CAPE Unit 1 Pure Math 2018 Paper 2 Solutions
CAPE 2017 Paper 2 SOLUTIONS
QUESTION 1
(a) (i) ~𝑝 ∨ 𝑞
(ii) 𝑝 → ~𝑞
(b) (i) ∗ is commutative since 2 ∗ 1 = 1 ∗ 2
The table is symmetric about the diagonal hence * is commutative or
showing that a*b =b*a for all a,b.
(ii) 1 ∗ 3 = 1
2∗3 =2
3*a=a*3 for all a in {1, 2,3,4} therefore 3 is the identity element of *.
3∗3 =3
4∗3 =4
(c) (i) 𝑓(𝑥) = 𝑎𝑥 + 9𝑥 − 11𝑥 + 𝑏
(ii) 𝑓(𝑥) = 2𝑥 + 9𝑥 − 11𝑥 − 30
𝑓(2) = 0
𝑎(2) + 9(2) − 11(2) + 𝑏 = 0
8𝑎 + 𝑏 = −14
𝑓(−2) = 12
𝑎(−2) + 9(−2) − 11(−2) + 𝑏 = 12
𝑓(𝑥) = (𝑥 − 2)(2𝑥 + 13𝑥 + 15)
𝑓(𝑥) = (𝑥 − 2)(2𝑥 + 3)(𝑥 + 5)
−8𝑎 + 𝑏 = −46
3
𝑥 = −5, − , 2
2
2𝑏 = −60
𝑏 = −30
𝑎=2
(d) Let 𝑃 : ∑
8𝑟 = 4𝑛(𝑛 + 1)
∀𝑛∈ℕ
𝑃 : 8(1) = 4(1)(1 + 1)
8=8
Therefore 𝑃 is true.
40
Assume 𝑃 is true for 𝑛 = 𝑘
𝑃 :
8𝑟 = 4𝑘(𝑘 + 1)
𝑃
:
8𝑟 = 4(𝑘 + 1)(𝑘 + 2)
𝑃
= 𝑃 + (𝑘 + 1)term
= 4𝑘(𝑘 + 1) + 8(𝑘 + 1)
= (𝑘 + 1)(4𝑘 + 8)
= 4(𝑘 + 1)(𝑘 + 2)
Therefore 𝑃
is true ∀ 𝑃 is true.
Hence by mathematical induction 8 + 11 + 24 + 32 + ⋯ + 8𝑛 = 4𝑛(𝑛 + 1) for all 𝑛 ∈ ℕ.
QUESTION 2
2.
(𝑎 + 𝑏) = 𝑎 + 2𝑎𝑏 + 𝑏
(a) (i)
ln
𝑎+𝑏
1
= (ln 𝑎 + ln 𝑏)
4
2
𝑎+𝑏
1
ln
= (ln 𝑎𝑏)
4
2
ln
𝑎+𝑏
= ln(𝑎𝑏)
4
𝑎+𝑏
= √𝑎𝑏
4
Therefore
𝑎 + 𝑏 = 14𝑎𝑏
𝑎 + 𝑏 − 14𝑎𝑏 = 0
𝑎 + 𝑏 + 2𝑎𝑏 − 16𝑎𝑏 = 0
(𝑎 + 𝑏) = 16𝑎𝑏
(𝑎 + 𝑏)
= 𝑎𝑏
16
𝑎 + 𝑏 = 4√𝑎𝑏
(𝑎 + 𝑏) = 16𝑎𝑏
𝑎 + 2𝑎𝑏 + 𝑏 = 16𝑎𝑏
𝑎+𝑏
4
ln
𝑎 + 𝑏 = 14𝑎𝑏
𝑎+𝑏
4
2 ln
Alternately
ln
= 𝑎𝑏
= ln 𝑎𝑏
𝑎+𝑏
= ln 𝑎 + ln 𝑏
4
𝑎+𝑏
1
= (ln 𝑎 + ln 𝑏)
4
2
41
(ii) 2
+ 3(2 ) = 4
2 =
1
+ 3(2 ) = 4
2
ln 2 = ln
1
+ 3𝑦 − 4 = 0
𝑦
1
3
1
3 = − ln 3
𝑥=
ln 2
ln 2
ln
3𝑦 − 4𝑦 + 1 = 0
(3𝑦 − 1)(𝑦 − 1) = 0
𝑦=
1
3
2 =1
𝑥=0
1
,1
3
(b)
(c) 𝛼 + 𝛽 + 𝛾 = − = 0
𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 =
𝛼𝛽𝛾 = −
𝑐
=3
𝑎
𝑑
= −2
𝑎
𝑥 − (𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾)𝑥 + (𝛼𝛽)(𝛼𝛾) + (𝛼𝛽)(𝛽𝛾) + (𝛼𝛾)(𝛽𝛾) 𝑥 − (𝛼𝛽)(𝛼𝛾)(𝛽𝛾) = 0
𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 = 3
(𝛼𝛽)(𝛼𝛾) + (𝛼𝛽)(𝛽𝛾) + (𝛼𝛾)(𝛽𝛾)
= 𝛼 𝛽𝛾 + 𝛽 𝛼𝛾 + 𝛾 𝛼𝛽
= (𝛼𝛽𝛾)(𝛼 + 𝛽 + 𝛾)
= (−2)(0)
42
=0
(𝛼𝛽)(𝛼𝛾)(𝛽𝛾)
= (𝛼𝛽𝛾)
= (−2)
=4
𝑥 − 3𝑥 − 4 = 0
QUESTION 3
3.
(ii) sin 𝐴 =
(a) (i) tan(𝐴 + 𝐵) =
tan 𝐴 =
sin(𝐴 + 𝐵)
cos(𝐴 + 𝐵)
sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
=
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
sin 𝐴 cos 𝐵 sin 𝐵 cos 𝐴
=
+
cos 𝐴 cos 𝐵 cos 𝐴 cos 𝐵
cos 𝐴 cos 𝐵 sin 𝐴 sin 𝐵
÷
−
cos 𝐴 cos 𝐵 cos 𝐴 cos 𝐵
tan 𝐴 + tan 𝐵
=
1 − tan 𝐴 tan 𝐵
3
4
cos 𝐵 = −
1
2
tan 𝐵 = −√3
tan(𝐴 + 𝐵) =
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
3
+ −√3
= 4
3
1−
−√3
4
ALTERNATELY
=
3
3√3
− √3 ÷ 1 +
4
4
tan 𝐴 + tan 𝐵
1 − tan 𝐴 tan 𝐵
=
3 − 4√3
4 + 3√3
×
4
4
=
sin 𝐴 sin 𝐵
sin 𝐴
+
÷ 1−
cos 𝐴 cos 𝐵
cos 𝐴
sin 𝐵
cos 𝐵
sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
=
÷
cos 𝐴 cos 𝐵
cos 𝐴 cos 𝐵
=
=
=
sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
sin(𝐴 + 𝐵)
=
cos(𝐴 + 𝐵)
4 + 3√3
3 − 4√3 4 − 3√3
4 + 3√3 4 − 3√3
=
12 − 9√3 − 16√3 + 36
16 − 27
=
48 − 25√3
−11
sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
cos 𝐴 cos 𝐵
×
cos 𝐴 cos 𝐵
cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
=
3 − 4√3
=−
48 25√3
+
11
11
= tan(𝐴 + 𝐵)
43
(b) sin 𝜃 − 2 cos 𝜃 + 3 cos 𝜃 + 5 = 0
(1 − cos 𝜃) − 2 cos 𝜃 + 3 cos 𝜃 + 5 = 0
−3 cos 𝜃 + 3 cos 𝜃 + 6 = 0
cos 𝜃 − cos 𝜃 − 2 = 0
(cos 𝜃 − 2)(cos 𝜃 + 1) = 0
cos 𝜃 = −1
𝜃=𝜋
𝜃 = 3𝜋
By using the graph of cos 𝜃.
(c) (i) 𝑓(𝜃) = 6 cos 𝜃 + 8 sin 𝜃
𝑟 = √6 + 8 = 10
6
𝛼 = tan
= 36.87°
8
𝑓(𝜃) = 10 sin(𝜃 + 36.87°)
(ii) 10 sin(𝜃 + 36.87°) = 2
sin(𝜃 + 36.87°) = 0.2
𝑅𝐴 = sin (0.2) = 11.54°
𝐼: 𝜃 + 36.87° = 11.54°
𝜃 = −25.33°
𝐼𝐼: 𝜃 + 36.87° = 180° − 11.54°
= 168.46°
𝜃 = 131.59°
General solution:
𝜃 = −25.33 + 360°𝑛, 𝑛 ∈ ℤ
𝜃 = 131.69° + 360°𝑛, 𝑛 ∈ ℤ
QUESTION 4
(a) (i) 𝑥 + 𝑦 − 4𝑥 + 2𝑦 − 2 = 0
(ii) 𝑥 + 𝑦 − 4𝑥 + 2𝑦 + 2 = 0
(𝑥 − 2) + (𝑦 + 1) = 2 + 4 + 1
𝑥 + 3𝑦 = 3
(𝑥 − 2) + (𝑦 + 1) = 7
(3 − 3𝑦) + 𝑦 − 4(3 − 3𝑦) + 2𝑦 − 2 = 0
Centre (2, −1)
+ −1 − (−2)
𝑥 = 3 − 3𝑦
9𝑦 − 18𝑦 + 9 + 𝑦 − 12 + 12𝑦 + 2𝑦 − 2 = 0
Radius of 𝐶 =
2 − (−1)
→
10𝑦 − 4𝑦 − 5 = 0
= √10
𝐶 : (𝑥 − 2) + (𝑦 + 1) = 10
𝑏 − 4𝑎𝑐
(−4) − 4(10)(−5) > 0
Since discriminant is positive the
quadratic has 2 distinct roots therefore
the 𝐿 cannot be a tangent to 𝐶 .
(b) (i) 𝑃𝑄⃗ = −2𝑖 − 3𝑗 + 2𝑘
(ii) 𝑟. 𝑛 = 𝑎. 𝑛
𝑥
−2
1
−2
𝑦 . −3 = −2 . −3
𝑧
2
4
2
−2𝑥 − 3𝑦 + 2𝑧 = −2 + 6 + 8
−2𝑥 − 3𝑦 + 2𝑧 = 12
44
(c) 𝐿 = −𝑖 + 𝑗 − 2𝑘 + 𝛼(−2𝑖 + 𝑗 − 3𝑘)
Solving (1) and (2) simultaneously
𝐿 = −2𝑖 + 𝑗 − 4𝑘 + 𝛽(𝑖 − 𝑗 + 𝑘)
2𝛼 + 𝛽 = 1
−𝑖 + 𝑗 − 2𝑘 − 2𝛼𝑖 + 𝛼𝑗 − 3𝛼𝑘
= −2𝑖 + 𝑗 − 4𝑘 + 𝛽𝑖 − 𝛽𝑗
+ 𝛽𝑘
𝛼+𝛽 =0
(−1 − 2𝛼)𝑖 = (−2 + 𝛽)𝑖
𝛽 = −1
−1 − 2𝛼 = −2 + 𝛽
Checking in (3)
2𝛼 + 𝛽 = 1
(1)
𝛼=1
3(1) + (−1) = 2
Therefore 𝐿 and 𝐿 intersect.
(ii) 𝐿 = −𝑖 + 𝑗 − 2𝑘 − 2𝑖 + 𝑗 − 3𝑘 since 𝛼 =
1
(1 + 𝛼)𝑗 = (1 − 𝛽)𝑗
1+𝛼 =1−𝛽
𝐿 = −3𝑖 + 2𝑗 − 5𝑘
𝛼+𝛽 =0
(2)
When 𝛽 = −1
𝐿 = −2𝑖 + 𝑗 − 4𝑘 − 𝑖 + 𝑗 − 𝑘
(−2 − 3𝛼)𝑘 = (−4 + 𝛽)𝑘
𝐿 = −3𝑖 + 2𝑗 − 5𝑘
−2 − 3𝛼 = −4 + 𝛽
Point of intersection is −3𝑖 + 2𝑗 − 5𝑘
3𝛼 + 𝛽 = 2
(3)
QUESTION 5
(a) lim
→
lim
→
=𝑘
(𝑥 − 1)(𝑥 + 𝑥 + 𝑥 + 𝑥 + 1)
=𝑘
𝑥−1
lim 𝑥 + 𝑥 + 𝑥 + 𝑥 + 1 = 𝑘
→
1 +1 +1 +1+1=𝑘
5=𝑘
45
(b) (i) 𝑥 = 5𝑡 + 3, 𝑦 = 𝑡 − 𝑡 + 2
When 𝑡 = 0
𝑑𝑥
=5
𝑑𝑡
𝑥 = 5(0) + 3 = 3
𝑦=0 −0 +2=2
𝑑𝑦
= 3𝑡 − 2𝑡
𝑑𝑡
(3, 2)
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
When 𝑡 =
𝑑𝑦 3𝑡 − 2𝑡
=
𝑑𝑥
5
(ii)
𝑥=5
=0
𝑦=
3𝑡 − 2𝑡 = 0
2
3
(c) (i) (a) 𝑦 = √2 + 2𝑥 = (2 + 2𝑥 )
+2=
50
27
+ 2𝑥 − (2 + 2𝑥 ) (4𝑥)
− 4𝑥 (2 + 2𝑥 )
𝑑 𝑦 4
−
𝑑𝑥
𝑦
𝑑𝑦
𝑦
− 2𝑥
𝑑𝑥
=0
2
3
𝑑 𝑦
= 2(2 + 2𝑥)
𝑑𝑥
𝑑𝑦
= 2𝑥(2 + 2𝑥 )
𝑑𝑥
= 2𝑥 − 2𝑥
−
= 2(2 + 2𝑥 )
(b)
𝑑𝑦 1
= (2 + 2𝑥 ) (4𝑥)
𝑑𝑥 2
= (2 + 2𝑥 ) 2𝑥(2 + 2𝑥 )
2
3
19 50
,
3 27
𝑡(3𝑡 − 2) = 0
𝑡 = 0,
2
19
+3=
3
3
= 2(2 + 2𝑥 )
− 4𝑥 (2 + 2𝑥 )
− 4 (2 + 2𝑥 )
= 2(2 + 2𝑥 )
− 4𝑥 (2 + 2𝑥 )
− 4(2 + 2𝑥 )
− 2𝑥
= 2(2 + 2𝑥 ) [2 + 2𝑥 − 2𝑥 − 2]
= 2(2 + 2𝑥 ) (0)
=0
(ii)
when 𝑥 = 0
= 2 2 + 2(0)
− 4(0) (2 + 2(0) )
= √2
46
QUESTION 6
Equation of 𝑄𝑅
(a) (i)
𝑚=
3−2
= −1
3−4
𝑦 = 𝑚𝑥 + 𝑐
(3, 3) 𝑚 = −1
3 = −1(3) + 𝑐
𝑐=6
𝑦 = −𝑥 + 6
Equation of 𝑃𝑅
(ii) Equation of 𝑃𝑄
𝑚=
3−1 2
=
3−0 3
𝑚=
2−1 1
=
4−0 4
𝑦=
2
𝑥+1
3
𝑦=
1
𝑥+1
4
(iii) ∫
=
𝑥 + 1 𝑑𝑥 + ∫ −𝑥 + 6 𝑑𝑥 − ∫
2 𝑥
3 2
+𝑥
3
𝑥
4
1 𝑥
+ − + 6𝑥 −
0
2
3
4 2
(3)
0
+3 −
+0
3
3
=
= (6 − 0) + 16 −
=
𝑥 + 1 𝑑𝑥
+
−
+𝑥
4
0
4
3
+ 6(4) − − + 6(3)
2
2
−
4
0
+4 −
+0
8
8
27
− (6 − 0)
2
5
2
47
(c) ∫ [3𝑓(𝑥) + 𝑔(𝑥)] 𝑑𝑥 = 5
and ∫ [5𝑓(𝑥) − 2𝑔(𝑥)] 𝑑𝑥 = 1
(1) × 2: 6𝑚 + 2𝑛 = 10
(2)
5𝑚 − 2𝑛 = 1
11𝑚 = 11
[3𝑓(𝑥) + 𝑔(𝑥)] 𝑑𝑥 = 5
𝑚=1
𝑛=2
3
𝑓(𝑥) 𝑑𝑥 +
𝑔(𝑥) 𝑑𝑥 = 5
𝑓(𝑥) 𝑑𝑥 = 1
5
𝑓(𝑥) 𝑑𝑥 − 2
𝑔(𝑥) 𝑑𝑥 = 1
𝑔(𝑥) 𝑑𝑥 = 2
Let ∫ 𝑓(𝑥) 𝑑𝑥 = 𝑚 and ∫ 𝑔(𝑥) 𝑑𝑥 = 𝑛
3𝑚 + 𝑛 = 5
(1)
5𝑚 − 2𝑛 = 1
(2)
48
MAY 2016 UNIT 1 PAPER 2
SECTION A
Module I
1. (a)
(i)
Data: f ( x ) = 2 x 3 − x 2 + px + q , f ( −3) = 0 and f ( −1) = 10
Required To Prove: p = −25 and q = −12
Proof:
Recall: The remainder and factor theorem
If a polynomial f ( x ) is divided by ( x − a ) the remainder is f ( a ) . If
f ( a ) = 0 , then ( x − a ) is a factor of f ( x ) .
f ( x ) = 2 x 3 − x 2 + px + q
f ( −3) = 0
(from data)
0 = 2 ( −3) − ( −3) − 3 p + q
3
2
0 = 2 ( −27 ) − 9 − 3 p + q
0 = −63 − 3 p + q
Let 63 = −3p + q
…
f ( −1) = 10
10 = 2 ( −1) − ( −1) + p ( −1) + q
3
2
10 = −2 − 1 − p + q
Let 13 = − p + q
…
Equation  − Equation 
−50 = 2 p
p = −25
Substitute p = −25 into equation 
13 = − ( −25) + q
13 − 35 = q
q = −12
Hence p = −25 and q = −12 .
Q.E.D.
(ii)
Required To Solve: f ( x ) = 0
Solution:
From the part (i), 𝑓(𝑥) = 2𝑥 3 − 𝑥 2 − 25𝑥 − 12
Since f ( −3) = 0 , then ( x + 3) is a factor of f ( x ) .
2 x2 − 7 x − 4
x + 3 2 x 3 − x 2 − 25 x − 12
− 2 x3 + 6x2
− 7 x 2 − 25 x
−
−7 x 2 − 21x
− 4 x − 12
−
− 4 x − 12
0
f ( x ) = ( x + 3) ( 2 x 2 − 7 x − 4 )
= ( x + 3)( 2 x + 1)( x − 4 )
When f ( x ) = 0
x+3=0
x = −3
 x = −3 or −
(b)
2x + 1 = 0
OR
x=−
x−4=0
1 OR
2
1
or 4
2
Required To Prove: 6n − 1 is divisible by 5 n  N .
Proof:
Assume the statement is true for n = k .
That is, 6k − 1 is divisible by 5.
Hence, 6k − 1 = 5 p, p  N .
Consider n = k + 1
6k + 1 − 1 = 6 ( 6 k ) − 1
Recall: 6k − 1 = 5 p  6k = 5 p + 1
x=4
6k + 1 − 1 = 6 ( 5 p + 1) − 1
= 30 p + 6 − 1
= 30 p + 5
= 5 ( 6 p + 1)
Since p is a natural number then 6p + 1 is also a natural number
 6k + 1 − 1 is divisible by 5.
 The statement is true for n = k + 1 .
When n = 1
61 − 1 = 5
= 5  1 which is divisible by 5.
The statement is true for n = 1 .
When n = 2
62 − 1 = 35
= 5  7 which is divisible by 5.
The statement is true for n = 2 .
Hence, by the principle of mathematical induction the statement is true n  N .
(c)
(i)
(ii)
Required To Complete: The truth table given.
Solution:
p
q
p→q
pq
pq
(p  q) → (p  q)
T
T
F
F
T
F
T
F
T
F
T
T
T
T
T
F
T
F
F
F
T
F
F
T
Required To State: Whether p → q and ( p  q ) → ( p  q ) are logically
equivalent.
Solution:
p → q is not logically equivalent to ( p  q ) → ( p  q ) since they do not
have the same truth values.
T
T
F 
F
T
F
F
T
2. (a)
Data: log 2 (10 − x ) + log 2 x = 4
Required To Find: x
Solution:
log 2 (10 − x ) + log 2 x = 4
log 2 (10 − x )( x )  = 4
Hence,
(Product law)
24 = (10 − x ) x (Definition of logs)
16 = 10 − x 2
x 2 − 10 x + 16 = 0
( x − 2 )( x − 8) = 0
x−2=0
OR
x=2
x=8
Test when x = 2
log 2 8 + log 2 = 4
Test when x = 8
log 2 2 + log 2 = 8
3+1 = 4
1+ 3 = 4
True
 x = 2 or 8, are both correct.
(b)
x −8 = 0
True
x+3
, x 1
x −1
Required To Determine: Whether f is bijective.
Solution:
x+3
f ( x) =
, x 1
x −1
Vertical asymptotes occur when denominator = 0
 x = 1 is a vertical asymptote of f ( x ) .
Data: f ( x ) =
x −1
−
1
x+3
x −1
4
4
x −1
As f ( x ) → 1 x →  , therefore, y = 1 is a horizontal asymptote.
 f ( x) = 1 +
When x = 2
5
f ( x) =
1
Consider f ( a ) = f ( b )
a+3 b+3
=
a −1 b −1
( a + 3)( b − 1) = ( b + 3)( a − 1)
ab − a + 3b − 3 = ab + 3a − b − 3
4b = 4 a
b=a
 f ( a ) = f ( b )  a = b , hence f is one to one or injective.
For f to be onto, y  Y , x  X such that y = f ( x ) .
When y = 1,  a value of x such that f ( x ) = 1 . Hence, f is not onto and thus not
surjective. Hence, f ( x ) is injective but not surjective and therefore not bijective
which requires both conditions to be true.
(c)
(i)
Data: The roots of 2 x 3 − 5x 2 + 4 x + 6 = 0 are  ,  and  .
Required To State: The values of  +  +  ,  +  +  and  .
Solution:
If
a
ax 3 + bx 2 + cx + d = 0
b 2 c
d
x + x+ =0
a
a
a
If  ,  and  are the roots of the equation, then
( x −  )( x −  )( x −  ) = 0
x3 +
x 3 − ( +  +  ) x 2 + ( +  +  ) x −  = 0
Equating coefficients of x 2 :
b
 +  + = −
a
Equating coefficients of x1 :
c
 +  +  =
a
Equating coefficients of x 0 which are the constant terms:
d
 = −
a
For 2 x 3 − 5x 2 + 4 x + 6 = 0
2
5 2
x + 2x + 3 = 0
2
Equating coeffcients of:
 5 5
x2 :  +  +  = −  −  =
 2 2
1
x :  +  +  = 2
x0 :  = −3
x3 −
(ii)
Required To Determine: An equation with integer coefficients which has
1 1
1
roots 2 , 2 and 2 , given

( )
2


+ ( ) + (  ) = ( +  +  ) − 2 ( +  +  )
2
2
2
 2 +  2 +  2 = ( +  +  ) − 2 ( +  +  )
2
Solution:
Any cubic equation can be expressed in the form:
x3 − ( sum of roots ) x 2 + ( sum of the product of the roots taken two at a time )
− ( product of roots ) = 0
Sum of roots:
1
1
1  2  2 +  2 2 +  2  2
+
+
=
2
2
2
2 2 2



  
( ) + (  ) + ( )
=
2
( )
2
 +  +  ) − 2 ( +  +  )
(
=
2
( )
2
=
( 2)
2
2
2
5
− 2 ( −3)  
2
2
( −3)
4 + 15
9
19
=
9
=
Sum of the product of the roots, taken two at a time:
1
1
1
 2 +2 +  2
+
+
=
2 2
2 2
2 2
2 2 2
 
 

  
=
( +  +  )
2
( )
2
5
  − 2 ( 2)
2
= 
2
( −3)
9
=4
9
1
=
4
Product of the roots:
1
1
1
1
 2 2 =
2


( )

=
=
1
( −3)
1
9
2
2
− 2 ( +  +  )
2
 The new equation whose roots are
x3 −
1

2
,
1

2
and
1
2
is
19 2 1
1
x + x − = 0.
9
4
9
36
36 x3 − 76 x 2 + 9 x − 4 = 0 (in integral form)
Alternative Method:
1
Let y = 2

2 =
1
y
=
1
y
But  is a root of 2 x 3 − 5x 2 + 4 x + 6 = 0
1
Substitute  =
y
3
2
 1
 1
 1
2
 − 5
 + 4
+6=0
 y
 y
 y
2
5
4
− +
+6=0
y y y
y
2 − 5 y + 4y + 6y y
=0
y y
y y
2 − 5 y + 4y + 6y y = 0
2 + 4 y = −6 y y + 5 y
2 + 4y =
Squaring both sides:
(2 + 4 y )
2
=
y (5 − 6 y )
( y ) (5 − 6 y )
2
2
4 + 16 y + 16 y 2 = y ( 25 − 60 y + 36 y 2 )
4 + 16 y + 16 y 2 = 25 y − 60 y 2 + 36 y 3
0 = 36 y 3 − 76 y 2 + 9 y − 4
SECTION B
Module 2
3. (a)
(i)
Required To Prove: sec 2  =
cosec 
cosec  − sin 
Proof:
Taking the R.H.S.:
cosec 
=
cosec  − sin 
1
sin 
1
− sin 
sin 
1
sin  
 1
=

−
sin   sin 
1 
1 − sin 2  
1

sin   sin  
Recall: sin 2  + cos2  = 1  cos2  = 1 − sin 2 
cosec 
1
cos 2 
=

cosec  − sin  sin  sin 
1
sin 
=

sin  cos2 
1
=
cos2 
= sec 2 
= L.H.S.
Q.E.D.
=
(ii)
cosec 
4
= for 0    2 .
cosec  − sin  3
Required To Find: 
Solution:
Data:
cosec 
4
=
cosec  − sin  3
4
3
1
4
=
2
cos  3
3
cos2  =
4
sec 2  =
cos  = 
3
4
cos  = 
3
2
When cos  =
3
2
When cos  = −
 


 ,  + 
6 
6

5 7
=
,
6 6
 3

 2 
 =  −
 = cos −1 

,  2 − 
6 
6
 11
= ,
6 6
=
 =
(b)
(i)
 
 5 7 11
6
,
6
,
6
,
6
3
2
for 0    2 .
Data: f ( ) = sin  + cos 
Required To Express: f ( ) = sin  + cos  in the form r sin ( +  ) ,
where r  0 and 0   
Solution:

2
.
sin  + cos  = r sin ( +  )
= r sin  cos  + sin  cos  
= r sin  cos  + r sin  cos 
Equating the terms in sin  :
r cos  = 1
…
Equating the terms in cos  :
r sin  = 1
…
Equation   Equation 
r sin  1
=
r cos  1
tan  = 1
 = tan −1 (1)
=

4
for 0   

2
Equation 2 + Equation 2
2
2
2
2
( r sin  ) + ( r cos  ) = (1) + (1)
r 2 sin 2  + r 2 cos2  = 2
r 2 ( sin 2  + cos2  ) = 2
Recall: sin 2  + cos2  = 1
 r2 = 2
r = 2, r  0


Hence, sin  + cos  = 2 sin   +  .
4

Alternative Method:
r=
(1)
2
+ (1)
2
= 2
 2
1
1
sin  +
cos  = sin  cos  + cos  sin 
2
2
Equating terms in sin  :
1
2
Equating terms in cos  :
1
sin  =
2
cos  =
Hence,  =

4
.


Therefore, sin  + cos  = 2 sin   +  .
4

(ii)
Required To Find: The maximum value of f and the smallest nonnegative value of  at which it occurs
Solution:
f ( ) = sin  + cos 


= 2 sin   + 
4



−1  sin   +   1
2



The maximum value of sin   +  = 1 .
4

 The maximum value of f ( ) = 2 (1)
= 2


This occurs when sin   +  = 1
4

+
+

4

4
= sin −1 (1)
=
=
=
(c)
Required to Prove: tan ( A + B + C ) =

2

2

4
−

4
(smallest non-negative value of  )
tan A + tan B + tan C − tan A tan B tan C
1 − tan A tan B − tan A tan C − tan B tan C
Proof:
Taking left hand side:
tan ( A + B + C ) = tan  A + ( B + C ) 
=
tan A + tan ( B + C )
...(1)
1 − tan A tan ( B + C )
Consider the numerator of equation (1):
 tan B + tan C 
tan A + tan ( B + C ) = tan A + 
 1 − tan B tan C 
=
tan A  tan B + tan C 
+
1
1 − tan B tan C 
tan A (1 − tan B tan C ) + tan B + tan C
1 − tan B tan C
tan A + tan B + tan C − tan A tan B tan C
=
1 − tan B tan C
=
Consider the denominator of equation (1):
 tan B + tan C 
1 − tan A tan ( B + C ) = 1 − tan A 
1 − tan B tan C 
=
(1 − tan B tan C ) − tan A ( tan B + tan C )
1 − tan B tan C
 tan A + tan B + tan C − tan A tan B tan C 


1 − tan B tan C
tan ( A + B + C ) = 
 1 − tan B tan C − tan A tan B − tan A tan C 


1 − tan B tan C
tan A + tan B + tan C − tan A tan B tan C
=
1 − tan A tan B − tan B tan C − tan A tan C
= Right hand side
Q.E.D.
4. (a)
(i)
Data: sin  = x
Required To Prove: tan  =
x
1− x
2
, where 0   

2
.
Proof:
sin  = x
sin  =
x
1
By Pythagoras’ Theorem:
adj =
(1)
2
− ( x)
2
= 1 − x2
tan  =
tan  =
opp
adj
x
1 − x2
Q.E.D.
(ii)
Data: A curve is defined by the parametric equation y = tan 2t and
x = sin t for 0  t 

2
.
Required To Determine: The Cartesian equation of the curve.
Solution:
y = tan 2t
2 tan t
1 − tan 2 t
 x 
2

1 − x2 

y=
2
 x 
1− 

2
 1− x 
y=
y=
y=
y=
y=
y=
y=
(b)
(i)

x2 
 1 −
2
1 − x2  1 − x 
2x
(1 − x ) − x

2
2x
1 − x2
2x
1 − x2
2x
1 − x2
2
1 − x2

1 − 2 x2
1 − x2

1 − x2
1 − 2x2
2 x 1 − x2 1 − x2
1 − x 2 (1 − 2 x 2 )
2x 1 − x2
1 − 2x2
 1
2


Data: u =  −3  and v =  1  are two position vectors in R 3 .
 2
 5
 
 
Required To Calculate: The lengths of u and v respectively
Calculation:
u=
(1)
2
+ ( −3) + ( 2 )
2
= 1+ 9 + 4
= 14 units
v =
( 2)
2
+ (1) + ( 5)
= 4 + 1 + 25
= 30 units
2
2
2
(ii)
Required To Find: cos  where  is the angle between u and v in R 3 .
Solution:
Recall: u.v = u v cos 
Hence, cos  =
u.v
u v
 1  2
 −3  .  1 
   
 2  5
=   
14 30
(1  2 ) + ( −3  1) + ( 2  5)
=
420
2 − 3 + 10
=
420
9
=
420
(c)
Data: A point P ( x, y ) moves such that its distance from the x – axis is half its
distance from the origin.
Required To Determine: The Cartesian equation of the locus of P
Solution:
Distance from P to the origin
( x − 0)
Distance of P from the x – axis is y.
1 2
x + y2 = y
2
x2 + y2 = 2 y
Square both sides:
2
+ ( y − 0) = x 2 + y 2
2
x2 + y 2 = 4 y 2
x2 = 3 y 2
x2 − 3 y 2 = 0
x2
− y2 = 0
3
(d)
Data: The line L has the equation 2 x + y + 3 = 0 and the circle C has the equation
x2 + y2 = 9 .
Required To Determine: The points of intersection of the circle C and the line L.
Solution:
…
L : 2x + y + 3 = 0
…
C : x2 + y 2 = 9
From equation 
…
y = −3 − 2 x
Substitute equation  into equation 
2
x 2 + ( −3 − 2 x ) = 9
x 2 + 4 x 2 + 12 x + 9 = 9
5 x 2 + 12 x = 0
x ( 5 x + 12 ) = 0
x = 0 or 5 x + 12 = 0
x=−
When x = 0
y = −3 − 2 ( 0 )
y = −3
12
5
12
5
 12 
y = −3 − 2  − 
 5
24
= −3 +
5
9
=
5
When x = −
 12 9 
 The points of intersection are ( 0, − 3) and  − ,  .
 5 5
SECTION C
Module 3
1
5. (a)
Required To Find:
 ( x + 1) 3 dx using an appropriate substitution.
Solution:
1
 ( x + 1) 3
dx
Let t = x + 1 
dt
=1
dx
dt = dx
1
1
  ( x + 1) 3 dx =  t 3 dt
4
t3
= + C (where C is a constant)
4
3
3 4
= t3 + C
4
Recall: t = x + 1
1
4
3
3 dx =
3 +C
x
+
1
x
+
1
(
)
(
)

4
(b)
Data: Diagram showing the finite region R which is enclosed by the curve
y = x 3 − 1 and the lines x = 0 and y = 0 .
Required To Calculate: The volume of the solid that results from rotating R
about the y – axis.
Calculation:
y = x3 − 1
y + 1 = x3
x=
y +1
3
2
x 2 = ( y + 1) 3
Vy =  
=
y2
y1
x 2 dy
2
( y + 1) 3
−1
0
dy
0
5
3

=   ( y + 1) 3 
5
 −1
5
3 
y
+
1
( ) 3 

5 
 −1
0
=
3
5
3
=
5
=
(c)
Data:

 53

(1) − 0 
units3
a
0
f ( x ) dx =  f ( a − x ) dx , a  0
Required To Prove:
a
0
ex
1
 0 e x + e1 − x dx = 2
1
Proof:
1
ex
ex
 0 e x + e1 − x dx =  0 x e dx
e + x
e
x
1
e
=  2x
dx
0 e
+e
ex
2x
1 e
=  2x
dx
0 e
+e
dt
Let t = e2 x + e 
= 2e2 x
dx
1
ex
e 2 x dt
 0 e x + e1 − x dx =  t 2e2 x
1 1
=  dt
2 t
1
= ln t + C (C is a constant)
2
1
1
1

=  ln ( e 2 x + e )
2
0


1
ln ( e 2 + e ) − ln (1 + e )
2
1  e2 + e 
= ln 

2  e +1 
=
=
1  e ( e + 1) 
ln 

2  e +1 
1
ln e
2
1
=
2
Q.E.D.
=
Alternative Method:
Given that

a
0
f ( x ) dx =  f ( a − x ) dx for a  0
a
0
Then for a = 1 and f ( x ) =

a
0
ex
consider:
e x + e1 − x
ex
dx
0 e x + e1 − x
1
e1 − x
=  1 − x 1 − (1 − x ) dx
0 e
+e
1− x
1
e
=  1− x
dx
0 e
+ ex
f ( x ) dx = 
1
=  f ( a − x ) dx
a
0
Now,
1
1
ex
ex
ex
2  x 1 − x dx =  x 1 − x dx +  x 1 − x dx
0 e +e
0 e +e
0 e +e
1
1
ex
e1 − x
dx
Recall:  x 1 − x dx =  1 − x
0 e +e
0 e
+ ex
1
1
1
ex
ex
e1 − x
2  x 1 − x dx =  x 1 − x dx +  1 − x
dx
0 e +e
0 e +e
0 e
+ ex
1
ex
e1 − x 
=   x 1− x + 1− x
 dx
0 e +e
e + ex 

1
x
1− x
1 e + e

=   1− x
dx
x 
0 e
+e 

1
=  1 dx
0
=  x 0
1
= (1 − 0 )
=1
ex
1
 0 e x + e1 − x dx = 2
1
Q.E.D.
Alternative Method:
Given that

a
0
f ( x ) dx =  f ( a − x ) dx for a  0
a
0
ex
consider:
e x + e1 − x
x
1− x
1e +e
− e1 − x
dx = 
dx
0
e x + e1 − x
x
1− x
1 e + e
e1 − x 
=   x 1 − x − x 1 − x  dx
0 e +e
e +e 

Then for a = 1 and f ( x ) =
ex
 0 e x + e1 − x
1
1
e1 − x 
=   1 − x 1 − x  dx
0
 e +e 
1
1
e1 − x
=  1 dx −  1 − x
dx
0
0 e
+ ex
1
1
ex
e1 − x
dx
Recall:  x 1 − x dx =  1 − x
0 e +e
0 e
+ ex
1
1
ex
ex
 0 e x + e1 − x dx =  0 1 dx −  0 e x + e1 − x dx
1
1
ex
dx +  x 1 − x dx =  1 dx
0 e +e
0
x
1
e
1
2  x 1 − x dx =  x  0
0 e +e
1
ex
2  x 1 − x dx = 1 − 0
0 e +e
1
ex
2  x 1 − x dx = 1
0 e +e
1
ex
1
 0 e x + e1 − x dx = 2
Q.E.D.
1
ex
 0 e x + e1 − x
1
(d)
(i)
Data: An initial population of 100, 000 bacteria grow exponentially at a
rate of 2% per hour where y = f ( t ) is the number of bacteria present t
hours later.
Required To Show: The number of bacteria at any time can be modelled
by the equation y = 10000e0.02t by solving an appropriate differential
equation.
Proof:
dy
Let
be the rate of growth of bacteria.
dt
dy
= 0.02 y
dt
dy
= 0.02dt
y
1
0.02dt = dy is of the form f ( t ) dt = g ( y ) dy , where
y
f ( t ) = 0.02 is a constant and
1
is function of y only.
y
The variables are in separable form and the solution can be found by
integrating:
dy
 y =  0.02 dt
ln y = 0.02t + C (C is constant)
g ( y) =
y = e0.02t + C
When t = 0, y = 10000
10000 = eC
ln10000 = C
y = eC e0.02 t
y = 10000e0.02 t
Q.E.D.
(ii)
Required To Determine: The time required for the bacteria population to
double in size.
Solution:
Initial population = 10000
New population = 2  10000
= 20000
 20000 = 10000e0.02 t
20000
= e0.02 t
10000
2 = e0.02 t
ln 2 = ln e0.02 t
ln 2 = 0.02t ln e
ln 2
=t
0.02
t = 34.66
t  34.7 hours (correct to 3 significant figures)
6. (a)
Data: f ( x ) = 2 x 3 + 5 x 2 − x + 12
Required To Find: The equation of the tangent to the curve at the point where
x = 3.
Solution:
f ( x ) = 2 x 3 + 5 x 2 − x + 12
The gradient function, f  ( x ) = 6 x 2 + 10 x − 1
f  ( 3) = 6 ( 3) + 10 ( 3) − 1
2
= 54 + 30 − 1
= 83
At x = 3
f ( 3) = 2 ( 3) + 5 ( 3) − ( 3) + 12
3
2
= 108
The tangent to the curve at the point ( 3, 108) .
The equation of the tangent is
y − 108 = 83 ( x − 3)
y = 83x − 249 + 108
y = 83x − 141
(b)
(i)
 x2 + 2 x + 3 x  0
Data: A function f is defined on R as f ( x ) = 
.
ax
+
b
x

0

Required To Calculate: lim− f ( x ) and lim+ f ( x ) .
x →0
x →0
Calculation:
lim+ f ( x ) = a ( 0 ) + b
x →0
=b
lim f ( x ) = ( 0 ) + 2 ( 0 ) + 3
2
x →0 −
=3
(ii)
Required To Calculate: The values of a and b such that f ( x ) is
continuous at x = 0 .
Calculation:
For f ( x ) to be continuous at x = 0 ,
lim f ( x ) = lim+ f ( x ) = f ( 0 )
x →0−
x →0
f ( 0) = 3
b=3
f ( x ) = 3x + a
a could assume any value, that is, a  R .
(iii)
Data: b = 3
Required To Determine: The value of a such that
f ( 0 + t ) − f ( 0)
f  ( 0 ) = lim
.
t →0
t
Solution:
(
)
at + b − ( 0 ) + 2 ( 0 ) + 3
f  ( 0 ) = lim
t →0
2
t
at + 3 − 3
t →0
t
at
= lim
t →0 t
= lim a
= lim
t →0
=a
But f  ( t ) = 2t + 2 at t = 0
a = 2
(c)
Required To Differentiate: f ( x ) = x with respect to x, using first principles.
Solution:
x+h − x
x+h−x
Gradient of PQ =
x+h − x
h
x+h − x
x+h + x

h
x+h + x
( x + h) − x
=
=
=
h
(
=
h
=
x+h + x
(
)
h
x+h + x
1
x+h + x
)
As h → 0 , the chord PQ → the tangent to the curve at P.
d
1

x =
dx
x+0 + x
1
=
2 x
( )
MAY 2015 UNIT 1 PAPER 2
SECTION A
Module 1
1. (a)
(b)
Data: p and q are two propositions.
(i)
Required To State: The inverse and contrapositive of p → q .
Solution:
The inverse of p → q is p → q .
The contrapositive of p → q is q → p .
(ii)
Required To Show: The table of truth values for p → q and
Solution:
q
p
p→q
q→ p
p
q
T
T
F
F
T
T
T
F
F
T
F
F
F
T
T
F
T
T
F
F
T
T
T
T
(iii)
Required To State: Whether p → q and q → p are logically
equivalent.
Solution:
Since their truth values are the same or equivalent, that is both are ‘T F T
T’ and p → q and q → p are logically equivalent.
q→ p.
Data: f ( x ) = x 3 + px 2 − x + q . ( x − 5) is a factor of f ( x ) and when f ( x ) is
divided by ( x − 1) it gives a remainder of 24.
(i)
Required To Find: The value of p and of q.
Solution:
f ( x ) = x 3 + px 2 − x + q
Recall: The Remainder and Factor Theorem
If f ( x ) is any polynomial and f ( x ) is divided by ( x − a ) , then
remainder is f ( a ) . If f ( a ) = 0 , then ( x − a ) is a factor of f ( x ) .
 f ( 5) = 0 and f (1) = 24
f ( 5) = ( 5) + p ( 5 ) − ( 5 ) + q
3
2
0 = 125 + 25 p − 5 + q
…
25 p + q = −120
f (1) = (1) + p (1) − (1) + q
3
2
24 = 1 + p − 1 + q
(From data)
…
p + q = 24
Equation  − Equation :
24 p = −144
−144
24
p = −6
p=
When p = −6 , substitute into Equation :
−6 + q = 24
q = 24 + 6
q = 30
 p = −6 and q = 30 and so, f ( x ) = x 3 − 6 x 2 − x + 30 .
(ii)
Required To Factorise: f ( x ) completely.
Solution:
Using long division and the fact that ( x − 5) is a factor of f ( x ) :
x2 − x − 6
x − 5 x 3 − 6 x 2 − x + 30
− x 3 − 5x 2
− x2 − x
− − x 2 + 5x
− 6 x + 30
− −6 x + 30
0
 f ( x ) = ( x − 5) ( x 2 − x − 6 )
f ( x ) = ( x − 5)( x − 3)( x + 2 )
Alternative Method:
By synthetic division:
( x − 5) ( ax 2 + bx + c ) = x 3 − 6 x 2 − x + 30
ax 3 + bx 2 + cx − 5ax 2 − 5bx − 5c = x 3 − 6 x 2 − x + 30
ax 3 + ( b − 5a ) x 2 + ( c − 5b ) x − 5c = x 3 − 6 x 2 − x + 30
Equating coefficients of:
x3  a = 1
x 2  b − 5a = −6
b − 5 = −6
b = −6 + 5
b = −1
x 0  −5c = 30
30
−5
c = −6
c=
 f ( x ) = ( x − 5) ( x 2 − x − 6 )
f ( x ) = ( x − 5)( x − 3)( x + 2 )
(c)
Data: S ( n ) = 5 + 52 + 53 + 54 + ... + 5n
Required To Prove: 4 S ( n ) = 5n + 1 − 5 for n  N by mathematical induction.
Proof:
Assuming statement is true for n = k
4 S ( k ) = 5k + 1 − 5
Proving true for n = k + 1
4 S ( k + 1) = 5( k +1)+1 − 5
th
4  S ( k + 1)  = 4  S ( k ) + ( k + 1) term 


= 5k + 1 − 5 + 4 ( 5k + 1 )
= 5 ( 5k + 1 ) − 5
= 5( ) − 5
 The statement is true for n = k + 1 .
k + 1 +1
Testing for n = 1 :
4 S (1) = 51 + 1 − 5
4 ( 5) = 25 − 5
20 = 20
Statement is true for n = 1 .
Hence, by the Principle of Mathematical Induction, the statement is true n  N .
2. (a)
Data: f : A → B and g : B → C are both one to one and onto.
(i), (ii) Required To Prove: g f is one to one.
Proof:
Each element of A is mapped onto one element in B and each element of B
is associated with only one element in A since f is one to one. The codomain of B is equal to the range since f is onto, similarly, for the function
B which is mapped onto C.
(b)
(i)
Required To Solve: 3 −
Solution:
4
4
3−
−
=0
x
x
( 9 ) (81)
3−
3−
4
(9)
x
4
(9)
x
−
−
4
( 92 )
x
4
(9x )
2
=0
=0
Let u = 9 x
4 4
3− − 2 = 0
u u
2
(u )
3u 2 − 4u − 4 = 0
( 3u + 2 )( u − 2 ) = 0
4
(9)
x
−
4
(81)
x
=0
u=−
u=9
2
3
x
2
9x = −
3
Take lg :
(ii)
 2
lg 9 x = lg  − 
 3
No real solutions
exist for x
Required To Solve: 5 x − 6 = x + 5
u=2
9x = 2
Take lg :
lg 9 x = lg 2
x lg 9 = lg 2
x=
lg 2
(in exact form)
lg 9
Solution:
y = 5x − 6
y = −5 x + 6
y = x+5
Equating left hand side:
5x − 6 = x + 5
y = x+5
Equating right hand side:
−5 x + 6 = x + 5
4 x = 11
11
3
x=
or 2
4
4
1
3
 x = or 2
6
4
Alternative Method:
+ (5x − 6) = x + 5
− (5x − 6) = x + 5
6x = 1
1
x=
6
5x − 6 = x + 5
4 x = 11
11
3
x=
or 2
4
4
3
1
 x = or 2
4
6
(c)
−5 x + 6 = x + 5
6x = 1
1
x=
6
Data: The growth of a bacteria population after t hours is given by N = 300 + 5t .
(i)
Required To Calculate: The number of bacteria present at t = 0 .
Calculation:
When t = 0
N = 300 + 50
= 300 + 1
= 301
(ii)
Required To Find: The time required to triple the number of bacteria.
Solution:
N = 301
 3N = 903
903 = 300 + 5t
5t = 603
Take lg:
lg 5t = lg 603
t lg 5 = lg 603
t=
lg 603
lg 5
t = 3.98 ( correct to 2 decimal places )
3. (a)
(i)
Required To Prove: cos3x = 4cos3 x − 3cos x
Proof:
cos3x = cos ( x + 2 x )
= cos x cos 2 x − sin x sin 2 x
Recall: cos 2 x = 2cos2 x − 1 and sin 2 x = 2sin x cos x
cos 3x = cos x ( 2 cos2 x − 1) − sin x ( 2sin x cos x )
= 2 cos3 x − cos x − 2 cos x ( sin 2 x )
Recall: cos2 x + sin 2 x = 1  sin 2 x = 1 − cos2 x
cos 3x = 2 cos3 x − cos x − 2 cos x (1 − cos2 x )
= 2 cos3 x − cos x − 2 cos x + 2 cos3 x
= 4 cos3 x − 3cos x
Q.E.D.
(ii)
Required To Solve: cos 6 x − cos 2 x = 0 for 0  x  2 .
Solution:
cos 6 x − cos 2 x = 0
cos 3 ( 2 x ) − cos 2 x = 0
4 cos3 2 x − 3cos 2 x − cos 2 x = 0
4 cos3 2 x − 4 cos 2 x = 0
4 cos 2 x ( cos 2 2 x − 1) = 0
4 cos 2 x = 0
cos 2 x = 0
2x =
=
 3 
2
,
2
,
2
+ 2 ,
 3 5 7
2
,
2
,
2
,
3
+ 2
2
2
cos2 2 x − 1 = 0
cos2 2 x = 1
cos 2 x = 1
cos 2 x = −1
2 x =  ,  + 2
=  , 3

cos 2 x = 1
2 x = 0, 2 , 2 + 2
= 0, 2 , 4
3
5
7
, 2 ,
, 3 ,
, 4
2
2
2
2
  3
5 3 7
 x = 0, , ,
, ,
,
,
, 2
4 2 4
4 2 4
2 x =, 0,
(b)
(i)
, ,
Required To Express: f ( 2 ) = 3sin 2 + 4 cos 2 in the form
r sin ( 2 +  ) , where r  0, 0   

2
.
Solution:
f ( 2 ) = 3sin 2 + 4 cos 2
= r ( sin 2 cos  + cos 2 sin  )
r = 32 + 42
=5
 3sin 2 + 4 cos 2 = 5 ( sin 2 cos  + cos 2 sin  )
4
3
or sin  =
5
5
= 0.927
cos  =
0 

2
 r ( sin 2 +  ) = 5 ( sin 2 + 0.927 )
(ii)
Required To Find: The maximum and minimum value of
1
.
7 − f ( )
Solution:
f ( ) = 5sin ( 2 +  )
− 1  sin 2  1
5 ( −1)  f ( )  5 (1)
− 5  f ( )  5
1
1
= is the maximum value.
7 − ( 5) 2
1
1
=
When f ( ) = −5 so
is the minimum value.
7 − ( −5 ) 12
When f ( ) = 5 so
4. (a)
Data: C1 : x = 10 cos  − 3; y = 10 sin  + 2
(i)
C2 : x = 4cos  + 3; y = 4sin  + 2
Required To Find: The Cartesian equation of C1 and C2 in the form
( x − a)
2
+ ( y − b) = r 2 .
2
Solution:
For C1 :
x+3
10
y−2
y = 10 sin  + 2  sin  =
10
2
2
Recall: cos  + sin  = 1
x = 10 cos  − 3  cos  =
2
2
 x + 3  y − 2 

 +
 =1
 10   10 
( x + 3)
2
+ ( y − 2 ) = 10 is of the form ( x − a ) + ( y − b ) = r 2 , where
2
2
2
a = −3 , b = 2 and r = 10 .
For C2 :
x−3
4
y−2
y = 4sin  + 2  sin  =
4
2
2
Recall: cos  + sin  = 1
2
2
 x − 3  y − 2 

+
 
 =1
 4   4 
x = 4 cos  + 3  cos  =
( x − 3)
2
+ ( y − 2 ) = 16 is of the form ( x − a ) + ( y − b ) = r 2 , where
2
2
2
a = 3 , b = 2 and r = 4 .
(ii)
Required To Find: The points of intersection of C1 and C2 .
Solution:
2
2
C1 : ( x + 3) + ( y − 2 ) = 10
x2 + 6x + 9 + y2 − 4 y + 4 = 0
x2 + y2 + 6x − 4 y + 3 = 0
C2 :
( x − 3)
2
+ ( y − 2 ) = 16
2
x 2 − 6 x + 9 + y 2 − 4 y + 4 = 16
x2 + y2 − 6x − 4 y − 3 = 0
Equating C1 and C2 :
x2 + y2 + 6x − 4 y + 3 = x2 + y 2 − 6x − 4 y − 3
12 x = −6
x=−
When x = −
1
2
1
2
2
 1  
2
  − 2  + 3  + ( y − 2 ) = 10
 

25
+ y 2 − 4 y + 4 − 10 = 0
4
1
y 2 − 4 y + = 0 is of the form ay 2 + by + c = 0 , where a = 1 ,
4
1
b = −4 and c = .
4
Using the quadratic formula:
−b  b 2 − 4ac
y=
2a
− ( −4 ) 
=
( −4 )
2
2 (1)
1
− 4 (1)  
4
4  15
2
15
= −2 
2
=
 1
 1
15 
15 
 The points of intersection are  − , − 2 +
.
 and  − , − 2 −
2 
2 
 2
 2
(b)
Data: The point P ( x, y ) moves so that its distance from ( 0, 3) is two times the
distance from ( 5, 2 ) .
Required To Prove: The equation of the locus to the point P ( x, y ) is a circle.
Proof:
PA = 2 PB
( x ) + ( y − 3)
2
2
=2
Square both sides:
( x − 5) + ( y − 2 )
2

x 2 + ( y − 3) = 4 ( x − 5 ) + ( y − 2 )
2
2
2
2

x 2 + y 2 − 6 y + 9 = 4  x 2 − 10 x + 25 + y 2 − 4 y + 4
x 2 + y 2 − 6 y + 9 = 4 x 2 − 40 x + 100 + 4 y 2 − 16 y + 16
3 x 2 + 3 y 2 − 40 x − 10 y + 107 = 0
( 3)
40
10
107
x− y+
= 0 is of the form x 2 + y 2 + 2 gx + 2 fy + c = 0 , where
3
3
3
107
20
5
.
g = − , f = − and c =
3
3
3
 20 5 
This is the equation of a circle with center  ,  and radius
 3 3
x2 + y 2 −
 20   −5  107 104
= −  +  −
=
units.
3
9
 3   3 
Q.E.D.
2
5. (a)
2
 sin ( ax )
if x  0, a  0

Data: f ( x ) =  x
.
4
if x = 0

Required To Find: The value of a if f is continuous at x = 0 .
Solution:
For f ( x ) to be continuous at x = 0 .
lim f ( x ) = lim− f ( x ) = f ( 0 )
x → 0+
x →0
sin ( ax )
=4
x →0
x
a sin ( ax )
lim
=4
x →0
ax
sin ( ax )
a lim
=4
x →0
ax
a.1 = 4
lim
a=4
(b)
Required To Find: The derivative of f ( x ) = sin ( 2 x ) by first principles.
Solution:
Consider part of the curve y = sin ( 2 x )
As h → 0 , the chord PQ → the tangent to y = sin 2 x at P.
sin 2 ( x + h ) + sin 2 x
Gradient of PQ =
( x + h) − x
sin ( 2 x + 2h ) − sin 2 x
h
sin 2 x cos 2h + cos 2 x sin 2h − sin 2 x
=
h
sin 2 x  cos 2h − 1 + cos 2 x sin 2h
=
h
sin 2 x  cos 2h − 1 cos 2 x sin 2h
=
+
h
h
sin 2h
 cos 2h − 1 
= lim sin 2 x 
+ lim cos 2 x

h →0
h
h

 h →0
sin 2h
 cos 2h − 1 
= sin 2 x lim 
+ cos 2 x lim

h →0
h →0
h
h


2.sin 2h
sin h
Recall: lim
= 2 (1) = 2
= 1 and lim
h →0
h →0
h
2h
Gradient of PQ = 0 + ( cos 2 x )( 2 )
=
= 2 cos 2x
(c)
Data: y =
(i)
2x
1 + x2
Required To Show: x
Proof:
From the data
dy
y
=
dx 1 + x 2
y=
2x
1 + x2
4 x2
y2 =
1 + x2
(1 + x 2 )( y 2 ) = 4 x 2
Differentiating implicitly with respect to x:
dy
y 2 .2 x + (1 + x 2 ) 2 y
= 8x
dx
dy
2 xy 2 + 2 (1 + x 2 ) y
= 8x
dx
dy
2 y (1 + x 2 ) = 8 x − 2 xy 2
dx
2
 4 x  dy
2 y  2  = 8 x − 2 xy 2
 y  dx
( 2 )
4 x 2 dy
= 4 x − xy 2
y dx
( x)
4 x dy
= 4 − y2
y dx
x
dy ( 4 − y
=
dx
4
4 x2
1 + x2
4 4 x2
−
2
dy
x
= y. 1 1 + x
dx
4
4 (1 + x 2 − 4 x 2 )
= y.
4 (1 + x 2 )
Recall: y 2 =
x
= y.
4 + 4 x2 − 4 x2
4 (1 + x 2 )
= y.
4
4 (1 + x 2 )
dy
y
=
dx 1 + x 2
Q.E.D.
2
)y
(ii)
Required To Prove:
d2y
3y
+
=0
2
2
dx
(1 + x 2 )
Proof:
dy
y
=
dx 1 + x 2
dy
x (1 + x 2 ) = y
dx
=y
( x + x3 ) dy
dx
2
dy
dy
2
2 d y
1 + 3 x ) + x (1 + x ) 2 = 1.
(
dx
dx
dx
2
dy
d y
3x 2
+ x (1 + x 2 ) 2 = 0
dx
dx
( x)
x
2
 dy 
2 d y
3  x  + (1 + x ) 2 = 0
dx
 dx 
2
 y 
2 d y
3
+ 1+ x ) 2 = 0
2  (
dx
 1+ x 
2
3y
2 d y
+
1
+
x
( ) dx 2 = 0
1 + x2
 (1 + x 2 )
3y
(1 + x )
2 2
d2y
+ 2 =0
dx
d2y
3y
+
=0
2
2 2
dx
1
+
x
( )
Q.E.D.
6. (a)
Data: The region bounded by the lines y = 3x − 7 , y + x = 9 and 3 y = x + 3 .
(i)
Required To Prove: The coordinates of A, B and C are ( 4, 5) , ( 3, 2 ) and
( 6, 3) .
Proof:
Drawing the lines on the same axes
Solving y + x = 9 and y = 3x − 7 simultaneously to get A :
…
y =9− x
…
y = 3x − 7
3x − 7 = 9 − x
4 x = 16
16
4
x=4
When x = 4
y =9−4
=5
x=
 A ( 4, 5)
Solving 3 y = x + 3 and y = 3x − 7 simultaneously to get B:
…
y = 3x − 7
…
3y = x + 3
Substitute equation  into equation :
3 ( 3x − 7 ) = x + 3
9 x − 21 = x + 3
8 x = 24
24
8
x=3
When x = 3
y = 3 ( 3) − 7
x=
= 9−7
=2
 B ( 3, 2 )
Solving 3 y = x + 3 and y + x = 9 simultaneously to get C:
…
3y = x + 3
…
y =9− x
Substitute equation  into equation :
3(9 − x ) = x + 3
27 − 3x = x + 3
4 x = 24
24
4
x=6
When x = 6
y =9−6
=3
x=
 C ( 6, 3)
Q.E.D.
(ii)
Required To Find: The area bounded by the lines using integration.
Solution:
A1 + A3 = 
4
3
( 3x − 7 ) dx
4
 3x 2

=
− 7x
 2
3
 3 ( 4 )2
  3 ( 3) 2

=
− 7 ( 4) − 
− 7 ( 3) 
 2
  2

1
= 3 square units
2
A2 + A4 = 
6
4
( 9 − x ) dx
6

x2 
= 9 x − 
2 4

2
2

( 6 )  − 9 4 − ( 4 ) 
= 9 ( 6 ) −
  ( )

2  
2 

= 54 − 18 − 36 + 8
= 62 − 54
= 8 square units
6 x

A1 + A2 =   + 1 dx
3
3 
6
 x2

=  + x
6
3
 ( 6)2
  ( 3) 2

=
+ 6 − 
+ 3
 6
  6

1
= 7 square units
2
Area bounded by the lines = ( A1 + A3 ) + ( A2 + A4 )  − ( A1 + A2 )
1
 1

=  3 + 8 − 7
2
 2

= 4 square units
(b)
Data: The gradient function of a curve y = f ( x ) is 3x 2 + 8 x − 3 . The curve
passes through ( 0, − 6 ) .
(i)
Required To Find: The equation of the curve.
Solution:
dy
Gradient function,
= 3x 2 + 8 x − 3
dx
y =  ( 3x 2 + 8 x − 6 ) dx
= x 3 + 4 x 2 − 3x + C (C is the constant of integration)
The point ( 0, − 6 ) lies on the curve.
−6 = ( 0 ) + 4 ( 0 ) − 3 ( 0 ) + C
3
2
C = −6
The equation of the curve is y = x 3 + 4 x 2 − 3x − 6 .
(ii)
Required To Find: The coordinates of the stationary points on the curve.
Solution:
dy
At a stationary point,
=0
dx
3x 2 + 8 x − 3 = 0
( 3x − 1)( x + 3) = 0
x=
When x =
1
or − 3
3
1
3
3
2
1
1
1
y =   + 4   − 3  − 6
 3
 3
 3
1 4
=
+ −1− 6
27 9
176
=−
27
When x = −3
3
2
y = ( −3) + 4 ( −3) − 3 ( −3) − 6
= −27 + 36 + 9 − 6
= 12
 1 176 
 The stationary points are  , −
 and ( −3, 12 ) .
27 
3
d2y
= 6x + 8
dx 2
 1 176 
At  , −
,
27 
3
d2y
1
= 6  + 8
2
dx
 3
= 10( 0)  Minimum point
 1 176 
 , −
 is a minimum point
27 
3
At ( −3, 12 ) ,
d2y
= 6 ( −3) + 8
dx 2
= −10
= −1( 0)  Maximum point
 ( −3, 12 ) is a maximum point.
(iii)
Required To Sketch: The curve by clearly labelling stationary points.
Sketch:
MAY 2014
CAPE PURE MATHEMATICS UNIT 1 PAPER 2
SECTION A
1. (a)
Data: p, q and r are three propositions.
Required To Draw: A truth table for ( p → q )  ( r → q ) .
Solution:
(b)
p
q
r
p→q
r →q
( p → q)  (r → q)
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
T
F
F
T
T
T
T
T
T
F
T
T
T
F
T
T
T
F
F
T
T
F
T
Data: y  x = y3 + x3 + ay 2 + ax 2 − 5 y − 5x + 16 , where a is a real number.
(i)
Required To State: If  is commutative, giving a reason.
Solution:
x  y = x 3 + y 3 + ax 2 + ay 2 − 5 x − 5 y + 16
= y 3 + x3 + ay 2 + ax 2 − 5 y − 5 x + 16
= yx
 is commutative.
(ii)
Data: f ( x ) = 2  x and ( x − 1) is a factor of f ( x ) .
a)
Required To Find: The value of a.
Solution:
f ( x) = 2  x
= ( 2 ) + x3 + a ( 2 ) + ax 2 − 5 ( 2 ) − 5 x + 16
3
2
= x3 + ax 2 − 5 x + 8 + 4a − 10 + 16
= x3 + ax 2 − 5 x + 4a + 14
Recall: The remainder and factor theorem
When a polynomial f ( x ) is divided by ( x − a ) , the remainder is
f ( a ) . If f ( a ) = 0 , then ( x − a ) is a factor of f ( x ) .
Hence,
f (1) = 0
0 = (1) + a (1) − 5 (1) + 4a + 14
3
3
0 = 1 + a − 5 + 4a + 14
0 = 5a + 10
a = −2
Required To Factorise: f ( x ) completely.
b)
Solution:
f ( x ) = x 3 − 2 x 2 − 5 x + 4 ( −2 ) + 14
= x3 − 2 x 2 − 5 x + 6
From the data ( x − 1) is a factor.
x2 − x − 6
x − 1 x3 − 2 x 2 − 5 x + 6
− x3 − x 2
− x2 − 5x
−
− x2 + x
− 6x + 6
− −6 x + 6
0
 f ( x ) = ( x − 1) ( x 2 − x − 6 )
= ( x − 1)( x − 3)( x + 2 )
(c)
Required To Prove: 12 + 32 + 52 + ... + ( 2n − 1) =
2
mathematical induction.
Proof:
n
12 + 32 + 52 + ( 2n − 1) =  ( 2r − 1)
2
2
r =1
n
4n 2 − 1)
(
3
r =1
Using the Principle of Mathematical Induction
Assume that the statement is true for n = k .
That is,
n
 ( 2r − 1)
2
=
n
4n2 − 1) for n  N , by
(
3
k
 ( 2r − 1)
2
=
r =1
k
4k 2 − 1)
(
3
Consider n = k + 1
k +1
 ( 2r − 1)
2
r =1
k
=  ( 2r − 1) + ( k + 1) term
2
th
r =1
2
k
4k 2 − 1) + ( 2 ( k + 1) − 1)
(
3
k
2
=  4k 2 − 1 + ( 2k + 1)
3
k ( 2k + 1)( 2k − 1)
2
=
+ ( 2k + 1)
3
 k ( 2k − 1)

= ( 2k + 1) 
+ ( 2k + 1) 
3


=
 k ( 2k − 1) + 3 ( 2k + 1) 
= ( 2k + 1) 

3


 2k + 1 
2
=
  2k − k + 6k + 3
 3 
= ( 2k + 1)  2k 2 + 5k + 3
 2k + 1 
=
 ( k + 1)( 2k + 3)
 3 
( k + 1) 2k + 3 2k + 1
=
(
)(
) 
3 
k +1
 4k 2 + 8k + 3
=
3
k +1
 4k 2 + 8k + 4 − 1
=
3
n
k +1
2
=
4 ( k + 1) − 1 which is of the form ( 4n 2 − 1) , where
3
3
n = k + 1 in this case
Hence, the statement is true for n = k + 1.
(
)
When n = 1 .
Left hand side:
1
 ( 2r − 1) = (1)
2
r =1
=1
Right hand side:
2
1
3
( 4 − 1) =
3
3
=1
Statement is true for n = 1 .
When n = 2
Left hand side:
1
 ( 2r − 1)
r =1
2
= 1 + ( 2 ( 2 ) − 1)
= 1 + ( 3)
2
2
= 10
Right hand side:
2
2
2
4 ( 2 ) − 1 = (16 − 1)
3
3
= 10
Statement is true for n = 2 .
(
)
Hence, by the Principal of Mathematical Induction, the statement is true
for n  N .
Q.E.D
2. (a)
x −1
, xR.
2
Required To Determine: f 2 ( x ) , in terms of x.
Solution:
f 2 ( x ) = ff ( x )
Data: f ( x ) = 2 x 2 + 1, 1  x   and g ( x ) =
(i)
a)
= f  2 x 2 + 1
= 2 ( 2 x 2 + 1) + 1
2
= 2  4 x 4 + 4 x 2 + 1 + 1
= 8x4 + 8x2 + 3
b)
Required To Determine: f  g ( x )  in terms of x.
Solution:
 x −1 
f  g ( x )  = f 

 2 
2
 x −1 
= 2 
 + 1
2


 x −1 
= 2
 +1
 2 
= x −1+1
=x
(ii)
Required To State: The relationship between f and g.
Solution:
Recall: ff −1 ( x ) = x
Since fg ( x ) = x , then g ( x ) = f −1 ( x ) .
(b)
Data: a3 + b3 + 3a 2b = 5ab2
a +b
Required To Prove: 3log 
= log a + 2 log b
 2 
Proof:
Assuming the base is the same throughout the equation,
Taking left hand side:
a + b
a+b
3log 
= log 


 2 
 2 
3
 a 3 + 3a 2b + 3ab 2 + b3 
= log 

8


 5ab2 + 3ab2 
= log 

8


 8ab2 
= log 

 8 
= log ( ab2 )
= log a + log b2
= log a + 2 log b (Right hand side)
Q.E.D.
(c)
(i)
Required To Solve: e x +
1
− 2 = 0.
ex
Solution:
1
ex + x − 2 = 0
e
x
(e )
e 2 x + 1 − 2e x = 0
Let u = e x
u 2 − 2u + 1 = 0
( u − 1)
2
=0
u =1
Recall: u = e x
ex = 1
ln e x = ln1
x=0
(ii)
Required To Solve: log 2 ( x + 1) − log 2 ( 3x + 1) = 2
Solution:
log 2 ( x + 1) − log 2 ( 3 x + 1) = 2
 x +1 
log 2 
=2
 3x + 1 
x +1
= 22 = 4
3x + 1
x + 1 = 4 ( 3 x + 1)
x + 1 = 12 x + 4
− 3 = 11x
x=−
(d)
Required To Prove:
Proof:
Taking left hand side:
3
11
3 −1
3 +1
2 −1
2 +1
+
+
+
= 10
3 +1
3 −1
2 +1
2 −1
3 −1
3 +1
2 −1
2 +1
+
+
+
3 +1
3 −1
2 +1
2 −1
(
=
) ( 3 + 1) + ( 2 − 1) + ( 2 + 1)
( 3 − 1)( 3 + 1) ( 2 + 1)( 2 −1)
2
3 −1 +
2
2
2
3 − 2 3 +1+ 3 + 2 3 +1 2 − 2 2 +1+ 2 + 2 2 +1
+
2
1
8
= +6
2
= 10 (Right hand side)
Q.E.D.
=
SECTION B
3. (a)
(i)
Required To Prove:
cot y − cot x sin ( x − y )
=
cot x + cot y sin ( x + y )
Proof:
Taking left hand side:
cos y cos x
−
cot y − cot x sin y sin x
=
cot x + cot y cos x + cos y
sin x sin y
cos y sin x − cos x sin y
sin y sin x
=
cos x sin y + cos y sin x
sin y sin x
=
sin ( x − y ) sin ( x + y )

sin y sin x sin y sin x
=
sin ( x − y )
(Right hand side)
sin ( x + y )
Q.E.D.
(ii)
Required To Find: The possible values of y in
cot y − cot x
1

= 1, 0  y  2 when sin x = , 0  x  .
cot x + cot y
2
2
Solution:
cos x =
3
2
cot y − cot x
=1
cot x + cot y
sin ( x − y )
=1
sin ( x + y )
sin ( x − y ) = sin ( x + y )
sin x cos y − sin y cos x = sin x cos y + sin y cos x
2sin y cos x = 0
 3
2 
 sin y = 0
 2 
3 sin y = 0
sin y = 0
y = 0,  , 2
(b)
(i)
Required to Express: f ( ) = 3sin 2 + 4 cos 2 in the form
r sin ( 2 +  ) , where r  0 and 0   

2
.
Solution:
3sin 2 + 4 cos 2 = r sin ( 2 +  )
= r sin 2 cos  + cos 2 sin  
Equating terms in sin 2 and cos 2
3 = r cos 
…(1)
4 = r sin 
…(2)
(2)  (1)
tan  =
4
3
 = 0.927 radians (3 significant figures) for 0  x 

2
(1)2 + (2)2
2
2
2
2
( r cos  ) + ( r sin  ) = ( 3) + ( 4 )
r 2 ( cos 2  + sin 2  ) = 25
r 2 = 25
r = 5 ( r  0)
 3sin 2 + 4 cos 2 = 5sin ( 2 + 0.927 )
(ii)
a)
Required To Find: The value of  at which f ( ) is a minimum.
Solution:
f ( ) = 5sin ( 2 + 0.927 )
f ( ) has a minimum value when sin ( 2 + 0.927 ) = −1 .
sin ( 2 + 0.927 ) = −1
3 7
,
2 2
2 = 3.785, 10.069
2 + 0.927 =
 = 1.89, 5.03 (3 significant figures)
b)
Required To Determine: The minimum value and the maximum
1
value of
.
7 − f ( )
Solution:
Minimum value of f ( ) = −5
 Minimum value of
1
1
=
7 − f ( ) 7 − ( −5 )
1
=
12
Maximum value of f ( ) = 5
 Maximum value of
1
1
=
7 − f ( ) 7 − 5
1
=
2
4. (a)
Data: L1 and L2 are two diameters of a circle C. L1 and L2 have equations
x − y + 1 = 0 and x + y − 5 = 0 respectively.
(i)
Required To Prove: The coordinates of the center of the circle are ( 2, 3) .
Proof:
Solving L 1 and L2 simultaneously.
x − y + 1 = 0 …(1)
x + y − 5 = 0 …(2)
(1) + (2)
2x − 4 = 0
2x = 4
x=2
When x = 2
2 − y +1 = 0
y=3
L 1 and L2 intersect at the center of the circle.
 The coordinates of the center of the circle are ( 2, 3) .
Q.E.D.
Alternative Method:
Substituting the center ( 2, 3) into the equations of L1 and L2 .
L1 : 2 − 3 + 1 = 0
3−3 = 0
0 = 0 which is true.
L2 : 2 + 3 − 5 = 0
5−5 = 0
0 = 0 which is true.
 L1 and L2 intersect at ( 2, 3) since L1 and L2 are diameters of the circle
C. Hence, the center of the circle is ( 2, 3) .
(ii)
Data: A and B are the endpoints of the diameter L 1 . The coordinates of A
are (1, 2 ) and the diameters of a circle bisect each other.
Required To Find: The coordinates of B.
Solution:
The midpoint of AB is the center, ( 2, 3)
A (1, 2 ) C ( 2, 3)
Let B ( x1 , y1 )
x1 + 1 y1 + 2 
,

2 
 2
Equating x and y coordinates:
( 2, 3) = 
x1 + 1
=2
2
x1 = 4 − 1
y1 + 2
=3
2
y1 = 6 − 2
x1 = 3
y1 = 4
 B ( 3, 4 )
(iii)
Data: A point p moves on the x – y plane such that the distance from
C ( 2, 3) is always 2 units.
Required To Determine: The locus of p.
Solution:
The locus of p is a circle with center ( 2, 3) and radius
That is,
( x − 2)
2
+ ( y − 3) = 2
2
x2 − 4 x + 4 + y2 − 6x + 9 = 2
x 2 + y 2 − 4 x − 6 x + 11 = 0
Alternative Method:
Let P have coordinates ( x, y ) .

( x − 2 ) + ( y − 3)
2
2
= 2
2 units.
( x − 2 ) + ( y − 3)
2
and radius
(b)
2
= 2 which is the equation of a circle with center ( 2, 3)
2 units.
1
1
and y =
.
1− t2
1+ t
Required To Determine: The Cartesian equation of S.
Solution:
1
x=
1+ t
x (1 + t ) = 1
Data: The parametric equations of a curve S are x =
x + xt = 1
xt = 1 − x
t=
t
2
1− x
x
(1 − x )
=
2
x2
1 − 2x + x2
=
x2
(1 − 2 x + x )
= 1−
2
1− t
2
=
x2
x 2 − (1 − 2 x + x 2 )
x2
2x −1
x2
1
x2
=
1 − t 2 2x −1
x2
y =
2x −1
=
(c)
Data: The points P ( 3, − 2, 1) , Q ( −1,  , 5 ) and R ( 2, 1, − 4 ) are three vertices
of a triangle PQR.
(i)
Required To Express: PQ, QR and RP in the form xi + yj + zk .
Solution:
PQ = OQ − OP
= ( −i +  j + 5k ) − ( 3i − 2 j + k )
= −4i + (  + 2 ) j + 4k
QR = OR − OQ
= ( 2i + j − 4k ) − ( −i +  j + 5k )
= 3i + (1 −  ) j − 9k
RP = OP − OR
= ( 3i − 2 j + k ) − ( 2i + j − 4k )
= i − 3j + 5k
(ii)
Data: Triangle PQR is right-angled with the side PQ as the hypotenuse.
Required To Find: The value of  .
Solution:
Re: If a . b = 0 , then a is perpendicular to b
RP . RQ = ( i − 3 j + 5k ) . ( −3i − (1 −  ) + 9k )
0 = −3 + 3 (1 −  ) + 45
0 = 42 + 3 (1 −  )
0 = 45 − 3
3 = 45
 = 15
SECTION C
5. (a)
ax + 2, x  3
Data: f ( x ) =  2
ax , x  3
(i)
Required To Find: The value of a if f ( x ) is continuous at x = 3 .
Solution:
f ( x ) is continuous at x = 3 , then
lim f ( x ) = lim+ f ( x )
x →3−
That is,
x →3
a ( 3) + 2 = a ( 3 )
2
 3a + 2 = 9a
2 = 6a
a=
(ii)
1
3
x2 + 2
Data: g ( x ) = 2
and lim 2 g ( x ) = lim g ( x ) .
x →1
x →0
bx + x + 4
Required To Find: The value of b.
Solution:
From data:
2 g (1) = g ( 0 )
 (1)2 + 2  2
2
=
 b (1)2 + (1) + 4  4


 3  1
2
=
b+5 2
3
1
=
b+5 4
12 = b + 5
b=7
(b)
1
x
dy
Required To Find:
, using first principles.
dx
Solution:
(i)
Data: y =
y=
1
x
1
1
−
x+h
x
h
dy
= lim
dx h→0
x − x+h
x+h x
h
= lim
h →0
 x − x+h  x + x+h



x
+
h
x
x + x+h



= lim
h →0
h
x − ( x + h)
x x + h + x ( x + h)
h →0
h
−h
= lim
h →0
h  x x + h + x ( x + h ) 
h
−1
= lim .
h →0 h x x + h +
x ( x + h)
= lim
= lim
h →0
1
x x + h + x ( x + h)
−1
x x+x x
−1
=
2 x
=
(ii)
Data: y =
x
1+ x
Required To Find:
dy
dx
Solution:
x
u
y=
is of the form y = , where
v
1+ x
(Rationalising)
u=x
du
=1
dx
v = 1+ x
and
1
= (1 + x ) 2
1
dv 1
−
= (1 + x ) 2
dx 2
1
=
2 1+ x
du
dv
−u
dy
= dx 2 dx
dx
v
v
(Quotient rule)
x
2 1+ x
=
1+ x
2 (1 + x ) − x
=
2 1 + x (1 + x )
1+ x −
=
(c)
x+2
2 (1 + x ) 1 + x
Data: The parametric equations of a curve are x = cos  and
y = sin  , 0    2 .
dy
Required To Find:
, in terms of  .
dx
Solution:
x = cos 
y = sin 
dx
dy
= − sin 
= cos 
d
d
dy dy d
=

(Chain rule)
dx d dx
1 

= cos    −

 sin  
cos 
=−
sin 
= − cot 
6. (a)
Data: The gradient of a curve which passes through the point ( −1, − 4 ) is
dy
= 3x 2 − 4 x + 1 .
dx
(i)
a)
Required To Find: The equation of the curve.
Solution:
dy
= 3x 2 − 4 x + 1
dx
y =  ( 3x 2 − 4 x + 1) dx
= x 3 − 2 x 2 + x + C (where C is the constant of integration)
When x = −1 and y = −4
−4 = ( −1) − 2 ( −1) + ( −1) + C
3
2
−4 = −1 − 2 − 1 + C
−4 = − 4 + C
C=0
The equation of the curve is y = x3 − 2 x 2 + x .
b)
Required To Find: The coordinates of the stationary points and
their nature.
Solution:
dy
= 3x 2 − 4 x + 1
dx
dy
Stationary points occur when
= 0.
dx
3x 2 − 4 x + 1 = 0
( 3x − 1)( x − 1) = 0
1
x= ,1
3
d2y
= 6x − 4
dx 2
When x =
3
1
3
2
1
1 1
y =   − 2  +
3
3 3
1 2 1
=
− +
27 9 3
4
=
27
d2y
1
= 6  − 4
2
dx
3
= −2 ( 0)
1 4 
 ,
 is a maximum point.
 3 27 
When x = 1
3
2
y = (1) − 2 (1) + 1
= 1− 2 +1
=0
d2y
= 6 (1) − 4
dx 2
= 2 (  0)
 (1, 0 ) is a minimum point.
(ii)
Required To Sketch: The curve showing all stationary points and
intercepts.
Solution:
The curve cuts the x – axis when y = 0 , that is, at
x3 − 2 x 2 + x = 0
x ( x 2 − 2 x + 1) = 0
x ( x − 1) = 0
2
x = 0 and x = 1
 The curve cuts the x – axis at ( 0, 0 ) and (1, 0 ) .
Since ( x − 1) is a repeated root, the curve touches the x – axis at (1, 0 ) ,
that is, the x – axis is a tangent at (1, 0 ) .
(b)
Data: f ( x ) = 2 x 1 + x 2
(i)
Required To Evaluate:

3
0
f ( x ) dx .
Solution:
Let
t = 1 + x2
dt
= 2x
dx
dt
= dx
2x
Converting limits:
When x = 3
When x = 0
t = 1 + ( 3)
t = 1 + ( 0)
2
= 10
2
=1
  f ( x ) dx =  2 x t
3
10
0
1
=
10
1
dt
2x
t dt
10
2 3 
=  t2 
 3 1
3
2
 2 3 
=  (10 ) 2  −  (1) 2 
3
 3

3
2
=  10 − 1 

3 
2
=
1000 − 1
3
(
(
(ii)
)
)
Required To Calculate: The volume generated by rotating the area
bounded by the curve in (b) (i), the x – axis and the lines x = 0 and x = 2
about the x – axis.
Calculation:
Volume =  
x2
x1
y 2 dx
2
=    2 x 1 + x 2  dx
0 

2
=   4 x 2 (1 + x 2 ) dx
2
0
=
2
0
(4x
2
+ 4 x 4 ) dx
2
4 
4
=   x3 + x5 
5 0
3
 32 128  
=   +
−0
5  
 3
544
=
units3
15
MAY 2013
CAPE PURE MATHEMATICS UNIT 1 PAPER 2
SECTION A
1. (a)
Required To Constuct: A truth table for p → q , where p and q are two
propositions.
Solution:
This statement is conditional meaning if p then q.
(i)
p
T
T
F
F
p→q
T
F
T
T
q
T
F
T
F
Required To Construct: A truth table for ~ ( p  q ) , where p and q are
(ii)
two propositions.
Solution:
This statement is the negation of the conjunction not (p and q).
(b)
p
q
( p  q)
~ ( p  q)
T
T
F
F
T
F
T
F
T
F
F
F
F
T
T
T
Data: A binary operator  is defined on set R + by y  x = y 2 + x 2 + 2 y + x − 5xy
Required To Solve: 2  x = 0
Solution:
2 x = 0
( 2)
2
+ x2 + 2 ( 2) + x − 5x ( 2) = 0
4 + x 2 + 4 + x − 10 x = 0
x2 − 9 x + 8 = 0
( x − 1)( x − 8) = 0
x = 1 or 8
(c)
Required To Prove: 5n + 3 is divisible by 2 for all values of n  N , by
mathematical induction.
Proof:
Let P ( n ) = 5n + 3
Assume the statement is true for n = k .
P ( k ) = 5k + 3 = 2 p …
pN
Consider n = k + 1
P ( k + 1) = 5k + 1 + 3
= 5k + 1 + 2 p − 5k
=5
k +1
(From )
−5 + 2p
k
= 5 ( 5 − 1) + 2 p
k
= 4  5k + 2 p
= 2 ( 2.5k + p ) which is a multiple of 2 and hence divisible by 2.
The statement is true for n = k + 1 .
Testing for n = 1
P (1) = 5 + 3
= 8 which is divisible by 2.
The statement is true for n = 1 .
Testing for n = 2
P ( 2 ) = 52 + 3
= 28 which is divisible by 2.
The statement is true for n = 2 .
Hence, by the Principle of Mathematical Induction the statement is true for
 nN .
(d)
Data: f ( x ) = x3 − 9 x 2 + px + 16 . ( x + 1) is a factor of f ( x ) .
(i)
Required To Prove: p = 6 .
Proof:
Recall: The remainder and factor theorem
If f ( x ) is any polynomial and f ( x ) is divided by ( x − a ) the remainder
is f ( a ) . If f ( a ) = 0 then ( x − a ) is a factor of f ( x ) .
( x + 1) is a factor of f ( x ) .
f ( −1) = 0
3
2
( −1) − 9 ( −1) + p ( −1) + 16 = 0
− 1 − 9 − p + 16 = 0
− p+6=0
p=6
Q.E.D.
(ii)
Required To Factorise: f ( x ) completely.
Solution:
x 2 − 10 x + 16
x + 1 x 3 − 9 x 2 + 6 x + 16
− x3 + x 2
− 10 x 2 + 6 x
− −10 x 2 − 10 x
+ 16 x + 16
− +16 x + 16
0
 f ( x ) = ( x 2 − 10 x + 16 ) ( x + 1)
= ( x + 1)( x − 8 )( x − 2 )
Alternative Method:
By synthetic division,
x 3 − 9 x 2 + 6 x + 16 = ( x + 1) ( ax 2 + bx + c )
= ax 3 + bx 2 + cx + ax 2 + bx + c
= ax 3 + ( b + a ) x 2 + ( c + b ) x + c
Equating coefficients of
x3 :1 = a
x 2 : −9 = b + a
− 9 = b +1
− 10 = b
x :16 = c
0
 f ( x ) = ( x + 1) ( x 2 − 10 x + 16 )
= ( x + 1)( x − 2 )( x − 8 )
(iii)
Required To Solve: f ( x ) = 0
Solution:
f ( x ) = ( x + 1)( x − 8 )( x − 2 )
( x + 1)( x − 8)( x − 2 ) = 0
 x = −1 or 8 or 2
2. (a)
Data: A =  x : x  R, x  1 , f ( x ) = x 2 − x
Required To Prove: f is one to one.
Solution:
f ( a ) = f (b )
a 2 − a = b2 − b
a 2 − b2 − a + b = 0
( a − b )( a + b ) − ( a − b ) = 0
( a − b ) ( a + b ) − 1 = 0
a −b = 0
a=b
 f ( x ) is one to one.
a +b =1
But a, b  1 .
 This is not possible.
Alternative Method:
f ( x ) = x2 − x
= x ( x − 1)
The minimum point of f ( x ) occurs when x =
1
.
2
As illustrated on the diagram, each element of the domain corresponds to one
element in the co-domain and each element of the co-domain is associated with
one element of the domain. A horizontal line cuts the graph at only one point.
Hence, f ( x ) is one to one for x greater than or equal to 1, as illustrated.
(b)
Data: f ( x ) = 3x + 2 and g ( x ) = e2 x .
(i)
a)
Required To Find: f −1 ( x ) and g −1 ( x )
Solution:
Let y = f ( x ) :
y = 3x + 2
Interchange x and y:
x = 3y + 2
Make y the subject of the formula:
x − 2 = 3y
x−2
=y
3
x−2
 f −1 ( x ) =
3
Let y = g ( x )
y = e2 x
Interchange x and y:
x = e2 y
Make y the subject of the formula:
Taking natural logs:
ln x = ln e 2 y
ln x = 2 y
1
ln x = y
2
1
 g −1 ( x ) = ln x
2
= ln x
b)
Required To Find: f  g ( x )  or f g ( x )
Solution:
fg ( x ) = f ( e 2 x )
= 3e 2 x + 2
(ii)
Required To Prove: ( f g )
−1
( x ) = g −1 ( x )
f −1 ( x )
Proof:
 x−2
g −1 f −1 ( x ) = g −1 

 3 
1  x−2
= ln 

2  3 
−1
 fg ( x ) 
Let y = fg ( x )
y = 3e 2 x + 2
Interchange x and y :
x = 3e 2 y + 2
Make y the subject of the formula:
x − 2 = 3e 2 y
x−2
= e2 y
3
Taking ln:
 x−2
2y
ln 
 = ln e
 3 
x−2
ln
= 2y
3
1  x−2
ln 
= y
2  3 
−1
1  x−2
  fg ( x )  = ln 

2  3 
Hence, ( f g )
(c)
(i)
−1
( x ) = g −1 ( x )
f −1 ( x ) .
Q.E.D.
Required To Solve: 3x 2 + 4 x + 1  5
Solution:
3x 2 + 4 x + 1  5
3x 2 + 4 x − 4  0
( 3x − 2 )( x + 2 )  0
2
 The graph of y = 3 x + 4 x − 4 meets the x – axis at x = −2 or
2
.
3

 x : −2  x 

(ii)
2
 shown shaded.
3
Required To Solve: x + 2 = 3x + 5
Solution:
x + 2 = 3x + 5
That is,
−3x − 5 = x + 2 = 3 x + 5
−3 x − 5 = x + 2
− 7 = 4x
7
x=−
4
x = −
7
3
or − .
4
2
(by definition)
x + 2 = 3x + 5
− 3 = 2x
3
x=−
2
x = −
3
2
SECTION B
3. (a)
(i)
Required To Prove: sin 2 =
2 tan 
.
1 + tan 2 
Proof:
Taking right hand side:
2 tan 
2 tan 
=
2
1 + tan  sec 2 
= 2 tan  .cos 2 
2sin 
=
.cos2 
cos 
= 2sin  cos 
= sin 2
(Left hand side)
Q.E.D.
Alternative Proof:
Simplifying the right hand side:
2 tan 
2 tan 
=
2
1 + tan  sec 2 
2sin 
= cos 
1
cos2 
2sin  cos2 
=

cos 
1
= 2sin  cos 
= sin 2
(Left hand side)
Q.E.D.
(ii)
Required To Solve: sin 2 − tan  = 0 for 0    2 .
Solution:
sin 2 − tan  = 0
2 tan 
− tan  = 0
1 + tan 2 
(1 + tan 2  )
2 tan  − (1 + tan 2  ) tan  = 0
2 tan  − tan  − tan 3  = 0
tan  − tan 3  = 0
tan  (1 − tan 2  ) = 0
tan  (1 − tan  )(1 + tan  ) = 0
tan  = 0
1 − tan  = 0
tan  = 1
 = 0,  , 2
=
 = 0,
(b)
 3
4
,
4
, ,
 5
4
,
4
1 + tan  = 0
tan  = −1
3 7
= ,
4 4
5 7
,
, 2 .
4 4
Data: f ( ) = 3cos  − 4sin  .
(i)
Required To Express: f ( ) in the form r cos ( +  ) , where r  0 and
0 

.
2
Solution:
3cos  − 4sin  = r cos ( +  )
= r  cos  cos  − sin  sin  
Equating terms in cos  and sin  .
3 = r cos 
…(1)
4 = r sin 
…(2)
(2)  (1)
tan  =
4
3
4
 
 = tan −1  
3
= 0.93 radians for 0   

2
(1)2 + (2)2
2
2
( r cos  ) + ( r sin  ) = 32 + 42
r 2 ( cos 2  + sin 2  ) = 25
r 2 = 25
r = 5 ( 0)
 3cos  − 4sin  = 5cos ( + 0.93) which is of the form r cos ( +  ) ,
where r = 5 and  = 0.93 .
(ii)
a)
Required To Find: The maximum value of f ( ) .
Solution:
f ( ) = 5cos ( + 0.93)
From 3 (b) (i), −1  cos ( + 0.93)  1
 −1( 5)  5cos ( + 0.93)  1 ( 5)
Maximum value of cos ( + 0.93) = +1
 Maximum value of f ( ) = 5 ( +1)
= +5
b)
Required To Find: The minimum value of
1
.
8 + f ( )
Solution:
The minimum value of
1
occurs when the denominator is
8 + f ( )
the largest.
Minimum value of
1
1
=
8 + f ( ) 8 + max f ( )
1
8+5
1
=
13
=
(iii)
Data: A + B + C =  , where A, B and C are the angles of a triangle.
a)
Required To Prove: sin A = sin ( B + C )
Proof:
A+ B +C =
A =  − (B + C)
sin A = sin  − ( B + C ) 
Recall: sin (180 −  ) = sin 
sin A = sin ( B + C )
Q.E.D
Alternative method of proof:
A+ B +C =
A =  − (B + C)
sin A = sin  − ( B + C ) 
Recall: sin ( A − B ) = sin A cos B − cos A sin B
sin ( − A ) = sin  cos A − cos  sin A
= 0 ( cos A ) − ( −1) sin A
= sin A
= sin  − ( B + C ) 
= sin ( B + C )
 sin A = sin ( B + C )
Q.E.D.
b)
Required To Prove:
sin A + sin B + sin C = sin ( A + B ) + sin ( B + C ) + sin ( A + C )
Proof:
B =  −( A+C)
sin B = sin  − ( A + C ) 
= sin ( A + C )
Similarly,
C =  − ( A + B)
sin C = sin  − ( A + B ) 
= sin ( A + B )
 sin A + sin B + sin C = sin ( A + B ) + sin ( B + C ) + sin ( A + C )
Q.E.D
4. (a)
Data: A circle, C, has equation x 2 + y 2 − 6 x − 4 y + 4 = 0 .
(i)
Required To Prove: Circle C has centre ( 3, 2 ) and radius 3.
Proof:
x2 + y 2 − 6 x − 4 y + 4 = 0
x2 + y 2 − 6 x + 9 − 4 y + 4 = 9
x2 − 6 x + 9 + y 2 − 4 y + 4 = 9
( x − 3) + ( y − 2 ) = 9
2
2
2
( x − 3) + ( y − 2 ) = ( 3)
2
2
This is the equation of a circle with centre
( 3, 2 )
and radius 3.
Alternative Method:
x2 + y 2 − 6 x − 4 y + 4 = 0 is of the form x2 + y 2 + 2 gx + 2 fy + c = 0 , which
is the equation of a circle with centre ( − g , − f ) and radius
= g2 + f 2 − c .
2 g = −6
2 f = −4
g = −3
f = −2
 Centre of the circle is ( 3, 2 ) .
Radius = g 2 + f 2 − c
=
( −3) + ( −2 )
2
2
−4
= 9+4−4
= 3 units
Q.E.D.
(ii)
a)
Required To Find: The equation of the normal to the circle, C, at
( 6, 2 ) .
Solution:
2−0
6−3
=0
Gradient of the normal =
Equation of the normal is ( y − 2 ) = 0 ( x − 6 )
y=2
b)
Required To Prove: The tangent to the circle, C, at the point
( 6, 2 ) is parallel to the y – axis.
Proof:
Equation of the normal is y = 2 .
Equation of the tangent is x = 6 , which is parallel to the y – axis.
Q.E.D.
(b)
Data: The parametric equations of a curve are x = t 2 + t, y = 2t − 4 .
Required To Prove: The Cartesian equation of the curve is 4 x = y 2 + 10 y + 24 .
Proof:
…(1)
x = t2 + t
…(2)
y = 2t − 4
From (2)
y+4
=t
2
Substitute into (1):
 y+4  y+4
x=
 +

 2   2 
2
x=
( 4 )
( y + 4)
4
2
+
( y + 4)
2
4 x = y 2 + 8 y + 16 + 2 ( y + 4 )
4 x = y 2 + 10 y + 24
Q.E.D.
(c)
Data: A ( 3, − 1, 2 ) , B (1, 2, − 4 ) and C ( −1, 1, − 2 ) are three vertices of a
parallelogram ABCD.
(i)
Required To Express: AB and BC in the form xi + yj + zk .
Solution:
AB = OB − OA
 1   3
=  2  −  −1 
 −4   2 
   
 −2 
=  3 
 −6 
 
= −2i + 3j − 6k is of the form xi + yj + zk , where x = −2, y = 3 and
z = −6 .
BC = OC − OB
 −1   1 
=  1  −  2 
 −2   −4 
   
 −2 
=  −1 
 2
 
= −2i − j + 2k is of the form xi + yj + zk , where x = −2, y = −1 and
z = 2.
(ii)
Required To Prove: The vector r = −16 j − 8k is perpendicular to the
plane through A, B and C.
Proof:
r . AB = ( −16 j − 8k )( −2i + 3j − 6k )
= −2 ( 0 ) + 3 ( −16 ) + ( −8 )( −6 )
=0
r . BC = ( −16 j − 8k )( −2i − j + 2k )
= −2 ( 0 ) − 16 ( −1) + 2 ( −8 )
=0
Since r is perpendicular to both AB and BC , then it is perpendicular to
the plane through A, B and C. r = −16 j − 18k is the normal to the plane
through A, B and C.
Q.E.D.
(iii)
Required To Find: The Cartesian equation of the plane through A, B and
C.
Solution:
Let
n = xi + yj + zk
n . r = a . r , where a = OA = 3i − j + 2k
( xi + yj + zk ) . ( −16 j − 8k ) = ( 3i − j + 2k ) . ( −16 j − 8k )
− 16 y − 8 z = 16 − 16
− 16 y − 8 z = 0
2y + z = 0
SECTION C
5. (a)
 x + 2, x  2
Data: f ( x ) =  2
x2
x ,
(i)
Required To Find: lim f ( x ) .
x →2
Solution:
lim f ( x ) = 2 + 2
x → 2−
=4
lim+ f ( x ) = ( 2 )
2
x →2
=4
 lim f ( x ) = 4
x →2
(ii)
Required To Determine: Whether f ( x ) is continuous at x = 2 .
Solution:
Since f ( x ) is not defined at x = 2 , f ( x ) is not continuous at x = 2 .
(b)
Data: y =
x2 + 2 x + 3
(x
2
+ 2)
3
dy −4 x3 − 10 x 2 − 14 x + 4
=
Required To Prove:
4
dx
( x2 + 2)
Proof:
x2 + 2 x + 3
u
is of the form y = , where
y=
3
v
( x2 + 2)
v = ( x2 + 2)
u = x2 + 2x + 3
du
= 2x + 2
dx
3
2
dv
= 3 ( 2 x ) ( x2 + 2)
dx
= 6 x ( x2 + 2)
dy
=
dx
v
du
dv
−u
dx
dx
v2
(x
=
=
(x
(x
=
=
=
2
(Quotient law)
+ 2 ) ( 2 x + 2 ) − ( x 2 + 2 x + 3) ( 6 x ) ( x 2 + 2 )
3
(x
2
2
2
2
+ 2)
2
6
+ 2 ) ( x 2 + 2 ) ( 2 x + 2 ) − 6 x ( x 2 + 2 x + 3) 
6
( x2 + 2)
2
+ 2 ) ( 2 x + 2 ) − 6 x ( x 2 + 2 x + 3)
(x
2
+ 4)
4
2 x 3 + 2 x 2 + 4 x + 4 − ( 6 x 3 + 12 x 2 + 18 x )
(x
2
+ 4)
4
−4 x 3 − 10 x 2 − 14 x + 4
(x
2
+ 4)
4
Q.E.D.
(c)
Data: The equation of an ellipse is given by x = 1 − 3cos  , y = 2sin  for
0    2 .
dy
Required To Find:
in terms of  .
dx
Solution:
x = 1 − 3cos 
y = 2sin 
dx
dy
= −3 ( − sin  )
= 2 cos 
d
d
= 3sin 
dy dy d
=

dx d dx
= 2 cos  
(Chain rule)
1
3sin 
2
= cot 
3
(d)
Data: Diagram showing the curve y = x 2 + 3 and the line y = 4 x intersecting at
two points P and Q.
(i)
Required To Determine: The coordinates of P and Q.
Solution:
y = x2 + 3
…(1)
…(2)
y = 4x
Equating (2) into (1):
4 x = x2 + 3
x2 − 4 x + 3 = 0
( x − 3)( x − 1) = 0
x = 1 or 3
When x = 1
y = 4 (1)
=4
When x = 3
y = 4 ( 3)
= 12
 Coordinates of P and Q are (1, 4 ) and ( 3, 12 ) respectively.
(ii)
Required To Calculate: The area of the shaded region.
Calculation:
Area of shaded region = Area under line − Area under curve
3
=  4 x dx − 
1
3
1
(x
3
2
+ 3) dx
3
 4 x 2   x3

=
−  + 3x 

 2 1  3
1
1

= 18 − 2 − 18 − 3 
3

2
= 16 − 4
3
4
= units 2
3
6. (a)
(i)
Data: u = 1 − x
Required To Find:
 x (1 − x )
2
dx
Solution:
u = 1− x  x = 1− u
du
= −1
dx
− du = dx
 x (1 − x )
2
dx =  (1 − u ) u 2 ( −du )
= −  u 2 − u 3 du
u3 u 4
= − + +C
3 4
(1 − x )
=
4
(ii)
4
(1 − x )
−
3
(where C is the constant of integration)
3
+C
Data: f ( t ) = 2 cos t and g ( t ) = 4sin 5t + 3cos t .
Required To Prove:
  f (t ) + g (t ) dt =  f (t ) dt +  g (t ) dt
Proof:
Consider the left hand side:
  f ( t ) + g ( t ) dt =  ( 2 cos t + 4sin 5t + 3cos t ) dt
=  ( 2 cos t ) dt +  ( 4sin 5t + 3cos t ) dt
=  f ( t ) dt +  g ( t ) dt
(Right hand side)
Q.E.D.
(b)
Data: A sports association is planning to construct a running track in the shape of
a rectangle surmounted by a semi-circle, as shown in the diagram below, where x
is the length of the rectangle and r is the radius of the circle. The perimeter of the
track is 600 metres.
(i)
Required To Prove: r =
600 − 2 x
.
2 +
Proof:
1
( 2 r )
2
600 = 2 x + 2r +  r
Perimeter of track = 2 x + 2r +
600 − 2 x = ( + 2 ) r
r=
600 − 2 x
2 +
Q.E.D.
(ii)
Required To Determine: The length x that maximises the area of the
track.
Solution:
Let A be the area of the track.
1
A = 2rx +  r 2
2
 600 − 2 x  1  600 − 2 x 
= 2x 
+ 
 2 +   2  2 +  
=
2 ( 600 x − 2 x 2 )
2+
+

2 ( + 2 )
2
2
( 600 − 2 x )
2
dA
2

=
 2 ( 600 − 2 x )( −2 )
( 600 − 4 x ) −
2
dx 2 + 
2(2 +  )
=
8
2
600 − 2 x )
(150 − x ) +
2 (
2+
(2 +  )
Stationary points occur when
0=
dA
=0.
dx
8
2
600 − 2 x )
(150 − x ) −
2 (
2+
(2 +  )
(2 +  )
2
0 = 8 ( 2 +  )(150 − x ) − 2 ( 600 − 2 x )
4 x + 8 ( 2 +  )(150 − x ) = 150 ( 8 )( 2 +  ) + ( 600 ) 2
(16 + 12 ) x = 2400 + 1200
2400 + 1200
16 + 10
600 ( 2 +  )
=
4 ( 4 + 3 )
x=
 2+ 
= 150 

 4 + 3 
Alternative Method:
dA
2
1
=
600 − 2 x )( 2 )( −2 )
( 600 − 4 x ) +
2 (
dx  + 2
2 ( + 2 )
=
2
4
600 − 2 x )
( 600 − 4 x ) −
2 (
 +2
2 ( + 2 )
Stationary points occur when
dA
=0
dx
That is,
0=
2
( + 2 )
( 2 )
2
( 600 − 2x ) =
2
4
600 − 2 x )
( 600 − 4 x ) −
2 (
 +2
2 ( + 2 )
2
( 600 − 4 x )
 +2
600 − 2 x
= 600 − 4 x
 +2
300 − x
= 300 − 2 x
 +2
300 − x = ( + 2 )( 300 − 2 x )
300 − x = 300 ( + 2 ) − 2 x ( + 2 )
2 x ( + 2 ) − x = 300 + 600 − 300
x  2 + 4 − 1 = 300 + 300
x ( 2 + 3) = 300 ( + 1)
x=
300 ( + 1)
2 + 3
d2A
8
4
=−
−
2
dx
2 +  ( 2 +  )2
 0 (Maximum)
(c)
Data: y = − x sin x − 2cos x + Ax + B , where A and B are constants.
(i)
Required To Prove: y = x sin x
Proof:
y = − x sin x − 2 cos x + Ax + B
y = − x cos x − sin x + 2sin x + A
y = − x cos x + sin x + A
y = − x ( − sin x ) − cos x + cos x
y = x sin x
Q.E.D.
(ii)
Required To Determine: The specific solution for the differential
equation y = x sin x , given that when x = 0, y = 1 and when x =  , y = 6 .
Solution:
 y dx =  x sin x dx
y = − x cos x + sin x + A
y =  y dx
y = − x sin x − 2 cos x + Ax + B
When x = 0, y = 1
1 = 0 ( sin 0 ) − 2 cos ( 0 ) + A ( 0 ) + B
1= B −2
B=3
When x =  , y = 6
6 = − sin ( ) − 2 cos ( ) +  ( A ) + 3
6 = 2 + 3 +  ( A)
1 =  ( A)
A=
1

 y = − x sin x − 2cos x +
1
x+3

(Specific solution OR Particular solution (P.S.))
JUNE 2012 UNIT 1 PAPER 2
1. (a)
Data: The expression 𝑓(𝑥) = 2𝑥 3 − 𝑝𝑥 2 + 𝑞𝑥 − 10 is divisible by 𝑥 − 1 and
has a remainder −6 when divided by 𝑥 + 1.
(i)
Required To Find: The value of 𝑝 and of 𝑞.
Solution:
Recall: The remainder and Factor Theorem
If a polynomial 𝑓(𝑥) is divided, by (𝑥 − 𝑎) the remainder is 𝑓(𝑎). If,
𝑓(𝑎) = 0, then (𝑥 − 𝑎) is a factor of 𝑓(𝑥).
𝑓(𝑥) is divisible by (𝑥 − 1)
∴ 𝑓(1) = 0
𝑓(1) = 2(1)3 − 𝑝(1)2 + 𝑞(1) − 10
0 = 2 − 𝑝 + 𝑞 − 10
8 = −𝑝 + 𝑞 … (1)
𝑓(𝑥) has a remainder of −6 when divided by 𝑥 + 1.
∴ 𝑓(−1) = −6
𝑓(−1) = 2(−1)3 − 𝑝(−1)2 + 𝑞(−1) − 10
−6 = −2 − 𝑝 − 𝑞 − 10
𝑝 + 𝑞 = −6 … (2)
Eq(1) + Eq(2)
2𝑞 = 2
𝑞=1
When, 𝑞 = 1
𝑝 = −6 − 1
𝑝 = −7
∴ 𝑝 = −7 and 𝑞 = 1.
(ii)
Required To Find: The factors of 𝑓(𝑥).
Solution:
2𝑥 2 + 9𝑥 + 10
(𝑥 − 1) 2𝑥 3 + 7𝑥 2 + 𝑥 − 10
−2𝑥 3 − 2𝑥 2
9𝑥 2 + 𝑥
− 9𝑥 2 − 9𝑥
10𝑥 − 10
− 10𝑥 − 10
0
2
𝑓(𝑥) = (𝑥 − 1)(2𝑥 + 9𝑥 + 10)
= (𝑥 − 1)(2𝑥 + 5)(𝑥 + 2)
Therefore the factors of 𝑓(𝑥) = 2𝑥 3 + 7𝑥 2 + 𝑥 − 10 are(𝑥 + 2), (𝑥 − 1)
and (2𝑥 + 5).
Alternative Method:
Using synthetic division:
(𝑥 − 1)(𝑎𝑥 2 + 𝑏𝑥 + 𝑐) = 2𝑥 3 + 7𝑥 2 + 𝑥 − 10
𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 − 𝑎𝑥 2 − 𝑏𝑥 − 𝑐 = 2𝑥 3 + 7𝑥 2 + 𝑥 − 10
𝑎𝑥 3 + (𝑏 − 𝑎)𝑥 2 + (𝑐 − 𝑏)𝑥 − 𝑐 = 2𝑥 3 + 7𝑥 2 + 𝑥 − 10
Expanding and comparing coefficients of 𝑥 3 :
𝑎=2
Comparing the constant term,
−𝑐 = −10
𝑐 = 10
Comparing coefficients of 𝑥 2 :
−𝑎 + 𝑏 = 7
−2 + 𝑏 = 7
𝑏=9
𝑓(𝑥) = (𝑥 − 1)(2𝑥 2 + 9𝑥 + 10)
= (𝑥 − 1)(2𝑥 + 5)(𝑥 + 2)
Hence the factors of 𝑓(𝑥) are (𝑥 − 1), (2𝑥 + 5) and (𝑥 + 2).
(b)
2
Data: (√𝑥 + √𝑦) = 16 + √240
Required To Find: The values of 𝑥 and 𝑦.
Solution:
2
(√𝑥 + √𝑦) = 16 + √240
𝑥 + 𝑦 + 2√𝑥𝑦 = 16 + √240
𝑥 + 𝑦 + √4𝑥𝑦 = 16 + √240
Equating terms:
𝑥 + 𝑦 = 16
… (1)
4𝑥𝑦 = 240 … (2)
From Equation (2):
𝑥𝑦 = 60
𝑦=
60
𝑥
Substitute into Equation (1):
𝑥+
60
𝑥
= 16
(× 𝑥)
𝑥 2 + 60 = 16𝑥
𝑥 2 − 16𝑥 + 60 = 0
(𝑥 − 10)(𝑥 − 6) = 0
𝑥 = 6 or 10
When 𝑥 = 6
𝑦=
60
6
= 10
When 𝑥 = 10
60
𝑦 = 10
=6
∴ 𝑥 = 6 and 𝑦 = 10 OR 𝑥 = 10 and 𝑦 = 6.
(c)
(i)
Required To Solve:|3𝑥 − 7| ≤ 5
Solution:
Consider 𝑦 = 3𝑥 − 7
when 𝑥 = 0, 𝑦 = −7
7
when 𝑦 = 0, 𝑥 = 3
−3𝑥 + 7 = 5
2 = 3𝑥
2
3
3𝑥 − 7 = 5
3𝑥 = 12
=𝑥
𝑥=4
2
Hence, {𝑥: 3 ≤ 𝑥 ≤ 4}
Alternative Method:
Recall |𝑥| ≤ 𝑎 implies that −𝑎 ≤ 𝑥 ≤ 𝑎.
|3𝑥 − 7| ≤ 5
−5 ≤ 3𝑥 − 7 ≤ 5
2 ≤ 3𝑥 ≤ 12
2
3
≤𝑥≤4
The answer is best written in set builder notation as {𝑥 ∶
(ii)
2
3
≤ 𝑥 ≤ 4}.
Required To Prove: There are no real solutions of 𝑥 for
|3𝑥 − 7| + 5 ≤ 0.
Proof:
|3𝑥 − 7| + 5 ≤ 0
|3𝑥 − 7| ≤ −5
but |3𝑥 − 7| is always positive or 0.
i.e.|3𝑥 − 7| ≥ 0
Hence, in the given inequality, no real solution exists for 𝑥 ∈ 𝑅.
Q.E.D
Alternative Method:
Consider the sketch of the graph 𝑦 = |3𝑥 − 7|.
0
𝑇=( )
5
𝑦 = |3𝑥 − 7| →
𝑦 = |3𝑥 − 7| +5 and |3𝑥 − 7| ≥ 0 ∀𝑥,
|3𝑥 − 7| + 5 ≥ 5 ∀𝑥
∴ |3𝑥 − 7| + 5 ≤ 0 has no real solutions.
2.
(a)
Data: The function 𝑓 on ℝ is defined by 𝑓: 𝑥 → 𝑥 2 − 3.
(i)
Required To Find: 𝑓(𝑓(𝑥)) in terms of 𝑥.
Solution:
𝑓(𝑓(𝑥)) = (𝑥 2 − 3)2 − 3
= 𝑥 4 − 6𝑥 2 + 9 − 3
= 𝑥 4 − 6𝑥 2 + 6
(ii)
Required To Find: The values of 𝑥 for which 𝑓(𝑓(𝑥)) = 𝑓(𝑥 + 3)
Solution:
𝑓(𝑥 + 3) = (𝑥 + 3)2 − 3
= 𝑥 2 + 6𝑥 + 9 − 3
= 𝑥 2 + 6𝑥 + 6
𝑓𝑓(𝑥) = 𝑓(𝑥 + 3)
𝑥 4 − 6𝑥 2 + 6 = 𝑥 2 + 6𝑥 + 6
𝑥 4 − 7𝑥 2 − 6𝑥 = 0
𝑥(𝑥 3 − 7𝑥 − 6) = 0
Let 𝑔(𝑥) = 𝑥 3 − 7𝑥 − 6
Using the remainder and factor theorem:
When a polynomial 𝑔(𝑥) is divided, by (𝑥 − 𝑎), the remainder is 𝑔(𝑎). If,
𝑔(𝑎) = 0, then (𝑥 − 𝑎) is a factor of 𝑔(𝑥).
𝑔(−1) = −1 + 7 − 6 = 0
(𝑥 + 1) is a factor of 𝑔(𝑥).
𝑥2 − 𝑥 − 6
(𝑥 + 1) 𝑥 3 − 7𝑥 − 6
𝑥3 + 𝑥2
− 𝑥 2 − 7𝑥
− −𝑥 2 − 𝑥
−6𝑥 − 6
− −6𝑥 − 6
0
𝑔(𝑥) = (𝑥 + 1)(𝑥 2 − 𝑥 − 6)
= (𝑥 + 1)(𝑥 − 3)(𝑥 + 2)
∴ 𝑥(𝑥 3 − 7𝑥 − 6) = 0
𝑥(𝑥 + 1)(𝑥 − 3)(𝑥 + 2) = 0
𝑥 = −2,−1, 0 or 3
(b)
Data: The roots of the equation 4𝑥 2 − 3𝑥 + 1 = 0 is 𝛼 and 𝛽.
(i)
Required To Find: The value of 𝛼 + 𝛽 and of 𝛼𝛽.
Solution:
Re: If 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 is a quadratic equation in x then
÷ (𝑎)
𝑏
𝑐
𝑥2 + 𝑎 𝑥 + 𝑎 = 0
If 𝛼 and 𝛽 are the roots of the equation then
(𝑥 − 𝛼)(𝑥 − 𝛽) = 0
𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 = 0
Equating coefficients
𝑏
𝛼 + 𝛽 = −𝑎
𝑐
𝛼𝛽 = 𝑎
The given equation is
4𝑥 2 − 3𝑥 + 1 = 0
3
1
(÷ 4) 𝑥 2 − 4 𝑥 + 4 = 0
−3
Hence, 𝛼 + 𝛽 = − ( 4 ) =
4
1
𝛼𝛽 = 4
and
(ii)
3
Required To Find: The value of 𝛼 2 + 𝛽 2.
Solution:
𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
3 2
1
= (4) − 2 (4)
1
= 16
(iii)
2
2
Required To Find: The quadratic equation whose roots are 𝛼2 and 𝛽2.
Solution:
A quadratic equation is of the form:
𝑥 2 − (sum of roots)𝑥 + (product of roots) = 0
The required equation is:
2
2
Re:  2  2 = ( )
2 2
𝑥 2 − (𝛼2 + 𝛽2 ) 𝑥 + (𝛼2 𝛽2 ) = 0
2
𝛼2
2
+ 𝛽2 =
2 2
𝛼2 𝛽 2
2(𝛽 2 +𝛼2 )
𝛼2 𝛽 2
4
= 𝛼2 𝛽 2 =
4
1 2
( )
4
=
1
16
1 2
( )
4
2( )
2
=2
= 64
2
2
The new equation whose roots are:𝛼2 and 𝛽2 is 𝑥 2 − 2𝑥 + 64 = 0.
(c)
(i)
1
3
5
5
7
9
Required To Evaluate: log10 (3) + log10 (5) + log10 (7) +
7
9
log10 (9) + log10 (10)
Solution:
Re: log 𝑎 + log 𝑏 = log 𝑎𝑏
1
3
log10 (3) + log10 (5) + log10 (7) + log10 (9) + log10 (10)
1
3
5
7
9
= log10 (3 × 5 × 7 × 9 × 10)
1
= log10 (10)
= log10 (10−1 )
= − log10 10
= −1
(ii)
𝑟
Required To Evaluate: ∑99
𝑟=1 log10 (𝑟+1)
Solution:
𝑟
∑99
𝑟=1 log10 (
1
2
3
98
) = log10 (2) + log10 (3) + log10 (4) + ⋯ + log10 (99) +
𝑟+1
99
+ log10 (100)
= log10 1 − log10 2 + log10 2 − log10 3 + log10 3
+ ⋯ − log10 99 + log10 99 − log10 100
= log10 1 − log10 100
1
= log10 (100)
= log10 (10−2 )
= −2 log10 10
= −2 × 1
= −2
Alternative Method:
𝑟
1
2
3
98
∑99
𝑟=1 log10 (𝑟+1) = log10 (2) + log10 (3) + log10 (4) + ⋯ + log10 (99)
99
+ log10 (100)
1
2
3
98
99
= log10 [2 × 3 × 4 × … × 99 × 100]
1
= log10 (100)
= log10 10−2
= −2
3.
(a)
Data: cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
cos 2𝜃 = 2 cos2 𝜃 − 1
(i)
1
Required To Prove: cos 3𝜃 = 2 cos 𝜃 (cos2 𝜃 − sin2 𝜃 − 2)
Proof:
Taking L.H.S.
cos 3𝜃 = cos(2𝜃 + 𝜃)
= cos 2𝜃 cos 𝜃 − sin 2𝜃 sin 𝜃
= (2 cos2 𝜃 − 1) cos 𝜃 − (2 sin 𝜃 cos 𝜃) sin 𝜃
= 2 cos3 𝜃 − cos 𝜃 − 2 sin2 𝜃 cos 𝜃
1
= 2 cos 𝜃 (cos2 𝜃 − 2 − sin2 𝜃)
1
= 2 cos 𝜃 (cos2 𝜃 − sin2 𝜃 − ) ≡R.H.S.
2
Q.E.D.
(ii)
1
Required To Prove: 2 [sin 6𝜃 − sin 2𝜃] ≡ (2 cos 2 2𝜃 − 1) sin 2𝜃
Proof
Taking L.H.S.
Recall: sin 𝑃 − sin 𝑄 = 2 cos (
1
𝑃+𝑄
1
[sin 6𝜃 − sin 2𝜃] = [2 cos (
2
2
) sin (
2
6𝜃+2𝜃
2
𝑃−𝑄
)
2
6𝜃−2𝜃
) sin (
2
(Factor formula)
)]
1
= 2 [2 cos 4𝜃 sin 2𝜃]
= cos 4𝜃 sin 2𝜃
= cos 2(2𝜃) sin 2𝜃
= (2 cos 2 2𝜃 − 1) sin 2𝜃
=R.H.S.
Q.E.D.
(iii)
Required To Solve: sin 6𝜃 − sin 2𝜃 = 0 for 0 ≤ 𝜃 ≤
𝜋
2
Solution:
From part (ii)
1
[sin 6𝜃 − sin 2𝜃] = (2cos2 2𝜃 − 1) sin 2𝜃
2
sin 6𝜃 − sin 2𝜃 = 2(2 cos2 2𝜃 − 1) sin 2𝜃 = 0
𝜋
(2 cos 2 2𝜃 − 1) = 0
OR
2sin 2𝜃 = 0 0 ≤ 𝜃 ≤ 2
(√2 cos 2𝜃 − 1)(√2 cos 2𝜃 + 1) = 0
𝜋
√2 cos 2𝜃 − 1 = 0 0 ≤ 𝜃 ≤ 2
cos 2𝜃 =
1
sin 2𝜃 = 0 0 ≤ 2𝜃 ≤ 𝜋
2𝜃 = 0, 𝜋
𝜋
0 ≤ 2𝜃 ≤ 𝜋
√2
𝜃 = 0 𝑜𝑟 2
1
2𝜃 = cos−1 ( )
𝜋 7𝜋
√2
2𝜃 = 4 ,
4
𝜋 7𝜋
𝜃 = 8,
𝜃=
𝜋
8
8
only (for the given range)
𝜋
√2 cos 2𝜃 + 1 = 0 0 ≤ 𝜃 ≤
cos 2𝜃 = −
2
1
0 ≤ 2𝜃 ≤ 𝜋
√2
−1
2𝜃 = cos −1 ( )
√2
2𝜃 =
𝜃=
𝜃=
3𝜋 5𝜋
,
4 4
3𝜋 5𝜋
8
3𝜋
8
,
8
only, for the given range
𝜋 3𝜋 𝜋
∴ 𝜃 = 0, 8 ,
(b)
Data: 2 cot 2 𝜃 + cos 𝜃 = 0
8
,2
0≤𝜃≤
𝜋
2
Required To Find: All possible values of cos 𝜃.
Solution:
2 cot 2 𝜃 + cos 𝜃 = 0
cos2 𝜃
2 ( sin2 𝜃 ) + cos 𝜃 = 0
2 cos2 𝜃+cos 𝜃 sin2 𝜃
sin2 𝜃
=0
2 cos 2 𝜃 + cos 𝜃 sin2 𝜃 = 0
(Recall: cos 2 𝜃 + sin2 𝜃 = 1)
2 cos 2 𝜃 + cos 𝜃 (1 − cos2 𝜃) = 0
2 cos 2 𝜃 + cos 𝜃 − cos 3 𝜃 = 0
∴ cos 3 𝜃 − 2 cos2 𝜃 − cos 𝜃 = 0
cos 𝜃 (cos 2 𝜃 − 2 cos 𝜃 − 1) = 0
cos 𝜃 = 0
OR
cos2 𝜃 − 2 cos 𝜃 − 1 = 0
2±√(−2)2 −4(1)(−1)
cos 𝜃 =
cos 𝜃 =
cos 𝜃 =
2(1)
2±√4+4
2
2±√8
2
= 1 ± √2
∴ cos 𝜃 = 0, 1 + √2 or 1 − √2
cos 𝜃 = 0, 1 − √2 for −1 ≤ cos 𝜃 ≤ 1
4.
(a)
Data: 𝑥 = 3 tan 𝜃, 𝑦 = 3 sec 𝜃 are parametric equations of the curve 𝐶.
(i)
Required To Find: The Cartesian equation of the curve 𝐶.
Solution:
𝑦
𝑥
sec 𝜃 = 3
and
tan 𝜃 = 3
Re: tan2 𝜃 + 1 = sec 2 𝜃
𝑥 2
𝑦 2
∴ (3 ) + 1 = ( 3 )
𝑥2
9
2
+1=
𝑦2
9
2
𝑥 +9=𝑦
(ii)
Data: The curve 𝑦 = √10𝑥 intersects the curve 𝐶.
Required To Find: The points of intersection.
Solution:
𝑥 2 − 𝑦 2 = −9
… (1)
𝑦 = √10𝑥 … (2)
Squaring equation (2):
𝑦 2 = 10𝑥
… (3)
Substituting (3) into (1):
𝑥 2 − 10𝑥 = −9
𝑥 2 − 10𝑥 + 9 = 0
(𝑥 − 1)(𝑥 − 9) = 0
𝑥 = 1 or 9
When 𝑥 = 1, 𝑦 = √10
When 𝑥 = 9, 𝑦 = √90 = 3√10
The points of intersection are (1, √10) and (9, 3√10).
(b)
−1
−3
), 𝑞 = ( ) are the position vectors of the points (−3, 4) and
6
4
Data: 𝑝 = (
(−1, 6).
(i)
Required To Express: 𝑝 and 𝑞 in the form 𝑥𝑖 + 𝑦𝑗.
Solution:
𝑝 = −3𝑖 + 4𝑗, which is of the form 𝑥𝑖 + 𝑦𝑗 where 𝑥 = −3 and 𝑦 = 4.
𝑞 = −𝑖 + 6𝑗, which is of the form 𝑥𝑖 + 𝑦𝑗 where 𝑥 = −1 and 𝑦 = 6.
(ii)
Required To Find: 𝑝 − 𝑞
Solution:
𝑝 − 𝑞 = (−3𝑖 + 4𝑗) − (−𝑖 + 6𝑗)
= −2𝑖 − 2𝑗, which is of the form 𝑥𝑖 + 𝑦𝑗 where 𝑥 = −2 and
𝑦 = −2
(iii)
Required To Find: 𝑝. 𝑞
Solution:
𝑝. 𝑞 = (−3)(−1) + 4(6)
= 3 + 24
= 27
(iv)
Required To Find: 𝜃, where 𝜃 is the angle between 𝑝 and 𝑞.
Solution:
Recall: 𝑝. 𝑞 = |𝑝||𝑞| cos 𝜃
𝑝.𝑞
cos 𝜃 = |𝑝||𝑞|
=
=
27
√(−3)2 +(42 )√(−1)2 +(6)2
27
√25√37
27
𝜃 = cos −1 (
)
√25√37
= 27.40 (to the nearest,0.10 )
5.
(a)
(i)
Required To Find: The values of 𝑥 for which
𝑥 3 +8
𝑥 2 −4
is discontinuous.
Solution:
𝑥 3 +8
Let 𝑓(𝑥) = 𝑥 2 −4
𝑓(𝑥) is discontinuous when denominator = 0
That is,
𝑥2 − 4 = 0
(𝑥 − 2)(𝑥 + 2) = 0
𝑥 = 2 or −2
As 𝑥 → 2, 𝑓(𝑥) → ∞ and
as 𝑥 → −2, 𝑓(𝑥) → ∞
Hence 𝑓(𝑥) is discontinuous at 𝑥 = 2 and at 𝑥 = −2.
(ii)
Required To Find: lim
𝑥 3 +8
𝑥→−2 𝑥 2 −4
Solution:
𝑥 3 +8
Let 𝑓(𝑥) = 𝑥 2 −4
(−2)3 +8
0
𝑓(−2) = (−2)2 −4 = 0 which is indeterminate.
Factorising and cancelling:
lim
𝑥→−2
𝑥 3 +8
𝑥 2 −4
= lim
(𝑥+2)(𝑥 2 −2𝑥+4)
= lim
=
(𝑥−2)(𝑥+2)
𝑥→−2
𝑥 2 −2𝑥+4
𝑥→−2 𝑥−2
(−2)2 −2(2)+4
−2−2
12
= −4
= −3
Alternative Method:
We could have used L’Hospital’s Rule
𝑓(𝑥)
lim 𝑔(𝑥) =
𝑔´(𝑎)
𝑥→𝑎
𝑓´(𝑎)
In this question, 𝑔(𝑥) = 𝑥 3 + 8 and 𝑔′(𝑥) = 3𝑥 2
And 𝑓(𝑥) = 𝑥 2 − 4 and 𝑓 ′ (𝑥) = 2𝑥
Hence lim
𝑥→−2
𝑥 3 +8
= lim
𝑥 2 −4
3𝑥 2
𝑥→−2 2𝑥
= lim
3𝑥
𝑥→−2 2
=
3(−2)
2
= −3
(iii)
Required To Find: lim
2𝑥 3 +4𝑥
𝑥→0 sin 2𝑥
Solution:
2𝑥 3 + 4𝑥
2𝑥(𝑥 2 + 2)
lim
= lim
𝑥→0 sin 2𝑥
𝑥→0
sin 2𝑥
2
=
=
lim (𝑥 +2)
𝑥→0
sin 2𝑥)
lim (
)
2𝑥
𝑥→0
2
(0 )+2
1
sin 2𝑥
lim (
)=1
𝑥→0
2𝑥
Let 𝑢 = 2𝑥
As 𝑥 → 0, 𝑢 →0
sin 2𝑥
sin 𝑢
∴ lim (
) = lim (
)
𝑥→0
𝑢→0
2𝑥
𝑢
=2
=1
Alternative Method:
lim
𝑓(𝑥)
𝑥→𝑎 𝑔(𝑥)
=
𝑔´(𝑎)
(L’Hospital’s Rule)
𝑓´(𝑎)
If 𝑓(𝑥) = 2𝑥 3 + 4𝑥 then 𝑓′(𝑥) = 6𝑥 2 + 4
and 𝑔(𝑥) = sin 2𝑥, then 𝑔′(𝑥) = 2 cos 2𝑥
lim
2𝑥 3 +4𝑥
𝑥→0 sin 2𝑥
6𝑥 2 +4
= lim 2 cos 2𝑥
𝑥→0
=
6(0)2 +4
2 cos 2(0)
4
=2
=2
(b)
𝑥2 + 1
𝑥>1
4 + 𝑝𝑥
𝑥<1
Required To Find: lim+ 𝑓(𝑥)
Data:𝑓(𝑥) = {
(i)
a)
𝑥→1
Solution:
lim+ 𝑓(𝑥) = (1)2 + 1
𝑥→1
=2
b)
Required To Find: The value of 𝑝, such that lim 𝑓(𝑥) exists.
𝑥→1
Solution:
For lim 𝑓(𝑥) to exist
𝑥→1
lim 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→1−
4 + 𝑝(1) =
𝑥→1
(1)2
+1
4+𝑝=2
𝑝 = −2
(ii)
Required To Find: The value of 𝑓(1) for 𝑓 to be continuous at the point
𝑥 = 1.
Solution:
lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(1)
𝑥→1
𝑥→1
∴ 𝑓(1) = (1)2 + 1
=2
𝑓(1) = 4 + (−2)(1)
=2
(c)
Data: A chemical process in a manufacturing plant is controlled by the function
𝑣
𝑀 = 𝑢𝑡 2 + 𝑡 2 , where 𝑢 and 𝑣 are constants.
𝑑𝑀
When 𝑡 = 1, 𝑀 = −1 and when 𝑡 = 2, 𝑑𝑡 =
Required To Find: The value of 𝑢 andof 𝑣.
Solution:
When 𝑡 = 1, 𝑀 = −1
−1 = 𝑢 + 𝑣 … (1)
𝑑𝑀
2𝑣
= 2𝑢𝑡 − 𝑡 3
𝑑𝑡
When 𝑡 = 2,
35
4
35
4
= 4𝑢 −
𝑑𝑀
2𝑣
𝑑𝑡
=
35
4
8
𝑣
= 4𝑢 − 4 … (2)
Equation (2) × 4:
35 = 16𝑢 − 𝑣 … (3)
Equation (1) + Equation (2):
34 = 17𝑢
2=𝑢
𝑣 = −2 − 1
𝑣 = −3
3
and 𝑀 = 2𝑡 2 − 𝑡 2
6.
(a)
(i)
Data: 𝑦 = √4𝑥 2 − 7
35
4
𝑑𝑦
Required To Prove: 𝑦 𝑑𝑥 = 4𝑥
Proof:
𝑦 = √4𝑥 2 − 7
Let t = (4𝑥 2 − 7) ⇒
1
𝑦 = 𝑡2 ⇒
𝑑𝑦
=
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑡
1
𝑑𝑡
𝑑𝑥
= 8x
1
= 2 𝑡 −2
𝑑𝑡
× 𝑑𝑥 (Chain Rule)
1
= 2 (8𝑥)(4𝑥 2 − 7)
1
2
−( )
4𝑥
= √4𝑥 2
=
−7
4𝑥
𝑦
𝑑𝑥
𝑦 𝑑𝑥 = 4𝑥
Q.E.D.
Alternative Method:
𝑦 = √4𝑥 2 − 7
Square both sides:
𝑦 2 = 4𝑥 2 − 7
Differentiating, implicitly w.r.t. 𝑥
𝑑𝑦
2𝑦 𝑑𝑥 = 8𝑥
𝑑𝑦
𝑦 𝑑𝑥 = 4𝑥
Q.E.D.
(ii)
𝑑2 𝑦
𝑑𝑦 2
Required To Prove: 𝑦 𝑑𝑥 2 + (𝑑𝑥 ) = 4
Proof:
𝑑𝑦
𝑦 𝑑𝑥 = 4𝑥
Differentiating implicitly w.r.t. 𝑥
𝑑𝑦
𝑑𝑦
𝑑2 𝑦
𝑑2 𝑦
𝑑𝑦 2
(𝑑𝑥 ) (𝑑𝑥 ) + 𝑦 𝑑𝑥 2 = 4
𝑦 𝑑𝑥 2 + (𝑑𝑥 ) = 4
Q.E.D.
Alternative Method:
𝑑𝑦
4𝑥
𝑢
= √4𝑥 2 is of the form 𝑣 , where
𝑑𝑥
−7
𝑣 = √4𝑥 2 − 7
𝑑𝑣
4𝑥
= 𝑦
𝑑𝑥
𝑢 = 4𝑥
𝑑𝑢
=4
𝑑𝑥
𝑑𝑦 2
𝑑𝑥 2
=
=
=
4√4𝑥 2 −7−4𝑥(
(√4𝑥 2 −7)
4√4𝑥 2 −7−
4𝑥
)
𝑦
2
16𝑥2
𝑦
4𝑥 2 −7
4𝑦√4𝑥 2 −7−16𝑥 2
𝑦(4𝑥 2 −7)
(× 𝑦)
𝑦
𝑑𝑦 2
𝑑𝑥 2
=
=
4𝑦√4𝑥 2 −7−16𝑥 2
4𝑥 2 −7
4𝑦(𝑦)
16𝑥 2
𝑦2
−
𝑦2
4𝑥 2
= 4−(𝑦)
𝑑𝑦 2
𝑑𝑦 2
𝑑𝑦 2
= 4 − (𝑑𝑥 )
∴ 𝑦 𝑑𝑥 2 + (𝑑𝑥 ) = 4
Q.E.D.
(b)
Data: The curve 𝐶 passes through the point (−1, 0) and its gradient at any point
(𝑥, 𝑦) is
(i)
𝑑𝑦
𝑑𝑥
= 3𝑥 2 − 6𝑥.
Required To Find: The equation of the curve.
Solution:
𝑦 = ∫(3𝑥 2 − 6𝑥) 𝑑𝑥
= 𝑥3 −
6𝑥 2
2
+ 𝑐 ( c is the constant of integration)
= 𝑥 3 − 3𝑥 2 + 𝑐
When 𝑥 = −1, 𝑦 = 0
0 = (−1)3 − 3(−1)2 + 𝑐
0 = −1 − 3 + 𝑐
4=𝑐
∴ 𝑦 = 𝑥 3 − 3𝑥 2 + 4 is the equation of the curve.
(ii)
Required To Find: The coordinates of the stationary points of the curve.
Solution:
Stationary points occur when
𝑑𝑦
𝑑𝑥
= 0.
Let 3𝑥 2 − 6𝑥 = 0
3𝑥(𝑥 − 2) = 0
𝑥 = 0, 2
When 𝑥 = 0, 𝑦 = (0)3 − 3(0)2 + 4 = 4
When 𝑥 = 2, 𝑦 = (2)3 − 3(2)2 + 4 = 8 − 12 + 4 = 0
Coordinates of the stationary points are (0, 4) and (2, 0).
(iii)
Required To Find: The nature of the stationary points.
Solution:
𝑑2 𝑦
𝑑𝑥 2
= 6𝑥 − 6
𝑑2 𝑦
When 𝑥 = 0, 𝑑𝑥 2 = −6 < 0
∴ (0, 4) is a maximum point
𝑑2 𝑦
When 𝑥 = 2, 𝑑𝑥 2 = 12 − 6 = 6 > 0
∴ (2,0) is a minimum point.
(iv)
Data: The curve meets the 𝑥- axis at 𝑃 and 𝑄.
Required To Find: The coordinates of the points 𝑃 and 𝑄.
Solution:
𝑦 = 𝑥 3 − 3𝑥 2 + 4
Let 𝑦 = 0
i.e. 𝑥 3 − 3𝑥 2 + 4 = 0
Let 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 4
𝑓(−1) = (−1)3 − 3(−1)2 + 4 = 0
(𝑥 + 1) is a factor of 𝑓(𝑥)
(𝑥 + 1)(𝑎𝑥 2 + 𝑏𝑥 + 𝑐) = 𝑥 3 − 3𝑥 2 + 4
𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑥 3 − 3𝑥 2 + 4
𝑎𝑥 3 + (𝑏 + 𝑎)𝑥 2 + (𝑐 + 𝑏)𝑥 + 𝑐 = 𝑥 3 − 3𝑥 2 + 4
Comparing coefficients of
𝑥3: 𝑎 = 1
𝑥 2 : 𝑎 + 𝑏 = −3
1 + 𝑏 = −3
𝑏 = −4
constant: 𝑐 = 4
𝑓(𝑥) = (𝑥 + 1)(𝑥 2 − 4𝑥 + 4)
= (𝑥 + 1)(𝑥 − 2)2
The curve cuts the 𝑥 − axis at y = 0 i. e at 𝑥 = −1 and x = 2.
Let 𝑃 = (−1,0) and 𝑄 = (2, 0)
When 𝑥 = 0, 𝑦 = 4
Hence the curve cuts the x- axis at – 1 and touches at 𝑥 = 2 AND cuts the
y- axis at y = 4. The point (0, 4) is a maximum point and (2, 0) is a
minimum point.
(v)
Required To Sketch: The curve of 𝑦.
Solution:
𝑦
(0,4) maximum point
P(−1,0) O
Q(2,0)
𝑥
minimum point
JUNE 2011 UNIT 1 PAPER 2
1.
(a)
(i)
2
2
Required To Find: The exact value of (√75 + √12) − (√75 − √12) .
Solution:
2
(√75 + √12) − (√75 − √12)
2
= (75 + 12 + 2√75√12) − (75 + 12 − 2√75√12)
= 4√75√12
= 4(5√3)(2√3)
= 4(5)(2)(3)
= 120
Alternative Method:
2
(√75 + √12) − (√75 − √12)
2
2
= (√3 × 25 + √3 × 4) − (√3 × 25 − √3 × 4)
2
= (5√3 + 2√3) − (5√3 − 2√3)
2
= (7√3) − (3√3)
2
2
2
= (49 × 3) − (9 × 3)
= 147 − 27
= 120
1
(ii)
3
1
Required To Find: The exact value of, 274 × 98 × 818 .
Solution:
1
3
1
274 × 98 × 818
4
8
8
= √27 × √(9)3 × √81
1
3
1
= (33 )4 × (32 )8 × (34 )8
3
6
1
= 34 × 38 × 32
3 6 1
= 34+8+2
= 32
=9
(b)
Data: 𝑓(𝑥) = 𝑥 3 + 𝑚𝑥 2 + 𝑛𝑥 + 𝑝, where 𝑚, 𝑛 and 𝑝 are constants.
(i)
Required To Find: The value of 𝑝.
Solution:
The curve cuts the 𝑦 −axis when 𝑥 = 0.
𝑓(𝑥) = (0)3 + 𝑚(0)2 + 𝑛(0) + 𝑝
=𝑝
However, the curve cuts the 𝑦-axis at (0,4) as illustrated on the diagram
∴𝑝=4
(ii)
Required To Find: The value of 𝑚 and of 𝑛.
Solution:
From the graph
𝑓(1) = 0 and 𝑓(2) = 0
since the graph cuts the x – axis at 𝑥 = 1 and at 𝑥 = 2.
Using 𝑓(1) = 0 we obtain
(1)3 + 𝑚(1)2 + 4 = 0
(1) + 𝑚 + 𝑛 + 4 = 0
𝑚 + 𝑛 = −5 … (1)
Using 𝑓(2) = 0 we obtain
(2)3 + (2)2 𝑚 + (2)𝑛 + 4 = 0
8 + 4𝑚 + 2𝑛 + 4 = 0
4𝑚 + 2𝑛 = −12 … (2)
Solving equations (1) and (2) simultaneously:
𝑚 + 𝑛 = −5
… (1)
4𝑚 + 2𝑛 = −12 … (2)
Equation (1) × 4
4𝑚 + 4𝑛 = −20 … (3)
Equation (3) − Equation (2)
2𝑛 = −8
𝑛 = −4
Sub 𝑛 = −4 into equation (1)
𝑚 + (−4) = −5
𝑚 = −1
∴ 𝑚 = −1 and 𝑛 = −4.
(iii)
Required To Find: The 𝑥- coordinate of the point 𝑄.
Solution:
𝑓(𝑥) = 𝑥 3 − 𝑥 2 − 4𝑥 + 4
Since the curve cuts the x -axis at 1 and at 2 then (x - 1) and (x - 2) must be factors
of f(x).
Hence f(x) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 𝑞) and where the curve cuts the negative x
axis at q.
Now, 𝑥 3 − 𝑥 2 − 4𝑥 + 4 = (𝑥 − 1)(𝑥 − 2)(𝑥 − 𝑞)
Equating coefficients of 𝑥:
And -1 x -2 x - q = 4
∴q = 2
And f(x) = (𝑥 − 1)(𝑥 − 2)(𝑥 + 2)
Hence, the 𝑥 −coordinate of 𝑄 is −2 .
(c)
Data: 𝑦 = log 2 𝑥 and √log 2 𝑥 = log 2 √𝑥.
(i)
Required To Find: The value of 𝑥.
Solution:
√log 2 𝑥 = log 2 √𝑥
1
1
(log 2 𝑥)2 = log 2 (𝑥)2
1
1
(log 2 𝑥)2 = 2 log 2 𝑥
Let 𝑦 = log 2 𝑥
1
1
𝑦2 = 2 𝑦
Squaring both sides:
1
𝑦 = 4 𝑦2
4𝑦 = 𝑦 2
0 = 𝑦 2 − 4𝑦
0 = 𝑦(𝑦 − 4)
𝑦 = 0 or 4
When 𝑦 = 0
and
log 2 𝑥 = 0
when 𝑦 = 4
log 2 𝑥 = 4
𝑥 = 20
𝑥 = 24
=1
𝑥 = 16
∴ 𝑥 = 1 and 16
(ii)
Data: 𝑥 2 − |𝑥| − 12 < 0
Required To Solve: For real values of 𝑥.
Solution:
𝑥 2 − |𝑥| − 12 < 0 has the same as solutions as 𝑥 2 − 12 < |𝑥|.
Consider 𝑥 2 − 12 = |𝑥|
Consider
The branches of 𝑦 = |𝑥| have equations y = x in quadrant 1 and y = - x in
quadrant 2.
Consider
𝑥 2 − 12 = 𝑥
𝑥 2 − 𝑥 − 12 = 0
(𝑥 − 4)(𝑥 + 3) = 0
𝑥 = −3 or 4
𝑥 = 4 (𝑥 > 0) for the point of intersection in quadrant 1
Now consider
𝑥 2 − 12 = −𝑥
𝑥 2 + 𝑥 − 12 = 0
(𝑥 − 3)(𝑥 + 4) = 0
𝑥 = −4 or 3
𝑥 = −4 (𝑥 < 0) for the point of intersection in quadrant 2
𝑂𝑅
After obtaining 𝑥 = 4, we could have deduced that 𝑥 = −4 by symmetry
𝑥 = −4 and 𝑥 = 4 are the critical values of the solution
The graph of 𝑦 = 𝑥 2 − 12 lies below (less than)the graph of 𝑦 = |𝑥| between x = -4 and
x=4
∴ Solution set = {𝑥: −4 < 𝑥 < 4}
2.
(a)
Data: 𝑥 2 − 𝑝𝑥 + 24 = 0 has roots 𝛼 and 𝛽.
(i)
a)
Required To Express: 𝛼 + 𝛽 in terms of 𝑝.
If 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
÷𝑎
𝑏
𝑐
𝑎
𝑎
𝑥2 + 𝑥 + = 0
If the roots are 𝛼 and 𝛽, then
(𝑥 − 𝛼)(𝑥 − 𝛽) = 0
𝑥 2 − 𝛼𝑥 − 𝛽𝑥 + 𝛼𝛽 = 0
Hence,
𝑏
𝑐
𝑥 2 + 𝑎 𝑥 + 𝑎 = 𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽
Equating coefficients:
𝑏
in x: (𝛼 + 𝛽) = − 𝑎
𝑐
and the constants: 𝛼𝛽 = 𝑎
𝑏
𝑐
𝑥 2 − 𝑝𝑥 + 24 = 0 is of the form 𝑥 2 + 𝑎 𝑥 + 𝑎 = 0
where 𝑎 = 1, 𝑏 = −𝑝 and 𝑐 = 24
∴ 𝛼 + 𝛽 = −(−𝑝)
𝛼+𝛽 =𝑝
b)
Required To Express: 𝛼 2 + 𝛽 2 in terms of 𝑝.
Solution:
𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
𝛼𝛽 =
24
1
= 24
𝛼 2 + 𝛽 2 = (−𝑝)2 − 2(24)
= 𝑝2 − 48
(ii)
Data: 𝛼 2 + 𝛽 2 = 33
Required To Find: The possible values of 𝑝.
Solution:
𝛼 2 + 𝛽 2 = 𝑝2 − 48
𝑝2 − 48 = 33
𝑝2 = 81
𝑝 = ±√81
𝑝 = ±9 𝜖 𝑅
(b)
Data: 𝑓(2𝑥 + 3) = 2𝑓(𝑥) + 3 for 𝑥 ∈ 𝑅 and 𝑓(0) = 6.
(i)
Required To Find: The value of 𝑓(3).
Solution:
𝑓(2𝑥 + 3) = 2𝑓(𝑥) + 3
Let 𝑥 = 0
𝑓(3) = 2𝑓(0) + 3
= 2(6) + 3
= 15
(ii)
Required To Find: The value of 𝑓(9).
Solution:
Let 𝑥 = 3
𝑓(9) = 2𝑓(3) + 3
= 2(15) + 3
= 33
(iii)
Required To Find: The value of 𝑓(−3).
Solution:
Let 𝑥 = −3
𝑓(−3) = 2𝑓(−3) + 3
−3 = 𝑓(−3)
−3 = 𝑓(−3)
(c)
Required To Prove: The product of any two consecutive integers 𝑘 and 𝑘 + 1 is
an even integer.
Proof:
Consider 𝑘 is even, that is, 𝑘 = 2𝑝
Then, 𝑘(𝑘 + 1) = 2𝑝(2𝑝 + 1)
= 2(2𝑝2 + 𝑝) which is a multiple of 2 and therefore divisible by
2 and is also even.
Hence, 𝑘(𝑘 + 1) is even.
Consider 𝑘 is odd, then 𝑘 + 1 is even, that is, 𝑘 + 1 = 2𝑝
∴ 𝑘(𝑘 + 1) = (2𝑝 − 1)(2𝑝)
= 2(2𝑝2 − 1) which is a multiple of 2, therefore divisible by 2.
Hence, 𝑘(𝑘 + 1) is even.
The product of any two consecutive integers 𝑘 and 𝑘 + 1 is an even
integer.
(d)
Required To Prove: By mathematical induction, that, 𝑛(𝑛2 + 5) is divisible by 6
∀𝑛𝜖𝑍 + .
Proof:
Let 𝑓(𝑛) = 𝑛(𝑛2 + 5)
Assume the statement true for 𝑛 = 𝑘
∴ 𝑓(𝑘) = 𝑘(𝑘 2 + 5)
𝑓(𝑘) = 6𝑝
𝑝 ∈ 𝑍+
Consider 𝑛 = 𝑘 + 1
𝑓(𝑘 + 1) = (𝑘 + 1)[(𝑘 + 1)2 + 5]
= (𝑘 + 1)[𝑘 2 + 2𝑘 + 6]
= 𝑘 3 + 2𝑘 2 + 6𝑘 + 𝑘 2 + 2𝑘 + 6
= 𝑘 3 + 3𝑘 2 + 8𝑘 + 6
𝑓(𝑘 + 1) = (𝑘 3 + 5𝑘) + (3𝑘 2 + 3𝑘 + 6)
= 𝑘(𝑘 2 + 5) + 3(𝑘 2 + 𝑘 + 2)
Recall: 𝑘(𝑘 2 + 5) = 6𝑝
𝑝 ∈ 𝑍+
𝑓(𝑘 + 1) = 6𝑝 + 3(𝑘 2 + 𝑘 + 2)
Now, we prove that 𝑘 2 + 𝑘 + 2 is divisible by 2 and so even, so that 3(𝑘 2 + 𝑘 +
2) will be shown to be divisible by 6.
Assume true:
𝑘 2 + 𝑘 + 2 = 2𝑟 𝑟 ∈ 𝑍 +
Consider for (𝑘 + 1):
(𝑘 + 1)2 + (𝑘 + 1) + 2 = 𝑘 2 + 2𝑘 + 1 + 𝑘 + 3
= (𝑘 2 + 𝑘 + 2) + 2𝑘 + 2
= 2𝑟 + 2(𝑘 + 1)
= 2(𝑟 + 𝑘 + 1)which is a multiple of 2 and so divisible by 2
Hence, 𝑘 2 + 𝑘 + 2 is divisible by 2.
Thus 𝑓(𝑘 + 1) = (𝑘 3 + 5𝑘) + 3(𝑘 2 + 𝑘 + 2)
= 6𝑝 + 6(𝑟 + 𝑘 + 1)
= 6(𝑝 + 𝑟 + 𝑘 + 1) which is multiple of 6 and so divisible by 6.
Hence, the statement is true for 𝑛 = 𝑘 + 1
Consider 𝑛 = 1
𝑓(1) = 2(22 + 5)
= 18 is a multiple of 6 and hence divisible by 6.
The statement is true for 𝑛 = 1
Hence, by the Principle of Mathematical Induction, the statement, 𝑛(𝑛2 + 5) is
divisible by 6 is true ∀𝑛 ∈ 𝑁.
3.
(a)
Data: 𝑎 = 𝑎1 𝑖 + 𝑎2 𝑗, 𝑏 = 𝑏1 𝑖 + 𝑏2 𝑗, |𝑎| = 13 and |𝑏| = 10.
(i)
Required To Find: The value of (𝑎 + 𝑏). (𝑎 − 𝑏).
Solution:
(𝑎 + 𝑏) = (𝑎1 𝑖 + 𝑎2 𝑗) + (𝑏1 𝑖 + 𝑏2 𝑗)
= (𝑎1 + 𝑏1 )𝑖 + (𝑎2 + 𝑏2 )𝑗
(𝑎 − 𝑏) = (𝑎1 𝑖 + 𝑎2 𝑗) − (𝑏1 𝑖 + 𝑏2 𝑗)
= (𝑎1 − 𝑏1 )𝑖 + (𝑎2 − 𝑏2 )𝑗
(𝑎 + 𝑏). (𝑎 − 𝑏) = [(𝑎1 + 𝑏1 )𝑖 + (𝑎2 + 𝑏2 )𝑗].[(𝑎1 − 𝑏1 )𝑖 + (𝑎2 − 𝑏2 )𝑗]
= (𝑎1 + 𝑏1 )(𝑎1 − 𝑏1 ) + (𝑎2 + 𝑏2 )(𝑎2 − 𝑏2 )
= (𝑎12 − 𝑏12 ) + (𝑎22 − 𝑏22 )
= (𝑎12 + 𝑎22 ) − (𝑏12 + 𝑏22 )
But |𝑎| = 13
i.e. √𝑎12 + 𝑎22 = 13
𝑎12 + 𝑎22 = 169
Also, |𝑏| = 10
i.e. 𝑏12 + 𝑏22 = 100
∴ (𝑎 + 𝑏)(𝑎 − 𝑏) = (𝑎12 + 𝑎22 ) − (𝑏12 + 𝑏22 )
= 169 − 100
= 69
(ii)
Data: 2𝑏 − 𝑎 = 11𝑖
Required To Determine: The possible values of 𝑎 and 𝑏.
Solution:
2𝑏 − 𝑎 = 11𝑖
2(𝑏1 𝑖 + 𝑏2 𝑗) − (𝑎1 𝑖 + 𝑎2 𝑗) = 11𝑖
(2𝑏1 − 𝑎1 )𝑖 + (2𝑏2 − 𝑎2 )𝑗 = 11𝑖 + 0𝑗
Equating the coefficients of the terms in 𝑖 and 𝑗 respectively, we obtain
2𝑏1 − 𝑎1 = 11 … (1)
2𝑏2 − 𝑎2 = 0 … (2)
𝑎2 = 2𝑏2
And 𝑎1 = 2𝑏1 − 11
Also |2𝑏 − 𝑎| = |11|
√(2𝑏1 − 𝑎1 )2 + (2𝑏2 − 𝑎2 )2 = 11
Squaring both sides:
(2𝑏1 − 𝑎1 )2 + (2𝑏2 − 𝑎2 )2 = 121
4𝑏12 − 4𝑏1 𝑎1 + 𝑎12 + 4𝑏22 − 4𝑏2 𝑎2 + 𝑎22 = 121
4(𝑏12 + 𝑏22 ) + (𝑎12 + 𝑎22 ) − 4(𝑏1 𝑎1 + 𝑏2 𝑎2 ) = 121
Recall: 𝑏12 + 𝑏22 = 100 and 𝑎12 + 𝑎22 = 169
4(100) + 169 − 4(𝑏1 𝑎1 + 𝑏2 𝑎2 ) = 121
400 + 169 − 121 = 4(𝑏1 𝑎1 + 𝑏2 𝑎2 )
448 = 4(𝑏1 𝑎1 + 𝑏2 𝑎2 )
112 = 𝑏1 𝑎1 + 𝑏2 𝑎2 … (3)
Substitute 𝑎2 = 2𝑏2 and 𝑎1 = 2𝑏1 − 11 into equation (3):
112 = (2𝑏1 − 11)𝑏1 + 𝑏2 (2𝑏2 )
112 = (2𝑏1 − 11)𝑏1 + 2𝑏22
112 = 2(𝑏12 + 𝑏22 ) − 11𝑏1
112 = 2(100) − 11𝑏1
11𝑏1 = 88
𝑏1 = 8
∴ 𝑎1 = 2(8) − 11
=5
Substitute for 𝑎1 and 𝑏1 into equation (3):
5(8) + 𝑎2 𝑏2 = 112
𝑎2 𝑏2 = 112 − 40
= 72
But 𝑎2 = 2𝑏2
2𝑏22 = 72
𝑏22 = 36
𝑏2 = ±6
𝑎2 = 2𝑏2
𝑎2 = ±12
𝑎 = (5𝑖 + 12𝑗) and 𝑏 = (8𝑖 + 12𝑗)
OR
𝑎 = (5𝑖 − 12𝑗) and 𝑏 = (8𝑖 − 12𝑗)
(b)
Data: 𝐿 has equation 𝑥 − 𝑦 + 1 = 0 and circle 𝐶 has equation
𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0.
(i)
Required To Prove: That 𝐿 passes through the center of the circle 𝐶.
Proof:
𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0 … (1)
𝑥 2 + 𝑦 2 − 2𝑦 + 1 − 1 − 15 = 0
𝑥 2 + (𝑦 − 1)2 − 1 − 15 = 0
(𝑥 − 0)2 + (𝑦 − 1)2 = 42
which is a circle with centre (0, 1) and radius 4 units.
𝐿: 𝑥 − 𝑦 + 1 = 0 … (2)
Sub 𝑥 = 0, 𝑦 = 1 into equation (2)
0−1+1= 0
0 = 0 and the equation is satisfied
∴ 𝐿 passes through 𝐶.
Alternative Method:
𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0
𝑥 2 + 𝑦 2 + 2(0)𝑥 + 2(−1)𝑦 + (−15) = 0 is of the form
𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0, where 𝑔 = 0 and 𝑓 = −1.
Centre is (−𝑔, −𝑓) = (−(0), −(−1))
= (0, 1)
Substitute (0, 1) into equation of line 𝐿:
𝑥−𝑦+1=0
0−1+1= 0
0=0
The point (0, 1) satisfies the equation, therefore the line 𝐿 passes through
the centre of the circle 𝐶.
(ii)
Data: The line 𝐿 intersects the circle 𝐶 at the points 𝑃 and 𝑄.
Required To Determine: The coordinates of 𝑃 and 𝑄.
Solution:
𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0 … (1)
𝑥−𝑦+1=0
… (2)
Solve equations (1) and (2) simultaneously:
From equation (2):
𝑦 =𝑥+1
… (3)
Substitute equation (3) into equation (1):
𝑥 2 + (𝑥 + 1)2 − 2(𝑥 + 1) − 15 = 0
𝑥 2 + 𝑥 2 + 2𝑥 + 1 − 2𝑥 − 2 − 15 = 0
2𝑥 2 − 16 = 0
2(𝑥 2 − 8) = 0
2(𝑥 + √8)(𝑥 − √8) = 0
𝑥 = ±√8
When 𝑥 = √8, 𝑦 = 1 + √8
When 𝑥 = −√8, 𝑦 = 1 − √8
∴ 𝑃(−√8, 1 − √8) and 𝑄(√8, 1 + √8).
(iii)
Data: 𝑥 = 𝑏 + 𝑎 cos 𝜃 , 𝑦 = 𝑐 + 𝑎 sin 𝜃
Required To Find: The value of the constants 𝑎, 𝑏 and 𝑐.
Solution:
𝑥 = 𝑏 + 𝑎 cos 𝜃
cos 𝜃 =
𝑥−𝑏
𝑦 = 𝑐 + asin 𝜃
sin 𝜃 =
𝑎
𝑦−𝑐
𝑎
Re: cos 2 𝜃 + sin2 𝜃 = 1
𝑥−𝑏 2
(
𝑎
𝑦−𝑐 2
) +(
𝑎
) =1
(× 𝑎2 )
(𝑥 − 𝑏)2 + (𝑦 − 𝑐)2 = 𝑎2 is of the form
(𝑥 − 0)2 + (𝑦 − 1)2 = 42
and where 𝑎 = 4, 𝑏 = 0, 𝑐 = 1.
(iv)
Required To Find: The possible equations of 𝐶2 .
Solution:
Let the coordinates of the centre of 𝐶2 be (𝑎, 𝑏)
Since 𝐶2 has the same radius as 𝐶, the equation of 𝐶2 is
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 42
The line joining the centre of 𝐶2 to the circumference of 𝐶 has gradient
−1.
𝑏−1
𝑎
= −1
𝑏 − 1 = −𝑎
𝑎 + 𝑏 = 1 … (1)
Also
√(𝑎 − 0)2 + (𝑏 − 1)2 = 4
𝑎2 + (𝑏 − 1)2 = 16
𝑎2 + 𝑏 2 − 2𝑏 + 1 = 16
From equation (1) 𝑏 = 1 − 𝑎
𝑎2 + (1 − 𝑎)2 − 2(1 − 𝑎) − 15 = 0
𝑎2 + 𝑎2 − 2𝑎 + 1 − 2 + 2𝑎 − 15 = 0
2𝑎2 − 16 = 0
𝑎2 − 8 = 0
𝑎 = ±2√2
𝑏 =1−𝑎
𝑏 = 1 ∓ 2√2
Equation of 𝐶2 is
2
2
(𝑥 − 2√2) + (𝑦 + (1 − 2√2)) = 42
OR
2
2
(𝑥 + 2√2) + (𝑦 − (1 + 2√2)) = 42
4.
(a)
Data: 𝑥 = cos2 𝜃
Required To Find: The values of 𝜃 such that 8cos4 𝜃 − 10cos 2 𝜃 + 3 = 0.
Solution:
8 cos 4 𝜃 − 10 cos2 𝜃 + 3 = 0
Let 𝑥 = cos 2 𝜃
8𝑥 2 − 10𝑥 + 3 = 0
(2𝑥 − 1)(4𝑥 − 3) = 0
1
3
𝑥 = 2 or 4
1
3
cos2 𝜃 = 2
cos 𝜃 = ±
cos 𝜃 =
cos 2 𝜃 = 4
1
cos 𝜃 = ±
√2
1
√2
𝜋 7𝜋
𝜃 = 4,
cos 𝜃 = −
𝜃=
4
𝜋 𝜋 3𝜋 5𝜋
𝜃 = 6,4,
4
,
6
1
√2
3𝜋 5𝜋
4
,
4
0≤𝜃≤𝜋
cos 𝜃 =
√3
2
√3
2
𝜋 11𝜋
𝜃 = 6,
6
√3
2
5𝜋 7𝜋
cos 𝜃 = −
𝜃=
6
,
6
(b)
Data: A
Q
B
6 𝑐𝑚
P
R
8 𝑐𝑚
𝜃
D
(i)
S
C
Required To Find: The length of the side 𝐵𝐶.
Solution:
𝐵𝐶 = 𝑅𝐶 + 𝐵𝑅
sin 𝜃 =
𝑅𝐶
8
8 sin 𝜃 = 𝑅𝐶
𝐵𝑅̂ 𝑄 = 𝜃
cos 𝜃 =
𝐵𝑅
6
6 cos 𝜃 = 𝐵𝑅
𝐵𝐶 = 8 sin 𝜃 + 6 cos 𝜃
(ii)
Data:|𝐵𝐶| = 7
Required To Find: The value of 𝜃.
Solution:
8 sin 𝜃 + 6 cos 𝜃 = 𝑅 sin(𝜃 + 𝛼)
= 𝑅[sin 𝜃 sin 𝛼 + cos 𝜃 cos 𝛼]
Equating terms in sin 𝜃 and cos 𝜃
𝑅 sin 𝛼 = 8
𝑅 cos 𝛼 = 6
8
4
tan 𝛼 = 6 = 3
4
𝛼 = tan−1 ( 3) = 53.10
𝑅 2 = 82 + 62
𝑅 2 = 100
𝑅 = 10(> 0)
8 sin 𝜃 + 6 cos 𝜃 = 10 sin(𝜃 + 53.10 )
|𝐵𝐶| = 8 sin 𝜃 + 6 cos 𝜃
Now,
∴ 10 sin(𝜃 + 53.10 ) = 7
7
sin(𝜃 + 53.10 ) = 10
7
(𝜃 + 53.10 ) = sin−1 ( )
10
(𝜃 + 53.10 ) = 44.40 , 135.60
𝜃 = 82.50 (correct to 1 decimal place)
(iii)
Required To Determine: If 15 is a possible value for |𝐵𝐶|.
Solution:
8 sin 𝜃 + 6 cos 𝜃 = 15
10 sin(𝜃 + 53.10 ) = 15
15
sin(𝜃 + 53.10 ) = 10 which is > 1
But −1 ≤ sin 𝜃 ≤ 1 for all values of θ
Hence |𝐵𝐶| ≠ 15.
Alternatively:
−1 ≤ sin(𝜃 + 53.1°) ≤ 1
−10 ≤ 10 sin(𝜃 + 53.1°) ≤ 10
∴ |𝐵𝐶| ≠ 15
(c)
(i)
Required To Prove:
Proof:
Taking L.H.S.
1−cos 2𝜃
sin 2𝜃
= tan 𝜃
Recall: sin 2𝜃 = 2 sin 𝜃 cos 𝜃 and cos 2𝜃 = 1 − 2sin2 𝜃.
1−cos 2𝜃
=
sin 2𝜃
1−(1−2 sin2 𝜃)
2 sin 𝜃 cos 𝜃
2 sin2 𝜃
= 2 sin 𝜃 cos 𝜃
sin 𝜃
= cos 𝜃
= tan 𝜃
= R.H.S.
Q.E.D.
(ii)
Required To Prove:
a)
Proof:
1−cos 4𝜃
sin 4𝜃
= tan2 𝜃
Taking L.H.S.
Let 𝜃 = 2𝐴
From (i)
1−cos 2𝜃
sin 2𝜃
1−cos 4𝐴
sin 4𝐴
∴
b)
1−cos 4𝜃
sin 4𝜃
= tan 𝜃
= tan 2𝐴
= tan 2𝜃
Required To Prove:
Proof:
Let 𝜃 = 3𝐴
From (i)
1−cos 2𝜃
sin 2𝜃
= tan 𝜃
1−cos 6𝜃
sin 6𝜃
= tan 3𝜃
1−cos 6𝐴
sin 6𝐴
∴
(iii)
1−cos 6𝜃
sin 6𝜃
= tan 3𝐴
= tan 3𝜃
Required To Evaluate: ∑𝑛𝑟=1 (tan 𝑟𝜃 sin 2𝑟𝜃 + cos 2𝑟𝜃), where 𝑛 ∈ 𝑍 + .
Solution:
From (i) tan 𝜃 sin 2𝜃 = 1 − cos 2𝜃
(ii) a) tan 2𝜃 sin 4𝜃 = 1 − cos 4 𝜃
b) tan 3𝜃 sin 6𝜃 = 1 − cos 6 𝜃
and
by extension tan 𝑟𝜃 sin 2𝑟𝜃 = 1 − cos 2𝑟 𝜃
Hence,
∑𝑛𝑟=1 (tan 𝑟𝜃 sin 2𝑟𝜃 + cos 2𝑟𝜃)
= ∑𝑛𝑟=1(1 − cos 2𝑟𝜃 + cos 2𝑟𝜃)
= ∑𝑛𝑟=1 1
= 1 + 1 + 1 + ⋯+1 (n times)
=𝑛
5.
(a)
∀𝑛 ∈ 𝑍 +
Required To Find: lim
𝑥 2 +5𝑥+6
𝑥→−2 𝑥 2 −𝑥−6
Solution:
Let 𝑓(𝑥) =
𝑓(−2) =
=
𝑥 2 +5𝑥+6
𝑥 2 −𝑥−6
(−2)2 +5(−2)+6
(−2)2 −(−2)−6
4−10+6
4+2−6
0
= 0 which is indeterminate
Factorising and cancelling for this indeterminate form
lim
𝑥 2 +5𝑥+6
𝑥→−2 𝑥 2 −𝑥−6
= lim
(𝑥+3)(𝑥+2)
𝑥→−2 (𝑥−3)(𝑥+2)
= lim
𝑥+3
𝑥→−2 𝑥−3
−2+3
= −2−3
1
= −5
Alternative Method:
Using L’Hospital’s Rule:
If
f ( x) =
g ( x)
h ( x)
then,
lim x →a f ( x ) =
g ( a )
h ( a )
Consider
x2 + 5x + 6
2x + 5
lim 2
= lim
x →−2 x − x − 6
x →−2 2 x − 1
2 ( −2 ) + 5
=
2 ( −2 ) − 1
=−
(b)
1
5
2
Data: 𝑓(𝑥) = { 𝑥 + 1 𝑥 ≥ 2
𝑏𝑥 + 1 𝑥 < 2
(i)
Required To Determine: 𝑓(2)
Solution:
𝑓(2) = (2)2 + 1
=5
(ii)
Required To Determine: lim+ 𝑓(𝑥)
𝑥→2
Solution:
From (b) (i)
lim 𝑓(𝑥) = 5
𝑥→2+
(iii)
Required To Determine: lim− 𝑓(𝑥) in terms of 𝑏.
𝑥→2
Solution:
lim 𝑓(𝑥) = 𝑏(2) + 1 = 2𝑏 + 1
𝑥→2−
(iv)
Required To Determine: The value of 𝑏 for which 𝑓(𝑥) is continuous.
Solution:
For 𝑓(𝑥) to be continuous at 𝑥 = 2
lim 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→2−
𝑥→2
hence 5 = 2𝑏 + 1
4 = 2𝑏
𝑏=2
(c)
Data: 𝑦 = 𝑝𝑥 3 + 𝑞𝑥 2 + 3𝑥 + 2 passes through the point 𝑇(1,2) and the gradient
at 𝑇 is 7. The line 𝑥 = 1 cuts the 𝑥 −axis at 𝑀 and the normal to the curve at 𝑇
cuts the 𝑥 −axis at 𝑁.
(i)
Required To Find: The value of 𝑝 and of 𝑞.
Solution:
𝑦 = 𝑝𝑥 3 + 𝑞𝑥 2 + 3𝑥 + 2
At 𝑇 (1, 2), 𝑥 = 1 and 𝑦 = 2
Hence,
2 = 𝑝(1)3 + 𝑞(1)2 + 3(1) + 2
2= 𝑝+𝑞+3+2
−3 = 𝑝 + 𝑞 … (1)
𝑑𝑦
𝑑𝑥
= 3𝑝𝑥 2 + 2𝑞𝑥 + 3
𝑑𝑦
At 𝑇, 𝑑𝑥 = 7
Hence,
7 = 3𝑝(1)2 + 2𝑞(1) + 3
4 = 3𝑝 + 2𝑞 … (2)
Solving equation (1) and equation (2) simultaneously:
Equation (1) × 2:
−6 = 2𝑝 + 2𝑞 … (3)
Equation (2) − Equation (3):
𝑝 = 10
Substitute 𝑝 = 10 into equation (1):
−3 = 10 + 𝑞
𝑞 = −13
𝑦 = 10𝑥 3 − 13𝑥 2 + 3𝑥 + 2
(ii)
Required To Find: The equation of the normal to the curve at 𝑇.
Solution:
Gradient of the normal at 𝑇 = −
1
7
(Since product of gradients of perpendicular lines = −1)
Equation of the normal at 𝑇 using
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
1
is 𝑦 − 2 = − 7 (𝑥 − 1)
1
1
1
15
𝑦 = −7𝑥 + 7 + 2
𝑦 = −7𝑥 +
7
7𝑦 = −𝑥 + 15
(iii)
Required To Find: The length of 𝑀𝑁.
Solution:
The line 𝑥 = 1 cuts the 𝑥 −axis at 𝑀. ∴ 𝑀 = (1, 0)
The normal to the curve at 𝑇 cuts the 𝑥 −axis at 𝑁.
When 𝑦 = 0
7(0) = 𝑥 + 15
0 = −𝑥 + 15
𝑥 = 15
Coordinates of 𝑁 = (15, 0)
Length of 𝑀𝑁 = √(0 − 0)2 + (15 − 1)2
= √(14)2
= 14 units
6.
(a)
𝑦
Data:
𝑓(𝑥) = 𝑥(𝑥 2 − 12)
𝐴
𝑂
𝑥
𝐵
𝑓(𝑥) = 𝑥(𝑥 2 − 12) 𝐴 and 𝐵 are stationary points.
(i)
Required To Find: The coordinates of 𝐴 and of 𝐵.
Solution:
𝑓(𝑥) = 𝑥(𝑥 2 − 12)
= 𝑥 3 − 12𝑥
𝑓 ′ (𝑥) = 3𝑥 2 − 12
= 3(𝑥 2 − 4)
= 3(𝑥 − 2)(𝑥 + 2)
Stationary points occur when 𝑓 ′ (𝑥) = 0
Let 0 = 3(𝑥 − 2)(𝑥 + 2)
𝑥 = −2 and 2
When 𝑥 = 2
𝑓(2) = 2(22 − 12) = −16
When 𝑥 = −2
𝑓(−2) = (2)((−2)2 − 12) = 16
At 𝐴, x is negative and at 𝐵, x is positive.
∴ 𝐴(−2, 16) and 𝐵(2, −16)
(ii)
Required To Find: The equation of the normal to the curve at the origin.
Solution:
At 𝑂, 𝑥 = 0
𝑓 ′ (0) = 3(0)2 − 12 = −12
Gradient of tangent at 𝑂 = −12
1
Gradient of normal at 𝑂 = 12
(Product of gradients of perpendicular lines = −1)
Equation of normal using:
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
1
is 𝑦 − 0 = 12 (𝑥 − 0)
1
𝑦 = 12 𝑥
(iii)
Required To Find: The area between the curve and the positive 𝑥 −axis.
Solution:
The curve cuts the 𝑥 −axis when 𝑦 = 0
i.e 0 = 𝑥(𝑥 2 − 12)
0 = 𝑥(𝑥 − √12)(𝑥 + √12)
𝑥 = −√12, 0, √12
Since the region denoted by 𝐴 lies below the x axis, then:
√12
𝐴 = |∫0
𝑥4
(𝑥 3 − 12𝑥) 𝑑𝑥|
= |[ 4 −
12𝑥 2 √12
2
]
0
|
=|
144
4
−
12(12)
2
|
= |36 − 72|
= |−36|
= 36 units2
(b)
(i)
𝑎
𝑎
Data: ∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥) 𝑑𝑥, 𝑎 > 0
𝜋
𝜋
Required To Prove: ∫0 𝑥 sin 𝑥 𝑑𝑥 = ∫0 ( 𝜋 − 𝑥) sin 𝑥 𝑑𝑥
Proof:
Let 𝑓(𝑥) = 𝑥 sin 𝑥
𝑎 =𝜋−𝑥
𝑥 =𝜋−𝑎
𝑑𝑥 = −𝑑𝑎
𝑓(𝜋 − 𝑎) = (𝜋 − 𝑎) sin(𝜋 − 𝑎)
= (𝜋 − 𝑎) sin 𝑎
When 𝑥 = 0, 𝑎 = 𝜋 and
When 𝑥 = 𝜋, 𝑎 = 0
𝜋
0
∫0 𝑥 sin 𝑥 𝑑𝑥 = − ∫𝜋 (𝜋 − 𝑎) sin 𝑎 𝑑𝑎
𝜋
= ∫0 (𝜋 − 𝑎) sin 𝑎 𝑑𝑎
𝜋
= ∫0 ( 𝜋 − 𝑥) sin 𝑥 𝑑𝑥
Q.E.D.
(ii)
a)
𝜋
𝜋
Required To Prove: ∫0 𝑥 sin 𝑥 𝑑𝑥 = ∫0 π sin 𝑥 𝑑𝑥 −
𝜋
∫0 𝑥 sin 𝑥 𝑑𝑥
Proof:
𝜋
𝜋
∫0 𝑥 sin 𝑥 𝑑𝑥 = ∫0 ( 𝜋 − 𝑥) sin 𝑥 𝑑𝑥
Separating:
𝜋
𝜋
= ∫0 π sin 𝑥 𝑑𝑥 − ∫0 𝑥 sin 𝑥 𝑑𝑥
Q.E.D.
b)
𝜋
Required To Prove: ∫0 𝑥 sin 𝑥 𝑑𝑥 = 𝜋
Proof:
Recall:
𝜋
𝜋
𝜋
∫0 𝑥 sin 𝑥 𝑑𝑥 = ∫0 π sin 𝑥 𝑑𝑥 − ∫0 𝑥 sin 𝑥 𝑑𝑥
𝜋
𝜋
∴ 2 ∫0 𝑥 sin 𝑥 𝑑𝑥 = 𝜋 ∫0 sin 𝑥 𝑑𝑥
= 𝜋[− cos 𝑥]
𝜋
0
= 𝜋[− cos 𝜋 − (− cos 0)]
= 𝜋[−(−1)— 1]
= 2𝜋
𝜋
∴ ∫0 𝑥 sin 𝑥 𝑑𝑥 =
2𝜋
2
=𝜋
Q.E.D.
JUNE 2010 CAPE UNIT 1 PAPER 2
1.
(a)
Data: 𝑓(𝑥) = 4𝑥 3 − (3𝑝 + 2)𝑥 2 − (𝑝2 − 1)𝑥 + 3 and (𝑥 − 𝑝) is a factor of
𝑓(𝑥).
Required To Find: The values of 𝑝.
Solution:
Recall the Remainder and Factor Theorem
If 𝑓(𝑥) is any polynomial, and 𝑓(𝑥) is divided b𝑦(𝑥 − 𝑎), then the remainder is
𝑓(𝑎) and if 𝑓(𝑎) = 0, then (𝑥 − 𝑎) is a factor of 𝑓(𝑥).
𝑓(𝑥) = 4𝑥 3 − (3𝑝 + 2)𝑥 2 − (𝑝2 − 1)𝑥 + 3 (data)
Hence, 𝑓(𝑝) = 0
∴ 4(𝑝)3 − (3𝑝 + 2)(𝑝)2 − (𝑝2 − 1)𝑝 + 3 = 0
4𝑝3 − 3𝑝3 − 2𝑝2 − 𝑝3 + 𝑝 + 3 = 0
−2𝑝2 + 𝑝 + 3 = 0
× (−1)
2𝑝2 − 𝑝 − 3 = 0
(2𝑝 − 3)(𝑝 + 1) = 0
∴ 𝑝 = −1 or
(b)
3
2
Data: log(𝑥 − 1) + 2 log 𝑦 = 2 log 3 and log 𝑥 + log 𝑦 = log 6
Required To Solve: For 𝑥 and for 𝑦.
Solution:
log(𝑥 − 1) + 2 log 𝑦 = 2 log 3 … (1)
log 𝑥 + log 𝑦 = log 6
… (2)
From equation (1):
log(𝑥 − 1) + log 𝑦 2 = log 52
log[(𝑥 − 1) × (𝑦)2 ] = log(3)2
Removing logs we obtain:
(𝑥 − 1)𝑦 2 = 9
𝑥𝑦 2 − 𝑦 2 = 9 … (3)
From equation (2):
log(𝑥𝑦) = log 6
Removing logs we obtain:
𝑥𝑦 = 6 … (4)
Make 𝑥 the subject in equation (4):
6
𝑥=𝑦
Substituting this expression into equation (3),we obtain:
6
(𝑦) 𝑦 2 − 𝑦 2 = 9
6𝑦 − 𝑦 2 = 9
𝑦 2 − 6𝑦 + 9 = 0
(𝑦 − 3)2 = 0
𝑦=3
6
Substituting, 𝑦 = 3 into 𝑥 = 𝑦
6
𝑥=3
=2
∴ 𝑥 = 2 and 𝑦 = 3
(c)
Data:
2𝑥−3
𝑥+1
−5>0
Required To Find: 𝑥 ∈ 𝑅
Solution:
2𝑥−3
𝑥+1
−5 >0
× (𝑥 + 1)2 and maintain the inequality
(2𝑥 − 3)(𝑥 + 1) − 5(𝑥 + 1)2 > 0
2𝑥 2 − 3𝑥 + 2𝑥 − 3 − 5(𝑥 2 + 2𝑥 + 1) > 0
2𝑥 2 − 3𝑥 + 2𝑥 − 3 − 5𝑥 2 − 10𝑥 − 5 > 0
−3𝑥 2 − 11𝑥 − 8 > 0
𝑇ℎ𝑒 𝑖𝑛𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 × (−1)
3𝑥 2 + 11𝑥 + 8 < 0
(3𝑥 + 8)(𝑥 + 1) < 0
Coefficient of 𝑥 2 > 0 ⇒ The quadratic graph of y = 3𝑥 2 + 11𝑥 + 8 has a minimum point . The
quadratic graph cuts the horizontal axis at
3𝑥 + 8 = 0
𝑥+1=0
or
8
𝑥 = −3
𝑥 = −1
When sketched the graph would look like:
𝑦 = 3𝑥 2 + 11𝑥 + 8
𝑥
2
−2 3
−1
8
3
8
3
𝑥 < −1 and 𝑥 > − or − < 𝑥 < −1
∴
2𝑥−3
𝑥+1
− 5 > 0 for
8
{𝑥: − 3 < 𝑥 < −1; 𝑥 ∈ 𝑅}
(d)
Data: 4𝑥 − 3(2𝑥+1 ) + 8 = 0
Required To Solve: For x.
Solution:
4𝑥 − 3(2𝑥+1 ) + 8 = 0
(22 )𝑥 − 3(2𝑥 ) × 21 + 23 = 0
(2𝑥 )2 − 6(2𝑥 ) + 8 = 0
Let 𝑦 = 2𝑥
𝑦 2 − 6𝑦 + 8 = 0
(𝑦 − 4)(𝑦 − 2) = 0
𝑦 = 2 or 4
Recall: 𝑦 = 2𝑥
∴ 2𝑥 = 21
2𝑥 = 4
Equating indices:
2𝑥 = 22
𝑥=1
Equating indices:
𝑥=2
Hence, 𝑥 = 1 or 2
2.
(a)
1
Data: 𝑆𝑛 = ∑𝑛𝑟=1 𝑟 = 2 𝑛(𝑛 + 1)
(i)
Required To Find: 𝑆2𝑛 = ∑2𝑛
𝑟=1 𝑟 , in terms of 𝑛.
Solution:
1
𝑆2𝑛 = ∑2𝑛
𝑟=1 𝑟 = 2 (2𝑛)((2𝑛) + 1)
= 𝑛(2𝑛 + 1)
Alternative Method:
∑𝑛𝑟=1 𝑟 = 1 + 2 + 3 + 4 + ⋯ + 𝑛 which is an arithmetic progression with
𝑎 = 1 and 𝑑 = 1.
For an arithmetic progression:
𝑛
𝑛
𝑆𝑛 = 2 (𝑎 + 𝑙)
𝑆𝑛 = 2 [2𝑎 + (𝑛 − 1)𝑑]
𝑛
= 2 (1 + 𝑛)
𝑛
= 2 [2(1) + (𝑛 − 1)1]
OR
1
1
= 2 𝑛(𝑛 + 1)
And 𝑆2𝑛 =
=
2𝑛
2
2𝑛
2
= 2 𝑛(𝑛 + 1)
(𝑎 + 𝑙)
(1 + 2𝑛)
= 𝑛(2𝑛 + 1)
(ii)
Data: 𝑆2𝑛 − 𝑆𝑛 = 𝑝𝑛2 + 𝑞𝑛
Required To Find: The value of p and of q.
Solution:
1
𝑆2𝑛 − 𝑆𝑛 = 𝑛(2𝑛 + 1) − 2 𝑛(𝑛 + 1)
1
1
= 2𝑛2 + 𝑛 − 2 𝑛2 − 2 𝑛
3
1
= 2 𝑛2 + 2 𝑛
which is of the form 𝑝𝑛2 + 𝑞𝑛 ,
3
1
Equating coefficients we obtain 𝑝 = 2 and 𝑞 = 2.
(iii)
Data: 𝑆2𝑛 − 𝑆𝑛 = 260
Required To Calculate: n
Calculation:
3
1
𝑛2 + 2 𝑛 = 260
2
(× 2)
3𝑛2 + 𝑛 = 520
3𝑛2 + 𝑛 − 520 = 0
(3𝑛 + 40)(𝑛 − 13) = 0
𝑛 = 13 or
−40
3
since, 𝑛 ∈ 𝑍 + then 𝑛 = 13 only.
(b)
Data: Diagram showing the graph of 𝑦 = 𝑥 2 (3 − 𝑥). The coordinates of 𝑃 and 𝑄
are (2, 4) and (3, 0) respectively.
(i)
Required To Calculate: The solution set of the inequality 𝑥 2 (3 − 𝑥) ≤ 0
Calculation:
y
O
3
𝑦 = 𝑥 2 (3 − 𝑥)
𝑥 2 (3 − 𝑥) ≤ 0
The graph is 0 at O and at 3 and negative for x greater than 3
Solution: 𝑥 = {0} ∪ {𝑥: 𝑥 ≥ 3}
(ii)
Data: 𝑥 2 (3 − 𝑥) = 𝑘 has 3 real solutions for 𝑥.
Required To Calculate: The possible values of 𝑘.
Calculation:
y
4
𝑦 = 𝑥 2 (3 − 𝑥)
𝑦=𝑘
𝛼1
O
𝛼2
𝛼3
x
The maximum value of the graph of 𝑦 = 𝑥 2 (3 − 𝑥) is 4.
Hence, for 3 real solutions
𝑘 < 4 and 𝑘 > 0 (𝑘 = 0 is the x-axis)
∴ {𝑘: 0 < 𝑘 < 4}
(iii)
Data: 𝑓: 𝑥 → 𝑥 2 (3 − 𝑥), 0 < 𝑥 < 2
𝑔: 𝑥 → 𝑥 2 (3 − 𝑥), 0 < 𝑥 < 3
a)
Required To Prove: 𝑓 has an inverse.
Proof: Consider the graph of 𝑓(𝑥) for 0 < 𝑥 < 2.
There exists 𝑓 −1 iff and only if injective, that is, both one to one and the
co-domain is equal to the range. For 𝑓, 𝐴 is the domain and 𝐵 is the range,
but for 𝑓 −1 , 𝐵 is the domain and 𝐴 is the co-domain.
For each element of 𝐴, there corresponds only one element of 𝐵 under 𝑓.
Similarly, for every element of the co-domain there corresponds only one
element of the domain, hence the function is one to one. The range is
equal to the co-domain for 0 < 𝑥 < 2, hence for 0 < 𝑥 < 2, the function
is both one to one (injective) and onto (surjective), hence 𝑓 −1 exists.
Q.E.D.
b)
Required To Prove: That 𝑔 does not have an inverse.
Proof: Consider the graph of 𝑔(𝑥) for 0 < 𝑥 < 3.
In the domain 0 < 𝑥 < 3 there can be two elements of the domain mapped
onto the same element of the co-domain as illustrated by the horizontal
line test on the above diagram. Therefore, it is not one to one and 𝑔−1 will
not exist.
Q.E.D.
3.
(a)
Data: 𝑝 = 6𝑖 + 4𝑗, 𝑞 = −8𝑖 − 9𝑗
(i)
Required To Calculate: The angle between 𝑝 and 𝑞.
Calculation:
Let 𝜃 be the angle between p and q.
6𝑖 + 4𝑗
p
𝑂
𝜃
q
−8𝑖 − 9𝑗
Using dot product of 𝑝 and 𝑞:
−8
6
𝑝. 𝑞 = ( ) . ( )
−9
4
𝑝. 𝑞 = (6 × −8) + (4 × −9)
= −48 − 36
= −84
|𝑝| = √(6)2 + (4)2
= √36 + 16
= √52
|𝑞| = √(−8)2 + (−9)2
= √64 + 81
= √145
Recall: 𝑝. 𝑞 = |𝑝||𝑞| cos 𝜃,
−84 = √52√145 cos 𝜃
cos 𝜃 = −
84
√52√145
𝜃 = 165.320
= 165.30 (to the nearest 0.10 )
(ii)
(a)
Required To Find: A non-zero vector, 𝑣, such that 𝑝. 𝑣 = 0
Solution:
Let 𝑣 = 𝑎𝑖 + 𝑏𝑗, 𝑎, 𝑏 ∈ 𝑍
𝑝. 𝑣 = 0
(6𝑖 + 4𝑗)(𝑎𝑖 + 𝑏𝑗) = 0
∴ 6𝑎 + 4𝑏 = 0
3𝑎 + 2𝑏 = 0
2
𝑎 = −3𝑏
∃ any number of possible values of a and b for which
3𝑎 + 2𝑏 = 0 and hence 𝑝. 𝑣 = 0
E.g. 𝑎 = −2 and 𝑏 = 3
i.e. 𝑣 = −2𝑖 + 3𝑗
2
𝑣 = 𝛼 (− 3 𝑖 + 𝑗) , 𝛼 ∈ 𝑅
(b)
Required To State: The relationship between 𝑝 and 𝑣.
Solution:
If 𝑝. 𝑣 = 0
and 𝑝. 𝑣 = |𝑝||𝑣| cos 𝜃
then cos 𝜃 = 0
and 𝜃 = 900
∴ 𝑝 must be perpendicular to 𝑣.
6𝑖 + 4𝑗
2
𝛼 (− 3 𝑖 + 𝑗) , 𝛼 ∈ 𝑅
(b)
Data: 𝐶1 has diameter with endpoints (−3, 4) and (1, 2).
(−3, 4)
(1, 2)
(i)
Required To Show: The equation of 𝐶1 is 𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 + 5 = 0.
Proof:
The center of 𝐶1 is the midpoint of the diameter.
−3+1 4+2
Centre of 𝐶1 is (
2
,
2
) = (−1,3)
2
Radius of 𝐶1 = √(1 − (−1)) + (2 − 3)2
= √22 + (−1)2
= √5 units
∴ Equation of 𝐶1 is (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 where the center of the
circle is (𝑎, 𝑏) and the radius is 𝑟.
i.e.
2
(𝑥 − (−1)) + (𝑦 − 3)2 = (√5)
2
(𝑥 + 1)2 + (𝑦 − 3)2 = 5
𝑥 2 + 2𝑥 + 1 + 𝑦 2 − 6𝑦 + 9 − 5 = 0
𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 + 5 = 0
Q.E.D.
(ii)
Data: 𝐶1 and 𝐶2 intersect.
𝐶2 : 𝑥 2 + 𝑦 2 + 𝑥 − 5𝑦 = 0
𝐶1 : 𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 + 5 = 0
Required To Calculate: Points of intersection of 𝐶1 and 𝐶2 .
Calculation:
Let
𝑥 2 + 𝑦 2 + 𝑥 − 5𝑦 = 0 … (1)
and 𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 + 5 = 0 … (2)
Equating equation (1) and equation (2):
𝑥 2 + 𝑦 2 + 𝑥 − 5𝑦 = 𝑥 2 + 𝑦 2 + 2𝑥 − 6𝑦 + 5
𝑥 − 5𝑦 = 2𝑥 − 6𝑦 + 5
0=𝑥−𝑦+5
𝑦 =𝑥+5
∴ 𝐶1 and 𝐶2 intersect when 𝑦 = 𝑥 + 5.
Substituting 𝑦 = 𝑥 + 5 into equation (1):
𝑥 2 + (𝑥 + 5)2 + 𝑥 − 5(𝑥 + 5) = 0
𝑥 2 + 𝑥 2 + 10𝑥 + 25 + 𝑥 − 5𝑥 − 25 = 0
2𝑥 2 + 6𝑥 = 0
𝑥 2 + 3𝑥 = 0
𝑥(𝑥 + 3) = 0
𝑥 = 0 and −3
When 𝑥 = 0, 𝑦 = 0 + 5 = 5
When 𝑥 = −3, 𝑦 = −3 + 5 = 2
∴ 𝐶1 and 𝐶2 intersect at (0, 5) and (−3, 2)
4.
(a)
(i)
Data: cos 3𝐴 = 0.5, 0 ≤ 𝐴 ≤ 𝜋
Required To Find: 𝐴
Solution:
cos 3𝐴 = 0.5
3𝐴 = cos−1 (0.5)
=
𝜋
3
If 0 ≤ 𝐴 ≤ 𝜋, then 0 ≤ 3𝐴 ≤ 3𝜋.
𝛼 = cos −1(0.5) =
𝜋
3
2𝜋 − 𝛼
The general solution for cosine is 𝐴, 2𝜋 − 𝐴, 2𝜋 + 𝐴, 4𝜋 − 𝐴
𝜋
𝜋
𝜋
∴ 3𝐴 = 3 , (2𝜋 − 3 ) , 2𝜋 + 3 ,…
𝜋 5𝜋 7𝜋
= 3,
3
,
3
𝜋 5𝜋 7𝜋
𝐴 = 9,
9
,
9
0≤𝐴≤𝜋
Other values that satisfy the equation are not within the given range.
(ii)
Required To Prove: cos 3𝐴 = 4 cos3 𝐴 − 3 cos 𝐴
Proof:
Taking L.H.S.
cos 3𝐴 ≡ cos(2𝐴 + 𝐴)
= cos 𝐴 cos 2𝐴 − sin 𝐴 sin 2𝐴
(Compound-angle formula)
Recall: cos 2𝐴 = 2 cos 2 𝐴 − 1
and
sin 2𝐴 = 2 sin 𝐴 cos 𝐴
Hence, cos 3𝐴 = cos 𝐴(2 cos 2 𝐴 − 1) − sin 𝐴(2 sin 𝐴 cos 𝐴)
= 2 cos3 𝐴 − cos 𝐴 − 2 sin2 𝐴 cos 𝐴
Recall: sin2 𝐴 + cos2 𝐴 = 1 and so sin2 𝐴 = 1 − cos2 𝐴
= 2 cos3 𝐴 − cos 𝐴 − 2(1 − cos2 𝐴) cos 𝐴
= 2 cos3 𝐴 − cos 𝐴 − 2 cos 𝐴 + 2 cos3 𝐴
= 4 cos3 𝐴 − 3 cos A
= R.H.S.
Q.E.D.
(iii)
Data: The 3 roots of the equation 4𝑝3 − 3𝑝 − 0.5 = 0 lie between −1 and
1.
Required To Calculate: The roots of the equation.
Calculation:
From (a) (i)
cos 3𝐴 = 0.5
From
cos 3𝐴 = 4 cos 3 𝐴 −3 cos 𝐴
(ii)
∴ 4 cos3 𝐴 − 3 cos 𝐴 = 0.5
and 4 cos3 𝐴 − 3 cos 𝐴 − 0.5 = 0 is of the form
4𝑝3 − 3𝑝 − 0.5 = 0 (where 𝑝 = cos 𝐴)
𝜋 5𝜋 7𝜋
From (a)(i) the solutions of A were 9 ,
9
,
9
radians.
Hence, if 𝑝 = cos 𝐴
𝜋
then when 𝐴 = 9 ,
𝜋
𝑝 = cos 9 = 0.940
when 𝐴 =
𝑝 = cos
5𝜋
9
when 𝐴 =
𝑝 = cos
7𝜋
9
5𝜋
9
,
= −0.174
7𝜋
9
,
= −0.766
∴ 𝑝 = 0.940, −0.174, −0.766 for the domain −1 ≤ 𝑝 ≤ 1.
(b)
Painting h m
𝛼
𝑂
dm
𝛽
xm
Data: Diagram showing the viewing angle of a painting on a vertical wall, with
given dimensions, where 𝛼 and 𝛽 are respective angles in inclination in radians.
ℎ𝑥
Required To Prove: tan(𝛼 − 𝛽) = 𝑥 2 +𝑑(𝑑+ℎ)
(i)
Proof:
hm
𝛼
𝑂
dm
𝛽
xm
tan 𝛼 =
ℎ+𝑑
𝑥
𝑑
tan 𝛽 = 𝑥
tan 𝛼 −tan 𝛽
tan(𝛼 − 𝛽) = 1+tan 𝛼 tan 𝛽
=
ℎ+𝑑
𝑑
−( )
𝑥
𝑥
ℎ+𝑑 𝑑
1+(
)( )
𝑥
𝑥
(Compound-angle formula)
=
ℎ
𝑥
𝑑(ℎ+𝑑)
1+ 2
𝑥
=
ℎ
𝑥
𝑥2 +𝑑(ℎ+𝑑)
𝑥2
( )
𝑥2
ℎ
= 𝑥 × 𝑥 2 +𝑑(ℎ+𝑑)
ℎ𝑥
= 𝑥 2 +𝑑(𝑑+ℎ)
Q.E.D.
(ii)
Data: Maximum viewing angle, (𝛼 − 𝛽) occurs at 𝑥 = √ℎ(𝑑 + ℎ).
Required To Calculate: The maximum viewing angle when 𝑑 = 3ℎ.
Calculation:
ℎ𝑥
tan(𝛼 − 𝛽) = 𝑥 2 +𝑑(𝑑+ℎ)
Let 𝐴 = tan(𝛼 − 𝛽)
ℎ𝑥
𝐴 = 𝑥 2 +𝑑(𝑑+ℎ)
Let 𝑥 = √ℎ(𝑑 + ℎ) and 𝑑 = 3ℎ
ℎ(√ℎ(𝑑+ℎ))
∴𝐴=
2
(√ℎ(𝑑+ℎ)) +𝑑(𝑑+ℎ)
ℎ√ℎ(3ℎ+ℎ)
= ℎ(3ℎ+ℎ)+3ℎ(3ℎ+ℎ)
ℎ√4ℎ2
= 4ℎ2 +12ℎ2
=
ℎ(2ℎ)
16ℎ2
2ℎ2
= 16ℎ2
2
= 16
1
=8
1
∴ tan(𝛼 − 𝛽) = 8
1
𝛼 − 𝛽 = tan−1 (8)
= 0.124 radians
5.
(a)
(i)
𝑥 2 −9
Required To Find: lim 𝑥 3 −27
𝑥→3
Solution:
𝑥 2 −9
Let 𝑓(𝑥) = 𝑥 3 −27
32 −9
𝑓(3) = 33 −27
0
= 0, which is indeterminate
Hence, factorising and cancelling is the expected method:
lim
𝑥→3
𝑥 2 −9
𝑥 3 −27
(𝑥−3)(𝑥+3)
= lim (𝑥−3)(𝑥 2
𝑥→3
+3𝑥+9)
𝑥+3
= lim 𝑥 2 +3𝑥+9
𝑥→3
3+3
= 32 +3(3)+9
6
= 27
2
=9
Alternative Method:
Using L’Hospital’s Rule:
If
g ( x)
f ( x) =
h ( x)
then,
lim x →a f ( x ) =
g ( a )
h ( a )
𝑓 ′ (𝑥) = 2𝑥 and 𝑔′ (𝑥) = 3𝑥 2
𝑥 2 −9
2𝑥
∴ lim 𝑥 3 −27 = lim 3𝑥 2
𝑥→3
𝑥→3
2(3)
= 3(3)2
6
= 27
2
=9
(ii)
tan 𝑥−5𝑥
Required To Find: lim sin 2𝑥−4𝑥
𝑥→0
Solution:
tan 𝑥−5𝑥
𝑓(𝑥)
Let 𝑓(𝑥) = sin 2𝑥−4𝑥, which is of the form 𝑔(𝑥).
tan 0−5(0)
𝑓(0) = sin 2(0)−4(0)
0
= 0 , which is indeterminate.
Using L’Hospital’s Rule, which is to differentiate both the numerator and
the denominator, then substitute the value of 𝑥 = 0 in both the numerator
and the denominator.
𝑓 ′(𝑥) = sec 2 𝑥 − 5 and 𝑔′ = 2 cos 2𝑥 − 4
sec2 𝑥−5
tan 𝑥−5𝑥
lim sin 2𝑥−4𝑥 = lim 2 cos 2𝑥−4
𝑥→0
𝑥→0
(
1
−5)
cos2 0
= 2 cos 2(0)−4
1−5
= 2−4
−4
= −2
=2
Alternative Method:
lim
tan 𝑥−5𝑥
𝑥→0 sin 2𝑥−4𝑥
= lim tan 𝑥 − 5𝑥 ÷ lim sin 2𝑥 − 4𝑥
𝑥→0
𝑥→0
𝑥
(÷ 𝑥)
=
tan 𝑥−5𝑥
lim
𝑥
𝑥→0
sin 2𝑥−4𝑥
lim
𝑥
𝑥→0
Consider numerator:
lim
tan 𝑥−5𝑥
𝑥
𝑥→0
= lim
sin 𝑥
cos 𝑥
𝑥
𝑥→0
= [[lim
−5
sin 𝑥
𝑥
𝑥→0
1
] ÷ [lim cos 𝑥]] − lim 5
𝑥→0
=1÷1−5
= −4
Consider denominator:
lim
sin 2𝑥−4𝑥
𝑥→0
𝑥
= lim
sin 2𝑥
𝑥
𝑥→0
= lim
𝑥→0
2 sin 𝑥 cos 𝑥
𝑥
𝑥→0
= lim
− lim 4
sin 𝑥
𝑥→0
𝑥
−4
[2 lim cos 𝑥] − 4
𝑥→0
= 1(2) − 4
= −2
tan 𝑥−5𝑥
−4
∴ lim sin 2𝑥−4𝑥 = −2
𝑥→0
=2
(b)
Data: 𝑓(𝑥) = {
(i)
(a)
3𝑥 − 7, if 𝑥 > 4
1 + 2𝑥, 𝑖𝑓 𝑥 ≤ 4
Required To Find: lim+ 𝑓(𝑥)
𝑥→4
𝑥→0
Solution:
lim 𝑓(𝑥) = lim+ (3𝑥 − 7)
𝑥→4+
𝑥→4
= 3(4) − 7
=5
(b)
Required To Find: lim− 𝑓(𝑥)
𝑥→4
Solution:
lim 𝑓(𝑥) = lim− (1 + 2𝑥)
𝑥→4−
𝑥→4
= 1 + 2(4)
=9
(ii)
Required To Prove: That 𝑓(𝑥) is discontinuous at 𝑥 = 4.
Proof:
Since
lim 𝑓(𝑥) ≠ lim− 𝑓(𝑥)
𝑥→4 +
𝑥→4
𝑓(𝑥) is not continuous at 𝑥 = 4 and hence it is discontinuous at 𝑥 = 4.
Q.E.D.
(c)
(i)
1 2
1
Required To Evaluate: ∫−1 [𝑥 − 𝑥] 𝑑𝑥
Solution:
1 2
Expand [𝑥 − 𝑥] :
1 2
1
1
1
∫−1 [𝑥 − 𝑥] 𝑑𝑥 = ∫−1 [𝑥 2 − 2 + 𝑥 2 ] 𝑑𝑥
1
= ∫−1[𝑥 2 − 2 + 𝑥 −2 ]𝑑𝑥
1
𝑥3
= [ 3 − 2𝑥 − 𝑥 −1 ]
𝑥3
1
1
= [ 3 − 2𝑥 − 𝑥]
=[
(1)3
3
−1
−1
1
− 2(1) − 1] − [
1
1
(−1)3
3
= [3 − 2 − 1] − [− 3 + 2 + 1]
8
8
= [− 3] − [3]
=−
(ii)
16
3
Data: 𝑢 = 𝑥 2 + 4
Required To Find: ∫ 𝑥√𝑥 2 + 4 𝑑𝑥
Solution:
𝑢 = 𝑥2 + 4
𝑑𝑢
𝑑𝑥
𝑑𝑢
2𝑥
= 2𝑥
= 𝑑𝑥
𝑑𝑢
∫ 𝑥√𝑥 2 + 4 𝑑𝑥 = ∫ 𝑥 √𝑢 2𝑥
1
1
= 2 ∫ 𝑢2 𝑑𝑢
1
− 2(−1) − −1]
3
1 𝑢2
= 2[
3
2
]+𝐶
(𝐶 – the constant of integration)
1
= 3 √𝑢3 + 𝐶
1
= 3 √(𝑥 2 + 4)3 + 𝐶
6.
(a)
(i)
R.T.Differentiate: 𝑦 = sin(3𝑥 + 2) + tan 5𝑥 w.r.t 𝑥.
Solution:
Let 𝑎 = sin(3𝑥 + 2)
Let 𝑡 = 3𝑥 + 2
𝑑𝑡
𝑑𝑥
=3
𝑎 = sin 𝑡
𝑑𝑎
𝑑𝑡
𝑑𝑎
𝑑𝑥
= cos 𝑡
=
𝑑𝑎
𝑑𝑡
𝑑𝑡
× 𝑑𝑥
(Chain rule)
= (cos 𝑡) × 3
= 3 cos(3𝑥 + 2)
Let 𝑏 = tan 5𝑥
Let 𝑡 = 5𝑥
𝑑𝑡
𝑑𝑥
=5
𝑏 = tan 𝑡
𝑑𝑏
𝑑𝑡
𝑑𝑏
𝑑𝑥
= sec 2 𝑡
=
𝑑𝑏
𝑑𝑡
𝑑𝑡
× 𝑑𝑥
= (sec 2 𝑡) × 5
(Chain rule)
= 5 sec 2 5𝑥
𝑑𝑦
∴ 𝑑𝑥 = 3 cos(3𝑥 + 2) + 5 sec 2 5𝑥
(ii)
𝑥 2 +1
𝐃𝐚𝐭𝐚: 𝑦 = 𝑥 3 −1
Required to: Differentiate 𝑦 w.r.t 𝑥.
Solution:
𝑥 2 +1
𝑦 = 𝑥 3 −1 is of the form 𝑦 =
𝑢
𝑣
𝑑𝑢
When 𝑢 = 𝑥 2 + 1 𝑑𝑥 = 2𝑥
𝑑𝑣
𝑣 = 𝑥 3 − 1, 𝑑𝑥 = 3𝑥 2
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
𝑑𝑢
𝑑𝑣
−𝑢
𝑑𝑥
𝑑𝑥
𝑣2
(Quotient Law)
(𝑥 3 −1)(2𝑥)−(𝑥 2 +1)(3𝑥 2 )
(𝑥 3 −1)2
=
2𝑥 4 −2𝑥−3𝑥 4 −3𝑥 2
(𝑥 3 −1)2
=
−𝑥 4 −3𝑥 2 −2𝑥
(𝑥 3 −1)2
=
(b)
𝑣
−𝑥(𝑥 3 +3𝑥+2)
(𝑥 3 −1)2
4
Data: the function 𝑓(𝑥) satisfies ∫1 𝑓(𝑥)𝑑𝑥 = 7.
(i)
4
Required To Find: ∫1 [3𝑓(𝑥) + 4]𝑑𝑥
Solution:
4
4
4
∫1 [3𝑓(𝑥) + 4]𝑑𝑥 = 3 ∫1 𝑓(𝑥) 𝑑𝑥 + ∫1 4 𝑑𝑥
= 3(7) + [4𝑥]14
= 21 + [4(4) − 4(1)]
= 21 + 16 − 4
= 33
(ii)
3
Required To Find: ∫0 2𝑓(𝑥 + 1) 𝑑𝑥
Solution:
Let 𝑢 = 𝑥 + 1
𝑑𝑢
𝑑𝑥
=1
𝑑𝑥 = 𝑑𝑢
3
?
∫0 2𝑓(𝑥 + 1) 𝑑𝑥 = 2 ∫? 𝑓(𝑢) 𝑑𝑢
Finding the new limits:
When 𝑥 = 3, 𝑢 = 4 and 𝑥 = 0, 𝑢 = 1
3
4
∴ ∫0 2𝑓(𝑥 + 1) 𝑑𝑥 = 2 ∫1 𝑓(𝑢) 𝑑𝑢
4
= 2 ∫1 𝑓(𝑥) 𝑑𝑥
= 2[7]
= 14
(c)
Data: The line 𝑥 + 𝑦 = 2 intersects the curve 𝑦 = 𝑥 2 at 𝑃 and 𝑄.
(i)
Required To Find: The coordinates of 𝑃 and 𝑄.
Solution:
Solving the equations simultaneously, to find the coordinates of 𝑃 and 𝑄.
𝑦 = 𝑥 2 … (1)
Let
and 𝑥 + 𝑦 = 2 … (2)
Substituting equation (1) into equation (2):
𝑥 + 𝑥2 = 2
𝑥2 + 𝑥 − 2 = 0
(𝑥 − 1)(𝑥 + 2) = 0,
∴ 𝑥 = 1 or −2
When 𝑥 = 1, 𝑦 = (1)2 = 1. When 𝑥 = −2 , 𝑦 = (−2)2 = 4
Since the x – coordinate of 𝑃 is – 𝑣𝑒 then 𝑃 is (−2, 4) and
since the 𝑥 – coordinate of 𝑄 is + 𝑣𝑒, then 𝑄
is (1, 1).
(ii)
Required To Calculate: The area of the shaded region.
Calculation:
Let the region 𝐴1 , 𝐴2 and 𝐴3 be defined as shown in the above diagram.
1
Area of (𝐴2 + 𝐴3 ) = ∫−2 𝑥 2 𝑑𝑥
1
𝑥3
= [3]
=[
−2
(1)3
−
3
1
(−2)3
3
]
8
=3+3
= 3 units2
1
2
Area of (𝐴1 + 𝐴2 + 𝐴3 ) = (3)(1 + 4)
=
15
∴ Area of shaded region =
2
units2
15
2
−3
1
= 4 2 units2
JUNE 2009 UNIT 1 PAPER 2
1.
(a)
Required To Express: √28 + √343 in the form 𝑘√7, where 𝑘 is an integer.
Solution:
√28 + √343 = √7 × 4 + √7 × 49
= 2√7 + 7√7
= 9√7
which is of the form 𝑘√7 where 𝑘 = 9 ∈ 𝑍.
(b)
Data: 𝑥 and 𝑦 are positive real numbers and 𝑥 ≠ 𝑦.
(i)
Required To Simplify:
𝑥 4 −𝑦 4
𝑥−𝑦
Solution:
𝑥 4 −𝑦 4
𝑥−𝑦
(𝑥 2 −𝑦 2 )(𝑥 2 +𝑦 2 )
=
𝑥−𝑦
=
(Difference of two squares, twice)
(𝑥−𝑦)(𝑥+𝑦)(𝑥 2 +𝑦 2 )
𝑥−𝑦
= (𝑥 + 𝑦)(𝑥 2 + 𝑦 2 )
(ii)
Required To Prove: (𝑦 + 1)4 − 𝑦 4 = (𝑦 + 1)3 + (𝑦 + 1)2 𝑦 +
(𝑦 + 1)𝑦 2 + 𝑦 3
Proof:
L.H.S.
Re: (i)
𝑥 4 −𝑦 4
𝑥−𝑦
= (𝑥 + 𝑦)(𝑥 2 + 𝑦 2 )
Let 𝑥 = 𝑦 + 1
i.e.
(𝑦+1)4 −𝑦 4
(𝑦+1)−𝑦
= ((𝑦 + 1) + 𝑦)((𝑦 + 1)2 + 𝑦 2 )
(𝑦 + 1)4 − 𝑦 4 = [(𝑦 + 1) + 𝑦][(𝑦 + 1) − 𝑦] [(𝑦 + 1)2 + 𝑦 2 ]
= 1[(𝑦 + 1) + 𝑦][(𝑦 + 1)2 + 𝑦 2 ]
= [(𝑦 + 1) + 𝑦][(𝑦 + 1)2 + 𝑦 2 ]
(𝑦 + 1)4 − 𝑦 4 = (𝑦 + 1)3 + (𝑦 + 1)2 𝑦 + (𝑦 + 1)𝑦 2 + 𝑦 3
Q.E.D.
(iii)
Required To Deduce: (𝑦 + 1)4 − 𝑦 4 < 4(𝑦 + 1)3
Solution:
Re: From (ii): (𝑦 + 1)4 − 𝑦 4 = (𝑦 + 1)3 + (𝑦 + 1)2 𝑦 + (𝑦 + 1)𝑦 2 + 𝑦 3
Now 𝑦 < 𝑦 + 1 (since y is a positive integer)
and 𝑦 2 < (𝑦 + 1)2
and 𝑦 3 < (𝑦 + 1)3
Hence,
(𝑦 + 1)4 − 𝑦 4 < (𝑦 + 1)3 + (𝑦 + 1)2 (𝑦 + 1) + (𝑦 + 1)(𝑦 + 1) 2 + (𝑦 + 1)3
i.e
< 4(𝑦 + 1)3
Q.E.D
(c)
Required To Solve: For x in log 4 𝑥 = 1 + log 2 2𝑥, 𝑥 > 0
Solution:
log 4 𝑥 = 1 + log 2 2𝑥
log2 𝑥
log2 4
log2 𝑥
log2 4
log2 𝑥
log2 (2)2
log2 𝑥
2log2 2
= 1 + log 2 2𝑥 (Change of base formula)
= log 2 2 + log 2 2𝑥
= log 2 (2 × 2𝑥)
= log 2 4𝑥
(log 2 2 = 1 )
log2 𝑥
2
1
2
= log 2 4𝑥
log 2 𝑥 = log 2 4𝑥
1
log 2 (𝑥 2 ) = log 2 4𝑥
Removing logs:
1
𝑥 2 = 4𝑥
√𝑥 = 4𝑥
𝑥 = 16𝑥 2
16𝑥 2 − 𝑥 = 0
𝑥(16𝑥 − 1) = 0
1
𝑥 = 0 or 𝑥 = 16
( If 𝑥 = 0 then one or more of the terms in the given equation will be the log of 0
or a negative number. Hence, x cannot be 0.)
∴𝑥=
1
16
only (𝑥 > 0)
Alternative Method:
log 4 𝑥 = 1 + log 2 2𝑥
log2 𝑥
log2 (2)2
log2 𝑥
2log2 2
1
2
= 1 + log 2 2 + log 2 𝑥
= 1 + 1 + log 2 𝑥
log 2 𝑥 = 2 + log 2 𝑥
1
− 2 log 2 𝑥 = 2
log 2 𝑥 = −4
2−4 = 𝑥
1
𝑥 = 24
(Definition of logs)
1
𝑥 = 16
2.
(a)
Data: The roots of the equation 2𝑥 2 + 4𝑥 + 5 = 0 are 𝛼 and 𝛽.
Required To Find: A quadratic equation with roots
2
𝛼
equation.
Solution:
Re:
If 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 is a quadratic in x,
(÷ 𝑎)
𝑏
𝑐
𝑥2 + 𝑎 𝑥 + 𝑎 = 0
If the roots are 𝛼 and 𝛽
then (𝑥 − 𝛼)(𝑥 − 𝛽) = 0
𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 = 0
Equating coefficients
𝑏
𝑐
𝛼 + 𝛽 = − 𝑎 and 𝛼𝛽 = 𝑎
4
𝛼 + 𝛽 = − (2)
= −2
5
𝛼𝛽 = 2
The required quadratic equation is
𝑥 2 − (sum of roots)𝑥 + (product of roots) = 0
2
2
2
2
i.e. 𝑥 2 − (𝛼 + 𝛽) + (𝛼 × 𝛽) = 0
2
𝛼
2
+𝛽 =
=
2(𝛼+𝛽)
𝛼𝛽
2(−2)
5
2
2
and 𝛽, without solving the
8
= −5
2
2
4
× 𝛽 = 𝛼𝛽
𝛼
=
4
5
2
8
=5
8
8
Required equation is 𝑥 2 − (− 5 𝑥) + 5 = 0
The equation is best expressed in integral form
(× 5) 5𝑥 2 + 8𝑥 + 8 = 0
(b)
Data: A coach trains six athletes 𝑢, 𝑣, 𝑤, 𝑥, 𝑦 and 𝑧. He makes and assignment 𝑓
of athletes 𝑢, 𝑣, 𝑥, 𝑦 and 𝑧 to activities 1, 2, 3 and 4 according to the diagram
where 𝐴 = {𝑢, 𝑣, 𝑤, 𝑥, 𝑦, 𝑧} and 𝐵 = {1, 2, 3, 4}.
(i)
Required To Express: 𝑓 as a set of ordered pairs.
Solution:
From the diagram, the set of ordered pairs of 𝑓 is
{(𝑢, 1), (𝑣, 2), (𝑣, 3), (𝑥, 1), (𝑦, 3), (𝑧, 4)}
(ii)
(a)
Required To Find: Two reasons why 𝑓 is not a function.
Solution:
𝑓 is a function ‘iff’ each element of the domain, A, is mapped onto
only one element of the co-domain, B. 𝑓 is NOT a function
because:
(i) 𝑣 ∈ 𝐴, is mapped onto two elements, 2 and 3 ∈ 𝐵, in the codomain.
(ii) 𝑤 ∈ 𝐴, is not mapped onto any element of 𝐵.
(b)
Required To Construct: 𝑔: 𝐴 → 𝐵 with minimum changes to 𝑓 as
a set of ordered pairs.
Solution:
A
𝑔
B
u
1
v
2
w
3
x
4
y
z
Each element of 𝐴 is to be mapped onto only one element of 𝐵,
under 𝑔.
Hence, 𝑔 will be a function, if we map w onto 3 as illustrated, so
that each element of 𝐴 is mapped onto one element of 𝐵, which is
the criterion for a function.
{(𝑢, 1), (𝑣, 2), (𝑤, 3), (𝑥, 1)(𝑦, 3), (𝑧, 4)}
(c)
Required To Determine: The number of different functions
possible for 𝑔.
Solution:
Any of the 6 elements of 𝐴 may be mapped onto any of the 4
elements of 𝐵.
∴ the number of different possible functions = 6 × 4 = 24
(c)
𝑥 − 3 if 𝑥 ≤ 3
Data: 𝑓 is defined on 𝑅 for 𝑓(𝑥) = {𝑥
if 𝑥 > 3
4
(i)
Required To Find: 𝑓[𝑓(20)]
Solution:
𝑓(20) =
20
4
=5
(As 20 > 3)
∴ 𝑓[𝑓(20)] = 𝑓(5)
5
=4
(ii)
Required To Find: 𝑓[𝑓(8)]
Solution:
8
𝑓(8) = 4 = 2
∴ 𝑓[𝑓(8)] = 𝑓(2)
(As 8 > 3)
(As 2 < 3)
=2−3
= −1
(iii)
Required To Find: 𝑓[𝑓(3)]
Solution:
𝑓(3) = 3 − 3 = 0
∴ 𝑓[𝑓(3)] = 𝑓(0)
=0−3
= −3
3.
(a)
Data: Circle, C, has equation (𝑥 − 3)2 + (𝑦 − 4)2 = 25.
(i)
Required To Find: The radius and coordinates of the centre of the circle.
Solution:
(𝑥 − 3)2 + (𝑦 − 4)2 = 25
(𝑥 − 3)2 + (𝑦 − 4)2 = (5)2 is of the form
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 which represents a circle of radius 𝑟 and
center (𝑎, 𝑏).
∴ Center of C is (3, 4) and radius = 5 units.
(3, 9)
(6,8)
5 units
(3,4)
(-1, 1)
𝑦 = 2𝑥 + 3
C
(ii)
Required To Find: The equation of tangent at the point (6, 8) on 𝐶.
Solution:
Gradient of the line connecting the centre (3, 4) and (6, 8)
8−4
= 6−3
4
=3
3
∴ Gradient of tangent = − 4.
(The angle made by the tangent and a radius, at the point of contact =
900
AND
the product of gradients of perpendicular lines = −1 )
ALTERNATIVE METHOD: for finding the gradient of the tangent
(𝑥 − 3)2 + (𝑦 − 4)2 = 25
𝑥 2 − 6𝑥 + 9 + 𝑦 2 − 8𝑦 + 16 = 25
𝑥 2 − 6𝑥 + 𝑦 2 − 8𝑦 = 0
Differentiating implicitly w.r.t x
𝑑𝑦
𝑑𝑦
2𝑥 − 6 + 2𝑦 𝑑𝑥 − 8 𝑑𝑥 = 0
𝑑𝑦
𝑥 − 3 + (𝑦 − 4) 𝑑𝑥 = 0
(÷ 2)
𝑑𝑦
𝑑𝑥
3−𝑥
= 𝑦−4
3−6
3
∴ the gradient of tangent at (6, 8) = 8−4 = − 4
3
Hence, equation of the tangent is (𝑦 − 8) = − 4 (𝑥 − 6)
4(𝑦 − 8) = −3(𝑥 − 6)
4𝑦 − 32 = −3𝑥 + 18
4𝑦 = −3𝑥 + 50
(iii)
Required To Calculate: The points of intersection of 𝐶 and 𝑦 = 2𝑥 + 3.
Solution:
Let 𝑦 = 2𝑥 + 3
… (1)
(𝑥 − 3)2 + (𝑦 − 4)2 = 25 … (2)
Substituting equation (1) into equation (2):
(𝑥 − 3)2 + (2𝑥 + 3 − 4)2 = 25
(𝑥 − 3)2 + (2𝑥 − 1)2 = 25
𝑥 2 − 6𝑥 + 9 + 4𝑥 2 − 4𝑥 + 1 = 25
5𝑥 2 − 10𝑥 − 15 = 0
𝑥 2 − 2𝑥 − 3 = 0
(÷ 5)
(𝑥 − 3)(𝑥 + 1) = 0
𝑥 = −1 or 3
When 𝑥 = 3, 𝑦 = 2(3) + 3 = 9
When 𝑥 = −1, 𝑦 = 2(−1) + 3 = 1
Points of intersection of 𝐶 and the straight line are (3, 9) and (−1, 1).
(b)
Data: The position vector of 𝑝 is given by 𝑝 = −𝑖 + 6𝑗 and the position vector
of 𝑞 is given by 𝑞 = 3𝑖 + 8𝑗.
(i)
(a)
Required To Calculate: The size of the acute angle, 𝜃, between p
and q, in degrees.
Solution:
j
Q
3𝑖 + 8𝑗
P
−𝑖 + 6𝑗
𝜃
O
i
𝑝. 𝑞 = |𝑝||𝑞| cos 𝜃
𝑝.𝑞
cos 𝜃 = |𝑝||𝑞|
𝑝. 𝑞 = (−𝑖 + 6𝑗). (3𝑖 + 8𝑗)
𝑝. 𝑞 = (−1)(3) + (6)(8)
= 45
|𝑝||𝑞| = √(−1)2 + (6)2 × √(3)2 + (8)2
= √37√73
cos 𝜃 =
45
√37√73
45
𝜃 = cos −1 (
√37√73
)
= 30.010
= 30.00 (to the nearest 0.1 degree)
(b)
Required To Find: The area of the triangle 𝑃𝑂𝑄.
Solution:
1
Area of triangle 𝑃𝑂𝑄 = 2 |𝑝||𝑞| sin 𝜃
1
= 2 (√37√73) sin 30.010
= 12.99 units2 (correct to 2 decimal places)
(ii)
(a)
Required To Find: The position vector of 𝑀, the midpoint of 𝑃𝑄,
in terms of 𝑖 and 𝑗.
Solution:
j
M
Q
3𝑖 + 8𝑗
P
−𝑖 + 6𝑗
O
i
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
𝑃𝑄 = 𝑂𝑄
𝑂𝑃
= (3𝑖 + 8𝑗) − (−𝑖 + 6𝑗)
= 4𝑖 + 2𝑗
1
⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
𝑃𝑀 = 2 (𝑃𝑄)
1
= 2 (4𝑖 + 2𝑗)
= 2𝑖 + 𝑗
⃗⃗⃗⃗⃗⃗
𝑂𝑀 = ⃗⃗⃗⃗⃗
𝑂𝑃 + ⃗⃗⃗⃗⃗⃗
𝑃𝑀
= (−𝑖 + 6𝑗) + (2𝑖 + 𝑗)
= 𝑖 + 7𝑗
(b)
Required To Find: The position vector of 𝑅, where 𝑅 is such that
𝑃𝑄𝑅𝑂 forms a parallelogram, when labelled in a clockwise
direction.
Solution:
j
M
Q
3𝑖 + 8𝑗
P
−𝑖 + 6𝑗
R
O
i
𝑇
𝑃→𝑄
4
( )
2
−1
3
)→( )
6
8
𝑃𝑄𝑅𝑂 is a parallelogram and opposite sides are equal and parallel.
Similarly
4
𝑇=( )
2
𝑂→𝑅
(
4
𝑇=( )
2
0
4
( )→( )
0
2
⃗⃗⃗⃗⃗ = 4𝑖 + 2𝑗
∴ 𝑂𝑅
Alternative Method:
1
( )
−6
𝑃→
1
( )
−6
𝑄→
𝑂
𝑅
3
If 𝑄 = ( )
8
3
1
𝑅 = ( )−( )
8
−6
4
=( )
2
= 4𝑖 + 2𝑗
4.
(a)
Data:
A
4cm
𝜃
B
P
9 cm
x cm
𝜃
C
(i)
Q
D
Required To Prove: 𝑥 = 4 cos 𝜃 + 9 sin 𝜃
Proof:
Considering triangle 𝐶𝐵𝑄
sin 𝜃 =
𝐵𝑄
9
∴ 𝐵𝑄 = 9 sin 𝜃
cos 𝜃 =
𝐴𝑃
4
Considering triangle 𝐴𝐵𝑃
∴ 𝐴𝑃 = 4 cos 𝜃
𝑥 = The length of 𝑃𝐷 + the length of 𝐴𝑃 (Since 𝑃𝐷 = 𝐵𝑄)
= 4 cos 𝜃 + 9 sin 𝜃
Q.E.D.
(ii)
Required To Find: The maximum possible value of 𝑥 by expressing 𝑥 in
the form 𝑟 cos(𝜃 − 𝛼), where 𝑟 is positive and 0 ≤ 𝛼 < 12𝜋.
Solution:
4 cos 𝜃 + 9 sin 𝜃 = 𝑟 cos(𝜃 − 𝛼)
𝑟 = √42 + 92
= √97
∴
4
√97
cos 𝜃 +
9
√97
sin 𝜃 = cos(𝜃 − 𝛼)
= cos 𝜃 cos 𝛼 + 𝑠𝑖𝑛𝜃 sin 𝛼
Equating coefficients
cos 𝛼 =
and sin 𝛼 =
4
√97
9
√97
y+
`
√97
+9
x+
𝛼
𝑂
+4
4
)
√97
𝛼 = cos −1(
= 1.152 𝑟𝑎𝑑
0≤𝛼<
𝑥 = √97 cos(𝜃 − 1.152)
−1 ≤ cos(𝜃 − 1.152) ≤ 1
𝜋
2
∴ 𝑥𝑚𝑎𝑥 = √97 × 1
= √97
(b)
3
5
Data: sin 𝐴 = 5 and cos 𝐵 = 13, where 𝐴 and 𝐵 are acute angles.
y+
y+
+5
+3
A
O
+13
+4
4
+12
B
x+
O
+5
x+
12
cos 𝐴 = 5 and sin 𝐴 = 13
(i)
Required To Find: The exact value of sin(𝐴 + 𝐵).
Solution:
sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
3
5
4
12
5
13
5
13
= ( )( ) + ( )( )
63
= 65
(ii)
Required To Find: The exact value of cos(𝐴 − 𝐵).
Solution:
cos(𝐴 − 𝐵) = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵
4
5
3
12
= (5) (13) + (5) (13)
56
= 65
(iii)
Required To Find: The exact value of cos 2𝐴.
Solution:
cos 2𝐴 = cos2 𝐴 − sin2 𝐴
4 2
3 2
= (5) − (5)
7
= 25
Alternative Method:
cos 2𝐴 = 2cos2 𝐴 − 1
4 2
= 2 (5) − 1
7
= 25
(c)
𝑥
𝜋
Required To Prove: tan [2 + 4 ] = sec 𝑥 + tan 𝑥
Proof:
Taking L.H.S.
𝑥
𝜋
tan [2 + 4 ] =
𝑥
2
𝜋
4
𝑥
𝜋
1−tan tan
2
4
(tan +tan )
𝜋
Recall: tan 4 = 1
L.H.S. =
𝑥
2
𝑥
1−tan(2)
1+tan( )
=
=
𝑥
2
𝑥
1−tan(2)
1+tan( )
×
𝑥
2
𝑥
1+tan(2)
1+tan( )
𝑥
2
𝑥
2
1+2 tan( )+tan2 ( )
𝑥
1−tan2 (2)
𝑥
𝑥
Re: (1 + tan2 (2) = sec 2 (2))
=
𝑥
2
𝑥
2
2 tan( )+sec2 ( )
𝑥
1−tan2 (2)
=
𝑥
2
𝑥
2
1−tan (2)
2 tan( )
+
𝑥
2
𝑥
2
1−tan (2)
sec2 ( )
1
Re: tan 𝑥 = tan 2 (2 𝑥)
=
𝑥
2
2 tan( )
𝑥
1−tan2 (2)
𝑥
L.H.S. = tan 𝑥 +
sec2 (2)
𝑥
1−tan2 (2)
1
(
𝑥
sec2 (2)
Consider
𝑥
1−tan2 (2)
=
)
𝑥
)
2
𝑥
sin2 ( )
2
1−
𝑥
𝑐𝑜𝑠2 ( )
2
cos2 (
=
1
𝑥 )
cos2( )
2
𝑥
sin2( )
1− 2 𝑥2
cos ( )
2
=
1
𝑥
𝑥
cos2 ( )−sin2 ( )
2
2
𝑥
2
cos (2)
(
=
𝑥
×
cos2(2)
𝑥
cos2 ( 2)
𝑥
× cos 2 ( )
2
1
𝑥
2
𝑥
2
(cos2 ( )−sin2 ( ))
𝑥
𝑥
𝑥
(Re: cos 𝑥 = cos 2 (2) = cos 2 (2) − sin2 (2))
1
= cos 𝑥
= sec 𝑥
𝑥
𝜋
tan (2 + 4 ) = tan 𝑥 + sec 𝑥
=R.H.S.
Q.E.D
5.
(a)
𝑥 3 −8
Required To Find: lim 𝑥 2 −6𝑥+8
𝑥→2
Solution:
𝑥 3 −8
Let 𝑓(𝑥) = 𝑥 2 −6𝑥+8
23 −8
𝑓(2) = 22 −6(2)+8
0
= 0 (which is indeterminate)
Hence, factorising and cancelling:
𝑓(𝑥) =
(𝑥−2)(𝑥 2 +2𝑥+4)
(𝑥−2)(𝑥−4)
𝑓(𝑥) =
𝑥 2 +2𝑥+4
𝑥−4
𝑥 3 −8
lim 𝑥 2 −6𝑥+8 = 𝑓(2)
𝑥→2
=
22 +2(2)+4
2−4
= −6
(Alternative method by using L’Hospital’s rule)
If
f ( x) =
lim x →a f ( x ) =
g ( x)
h ( x)
g ( a )
h ( a )
𝑔′(𝑥) = 3𝑥 2 and 𝑓 ′ (𝑥) = 2𝑥 − 6
Differentiate both numerator and denominator and substitute x = 2
𝑥 3 −8
3𝑥 2
lim 𝑥 2 −6𝑥+8 = lim 2𝑥−6
𝑥→2
𝑥→2
3(2)2
= 2(2)−4
= −6
(b)
Data: 𝑓(𝑥) = {
(i)
3 − 𝑥 if 𝑥 ≥ 1
1 + 𝑥 if 𝑥 < 1
Required To Sketch: 𝑓(𝑥)
Solution:
y
𝑓(𝑥) = 3 − 𝑥
2
𝑓(𝑥) = 1 + 𝑥
x
-1
(ii)
(a)
O
1
2
Required To Find: lim+ 𝑓(𝑥)
𝑥→1
Solution:
lim 𝑓(𝑥) = 1 + 1
𝑥→1+
=2
(b)
Required To Find: lim− 𝑓(𝑥)
𝑥→1
Solution:
lim 𝑓(𝑥) = 3 − 1
𝑥→1−
=2
(iii)
Required To Deduce: That 𝑓(𝑥) is continuous at 𝑥 = 1.
Solution:
Since lim+ 𝑓(𝑥) = lim− 𝑓(𝑥) = 𝑓(1) = 2
𝑥→1
𝑥→1
𝑓(𝑥) is continuous at 𝑥 = 1.
Q.E.D.
(c)
1
Required To Differentiate: 𝑦 = 𝑥 2 from first principles.
Solution:
1
1
(𝑥+ℎ , (𝑥+ℎ)2 )
1
(𝑥, 𝑥 2 )
𝑓
′ (𝑥)
h
= lim
1
1
−
(𝑥+ℎ)2 𝑥2
ℎ→0 (𝑥+ℎ)−𝑥
𝑥2 −(𝑥+ℎ)2
= lim
( 2
)
𝑥 (𝑥+ℎ)2
ℎ
ℎ→0
𝑥 2 −(𝑥 2 +2ℎ𝑥+ℎ2 )
= lim ℎ(𝑥 2 (𝑥 2 +2ℎ𝑥+ℎ2 ))
ℎ→0
−2ℎ𝑥−ℎ2
= lim 𝑥 2 (𝑥+ℎ)2 .ℎ
ℎ→0
(−2𝑥−ℎ)
= lim 𝑥 2 (𝑥+ℎ)2
ℎ→0
(−2𝑥−0)
= 𝑥 2 (𝑥+0)2
2𝑥
= − 𝑥4
2
= − 𝑥3
(d)
Data: 𝑓 ′ (𝑥) = 3𝑥 2 + 6𝑥 + 𝑘, 𝑓(0) = −6, 𝑓(1) = −3. 𝑘 is a constant.
Required To Find: 𝑓(𝑥)
Solution:
𝑓(𝑥) = ∫(3𝑥 2 + 6𝑥 + 𝑘) 𝑑𝑥
= 𝑥 3 + 3𝑥 2 + 𝑘𝑥 + 𝑐 (c = constant)
𝑓(0) = −6
−6 = (0)3 + 3(0)2 + 𝑘(0) + 𝑐
−6 = 𝑐
𝑓(𝑥) = 𝑥 3 + 3𝑥 2 + 𝑘𝑥 − 6
𝑓(1) = −3
−3 = (1)3 + 3(1)2 + 𝑘(1) − 6
−3 = −2 + 𝑘
−1 = 𝑘
∴ 𝑓(𝑥) = 𝑥 3 + 3𝑥 2 − 𝑥 − 6
6.
(a)
Data: 𝑦 = sin 2𝑥 + cos 2𝑥
𝑑2 𝑦
Required To Prove: 𝑑𝑥 2 + 4𝑦 = 0
Proof:
𝑦 = sin 2𝑥 + cos 2𝑥
𝑑𝑦
𝑑𝑥
= 2(cos 2𝑥) + 2(− sin 2𝑥)
= 2 cos 2𝑥 − 2 sin 2𝑥
𝑑2 𝑦
𝑑𝑥 2
= 2(−2 sin 2𝑥) − 2(2 cos 2𝑥)
= −4 sin 2𝑥 − 4 cos 2𝑥
= −4(sin 2𝑥 + cos 2𝑥)
= −4𝑦
𝑑2 𝑦
𝑑𝑥 2
+ 4𝑦 = 0
Q.E.D.
(b)
𝑎
𝑎
Data: ∫0 (𝑥 + 1)𝑑𝑥 = 3 ∫0 (𝑥 − 1)𝑑𝑥 , 𝑎 > 0
Required To Find: The value of a.
Solution:
𝑎
𝑎
∫0 (𝑥 + 1)𝑑𝑥 = 3 ∫0 (𝑥 − 1)𝑑𝑥
𝑎
𝑥2
𝑥2
[ 2 + 𝑥] = 3 [ 2 − 𝑥]
0
𝑎2
02
𝑎
0
𝑎2
02
[( 2 + 𝑎) − ( 2 + 0)] = 3 [( 2 − 𝑎) − ( 2 − 0)]
𝑎2
2
+𝑎 =
3𝑎2
2
− 3𝑎
∴ 𝑎2 − 4𝑎 = 0
𝑎(𝑎 − 4) = 0
𝑎 = 0, 4
𝑎 = 4 (𝑎 > 0)
(c)
Data: A cardboard 16 cm by 10 cm has squares 𝑥 cm by 𝑥 cm removed from
each corner. The remainder of the cardboard is used to fold a tray.
(10 − 2𝑥 )cm
x cm
x cm
(16 − 2𝑥) cm
𝑥 cm
16 − 2𝑥 cm
10 − 2𝑥 cm
(i)
Required To Prove: 𝑉 = 4(𝑥 3 − 13𝑥 2 + 40𝑥)
Proof:
Volume of tray = 𝑥(16 − 2𝑥)(10 − 2𝑥)
= 𝑥(160 − 32𝑥 − 20𝑥 + 4𝑥 2 )
= 𝑥(160 − 52𝑥 + 4𝑥 2 )
= 160𝑥 − 52𝑥 2 + 4𝑥 3
= 4(𝑥 3 − 13𝑥 + 40𝑥)
Q.E.D.
(ii)
Required To Find: The value of 𝑥 such that 𝑉 is maximum.
Solution:
𝑉 = 4𝑥 3 − 52𝑥 2 + 160𝑥
𝑑𝑉
𝑑𝑥
(÷ 4)
= 12𝑥 2 − 104𝑥 + 160
= 3𝑥 2 − 26𝑥 + 40
𝑑𝑉
At a stationary value of V, 𝑑𝑥 = 0
∴ 3𝑥 2 − 26𝑥 + 40 = 0
(3𝑥 − 20)(𝑥 − 2) = 0
𝑥=
When 𝑥 =
∴𝑥≠
20
3
or 2
20
3
, the width of the rectangle will be negative.
20
3
Hence, 𝑥 = 2 only.
𝑑2 𝑉
𝑑𝑥 2
= 6𝑥 − 104
𝑑2 𝑉
When = 2, 𝑑𝑥 2 < 0.
Hence, V is a maximum at 𝑥 = 2.
JUNE 2008 UNIT 1 PAPER 2
THE REGION OF TRINIDAD AND TOBAGO ONLY
1. (a)
(i)
Data: 4𝑥 2 − ℎ𝑥 + (8 − ℎ) = 0
R.T.F. The real values of h for which the quadratic equation has real
roots.
Solution:
If 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 has real roots, then 𝑏 2 ≥ 4𝑎𝑐
Hence, if 4𝑥 2 − 2ℎ𝑥 + (8 − ℎ) = 0 has real roots, then
(−2ℎ)2 ≥ 4(4)(8 − ℎ)
4ℎ2 ≥ 128 − 16ℎ
(÷ 4) ℎ2 + 4ℎ − 32 ≥ 0
(ℎ + 8)(ℎ − 4) ≥ 0
+ ve
+ ve
h
-8
4
∴ The given quadratic has real roots for {ℎ: ℎ ≥ 4} ∪ {ℎ: ℎ ≤ −8}
(ii)
Data: The roots of the cubic equation 𝑥 3 − 15𝑥 2 + 𝑝𝑥 − 105 = 0 are
(5 − 𝑘), 5 and (5 + 𝑘).
R.T.F. The value of the constant p and of k
Solution:
The polynomial can be expressed as
(𝑥 − (5 − 𝑘))(𝑥 − 5)(𝑥 − (5 + 𝑘)) = 0
When we multiply the last term of each factor we should obtain, -105
Therefore − (5 − 𝑘)(−5)(−)(5 + 𝑘) = −105
(25 − 𝑘 2 )(−5) = −105
25 − 𝑘 2 = 21
𝑘2 − 4 = 0
(𝑘 − 2)(𝑘 + 2) = 0
Hence, 𝑘 = ±2
When 𝑘 = 2 the polynomial becomes
(𝑥 − 3)(𝑥 − 5)(𝑥 − 7) = 0
(𝑥 2 − 8𝑥 + 15)(𝑥 − 7) = 0
𝑥 3 − 8𝑥 2 + 15𝑥 − 7𝑥 2 + 56𝑥 − 105 = 0
𝑥 3 − 15𝑥 2 + 71𝑥 − 105 = 0
Equating coefficients ⇒ 𝑝 = 71
When 𝑘 = −2, the polynomial becomes
(𝑥 − (5 − (−2))(𝑥 − 5)(𝑥 − (5 + (−2))) = 0
(𝑥 − 3)(𝑥 − 5)(𝑥 − 7) = 0 (the same as when k = 2)
∴ 𝑝 = 71, also
Hence, 𝑘 = ±2 𝑎𝑛𝑑 𝑝 = 71
(b)
Data:𝑓(𝑥) = |𝑥 + 2| 𝑎𝑛𝑑 𝑔(𝑥) = 2|𝑥 − 1|
(i)
R.T.Complete: the given table
Solution:
x
f(x)
g(x)
-3
1
8
-2
0
6
-1
1
4
0
2
2
1
3
0
2
4
2
3
5
4
4
6
6
5
7
8
(ii)
(iii)
R.T.F. The values of x for which 𝑓(𝑥) = 𝑔(𝑥)
Solution:
From the graph the values of 𝑥 for which 𝑓(𝑥) = 𝑔(𝑥) are x = 0 and x = 4.
( The x – coordinates of the points of intersection of 𝑓(𝑥) 𝑎𝑛𝑑 𝑔(𝑥). )
2. (a)
2710 +910
R.T.Evaluate√ 274 +911
Solution:
2710 +910
√
274 +911
=√
(33 )10 +(32 )10
= √ (33 )4 +(32 )11
(330 +320 )
312 +322
320 (310 +1)
= √312 (1+310 )
320
= √312
= √38
= 34
= 81
(b)
(i)
R.T.P. log n m =
log10 m
, for 𝑚, 𝑛 ∈ 𝑁
log10 n
Proof:
If log n m = x then 𝑛 𝑥 = 𝑚 (definition of logs)
Taking lg
𝑙𝑔(𝑛 𝑥 ) = 𝑙𝑔𝑚
𝑥𝑙𝑔 𝑛 = 𝑙𝑔 𝑚
𝑥=
𝑙𝑔𝑚
𝑙𝑔𝑛
=
𝑙𝑜𝑔10 𝑚
𝑙𝑜𝑔10 𝑛
Hence, log n m =
log10 m
log10 n
Q.E.D.
(ii)
Data: y = (log 2 3)(log 3 4)(log 4 5)...(log 31 32)
R.T.C. The exact value of y
Calculation:
y = (log 2 3)(log 3 4)(log 4 5)...(log 31 32)
Re: log n m =
Hence, y =
log10 m
log10 n
log10 3 log10 4 log10 5
log10 32


 ... 
log10 2 log10 3 log10 4
log10 31
After cancelling
y=
log10 32
log10 2
=
log10 2 5
log10 2
=
5 log10 2
log10 2
=5x1
=5
(c)
R.T.P. By the Principle of Mathematical Induction, 𝑓(𝑛) = 7𝑛 − 1 is divisible by
6 , ∀𝑛 ∈ 𝑁.
Proof:
Assume that the statement true for 𝑛 = 𝑘
Then 𝑓(𝑘) = 7𝑘 − 1 is divisible by 6.
Therefore 7𝑘 − 1 = 6 × 𝑎 where 𝑎 ∈ 𝑁
For 𝑛 = 𝑘 + 1
𝑓(𝑘 + 1) = 7𝑘+1 − 1
= (7.7𝑘 ) − 1
Recall: 7𝑘 − 1 = 6𝑎
7𝑘 = 6𝑎 + 1
= 7(6𝑎 + 1) − 1
= 42𝑎 + 7 − 1
𝑓(𝑘 + 1) = 6(7𝑎 + 1)
(𝐼𝑓 𝑎 ∈ 𝑁, 𝑡ℎ𝑒𝑛 7𝑎 + 1 ∈ 𝑁)
Hence, 𝑓(𝑘 + 1), which is a multiple of 6 is therefore divisible by 6.
When 𝑛 = 1: 𝑓(1) = 71 − 1 = 6(1) and which is divisible by 6.
When 𝑛 = 2: 𝑓(2) = 72 − 1 = 6(8) which is divisible by 6.
Hence, by the Principle of Mathematical Induction, 𝑓(𝑛) = 7𝑛 − 1 is divisible by
6 ∀𝑛 ∈ 𝑁.
3.
(a)
Data: p and q are 2 vectors such that 𝑝 = 𝑖 − 𝑗 𝑎𝑛𝑑 𝑞 = 𝜆𝑖 + 2𝑗
R.T.F. The values of 𝜆 such that
(i)
q is parallel to p
Solution:
If p and q are parallel, then 𝑞 = 𝛼 × 𝑝 (𝛼 = 𝑎 𝑠𝑐𝑎𝑙𝑎𝑟)
∴ 𝜆𝑖 + 2𝑗 = 𝛼(𝑖 − 𝑗) = 𝛼𝑖 − 𝛼𝑗
Equate components
−𝛼 = 2
𝛼 = −2
and 𝜆 = 𝛼
∴ 𝜆 = −2
(ii)
q is perpendicular to p.
Solution:
If q is perpendicular to p then 𝑞. 𝑝 = 0
∴ (1 × 𝜆) + (−1 × 2) = 0
𝜆−2=0
∴𝜆=2
(iii)
𝜋
The angle between p and q is 3 .
Solution:
𝜋
Re: 𝑝. 𝑞 = |𝑝||𝑞| cos 3
1
(1 × 𝜆) + (−1 × 2) = √(1)2 + (−1)2 × √(𝜆)2 + (2)2 ×
2
1
𝜆 − 2 = √2√𝜆2 + 4 × 2
2𝜆 − 4 = √2(𝜆2 + 4)
Squaring both sides
4𝜆2 − 16𝜆 + 16 = 2𝜆2 + 8
2𝜆2 − 16𝜆 + 8 = 0
(÷ 2) 𝜆2 − 8𝜆 + 4 = 0
𝜆=
𝜆=
𝜆=
𝜆=
8±√64−16
2
8±√48
2
8±√16×3
2
8±4√3
2
𝜆 = 4 ± 2√3
(b)
1−𝑐𝑜𝑠2𝐴+𝑠𝑖𝑛2𝐴
R.T.P. 1+𝑐𝑜𝑠2𝐴+𝑠𝑖𝑛2𝐴 ≡ 𝑡𝑎𝑛𝐴
Proof:
Re: 𝑐𝑜𝑠2𝐴 = 2 cos 2 𝐴 − 1 = 1 − 2 sin2 𝐴
𝑠𝑖𝑛2𝐴 = 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴
Taking L.H.S.
1−𝑐𝑜𝑠2𝐴+𝑠𝑖𝑛2𝐴
1−(1−2 sin2 𝐴)+2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴
= 1+(2 cos2 𝐴−1)+2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴
1+𝑐𝑜𝑠2𝐴+𝑠𝑖𝑛2𝐴
2 sin2 𝐴+2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴
= 2 cos2 𝐴 +2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴
(2𝑠𝑖𝑛𝐴(𝑠𝑖𝑛𝐴+𝑐𝑜𝑠𝐴))
= (2𝑐𝑜𝑠𝐴(𝑐𝑜𝑠𝐴+𝑠𝑖𝑛𝐴))
𝑠𝑖𝑛𝐴
= 𝑐𝑜𝑠𝐴
= 𝑡𝑎𝑛𝐴
= R.H.S.
Q.E.D.
(c)
(i)
R.T.P. sin(𝑛 + 1) 𝜃 = 𝑡𝑠𝑖𝑛 𝑛𝜃 − sin(𝑛 − 1) 𝜃 𝑤ℎ𝑒𝑟𝑒 𝑡 = 2𝑐𝑜𝑠𝜃
Proof:
 (n + 1) + (n − 1)   (n + 1) − (n − 1) 
sin(𝑛 + 1) 𝜃 + sin(𝑛 − 1) 𝜃 = 2 sin
 cos

2
2

 

= 2 sin 𝑛𝜃 𝑐𝑜𝑠𝜃
(Factor Formula)
∴ sin(𝑛 + 1) 𝜃 + sin(𝑛 − 1) 𝜃 = 2 cos 𝜃 sin 𝑛𝜃
= 𝑡𝑠𝑖𝑛 𝑛𝜃where𝑡 = 2 cos 𝜃
and sin(𝑛 + 1) 𝜃 = 𝑡sin 𝑛𝜃 − sin(𝑛 − 1) 𝜃
Q.E.D.
(ii)
R.T.P. 𝑠𝑖𝑛3𝜃 = (𝑡 2 − 1)𝑠𝑖𝑛𝜃
Proof:
Re: sin(𝑛 + 1) 𝜃 = 𝑡𝑠𝑖𝑛 𝑛𝜃 − sin(𝑛 − 1) 𝜃
When 𝑛 = 2
𝑠𝑖𝑛3𝜃 = 𝑡𝑠𝑖𝑛2𝜃 − 𝑠𝑖𝑛𝜃
(𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃)
= 2𝑡𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃
= (2𝑐𝑜𝑠𝜃)𝑡(𝑠𝑖𝑛𝜃) − 𝑠𝑖𝑛𝜃
= 𝑡 2 𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛𝜃
(𝑡 = 2 cos 𝜃)
= (𝑡 2 − 1)𝑠𝑖𝑛𝜃
Q.E.D.
(iii)
R.T.F. All the solutions of 𝑠𝑖𝑛3𝜃 = 𝑠𝑖𝑛𝜃, 0 ≤ 𝜃 ≤ 𝜋
Solution:
𝑠𝑖𝑛3𝜃 = 𝑠𝑖𝑛𝜃
Re: 𝑠𝑖𝑛3𝜃 = 3𝑠𝑖𝑛𝜃 − 4 sin3 𝜃
Hence, 3𝑠𝑖𝑛𝜃 − 4 sin3 𝜃 = 𝑠𝑖𝑛𝜃
4 sin3 𝜃 − 2𝑠𝑖𝑛𝜃 = 0
2𝑠𝑖𝑛𝜃(2 sin2 𝜃 − 1) = 0
∴ 𝑠𝑖𝑛𝜃 = 0or±
1
√2
For 0 ≤ 𝜃 ≤ 𝜋
If 𝑠𝑖𝑛𝜃 = 0
then𝜃 = 0 𝑜𝑟 𝜋
If 𝑠𝑖𝑛𝜃 =
then𝜃 =
𝜋
4
If 𝑠𝑖𝑛𝜃 =
1
√2
or
3𝜋
4
−1
√2
then 𝜃 has NO solutions for the given range.
𝜋 3𝜋
Hence, 𝜃 = 0, 4 ,
4
,𝜋
Alternative Method
Re: 𝑠𝑖𝑛3𝜃 = (𝑡 2 − 1)𝑠𝑖𝑛𝜃
𝑠𝑖𝑛3𝜃 = 𝑠𝑖𝑛𝜃
∴ (𝑡 2 − 1)𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛𝜃 = 0
𝑠𝑖𝑛𝜃(𝑡 2 − 1 − 1) = 0
(𝑡 = 2 cos 𝜃)
𝑠𝑖𝑛𝜃((2𝑐𝑜𝑠𝜃)2 − 2) = 0
𝑠𝑖𝑛𝜃(4 cos2 𝜃 − 2) = 0
2𝑠𝑖𝑛𝜃(2 cos2 𝜃 − 1) = 0
∴ 2𝑠𝑖𝑛𝜃 = 0 or cos 𝜃 = ±
1
√2
For 0 ≤ 𝜃 ≤ 𝜋
when 2𝑠𝑖𝑛𝜃 = 0
𝜃 = 0, 𝜋
𝑐𝑜𝑠𝜃 =
𝜃=
√2
𝜋
4
𝑐𝑜𝑠𝜃 =
𝜃=
1
−1
√2
3𝜋
4
𝜋 3𝜋
Hence, 𝜃 = 0, 4 ,
4.
(a)
(i)
4
,𝜋
Data: The line 𝑥 − 2𝑦 + 4 = 0 cuts the circle
𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 = 0, with centre, P at A and B
R.T.F. the coordinates of P, A and B
Solution:
𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 = 0
(𝑥 − 1)2 + (𝑦 − 10)2 − 1 − 100 + 51 = 0
(𝑥 − 1)2 + (𝑦 − 10)2 = 50 = (5√2)
2
is of the form (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 which is a circle with centre
(a, b) and radius r.
Hence, (𝑥 − 1)2 + (𝑦 − 10)2 = 50 represents the equation of a circle
with centre (1, 10).
∴ P has coordinates (1, 10).
Solving 𝑥 − 2𝑦 + 4 = 0 and 𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 = 0
simultaneously, to find A and B.
𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 = 0
P
𝑥 − 2𝑦 + 4 = 0
B
A
Let 𝑥 − 2𝑦 + 4 = 0 … (1)
and 𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 = 0 … (2)
From (1) 𝑥 = 2𝑦 − 4
Substituting (1) into (2)
(2𝑦 − 4)2 + 𝑦 2 − 2(2𝑦 − 4) − 20𝑦 + 51 = 0
4𝑦 2 − 16𝑦 + 16 + 𝑦 2 − 4𝑦 + 8 − 20𝑦 + 51 = 0
5𝑦 2 − 40𝑦 + 75 = 0
𝑦 2 − 8𝑦 + 15 = 0
(𝑦 − 3)(𝑦 − 5) = 0
𝑦 = 3 𝑜𝑟 5
When 𝑦 = 3, 𝑥 = 2(3) − 4 = 2
When 𝑦 = 5, 𝑥 = 2(5) − 4 = 6
Points of intersection are (2, 3) and (6, 5)
Let the coordinates of A be (2, 3) and B (6, 5).
(ii)
𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 = 0
P (1, 10)
B (6, 5)
M
A (2, 3)
Q
𝑥 − 2𝑦 + 4 = 0
𝑥 2 + 𝑦 2 − 2𝑥 − 0𝑦 + 51 + 𝜆(𝑥 − 2𝑦 + 4) = 0
Data: 𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 + 𝜆(𝑥 − 2𝑦 + 4) = 0
P, A and B lie on the circle.
(a)
R.T.F. The value of 𝜆.
Solution
𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 + 𝜆(𝑥 − 2𝑦 + 4) = 0 … (3)
Since P (1, 10) lies on the circle, then 𝑥 = 1 and 𝑦 = 10 satisfies
the equation.
Sub 𝑥 = 1, 𝑦 = 10 𝑖𝑛 (3)
(1)2 + (10)2 − 2(1) − 20(10) + 51 + 𝜆(1 − 2(10) + 4) = 0
1 + 100 − 2 − 200 + 51 + 𝜆(−15) = 0
−50 − 15𝜆 = 0
50
𝜆 = − 15 = −
(b)
10
3
R.T.F. The equation of the circle C.
Solution:
50
10
Substituting 𝜆 = − 15 = − 3 , the equation of C is
10
𝑥 2 + 𝑦 2 − 2𝑥 − 20𝑦 + 51 −
3
(𝑥 − 2𝑦 + 4) = 0
3𝑥 2 + 3𝑦 2 − 6𝑥 − 60𝑦 + 153 − 10𝑥 + 20𝑦 − 40 = 0
3𝑥 2 + 3𝑦 2 − 16𝑥 − 40𝑦 + 113 = 0
𝑥2 + 𝑦2 −
(c)
16
3
𝑥−
40
3
𝑦+
113
=0
3
R.T.F. The distance |𝑃𝑄|
Solution:
𝑥2 + 𝑦2 −
16
3
𝑥−
8 2
(𝑥 − 3) + (𝑦 −
8 2
(𝑥 − 3) + (𝑦 −
40
3
𝑦+
20 2
3
) +
113
=0
3
113
20 2
3
5√5
) =(
3
64
−
9
−
400
9
=0
2
3
)
8 20
∴Centre of C, Q is(3 , 3 )
2
2
8
20
|PQ|=√( − 1) + ( − 1)
3
3
5 2
10 2
= √(− 3) + (− 3 )
25
=√9 +
=√
(d)
125
9
100
=
9
5√5
3
𝑢𝑛𝑖𝑡𝑠
R.T.F. The distance |PM|if PQ cuts AB at M.
Solution:
A is (2, 3) and B is (6, 5).
5−3
2
𝑦−3
1
1
Gradient of AB = 6−2 = 4 = 2
Equation of AB is 𝑥−2 = 2
2𝑦 − 6 = 𝑥 − 2
2𝑦 = 𝑥 + 4
8 20
P = (1, 10) and Q = (3 , 3 )
Gradient of PQ =
20
−10
3
8
−1
3
=
10
3
5
3
−
= −2
𝑦−10
Equation of PQ is 𝑥−1 = −2
𝑦 − 10 = −2𝑥 + 2
𝑦 = −2𝑥 + 12
Let 2𝑦 = 𝑥 + 4 … (1) and 𝑦 = −2𝑥 + 12 … (2)
Solving simultaneously to find M, the point of intersection of PQ
and AB.
Sub (2) into (1)
2(−2𝑥 + 12) = 𝑥 + 4
−4𝑥 + 24 = 𝑥 + 4
20 = 5𝑥
4=𝑥
𝑦 = −2(4) + 12
𝑦=4
M has coordinates (4, 4)
|𝑃𝑀| = √(4 − 1)2 + (4 − 10)2
= √(3)2 + (−6)2
= √9 + 36
= √45
= 3√5 units
(b)
Data: 𝑥 = 2 + 3𝑠𝑖𝑛𝑡 𝑦 = 3 + 4𝑐𝑜𝑠𝑡
(i)
R.T.P. The Cartesian equation of the curve is
(𝑥−2)2
9
+
(𝑦−3)2
16
=1
Solution:
𝑥 = 2 + 3𝑠𝑖𝑛𝑡 𝑦 = 3 + 4𝑐𝑜𝑠𝑡
Hence, 𝑠𝑖𝑛𝑡 =
𝑥−2
3
and cos 𝑡 =
Re: sin2 𝑡 + cos2 𝑡 = 1
∴(
𝑥−2 2
3
4
∀𝑡
𝑦−3 2
) +(
(𝑥−2)2
Hence,
𝑦−3
9
4
+
) =1
(𝑦−3)2
16
=1
Q.E.D.
(ii)
R.T.P. That every point on the curve
(𝑥−2)2
9
+
(𝑦−3)2
16
= 1, lies on or within
the circle(𝑥 − 2)2 + (𝑦 − 3)2 = 25
Proof:
(𝑥 − 2)2 + (𝑦 − 3)2 = 25 is a circle and is of the form
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑐 2
∴ Centre of this circle is (2, 3) and radius √25 = 5units.
(𝑥)2
9
𝑥2
9
+
(𝑦)2
𝑦2
16
= 1 is an ellipse
+ 16 = 1 𝑜𝑟
𝑥2
𝑦2
+ 42 = 1 is an ellipse with centre (0, 0) cutting the
32
x – axis at (3,0) and (-3 ,0) and y- axis at (4, 0) and ( -4, 0)
y
4
x
3
-3
-4
2
𝑇= ( )
3
𝑥2
9
(𝑥−2)2
32
+
𝑦2
(𝑥−2)2
+ 16 = 1
(𝑦−3)2
42
= 1 𝑖𝑠
9
𝑥2
32
+
(𝑦−3)2
16
=1
𝑦2
2
+ 42 = 1,translated by 𝑇 = ( ).
3
(Note that the centre of the ellipse is now the same as the centre of the circle and
is the point, (2, 3).
We draw both the circle and the ellipse on the same axes
(2, 8)
(𝑥−2)2
(𝑥 − 2)2 + (𝑦 − 3)2 = 25
9
(2, 7)
4
(-3, 3)
3
(-1, 3)
(7, 3)
(5, 3)
(2, 3)
(2,-1)
(2, -2)
+
(𝑦−3)2
16
=1
Hence, all the points of
(𝑥−2)2
9
+
(𝑦−3)2
= 1lie within (𝑥 − 2)2 + (𝑦 − 3)2 = 25 only, as clearly
16
illustrated in the above diagram.
Q.E.D.
5.
(a)
sin 4 x
x →0 sin 5 x
R.T.F. lim
Solution:
lim
x →0
sin 4 x sin 4(0) 0
=
= and which is indeterminate
sin 5 x sin 5(0) 0
Re: L’Hospital’s Rule
lim
x →a
f ( x)
f ( x)
= lim
g ( x) x→0 g ( x)
Hence, lim
x →0
sin 4 x
4 cos 4 x
= lim
x
→
0
sin 5 x
5 cos 5 x
=
4
cos 4 x
lim
5 x→0 cos 5 x
=
4  cos 4(0) 


5  cos 5(0) 
4
1
=5×1
4
=5
(b)
𝑥
Data: 𝑦 = 1−4𝑥
(i)
(a)
𝑑𝑦
R.T.F. 𝑑𝑥
Solution:
𝑥
𝑢
𝑦 = 1−4𝑥 is of the form 𝑦 = 𝑣 where
𝑢 = 𝑥,
𝑑𝑢
𝑑𝑥
=1
𝑑𝑣
𝑣 = 1 − 4𝑥,
and
𝑑𝑦
𝑑𝑥
𝑑𝑦
and 𝑑𝑥 =
=
𝑑𝑥
= −4
(1−4𝑥).1−(−4)(−𝑥)
(Quotient Law)
(1−4𝑥)2
1−4𝑥+4𝑥
(1−4𝑥)2
1
= (1−4𝑥)2
𝑑𝑦
R.T.P.𝑥 2 (𝑑𝑥 ) = 𝑦 2
(b)
Proof:
Taking L.H.S.
𝑑𝑦
1
𝑥 2 (𝑑𝑥 ) = 𝑥 2 × ((1−4𝑥)2 )
𝑥2
= (1−4𝑥)2
𝑥
2
= (1−4𝑥)
= 𝑦2
= R.H.S.
Q.E.D.
(ii)
𝑑2 𝑦
𝑑𝑦
R.T.P. 𝑥 2 (𝑑𝑥 2 ) + 2(𝑥 − 𝑦) (𝑑𝑥 ) = 0
Proof:
𝑑𝑦
Re: 𝑥 2 (𝑑𝑥 ) = 𝑦 2 (from (b))
Differentiating implicitly w.r.t x
𝑑2 𝑦
𝑑𝑦
𝑑𝑦
2𝑥 (𝑑𝑥 ) + 𝑥 2 (𝑑𝑥 2 ) = 2𝑦 (𝑑𝑥 )
𝑑2 𝑦
𝑑𝑦
𝑑𝑦
∴ 𝑥 2 (𝑑𝑥 2 ) + 2𝑥 (𝑑𝑥 ) − 2𝑦 (𝑑𝑥 ) = 0
𝑑2 𝑦
𝑑𝑦
and 𝑥 2 (𝑑𝑥 2 ) + (2𝑥 − 2𝑦) (𝑑𝑥 ) = 0
𝑑2 𝑦
𝑑𝑦
and 𝑥 2 (𝑑𝑥 2 ) + 2(𝑥 − 𝑦) (𝑑𝑥 ) = 0
Q.E.D.
(Product Law)
ALTERNATIVE METHOD
𝑑𝑦
𝑑𝑥
1
= (1−4𝑥)2 = (1 − 4𝑥)−2
𝑑2 𝑦
𝑑𝑥 2
= (−2)(1 − 4𝑥)−3 (−4)
(Chain Rule)
8
= (1−4𝑥)3
𝑑2 𝑦
𝑑𝑦
∴ 𝑥 2 (𝑑𝑥 2 ) + 2(𝑥 − 𝑦) (𝑑𝑥 )
8
= 𝑥 2 × (1−4𝑥)3 + 2 (𝑥 −
8𝑥 2
2𝑥
1
1−4𝑥
1
) ((1−4𝑥)2 )
2𝑥
= (1−4𝑥)3 + (1−4𝑥)2 − (1−4𝑥)3
8𝑥 2 +2𝑥(1−4𝑥)−2𝑥
(1−4𝑥)3
=
8𝑥 2 +2𝑥−8𝑥 2 −2𝑥
(1−4𝑥)3
0
= (1−4𝑥)3
=0
Q.E.D
(c)
Open lid
h cm
3x cm
2x cm
Data: Total surface area of the box is 200𝑐𝑚3
(i)
R.T.P. ℎ =
20
𝑥
−
3𝑥
5
Proof:
Surface area of box = (Area of base) + (Area of front and back faces) +
+ (Area of right and left faces)
= 1. (2𝑥)(3𝑥) + 2. (2𝑥)(ℎ) + 2. (3𝑥)(ℎ)
= 6𝑥 2 + 4ℎ𝑥 + 6ℎ𝑥
= 6𝑥 2 + 10ℎ𝑥
Hence, 6𝑥 2 + 10ℎ𝑥 = 200
(200−6𝑥 2 )
ℎ=
10𝑥
6𝑥 2
200
ℎ = 10𝑥 − 10𝑥
∴ℎ=
20
𝑥
−
3𝑥
5
Q.E.D.
(ii)
R.T.F. h when its volume is a maximum.
Solution:
Volume of box = (2𝑥)(3𝑥)ℎ
20
= 6𝑥 2 ( 𝑥 −
𝑉 = 120𝑥 −
18
5
𝑥3
V is a function of x only.
𝑑𝑉
At stationary value 𝑑𝑥 = 0
𝑑𝑉
𝑑𝑥
= 120 −
18
5
(3𝑥 2 )
𝑑𝑉
Let 𝑑𝑥 = 0
120 =
18
5
(3𝑥 2 )
5
1
𝑥 2 = 120 (18) (3)
3𝑥
5
)
(in terms of x only).
=
100
9
𝑥=±
10
3
Since x > 0, 𝑥 =
𝑑2 𝑉
𝑑𝑥 2
=−
18
5
When 𝑥 =
andℎ =
6.
(a)
20
10
3
10
3
, only
(3)(2𝑥)
10
,
3
𝑑2 𝑉
𝑑𝑥 2
10
3
3( )
−
5
< 0  V is maximum at 𝑥 =
10
3
.
= 6 − 2 = 4 𝑐𝑚
Data: 𝑢 = 3𝑥 2 + 1
R.T.F.∫
𝑥
(√3𝑥 2 +1)
𝑑𝑥
Solution:
When 𝑢 = 3𝑥 2 + 1
𝑑𝑢
𝑑𝑥
= 6𝑥 and 𝑑𝑥 =
∴∫
𝑥
(√3𝑥 2 +1)
𝑑𝑢
6𝑥
𝑑𝑥 ≡ ∫
𝑥 𝑑𝑢
√𝑢 6𝑥
1
=∫
−
𝑢 2
6
𝑑𝑢
1
=
=
𝑢2
1
2
6( )
√𝑢
3
+𝑐
(c = the constant of integration)
+𝑐
Re-substituting
𝑥
∫ √3𝑥 2 +1 =
(b)
√3𝑥 2 +1
3
+𝑐
Data: A curve, C, passes through ( 3, -1) and has gradient function 𝑥 2 − 4𝑥 + 3
R.T.F. The equation of C.
Solution:
𝑑𝑦
Let the gradient function be 𝑑𝑥
𝑑𝑦
So 𝑑𝑥 = 𝑥 2 − 4𝑥 + 3
Equation of the curve is 𝑦 = ∫(𝑥 2 − 4𝑥 + 3)𝑑𝑥
=
𝑥3
−
3
4𝑥 2
2
+ 3𝑥 + 𝑘 (k = constant)
The point, (3, -1) lies on the curve.
∴ 𝑥 = 3 𝑎𝑛𝑑 𝑦 = −1 satisfies the equation and
and−1 =
(3)3
3
−
4(3)2
2
+ 3(3) + 𝑘
k= −1
Hence, equation of C is =
(c)
(i)
𝑥3
3
−
4𝑥 2
2
+ 3𝑥 − 1
R.T.F. the coordinates of A, B and C.
Solution:
At B, 𝑥 = 0 in the equation of the straight line 𝑦 + 2𝑥 = 5
i.e. y = 5. Hence, B has coordinates (0, 5)
At C, y = 0 in the equation of the curve𝑦 = 𝑥(4 − 𝑥)
𝑥 = 0 𝑜𝑟 4
𝑥 = 0at the origin, O. ∴ 𝑥 = 4 at C.
C has coordinates (4, 0).
Solving simultaneously, to find A
Let 𝑦 + 2𝑥 = 5 … (1)
And 𝑦 = 𝑥(4 − 𝑥) … (2)
From (1) 𝑦 = 5 − 2𝑥
Substituting into (2)
5 − 2𝑥 = 𝑥(4 − 𝑥)
5 − 2𝑥 = 4𝑥 − 𝑥 2
𝑥 2 − 6𝑥 + 5 = 0
(𝑥 − 1)(𝑥 − 5) = 0
𝑥 = 1 𝑜𝑟 5
A lies between 𝑥 = 0 𝑎𝑛𝑑 𝑥 = 4
∴ 𝑥 = 1 at A.
When 𝑥 = 1, 𝑦 = 5 − 2(1) = 3
Hence, A has coordinates (1, 3).
(ii)
R.T.F. The exact value of the shaded region.
Solution:
y
5
𝑦 + 2𝑥 = 5
(1, 3)
A1
𝑦 = 𝑥(4 − 𝑥)
A2
x
O
1
4
The shaded region is divided into regions A1 and A2 as shown on the
diagram.
1
The area of A1= 2 (5 + 3)(1)
= 4 units2
4
The area of A2= ∫1 {𝑥(4 − 𝑥)} 𝑑𝑥
4

x3 
= 2 x 2 − 
3 1

= (2(4)2 −
1
(4)3
3
) − (2(1)2 − (1)2 −
(1)3
3
1
= 32 − 21 3 − 2 + 3
= 9 units2
∴ Area of the shaded region in the given diagram= 4 + 9
= 13 units2 (exact)
)
JUNE 2008 UNIT 1 PAPER 2
(REST OF THE CARIBBEAN ONLY)
1.
(a)
Data: The roots of the cubic equation 𝑥 3 + 3𝑝𝑥 2 + 𝑞𝑥 + 𝑟 = 0 are -1, 1 and 3.
R.T.F. The values of p, q and r
Solution:
Remainder and Factor Theorem – When a polynomial f(x) is divided by
(x - a), the remainder is f(a). If f(a) = 0 then (x – a) is a factor of f(x).
Let 𝑓(𝑥) = 𝑥 3 + 3𝑝𝑥 2 + 𝑞𝑥 + 𝑟
Since the roots are – 1, 1 and 3 then 𝑓(𝑥) = (𝑥 + 1)(𝑥 − 1)(𝑥 − 3)
Expanding, 𝑓(𝑥) = (𝑥 2 − 1)(𝑥 − 3)
= 𝑥 3 − 3𝑥 2 − 𝑥 + 3
Equating constants, r = 3
Equating coefficient of 𝑥: q = - 1
Equating coefficient of 𝑥 2 : 3𝑝 = −3
∴ 𝑝 = −1
Hence, 𝑝 = −1 ∈ 𝑅, 𝑞 = −1 ∈ 𝑅 𝑎𝑛𝑑 𝑟 = 3 ∈ 𝑅
(b)
(i)
R.T.P:
√6+√2
√6−√2
= 2 + √3
Proof:
Taking, L.H.S
Rationalising to get
√6+√2
√6−√2
=
=
√6+√2
√6+√2
×
√6−√2
√6+√2
6+2√12+2
6−2
=
8+2√12
4
=
8+4√3
4
= 2 + √3
Q.E.D
(ii)
R.T.P:
√6+√2
√6−√2
+
√6−√2
√6+√2
=4
Proof:
√6+√2
√6−√2
+
√6−√2
√6+√2
2
2
(√6 + √2) + (√6 − √2)
=
4
=
6+2+2√12+6+2−2√12
4
=
16
=4
4
Q.E.D
n
(c)
(i)
R.T.P:
1
 r (r + 1) = 3 n(n + 1)(n + 2)
r =1
Proof:
Using the method of induction
Assume the statement true for n = k
k
1
i.e.  r (r + 1) = k (k + 1)(k + 2)
3
r =1
Consider n = k + 1
k +1
k
r =1
r =1
 r (r + 1) =  r (r + 1) + (k + 1) th term
1
= k (k + 1)(k + 2) + (k + 1)(k + 2)
3
1

= (k + 1)( k + 2) k + 1
3

 k + 3
= (k + 1)( k + 2)

 3 
1
= (k + 1)(k + 2)(k + 3)
3
We re-write the expression to note that k is replaced by k+1
=
1
( k + 1)([k + 1] + 1)([k + 1] + 2) )
3
Hence, the statement is true for n = k + 1
Consider n = 1
1
L.H.S. :  r (r + 1) = 1(1 + 1) = 2
r =1
R.H.S.:
1
1
(1)(1 + 1)(1 + 2) = (6) = 2
3
3
The statement is true for n = 1
Consider, n = 2
2
L.H.S. :  r (r + 1) = 2(2 + 1) = 6
r =1
And 6 + 2 = 8
R.H.S.:
1
1
(2)(2 + 1)(2 + 2) = (24) = 8
3
3
The statement is true for n = 2
Hence, by the Principle of Mathematical Induction
n
1
 r (r + 1) = 3 n(n + 1)(n + 2) is true n  Z
+
r =1
Alternative Method
Taking, L.H.S
n
n
n
r =1
r =1
r =1
 r (r + 1) =  r 2 +  r
n
Recall:
r
2
=
r =1
n
(n + 1)(2n + 1) and
6
n
Hence,
n
n
 r = 2 (n + 1)
r =1
n
n
 r (r + 1) = 6 (n + 1)(2n + 1) + 2 (n + 1)
r =1
=
n
1

(n + 1)  (2n + 1) + 1
2
3

=
n
 2n + 1 + 3 
(n + 1) 

2
3


=
n
 2n + 4 
( n + 1) 

2
 3 
=
n
(2)(n + 1)(n + 2)
6
=
n(n + 1)(n + 2)
3
Q.E.D.
50
(ii)
R.T.C:  r (r + 1)
r =31
Calculation:
50
50
30
r = 31
r =1
r =1
 r (r + 1) =  r (r + 1) −  r (r + 1)
1
1
= (50)(51)(52) − (30)(31)(32)
3
3
= 44 200 – 9 920
= 34 280
2.
(a)
Data: The roots of the equation 2𝑥 2 + 4𝑥 + 5 = 0 are 𝛼 and 𝛽.
(i)
R.T.F: the values of 𝛼 + 𝛽and 𝛼𝛽.
Solution:
Recall: If 𝛼 and 𝛽 are the roots of 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 then
𝑏
𝑐
(𝑥 − 𝛼)(𝑥 − 𝛽) = 𝑥 2 + 𝑥 +
𝑎
𝑎
Equating coefficients:  +  =
−b
c
and  =
a
a
Hence, if 𝛼 and 𝛽 are the roots of 2𝑥 2 + 4𝑥 + 5 = 0 then
 + =
− (4)
= −2
2
and  =
(ii)
(a)
5
2
R.T.C. The value of 𝛼 2 + 𝛽 2
Calculation:
(𝛼 + 𝛽)2 = 𝛼 2 + 2𝛼𝛽 + 𝛽 2
∴ 𝛼 2 + 𝛽 2 = (𝛼 + 𝛽 )2 − 2𝛼𝛽
5
= ( −2) 2 − 2 
2
=4–5
=-1
(b)
R.T.C. The value of 𝛼 3 + 𝛽 3
Calculation:
(𝛼 + 𝛽)3 = 𝛼 3 + 3𝛼 2 𝛽 + 3𝛼𝛽 2 + 𝛽 3
∴ 𝛼 3 + 𝛽 3 = (𝛼 + 𝛽)3 − 3𝛼 2 𝛽 − 3𝛼𝛽 2
∴ 𝛼 3 + 𝛽 3 = (𝛼 + 𝛽)3 − 3𝛼𝛽(𝛼 + 𝛽)
5
= (−2) 3 − 3 (−2)
2
= - 8 +15
= 7
(iii)
R.T.F. The quadratic equation whose roots are 𝛼 3 𝑎𝑛𝑑 𝛽 3 .
Solution:
A quadratic is of the form
x 2 - (The sum of the roots) x + (the product of the roots) = 0
Hence, the required equation is
𝑥 2 − (𝛼 3 + 𝛽 3 )𝑥 + 𝛼 3 𝛽 3 = 0
3
5
x − 7x +   = 0
2
2
 125 
x2 − 7x + 
=0
 8 
Such an equation is best expressed in integral form
(x 8)
8 x 2 − 56 x + 125 = 0 (in integral form)
(b)
(i)
1
3
Data: x − 4 x
−
1
3
=3
R.T.F.The value of x.
Solution:
1
x 3 − 4x
−
1
3
1
=3
4
and x 3 −
x
1
3
=3
1
Let u = x 3  u −
(x u)
4
=3
u
u 2 − 4 = 3u
u 2 − 3u − 4 = 0
(u − 4)(u + 1) = 0
𝑢 = 4 𝑜𝑟 𝑢 = −1
When u = 4,
1
x3 = 4
𝑥 = (4)3
= 64
And when u = -1
1
x 3 = −1
𝑥 = (−1)3
= -1
Hence, x = -1 or 64
(It is always wise, in questions of this type, to substitute these values obtained, to ensure
that they are valid for the given equation)
(ii)
Data: log 5 ( x + 3) + log 5 ( x − 1) = 1
R.T.F. The value of x
Solution:
log 5 ( x + 3) + log 5 ( x − 1) = 1
log 5 {( x + 3)( x − 1)} = 1
(Product Law)
From the definition of logs
( x + 3)( x − 1) = 51
∴ 𝑥 2 + 2𝑥 − 3 = 5
𝑥 2 + 2𝑥 − 8 = 0
(𝑥 + 4)(𝑥 − 2) = 0
𝑥 = −4 𝑜𝑟 𝑥 = 2
When x = - 4, then x is log 5 (-4), which does not exist.
Hence x  −4 .
and x = 2 only.
(iii)
R.T.C. The value of
1
2
3
8
9
log10   + log10   + log10   + ... + log10   + log10  
2
3
4
9
 10 
Solution:
1
2
3
8
9
log10   + log10   + log10   + ... + log10   + log10  
2
3
4
9
 10 
1 2 3 4 5 6 7 8 9 
= log10  . . . . . . . . 
 2 3 4 5 6 7 8 9 10 
1
= log10  
 10 
= log10 10 −1
(Product Law)
= −1(log10 10) (Power Law)
= -1
3.
(a)
Data:𝑦 = 3𝑥 + 4 and 4𝑦 = 3𝑥 + 5 are inclined at angles 𝛼 and 𝛽.
(i)
R.T.State The values of tan 𝛼 and tan 𝛽.
Solution:
𝑦 = 3𝑥 + 4 is of the form y = mx + c where m1 = 3 is the gradient.
 tan  = 3
4𝑦 = 3𝑥 + 5
(÷ 4) y =
3
3
5
is the gradient.
x + is of the form y = mx + c where m2 =
4
4
4
 tan  =
(ii)
3
4
R.T.F. The angle between the two lines.
Solution:
tan 𝛼 = m1
tan 𝛽= m2
𝑚1 = 3
𝛼
𝑚2 =
3
4
𝜃
𝛽
Let  be the difference between the 2 angles.
𝑥
 = −
 tan  = tan( −  )
(Using the compound angle formula)
=
m1 − m2
1 + m1 m2
9
3
4 = 4 = 9
=
9 13
3
1+
1 + 3 
4
4
3−
(b)
(i)
R.T.P.𝑠𝑖𝑛2𝜃 − 𝑡𝑎𝑛𝜃𝑐𝑜𝑠2𝜃 = 𝑡𝑎𝑛𝜃
Proof:
Taking L.H.S
Re: 𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 (double angle formula)
𝑐𝑜𝑠2𝜃 = 2 cos2 𝜃 − 1
and tan  =
sin 
cos 
𝑠𝑖𝑛2𝜃 − 𝑡𝑎𝑛𝜃𝑐𝑜𝑠2𝜃 = 2 sin  cos  −
sin 
(2 cos 2  − 1)
cos 
= 2 sin  cos  − 2 sin  cos  +
=
sin 
cos 
= tan 
= R.H.S.
And L. H. S. = R.H.S.
Q.E.D
sin 
cos 
(ii)
R.T.Express tan  in terms of 𝑠𝑖𝑛2𝜃 𝑎𝑛𝑑 𝑐𝑜𝑠2𝜃
Solution:
𝑠𝑖𝑛2𝜃 − 𝑡𝑎𝑛𝜃𝑐𝑜𝑠2𝜃 = 𝑡𝑎𝑛𝜃 (b(i))
𝑠𝑖𝑛2𝜃 = 𝑡𝑎𝑛𝜃𝑐𝑜𝑠2𝜃 + 𝑡𝑎𝑛𝜃
𝑠𝑖𝑛2𝜃 = 𝑡𝑎𝑛𝜃(1 + 𝑐𝑜𝑠2𝜃)
sin 2
1 + cos 2
∴ 𝑡𝑎𝑛𝜃 =
(iii)
1
R.T.P. tan 22 2 ° = √2 − 1
Proof:Using the data from the previous part of the question
Re: 𝑡𝑎𝑛𝜃 =
Let  = 22
1
2
sin 2
1 + cos 2
0
1
sin 2(22 ) 0
1
2
𝑡𝑎𝑛22 2 ° =
1
1 + cos 2(22 ) 0
2
=
sin 45 0
1 + cos 45 0
Re: cos 45 0 = sin 45 0 =
1
and𝑡𝑎𝑛22 2 ° =
1
=
1+
2
1
2
1
2
sin 45 0
1 + cos 45 0
1
=
2
2 +1
2
 2

 

2


=
1
2 +1
=
(Rationalising)
=
1
2 +1

2 −1
2 −1
2 −1
1
= 2 −1
Q.E.D
(c)
(i)
(a)
R.T.P. sin
A+ B
C
= cos
2
2
Proof:
A
B
C
A + B + C = 1800 (sum of the angles of a triangle = 1800)
A + B = 1800 – C
A+ B
C
= 90 0 −
2
2
 A+ B
 0 C
sin
 = sin 90 − 
2
 2 

Re: sin(90 0 −  ) = cos 
C
C

 sin 90 0 −  = cos
2
2

Hence, sin
A+ B
C
= cos
2
2
Q.E.D.
R.T.P. sin B + sin C = 2 cos
(b)
A
B−C
cos
2
2
Proof:
sin B + sin C = 2 sin
B+C
B−C
cos
2
2
(Factor Formula)
B + C = 1800 – A
B+C
A
= 90 0 −
2
2
A
B+C
 0 A
sin
 = sin 90 −  = cos
2
2
 2 

Hence, sin B + sin C = 2 cos
A
B−C
cos
2
2
Q.E.D.
(ii)
R.T.P. sin A + sin B + sin C = 4 cos
A
B
C
cos cos
2
2
2
Proof:
Taking L.H.S
sin A + sin B + sin C = 2 sin
= 2 cos
A
A
A
B−C
cos + 2 cos cos
2
2
2
2
A
A
B−C 
 sin + cos

2
2
2 
A + B + C = 1800
A = 1800 – (B + C)
(B + C )
A
= 90 0 −
2
2
(B + C ) 
 A

sin  = sin 90 0 −

2 
2

B+C
= cos

 2 
Hence, sin A + sin B + sin C = 2 cos
(By the Factor Formula)
A
B+C
B−C 
+ cos
 cos

2
2
2 
B+C B−C
B+C B−C

+
−

A
2 cos 2
2
= 2 cos  2 cos 2
2
2
2


= 2 cos






A
B
C
cos cos
2
2
2
Q.E.D
4.
(a)
Data: P (-2, 0) and Q( 8, 8). M is the midpoint of PQ.
(i)
R.T.F. The equation of the line through M and perpendicular to PQ
Solution:
8− 2 0+8
The midpoint of PQ = 
,

2 
 2
= (3, 4)
Gradient of PQ =
8−0
8 4
=
=
8 − (−2) 10 5
Gradient of line perpendicular to PQ = −
5
4
(Product of gradients of perpendicular lines = -1)
 The equation of the required line is
y−4=
−5
( x − 3)
4
y−4=
−5
15
x+
4
4
y=
−5
31
x+
4
4
( × 4) 4 y = −5 x + 31
(ii)
R.T.F. The coordinates of the centre of the circle passing through P, O
and Q.
Solution:
Let the equation of the circle be
𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0 with centre (-g, -f) and where g, f and c are
constants.
If (-2, 0) lies on the circle then
(−2)2 + (0)2 + 2𝑔(−2) + 2𝑓(0) + 𝑐 = 0
4 − 4𝑔 + 𝑐 = 0 ...(1)
Also (0, 0) lies on the circle
Hence, (0)2 + (0)2 + 2𝑔(0) + 2𝑓(0) + 𝑐 = 0
c = 0 ...(2)
Substitute (2) into (1)
4 − 4𝑔 = 0
g=1
(8, 8) lies on the circle
(8)2 + (8)2 + 2𝑔(8) + 2𝑓(8) + 𝑐 = 0
128 + 16+ 16f = 0
16f = -144
f = -9
Hence, centre of the circle is (-1, 9).
ALTERNATIVE METHOD
y
C (-g, -f)
Q (8,8)
M
x
P (-2,0)
O
4𝑦 = −5𝑥 + 1
Gradient of CM =
4+ f −5
=
3+ g
4
Equating numerator
4 + 𝑓 = −5
f = −9
Equating denominator
3+𝑔=4
𝑔=1
Hence, the centre is (-(1), -(-9)) = (-1, 9)
(b)
(i)
R.T.P. 𝑦 = 𝑥 + 1 is a tangent to the circle,
𝑥 2 + 𝑦 2 + 10𝑥 − 12𝑦 + 11 = 0
Proof:
Let 𝑦 = 𝑥 + 1...(1)
and 𝑥 2 + 𝑦 2 + 10𝑥 − 12𝑦 + 11 = 0 ...(2)
If the line (1) is a tangent to the circle (2) then if solved simultaneously
there should be ONLY 1 real solution.
Substituting (1) into (2)
𝑥 2 + (𝑥 + 1)2 + 10𝑥 − 12(𝑥 + 1) + 11 = 0
𝑥 2 + 𝑥 2 + 2𝑥 + 1 + 10𝑥 − 12𝑥 − 12 + 11 = 0
2𝑥 2 + 12𝑥 − 12𝑥 − 12 + 12 = 0
𝑥2 = 0
and 𝑥 = 0 only
Hence, 𝑦 = 𝑥 + 1is a tangent to the circle
𝑥 2 + 𝑦 2 + 10𝑥 − 12𝑦 + 11 = 0 at x = 0.
(ii)
R.T.F. The coordinates of the point of contact.
Solution:
When x = 0, y = 1
Hence, point of contact is (0, 1).
5.
(a)
R.T.F. lim
x →3
Solution:
x 3 − 27
x 2 + x − 12
Let f ( x) =
f (3) =
x 3 − 27
x 2 + x − 12
33 − 27
0
= (Which is indeterminate)
2
3 + 3 − 12 0
Using factorising and cancelling method for this indeterminate form
𝑥 2 + 𝑥 − 12 = (𝑥 + 4)(𝑥 − 3).
Clearly (x + 4) is clearly not factor of 𝑥 3 − 27 ( whose factors cannot be readily
obtained)
We deduce that since the indeterminate form is
0
and factorising and cancelling is
0
the method to be used. We suggest that (x - 3) is a likely factor of 𝑥 3 − 27.
𝑥 2 + 3𝑥 + 9
(x – 3)
𝑥 3 − 27
-𝑥 3 − 3𝑥 2
3𝑥 2 − 27
3𝑥 2 − 9𝑥
9𝑥 − 27
9𝑥 − 27
0
Hence, lim
x →3
x 3 − 27
( x − 3)( x 2 + 3x + 9)
=
lim
x 2 + x − 12 x→3 ( x − 3)( x + 4)
= lim
x →3
=
x 2 + 3x + 9
x+4
3 2 + 3(3) + 9 27
=
3+ 4
7
ALTERNATIVE METHOD:
L’Hospital’s Rule:
If
g ( x)
f ( x) =
h ( x)
then,
g ( a )
h ( a )
lim x →a f ( x ) =
f ( x) =
x 3 − 27
x 2 + x − 12
g ( x ) = x 3 − 27
h ( x ) = x 2 + x − 12
g  ( x ) = 3x 2
h ( x ) = 2 x + 1
g  ( 3) = 3 ( 2 )
h ( 3 ) = 2 ( 3 ) + 1
2
=7
= 27
lim x→3
(b)
x3 − 27
27
=
2
x + x − 12 7
u
+ vt 2 , where u and v are constants.
t
1
dP
When t = 1, P = - 1.
= −5 at t =
2
dt
R.T.F. The value of u and of v
Solution:
u
P = + vt 2
t
Data: P =
P = -1 at t = 1
i.e. − 1 =
u
+ v(1) 2
1
u + v = −1
...(1)
𝑃 = 𝑢𝑡 −1 + 𝑣𝑡 2
dP
= u (−t −2 ) + v(2t )
dt
dP − u
= 2 + 2vt
dt
t
1
dP
= −5 at t =
2
dt
−5 =
−u
1
+ 2 v
2
1
 
2
2
− 5 = −4u + v ...(2)
Solving (1) and (2) simultaneously
(1) + (2) 5𝑢 = 4
4
u=
5
Substituting into (1)
v=−
Hence, P =
(c)
Data:
(i)
4
+ v = −1
5
9
5
4 9 2
+ t
5t 5
dy
= 3x 2 − 6 x and (-1, 0) is a point on the curve.
dx
R.T.F. The equation of the curve
Solution:
dy
= 3x 2 − 6 x
dx
y =  (3x 2 − 6 x)dx
y = x 3 − 3x 2 + c
When x = -1, y = 0
 0 = (−1) 3 − 3(−1) 2 + c
0 = −1 − 3 + c
c=4
y = x 3 − 3x 2 + 4
(ii)
R.T.F. The coordinates and nature of the stationary points.
Solution:
y = x 3 − 3x 2 + 4
dy
= 3x 2 − 6 x
dx
Stationary points occur when
dy
=0
dx
Let 3𝑥 2 − 6𝑥 = 0
3𝑥(𝑥 − 2) = 0
𝑥 = 0 𝑜𝑟 𝑥 = 2
When 𝑥 = 0, 𝑦 = (0)3 − 3(0)2 + 4 = 4
When 𝑥 = 2, 𝑦 = (2)3 − 3(2)2 + 4 = 0
Stationary points are (0, 4) and (2, 0)
d2y
= 6x − 6
dx 2
At x = 0
d2y
= 6(0) − 6 = −6  0 ⇒ (0, 4) is a maximum point
dx 2
At x = 2
d2y
= 6(2) − 6 = 6  0 ⇒ (2, 0) is a minimum point
dx 2
(iii)
R.T.Sketch the curve
Solution:
When x = 0, y = 4 ∴the curve cuts the y-axis at (0, 4).
Since (2, 0) is a minimum point (x – 2) is a repeated factor.
(
)
x 3 − 3x 2 + 4 = x 2 − 4 x + 4 (x + 1)
∴ the curve cuts the x-axis at (-1, 0)
y
Maximum point
(0,4)
6.
(a)
x
O
(-1, 0)
y = x 3 − 3x 2 + 4
(2, 0)
Minimum point
R.T.Differentiate: y w.r.t. x
(i)
y = x 2x − 1
= x(2 x + 1)
1
2
du
=1
dx
Let u = x
1
v = (2 x − 1) 2
1
dv 1
−
= (2)(2 x − 1) 2
dx 2
= (2 x − 1)
Re: If y = uv , then
1
−
1
2
dy
dv
du
(Product Law)
=u +v
dx
dx
dx
1
−
dy

= x(2 x − 1) 2 + (2 x − 1) 2
dx
x
=
=
2x − 1
+ 2x − 1
x + (2 x − 1)
2x − 1
=
3x − 1
2x − 1
ALTERNATIVE METHOD
y = x 2x − 1
y 2 = x 2 (2 x − 1)
Differentiating implicitly w.r.t x
2y
dy
= 6x 2 − 2x
dx
dy 6 x 2 − 2 x
=
dx
2y
dy 2 x(3x − 1)
=
dx 2 x 2 x − 1
dy
3x − 1
=
dx
2x − 1
(ii)
y = sin 2 ( x 3 + 4)
Let t = x 3 + 4
dt
= 3x 2
dx
 y = sin 2 t
Let v = sin t
y = v2
dy
= 2v
dv
dv
= cos t
dt
dy dy dv dt
(chain rule)
=
 
dx dv dt dx
= (2v)  (cos t )  (3x 2 )
= 6 x 2 sin t cos t
= 3x 2  2 sin t cos t
= 3x 2  sin 2t
(
= 3 x 2 sin 2( x 3 + 4)
)
6
(b)
(i)
Data:  f ( x)dx = 7
1
6
R.T.C.  [2 − f ( x)]dx
1
Calculation:
6
6
6
1
1
1
 [2 − f ( x)]dx   2dx −  f ( x)dx
= 2 x1 − 7
6
= [2(6) − 2(1)] − 7
= (12 − 2) − 7
=3
(ii)
Data: The area under the curve 𝑦 = 𝑥 2 + 𝑘𝑥 − 5 bounded by the x-axis
2
and 𝑥 = 1 𝑎𝑛𝑑 𝑥 = 3 is 14 units 2 .
3
R.T.F. The value of k.
Solution:
3
A =  ( x 2 + kx − 5)dx = 14
1
2
(Data)
3
3
 x 3 kx 2

2
 +
− 5 x  = 14
2
3
3
1
 (3) 3 k (3) 2

  (1) 3 k (1) 2
2
+
+
− 5(1) = 14
−
5
(
3
)

 −
2
2
3
 3

  3
2
 27 9k
 1 k

 3 + 2 − 15 −  3 + 2 − 5 = 14 3

 

9k
1 k
2


9 + 2 − 15 − 3 − 2 + 5 = 14 3


4k −
4
2
= 14
3
3
4k = 16
k=4
Hence, 𝑦 = 𝑥 2 + 4𝑥 − 5
(c)
Data: Volume of figure = 45𝜋 𝑢𝑛𝑖𝑡𝑠 3
(i)
R.T.P. h =
45 2r
−
r2 3
14

Proof: Volume of figure = r 2 h +  r 3  = 45
23

2
 r 2 h + r 3 = 45
3
(÷ 𝜋)
r 2h +
2 3
r = 45
3
2
r 2 h = 45 − r 3
3
h=
45 −
2 3
r
3
r2
h=
45 2
− r
r2 3
Q.E.D.
(b)
R.T.P. A =
5r 2 90
+
3
r
Proof :
Area of external surface = Area of curved surface of cylinder +
Area of hemisphere + Area of circular base
1
= 2rh + (4r 2 ) + r 2
2
A
= 2rh + 3r 2
( h=
45 2
− r)
r2 3
 45 2 
= 2r  2 − r  + 3r 2
3 
r
=
90 4r 2
−
+ 3r 2
r
3
A=
90 5r 2
+
r
3
Q.E.D.
(ii)
R.T.F. The value of r for which A is a minimum and the corresponding
value of A.
Solution:
5
A = 90r −1 + r 2
3
dA
10
= −90r −2 + r
dr
3
At a minimum value of A,
i.e.
− 90r −2 +
dA
=0
dr
10
r = 0
3
− 90 10
+ r = 0
3
r2
r 9
=
3 r2
r 3 = 27
And r = 3
d 2 A 10
180
= + 3
2
3
dr
r
When r = 3 ,
d 2 A 10
180
=  + 3 0
2
3
dr
3
 A has a minimum value at r = 3
When r = 3
5 (3) 2 90
A=
+
= 15 + 30 = 45 units2 (in exact form)
3
(3)
JUNE 2007 UNIT 1 PAPER 2
1.
(a)
Data: g ( x) = x 4 − 9  R
(i)
R.T.F all the real factors of g(x).
Solution:
g ( x) = x 4 − 9
( ) − (3)
= x2
2
2
= ( x 2 − 3)( x 2 + 3) (the difference of two squares)
= ( x − 3 )( x + 3) ( x 2 + 3) (by further factorising)
(ii)
R.T.F all the real roots of g ( x) = 0
Solution:
g ( x) = x 4 − 9 = 0
= ( x − 3 )( x + 3) ( x 2 + 3) = 0 (from part (i))
Either ( x − 3 ) = 0 or ( x + 3) = 0 or ( x 2 + 3) = 0
x2 + 3 = 0
When x = − 3 , there are no real solutions or roots.
 Real roots occur when ( x − 3 ) = 0 and ( x + 3) = 0
Hence, − 3 and
3 are the real roots.
(b)
Data: f ( x) = x 4 − 9 x 3 + 28x 2 − 36 x + 16 and u = x +
(i)
4
x
R.T.Express u 2 in terms of x.
Solution:
4
u = x+
x
2
4

u = x + 
x

4 
4

u 2 =  x +  x + 
x 
x

16
u2 = x2 + 4 + 4 + 2
x
16
u2 = x2 + 8 + 2
x
2
(ii)
36 16 

R.T.P. By writing f ( x) = x 2  x 2 − 9 x + 28 −
, to show
+
x x 2 

that if f ( x) = 0 , then u 2 − 9u + 20 = 0
Proof:
36 16 

f ( x) = x 2  x 2 − 9 x + 28 −
+
x x 2 


16
36 


= x 2  x 2 + 2 + 8  + 20 −  9 x + 
x 
x




16


= x 2  x 2 + 2 + 8  + 20 − 9 x +
x



Recall: u = x +
4 

x 
16
4
and u 2 = x 2 + 8 + 2
x
x
Hence, f ( x) = x 2 [u 2 + 20 − 9u]
If f ( x) = 0 , then x 2 [u 2 + 20 − 9u] = 0
x  0 (data)
Hence, u 2 + 20 − 9u = 0 .
Q.E.D.
(iii)
R.T.Determine the values of x  R , for which f ( x) = 0
Solution:
u 2 − 9u + 20 = 0
(u − 5)(u − 4) = 0
u = 4 or u = 5
When u = 4
x+
x
4
=4
x
x 2 + 4 = 4x
x 2 − 4x + 4 = 0
( x − 2) 2 = 0
x = 2
When u = 5
x+
x
4
=5
x
x 2 + 4 = 5x
x 2 − 5x + 4 = 0
( x − 4)( x − 1) = 0
 x = 4 or x = 1
Hence, x = 1  R or x = 2  R or x = 4,  R
n
2.
(a)
Data: S n =  r , r  N ;
r =1
(i)
n
1
 r = 2 n(n + 1)
r =1
R.T.F. the value of n for which 3S 2 n = 11S n
Solution:
n
Sn =  r =
r =1
2n
1
n(n + 1)
2
S 2n =  r =
r =1
1
(2n)( 2n + 1)
2
= n(2n + 1)
3S 2n = 11S n
1

 3n(2n + 1) = 11 n(n + 1)
2

2
6n(2n + 1) = 11n(n + 1)
12n 2 + 6n = 11n 2 + 11n
n 2 − 5n = 0
n(n − 5) = 0
Either n = 0 or n = 5
n = 0 N
Hence, n = 5  N only
(b)
Data: x 2 − px + 24 = 0 , p  R , has roots  & 
x 2 − 8 x + q = 0 , q  R , has roots 2 +  & 2 − 
(i)
R.T.Express p and q in terms of  & 
Solution:
Let ax 2 + bx + c = 0 be any quadratic, where a, b and c are constants
 (a)
x2 +
b
c
x+ =0
a
a
If the roots are  and 
Then ( x −  )( x −  ) = 0
Hence, x 2 +
b
c
x + = x 2 − ( +  ) x + 
a
a
x 2 − (The sum of the roots) x + (product of the roots) = 0
Equating coefficients:
In x: ( +  ) =
−b
a
The constant:  =
c
a
x 2 − px + 24 = 0 is of the form ax 2 + bx + c = 0
Where a = 1, b = - p and c = 24
Hence,  +  =
− (− p)
1
 + = p
and  = 24
x 2 − 8 x + q = 0 is of the form ax 2 + bx + c = 0
Where a = 1, b = - 8 and c = q
Roots are (2 +  ) and (2 −  )
 (2 +  ) + (2 −  ) =
− (−8)
1
4 = 8
 =2
(2 +  )(2 −  ) = q
4 2 −  2 = q
4( 2) 2 −  2 = q
 q = 16 −  2
(ii)
R.T.F. the value of  and of  .
Solution:
Recall:  = 2 ( From (b) (i))
and  = 24
 2 = 24
and  = 12
(iii)
R.T.Calculate the value of p and of q.
Calculation:
Recall:  +  = p
Hence, p = 2 + 12
= 14  R
Recall: 4 2 −  2 = q
Hence, q = 4(2) 2 − (12) 2
q = 16 – 144
= - 128  R
 p = 14  R and q = - 128  R
(c)
R.T.P. By Mathematical Induction, that n 2  2n  integers n  3
Proof:
If n 2  2n , then n 2 − 2n  0
Assume the statement is true for n = k
i.e. k 2 − 2k  0 for k  3
Consider the case when n = k + 1
(k + 1) 2 − 2(k + 1)
= k 2 + 2k + 1 − 2k − 2
(
)
= k 2 − 2k + (2k − 1)
But k 2 − 2k  0 for k  3
And (2k − 1)  0 for k  3
(
)
Hence, k 2 − 2k + (2k − 1)  0 k  3
 The statement is true for n = k + 1
Also, we may test to illustrate that the statement is true for a value of n
Let n = 3
n2 = 9
2n = 2(3) = 6
9  6 (True)
Let n = 4
n 2 = 16
2n = 2(4) = 8
16  8 True
Hence, by the Principle of Mathematical Induction, the
statement, n 2 − 2n  0 is true  integers n  3 .
and so n 2  2n is true  integers n  3 .
3.
(a)
(i)
R.T.F. The length of the radius of the circle.
Solution:
CDˆ O = 90 0 ( The angle made by a tangent to a circle and a radius, at the
point of contact, is a right angle.)
Hence, D is the point (0, -4).
 Radius of the circle, CD = 5 – 0
= 5 units.
(ii)
R.T.F The equation of the circle.
Solution:
Recall: For center (a, b) and radius r, the equation of a circle is given by
( x − a) 2 + ( y − b) 2 = r 2
For a circle with center (5, -4) and radius 5 units, the equation of the circle
is ( x − 5) 2 + ( y − (−4)) 2 = 5 2 .
( x − 5) 2 + ( y + 4) 2 = 25
or x 2 − 10 x + 25 + y 2 + 8 y + 16 = 25
i.e. x 2 − 10 x + y 2 + 8 y + 16 = 0
(iii)
R.T.C the coordinates of A and B
Solution:
Recall: x 2 + y 2 − 10 x + 8 y + 16 = 0
The circle cuts the x-axis at A and at B, when y = 0.
 x 2 + (0) 2 − 10 x + 8(0) + 16 = 0
x 2 − 10 x + 16 = 0
( x − 8)( x − 2) = 0
Either x = 2 or x = 8
The circle cuts at A first and after at B
Hence, A has coordinates (2, 0) and B has coordinates (8, 0).
(iv)
R.T.C The equation of the tangent at B.
Solution:
Differentiating the equation of the circle, implicitly w.r.t. x,
x 2 + y 2 − 10 x + 8 y + 16 = 0
 2x + 2 y
(2 y + 8)
dy
dy
− 10 + 8
=0
dx
dx
dy
= 10 − 2 x
dx
dy 10 − 2 x
=
dx 2 y + 8
The gradient of the tangent at B, (8, 0) =
Equation of the tangent at B is y − 0 =
y=
−3
x + 6 or 4 y = −3x + 24
4
10 − 2(8)
2( 0) + 8
=
−6
8
=
−3
4
−3
( x − 8)
4
(v)
R.T.F The coordinates of P
Solution:
P is a point on the line PB.
P is also the y-intercept on the line PB.
4 y = −3x + 24
When x = 0 , y = 6
Hence, the point P has coordinates (0, 6).
(b)
R.T.P that PD = PB
Proof:
Length of PB
=
(8 − 0)2 + (0 − 6)2
= 82 + 62
= 100
= 10 units
Length of PD
=
(0 − 0)2 + (− 4 − 6)2
= 100
= 10 units
Hence, PD = PB.
Q.E.D
4.
(a)
(i)
R.T.P. cos 2 
1 − tan 2 
1 + tan 2 
Proof:
Taking, R.H.S
Recall: tan  =
sin 
cos 
sin 2 
1−
1 − tan 2 
cos 2 
=
Hence,
2
1 + tan 
sin 2 
1+
cos 2 


sin 2  
1 −
  cos 2 
2
cos  
=

sin 2  
1 +
  cos 2 
2
cos



cos 2 
cos 2 
=
cos 2  − sin 2 
cos 2  + sin 2 
Recall: cos 2  + sin 2  = 1
& cos 2 = cos 2  − sin 2 
 R.H.S
=
cos 2
1
= cos 2
= L.H.S
 L.H.S = R.H.S
Q.E.D
0
(ii)
R.T.P. tan 67
1
= 1+ 2
2
Proof:
1
Let  = 67
2
0
0
1
2
Then cos 135 0 =
0
1
1 + tan 2 67
2
1 − tan 2 67
(
cos1350 = cos 180 0 − 450
= − cos 450
−1
=
1
Let t = tan 67
2
−1
2
=
2
0
1− t2
1+ t2
− 1(1 + t 2 ) = 2 (1 − t 2 )
− 1 − t 2 = 2 − 2t 2
(
)
2 −1 t 2 = 2 +1
t2 =
2 +1
2 −1
Rationalising to get
t2 =
(
=
2 +1
2 −1

)
2 +1
2 −1
2
2 +1
2 +1
)
t2 =
(
)
2
2 +1
Tan 67 ½ 0 is positive and so we take he positive value only
t =
(
)
2 +1
0
1
Hence, tan 67
=
2
(
)
2 +1
Q.E.D.
(b)
Data: sin q =
5
3
and cos p =
13
5
Assuming that the angles marked p, q, r, s, and t are acute
(i)
R.T.F the exact value of cos q
Solution:
y+
+5
+3
q
x+
O
adj+
(adj)2 + 32 = 5 2
(adj)2 = 5 2 − 32
adj = 5 2 − 3 2
adj = 4 or − 4
Since the measure is along Ox+ , then the adjacent is +4 units
Hence, cos q =
(ii)
4
(exactly)
5
R.T.F. The exact value of sin p
Solution:
y+
+13
+opp
p
x+
O
+5
(opp )2 + 5 2 = 132
(opp)2 = 132 − 5 2
opp = 13 2 − 5 2
opp = 12
Opposite side is in the direction of Oy+
opp = 12
Hence, sin p =
12
(exactly)
13
(iii)
R.T.F. The exact value of sin r
Solution:
r=p+q
(the exterior angle of a triangle is equal to the sum of the
interior opposite angles).
 sin r = sin( p + q)
= sin p cos q + sin q cos p (Compound angle formula)
 12 4   5 3 
=  +  
 13 5   13 5 
48 3 63
(Exactly)
+ =
65 13 55
=
(iv)
R.T.F the exact value of cos (p + t)
Solution:
p + t + q + 60 0 = 180 0
 p + t = 120 0 − q
Hence, cos( p + t ) = cos(120 0 − q)
= cos1200 cos q + sin120 0 sin q (compound angle formula)
cos120 0 = cos(180 0 − 60 0 ) = − cos 60 0 =
sin 120 0 = sin(180 0 − 60 0 ) = sin 60 0 =
 − 1 4   3 3 
cos( p + t ) =    + 

 2 5   2 5 
=
−4 3 3
+
10
10
−1
2
3
2
Hence, cos( p + t ) =
5.
(a)
3 3−4
(exactly)
10
Data: y = 5 x 2 + 3
(i)
R.T.F an expression for
dy
.
dx
Solution:
y = 5x 2 + 3
(
)
y = 5x 2 + 3
1
2
Let t = 5 x 2 + 3
dt
= 10 x
dx
1
y =t2
1
dy 1 − 2
1
= t =
dt 2
2 t
dy dy dt
(Chain Rule)
=

dx dt dx
=
=
=
1
2 t
 10 x
1
2 5x 2 + 3
 10 x
5x
5x 2 + 3
Alternative Method
y = 5x 2 + 3
Squaring, y 2 = 5 x 2 + 3
Differentiating implicitly w.r.t. x
2y
dy
= 10 x
dx
dy 10 x
=
dx 2 y
dy
5x
=
dx
5x 2 + 3
(ii)
R.T.S. y
dy
= 5x
dx
Proof:
Taking L.H.S
Recall:
dy
5x
5x
=
=
2
dx
y
5x + 3
Hence, y
 5x 
dy
= y 
dx
 y 
= 5x
= R.H.S
L.H.S. = R.H.S.
Q.E.D
2
d 2 y  dy 
R.T.S. y 2 +   = 5
dx
 dx 
(iii)
Proof:
Taking L.H.S
Now
dy
5x
=
dx
5x 2 + 3
=
5x
= 5 xy −1
y
Differentiating implicitly w.r.t x
d2y
− 2 dy
= 5 x(− y )
+ 5 y −1 (Product Law)
2
dx
dx
d2y
5x 5x 5
=

+
2
dx
− y2 y y
d 2 y − 25 x 2
=
+ 5 y −1
2
3
dx
y
 5x 
 dy 
  =  
 dx 
 y 
2
=
Hence, y
2
25 x 2
y2
d 2 y  dy 
+  =
dx 2  dx 
 − 25 x 2 5  25 x 2
y
+  + 2
3
y
y
 y
− 25 x 2
25 x 2
=
+5+ 2
y2
y
= 5 = R.H.S.
Q.E.D.
(b)
t 

Data: h = 21 + cos
, t  0
450 

(i)
R.T.Calculate: The height of the tide when high tide occurs, for the first
time.
Calculation:
t 

h = 21 + cos

450 

t 

Recall: − 1   cos
 1
450 

t 

− 2   2 cos
2
450 

 hmax = 2 + 2(1)
=4m
(ii)
R.T.C. The time elapsed between first high tide and first low tide.
Solution:
t 

h = 21 + cos

450 

t 

− 1   cos
 1
450 

 ( 0) 

When t = 0,  cos
 =1
450 

And h = 2 + 2(1)
= 4m
t = a maximum
When
t
=
450
Then cos
t
= cos 
450
= -1
h = 2[1+(-1)]
=0
t =a minimum

t
=
450
1
 t = 450 minutes or 7 hours
2
 time elapsed = tmin - tmax
= 450 – 0
= 450 minutes.
(iii)
R.T.C
dh
at t = 75
dt
Calculation:
h = 2 + 2 cos
Let a =
t
450
t
450
da

=
dt 450
h = 2 + 2 cos a
dh
= −2 sin a
da
dh dh da
(Chain Rule)
=

dt da dt
  
= −2(sin a )

 450 
t   

=  − 2 sin


450  450 

When t = 75
dh
    75 
= −2
 sin

dt
 450   450 
= − 6.98  10 −3 metres per minute
6.
(a)
Data:
(i)
a
a
0
0
 f ( x)dx =  f (a − x)dx, a  0


0
0
R.T.S that if I =  2 sin 2 x dx then I =  2 cos 2 x dx
Proof
Taking L.H.S
Let f ( x) = sin 2 x
f(



− x) = sin 2  − x 
2
2


I =  2 sin 2 x dx
0



=  2 sin 2  − x  dx
0
2

=

2
0
2
 

sin 2 − x  dx

 


But sin − x  = cos x
2


Hence, I =  2 (cos x ) dx
2
0
(using the above result)

=  2 cos 2 xdx
0
= R.H.S.
L.H.S. = R.H.S
Q.E.D
(ii)
R.T.S. I =

4
Proof:

Recall: I =  2 cos 2 xdx & cos 2 x = 2 cos 2 x − 1
0

 cos 2 x + 1 
 I = 2 
dx
0
2


=

1 2
(cos 2 x + 1)dx
2 0

1  1
 2
=   sin 2 x + x 
 0
2  2
1  1
        1  1

=   sin 2  +    -   sin 2(0) + (0)

 2   2    2  2
2  2
1  

=   − 0 

2  2
I =

4
Q.E.D
(b)
(i)
R.T.Sketch the curve y = x 2 + 4
Solution:
y = x2 + 4
y = x 2 , is a standard graph.
0
T =  
 4
2
x → x2 + 4
Alternative Method
The curve cuts the y-axis when x = 0
When x = 0, y = 4
Therefore the curve cuts the vertical axis at (0, 4)
The curve cuts the x-axis when y = 0
i.e. x 2 + 4 = 0
x 2 = −4 and x = − 4 has no real solutions.
Hence, The curve does not cut the x-axis.
dy
= 2x
dx
dy
= 0 at x = 0
dx
(0, 4) is a stationary point.
d2y
= 2  0  (0, 4) is a minimum point.
dx 2
y
y = x2 + 4
O
4
x
(ii)
R.T.F The volume of the shaded region
Solution:
y
y = x2 + 4
5
y2
V y =   x 2 dy
y1
=   ( y − 4)dy
5
4
5
 y2

=   − 4 y
 2
4
 5 2
  42

=   − 4(5)  −  − 4(4) 
  2

 2
 25

=   − 20 − 8 + 16 
 2

 25

=   − 12 
 2

=

2
cubic units
JUNE 2007 UNIT 1 PAPER 1
1.
(a)
Data: (x – 1) is a factor of f ( x) = x 3 + px 2 − x − 2 .
R.T F. The value of p.
Solution:
Recall the‘Remainder and Factor Theorem'–
If f(x) is any polynomial and f(x) is divided by (x – a) then f(a) is the remainder. If
f(a) = 0, then (x – a) is a factor of f(x).
f ( x) = x 3 + px 2 − x − 2 , p  R
From data, f(1) = 0
 f (1) = (1) 3 + p(1) 2 − (1) − 2 = 0
1+ p −1− 2 = 0
p−2=0
 p = 2R
(b)
R.T.F. The remaining factors of f(x).
Solution:
f ( x) = x 3 + 2 x 2 − x − 2 (From (a))
x 2 + 3x + 2
x − 1 x3 + 2x 2 − x − 2
_
x3 − x2
3x 2 − x − 2
_
3x 2 − 3x
2x − 2
_
2x − 2
0
Therefore f ( x) = ( x − 1)( x 2 + 3x + 2)
f ( x) = ( x − 1)( x + 1)( x + 2)
Hence, the remaining factors of f(x) are (x + 1) and (x + 2).
2.
(a)
Data: (3 x ) = 27 ( x − 2 )
2
R.T.C. The value of x
Calculation:
(3 )
x 2
= 27 ( x − 2 )
(3 ) = (3 )( )
(3 ) = (3 )
x−2
2x
3
2x
3 x −6
Equating indices (since the bases are equal)
2 x = 3x − 6
6 = 3x − 2 x
6=x
Hence, x = 6 .
(b)
R.T.Express.
5− 3
2+ 3
in the form x + y 3
Solution:
Rationalising :
5− 3
2+ 3

2− 3
2− 3
=
10 − 5 3 − 2 3 + 3
4−3
=
10 − 7 3 + 3
1
= 13 − 7 3
Which is of the form x + y 3 where x = 13  Z and y = -7  Z .
3.
(a)
Data:
y
𝑦 = 𝑓(𝑥)
A (1, 3)
3
x
0
1
R.T.Determine: The maximum points of the following graphs
(i)
y = f ( x) − 2
Solution:
 0 
T  
 − 2
y = f ( x) − 2
y = f (x)
i.e. f (x) is mapped onto f ( x) − 2 by a vertical shift of 2 units
downwards. This is a congruent transformation, as expected for a
translation. Hence, the maximum point of y = f ( x) − 2 is A(1,3 − 2)
and A (1, 1).
y
𝑦 = 𝑓(𝑥) − 2
𝐴′ (1, 3 − 2)= (1,1)
1
x
0
1
-2
(ii)
y = f ( x + 3)
 − 3
T  
 0 
Solution.
y = f (x)
y = f ( x + 3)
i.e. f (x) is mapped onto y = f ( x + 3) by a horizontal shift of 3 units to
the left. This is a congruent transformation, as expected for a translation.
Hence, the maximum point of y = f ( x + 3) is A (1-3, 2) and A = (-2, 3).
(Illustrated in the diagram shown below)
𝑦 = 𝑓(𝑥) + 3
′′
𝐴 (−2,3)
x
-3
-2
(The y-axis is excluded from this sketch because its position, with respect to the given
points on the graph, after the shift, cannot be ascertained based on the given information).
(b)
(i)
Data: f : x → 3x − 2 , x  R
R.T.P. f (x) is one to one.
Proof:
Let f ( x) = 3x − 2
x
f(x)
a
f(a)
2
3
0
b
f(b)
For every a  A ,  only one value of f (a)  R . If f (a) = f (b) then a = b .
Hence, f is a one to one function, as expected for a linear relationship.
Q.E.D
y
f(a)
f(x) = 3x+ 2
x
O
(ii)
a
Data: f ( f ( x + 3)) = f ( x − 3)
R.T.C. The value of x  R
Calculation:
f ( x + 3) = 3( x + 3) − 2
L.H.S.:  f [ f ( x + 3)] = 33( x + 3) − 2 − 2
= 33x + 9 − 2 − 2
= 33x + 7 − 2
= 9 x + 21 − 2
= 9 x + 19
f ( x − 3) = 3( x − 3) − 2
R.H.S.:
= 3x − 9 − 2
= 3x − 11
When L.H.S. = R.H.S.
 9 x + 19 = 3x − 11
9 x − 3x = −19 − 11
6 x = −30
x = −5 ,  R
4.
(a)
R.T.Solve x − 4 − 6  0 , x  R
Solution:
x−4 −6  0
shall have the same set of solutions as,
x−4 6
Let y = x − 4
Sketching y = x − 4 and y = 6 are drawn on the same axes
y
y = x−4
y = x−4
y = 4− x
4
x
O
4
The graphs cut at x = 10 and x = -2. The shaded region illustrates x − 4  6
y
y = x−4
y = x−4
6
y = 4− x
y=6
4
x
-2
O
4
10
x − 4  6 for the solution set of x : x  10  x : x  −2
as illustrated on the above diagram.
(b)
R.T.Express − 3x 2 − x + 2 in the form u ( x + v) 2 + w
Solution:
Completing the square
1 

− 3 x 2 − x + 2 = −3  x 2 + x  − 2
3 

2
1
25

= −3 x +  +
6  12

This is of the form u ( x + v) 2 + w where u = −3  R , v =
w=
25
R.
12
Alternatively
u( x + v) 2 + w = u( x 2 + 2 xv + v 2 ) + w
= ux 2 + 2uvx + uv 2 + w
1
 R and
6
Equating coefficients of:
x 2 : u = −3
x : 2vu = −1
i.e. 2(−3)v = −1
v=
1
6
Constant: uv 2 + w = 2
1
− 3( ) 2 + w = 2
6
−
1
+w=2
12
w=2
1 25
=
12 12
2
1
25

 −3 x − x + 2 = −3 x +  +
6
12

2
R.T. Solve x 2 + xy = 2 and y + 3x = 5 simultaneously.
5.
Solution:
Let x 2 + xy = 2 ...(1)
and y + 3x = 5 ...(2)
From (2) y = 5 − 3x
Substituting y = 5 − 3x into (1)
x 2 + x(5 − 3x) = 2
x 2 + 5 x − 3x 2 = 2
− 2 x 2 + 5x − 2 = 0
( −1)
2 x 2 − 5x + 2 = 0
(2 x − 1)(x − 2) = 0
Either x =
1
or x = 2
2
When x =
1
2
When x = 2
AND
1
y = 5 − 3 
2
=3
1
2
= −1
Hence, x =
OR
6.
y = 5 − 3(2)
1
1
and y = 3
2
2
x = 2 and y = −1
Data: A (7, 3), B (1, -4) and C (-5, -1).
(a)
(i)
R.T.Calculate the equation of the line AC
Calculation:
Gradient of AC =
−1− 3
−5−7
=
−4
− 12
=
1
3
Using the point A, (7, 3)
Equation of AC is given by y − 3 =
OR
(ii)
y=
1
7
x− +3
3
3
y=
1
2
x+
3
3
3y = x + 2
R.T.F. the equation of the line BD
Solution:
1
(x − 7 )
3
Gradient of BC =
=
=−
− 1 − (−4)
− 5 −1
3
−6
1
2
Line BD is perpendicular BC (from the diagram)
 Gradient of BD = 2 (product of gradients of perpendicular lines = -1)
Using the point B, (1, - 4)
Equation of BD is given by y − (−4) = 2(x − 1)
y = 2x − 2 − 4
y = 2x − 6
(b)
R.T.F. The coordinates of M.
Solution:
M is the point of intersection of the lines BD and AC.
Hence, solving the equations of AC and BD simultaneously to obtain M
Let 3 y = x + 2 ...(1)
and y = 2 x − 6 ...(2)
Subst. (2) into (1)
3(2 x − 6) = x + 2
6 x − 18 = x + 2
6 x − x = 18 + 2
5 x = 20
x=4
When x = 4
y = 2(4) − 6
y=2
Hence, the coordinates of M is (4, 2).
7.
R.T.Express cos  − sin  in terms of R cos( +  )
(a)
Solution:
Recall: cos( +  )  cos  cos  − sin  sin  (compound angle formula)
R = 12 + 12
R= 2
cos  − sin  = 2 cos( +  )
1
2
1
cos  −
2
sin  = cos  cos  − sin  sin 
y
R= 2
1

x
O
1
Hence cos  =
1
2
1 

 sin  =

2

or
 1 

 2
 = cos −1 
=

4
radians
Hence, cos  − sin  = 2 cos( +

4
)
which is of the form R cos( +  )
where, R = 2 , ( R  0)
and  =

4
rads. 0   

2
(b)
R.T.F. The general solution of cos  − sin  = 1
Solution:
cos  − sin  = 1
From (a)
 2 cos( +
cos( +

4
)=


4
y
) =1
1

4
2
 1 
( + ) = cos 

4
 2
−1
O
 

 +  =
4 4

 =0
The general solution for cos  = 2n  A ,
In this case, A = 0
 = 2n  0 ,
 = 2 n , n  Z
8.
A
6 cm
C
3cm
O

6
3cm
D
B
x
y
(a)
R.T.F. The length of the arc AB
Solution:
Length of arc = r (definition)
Length of arc AB = 6 

6
=  cm
(b)
R.T.F. The area of the shaded region
Solution:
The area of shaded region ABCD = The area of sector OAB – The area of
=
1 2  1 2  
(6)   − (3) sin 
2
6 2
6
1

=  3 − 2  cm2
4

9.
(a)
Data: z = 4 + 3i
R.T.Express
z
in the form a + ib
z
Solution:
z 4 − 3i
=
, where z = 4 − 3i is the complex conjugate of z.
z 4 + 3i
By rationalising, we get
z 4 − 3i 4 − 3i 4 − 3i
=
=

z 4 + 3i 4 + 3i 4 − 3i
=
16 − 12i − 12i + 9i 2
16 − 12i + 12i − 9i 2
(Recall i 2 = −1 )
COD
=
16 − 24i − 9
25
=
7 − 24i
25
=
7 24
− i
25 25
Which is of the form a + ib where a =
(b)
R.T.F.
24
7
 R and b = −
R.
25
25
z
z
Solution:
2
z
7 24
 7   24 
=
− i =   + 
z 25 25
 25   25 
2
 49   576 
= 
+
 =1
 625   625 
Alternative Method
Recall,
z
z
=
z
z
z = 4 + 3i = 4 2 + 3 2
=5
z = 4 − 3i = 4 2 + (−3) 2 = 5
Hence,
z 5
= =1
z 5
(NOTE THAT COMPLEX NUMBERS IN NOW TESTED IN UNIT 2)
10.
(a)
(i)


Data: OA = 3i + 2 j OB = 2i − 4 j
j

R.T.F AB
j
Solution:
A
3i + 2j



AB = AO + OB
= − (3i + 2 j ) + (2i − 4 j )
= (2 − 3)i + (− 4 − 2) j
i
M
M lies on the
O
= −i −6j
(ii)
i axis

R.T.F AB
Solution:

AB =
(−1) 2 + (−6) 2
= 1+ 36
=
(iii)
37 units
Data: M divides AB internally in the ratio 1 : 2.
R.T.F. The position vector of M
Solution:
2i - 4j
B
A (3, 2)
1
M
P
6
2
B (2, -4)
R
Q
1
BR =
2
2
(1) =
3
3
 M has the i position of = 2i +
PQ =
2
2
i=2 i
3
3
2
(6) = 4
3
 M has the j position of = − 4 j + 4 j = 0 j

2
 OM = 2 i
3
(b)


R.T.Determine whether or not OA is perpendicular to OB
Solution:


If OA is perpendicular to OB then their dot product = 0.
   3  2 
OA  OB =   
 2  − 4 
 
OA  OB = (3  2) + (2  −4)
=6–8
=-2 0


Hence, OA is NOT perpendicular to OB .
11.
(a)
x3 + 8
R.T.F lim 3
x → −2 x − 4 x
Solution:
Let f ( x) =
f (−2) =
x3 + 8
x3 − 4x
(−2) 3 + 8
(−2) 3 − 4(−2)
=
−8+8
−8+8
=
0
(which is indeterminate)
0
Hence, the factorising and cancelling method is a good choice for finding
x3 + 8
x → −2 x 3 − 4 x
lim
Note: The factorisation of the numerator is not clearly seen in this case.
The denominator, x( x 2 − 4) = x( x − 2)( x + 2) . It is clear from the
indeterminate form, that both numerator and denominator should have a
common factor.
It is not x or x – 2 since the numerator has no negative
sign.
 It has to be (x+ 2).
We divide by (x+ 2)
x 2 − 2x + 4
x + 2 x3 + 8
_
x 3 + 2x 2
− 2x 2 + 8
_
− 2x 2 − 4x
4x + 8
_
4x + 8
0
( x + 2)( x 2 − 2 x + 4)
x3 + 8
Hence, lim 3
= lim
x → −2 x − 4 x
x → −2
x( x + 2)( x − 2)
( x 2 − 2 x + 4)
x →−2
x( x − 2)
= lim
=
((−2) 2 − 2(−2) + 4)
(−2)(−2 − 2)
=
4+4+4
(−2)( −4)
=
12
8
=
3
2
x3 + 8 3
=
x → −2 x 3 − 4 x
2
Hence, lim
Alternate method:
Using L’Hospital’s rule: If f ( x ) =
Then lim f ( x ) =
x →a
Let f ( x ) =
= 3( 2)
g ( a )
h ( a )
h ( x ) = 3x 2 − 4
= 3( 2) − 4
2
2
=8
= 12
x →2
(b)
Data:
h ( x)
x3 + 8
x3 − 4 x
g  ( x ) = 3x 2
 lim
g ( x)
x3 + 8 12
=
x3 − 4 x 8
3
=
2
f ( x) =
x2 +1
( 2 x − 3 − 9)
R.T.F. The values of x  R such that f (x) are continuous.
Solution: f(x) is NOT continuous when ( 2 x − 3 − 9) = 0  2 x − 3 = 9
Let y = 2 x − 3 and y = 9
y = 2x − 3
y
y = 2x-3
y = 3-2x
9
3 − 2x = 9
− 2x = 6
x = −3
y=9
2x − 3 = 9
2 x = 12
x=6
3
-3
O
x
3
2
6
When x = 0
y = 3 - 2(0) = 3
and when y = 0, x =
3
.
2
As x → −3 f (−3) → 
And as x → 6 f (6) → 
 f (x) is continuous for x  R and x  −3 or x  6 .
Alternative Method
x = A  −A = x = A
2x − 3 − 9 = 0
2x − 3 = 9
− 9 = 2x − 3 = 9
2 x − 3 = −9 or 2 x − 3 = 9
x = −3 or x = 6
 f (x) is continuous for x  R and x  −3 or x  6 .
This is best expressed as { x: x  R , x  −3 or x  6 }
12.
(a)
Data: f ( x) =
x2 − 4
x3 + 1
R.T.C. f (x)
Solution:
x2 − 4
u
f ( x) = 3
is of the form
x +1
v
du
= 2x
dx
where u = x 2 − 4
dv
= 3x 2
dx
and v = x 3 + 1
 f ( x) =
=
v
du
dv
−u
dx
dx
2
v
(Quotient Law)
( x 3 + 1)(2 x) − (3x 2 )( x 2 − 4)
( x 3 + 1) 2
(2 x 4 + 2 x) − (3x 4 − 12 x 2 )
=
( x 3 + 1) 2
 f ( x) =
− x 4 + 12 x 2 + 2 x
( x3 + 1) 2

(b)
R.T.Evaluate  4 sin 2 x cos 2 xdx using the substitution u = sin 2x or otherwise.
0
Solution:
 sin 2 x cos 2 xdx
Let u = sin 2x
du
= 2 cos 2 x
dx
du
= dx
2 cos 2 x
  sin 2 x cos 2 xdx =  u cos 2 x
=
1
udu
2
du
2 cos 2 x
u 2

=  + c  c = a constant
4

 (sin 2 x) 2

+ c
=
4




 (sin 2 x) 2  4
  sin 2 x cos 2 xdx = 

4

0
0
4
 12
= 
4
=
Hence,


4
0
 0
 −  
 4
2
1
4
sin 2 x cos 2 xdx =
1
4
Alternative Method
Recall: sin 4 x = 2 sin 2 x cos 2 x (double angle formula)


4
0

1
sin 2 x cos 2 xdx =  4 sin 4 xdx
2 0

 − cos 4 x  4
=

 4 0
 −1  −1
= −  −  
 8   8 
=
13.
(a)
1
4
Data: y = px 3 + qx + r
R.T.C. the values of p, q and r
Solution:
(0, 0) lies on y
0 = p(0) 3 + q(0) + r
r = 0
Similarly, P (1, 2) lies on the curve.
 2 = p(1) 3 + q(1) + 0
p + q = 2 ...(1)
If y = px 3 + qx + r
Then the gradient function,
dy
= 3 px 2 + q
dx
At P(1, 2), the gradient = 8
 8 = 3 p (1) 2 + q
 3 p + q = 8 ...(2)
Solving (1) & (2) simultaneously
From (1) p = 2 − q
Substituting into (2)
3(2 − q) + q = 8
6 − 3q + q = 8
− 2q = 2
q = −1
When q = −1
p = 2 − (−1)
p=3
Hence, p = 3 , q = −1 and r = 0 .
(b)
R.T.F. The equation of the normal to the curve at P.
Solution:
Gradient of the normal to the curve at P =
−1
(the product of gradients of
8
perpendicular lines = -1).
Equation of normal to the curve is y − 2 =
OR
14.
(a)
−1
( x − 1)
8
y=
−1
( x − 1) + 2
8
y=
−1
17
x+
8
8
8 y = − x + 17
Data: f : x → 12 x − x 3
R.T.F The stationary points
Solution:
Let f ( x) = 2 x − x 3
f ( x) = 2 − 3x 2
At stationary points, f ( x) = 0
 0 = 12 − 3x 2
3x 2 = 12
x2 = 4
x = 2
When x = 2 , f (2) = 16
When x = −2 , f (−2) = −16
 The stationary points are (2, 16) and (-2, -16)
(b)
R.T.Determine the nature of the stationary points.
Solution:
f ( x) = −6 x
When x = 2, f (2) = −6(2) = −12  0  (2, 16) is a maximum point.
When x = −2, f (−2) = −6(−2) = 12  0  (-2, -16) is a minimum point.
15.
(a)
R.T.F the coordinates of P and of Q
Solution:
Let y = x 2 ...(1)
Let y = 2 x + 3 ...(2)
Solving simultaneously, to obtain the points of intersection, P and Q
Equating (2) and (1) to get
2x + 3 = x 2
0 = x 2 − 2x − 3
0 = ( x − 3)( x + 1)
Either x = 3 or x = -1
When x = 3, y = 2(3) + 3 = 9
And when x = -1, y = 2(−1) + 3 = 1
At P, x is negative. Hence, the point P is (-1, 1).
At Q, x is positive. Hence, the point Q is (3, 9).
(b)
R.T.Calculate: The area of the shaded portion, POQ.
Calculation:
The regions A, B and C are as shown in the diagram.
The entire region(A + B + C) is a trapezium
Area of trapezium =
1
(the sum of the parallel sides)x(the perpendicular height)
2
1
(1 + 9)4
2
Area of (A + B + C) =
= 20 sq. units
Area of (B + C) =

3
−1
x 2 dx
3
 x3 
=  
 3  −1
 (3) 3 (−1) 3 
−
=

3 
 3
1
= 9 sq. units
3
Hence, the area of the shaded region, A = 20 − 9
2
= 10 sq. units
3
1
3
JUNE 2006 UNIT 1 PAPER 2
1.
(a)
R.T.S the simultaneous equations x 2 + xy = 6 and x − 3 y + 1 = 0
Solution:
Let x 2 + xy = 6 ... (1)
Let x − 3 y + 1 = 0 ... (2)
From equation (2), x = 3 y − 1
Subst. into equation (1)
(3 y − 1) 2 + (3 y − 1) y = 6
9 y 2 − 6 y + 1 + 3y 2 − y = 6
12 y 2 − 7 y − 5 = 0
(12 y + 5)( y − 1) = 0
y=
−5
or 1
12
When y =
−5
12
 −5
x = 3
 −1
 12 
x = −2
1
4
When y = 1
x = 3(1) − 1
x=2
Hence x = 2 and y = 1 OR x = −2
−5
1
and y =
12
4
If ax 2 + bx + c = 0
(b)
 (a)
x2 +
b
c
x+ =0
a
a
If the roots are  and 
Then
( x −  )( x −  ) = 0
x 2 − x − x +  = 0
Hence, x 2 +
b
c
x + = x 2 − ( +  ) x + 
a
a
Equating terms:
in x: ( +  ) =
−b
a
and constant:  =
(i)
c
a
R.T.F the value of ( +  ) and of 
Solution:
In x 2 + 4 x + 1 = 0 , x 2 + 4 x + 1
is of the form ax 2 + bx + c
where a = 1, b = 4 and c =1
( +  ) = −
b
a
 ( +  ) =
−4
= −4
1
 =
c
a
1
  = = 1
1
(ii)
R.T.C. the value of  2 +  2
Calculation:
( +  )2 =  2 + 2 +  2
( +  )2 − 2 =  2 +  2
 2 +  2 = ( +  ) 2 − 2
= (4) 2 − 2(1)
= 16 − 2
= 14
(iii)
R.T.F the equation whose roots are 1 +
1

and 1 +
1

Solution:
x2 +
b
c
x + = x 2 − (sum of the roots)x + (product of the roots) = 0
a
a
 Required equation is

1
1 
1 
1
x 2 − 1 + + 1 +  x + 1 + 1 +  = 0
      
 

1 1 
1 1
1 
=0
x 2 −  2 + +  x + 1 + + +
       


 +   +
1 
 x + 1 +
=0
x 2 −  2 +
+
  

 

( −4 )   ( − 4 ) 1 

x2 − 2 +
+ =0
 x + 1 +
1  
1
1

x 2 − (−2) x − 2 = 0
x 2 + 2x − 2 = 0
Required equation is x 2 + 2 x − 2 = 0
n
2.
(a)
R.T.P. by Mathematical Induction that
 r = 2 n(n + 1)
1
r =1
Proof:
Assume statement true for n = k
k
1
  r = k (k + 1)
2
r =1
Consider n = k + 1
k +1
k
r =1
r =1
  r = r + (k + 1) th term
=
1
k (k + 1) + (k + 1)
2
=
1
k (k + 1) + 2(k + 1)
2
=
1
(k + 1)(k + 1) + 1
2
1
= (k + 1){( k + 1) + 1}
2
which is of the form
1
n(n + 1) where n = (k + 1)
2
Hence, statement is true for n = k + 1
Also consider when n = 1
L.H.S.:
r =1
r =1
R.H.S.:
1
1(1 + 1) = 1
2
i.e. statement is true for n = 1
Consider n = 2
2
r = 1+ 2 = 3
L.H.S.:
r =1
1
(2)(2 + 1) = 3
2
R.H.S.:
i.e. statement is true for n = 2
Consider n = 3
3
r = 1+ 2 + 3 = 6
L.H.S.:
r =1
1
(3)(3 + 1) = 6
2
R.H.S.:
i.e. statement is true for n = 3
Hence by the Principle of Mathematical Induction statement is true for
n  Z + .
Q.E.D
2n
(b)
(i)
R.T.E.
 r in terms of n.
r =1
Solution:
n
Recall:
 r = 2 n(n + 1)
1
r =1
Replacing n by 2n
2n
 r = 2 (2n )(2n + 1)
1
r =1
2n
 r = n(2n + 1)
r =1
2n
(ii)
 r in terms of n.
R.T.E.
r = n +1
Solution:
2n
r =
r = n +1
2n
r r =1
n
r
r =1
1
= n(2n + 1) − n(n + 1)
2
1
1
= 2n 2 + n − n 2 − n
2
2
(c)
=
3 2 1
n + n
2
2
=
n
(3n + 1)
2
R.T.F. the value of n
Solution:
2n
Given
 r = 100
r = n +1

1
n(3n + 1) = 100 (from b (ii))
2
3 2 1
n + n = 100
2
2
x(2)
3n 2 + n − 200 = 0
(3n + 25)(n − 8) = 0
 n = 8 and n = 8
3
4
Since n  Z + then n = 8 only.
3.
(a)
(i)
R.T.F coordinates of centre and radius of the circle with equation
x 2 + y 2 + 2x − 4 y = 4
Solution;
x 2 + y 2 + 2 x − 4 y − 4 = 0 = x 2 + y 2 + 2(1) x + 2(−2) y + (−4)
is of the form x 2 + y 2 + 2 gx + 2 fy + c = 0
which is the equation of a circle with centre (− g ,− f ) and
radius =
g2 + f 2 −c
where g = 1  R
f = −2  R
c = −4  R
Centre = (− g ,− f )
= (−1,−(−2))
= (-1, 2)
Radius =
g2 + f 2 −c
= 12 + (−2) 2 − (−4)
= 1+ 4 + 4
= 9
= 3 units
Hence, x 2 + y 2 + 2 x − 4 y = 4 represents the equation of a circle with
center (-1, 2) and radius 3 units.
Data: x = −1 + 3 sin 
y = 2 + 3 cos 
(b) R.T.P. the parametric equations of the circle are x = −1 + 3 sin  and
y = 2 + 3 cos 
Proof:
x = −1 + 3 sin 
y = 2 + 3 cos 
x 2 + y 2 + 2x − 4 y = 4
Taking L.H.S.
x 2 + y 2 + 2x − 4 y
= (−1 + 3 sin  ) 2 + (2 + 3 cos  ) 2 + 2(−1 + 3 sin  ) − 4(2 + 3 cos  )
= 1 + 9 sin 2  − 6 sin  + 4 + 12 cos  + 9 cos 2  − 2 + 6 sin  − 8 − 12 cos 
= 1 + 9(sin 2  + cos 2  ) + 4 + 2 − 8
= 1+ 9 + 4 + 2 − 8
=4
= R.H.S.
L.H.S. = R.H.S.
Q.E.D.
(sin
2
 + cos 2  = 1)
(b) R.T.S that the x-coordinates of the points of intersection of the circle with the line
3
x + y = 1 are − 1
2.
2
Proof:
Solving simultaneously to find the points of intersection
Let x + y = 1 …(1)
Let x 2 + 2 x + y 2 − 4 y = 4 …(2)
Subst. y = 1 − x into (2)
x 2 + 2 x + (1 − x) 2 − 4(1 − x) = 4
x 2 + 2x + 1 − 2x + x 2 − 4 + 4x − 4 = 0
2x 2 + 4x − 7 = 0
is of the form ax 2 + bx + c = 0 , where a = 2, b = 4 and c = -7.
Recall: x =
− b  b 2 − 4ac
(Quadratic Formula)
2a
x=
− 4  4 2 − 4(2)(−7)
2(2)
x=
− 4  16 + 56
4
x=
− 4  72
4
x=
−46 2
4
x = −1 
Q.E.D.
3 2
2
(b)
R.T.F. the general solution of cos  = 2 sin 2  − 1
Solution:
cos  = 2 sin 2  − 1
Recall: cos 2  + sin 2  = 1
cos  = 2(1 − cos 2  ) − 1
2 cos 2  + cos  − 1 = 0
(2 cos  − 1)(cos  + 1) = 0
2 cos  − 1 = 0 or cos  + 1 = 0
When cos  =
1
, cos  +ve in the 1st. and 4th. quadrants.
2

2 − 
1
2
 is acute.  = cos −1   =

3
rads


Hence, general solution:  =  2n  , n  Z
3

When cos  = −1 , cos  in the 2nd and 3rd. quadrants.
 −
 +
 = cos −1 (−1) =  rads
Hence, general solution :  = (2n   ) n  Z
4.
(a)
R.T.C. the values of R and  .
(i)
Calculation:
Recall: sin( x −  )  sin x cos  − sin  cos x (compound angle formula)
4 sin x − cos x = R sin(x −  )
R = 4 2 + 12
R = 17
R = 4.12
R = 4.1 to 1d.p.
4
17
1
sin x −
17
cos x = sin x cos  − cos x sin 
y
R = 17
1

x
O
4
cos  =
4
17
 4 

 17 
 = cos −1 
 = 14.04 0
 = 14.0 0 to 1 d.p.
(iii)
R.T.F. one value of x between 0 0 and 360 0 where there exists a stationary
point.
Solution:
Let y = 4 sin x − cos x
 y = 17 sin( x − 14.0 0 )
y max = 17  1
y min = 17  (−1) , since − 1  sin( x − 14.0 0 )  1
Occurs when (x − 14.0 0 ) = 90 0
y max = 17  1
Since sin 90 0 = sin( x − 14.0 0 ) = 1
Equating angles,
90 0 = x − 14.0 0
x = 104.0 0 at stationary point when y = 4 sin x − cos x and is a maximum.
(b)
Data : Let z1 = 2 − 3i and z 2 = 3 + 4i
(i)
R.T.F. in the form a + bi
(a)
z1 + z 2 = (2 − 3i) + (3 + 4i)
= (2 + 3) + (−3 + 4)i
= 5 + i is of the form a + bi where a = 5  R and b = 1  R
(b)
z1  z 2 = (2 − 3i )  (3 + 4i )
= 6 + 8i − 9i − 12i 2
= 6 + 12 − i
 i 2 = −1
= 18 − i is of the form a + bi where a = 18  R and
b = - 1 R
©
z1 2 − 3i
=
z 2 3 + 4i
=
2 − 3i 3 − 4i

3 + 4i 3 − 4i
=
6 − 9i − 8i + 12i 2
3 2 − 16i 2
=
6 − 17i − 12
 i 2 = −1
9 + 16
=
− 6 − 17i
25
=
−6
− 6 17
 R and
− i is of the form a + bi where a =
25
25 25
b=
(iii)
− 17
R
25
R.T.F the quadratic equation whose roots are z1 and z 2 .
Solution:
2 − 3i is a root  z - (2 - 3i) is a factor
3 + 4i is a root  z - (3 + 4i) is a factor
z − (2 − 3i) z − (3 + 4i) = 0
z 2 − z (2 − 3i) − z(3 + 4i) + (2 − 3i)(3 + 4i) = 0
z 2 − z[(2 − 3i) + (3 + 4i)] + [6 − 9i + 8i − 12i 2 ] = 0
z 2 − (5 + i) z + (18 − i) = 0
the quadratic equation whose roots
are z1 and z 2 is z 2 − (5 + i) z + (18 − i) = 0
5.
(a)
(i)
sin x
x →0 x
R.T.State the value of lim
Solution:
As x is very small and in radians sin x  x (Maclaurin’s expansion)
 lim
x →0
sin x
x
= lim

x
→
0
x
x
= lim 1
x →0
= 1 (constant law)
Alternative Solution
lim
x →0
(ii)
sin x
= 1 (basic trigonometric limit)
x
R.T.Express A and B in terms of x and/or x
Solution:
Data : sin 2( x + x) − sin 2 x = 2 cos A sin A
Taking L.H.S.
Using Factor Formula
 4 x + 2x   2x 
sin 2( x + x) − sin 2 x  2 cos
 sin

2

  2 
= 2 cos(2 x + x )sin(x )
Equating coefficients
A = (2 x + x )
B = x
(iii)
R.T. Differentiate from 1st. principles y = sin 2 x
y = sin x
P
Solution
Recall:
f ( x) =
sin 2( x + x)
)
f ( x + x) − f ( x)
( x + x) − x
f ( x) = sin 2 x
O
f ( x + x) = sin 2( x + x)
f ( x) = lim
x →0
sin 2( x + x) − sin 2 x
( x + x) − x
Recall from (ii)
sin 2( x + x) − sin 2 x = 2 cos(2 x + x) sin(x)
Subst. into numerator
f ( x) = lim
x →0
2 cos(2 x + x) sin(x)
x
sin x
x →0 x
= lim 2 cos(2 x + x)  lim
x →0
sin x
=1
x →0 x
From (i) lim
f ( x) = lim 2 cos(2 x + x)  1
x →0
= lim 2 cos(2 x + x)
x →0
= 2 cos(2 x + 0)
= 2 cos 2 x
f ( x) = 2 cos 2 x
Q
sin x
x
x + x
x
Data: y = hx 2 +
(b)
(i)
k
x
R.T.F. the values of h and k.
Solution:
y = hx 2 +
k
x
The curve passes through the point P (1, 1). (Data)
Subst. x =1 and y = 1
1 = h(1) 2 +
k
(1)
1 = h + k ...(1)
Gradient of curve at P is 5.
y = hx 2 +
k
x
y = hx 2 + kx −1
dy
= 2hx − kx −2
dx
= 2hx −
At P, x = 1,
k
x2
dy
=5
dx
 5 = 2h(1) −
(Data)
k
(1) 2
2h − k = 5 ...(2)
Solving equations (1) & (2) simultaneously
From (1) h = 1 − k
Subst. into (2)
2(1 − k ) − k = 5
2 − 2k − k = 5
− 3k = 3
k = −1
Subst. k = −1 into h = 1 − k
h = 1 − (−1)
h=2
 y = 2x 2 −
(ii)
1
x
R.T.F. the equation of the tangent to the curve at x =
Solution:
y = 2x 2 −
1
x
When x =
1
2
2
1
1
y = 2  −
1
2
2
y = −1
1
2
dy
1
= 4x − 2
dx
x
At x =
1
2
dy
1
1
= 4  +
2
dx
2 1
 
2
dy
= 2+4
dx
=6
1
.
2
= gradient of tangent at x =
Equation of tangent at x =
1
2 =6
1
x−
2
y −1
1
1

y − 1 = 6 x − 
2
2

1 1

y = 6 x −  + 1
2 2

y = 6x − 1
1
2
or 2 y = 12 x − 3
1
is
2
1
2
y
6.
y=x
x
O
(a)
1
n
(i)
2
n
3
n
(1,0)
n −1 n
n n
4
n
R.T.S that the area S is approximately
1
2
3
n −1
+ 2 + 2 + ... + 2
2
n
n
n
n
Proof:
Area S = Sum of the area of the rectangles under the line y = x
Length of Strip
Width of Strip
Area of Strip
(constant)
1
n
2
n
3
n
4
n
n −1
n
1
2
3
n −1
Area S  2 + 2 + 2 + ... + 2
n
n
n
n
1
n
1
n2
1
n
1
n
1
n
1
n
2
n2
3
n2
4
n2
n −1
n2
Q.E.D
n −1
(ii)
Data:
1
 r = 2 n(n − 1)
r =1
R.T.S. S 
1 1
1 − 
2 n
Proof:
S
1
2
3
n −1
+ 2 + 2 + ... + 2
2
n
n
n
n
S
1
(1 + 2 + 3 + ... + n − 1)
n2
S
1
n2
n −1
r
r =1
S
1 1

n(n − 1) 
2 
n 2

S
1  n2 − n 


2  n 2 
S
1 1
1 − 
2 n
Q.E.D.
(b)
(i)
R.T.S. f ( x) =
8 − 2x 2
( x 2 + 4) 2
Proof:
f ( x) =
2x
u
is of the form
v
x +4
2
du
=2
dx
Where u = 2 x
and
v = x2 + 4
f ( x) =
v
dv
= 2x
dx
du
dv
−u
dx
dx
2
v
(Quotient Law)
f ( x) =
( x 2 + 4)(2) − (2 x)(2 x)
( x 2 + 4) 2
f ( x) =
2x 2 + 8 − 4x 2
( x 2 + 4) 2
Hence, f ( x) =
8 − 2x 2
( x 2 + 4) 2
Q.E.D.
24 − 6 x 2
0 ( x 2 + 4) 2 dx
1
(ii)
R.T.Evaluate
Solution:
If f ( x) =
8 − 2x 2
2x

f
(
x
)
=
and
( x 2 + 4) 2
x2 + 4
8 − 2x 2
 ( x 2 + 4) 2 dx
=
2x
+c
x +4
c = constant
2
Hence,
24 − 6 x 2
(8 − 2 x 2 )
3
dx
=
0 ( x 2 + 4) 2 dx
0 ( x 2 + 4) 2
1
1
1
 2x 
= 3 2

 x + 40
 2(1)
2(0) 
−
= 3 2

 (1) + 4 0 + 4 
=
6
5
(c)

Data:
2u
u
1
7
dx =
4
192
x
R.T.C. the value of u.
Calculation:
2u

2u
u
 x −3 
x dx = −

 3 u
−4
2u
=
 1 
− 3 x 3 

u
 −1   −1 
− 3 
=
3 
 24u   3u 
=
−1
1
7
+ 3 =
3
24u
3u
24u 3
Hence,
7
7
=
3
192
24u
 24u 3 = 192
u3 = 8
u = 2, u  0
JUNE 2006 Unit 1 Paper 1
1.
(a)
(i)
Recall: Remainder & Factor Theorem - If f(x) is any polynomial and f(x)
is divided by (x - a) , then the remainder is f(a). If f(a) = 0, then (x - a) is
a factor of f(x).
R.T.P. (x - 1) is a factor of f ( x) = x 4 − ( p + 1) x 2 + p for all values of p;
pN .
Proof:
f ( x) = x 4 − ( p + 1) x 2 + p
and
f (1) = (1) 4 − ( p + 1)(1) 2 + p
= 1 − ( p + 1) + p
=0
Hence (x-1) is a factor of f(x)  p .
Q.E.D.
(ii)If (x- 2) is a factor of f(x)
Then f(2) = 0
Similarly,
(2) 4 − ( p + 1)(2) 2 + p = 0
16 – 4p – 4 + p = 0
12 = 3p
and
p=4 N
n
(b)
Data:
n
 r = 2 (n + 1)
r =1
n
R.T.P.
1
 (3r + 1) = 2 n(3n + 5)
r =1
Proof:
n
n
n
r =1
r =1
r =1
 (3r + 1)   3r + 1
n
n
r =1
r =1
= 3 r +  1
n
1 = (1 + 1 + 1 + ... + 1) = 1 x n = n
r =1
(n 1’s)
n

and 3 r +  1 = 3 (n + 1) + n
2

r =1
r =1
n
n
=
3n 2 3n
+
+n
2
2
=
1
n(3n + 3 + 2)
2
=
1
n(3n + 5)
2
Q.E.D.
2.
(a)
A = {x : 2  x  7}
B = {x : x − 4  h , h  R}
R.T.C. Largest h for which B  A
Calculation:
y = x − 4, x  R
y=-x+4
y=x-4
4
O
4
x
When x = 2, y = x − 4 = 2
and
x = 7, y = x − 4 = 3
y
y = x−4
2 x7
3
(7, 3)
(2, 2)
2
x
O
2
4
7
Hence, largest h for which B  A is 3 (as illustrated in the diagram).
(b)
Data: ( x +
1 2
y) + ky 2 = x 2 + xy + y 2 , x, y, k  R
2
R.T.C. k
Calculation:
(x +
1 2
y) + ky 2
2
= x2 +
1
1
1
xy + xy + y 2 + ky 2
2
2
4
1
= x 2 + xy + (k + ) y 2
4
1
Hence x 2 + xy + (k + ) y 2 = x 2 + xy + y 2
4
Equating coefficients of the term in
k+
1
=1
4

k=
3
R
4
y2
3.
(a)
(i)
Data:
ax + b
3x
, x  −1
−2 
x +1
x +1
R.T.C. a, b  R
Calculation:
3x
3x
2
−
−2 =
x +1 1
x +1
=
3x − 2 x − 2
x +1
=
x−2
x +1
is of the form
ax + b
x +1
where a = 1  R and b = -2  R
(ii)
Data:
3x
 2, xR
x +1
R.T.F range of x
Calculation:
3x
2
x +1
=
3x
−20
x +1
x−2
0
x +1
 ( x + 1) 2
 ( x − 2)( x + 1)  0
y = ( x − 2)( x + 1) cuts the x-axis at -1 and 2.
Coefficient of x 2  0  y has a minimum point.
Sketch of y = ( x − 2)( x + 1)
> 0
> 0
-1

(b)
2
3x
 2 {x : x  2}  {x : x  −1} as illustrated
x +1
42
R.T.P.
2 8
−
= 24 2
1
3
Proof:
Taking L.H.S.
42
2 8
−
1
3
(2 2 ) 2
=
2  (2 )
3
=
=
24
2  2 −1
24
1
2 2  2 −1
=
24
2
−
1
2
1
= 24  2 2
= 24 2
Q.E.D.
−
1
3
x
f(x)
4.
(-1, p)
O
-1
(a)
(i)
(q, 2)
2
x
+1
f ( x) = x 2 + 1 . If (-1, p) lies on f(x) then
f (−1) = p
f(-1) = ( −1) 2 + 1 = 2
 p = 2 R
Similarly, if (q, 2) lies on f(x)
then f (q) = 2

2 = (q) 2 + 1
 q2 = 1
q = 1
From the diagram q is + ve.
 q = 1 R
(ii)
f(0) = (0) 2 + 1 = 1
 f(x) cuts the vertical axis at (1, 0), the minimum point of f(x).
Hence, for the given domain − 1  x  1 , the range of the function f(x) is
1  f ( x)  2 , as illustrated on the diagram.
(b)
(i)
Domain
Co domain = Range
Values of x
Values of f( x)
−1  x  1
1  f ( x)  2
f
Assuming (not indicated in the question) that the co-domain
is 1  f ( x)  2 , then the co-domain
is equal to the range. Hence the
function f(x) is surjective (onto).
(ii)
There may be more than one value of x that are mapped onto one value of
f(x).
x
E.g.
f(1) = 2
f(-1) = 2
f(x)
f
1
2
-1
Hence, the function f(x) is not one to one (injective).
(iii)
The function, f(x), is not both surjective and injective i.e. not bijective,
and hence does not have an inverse.
5.
Let x + 2 y = 1 ...(1) and
2 x + my = n ...(2)
From (1)
From (2)
1
1
y =− x+
2
2
y=−
2
n
x+
m
m
These represent equations of straight lines.
(a)
For a unique solution the straight lines intersect at exactly one point.
This occurs if and only if the two lines are not parallel, i.e. their
gradients are not the same.
1
2
Hence, −  −
2
m
Grad.= − 2
m
Grad.= − 1
2
− m  −4
and m  4 , m  R
(b)
For the system of equations to be inconsistent, then there are no solutions.
The two straight lines are therefore parallel and do not meet. Hence, their
gradients are the same and their intercepts on the vertical axis must be different.
y
1
2
Hence, − = −
2
m
1
2
− m = −4
m=4
If intercepts on the vertical axis are not the
same then for the equations to be inconsistent
grad = - 1
2
n
m
grad = − 2
m
1 n

2 4
Subst. m = 4
1 n

2 4
n2
 m = 4 and n  2
(c)
If the system of equations have an infinite number of solution, then the two
(2)lines are the same. Hence, then gradients are the same and intercepts on the yaxis are also the same.
When gradients are the same
Lines (1) & (2)
1
2
− =−
2
m
and − m = −4
1 n
=
2 m
m=4
When the intercepts on the vertical axes
Gradients - 1 = − 2
are the same, subst. m = 4
1 n
=
2 m
1 n
=
2 4
and n = 2
 For an infinite number of solutions m = 4 and n = 2
2
m
6.
(a)
11
−0
1
11
Gradient of the line joining (0, ) to (11, 0) = 2
= −
2
0 − 11
2
Hence gradient of the line PQ is 2. (product of gradients of perpendicular
lines = -1)
Using the point (2, 7), the equation of PQ is
y−7
=2
x−2
y − 7 = 2x − 4
y = 2x + 3
(b)
Solving the equations x + 2 y = 11 ...(1) and y = 2 x + 3 ...(2) simultaneously to find
point Q.
Subst. (2) into (1)
x + 2(2 x + 3) = 11
5x = 5
x =1
Subst. x = 1 into equation (2)
y = 2(1) + 3
y=5
Hence, point of intersection of the two lines is Q (1, 5).
(c)
P = (2, 7) and Q = (1, 5)
Length of the line PQ =
(2 − 1) 2 + (7 − 5) 2
= 1+ 4
=
5 units.
7.
R.T.C (a) AC
(b) AB
Calculation
(a)
 2 
AC 2 = (7) 2 + (8) 2 − 2(7)(8) cos

 3 
(cosine rule)
1
AC 2 = 49 + 64 − 112(− )
2
AC 2 = 169
AC = 13 units
(b)
AB 2 = (13) 2 + (13) 2 (Pythagoras’ Theorem)
AB = 338
AB = 13  13  2
AB = 13 2 units
8.
(a)
Data: 4 cos 2  − 4 sin  − 1 = 0
R.T.C.  for 0    
Calculation
4 cos 2  − 4 sin  − 1 = 0
Re: cos 2  + sin 2  = 1
cos 2  = 1 − sin 2 

4(1 − sin 2  ) − 4 sin  − 1 = 0
and
4 − 4 sin 2  − 4 sin  − 1 = 0
4 sin 2  + 4 sin  − 3 = 0
x(-1)
(2 sin  − 1)(2 sin  + 3) = 0
sin  =
and
When sin  =
−3
1
or sin  =
2
2
−3
,  has no real solutions since sin  < -1 
2
or sin  is not less than -1  .
When sin  =
1
2
−A
1

A = sin −1 ( ) =
2
6
0  
=
=

6

6
, −
and

6
5
6
(b)
R.T.P
1 − cos 2 x
 tan 2 x
1 + cos 2 x
Proof:
Recall: cos 2 x = 2 cos 2 x − 1 = 1 − 2 sin 2 x
Taking L.H.S.
1 − cos 2 x 1 − (1 − 2 sin 2 x)
=
1 + cos 2 x 1 + (2 cos 2 x − 1)
=
2 sin 2 x
2 cos 2 x
=
sin x  sin x
cos x  cos x
Recall: (tan x =
sin x
)
cos x
= tan 2 x = R.H.S.
Q.E.D.
9.
(a)
Data: -3 + 2i and -3 -2i are the roots of x 2 + 6 x + k
R.T.C the value of k.
Calculation:
If -3 + 2i is a root of the equation x 2 + 6 x + k = 0
then (−3 + 2i) 2 + 6(−3 + 2i) + k = 0

9 − 12i + 4i 2 − 18 + 12i + k = 0
Recall: i 2 = −1
 9 − 12i + 4(−1) − 18 + 12i + k = 0

and
− 13 + k = 0
k = 13.
Alternative Method:
Similarly (−3 − 2i) 2 + 6(−3 − 2i) + k = 0
 k = 13.
Alternative Method:
( [ x − (−3 − 2i)][ x − (−3 − 2i)] = x 2 + 6 x + k
Equating coefficients  k = 13.
(b)
Data:
u + 2i
 1 + vi
3 − 4i
R.T.C u, v  R
Calculation:
Taking L.H.S.
u + 2i 3 + 4i

 1 + vi
3 − 4i 3 + 4i
=
3u + 6i + 4ui + 8i 2
 1 + vi
9 − 12i + 12i − 16i 2
Recall: i 2 = −1

(3u − 8) + i(6 + 4u )
 1 + vi
25
 3u − 8   6 + 4u 

+
i  1 + vi
 25   25 
Equating real part and imaginary part respectively
 3u − 8 

 =1
 25 
and
u = 11
 6 + 4u 

=v
 25 
 6 + 4 (11) = 25v
and
v=2
Hence, u = 11  R and
v = 2 R
10.
Data: p = 2i + 3j and q = 3i – 2j
x, y  R such that xp + yq = -3i -11j
R.T.C.
Calculation:
xp + yq = -3i -11j
 x(2i + 3j) + y(3i – 2j) = -3i -11j
2x i + 3x j + 3yi -2yj = -3i -11j
(2x + 3y)i + (3x – 2y)j = -3i – 11j
Equating i and j components
Let
2x + 3y = -3 ...(1)
Let
3x -2y = -11 ...(2)
− 3 − 2x
3
From (1)
y=
Subst. into (2)
 − 3 − 2x 
3 x − 2
 = −11
3


( 3 )
9x + 6 +4x = -33
13x = -39
x = -3
Subst. into (1)
2(-3) + 3y = -3
3y = 3
y=1
Hence, x = -3 and y = 1
(b)
Data:
p = 2i + 3j and q = 3i – 2j
R.T.P. p and q are perpendicular.
Proof
Recall: If a and b are 2 vectors a.b = a b cos 
If  = 90 0 cos  = 0 and a.b = 0
p.q = (2  3) + (3  −2)
= 6−6 = 0
Hence, p and q are perpendicular.
Q.E.D.
11.
(a)
R.T.F. lim
x→1
x2 + x − 2
x 2 − 3x + 2
Solution
Let f(x) =
x2 + x − 2
x 2 − 3x + 2
(1) 2 + 1 − 2
(1) 2 − 3(1) + 2
f(1) =
=
0
0
(indeterminate)
Hence factorising and cancelling
lim
x →1
x2 + x − 2
x 2 − 3x + 2
( x + 2)( x − 1)
( x − 1)( x − 2)
= lim
x →1
= lim
x →1
=
( x + 2)
( x − 2)
1+ 2
1− 2
=-3
Hence, lim
x →1
(b)
x2 + x − 2
= -3
x 2 − 3x + 2
R.T.C. values of x  R for which f(x) =
9 − x2
is discontinuous
( x 2 − 3)( x − 3)
Calculation
As x →  3 , (x 2 − 3) → 0 and f (x) →  .
Hence f(x) is discontinuous at x = + 3 and x = − 3
As x → 3 , f ( x) →
0
which is indeterminate.
0
 f(x) is discontinuous at x =  3 .
12.
(a)
f (x) =
2− x
, xR, x  R
x2
R.T.F. the nature of the stationary points of f(x).
Solution
f (x) =
2− x
u
is of the form f(x) = where
2
v
x
u = 2− x,
du
= −1
dx
v = x2 ,
dv
= 2x
dx
and
f ( x) =
x 2 (−1) − (2 − x)( 2 x)
(x2 )2
f ( x) =
− x 2 − 4x + 2x 2
x4
f ( x) =
x 2 − 4x
x4
At a critical value f ( x) = 0

 (x4 )
x 2 − 4x
=0
x4
x( x − 4) = 0
x = 0 or 4
x  0 (data)
 x = 4 only
f ( x) =
x 2 − 4x
x4
f ( x) = x −2 − 4 x −3
f ( x) = −2 x −3 − 4(−3x −4 )
f ( x) =
− 2 12
+
x3 x4
(quotient rule)
When x = 4, f (4) =
When x =4, f(4) =
2 − 4 − 2 −1
=
=
( 4) 2 16
8
When x = 4, f(4) =
(b)
−2
12
+
 0  minimum value
3
( 4)
( 4) 4
−1
and is a minimum value.
8
R.T.Differentiate w.r.t. x f ( x) = sin 2 ( x 2 )
Solution
f ( x) = sin 2 ( x 2 )
Let y = f(x), t = x 2 and v = sin t (= sin x 2 )
 y = v2
dv
dy
dt
= cos t
= 2x
= 2v
dx
dv
dt
dy dy dv dt
=
 
dx dv dt dx
(Chain Rule)
= 2v  (cos t )  2 x
= (2 sin x 2 ) (cos x 2 ) 2 x
= 4 x sin x 2 cos x 2
Data: f ( x) = x( x 2 − 12)
13.
(a)
R.T.F. the coordinates of A and B
Solution
At stationary points f ( x) = 0
f ( x) = x 3 − 12 x
f ( x) = 3 x 2 − 12
When 3x 2 − 12 = 0
x2 = 4
and
x = 2
 Stationary points occur at x = 2 and x = −2
When x = 2 , f (2) = (2) 3 − 12(2) = −16
At B, x is + ve
 B = (2, -16)
When x = −2 , f (−2) = (−2) 3 − 12(−2) = 16
At A, x is - ve
 A = (-2, 16)
(b)
y
f ( x) = x( x 2 − 12)
A
O
x
B
R.T.F. the equation of the normal at the origin.
Solution
f (0) = 3(0) 2 − 12 = -12
 gradient of the tangent at O is -12.
Hence, gradient of the normal at O is
1
.
12
(product of gradient of perpendicular lines = -1)
Equation of normal at O is
i.e.
y−0 1
=
x − 0 12
12y = x
14
(a)
y
y=
A
O
y=
A
3
2
1
x −1
2
16
x2
R.T.E. the shaded area, A as a difference of two integrals.
Solution
Let region A be as shown on the diagram.
Area of ( A + A )=
3
16
x
2
dx
2
Area of A
1

=   x − 1dx
2

2
3
16
1

Hence area of shaded region, A =  2 dx -   x − 1dx
2

2
2 x
3
3
and is expressed as the difference of two definite integrals
x
3
(b)
R.T.P.
A = 16  x − 2 dx 2
3
3
1
xdx +  1dx
2 2
2
Proof
16
1

dx -   x − 1dx
2
2

2 x
2
3
3
Area of A = 
3
3
3
1
=  16 x dx -  ( x)dx +  dx
2
2
2
2
−2
3
= 16 x − 2 dx 2
3
3
1
xdx +  dx
2 2
2
Q.E.D.
(c)
R.T.F. the value of A
Solution
3
Area of A = 16 x − 2 dx 2
3
3
1
xdx +  dx
2 2
2
3
3
 x −1 
1 x2 
3
−
= 16

   + x 2
 −1 2 2 2 2
3
 − 16 x 2

−
+ x
= 
4
 x
2
1
 1

= − 5 − 2 + 3 − − 8 − 1 + 2
4
 3

=2
5
units2
12
= 2.417
= 2.42units2 (2 d.p.)
15.
(a)
Data:

R.T.P
a
0
a
f ( x)dx =  f (a − x)dx , a  0
0


0
0
 x sin xdx =  ( − x) sin xdx
Proof:
Replacing x by  − x in

 x sin xdx
0
Upper limit:  =  − x  x = 0
Lower limit: 0 =  − x  x = 
d
( − x) = −1
dx

0
0

 x sin xdx =  ( − x) sin( − x)(−1)dx
0
= −  ( − x) sin( − x)dx

0
a
a
0
Recall: sin( − x) = sin x and −  f ( x)dx =  f ( x)dx


0
0
  x sin xdx   ( − x) sin xdx
Q.E.D.
(b)
(i)
R.T.P



0
0
0
 x sin xdx    sin xdx −  x sin xdx
Proof


0
0
 x sin xdx =  ( − x) sin xdx

=  ( sin x − x sin x)dx
0


0
0
=   sin xdx −  x sin xdx


0
0
=   sin xdx −  x sin xdx
Q.E.D

(ii)
R.T.P.
 x sin xdx = 
0
Proof



0
0
0
 x sin xdx    sin xdx −  x sin xdx


0
0
 2 x sin xdx =   sin xdx

  x sin xdx =
0
=
=
=

2 0

2

2

2
=
Q.E.D

sin xdx
[− cos x]0
(− cos  ) − (− cos 0)
(− (−1)) − (−1)
JUNE 2005 Unit 1 Paper 2
1.
(a)
(i)
R.T.Complete the table
Solution
f ( x) = x( 2 − x)
f ( x) = 2 x − x 2
f ( x) = 2 x − x 2
f (−1) = 2(−1) − (−1) 2 = 3
f (−1) = − 3 = 3
f (0) = 2(0) − (0) 2 = 0
f ( 0) = 0
f (1) = 2(1) − (1) 2 = 1
f (−1) = 1
f (3) = 2(3) − (3) 2 = 3
f (3) = 3
x
f (x)
-2
8
-1
3
0
0
1
1
2
0
3
3
4
8
(ii)
R.T.Draw f (x)
Solution
f ( x) = x( 2 − x)
Coefficient of x 2 is –ve  a maximum point.
The curve cuts the x-axis at 0 and 2 and it cuts the y-axis at 0.
y
(1, 1)
x
O
1
2
y
f ( x) = − x( 2 − x)
8
f ( x) = − x( 2 − x)
f ( x) = x( 2 − x)
(1, 1) max. pt.
-2
O
2
x
4
(b)
R.T.C. the value of k
Solution
k ( x 2 + 5) = 6 + 12 x − x 2
kx 2 + 5k = 6 + 12 x − x 2
x 2 (k + 1) − 12 x − 6 + 5k = 0
Which is of the form ax 2 + bx + c = 0
Where a = (k + 1), b = -12 and c = 5k - 6
Equation has equal roots iff b 2 − 4ac = 0
i.e. (−12) 2 − 4(k + 1)(5k − 6) = 0
20k 2 − 4k − 168 = 0
5k 2 − k − 42 = 0
4
(5k + 14)(k − 3) = 0
k=
(c)
(i)
− 14
 R or k = 3  R
5
R.T.C. the value of x
Solution
Data: 2 x = 16 x −1
2
2 x = (2 4 ) x −1
2
2 x = 2 4 x−4
2
Equating powers of x (bases are equal)
x 2 = 4x − 4
x 2 − 4x + 4 = 0
( x − 2) 2 = 0
x = 2
(ii)
(
R.T.C. the value of
) (
3
2 +1 −
)
2 −1
3
Calculation
(
) (
3
2 +1 −
(
)(
=  2 +1

(
= 2
)
2 −1
2
) (
)(
2
) (
)
2 −1 

)(
2 +1 − 2 − 2 2 +1
 
)
2 −1

2 + 2 + 4 + 2 2 + 2 +1 − 2 2 − 2 − 4 + 2 2 + 2 −1
 
= 5 2 +7 − 5 2 −7
= 14
)(
2 +1  −  2 −1
 
= 2 + 2 2 +1

3

2.
R.T.P. 10 n − 1 is divisible by 9, n  Z +
(a)
Proof
Assume statement true for n = k
i.e. 10 k − 1 is divisible by 9
 10 k − 1 = 9 p , p  Z
10 k = 9 p + 1
Let n = k + 1
(
)
10 k +1 − 1 = 10 k  10 − 1
= 10(9 p + 1) − 1
= 90 p + 10 − 1
= 90 p + 9
= 9(10 p + 1) ,  Z
which is divisible by 9
Testing for n = 1
101 − 1 = 9 which is divisible by 9
Testing for n = 2
10 2 − 1 = 99 which is divisible by 9
Hence by the Principle of Mathematical Induction 10 n − 1 is divisible by 9
n  Z + .
(b)
(i)
R.T.F. the values of p
Solution
Let px + 2 y = 8 ...(1)
and − 4 x + p 2 y = 16 ...(2)
Recall:  an infinite number of solutions iff (1) = (2)
(1) x 2
2 px + 4 y = 16
Equating corresponding coefficients
2 p = −4
and
p2 = 4
p = 2
p = −2
Hence p = −2
(ii)
When p = −2
(1):
− 2x + 2 y = 8
y = x+4
 ( x, x + 4) x  R
(c)
R.T.F the range of values of x.
Solution
Data:
x+4
5
x−2
x+4
−5  0
x−2
x + 4 − 5( x − 2)
0
x−2
− 4 x + 14
0
x−2
 ( x − 2) 2
− 4 x + 14
 ( x − 2) 2  0
x−2
(−4 x + 14)( x − 2)  0
− 2(7 − 2 x)( x − 2)  0
(2 x − 7)( x − 2)  0
Coefficient of x 2 is +ve  a minimum point.
y
O
Solution set
x
2
7
2
7

x : 2  x   x  R
2

3.
R.T.E the equation of a circle, Q, in the form (x − a )2 + ( y − b )2 = c
(a)
Data: Equation of Q is x 2 + y 2 − 2 x + 2 y = 23
Solution
x 2 − 2 x + y 2 + 2 y = 23
x 2 − 2 x + 1 + y 2 + 2 y + 1 = 23 + 1 + 1
x 2 − 2 x + 1 + y 2 + 2 y + 1 = 25
(x − 1)2 + ( y − (−1))2 = 5 2
which is of the form (x − a )2 + ( y − b )2 = c
where a = 1, b = -1 and c = 25
(b)
(i)
R.T.F the coordinates of the centre of the circle, Q.
Solution
(x − a )2 + ( y − b)2 = c is the equation of a circle with centre (a, b)
Hence the coordinates of the centre of the circle is (1, -1)
(ii)
R.T.F the radius of the circle Q.
Solution
(x − a )2 + ( y − b)2 = c 2 is the equation of a circle with radius
Hence, the radius of Q is 25 = 5 units.
c.
(c)
R.T.P. that the point A (4, 3) lies on Q
Proof
Subst. x = 4 and y = 3 into the equation of the circle,
L.H.S : 4 2 + 32 − 2(4) + 2(3)
= 16 + 9 – 8 + 6
= 23
= R.H.S
Hence, the point A lies on the circle.
Q.E.D
(d)
R.T.F the equations of the tangent to the circle, Q, at the point A.
Solution
Recall: x 2 + y 2 − 2 x + 2 y = 23
Differentiating implicitly with respect to x
2x + 2 y
dy
dy
−2+2
=0
dx
dx
(2 y + 2)
dy
= 2 − 2x
dx
dy 2 − 2 x
=
dx 2 y + 2
Subst. x = 4 and y = 3
dy 2 − 2(4)
=
dx 2(3) + 2
dy − 6
=
dx
8
dy − 3
=
dx
4
 Gradient of tangent at A is
−3
4
The equation of the tangent at A is given by
y −3 =
−3
(x − 4)
4
y=
−3
x +3+3
4
y=
−3
x+6
4
4 y + 3x = 24
A (4, 3)
(e)
O (1,-1)
B
R.T.F the coordinates of B
Solution
Let the coordinates of B be (a, b)
AB is the diameter of the circle Q. O is the centre of the circle which is also the
midpoint of AB.
Using the midpoint formula:
(1,−1) =  4 + a , 3 + b 
 2
2 
Equating corresponding coordinates
x:
1=
4+a
2
&
y:
2 = 4+a
−2= a
Hence, the coordinates of B is (-2, -5)
−1 =
3+b
2
−2 = 3+b
−5 = b
4.
(a)
(i)
R.T.F the length of the arc ABC
Solution
Length of arc ABC = R
= 7
=
(ii)
R.T.P. r =
(where R is the radius of sector OABC)

3
7
cm
3
7
cm
6
Proof
When sector OABC is folded, the arc ABC becomes the circular base of
the cone.
Hence, length of arc ABC = circumference of circular base of cone
 2r =
7
3
6r = 7
r =
7
cm
6
Q.E.D
R.T.P h =
(b)
7 35
cm
6
Proof
O
7 cm
h
r
A, C
By Pythagoras’ Theorem
72 = h2 + r 2
h2 = 72 − r 2
7
h = 7 − 
6
2
2
2
 49 
h 2 = 49 −  
 36 
h2 =
49(36) − 49
36
h2 =
49(36 − 1)
36
h2 =
49(35)
36
h=
49(35)
36
h=
7
35 cm
6
Q.E.D.
(b)
(i)
R.T.P cos 3  4 cos 3  − 3 cos 
Proof
Taking L.H.S.
cos 3 = cos(2 +  )
cos 3 = cos 2 cos  − sin 2 sin 
(Compound Angle Formula)
Recall: cos 2 = 2 cos 2  − 1
sin 2 = 2 sin  cos 
cos 3 = (2 cos 2  − 1) cos  − (2 sin  cos  ) sin 
= 2 cos 3  − cos  − 2 sin 2  cos 
Recall: sin 2  = 1 − cos 2 
Hence, cos 3 = 2 cos 3  − cos  − 2(1 − cos 2  ) cos 
= 2 cos 3  − cos  − 2 cos  − 2 cos 3 
= 4 cos 3  − 3 cos 
= R.H.S.
Q.E.D.
(ii)
R.T.F the value of 
Solution
Data: a = (4 cos 2  )i + (6 cos  − 1) j
b = (2 cos  )i − j
If a and b are perpendicular then a  b = 0
a  b = (4 cos 2  )(2 cos  ) + (6 cos  − 1)(−1)
= 8 cos 3  − 6 cos  + 1
Hence, 8 cos 3  − 6 cos  + 1 = 0
2(4 cos 3  − 3 cos  ) + 1 = 0
But cos 3 = 4 cos 3  − 3 cos 
 2 cos 3 + 1 = 0
cos 3 =
If cos 3 =
1
2
−A
−1
2
0  

4
, 0  3 
3
4
1
3 = cos −1  
2
=

3


3 =   − 
3

3 =
2
3
3 =
2
3
Hence,  =
(c)
2
, all other values are out of range.
9
R.T.F. the modulus of the complex number z.
Solution
Data:
z=
25(2 + 3i )
(4 + 3i )
Let z1 = 25(2 + 3i ) and z 2 = (4 + 3i )
z1 = 25(2 + 3i ) = 25 2 2 + 3 2
= 25 4 + 9
= 25 13
z 2 = (4 + 3i ) = 4 2 + 3 2
A=

3
=
25
=5
z =
z1
z1
25 13
=
=
z2
z2
5
= 5 13 units
5
(a)
(i)
sin u
u →0
u
R.T.State the value of lim
Solution
lim
u →0
(ii)
sin u
= 1 (Basic Trigonometric Limit)
u
sin 3x
=3
x →0
x
R.T.Show that lim
Proof
Taking L.H.S.
sin 3x
3 sin 3x
= lim
x →0
x →0
x
3x
lim
= 3 lim
x →0
sin 3x
3x
Let u = 3 x
as x → 0, u → 0,
sin 3x
sin u
= 3 lim
x →0
u →0
x
u
Hence, lim
= 3(1)
=3
L.H.S = R.H.S
Q.E.D
sin 3x
x →0 sin 5 x
(iii)
R.T.Evaluate lim
Solution
lim
x →0

x
x
sin 3x
sin 5 x
= lim
x →0
x sin 3x
x sin 5 x
 sin 3 x sin 5 x 
= lim

x →0
x 
 x
sin 3 x
x
=
sin 5 x
lim
x →0
x
lim
x →0
sin 3x
=3
x →0
x
Recall: lim
Similarly lim
x →0
sin 5 x
=5
x
3
sin 3x
=
x →0 sin 5 x
5
Hence, lim
(b)
R.T.Show that x 2
d2y
dy
+x
=y
2
dx
dx
Proof
y=
A
+ Bx , where A and B are constants
x
y = Ax −1 + Bx
dy
= − Ax −2 + B
dx
dy
A
=− 2 +B
dx
x
d2y
= − A(−2 x −3 )
2
dx
d 2 y 2A
= 3
dx 2
x
Taking L.H.S.
x2
d2y
dy
 2A   A

+x
= x 2  3  + x − 2 + B 
2
dx
dx
x   x

=
2A A
− + Bx
x
x
=
A
+ Bx
x
= y
L.H.S. = R.H.S.
Q.E.D.
y
(c)
P
y 2 = 4x
P
R1
y = 2x
R2
x
O
(i)
R.T.F the coordinates of P
Solution
Let y 2 = 4 x ...(1)
Let y = 2 x ...(2)
Subst. (2) into (1)
( 2 x) 2 = 4 x
4x 2 = 4x
4x 2 − 4x = 0
4 x( x − 1) = 0
either 4 x = 0 or ( x − 1) = 0
x=0
x =1
or
x = 0 at O
When x = 1 , y = 2
 x = 1 at P.
Hence, coordinates of P is (1, 2)
(ii)
R.T.F the volume of the solid generated when the shaded region is rotated
2 radians about the x-axis.
Solution
Regions R1 and R2 are as shown in the diagram
Volume of the required region, R1
= Volume when (R1+R2) are rotated 2 radians about the x – axis Volume when R2 are rotated 2 radians about the x – axis
1
1
0
0
=   4 xdx −   (2 x) 2 dx
 
=  2x
2 1
0
1
 4x3 
− 

 3 0
4

=  2 − 0 −   − 0
3

=
2
 cubic units.
3
6.
(a)
R.T.Differentiate (x 2 + 7 ) + sin 3 x with respect to x.
5
Solution
Let y = (x 2 + 7 ) + sin 3 x
5
Let u = (x 2 + 7 ) and v = sin 3x
5
dy du dv
=
+
dx dx dx
(
u = x2 + 7
(
)
5
Let a = x 2 + 7
)
da
= 2x
dx
u = (a )
5
du
= 5a 4
da
du du da
(Chain Rule)
=

dx da dx
= 5a 4  (2 x)
(
= 10 x x 2 + 7
v = sin 3x
Let t = 3 x
dt
=3
dx
v = sin t
dv
= cos t
dt
)
5
dv dv dt
=

dx dt dx
(Chain Rule)
= (cos t )3
= 3 cos 3x

(b)
(i)
(
)
4
dy
= 10 x x 2 + 7 + 3 cos 3x
dx
R.T.F the values of x for which y = x 3 − 9 x 2 + 15x + 4 has stationary
points.
Data: y = x 3 − 9 x 2 + 15x + 4
Solution
dy
= 3x 2 − 18 x + 15
dx
Stationary points occur when
dy
=0
dx
dy
= 3x 2 − 18 x + 15 = 0
dx
3( x 2 − 6 x + 5) = 0
x 2 − 6x + 5 = 0
( x − 5)( x − 1) = 0
(ii)
Either ( x − 5) = 0
or
( x − 1) = 0
i.e.
or
x=1
x=5
R.T.F the values of x for which y = x 3 − 9 x 2 + 15x + 4 is increasing.
Solution
Recall: A function f (x) increases provided that f (x) increases as x
increases.
y = x 3 − 9 x 2 + 15x + 4
dy
= 3x 2 − 18 x + 15
dx
If y increases then
dy
0
dx
3( x 2 − 6 x + 5)  0
x 2 − 6x + 5  0
( x − 5)( x − 1)  0
3x 2 − 18x + 15  0
1
5
3x 2 − 18x + 15  0
Solution set = x : x  1}  {x : x  5
(iii)
R.T.F the values of x for which y = x 3 − 9 x 2 + 15x + 4 is decreasing.
Solution
Recall: A function f (x) decreases provided that f (x) decreases as x
increases.
y = x 3 − 9 x 2 + 15x + 4
dy
= 3x 2 − 18 x + 15
dx
If y decreases then
dy
0
dx
Solution set = x : 1  x  5
(c)
(i)
R.T.Show

a
0
a
f ( x)dx =  f (a − x)dx
0
Proof
Taking L.H.S

a
0
f ( x)dx
Let t = a − x
x = a −t
dt
= −1
dx
dx = -dt
When x = a, t = 0
And when x = 0, t = a
a
0
0
a
  f ( x)dx = −  f (a − t )dt
=

a
=

a
0
0
f (a − t )dt
f (a − x)dx
L.H.S. = R.H.S.
Q.E.D
(ii)
R.T.Evaluate
5
 3 f ( x − 1)dx
1
Solution
Data:

4
0
f ( x)dx = 12
Let t = x – 1
dt = dx
When x = 5, t = 4
And when x = 1, t = 0
5
4
 3 f ( x − 1)dx =  3 f (t )dt
1
0
4
= 3  f (t )dt
0
4
= 3 f ( x)dx
0
= 3 x 12
= 36
JUNE 2005 UNIT 1 PAPER 1
f(x)
1.
(0, 12)
O
(a)
f(x)= x 3 + hx 2 − 8 x + k
x 3 + hx 2 − 8 x + k
x
(2, 0)
R.T. State the value of f(0) and f(2).
Solution:
f(x) cuts the vertical axis at f(0) = 12
When f(x) touches the horizontal axis at f(2), where f(2) = 0.
(b)
R.T.F the value of the constants h and k.
Solution:
f(0) and f(2) are the points on the graph where f(0) = 12 and f(2) = 0.
Hence (0) 3 + h(0) 2 − 8(0) + k = 12
 k = 12.
f (2) = 0
and 0 = (2) 3 + h(2) 2 − 8(2) + k
= 8 + h(4) − 16 + 12
When k = 12
0 = 4h + 4
 − 4 = 4h
and h = −1
The values of h and k are -1 and 12 respectively.
(c)
R.T.F factorise f(x) completely
Solution : f ( x) = x 3 − x 2 − 8x + 12
Since the curve touches the x-axis at (2, 0), then (x - 2) is a repeated factor.
By synthetic division
x  x  x = x 3 and − 2  −2  −3 = −12
Hence, the 3rd. factor is (x + 3)
f ( x) = ( x − 2)( x − 2)( x + 3)
2.
(a)
R.T.F. range of values of x  0 for which x 2 − 2 x − 3  0
Solution:
x2 − 2 x − 3  0
Rearranging the above inequality
x2 − 3  2 x
y
y
y = x2 − 3
y = 2x
y = 2x
y = -2x
O
x
O
-3
See diagram on overleaf
Consider x 2 − 3 = 2 x
Consider x 2 − 3 = −2 x
x
x 2 + 2x − 3 = 0
x 2 − 2x − 3 = 0
( x + 3)( x − 1) = 0
( x − 3)( x + 1) = 0
x = 1 or x = −3
x = −1 or x = 3
y
x
-3
-1
-3
Hence, x 2 − 3  2 x for x : −3  x  0
(b)
Data x  y and k  0
R.T.P. kx  ky
Proof:
x y
+ (-y) x + (− y)  y + (− y)
x− y 0
k  −ve
k
k ( x − y)  0
kx − ky  0
+ ky
kx − ky + ky  +ky
kx  ky
Q.E.D
3.
(a)
R.T.P. 11 + 7 =
4
11 − 7
Proof:
Consider R.H.S.
Rationalize denominator on R.H.S.
R.H.S. =
4
11 − 7

=
4( 11 + 7)
11 − 7
=
4( 11 + 7)
4
11 + 7
11 + 7
= 11 + 7 = L.H.S.
 L.H.S = R.H.S.
Q.E.D.
(b)
(i)
Data : x +
1
=1
x
R.T.S. x 2 +
1
= −1
x2
Proof:
2
1

2
 x +  = (1)
x


x2 + 2 +
and
Q.E.D.
x2 +
1
=1
x2
1
= −1
x2
(ii)
R.T.F. the value of x 3 +
1
x3
Solution
1
 2 1 
 x + 2  x +  = −1  1
x
x 

1
 3 1  
 x + 3  +  x +  = −1
x
x  

Recall: x +
1
=1
x
1 

Hence,  x 3 + 3  + 1 = −1
x 

and
Alternative method
 3 1 
 x + 3  = −2
x 

4.
R.T.Solve simultaneously x − 2 y = −3 and x 2 + 3 y = 7
Solution
Let x − 2 y = −3 …(1)
and x 2 + 3 y = 7 …(2)
From (1) x = 2 y − 3
Subst. into (2)
(2 y − 3) 2 + 3 y = 7
4 y 2 − 12 y + 9 + 3 y = 7
4y2 − 9y + 2 = 0
(4 y − 1)( y − 2) = 0
or
( y − 2) = 0
1
4
or
y=2
Subst. y =
1
into (1)
4
(4 y − 1) = 0
y=
1
x = 2  − 3
4
x=
x = 2(2) − 3
1
−3
2
x=−
Subst. y = 2 into (1)
x = 4−3
5
2
x =1
Solution: at x = −
5
1
and y = or
2
4
x = 1 and y = 2
Data: f :→ −2 x + 3
5.
(a)
R.T.S. that f is one-to-one
Solution
Let f ( x) = −2 x + 3
y
f(x)= - 2x + 3
3
A straight line
x
O
E.g.
3
2
When x = 3, f(x) = -3
When x = 8, f(x) = -13
A
B
x
f(x)
3
-3
8
-13
For every a  A there corresponds one value of f (a)  B (i.e. one to one
relationship) and for every f (a)  B there exists only one a  A
If f(a) = f(b), then a = b. Hence, f is one to one or injective.
(b)
R.T.F. x  R : f ( f ( x)) = f ( x) + 6
Solution
f ( x) = −2 x + 3
f ( f ( x)) = −2(−2 x + 3) + 3
= 4x − 3
Hence, 4 x − 3 = (− 2 x + 3) + 6
4 x − 3 = −2 x + 9
6 x = 12
x = 2, xR
6.
(a)
(i)
R.T.F. the coordinates of M
Solution
M lies on AB such that M is the midpoint of AB
 6 + (−4) − 5 + 3 
,
Midpoint of AB, M = 

2
2 

= (1, -1)
(ii)
R.T.F the gradient of the line through AB
Solution
Gradient of line AB =
=
3 − (−5)
−4−6
8
4
=−
− 10
5
(iii)
R.T.F. the equation of the line through M and N.
Solution
From the diagram, MN is the perpendicular bisector of AB.
 the gradient of MN =
5
4
( product of gradient of perpendicular lines = -1)
Equation of MN is y − (−1) =
5
( x − 1)
4
4( y + 1) = 5( x − 1)
4 y + 4 = 5x − 5
4 y = 5x − 9
(b)
R.T.F the coordinates of P
Solution
B (-4, 3)
P (x, y)
1
of 8 8 units
4
1
of 10
4
= 2
= 2.5
A (6, -5)
K(4, 3)
10 units
P lies on AB such that AP : PB = 3 : 1
Using similar traingles
Hence x-coordinate of P = − 4 + 2.5
= -1.5
And the y-coordinate of P = 3 – 2
=1
 the coordinates of P is ( -1.5, 1)
7.
R.T.Express f ( ) in terms of R cos( +  )
(a)
Solution
f ( ) = 2 cos  − sin 
Recall: cos( +  )  cos  cos  − sin  sin  (compound angle formula)
2 sin  − cos  = R(sin  −  )
R=
( 2)
2
sin  −
3
2
+ 12
1
3
cos  = cos  cos  − sin  sin 
y
R= 3
1

x
O
2
4
cos  =
4
17
 1 

 3
 = cos −1 
 = 0.615 radians
Hence,
2 cos  − sin  = 3 cos( + 0.615)
(b)
R.T.F. the minimum value of f ( )
Solution
2 cos  − sin  = 3 cos( + 0.615)
− 1  cos( + 0.615)  1
Hence, the minimum value of 3 cos( + 0.615) = 3  −1 = − 3
(c)
R.T.F. the value of  for which the minimum value occurs.
Solution
3 cos( + 0.615) is − 3
Recall: the minimum value of
Thus
3 cos( + 0.615) = − 3
cos( + 0.615) = −1
( + 0.615) = cos −1 (−1)
( + 0.615) = 
 =  − 0.615
 = 2.526
 = 2.53 rads. (to 3 s.f.)
0    2
8.
(a)
R.T.F. the range of values of k
Solution
x 2 + 2kx + 9 = 0 ...(1)
is of the form ax 2 + bx + c = 0
where a = 1, b = 2k and c = 9
For complex roots b2 – 4ac < 0
i.e. (2k)2 - 4(1)(9) < 0
4k2 - 36 < 0
4(k2 - 9) < 0
(k - 3)(k + 3) < 0
k
-3
Solution {k : −3  k  3}
3
(b)
R.T.Express
2 + 3i
in the form x + iy
3 + 4i
Solution
Rationalising the denominator
2 + 3i 2 + 3i 3 − 4i
=

3 + 4i 3 + 4i 3 − 4i
=
6 − 8i + 9i − 12i 2
9 − 16i 2
(Recall i 2 = −1 )
=
18 + i
25
=
18 1
+ i
25 25
Which is of the form x + iy where x =
18
1
 R and y =
R
25
25
9.
(a)
R.T.Express the points A (1, 2), B (2, 5) and C (0,- 4) as position vectors
Solution
(b)
A = (1, 2)

 OA = i + 2 j
B = (2, 5)

 OB = 2i + 5 j
C = (0, -4)

 OC = −4 j
R.T.F. the position vector of D
Solution

Let OD = ai + bj a, b  R


AB = CD



AB = AO + OB and



CD = CO + OD




 AO + OB = CO + OD

AB = −(i + 2 j ) + (2i + 5 j )
= i +3j

CD = −(−4 j ) + (ai + bj )
= ai + (b + 4) j
 i + 3 j = ai + (b + 4) j
Equating components
a = −1 , 3 = b + 4
b = −1

Hence, OD = i − j
10.
R.T.F the values of  for which the vectors (cos  )i + 3 j and
1
i + (sin  ) j are parallel.
4
Solution
Let a = (cos  )i + 3 j and b =
1
i + (sin  ) j
4
If a and b are parallel then b can be expressed as a scalar multiple of a
Let a = b where  is a scalar.
1

 (cos  )i + 3 j =   i + (sin  ) j 
4

Equating components of i and j respectively.
1
Let cos  = 
4
Let
3 =  sin 
…(1)
…(2)
3
sin 
From (2)  =
Subst. into (1)
cos  =
1
3

4 sin 
sin  cos  =
3
4
Recall: sin 2 = 2 sin  cos 
Hence,
1
3
sin 2 =
2
4
sin 2 =
3
2
 3

2 = sin −1 

2


0    2
0  2  4
−A
A
2 =
2 =
=
 3 
=
A = sin −1 

 2  3
 



,   − ,  2 + ,  2 +  − 
3 
3 
3 
3
 2 7 8
3
,
3
,
3
,
3
  7 4
, ,
,
6 3 6 3
Hence, the vectors are parallel for  =
  7 4
radians.
, ,
,
6 3 6 3
11.
(a)
R.T.P lim
h →0
Data:
(
x+h − x
1
=
h
2 x
x+h + x
)(
)
x+h − x =h
Proof
Taking L.H.S
lim
h →0
x+h − x
= lim
h →0
h
(
x+h + x
1
= lim
x+h + x
h →0
1
=
=
x+h − x
x+ x
1
2 x
Q.E.D
)(
x+h − x
)
y
(b)
y= x
Q
x+h
x+h
x
+h
P
R
x
O
x
x+h
R.T.Deduce, from first principles, the derivative of y = x
Solution
f ( x) = lim
h →0
f ( x + h) − f ( x )
h
Let f ( x) = x
f ( x) = lim
h →0
=
OR
1
2 x
x+h − x
h
(from (a))
Gradient of PQ =
QR
PR
=
x+h − x
( x + h) − x
=
x+h − x
h
As h → 0 , the chord PQ → tangent to the curve at P =
= gradient function.
 when y = x
dy
1
=
dx 2 x
1
2 x
(from (a))
12.
(a)
R.T.F. the values of for which f(x) is discontinuous.
Solution
Data: f ( x) =
=
x
x − 2x − 8
2
x
( x − 4)( x + 2)
As x → 4 , f (x) →  , i.e. f(x) is undefined.
As x → −2 , f (x) →  , i.e. f(x) is undefined.
Hence, f(x) is discontinuous at x = 4 and x = -2.
(b)
R.T.S x 3 = 8 + 4 x has a root in the closed interval [2, 3].
Proof
Let f ( x) = x 3 − 4 x − 8
f (2) = 2 3 − 4(2) − 8 = −8  0
f (3) = 33 − 4(3) − 8 = 7  0
 f (x) is continuous,  x as expected for a polynomial, and hence, continuous in
the interval 2,3 .Thus, by the Intermediate Value Theorem,  a root c, that lies
in [2, 3].
y
f(x)
7
O
-8
2
3
x
13.
(a)
R.T.F the value of k.
Data: y = 2 x 3 + kx − 5
Solution
dy
= 6x 2 + k
dx
When x = 1,
dy
= −2
dx
− 2 = 6(1) 2 + k
 k = −8
Hence, y = 2 x 3 − 8x − 5
(b)
R.T.F. the value of
d2y
at P
dx 2
Solution
d2y
= 12 x
dx 2
At x =1
d2y
= 12(1)
dx 2
= 12
(c)
R.T.F. the equation of the normal to the curve at P.
Solution
At P, the gradient of the curve = -2.
i.e. the gradient of the tangent to the curve at P = -2.
Hence, the gradient of the normal to the curve at P =
1
( the product of the
2
gradients of perpendicular lines = -1).
At P, when x = 1
y = 2(1) 3 − 8(1) − 5
= -11
The normal to the curve at P passes through (1, -11) and has a gradient =
Hence, the equation of the normal to the curve at P is given by
y − (−11) =
y + 11 =
1
(x − 1)
2
1
1
x−
2
2
y=
1
1
x − − 11
2
2
y=
1
23
x−
2
2
i.e. 2 y = x − 23
1
.
2
14.
(a)
Data: f : x → x 3 − 3x 2 − 9 x + 6
R.T.F the coordinates of the stationary points.
Solution
Let f ( x) = x 3 − 3x 2 − 9 x + 6
f ( x) = 3x 2 − 6 x − 9
At stationary points, f ( x) = 0
 3x 2 − 6 x − 9 = 0
 3 x 2 − 2x − 3 = 0
( x − 3)( x + 1) = 0
Either x = -1 or x = 3
When x = 3
f : (3) = 33 − 3(3) 2 − 9(3) + 6
= - 21
 stationary point is (3, -21)
When x = -1
f : (−1) = (−1) 3 − 3(−1) 2 − 9(−1) + 6
= 11
 stationary point is (-1, 11)
(b)
R.T.Determine the nature of the stationary points of f.
Solution
When f ( x) = 3x 2 − 6 x − 9
then f ( x) = 6 x − 6
For the stationary point at ( -1, 11)
f (−1) = 6(−1) − 6 = −12  0
Hence, (-1,11) is a maximum point.
For the stationary point at ( 3, -21)
f (3) = 6(3) − 6 = 12  0
Hence, (3, -21) is a minimum point.
y
15.
y = x 2 − 2x
-1
P
x
-1
O
R
Q
(a)
R.T.F the coordinates of the points P, Q and R
Solution
y = x 2 − 2x
Assuming that dotted line from P is vertical
At P, x = -1
Subst. x = -1
y = (−1) 2 − 2(−1)
=3
Coordinates of P is (-1, 3)
Assuming that Q is the minimum point of the curve y = x 2 − 2 x
dy
= 2x − 2
dx
At the minimum point,
2x − 2 = 0
x =1
dy
=0
dx
Subst. x = 1 into y
y = 12 − 2(1)
y = −1
Coordinates of Q is (1, -1)
Assuming that R lies on the x-axis
At R, y = 0
Subst. y = 0
x 2 − 2x = 0
x( x − 2) = 0
Either x = 0 or x = 2
x = 0 at O and x = 2 at R
Coordinates of R is (2, 0)
(b)
R.T. F area bounded by the curve and the lines x = -1 and x = 2
Solution
x = -1
(-1, 1)
x=2
y = x 2 − 2x
-1
P
x
-1
O
R
Q
(1, -1)
(2, 0)
Area of shaded region =

0
−1
( x 2 − 2 x)dx +
0
 (x
2
0
2
)
− 2 x dx
2
 x3

 x3

=  − x2  +  − x2 
3
 −1  3
0
3

−1  2
= 0 −  − 1 +  − 2 2  − 0
3
 3

=
4 4
+
3 3
=
8
square units.
3
June 2004 Unit 1 Paper 2
1.
(a)
Data: ( x − 1) and ( x − 2 ) are factors of f (x ) = x 3 + mx + n .
R.T.F. the value of m and n and the third factor.
Solution
f (x ) = x 3 + mx + n
Recall: Remainder and Factor theorem - If f ( x ) is any polynomial and f ( x ) is
divided by ( x − a ) , the remainder is f (a ) . If f (a ) = 0 then ( x − a ) is a factor of
f (x ) .
(x − 1) and (x − 2) are factors of
 f (1) = 0
(1)3 + m(1) + n = 0
m + n +1 = 0
Let m + n = −1 ...(1)
f (2) = 0
(2)3 + m(2) + n = 0
8 + 2m + n = 0
Let 2m + n = −8 ...(2)
From (1):
n = −1 − m
Substitute in (2)
2m + (−1 − m) = −8
2m + (−1 − m) = −8
2m − m = −8 + 1
m = −7
Substitute m = −7 in n = −1 − m
n = −1 − (− 7 )
n=6
 m = −7 and n = 6
f ( x ) (data)
 f (x ) = x 3 − 7 x + 6
(x − 1)(x − 2)(ax + b) = x 3 − 7 x + 6

ax 3 = x 3

a =1
And
− 1 −2  b = 6

b=3
ax + b = (x + 3)
 The third factor of f ( x ) is (x + 3)
(b)
Data: 2 y 2 − 9 y + 14  p( y − 1)( y − 2) + q( y − 1) + r
R.T.F p, q and r.
Solution
Let
y =1

2(1) − 9(1) + 14 = r
2
r = 7
Let
y=2

2(2) − 9(2) + 14 = q + 7
2
q = −3  
Similarly
Let
y=0

p(− 1)(− 2) − (− 3)(− 1) + 7 = 14
2 p + 3 + 7 = 14
p = 2
Hence: p = 2 , q = −3 and r = 7 .
(c)
Data: 2 x − 3  5
(i)
R.T.F. the range of values of x
Solution
y = 2x − 3
Let
y = 3 − 2 x …(1)
And
y = 5 …(2)
 5 = 3 − 2x
x = −1
Let
y = 2 x − 3 ...(3)
And
y = 5 …(4)

5 = 2x − 3
x=4
For the interval − 1  x  4 , the graph of y = 2 x − 3 lies under the line
y = 5 and so the range of x   for 2 x − 3  5 is x : −1  x  4.
(ii)
R.T.F. the least possible value of x + 1.
Solution
xA
 −A  x  A

(form definition)
2x − 3  5
 −5  2 x − 3  5
(+ 5)
0  2 x + 2  10
( 2)
0  x +1  5
 The least possible value of
(iii)
(x + 1)
is 0
R.T.F. the greatest possible value of x + 1.
Solution
0  x +1  5
The greatest possible value of
(x + 1)
is 5.
2.
(a)
(i)
R.T.E. f ( x ) = 12 x − 2 x 2 in the form A + B(x + p ) and also find the
2
maximum value.
Solution
f ( x ) = 12 x − 2 x 2
= −2 x 2 + 12 x
(
= −2 x 2 − 6 x

= −2(x − 3)
)
f (x ) = −2 x 2 − 6 x + (− 3) − (− 3)
2
2
2


−9
 f (x ) = 18 − 2(x − 3)
2
This is of the form A + B(x + p )
2
Where A = 18   , B = −2   and p = −3   .
From the completed square form
(x − 3)2  0
x , the maximum value of
f (x ) = 18 .
Maximum value occurs when
(x − 3)2 = 0
x = 3
 Coordinates of Max = (3,18)
(ii)
R.T.Sketch
f ( x ) = 12 x − 2 x 2
Solution
The curve cuts the y-axis when x = 0
y = 12(0) − 2(0) = 0
2
 Curve passes through origin.
The curve cuts the x - axis when y = 0 .
 18 − 2(x − 3) = 0
2
18 = 2(x − 3)
2
(x − 3) = 3
or
x=6
or
(x − 3) = −3
x=0
Hence, the curve cuts x - axis at (6,0 ) and (0,0 ) .
Coefficient of x 2 , − 2 is –ve indicates a max. point for the quadratic curve.
max pt.
(iii)
R.T.P. f (x ) = −2 x 2 + 12 x is NOT one-to-one
Proof: There may be more than one value of x mapped onto the same
value of f ( x )
e.g.
y
f (0) = 0
f (6) = 0
0
0
6
0
6
x
(i)
(a)
R.T.Complete the table for f ( x) = sin x .
Solution
(b)
x
0

2

3
2
2
f (x )
0
1
0
-1
0
R.T.Sketch the graph of f.
Solution
(ii)
R.T.Sketch f ( x) = sin x
Solution
This region is to be reflected in the x-axis.
(iii)
R.T.F. the solution set of the equation sin x = sin x .
Solution
The solution set for sin x = sin x is x : 0  x     x = 2 
3.
(a)
(i)
R.T.F The equation of line AB
Solution
Gradient of AB =
4 − (− 8)
4 − (− 4 )
=
12
8
=
3
2
Equation of line AB is
y − (− 8) 3
=
x − (− 4 ) 2
y +8 3
=
x+4 2
2 y + 16 = 3x + 12
2 y = 3x − 4
Equation of AB is y =
(ii)
3
x−2
2
R.T.F The equation of line PQ
Solution
PQ is perpendicular to AQB
Hence gradient of PQ = −
2
(the product of the gradients of
3
perpendicular lines is = -1)
Equation of line PQ is
y −3
2
=−
x − (− 1)
3
3 y − 9 = −2 x − 2
3 y = −2 x + 7
(iii)
R.T.F. the coordinates of Q.
Solution
Let y =
3
x − 2 ...(1)
2
Let 3 y = −2 x + 7 …(2)
Solving simultaneously to obtain the point of intersection i.e. co-ordinates
of Q .
Substitute (1) into (2)
3

3 x − 2  = −2 x + 7
2

9
x − 6 = −2 x + 7
2
( 2)
9 x − 12 = −4 x + 14
13x = 26
x=2
When x = 2
y=
3
(2) − 2
2
y =1
Co-ordinates of Q is (2,1)
(b)
R.T.S. 6 cos 2  + sin  = 4 for 0    180
Solution
6 cos 2  + sin  = 4
Recall: cos 2  = 1 − sin 2 
(
)
6 1 − sin 2  + sin  − 4 = 0
6 − 6 sin 2  + sin  − 4 = 0
6 sin 2  − sin  − 2 = 0
(3 sin  − 2)(2 sin  + 1) = 0
sin  =
2
or
3
For sin  =
sin  = −
1
2
2
3
180 - A
A
sin  is + ve in the 1st and 2nd Quadrant
2
3
 = sin −1  
 = 41.81 (Acute, 1st quadrant)
 = 180  − 41.81 (2nd quadrant)
 = 138.19 
OR
sin  = −
1
2
 1
 2
 = sin −1  − 
 = 30  (Acute)
sin  is – ve, hence  lies in the 3rd and 4th Quadrants. (Out of Range)
A= 30 
 = 180 + A
 = 360 − A
There are no values of    for given range.
 = 41.8  ,138.2  (to 1 d.p.)
(c)
R.T.S sin x + sin 3x = 0 for x : 0  x  
Solution:
 P +Q   P −Q 
Recall: sin P + sin Q = 2 sin
 cos

 2   2 
 3x + x   3x − x 
sin x + sin 3 x = 2 sin
 cos

 2   2 
= 2 sin 2 x cos x
Now 2 sin 2 x cos x = 0 where 0  x  
cos x = 0
x=
where 0  x  

sin 2 x = 0 where x : 0  2 x  2
2 x = 0,  ,2
2
x = 0,
Solution x = 0 ,

2
, 0  x   .

2
,
4)
(a)
R.T.Express w =
z −1
in the form x + iy x, y  
z+2
Solution:
w=
(x − 1) + iy
(x + 2) + iy
Rationalizing by x complex conjugate of denominator.
=
(x − 1) + iy  (x + 2) − iy
(x + 2) + iy (x + 2) − iy
=
(x − 1)(x − 2) − iy(x − 1) + iy(x + 2) − i 2 y 2
( x + 2 )2 − i 2 y 2
Recall: i 2 = −1
=
(x − 1)(x + 2) + y 2 + i( y(x + 2) − y(x − 1))
( x + 2 )2 + y 2
=
(x − 1)(x + 2) + y 2 +
3y
i
2
2
(x + 2) + y
( x + 2 )2 + y 2
(x + iy )
This is of the form
where x =
and
(b)
Data: arg(w)=
(i)
(x − 1)(x + 2) + y 2  
( x + 2 )2 + y 2
y=
(x + 2)2 + y 2

4
R.T.F equation connecting x and y
Calculation:
arg(w)=

4
Let the argument of w =  .
i.e. tan  =
y
x
3y

3y
 x 2 + 4x + 4 + y 2
Hence, tan = 2
4
x + x − 2 + y2
x 2 + 4x + 4 + y 2
1=
3y
x 2 + 4x + 4 + y 2

x 2 + 4x + 4 + y 2 x 2 + x − 2 + y 2
1=
3y
x + x − 2 + y2
2
x 2 + x − 2 + y 2 = 3y
x 2 + x − 2 + y 2 − 3y = 0
x 2 + y 2 + x − 3y − 2 = 0
which is of the form ax 2 + by 2 + cx + dy + f = 0
where a = 1, b = 1, c = 1, d = −3, f = −2
and a, b, c, d , f are integers
(ii)
R.T.P. that the equation in (i) represents a circle, C
Proof
x 2 + y 2 + x − 3y − 2 = 0
is of the form x 2 + y 2 + 2 gx + 2 fy + c = 0
(where g, f & c are constants) & 2 g = 1,2 f = −3 and c = −2
g =
1
−3
and c = −2
,f =
2
2
 x 2 + y 2 + x − 3 y − 2 = 0 is the equation of a circle
(iii)
R.T.F the centre and radius of the circle, C
Solution
for the circle, x 2 + y 2 + 2 gx + 2 fy + c = 0
the centre = (− g ,− f )
and radius = g 2 + f 2 − c
Hence, for the circle, x 2 + y 2 + x − 3 y − 2 = 0
2g = 1
g=
2 f = −3
and
1
2
−g =−
f =−
1
2
−f =
 1 3
 Centre =  − , 
 2 2
Radius = g 2 + f 2 − c
1
 − 3
=   +
 − (− 2)
2
 2 
2
2
=
1 9
+ +2
4 4
=
5
+2
2
=
9
2
 Radius =
3
2
units
3
2
3
2
(c)
R.T.F the position vector of P .
Solution
The sketch shows a parallelogram OLMN whose diagonals OM and LN intersect at P
Position vectors of L and N relative to origin O , are
− 3i + 6 j and 2i + 3 j respectively
OL = −3i + 6 j
ON = 2i + 3 j
NM = OL (opposite sides of parallelogram are parallel and equal)
OM = ON + NM
The diagonals of a parallelogram bisect each other
 OP =
1
OM
2
OM = ON + NM
= (2i + 3 j ) + (− 3i + 6 j )
= −i + 9 j
 OP =
1
(− i + 9 j )
2
=
−i 9
+ j
2 2

1
9
Position vector of P, OP = − i + j
2
2
5.
(a)
R.T.E lim
x →3
x 2 − 2x − 3
x 2 − 4x + 3
Solution
Let f (x ) =
f (3) =
x 2 − 2x − 3
x 2 − 4x + 3
9−6−3 0
(indeterminate)
=
9 − 12 + 3 0
 Factorising and cancelling
x 2 − 2x − 3
(x + 1)(x − 3)
lim 2
= lim
x →3 x − 4 x + 3
x →3 ( x − 1)( x − 3)
(x + 1)
(x − 1)
(3 + 1)
=
(3 − 1)
= lim
x →3
=
4
2
=2

(b)
x 2 − 2x − 3
lim 2
=2
x →3 x − 4 x + 3
R.T.C. the values of x   for which
(x + 2)
x(x + 1)
is not continuous.
Calculation
Let f ( x ) =
(x + 2)
x(x + 1)
f ( x ) is discontinuous when denominator = 0
i.e. x(x + 1) = 0
x = 0 or x = −1
Hence, f ( x ) is discontinuous at x = 0 or -1.
(The lines x = 0 and x = −1 are vertical asymptotes of f ( x ) )
(c)
Data: y =
R.T.F.
x2 −1
x2 +1
dy
dx
Solution
y=
u
x2 −1
of the form y =
2
v
x +1
Where u = x 2 − 1
v = x2 +1
and
du
= 2x
dx
dv
= 2x
dx
dy
=
dx
Recall:
v
du
dv
−u
dx
dx
v2
(
)
(
)
dy
x 2 + 1 (2 x ) − x 2 − 1 (2 x )
=
2
dx
x2 +1
( )
(2 x + 2 x ) − (2 x − 2 x )
=
(x + 1)
3
3
2
2
=
2x3 + 2x − 2x3 + 2x
(x
2
)
+1
2

dy
4x
=
in terms of x
dx x 2 + 1 2
(ii)
R.T.P
(
)
(
4x
) dy
− 4y =
dx
x +1
x x2 +1
2
Proof:
(
) dy
− 4y
dx
Taking L.H.S = x x 2 + 1
Recall:
x2 −1
dy
4x
=
& y= 2
x +1
dx x 2 + 1 2
(
)
 4x
dy
x x +1
− 4 y = (x 3 + x )
 (x 2 + 1)2
dx

(
)
2
(
4x x3 + x
=
(x
2
)
+1
2
) − 4(x
  x2 −1
 − 4

  x 2 + 1 

)
−1
x +1
2
2
(
) ( )(
(x + 1)
+ 4 x − 4(x − 1)
(x + 1)
)
4x 2 x 2 + 1 − 4 x 2 −1 x 2 +1
=
2
2
4x 4
=
2
4
2
2
4x 4 + 4x 2 − 4x 4 + 4
=
(x
2
)
+1
2
4x 2 + 4
=
(x + 1)
4(x + 1)
=
(x + 1)
2
2
2
2
2
=
4
x +1
(
2
)
= R.H.S
L.H.S. = R.H.S.
Q.E.D.
(d)
R.T.F. the range of values of x for which x 5 − 5 x + 3 is decreasing.
Solution
Let f (x ) = x 5 − 5x + 3
 f (x ) = 5x 4 − 5
( )
= 5(x − 1)(x + 1)
5  (x + 1) > 0 , x
= 5 x 4 −1
2
2
2
f ' (x ) = −ve iff x 2 − 1  0
(x − 1)(x + 1)  0
Range of values of x , for which f ' (x )  0
Hence, the solution set is x : −1  x  1
6.
(a)
f ( x ) = x 3 − 3x + 2
R.T.F the stationary points of the curve
Solution:
Recall: Stationary points occur when f ' ( x ) = 0
f ( x ) = x 3 − 3x + 2
f ' ( x ) = 3x 2 − 3 = 0
3x 2 = 3
x2 = 1
 x = 1
When x = +1
f (1) = 1 − 3 + 2 = 0
When x = −1
f (− 1) = −1 + 3 + 2 = 4
Stationary points occur at the pts (1,0 ) and (− 1,4)
(b)
R.T.F the nature of these stationary points
Solution:
f ' ' (x ) = 6 x
When x = 1
f ' ' (1) = 6  min. pt at (1,0 )
When x = −1
f ' ' (− 1) = −6  max. pt at (− 1,4 )
(c)
R.T.P
f ( x ) touches the x axis at x = 1
Proof:
Dividing
f ( x ) by ( x − 1)
x2 + x − 2
(x − 1) x 3 − 3x + 2
−
x3 − x 2
x 2 − 3x + 2
−
x2 − x
− 2x + 2
−
− 2x + 2
0
(
f (x ) = (x − 1) x 2 + x − 2
)
= (x − 1)(x − 1)(x + 2)
= (x − 1) (x + 2)
2
Whenever there is a repeated root, the x-axis is a tangent at that point.
 The x-axis is a tangent at the point (1,0 ) showing that f ( x ) touches the x
axis at x = 1 .
(d)
O
(e)
R.T.F Area bounded by curve and x axis for − 2  x  1
Solution:
Area =
 f (x )dx
1
−2
=  (x 3 − 3x + 2)dx
1
−2
1
 x 4 3x 2

= −
+ 2 x
2
4
 −2
1 3

=  − + 2  − (4 − 6 − 4 )
4 2

=
3
+6
4
3
= 6 sq.units
4
JUNE 2004 Unit 1 Paper 1
1.
Data: f (x ) = x 3 − p 2 x 2 + 2 x − p
R.T.F. the value of p.
Solution
f (x ) = x 3 − p 2 x 2 + 2 x − p
Recall: Remainder and Factor Theorem - If f ( x ) is any polynomial and f ( x ) is divided
by ( x − a ) . Then the remainder is f (a ) . If f (a ) = 0 , then ( x − a ) is a factor of f ( x ) .
Hence f (− 1) = (− 1) − p 2 (− 1) + 2(− 1) − p = −5
3
2
−1 − p 2 − 2 − p + 5 = 0
− p2 − p + 2 = 0
p2 + p − 2 = 0
( p − 1)( p + 2) = 0
p = 1 or p = −2
 p = 1 or -2
2.
(a)
Given x  y and k  0
R.T.P. kx  ky
Proof:
x y
(-y)

x− y  y− y
x− y 0
Multiplying by –ve changes the sign of an inequality.
(= k )
kx − ky  0
+ ky
kx − ky + ky  ky

kx  ky
Q.E.D.
(b)
R.T.S. x 2 − 6 x + 8 = 0
Solution
x2 − 6 x + 8 = 0
x2 + 8 = 6 x
y
y = x2 + 8
y = −6 x
y = 6x
8
x
-4
x 2 + 8 = 6x
x 2 − 6x + 8 = 0
(x − 4)(x − 2) = 0
 x = 4 or x = 2
By Symmetry
x = −4 or x = −2
 x = −4 or -2 or 2 or 4
-2 O
2
4
3.
(a)
4
= 2 2− x
x
2
R.T.P.
Proof:
4
2x
=
22
2x
= 2 2− x
Q.E.D.
(b)
R.T.S. 2 x + 2 2− x = 5
Solution:
2 x + 2 2− x = 5
2 x + 22 2−x = 5
Let t = 2 x

2−x =
1
t
1
 t + 4  = 5
t 
t 2 − 5t + 4 = 0
(t − 1)(t − 4) = 0
t = 1 or 4
When t = 1
2x =1
2 x = 20
x=0
When t = 4
2x = 4
2x = 22
x=2
Hence, x = 0 or 2.
4.
Data:
f : x → −3x + 6
g :x → x+7
R.T.S. f ( g (2 x + 1) = 30
Solution
f ( g (2 x + 1) = 30
f ( g (2 x + 1) = f (2 x + 1 + 7)
= f (2 x + 8)
= −3(2 x + 8) + 6
= −6 x − 18
Hence, − 6 x − 18 = 30
− 6 x = −48
x=8
5.
(a)
R.T.C. the value of 
Calculation
( AB )2 = (OA)2 + (OB )2 − 2(OA)(OB ) cos(Cosine

rule)
2
2
2
(8) = (5) + (5) − 2(5)(5) cos 
64 = 50 − 50 cos 
2
14 = −50 cos 
14
50
 7 
 = cos −1  − 
 25 
cos  = −
5
4
 7 
 =  − cos −1  
 25 
= 1.85 rad (to 2 d.p)
Alternative Method
  4
sin  =
2 5

4
= sin −1  
2
5
4
  = 2 sin −1  
5
 = 1.85 rad (to 2 d.p)

(b)
R.T.C. the length of the arc ACB.
Calculation
Reflex AOˆ B =  − 1.85
Length of arc = r
 in radians
 7 
= 5

 5 
= 7
= 22m
6.
Data: x = 2t 2 + 3 and y = 3t 4 + 2
R.T.F. the equation of a curve in the form y = Ax 2 + Bx + C
Solution
Using x = 2t 2 + 3
t2 =
x −3
2
Substitute t 2 into y
 x −3
y = 3
 +2
 2 
3
2
y = ( x − 3) + 2
4
3
y = x 2 − 6x + 9 + 2
4
3
9
35
y = x2 − x +
4
2
4
2
(
)
is of the form y = Ax 2 + Bx + C where
A=
3
 ,
4
9
B = − 
2
,
C=
35

4
7.
x−2
0
x+3
R.T.F. the range of values of
Solution
x−2
0
x+3
x  −3
Multiply both sides by (x − 3)
2
(x − 2)(x + 3)  0
Coefficient of x 2 = +ve

y = (x − 2)(x + 3)
+
+
-3
x = x : x  2 x : x  −3
2
8.
(a)
R.T.P
sin 2 A
 cot A
1 − cos 2 A
Proof:
Taking L.H.S
Recall: sin 2 A = 2 sin A cos A
cos 2 A = 1 − 2 sin 2 A

2 sin A cos A
sin 2 A
=
1 − cos 2 A 1 − (1 − 2 sin 2 A)
=
2 sin A cos A
2 sin 2 A
=
cos A
sin A
= cot A
= RHS
L.H.S = R.H.S.
Q.E.D.
(b)
R.T.S. cos 2 = 3 cos  − 2 for 0    
Solution
cos 2 = 3 cos  − 2
Recall: cos 2 = 2 cos 2  − 1
 2 cos 2  − 1 = 3 cos  − 2
2 cos 2  − 3 cos  + 1 = 0
(2 cos  − 1)(cos  − 1) = 0
 cos  =
cos  =
=
1
2
or
1
1
2

3
0  
or cos  = 1
 = 0 0  
Hence,  = 0 or

3
for 0    
9.
Data: Given that x 2 − 3x − 1 = 0 has roots  and  .
R.T.F. the equation whose roots are 1 +  and 1 +  .
Solution
If ax 2 + bx + c = 0
x2 +
b
c
x+ =0
a
a
If  ,  are the roots,
Then
(x −  )(x −  ) = 0

x 2 = ( +  ) +  = 0
Equate coefficients
b
a

 + =

 =
 in
x 2 − 3x − 1 = 0

 + =

 =
c
a
− (− 3)
=3
1
−1
= −1
1
Equation is x 2 − (sum of roots) x + product of roots = 0
x 2 − x(( +  ) + 2) + ( + ( +  ) + 1) = 0
Equation whose roots are 1 +  and 1 +  is
x 2 − x(1 +  + 1 +  ) + (1 +  )(1 +  ) = 0
x 2 − x( +  + 2) + ( +  +  + 1) = 0
 x 2 − x(3 + 2) + (− 1 + 3 + 1) = 0
x 2 − 5x + 3 = 0
10.
(a)
Data: The position vector of the point P is i + 3 j .

R.T.F. the unit vector in the direction OP .
Solution
The unit vector in the direction of OP = i + 3 j is  (i + 3 j )
Since it is a unit vector, its modulus is = 1 and  = a scalar
i.e.
 (i + 3 j ) = 1
i + 3j = 1

( )2 + (3 )2
=1
 2 + 9 2 = 1
10 2 = 1
=
1
10
j
i + 3j
0
i
 The unit vector in the direction of OP is
1
(i + 3 j )
10
(b)
Data: OQ = 5

R.T.F. the position vector of OQ
Solution
OQ = 5

( ) + (3 ) = 5
2
2
10 2 = 5
=

(c)
5
5 5
5
5
=
=
=
2
10
5 2
2
OQ =
5
(i + 3 j )
2
Data: 3ti + 4 j is perpendicular to OP
R.T.F the value of t
Solution
Since 3ti + 4 j is perpendicular to OP , their dot product = 0.
. i + 3 j) = 0
i.e. (3ti + 4 j )(
(3t  1) + (4  3) = 0
3t + 12 = 0
3t = −12
t = −4
11.
Data: lim 4 f (x ) = 5
x → −2
R.T.C lim  f (x ) + 2 x 
x → −2
Calculation:
lim 4 f (x ) = 5
x → −2
4 lim f ( x ) = 5
x → −2
lim f (x ) =
x → −2
5
4
R.T.Evaluate lim  f (x ) + 2 x
x →−2
Solution
lim  f (x ) + 2 x = lim f (x ) + lim 2 x (Sum Law)
x →−2
x →−2
x →−2
= lim f (x ) + 2 lim x
x →−2
=
5
+ 2(− 2)
4
1
=1 − 4
4
=
− 11
4
x →−2
R.T.Differentiate f ( x ) = x 3 from first principles.
12.
Solution
y
y = x3
Q
(x + h)3
x3
(x + h, (x + h)3)
P
R
x
x+h
x
QR ( x + h ) − x 3
=
(x + h ) − x
PR
3
Grad of PQ =
=
x 3 + 3x 2 h + 3xh 2 + h 3 − x 3
h
= 3x 2 + 3xh + h 2
As h → 0 , gradient of PQ → gradient of the tangent of the curve
f ( x) = lim 3x 2 + 3xh + h 2
h →0
= 3x 2
13.
Data: f (x ) = rx 2 + sx + t
(a)
(i)
r0
R.T.F. f ' ( x )
Solution
f ' (x ) = r (2 x ) + s
f ' ' ( x ) = 2r
(ii)
R.T.F. f ' ' ( x ) = 2r
Solution
f (x ) = 2rx + s
f ' ' ( x ) = 2r
(b)
R.T.F. the conditions under which f ( x ) has a maximum point.
Solution
The stationary point of f ( x ) occurs at f ' ( x ) .
i.e.
when 2rx + s = 0
2rx = − s
x=
−s
2r
For this point to be a maximum when substituted into f ' ' (x ) result must be
negative.
i.e.
2r  0
(assuming s  0 )
r0
(c)
R.T.F. the maximum value.
Solution
−s
The maximum value of f ( x ) occurs at x = 

 2r 
−s −s
−s
f
 = r
 + s
+t
 2r   2r 
 2r 
2
 s2
= r  2
 4r
=
 s2
 − + t
 2r
s2 s2
− +t
4r 2r
s2  1 
=  − 1 + t
2r  2 
=t−
s2
4r
14.
Data: y = px 3 + qx 2 + 3x + 2 passes through T (1,2 ) where gradient is 7 ,
R.T.F. the values of p and q.
Solution
y = px 3 + qx 2 + 3x + 2
dy
= 3 px 3 + 2qx 2 + 3
dx
When x = 1 ,
dy
=7
dx
7 = 3 p(1) + 2q(1) + 3
3
7 = 3 p + 2q + 3
3 p + 2q = 4 ...(1)
Sub (1,2) into given equation
2 = p(1) + q(1) + 3(1) + 2
3
2
2 = p + q +3+ 2
p + q = −3 ...(2)
Solving simultaneously to find p and q
From (2), q = −3 − p
Sub into (1)
3 p + 2(− 3 − p ) = 4
3p + 6 − 2p = 4
p = 10
Sub p = 10 in (2)
10 + q = −3
q = −13
 p = 10 and q = −13
15.
y
(4, 2)
A
y=-x+2
y=x-2
x
0
(a)
B
4
R.T.F. the coordinates of A and B
Solution
y = x − 2 cuts the x-axis at y = 0
x=2

B = (2,0)
y = − x + 2 cuts the x-axis at x = 0
y=2

(b)
A = (0,2)
R.T.F. the volume generated.
Solution
When
AOB is rotated about the x-axis through 360  , the figures generated
is a cone of radius OA = 2 units and height OB = 2 units
1
Volume of cone = r 2 h
3
1
2
=  (2) (2)
3
=
8
units 3
3
where h = length of OB
= 2units
where r = length of OA
= 2units
Alternative Solution
x2
V =   y 2 dx
x1
=   (− x + 2) dx
2
2
0
=   (x 2 − 4 x + 4)d x
2
0
2
 x3

=   − 2 x 2 + 4 x
3
0
 (2)3
 0

2
=  
− 2(2) + 2(2)  −  − 2(0) + 0 

 3
 4
 2

=   2 − 8 + 8
 3

8
=  units 3
3
June 2003 Unit 1 Paper 2
1)
a)
Data : (x – 1) and (x + 2) are factors of f (x ) = x 3 + px + q .
R.T.F. the values of p and q.
Recall: Remainder and Factor Theorem - If f ( x ) is any polynomial and f ( x ) is
divided by (x – a), then the remainder is f (a ) . If f (a ) = 0, then ( x − a ) is a factor
of f ( x ) .
From data f (1) = 0
 f (1) = (1) + p(1) + q = 0
3
1+ p + q = 0
p + q = −1 ...(1)
f (− 2) = 0
f (− 2) = (− 2) + p(− 2) + q = 0
3
−8− 2p + q = 0
− 2 p + q = 8 ...(2)
Solving simultaneously
(2) - (1)
− 3p = 9
 p = −3  
Sub into (1)
1 + (− 3) + q = 0

q = 2
Solution p = −3 and q = 2
R.T.F. the remainder when f ( x ) is divided by (x – 1)
Solution
f ( x ) = x 3 − 3x + 2
When f ( x ) is divided by (x – 1) remainder = f (− 1)
f (− 1) = (− 1) − 3(− 1) + 2
3
= −1 + 3 + 2
=4
 Remainder = 4
(b)
A
3 cm
120
2 cm
0
C
B
(i)
R.T.F. the length of BC
Solution
BC 2 = (2) + (3) − 2(2)(3) cos 120 
2
2
 1
BC 2 = 4 + 9 − 2(6 ) − 
 2
BC 2 = 4 + 9 + 6
BC 2 = 19
BC = 19 cm
BC = 4.36 cm
(to 2 d.p)
(Cosine Law)
(ii)
R.T.F. the value of sin C.
Solution
2
19
=
sin C sin 120 
(Sine Rule)
 2 sin 120 0 = 19 sin C
2 sin 120 0
19
= sin C
0.3973 = sin C
0.397 = sin C (3 s.f.)
(c)
(i)
R.T.P. area of shaded region = 2( − 2 2 ) .
Proof
Area of shaded region = Area of Sector AOB - Area of ABO
Area of Sector AOB =
1 2 
( 4)  
2
4
= 2 cm 2
Area of
ABO =
1
(4)(4)sin 
2
4
= 8 sin
=

4
8
2
= 4 2 cm 2
 Area of shaded region = 2 − 4 2
(
)
= 2  − 2 2 cm 2
Q.E.D
(ii)
(
)
R.T.P. the length of chord AB = 4 2 − 2 .
Proof
Using Cosine Rule
( AB )2 = (OA)2 + (OB )2 − 2(OA)(OB ) cos 
4
( AB )2 = (4)2 + (4)2 − 2(4)(4)
( AB )2 = 32 − 16
( AB )2 = 16(2 −
(
2
2
AB = 16 2 − 2
)
(2 − 2 )
(2 − 2 )
AB = 16 
AB = 4
)
Q.E.D
1
2
2.
(a)
R.T.Sketch y = f (x − 1)
(i)
y
y = f ( x − 1)
y = f (x)
x
O
(2,-2)
(3,-2)
f (x − 1) is mapped onto f ( x ) by a horizontal shift of one unit to the right
1
defined by the vector   . This is a congruent transformation.
 0
1
 2  T  0   2 + 1   3 
  ⎯⎯
 =  
⎯→
 − 2
 − 2 + 0  − 2
(ii)
R.T.Sketch y = f (x ) + 3
y
y = f ( x) + 3
(2,1)
x
y = f (x)
(2,-2)
f (x ) + 3 is mapped onto f ( x ) by a vertical shift of 3 units upwards
 0
defined by the vector   . This is a congruent transformation.
 3
0
 2  T  3   2 + 0   2 
  ⎯⎯
 =  
⎯→
 − 2
 − 2 + 3  1 
(iii)
R.T.Sketch y = f ( x )
Solution
y
y = f (x)
Max pt. (2, 2)
2
x
O
-2
The region only below the x-axis is reflected in the x-axis for the modulus
graph.
b)
Data: A = {x : 0  x  4}
B = {x : 0  x  8}
f : x → x( 4 − x)
(i)
R.T.Sketch f : x → x(4 − x)
Solution
f (x )
0 x4
y
4
(2,4) max. pt.
x
O
f (x ) = x(4 − x )
2
4
f (x ) = 4 x − x 2 = 0
4x − x 2 = 0

x = 0 or 4
f ' (x ) = 4 − 2 x = 0

x=2
f ' ' (x ) = −ve
f (2 ) = 4
 (2,4) is a maximum pt.
(ii)
R.T.F. C such that C  A and f : C → B is one to one.
Solution
f (x )
y
4
x
O
2
f :A→B
CA
For 0  x  2
For x  C  only 1 f (x )  B
For a, b  C if f (a ) = f (b )

a=b
f is 1-1 for 0  x  2 as shown in the graph
Similarly
y
4
2
O
4
x
For 2  x  4
CA
(iii)
R.T.P. f is not onto.
Proof
x A 0  x  4
B 0  f (x )  8
4  f (x )  8
There are no elements of f ( x ) that are mapped onto 4  x  8

Range  Co-domain
Hence f ( x ) is not onto.
Q.E.D.
(iv)
R.T.P. f : A → B is not one to one.
Proof
f (x ) = x(4 − x )
x=0
f (0 ) = 0
x=4
f (4 ) = 0
Solution is x = 0 or 4
A
B
0
0
4
Hence, f is not one to one.
(v)
R.T.F. the range of values of y for which f ( x) = y has a solution.
Solution
y
8
()
y = f x  4
4
x
4
0  f (x )  4
 The equation possesses a solution when 0  f (x )  4 as illustrated.
3.
(a)
R.T.F The co-ordinates of A , B and C
Solution
At A , x = 0 in eq’n 2 x + 3 y = 6
0 + 3y = 6

3y = 6
y=2
A = (0,2)

At B , y = 0 in eq’n 2 x + 3 y = 6
2x + 0 = 6
2x = 6
x=3
B = (3,0)

Let C = (x 2 , y 2 )
A
(0,2)
(0 + x2 )
2
B
(3,0)
=3
 x2 = 6
(2 + y 2 )
2
=3
 y 2 = −2
Hence, C = (6,−2)
C
(x2,y2)
(b)
R.T.C the equations of CD and AD
Calculation:
Equation of AC is:
3 y = −2 x + 6
y=−
2
x+2
3
 Gradient of AC is = −
2
3
Hence Gradient of CD =
3
(product of the gradient of
2
Using pt C, (6, -2)
Equation of CD is:
y+2 3
=
x−6 2
2( y + 2) = 3(x − 6)
2 y + 4 = 3x − 18
2 y = 3x − 22
y=
3
x − 11
2
AD is parallel to the line y = −5 x + 7
 grad AD = -5 (parallel lines have the same gradient)
Using A(0,2) with gradient = -5.
Equation of AD is:
y−2
= −5
x−0
y − 2 = −5 x
y = −5 x + 2
lines = -1)
(c)
R.T.F co-ordinates of D
Solution
Solving the equations of AD and CD simultaneously to obtain D
Let
y = −5 x + 2 …(1)
Let
y=
3
x − 11 …(2)
2
Subst. (1) into (2)
3
x −11 = − 5 x + 2
2
1
6 x = 13
2
x=2
Substitute x = 2 into (1)
y = −5(2) + 2
y = −10 + 2
y = −8
 D = (2,−8)
d)
R.T.F Area of Triangle ACD
Solution
Length of AC =
=
(− 2 − 2)2 + (6 − 0)2
(16) + (36)
= 52
Length of CD =
(− 2 + 8)2 + (6 − 2)2
=
(6)2 + (4)2
= 52

Area of Triangle ACD =
=
L B
2
52  52 52
=
= 26 units 2
2
2
4.
(a)
R.T.S cos 2 − 3 cos  = 1
for
0    2
and
cos  = 2  no real solution since
Solution:
Recall: cos 2 = 2 cos 2  − 1

2 cos 2  − 1 − 3 cos  = 1
2 cos 2  − 3 cos  − 2 = 0

(2 cos  + 1)(cos  − 2) = 0
cos  = −

1
2
− 1  cos   1 for 
A = cos
 −A
−1  1 
 =
2
+A

 = −
=

3
,+
2
4
or
3
3

3
0  x  2

3
(b)
Data: cos A =
R.T.F. tan
3
5
A
2
Solution
Assuming the angle is acute
y
+5
x = +4
A
x+
O
+3
5 2 = 3 2 + x 2 (Pythagoras’ Theorem)
x 2 = 5 2 − 32
If A is acute then x = +4
x = +4
tan A =
+4 4
=
+3 3
tan 2 =
2 tan 
1 − tan 2 
A
2
tan A =
2 A
1 − tan
2
2 tan
Let
t = tan
A
2
4
2t
=
3 1− t2
4t 2 + 6t − 4 = 0
()
2t 2 + 3t − 2 = 0
(2t − 1)(t + 2) = 0
 Now tan
1
A
= −2 or
2
2
Since A is acute then
 tan
A
is acute
2
1
A
only
= +ve and
2
2
But A could have been in quadrant 4.
+3
O
x+
+5
-4
yIn this diagram adj = -4
 tan A =
−4
3
−4
2t
=
3 1− t 2
− 4 + 4t 2 = 6t
2t 2 − 3t − 2 = 0
(2t + 1)(t − 2) = 0
t = 2 or −
1
2
 Now tan
1
A
= 2 or −
2
2
In quadrant 4, tan
A
1
=−
2
2
(c)
R.T.P cos 4 4 A − sin 4 4 A + 1 = 2 cos 2 A
Proof: L.H.S
(
= (cos
)(
A) + 1 (cos
)
cos 4 4 A − sin 4 4 A + 1 = cos 2 A − sin 2 A cos 2 A + sin 2 A + 1
2
A − sin 2
= cos 2 A + 1 − sin 2 A
= 2 cos 2 A
= R.H.S.
L.H.S. = R.H.S.
Q.E.D
(d)
Data: sin A =
12
4
and sin B =
5
13
R.T.F. cos( A − B) and sin( A + B)
Solution
Taking sin A =
12
13
y+
+13
+12
A
x+
O
adj = +5
132 = 12 2 + (adj ) (Pythagoras’ Theorem)
2
(adj)2 = 132 − 12 2
(adj)2 = 25
adj = 5
Since A is acute than adj = 5
2
) 
A + sin 2 A = 1
sin B =
4
5
y+
+5
+4
B
x+
O
(adj)2 + 4 2 = 5 2
adj = +3
(Pythagoras’ Theorem)
adj = 3
Since B is acute than adj = 3
Recall: cos( A − B ) = cos A cos B + sin A sin B
Substitute values from the diagrams

 5  3   12  4 
cos( A − B ) =    +   
 13  5   13  5 
=

cos( A − B ) =
15 48
+
65 65
63
65
Recall: sin( A + B ) = sin A cos B + cos A sin B
Substitute values from the diagrams

 12  3   4  5 
sin ( A + B ) =    +   
 13  5   5  13 
=

sin( A + B ) =
36 20
+
65 65
56
65
5.
(a)
Data: f (x ) = x 3 − 5x 2 + 3x
1 
R.T.P. f ( x ) = 0 possesses a root in the interval  ,1 and there is another root > 1.
2 
Proof
f ( x ) = x 3 − 5 x 2 + 3x
Since f ( x ) is a polynomial, f ( x ) is continuous everywhere and hence, in the
1 
interval  ,1 for x   .
2 
f (1) = (1) − 5(1) + 3(1)
3
2
= −1
3
2
1 1
1
1
f   =   − 5  + 3 
2 2
2
2
=
3
8
1
Since f (1)  0 and f    0 differ in sign, then a root exists in the
2
1 
interval  ,1 according to the Intermediate Value Theorem.
2 
x 1
f (x )
2
-6
3
-9
4
-4
5
15
6
54
Similarly f (4)  0 and f (5)  0 differ in sign in the closed interval 4,5 , where
f ( x ) is continuous, there exists a root between x = 4 and x = 5 according to the
Intermediate Value Theorem.
Q.E.D.
b)
i)
R.T.F. the stationary points of f(x).
Solution
Recall: Stationary points are found when f ' ( x ) = 0
f ( x ) = x 3 − 5 x 2 + 3x
f ' (x ) = 3x 2 − 10 x + 3
f ' (x ) = (x − 3)(3x − 1)
when f ' ( x ) = 0
(x − 3)(3x − 1) = 0
 x = 3 or
1
3
f (3) = (3) − 5(3) + 3(3) = −9
3
2
3
2
1 1
1
 1  13
f   =   − 5  + 3  =
 3  3
 3
 3  27
 1 13 
Hence Stationary points on f ( x ) occur at  ,  and (3,−9 )
 3 27 
(ii)
R.T.F. nature of the stationary points.
Solution
f ' (x ) = 3x 2 − 10 x + 3
f ' ' (x ) = 6 x − 10
f ' ' (3) = 6(3) − 10 = 8 = +ve
 (3,−9 ) is a local minimum pt.
1
1
f ' '   = 6  − 10 = −8 = −ve
3
3
 1 13 
  ,  is a local maximum pt.
 3 27 
 1 13 
Hence, the stationary points occurring at (3,−9 ) and  ,  are
 3 27 
local minimum and maximum respectively.
(c)
Data: y =
R.T.P.
1
x +2
2
d2y
= 2(3 x 2 − 2) y 3
2
dx
Proof
Given y =
(
1
= x2 + 2
x +2
2
)
−1
Let t = x 2 + 2
y = t −1
dy
= −t − 2
dt
dt
= 2x
dx
dy dy dt
=

dx dt dx
(Chain Rule)
dy
= −t −2  2 x
dx
=
(x
− 2x
2
+2
)
2
 1 
= −2 x  2

 x + 2
2
= −2xy 2
Differentiating
(
dy
implicitly w.r.t x
dx
)
d2y
dy 

= y 2  −2 +  (− 2 x ).2 y 
2
dx 
dx

= −2 y 2 − 4 xy
Substituting
dy
dx
dy
= −2xy 2
dx
(
d2y
= −2 y 2 − 4 xy − 2 xy 2
2
dx
= 8x 2 y 3 − 2 y 2
)
(Product Law)

1
= 2 y 3  4 x 2 − 
y

But y =
1
1
= x2 + 2

y
x +2
2
Substituting
Hence,
1
= x2 + 2
y
(
d2y
= 2 y 3 4x 2 − x 2 − 2
2
dx
= 2 y 3 3x 2 − 2
(
)
(
)
d2y
= 2 y 3 3x 2 − 2
dx 2
Q.E.D.
)
6.
(a)
R.T.P. S n =
1
1
1
+
+ ... +
n +1 n + 2
2n
Proof
Base of each rectangle =
1
n
Height of each rectangle given by y =
1
1+ x
Area under curve between x = 0  
Area of first rectangle at x =
1
n
= Base  Height




1 1 
=
n 1
1+ 
 n
=
2
Area of second rectangle at x =
n




1 1 
=
n 2
1+ 
 n
=
3
Area of third rectangle at x =
n
1 n 


n  n +1
1 n 


n n+2




1 1 
=
n 3
1+ 
 n
=
1 n 


n  n +3




1 1 
=
n n
1+ 
 n
n
 Area of nth rectangle at x =
n
=
11
 
n 2




1 1
1
1
1
 S n areas of rectangular strips =
+
+
+ ... +
n 1  2  3
2
 1 +  1 +   1 + 

 n  n  n

=
Hence, S n =
1 n
n
n
1
+
+
+ ... + 

n  n +1 n + 2 n + 3
2
1
1
1
1
+
+
+ ... +
n +1 n + 2 n + 3
2n
Q.E.D.
b)
(i)
Data: f (x ) =
x
x +4
2
R.T.P.
Proof
f (x ) =
x
u
is of the form f ' (x ) =
v
x +4
2
du
Where u = x
And v = x 2 + 4 ,
f ' (x ) =
v
dv
= 2x
dx
du
dv
−u
dx
dx
2
v
(x
f ' (x ) =
=
=1
dx
2
)
(Quotient Rule)
+ 4 − x(2 x )
(x
2
+4
)
2
x 2 + 4 − 2x 2
(x
2
+4
)
2
4 − x2
=
(x
+4
2
)
2
Q.E.D.
ii)
R.T.Evaluate 
2
0
12 − 3x 2
(x
2
+4
)
2
dx
Solution
12 − 3x 2
 (x
2
+4
(
)dx
3 4 − x2
dx = 
2
(x + 4)
(4 − x ) dx
= 3
(x + 4)
)
2
2
2
2
2
(4 − x )  x

 (x + 4) dx =  x + 4 + c c = constant
(4 − x ) dx =  3x + A
3
x + 4



(x + 4)
2
2
2
2
2
2
2
2
where A = 3c
Hence, 
2
0
12 − 3x 2
(x
2
+4
dx = 3
2
)
2
0
(4 − x ) dx
(x + 4)
2
2
2
2
 3x 
= 2

 x + 40
3
=   − 0
4
=
3
4
(c)
(i)
R.T.Sketch the curve y = x 2 + 1
Solution
x = 0 , y =1
Curve is symmetrical about y − axis
Minimum point is (0,1)
0
T = 
1
x ⎯⎯⎯→ x 2 + 1
2
y = x 2 is a standard curve
y = x2 + 1
2
1
O
ii)
A
B
1
R.T.F The volume of region B when rotated 2 radians about the
y − axis
Solution
When x = 0 , y = 1
When x = 1 , y = 2
(A+B) rotated through 2 radians about y-axis is a cylinder.
V =  (1) (2) (Volume of cylinder = r 2 h
2
V = 2 units3
When A rotated through 2 radians about y-axis
V =   ( y − 1)dy
2
1
2
 y2

=   − y
2
1
1

=  0 − − 
2

1
=  units3
2
1 

Volume when B is rotated =  2 −  
2 

=
3
units3
2
JUNE 2003 Unit 1 Paper 1
1.
(a)
Data: hx 3 − 12 x 2 − x + 3  (2 x − 1)(2 x + 1)(x − k )
R.T.F. the value of h and k.
Solution
hx 3 − 12 x 2 − x + 3  (2 x − 1)(2 x + 1)(x − k )


 (2 x ) − 12 x − k
2
(
)
 4 x 2 − 1 (x − k )
hx 3 − 12 x 2 − x + 3  4 x 3 − 4 x 2 − x + k
Equating corresponding coefficient

h = 4 and
Alternative Method
2 x  2 x  2 x = hx 3

h=4
And
− 1 1 − k = 3

k =3
k =3
2
(b)
3x
= 9x
R.T.S.
27
Solution
2
3x
= 9x
27
( )
3 x  33 = 3 2
2
3x
2
−3
x
= 32 x
Equating Indices (bases are equal)
x 2 − 3 = 2x
x 2 − 2x − 3 = 0
(x + 1)(x − 3) = 0

x=3
or
x = −1
2.
R.T.F. the real values of x that satisfy 2 x − 3 − 6 2 x − 3 + 5 = 0
2
Solution
2x − 3 − 6 2x − 3 + 5 = 0
2
Let
t = 2x − 3
t 2 − 6t + 5 = 0
(t − 1)(t − 5) = 0
 t = 1 or t = 5
We substitute t = 2 x − 3
Recall: x  a 
−axa
x =a 
−a = x = a
2x − 3 = 1
Hence
− 1 = 2x − 3 = 1
Where 2 x − 3 = 1
or
2 x − 3 = −1
2x = 4
2x = 2
x=2
x =1
2x − 3 = 5
Also
− 5 = 2x − 3 = 5
Where 2 x − 3 = 5

or
2 x − 3 = −5
2x = 8
2 x = −2
x=4
x = −1
The real values of x which satisfy 2 x − 3 − 6 2 x − 3 + 5 = 0
2
are -1, 1, 2 or 4
3.
(a)
Data: − x 2 − 2 x + 3  a(x + h ) + k
2
R.T.F The values of the constants a , h and k
Solution


− x 2 − 2 x + 3  −1 x 2 + 2 x − 3

= −1(x + 1)

= −1 (x + 1) − 1 − 3
2
2

−4
= −(x + 1) + 4
2
is of the form a(x + h ) + k
2
Where a = −1 , h = 1 , k = 4
Alternative Method
− x 2 − 2 x + 3  ax 2 + 2ahx + ah 2 + k
Equating Coefficients
x2 :
a = −1
x:
2(− 1)h = −2

h =1
Constant: (− 1)(1) + k = 3
2

(b)
k =4
R.T.F. the maximum value of 3 − 2 x − x 2 .
Solution
Let
f (x ) = 4 − (x + 1)
2
The maximum value of f(x) is 4 and it occurs when
(x + 1)2 = 0
i.e. at x = −1 . Hence, the maximum point is (-1, 4).
4.
(a)
R.T.F. area of shaded region.
Solution
C
D
B
300
A
O
Area of sector OAD =
30 
 r 2

360
=
1
2
  (35)
12
= 320.7m 2
Area of sector OCB =
=
1
2
  (42)
12
1
 (5541.8)
2
= 461.8m 2
Area of CBAD = Area of sector OCB - Area of sector OAD
= 461.8 - 320.7
= 141.1m 2
(b)
R.T.F. the length of the chord AD
Solution
AD 2 = (35) + (35) − 2(35)(35)cos 30 (cosine rule)
2
2
AD = 18.12m
= 18.1 m (3 s.f.)
5.
Data: f (x ) = 2 x and g (x ) = 4 x + 6
R.T.F. the value(s) of x such that xg (x ) = x(4 x + 6)
Solution
xg (x ) = x(4 x + 6)
= 4x 2 + 6x
g ( f (x )) = g (2 x )
= 4(2 x ) + 6
= 8x + 6
xg (x ) = g ( f (x ))
4x 2 + 6x = 8x + 6
4x 2 − 2x − 6 = 0
(4 x − 6)(x + 1) = 0
4x − 6 = 0
4x = 6
x=
3
2
or
x +1 = 0
x = −1
6.
(a)
R.T.F the equation of the line perpendicular y = 3x + 2 and passing through (0, 1).
Solution
y = 3x + 2
Gradient of perpendicular line =
−1
3
(product of gradients of perpendicular lines = −1 )
 Equation of line is
y −1
1
=−
x−0
3
1
y −1 = − x
3
1
y = − x +1
3
3 y = −1x + 1
or
(b)
y
1
x-
2
x
0
-2 r = 2
(1,-2)
-2
y-
(3, -2)
x=3
(i)
R.T.F. the equation of the circle with centre (1,−2) and radius 2 units .
Solution
Centre (1,−2) , Radius 2 units
Recall: Eq’n of a circle (x − a ) + ( y − b ) = r 2
2
2
Where (a, b ) is the centre and r = radius

Eq’n is (x − (1)) + ( y − (− 2)) = (2)
2
2
2
− 2x + x 2 + 1 + y 2 + 4 y + 4 = 4
x 2 + y 2 − 2x + 4 y + 1 = 0
(ii)
R.T.P. line = 3 touches circle at (3,−2)
Proof
When x = 3
− 2(3) + (3) + y 2 + 4 y + 1 = 0
2
y 2 + 4y + 4 = 0
y 2 + 2y + 2y + 4 = 0
y ( y + 2) + 2( y + 2) = 0
( y + 2)( y + 2) = 0
i.e.
y = −2 only
 The line touches circle at (3,−2) .
7.
(a)
Data:
x
0
x +1
R.T.F the range of values of x for which
x
0
x +1
Solution
(  (x + 1) )
2
x
2
2
 (x + 1)  0  (x + 1)
x +1
x( x + 1)
0
x +1
2
x( x + 1)  0
Critical values are x = 0 and x = −1
Coefficient of x 2 is + ve 
y
x
-1

0
x : x  −1 x : x  0, x  R
(b)
Data: (2 x + 1)  9
2
R.T.F the range of values of x for which (2 x + 1)  9
2
Solution
(2 x + 1)2  9
4x 2 + 4x +1  9
4x 2 + 4x − 8  0
(
)
4 x2 + x − 2  0
4(x + 2)(x − 1)  0
Critical values are x = −2 and x = 1
Coefficient of x 2 is + ve  ”Minimum point”
y
-2

0
1
x : −2  x  1
x
8.
R.T.E sin  − cos  in the form R sin ( −  )
Solution
y

O
x
1
R = 12 + 12
R= 2
sin − cos = 2 sin( −  )
1
1
sin  −
cos  = sin  cos  − cos  sin 
2
2
cos  =
=
1
2

4


sin  − cos  = 2 sin − 
4

is of the form R sin ( −  )
where R = 2
and
=

4
R.T.S. sin  − cos  = 1 0    
Solution


2 sin −  = 1
4

 1

sin −  =
4

2
−
−

 1 
= sin −1 

4
 2

4
=

4
or
 −
3
4
But sin is + ve in the 1st and 2nd Quadrant


4

2
= −

4
=
Hence,  =

2
3

= −
4
4
 =
or  0    

9.
(a)
R.T.E.
4 − 2i
in the form a + ib
1 − 3i
Solution
Multiplying by the complex conjugate of the denominator = 1+ 3i
4 − 2i 4 − 2i 1 + 3i
=

1 − 3i 1 − 3i 1 + 3i
=
4 − 2i + 12i − 6i 2
(1)2 − (3i )2
Recall: i 2 = −1
=
10 + 10i
10
= 1 + 1i
is of the form a + bi where a = 1 and b = 1 and a , b  
(b)
 4 − 2i  
R.T.P. arg
=
 1 − 3i  4
Proof
from (a):
 4 − 2i 
arg
 = arg (1 + i )
 1 − 3i 
1
= tan −1  
1
=

4
radians
Im(z)
1

4
O
Re(z)
1
10.

Data: OA = −2i + j
(a)
(i)

and OB = i + j
R.T.F. the unit vector in the direction of OB .
The unit vector has magnitude of 1 and is parallel to OB
j
-2i + j
i+j
A
B
i
-2
0
Unit vector =  (i + j ) = i + j
 = a scalar
i + j = 1

 2 + 2 =1

2 = 1

=
1
2
Hence the unit vector parallel to OB is
(ii)
1
2
(i + j )
R.T.F. the position vector of C.
Solution
If OB is produced, a point C along the line will be of the form
OC =  (i + j )
 = a scalar
Given: OC = OA

OC = i + j

2 +2 =
2 2 = 5
(− 2)2 + (1)2
(b)

=

OC =
5
2
5
(i + j )
2
R.T.P. ai + bj and − bi + aj are perpendicular.
Proof
Recall: If a and b are two vectors and then the angle between them is
p.q = p q cos 
Let
p = ai + bj
q = −bi + aj
p.q = (a  −b ) + (b  a )
p.q = 0
If p.q = 0
then p is perpendicular to q (since cos 90  = 0 ).
11.
(a)
x2 + x − 2
x 2 + 5x + 6
R.T.F lim
x → −2
Solution
Let f ( x ) =
x2 + x − 2
x 2 + 5x + 6
2
(
− 2) + (− 2) − 2 0
f (− 2) =
=
(− 2)2 + 5(− 2) + 6 0
(is indeterminate)
Using Factor and Cancelling Method
(x + 2)(x − 1)
x2 + x − 2
= lim
2
x → −2 x + 5 x + 6
x → −2 ( x + 3)( x + 2 )
lim
(x − 1)
(x + 3)
(− 2 − 1) = − 3 = −3
=
(− 2 + 3) 1
= lim
x → −2
(b)
Data: f ( x ) =
(x
x
2
−9
)
R.T.C. the real values of x for which the function is continuous
Calculation:
f (x ) =

(x
x
2
−9
)
is discontinuous when the denominator is 0
x −9  0
2
( x − 3)( x + 3)  0
x −3 = 0
x =3
−3 = x = 3

x = 3
x +3= 0

No real values of x
Solution x = 3
12.
Data: y = 2x 3
R.T.F. the gradient of the curve
Solution
Gradient function 
dy
= 6x 2
dx
Substitute y = 16 into equation of the curve
16 = 2 x 3
8 = x3
x=3 8
x=2
As x = 2 , when y = 16
dy
2
= 6(2)
dx
dy
= 24
dx
The gradient of tangent to the curve = 24
13.
(a)
Data: g : x → 2 x 3 − 3x 2 + 4
R.T.F. the values of x for which the stationary point occurs.
Solution
Let g ( x) = 2 x 3 − 3x 2 + 4
g ' (x ) = 6 x 2 − 6 x
There exists a stationary point when g ' (x ) = 0
6x 2 − 6x = 0
6 x(x − 1) = 0
x = 0 or 1
(b)
R.T.D. the nature of the stationary points.
Solution
g (x ) = 12 x − 6
When x = 0
g (0) = 12(0) − 6 = −6  (−ve)
 Max point at x = 0
When x = 1
g (1) = 12(1) − 6 = 6  (+ve)  Min point at x = 1
14.
Data: f (x ) =
x
x +7
2
R.T.F. f (x)
Solution
y=
u
v
Where u = x
v = x2 + 7
and
du
=1
dx
dy
=
dx
v
dv
= 2x
dx
du
dv
−u
dx
dx
2
v
(
)
(Quotient Law)
dy
x 2 + 7 .1 − x(2 x )
=
2
dx
x2 + 7
=
=

(
(x
2
)
)
+ 7 − 2x 2
(x
+7
2
)
2
7 − x2
(x
2
+7
)
7 − x2
(x
2
+7
)
2
14 − 2 x 2
 (x
2
+7
)
2
2
dx =
x
+c
x +7
dx  
c = constant
2
(
2 7 − x2
(x
= 2
2
+7
)
)dx
2
7 − x2
(x
2
+7
)
2
dx
 x

= 2 2
+ c
x +7

=
2x
+A
x +7
2
( A = 2c )
1
14 − 2 x 2
 (
−1
1
 2x 
dx =  2
2

 x + 7  −1
x2 + 7
)
 2(1)   2(− 1) 
=
−
 8   8 
=
1
2
15.
Data: Equation of line is y = 3x
Equation of curve y = x 2
(a)
R.T.F. the coordinates of P.
Solution
P is the point of intersection of y = 3x and y = x 2
Solving simultaneously
3x − x 2 = 0
x(x − 3) = 0
x = 0 or 3
From the graph, the x -coordinate of P is > 0
Hence, x -coordinate of P is 3
Sub x = 3
y = 3(3) = 9
 P is the point (3,9 )
(b)
y = x2
P
y = 3x
(3, 9)
y
A1
A2
x
0
3
Regions A1 and A2 are as shown in the diagram.
Area of A1 + A2 =
Area of A2 =

3
1
3 9
1
= 13
2
2
x 2 dx
3
 x3 
= 
 3 0
=9−0 =9

1
1
Area of shaded region = 13 − 9 = 4 sq . units
2
2
CAPE - 2002 - Unit 1
Pure Mathematics
Paper 02
Section A (Module 1)
1
Let A = { x ε R : x > 2} and B = { x ε R : x > 0} Let f : A maps to B be the function given by
2x
x 2
f. ( x )
q
( a)
Find p, q ε R such that f (x) =
(b)
Show that f is one-to-one
[6 marks]
( c)
Determine whether there is an x ε A such that f (x) = 1
[7 marks]
(d)
Use Part (c) above to determine
p
x
[3 marks]
2
( i)
the range of f
[5 marks]
( ii )
whether or not f is onto
[4 marks]
( a)
f. ( x )
(b)
f . x1
2
x1
x
2
f . x2
4
2
4
2
2
4
x2
2
x1
( c)
2
4
x
2
1
1
x
4 1
4
x
4
x
2
hence n x ε R such that f (x) = 1
1
1
x
4 2
f is one-to-one
x2
since x > 2
hence
2
2
1
>0
2
(d)
( i)
y
2
y
4
2
x
4
x
for y ε B and x ε A
2
then y - 2 > 0
2
hence y > 2
1 ε B and from Part (c) there is no x ε A such that f (x) = 1
( ii )
hence f is not onto
2
( a)
If
t
tan .
θ
tan .
Hence find
(b)
express cos θ and sin θ in terms of t
2
θ
2
[6 marks]
11
5
when cos θ + 2 sin θ =
[7 marks]
In triangel ABC angle ACB = 90 degrees and D is the point on BC such that angle
ABD = angle BAD = θ. Given that BD = 5 cm and AB = 8 cm find
( i)
cos θ
[3 marks]
( ii )
sin θ
[3 marks]
( iii )
the length of BC
[3 marks]
( iv )
the length of AC
[3 marks]
( a)
θ
2
sec2
( i)
2
sin θ
t2
1
cos θ
1
2
2
t
2t
2
1
2
t
θ
2
1
( cos . θ
2
cos . θ
2
1
1
t2
1
t2
1)
2 2
1
t
1
t
2
4t
2
1
t2
2
2t
1
t
1
1
sin . θ
1
cos2
2
cos . θ
( ii )
t2
1
11
5
2
2
t
(4 t
3) ( 2 t
1)
0
tan .
θ
2
3 1
,
4 2
(b)
A
q
4
F
4
q
B
5
5
D
C
8 cos . θ
BC
4
5
cos . θ
BF = FA = 4
sin . θ
32
cm
5
BC
3
5
AC
8 sin . θ
AC
24
cm
5
Section B (Module 2)
3
( a)
A curve is given by the parametric equations
x
dy
dx
4 t2
y
8t
( i)
Find
in terms of t
( ii )
Show that the Cartesian equation of the tangent to the curve at the point P
with parameter T is
Ty
( iii )
[4 marks]
4 T2
x
The tangent in (ii) above meets the y-axis at the point Q and the x-axis at the
point R. If O is the origin show that the area of the triangle OQR is
8 T3
(b)
[5 marks]
square units
[6 marks]
Solve for x the inequality
2x
3x
3
3
<1
4
[10 marks]
( a)
( i)
dy
dx
( ii )
8
8t
y_8T
1
x
T
1
t
4 T2
Ty
Q . 4 T2 , 0
( iii )
area
(b)
dy
dx
2x
3
3x
4
R . ( 0 , 4 T)
1
4 T2 ( 4 T)
2
1< 0
4 T2
x
x
1
3x
4
<0
area
8 T3
(x
1) > 0
(3 x
f. ( x )
<0
g.(x)
4
( a)
x<
hence
4
3
4)< 0
units square
x> 1
x<
4
3
x> 1
Find the two square roots of the complex number 5 - 12i in the form x + iy where x,y
εR
[8 marks]
(b)
( i)
If z = x + iy where x,y ε R
(y 0)
z
( ii )
1
z
[5 marks]
Find and identify the locus of the points for which the imaginary part of
1
z
z
( c)
find the real and imaginary parts of
is zero
[5 marks]
If the position vector of the point A is i - 3j and the position vector of the point B is 2i
+ 5j find
( i)
( ii )
[4 marks]
AB
the position vector of the midpoint of
4
AB
[3 marks]
( a)
5
12i
x2
5
x2
y2
( i)
x
( ii )
x
x
iy x
x
x . x2
x2
y2
y2
1
2
iy
iy
x
y
y
x2
y2
zImag
y
x2
0
y2
1
0
x2
9
z1
2
x
zImag
1
iy
x2
xy
2, 2
y
x
zReal
( 2 xy) i
6
3, 3
x
(b)
y2
3
iy
4
2i
0
z2
x
iy
x2
y2
3
2i
i
y. x2
2
x
y2
1
2
y
circle with centre (0, 0)
radius = 1 unit
( c)
( i)
( ii )
AB
82
1
3
i
2
M
65
j
Section C (Module 3)
5
( a)
By expressing x - 4 as
x
lim
x
Hence find
2
2
find
x 2
x
4
4
x
lim
x
x
2
4x
2
5x
5
[5 marks]
4
(b)
f. ( x )
The equation of a curve is given by
x. ( x
2 )2
d .
f ( x)
dx
( i)
Obtain an expression for
( ii )
Find the stationary point(s) of f
[2 marks]
( iii )
Determine the nature of the stationary point(s) of f
[5 marks]
( iv )
Sketch the curve
[5 marks]
(v)
Find the area bounded by the curve and the interval of the x-axis -2 < x < 0
[3 marks]
[5 marks]
( a)
( i)
x
4
( ii )
x
2
x
4 (x
d .
f ( x)
dx
( ii )
stat points:
d
(x
dx
for x - -2
1)
(x
2) ( 3 x
( 2 , 0 ) max
6
1
4
2
x
( i)
( iii )
2
lim
x
(b)
x
lim
2
x
2
x
2) ( 3 x
2)
2)
6x
2
2
3
( 2 , 0)
1
12
32
27
8
for x =
2
3
2 32
,
3 27
min
( iv )
10
t
3
4t
2
4t
5
0
10
t
0
(v)
x3
area
4 x2
yields
4 x dx
2
6
( a)
Using the substitution u = x + 3 or otherwise evaluate
x. ( x
(b)
4
3
The section of the curve
y
[6 marks]
3) dx
x. ( x
3)
is rotated about the x-axis
0 x 2
through 360 deg. Find the volume generated by
( i)
direct integration
( ii )
the trapezium rule using five ordinates
( a)
[8 marks]
u=x+3
Vol
π.
du = dx
[11 marks]
limits: 3, 5
f. ( x )
5
(u
2
3 ) u du
yields
2
2
x (x
3)
12 . π
3
(b)
Vol
π ( 2 0) .
(0
2
4
20
2 ( 0.875
4
10.125 ) )
yields
7
12.5 . π
June 2002 Unit 1 Paper 1
1.
Data: x = 2 is a root of 6 x 3 − px 2 − 14 x + 24 = 0 .
R.T.F. the value of p and the other roots of the equation.
Solution
Let f (x ) = 6 x3 − px 2 − 14 x + 24
Recall: Remainder and Factor Theorem - If f ( x ) is any polynomial and f ( x ) is
divided by ( x − a ) , the remainder is f (a ) .If f (a ) = 0 , then ( x − a ) is a factor of
f (x ) .
(x − 2) is a factor of f (x )
f (2) = 6(2) − p(2) − 14(2) + 24 = 0
3
2
48 − 4 p − 28 + 24 = 0
44 − 4 p = 0
4 p = 44
and
p = 11
(
)
 6 x 3 − 11x 2 − 14 x + 24 = (x − 2) ax 2 + bx + c where a, b and c are constants.
ax 2  x = 6x 3
 a=6
− 2 + c = 24
c = 12
(bx  x) + (−2 + ax 2 ) = −11x 2
b − 2(6) = −11
b =1
 ax 2 + bx + c = 6 x 2 + x − 12
= (x − 2)(3x − 4)(2 x + 3)
 The remaining roots are +
4
3
and − .
3
2
2.
Data: 2 2 x − 3  2 x +1 + 8 = 0 .
R.T.C. the value of x.
Calculation
2 2 x − 3  2 x +1 + 8 = 0
(2 )
x 2
( )
− 3 2 2 x + 8 = 0
y = 2x
Let
y2 − 6y + 8 = 0
( y − 2)( y − 4) = 0
where y − 2 = 0
or
y=2
and
When y = 2
y−4=0
y=4
or
2 x = 21
When y = 4
2 x = 22
By equating powers (bases are equal)
x =1
or
x = 2.
3.
Data: f : x → x 2 − 3 and f  f (x ) = f (x − 3)
R.T.F the set of values of x .
Solution
Let f (x ) = x 2 − 3
(
)
f  f ( x ) = x 2 − 3 − 3
2
f (x + 3) = (x + 3) − 3
2
Hence f  f (x ) = f (x + 3)
(x
2
)
− 3 − 3 = ( x + 3) − 3
2
(x
2
2
−3
) = ( x + 3)
2
2
x 4 − 6x 2 + 9 − x 2 − 6x − 9 = 0
x 4 − 7 x 2 − 6x = 0
(
)
x x3 − 7x − 6 = 0
Let
g (x ) = x 3 − 7 x − 6
Recall: Remainder and Factor Theorem
If f ( x ) is any polynomial and f ( x ) is divided by ( x − a ) then the remainder is
given by f (a ) .If f (a ) = 0 then ( x − a ) is a factor .
f (− 1) = −1 + 7 − 6 = 0
 ( x + 1) is a factor

x2 − x − 6
x +1 x 3 − 7x − 6
- x3 + x2
0 − x 2 − 7x
− x2 − x
− 6x − 6
- − 6x − 6
0–0

f ( x) = x(x + 1) x 2 − x − 6
(
Hence, x(x + 1)( x + 2)( x − 3) = 0
 x = − 2, − 1, 0, 3, x  
)
4.
R.T.S. 2 x − y = 5 and x 2 − 6 y = xy simultaneously.
Solution
2 x − y = 5 ...(1)
x 2 − 6 y = xy ...(2)
From (1)
y = 2x − 5
Sub y = 2 x − 5 into (2)
x 2 − 6(2 x − 5) = x(2 x − 5)
x 2 − 12 x + 30 = 2 x 2 − 5 x
x 2 + 7 x − 30 = 0
x 2 − 3x + 10 x − 30 = 0
x(x − 3) + 10(x − 3) = 0
(x + 10)(x − 3) = 0
x + 10 = 0
or
x −3 = 0
x = −10
x=3
Sub x = −10 into (1)

2(− 10) − y = 5
− 20 − y = 5
y = −25
Sub x = 3 into (1)

2(3) − y = 5
6− y = 5
y =1
 Solutions are:
or
x = −10
and
y = −25
x=3
and
y =1
A
5.
A
2

3
2
O
2
B
R.T.F. area of pendant.
Solution
Area of shaded segment = Area of sector AOB - Area of Triangle AOB
Area of sector AOB =
1 2
r 
2
=
1 2 
(2)  
2
3
=
2
cm 2
3
Area of Triangle AOB =
1
(2)(2) sin 60 
2
=
1
(2)(2) sin 60 
2
= 3 cm 2
2
3

Area of shaded segment =

Area of pendant = 2 (Area of shaded segment)
3 cm 2
 2

= 2
− 3  cm 2
 3

y
(By Symmetry)
y
6.
A (-1, 1)
B (2, 0)
2
O
(a)
x
C ( x2 , y 2 )
R.T.F. the coordinates of C.
Solution
Let C = ( x 2 , y 2 )
Midpoint of AC is B (2,0 )
2=
−1 + x2
2
0=
4 = −1 + x 2
1+ y2
2
0 = 1+ y2
5 = x2
− 1 = y2
 C (5,−1)
(b)
R.T.F. the equation of the straight line perpendicular to OC.
Solution
Gradient of AC =
1 − (−1)
−1− 5
=
2
−6
=−
1
3
Gradient of line perpendicular to AC = 3 (product of gradients of
perpendicular lines = -1).
The line passes through the origin.
Hence, equation of line is y − 0 = 3( x − 0)
y = 3x
7.
Data:
p 2 x − 4 y = 8 and 8 x − 2 y = p
(a)
R.T.F. the value of p for which the equations have an infinite number of
solutions.
Solution
Let p 2 x − 4 y = 8
…(1)
And 8 x − 2 y = p
…(2)
p=
Equations will have an infinite number of solutions when both equations
are the same.
Multiply (2) by 2
i.e., 2(8 x − 2 y = p )
16 x − 4 y = 2 p
Equating corresponding coefficients of both equations.
x: p 2 = 16
and
constant :
p 2 = 16
p=4
p = 16
p = 4

(b)
The value of p is 4 only.
R.T.F. the solutions for this value of p.
Solution
Subst. p = 4 into (2)
8x − 2 y = 4
8x − 4 = 2 y

y=
8 = 2p
8x − 4
2
y = 4x − 2
Solutions are
(x,4 x − 2);
x  
8.
R.T.P. cos 2 =
1 − tan 2 
1 + tan 2 
Proof
Taking R.H.S.
sin 2 
1 − tan 2 
cos 2 
=
1 + tan 2 
sin 2 
1+
cos 2 
1−
sin  

 tan  =

cos  

cos 2  − sin 2 
cos 2 
=
cos 2  + sin 2 
cos 2 
=
cos 2  − sin 2 
cos 2  + sin 2 
=
cos 2
1
= cos 2
= L.H.S.
L.H.S. = R.H.S.
Q.E.D.
(cos 2 = cos 2  − sin 2  ) and (cos 2  + sin 2  = 1)
9.
(a)
Data: x 2 − 3x − 1 = 0 has roots  and  .
R.T.F. the equation whose roots are
2

and
2

Solution
If ax 2 + bx + c = 0
x2 +
b
c
x+ =0
a
a
(x −  )(x −  ) = 0
 ,  are roots
x 2 − ( +  ) +  = 0
Equating coefficients

+ =
 =
−b
a
c
a
x 2 − 3x − 1 = 0
Let
y = x 2 − 3x − 1 = 0
Hence
 − 3

 1 
 +  = −
and
 =
 = −1
 + =3
Required equation:
2 2 2 2
x 2 + (− ) +  +    = 0
     
1 1
 1 
 = 0
x 2 − 2 +  + 4
  
  
 +  
 1 
 + 4
 = 0
x 2 − 2
  
  
Re  +  = −3
−1
1
and

 3 
 1 
x 2 − 2  + 4  = 0
 − 1
 − 1
and
x 2 + 6x − 4 = 0
 = −1
(b)
Data: x 2 + 12 x − a = 0 . One root is three times the other.
R.T.F. the value of a.
Solution
Let  and  be the roots of x 2 + 12 x − a = 0 .
Let
 = 3
Equation is of the form Ax 2 + Bx + C = 0
Where A = 1 , B = 12 , C = −a

 + =−
B
A
 + =−
− 12
1
 +  = −12
and
 = 3
3 +  = −12
i.e. 4 = −12
and
 = −3
Hence  = −12 + 3
 = −9
and
 =
C
A
(− 3)(− 9) = C
C = 27
Hence, a = −27 .
10.
(a)
Data: r = (cos  + 2 sin  )i + (sin  − 2 cos  ) j
R.T.P. r is independent of  .
Proof
r =
(cos  + 2 sin  )2 + (sin  − 2 cos  )2
= cos 2  + 4 sin  cos  + 4 sin 2  + sin 2  − 4 sin  cos  + 4 cos 2 
(
= cos 2  + sin 2  + 4 sin 2  + cos 2 
)
(sin 2  + cos 2  = 1)
= 1+ 4
= 5 which is independent of  .
Q.E.D.
(b)
R.T.F. the vector parallel to i + 3 j and has the same magnitude as 2i − j
Solution
If two vectors are parallel, then one is a scalar multiple of the other.
Let required vector be =  (i + 3 j )
= i + 3j
Vector has the same magnitude as 2i − j

(i )2 + (3j )2
=
(2)2 + (− 1)2
10 2 = 5
10 2 = 5
2 =
5
10
=
1
2
Hence, the required vector is
1
2
i+
3
2
j
11.
(a)
x+2
x →−2 2 x 3 − 8 x
R.T.F lim
Solution:
Let f (x ) =
f (− 2 ) =
x+2
2 x 3 − 8x
−2+2
2(− 2 ) − 8(− 2 )
3
=
0
0
i.e. indeterminate
Factorising and cancelling
x+2
x+2
= lim
3
x → −2 2 x ( x 2 − 4 )
x →−2 2 x − 8 x
lim
x+2
x → −2 2 x ( x + 2 )( x − 2 )
= lim
x+2
x → −2 2 x ( x − 2 )
= lim
=
1
2(− 2 )(− 2 − 2 )
=
1
16
Hence, lim f (x ) = lim
x → −2
x → −2
x+2
1
=
3
2 x − 8 x 16
(b)
R.T.F values of x for f (x ) =
2x + 1
is continuous.
x + x−2
2
Solution:
f (x ) =
2x + 1
x + x−2
f (x ) =
2x + 1
(x − 1)(x + 2)
2
As x → 1 , f (1) → 
(x − 1)(x + 2) = 0
And as x → −2 , f (− 2 ) → 
x = -2 and x = 1 are vertical asymptotes of f ( x )
 f ( x ) is continuous for x   and x  1 & x  −2
12.
x2
x3 + 2
Data: f ( x ) =
R.T.F. f (x) and evaluate
16 x − x 4
1
 (x
3
0
+2
)
2
dx
Solution
x2
u
is of the form
3
x +2
v
f (x ) =
Where u = x 2
du
= 2x
dx
and v = x 3 + 2
dv
= 3x 2
dx
v
f ' (x ) =
du
dv
−u
dx
dx
2
v
(x
f ' (x ) =
=
=
Hence,
)
(x
3
2
3
2 x 4 + 4 x − 3x 4
(x
3
+2
)
2
4x − x 4
(x
3
+2
)
2
4x − x 4
 (x
16 x − 4 x 4
 (x
( )( )
+ 2)
+ 2 (2 x ) − x 2 3x 2
3
+2
)
2
+2
3
)
2
=
x2
+c
x3 + 2
c = constant
(
)dx
(x + 2)
(4 x − x )dx = 4 x + A 
= 4
x +2

(x + 2)


dx  
4 4x − x 4
3
2
4
3
1

0
16 x 3 − 4 x 4
(x
3
+2
)
2
2
2
3
1
 4x 2 
dx =  3

 x + 20
 4(1)2   4(0 )2 
1
= 3
− 3
 =1
3
 (1) + 2   (0 ) + 2 
A = 4c
13.
Data: f : x → 27 x − x 3
(a)
R.T.F. the stationary points.
Solution
Let f ( x) = 27 x − x 3
f ' (x ) = 27 − 3x 2
Stationary points occur when f ( x ) = 0
i.e. 27 − 3x 2 = 0
3x 2 = 27
x2 = 9
x= 9
x = 3
When x = 3 and
f (x ) = 54
b)
when x = −3
f (x ) = −54
R.T.F. the nature of the stationary points
Solution
f ' (x ) = 27 − 3x 2
f ' ': x = −6 x
At x = 3
f ' ' (3) = −18
Hence, (3, 54) is a maximum pt.
At x = −3
f ' ' (− 3) = 18
Hence (-3, -54) is a minimum pt.
14.
R.T.C.
 cos
3
xdx
Solution
Let u = sin x
du
= cos x
dx
dx =
du
cos x
  cos 3 xdx =  cos 3 x
du
cos x
=  cos 2 xdu
(
( sin 2 x + cos 2 x = 1 )
)
=  1 − sin 2 x d u
(
)
=  1 − u 2 du
u3
=u−
+c
3
c = constant
sin 3 x
= sin x −
+c
3
sin 3 x
+c
Hence,  cos xdx = sin x −
3
3
15
(a)
R.T.Sketch y = x 2 + x
Solution
y = x 2 + x = x(x + 1)

The graph cuts the x-axis at -1 and 0
At the minimum point
dy
=0
dx
dy
= 2x + 1 = 0
dx
2 x = −1
x=
−1
2
When x =
−1
2
2
 1  1
y = −  + − 
 2  2
y=
1 1
−
4 2
y=−
1
4
 −1 −1
Coordinates of stationary point is  , 
 2 4 
y
y = x2 + x
-1
−1
2
O
−1
4
Minimum point
 −1 −1 
 , 
 2 4 
x
(b)
R.T.F. the area bounded by the curve, the x-axis and the lines x = -1
and x = 3.
Solution
y
y = x2 + x
x
-1
O
3
Minimum point
 −1 −1 
 , 
 2 4 

Total Area =
0
−1
3
ydx +  ydx
0
The modulus is needed because the region lies below the curve and the
area when computed will be negative.
=

0
−1
3
x 2 + xdx +  x 2 + xdx
0
0
3
 x3 x2 
 x3 x2 
= +  + + 
2  −1  3
2 0
3

1
 −1 1   
= (0) −  +  +   9 + 4  − (0)
2
 3 2  

=
1
9
+9+
6
2
= 13
2
sq units
3
CAPE - 2001
Pure Mathematics - Unit 1
Paper 02
Section A (Module 1)
1
( a)
Show that the inequality
(b)
Factorize completely the polynomial
( c)
Find the value of r such that
x
3
x
x4
x3
7x
holds for any x ε R
3
2 x2
x
[4 marks]
[4 marks]
2
has a remainder -2 when is is
4r
divided by x + 3
(d)
[4 marks]
The functions f and g are defined by
x2
f : x maps to
1
( e)
2
x2 1
x
2
g : x maps to
x
3
( i)
Explain clearly why f is not one-to-one
[3 marks]
( ii )
Calculate and simplify gf (x)
[5 marks]
The table below shows the growth in bacterial numbers n with time t in minutes as
recorded during a laboratory experiment. After treatment with an antibiotic the
number of bacteria is reduced at the same rate for the next 50 minutes
Time (t mins)
0
10
20
30
40
50
No of bacteria (n)
5
10
20
40
80
160
( i)
Using a scale of 2 cm to represent 10 minutes on the x-axis and 2 cm to
represent 20 bacteria on the y-axis draw the growth curve for the treatment
stage
[4 marks]
( ii )
Hence estimate the number of bacteria present after the first 35 minutes of
treatment
[1 mark]
1
( a)
(t
3) ( t)
( 3)
(t
3)
( 3)
( t)
for all real t
for all real t
for
t 3
t
3
for
t 3
t
3
t
3
t
3
10
t
3
t
3
12
9
6
3
0
3
6
9
12
t
(t
3)2 ( t
3 )2
t2
9
6t
(b)
x3
2 x2
( c)
f (-3) = -2
(d)
( i)
f. ( 2)
( ii )
1
x2
2 1 x2
x
2
21
4
f. ( 2)
3
81
r
f. ( a )
(x
1) . ( x
26
f. ( b)
simplifies to
a b
7 . x2
2 1
2
2. 3 . t
( 3 )2
6. t
by factoring, yields
2
r
6 t ( t )2
6
x2
2) . ( x
1)
( e)
( i)
Growth Curve
No of Bacteria
200
150
100
50
0
0
10
20
30
40
50
60
Minutes (t)
( ii )
2
( a)
approx 80 bacteria
The function f is given by f (x) =
2 x2
11 x
( i)
Express f (x) in the form
p. ( x
q )2
( ii )
Sketch the graph of f (x)
( iii )
Hence or otherwise find the turning point on the graph and determine whether
it is a maximum or a minimum
3
where p, q, r ε R
r
[3 marks]
[2 marks]
[3 marks]
(b)
Given that δ is an acute angle such that cos δ =
1
x
4
find an expression for sin 2δ in terms of x
( c)
[4 marks]
The parametic equations of a curve C are
2x
cot . θ
cosec. θ
4y
cosec. θ
2 cot . θ
( i)
Express cot θ and cosec θ in terms of x and y
[3 marks]
( ii )
Find an equation connecting x and y
[2 marks]
3
(d)
( i)
Sketch in separate diagrams the graphs of y = sin x and y = cos x for
2π x 2π
( ii )
6 marks
State clearly the transformation which maps y = cos x onto y = sin x
[2 marks]
( a)
2 x2
( i)
11 x
3
11
4
2. x
2
97
8
( ii )
10
2t
2
11 t
3
5
0
10
t
11
( iii )
(b)
4
cos . δ
97
,
8
1
x
4
sin . 2 δ
( c)
( i)
min
sin . δ
2 sin . δ . cos . δ
x2
16
sin . 2 δ
cot . θ
2x
cosec. θ
cot . θ
2x
4 y 2 cot . θ
cosec. θ
( ii )
1
cot2 θ
(2 x
1
4 y) 2
4 y 2. ( 2 x
cosec. θ
16 . ( x
4
(2 x
y) 2
16
x2
4 y 2 cot . θ
cot . θ
4 y)
cosec2 θ
1
1 .
x
8
(2 x
cosec. θ
4 y) 2
1
4 y)
4 (x
y)
16 . ( x
y) 2
simplifies to
16 . ( x
y) 2
(d)
( i)
1
sin ( t )
cos( t )
5
0
5
1
t
f (x) = sin (x)
( ii )
cos x = f (x +π/2)
Section B (Module 2)
3
( a)
A pair of simultaneous equations is given by
2x + py = 13 and px + 32y = 52 where p ε R
Find the value(s) of p for which the system of equations above has
(b)
( i)
a unique solution
[3 marks]
( ii )
an infinite number of solutions
[3 marks]
( iii )
no solution
[3 marks]
Solve the following equations for
( i)
( ii )
( c)
6 sin x
2
cos . x
3 sin . x
cos . x
Show that for
A
π
2
4
0 x
π
2
[5 marks]
0
[5 marks]
2
sec . A
tan . A
5
tan .
π
4
A
2
[6 marks]
( a)
2 p
p 32
( i)
p=8
( ii )
2x
for p = 8
8
for a unique sol
p 8
4.(2 x
8 y 13
8 y)
an infinite number of sol
p = -8
( iii )
p
2x
8x
8y
13
32 y
....... ( 1 )
52
....... ( 2 )
4 (1) + 2 gives 0 = 104 inconsistent - no sol
(b)
( i)
6 cos 2 x
cos . x
( 2 cos . x
1 ) ( 3 cos . x
( c)
1
π
6
2 sin . x
tan .
π
4
0
2)
0
x
π
3
π
3
x
( ii )
2
tan .
A
2
π
A
tan . . tan .
4
2
A
2
A
.
2
1
tan .
A
2
1
tan .
A
2
sin .
1
cos
sin
1
cos 2
2
.A
2
A
cos .
2
A
2
sin2
cos 2
cos .
A
2
A
cos .
2
A
2
A
sin .
2
6
cos .
A
2.
2
A
A
sin .
cos .
2
2
2 sin .
sin2
A
A
2
A
2
cos .
sin .
A
2
A
sin .
2
A
2
1
sin . A
cos . A
4 ( 13 )
4
( a)
The complex number z is expressed in the form x + iy where x,y, ε R
Express the complex number
z
1
z
1
in a similar form
The argument of the complex number
(b)
z
1
z
1
[7 marks]
π
is
4
( i)
Find the equation connecting x and y
[3 marks]
( ii )
Show that the equation represents a circle C
[3 marks]
( iii )
Determine the centre and radius of C
[2 marks]
( i)
r1
3i
r2
i
j
r3
are three vectors and t is a scalar
4j
Find the values of t such that the vector
to the vector
tr2
tr1
is perpendicular
r2
r3
[5 marks]
The position vectors of points A and B relative to the origin O are
( ii )
4i + j and i + 7j respectively. The point C lies on AB such that
AC : CB = 2 : 1. Find the position vector of C relative to O
[3 marks]
Determine angle cos AOC
( iii )
( a)
(x
(x
1)
1)
2
x
(x
iy . ( x
iy ( x
1
1)
2
( ii )
( iii )
( i)
1)
1)
iy
iy
(x
2 . iy
2
2
(x
1)
2
x
C (1, 0)
1) ( x
y2
1)
(x
2
y
y
[2 marks]
1
(x
1 ) iy
2
2
1)
y
2
2
2
y
2
x
y
2
rad
2
7
1
2
y
1
(x
1 ) iy
(b)
([3t + 1)i + j] . [(ti + (t - 4)j] = 0
( i)
(3 t
3 t2
2t
4
1) t
(1) ( t
4)
0
has solution(s)
0
( 4i
j) 2 ( i
2 1
( ii )
OC
( iii )
cos . ( AOC )
( 4i
7j)
2t
1
3
1.
13
3
1
3
1.
13
3
OC
j ) . ( 2i
17 .
3 t2
5j )
2i
4
0
=
1.535
5j
cos . ( AOC )
29
0.869
13
( 493 )
Section C (Module 3)
5
( a)
Using differentiation determine the range of real values of x for which the function
f. ( x )
12
6 x2
x3
is decreasing
[5 marks]
x3 with respect to x from first principles
(b)
Differentiate
( c)
A farmer plans to construct an enclosure for his sheep making the use of one side of
a barn that is 150 metres in length. Using 500 metres of fencing material the farmer
will build a fence QRSTP which along with the existing barn wall PQ will form a
rectangular enclosure PRST
150 m
Barn
P
Q
T
xm
R
S
Fence
8
[6 marks]
Let the length of QR be x metres. Simplifying your answers where possible find
expressions in terms of x for
( i)
the length of TS
[1 mark]
( ii )
the length of RS
[4 marks]
( iii )
the area A of PRST
[2 marks]
( iv )
the stationary value of x and show that it is a maximum
[5 marks]
( vi )
the maximum area of the sheep enclosure
[2 marks]
( a)
(b)
lim
δx 0
3
δx )
(x
x3
3 x . ( δx )
( δx )
( i)
TS
150
x
( ii )
RS
175
x
( iii )
( iv )
A
25
( 150
2x
0
dx2
A
( 150
x3
x
3 x2 δx
x< 0
x> 4
2
3 x . ( δx )
( δx )
3
δx
2
x ) ( 175
d2 A
(v)
x)< 0
lim
δx 0
δx
lim 3 x2
δx 0
( c)
x. ( 4
3 x2 < 0
12 x
3 x2
x)
A
26250
25 x
x2
12.5
<0
12.5 ) ( 175
9
12.5 )
A
26406.25 m2
x3
6
( a)
Calculate the volume generated when the finite region in the first quadrant bounded
by the curve y = 2x2, the y-axis and the line y = 2 is rotated completely about the
y-axis.
[6 marks]
(b)
The region R is bounded by the curve
x2
y
the x-axis and the lines
1
x = 0 and x = 2
( i)
Calculate the area of R
[5 marks]
( ii )
The area of R is estimated using the trapezium rule with 2 intervals of equal
width. Show that this trapezium rule estimate differs by 1/3 from the exact
value for the area of R found in (b) (i)
[5 marks]
( iii )
On a carefully labelled sketch of
y
x2
shade in the 2 trapezia
1
of unequal width. The first trapezium has width h and the second trapezium
width (2 - h) with the three ordinates occurring where x = 0 x = h and x = 2
Show that the total area of these 2 trapezia of unequal width is given by
h2
2
( a)
Vol
2h
y
dy
2
π.
[5 marks]
6
π
yields
0
2
(b)
( i)
x2
A
1 dx
yields
0
( ii )
A
( ii )
( iii )
(2
0) .
4
( i)
sketch
10
(1
1
3
5
2 (2))
14
3
yields
5
( iv )
A
1
1
2
h2
1 h
1 2
h
2
simplifies to
A
1
( 2)2
2.h
h2
6
h2
2h
6
11
1 (2
h)
10
CAPE - 2001 - Unit 1
Pure Mathematics - Paper 01
Section A (Module 1)
1
Solve for x
2x
1
3
xεR
x
[5 marks]
4
(2 x
2
1)
2
(x
3)
5
2
42
Solve the equation
( 2)
2
1
4
(x
a
5
2
5x
13
2
a )2
b
2
3
[5 marks]
b
13
2
x
[6 marks]
32
2
x
1
2
22
has solution(s)
0
x2
Find real numbers a and b such that
x
3
2
2x
25
2
x
x2
4x
5
0
(x
xy
2
and
1) ( x
5)
0
x = 1, -5
4
2
Solve the pair of simultaneous equations
(1
2 x )2
x( 1
2 x)
2
y
0
1
2
6x
5x
1
0
has solution(s)
1
6
1 4
,
6 3
sol : x,y : (1, -1)
1
2x
y
1
[7 marks]
5
A
15 km
O
f
15 km
B
The diagram above (not drawn to scale) shows part of the radar system of a coast guard
boat. O represents the position of the coast guard station. A represents the position of the
coast guard boat at sea and B represents the position of a fishing vessel from which
emanates a distress signal. OA = 15 km and angle AOB = φ deg
( a)
(b)
Find the expression in terms of π and φ for
( i)
the length of the minor arc AB
[2 marks]
( ii )
the area of the corresponding sector OAB
[2 marks]
Given that φ = 40 calculate the area of the triangular region mapped out between the
coast guard station the boat and the vessel
[3 marks]
( a)
(b)
( i)
arcAB
( ii )
areasector
area
15 .
π
φ
180
π φ
km
12
1
π
( 15 ) 2
φ
2
180
1
2π
2
( 15 ) sin .
2
9
2
5π φ
km2
8
area
2
73.3 km
Section B (Module 2)
6
Given the points A (3, -5) and B (-4, 2) find
( i)
the coordinates of the midpoint M of AB
[1 mark]
( ii )
the gradient of AB
[1 mark]
( iii )
the equation of the straight line through M that is normal to AB
7
1 3
,
2 2
( i)
M.
( ii )
GradAB
( iii )
y
[3 marks]
1
3
2
1
2
x
2
x. ( x
5
f. ( x )
Sketch the graph of the function
x. ( x
Hence solve the inequality
2y 2x
4)
y-x+1=0
0
4)> 5
[6 marks]
10
t
2
4t
5
10
5
0
5
10
t
(x
8
f. ( θ )
1) ( x
2 cos . θ
5) > 0
sin . θ
x < -5 x > 1
in the form R cos (θ + α)
( a)
Express
(b)
Hence find the values of θ for which f (θ) is a maximum or a minimum where
00 θ 3600
3
[5 marks]
[2 marks]
3 cos . θ
( a)
(b)
9
35.260
f . θ max
θ
cos
1
( 1)
f . θ min
θ
cos
1
( 1)
35.260
35.260
Given that α and β are roots of the equation
α
2
3x
β
2
(α
α
β)
α
10
Given that
AB
(α
2
0
α
β ) (α
β)
4
2
α
2
(α
β
2
β
2i
2
2 αβ
β ).
(α
β)
2
3 .
2
9
4
8
3j
CB
(α
5i
2
find the unit vector in the direction of
BC
AB . AC
(b)
( 2i
unit . vector
AB
AC
3j ) . ( 3i
1
( 2i
13
4
2
[6 marks]
2
2
(α
β)
2
4 αβ
41
j
is perpendicular to
AB
β
4
(b)
AC
2
3.
show that
( a)
β)
and α > β
4 αβ
( a)
AB
215.260
αβ
2
2
β
θ
3x
α
3
β
35.260
2 x2
4
find without solving the equation the exact value of
2 x2
θ
[3 marks]
AC
[3 marks]
AB
2i
2j)
3j )
3j
( 5i
j)
AB . AC
AC
0
3i
2j
Section C (Module 3)
11
lim f . ( x )
3
Given that
3x
lim 9 . f . ( x )
3
evaluate
1
x
lim f . ( x )
x 3
12
f. ( x )
3.(3)
1
x3
Show that the equation
[5 marks]
x
x3
10
3x
lim 9 . f . ( x )
x 3
72
has a root between 1 and 2
3x
f. ( 1 )
10
f. ( 2 )
6
[5 marks]
4
By Intermediate Value Theorem - result proven
13
The curve
ax2
y
where a and b are real constants has a stationary point at
bx
(1, 2). Calculate the respective values of a and b
x
1
a
b
2
dy
dx
2 ax
b
[6 marks]
2a
b
0
a = -2 b = 4
14
x2
Differentiate
2
2 x2
with respect to x
50 x
Hence or otherwise find
2
2
2x
2
2
d x
2
dx 2 x2 1
5.
[3 marks]
1
2x
1
2
1 ( 2 x)
x
2
2 ( 4 x)
10 x
2
2
2x
d x
2
dx
dx 2 x2 1
[3 marks]
dx
2
1
5. x
2
2 x2
1
2
5
2x
K
2
1
15
Initially the depth of water in a tank is 32 metres. Water drains from the tank through a hole
cut in the bottom. At t minutes after the water begins draining the depth of water in the tank
is x metres. The water level changes at a rate equal to (-2t - 4)
( a)
Find an expression for x in terms of t
(b)
Hence determine how long it takes for the water to completely drain from the tank
[5 marks]
[3 marks]
( a)
dx
dt
x
2t
t
4
1 dx
32
t2
x
( 2t
4 ) dt
0
4t . t
32
x
32
0
x=0
(t
4) ( t
8)
6
0
t = 4 minutes
4t
t2
CAPE - 2000
Pure Mathematics - Unit 1
Paper 01
Section A (Module 1)
1
x3
2 x2
( a)
If
(b)
Find the real values of x which satisfy the equation
5x
6
(x
2) ( x
2x
2
( a)
3
3) ( x
[3 marks]
comparing constants: k = 1
(b)
(2 x
Express
1
2x
25
has solution(s)
0
3 x2
[2 marks]
5
( a)
3 )2
find the value of k
k)
a. ( x
in the form
h)2
1
4
k
stating explicitly the
values of a, h and k
(b)
23
Solve for x the equation
( a)
(b)
3
[3 marks]
3 . x2
2x
3
5x
2
23
6 x
1
64
5x
1
3
x
1
[3 marks]
3. x
1
3
3
5x
2
4
9
3. x
1
3
2
4
3
1
6
6x
x=3
The diagram below shows the graph of y = f (x) which has a minimum point at (1, -2)
y
x
0
(1, -2)
1
State the coordinates of the minimum point on the graph with equation
( i)
y = f (x) + 2
[2 marks]
( ii )
y = f (x - 3)
[2 marks]
( iii )
y = 4 f (x)
[2 marks]
( i)
( 1 , 0 ) min
0
translation of
4
( 4 , 2 ) min
( ii )
translation of
2
( iii )
3
( 1 , 8 ) min
y-stretch factor 4, x-axis invariant
0
In triangle ABC angle ABC = 120 deg, AB = 2 m AC = 2x m and BC = (x + 3) m
A
2x
2
C
120 0
(x + 3)
B
Show that
3 x2
8x
( 2 x)2
19
0
22
(x
[6 marks]
3)2
2
3x
5
2 (2) ( x
2
x
8x
3 ) cos . 1200
19
0
g : x maps to
x
The functions f and g are defined on R by
f : x maps to 2x
2
3
Determine the set of values of x for which f [f (x)] = g [f (x)]
2 ( 2 x)
( 2 x)
2
3
2
4x
2
4x
3
0
[7 marks]
has solution(s)
3
2
1
2
Section B (Module 2)
6
4 x2
Let α and β be the roots of the equation
( a)
3x
0
Without solving the equation write down the values of
( i)
α
( ii )
αβ
β
[1 mark]
α
2
β
2
(b)
Find the value of
[2 marks]
( c)
Find the equation whose roots are
2
α
( a)
( i)
(b)
α
α
2
β
2
3
4
β
3
4
2
1
4
2 α
Prove that
cos . 2 A
cos . 2 A
1
1
β
1
4
2
x2
β
2
2
2x
prod of roots = 64
2
64
0
sin . 2 A
tan . A
sin . 2 A
[5 marks]
2 sin . A . cos . A 2 sin . A . ( sin . A
. .
.
1 2 sin . A . cos . A 2 cos A ( sin A
2 sin2 A
1
1
1
2 cos 2 A
LHS
8
( a)
Express
sin . θ
[3 marks]
2
1
16
( αβ )
equation:
2
and
2
αβ
( ii )
2
sum of roots =
( c)
7
1
3 cos . θ
cos . A )
cos . A )
tan . A
in the form R sin (θ + α) where R > 0 and α is acute
[4 marks]
(b)
Hence deduce the minimum value of
3
sin . θ
3 cos . θ
6
[3 marks]
( a)
π
3
2 sin . θ
sin . θ
10
π
3
4 2 sin . θ
(b)
9
3 cos . θ
6 8
min value = 4
Let A (1, 2) be a point in the coordinate plane with origin O. Find
( a)
the equation of the straight line OA
[3 marks]
(b)
the equation of the straight line AB through A perpendicular to OA
[2 marks]
( c)
the coordinates of the point B at which the line AB crosses the x-axis
( a)
( a)
y - 2 = 2 (x - 1)
(b)
y
( c)
(5 , 0)
2
1
2
(x
[1 mark]
y = 2x
1)
1
y
2
x
5
2
In a triangle ABC the position vectors of A, B and C are respectively
i + j, 3i + 4j, and 4i - j
(b)
( i)
Find
( ii )
Show that angle BAC = 90 deg
BA
and
[2 marks]
AC
[2 marks]
A pair of simultaneous equations is given by
px + 4y = 8 and 6x + 2y = q where p,q, ε R
State the values of p and q for which the simultaneous equations have an infinite
number of solutions
[2 marks]
( a)
( i)
BA
( ii )
(b)
( 2i
p
4
6
2
2i
3j
AC
3j ) . ( 3i
2j )
p = 12
4
3i
2j
0
4
8
2
q
q=4
Section C (Module 3)
11
( a)
Find
lim
x
(b)
x2
3 3 x2
( a)
lim
x
(b)
( a)
[3 marks]
9x
Determine the real values of x for which the function
x
2x
f. ( x )
12
9
( x 3) ( x 3)
.
3 3 x ( x 3)
2
3
7
2
x
2x
7 0
x
13
[3 marks]
7
2
Find the value of x at the stationary point(s) of the function
f. ( x )
(b)
is continuous
7
3 x3
4x
[3 marks]
Determine the nature of the stationary point(s)
9 x2
( a)
4
(b)
d
4
dx
The function f is defined by
[3 marks]
2
3
has solution(s)
0
9 x2
2
18 x
6
x
f. ( x )
3
4
x
4
2
x
,
2
3
16
9
2 16
,
3 9
min
xεR
max
x 0
x
( a)
Find
(b)
Evaluate
f. ( x )
3
[3 marks]
3 dx
f. ( x)
[3 marks]
3 dx
1
5
x2
( a)
1
1
2
x3
3 dx
3
x
F. ( 3)
(b)
F . ( 1)
2
14
Given that
( a)
4x
1
x
K
16
4
f. ( x) d x
f. ( x) d x
and
17
0
where f (x) is
22
0
a real and continuous function in the closed interval [0, 4] evaluate
4
f. ( x) d x
[2 marks]
2
(b)
Differentiate with respect to x from first principles the function
x2
y
( a)
I
22
17
I
[4 marks]
2
5
(b)
lim
δx 0
(x
δx )
2
2
δx
x2
2
lim
δx 0
2
x
2 x . δx
( δx )
2
2
x2
2
δx
d 2
x
dx
15
2
2x
Let f be a cubic function in x. Suppose that f (x) = 0 has roots at x = 0 and x = 3 abnd f has a
maximum point at (1, 4) and a minimum point at (3, 0). Sketch the graph of y = f (x)
indicating clearly its maximum point, its minimum point and its intercepts with the axes
[6 marks]
6
y
(1, 4)
x
0
(3, 0)
7
CAPE - 1999
Pure Mathematics - Unit 1
Paper 02
Section A (Module 1)
1
Solve the simultaneous equations
( a)
2)2
(x
( y 2)2
4
y+x-2=0
[8 marks]
(b)
P
R
S
Q
The diagram above represents the logo of a company. The logo consistes of three
circles each of radius r which touch one another externally. P Q and R are the centres
of these three large circles. A fourth circle with centre S and radius α touches each of
these three large circles.
( i)
Write down the size of angle PQR
( ii )
Calculate in terms of r the area of the triangle PQR
( iii )
Write down the size of angle PSQ
[1 mark]
[2 marks]
[1 mark]
By considering triangle PSR or otherwise
( iv )
r
show that
r
(v)
α
3
2
[2 marks]
calculate in terms of r the area of the region enclosed by the arcs of the three
large circles and the circumference of the circle with centre S
[6 marks ]
1
( a)
subs y = 2 - x
(x
2)2
y4
(b)
( i)
( iii )
(v)
x )2
(4
2
y2
π
4
(2 , 0)
0
2π
r
( iv )
3
areatriangle . ( PRQ )
r2 . 3
π α
sin .
α
r
2
( 4 , 2)
1
π
( 2 r ) 2 r sin
2
3
( ii )
6
4
has solution(s)
0
r2 . 3
π
r
3
3
α
r
2
areacircle .( centre . S )
2
subs
α
2
3 r
3
2.
r
2
The function f is defined by
( a)
3
f. ( x )
2
π. 2
3
r2
3
1
6x
3 x2
xεR
x > -1
( i)
Evaluate ff (-2)
[2 marks]
( ii )
Calculate the exact values of x which map onto themselves under the function
f
[4 marks]
u. ( x
2
where u, v, w ε R
( iii )
Express f (x) in the form
( iv )
By sketching the graph of y = f (x) or otherwise state the turning point P of the
graph and indicate whether P is a maximum or a minimum
v)
w
[3 marks]
[3 marks]
2
(b)
( i)
Determine the range of f
[1 mark]
( ii )
Explain why the function f has an inverse
[1 mark]
( iii )
Find an expression for f -1 (x)
[4 marks]
( iv )
Describe the geometrical relationship between the graphs
y = f (x) an y = f -1 (x)
( a)
[2 marks]
ff (-2) = f (1) = -8
( i)
( ii )
1
6x
3 x2
x
0
has solution(s)
7
6
( iii )
completing the square:
f. ( x )
4
3.(x
1)2
( iv )
f( x )
1
3 x2
6x
5
f( x )
1
0
1
5
x
P . ( 1 , 4 ) max
3
2
1.
61
6
(b)
( i)
f. ( x ) 4
( ii )
f has an inverse since x > -1 and is one-to-one (bijectve)
1
(4
( x)
x)
( iii )
f
1
( iv )
f (x) is a reflection of f -1 (x) in the line y = x
3
x 4
Section B (Module 2)
3
( a)
Find the equation of the line which passes through the point (4, -1) and is
perpendicular to the line y = 2 - 2x
( i)
[3 marks]
( ii )
Calculate the coordinates of the point of intersection of these two straight lines
[3 marks]
(b)
The function f is defined on the set R of real numbers by
f. ( x )
3 ) x2
(k
kx
1
where k ε R
Calculate the range of values of k for which the equation f (x) = 0 has no real
roots
( i)
[5 marks]
( ii )
Solve the equation f (x) = 0 for k = 1 giving your answer in the form a + bi and
a - bi where a, b ε R
[3 marks]
( iii )
Let α and β be roots of f (x) = 0 when k = 8. Without first solving f (x) = 0
determine the equation whose roots are respectively
1
( a)
( i)
y 1
( ii )
1
x
2
1
2
3
1
and
α
(x
4)
2
2x
4
[6 marks]
β
y
1
x
2
3
x = 2 y = -2
(b)
4. ( k
k2
( i)
3 ) ( 1) < 0
( ii )
4 x2
( iii )
11 x2
8x
α
8
( 11 )
11
x
β
αβ
1
1
1. .
i 15
8
1
8
1. .
i 15
8
sum of roots = -8
product of roots = 11
x2
( a)
1
8
0
11
αβ
4
has solution(s)
0
1
2 < k< 6
has solution(s)
8x
Given that θ is an obtuse angle such that
11
0
sin . θ
2
3
find the value of cos 2θ
[2 marks]
(b)
The diagram below shows a triangle ABC in which AC = 4 units, CD = 3 units and
angle CAB = angle DCB = θ. AB is perpendicular to CB
A
q
4
D
3
q
C
B
( i)
Obtain an expression for AD in terms of θ
( ii )
Express AD in the form R cos (θ + α) where R is positive and α is an acute
angle
[3 marks]
[4 marks]
5
( c)
( i)
Express cos 3θ in terms of cos θ
( ii )
Hence solve for 0 < θ < 2π the equation
[5 marks]
cos 3θ + 2 cos 2θ + 4 cos θ + 2 = 0
( a)
cos . 2 θ
(b)
( i)
1
2 cos2 θ
( i)
1 cos . θ
5 cos . θ
cos . ( 2 θ
36.870
θ)
( 2 sin . θ . cos . θ ) sin . θ
cos . 3 θ
( ii )
4 cos 3 θ
1
2
3
2
2
1
9
cos . 2 θ
AD = 4 cos θ - 3 sin θ (AB - DB)
( ii )
( c)
cos . 2 θ
2 sin2 θ
[6 marks]
cos . θ . ( 2 cos . θ
1)
2
cos . θ
tan
3
4
1
sin . 2 θ . sin . θ
2. 1
cos 2 θ cos . θ
3 cos . θ
3 cos . θ
cos . θ
4 cos2 θ
cos . 2 θ . cos . θ
2 cos 3 θ
4 cos3 θ
4 cos 3 θ
α
2 . 2 cos2 θ
4 cos . θ
cos . θ . 4 cos 2 θ
0
cos . θ
0
θ
6
1
0
cos . θ
π 3π 2π 4π
,
,
,
2 2 3 3
2
4 cos . θ
1
2
0
1
0
Section C (Module 3)
5
A groove cutter is operated by a robot to produce small discs from a flat sheet of metal. The
path of the cutter blade traces out the curve
0 t π
y = sin t
after t seconds
( a)
Sketch the curve traced out by the cutter
[2 marks]
(b)
Use the Trapezium rule to find the approximate area of a flat side of each disc by
using eight subintervals
[6 marks]
( c)
Compare this approximate area with the exact area of a side of each disc
[4 marks]
(d)
Each disc is fed automatically into another machine that rotates it through 360 deg
about the t-axis to sand the edge
( i)
Sketch the solid that is generated by the rotation
[2 marks]
( ii )
Find the volume of the solid that is generated
[6 marks]
( a)
f( t)
sin ( t )
1
f( t )
0.5
0
1
2
3
t
(b)
π
8
h
t0
0
tπ
0.38268
tπ
8
t3
π
0.92388
8
t7
π
tπ
4
1
2
0.38268
tπ
0.70711
t5
π
8
0
8
7
0.92388
t3
π
4
0.70711
area
π
(2
16
4 ( 0.38268 )
simplifies to
4 ( 0.70711 )
1.974 units
π
( ii )
area
4 ( 0.92388 ) )
cos . t . π
sin ( t) d t
0
2 units
0
difference = 0.026 units
( c)
( i)
y
x
p
0
π
( ii )
Vol
π . sin2 t d t
0
π
π.
2 0
(1
cos . 2 t ) d t
π
t
2
8
1
2
sin . 2 t . π
0
2
Vol
π
units3
2
6
The function g is given by
g. ( x)
x2
1 (x
xεR
1)
( a)
Find g (0) g (1) and g (-1)
[3 marks]
(b)
Obtain an expression for the derivative d/dx (g) at x ε R
[2 marks]
( c)
Find the stationary points of g
[4 marks]
(d)
Determine the value(s) of x where g has
( e)
( i)
a local minimum
[2 marks]
( ii )
a local maximum
[2 marks]
Using the above and any other information sketch the graph of g
( a)
g. ( 0)
1
g. ( 1)
(b)
d
x2
dx
1 (x
1)
( c)
(3 x
(d)
d2
dx2
1) ( x
6x
( i)
1)
g. ( 1 )
0
3 x2
0
2x
d2 g
dx2
32
,
3 27
1> 0
3
1
( e)
9
1
1
, 1
3
x
2
0
d2 g
dx2
( ii )
min
<0
1
( 1 , 0 ) max
[7 marks]
( e)
x2
f( x )
1 (x
1)
3
2
1
f( x )
3
2
1
0
1
1
2
3
x
.
10
2
3
1
5
8
CAPE - 1999
Pure Mathematics - Unit 1
Paper 01
Section A (Module 1)
1
f (2) = 0
2
k
Express
32
x5
2x
calculate the value of k
k
5
x
5
3x
3x
2x
2x
81x
Solve the equation
34
x
1
x ) x4
1
5
x
1
32
x5
2 x3
f. ( 2 )
0
8x
16
4 x2
27x
33
[2 marks]
[3 marks]
5
in the form (2 - x) g (x) where g (x) is a polynomial in x
(2
5
5x
6
f. ( x )
4
kx2
Find the real values of x which satisfy the equation
3x
3
x3
Given that x - 2 is a factor of
[3 marks]
[4 marks]
x
4x
4
3x
x
4
In triangle ABC below angle ABC = 1200, BA = (y - 3) cm, BC = y cm and AC= 6 cm
Show that
y2
3y
9
[3 marks]
0
1
A
(y – 3) cm
1200
B
6 cm
y cm
C
62
using the cosine rule:
6
The function f is given by
( a)
(b)
y2
3y 9
f. ( x )
2x
( y 3)2
y2
2 ( y 3 ) ( y) cos . 1200
0
3
xεZ
Show that f is
( i)
injective (one-to-one)
[1 mark]
( ii )
not surjective (not onto Z)
[2 marks]
Determine the value(s) of x ε Z for which f -1 (x) ε Z
( a)
(b)
( i)
a b
( ii )
2. ( a)
f -1 (x) =
1
(x
2
f. ( a ) f. ( b )
3 2
3)
a, b ε Z
for any a ε Z
x = 2n + 1 n ε Z
2
[2 marks]
Section B (Module 2)
2 x2
Given that α and β are roots of the equation
7
4x
α
find without solving the equation the exact value of
(α
consider
α
β
(α
α
Prove that
8
β
β)
2
β)
αβ
θ
2
2
sin2
In a triangle ABC show that
β
2
2
β
2 αβ
2
5
sin2
θ
2
(α
β ). (α
(α
α
2
cos . θ
1
9
α
α
2
cos2
2
2
β
β)
β)
2
2
β
0
2
(α
2
2 . ( 2)2
2
5
[3 marks]
β)
2
4 αβ
4 αβ
4.
5
2
2 . 14
θ
2
[3 marks]
2 sin2
1
AB
AC
θ
2
sin2
θ
2
where D is the midpoint of BC
2 AD
[3 marks]
AB
BC
AC
AB
AB
BC
AC
AB
2 AD
2 AD
AB
AC
BC
2 AD
2 AD
AB
AC
2 AD
Sketch a graph to represent the range of values of x which satisfy the inequality
10
x2
(x
-4
0
2) ( x
8
2x
4) 0
4 x 2
2
3
[3 marks]
11
Given that points C (-1, 1) and D (2, -2) find the equation of
( a)
the line CD
[2 marks]
(b)
the line through E (-1, -2) parallel to the line CD
[2 marks]
( a)
y
1
2 1
(x
2 1
(b)
y
2
(x
y+x=0
1)
y+x+3=0
1)
Section C (Module 3)
13
Given that
lim
x
2a
14
a. x
x
9
a
calculate the value of a
9
[2 marks]
4
4
5
2
f. ( x )
The function f is defined by
(x
1
2).( x
1)
and is continuous for all values of x except a and b where a < b.
Find the value of a and of b
a = -2
15
( a)
b=1
Find the value(s) of x at the stationary point(s) of the function
f. ( x )
(b)
[2 marks]
2 x3
9 x2
Determine the nature of the points
( a)
(b)
6 x2
d2
2
dx
18 x
has solution(s)
0
f. ( x)
0
( 0 , 0 ) max
0
4
[4 marks]
0
3
d2
dx
2
f. ( x) > 0
3
( 3 , 27 ) min
16
3 (x
lim
h
17
3 x2
Find from first principles the derivative of
0
h)2
h
3 x2
lim 6 x
h 0
f. ( x )
The function f is defined by
x3
x2
with respect to x
d
3 x2
dx
3h
2
x5
where x ε R
[4 marks]
6x
x 0
f. ( x ) dx
( a)
Find
(b)
Evaluate
2
f. ( x) d x
[5 marks]
1
( a)
1
(b)
x
5
18
2
x
Given that
1
2
2x
x
3
2x
5
1 .
2
2 x4 1
f. ( x) d x
25
2 . ( f. ( x)
x) d x
dx
1
1
1
x
2 x2
2 x4
C
13
32
where f is a real continuous function evaluate
2
5
[3 marks]
2
5
2
2 . f. ( x ) dx
5
2 x dx
2
5
50
25
4
I
71
2
CAPE - 1998
Pure Mathematics - Unit 1
Paper 02
Section A (Module 1)
1
Solve the simultaneous equations
( a)
y - 2x + 3 = 0
x2
y2
5x
y
[8 marks]
26
The diagram below not drawn to scale shows a plan view of a flower bed covered in
grass ferns and hibiscus
(b)
ACB is an arc of a circle with centre O and radius 10 cm
XZY is also an arc of a circle with centre O and radius 11 cm
AB is a straight line and the angle at O is θ
Z
Grass
C
Ferns
Y
X
1m
B
A
Hibiscus
10 m
q
( i)
Write down an expression in terms of θ for the area covered by the grass
[3 marks]
( ii )
Write down an expression in terms of θ for the area covered by the hibiscus
[2 marks]
( iii )
Given that the area covered by hibiscus is one-third of the area of the circle of
radius 6 m calculate sin θ
[3 marks]
( iv )
Calculate the area covered by the ferns
1
[4 marks]
( a)
y = 2x - 3
x2
(2 x
3 )2
x=4
(b)
5x
(2 x
y=5
1
112
2
3)
26
x = -1
( a)
1
y = -5
102 θ
( i)
areagrass
( ii )
areahibiscus
( iii )
50 sin . θ
1 . 2
π (6)
3
( iv )
areaferns
1.
6π
( 10 ) 2 sin 1 .
2
25
21
θ
2
areagrass
1
( 10 ) 2 sin . θ
2
areahibiscus
50 sin . θ
6π
25
sin . θ
areaferns
2
4
has solution(s)
0
50
6π
25
50 . ( θ
50 . ( 0.8540
sin . θ )
0.7540 ) m2
5.0 m2
The function f is defined on the set R of real numbers by f (x) = 2x + k, where k is a
real constant. Given that ff (2) = 5 calculate the value of k
[3 marks]
(b)
The function g is defined on R for all
( i)
by
x 1
g. ( x)
x2
2x
2
Calculate the values of x which map onto themselves under the function g
[4 marks]
2
where a, b ε R
( ii )
Express g (x) in the form
( iii )
Sketch the graph of y = g (x) showing clearly the coordinates of the minimum
point
(x
a)
b
[3 marks]
[3 marks]
( iv )
Explain why g has an inverse
2
[1 mark]
The inverse of g is denoted by g -1
Describe the geometrical relationship between the graphs
(v)
y = g (x) and y = g -1 (x)
Find an expression for g -1 (x)
( vi )
f (2) = 2(2) + k = 4 + k
( a)
( i)
( ii )
x2
2x
g. ( x)
2
x
k = -1
(x
1 )2
(x
[4 marks]
f (4 + k) = 2(4 + k) + k = 8 + 3k
8 + 3k = 5
(b)
[2 marks]
1) ( x
2)
0
x = 1, 2
1
( iii )
5
4
2
x
3
2x
2
2
1
1
2
3
4
x
( 1 , 1 ) min
( iv )
f. ( a )
(v)
g -1 (x) is a reflection of g (x) in the line y = x
( vi )
x
b
( y 1 )2
bijective - one-to-one and onto
a b
1
y
3
(x
1)
1
g -1 (x) =
(x
1)
1
Section B (Module 2)
3
z1 and
( a)
z2
are the complex numbers 4 + 2i and 3 + i respectively
( i)
Express the product
( ii )
Find the modulus and principal value of the argument of w
( iii )
Write down the complex numbers iw and
w
in the form a + bi
z1 z2
w
[2 marks]
[3 marks]
(the conjugate of w)
[2 marks]
( iv )
Find the distance between the points on the Argand diagram represented by
iw and
(b)
[2 marks]
w
Find the cartesian equation of the curve with parametric representation
( i)
x = 3 cos θ
y = sin θ
[2 marks]
( ii )
Sketch the curve
( iii )
Find the equation of the tangent to this curve at the point
( iv )
Express
x2
[2 marks]
y2
cos2 θ
in terms of
( i)
( ii )
z1 z2
(4
102
z1 z2
arg. z1 z2
( iii )
( iv )
iw
iw . w
2i ) ( 3
10
tan
10 + 10i
i)
10 . 2
102
1
π
(1)
10i
w
10 . 2
10 . 2
4
4
10
10i
20 . 2
π
4
[4 marks]
and hence find the greatest
and least distances from the origin to any point on the curve
( a)
θ
[3 marks]
(b)
2
x
3
( i)
f( t)
y2
x2
1
y2
9
3 cos ( t)
1
g ( t)
sin ( t)
( ii )
1
g( t )
4
3
2
1
0
1
2
3
4
1
f( t )
( iii )
cos . θ
3 sin . θ
dy
dx
tangent:
dy
dx
1
9 cos 2 θ
θ
4
( a)
( i)
0
cos 2 θ
1
x
3
2
( iv )
3
4
1
y
1
π
2
x2
distancemax
1
y2
θ
3
π
2
distancemin
1
Given that for all values of θ
3 cos . θ
0
45
k . sin . θ
0
45
b . cos . θ
evaluate k and b
[4 marks]
( ii )
Express 4 cos θ + 2 sin θ in the form R cos (θ - α) where R is positive and α is
acute
[4 marks]
5
(b)
A
E
B
D
C
In the above figure E is the foot of the perpendicular from B to AC. D is the foot of the
perpendicular from A to BC.
If 2 BD = 3 AE θ is the angle ABE and 2θ is the angle BAD
( i)
show that 2 sin 2θ = 3 sin θ
( ii )
solve the equation 2 sin 2θ = 3 sin θ for
( iii )
find the general solution(s) of the equation 2 sin 2θ = 3 sin θ
( a)
( i)
3 cos . θ .
3. 2
2
2
2
[4 marks]
3 sin . θ .
k. 2
2
2
2
00 θ 900
2
2
k . cos . θ .
k. 2
2
3. 2
2
k . sin . θ .
b
[4 marks]
[4 marks]
2
2
b . cos . θ
0
k=3
b
( ii )
R
2. 5
2 . 5 cos . θ
tan . α
26.60
6
1
2
α
26.60
3. 2
( i)
3
AE
2
BD
( ii )
3 sin . θ
( iii )
θ
BD
AB
sin . 2 θ
AE
AB
sin . θ
4 . sin . θ . cos . θ
( 180 n ) 0
θ
3 AE
AB
2 sin . 2 θ
3 sin . θ
2 sin . 2 θ
sin . θ . ( 4 cos . θ
( 360 n ) 0
3)
41.410
0
θ
( 360 n ) 0
θ
00 , 41.410
41.410
Section C (Module 3)
5
Oil flows into a cylindrical drum of radius 10 metres. At time t = 0 the drum was empty and at
time t the oil level in the drum is rising at the rate
10 .
1
1
1
4
t
metres per second
( a)
What is the rate of increase in the volume V of oil in the drum?
(b)
Sketch the graph of f (t) = 10 .
( c)
At what time is the rate of increase in oil level zero?
[2 marks]
(d)
Write down an expression for V at time t = 3
[3 marks]
( e)
Evaluate V by ordinary integration
[3 marks]
( f)
Evaluate V by the trapezium rule using seven ordinates.
[5 marks]
( i)
V
π . ( 10 ) h
2
dV
dt
1000 π
1
1
dV
dh
100 π
1
1
t
( ii )
7
1
4
t
1
4
for
0 t 3
dV
dt
dV . dh
dh dt
[5 marks]
[2 marks]
1
1
10
1
t
5
4
0
1
2
3
t
( c)
10
1
1
t
1
4
has solution(s)
0
3
(d)
V
1000 π .
1
1
t
t
3 secs
1
dt
4
0
( e)
V
1000 π . ln. ( 1
1 .
t 3
4
0
t)
V
1000 π . ln. 4
V
636.29 π . m3
( f)
Vol
1
1000 π .
( 0.75
4
2 ( 0.41667
0.25
0.15
0.08333
yields
8
0.03571 ) )
655.355 . π m3
3
m3
4
6
Explain what is meant by the derivative of a function at a point of its domain.
f (x) =
(x
2)2
1) ( x
xεR
( a)
Find f (0) and the value(s) of x for which f (x) = 0
(b)
Obtain an expression for
( c)
Determine the intervals over which f
(d)
( e)
d .
f ( x)
dx
[2 marks]
[2 marks]
and find the stationary points of f
[2 marks]
( i)
is increasing
[2 marks]
( ii )
has negative second derivative
[2 marks]
Determine the value(s) of x where f has
( i)
a local minimum
[2 marks]
( ii )
a local maximum
[2 marks]
Using the above and any other information sketch the graph of f
d .
f ( x)
dx
derivation of f at point x is
lim
h 0
f. ( x
h)
h
[6 marks]
f. ( x )
the limit exists
( a)
f (0) = 4
(b)
d
(x
dx
d
dx
f. ( x)
f. ( x )
1) ( x
0
2 )2
2 (x
0
2.(x
2) ( x
2) ( x
1)
( i)
d
dx
1)
(x
(0 , 4)
Stationary points:
( c)
1, 2
x
f. ( x) > 0
3 x.(x
9
(x
2)2
2 )2
0
has solution(s)
(2 , 0)
2) > 0
has solution(s)
x< 0
2< x
0
2
d2
( ii )
dx2
f. ( x )
3x
( 0 , 4 ) max
2)
1)< 0
6 (x
(d)
3 (x
has solution(s)
( 2 , 0 ) min
( e)
5
4
3
(x
1) (x
2)
2
2
1
2
1
1
0
1
2
2
x
10
3
4
5
( x < 1)
6
CAPE - 1998
Pure Mathematics - Unit 1
Paper 01
Section A (Module 1)
1
Use the remainder theorem to find the remainder when
3
2
2x
f. ( 2 )
2
3x
9
is divided by x + 2
[2 marks]
5
Given that x is a real number solve the equation x + 3 =
[2 marks]
4x
3
(x
3)
2
( 4 x)
2
5
has solution(s)
0
1
3
2
Express
3x
6x
4
a. ( x
in the form
h)
2
k
stating clearly the values of
a, h and k
[3 marks]
3. ( x
4
1)
2
7
The diagram below shows the graph of y = A + B sin (2x - C) for
0 x π
where A, B
and C are positive real constants
y
3
2
1
0
p/4
p/2
Obtain the values of A, B and C
3p/4
p
x
[4 marks]
ymax
A+B=3
3
ymin
A-B=1
1
A = 2 B = 1 C =π /2
5
In triangle ABC angle ABC = 60 deg AB = (x + 2) cm AC = 5 cm and BC = x cm
B
600
(x + 2) cm
x cm
A
5 cm
2
Show that
x
2x
C
21
cosine rule:
(x
6
2)
2
2
x
[4 marks]
0
(x
2)
2. ( x
2
2
x
2 ) ( x)
2. ( x
1
2
2 ) ( x)
2
1
2
5
2
simplifies to
5
2
x
2. x
21
The function f is defined by f : x maps to 3x + 1, xε R Show that f is
( a)
injective (one-to-one)
[2 marks]
( b)
surjective (onto R)
[2 marks]
( c)
find the inverse f -1 of f
[1 mark]
( a)
f. ( a ) f. ( b )
3 (a)
1 3 (b)
a b
1
3 (a)
a,b 0
1
3 ( b)
1 0
a and b are distinct and hence a maps to f (a), b maps to f (b)
f.
for any y ε R
( b)
1
(y
3
1
(y
3
3.
1)
every y ε R is the image under f of x =
1
(y
3
f -1
( c)
1)
1)
1
1
(y
3
1)
yεR
Section B (Module 2)
7
2
Given that α and β are the roots of the equation
2x
(α
find without solving the equation the value of
8
(α
β)
2
α
(α
β)
2
(α
Prove that
1
1
9
2
β
2
β)
2 αβ
2
2
β
2
5
2
4 αβ
7
0
2
(α
[3 marks]
β)
2
2
7
4
2
2 αβ
=
cos . 4 θ
tan. 2 θ
sin . 4 θ
2
[3 marks]
2
2
cos 2 θ sin 2 θ
2 sin . 2 θ . cos . 2 θ
2
sin 2 θ sin 2 θ
2 sin . 2 θ . cos . 2 θ
tan 2θ
Find the magnitude of the vector (tanθ i + j) in its simplest form
2
( tan. θ )
10
α
β)
5x
sec . θ
1
Find the range of values of x which satisfy the inequality (3x + 1)(2x - 3) > 0
(3 x
1) ( 2 x
3 )> 0
x<
[2 marks]
1
3
x>
3
2
[4 marks]
11
12
Given the points A (1, 2) and B (-1, 3) find
(i)
the coordinates of M the mid-point of AB
( ii )
the equation of the line through the origin parallel to AB
(1)
M (0, 5/2)
( ii )
y
1
0
(x
2
0)
x
2y
[4 marks]
0
The vectors x and y are given by
x = (t + 1) i + 3j and y = i + (t - 1) j where t is a real number
Find the value(s) of t for which x and y are parallel
3
comparing ratio of components:
2
t
1
3
t
[4 marks]
t
t
1
1
1
2,2
Section C (Module 3)
2
13
Find
x
lim
x
2
x
x
2
(x
lim
x
14
Given the function
(i)
undefined
( ii )
continuous
(i)
[2 marks]
2
2
f. ( x )
1
x
1) (x
x 2
2)
3
state the values of x for which f (x) is
[2 marks]
x=0
( ii )
all real x except x = 0
Determine the value(s) of x at the stationary points of the function
15
f. ( x )
3
2
x
d .
f ( x)
dx
d
2
2
and determine the nature of these points
6x
2
3x
f. ( x )
3 x. ( x
12 x
6x
d
12
2
2
dx
4)
f. ( x )
dx
d
2
2
3
(i)
Find
( ii )
Hence evaluate
2x
5x
< 0
x
0,4
f (x) is max
0
f. ( x )
dx
16
0
[4 marks]
> 0
f (x) is min
4
3 dx
3
3
2x
5x
[4 marks]
3 dx
2
1 4
x
2
(i)
1
4
( 3)
2
( ii )
5.
2
(3)
2
2
Given that
17
(3
5 2
x
2
C
1
4
( 2)
2
3 ( 3)
f. ( x ) ) d x
3x
5
2
( 2)
2
3 ( 2)
35
where f (x) is real and continuous evaluate
1
0
2
f. ( x ) d x
[4 marks]
0
2
2
3 dx
0
0
f. ( x ) d x
2
1
0
f. ( x ) d x
1
( 3 x) .
2
2
0
0
f. ( x ) d x
5
18
The function f is defined by f : x maps to f (x), εx R where for each x ε R f (x) is a positive real
number
The area under the graph of f from x = 0 to x = t
F. ( t)
3
t
t 0
is given by
3t
Find f (1)
[4 marks]
f. ( t )
2
3t
3
t
1
End of Test
f. ( 1 )
6