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2013 Revision Package BT1 Solutions
Chapter 2: Binomial Expansion Solution
1 [AJC/2005/Prelim/P1/Q3]
Solution
( )( )
( )( ) ( )
( )
( )
�
�
�
�
�
++++=
++++
++++=
+++++=
+−
−−
−−
−
+−
−−
−
+−
−+
+=
−+=−
+ −
32
32
32
32
3
2
2
1
42
521
2
32
5
2
31
2
5
2
311
2!3
22
11
2
1
2
1
2!2
12
1
2
1
22
11
1
21121
1
xxx
xxx
xxx
xxxx
x
xx
x
xxx
x
The series is valid for 2
112 <⇒< xx
When 11
1=x ,
11
4
3
11
11
12
11
9
11
12
11
121
111
1
21
1=×==
−
+=
−
+
x
x
32.3
3209
10648
2662
3209411
11
14
11
1
2
5
11
121
11
432
≈
≈
÷≈
+
+
+
+= �
2 [TJC/2005/Prelim/P1/Q7]
Solution
2013 Revision Package BT1 Solutions
( )( )
( ) ( )( ) ( )22
22
2
12111232
1
2112112
1123
xCxBAxxx
x
C
x
BAx
xx
xx
++−+=−+
−+
+
+=
−+
−+
When 2
1=x ,
2
1
4
11
4
111
2
123
2
1
=
++=
−
+
C
C
When 0=x , ( ) ( )
1
12
11
2
3
=
+=
B
B
When 1=x , ( ) ( )( ) ( )
3
112
12111123
2
1
=
++−+=−+
A
A
( )( ) ( )xx
x
xx
xx
212
1
1
13
2112
112322
2
−+
+
+=
−+
−+
( )( ) ( )
( )( ) ( )
( )( ) ( )
�
��
��
+++=
+++++−+=
+++++−+=
−+++=
−+
+
+=
−+
−+
−−
2
22
22
112
22
2
42
3
22
113
4212
1113
212
1113
212
1
1
13
2112
1123
xx
xxxx
xxxx
xxx
xx
x
xx
xx
The expansion is valid for 2
112x and 12 <⇒<< xx
3 [RJC/2005/Prelim/P1/Q5]
Solution
2013 Revision Package BT1 Solutions
( )
�
�
�
++−+=
++−+=
+
−
−
+
−
+
+=
+=+
32
32
32
2
1
2
1
512
1
64
1
4
12
1024
1
128
1
8
112
4!3
22
11
2
1
2
1
4!2
12
1
2
1
42
112
4124
yyy
yyy
yyy
yy
The expansion is valid for 414
<⇒< yy
Let 28 kxxy +=
( ) ( )
( ) ( ) ( )
( ) ( )
��
���
�
�
++−=
+++++−=
++++−++=
++−+=
+=++
33
34232
32222
32
2
1
2
12
4
512512
11664
64
1
8512
18
64
18
4
12
512
1
64
1
4
12
484
xxk
xxkkxx
kxxkxxkxx
yyy
ykxx
4
014
=
=+−
k
k
4 [NYJC/2005/Prelim/P1/Q1]
Solution
( ) �++++=−− 322
43211 xxxx
The expansion is valid for 1<x .
2013 Revision Package BT1 Solutions
12
2
113
expansion into 2
1subst ,
2
14
2
13
2
1213
23
2
3
2
32
11
11
=
−=
=
+
+
+
+=
=
−
∞
=−
∞
=− ∑∑
x
rr
rr
rr
�
5 [HCI/2005/Prelim/P1/Q2]
Solution
( )
( ) ( )( )
( )( )( )
�
�
�
+++=
++++=
+
−
−−−−−
+
−
−−−+
−−+
=
−=−
−
−−
32
32
3
21
2
22
8
1
16
3
4
1
2
1
4
31
4
1
2!3
22122
2!2
122
221
4
1
2122
xx
xxx
x
xx
xx
The series is valid for 212
<⇒< xx
.
The term of rx is the ( )th
r 1+ term in the series.
( ) ( )( )( ) ( )
( ) ( )( )( ) ( )( ) ( )
( )22
1
2
1
!
14321
4
1
2!
1222122
4
1 term1
+
+=
−
+−=
−
+−−−−−−−=+
r
r
r
rr
r
th
r
xr
rr
x
r
rr
�
�
6(i) [CJC/2005/Prelim/P1/Q2]
Solution
2013 Revision Package BT1 Solutions
( )( )( )
( ) ( )( )( )11
11
11
1
11
1f
2
2
23
2
23
3
2
3
+−
−+=
−−+
−+=
−−+
−−=
+−
−−=
xx
x
xxx
x
xxx
xx
xx
xxx
( )( ) ( )( ) ( )22
2
2
2
1111111 ++
++
−=
+−
−=
+−
−
x
D
x
C
x
B
xx
x
xx
x
( ) ( )( ) ( )111122 −++−++=− xDxxCxBx
When 1=x , ( )4
121
2−=⇒=− BB
When 1−=x , ( )2
1111 =⇒−−=− BD
When 0=x , ( ) ( ) ( )4
31
2
111
2
10
2−=⇒−+−+= CC
( )( )( ) ( ) ( ) ( )22
3
12
1
14
3
14
11
11
1f
++
+−
−−=
+−
−−=
xxxxx
xxx
6(ii) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
2
1 1 2
2 3 2 3 2 3
2 3 2 3 2 3
2 3
f
1 3 11
4 1 4 1 2 1
1 3 11 1 1 1
4 4 2
1 3 11 1 1 1 2 3 4
4 4 2
1 1 1 1 3 3 3 3 1 31 ... ... 2 ...
4 4 4 4 4 4 4 4 2 2
1
x
x x x
x x x
x x x x x x x x x
x x x x x x x x x
x x
− − −
= − − +− + +
= + − − + + +
= + + + + + − − + − + + + − + + − +
= + + + + + − + − + + + − + − +
= + − +
� � �
�
The series is valid for 1<x .
2005x of ( )xf is the 2006th term of ( ) 11
4
1 −− x + 2006th term of ( ) 1
14
3 −+− x + 2006th term of
( ) 21
2
1 −+ x
( ) ( ) ( ) 200520052005120062005120062005 100210034
3
4
120061
2
11
4
3
4
1xxxxx −=
−+=−+−−
++
Therefore coefficient of 2005x is -1002.
7 [ACJC/2005/Prelim/P1/Q2]
2013 Revision Package BT1 Solutions
Solution
Using sine rule
C
AB
x
ACBC
sin
6sin
4sin
=
+
=ππ
( )
( )( )
( )
+−≈
++−≈
−+
−−≈
+
+−
−−−+
+−−=
+−+=
+−+=
++
+−
=
+
=
+
=
+
=
−
−
2
22
222
22
12
12
12
2
2
7312
32
312
23
2312
23
!2
111
2312
2312
2312
32
1
2
sin2
3cos
2
12
1
sin6
coscos6
sin2
1
6sin
4sin
xx
xx
x
xx
xx
xx
xx
xx
xx
xx
xx
xx
xAC
BC
���
�
�
��
ππ
π
π
A
B
C4
π x+6
π
2013 Revision Package BT1 Solutions
Chapter 3: Arithmetic & Geometric Progressions Solution
1
(i)
(ii)
(iii)
[RJC/2007/Prelim/P1/Q13]
Solution
Let Sn and Hn denote the amount of savings that Selina and Hebe have at the end of the nth year
respectively.
1000 200n
S n= +
( ) ( ) ( )
( )
( )
2 6
5
5
1000 100 100 1.5 100 1.5 ... 100 1.5
100 1.5 11000
1.5 1
1000 200 1.5 1
n
n
n
n
H−
−
−
= + + + + +
−= +
−
= + −
where 5n >
When n n
H S> ,
( )51000 200 1.5 1 1000 200
nn
−+ − > +
i.e. ( )5200 1.5 1 0
nn
−− − >
From the GC, the value of k is 12.
Let k
u denote the interest that Hebe gets in the kth
year.
Then ( )12 5 1
12100 1.5 1139.0625u
− −= = ,
so Hebe receives $1140 in interest at the end of the 12th
year.
Then for 12n > ,
12
12
1
3
n
nu u
−
=
.
The total investment yield from the 13th year onwards therefore
cannot exceed 12
12
13
13
1
1 2
uu=
−.
Hence, the maximum value of Hebe’s investment is
12 12
1$4790
2H u+ = (to 3 s.f.).
2 [AJC/2008/Prelim/P2/Q2]
Solution
Let T7, T3, T1 be the seventh, third and first term of an arithmetic series with first term a and
common difference d.
7 3 16 , 2 ,T a d T a d T a= + = + =
2013 Revision Package BT1 Solutions
( ) ( )2
2 2 2
2
2
6 2
6 2
6 4 4
4 2 0
a d a
a d a d
a a d a d
a ad a ad d
d ad
+=
+ +
+ = +
+ = + +
− =
1
3
Since 02
1Common ratio
2 2
ad d
T a ar
T a d a a
≠ ⇒ =
= = = =+ +
Since common ratio < 1, the geometric progression is convergent.
( )
13 1
23 32 1 100
12 4 8 12
9 3 16 1 100 0
2 8 8 2
n
n
nn
nn
− + − − ≥
−
+ − − − ≥
Using GC, 22.3n ≥ ∴ Least n = 23.
3
(i)
(ii)
[DHS/2008/Prelim/P2/Q4]
Solution
Amount saved in 1st
year = ( )1
36000 $180002
=
9
2904.1
9
20104.1
1000000104.1
)104.1(18000
≥
≥−
≥−
−=
n
n
n
nS
7 6 3
3 3Since 3 3 and
2 4 8
T a d
ad a a a d
= + =
= ⇒ + = ⇒ = =
2013 Revision Package BT1 Solutions
(iii)
30
8.2904.1ln
9
29ln
=∴
=≥
n
n
9 years = 108 months
108 = 21(5) + 3 ∴n = 21
total amount = ( ) ( )21
10000 2 25 20 32
+ +
= $11155
4 [JJC/2008/Prelim/P2/Q1]
Solution
Let the first term of the AP be a and common difference be d.
( )1 2 9 10 11 18
2 T T ... T T T ... T+ + + = + + +
( ) ( )9 9
2 8 9 172 2
a a d a a a d+ + = + + +
[ ]2 2 8 2 26a d a d+ = +
5a d=
16
1
20
20
T
T=
20 15
20
a d
a
+=
20 20
5 20
d
d=
24d =
2d = (reject 2d = − , since a is positive)
Hence, 10a =
5
(i)
(ii)
[NYJC/2008/Prelim/P1/Q4]
Solution
11 1ln( ) ln( ) ln n
n n n n
n
ab b a a
a
++ +
− = − =
.
Since na is a geometric progression, thus 1n
n
ar
a
+ = for all n+∈� . Thus 1 ln( )n n rb b+ − = for all
n+∈� . Thus
nb is an arithmetic progression with common difference ln( )r .
Since nb is an arithmetic progression
2013 Revision Package BT1 Solutions
(iii)
1
1 1
1
1 1
1
1( )
2
1[ln( ) ln( )]
2
1ln( )
2
N
n N
n
N
N
Nb b b
Na a
Naa
+
+=
+
+
+= +
+= +
+=
∑
Since ( )1
1 2 11
lnN
n N
n
b a a a+
+=
= × × ×∑ � , thus
( )
1
1 2 1
1
1
1
21
1exp ln( )
exp
2
N
N n
n
N
N
N
a a a b
N
a
aa
a
+
+=
+
+
+
× × ×
+ =
=
=
∑�
6
(a)
(b)
[TJC/2008/Prelim/P1/Q11]
Solution
(i) S8 = T29
4(2a + 7d) = a + 28d + 98
7a = 98
a = 14
(ii) Given that u14 ≤ 196 and u15 > 196,
14 + 13d ≤ 196 and 14 + 14d > 196
Hence d ≤ 14 and d > 13
i.e. 13 < d ≤ 14.
(i) wn
wn − 1 =
v2n − 1 + v2n
v2n − 3 + v2n − 2 =
ar2n − 2
+ ar2n − 1
ar2n − 4
+ ar2n − 3
= r2
Hence the sequence {wn} is a g.p. with common ratio r2.
(ii) w1 + w3 + w5 + … = 4 + 4r
1 − r4 =
32
15
From GC, r = − 1
2 (since r ≠ −1)
Hence ∑r = 1
∞ vn =
4
1 −
−1
2
= 8
3 .
7
[TPJC/2008/Prelim/P2/Q2]
Solution
2013 Revision Package BT1 Solutions
(a)
(b)
7
( 1) 70
7 ( 1) 70
( 1) 63.......(1)
a
a n d
n d
n d
=
+ − =
+ − =
− =
[2 ( 1) ] 3852
[2(7) ( 1) ] 385.......(2)2
na n d
nn d
+ − =
+ − =
Subt (1) into (2)
(14 63) 770
10
n
n
+ =
=
number of terms = 10
Subt into (1)
9 63
7
d
d
=
=
common difference = 7
Given GP. Common ratio = 1
2
x +− .
If the sum exists, 1
2
x +− < 1
3 1 x− < ≤
S∞ = 4
3
2 4
1 31
2
4 4
3 3
0
x
x
x
=+
− −
=+
=
8 [MJC/2006/Prelim/P1/Q10(a)]
Solution
2013 Revision Package BT1 Solutions
217 3
nS n n= −= −= −= −
2 2
117( 1) 3( 1) 3 23 20
nS n n n n
−−−−= − − − = − + −= − − − = − + −= − − − = − + −= − − − = − + −
2 2
1(17 3 ) ( 3 23 20)
n n nT S S n n n n
−−−−= − = − − − + −= − = − − − + −= − = − − − + −= − = − − − + −
20 6n= −= −= −= −
1
(20 6( 1)) (20 6 ) 6n n
T T n n++++
− = − + − − = −− = − + − − = −− = − + − − = −− = − + − − = −
∴∴∴∴ the series is an A.P. with common difference 6d = −= −= −= −
9 [SRJC/2006/Prelim/P1/Q8]
Solution
Let the first term of the AP be a = 21 h, and the common difference d =
121
605 = h.
(i) ( ) 166121
21
7 =+=+= daT h
(ii) [ ]=+= 12
721
7S 415 hours
(iii) Let ( ) ( )[ ] 60)1(22 12
121 =−+ n
n
( )[ ] 60)1(12 12
1 =−+ nn
( )[ ] 120)1(1121 =−+ nn
1440)1(12 =−+ nnn
01440112 =−+ nn
=±−
=+±−
=2
588111
2
576012111n 32.8 or – 43.8
Hence during the 33rd lesson, Betty will have completed a total of 60 hours.
10 [SAJC/2006/Prelim/P1/Q2]
Solution
GP 1: 2, , ,a ar ar …
( ) ( )
3
31
3 1 1
S
a
r
a r
∞ =
=−
= − �
GP 2: 2 2 2 2 4, , ,a a r a r …
( )
'
2
2
2
9 2
21
aS
a
r
∞ =
=−
�
Substitute (1) into (2)
2013 Revision Package BT1 Solutions
( )( )( )
( )
29 1 1
1 1 2
2 1 1
1
3
13 1
3
2
r
r r
r r
r
a
−=
+ −
− = +
=
= −
=
11 [TJC/2006/Promo/Q6]
Solution
Given that 2T5 = S2, we have 2ar4 = a + ar, i.e., 2r
4 – r – 1 = 0.
From G.C., r = - 0 .648 or r = 1. (rejected as |r| < 1 for S to exist).
For Sn is within 5% of S, i.e., |Sn – S| < 0.05S ,
we have (1 ) 0.05
1 1 1
na r a a
r r r
−− <
− − − --------------(1)
Since 1
a
r− > 0 (as a > 0 and r = −0 .648), (1) simplifies to
(1 ( 0.648) ) 1 0.05n− − − <
(0.648) 0.05n <
( )lg(0.648) lg 0.05n <
lg(0.05)6.90
lg(0.648)n > = (correct to 3 significant figures)
Hence, minimum value of n is 7.
12
(i)
(ii)
[RJC/2005/Prelim/P1/Q4]
Solution
The first element in the thn row
= 1 2 ... ( 1)2 n+ + + −
=
( )1
22
n n−
The first element in the 10th
row =452 and the first element in the 21
st row =
2102
The sum of all the elements from the 10th to 20th row = 45 46 47 2092 2 2 ... 2+ + + + = ( )45 1652 2 1−
2013 Revision Package BT1 Solutions
13
(a)
(i)
(ii)
(b)
[SAJC/2005/Prelim/P1/Q9]
Solution
Row 1 2 3 … n n+1 n+2 … n+m
No. of tiles 1 3 5 … 2n - 1 2n + 1 2n + 3 …
Total number of tiles
( )
( )( )
2
terms,2,1:
1212
12531
n
nn
n
ndaAP
=
−+=
−++++=
==
��� ���� ���
Extra tiles needed
( ) ( )( )[ ]
[ ]( )mnm
mnm
mnm
+=
−++=
−++=
2
112
211222
square 1 2 3 … n
length
10
5
410
2
5
410
…
1
5
410
−
n
Area of nth square
222
1
5
4100
5
410
−−
=
=
nn
( )
2 2 2 2
4 4100 17 0.17
5 5
1 ln 0.172 2 ln 0.8 ln 0.17 2
2 ln 0.8
4.97
n n
n n
n
− −
< ⇒ <
− < ⇒ > +
>
5n =
2013 Revision Package BT1 Solutions
Chapter 4: Series, Sequences, Mathematical Induction
1 [DHS/Prelim/P1/2]
Prove by induction that1
1
(2 1)(2 3) 3(2 3)
n
r
n
r r n=
=+ + +
∑ . [5]
Hence state the value of 1
1
(2 1)(2 3)r r r
∞
= + +∑ . [1]
[Ans: 1/6]
2 [HCI/Prelim/P1/8]
(i) Express 4
( 1) ( 2)
r
r r r
−
− + in the form
1 2
A B C
r r r+ +
− +. [2]
(ii) Hence find 2
4
( 1) ( 2)
n
r
r
r r r=
−
− +∑ . [3]
Give a reason why the series is convergent, and state its limit. [2]
Use your answer to part (ii) to find 2
3
( 1)( 3)
n
r
r
r r r=
−
+ +∑ . [2]
[Ans: (i) 1, 2, 1A B C= = − = ; (ii)1 1 1 1
6 1 2n n n− + +
+ +;
1
6;
1 1 1 1
12 1 2 3n n n− − + +
+ + +]
3 [NJC/2010/Prelim/P2/2(b)]
(i) By expressing ( )
2 7 11
4 !
r r
r
+ +
+ in the form
( ) ( )2 ! 4 !
A B
r r+
+ + where A and B are
real constants, show that ( ) ( )
2
1
7 11 5 5
4 ! 4! 4 !
n
r
r r n
r n=
+ + += −
+ +∑ . [3]
(ii) Use the method of mathematical induction to prove your result in (i). [5]
(iii) Hence, find ( )
2
1
9 19
5 !r
r r
r=
∞ + +
+∑ . [3]
2013 Revision Package BT1 Solutions
[Ans: (i) ( ) ( )
1 1
2 ! 4 !r r−
+ +; (iii)
1
20]
4 The sequence of numbers x1, x2, x3, ... is such that x1 = −2 and xn + 1 =
5
xn − 1.
(i) Find the value of x5 x10 correct to 2 decimal places. [2]
(ii) As n → ∞, xn → θ . Find the exact value of θ . [3]
(ans : (i) 7.40 (ii) 1 1
212 2
− − )
5 (VJC Promo 2008) A sequence of real numbers x1 , x2 , x3 , ... satisfies the relation
xn + 1 = 4 + 15xn − 10
4xn − 4
for n ≥ 1. As n → ∞, xn → l.
(i) By solving x = 4 + 15x − 10
4x − 4 , find the value(s) of l. [3]
(ii) Prove that (xn + 1 − 4)2 − (l − 4)2 = 6 − xn
4xn − 4 . [2]
(iii) Hence show that if xn > l, then xn + 1 < l. [4]
(ans : (i) 6 )
6 (JJC/CT/09)
The rth term of a sequence is given by ur = ( )
1
1r r + for r = 1, 2, 3, … .
(i) Evaluate 1
n
r
r
u=
∑ for n = 1, 2 and 3. By observing the pattern of values, make a
conjecture for a formula for 1
n
r
r
u=
∑ in terms of n. [3]
(ii) Prove, using mathematical induction, your conjecture for 1
n
r
r
u=
∑ in (i), in terms of n.
2013 Revision Package BT1 Solutions
[5]
(iii) Using the formula, evaluate 2
1
1nr
r
u
n=
−∑ . [2]
[Ans: (iii) 1
( 1)n +]
7 [RVHS/2010/Prelim/P1/5]
The sequence of numbers u u u1 2 3, , ,… is given by 1 2u = and 1
2n
n
nu
nu+
+= for all positive
integers.
(i) By writing down the terms 2u and 3u , make a conjecture for nu in terms of n . [2]
(ii) Prove your conjecture by mathematical induction. [4]
(iii) Write down the limit of 1n nu u − as n tends to infinity. [1]
[Ans: (iii) 1]
8 [HCI/Prelim/2010]
The sequence of numbers nu , where n = 0, 1, 2, 3, …, is such that 0u = 2− and 1
1
( 2)
2 1n
n
n
n uu
u n
−
−
+=
+ +.
Prove by induction that, for 0,n ≥
2
2 1n
nu
n
+=
−. [5]
9 [NYJC/2010/Prelim/P2/2]
The sequence of numbers {ur} where r = 1, 2, 3, …, is such that it satisfies the recurrence relation
21
1r
r
ruu r
r
+ − =+
and 1u = 1.
(i) By dividing the above recurrence relation by r and using the method of difference, show that
2013 Revision Package BT1 Solutions
( )2 22
n
nu n n= − + for n = 1, 2, 3, …. [5]
(ii) Prove the result in (i) using mathematical induction. [4]
(iii) Find the exact value of 3
nu
n as n → ∞ . [1]
[Ans: (iii) ½]
10 [CJC/Prelim/P1/4]
The diagram below shows the graph of xxy −+= 2ln2 . The two roots of the equation
2ln2 =− xx are denoted by α and β , where βα < .
α βx
y
α βx
y
(i) Find the values of α and β , correct to 3 decimal places. [2]
A sequence of real numbers x1, x2, x3, … satisfies the recurrence relation
2)ln( 2
1 ++++====++++ nn xx for 1≥n .
(ii) Prove that if the sequence converges, it must converge to either α or β . [2]
(iii) By considering nn xx −+1 and the graph above, prove that
βα <<>+ nnn xxx if1 ,
βα ><<+ nnnn xxxx or if1 . [2]
2013 Revision Package BT1 Solutions
(iv) Hence deduce the value that xn converges to for 21 =x , giving your answer correct to 3 decimal
places. [2]
[Ans: (i) 357.5,464.0 == βα ; (iv) 357.5=→ βnx ]
11 [DHS/2010/Prelim/P1/3]
(i)Show that 2 24 5 ( )n n n a b− + = − + , where a and b are constants to be determined. [1]
(ii) Show that ( )2 2 2 2
3
1 4 5 1 ( 1) 1 5 2.N
n
n n n N N=
+ − − + = + + − + − −∑ [3]
(iii) Without the use of a graphic calculator, deduce that the sum in (ii) is strictly less than 2 1.N +
[2]
[Ans: (i) 2( 2) 1n − + ]
2013 Revision Package BT1 Solutions
Chapter 5, 7: Graphing Techniques & Transformations Solutions
1
(i)
[RJC/2010/Prelim/P1/Q7]
2 11
ay x
bx= + +
+
( )2
d2
d 1
y ab
x bx= −
+
When
d0,
d
y
x=
( )2
21
ab
bx=
+
( )2
12
1 1
2
abbx
abx
b
+ =
= − ±
Since a and b are positive constants, ab > 0 and thus there are 2 distinct real solutions for x.
Hence, C has exactly two stationary points. (shown)
(ii) Given that C passes through the point (0,3) ,
3 1
2
a
a
= +
=
x
y
2y =2
1x
b= −
'f ( )y x=
( )1
1b
b
−O ( )
11b
b
− +
||
__2 2b−
2013 Revision Package BT1 Solutions
2(i)
(ii)
[MJC/2010/Prelim/P1/Q8]
Let the curve be 3 2y ax bx cx d= + + + .
Since the points ( )2,75− , ( )0,3 and ( )1,12 lie on the curve.
Using ( )0,3 , 3d =
Using ( )2,75− ,
( ) ( ) ( )3 2
2 2 2 75a b c d− + − + − + =
8 4 2 72 (1)a b c− + − = ���
Using ( )1,12 ,
12a b c d+ + + =
9 (2)a b c+ + = ���
Since ( )1,12 is a maximum point, d
0d
y
x= .
x
y
( )f 1 2y x= −
1' , 3
2B
3' , 75
2A
( )' 0,12C
0
x
y
( )" 1, 3C
( )" 0, 0B
( )" 2, 72A −
( )"' 1, 3C −
α β
( )"' 2, 72A − −
0
2013 Revision Package BT1 Solutions
23 2 0ax bx c+ + =
3 2 0 (3)a b c+ + = ���
Using GC to solve (1),(2) and (3),
8, 7, 10a b c= − = =
Thus the equation of the curve is 3 28 7 10 3y x x x= − + + + .
3
(i)
[DHS/2010/Prelim/P1/10b]
(ii)
4
[TJC/2010/Prelim/P2/Q4]
(i)
y
x
1x = −
6y =
( )fy x′=
y
f '( )y x=
O x
x = 2
-2
O x
2
y
A’(-2, 0.5)
1
f( )y
x=
2013 Revision Package BT1 Solutions
(ii)
(b) Note:
2 2 5 23
1 1
x xy x
x x
− − += = − − +
− −
24y x
x= − − A→
2 24 4y x x
x x= − − − = − − +
−
B→ ( )
( )2 2
1 4 31 1
y x xx x
= − − − + = − − +− −
A: a reflection about y-axis
B: a translation of 1 unit in the positive direction of x-axis.
(ii)
2 2 5 23
1 1
x xy x
x x
− − += = − − +
− − ,
Asymptotes are: x = 1 and y = −x−3
Adding graph of ( )
( )2
2
2
14 1
2
xy
−+ + =
to part (ii), there are 4 roots for the
equation ( )
22 2
2
1 2 54 1
2 1
x x x
x
− − − ++ + =
− .
5(i) [RVHS/2010/Prelim/P1/Q10]
2d =
y
x
1x = −
1 2−
( )2 fy x=
0
2 2 5
1
x xy
x
− − +=
− ( )
( )2
2
2
14 1
2
xy
−+ + =
x =
y x= − −
2013 Revision Package BT1 Solutions
Method 1:
Using long division:
( )( )2 2 2
22
c b aax bx cy ax b a
x d x
− −+ += = + − +
+ +
1a∴ = , 2b = −
(ii) 2 2
2
x x cy
x
− +=
+
At ( )0, 4.5− : 4.52
c− =
9c∴ = −
(iii)
(0,-4.5)
( )2 2 24x y k− + = is a circle centre (4, 0), radius k
2 24 5 4
145 145 or
2 2
k .
k k
> +
⇒ < − >
y
x O 4
4−
4
(4, k)
2013 Revision Package BT1 Solutions
6
(a)
[HCI/2010/Prelim/P1/Q12]
2 2 2( 2) (1 )x a y− = −
22
2
( 2)1
xy
a
−⇒ + = reps an ellipse.
Method 1:
Sequence of transformations:
Scale // to x-axis by factor a.
Translate in the positive x-direction by 2 units.
Method 2:
Sequence of transformations:
Translate in the positive x-direction by 2
a units.
Scale // to x-axis by factor a.
(bi) 2
2
1
1 4
( )
4
( 1)
3
−
+ −
− +
− −
− − −
−
x
x x
x x
x
x
x
y
(bii) Sub
2 4
1
−=
+
xy
x into
2 2 2( 2) (1 )x a y− = − :
22
2 2 4( 2) 1
1
− − = − +
xx a
x
( ) ( ) ( )22 22 2 2 21 ( 2) 1 4⇒ + − = + − −x x a x a x --- (*)
(shown)
Hence the x-coordinate of the points of intersection of 1C and 2C satisfy equation (*).
(b)
(iii) From (ii), number of intersection points between 1C and 2C gives the number of real roots of the
1= −x
1= −y x
2
2 4
1
−=
+
xy
x
4− 2
2
2
( 2)1
xy
a
−+ =
2 − a 2 a+ 2−
2013 Revision Package BT1 Solutions
equation (*). From the graphs, there are 2 points of intersection between 1C and 2C . Hence 2
real roots.
7(i) [NYJC/2010/Prelim/P1/Q6]
y = f(x)
(ii)
(iii)
8a [AJC/2010/Prelim/P1/Q10]
x
y
2013 Revision Package BT1 Solutions
b ( )
2
2
1ln ln 3 2ln(3 )
6 9y x x
x x
− = = − = − −
− +
The graph can be obtained from lny x= by
1. translate of 3 unit in the negative x direction ln( 3)y x= +
2. reflect in the y-axis ln( 3)y x= − +
3. scale by factor 2, parallel to the y-axis
4. reflect in the x-axis
2013 Revision Package BT1 Solutions
Chapter 6: Functions Solution
1 (i)
2
2 2
2
2 2
(1 ) 2 ( 1)f ( )
(1 )
2 1
(1 )
x x xx
x
x x
x
+ − −′ =
+
− + +=
+
For stationary point, f ( ) 0 1 2,1 2x x′ = ⇒ = − +
Coordinates of stationary points 1 2
1 2,2 2
− − −
and
1 21 2,
2 2
+ − +
(ii)
Range of f is 1 2 1 2
,2 2 2 2
− − − +
(iii) 1
fg( )1
xx
x
−=
+
Domain of fg is { }: 0x x∈ ≥�
Range of g is [0, )∞
Thus, range of fg is 1 2
1,2 2
− − +
2 Let
2 1e , 0xy x
+= < .
2013 Revision Package BT1 Solutions
1ln
ln12
−−=
=+
yx
yx
Range of f = ( )e,∞
( )1f ln 1, ex x x−∴ = − − >
( )
( )
1g f fg( )
ln e 1 1x
x x
x
−=
= − − = − −
Alternative:
( )
( )
( )
( )
( )
2g 1
2
2
fg( ) e e , 1
g 1
g 1
g 1
since domain of f , 0
x xx x
x x
x x
x x
+ = = >
+ =
= −
= − −
= −∞
3 (i)
fR is ( )∞− ,1
(ii) gD is ( ) }2{\,1 ∞
since gf DR ⊄
as ( )∞− ,1 ⊄ ( ) }2{\,1 ∞
gf∴ does not exist
(iii) gf DR ⊂
f(x) > 1
122 >+ xx
2013 Revision Package BT1 Solutions
( )
12
212
−>
>+
x
x
12 −=∴k
222 ≠+ xx
0222 ≠−+ xx
Solving 31 ±−≠x i.e. 1 3a = − +
4 (i) Range of f is (0, 1]
(ii) gf(x) = g(cos x) =1
cos x, 0
2x≤ <
π
(iii) Area = 3 3
0 0
1sec
cos dx x dx
x
π π
=∫ ∫
= 30
ln sec tanx xπ
+
12
1ln 3 ln(1 0)
= + − +
ln( 3 2)= +
5 (i) An inverse function exists if and only if the function is a one-one function.
Let 2
1
)24( xy +=
)4(
2
124 22 −=⇒+= yxxy
1
x
y
y = cos x
2π
1
x
y y = sec x
3π 2
π
1
x
y y = sec x
2013 Revision Package BT1 Solutions
62),4(
2
1: 21 ≤≤−→∴ −
xxxf
(ii)
(iii)
...2!2
12
1
2
1
22
11
21
22
1
+
−
+
+=
+
xxx
= ...
32
1
41 2 +−+ x
x
Expansion is valid for
12
<x
2<⇒ x
x3)
4
x1( 2 x3)x(f −=+⇒−=
3
2x =⇒
6 ( )
( )2 2
2
2 2
2f 3
2 32 2 2
32 4
x x kx
k k kx x
k kx
= − +
= − + − +
= − + −
The minimum point is at
2
,32 4
k k −
.
x
f-1(x)
f(x) y = x y
2013 Revision Package BT1 Solutions
2
3 ,4
f
kR
= − ∞
From the graph of f , it is not one-one and hence 1f −
doesn’t exist. For example ( ) ( )f 0 f 3k= =
.
2
2
[ 1, )
3 14
4
4
4
fR
k
k
k
= − ∞
= −
=
= ±
−
When domain of f is restricted to the set of negative real numbers, 1f −
exist. The minimum point
at 0 0x k≥ ⇒ ≥ . Therefore 4k = .
2f : 4 3, x x x x−− + ∈ �
( )3,fR = ∞
Let ( ) 2f 4 3x x x y= − + =
( )
( )
2
2
2
4 3
2 1
2 1
2
2
1
1
2 1 since 0
x x y
x y
x y
y
x y
x y x
x
− + =
− − =
− = +
− = ± +
= ± +
= − + <
( )-1 1 32 ,f x xx += − >
y
x
( )fy x=
2
,32 4
k k −
3
2013 Revision Package BT1 Solutions
7
( )1
g 21
xx
= ++
The vertical asymptote is 1x = − .
The horizontal asymptote is 2y = .
When ( )g 0x = ,
12 0
1
11
2
3
2
x
x
x
+ =+
+ = −
= −
(ii)
( )
( )
1fg ln 2 1
1
1 4fg ln 3 ,
1 3
xx
x xx
= + +
+
= + < −
+
y
( )gy x=
x1−
2y =
3
2−
2013 Revision Package BT1 Solutions
( ), ln 3fgR = −∞
iii)
( )
( ) ( )
( )
( )
2
2
2
2
h 3 14
h 3 14
3 9h 14
2 4
65 3h
4 2
x x x
x x x
x x
x x
= − + +
= − − +
= − − − +
= − −
Maximum point is 3 65
,2 4
.
Greatest 3
2b = .
y
( )fy x=
x1−
( )1,2gR = −
ln 3
fgR
2013 Revision Package BT1 Solutions
65,
4hR
= −∞
Let ( )h x y=
2
2
65 3
4 2
3 65
2
3, since
4
3 65
2 4
3 65
2 4
3
2
65
2 4
x y
x y
x y
x y
yx x
− − =
− = −
− = ±
±
=
= −
− <
−
−
( )-1 3 65h
2 4
65,
4x y x− − <=
8 (i)
( )2 1 1
f 21 1
xx
x x
+= = −
+ +
The horizontal asymptote is 2y = .
( )f 0 1=
y
( )hy x=
x
3 65,
2 4
2013 Revision Package BT1 Solutions
( ) ( )1 ln 2g x x= − + +
( )g 0 1 ln 2= − +
When ( )g 0x = ,
( )
( )
1 ln 2 0
ln 2 1
2
2
x
x
x e
x e
− + + =
+ =
+ =
= −
ii) ( )1, 2fR =
Let ( )2 1
f1
xx y
x
+= =
+
y
x
( )fy x=
2y =
1 ln 2− +
2e −
y
x
( )fy x=
2y =
1
2013 Revision Package BT1 Solutions
( )
2 1
1
2 1 1
2 1
1
2
xy
x
x y x
x xy y
yx
y
+=
+
+ = +
− = −
−=
−
( )-1 1f , 1 2
2
xx x
x
−= < <
−
iii)
( ) ( )
( )( ) ( )
( )( ) ( )
( )
1
ln 2
ln
1
1 2
hg 2
h 1 ln 2
h 1 ln 2
h
x
x
x
x e x
x e e
x e
x e
+
− + +
−
−
= +
− + + =
− + + =
=
Alternative
Let ( ) ( )g 1 ln 2x x y= − + + =
( )
( )1
1
1 ln 2
ln 2 1
2
2
y
y
x y
x y
x e
x e
+
+
− + + =
+ = +
+ =
= −
( )-1 1g 2, 1 ln 2xx e x
+= − > − +
( )( ) ( )
( ) ( )( )
( )
1
1 1
1 1
hg h
h 2 2
h
h
x
x
x
g x x
x e e
x e e
x e
−
− +
− +
=
= −
×
+
=
=
iv) ( ) ( )1 ln 2, 0,g fR D= − + ∞ ⊄ = ∞ therefore fg doesn’t exist.
2013 Revision Package BT1 Solutions
Chapter 8: Inequalities Answers
1 22 3x x+ <
23 2 3x x− < + <
23 2x x− < + and 22 3x x+ <
22 3 0x x+ + > and 22 3 0x x+ − <
21 23
04 16
x
+ + > and ( )( )2 3 1 0x x+ − <
x∈� and 3
12
x− < <
31
2x∴− < <
2
1 1 1
2 2 22 3 2 3x x x
xe e e e
+ < ⇒ + <
∴replace x with
1
2x
e
1 1
2 23
1 0 12
x x
e e− < < ⇒ < <
1ln1
2x <
0x∴ <
2
From graph, to solve for values of x at the points of intersection:
x
y
5 -1 0
-5
y = 5
542 −−= xxy
( )542 −−−= xxy
2013 Revision Package BT1 Solutions
5542 =−− xx or ( ) 5542 =−−− xx
01042 =−− xx or 042 =− xx
142 ±=x or
0=x or 4=x
142 −≤∴ x or 142 +≥x or 40 ≤≤ x
Otherwise method :
.|1||5|
5+≤
−x
x
( )( )515 −+≤ xx
( ) ( ) 222551 ≥−+ xx
( )( )[ ] ( )( )[ ] 0551551 ≥−−++−+ xxxx
Consider
( )( ) 0551 =+−+ xx or ( )( ) 0551 =−−+ xx
142 ±=x or
0=x or 4=x
142 −≤∴ x or 142 +≥∴ x or 40 ≤≤ x
3
2
71
4 3
x
x x
+≤
+ −
2
71 0
4 3
x
x x
+− ≤
+ −
+ + + - -
142 − 142 + 0 4
x
2013 Revision Package BT1 Solutions
( )2
2
7 4 30
4 3
x x x
x x
+ − + −≤
+ −
( )
2
2
7 4 30
3 4
x x x
x x
+ − − +≤
− − −
2
2
2 30
3 4
x x
x x
− +≥
− −
Since 2 22 3 ( 1) 2 0 for ,x x x x− + = − + > ∈�
2 3 4 0x x− − >
( ) ( )4 1 0x x− + >
1 or 4x x< − >
4
2
22 5 2exx
x+ − ≤
2
25 2e 2x
xx
⇒ − ≤ −
2
5 1e
2
xx
x⇒ − ≤ −
From the diagram,
or 0 or 0x x xα β≤ ≤ < >
Using G.C.,
1− 4
− + +
α β x
y
O
exy x= −
2
5 1
2y
x= −
2013 Revision Package BT1 Solutions
2.18 or 0.920 0 or 0x x x≤ − − ≤ < > (3s.f.)
Therefore solution set is ( , 2.18] [ 0.920,0) (0, )−∞ − ∪ − ∪ ∞ .
5 2 21 34 2 1 4( )
4 4x x x+ + = + +
Since x is real, 21
( ) 04
x + ≥
Thus 24 2 1x x+ + is always positive.
2
2
12 0
1 2
4 2 10
1 2
Since 4 2 1 0,
1 2 0
1
2
xx
x x
x
x x
x
x
+ >+
+ +>
+
+ + >
+ >
> −
2 2 11 1 1
22 1
x x
x x x
x
++ > ⇒ + + >
+ +
i.e. 1 1
2 01
1 2x
x
+ > +
Replace x by 1
x, we have
12
x> −
2 or 0x x∴ < − >
6
2
12 294
5
x
x
+≤
−⇒
2
2
(12 29) 4(5 )0
5
x x
x
+ − −≤
−
⇒ 2( 5 )( 5 )(2 3) 0x x x− + + ≤
⇒ x= -3/2, 5x ≤ − or 5x ≥ x= -3/2, 5x < − or 5x > ( 5x ≠ ± )
2013 Revision Package BT1 Solutions
2
(12 29 )4
5 1
x x
x
+≤
−
Replace x by 1/x, ⇒ x= -2/3, 1
5x
< − or 1
5x
>
⇒ x= -2/3, 1
05
x− < < or 1
05
x< <
⇒ x= -2/3, 1 1
5 5x− < < as x = 0 is a solution to the inequality
7 ln 2 0
ax
x
− ≥
where 1a > .
2
2 1
20
1 1 8 ) 1 1 8 )2
4 40
1 1 8 1 1 80
4 4
ax
x
x x a
x
a ax x
x
a ax or x
− ≥
− −≥
+ + − +− −
≥
− + + +≤ < ≥
2013 Revision Package BT1 Solutions
Chapter 9: System of Linear Equations
1. [CJC/2010/Prelim/P1/Q1]
Let x be the price of high heels, y be the price of facial mask and z be the price of
handbag in dollars.
We have,
70.18375153
30.1158872
20.12983105
=++
=++
=++
zyx
zyx
zyx
Solving, x = 29.9, y = 99.9, z = 49.9
Total cost of gift 20.629$90.49290.99590.29 =×+×+=
2. [NJC/2010/Prelim/P1/Q1]
Let the unit digit be z.
Let the tenth digit be y.
Let the hundredth digit be x.
15 (1)x y z+ + = −
(100 10 ) (100 10 ) 594
99 99 594 (2)
x y z z y x
x z
+ + − + + =
⇒ − = −
4 5
4 5 (3)
y z x
x y z
+ = +
⇒ − + + = −
Using GC to solve the equations simultaneously,
8, 5, 2x y z= = = .
Thus the number is 852.
3. [IJC/2010/Prelim/P2/Q1]
2013 Revision Package BT1 Solutions
Let V 2ms− , M 2ms− , S 2ms− be the gravitational pull on each planet Venus, Mars
and Saturn respectively.
630S + 630V + 630M = 13860 ---------- (1)
900S − 600V = 2880 ---------- (2)
600S + 630V + 900M = 3(4870) = 14610 ---------- (3)
630 630 630 13860
900 600 0 2880
600 630 900 14610
S
V
M
− =
From GC,
M = 3.8, V = 9, S = 9.2
Hence the weight of Probe D on Saturn is ( )500 9.2 N× = 4600 N
4. [PJC/2010Prelim/P1/Q1]
Let the number of diagonals be 2d An Bn C= + +
No of diagonals in a triangle = 0
No of diagonals in a quadrilateral = 2
No of diagonals in a pentagon = 5
Therefore,
9 3 0
16 4 2
25 5 5
A B C
A B C
A B C
+ + =
+ + =
+ + =
Solving using GC, 1 3
, , 02 2
A B C= = − =
Triangle
(3 sides)
Quadrilateral
(4 sides)
Pentagon
(5 sides)
2013 Revision Package BT1 Solutions
Thus, 21 3
2 2d n n= −
For a polygon of 200 sides,
The number of diagonals = 21 3(200) (200) 19700
2 2− =
5. [RI/2010/Prelim/P1/Q3]
Let x, y, z be the exchange rate quoted for Sterling Pound, Euro Dollar and Swiss
Franc, respectively (i.e. 1 Sterling Pound = x Singapore Dollars, 1 Euro Dollar = y
Singapore Dollars and 1 Swiss Franc = z Singapore Dollars).
36 77 42 269.9x y z+ + =
55 18 63 233.45x y z+ + =
40 31 26 175.5x y z+ + =
Using the GC, x = 2.15, y = 1.78, z = 1.32
59 24 313kx y z+ + =
313 59(1.78) 24(1.32)82
2.15k
− −∴ = =
6. [RVHS/2010/Prelim/P1/Q1]
Let the no. of small, medium and large bottles manufactured be denoted by s, m and l
respectively.
So, 150 335 475 280400s m l+ + = ,
200 450 600 370700s m l+ + = and
3.8 2(2.5 4.2 )m s l= +
Using GC, solving the augmented matrix:
150 335 475 280400
200 450 600 370700
5 3.8 8.4 0
−
2013 Revision Package BT1 Solutions
Ans: s = 166, m = 550 and l = 150
Assumption:
The plastic bottles are of negligible thickness. OR
The plastic bottles are of the same thickness.
7. [SAJC/2010/Prelim/P1/Q1]
Let 3 2
nu an bn cn d= + + +
1 63u = : ( ) ( )
231 1 (1) 63
63 (1)
a b c d
a b c d
+ + + =
+ + + = −
2 116u = : ( ) ( )
232 2 (2) 116
8 4 2 116 (2)
a b c d
a b c d
+ + + =
+ + + = −
3 171u = : ( ) ( )
233 3 (3) 171
27 9 3 171 (3)
a b c d
a b c d
+ + + =
+ + + = −
4 234u = : ( ) ( )
234 4 (4) 234
64 16 4 234 (4)
a b c d
a b c d
+ + + =
+ + + = −
Using the GC APPL to solve (1), (2), (3), (4) simultaneously, we get:
1, 5, 61, 6a b c d= = − = =
3 25 61 6n
u n n n= − + +
Hence 50u = ( ) ( ) ( )3 2
50 5 50 61 50 6 115556− + + =
8. [TJC/2010/Prelim/P1/Q6]
(a) ( )( )
( )( )( )
2
2
f 5 84 61 1 0 0
g 2 2 1
x xx x
x x x x x
++ +≤ ⇒ − ≤ ⇒ ≤
− − − +
2013 Revision Package BT1 Solutions
( )( )( ) [ ]5 8 2 1 0 1,2x x x x⇒ + − + ≤ ≠ −
8 or 1 2
5x x⇒ ≤ − − < <
( )( )
f 81 or 1 2
5g
x
x x
x
ee e
e
−
− −
−≤ ⇒ ≤ − − < <
0 2xe−⇒ < <
1
ln2
x⇒ >
(b) ( ) ( )h fy x x= +
( ) ( )3 2 2 3 24 6 1 4 6y ax bx c x x ax b x x c⇒ = + + + + + = + + + + +
Since it passes through the points ( ) ( )1, 18 , 1, 14− − − and ( )3,30
( ) ( )( ) ( ) ( )3 2
1 1 1 4 1 6 18a b c− + + − + − + + = −
21 (1)a b c⇒ − + + = − − − −
( ) ( )( ) ( ) ( )3 2
1 1 1 4 1 6 14a b c+ + + + + = −
25 (2)a b c⇒ + + = − − − −
( ) ( )( ) ( ) ( )3 2
3 1 3 4 3 6 30a b c+ + + + + =
27 9 3 (3)a b c⇒ + + = − − − [M2 – -1 for any one wrong equation]
Solving (1),(2) and (3) gives 2, 10, 33a b c= − = = −
( ) ( )3 2h 2 10 33 h 101 1958625x x x= − + − ⇒ = −
2013 Revision Package BT1 Solutions
Chapter 12: Applications of Differentiation Solution
1 [TPJC Promo 08]
2013 Revision Package BT1 Solutions
2 [NJC promo 08]
2V x y=
Surface Area: 24 48xy x+ =
248
4
xy
x
−∴ =
21
(48 )4
V x x⇒∴ = −
Using GC, max volume occurs when 4x = and hence 2y = .
Dimension of box is 4 cm x 4 cm x 2 cm.
3 [JJC promos 08]
2
2102
xQR
= − ⇒
2
22( 10 )2
xp x
= + −
x
x
4
V
x
.
y
2013 Revision Package BT1 Solutions
22 400x x= + −
2
2102
xA x
= −
24002
xx= − [shown]
( )1
2 221 1
400 ( 2 ) 4002 2 2
dA xx x x
dx
− = − − + −
22
2
1400
22 400
xx
x= − + −
−
For stationary value of A, 0dA
dx=
22
2
1 400
22 400
xx
x⇒ = −
−
2 2 400x x⇒ = −
10 2 ( 0)x x⇒ = >∵
2(10 2) 400 200 30 2p∴ = + − = (cm)
4 [RJC promos 08]
By Pythagoras’ Theorem,
2 2 2 2
2 2 2 2
f ( )
( ) ( ) ( ) ( )
3 (5 ) 4 (Shown)
x AP PB
OA OP NP NB
x x
= +
= + + +
= + + − +
.
2 2 2 2 2 2 2 2
2 2
2 2
df( ) 0.5(2 ) 0.5(2)(5 ) 5
d 3 (5 ) 4 3 (5 ) 4
(5 ) 16 ( 5) 9
9 (5 ) 16
x x x x x
x x x x x
x x x x
x x
− − −= + = +
+ − + + − +
− + + − +=
+ − +
Let df( )
0d
x
x= , and hence we are solving
2013 Revision Package BT1 Solutions
2 2
2 2 2 2
(5 ) 16 ( 5) 9
((5 ) 16)) (5 ) ( 9)
x x x x
x x x x
− + = − − +
⇒ − + = − +
On simplification, we have
2 2
2
16 9(5 )
7 90 225 0
(7 15)( 15) 0
x x
x x
x x
= −
+ − =
− + =
When df ( )
0,d
x
x= x = 15/7 (since x > 0) satisfies the quadratic equation given.
Using the 1st
derivative test,
x 15
7
−
15
7
15
7
+
( )df
d
x
x −ve 0 +ve
Or alternatively using the second derivative test,
22
2 3 32 2 2 22 2 2 22 2
2 2 2 2 2
3 3
2 2 2 22 2
3 3
2 2 2 22 2
1 1( )(2 ) 2( 5)
d f( ) 1 12 2
d 3 (5 ) 4( 3 ) ((5 ) 4 )
9 ((5 ) 4 ) ( 5)
( 3 ) ((5 ) 4 )
9 16
( 3 ) ((5 ) 4 )
x x xx
x x xx x
x x x x
x x
x x
− −−
= + + ++ − +
+ − +
+ − − + − −= +
+ − +
= +
+ − +
which is greater than zero when x = 15/7.
Hence f(x) is minimum when x = 15/7, and on substituting x gives f(x) = 74 or 8.60 (to 3.s.f)
2013 Revision Package BT1 Solutions
5 [SAJC 2006 Prelims/P1/Q11]
2 1
2
x t
dx
dt
= −
=
( ) ( )( )
22
22
2
1
1
21 1 2
1
yt
dy tt t
dt t
−
=+
= − + = −+
( ) ( )2 2
2 2
2 1
1 12
dt tdy dy
dx d
t
d t tt x× = − × = −=
+ +
Equation of tangent at 2
12 1,
1t
t
−
+
( )( )
( ) ( )
( ) ( )
222
22 2 2
22 2
12 1
1 1
1 1 2
1 3 +1 shown
ty x t
t t
t y t xt t t
t y xt t t
− = − − − + +
+ − + = − + −
+ + = −
(ii) When 3t = , equation of tangent
( ) ( )2 223 1 3 3 3 3 1
100 3 25
y x
y x
+ + = − +
+ =
When the tangent meets the curve at 2
12 1,
1q
q
−
+
( )
( )( ) ( )
( )
2
2 2
3 2 2
3
3
2 2
2
100 3 25
1100 3 2 1 25
1
100 3 2 1 1 25 1
100 3 2 2 1 25 25
100 6 3 6 3 25 25
6 28 6 72 0
4(From GC) 3,
3
y x
q q q
q q q q
q q q q
q q q
x
+ =
+ − =
+
+ − + = +
+ + − − = +
+ − + − = +
− + + =
= −
Therefore the point of intersection Q is
2013 Revision Package BT1 Solutions
2
4 1 8 92 1, ,
3 3 2541
3
− − = −
− +
6 [CJC 2006 Prelims/P1/Q11]
i)
5 se
5 sec
c tandx
ad
x a θ
θ θθ
=
=
2
3 tan
3 secdy
ad
y a θ
θθ
=
=
23 sec 3 3
sin5 sec tan 5sin5cos
cos
d d a
dx a
y dy
dx d
θ θ
θθ θ θ θθθ
× = = ==
ii) When the normal’s gradient = 1−
1
1
3
5sin1
3sin
5
dy
dx
θ
θ
=
−
=
=−
4cos
5
5sec
4
3tan
4
θ
θ
θ
=
=
=
The point is 25 9
5 ,4
5 3,3
4 4 4
aa
aa
=
×
×
When the tangent is parallel to the y -axis,
0
sin 05 θ
θ
=
=
The point is ( ) ( )sec0,3 tan 05 5 ,0aa a=
θ3
5
2 25 43− =
2013 Revision Package BT1 Solutions
7 [AJC 2007 Prelims P1Q13b]
(i)
By similar triangles
2 2
6
and 63
r
h d
hr d
= =
= =
( ) ( ) ( )
( ) ( ) ( )
( )
2 2
22
3
1 1
3 3
1 16 6
3 3
8
2 1
2 23
6 (shown)27
r
h
V h d
h
h
π
π π
ππ
π +
+
+ −
= −+
= −
=
(ii)
( )
( )
3
2
6 8
9
27
6
h
hh
V
dV
d
ππ
π=
−+
+
=
( ) ( )2 2
9 18020
6 6
dV
d
dh dh
dt d t hV hπ π× = × =
+ +=
When 3h = , ( )3
1
2
180 20 cms
93 6h
dh
dt ππ=
−=+
=
6 cm
2 cm
cmr
cmh
2 cm
2 cm
d cm
2013 Revision Package BT1 Solutions
8 [HCI 2007 Prelims P1Q2]
i)
2
3
12
4
xV
dV
dx
x
π
π
=
=
2 2
4 123
dV
dx
dx d
x
x
t dV xd π π= × = × =
When 3x = , ( )
-1
2
12 4m s
33
dx
dt ππ==
ii)
2
2
3
3
12
123
36
12dx
dt
dx dt
C t
t
x
x
x
xC
π
π
π
π=
=
=
+ =
+
∫ ∫
When 5x = , 36
125t C
π= +
When 10x = , 1000
36t C
π= +
Time taken 1000 125
s36 36 36
875C C
π π π = + − + =
2013 Revision Package BT1 Solutions
Chapter 13: Maclaurin’s Series Solutions
1
(i)
(ii)
(iii)
[ACJC/2008/Promo]
Solution
3 2
3 22 0
d y dy d y
dx dx dx+ =
24 2 3
4 2 32 2 0
d y d y dy d y
dx dx dx dx+ + =
When 0x = , 0y = 0dy
dx=
2
21
d y
dx= −
3
30
d y
dx=
4
42
d y
dx= −
Using Maclaurin’s series,
( ) ( )2 42 41 2 1 1
... ...2! 4! 2 12
x xy x x
− −= + + = − − +
( )
2 4
2 4 2 4
1 1ln cos ...
4 2 4 12 4
1 ln 2 ln 2
2 32 3072 16 1536
π π π
π π π π
= − − +
− ≈ − − ∴ ≈ +
( ) 2 41 1ln cos ... 2
2 12y x x x= = − − +
Differentiate y wrt x,
3
3
1tan ...
3
8 tan 2 2 ...
3
x x x
x x x
− = − − +
∴ = + +
2
(i)
(ii)
[MI/2008/Promo]
Solution
21y x x= + +
2
21
2 1
dy x
dx x= +
+ =
2
2
1
1
x x
x
+ +
+
2 2
1 1dy
x x xdx
+ = + +
2
1dy
x ydx
+ = (shown).
( )dx
dy x x
dx
dy
dx
y d x =⋅+ ⋅+ +
−)2(1
2
11 2
12
2
22
2013 Revision Package BT1 Solutions
2
2 2
2(1 ) 1
d y dy dyx x x
dx dx dx+ + = +
ydx
dyx
dx
ydx =++
2
22 )1( (shown).
( )3 2 2
2
3 2 21 (2 ) .1
d y d y d y dy dyx x x
dx dx dx dx dx+ + + + =
( )3 2
2
3 21 3 . 0
d y d yx x
dx dx+ ⋅ + =
( )4 3 3 2
2
4 3 3 21 (2 ) 3 .3
d y d y d y d y dyx x x
dx dx dx dx dx+ + + ⋅ + =
( ) 03.512
2
3
3
4
42 =++⋅+
dx
yd
dx
ydx
dx
ydx
When ,0=x ,1=y ,1=dx
dy ,1
2
2
=dx
yd
3
30,
d y
dx=
4
43
d y
dx= − .
Maclaurin’s Series: ....8
1
2
11 42 +−++= xxxy
3
1=x is suitable as the terms 0
3
1→
n
for all n ≥ 4.
3 [RJC/2008/Promo]
Solution
Let1e x
y− +=
ln 1y x= − + --- (1)
Differentiate (1) with respect to x:
1 d 1 d
2 1d d2 1
y yx y
y x xx
−= ⇒ + = −
+ --- (2)
Differentiate (2) with respect to x:
( )
( )
( )
2
2
2
2
2
2
2
2
d 2 d d2 1
d d d2 1
d d d2 1 1
d d d
d d d4 1 2 2 1
d d d
d d4 1 2
d d
y y yx
x x xx
y y yx x
x x x
y y yx x
x x x
y yx y
x x
+ + = −+
+ + = − +
+ + = − +
+ + =
2013 Revision Package BT1 Solutions
( )2
2
d d4 1 2 0
d d
y yx y
x x+ + − = (shown) --- (3)
Differentiate (3) with respect to x:
( )3 2 2
3 2 2
d d d d4 1 4 2 0
d d d d
y y y yx
x x x x+ + + − =
( )3 2
3 2
d d d4 1 6 0
d d d
y y yx
x x x+ + − =
At 0x = ,
2
1 1 1
2
d 1 d 1e , e , e
d 2 d 2
y yy
x x
− − −= = − = and
3
1
3
d 7e
d 8
y
x
−= −
By Maclaurin’s series,
1 1
1 1 2 3
1 7e e
1 2 8e e
2 2! 3!y x x x
− −
− −
−
= + − + + +
…
1 2 31 1 7
e 12 4 48
x x x−
≈ − + −
( ) ( )
2 1 2 1
2 3
1 2 3
2 3
.
2 2 1 1 7 = 1 2 ... 1 ...
2! 3! 2 4 48
1 = (16 24 20 11 ...)
16
x x x xe e e
x xe x x x x
x x xe
− + − +
−
=
+ + + + − + − +
+ + + +
4
(i)
[TJC/2008/Promo]
Solution
4 sin 2y x= +
2
4 sin 2y x⇒ = +
cos 2dy
y xdx
⇒ =
22
22sin 2
d y dyy x
dx dx⇒ + = −
2 3 2
2 3 2 2 4 cos 2
dy d y d y dy d yy x
dx dx dx dx dx⇒ + + = −
3 2
3 23 4 cos 2
d y dy d yy x
dx dx dx⇒ + = −
When x = 0, y = 2, 1
2
dy
dx= ,
2
2
1
8
d y
dx= − ,
3
3
61
32
d y
dx= −
Maclaurin’s series of y:
2013 Revision Package BT1 Solutions
(ii)
2 3
2
3
1 1 612 ......
2 8 2! 32 3!
612
2 16 192
x xy x
x xx
= + + − + − +
≈ + − −
When x = 0.5, by part (i) result, y ≈ 2.19466 (to 5 d.p.)
This gives an estimated error = 2.20033 2.19466
100% 0.258%2.20033
−× =
5 [ACJC/2008/Prelim/P1/Q9]
Solution
1siny x
−= ⇒
2
1
1
dy
dx x
=−
.
⇒ 2
1 1dy
xdx
− =
( )
( ) ( )
( )
22
2 2
22
2
3 2 22
3 2 2
3 22
3 2
diff. w.r.t ,
21 0
2 1
1 0
diff. w.r.t ,
1 2 0
1 3 0 ( )
x
d y dy xx
dxdx x
d y dyx x
dxdx
x
d y d y d y dyx x x
dxdx dx dx
d y d y dyx x Shown
dxdx dx
−− + =
−
− − =
− + − − + =
− − − =
( )
( )
4 3 22
4 3 2
5 4 32
5 4 3
diff. w.r.t ,
1 5 4 0
diff. w.r.t ,
1 7 9 0
x
d y d y d yx x
dx dx dx
x
d y d y d yx x
dx dx dx
− − − =
− − − =
(0) 0, '(0) 1, ''(0) 0, '''(0) 1, ''''(0) 0, '''''(0) 9f f f f f f= = = = = =
3 53
...6 40
x xy x= + + +
2013 Revision Package BT1 Solutions
1
2
3 5
2 4
1sin
1
3 ...
6 40
3 1 ...
2 8
dx
dxx
d x xx
dx
x x
−=
−
= + + +
= + + +
Using 1
2x = ,
2 4
2
1 13
1 2 21 ...
2 811
2
= + + +
−
2 4
2
1 13
1 2 21 ...
2 811
2
= + + +
−
≈ 147
128
⇒ 256
3147
≈
6
(a)
(b)
(c)
[CJC/2008/Prelim/P2/Q1]
Solution
1 sin 1x x+ ≈ +
Since x is small.
By standard Maclaurin’s Series expansion,
( )1
22
1 11
1 2 21 1 1
2 2!x x x x
−
+ = + ≈ + +
21 11
2 8x x≈ + − (Shown)
Given 3x − e
x ln a = 0
⇒ 3x = e
x ln a
⇒ 3x = a
x
⇒ 3 = a
Given that ex ≈ 1 + x +
x2
2 +
x3
3!
3x = e
x ln a
⇒ 3x ≈ 1 + xln3 + (xln3)
2
2 +
(xln3)3
3!
( ) (0) 1y f x f= ⇒ =
2013 Revision Package BT1 Solutions
2
2
1 sin 12 1 sin '(0)
2 2
dy xy x f
dx y
+= + ⇒ = =
By implicit differentiation, 2 2
22
2
2 2
1cos 4 1 4
22 4 cos "(0) 0
2 4
dyx y
d y dy dxy y x f
dx dx y
− −
+ = ⇒ = = =
By implicit differentiation, 3 2
33 2 2
3 2 2
2
sin 4 121
12 4 sin '"(0)2 4
2
dy dy d yx y
d y dy d y dy dx dx dxy x f
dx dx dx dx yy
− − − ⋅
+ ⋅ + = − ⇒ = = −
By Maclaurin’s Expansion,
2 3
3
"(0) '"(0)(0) '(0) .....
2! 3!
1 11
2 24
f fy f xf x x
x x
= + + + +
≈ + −
7 [JJC/2008/Prelim/P1/Q11]
Solution
1 sin x
y e−
=
1
sin
2
d 1.
d 1
xye
x x
−
=−
2 d
1-d
yx y
x=
2
2 2d(1 )
d
yx y
x− =
2 2
2
2
d d d d2 (1 ) 2 2
d d d d
y y y yx x y
x x x x− + − =
2
2
2
d d (1 )
d d
y yx x y
x x∴ − − =
Diff wrt x,
2 3 2
2
2 3 2
d d d d d2 (1 )
d d d d d
y y y y yx x x
x x x x x− + − − − =
2 3
2
2 3
d d d3 (1 ) 2
d d d
y y yx x
x x x− + − =
When x = 0,
y =1, d
d
y
x
= 1,
2
2
d
d
y
x
=1 ,
3
3
d
d
y
x
=2
12 3
sin.1 ..
2 3
x x xe x
−
= + + + +
2013 Revision Package BT1 Solutions
1sin 2 3 21(1 ...)(1 )
cos 2 3 2
xe x x x
xx
−
−= + + + + −
2 3 2
(1 ...)(1 ...)2 3 2
x x xx= + + + + + +
2 351 ...
6x x x= + + + +
2013 Revision Package BT1 Solutions
Chapter 14: Integration Techniques Solutions
1 [AJC 2007 Common Test]
2
22
2 1
2 1
1
1 4
1 8 1 1
8 2 11 4
4
1 1.2 1 4 sin (2 )
8 2
1 11 4 sin (2 )
4 2
xdx
x
xdx dx
xx
x x c
x x c
−
−
−
−
−= − −
−−
= − − − +
= − − − +
∫
∫ ∫
2 [AJC 2007 Common Test]
2
2
2
1
2 4
02
24
2
cos 2sin cos
10, , ,
2 2 4
cos cos
1 sin sin
cos2sin cos
1 sin
2cos
x dx d
x x
x
x
xdx d
x
d
π
π
π
π
θ θ θ θ
π πθ θ
θ θ
θ θ
θθ θ θ
θ
θ θ
= ⇒ = −
= = = =
= =−
= −−
= −
∫ ∫
∫
2
4
2
4
(1 cos 2 )
sin 2
2
1 1
2 4 2 4 2
d
π
π
π
π
θ θ
θθ
π π π
= +
= +
= − + = −
∫
3 (AJ 06Prelim P2/Qn 3)
(i) ( ) ( )1
2 2 221
1 1 2 12
dx x x x x x
dx
− − = − − + −
2013 Revision Package BT1 Solutions
( ) ( )
( )
12 22
22
2
2 2
2
2
2
11 2 1
2
11
1
1
1 2
1
x x x x
xx
x
x x
x
x
x
− = − − + −
−= + −
−
− + −=
−
−=
−
(ii) ( )1 1 2 2
2
14 sin sin 2 2
1x xdx x x x dx
x
− −⌠⌡
= − ×−∫
( )
( )
21 2
2 2
1 2 2 1
1 2 1sin 2
1 1
sin 2 1 sin
xx x dx dx
x x
x x x x x C
−
− −
⌠ ⌠
⌡⌡
−= + −
− −
= + − − +
4 (VJ 06FE Qn 6)
( ) ( )sin 3cos 3sin 2cos 3cos 2sinx x A x x B x x+ = + + −
sin 3cos 3 sin 2 cos 3 cos 2 sinx x A x A x B x B x+ = + + −
Comparing the coefficients for xsin : 123 =− BA (1)
Comparing the coefficients for xcos : 232 =+ BA (2)
Solving 13
9=A ,
13
7=B
( ) ( )
( ) ( )
( )⌡⌠
+
−+=
⌡⌠
+
−+
⌡⌠
+
+=
⌡
⌠
+
−++=
⌡⌠
+
+
dxxx
xxx
dxxx
xxdx
xx
xx
dxxx
xxxx
dxxx
xx
cos2sin3
sin2cos3
13
7
13
9
cos2sin3
sin2cos3
13
7
cos2sin3
cos2sin3
13
9
cos2sin3
sin2cos313
7cos2sin3
13
9
cos2sin2
cos3sin
2013 Revision Package BT1 Solutions
Let 3sin 2cosu x x= +
3cos 2sindu
x xdx
= −
( )
Cxxx
Cux
duxxu
xxx
dxxx
xxx
+−+=
++=
−
−+=
⌡⌠
+
−+
∫
sin2cos3ln13
7
13
9
ln13
7
13
9
sin2cos3
1sin2cos3
13
7
13
9
cos2sin3
sin2cos3
13
7
13
9
5 (RJ 06Prelim P1/Qn 14)
(a)
( )
BAAxx
BxAx
++−=
++−=
62
62
Comparing coefficients
3,2
1=−= BA
( )
( )
( )( )
( )
( ) Cxxx
dxx
xx
dxxx
xx
dxxx
dxxx
x
dxxx
x
dxxx
x
+−+−−−=
⌡
⌠
−−+−−−=
⌡
⌠
+−+−−+
−−
−=
⌡
⌠
−−+
⌡
⌠
−−
+−−=
⌡
⌠
−−
++−−=
⌡
⌠
−−
3sin386
31
1386
89932
13
2
1
86
2
1
86
13
86
62
2
1
86
3622
1
86
2
2
2
2
2
22
22
6 (NJ 06FE Qn 8)
(a)
2013 Revision Package BT1 Solutions
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )( )
( ) ( ) ( )( )xxx
dxx
xxxdxx
dxxxxxxdxx
dxx
xxxxxxdxx
dxxxxdxx
dxx
xxxxdxx
lnsinlncos2
lncos
lnsinlncoslncos 2
lncos lnsinlncoslncos
1lncoslnsinlncoslncos
lnsinlncoslncos
1lnsinlncoslncos
+=
+=
−+=
⌡
⌠
−+=
+=
⌡
⌠
−−=
∫
∫∫∫
∫
∫∫
∫
( ) ( ) ( )( )
( )
( )2
113
2
1
0sin0cos2
1
6sin
6cos
2
1
lnsinlncos2
lncos
6
6
11
66
−+=
+−
+=
+=∫
π
πππ
ππ
e
e
xxx
dxx
ee
(b)
( ) ( ) ( )
( ) ( )( )2223
2222
23
1432811123
2
43
121
811123
++++=+++
+
++
+=
++
+++
xxxAxxx
x
x
x
A
xx
xxx
Comparing coefficient of 2
x , 2=A
( ) ( )
3 2
2 2
3 12 11 8
1 2
x x xdx
x x
⌠⌡
+ + +
+ +
2013 Revision Package BT1 Solutions
( )
( )
( )( )
( )
2 2
2 2 2
1
2 1
2 1
2 3 4
21
1 3 2 12 4
2 2 21
1 3 12 ln 2 4 tan
1 2 2 2
2 3 4ln 2 tan
1 2 2 2
xdx
xx
xdx dx dx
x xx
x xx C
xx C
x
−
−
−
⌠⌡
⌠ ⌠ ⌠
⌡ ⌡⌡
+= +
++
= + ++ ++
+ = + + + + −
= − + + + ++
7 (SAJC 07 Prelim P1/Qn 14a)
dxaxa
∫ −
3
0
2)(9
= θθ
π
da ∫
2
0
2cos33
= θθ
π
da ∫
+2
02
12cos9
=2
0
2sin2
1
2
9π
θθ
+
a
=a4
9π
So a = 2.
Let θsin3=ax .
θθ
cos3
ad
dx=
2013 Revision Package BT1 Solutions
8 1
222 3
23 2
1
23
2
1
1
3
2
1 1
6 611 1
1
61
sin6
kk
k
k
ydx dy
yx x
y
dyy
y
π π
π
π−
= ⇒ − =
− −
⇒ − =−
⇒ − =
∫ ∫
∫
1
1
1 sin
3 6
1 sin
6
k
k
π π
π
−
−
⇒ − − =
⇒ =
∴k = 2
2013 Revision Package BT1 Solutions
Chapter 15: Areas and Volumes Solutions
1a [AJC 2007 Common Test]
1
2
0
11
2
200
1 2
2
0
1
2
0
11
0
Area = ln(1+x )
2ln(1 ) .
1
2(1 ) 2ln 2
1
2ln 2 2
1
ln 2 2 2 tan
ln 2 2 2 ln 2 24 2
dx
xx x x dx
x
xdx
x
dxx
x x
π π
−
= + − +
+ −= −
+
= − −+
= − −
= − − = − +
∫
∫
∫
∫
b ( )
ln2
2
0
ln 2
0
ln 2
Volume = (1) ln 2 1
ln 2
ln 2 ln 2 1
2 ln 2 (2 1) (2 ln 2 1)
y
y
e dy
e y
e
π π
π π
π π
π π π
− −
= − −
= − − − = − − = −
∫
2 (SRJC 06FE Qn 10b)
When exy == ,1
2013 Revision Package BT1 Solutions
(i) Area of R
( )
( )
( ) [ ]( ) ( )[ ]
2
2
2
2
units 42ln2
22ln2ln2
ln2
1ln2
ln2
−+=
−−−−−=
−−−=
−−−=
−−=
∫
∫
e
eeee
xxxe
dxx
xxxe
xdxe
e
e
e
(ii) ln yy x x e= ⇒ =
When 2ln,2 == yx
Volume
( )( )
( )
( )
[ ] ( )
[ ] ( )
32
2
2ln22
1
2ln
2
1
2ln
2
21
2ln
2
units 2ln462
1
2ln1442
1
2ln142
1
2ln142
1
2ln14
22ln1
+−=
−−−=
−−−=
−−
=
−−=
−−=
∫
∫
e
e
ee
e
dxe
dxy
y
y
ππ
ππ
ππ
ππ
ππ
3 θcos4=x , θsin5=y
Using the GC, we can see that the curve is an ellipse.
2013 Revision Package BT1 Solutions
θθ
sin4−=d
dx, θ
θcos5=
d
dy
θθ
cot4
5−=
d
dy
When 3
πθ = , 2
3cos4 ==
πx ,
2
35
3cos5 ==
πy ,
34
5
3cot
4
5−=−=
π
dx
dy.
Equation of tangent:
( ) 32
52
34
5
34
5
2
32
5
+−−=
−=−
−
xy
x
y
When 0=y ,
( )
( )
8
32
52
34
5
32
52
34
50
=
=−
+−−=
x
x
x
The tangent cuts the x-axis at 8=x .
2013 Revision Package BT1 Solutions
When 0=y ,
0
sin50
=
=
θ
θ
4
0cos4
=
=
x
x
( )
( )
( )
4
2
0
3
02
3
0
3
0
3
Area under curve
1 5 38 2
2 2
15 35sin 4sin
2
15 320 sin
2
15 3 1 cos 220
2 2
15 3 sin 210
2 2
15 3 1 210 0 0 sin
2 3 2 3
15 3 310
2 4 3
10 3 uni3
ydx
d
d
d
π
π
π
π
θ θ θ
θ θ
θθ
θθ
π π
π
π
= − −
= − −
= +
−= +
= + −
= + − − −
= + −
= −
∫
∫
∫
∫
2ts
2013 Revision Package BT1 Solutions
( )
( )
( )
( )
2
42
2
02
3
02
3
02
3
02
3
03
3
Volume
1 5 38 6
3 2
75 25sin 4sin
75 100 sin sin
75 100 sin 1 cos
75 100 sin sin cos
cos75 100 cos
3
1 175 100 1 cos co
3 3 3
y dx
d
d
d
d
π
π
π
π
π
π π
π π θ θ θ
π π θ θ θ
π π θ θ θ
π π θ θ θ θ
θπ π θ
ππ π
= − −
= − −
= +
= + −
= + −
= + − +
= + − + − − +
∫
∫
∫
∫
∫
3
3
s3
2 1 175 100
3 2 24
575 100
24
325 units
6
π
π π
π π
π
= + − − − +
− = +
=
4 (NJ 06FE Qn 10)
a) Using the GC
tdt
dx+= 1
2
3
2013 Revision Package BT1 Solutions
3
1
0
2
1
0
units 8
3
42
3
12
3
12
31
2
1
π
π
=
=
−=
⌡
⌠
+−=
=
∫
∫
dtt
dttt
ydxx
x
b)
From GC
When 1=y
3
42
14
2 4
1
=
−=−
=
−
−
x
xx
x
x
Volume⌡
⌠
−
−−=
3
2
2
1
4
2dx
x
xππ
Let ( )22 1 cosx θ= +
4sin cosdx
dθ θ
θ= −
When 2=x ,
1=y
2 3
2013 Revision Package BT1 Solutions
( )
2
0cos
cos122
2
2
πθ
θ
θ
=
=
+=
When 3=x ,
( )
4
2
1cos
cos123
2
2
πθ
θ
θ
=
=
+=
( )( )
( )
( )
( )
22
4
2
4
2
4
2
2
4
2
4
2
2
2
4
2
2
2
3
2
2
1
units 4
2
42
12
2sin
2
1
42sin
2
12
2
2sin2
2
12cos4
cos4
cossin4sin
cos
cossin4cos1
cos
cossin4cos124
2cos12
4
2
ππ
πππ
ππ
ππππ
θθ
ππ
θθ
ππ
θθππ
θθθθ
θππ
θθθθ
θππ
θθθθ
θππ
ππ
π
π
π
π
π
π
π
π
π
π
π
π
−=
−+=
+−
++=
++=
⌡⌠ +
+=
+=
⌡⌠ −−=
⌡
⌠−
−−=
⌡
⌠−
+−
−+−=
⌡
⌠
−
−−=
∫
d
d
d
d
d
dxx
xV
2013 Revision Package BT1 Solutions
5 (CJ 06Prelim P1/Qn 14)
(i)
( )
( ) ( )
2
0
2
0
2Area of
1
2 ln 1
2ln 2 1 2ln 0 1
2ln 3
R dxx
x
=+
= +
= + − +
=
∫
(ii)
Volume ⌡
⌠
+
+==−
−∫0
5.0
20
5.0
2
1
21 dx
xdxy
(iii) When 5.0−=x , 515.0
21 =
+−+=y
2013 Revision Package BT1 Solutions
( )1
1
4
1
4
11
2
11
2
1
21
1
21
1
21
2
2
2
2
+−
−−
=
−
−=
−−
=
−=+
+=−
++=
yyx
yx
yx
yx
xy
xy
Volume
( )
( )
3
5
3
5
3
2
5
3
2
2
units 02.1
32ln42
454ln4
4
425.1
1ln41
425.1
11
4
1
425.1
52
1
π
ππ
ππ
ππ
ππ
=
+−
−−
+−
−−=
+−−
−
−−=
⌡
⌠+
−−
−−=
−
= ∫
yyy
dxyy
dxx
6i [ACJC 2007 Common Test]
( ) ( )( )( ) ( )( )
( ) ( )
( ) ( )( )( ) ( )( )
( ) ( ) ( )
( ) ( )
sin lncos ln cos ln
cos ln sin ln
cos ln cos ln sin ln
cos ln sin ln cos ln
2 cos ln cos ln sin ln
xx dx x x x dx
x
x x x dx
xx x x x x dx
x
x x x x x dx
x dx x x x
− = −
= +
= + −
= + −
∴ = +
∫ ∫
∫
∫
∫∫ ( )
( ) ( ) ( ) cos ln
x C
x dx
+
∴ =∫x x
cos ln x + sin ln x + K2 2
2013 Revision Package BT1 Solutions
ii
Turning point: (1, 1)
x-intercepts: ( )
π-
2e ,0 = 0.208,0 and ( )
π
2e ,0 = 4.81,0
( ) ( )2
2
5
3 Required area cos ln cos ln
0.3871 0.0037
e
ex dx x dx
π
π= +
= +
=
∫ ∫
0.391
7
2
2 2
1 secd d
1 tan 1x
x
θθ
θ=
+ +∫ ∫
( )1
sec d
ln sec tan
ln sec tan (Shown)
c
x x c
θ θ
θ θ
−
=
= + +
= + +
∫
(i)
x
y
O
R
1
1
2 3y x= 2
1
1y
x=
+
2013 Revision Package BT1 Solutions
.
2
22
1 1 1
1 122 31
xx
xx= ⇒ =
++
( )( )
4 2
2 2
12 0
3 4 0
3
1
2
x x
x x
x
y
⇒ + − =
⇒ − + =
⇒ =
⇒ =
(ii) Area of Region R 2
0
31 1 1
d 32 21
xx
= −
+∫
( )
( )
( )
1
0
1
3 3ln sec tan
4
3ln sec tan 3 3
4
3ln 2 3 (Ans)
4
x x−
−
= + −
= + −
= + −
(iii) Volume of solid
2
0
3 1d
2y xπ
= −
∫
2
20
3 1 1d 0.513 (Ans)
21x
xπ
= − =
+ ∫
8 [IJC 2007 Prelim]
(a) 1
ln 1dx
x t tdt t
= + ⇒ = +
Area required 2 2
11 1
1(2 ) 2 2 2 2 3
e e ett dt t dt t t e e
t
+ = = + = + = + −∫ ∫ units
(b) 2 4
12 2
y
y y
−= − +
+ +
2013 Revision Package BT1 Solutions
2 2
2
2 4 8 16d 1 1
2 2 2 (2 )
168ln 2
2
yy dy dy
y y y y
y y Cy
−= − + = − +
+ + + +
= − + − ++
∫ ∫ ∫
Volume required
[ ]
2 2 21 1 0
1 0 1
1
0
2 | | 2 2
2 2 2
8ln | 2 | 16( 2)
161 8ln 3 8ln 2 8
3
14 28 ln
3 3
y y ydy dy dy
y y y
y y y
π π π
π π
π π
π π
− −
− − += = +∫ ∫ ∫
+ + +
= − + − + +
= − − + + +
= +
9 [PJC/2007/Block Test]
area of PQRS < A < area of PQUT
Using GC, min point is (1, 2) 2 1.5 2.5 1.5A∴ × < < ×
3 < A < 3.75.
222
0.50.5
1 15( ) d ln 2 ln 2
2 8
xA x x x
x
= + = + = +
∫
Thus 15
3 2ln 2 3.758
< + <
9 15
ln 216 16
⇒ < <
232
2
0.50.5
1 1 57( ) d 2
3 8
xV x x x
x x
π= π + = + − =
∫
10 [TPJC 2007 Common Test]
0.5 2
2.5
P Q
R S
U T
2013 Revision Package BT1 Solutions
(a) Area of region required = 2
1
(ln )x e
x
x dx=
=∫
When x = eu, 1 0 ; 1x u x e u= ⇒ = = ⇒ = and
udxe
du=
Area of region required = 1 1
2 2
0 0
uu
u
dxu du u e du
du
=
=
=∫ ∫ (shown)
Using integration by parts,
The exact area
1 112 2
00 0
11
00
2
2 2 2 2 2 2
u u u
u u
u e du u e ue du
e ue e du e e e e
= −∫ ∫
= − − = − + − = −∫
(b) Using the graphic calculator, find the volume of the solid formed when the region bounded by
the curves cosy x= , the lines 3
22
y x= − + and 1y = is rotated 2π radians about the y-axis, giving
your answer correct to 3 significant figures.
When 3
22
y x= − + and cosy x= intersect, from GC, 0.9403x = and 0.5895y =
Volume required
( )2
1 21
0.5895
4 2cos
3
0.285 to 3sf
yx dxπ −−
= −∫
=
2013 Revision Package BT1 Solutions
Chapter 16: Differential Equations
1 2
2sec tan
d yx x
dx=
Integrating both sides with respect to x: sec tan secdy
x x dx x Cdx
= = +∫
Integrating both sides with respect to x again: ( )
( )
sec
ln sec tan
y x C dx
x x Cx D
= +
= + + +
∫
The general solution is ( )ln sec tany x x Cx D= + + +
2(i)
dx
duxee
dx
dy uu +=
xdx
dy4=
2(ii) ∫ ∫= xdxdy 4
Cxy += 22
Cxxeu += 22
x
Cxe
u += 2
Cx
Cxu ,2ln
+= is an arbitrary constant.
2 (iii)
3 0.4 (10 )
dyy y
dx= −
12
− 1 x
1
2x = − 1
2x = y
2013 Revision Package BT1 Solutions
( )( )
10.4
(10 )
1 1 10.4
10 (10 )
1ln ln 10 0.4
10
dy dxy y
dy dxy y
y y x C
=−
+ =−
− − = +
∫ ∫
∫ ∫
4 10
ln 4 1010
, 10
x C
yx C
y
yAe A e
y
= +
−
= =−
When x = −1, y = 5 ⇒ 4
A e=
4 44 4 4 4
4 4
10 (10 )
10 1
xx x
x
y ee y e y y
y e
++ +
+= ⇒ = − ⇒ =
− +
4 1
dy dzy z x
dx dx= − ⇒ = −
Substituting into 1 2
1dy x
y dx y
−− = :
( ) ( )1 2
1 1
1 2
1
dz x
z x dx z x
dzz x x
dx
dzz
dx
− − − =
− −
− − + = −
= −
Integrating w.r.t. x:
( )
( )
1 1
1
ln 1 , where is a constant
1
1, where
1 1 (Shown)
x c
x c
x x
dz dxz
z x c c
z e
z Ae A e
y x Ae y x Ae
+
=−
− = +
− =
= + =
+ = + ⇒ = − +
∫ ∫
When curve passes through ( )1,0 , A = 0 1y x⇒ = −
When curve passes through ( )0,2 , A = 1 1 xy x e⇒ = − +
2013 Revision Package BT1 Solutions
(Notice that 1y x= − is an oblique asymptote for 1 xy x e= − + )
5
( )( )
2
1
2
4
( 2)
2 41 4 2 4
1 2
dy
dx x
xdy x dx y c y c
x
−
−
=+
+= + ⇒ = + ⇒ = − +
− + ∫ ∫
At ( )2,7A = , 4
7 82 2
c = + =+
Thus, equation of the curve is 4
82
yx
= − ++
Gradient of tangent at A
( ) ( )2
2,7
4 1
42 2
dy
dx= = =
+
Gradient of normal at A 4= −
Equation of normal at A: ( )7 4 2 4 15y x y x− = − − ⇒ = − +
6 ( )1
dxkx x
dt= −
1y x= −
1x
y x e= − +
x
y
2
1
1
2013 Revision Package BT1 Solutions
( )
( ) ( )
1 1 1
1 1
ln ln 1 ln
ln (shown)1
dx k dt dx k dtx x x x
x x C kt
Cxkt
x
= ⇒ + =− −
⇒ − − + =
=
−
∫ ∫ ∫ ∫
1When 0, ln 0 9
10 9
1When 1, ln ln 3
4 3
Ct x C
Ct x k k
= = ∴ = ⇒ =
= = ∴ = ⇒ =
( )
9 9 9ln ln 3 ln ln 3 3
1 1 1
3 9 3 1
9 3
t t
tt
t
x x xt
x x x
x x x
∴ = ⇒ = ⇒ =
− − −
⇒ = − ⇒ =+
3 11 as (shown)
99 3 13
t
t
t
x t= = → → ∞+ +
2013 Revision Package BT1 Solutions
7 i)
22death of rate birth of rate dt
daxx
x−=−=
At x = 10, 20101020 2 .a)(a)( =⇒−=
5
2dt
d 2x
xx
−=∴
ii) Selling away 1800 prawns daily, D.E. becomes 8.15
2dt
d 2
−−=x
xx
( ) ( )( )195
1910
5
1
dt
d
5
9
52
dt
d
2
2
−−−=+−−=
−−=
xxxxx
xx
x
iii) Separating the variables, ( )( ) 5
1
dt
d
19
1−=
−−
x
xx
Integrating both sides w.r.t. t: ( )( )
⌡⌠ −=
⌡
⌠−−
txxx
d 5
1d
19
1
( ) ( )
( ) ( )
c
t
eA,Aex
x
cct
xlnxln
txxx-
==−
−
+−=−−−
⌡⌠ −=
⌡
⌠−
−
−
1
9
constantarbitrary an is where,5
819
d 5
1d
18
1
98
1
5
8
At 0=t , x = 13: 3
1
113
913=
−
−=A 1
24
31
98
5
8
+=⇒=−
−∴
−
t
t
xe
x
x
2013 Revision Package BT1 Solutions
Chapter 10 and 17: Vectors
1 (SRJC/2009/Prelim/P1/Q2) 1. (a)
14 2
0 414 2
1 1 14
1 1 11
���� �������� OB OAOP
−
+ λ − + λ + λ = = = λ + λ + λ + λ + λ
14 22
114 2 21 4
4 5 51 1
1 12
����CP
− + λ − + λ− + λ
λ = λ − − = + + λ + λ + λ −
14 22
1
45
1
2
CP
− + λ − + λ
λ = + + λ
���� 2
4
1
k
−
=
2k⇒ =
4 14 25 4 8 2 2 4
1 1
3
k or kλ − + λ
+ = = − = − = −+ λ + λ
⇒ λ =
(b) 2
1 0
0 1
2cos150
5 1
µ
µ
• =
+
⇒2
3 2
2 5 1
µ
µ− =
+
24 15 1µ µ⇒ = − +
2013 Revision Package BT1 Solutions
2 216 15( 1)µ µ⇒ = +
2 15µ⇒ =
15µ = − (reject positive value of µ)
2
(i)
(ii)
Equation of line l1 : ~
1 3
2 2 ,
5 0
r
= + λ
λ ∈� .
Equation of line l2 : ~
1 1
0 1 ,
1 1
r
= + β −
β∈� .
When the lines intersect,
+
−
=
+
1
1
1
1
0
1
0
2
3
5
2
1
βλ
βλ +=+ 131 ------ (1)
βλ =+ 22 ------ (2)
β+−= 15 ------ (3)
From (3), 6=β
Substitute into (2), 2=λ
Check (1) : LHS = 1 + 3(2) = 7
RHS = 1 + 6 = 7 = LHS
Therefore the lines intersect.
=
+
−
=
5
6
7
1
1
1
6
1
0
1
OX
2013 Revision Package BT1 Solutions
0230
0
2
3
1
=+⇒=
•
bab
a
------ (1)
010
1
1
1
1
=++⇒=
•
bab
a
------ (2)
Solving, 3,2 −== ba .
Equation of line l3 : ,
1
3
2
5
6
7
~
−+
= sr s ∈� .
Since V lies on l3,
−+
=
1
3
2
5
6
7
sOV , for some s ∈� .
7 7 2 2
5 14 6 6 3 5 14 3 5 14
5 5 1 1
VX s s
= ⇒ − − − = ⇒ − =
����
14 5 14 5 5s s s⇒ = ⇒ = ⇒ = ±
Therefore V = ( )17, 9, 10− or ( )0 ,21 ,3− .
Alternative Solution:
( )2
15 14 3
141
OV OX
= + −
���� ���� or ( )
21
5 14 314
1
OV OX
= − −
���� ����
⇒
7 2
6 5 3
5 1
OV
= + −
���� or
7 2
6 5 3
5 1
OV
= − −
����
⇒
17
9
10
OV
= −
���� or
3
21
0
OV
−
=
����
2013 Revision Package BT1 Solutions
Therefore V = ( )17, 9, 10− or ( )0 ,21 ,3− .
3
1
5
: 4 15
3
Π
⋅ − =
r
Distance from A to 1Π
6 5
2 4
6 3 15 4 15 11 11 50
5050 50 50 50 50
⋅ − − = − = − = =
1
15 4 and are on the same side of
50 50A O Π> ⇒
Vector parallel to 2Π
6 5 1
2 2 0
6 8 2
= − = − −
Normal vector of 2Π
1 1 4 2
0 2 12 2 6
2 10 2 1
− −
= × = = −
Acute angle between 1Π & 2Π
5 2
4 6
3 11 1
5 2
4 6
3 1
31cos cos
50 41
−
− ⋅ − −
−
−
= =×
46.8= ° (to 3 s.f.)
1 2 3: 5 4 3 15; : 2 6 6; : 8x y z x y z x y az bΠ Π Π− + = − + + = − + + =
For line of intersection of 1 2 and Π Π (also the intersection of the 3 planes):
5 4 3 15 1 0 1 3
2 6 1 6 0 1 0.5 0
x x
y y
z z
− = ⇒ = − −
2013 Revision Package BT1 Solutions
Let , z t t= ∈� .
Then 2
3 3 2
0 12
0 2
t
x tt
y
z t
−
= − = − −
3 2
0 1 ,
0 2
µ µ
⇒ = + ∈ −
r �
Take 0 and 1µ µ= = , two points lying on 3Π are ( )3,0,0 and ( )5,1, 2− .
Sub. into 8x y az b+ + = : ( ) ( )
( ) ( )
3 8 0 0 3
5 8 1 2 3 5
a b b
a a
+ + = ⇒ =
+ + − = ⇒ =
4
(MJC/2009/Prelim/P1/Q5)
(i) Since AB = −b a����
and A, B and C are collinear,
2k∴ =
2 2
3 2
OC OB BC= +
= −
−
b+ b a
= b a
���� ���� ����
(ii) 3
4.
aa b
a=
2
2
34
4
16
3
4 4 3
33
a
a
a
=
=
= =
(iii) 2 2 2b 2 ( 2) 2 2 3= + − + =
In this case, the length of projection is ON����
which
is 3
a4
and is equal to 4 in this question.
2013 Revision Package BT1 Solutions
( )
0
a.bcos
a b
4
4 32 3
3
1
2
or 603
θ
πθ
=
=
=
=
(iv)
( ) ( )
2
3
2 2 3
3
2
2
2
DC DADE
+=
− + −=
= −
− −
a a b
b
=
���� ��������
5
(YJC/2009/Prelim/P1/Q9)
(a) 1l passes through A(0, 3, 1) and is parallel to
−=
1
2
1
1m .
A
B
C O
N
D
E
2013 Revision Package BT1 Solutions
)
2l passes through B(1, 0, 2) and is parallel to
−
−
=
2
4
2
2m .
Since 12 2mm −= . 1l and 2l are parallel.
(b)(i) Since P lies on 1l , ie. =→
OP
+
−
λ
λ
λ
1
23 , for some λ∈ℜ.
=→
BP
−
−
−
=
−
+
−
1
23
1
2
0
1
1
23
λ
λ
λ
λ
λ
λ
.
If APB is a right angle, 0
1
2
1
=
−•→
BP .
0
1
2
1
1
23
1
=
−•
−
−
−
λ
λ
λ
0)1()23(2)1( =−+−−− λλλ ⇒ 3
4=λ
=→
OP
=
=
+
−
7
1
4
3
1
3
73
13
4
3
41
3
423
3
4
.
Hence the coordinates of P is
3
7,
3
1,
3
4
Shortest distance between 1l and 2l
2013 Revision Package BT1 Solutions
=
4 11
3 3
4 1 13 2 units
3 3 3
141
33
BP→
−
= − = = −
6 6.
(i) Let AB→
= k
1
2
−2 . b =
k
2k
−2k
+
−2
1
0
=
k − 2
2k + 1
−2k
(k − 2) + 2(2k + 1) − 2(−2k) = 18
9k = 18 ⇒ k = 2
∴ b =
0
5
−4
(ii) AB→
= 2 1 + 4 + 4 = 6.
(iii) BA→
= 2 AC→
⇒ a − b = 2( c − a)
2c = 3a − b
2c = 3
−2
1
0 −
0
5
−4 ⇒ c =
−3
−1
2 .
(iv) CD→
=
2
5
−3 −
−3
−1
2
=
5
6
−5
2013 Revision Package BT1 Solutions
CB→
=
0
5
−4 −
−3
−1
2
=
3
6
−6 = 3
1
2
−2
∴ CD→
×××× CB→
=
5
6
−5 ×××× 3
1
2
−2 = 3
−2
5
4
Therefore equation of plane is
2 2 2
5 5 5 9
4 3 4
r
− −
⋅ = ⋅ = −
(v)
=
1
0
2
BD
Since 0
2
2
1
=
−
⋅BD , the length of projection of line BD on the plane = 5122 =+ units
7 7.
1
2
4
OA
−
=
����,
0
1
5
OB
=
����
1
1
1
AB OB OA
= − = −
���� ���� ����
(i) 1l :
1 1
2 1 ,
4 1
λ λ
−
= + − ∈
r �
2013 Revision Package BT1 Solutions
(ii)
1 1
5 1
3 1cos
35 3θ
− • − =
28.56θ = °
Acute angle between 1l and 1n is 28.56° .
Therefore, angle between 1Π and 1l is 61.4° .
(iii) Since A lies on 1Π , perpendicular distance required is 1
1
AB n
n
•����
.
1 1
1 5
1 3 9
35 35
− • − =
Perpendicular distance = 1.52 units
(iv) 2 1n AB n= ����
1 1
1 5
1 3
= − × −
2 1
2 2 1
4 2
= − = − − −
2Π : ( ) ( ) ( )2 2 4 2− − = − + + − −r i j k i j k i j ki i = – 11
2013 Revision Package BT1 Solutions
Cartesian Eqn: x – y – 2z = – 11
(v) Direction vector of 2 1 2l n n= ×
1 1
5 1
3 2
= − × − −
13
5
4
=
2l :
1 13
2 5 ,
4 4
µ µ
−
= + ∈
r �
8
8. (i) Since P lies on l ,
4 3
2
5
OP
λ
λ
λ
− + → = − +
for some λ ∈R .
Since
3
, 2 0.
1
OP l OP
→ → ⊥ ⋅ − =
Now
4 3 3
2 2 0
5 1
λ
λ
λ
− +
− ⋅ − = +
12 9 4 5 0
14 7
λ λ λ
λ
∴ − + + + + =
=
1
2λ =
2013 Revision Package BT1 Solutions
14 3
25
1 12 2
2 211
15
2
OP
− +
− → ∴ = − = −
+
(ii)
3 5 3 20 101 1
2 2 2 38 192 2
1 11 1 16 8
OP
− → × − = − × − = =
10
19 0
8
∴ ⋅ =
r
The Cartesian equation is 10 19 8 0x y z+ + = .
Remark: Better to use
4
0
5
−
instead of OP→
.
(iii)
4 3 2
2 3
5 1
k
λ
λ
λ
− +
− ⋅ = − +
8 6 2 5 3kλ λ λ− + − + + = −
3 (7 2 ) 3k λ− + − = −
7 2 0k⇒ − =
7
2k⇒ =
OR Use point P (or any point on l).
2013 Revision Package BT1 Solutions
5
22
1 3
11 1
2
k
−
− ⋅ = −
7
2k⇒ =
Alt. Since l lies in Π2,
2
1
l k
⊥
.
∴
2 3
2 0
1 1
k
⋅ − =
7
2k⇒ =
(iv) Let θ be the angle between planes 1Π and 2Π .
72
10 2
19
8 1cos
525 17.25θ
⋅
=⋅
6.8θ∴ =
9
9. (a) The augmented matrix
2 2 1 4
2 3 4 1
4 3 1 2
− − = − −
The RREF of the augmented matrix
1 0 0.5 0
0 1 1 0
0 0 0 1
− = −
The final row of the RREF shows that 0 0 0 1x y z+ + = , which implies that the
system of equations is inconsistent. ∴ The equations do not have a solution.
The 3 planes represented by the 3 equations do not intersect in a point or line.
Furthermore, since neither of the planes is parallel to any other (∵ neither of the
normals is parallel to any other), the 3 planes form a triangular prism.
2013 Revision Package BT1 Solutions
(b)
2 2 5
2 3 10
1 4 10
− × =
−
1 2
1
A vector to both normals of and is 2
2
π π ∴ ⊥
.
(i)
( ) 1 2
2 2 4 ---------(1)
2 3 4 1 ---------(2)
When 0, 2 2 4 -------(4)
and 2 3 1 --------(5)
(5) (4) gives 1 and 1.
1, 1, 0 is a point on both and .
12 is para2
x y z
x y z
z x y
x y
y x
π π
− + = −
+ − =
= − = −
+ =
− = = −
∴ −
1 2llel to both and .
1 1a vector equation of is 1 + 2 , .
0 2l
π π
λ λ− ∴ = ∈
r �
(ii) A vector normal to π is
1
2 .
2
A vector equation of π is
1 1 11
2 1 232 2 3 2
= − =
r . .
3
6 1
6
∴ =
r . .
10 10.
2013 Revision Package BT1 Solutions
1
2 1
: 2 0l r
p q
λ
= +
� and 1
2
: 2 2r
p
Π ⋅ = �
Since 1l lies on 1Π ⇒
2 1
2 1 2 2 (shown)
1
p
p
⋅ = ⇒ = −
Also since 1
1
1
1
l
⊥
⇒
1 1
0 1 0 1
1
q
q
⋅ = ⇒ = −
Therefore, 1
2 1
: 2 0
2 1
l r λ
= + − −
�
When
1
3, 2 (verified)
1
r OAλ→
−
= − = =
�
2 2
2 1 2 1
Normal of is 2 0 0 2 0
2 1 2 1
n
−
Π = × = = − − − −
�
Thus 2
1
: 0 0
1
r
Π ⋅ = �
And 3
1 2 1
: 5 1 5 10
1 3 1
r
Π ⋅ = ⋅ = �
1 1 0
3 2 1
2 1 1
AB OB OA
− −
= − = − =
→ → →
Thus length of projection of 3 on AB Π→
= 3
3
nAB
n×
→�
�
2013 Revision Package BT1 Solutions
0 11
1 527
1 1
41
127
1
18 2
27 3
= ×
−
= −
= =
x + y + z = 2, x + z = 0 and x + 5y + z = 10, intersects at 1l .
Otherwise method
3 3
2 1 2
For , 2 5 10 2 lies on
2 1 2
Π ⋅ = ⇒ Π − −
.
When 3
3 1 3 3
1, 2 5 10 2 also l 2 ies on
3 1 3 3
λ
= ⋅ = ⇒ Π − − −
.
Since
2
2
2
−
,
3
2
3
−
lies both on 1 3 and l Π
⇒ 1 2 3, and Π Π Π all intersects on the line 1.l
Solving method
2013 Revision Package BT1 Solutions
1 1 1 2
1 0 1 0
1 5 1 10
1 1 1 2 1 0 1 0
1 0 1 0 0 1 0 2
1 5 1 10 0 0 0 0
rref
x
y
z
=
⇒ →
Thus y = 2, x + z = 0.
Let ,z xµ µ= = −
Thus
0 1
2 0
0 1
r µ
−
= +
� which is line 1.l
⇒ 1 2 3, and Π Π Π all intersects on the line 1.l
11
(i)
(ii)
(iii)
−
−
=→
0
3
6
OA and
=→
6
0
0
OV
=∴→
2
1
2
3AV
Equation of AV is
r
+
=
2
1
2
6
0
0
λ , ℜ∈λ or r
+
−
−
=
2
1
2
0
3
6
λ , ℜ∈λ
Equation of l is r
−
+
−
−
=
2
1
10
2
4
µ
t
If AV and l intersect, let
+
+−
−−
=
+ µ
µ
µ
λ
λ
λ
2
2
104
26
2
t
⇒ )1.....(..........2−=− µλ and 25 −=+ µλ ……….(2)
Solving eq (1) & (2) we have 0=µ and 2−=λ .
2013 Revision Package BT1 Solutions
∴
−
−
=
2
2
4
MO�
. Also, 2262 =⇒+=+ tt λµ
Acute angle between AV and line l = 1
2 10
cos 1 . 1 / 9 105 60.8
2 2
−
−
= °
→
AM =
2
1
2
Perpendicular distance required = 079.60sin
→
AM = 2.62 units
12 or OA
8 4
OZ OA AZ OB= + +
= +i k
���� ���� ���� ���� ����
By ratio Theorem
8 1 8 11 1 1
0 4 41 1 1 1 1
4 0 4
OP OZ OC
λλ λ
λ λ λ λ λλ
+
= + = + = + + + + +
���� ���� ����
Since OP ⊥⊥⊥⊥ CZ,
0
7
4
4
OP CZ
CZ OZ OC
=
= − = −
���� ����i
���� ���� ����
8 1 71
4 4 0 56 7 16 16 0 8
4 4
λ
λ λ λ
λ
+
− = ⇒ + − + = ⇒ =
i
1 1 41 4
4 81 1 / 8 9
1 / 2 1
OP
+
= = +
����
2013 Revision Package BT1 Solutions
2 2 244 8 1 4
9OP = + + =����
7 / 9 71
32 / 9 329
-32 / 9 -32
BP OP OB
= − = =
���� ���� ����
The direction vector of line BP is
7
32
32
−
.
Cartesian Equation of line BP:
1 4
7 32 32
x y z− −= =
− or equivalent
(i)
(ii)
(iii)
Write down the position vector of Z in terms of i, j and k.
The point P divides CZ such that 1
CP
PZ
λ==== . Given that OP
���� is perpendicular to CZ
����, find the
value of λ and evaluate OP����
.
Find a cartesian equation of the line BP.
Ans: (i) 8 4OZ = +i k����
; (ii) 1
8λ = ; 4OP =
����; (iii)
1 4
7 32 32
x y z− −= =
−
13
3 1
4 2 , for some
1 0
OP s s
−
= + ∈
→� .
2013 Revision Package BT1 Solutions
(i)
(ii)
(iii)
Since OP is perpendicular to l,
1
2 0
0
OP•
−
∴ =
→
3 1
4 2 2 0 3 8 4 0
1 0
s
s s s•
− − ⇒ + = ⇒ − + + + =
1s⇒ = −
So,
3 1 4
4 2 2
1 0 1
OP
= + − =
→ #
Let
3 1 6 1
4 2 3
1 0 0
s a t
a
−
+ = +
3 6 (1)
4 2 3 (2)
1 (3)
s t
s a t
at
− = + − − − − − − − − − − −
⇒ + = + − − − − − − − − − − − = − − − − − − − − − − −
From (1) : 3s t= − − ------------------- (4)
Sub (4) into (2) : 4 2( 3 ) 3t a t+ − − = +
5 2t a⇒ = − − ----------- (5)
Sub (5) into (3) : 5 ( 2 )a a= − −
2 2 5 0a a⇒ + + =
2( 1) 4 0a⇒ + + =
Since, there are no real solutions for the equation 2( 1) 4 0a + + = , therefore there does not exist
real values of a such that the two lines l and m intersect. Hence, the 2 lines do not have a common
point.
2013 Revision Package BT1 Solutions
o 2 2
2 2
1 0
3 0
1 1 10cos60 4 10
2 310 .1 10
a aa a a
a a
•
= ⇒ = ⇒ = + ⇒ = ±
+ +
14
(i)
(ii)
(iii)
(iv)
(v)
Since AB ⊥����
1Р, AB����
//
1
2
2
−
i.e.
1
2 .
2
AB k
= −
����
2 2
2 1 2 1
2 0 2
k k
OB AB OA k k
k k
− −
= + = + = + − −
���� ���� ����
B lies on 1Π , so
1 2 1
2 18 2 1 2 18 9 18 2
2 2 2
k
OB k k k
k
−
⋅ = ⇒ + ⋅ = ⇒ = ⇒ = − − −
����
0
5
4
OB
∴ = −
����
Perpendicular distance from A to 1Π ( )22 22 4 4 6AB= = + + − =
����
By Ratio Theorem,
( )2 13
2 1 2
2 0 31
3 1 5 12
0 4 2
OC OBOA OC OA OB
OC
+= ⇒ = −
+
− −
∴ = − = − −
���� �������� ���� ���� ����
����
2 3 5 0 3 3
5 1 6 , 5 1 6
3 2 5 4 2 6
CD CB
− −
= − − = = − − = − − − −
���� ����
2013 Revision Package BT1 Solutions
5 3 6 2
6 6 15 3 5
5 6 12 4
CD CB
− −
× = × = = − −
���� ����
Therefore equation of plane DBC is
2 2 2
5 5 5 9
4 3 4
− −
⋅ = ⋅ = −
r
2
0
1
BD OD OB
= − =
���� ���� ����
Length of projection of line BD on 1Π
2 1 2
0 2 5
1 2 4 4 25 165
1 1 4 4 9
2
2
−
× − + +
= = = =+ +
−
15 [HCI 2007 Prelims]
Sub. equation of line into plane:
1 0 1
1 1 . 0 3
0 1 2
+ =
λ µ and get 3 2λ µ= − .
Sub. 3 2λ µ= − into 2Π :
1 0 3 2
(3 2 ) 1 1 3 1
0 1 0 1
µ µ µ
−
= − + = + −
r
To show the two lines are parallel:
The direction vector of 2l is scalar multiple of direction vector of 1l , thus 1l and 2l are parallel.
Note that B lies on 2l , thus
Shortest distance
2 0 2 21 1 1 30
1 2 1 036 6 6
1 0 1 4
AB
−
= × = × = = − − −
����
(i)
(ii)
(iii)
2013 Revision Package BT1 Solutions
Let
3 2
1 1
0 1
OC α
= + −
���� for some α ∈�
0 0 2
0 2 2 1 0 2
0 0 1
BA BC α α
⋅ = ⇒ − ⋅ − + = ⇒ = −
���� ����.
7
3
2
OC
∴ = −
����
7 0 7
Since , 3 2 . 5
2 0 2
CD BA OD OD
= − = ∴ = − −
���� ���� ���� ����
Observe that O lies on 2Π . Midpoint of OA which is ( )3 12 2, , 0 then lies on 3Π .
Since
1 0 1
1 1 1
0 1 1
= × = −
2n ,
32
12
1 1 1
1 1 1 1
1 0 1 1
⋅ − = ⋅ − ⇒ ⋅ − =
r r (shown)
Let F be foot of perpendicular from A to 3Π and pick a point E (1, 0, 0) on 3Π .
1 1 11 1 1
1 1 133 3
1 1 1
AF AE
= ⋅ − − = − −
���� ����
81
43
1
OF OA AF
∴ = + = −
���� ���� ����
16
(i)
(ii)
[NJC 2007 Prelims]
BN ⊥ 1Π ⇒ BN // n ⇒ BN k= n
Since N lies on 1Π ,
1
2 4
1
ON
⋅ − =
����
But
7 1
18 2
1 1
ON OB BN k
−
= + = + − −
���� ���� ����, so
7 1
18 2 2 4 8
1 1
k
k k
k
− +
− ⋅ − = ⇒ = − +
2013 Revision Package BT1 Solutions
7 8 1
18 16 2
1 8 7
ON
− +
∴ = − = − +
����
⊥ distance from B to 1Π ( )( )22 2
1
8 2 8 1 2 1 8 6
1
BN
= = − = + − + =
����
To verify that point A lies on 1Π :
LHS
2 1
0 2 2 2 4
2 1
= ⋅ − = + = =
RHS (Verified)
Let B′ be the reflection of B in 1Π .
By midpoint theorem,
( )1 2
2AN AB AB AB AN AB′ ′= + ⇒ = −���� ���� ����� ����� ���� ����
1 2 7 2 7
2 2 0 18 0 14
7 2 1 2 13
AB
− − − ′∴ = − − − = − − − −
����� , which is parallel to
7
14
13
− −
.
2 1Π Π⊥ ⇒ 2Π //
1
2
1
−
Direction vector of line l =
7 2 9 3
18 0 18 3 6
1 2 3 1
AB
− − −
= − = = − − − − −
����
Normal vector of 2Π
1 3 4
2 6 2
1 1 0
= − × − =
Equation of 2Π :
4 2 4 4
2 0 2 2 8
0 2 0 0
⋅ = ⋅ ⇒ ⋅ =
r r
Note that the triangle ABN is a right-angle triangle with 90ANB∠ = ° .
2013 Revision Package BT1 Solutions
( )2 2 2
3
3 6 3 3 6 1 3 46
1
AB
= − − = + + =
����
By Pythagoras’ Theorem,
Distance of A from 3Π ( ) ( )( )2 2 9 46 64 6 30AN AB BN= = − = − =
Acute angle between l and 3Π
= 1 1 8 6
cos cos 15.63 46
BNABN
AB
− − ∠ = = = °
(to 3 s.f.)
17
a
b
[MI 2007 Prelims]
Let 1
3, 23
xz yµ
+= = + = . Then 3 1, 3, 2x z yµ µ= − = − = .
2
1 3
: 3 1 ,
2 0
l µ µ
−
= − + ∈
r �
To check if lines intersect:
1 2 1 3
3
1 2
λ µ
λ µ
λ
+ − +
= − + −
2 3 2 -----(1)
3 -----(2)
1 -----(3)
λ µ
λ µ
λ
− = −
⇒ − = −− =
Solving equations (2) & (3): 1λ = − and 1 3 2µ = − + =
Substitute into equation (1): LHS ( ) ( )2 1 3 2 8 RHS= − − = − ≠
Therefore the lines do not intersect.
Let the angle between a and b be θ .
( )
( )( )
2
2
cos
3 2 3 cos5
4.15 (to 3 s.f.)
θ
π
− ⋅ = ⋅ − ⋅
= −
= −
=
a b a a a b a
a b a
2013 Revision Package BT1 Solutions
Chapter 18: Complex Numbers 1
1 [SAJC/2010/Prelim/P1/Q4]
(a) 2 3 4w i= +
Let w x iy= +
2( ) 3 4x iy i+ = +
2 22 3 4x xyi y i+ − = +
2 2 3x y− = ----- (1)
2 4xy = ----- (2)
From eq (2): 2
yx
=
Sub into eq (1):
22 4 22
3 3 4 0x x xx
− = ⇒ − − =
Solving, we get 2x = ± , 1y = ±
Hence (2 )w i= ± +
(b) Let 4 16z = −
14
4
4 ( 2 )
( 2 )
16
16
2 , 2, 1,0,1
i
i k
i k
z e
z e
z e k
+
+
=
=
= = − −
π
π π
π π
3 34 4 4 42 , 2 , 2 ,2
i i i iz e e e e
π π π π− −=
2 [MI/2010/Prelim/P2/Q2]
(i) 5 (2 )
2
5
32 32
2 where 2, 1,0
i k
ki
z e
z e k
π
π
= =
= = ± ±
(ii)
(iii)
5
5 32 1 02
ww
− − =
1z 2z
3z 4z
Re(z)
Im(z)
2
4π
O
Re (z)
Im (z)
2
5
π
2
5
π−
2
5
π
2
5
π
2
5
π
2Z
3Z
4Z
1Z
2013 Revision Package BT1 Solutions
5
32
12
w
w
= −
2
5
2 2
5 5
2
5
2
5
2
5
5 5 5
2
12
2
2
1
2
ki
k ki i
ki
ki
ki
k k ki i i
we
w
w we e
ew
e
ew
e e e
π
π π
π
π
π
π π π
−
=
−
= −
=
−
=
−
5
5
2
2 Im
ki
ki
ew
e
π
π
=
cos sin5 5
sin5
cot5
1
1 cot (shown)5
k ki
wk
i
k
wi
kw i
π π
π
π
π
+
=
= +
= −
3 [ACJC/2010/Prelim/P1/Q4]
1
* 3
i
w
−=
2arg
* 2 3 6
i
w
π π π− = − − − =
1 1 3 1 3 1cos sin
* 3 6 6 3 2 2 6 6
ii i i
w
π π − = + = + = +
*
ni
w
−
is purely imaginary, ( )2 1 , 6 2
n k kπ π
= + ∈� ,
2013 Revision Package BT1 Solutions
( )3 2 1 ,n k k∴ = + ∈� .
4 [ACJC/2010/Prelim/P1/Q5]
24 4 20 ,
i k
z i z i e k
ππ
+
− = ⇒ = = ∈�
12
4 2
7 3 5
8 8 8 8
, 2, 1,0,1
, , ,
i k
i i i i
z e k
z e e e e
ππ
π π π π
+
− −
= = − −
=
5
81
82
i
i
z e
z e
π
π
=
=
( )1 2
1 4 3arg
8 2 8 8z z
π π π + = + =
5 [RI/2010/Prelim/P1/Q1]
Let z = x + iy. Substitute the second equation into the first.
2
2 2 2
2 2 2
( * (1 i)) 2 0
( i ) ( )(1 i) 2 0
2 2i ( )i 2 0
z z z
x y x y
x xy x y
+ + − =+ + − =+ + − =+ + − =
+ + + + − =+ + + + − =+ + + + − =+ + + + − =
+ + + − =+ + + − =+ + + − =+ + + − =
On comparing real and imaginary parts, 2 2 2 22 2 0, 2 ( ) 0x xy x y x y− = + + = + =− = + + = + =− = + + = + =− = + + = + =
1, 1x y= ± == ± == ± == ± = ∓ .
When 1 iz = −= −= −= − , w = 2i
When 1 iz = − += − += − += − + , 2iw = −= −= −= −
6 [NYJC/2010/Prelim/P1/Q2]
(1+ai)(b+2i) = 8i
=> (b-2a)+(ab+2)i = 8i
Comparing real/imaginary parts, b-2a = 0 and ab+2 = 8
i.e. b = 2a and a(2a) + 2 = 8
i.e. 2
a = 3
i.e. a = - 3 (since a < 0) and b = - 2 3
w = - 2 3 - 2i => arg(w) = 5
6
π−
z1
z2
z2 + z2
x
y
8
π
5
8
π
O
2013 Revision Package BT1 Solutions
Hence, arg(wn) = 5
6
nπ−
∴Least n = 3
(then argument is 5
2
π−, so that wn is of form – ki with k > 0).
7 [DH/2010/Prelim/P1/Q7]
(i) No. The statement is not always true. It applies only for (polynomial) equation in z with real
coefficients.
(ii) 4 3+ i 0z + = ⇒
4 3 iz = − −
⇒
5i
4 62ez
π − =
1 1
4 4
1 5 (12 5)i ( 2 ) i4 6 242 e 2 e
kk
z
π ππ
−− +
= = , 0,1, 2,3k =
1 1 1 1
4 4 4 4
5 7 19 17-i i i -i
24 24 24 242 e or 2 e or 2 e or 2 ezπ π π π
∴ =
(iii)
(iv) The quadrilateral is a square.
Let the length of each side be L
Pythagoras Theorem:
12 2 242| | =2(2 ) 2 2L z= =
8 [YJC/2010/Prelim/P1/Q9]
(a) z = 1 + ip , w = 1 + iq
zw = (1 + ip)(1 + iq)
3 – 4i = 1 – pq + i(p + q)
∴ 3 = 1 − pq … … (1)
& � 4 = p + q ⇒ q = – 4 – p
Substitute into (1) ⇒ 3 = 1 − p(– 4 – p)
p2 + 4p – 2 = 0
p = 62)1(2
)2)(1(4164±−=
−−±−
Since p > 0 ∴ p = 62 +− //
Im
L
Re
L
2Z
3Z
4Z
O
1Z
2013 Revision Package BT1 Solutions
q = – 4 – ( 62 +− ) = 62 −− //
(b) a = 1 + i 3 =
3
i
2
π
e
∴ 1 + a + a2 + a3 + … + a9 = a
a
−
−
1
1 10
= )3i1(1
21
10
3 i
+−
−
π
e
= 3i
21 3
2 i
10
−
−
−
π
e
= 3i
)i(2123
2110
−
−−−
=3i
i 3512513
−
+
3i
3i×
= i 3171512 +− //
9 [SRJC/2010/Prelim/P1/Q11]
(i) z5 = 32eπ + 2kπ( )i
, k = 0,±1,±2
2
52 , 0, 1, 2
ki
z e k
π π+ = = ± ±
52i
z e
π
= ,
3
52i
e
π
, 52i
e
π−
, 2 ie
π,
3
52i
e
π−
(ii) w5 = 32i ( )
55 32 32iw iw⇒ = − ⇒ = −
Let z iw= ⇒ w iz= −
Area of triangle OW1W21 2
2 2 sin 1.902 5
π= × × × = units2.
(iii) ( )3 3
5 5 5 52 2 2 2 2i i i i
iz e z e z e z e z e
π π π ππ
− − − − − − −
2 R
e
I
m
2013 Revision Package BT1 Solutions
Re( )z
Im( )z
1
1
–1
–1 ××××
××××
××××
××××
××××
O
( )3 3
2 25 5 5 52 4 2 4 2i i i i
z z e e z z e e z
π π π π− −
− + + − + + +
( )2 2 34cos 4 4cos 4 2
5 5z z z z z
π π − + − + +
p = 4,α =π
5,β =
3π
5,k = 2
10 [NJC/2010/Prelim/P1/Q10]
(a) 2 *( i) 16 i 0z a z b+ − + + =
( ) ( )
( )
22 3i ( i) 2 3i 16 i 0
5 12i 2 3 i 2i 3 16 i 0
8 2 10 3 i 0
a b
a a b
a a b
+ + − − + + =
⇒ − + + − − − + + =
⇒ + + − + =
By comparing real and imaginary coefficient,
Real: 8 2 0 4 (ans)
Im :10 3 0 22 (ans)
a a
a b b
+ = ⇒ = −
− + = ⇒ = −
(b) 5 1 0z + =
( )
( )
5
5
5 iπ
i 2 1 π5
i 2 1 π
5
1 0
1
e
e
e , 0, 1, 2
k
k
z
z
z
z
z k
+
+
+ =
= −
=
=
= = ± ±
iπ iπ i3π i3π
5 5 5 51, e , e ,e ,ez− −
= −
(i)
2 2 2
2
2
i i ii
i
i
1 e e e e
e cos isin + cos i sin2 2 2 2
2cos e (shown)2
θ θ θ
θ
θ
θ
θ θ θ θ
θ
− + = +
= − + − +
=
(ii) Replace complex number z with 1w − ,
( )
( )
( )
i 2 1 π
5
i 2 1 π
5
i 2 1 π
5
e
1 e
1 e
k
k
k
z
w
w
+
+
+
=
⇒ − =
∴ = +
2013 Revision Package BT1 Solutions
From (i),
( ) ( ) ( )i 2 1 π
5
2 1 πi
102 1 π
1 e 2cos e 10
k kk
w
+ ++
∴ = + =
for 0, 1, 2k = ± ± .
2013 Revision Package BT1 Solutions
Chapter 19: Complex Numbers II
1 [TPJC/2010/Prelim/P2/Q2]
( ) ( ) ( ) ( )π
arg i 0 arg arg i 0 arg2
z z z= ⇒ + = ⇒ = −
(i) maximum value of 14 3 8 11z AP+ = = + =
(ii) 2 2
2
8 4 3 4 3 3tan
4 4OAP
− − −= =
( )1 4 3 3tan arg 4 π
4z
− −
∴− ≤ + ≤
or ( )0.776 rad arg 4 πz− ≤ + ≤
2 [YJC/2010/Prelim/P2/Q3(b)]
C (–1, 1)
51 −=− zz
3
4
π−
O B
2
2i1 =−+w
5 1
2013 Revision Package BT1 Solutions
(i) Min value of 2=− wz
(ii) Intersection point at B = w
22 −=−== OCCBOBw
4
argπ
−=w
( )
−
−=∴ 4 i
22
π
ew
( )( )i112 −−=
( ) ( )i1212 −−−=
3 [ACJC/2010/Prelim/P2/Q2]
(i)
(ii) Least value of z w− = 1
(iii) Greatest ( )arg 3z + = 1 11 3
tan sin 0.826 (3 dp)5 26
− − + =
Least ( )arg 3z + = 1 11 3
tan sin 0.432 (3 dp)5 26
− − − = −
4 [CJC/2010/Prelim/P1/Q11]
(a)(i)
[k=0]
[k=1]
[k= –1]
[k=2]
[k= –2]
[k= –3]
x
y
i (2,1)
O −3
(5,−3)
o
2
3
26
3
3
z
w
2013 Revision Package BT1 Solutions
(ii)
Equation of circle:
(b)(i) iiz +−≤+− 3223
( ) 11223 +≤−− iz
( ) 1323 ≤−− iz
Circle, center (3, -2), radius
half-line from point (6, -5), arg =
Area of shaded region = = =
5 [JJC/2010/Prelim/P1/Q8]
(a) 13 13pq i= +
( )( )2 13 13
2 2 13 13
( 2 ) ( 2) 13 13
ia b i i
b i abi a i
a b i ab i
+ − = +
− + + = +
+ + − = +
Comparing real and imaginary parts,
2 13 - (1)
2 13 - (2)
a b
ab
+ =
− =
(2): 15
ba
=
Subst. 15
ba
= into (1):
Re(z)
Im(z)
z1
z6
z5
z4
z3
z2
(3, -2)
(6, -5)
2013 Revision Package BT1 Solutions
( )( )
2
152 13
13 30 0
3 10 0
3 or 10
35 or
2
aa
a a
a a
a
b
+ =
− + =
− − =
=
=
(i)
2 2
Least
4 1
17
z OP=
= +
=
2 2
Greatest
4 3 2
7
z OQ=
= + +
=
(ii)
1
Max arg( )
4tan
4
4
z i PAR
π
−
− = ∠
=
=
6 [MI/2010/Prelim/P1/Q10]
(a) Given that 1 iz k= + is a root, so substitute into the given equation
( ) ( ) ( ) ( )4 3 2
1 i 1 i 9 1 i 29 1 i 60 0k k k k+ − + − + + + − =
( ) ( )2 3 4 2 3 21 4 i 6 4 i 1 3 i 3 i 9 18 i 9 29 29 i 60 0k k k k k k k k k k+ − − + − + − − − + − + + − =
(4, 3)
y
O 4
3
Locus of z
P (4, 1)
Q
A(0, 1)
R (4,5)
2013 Revision Package BT1 Solutions
Comparing the real or imaginary parts on both sides,
3 3 34 4 3 18 29 3 12 0
2 or 0 (N.A.)
k k k k k k k k
k k
− − + − + = − + =
⇒ = ± =
OR, 2 4 2 2 4 26 3 9 9 31 6 40 0
By GC, 2
k k k k k k
k
− + + − + − = + − =
⇒ = ±
Hence, ( )( ) ( )( ) 21 2i 1 2i 2 5z z z z− − − + = − + is a factor of the given equation
( )( )4 3 2 2 29 29 60 2 5 12 0z z z z z z z z− − + − = − + + − =
( ) ( )2 22 5 0 or 12 0
1 2i , 4 or 3
z z z z
z z z
− + = + − =
∴ = ± = − =
(b) (i)
(ii) min. (arg z) = argument of any complex numbers along CD
1 2tan
2
4
π
−=
=
At A, 2 2 and 2 2x y= − = + . -----(1)
max. (arg z) = argument of a, where A ≡ a
1
1
tan -----(2)
2 2tan
2 2
AB
OB
−
−
=
+= −
Hence, ( ) 1 2 2arg tan
4 2 2z
π − +
≤ ≤ −
(Shown)
7 [PJC/2010/Prelim/P2/Q5]
i 2i 2 2z − − ≤ and ( )Re 1 3iz > +
( )3
arg 2 24
z iπ
− − =
( )arg 2 24
z iπ
− − =
2
2 0 B
D A
x
y
C α α Note : α = π/4
2013 Revision Package BT1 Solutions
i 2+2i 2z − ≤ and ( )Re 2z >
( )2 2i 2z − − ≤ and ( )Re 2z >
( )arg 2 2i2 2
zπ π
α
− < − − ≤ − −
( ) 1 2arg 2 2i sin
2 2 4z
π π − − < − − ≤ − −
( )arg 2 2i2 3
zπ π
− < − − ≤ −
At maximum ( )arg 2 2iz − − ,
( )2 2cos ( 2 2sin )iz α α= + + − +
2 2cos ( 2 2sin )i6 6
zπ π
= + + − +
( )2 3 iz = + −
1 2d k d≤ <
2 21 2 3k≤ < +
1 13k≤ <
8 [SAJC/2010/Prelim/P2/Q3]
•(5,−2)
Im (z)
(2,2)
α
(2,−2)×
Re (z)
locus of z
2
2
πα−
α d1
d2
2013 Revision Package BT1 Solutions
(a) 0 arg( 1 )2
≤ + − ≤z iπ
and 2z i− =
Let P represent the complex number z x iy= + .
2cos 24
a x= = =π
2sin 24
b = =π
2 1y = +
( )2 2 1z i∴ = + +
(b) (i)
33 3
2 cos sin cos sin4 4 6 6
w i iπ π π π
= + −
Let 13 3
cos sin4 4
w iπ π
= +
2 cos sin6 6
w iπ π
= −
( )( )31 22∴ =w w w
3
1 22 2(1)(1) 2= = =w w w
( )( )( )3
1 2
1 2
arg( ) arg 2
arg(2) arg( ) 3arg( )
w w w
w w
=
= + +
30 3
4 6
4
= + + −
=
π π
π
2 cos sin4 4
w iπ π
∴ = +
(ii) 2 cos sin4 4
n n n nw i
π π = +
2
(0,1)
4π b
a
Re(z)
1
-1
P(x,y)
Im(z)
4π
0
arg( 1 )z i+ − = 0
arg( 1 )z i+ − = 2π
2013 Revision Package BT1 Solutions
Since 4 , n k k= ∈�
4
4
4
4 42 cos sin
4 4
2 cos
( 1) 2
n k
k
k k
k kw i
k
π π
π
= +
=
= −
9 [SRJC/2010/Prelim/P1/Q7]
(a)(i) 64 3 16 0ww i iw∗ + + =
( )( ) ( )64 3 16 0x yi x yi i i x yi+ − + + + =
2 2 64 3 16 16 0x y i ix y+ + + − =
( ) ( )2 2 16 16 64 3 0x y y x i+ − + + =
Comparing coefficients,
( )16 64 3 0x + = and ( )2 2 16 0x y y+ − =
4 3x = − 2 16 48 0y y− + =
( )( )4 12 0y y− − =
y = 4 since y < 5
4 3 4w i∴ = − +
(ii) 8w = , ( )5
arg6
wπ
= , thus 5 5
8 cos sin6 6
n n n nw i
π π = +
Since n
w is to real, Im ( ) 0nw = , 5
,6
nk k
ππ= ∈�
6
,5
kn k⇒ = ∈�
(b) 1 3 1z i iz− − = +
Let z x iy= + , then ( )1 3 1x iy i i x iy+ − − = + +
( )1 1 3 1x i y ix y− + − = + −
( ) ( ) ( )1 1 3 1x i y y ix− + − = − +
( ) ( ) ( )2 2 221 1 3 1x y x y− + − = + −
( )2 2 2 22 1 2 1 9 1 2x x y y x y y− + + − + = + − +
2 28 2 8 16 7x x y y+ + − = −
2013 Revision Package BT1 Solutions
( )2
21 91
8 64x y
+ + − =
The locus is a circle of centre 1
,18
−
of radius 3
8units.
10 [VJC/2010/Prelim/P2/Q1]
y
x 0 2
8
1
8−
4
8−
31
8
1
5
8
×