + c
2
4
2
c= 1 −
4
1
π
2

∴ x
=
sin  2t −  + 1 −

4
2
4
dy
6
= x (3 + 4x)
dx
=
y ∫ 3x + 4x 2 dx
y=
3 2 4 3
x + x +c
2
3
x = 1, y = 2 ⇒ 2 =
c=
−5
6
3 4
+ +c
2 3
3 2 4 3 5
x + x −
2
3
6
3
4
5
x = – 2, y = a ⇒ a = ( − 2)2 + ( − 2)3 −
2
3
6
32 5
a =6 −
−
3 6
11
= −
2
∴ y=
Try these 15.3
∫ sin x dx = ∫ sin x (sin x) dx
= ∫ sin x (1 − cos x) dx
= ∫ sin x (1 − 2 cos x + cos x) dx
=
∫ sin x − 2 sin x cos x + sin x cos x dx
(a)
5
2
2
2
2
2
4
2
=
− cos x +
(b)
4
2
1
cos3 x − cos5 x + c
3
5
∫ cos x dx = ∫ cos x cos x dx
5
Unit 1 Answers: Chapter 15
4
© Macmillan Publishers Limited 2013
Page 9 of 32
∫ cos x (1 − sin x) dx
= ∫ cos x (1 − 2 sin x + sin x) dx
=
∫ cos x − 2 cos x sin x + cos x sin x dx
=
2
2
2
4
2
4
2 3
1
sin x + sin 5 x + c
3
5
2
1 + cos2x
 1 + cos2x 
4
2
(c)
=
=
cos
x
dx
∫
∫  2  dx, cos x
2
1
= ∫ (1 + 2 cos2x + cos2 2x) dx
4
1
1 + cos4x
=
1 + 2 cos 2x +
dx
∫
4
2
1 3
1
=∫ + 2 cos 2x + cos 4x dx
4 2
2
1 3
1

=
x + sin2x + sin 4x  + c
4  2
8

=
sin x −
Try these 15.4
∫ tan x dx = ∫ tan x tan x dx
= ∫ tan x (sec x − 1) dx
= ∫ tan x sec x − tan x dx
= ∫ tan x sec x − (sec x − 1)  dx
4
(a)
2
2
2
2
2
2
2
2
2
2
tan 3 x
− tan x − x + c
3
5
3
2
(b) =
∫ tan x dx ∫ tan x (sec x − 1) dx
=
∫ (tan x sec x − tan x) dx
= ∫ tan x sec x − tan x (sec x − 1) dx
= ∫ tan x sec x − sec x tan x + tan x dx
=
=
3
2
3
2
3
2
3
2
2
1
1
tan 4 x − tan 2 x + ln sec x + c
4
2
Try these 15.5
(a)
(b)
1
=
3x) dx
(sin 9x − sin3x) dx
∫ (cos 6x sin
2∫
1 1
1

=
− cos9x + cos 3x  + c

2 9
3

1
cos2x) dx
(cos 10x + cos 6x) dx
∫ (cos8x=
2∫
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 10 of 32
11
1

sin 10 x + sin 6x  + c

2 10
6

1
− ∫ (cos 11x − cos 9x) dx
∫ (sin 10x sin x) dx =
2
11
1

=
−  sin (11x) − sin (9x)  + c
2 11
9

=
(c)
Exercise 15 C
1
4
∫ f (x) dx = 8
(a)
− ∫ f (x) dx =
−8
∫ f (x) dx =
(b)
dx 5=
(8) 40
∫ 5f(x)=
(c)
∫ f(x) dx + 4 ∫ x dx + ∫ f(x) dx
1
1
4
4
4
1
1
3
3
1
4
1
3
∫ f(x) dx + [ 2x 
4
=
2
3
1
1
= 8 + (18 – 2)
= 8 + 16
= 24
2
6
∫ g (x) dx = 12
(a)
2 ∫ g(x)
=
dx 2=
(12) 24
(b)
∫ 3 g (x) + 5 dx
= 3 ∫ g (x) dx + ∫ 5 dx
1
6
1
6
1
6
6
1
1
6
= 3[12] + [ 5x 1
= 36 + (30 – 5)
= 36 + 25
= 61
1
6
6
1
(c)
− ∫ g (x) dx =
− 12
∫ g (x) dx =
(d)
47
∫ [g (x) + Kx] dx =
47
∫ g (x) dx + K ∫ x dx =
6
1
6
6
1
1
6
 Kx 2 
12 + 
47
 =
 2 1
1
18K − K = 35
2
35K
= 35
2
K=2
4
4
0
0
3=
=
∫ f (x) dx 10,
∫ g (x) dx 6
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 11 of 32
∫ [f (x) + 3g (x)] dx
4
(a)
0
∫ f (x) dx + 3 ∫ g (x) dx
=
4
4
0
0
= 10 + 3(6) = 28
(b)
∫ f(x) f(x) dx cannot be evaluated
(c)
∫ f(x) dx + ∫ g(x) dx = ∫ f (x)dx − ∫ g (x)dx
4
0
4
0
4
4
0
4
0
0
= 10 – 6
=4
4
4
4
0
0
3 dx 2 ∫ g (x) dx + ∫ 3 dx
∫ 2g (x) +=
(d)
0
= 2 (6) + [3x ]0
4
= 12 + 3(4) = 24
∫ g(x) dx cannot be evaluated
5
(e)
4
0
y = x (1 + x2) ½
dy
 1
=
(1 + x 2 )1/ 2 + x   (2x) (1 + x 2 )−1/ 2
 2
dx
x2
=
(1 + x 2 )1/ 2 +
=
=
1 + x2
1 + x2 + x2
1 + x2
1 + 2x 2
1 + x2
Since
1
d
1 + 2x 2
[x (1 + x 2 ) 2 ] =
dx
1 + x2
Integrating both sides wrt x from 0 to 1
1
1
2
1 1 + 2x


⇒  x (1 + x 2 ) 2  =
∫0 1 + x 2 dx

0
2
1 1 + 2x
⇒ 21/ 2 =
∫0 1 + x 2 dx
3 + 6x 2
dx
0 (1 + x 2 )1/ 2
1
×3 ⇒ 3 2 =
∫
5
y = (1 + 4x) 3/2
dy 3
=
(1 + 4x)1/ 2 (4)
dx 2
= 6 (1 + 4x) 1/2
d
Since
[(1 + 4x)3/ 2 ] =
6 (1 + 4x)1/ 2
dx
Integrating both sides wrt x from 0 to 1
1
3


⇒ (1 + 4x) 2  =

0
1
∫ 6 (1 + 4x)
0
Unit 1 Answers: Chapter 15
1/ 2
dx
© Macmillan Publishers Limited 2013
Page 12 of 32
3
⇒ [5 2 − 1]
6
1
1
1
= ∫ (1 + 4x) 2 dx
6 0
1
1
1
(5 5 − 1)= ∫ (1 + 4x) 2 dx
0
6
sin x
y=
1 + cos x
dy (1 + cos x) cos x − sin x ( − sin x)
=
dx
(1 + cos x) 2
cos x + cos2 x + sin 2 x
(1 + cos x)2
1 + cos x
=
(1 + cos x)2
1
=
1 + cos x
=
d  sin x 
1
=


dx  1 + cos x  1 + cos x
⇒∫
π/ 4
0
π/ 4
 sin x 
1
dx =


1 + cos x
1 + cos x  0
π
sin
1
4
dx =
⇒∫
0 1 + cos x
π
1 + cos
4
2
2
2
2− 2
= =
= 2
×
2
2+ 2 2+ 2 2− 2
1+
2
2 2 −2
= 2 −1
=
2
x +1
y=
2 − 3x
dy (2 − 3x) (1) − (x + 1) (− 3)
=
dx
(2 − 3x) 2
2 − 3x + 3x + 3
=
(2 − 3x) 2
5
=
(2 − 3x) 2
π/ 4
7
d  x +1 
5

=
dx  2 − 3x  (2 − 3x) 2
1/ 2
1/ 2
 x +1 
1
5∫
dx
⇒
 =
0
(2 − 3x) 2
 2 − 3x  0
1/ 2
 3 / 2  1
1
⇒
5∫
dx
− =
0 (2 − 3x) 2
2
−
3
/
2
2

  
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 13 of 32
1/ 2
5 1
1
1
× =
dx =
2
∫
2 5 0 (2 − 3x)
2
2 sin 5θ cosθ = sin (5θ + θ) + sin (5θ – θ)
= sin 6θ + sin 4θ
⇒
8
9
π/4
1
 1

(sin 6θ + sin 4θ) dθ =  − cos 6θ − cos 4θ
∫0
0
4
 6
0
6π 1
4π   1
1
 1

=  − cos
− cos  −  − cos 0 − cos 0
 6

4 4
4  6
4
1 1 1 2
= + + =
4 6 4 3
2 sin 7θ cos 3θ = sin (7θ + 3θ) + sin (7θ – 3θ)
= sin (10θ) + sin (4θ)
π/4
2 sin 5θ cos θdθ = ∫
π/4
π
6
π/ 6
0
0
=
3θ dθ ∫ (sin 10θ + sin 4θ) dθ
∫ 2 sin 7θ cos
π
1
 1
6
=  − cos 10θ − cos 4θ 
4
 10
0
10π 1
4π   1
1
 1

=  − cos
− cos  −  − cos 0 − cos 0 
6
4
6   10
4
 10

− 1 1 1 1 17
=
+ +
+ =
20 8 10 4 40
10
∫
π/2
π
4
(cos 6θ cos 2θ) dθ
2 cos 6θ cos 2θ = cos (6θ + 2θ) + cos (6θ – 2θ)
= cos 8θ + cos 4θ
π
π
2
π
4
1 π2
1 1
1
2
θ
=
θ
θ
cos
6
cos
2
d
sin 8θ + sin 4θ 
π cos 8θ + cos 4θ dθ =
∫
∫

2 4
2 8
4
π
4
1  1
1
1
 1

 sin 4π + sin 2π  −  sin 2π + sin π  

2  8
4
4
 8

=0
=
11
π
2
∫ 2 sin 5θ sin θ dθ
0
– 2 sin 5θ sinθ = cos (5θ + θ) – cos (5θ – θ)
= cos6θ – cos 4θ
2 sin 5θ sinθ = cos 4θ – cos 6θ
π
π
2
0
π/ 2
1
1

θ ) dθ  sin 4θ − sin 6θ 
∫ cos 4θ − cos 6=
0
6
4
0
1
1
1
 1

=  sin 2π − sin 3π  −  sin 0 − sin 0 
6
6
4
 4

=0
12
– 2 sin 5θ sin 3θ = cos (5θ + 3θ) – cos (5θ – 3θ)
= cos 8θ – cos 2θ
∴ 2 sin 5θ sin 3θ = – cos 8θ + cos 2θ
∴ ∫ 2 2 sin 5θ sin
=
θ dθ
∫
π/ 2
π /3
Unit 1 Answers: Chapter 15
π/ 2
1
 1

− cos 8θ + cos 2θ dθ =  − sin 8θ + sin 2θ 
π /3
2
 8
 π /3
2 sin 5θ sin 3θ dθ = ∫
π/ 2
© Macmillan Publishers Limited 2013
Page 14 of 32
1
8π 1
2 π
 1
  1
=  − sin 4π + sin π −  − sin
+ sin

 8


2
8
3
2
3 
3
3 −3 3
−
=
16
4
16
2 cos7θ cos3θ = cos (7θ + 3θ) + cos (7θ – 3θ) = cos10θ + cos4θ
1
1
cos 7θ cos 3θ = cos 10θ + cos 4θ
2
2
π /12
π /12
π /12  1
1
1

1

θ
θ
θ
θ
+
θ
θ
θ
+
θ
(cos
7
cos
3
)
d
=
cos
10
cos
4
d
=
sin
10
sin
4

∫0
∫0  2
 20

2
8

0
=
13
1
10 π 1
4π 1
3
=
sin
+ sin
=
+
20
12 8
12 40 16
π/ 2
π/ 2
sin x dx = [ − cos x ]
14
∫
15
π
=
− cos + cos 0 =
1
2
3π / 2
3π / 2
 1

=
−
sin
3x
dx
cos
3x
∫0
 3


0
1
9π 1 1
=
− cos
+ =
3
2 3 3
16
∫
0
0
π/4
0
[ x + sin x ]0
π/ 4
(1 + cos x) dx.=
π
π π
2
=+ sin =+
4
4 4
2
π/ 4
π/4
1

17
cos
2x
dx
=
sin
2x
∫0
2


0
1
π 1
= =
sin
2
2 2
18
∫
π/2
0
=
π/ 2
1

cos 4x dx =  sin 4x 
4
0
1
sin 2π = 0
4
π/ 2
π
1
π 


+  dx  sin  7x +  
∫0 cos  7x=
2
2 0

7
1
8π 1
π
=
sin
− sin
7
2
7
2
1
= −
7
π/ 2
1 π/ 2
2
20
2x dx
1 − cos 4x dx
∫0 sin =
2 ∫0
π/2
19
using sin 2 x =
1 − cos 2x
2
π/2
=
1
1

x − sin 4x 

2
4
0
1 π 1

=  − sin 2 π 
2 2 4

Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 15 of 32
=
∫
21
π
4
π/ 4
0
cos 2 4x
=
dx
1 π/ 4
1 + cos 8x dx
2 ∫0
π/4
1
1

x + sin 8x 

2
8
0
=
1 + cos 2x
2
1 + cos 8x
using cos 2 4x =
2
using cos 2 x =
1 π 1

=  + sin 2π 
2 4 8

π
=
8
∫
22
π/ 2
0
cos 2 6x dx
1 π/ 2
1 + cos 12x dx
2 ∫0
π/2
1
1

=
+
x
sin
12x

2 
12
0
=
1 π 1

=  + sin 6π 
2  2 12

π
=
4
π/2
2  x
23
∫0 sin  2  dx
1 π/ 2
=
1 − cos x dx
2 ∫0
1
π/2
=
x − sin x ] 0
[
2
1 π
π
=
− sin 

2 2
2
1 π 
=
−1
2  2 
24
∫
∫
π /3
π
6
π/3
π
6
tan 2 (2x) dx
[sec 2 (2x) − 1] dx
π/3
1

=  tan (2x) − x 
2

π
6
1
 2π  π   1
 2π  π 
=  tan   −  −  tan   − 
 3  3 2
 6  6
2
− 3 π
3 π
=
− −
+
3 2
6
2
π
=
− 3−
6
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 16 of 32
25
4
∫ 4x − 6 x dx
= ∫ 4x − 6x dx
(a)
1
4
1/2
1
4
=  2x 2 − 4x 3/2 
1
(b)
= [32 − 4 ( 4)3 ] − [2 − 4]
=2
4
4
∫1 2 x − x dx
4
∫ 2x
=
1
1/2
− 4x −1/2 dx
4
26
4

=  x 3/2 − 8x1/2 
3

1
4
 4

=  (8) − 16  −  − 8
3
3

 

32 48 4 24
=
−
− +
3
3 3 3
4
=
3
0
4
(a)
∫−1 1 − 3x dx
0
∫ 4 (1 − 3x)
=
−1/ 2
−1
dx
0
4

=  × 2 (1 − 3x)1/ 2 
 −3
 −1
−8
=
[1 − 2]
3
8
=
3
(b)
∫
2
0
1 + 4x dx = ∫ (1 + 4x)1/ 2 dx
2
0
2
1 2

= × (1 + 4x)3/ 2 
4
3

0
1 3/ 2
1
13
=
[9 −=
1]
(26)
=
6
6
3
27
cos 3x = cos (2x + x)
= cos 2x cos x – sin 2x sin x
= (2 cos2x – 1) cos x – 2 sin2x cos x
= 2 cos3x – cos x – 2 cos x (1 – cos2x)
= 2 cos3x – cos x – 2 cos x + 2 cos3x
= 4 cos3x – 3 cos x
4 cos3x = cos3x + 3 cos x
1
3
cos3x
=
cos3x + cos x
4
4
π/ 2
π/ 2 1
3
3
∫π/ 4 cos x dx = ∫π/ 4 4 cos 3x + 4 cos x dx
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 17 of 32
π/ 2
3
1

=  sin 3x + sin x 
4
12
 π/ 4
3π 3
π  1
3π 3
π
1
=  sin
+ sin  −  sin
+ sin 
2 4
2  12
4 4
4
12
=
28
−1 3
2 3 2 2
2 1
+ −
−
= +
= (2 + 2)
12 4 24
8
3
3 3
(a)
5
∫ (3x + 1)
−1/ 2
1
dx
5
2

=  (3x + 1)1/ 2 
3
1
2 1/ 2
=
[16 − 41/ 2 ]
3
2
4
=
[4 − 2]=
3
3
(b)
∫
π/2
0
(sin 3x − cos 2x) dx
π/ 2
1
 1

=
 − 3 cos 3x − 2 sin 2x 

0
3π 1
 1
  1
=  − cos
− sin π  −  − 
2 2
 3
  3
1
=
3
29
(a)
9
∫ (1 + 7x)
−1/3
1
dx
9
 1 (1 + 7x) 2/3 
=

2 / 3 1
7
3
=
[642/3 − 82/3 ]
14
3
=
[16 − 4]
14
18
7
=
(b)
15
∫ (x − 3)
4
−3/ 2
dx
15
 − 2 (x − 3) −1/ 2 
=
4
 1

=
−2
− 1
 12 
12
3
=
2−
6
3
sin 3x = sin (2x + x)
= sin 2x cos x + cos 2x sin x
= 2 sin x cos2x + (1 – 2 sin2x) sin x
= 2 sin x [1 – sin2x] + sin x – 2 sin3x
= 2 sin x – 2 sin3x + sin x – 2 sin3x
= 3 sin x – 4 sin3x
=
2−
30
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 18 of 32
4 sin3x = 3 sin x – sin 3x
∫
π/ 4
0
4 sin 3 x dx = ∫
π/ 4
0
3 sin x − sin 3x dx
π/ 4
1


=
 − 3 cos x + 3 cos 3x 

0
π 1
3π  
1

=  − 3 cos + cos  −  − 3 + 



4 3
4
3
=
−3 2
2
1 −5
8 1
−
+3− =
2+ =
(8 − 5 2)
2
6
3
3
3 3
Exercise 15 D
∫ x (x + 1) dx
2
1
3
u = x3 + 1
du = 3x2 dx
1
du = x 2 dx
3
1
∫ x (x + 1) dx = ∫ 3 u du
2
3
3
1 2
u +c
6
1 3
=
(x + 1) 2 + c
6
x +1
2
∫ (6x 2 + 12x + 5)4 dx
u = 6x2 + 12x + 5
du
= 12x + 12
dx
1
du= (x + 1) dx
12
(6x2 + 12x + 5)4 = u4
x +1
1 1
∫ (6x 2 + 12x + 5)4 dx = 12 ∫ u4 du
1
=
u − 4 du
12 ∫
1  u− 3 
=

+c
12  −3 
=
1
=
−
+c
12u3
1
=
−
+c
2
12 (6x + 12x + 5)3
3
∫ cos x sin x dx
4
u = cos x
du = – sin x dx
– du = sin x dx
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 19 of 32
cos4x = u4
4
4
∫ cos x sin x dx = − ∫ u du
u5
=
−
+c
5
cos5 x
=
−
+c
5
3
∫ cos 4x sin 4x dx
4
u = sin 4x
du = 4 cos4x dx
1
du = cos 4x dx
4
1
∫ cos 4x sin 4x dx = 4 ∫ u du
3
3
1 4
u +c
16
1
=
sin 4 (4x) + c
16
6x
5
∫ 3x + 1 dx
u = 3x + 1
du = 3 dx
1
du = dx
3
Since 3x = u – 1
6x = 2u – 2
6x
∫ 3x + 1 dx
1 2u − 2
= ∫
du
3
u
2 1/2
=
u − u −1/2 du
3∫
2  2 3/2

=
u − 2u1/2  + c
3  3

4
4
= (3x + 1)3 / 2 − (3x + 1)1/ 2 + c
9
3
1
x
6
∫0 (7x + 2)3 dx
u = 7x + 2
du = 7dx
1
du = dx
7
7x = u – 2
u−2
x=
7
(7x + 2)3 = u3
When x = 0, u = 2
=
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 20 of 32
x = 1, u = 9
u−2
x
1 9 7
∫0 (7x + 2)3 dx = 7 ∫2 u3 du
1 9 −2
=
u − 2u −3 ) du
49 ∫2
9
1  1 1
=
−
+
49  u u 2  2
1
1  1 1   1 1  
− +  − − + 
49  9 81  2 4  
=
=
7
1 − 8 1
+
49  81 4 
1
=
324
∫
π/ 4
0
tan 3 x sec 2 x dx
u = tan x
du = sec2x dx
tan3x = u3
π
π
x = , u = tan = 1
4
4
x = 0, u = tan 0 = 0
∫
π/ 4
0
8
π/4
1
3
tan 3 xsec 2 x dx ==
∫ u du
0
1
1
1 4 
u 
=
 4 0 4
∫ 2 sin x cos x dx
4
0
u = sin x
du = cos x dx
sin4x = u4
π
π
2
x = , u = sin =
4
4
2
x = 0, u = sin 0 = 0
∫
π/4
0
2 sin 4 x cos x dx = 2 ∫
2 
=  u5 
 5 0
2 /2
0
u 4 du
2 /2
5
2  2
4 2 ×2
2
= =
=


5 2 
5 × 32
20
9
x+2
dx
0
x +1
u=x+1
du = dx
u+1=x+2
x +=
1 =
u u1/2
x = 1, u = 2
x = 0, u = 1
∫
1
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 21 of 32
∫
x+2
1
dx
x +1
2 u +1
=∫
du
1
u
0
2
∫u
=
1/2
1
+ u -1/2 du
2
2

=  u 3/2 + 2u1/2 
3
1
4
 2

= 
2 + 2 2  −  + 2
3
 3

10 2 8
−
3
3
=
10
∫
π/3
π/4
sin 2 x cos x dx
u = sin x
du = cos x dx
sin2x = u2
π
π
x= =
, u sin
=
3
3
π
π
x= =
, u sin
=
4
4
π /3
3
2
2
2
∫ sin x cos x dx = ∫
2
π/ 4
3
1  3
1  2
=
−
3  2 
3  2 
3
2
2
2
3/2
1 
u du =  u 3 
 3  2 /2
2
3
13 3
2
3
2
−
=
−
3 8
12
8
12
3 3−2 2
=
24
(3 3 − 2 2) (3 3 + 2 2)
=
24(3 3 + 2 2)
27 − 8
19
=
=
24 (3 3 + 2 2) 24 (3 3 + 2 2)
=
11
∫ x cos x du
2
(a)
u = x2
du = 2x dx
1
du = x dx
2
cos x2 = cos u
1
1
∫ x cos x dx = 2 ∫ cos u du = 2 sin u + c
2
=
1
sin x 2 + c
2
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 22 of 32
(b)
∫ x sin (x ) dx
2
3
u = x3
du = 3x2 dx
1
du = x 2 dx
3
1
∫ x sin x dx = 3 ∫ sin u du
2
3
1
cos u + c
3
1
= − cos x 3 + c
3
2 − 3x
∫ (4 + x)4 dx
u=x+4
du = dx
x=u–4
2 – 3x = 2 – 3 (u – 4) = 14 – 3u
2 − 3x
14 − 3u
∫ (4 + x)4 dx = ∫ u4 du
= −
(c)
∫14u − 3u du
=
-4
-3
14
3
+ 2 +c
3
−3u
2u
14
3
=
−
+
+c
3
3 (x + 4)
2 (x + 4)2
=
Exercise 15 E
1
2
dy
= 6x 2 − 4
dx
=
y ∫ 6x 2 − 4 dx
y = 2x3 – 4x + c
x = 1, y = 2 ⇒ 2 = 2 – 4 + c
c=4
y = 2x3 – 4x + 4
dy
(a)
= px − 5
dx
px – 5 = 4
x=–3⟹–3p=9
p=–3
dy
=
− 3x − 5
dx
(b)
y = ∫ − 3x − 5 dx
y=
−3 2
x − 5x + c
2
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 23 of 32
x = – 3, y = 2 ⇒ 2=
c=2−
− 27
+ 15 + c
2
3 1
=
2 2
−3 2
1
x − 5x +
2
2
dy
6
=
dx
4x + 1
y=
3
y = ∫ 6 (4x + 1)-1/2 dx
6 (4x + 1)1/ 2
+c
4
12
y = 3 (4x + 1)1/2 + c
x = 2, y = 6 ⇒ 6 = 3(9) 1/2 + c
c=–3
y = 3 (4x + 1)1/2 – 3
x = 6, y = h ⇒ h = 3 (25)1/2 – 3
= 15 – 3 = 12
dy
2
= 3x 2 + 2
dx
x
y = ∫ 3x 2 + 2x −2 dx
y=
4
2
+c
x
x = 1, y = 4 ⇒ 4 = 1 – 2 + c
c=5
2
y = x3 − + 5
x
dy
= 1 + 8x
dx
y = ∫ (1 + 8x)1/2 dx
y = x3 −
5
1 (1 + 8x)3/ 2
+c
8
32
1
y = (1 + 8x)3 / 2 + c
12
1
x = 3, y =
=8⇒ 8
(25)3/ 2 + c
12
125
=
8
+c
12
29
c=−
12
1
29
y = (1 + 8x)3/ 2 −
12
12
dy
= (4x − 6)3
dx
y = ∫ (4x − 6)3 dx
y=
6
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 24 of 32
1 (4x − 6) 4
+c
4
4
1
y = (4x − 6) 4 + c
16
1
x = 1, y = 2 ⇒ 2 = (16) + c
16
c=1
1
y = (4x − 6) 4 + 1
16
dy
4
= 1− 2
dx
x
y = ∫ 1 − 4x −2 dx
y=
7
4
+c
x
x = 4, y = 6 ⇒ 6 = 4 + 1 + c
c=1
4
y = x + +1
x
dy
= 4x 3 − 2x + 1
dx
y = ∫ 4x 3 − 2x + 1 dx
y= x+
8
9
y = x4 − x2 + x + c
x = 4, y = 5 ⇒ 5 = 44 – 42 + 4 + c
c = – 239
y = x4 – x2 + x – 239
f ′(x) = 3x 3 + 6x 2 - 2x + k
f (x) =
∫ 3x + 6x − 2x + k dx
3
2
3 4
x + 2x 3 − x 2 + kx + c
4
x = 0, f (x) = 4 ⇒ 4 = c
3
x = 1, f (1) = – 2 ⇒ − 2 = + 2 − 1 + k + 4
4
3 − 31
k =−2−5 =
4
4
3 4
31
f (x) = x + 2x 3 − x 2 − x + 4
4
4
dy
= 2x 4 + 3x 3
dx
y = ∫ 2x 4 + 3x 3 dx
f (x) =
10
y=
2 5 3 4
x + x +c
5
4
x = 2, y = – 2 ⇒ −2=
c=
− 10 64 60
−
−
5
5
5
Unit 1 Answers: Chapter 15
64
+ 12 + c
5
© Macmillan Publishers Limited 2013
Page 25 of 32
134
5
2
3
134
y = x5 + x 4 −
5
4
5
= −
Review Exercise 15
1
∫ f (x) dx = 42
(a)
∫ 6f (x) dx = (42) (6) = 252
(b)
∫ (f (x) − 4) dx = ∫ f (x) dx − ∫ 4 dx
9
1
9
1
9
9
9
1
1
1
= 42 − [4x]19
= 42 – [36 – 4] = 10
2
∫ g (x) dx = 12
3
0
(a)
∫ 5g (x) dx = 5 (12) = 60
(b)
∫ g (x) + 2] dx = ∫ g (x) dx + ∫ 2 dx
3
0
3
3
3
0
0
0
= 12 + [2x]30 = 12 + 6 = 18
∫ [g (x) + x] dx = − ( ∫ g (x) dx + ∫ x dx )
0
(c)
3
3
3
0
0
3
1 
=
− 12 −  x 2 
 2 0
9
33
=
− 12 − =
−
2
2
3
2
2 t + 4t
2
-3
-4
3
∫1 t 6 dt = ∫1 t + 4t dt
2
4 
 −1
=  2 − 3
3t 1
 2t
 1 1  1 4
= − −  − − − 
 8 6  2 3
=
4
37
24
∫ (x + 1) (2x − 3) dx
1
0
=
1
∫ 2x − x − 3 dx
2
0
1
1
2

=  x 3 − x 2 − 3x 
2
3
0
2 1
= − −3
3 2
− 17
=
6
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 26 of 32
5
∫ (4x − 1)
1/ 2
(a)
dx
1 (4x -1)3/ 2
+c
4
32
1
=
(4x - 1)3 / 2 + c
6
−1/ 2
(b)
∫ 4 (2 − 3t) dt
=
1/ 2
 1  (2 − 3t)
= 4 
+c
12
 −3 
8
=
−
2 − 3t + c
3
π/ 2
1 π/ 2
2
6
dx
=
1 + cos ( 4x ) dx
∫π/ 4 cos (2x)
2 ∫π / 4
π/2
1
1

=  x + sin 4x 
2
4
π / 4
=
1  π 1
 π 1

 + sin 2 π −  + sin π 

2 2 4
4 4

=
1 π π
=
2  4  8
7
8
1
1
using 2 sin 6x cos 4x = sin 10x + sin 2x
−1
1
1
1
(b)
∫ sin 8x sin 4x dx = ∫ 2 cos 12x + 2 cos 4x dx = − 24 sin 12x + 8 sin 4x + c
using – 2 sin 8x sin 4x = cos 12x – cos 4x
−1
1
sin 8x sin 4x =
cos 12x + cos 4x
2
2
x
y=
1 + 2x
dy (1 + 2x) − (x) (2)
=
dx
(1 + 2x) 2
1
=
(1 + 2x) 2
Since
⇒
9
1
∫ sin 6x cos 4x dx = 2 ∫ sin 10x + sin 2x dx = − 20 cos 10x − 4 cos 2x + c
(a)
d  x 
1

=
dx 1 + 2x  (1 + 2x) 2
x
1
+c=∫
dx
1 + 2x
(1 + 2x) 2
d
[cot x]
dx
=
d  cos x 


dx  sin x 
=
sin x ( − sin x) − (cos x) cos x
sin 2 x
=
− [sin 2 x + cos2 x]
sin 2 x
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 27 of 32
−1
= − cosec 2 x
sin 2 x
=
d
[cot 2x] = − 2 cosec 2 (2x)
dx
cot2 2x = cosec22x – 1
1
∫ cot 2x dx = ∫ cosec 2x − 1 dx = − 2 cot 2x − x + c
2
10
2
π/2
2
=
∫ (sin x + cos x) dx
0
∫
π/ 2
0
sin 2 x + 2 sin x cos x + cos 2 x dx
π/ 2
∫ 1 + sin 2x dx
=
0
π/ 2
1


=  x − cos 2x 
2

0
π 1
  1
=  − cos π  −  − 
2 2
  2
π 1 1 π
= + + = +1
2 2 2 2
x
11
y=
1 + x2
dy (1 + x 2 ) − (x) (2x)
=
dx
(1 + x 2 ) 2
=
1 − x2
(1 + x 2 )2
Since
⇒
12
d  x 
1 − x2
=

2 
dx 1 + x  (1 + x 2 ) 2
4x
4 − 4x 2
+
c
=
∫ (1 + x 2 )2 dx
1 + x2
d
[x sin x] = x cos x + sin x
dx
Integrate both sides from 0 to
π
4
π /4
⇒ [ x sin x ]0 =
∫ x cos x + sin x dx
π /4
0
π/4
π/4
π
π
sin =
x cos x dx + ∫ sin x dx
∫
0
0
4
4
π/4
2π
=
⇒
x cos x dx + [ − cos x]0π/4
∫
0
8
π/ 4
2
π


=
π ∫ x cos x dx +  − cos + cos 0 
0
8
4


π/ 4
2
2
∫0 x cosx dx = 8 π + 2 − 1
x
13
(a)
y= 2
x + 32
⇒
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 28 of 32
dy (x 2 + 32) − x (2x)
=
dx
(x 2 + 32) 2
=
32 − x 2
(x 2 + 32) 2
1
 x 
32 − x 2
∫0 (x 2 + 32)2 dx =  x 2 + 32 
0
1
(b)
1
33
2x
f (x) =
1 − 3x
(1 − 3x) (2) − 2x ( − 3)
f ′(x) =
(1 − 3x) 2
1
=
(1 − 3x) 2
=
14
d  2x 
1

=
dx 1 − 3x  (1 − 3x) 2
1
15
1
 2x 
1
⇒
dx
 =
∫
0
(1 − 3x) 2
1 − 3x  0
1
1
⇒ − 1 =∫
dx
0 (1 − 3x)2
1
6
⇒ − 6 =∫
dx
0 (1 − 3x)2
x
y=
2
(2x + 3)1/2
1
(2x 2 + 3)1/ 2 − x   (4x) (2x 2 + 3) −1/ 2
dy
2
=
dx
2x 2 + 3
2x 2 + 3 −
2x 2 + 3
2x 2 + 3
=
3
(2x + 3)3 / 2
2x
6
+c=∫
dx
2
1/ 2
2
(2x + 3)
(2x + 3)3/2
5x
y= 2
3x + 1
dy (3x 2 + 1) (5) − 5x (6x)
=
dx
(3x 2 + 1) 2
=
16
2x 2
=
2
5 − 15x 2
(3x 2 + 1)2
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 29 of 32
d  5x  5 (1 − 3x 2 )

=
dx  3x 2 + 1  (3x 2 + 1) 2
1
17
18
2
1 1 − 3x
 x 
dx
⇒ 2
=
 ∫
2 2
 3x + 1  0 0 (1 + 3x )
2
1 1 − 3x
1
⇒ =
dx
4 ∫0 (1 + 3x 2 ) 2
dy
(a)
= 4x 3 − 6x + 2
dx
y = ∫ 4x 3 − 6x + 2 dx
y = x4 – 3x2 + 2x + c
x = 1, y = – 5 ⇒ – 5 = 1 – 3 + 2 + c
c=–5
y = x4 – 3x2 + 2x – 5
(b) when x = 2, y = 24 – 3 (2)2 + 2 (2) – 5
= 16 – 12 + 4 – 5
=3
dy
= 4 (2)3 − 6 (2) +=
2 22
dx
Equation of the tangent at (2, 3) is
y – 3 = 22 (x – 2)
y = 22x – 41
3
x
∫124 1 − x 2 dx
u = 1 – x2
du = – 2x dx
1
− du = x dx
2
3
9
7
x = , u =1− =
4
16 16
1
1 3
x = , u =1− =
2
4 4
7
/16
1
− ∫ u -1/2 du
2 3/4
7/16
=  − u 
3/ 4
19
7
=
−
+
16
= 0.205
1
x2
∫
0
4 − x2
3
4
dx
x = 2 sinθ
dx = 2 cosθ dθ
4 − x 2 = 4 − 4 sin 2 θ
=
4(1 − sin 2 θ)
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 30 of 32
=
2
4 cos=
θ 2 cos θ
x2 = 4 sin2 θ
when x = 0 ⇒ sinθ = 0, θ = 0
1
π
x = 1, sinθ=
, θ=
2
6
2
1
x
∴∫
dx
0
4 − x2
∫
π/6
∫
π/4
0
0
4 sin 2 θ
(2 cos θ) dθ
2 cos θ
4 sin 2 θ dθ
π/ 4
1 − cos 2θ 
=
∫0 4  2  dθ
= 2∫
π/ 4
0
1 − cos 2 θ dθ
π/ 4
1


= 2 θ − sin 2θ 
2

0
π 1
= 2 − 
4 2
π
=
−1
2
dy
8
20
=
dx (4x − 5) 2
y = ∫ 8 (4x − 5) −2 dx
y=−
2
+c
4x − 5
x = 2, y = 6 ⇒=
6
c=
−2
+c
3
20
3
20
2
−
3 4x − 5
2
20
y=0⇒
=
4x − 5 3
3
= 4x − 5
10
3
4x = 5
10
53
x=
40
 53 
 , 0
 40 
y=
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 31 of 32
21
sin 3 x
∫ (1 + cos x)4 dx
u = 1 + cos x
du = – sin x dx
sin2x = 1 – cos2x
= 1 – (u – 1)2
= 1 – u2 + 2u – 1
= 2u – u2
sin 3 x
2u − u 2
dx
=
−
∫ (1 + cos x)4
∫ u4 du
u 2 − 2u
du
u4
u−2
= ∫ 3 du
u
1
2
= ∫ 2 − 3 du
u
u
=∫
1
−2
−3
2
Since u = 1 + cos x ⇒ ∫
22
1
∫ u − 2u du = − u + u + c
=
sin 3 x
1
1
+
+c
dx = −
4
(1 + cos x)
(1 + cos x) (1 + cos x) 2
x −1
∫ (3x − 6x + 5) dx
1
2
0
4
u = 3x2 – 6x + 5
du = (6x – 6) dx
x = 1, u = 3 – 6 + 5 = 2
x = 0, u = 5
1
x -1
1 2 1
∫0 (3x 2 − 6x + 5)4 dx = 6 ∫5 u 4 du
1 5 1
= − ∫ 4 du
6 2u
5
1 1
=
18  u3  2
1  1
1
− 

18  125 8 
= – 0.0065
π /6 sin θ
∫0 cos3θ dθ
u = cosθ, du = – sinθ dθ
=
23
θ = π 6, u = cos π 6 =
=
θ 0,=
u cos
=
0 1
3
2
3
∫
π /6
0
=
3
sin θ
1
 1 2 2 1
2
d
θ
=
−
du =  2  = −
3
3
∫
1
cos θ
u
3 2
 2u 1
1
6
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 32 of 32
24
25
x
1/ 2
∫ (1 − x ) dx
2 2
0
u = 1 − x2
du = – 2x dx
1
− du = x dx
2
1
1 3
x = , u =1− =
2
4 4
x = 0, u = 1
1 3/4 1
− ∫
du
2 1 u2
3/ 4
2 1 1
1 
=  = − =
2u
3 2 6
 1
3
1
∫0 (1 + x 2 )3/ 2 dx
x = tan u
⇒ dx = sec2u du
(1 + x 2 ) =
(1 + tan 2 u)3 / 2
= (sec2 u) 3/2
= sec3 u
3/2
When x = 3, u = tan -1 ( 3) =
π
3
x = 0, u = tan–1 (0) = 0
2
π /3 sec u
3
1
dx
=
∫0 (1 + x 2 )3/ 2
∫0 sec3 u du
π /3
1
=∫
du
0 sec u
=∫
π/3
0
cos u du
= [sin u ]0
π /3
π
= sin
=
3
3
2
Unit 1 Answers: Chapter 15
© Macmillan Publishers Limited 2013
Page 1 of 22
Chapter 16 Applications of Integration
Exercise 16A
1
(a)
x2 – 4x + 7 = 3x – 5
x2 – 7x + 12 = 0
(x – 3) (x – 4) = 0
x = 3, 4
x = 3, y = 9 – 5 = 4
x = 4, y = 12 – 5 = 7
P (3, 4)
Q (4, 7)
(b)
Area under curve
=
4
∫ x − 4x + 7 dx
2
3
4
2
1

=  x 3 − 2x 2 + 7x 
3
3
 64

= 
− 32 + 28 − (9 − 18 + 21)
 3

64 12 36 16
=
− −
=
3
3
3
3
1
16
Required area
=
(1) (7 + 4) −
2
3
11 16
=
−
2
3
33 − 32 1
= =
6
6
y = sin x, y = cos x
sin x = cos x
tan x = 1
π
x=
4
Calculate as 2 regions – A: area under y = sin x for x = 0 to x =
B: area under y = cos x for x =
Shaded
area
=
∫
π/ 4
0
sin x dx + ∫
π
π
to
4
2
π/ 2
π/ 4
π
4
cos x dx
=
[ − cos x ]0 + [sin x ]π/ 4
π/ 4
Unit 1 Answers: Chapter 16
π/ 2
© Macmillan Publishers Limited 2013
Page 2 of 22
2
2
+ 1+1−
2
2
= 2− 2
=−
3
∫ 4 − x dx
2
2
−1
2
1 

=  4x − x 3 
3  −1

8 
1

= 8 −  −  − 4 + 



3
3
=12 – 3
=9
4
∫ (x + 2) (3 − x) dx
= ∫ x − x + 6 dx
3
−2
3
2
−2
3
1
1 

= 6x + x 2 − x 3 
2
3  −2

9
8

 
= 18 + − 9  −  − 12 + 2 + 
2
3

 
9 8 125
= 19 + − =
2 3
6
5
y = 7x
y = 9x – x2
9x – x2 = 7x
x2 – 2x = 0
x (x – 2) = 0
x = 0, 2
when x = 0, y = 0
P (2, 14)
x = 2, y = 14
Total area
=
9
∫ 9x − x dx
2
0
9
1 
9
=  x 2 − x3 
3 0
2
729 729
=
−
2
3
243
=
2
2
Area of A =∫ 9x − x 2 dx −
0
Unit 1 Answers: Chapter 16
2 × 14
2
© Macmillan Publishers Limited 2013
Page 3 of 22
2
6
1 
9
= x 2 − x 3  − 14
3 0
2
8
= 18 − − 14
3
4
=
3
243 4
Area of=
B
−
2
3
721
=
6
4
8
3
= =
Ratio
721 721
6
Ratio 8 : 271
4x – x2 = 0
x (4 – x) = 0
x = 0, 4
4
 2 1 3
2
∫0 4x − x dx =2x − 3 x  0
64
= 32 −
3
96 − 64 32
= =
units 2
3
2
4
7
y= x + 2
x
3
4
Required area = ∫ x + 2 dx
1
x
4
3
4
1
=  x2 − 
x 1
2
9 4 1

=  −  −  − 4
2
3
2

 

20
=
3
8
x2 + 2x – 8 = x + 4
x2 + x – 12 = 0
(x + 4) (x – 3) = 0
x = – 4, 3
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 4 of 22
2
∫ x + 2x − 8 dx
Area under the curve from – 4 to 2=
2
−4
2
1

=  x 3 + x 2 − 8x 
3
 −4
8
  −64

=  + 4 − 16 − 
+ 16 + 32
3
  3

= |– 36 | = 36
Area under the curve from 2 to 3 =
3
∫ x + 2x − 8 dx
2
2
3
9
1

=  x 3 + x 2 − 8x 
3
2
8

= (9 + 9 − 24) −  + 4 − 16
3

8 10
=− 6 + 12 − =
3 3
7 × 7 49
Area of=
∆
=
2
2
49 10 343
Required area = 36 +
− =
2
3
6
(a) y = 2 + sin 3x
dy
= 3 cos 3x
dx
3 cos 3x = 0
⇒ cos 3x = 0
π 3π 5π 7 π 9π
3x = , , ,
,
2 2 2 2 2
π π 5π 7 π 9 π
x= , , ,
,
6 2 6 6 6
(b)
Area
=
π
3
0
∫ 2 + sin 3x dx
π /3
1


=  2x − cos 3x 
3

0
 2π 1
  1
= 
− cos π −  − 
 3 3
  3
=
2π 2
+
3 3
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 5 of 22
10
Area = ∫
π/ 4
0
2 cos 2x dx
= [sin 2x ]0
π/ 4
=1
11
1
+ 16x
x2
dy − 2
=
+ 16
dx x 3
dy
2
=0 ⇒ 3 =
16
dx
x
1
x3 =
8
1
x=
2
(a)
=
y
(b)
1 1
Area under the curve from =
to
4 2
1
2
1
4
1
∫ x + 16x dx
2
1/ 2
 1

= − + 8x 2 
x

1/ 4
1

=(− 2 + 2) −  − 4 + 
2

1
=3
2
1
2


12
A =  ,12  , gradient of OA =
= 24
1
2
equation of line OA: y = 24 x
When x =
1
,y=6
4
Area of=
trapezia
=
9
.
4
Shaded Area = 3
12
Required area = ∫
π /8
0
=6∫
π /8
= 3∫
π /8
0
0
1 1
  (12 + 6)
2 4
1 9 5
− =
2 4 4
6 cos 2 (2x) dx
1 + cos 4x
dx
2
(1 + cos 4x) dx
π /8
1


= 3  x + sin 4x 
4

0
π 1
 π 
= 3  + sin   
 2 
8 4
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 6 of 22
π 1
= 3 + 
8 4
13
x3 – 6x2 + 11x – 6 = (x – 1) (x2 – 5x + 6)
= (x – 1) (x – 2) (x – 3)
dy
= 3x 2 − 12x + 11
dx
3x2 – 12x + 11 = 0
12 ± 144 − 132
x=
6
= 1.43, 2.58
x = 0, y = –6
Area =
2
3
∫ x − 6x + 11x − 6 dx + ∫ x − 6x + 11x − 6) dx
3
2
1
3
2
2
2
3
11
11
1

1

=  x 4 − 2x 3 + x 2 − 6x  +  x 4 − 2x 3 + x 2 − 6x 
2
2
4
1
4
2
14
11
1

= (4 − 16 + 22 − 12) −  − 2 + − 6 
4
2


99
 81

+  − 54 +
− 18 − [4 − 16 + 22 − 12]
2
4

= – 2 – (– 2.25) + | – 2.25 – (– 2)|
1
=
2
y = 2x3 + 3x2 – 23x – 12
2x3 + 3x2 – 23x – 12
= (2x + 1) (x2 + x – 12)
= (2x + 1) (x + 4) (x – 3)
dy
= 6x 2 + 6x − 23 = 0 −2 + 3 + 23 − 12
dx
− 6 ± 36 − 4 (6) ( − 23)
x=
2 (6)
− 6 ± 14 3
12
= 1.521 , – 2.521
x = 1.521, y = 2 (1.521)3 + 3 (1. 521)2 – 23 (1.521) – 12
= – 33. 01
=
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 7 of 22
Area
=
∫
− 1/ 2
−4
3
2x 3 + 3x 2 − 23x − 12 dx + ∫ 1 2x 3 + 3x 2 − 23x − 12 dx
−
2
−1/ 2
3
23
23
1

1

=  x 4 + x 3 − x 2 − 12x  +  x 4 + x 3 − x 2 − 12x 
2
2
2
 −4
2
 −1/ 2
 1  1  4  1 3 23  1  2
23
 1   1

4
3
=  −  + −  −
(−4) 2 − 12 (− 4) 
 −  − 12  −   −  (− 4) + (− 4) −
2  2
2
 2    2

 2  2   2 
4
3
2
23 2
1 4
  1  1   1  23  1 
 1 
3
(3) − 12 (3)  −   −  +  −  −
+  (3) + 3 −
−  − 12  −  

2
2  2
2
  2  2   2 
 2  
97
 97

=  − (−72)  + − 72 −
32
 32

= 150.0625
15
−1
0
Area = ∫ x 3 + 3x 2 + 2x dx + ∫ x 3 + 3x 2 + 2x dx
−2
−1
−1
0
1

1

=  x 4 + x3 + x 2  +  x 4 + x3 + x 2 
4
 −2
4
 −1
1
 1

=  ( −1)4 + ( −1)3 + ( −1)2  −  ( −2)4 + ( − 2)3 + ( −2)2 
4
 4

 1

+ 0 −    (−1) 4 + (− 1)3 + (−1) 2 
 4

Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 8 of 22
1

 1
=  − 1 + 1 − (4 − 8 + 4) +  − 
4

 4
=
1 1 1
+ =
4 4 2
Exercise 16 B
b
1
V = π ∫ y 2 dx
a
2
 1 
= π∫ 
 dx

0  3x + 1 
1
1
= π∫
dx
3x + 1
0
1
π

=  1n 3x + 1 
3

0
π
= 1n 4
3
(a)
(1 − 2 sin x) 2 =
1 − 2 2 sin x + 2 sin 2 x
1
2
1 − cos2x 
=
1 − 2 2 sin x + 2 

2


=
2 − 2 2 sin x − cos 2x
π/4
V = π ∫ (1 − 2 sin x)2 dx
(b)
0
π/ 4
= π ∫ 2 − 2 2 sin x − cos 2x dx
0
π/4
1


= π  2x + 2 2 cos x − sin 2x 
2

0
3
 2π

π 1
π
= π 
+ 2 2 cos − sin  − (2 2 cos 0) 
4 2
2
 4

1
π

=π  + 2 − − 2 2 
2
2


3
π
=
π −2 2 + 
2
2
2
y = 3x – x
V = π ∫ y 2 dx
2
V = π ∫ (3x − x 2 ) 2 dx
1
2
= π ∫ 9x 2 − 6x 3 + x 4 dx
1
2
3
1 

= π 3x 3 − x 4 + x 5 
2
5 1

Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 9 of 22

32  
3 1 
= π  24 − 24 +  −  3 − +  
5  
2 5 

 32 3 1 
=π − − 
 5 2 5
47
=
π
10
π/4
4
V = π ∫ sin 2 x dx
0
π/ 4
=
π
1 − cos 2x dx
2 ∫0
π/4
π
1

=  x − sin 2x 
2
2
0
π π 1
π
− sin 

2 4 2
2
π π 1
=
−
2  4 2 
=
3π/2
5
V = π ∫ y 2 dx
0
3π/2
= π ∫ (1 + sin x)2 dx
0
3π / 2
= π ∫ 1 + 2 sin x + sin 2 x dx
0
3π / 2
= π ∫ 1 + 2 sin x +
0
1 1
− cos 2x dx
2 2
3π / 2
3
1
+ 2 sin x − cos 2x dx
2
2
0
=π ∫
3π / 2
1
3

= π  x − 2 cos x − sin 2x 
4
2
0
 9π

3π 1

= π 
− 2 cos
− sin 3π  − ( −2) 
2
4

 4

 9π

=
π  + 2
4

π/ 4
6
V = π ∫ sin 2 x dx + π ∫
π/ 2
π/ 4
cos 2 x dx
0
π/ 4
π/ 2 1 + cos 2x
1 − cos2x
dx + π ∫
dx
π/ 4
2
2
0
=π∫
π/ 4
π/ 2
π 1
π 1


x − sin 2x  +  x + sin 2x 

2 2
2
2
0

 π/ 4
π π 1 π π π 1
=
−
+
− −
2  4 2  2  2 4 2 
=
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 10 of 22
π π 
−1
2  2 
=
7
1
8
−3 −
4
Gradient of AB = 3
= 3= −
3 −1
2
3
Eq of AB:
4
y − 3 = − (x − 1)
3
4
4
y=− x+ +3
3
3
4
13
y=− x+
3
3
3
9
Volume under the curve = π ∫ 4 dx
x
1
3
 − 3
= π 3 
 x 1
 1
 26
=π  − + 3 = π
 9
 9
b
Volume under the line = π ∫ y 2 dx
a
13 
−4
= π ∫
x +  dx
3
3
1
2
3
π
(13 − 4x) 2 dx
9 ∫1
3
=
3
=
=
π  − (13 − 4x)3 


9
12
1
=
π  1 93 
− + 
9  12 12 
182
π
27
Required volume
=
=
8
182
26
π−
π
27
9
104
π
27
(a)
1
24
−5 −
5 = −6
Gradient of AB = 5
=
5 −1
4
5
6
Equation of AB = y – 5 =
− (x − 1)
5
−6
6
y=
x+ +5
5
5
−6
31
y=
x+
5
5
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 11 of 22
2
5
6
31 
Volume =
π ∫  − x +  dx
1
5
 5
π 5
=
( −6x + 31)2 dx
∫
1
25
(b)
5
π  (−6x + 31)3 
π  1 253 
=
=


− +

25 
− 18
1 25  18 18 
= 34.72 π
5
25
−25π  1 
dx =
4
x
3  x 3 1
1
Volume curve = π ∫
5
− 25π  1

− 1

3 125 
= 8.27 π
Required volume = 34.72π – 8.27π
= 26π
π

9
y = − cos  2x +  =
0

6
π π
2x + =
6 2
π
x=
6
π/6
π

volume below x-axis = π ∫ cos 2  2x +  dx
0
6

π π/6
π

=
1 + cos  4x +  dx
∫
3
2 0

=
π/6
1
π
π 

=  x + sin  4x +  
4
2
3 0

π  π 1
 1
 π 
 + sin π  − sin   

2  6 4
 4
 3 
=
=
π π
3
 −
 = 0.4824
2 6
8 
π /3
1
π
π 

volume above x-axis =  x + sin  4x +  
4
2
3   π /6

=
5π   π 1
π  π 1

 + sin
 −  + sin π  

2  3 4
3  6 4

=
π π
3 π
− 
 −
2 3 8
6
=
π π
3
 −
 = 0.4828
2 6
8 
Required volume = 0.965
π/ 4
10
V = π ∫ tan 2 x dx
0
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 12 of 22
π/ 4
=
π ∫ sec 2 x − 1 dx
0
=
π [tan x − x]0π/4
π π

=
π  tan − 
4 4

 π
=
π 1 − 
 4
3π / 8
11
π

V = π ∫ 4 sin 2  2x +  dx

4
0
3 π /8
π 

2π ∫ 1 − cos  4x +  dx
=
2 

0
3 π /8

1
π 

2  x − sin  4x +  
=π
4
2 0


12
  3π 1
π 
 1
=2π   − sin (2π)  −  sin 
2 
 4
 8 4
 3π 1 
=
2π  − 
 8 4
y = 2 sin x + 4 cos x
a
V = π ∫ (2 sin x + 4 cos x) 2 dx
0
a
=
π ∫ 4 sin 2 x + 16 sin x cos x + 16 cos 2 x dx
0
a
1 + cos 2x 
=
π ∫ 4 + 8 sin 2x + 12 
 dx
0
2


a
=
π ∫ 10 + 8 sin 2x + 6 cos 2x dx
0
=
π [10x − 4 cos 2x + 3 sin 2x]a0
=
π [10a − 4 cos 2a + 3 sin 2a + 4]
13
8
V = π ∫ y 2/3 dx
1
8
 3y5/3 
= π

 5 1
3π 5/3
=
[8 − 1]
5
3π
93
=
[32 − 1] =
π
5
5
2
14
x=
y
4
x2 = 2
y
5
4
dy
y2
2
V=π∫
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 13 of 22
5
 4
= π − 
 y 2
 4

= π  − + 2
 5

6
=
π
5
Review exercise 16
π/ 4
1
∫ sin x dx = [− cos x]
π /4
0
0
π
=
− cos + cos 0
4
2
=
−
+1
2
π/ 4
π
= sin − sin 0
∫ cos x dx = [sin x]
4
π/ 4
0
0
=
2
2
Required area =
2 
2
− 1 −
2 
2 
= ( 2 − 1) Square units
2
y = x2 – 2x, y = 6x – x2
2
x – 2x = 6x – x2
2x2 – 8x = 0
2x (x – 4) = 0
x = 0, 4
For y = x2 – 2x
if y = 0, x = 0, 2
so curve is below x-axis from x = 0 to x = 2, need to find the area in two sections,
0 < x < 2, 2 < x < 4
Area above y = x2 =
– 2x
2
∫ (x − 2x) dx
2
0
2
1

=  x3 − x 2 
3

0
=
8
4
−4 =
3
3
– 2x
Area under y = x2 =
4
∫ (x − 2x) dx
2
2
4
1

=  x3 − x 2 
3
2
64
8
20
=
− 16 − + 4 =
3
3
3
Area under y = 6x=
– x2
Unit 1 Answers: Chapter 16
∫ (6x − x ) dx
4
2
0
© Macmillan Publishers Limited 2013
Page 14 of 22
4
1 

= 3x 2 − x 3 
3 0

64 80
= 48 −
=
3
3
Required area = area under y = 6x – x2 – area under y = x2 – 2x + area above y = x2 – 2x
80 20 4 56
Square units
=
−
+ =
3
3 3 3
3
y = x3 – 6x2 + 8x.
dy
= 3x 2 − 12x + 8
dx
d2 y
= 6x − 12
dx 2
dy
=
0 ⇒ 3x 2 − 12x + 8 = 0
dx
12 ± 48
x=
6
12 ± 4 3
=
6
x = 3.15, 0.85
d2 y
when x = 0.85, =
6 (0.85) − 12 < 0 ⇒ max point
dx 2
d2 y
x = 3.15, =
6 (3.15) − 12 > 0 ⇒ min point
dx 2
x = 0.85, y = 3.08
x = 3.15, y = – 3.08
Area =
2
4
∫ x − 6x + 8x dx + ∫ (x − 6x + 8x) dx
3
2
0
3
2
2
2
4
1

1

=  x 4 − 2x 3 + 4x 2  +  x 4 − 2x 3 + 4x 2 
4
0  4
2
=(4 − 16 + 16) + (64 − 128 + 64) − (4 − 16 + 16)
4
= 4 + −4
= 8 Square units
y = x2 – 4
y = – 2x2
– 2x2 = x2 – 4
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 15 of 22
3x2 = 4
x=±
4
3
x=±
2 3
3
Area
=
For y = x2 – 4
2 3
3
∫ x − 4 dx
2
−2 3
3
2 3
1
 3
=  x 3 − 4x 
3
− 2 3
3
 64 3 
− 64
=
3−

27
 27 
128
= −
3
27
For y = – 2x2
Area
=
2 3
3
∫ − 2x dx
2
−2 3
3
2 3
 −2  3
=  x3 
3
− 2 3
3
=
− 32 3
27
∴ Required
area
=
128
32
3−
3
27
27
= 6.16 Square units
5
(a)
(b)
2−0
= 1
3 −1
Eq of AB: y – 0 = x – 1
y=x–1
y = 0 ⇒ x2 – 3x + 2 = 0
(x – 2)(x – 1) = 0
curve is above the x-axis from x = 2
Gradient of =
AB
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 16 of 22
Volume under the curve from x = 2:
3
3
2
2
π ∫ (x 2 − 3x + 2) 2 dx = π ∫ (x 4 − 6x 3 + 13x 2 − 12x + 4) dx
3
13
1

 21 16  31
=
π  x 5 − x 4 + x 3 − 6x 2 + 4x  =
π  −  =π
2
3
5
2
10 15  30
3
1

Volume under the line: π ∫ (x – 1) dx = π  (x − 1)3 
1
3
1
8
=
π
3
8
31
49
Required volume π −
π
=
π
3
30
30
3
6
(a)
3
2
V = π ∫ y 2 dx
b
a
2
 1 
V=π∫ 
 dx
1
 4x − 1 
2
2
= π ∫ (4x − 1) −2 dx
1
2
 1

= π  − (4x − 1)−1 
 4
1
1
 1
= π −
+ 
 28 12 
1
=
π
21
(b) y = 6
y = 5x – x2
5x – x2 = 6 ⇒ x2 – 5x + 6 = 0
(x – 2) (x – 3) = 0
x = 2, 3
3
3
2
2
Required area =∫ (5x − x 2 ) dx − ∫ 6 dx
3
1 
5
= x 2 − x 3  − [6x]32
3 2
2
8
 45
 
=  − 9  − 10 −  − [18 − 12]
2
3

 
1
= .
6
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 17 of 22
7
2
Area of region A = ∫ x 3 − 5x 2 + 6x dx
0
2
5
40
1
 
 8
=  x 4 − x 3 + 3x 2  = 4 −
+ 12  =
3
3
4
0 
 3
Area of region B =
3
1 4 5 3
 81
 8 5
3
2
2
∫2 ( x − 5x + 6x ) dx =  4 x − 3 x + 3x  2 =  4 − 45 + 27  −  3  = 12
3
83
= 32 : 5
5 12
y = x3 + 4x2 + 3x
= x (x2 + 4x + 3)
= x (x + 1) (x + 3)
y = 0, x = 0, – 1, – 3
dy
= 3x 2 + 8x + 3
dx
dy
=
0 ⇒ 3x 2 + 8x + 3 = 0
dx
− 8 ± 28
x=
6
= – 0.45, – 2.22
d2 y
x = –=
0.45,
6x + 8 = 6 ( − 0.45) + 8 > 0 min point
dx 2
d2 y
x = – 2.22, =
6 ( − 2.22) + 8 < 0 max point
dx 2
Ratio of A : B
8
(a)
(b)
y = x3 + 4x2 + 3x
When y = 0 ⇒ x3 + 4x2 + 3x = 0
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 18 of 22
x (x2 + 4x + 3) = 0
x (x + 3) (x + 1) = 0
x = 0, – 1, – 3
dy
Now = 3x 2 + 8x + 3
dx
dy
When x = 0,
=3
dx
dy
x = – 1,
=3 (− 1) 2 + 8 (− 1) + 3 =− 2
dx
dy
x = – 3,
= 3 (− 3) 2 + 8 (− 3) + 3 = 27 − 24 + 3 = 6
dx
dy
∴ When x = 0,
=3
dx
dy
x = – 1,
= −2
dx
dy
x = – 3,
=6
dx
−1
Area from x = – 1 to x = – 3 = ∫ (x 3 + 4x 2 + 3x) dx
(c)
−3
−1
4
3 
1
=  x 4 + x3 + x 2 
3
2  −3
4
27 
 1 4 3   81
=  − +  −  − 36 + 
 4 3 2  4
2
8
= Square units
3
b
9
V = π ∫ y 2 dx
a
2
4
6 
⇒V=
π∫ 
 dx
1
 x +2
= π ∫ 36 (x + 2)−2 dx
4
1
4
10

1 
= 36π  −

 x + 2 1
 1 1
= 36π  − + 
 6 3
= 6π.
Rotation about the y-axis:
d
V = π ∫ x 2 dy
c
Since x – y2 – 4 = 0
⇒ x = y2 + 4
x2 = (y2 + 4)2 = y4 + 8y2 + 16
3
V = π ∫ (y 4 + 8y 2 + 16) dy
0
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 19 of 22
3
8
1

= π  y 5 + y 3 + 16y 
3
5
0
 243

+ 72 + 48 
=π
5


3
= 168 π
5
11
d
V = π ∫ x 2 dy
c
y = 4x
y2
1
=x ⇒ x 2 = y 4
4
16
3
1
V = π ∫ y 4 dy
16
0
3
1

= π  y5 
 80  0
243
=
π
80
3
V = π ∫ (9 − x 2 )2 dx
12
2
3
= π ∫ (81 − 18x 2 + x 4 ) dx
2
3
1 

= π 81x − 6x 3 + x 5 
5 2


1
1


= π 81(3) − 6 (3)3 + (3)5 − 81 (2) − 6 (2)3 + (2)5  
5
5



46
=
π
5
13
Equation of the line OP:
y = 3x
1
1
0
0
Required volume = π ∫ 9x dx − π ∫ 9x 2 dx
=
9π 2 1
[x ]0 − 3π [x 3 ]10
2
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 20 of 22
9
π − 3π
2
3
=
π
2
=
14
(a)
Gradient of AB
=
Equation y =
(b)
2
2 2
2
=
π
π
4
2 2
x
π
Shaded volume = π ∫
π/ 4
0
sin x dx − π ∫
2
π/ 4
0
2
2 2 
x  dx

 π

π/ 4
π π/ 4
8  x3 
= ∫ 1 − cos 2x dx −  
2 0
π  3 0
π/4
1
π

x − sin 2x 
=

2
2
0
15
−
8  π3   1 
 
π  64   3 
π π 1
π  π2
= − sin  −
2 4 2
2  24
2
2
π
π π
π2 π
=
− −
=
−
8 4 24 12 4
For y = 9 – x2
Area
=
∫ (9 − x ) dx
3
2
0
3
1 

= 9x − x 3  = 27 − 9 = 18
3 0

For y = x2 + 1:
3
1

Area = ∫ (x + 1) dx =  x 3 + x  = 9 + 3 = 12
0
3
0
∴ Required area = 18 – 12 = 6
3
2
16
x2 = x
⇒ x4 = x
x = 0, 1
(a)
1
1
0
0
V = π ∫ ( x )2 dx − π ∫ (x 2 )2 dx
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 21 of 22
1
1
0
0
= π ∫ x dx − π ∫ x 4 dx
π 21 π 51
[x ]0 − [x ]0
2
5
π π 3π
= − =
2 5 10
when x = 1, y = 1
=
(b)
y = x2,
y=
1
x ⇒ y4 = x2
1
V = π ∫ y dy − π ∫ y 4 dy
0
=
17
π 21 π 51
[y ]0 − [y ]0
2
5
π π 3π
= − =
2 5 10
(a)
(b)
(c)
y = x2
y = 18 – x2
18 – x2 = x2
2x2 = 18
x2 = 9 ⇒ x = ± 3
when x = 3, y = 9
x = –3, y = 9
Points of intersections (3, 9) and (–3, 9)
(a)
3
1 3

2
∫−318 − x dx = 18x − 3 x  −3
= (54 – 9) – (– 54 + 9)
= 90
3
3
1 3 
∫−3 x dx =  3 x  −3 = 9 + 9 = 18
Required area = 90 – 18 = 72
sin 2x = cos x
⇒ 2 sin x cos x – cos x = 0
⇒ cos x (2 sin x – 1) = 0
1
cos x = 0, sin x =
2
π
π
x= , x=
2
6
π 3 π 
Points of intersections:  ,
 ,  , 0 
6 2  2 
3
18
0
2
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 22 of 22
(b)
π/ 2
−1

∫ sin 2x dx =  2 cos 2x  π/6
1
π
1
= − cos π + cos
2
2
3
1 1 3
= + =
2 4 4
π
π
π
1 1
π/ 2
∫π62 cos x dx = [sin x ]π/6 =sin 2 − sin 6 =1 − 2 =2
3 1 1
Required area = − =
4 2 4
π/ 2
π
6
Unit 1 Answers: Chapter 16
© Macmillan Publishers Limited 2013
Page 1 of 9
Chapter 17 Differential Equations
Exercise 17
1
dy
= 4x 3 − 2x + 1
dx
y= ∫ 4x 3 − 2x + 1 dx
y = x4 − x2 + x + c
2
3
dy
1 
= sin  x
2 
dx
1 
y = ∫ sin  x dx
2 
1 
y = –2cos  x + c
2 
dy
= cos2 x
dx
y = ∫ cos2 x dx
1
1 + cos 2x dx
2∫
1
1

y =  x + sin 2x  + c
2
2

dy
= x(x 2 + 2)
dx
y = ∫ x 3 +2x dx
y=
4
1 4
x + x2 + c
4
dy
y2
=
dx
3x + 1
1
dy ∫ (3x + 1)−1/ 2 dx
∫ y2=
y=
5
1 (3x + 1)1/ 2
+c
3 1/ 2
1 2
⇒=
−
(3x + 1)1/ 2 + c
y 3
1
y=
2
− (3x + 1)1/ 2 − c
3
dy
x2
= y2
dx
1
1
∫ y 2 dy = ∫ x 2 dx
=
⇒ − y −1
6
∫ y dy = ∫ x dx
−2
−2
Unit 1 Answers: Chapter 17
© Macmillan Publishers Limited 2013
Page 2 of 9
7
y −1 x −1
=
+c
−1 −1
1
1
− =
− +c
y
x
−1 + cx
=
x
 x 
y = −
 −1 + cx 
dy
sec2x
= cos2 x
dx
dy
= cos 4 x
dx
y = ∫ cos4 x dx
 1 + cos 2x 
y = ∫
 dx
2


1
y = ∫ 1+ 2 cos 2x + cos 2 2x dx
4
1
1 + cos 4x
y = ∫ 1 + 2 cos2x +
dx
4
2
1 3
1
y = ∫ + 2 cos 2x + cos 4x dx
4 2
2
1 3
1

y =  x + sin 2x + sin 4x  + c
4 2
8

dy
= y 2 + 2xy 2
dx
1
= ∫ (1 + 2x) dx
∫ y 2 dy
1
− = x + x2 + c
y
1
y=
2
−x − x − c
dx
= cos t + sin t
dt
x = ∫ cos t + sin t dt
2
8
9
10
x = sin t – cos t + c
x = 0, t = 0 ⇒ 0 = –1 + c, c = 1
x = sin t – cos t + 1
dy
= (x + 1)3
dx
y = ∫ (x + 1)3 dx
y =
1
(x + 1) 4 + c
4
y = 0, x = 2 ⇒ 0 =
Unit 1 Answers: Chapter 17
1 4
(3) + c
4
© Macmillan Publishers Limited 2013
Page 3 of 9
c= −
y=
11
13
1
81
(x + 1) 4 −
4
4
dy
= sin 2t
dt
y = ∫ sin 2t dt
y=
12
81
4
−1
cos2t + c
2
1π
y = 1, t = π / 4 ⇒ 1= − cos + c
2
2
c=1
1
y = − cos2t + 1
2
dy
= 2xy 2 − y 2
dx
1
∫ y2 dy = ∫ 2x − 1 dx
1
− = x2 − x + c
y
1
y= 2
x −x+c
dy (4x + 1)3
=
dx (y − 1)2
∫ (y − 1) dy = ∫ (4x + 1) dx
2
3
(y − 1)3 1
=
(4x + 1)4 + c
3
16
3
y – 1 = 3 (4x + 1) 4 + c
16
⇒
3
(4x + 1)4 + c
16
dy 3x 2 + 4x − 4
=
dx
2y − 4
y = 1+ 3
14
∫ (2y − 4) dy = ∫ 3x + 4x − 4 dx
2
1
(2y − 4) 2 = x 3 + 2x 2 − 4x + c
4
x = 3, y = 1 ⇒ 1 = 27 + 18 – 12 + c
c = –32
1
(2y − 4)2 = x 3 + 2x 2 − 4x − 32
4
(2y – 4)2 = 4x3 + 8x2 – 16x – 128
⇒
2y –4 = 4x 3 + 8x 2 − 16x − 128
1
y=
4x 3 + 8x 2 − 16x − 128 + 2
2
Unit 1 Answers: Chapter 17
© Macmillan Publishers Limited 2013
Page 4 of 9
15
16
17
dw
Iα 0+ =
dt
∫ I dwα= dt∫ −
Iw = –αt + c
t = 0, w = w0 ⇒ I w0 = c
I w = –αt + I w0
−α
w =
t + w0
I
dv
= − kv 2
dt
1
∫ v2 dv= ∫ −k dt
1
− =− kt + c
v
1
t = 0, v = v1 ⇒ − =
c
v1
1
1
− =− kt −
v
v1
1 ktv1 + 1
=
v
v1
v1
v=
ktv1 + 1
dx
∝ (9 − x)1/3
dt
dx
= k(9 − x)1/3
dt
dx
x = 1,
=1
dt
⇒ 1 = k (8)1/3
1
k=
2
dx 1
=
(9 − x)1/ 3
dt 2
2
∫ (9 − x)1/ 3 dx = ∫ dt
⇒ ∫ 2 (9 − x)−1/ 3 dx =
∫ dt
(9 − x) 3
= t+c
2
3
2
⇒−2
2
⇒ − 3(9 − x) 3 = t + c
t = 0, x = 1 ⇒ –3(8)2/3 = c
–12 = c
–3(9 – x)2/3 = t – 12
t − 12
(9 – x)2/3 =
−3
Unit 1 Answers: Chapter 17
© Macmillan Publishers Limited 2013
Page 5 of 9
3
 12 − t  2
9–x = 

 3 
3/2
x=
18
(a)
(b)
 12 − t 
9−

 3 
dp
−
∝ p − p0
dt
dp
− = k p − p0
dt
1
∫ − p − p dp =
∫ k dt
0
∫ −(p − p )
−1/ 2
0
dp =
∫ k dt
–2(p – p0) = kt + c
t = 0, p = 5p0
–2 (4p0)1/2 = c
c = −4 p 0
1/2
(c)
–2 (p – p0)1/2 = kt – 4 p 0
t = 2, p = 2p0 ⇒ –2
p 0 = 2k – 4
p0
2 p 0 = 2k
k=
(d)
p0
–2(p – p0)1/2 = t p 0 − 4 p 0
1
(p – p0)1/2 =
− t p0 + 2 p0
2
2
19

1 

p – p0 =  p 0  2 − t  
2 


1 

p = p0 + p0  2 − t  2
2 

dy
(a)
∝ (40 − y) 2
dt
dy
= k(40 − y) 2
dt
dy
(b)
y = 0,
= 4 ⇒ 4 = k(40) 2
dt
4
1
=
k =
1600 400
dy
1
=
( 40 – y ) 2
dt 400
1
1
dy =
dt
2
(40 − y)
400
1
1
∫ (40 − y)2 dy = ∫ 400 dt
1
1
⇒
=
t+c
40 − y 400
Unit 1 Answers: Chapter 17
© Macmillan Publishers Limited 2013
Page 6 of 9
40 – y =
1
t
+c
400
1
y = 40 –
t
+c
400
(c)
t = 0, y = 0 ⇒ 0 = 40 –
1
c
1
= 40
c
1
c=
40
1
400
= 40 −
t
1
t + 10
+
400 40
400
y = 35 ⇒ 35 = 40 –
t + 10
400
5=
t + 10
400
t + 10 =
= 80 cm
5
t = 70 minutes
dx
1
∝ 2
dt x
dx k
=
dt x 2
2
∫ x dx = ∫ k dt
y = 40 –
20
(a)
(b)
21
(a)
1 3
x = kt + c
3
t = 0, x = 0 ⇒ c = 0
1 3
x = kt
3
183
t = 2, x = 18 ⇒
= 2k
3
183 1
k=
× =
972
3 2
1
∴ x 3 = 972t
3
x3 = 2916t
303
x = 30 ⇒
= 2916t
3
t = 3.09
d2 y
= (3x + 2)2
dx 2
d2 y
⇒ ∫ 2 dx = ∫ (3x + 2) 2 dx
dx
Unit 1 Answers: Chapter 17
© Macmillan Publishers Limited 2013
Page 7 of 9
dy 1
=
(3x + 2)3 + A
dx 9
dy
1
3
∫ dx dx = ∫ 9 (3x + 2) +A dx
1
y=
(3x + 2) 4 + Ax + B
108
2
d y
= sin 2x cos2x
dx 2
d2y 1
⇒ 2 = sin 4x , since sin 4x = 2sin 2x cos 2x
2
dx
2
d y
1

∫ dx2 dx = ∫  2 sin 4x dx
dy
1
=
− cos 4x + A
dx
8
dy
 1

∫ dx dx =
∫  − 8 cos 4x + A dx
1
y = − sin 4x + Ax + B
32
⇒
(b)
22
23
d2 y
=
− 4t 2 + 5t + 3
2
dt
d2 y
⇒ ∫ 2 dt = ∫ (−4t 2 + 5t + 3) dt
dt
dy
4
5
⇒
=
− t 3 + t 2 + 3t + A
dt
3
2
dy
 4 3 5 2

∫ dt dt = ∫  − 3 t + 2 t + 3t + A dt
1
5
3
⇒y=
− t 4 + t 3 + t 2 + At + B
3
6
2
When t = 0, y = 1 ⇒ 1 = B
1
5
3
y=
− t 4 + t 3 + t 2 + At + 1
3
6
2
dy −4 3 5 2
=
t + t + 3t + A
dt
3
2
dy
t = 0,
=1⇒ 1 = A
dt
1
5
3
∴ y =− t 4 + t 3 + t 2 + t + 1
3
6
2
d2 y
=x+2
dx 2
d2 y
dx ∫ (x + 2) dx
∫ dx 2 =
dy 1
=
(x + 2)2 + A
dx 2
dy
1
2
∫ dx = ∫ 2 (x + 2) +A dx
Unit 1 Answers: Chapter 17
© Macmillan Publishers Limited 2013
Page 8 of 9
1
(x + 2)3 + Ax + B
6
8
x = 0, y = 0 ⇒ 0 = + B
6
−8 −4
=
B =
6
3
1
4
∴ y = (x + 2)3 + Ax −
6
3
dy 1
= (x + 2) 2 + A
dx 2
dy
x = 0,
=1⇒ 1 = 2 + A
dx
A = –1
1
4
∴=
y
(x + 2)3 − x −
6
3
dy 1
=
(x + 2)2 − =
1 0
dx 2
1
⇒ (x + 2)2 =
1
2
x+2= ± 2
x = –2 ± 2
d2 y
when x = – 2 + 2 ,
= – 2 + 2 + 2 > 0 min point
dx 2
d2 y
x =− 2 − 2, 2 =− 2 − 2 + 2 < 0 max point
dx
⇒ y=
24
dx
2
=
dt (1 − 2x)3
∫ (1 − 2x) dx =
∫ 2 dt
3
1
− (1 − 2x) 4 =
2t + c
8
(1 – 2x)4 = –16 t + c
1 – 2x = 4 −16t + c
x=
1 4
1 − −16t + c 
2
Unit 1 Answers: Chapter 17
© Macmillan Publishers Limited 2013
Page 9 of 9
25
(a)
d2 x
= 13 − 6t
dt 2
d2 x
dt ∫ (13 − 6t) dt
∫ dt 2=
dx
⇒
= 13t − 3t 2 + A
dt
dx
when t = 0,
= 30
dt
⇒ 30 =
A
dx
∴ =
− 3t 2 + 13t + 30
dt
When t = 3,
dx
= –3(3)2 + 13(3) + 30 = 42
dt
dx
= 0 ⇒ − 3t 2 + 13t + 30 = 0
dt
3t 2 − 13t − 30 =
0
(3t + 5) (t − 6) =
0
−5
=
t =
,t 6
3
Since t > 0, t = 6s
dx
(c) Now
=
− 3t 2 + 13t + 30
dt
dx
⇒∫
dt = ∫ ( −3t 2 + 13t + 30) dt
dt
13
⇒x=
− t 3 + t 2 + 30t + B
2
t = 0, x = 0 ⇒ 0 = B
13
∴ x =− t 3 + t 2 + 30t
2
t = 6 ⇒ x = –216 + 234 + 180 = 198m
(b)
when
Unit 1 Answers: Chapter 17
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
Module Tests
Module 1 Test 1:
___
____
____
√
1
(a)​​√ 80 ​​______
+ √​​ 245 ​​ +______
​​ 320 ​​
______
√ 15 × 5 ​​+ √
​​
49
×__5 ​​+ √​​ 64 × 5 ​​
= ​​
__
__
√ 5 ​​ + 7​​√ 5 ​​ + 8​​√ 5 ​​
= 4​​
__
√ 5 ​​
= 19​​
x = 19
(b) x3 + 3ax2 − bx + 4c = 0
Roots are 2, −3, 4
Sum of the roots = 2 + (−3) + 4
=3
Sum of the product of two roots =
(2)(−3) + (2)(4) + (−3)(4)
= −6 + 8 − 12 = −10
Product of the roots = (2)(−3) × (4)
= −24
Eq 11
x3 − 3x2 − 10x + 24 = 0
3a = −3 ⇒ a = −1
−b = −10 ⇒ b = 10
4c = 24 ⇒ c = 6
∴ a = −1, b = 10, c = 6
(c)
n
(i)​​∑​  ​​​r (2r − 1)
r=1
n
​  ​​​(2r2 − r)
= ∑
​​
r=1
n
∑
r=1
n
∑​  ​​​r
​  ​​​r2 − ​​
= 2 ​​
r=1
n(n + 1)
2n(n + 1)(2n + 1) _______
_______________
​​− ​​
​​
= ​​
2
6
n(n + 1)
= _______
​​
​​ [2(2n + 1) − 3]
6
n(n + 1)(4n − 1)
______________
= ​​
​​
6
40
(ii)​​∑​​​​r (2r − 1)
r = 18
40
17
r=1
r=1
∑​​​​r(2r − 1) − ∑
​​  ​​​​r (2r − 1)
​​
40(41)(159) 17(18)(67)
​​
​​
= __________
​​   ​​− _________
6
6
= 43460 − 3417 = 40043
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
2x + 1
(a)​​ ______ ​​ > 2
4x − 3
(2x + 1)(4x − 3) > 2(4x − 3)2
(2x + 1)(4x − 3) − 2(4x − 3)2 > 0
(4x − 3)[(2x + 1) − 2(4x − 3)] > 0
(4x − 3)(2x + 1 − 8x + 6) > 0
(4x − 3)(7 − 6x) > 0
2
3
4
​​ _34 ​​ < x < _​​  76 ​​
7
6
(b) (i)
3|x| − 2x − 1 = 20
Let |x| = x2
3x2 − 2x − 1 = 0
(3x + 1)(x − 1) = 0
x = −​​ _13 ​​, x = 1
Substitute x = 1,
x = −​​ _13 ​​
_1
_2
_2
3​​
(​  3 ​)​​ + ​​  3 ​​− 1 = ​​ 3 ​​
∴ x = 1
(ii) 2x2 − px + 2p + 1 = 0
a = 2, b = −p, c = 2p + 1
b2 − 4ac < 0
(−P)2 − 4(2)(2p + 1) < 0
P2 − 16P − 8 < 0
_______________
16
±
​
(−16)2 − 4(1)(−8) ​
√
____________________
​​
P = ​​
2
____
16 ± √​ 288 ​
_________
​​
P = ​​
2
__
16 ± 12​√ 2 ​
P = _________
​​  ​​
2 __
= 8 ± 4​​√ 2 ​​
__
__
(P − (8 + 4​​√ 2 ​​))(P + 8 + 4​​√ 2 ​​) < 0
8+4 2
8 + 4 2
__
__
8 − 4​​√ 2 ​​< P < 8 + 4​​√ 2 ​​
(c) P q
Nq (P ∩ Nq) N(P ∩ Nq)
T T F
F
T
T F T
T
F
F T F
F
T
F F T
F
T
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
3
(a) R.T.P.:
15r − 1 ≡ 7A
PROOF:
r = 0, 152 − 1 = 0
True for r = 0
Assume true for r = n
i.e. 15n − 1 = 7A
R.T.P. true for r = n + 1
i.e. 15n + 1 − 1 = 7B
PROOF:
15n + 1 −1 = 15n + 1 − 15n + 7A
= 15n(15 − 1) + 7A
= 15n(14) + 7A
= 7[(2)(15n) + 7]
= 7B
∴ divisible by 7
(b) (i)
R.T.P.
log2 b
​​
​​
loga b = _____
log2 a
PROOF:
Let y = loga b
⇒ ay = b
Taking log2 to base 2:
log2 ay = log2 b
y log2 a = log2 b
log2 b
y = ​​ _____ ​​
log2 a
log2 b
​​
​​
∴ loga b = _____
log2 a
(ii)
3 log2 x + 2 logx 2 = 7
2
3 log2 x + _____
​​ = 7
​​
log2 x
3(log2 x) − 7 log2 x + 2 = 0
y = log2 x
3y2 − 7y + 2 = 0
y = _​​ 13 ​​, y = 2
log2 x = _​​  13 ​​, log2 x = 2
_1
x = ​​2​​  ​ 3​​​, x = 22
__
√ 2 ​​, x = 4
x = 3​​
(iii) 22x + 1 = 34 − x
log 22x + 1 = log 34 − x
(2x + 1) log 2 = (4 − x) log 3
2x log 2 + log 2 = 4 log 3 − log 3
2x log 2 + x log 3 = 4 log 3 − log 2
x(2 log 2 + log 3 = 4 log 3 − log 2
4 log 3 − log 2
____________
x = ​​
​​
2 log 2 + log 3
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
(c) 32x − 2(3x) − 3 = 0
(3x)2 − 2(3x) − 3 = 0
y = 3x
y2 − 2y − 3 = 0
(y − 3)(y + 1) = 0
y = 3, y = −1
3x = 3 ⇒ x = 1
3x = −1 ⇒ No solution
4
f(x) = x2 + 5x + 6
2
5
25
​   ​)​​​  ​​+ 6 − ___
​​   ​​
= (​​​ x + __
4
2
2
5
1
​   ​)​​​  ​​− __
​​   ​​
= (​​​ x + __
4
2
1
Range: f(x) > − __
​​   ​​
4
1
(ii)​​ __________
​​ : Range: {−∞ to −4} ∪ {0, ∞}
x2 + 5x + 6
_____________
4
2
2 + 16x + 19 ​​
√
(b) x + 4x + 4 ​​ 4x
(4x2 + 16x + 16)
______________
​​
− ​​
3
3
​​
∴ 4 + _______
​​
(x + 2)2
A = 3, B = 2, C = 4
1
f(x) = ​​ __2 ​​
x
1
f(x + 2) = ​​ _______2 ​​ Shift to the left by 2 units
(x + 2)
3
​​ A stretch along the y-axis by factor 3
3f(x + 2) = _______
​​
(x + 2)2
3
3f(x + 2) + 4 = ​​ _______2 ​​+ 4 Move upwards by 4 units
(x + 2)
(c) (i)
h = {(a, 4), (a, 5), (b, 6), (c, 7), (d, 8), (e, 6), (f,8)}
(ii) a maps onto two different value
g has no corresponding mapping
(iii) a → 4
b → 6
c→7
d → 8
e → 6
f → 8
g→5
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
Module 1 Test 2:
1
(a)
1
1
3
5
7
1
3
5
7
3
3
1
7
5
5
5
7
1
3
7
7
5
3
1
Since all elements ∈ 5
⇒ Sin closed under multiplication module B
(i)
Identity is 1
(ii)Element
Inverse
1
1
3
3
5
5
7
7
(b) log x + log x2 + log x3 + … log xn
= log x + 2 log x + 3 log x + …n log x
= log x [1 + 2 + 3 + … n]
n(n + 1)
log x ​​[​ _______
​
​​
2 ]
(c) f(−1) = 4(−1) + 1 = −3
f(2) = 2(2)2 + 1 = 9
f(5) = 53 = 125
(iv) f(3) = 2(3)2 + 1 = 19
ff(3) = f(19) = 193 = 6859
2
(a) R.T.P.
n
n2
r2 + r − 1 _____
​​ = ​​
​​
∑​  ​​​ ________
​​
​​
n+1
r = 1 r(r + 1)
1
r2 + r − 1 1 + 1 − 1 __
PROOF: When n = 1, L.H.S. = ∑
​​ ​  ​​​ ​​ ________ ​​ = ​​ ________
​​ = ​​   ​​
2
r(r
+
1)
1(2)
r=1
∴ L.H.S. = R.H.S.
true for n = 1
Assume true for n = k
1
12
1
_____
R.H.S. = ​​
​​ = __
​​   ​​
1+1 2
k
k2
r2 + r − 1 _____
i.e. ​​∑​  ​​​ ________
​​
​​ = ​​
​​
r(r + 1)
k+1
r=1
R.T.P. true for n = k + 1
(k + 1)2
r2 + r − 1 _________
​​
​​ = ​​
​​
i.e. ​​∑​​​​ ________
r(r + 1)
(k + 1) + 1
r=1
k+1
k+1
k
(k + 1) + (k + 1) − 1
r + r − 1 __________________
​​ + ​​
​​
∑ r(r + 1) ∑​  ​​​ ________
​​
r(r + 1)
(k + 1)(k + 2)
r +r−1
​​ = ​​
PROOF: ​​
​​​​ ________
​​
2
r=1
Unit 1 Answers: Module Tests
2
2
r=1
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
k2 + 2k + 1 + k + 1 − 1
k2
​​
​​
= ​​ _____ ​​ + ___________________
k+1
(k + 1)(k + 2)
k2 + 3k + 1
k2
= _____
​​
​​ + ____________
​​     ​​
k + 1 (k + 1)(k + 2)
k2(k + 2) + k2 + 3k + 1
___________________
= ​​
​​
(k + 1)(k + 2)
(k + 1)3
k3 + 3k2 + 3k + 1 ____________
= ​​ ______________
​​ = ​​     ​​
(k + 1)(k + 2)
(k + 1)(k + 2)
(k + 1)2 _________
(k + 1)2
= ​​ _______
​​ = ​​
​​
k+2
(k + 1) + 1
∴ true for n = k + 1
Hence by P.M.I.
n
n2
r2 + r − 1 _____
​​ = ​​
​​
​​
∑​  ​​​ ________
​​
n+1
r = 1 r(r + 1)
R.T.P.:
n4 + 3n2 ≡ 4A
PROOF:
when n = 1, 12 + 3(1)2 = 4 = 4(1)
∴ true for n = 1
Assume true for n = k
i.e. k4 + 3k2 ≡ 4A
R.T.P. true for n = k + 1
(k + 1)4 + 3(k + 1)2 = 4B
PROOF:
(k + 1)4 + 3(k + 1)2
= k4 + 4k3 + 6k2 + 4k + 1 + 3k2 + 6k + 3
= k4 + 4k3 + 9k2 + 10k + 4
= 4A − 3k2 + 4k3 + 9k2 + 10k + 4????
= 4A + 4k3 + 6k2 + 6k + 4k + 4
= 4A + 4k3 + 6k(k + 1) + 4k + 4
= 4A + 4k3 + 3 × 2k(k + 1) + 4k + 4
= 4A + 4k3 + 3(4C) + 4k + 4
= 4[A + k3 + 3C + k + 1] = 4B
(c) (i)
If the pass mark is 50 then I will pass.
(ii) The examination is difficult and I will not pass.
(d) (i)
v(t ∩ ∼z)
= ∼t ∪ ∼z = ∼t vz
Tobago is NOT beautiful or Zico likes Tobago.
(ii) ∼t ∩ ∼z
Tobago is NOT beautiful and Zico does not like Tobago.
3
(a) (i)
f(x) = x2 − 4x + 7
= (x − 2)2 + 7 − 4
= (x − 2)2 + 3
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
(ii) f(x) is NOT 1 − 1 since c
f(1) = 1 − 4 + 7 = 4
f(−1) = 1 + 4 + 7 = 12
Since 1
d
A horizontal line drawn at
2
y = c cuts the graph above
more than once.
∴ f(x) is NOT r − 1
(iii) y = d does NOT cut the graph of y = f(x)
∴ NOT onto
(iv) Domain: x > 2
Range: y > 3
(v) Let y = (x − 2)2 + 3
x = (y − 2)2 + 3
x − 3 = (y −____
2)2
√
(y − 2) = ____
​​ x −3 ​​
y = 2 + ​​√ x −3 ​​ _____
∴ f−1(x) = 2 + √​​ x − 3 ​​, x > 3
(b) 2x3 − 5x2 + 6x + 2 = 0
5
α + β + γ = __
​​   ​​
2
6
αβ + αγ + βγ = ​​ __ ​​ = 3
2
2
__
αβγ = − ​​   ​​= −1
2
(i)
Σα2 = (Σα)2 − 2(Σαβ)
2
1
__
​   ​)​​​  ​​− 2(3)
= (​​​
2
25
1
​​− 6 = __
​​   ​​
= ___
​​
4
4
(ii)
Σα3 = (Σα)3 − 3Σα Σβ + 3Σαβγ
3
5
5
__
= (​​​
​   ​)​​​  ​​− 3​​(__
​   ​)​​(3) + 3(−1)
2
2
79
45
125 ___
​​− ​​   ​​− 3 = − ​​ ___ ​​
= ____
​​
8
2
8
(iii)
1
1 1 __
​​   ​​, ​​  γ ​​
Roots __
​​ α ​​, __
β
1
1 1 __
​​   ​​ + ​​  γ ​​
Sum of the roots = __
​​ α ​​ + __
β
βγ + αγ + αβ 3
= ____________
​​   ​​ = ​​ ___ ​​= −3
−1
αβγ
Sum of the product of 2 roots
1 ___
1
1
= ___
​​
​​ + ​​  αγ ​​ + ___
​​   ​​
αβ
βγ
γ + β + α ___
5
5/ 2
________
= ​​
​​ = ​​   ​​= − ​​ __ ​​
−1
2
αβγ
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
1
1
​​   ​​= −1
Product of roots = ____
​​   ​​ = ___
αβγ −1
1
1 1 __
​​   ​​, ​​  γ ​​is
Eqn with roots __
​​ α ​​, __
β
5
x3 + 3x2 + ​​(− ​ __ ​)​​x + 1 = 0
2
2x3 + 6x2 − 5x + 2 = 0
4
4x + 1
4x + 1
(a)​​ ______ ​​ < 1​​ ______ ​​> −1
x−2
x−2
(4x + 1)(x − 2) < (x − 2)2
(4x + 1)(x − 2) − (x − 2)2 < 0
(x − 2)[4x + 1 − (x − 2)] < 0
(x − 2)(3x + 3) < 0
–1
2
(x − 2)(4x + 1) > −(x − 2)2
(x − 2)(4x + 1) + (x − 2)2 > 0
(x − 2)[4x + 1 + x − 2] > 0
(x − 2)(5x − 1) > 0
1
5
2
1
−1 < x < 2
x < ​​ __ ​​, x > 2
5
1
Solution set: x: −1 < x < __
​​   ​​
5
(b) |zx − 1|2 − 6|2x − 1| + 8 = 0
y = |2x − 1|
y2 − 6y + 8 = 0
(y − 2)(y − 4) = 0
y = 2, 4
|2x − 1| = 2, |2x − 1| = 4
2x − 1 = 2, 2x − 1 = −2, 2x − 1 = 4, 2x − 1 = −4
5
3
3
1
x = __
​​   ​​, x = − ​​ __ ​​, x = __
​​   ​​, x = − ​​ __ ​​
2
2
2
2
(c) 3x − 9(32x) + 26(3x) − 24 = 0
y = 3x, y3 − 9y2 + 26y − 24 = 0
y = 2 is a root
y2 −7y + 12
________________
y3 − 9y2 + 26y − 24 ​​
y − 2​​√
− (y3 − 2y2)
−7y2 + 26y − 24
−(−7y2 + 14y)
1 + y − 24
1 + y − 24
0
2
∴ (y − 2)(y − 7y + 12) = 0
(y − 2) (y − 3) (y − 4) = 0
y = 2, 3, 4
3x = 2, 3x = 3, 3x = 4
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
PURE MATHEMATICS Unit 1
FOR CAPE® EXAMINATIONS
x lg 3 = lg 2, x = 1, x lg 3 = lg 4
lg 4
lg 2
x = ​​ ____ ​​ = x = ​​ ____ ​​ =
lg 3
lg 3
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 1 of 20
Module Tests
Module 2 Test 1:
1
(a)
sin 2x + sin 3x + sin 4x = 0.
sin 4x + sin 2x + sin 3x = 0.
 4x + 2x 
 4x − 2x 
⇒ 2 sin 
0
 cos 
 + sin 3x =

2
2 
⇒ 2 sin 3x cos x + sin 3x = 0
⇒ sin 3x [2 cos x + 1] = 0
1
sin 3x = 0, cos x = −
2
2π
3x = n
π x = 2nπ ±
, n ∈
3
nπ
x=
, n ∈
3
nπ

∴x =

3
 n ∈
π 

x = 2π n±

3  

(b) sin A + sin B + sin C
A+B
A−B
since C = 180 – (A + B)
= 2 sin
cos
+ sin (180 − (A + B))
2
2
 A + B
 A − B
= 2 sin 
cos 
+ sin(A + B)

 2 
 2 
 A + B
 A − B
 A + B
 A + B
2 sin 
cos 
+ 2 sin 
cos 



 2 
 2 
 2 
 2 
 A + B 
 A − B
 A + B 
2 sin 
cos 
+ cos 



 2 
 2 
 2  
 A + B 
A
 B 
= 2 sin 
  2 cos   cos   
 2 
2
 2 
C

 A
 B
= 4 sin  90 −  cos   cos  

 2
 2
2
(c)
A
 B
C
= 4 cos   cos   cos  
2
2
2
(i)
2sin θ + cos θ = R sin (θ + α )
= R sin θ cos α + R cos θ sin α
R cos α =2
1
, α= 26.6°
 ⇒ tan α=
R sin α =1 
2
A+B 
C
=  90 − 
2
2

C

C
since sin  90 −  =
cos  
2

2
since
R 2 = 22 + 12 ⇒ R = 5
(ii)
∴ 2 sin θ +=
cos θ
5 sin(θ + 26.6°)
2 sin θ + cos θ = 2
⇒ 5 sin(θ + 26.6) =2
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 2 of 20
 2 
θ + 26.6 =sin −1  
 5
=
θ + 26.6° 180n + (-1) n (63.4), n ∈ 
=
θ 360°n + 36.8°

 n ∈
=
θ 180°(2n + 1) − 90° 
1
4− 5
P(2, 3), Q(4, −1) R(3, −1)
Equation: (x − a) 2 + (y − b) 2 =
r2
[1]
(2 − a) 2 + (3 − b) 2 = r 2
2
2
2
[2]
(4 − a) + (−1 − b) =r
2
2
2
[3]
(3 − a) + (−1 − b) =r
2
2
[2] − [3] ⇒ (4 − a) − (3 − a) = 0
16 − 8a + a 2 − 9 + 6a − a 2 = 0
7 − 2a = 0
7
a=
2
2
2
7 
7
7

Substitute a = ,  2 −  + (3 − b) 2 = 4 −  + (−1 − b) 2
2 
2
2

9
1
+ 9 − 6b + b 2 = + 1 + 2b + b 2
4
4
10 = 8b
10 5
=
b =
8 4
7 5
centre  , 
2 4
(iii)
2
(a)
Max =
2
r2 =
9 
5  85
+ 3 −  =
⇒r=
4 
4  16
2
(b)
(c)
85
4
2
7 
5  65

Equation is  x −  +  y −  =
2 
4  16

x = 1 + 4 cos θ
x −1
cos θ =
4
y = −2 + 4 sin θ
y+2
sin θ =
4
(y + 2) 2 (x − 1) 2
sin 2 θ +=
cos 2 θ
+
16
16
2
2
(x − 1)
(y + 2)
⇒
+
=
1
16
16
⇒ (x − 1) 2 + (y + 2) 2 = 16 = 42
Circle centre (1, −2) radius 4
x = 2 + cos t,
y = 3 + 2 sin t
cos t = x − 2
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 3 of 20
y−3
2
sin t =
(y − 3) 2
+ (x − 2) 2
2
2
(y
− 3) 2
⇒ 1 = (x − 2) 2 +
22
Which is an ellipse with centre (2, 3)
 1  −2
 2 .  1 
   
 1  1 
(i)
cos θ =
 1  −2
 2  1 
   
 1  1 
2
sin 2 t + cos
=
t
3
(a)
1
1
=
6 6 6
⇒=
θ 80.4°
∴ angle AOB = 80.4°
 −2   1   −3 
→ → →      
AB =
OB − OA =
 1  −  2 =
 −1 
 1  1  0 
     
=
(ii)
 p   −2   p + 2 
→ → →     

BC = OC − OB =  2  −  1  =  1 
5  1   4 
    

→
→
Since AB is perpendicular to BC
→ →
⇒ AB . BC =
0
(b)
(i)
 −3  p + 2
 −1 .  1  =
0
  

 0  4 
⇒ −3p − 6 −1 = 0
3p = −7
−7
p=
3
2
x + y 2 − 8y − 9 =
0
y = 11 − x
x 2 + (11 − x) 2 − 8(11 − x) − 9 =
0
x 2 + 121 − 22x + x 2 − 88 + 8x − 9 =
0
2x 2 − 14x + 24 =
0
x 2 − 7x + 12 =
0
(x − 3) (x − 4) = 0
x= 3, 4
when x = 3, y =8
x = 4, y = 7
A = (3, 8), B = (4, 7)
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 4 of 20
4
 7 15 
Mid point of AB =  , 
2 2 
7−8
(ii) Gradient of AB =
= −1
4−3
Gradient of the ⊥ =1
Equation of ⊥ bisector is :
15
7
y − =x −
2
2
7 15
y=x − +
2 2
y=x+4
(c) 25x2 + 9y2 = 225
25x 2 9y 2
÷ 225 ⇒
+
=
1
225 225
x 2 y2
⇒
+
=
1
9 25
Centre (0, 0)
Length of the major axis is 10 units
Parametric equation:
x = 3 cos θ, y = 5 sin θ
1
 −2 
 
 
(a) =
r  2 + t  1 
 4
1
 
 
 x  1 − 2t 
⇒  y = 2 + t 
  

 z  4 + t 
⇒ x =1 − 2t 

y = 2 + t  t ∈  parametric equations
z = 4 + t 
(b)
(c)
(d)
 2
r .  3 = 8
 
 2
 x   2
⇒  y .  3 =8 ⇒ 2x + 3y + 2z =8 is the Cartesian equation of the plane
   
 z   2
Substitute x = 1 − 2t, y = 2 + t, z = 4 + t into the equation of the plane:
2(1 − 2t) + 3(2 + t) + 2(4 + t) = 8
2 − 4t + 6 + 3t + 8 + 2t =
8
t = −8.
Substituting t = −8 into the line we get
x = 1 + 16 = 17
y = 2 − 8 = −6
z = 4 − 8 = −4
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 5 of 20
5
(a)
(b)
 17 
 
intersection is  −6 
 −4 
 
cos 5x + cos x + cos 3x + cos 7x = 0
⇒ 2 cos 3x cos 2x + 2 cos 5x cos 2x = 0
2 cos 2x (cos 3x + cos 5x) = 0
4 cos 2x cos 4x cos x = 0
cos x = 0, cos 2x = 0, cos 4x = 0
π
π
π
x = 2n
π ± , 2x = 2nπ ± , 4x = 2nπ ±
2
2
2
π
x = 2n
π±
2

π 
x = nπ ±
n ∈ 
4 
1π

x=π±
n
2
8 
x = 1 + tan t
tan t = x − 1
Drawing a triangle:
cos t =
1
.
1 + (x − 1) 2
y = 2 + cos t.
1
y= 2 +
1 + (x − 2) 2
Module 2 Test 2
1
(a)
(i)
x 2 − 4x + y 2 − 9y − 12 =
0
(x − 2) 2 − 4 + (y − 3) 2 − 9 − 12 =
0
(ii)
(x − 2) 2 + (y − 3) 2 =
25
Circle centre (2, 3) radius 5
2−6 4
Gradient =
−
=
3−0 3
Equation: y − =
0
(iii)
3y = 4x − 24
3y − 4x + 24 = 0
y=x+8
Unit 1 Answers: Module Tests
4
(x − 6)
3
© Macmillan Publishers Limited 2013
Page 6 of 20
(x − 2) 2 + (x + 8 − 3) 2 =
25
x 2 − 4x + 4 + x 2 − 10x + 25 =
25
2x 2 − 14x + 4 =
0
x 2 − 7x + 2 =
0
x=
(b)
2
(a)
7 ± 41
2
x=1−t
y = 2 + 4t
z = 2 +t
2 + 3s = 1 − t
1 − s = 2 + 4t
s=2+t
[1] + [3] ⇒ 2 + 4s = 3
1
s=
4
[1]
[2]
[3]
1
7
Substitute into [3] ⇒ t = − 2 =−
4
4
1
7
Substitute s = , t = − into [2]
4
4
1
28
⇒1− = 2 −
4
4
3 −20
=
inconsistent
4
4
Since the lines are not parallel and do not intersect, they are skew lines
2 tan x
tan 2x =
1 − tan 2 x
2 tan 22.5
tan 2(22.5) =
1 − tan 2 22.5
2 tan 22.5
⇒1=
1 − tan 2 22.5
⇒ 1 − tan 2 (22.5) =
2 tan(22.5)
tan 2 (22.5) + 2 tan (22.5) − 1 =
0
tan(22.5) =
−2 ± 8
2
−2 ± 2 2
2
=− 1 ± 2
Since 22.5° is in the first quadrant
tan(22.5°) =− 1 + 2
=
(b)
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 7 of 20
3
5
4
tan A =
3
cos A =
5
13
5
tan B =
12
(i)
sin (A +B) = sin A cos B + cos A sin B
 4   12   3   5 
=     +   
 5   13   5   13 
sin B =
48 15 63
+
=
65 65 65
cos(A + B) = cos A cos B + sin A sin B
=
(ii)
 3  12   4  5  36 20 56
=
 +    = +
 5  13   5  13  65 65 65
=  
sin θ + sin 3θ − sin 2θ
cos θ + cos3θ − cos 2θ
(c)
=
=
2 sin 2θ cos θ − sin 2θ
2 cos 2θ cos θ − cos 2θ
sin 2θ [2 cos − 1]
cos 2θ [2 cos − 1]
sin 2θ
= tan 2θ
cos 2θ
x = 4 + 2 tan t
x−4
tan t =
2
(x
− 4) 2
tan 2 t =
22
y = 3 + sec t
sec t = y − 3
sec2 t = (y − 3)2
=
3
(a)
Unit 1 Answers: Module Tests
[1]
[2]
© Macmillan Publishers Limited 2013
Page 8 of 20
[2] − [1] ⇒ sec 2 t − tan 2 t = (y − 3) 2 −
1 =(y − 3) 2 −
(x − 4) 2
4
∴ Cartesian equation is (y − 3) 2 −
(b)
4
(a)
(x − 4) 2
22
x + 2y + z = 4
−x + y + 2z = 6
 1  −1
 2 .  1 
   
 1  2 
cos θ=
=
 1  −1
 2  1 
   
 1  2 
3
6 6
=
(x − 4) 2
=
1
4
3 1
=
6 2
θ = 60°
(i)
y2 = 16x
(t2, 4t) ⇒ Gradient of the tangent is
2
t
Equation of the tangent:
2
y − 4t=
(x − t 2 )
t
ty − 4t 2 = 2x − 2t 2
ty − 2x = 4t 2 − 2t 2
(b)
ty − 2x = 2t 2
(ii) At t = 3, 3y − 2x = 18
1 1
2
=
t
, y − 2x
=
3 3
9
8
160
[1] − [2] ⇒ y =
3
9
20
1
, x=
y=
3
2
(i)
cos 5θ + cos θ + 2 cos 3θ
= 2 cos 3θ cos 2θ + 2 cos 3θ
= 2 cos 3θ (cos 2θ + 1)
= 2 cos 3θ (2 cos2 θ)
= 4 cos2 θ cos 3 θ
(ii) 4 cos2 θ cos 3θ = 0
2
⇒ cos
=
θ 0, cos
=
3θ 0
cos θ = 0 cos 3θ = 0
π
π
θ= 2nπ ± , 3θ= 2nπ ±
2
2
2nπ π
=
θ
±
3
6
Unit 1 Answers: Module Tests
[1]
[2]
© Macmillan Publishers Limited 2013
Page 9 of 20
π
2 
 n ∈
2nπ π 
±
3
6 
x2 + y2 − 7x + 2y + a = 0
x = 7, y = 1
⇒ 49 + 1 − 49 + 2 + a =
0
a = −3
x2 − 7x + y2 + 2y − 3 = 0
∴θ= 2nπ ±
(c)
2
7
49

+ (y + 1) 2 − 1 − 3 =
0
 x −  −
2
4
2
7
49
65

2
+ 4=
 x −  + (y + 1) =
2
4
4

7

Centre  , − 1
2

1 − (−1) 2 4
Gradient=
= =
7
7 7
7−
2
2
4
Equation: y −=
1
(x − 7)
7
7y − 7 = 4x − 28
7y = 4x − 21
Let the coordinates be (x1, y1)
x1 + 7 7
= ⇒ x1 = 0
2
2
y1 + 1
=− 1 ⇒ y1 =− 3
2
(0, −3)
Module 3 Test 1
1
(a)
(i)
d
(x + 1)(4x 2 + 1)1/ 2 
dx 
1
=
(x + 1) (8x)(4x 2 + 1)−1/ 2 + (4x 2 + 1)1/ 2
2
4x(x + 1)
=
+ 4x 2 + 1
2
4x + 1
=
=
(ii)
4x(x + 1) + 4x 2 + 1
4x 2 + 1
8x 2 + 4x + 1
4x 2 + 1
d
cos3 (3x −=
2)  3cos 2 (3x − 2) [3 (− sin(3x − 2)) ]
dx
= –9 cos2 (3x–2) sin(3x–2)
Unit 1 Answers: Module Tests
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Page 10 of 20
2
(b)
(i)
∫ f(x) dx = 8
0
2
1 3 
∫0 [x − f (x)] dx =  3 x  − ∫0 f (x)dx
0
8
−16
= −8 =
3
3
y = x2 + 2
y = 14 – x
x2 + 2 = 14 – x
x2 + x –12 = 0
(x + 4) (x – 3) = 0
x = 3, –4
2
2
2
(ii)
3
Area under the curve = ∫ x 2 + 2 dx
−4
3
1

=  x 3 + 2x 
3
 −4
 −64

− 8
= (9 + 6) – 
 3

88 133
square units
= 15 +
=
3
3
3
=
Area under the line
∫ (14 − x) dx
−4
3
(c)
 (14 − x) 2 
−121
203
= −
+ 162=
square units
 =
2
2
2

 −4
203 133
Required area
=
−
2
3
1
= 57
6
Total surface area = x2 + 3xh.
v = x2h
x2h = 0.064
0.064
h=
x2
 0.064 
∴ A = x 2 + 3x  2 
 x 
0.192
= x2 +
x
dA
0.192
= 2x − 2
dx
x
dA
0.192
= 0 ⇒ 2x =
dx
x2
3
x = 0.096
x = 0.458 cm
0.192
When x = 0.458, A = (0.458)2 +
0.458
Unit 1 Answers: Module Tests
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Page 11 of 20
= 0.629 m2
2
(a)
(i)
64x 3 − 125
2
5
x → 12x − 11x − 5
lim
4
= lim
x→
(4x − 5) (16x 2 + 20x + 25)
(4x − 5) (3x + 1)
5
4
16x + 20x + 25 
= lim 

5
3x + 1
x→ 

4
2
2
 5
 5
16   + 20   + 25
 4
 4
=
 5
3  + 1
 4
= 15.79
(ii)
 sin 9x 
lim 

x→0 
x 
 sin 9x 
= lim 9 
x→0
 9x 
(b)
(i)
(ii)
 sin 9x 
= 9 lim 

x → 0  9x 
=9(1)
=9
dy
4
= 3x − 2
dx
x
4

y = ∫  3x − 2  dx

x 
3
4
y = x2 + + c
2
x
5
5 3
x=1, y = ⇒ = + 4 + c
2
2 2
c = –3
3
4
∴ y = x2 + − 3
2
x
dy
At stationary points
=0
dx
4
⇒ 3x − 2 =
0
x
3x3 = 4
4
x3 =
3
4
x = 3 = 1.1
3
3
4
=
y
(1.1)2 +
−3
2
1.1
= 2.45
Unit 1 Answers: Module Tests
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Page 12 of 20
(c)
(i)
d2 y
8
=3+ 3
2
dx
x
d2y
=3+ 8
= 6 + 3=9 > 0
x 3 = 4 / 3,
4/3
dx 2
∴ min point at (1.1, 2.45)
d2 y
Find =
4x 3 + 3x 2
dx 2
dy
⇒
= 4x 3 + 3x 2 dx
dx ∫
dy
= x 4 + x3 + A
dx
Integrating again wrt x:
y = ∫ x 4 + x 3 + A dx
1 5 1 4
x + x + Ax + B
5
4
When x = 0, y = 1 ⇒ 1 = B
dy
= x 4 +x 3 + A
dx
dy
x = 0,
=0⇒0=A
dx
∴ solution is
1
1
y = x 5 + x 4 +1
5
4
dy
∝ (9 − y)1/3
dt
dy
⇒
= k (9 − y)1/ 3
dt
dy
when y = 1,
= 0.2
dt
⇒ 0.2 =
k(8)1/3
0.2
k=
2
= 0.1
dy
∴ = 0.1 (9 − y)1/3
dt
dy
= 0.1(9 − y)1/ 3
dt
1
⇒∫
=
dy ∫ 0.1 dt
(9 − y)1/3
y =
(ii)
3
(a)
(i)
(ii)
⇒ ∫ (9 − y) −1/3 =
dy ∫ 0.1 dt
⇒
(9 − y)2 / 3
=
0.1t + c
−2 / 3
3
when t = 0, y = 1 ⇒ − (8) 2/3 =
c
2
c = –6
Unit 1 Answers: Module Tests
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Page 13 of 20
3
2/3
∴− (9 − y)=
0.1t − 6
2
3/ 2
1
1 

t + 4⇒ 9− y = 4− t
15
15


1 3/ 2
y =9 − (4 − t)
15
dy
(iii) At max height
=0⇒9−y=0⇒ y=9
dt
When y = 9, t = 15(4) = 60 yrs
f '(x) = 4x3 + 6x2 + 2x + k
f(x) = ∫ 4x 3 + 6x 2 + 2x + k dx
(9 – y)2/3 = −
(b)
(c)
4
(a)
f(x) = x4 + 2x3 + x2 + kx + c
f(0) = 5 ⇒ 5 = c
f(1) = 10 ⇒ 10 = 1 + 2 + 1 + k + 5
k=1
∴ f(x) = x4 + 2x3 + x2 + x + 5
y = Acos3x + Bsin3x
dy
=
− 3A sin 3x + 3Bcos3x
dx
d2 y
=
− 9Acos3x − 9Bsin3x
dx 2
d2 y
+ 9y =
−9A cos3x − 9Bsin 3x + 9A cos3x + 9Bsin 3x
dx 2
=0
d2 y
Hence
+ 9y = 0
dx 2
(i)
y = x3 + bx2 + x + c
dy
= 3x 2 + 2bx
dx
d2 y
= 6x + 2b
dx 2
d2 y
when x = 1,
= 0 ⇒ 0 = 6 + 2b
dx 2
b = –3
when x = 1, y = 4
⇒ 4 =1 − 3 + 1 + c
c=5
b = –3, c = 5
dy
= 3 − 6 = −3
(ii) when x = 1,
dx
Equation of the tangent: y – 4 = − 3(x – 1)
y + 3x = 7
1
(b)
x2 + 2
∫0 (3x 3 + 18x + 1)3 dx
Unit 1 Answers: Module Tests
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Page 14 of 20
(c)
u = 3x 3 + 18x + 1
du
= 9x 2 + 18
dx
du = 9(x2 + 2)dx
1
du = (x 2 + 2) dx
9
When x = 1, u = 3 + 18 + 1 = 22
x = 0, u = 1
1
22
du
1
22
22
x2 + 2
1
1 1
−3
9
dx =
u du =
∴∫
−  2
∫1 u 3 =
(3x 3 + 18x + 1)3
9 ∫1
18  u 1
0
1  1

=
−  2 − 1 =
0.055
18  22

π /6
π /6
1
=
θ dθ
cos 6θ + cos
θ 2θ d
(i)
∫0 cos 4θ cos2
2 ∫0
π /6
1 1
1

sin 6θ + sin=
2θ 

2 6
2
0
=
3
8
π /3
π /3
(ii)
1 1
1
π
sin π + =
sin 

2 6
2
3
1
1 + cos 6x dx
2 ∫0
2
dx
∫ cos 3x=
0
π/3
1
1

=
x + sin 6x 

2
6
0
=
1 π 1

+ sin 2 π − 0 
2  3 6

π
=
6
Module 3 Test 2
1
(a)
point of intersection:
–x2 + 6x + 3 = 2x + 6
x2 – 4x + 3 = 0
(x – 1) (x – 3) = 0
x = 1, 3
3
Area under the curve = ∫ -x 2 + 6x + 3 dx
1
3
 1

=  − x 3 + 3x 2 + 3x 
3

1
 1

= (–9 + 27 + 9) –  − + 3 + 3
 3

1
= 21 square units
3
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 15 of 20
3
Area under the line = ∫ (2x + 6) dx
1
=  x + 6x 
2
3
1
= (9 + 18) – (1 + 6)
= 20 square units
1
1
∴ shaded area = 21 − 20 = 1 square units
3
3
b
(b)
V = π ∫ y 2 dx
a
π /8
V = π ∫ cos 2 (2x)dx
0
π /8
1 + cos 4x
dx
2
0
= π∫
π /8
=
π
1

x + sin 4x 
2 
4
0
π π 1
π
+ sin 

2 8 4
2
π π 1
=
+
2  8 4 
d  x 
(c) (i)
dx  x 2 + 4 
=
=
(x 2 + 4)(1) – x(2x)
(x 2 + 4)2
=
−x 2 + 4
(x 2 + 4) 2
d  x 
4 - x2
 2
= 2
dx  x + 4  (x + 4) 2
Since
4 − x2
 x 
⇒  2
=
∫ 2 2 dx

 x + 4  0 0 (x + 4)
2
2
2
⇒
2
4-x 2
=
∫0 (x 2 +1)2 dx
8
6
12 − 3x 2
3
=
dx =
8 ∫0 (x 2 + 4) 2
4
2
×3⇒
2
1
k 4
3
∫0  kx − 2f (x)  dx = 4 x  − 2 (12)
0
2
(ii)
= 4k – 6
4k – 6 = 1
4k = 7
7
k=
4
Unit 1 Answers: Module Tests
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Page 16 of 20
2
(a)
(b)
dv
= 300 π cm 3s−1
dt
dr
(i)
RTF
when r = 25cm
dt
dv dv dr
=
×
dt dr dt
4
v = πr 3
3
dv
= 4πr 2
dr
dv
when r = 25,
= 4π(25) 2
dr
dr
∴ 300π = 4π(25) 2 ×
dt
3
dr π300
=
cms -1
=
dt 4π (25) 2 25
dA
(ii) RTF
when r = 25cm
dt
Solution:
A = 4πr2
dA
=8πr
dr
dA
when r = 25, = 25(8π=
) 200π
dr
dA dA dr
=
×
dt
dr dt
3
= 200π × = 24π cm 2s −1
25
6
(i)
y= x+ 2
x
dy
12
= 1− 3
dx
x
2
d y 36
=
dx 2 x 4
d2 y
dy
 36 
 12 
x 2 + 3 = x  4  + 3 1 − 3 
x 
dx
dx
 x 
36 36
= 3 − 3 +3
x
x
2
d y
dy
⇒x 2 +3 =
3
dx
dx
dy
(ii) when x = 1,
=
1 − 12 =
−11
dx
1
Gradient of the normal =
11
x = 1, y = 1 + 6 = 7.
Equation of the normal at (1, 7) is
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 17 of 20
1
(x − 1)
11
11y – 77 = x–1
11y – x = 76
y = sin x2
dy
= 2x cos x2
dx
d
sin x 2  = 2x cos x 2
Since
dx 
y–7=
(c)
(i)
⇒ sin x 2 
π/ 2
0
0
π/2
π 
1
sin   =
x cos x 2 dx
∫
2
 4  0
= 0.312
dy
(ii)
x3 = x + 1
dx
dy x + 1
= 3
dx
x
dy 1
1
= 2 + 3
dx x
x
1
1
⇒ y = ∫ 2 + 3 dx
x
x
1
1
y= − − 2 +c
x 2x
d
(x 2 +2) tan x  = (x 2 + 2)sec 2 x + 2x tan x
(a) (i)
dx 
d
1
cos(5x 3 − 2x)1/ 2 =

(ii)
(5x 3 − 2x)−1/ 2 (15x 2 − 2) (− sin(5x 3 − 2x)1/ 2 )
dx
2
2 − 15x 2
sin ( 5x 3 − 2x )
=
3
2 5x − 2x
(b) f(–2) = –2a + b
–2a + b = –2
[1]
f(2) = 2a + b
1
2a + b = (2) − 1
2
2a + b = 0
[2]
[1] + [2] ⇒ 2b = –2, b = –1
1
a=
2
1
a = , b = −1
2
x+5 −3
(c) (i)
lim
x→4
x−4
x+5 −3
x+5 +3
= lim
×
x→4
x−4
x+5 +3
⇒
3
π/ 2
2
=
∫ 2x cos x dx
Unit 1 Answers: Module Tests
2
© Macmillan Publishers Limited 2013
Page 18 of 20
= lim
x→4
x+5−9
(x − 4) x + 5 + 3
x−4
= lim
x − 4 ( x + 5 + 3)
1
= lim
x →4
x+5 +3
1
1
= =
9 +3 6
 tan 2x 
lim 
(ii)

x→0 
x 
sin 2x
= lim
x → 0 cos 2x
x
 1 
 sin 2x 
= lim 
 × lim


x → 0  cos2x 
x→0 
x 
= 1 × (2)
=2
(a) y = x3 – 6x2 + 9x + 1
dy
= 3x 2 − 12x + 9
dx
d2 y
= 6x − 12
dx 2
dy
= 0 ⇒ 3x 2 − 12x + 9 = 0
dx
x2 – 4x + 3 = 0
(x – 1) (x – 3) = 0
x = 1, 3
d2 y
=0 ⇒ 6x − 12 =0
dx 2
x=2
d2 y
when x = 1,
=6(1) − 12 =− 6 < 0 Max
dx 2
d2 y
x =3, 2 =6(3) − 12 =6 > 0 Min
dx
x = 1, y = 1 – 6 + 9 + 1 = 5
x = 2, y = 8 – 24 + 18 + 1 = 3
x = 3, y = 27 – 54 + 27 + 1 = 1
∴ (1, 5) Maximum point
(2, 3) point of inflexion
(3, 1) Minimum point
x→4
4
p
(b)
0
∫ (x − 2) dx =
3
0
p
 (x − 2) 4 
⇒
0
 =
 4 0
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 19 of 20
(p − 2) 4
−4=
0
4
4
(p – 2) = 16
p – 2 = 4 16
p = 2+2
=4
2x + 1
(i)
y=
x−4
Vertical asymptote : x – 4 = 0
x=4
Horizontal asymptote :
 2x + 1
lim y = lim 

x →∞
x →∞  x − 4 
⇒
(c)
1
x
= lim
x →∞
4
1−
x
2+0
= = 2
1− 0
∴ y = 2 is a horizontal asymptote
Asymptotes are: x = 4, vertical
y = 2, horizontal
dy (x − 4)(2) − (2x + 1)(1)
(ii)
=
dx
(x − 4)2
2x − 8 − 2x − 1
=
(x − 4)2
−9
=
(x − 4) 2
dy
−9
= 0⇒
= 0 ⇒ − 9 = 0, inconsistent
dx
(x − 4)2
2+
∴ There are no turning points
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 20 of 20
Unit 1 Answers: Module Tests
© Macmillan Publishers Limited 2013
Page 1 of 27
Multiple Choice Tests
Multiple Choice Test 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
A
B
A
A
A
A
D
A
B
C
A
C
A
C
D
D
D
B
B
B
A
D
C
B
C
B
C
A
B
C
B
C
B
C
A
D
B
C
C
C
A
B
B
A
D
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 2 of 27
1
A
1
1
=
2
( 3 + 2)
3+ 2 3 2 + 2
1
=
5+2 6
1
=
5+2 6
×
5−2 6
5−2 6
2
5−2 6
25 − 24
= 5−2 6
B
3x2 − 13x + 14 < 0
(3x − 7) (x − 2) < 0
7

 x: z < x < 
3

3
4
A
A
=
50
50
9
∑ ∑ ∑r
=
r
r−
=r 10 =r 1 =r 1
50(51) 10(9)
−
2
2
= 1230
A
f(x) = 4x3 + ax2 + 7x + 2
f(2) = 0
=
> 4(2)3 + a(2) 2 + 7(2) + 2 =
0
32 + 4a + 14 + 2 =
0
4a = −48
a = −12
A
2 log e (5p) − 3 log e (2p)+ 2
=
5
6
=log e (5p) 2 − log e (2p)3 + 2log e e
 25p 2

=log e  3 × e 2 
 8p

2
 25 e 
=log e 

 8p 
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 3 of 27
7
8
D
P=240(1.06) n
2500=240(1.06) n
2500
(1.06) n =
240
 250 
nlog (1.06)=log 

 24 
 250 
log 

 24 
n=
log (1.06)
=40.22
The year:1841
A
y = 4x − 7
x = 4y − 7
x+7
4
x+7
-1
f (x) =
4
fg(x) = x + 1
y=
9
10
g(x)=f −1 (x + 1)
x+1+7
=
4
x+8
=
4
1
−1
∴ g (x) = x + 2, x ∈ 
4
B
x+1
y+1
y=
,x =
x−2
y−2
xy - 2x = y + 1
xy - y = 1 + 2x
1 + 2x
y=
x-1
1 + 2(4)
9
g -1 (4) =
= =3
4-1
3
3a
+
3
hg -1 (4) =
=6
3
3a = 15
a=5
C
x 3 − 6x 2 + 11x − 6 =
0
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 4 of 27
∑α =
6
∑ αβ =
11
αβγ = 6
∑ α 2 =( ∑ α ) − 2 ∑ αβ
2
= (6) 2 − 2(11)
= 36 − 22 = 14
11
A
2x3 – x2 + 3x – 4 = 0
1
1
y= , x=
x
y
2 1 3
− + −4=
0
y3 y 2 y
12
⇒ 2 – y + 3y2 – 4y3 = 0
4y3 – 3y2 + y – 2 = 0
C
2x + 1 =
3
13
2x + 1 = 3, 2x + 1 = –3
x = 1, z = –2C
A
16 x +1 + 42 x
2 x −3 8 x + 2
(2 )
=
4 x +1
+ ( 22 )
2 x − 3 ( 23 )
=
=
2x
x+2
24 x + 4 + 24 x
24 x +3
2
4x
( 2 + 1)
A
4
2 4 x × 23
17
=
8
14
C
2x + 3 =
1 – 2x
⇒ (2x + 3)2 = (1 – 2x)2
⇒ 4x2 + 12x + 9 = 1 – 4x + 4x2
16x = – 8
x= −
15
1
2
D
x = 1, x = –2, x = 3
(x – 1) (x + 2) (x – 3) = 0
(x2 + x – 2) (x – 3) = 0
x3 – 3x2 + x2 – 3x – 2x + 6 = 0
x3 – 2x2 – 5x + 6 = 0
p = –2, q = –5, r = 6
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 5 of 27
16
D
sin θ =
1
2
n  π
=
θ nπ + ( -1)   n ∈ 
6
17
D
 11θ 
 3θ 
cos 7θ + cos 4θ =
2 cos 
 cos  
 2 
 2 
18
B
4cos θ + 3 sin θ = r cos (θ – α)
= r cos θ cos α + r sin θ sin α
r cos α = 4 
3
 tan α =
r sin α = 3 
4
r=
32 + 42 = 5

 3 
∴ 4 cos θ + 3sin θ = 5cos  θ − tan −1   
 4 

19
20
21
Min value – 5
B
An ellipse
B
4x2 + y2 – 8x + 4y + 6 = 0
4x2 – 8x + y2 + 4y + 6 = 0
4(x – 1)2 – 4 + (y + 2)2 – 4 + 6 = 0
Ellipse centre (1, –2)
A
 2   2
   
 2  ⋅ 1 
 −1  2 
4
   
cos θ =
=
 2   2 9
   
 2  1 
 −1  2 
   
22
D
 2
  
AB =  2 
3
 
 −2 
 

BC =  −2 
 k − 2


 
AB ⋅ AB =
0
 2   −2 
 

⇒  2  .  −2  =
0
3   k − 2
 

Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 6 of 27
⇒ –4 –4 + 3k – 6 = 0
3k = 14
k=
23
14
3
C
1 
2 
     
OA  4  OB  5 
 2
 −1
 
 
1 
1 
 
 
r=
 4  + λ 1  , λ ∈ 
 2
 −3 
 
 
24
B
cos (A − B) 7
=
cos (A + B) 3
cos A cos B + sin A sin B 7
=
cos A cos B − sin A sin B 3
3 cos A cos B + 3 sin A sin B = 7 cos A cos B – 7 sin A sin B
10 sin A sin B = 4 cos A cos B
sin A 4
=
cos B 10
2
tan A = cot B
5
25
5 tan A = 2 cot B
C
cot (θ + 45)
1
tan(θ + 45)
1
= tan θ + tan 45
1 − tan θ tan 45
1
1− t
= =
t +1 1+ t
1− t
=
26
B
tan 2θ
=
1 + sec 2θ
=
sin 2θ
1 + cos 2θ
sin 2θ
cos 2θ
1
1+
cos 2θ
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 7 of 27
2sin θ cos θ
2 cos 2 θ
sin θ
= = tan θ
cos θ
=
27
C
cos A =
3
4
cos 4A = 2 cos22A – 1
= 2[2cos2 A – 1]2 –1
2
 9 
= 2  2   − 1 − 1
  16  
1
−31
=
− 1=
32
32
28
A
x = 2 cos t
y = sin t + 1
cos t =
x
2
sin t = y – 1
cos 2 t =
x2
4
sin2t=(y–1)2
sin 2 t + cos 2 t = (y − 1) 2 +
x2
4
x2
1 = (y − 1) +
4
2
29
4(y – 1)2 + x2 = 4
B
5(x2 + y2) – 4x – 22y + 20 = 0
4
22
x 2 + y2 − x − y + 4 = 0
5
5
 2 11 
centre  , 
5 5 
r=
30
4 121
+
−4 = 1
25 25
C
11 8
3
−
−3
5 =
5
Gradient of normal
= 5=
2 6 −4 5 4
−
5 5
4
Gradient of the tangent =
3
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 8 of 27
31
B
lim
x →0
32
sin3x
3 sin3x
= lim
→
x
0
2x
2 3x
3
sin3x
= lim
→
x
0
2
3x
3
=
2
C
7
lim 4f(z) =
7 ⇒ lim f(x) =
x →2
x →2
4
7
23
lim(f(x) + 2x) = + 2(2) =
x →2
4
4
33
34
B
x(x + 1) = 0
x = 0, x = – 1
C
d
[sin(x 3 )]=3x 2 cos(x 3 )
dx
35
36
A
f(x) = x2ex
f′(x) = 2xex + x2ex
f′(0) = 0
D
P = 40 cm
2x + 2l = 40
l = 20 – x
dx
= 0.5
dt
A = lx = (20 – x) x = 20x – x2
dA
= 20 − 2x
dx
dA
= 20 − 6 = 14
when x = 3,
dx
dA dA dx
=
× =14×0.5 = 7cm 2s -1
dt dx dt
37
B
2x2 + 4xh = 150
4xh = 150 – 2x2
150 − 2x 2
4x
75 − x 2
=
2x
h=
38
C
V = x2 h
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 9 of 27
75 − x 2
× x2
2x
75
1
=
x − x3
2
2
dv 75 3 2
=
− x
dx 2 2
dv
75 3 2
=0 ⇒
=x
dx
2 2
=
x2 = 25
x=5
x = 5, V
=
39
75
53
(5) − = 125
2
2
C
dy
1
= 8− 3
dx
x
dy
=⇒
0 8x 3 =
1
dx
1
x3 =
8
1
x=
2
1
when x = , y = 4 + 2 = 6
2
40
C
 cos 4x − 1 
lim 

x →0
x


 cos 4x − 1 
= lim 4 
x →0
 4x 
41
= 4(0) = 0
A
∫ 4 cos 6θ cos 2θ dθ
2 cos 6θ cos 2θ = cos 8θ + cos4θ
∫ 4 cos 6θ cos 2θ dθ
= 2 ∫ cos8θ cos 4θ dθ
2
2
= sin 8θ + sin 4θ + c
8
4
1
1
= sin 8θ + sin 4θ + c
4
2
42
C
π
π
2
2
(2x) dx ∫ sec (2x) − 1 dx
∫ tan=
2
0
2
0
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 10 of 27
π
 1π

=
−
 2 tan(2x) − x  =
2
0
43
2
B
2x2 + 5 = 3x2 + 1
x2 = 4
x=±2
x = 12, y = 13
2
2
∫ (2x + 5)dx − ∫ (3x + 1)dx
shaded area=
2
0
2
0
2
= ∫ (− x 2 + 4)dx
0
2
 1 3

=
 − 3 x + 4x 
0
−8
16
=
+ 8=
3
3
44
A
13  y − 1 
13  y − 5 
dy-π ∫ 
π ∫ x 2 dy = π ∫ 

dy
1
5
 5 
 2 
13
13
 π (y − 1) 2 
 π (y − 5) 2 
−
=


2  5
 3 2 1  2
π
π
= [144] − [64]
6
4
= 24π – 16π = 8π = 25.1
45
D
5
∫ f(x)dx+[4x]
2
4
2
= 12 + 16 – 8 = 20
Multiple Choice Test 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
B
C
A
B
C
B
A
D
C
B
D
A
C
C
B
D
A
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 11 of 27
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
B
A
C
C
A
B
B
A
A
B
A
B
A
B
A
C
C
B
A
B
B
C
A
B
A
A
C
A
1
B
x2 – 3 < 1
⇒ –1< x2 – 3 < 1
2 < x2 < 4
For x2 < 4
x2 – 4 < 0
(x – 2) (x + 2) < 0
∴–2<x<2
For x2 > 2
x2 – 2 > 0
(x − 2 ) (x − 2 ) > 0
x < − 2, x > 2
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 12 of 27
{
}
∴{x : 2 < x < − 2}  x : 2 < x < 2
2
C
 2
−5
−5
1 −3
2
log=
log 2 2
=
=
2
2 2
 8  log
2
2


3
4
5
6
A
f(x) = x3 – 7x2 + kx – 12
f(2) = 23 – 7(2)2 + 2k – 12 = 0
2k = 32, k = 16
B
“If presentation college chaguanas do not win, then it is not raining.”
C
Truth table:
p
q
p⇒q
∼(p ⇒ q)
p ∩(∼q)
T
T
T
F
F
T
F
F
T
T
F
T
T
F
F
F
F
T
F
F
B
2
3+ x
2
 2 
f
=
3 + x 3 + 2
3+x
2(3 + x)
=
11 + 3x
f(x) =
∴ ff(x) is undefined at x = –3, x =
7
−11
3
A
|x + 1| = 3|2x + 1|
(x + 1)2 = 32 (2x + 1)2
x2 + 2x + 1 = 9 (4x2 + 4x + 1)
35x2 + 34x + 8 = 0.
(5x + 2) (7x + 4) = 0
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 13 of 27
=
x
8
−2 −4
,−
5
7
D
 1 
log 3   = log 3 3−3 = −3
 27 
9
10
C
(x – 2) (x + 3) (4x – 3) = 0
(x2 + x – 6) (4x – 3) = 0
4x3 – 3x2 + 4x2 – 3x – 24x + 18 = 0
4x3 + x2 – 27x + 18 = 0
B
x
<0
x−2
11
x
x–2
x
x−2
x<0
–ve
–ve
+ve
0<x<
2
+ve
–ve
–ve
x>2
+ve
+ve
+ve
∴ {x : 0 < x < 2}
D
1
1
x+2+2
=
×
x+2 −2
x+2 −2
x+2+2
x+2+2
x+2+2
=
=
x+2−4
x−2
12
A
2
8 
(3x) 2 +  x  =
31
3 
3
2
x = 3, 9 2 + 8 3 = 27 + 4 = 31
3
3
13
14
15
C
x3 + 6x2 + 11x + 6 = 0
(–1)3 + 6 (–1)2 + 11(–1) + 6 = 0.
x3 + 6x2 + 11x + 6 = (x + 1) (x2 + 5x + 6)
= (x + 1) (x + 2) (x + 3)
x = –1, –2, –3
C
x3 – 2x2 + 4x – 7 = 0
∑α = 2
∑αβ = 4
αβ𝛾 = 7
∑α2 = (∑α)2 – 2∑αβ
=(2)2 – 2(4)
= –4
B
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 14 of 27
a ∗ b = 2ab + a – 3
a∗e=a
⇒ 2ae + a – 3 = a
e=
16
17
3
2a
D
sin 9θ – sin 3θ
= 2cos 6θ sin 3θ
A
tan θ = 2
sin θ =
18
2
5
B
f(x) = 1 – 6 sin x cos x + 4cos2 x
= 1 – 3sin 2x + 2 [cos 2x + 1], cos 2x = 2cos2 x - 1
= 3 – [3sin 2x – 2cos 2x]
3 sin 2x – 2 cos 2x = R sin (2x – α)
= R sin 2x cos α – R cos 2x sin α
R cos α = 3
R sin α = 2
tan α = 2 3 ⇒ α = 33.7°
R
=
2
32 + (−2)=
13
∴ f(x) =−
3 13 sin(2x − 33.7°)
19
A
3 + 13 when sin (2x – 33.7°) = –1
20
21
C
sin (2x –33.7°) = –1
2x = –90 + 33.7°, 270 + 33.7°
x = 151.9°
C
 −2 
 −3 
 −2 
    
   
OA =
 2  , OB =
 m + 2  , OC =
 4
 −1 
 −1 
 −5 
 


 
 −1 
 1 
    

AB
=  m  , BC
= 2 − m
 0 
 −4 
 


 −1   1 
 
 

AB.BC =
0 ⇒  m  . 2 − m  =
0
 0   −4 
 

– 1+ (2 – m) m= 0
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 15 of 27
22
m2 – 2m + 1 = 0
(m – 1)2
= 0 ⇒m = 1
A
 2
 0
     
=
AB =
2  , AC  3 
1 
 −1 
 
 
 2  0
     
=
AB . AC =
2  . 3  5
1   −1 
  
23
B
5
cos θ =
9 10
θ = 58.2°
24
B
1 
 2
 
 
r =+
 2 λ 3 , λ ∈ 
 4
1 
 
 
x −1 y − 2
=
=
2
3
25
z−4
A
 4  2  4
    
r.  2  = 1  .  2 
0 3 0
    
26
= 10
4x + 2y = 10
2x + y = 5
A
x2 + 4x + y2 – y + 2 = 0
(x + 2)2 – 4 + (y – 1)2 – 1 + 2 = 0
(x + 2) 2 + (y − 1) 2 =
( 3) 2
r= 3
27
B
x = 2 sec t
sec 2 t =
x2
4
tan t = y – 1
tan2 t = (y – 1)2
1=
x2
− (y − 1) 2
4
(y − 1) 2 =
x2
−1
4
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 16 of 27
y −1 =
x2 − 4
4
y= 1+
x2 − 4
2
=
28
29
2 + x2 − 4
2
A
cos2 θ = (x – 2)2
sin2 θ = (y – 3)2
(y – 3)2 + (x – 2)2 = 1
Circle centre (2, 3)
B
 2
 4
 
 
r=
 1  + λ 1  , λ ∈ 
 −4 
5
 
 
x= 2 + 4λ 

y=
1+ λ λ ∈ 
z =−4 + 5λ 
30
31
A
x2 + y2 = 49
Circle centre (0, 0), r = 7
x2 – 6x + y2 – 8y + 21 = 0
(x – 3)2 – 9 + (y – 4)2 – 16 + 21 = 0
(x – 3)2 + (y – 4)2 = 4
A
Circle centre (3, 4), r = 2
B
 sin 4x 
 sin 4x 
lim
lim 
=

 4=
 4
x →0
x
→
0
 x 
 4x 
32
A
4x 2 + 10x + 4
lim
x →−2
3x + 6
2(2x + 1)(x + 2 )
= lim
x →−2
3 (x + 2)
2
= (−4 + 1) =−2
3
33
C
x2 – 5x + 5 = 0
x=
34
5± 5
2
= 3.62, 1.38
C
x3 – 3x2 + 2x = 0
x(x2 – 3x + 2) = 0
x(x – 2) (x – 1) = 0
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 17 of 27
35
x=0,1,2
B
 x 2 − 3x  0 − 0
=
lim 
 = 0
x →0
2
 2 
36
37
A
f(x) = x cos 3x
f′(x) = cos 3x – 3x sin 3x
B
y = ax2 + bx – 3
dy
= 2ax + b
dx
dy
x = −4,
= 9 ⇒ −8a + b = 9
dx
x = –4, y = –31 ⇒ –31 = 16a – 4b – 3
16a – 4b = –28
2 × [1] + [2] ⇒ –2b = 10
b=5
a=
38
[1]
[2]
−1
2
B
y = 2x 3 − 9x 2 +12x
dy
= 6x 2 − 18x +12 = 0
dx
39
x2 – 3x + 2 = 0
(x – 1) (x – 2) = 0
x = 1, 2
x = 1, y = 2 – 9 + 12 = 5
x = 2, y = 16 – 36 + 24 = 4
(1, 5) (2, 4)
C
f(x) = 4(2x + 1) −3
= − 24(2x + 1) −4
−24
=
(2x + 1) 4
40
41
A
f (x) =2x2 + 4x – 3
f′(x) =4x + 4 < 0
x < –1
B
x2 = 2x – x2
2x2 – 2x = 0
2x (x – 1) = 0
x = 0, x = 1
1
1 3  1
dx
x
∫0 x=
=
3  0 3
1
2
1
2
 2 1 3
∫0 2x − x dx = x − 3 x  0 =3
1
2
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 18 of 27
∴ Area =
42
2 1 1
− =
3 3 3
A
dy
= 2x + 1
dx
y = ∫ (2x + 1)dx
1
y = (2x + 1) 2 + C
4
x = 0, y =3 ⇒ C =2
3
4
1
3
y = (2x + 1) 2 + 2
4
4
43
A
∫ x (5 − x )
= ∫ 5x − x dx
1
2
0
1
2
5/ 2
0
1
2
5

=  x 3 − x 7/ 2 
7
3
0
5 2 29
= − =
3 7 21
44
C
V = π ∫ y 2 dx
V = π ∫ (5x + 1)dx
2
0
2
 5x 2

= π
+ x
 2
0
= π [10 + 2] = 12π
45
A
dp
∝P
dt
dp
⇒
= KP, k > 0
dt
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 19 of 27
Multiple Choice Test 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
B
C
C
A
B
A
D
C
A
B
D
B
A
C
C
D
A
D
B
B
C
A
C
B
D
A
B
C
A
D
B
C
B
D
A
D
B
A
D
C
D
B
C
C
D
1
B
f(x) = 4x – 2
f(2) = 4(2) – 2 = 6
f 2(2) = f(6) = 4(6) – 2 = 22
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 20 of 27
2
3
4
C
C
p ⇔ q is logically equivalent to
∼ p ⇔ ∼q
A
x+3
>0
2x −1
− ve
− ve
x+3
2x −1
+ ve
+ ve
− ve
− ve
+ ve
+ ve
+ ve
x + 3 2x −1
x < −3
−3 < x <
x>
6
1
2
{ x : x < −3} ∪  x : x >
∴
5
1
2

1

2
B
a*e = a + e – 2ae = a
e(1– 2a) = 0
e=0
A
20
20
20
∑ (3r + 2)= 3∑ r + ∑ 2
=r 1
=r 1 =r 1
3(20)(21)
+ 2(20)
2
= 670
=
7
D
log (x − 3)(x + 3) + log
=
=
8
log
x +3
x −3
( x − 3 x + 3 ) × xx +− 33
log (x + 3)
C
2x2 – 5x – 3 ≥ 0
(2x + 1) (x – 3)
≥0
1
{x : x ≤ − } ∪ {x : x ≥ 3}
2
9
10
A
B
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 21 of 27
1
1
y= , x=
x
y
1 4 6
− + −8 =
0
y3 y 2 y
1 − 4y + 6y 2 − 8y3 =
0
8y3 − 6y 2 + 4y − 1 =0
11
D
6x 3 − 3x 2 − 3x + 2 =
0
1
∑α = 2
1
∑ αβ = − 2
=
∑α
2
( ∑ α ) − 2∑ αβ
2
2
1
 1
=   − 2 − 
2
 2
1
5
= +1 =
4
4
12
B
2x − 1 > 5
13
2x – 1 >5, 2x – 1 < –5
x > 3,
x < –2
{x: x < –2} ∪ {x: x > 3}
A
4
−3
x
4
y=
−3
x
4
4
y+3= ⇒ x =
x
y+3
f(x)=
y = 1, x = 1,
y = –1x = 2
11
, x=5
5
5
y=– , x=8
2
y= −
14
C
2 + 36 + 72
=
2 +6+6 2
= 6+7 2
15
C
f(x) = x 3 − ax 2 + 2x + 5
f(2) = 8 − 4a + 4 + 5 = 0
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 22 of 27
4a=17
17
a=
4
16
D
1 − cos 2θ
2sin 2 θ
=
sin 2θ
2sin θ cos θ
sin θ
=
= tan θ
cos θ
17
A
4
5
5
cos B =
13
cos A =
cos(A
=
+ B) cos A cos B − sin A sin B
 4  5   3  12 
=    −   
 5  13   5  13 
20 36 −16
=
−
=
65 65 65
18
D
3cos θ + 4sin
=
θ r cos θ cos α + r sin θ sin α
r sin α = 4
r cos α = 3
tan
=
α
r=
4
4
=
, α arctan  
3
3
32 + 42 = 5

 4 
∴ 3cos θ + 4sin
=
θ 5cos  θ − arctan   
 3 

19
20
B
Max = 9, Min = 6 – 3 = 3
B
x−2
(x − 2) 2
2
⇒ cos
=
θ
3
9
y−3
(y − 3) 2
2
sin
=
θ
⇒ sin=
θ
2
4
cos
=
θ
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 23 of 27
(x − 2) 2 (y − 3) 2
∴
+
=
1
9
4
⇒ 4(x 2 − 4x + 4) + 9(y 2 − 6y + 9) =
36
4x 2 + 9y 2 − 16x − 54y + 61 =
0
21
22
23
24
25
C
A
x2 – 2x + y2 – 4y – 4 = 0.
(x – 1)2 – 1 + (y – 2)2 – 4 – 4 = 0
(x – 1)2 + (y – 2)2 = 9
Circle centre (1, 2), r = 3
C
cos4 x – sin4 x
= (cos2 x – sin2 x) (cos2 x + sin2 x)
= cos2x
B
cos 6θ + cos 4θ
= 2 cos 5θ cos θ
D
a ⋅b =
0
 1   −2 
  

0
 4⋅ 1  =
 5   P + 3
  

⇒ −2 + 4 + 5P + 15 =0
5P = −17
−17
P=
5
26
A

Mid point of PQ
4
 
==
 9  4i + 9j − 3k
 −3 
 
27
B
1
 
r ⋅ 2 =
12
 2
 
1 
 3
12
r ⋅ 2  = =
4
 3 3
 2 
 3
Distance = 4
28
C
x 2 − 4x + y 2 − 2y + 1 =
0
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 24 of 27
centre (2,1)
4+a
=2⇒a =0
2
b +1
= 1,=
b 1
2
(0,1)
29
A
1
2
=
2θ 2n(180) ± 60°, n ∈ 
cos 2θ =
θ= 180°n ± 30°, n ∈ 
30
31
D
cos 75 cos15 – sin 75 sin 15
= cos (75 + 15) = cos 90 = 0
B
sin 6x
x →0
x
sin 6x
= 6 lim
x →0
6x
=6
lim
32
C
x2 − x − 6
x →3
2x 2 − 5x − 3
(x − 3) (x + 2)
= lim
x →3
(2x + 1) (x − 3)
lim
=
33
3+ 2 5
=
6 +1 7
B
x +1 − 2
x →3
x −3
x +1 − 2
x +1 + 2
= lim
×
x →3
x −3
x +1 + 2
(x + 1) − 4
= lim
x →3
(x − 3)( x + 1 + 2)
1
= lim
x →3
x +1 + 2
1
1
= =
4+2 4
lim
34
D
x7 2
∫ x 4 − x 4 dx
= ∫ (x 3 − 2x -4 )dx
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 25 of 27
=
1 4 2
x + 3+c
4
3x
35
A
f (x) = x3 – 3x2 + 2
f′(x) = 3x2 – 6x < 0
x (x – 2) < 0
0<x<2
36
D
f(x) = cos3 x
f′(x) = –3 cos2 x sin x
B
37
∫ 2sin 8θ cos 4θ dθ
= ∫ (sin12θ + sin 4θ) dθ
=−
38
39
1
1
cos12θ − cos 4θ + C
12
4
A
| x |2 – 4 = 0
x2 = 4
x = 2, –2
D
V = π ∫ y 2 dx = π ∫ (x 2 +1) 2 dx
2
0
2
2
1

= π ∫ (x + 2x + 1)dx = π  x 5 + x 3 + x 
0
3
5
0
 32 16
 206
= π  + +=
2
π
5 3
 15
2
40
4
2
C
20 cos10x
cos 6x sin 6x
20sin10x
= lim
x →0 1
sin12x
2
sin10x
10
10x
= 40 lim
x→ 0
sin12x
12
12x
400 100
= =
12
3
lim
x →0
41
D
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 26 of 27
π


ϕ (x)= sec 4  x + 
4

π 
π
π


ϕ ′(x)= 4sec3  x +  sec  x +  tan  x + 
4 
4
4


π
π


= 4sec 4  x +  tan  x + 
4
4


π




π


π


π 
π



ϕ ²(x)= 4sec 4  x +  sec 2  x +  +16sec3  x +  sec  x +   tan 2  + x  
4
4
4
4
4




π
π
π



= 4sec6  x +  +16sec 4  x +  tan 2  x + 
4
4
4



2
π
4
16
when x =
, ϕ ″( x) =
3 = 1024
+
6
4
12
1 1
   
2 2
( )
42
B
y
=π–x
dy
= –dx
cos y = cos(π – x) = –cos x
x = 0, y = π
π
π
x= ,y=
2
2
π 2
π 2
π 2
π
π
π
cos y dy ∫=
cos x dx [sin x ]
∫=
43
C
∫ x dx
¥
-3/2
1
∞
 2 
= −

x 1

=2
44
C
3
15
4(2x
1)
dx =
+
∫0
2
c
1
15
(2x + 1) 4  0 =
2
2
4
(2c + 1) − 1 =
15
c
2c +=
1
1
c=
2
45
4
16
= 2
D
dy x 2
=
dx y 2
∫ y dy= ∫ x dx
2
2
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013
Page 27 of 27
1 3 1 3
=
y
x +c
3
3
x =2, y =0 ⇒ c =−8 / 3
3
y=
x3 − 8
Unit 1 Answers: Multiple Choice Tests
© Macmillan Publishers Limited 2013