1.
Let x be the no. of toy cars of type A
Let y be the no. of toy cars of type B
Let z be the no. of toy cars of type C
2x + 3y + z = 100
x + 4y +3z = 150
x + 5y + 2z = 130
… (1)
… (2)
… (3)
Using GC, x = 20, y = 10, z = 30
No. of toy cars of type A is 20
No. of toy cars of type B is 10
No. of toy cars of type C is 30
Total profit = 20($1) +10($2)+30($3) = $130
2.
x − 10
− ( x − 5) > 0
x+2
x − 10 − ( x + 2)( x − 5)
>0
x+2
− x 2 + 4x
>0
x+2
x ( x − 4)
<0
x+2
x < - 2,
or
0<x<4
For the inequality
x − 11
x +1
x − 1 < −2
Hence,
x −1
0 < x −1 < 4
or
x < −1 (n.a.) or
3.
> x − 6, replace x above by
1< x <5
∴ 1 < x < 25
2
(i) Let the first three terms of the GP be a , ar , ar .
( )
a (ar ) ar 2 = a 3 r 3 = (ar ) = 1728
ar = 1728 = 12
12
r=
"
a
3
[M1]
3
(1)
(
2
When the third term is reduced by 2, the resulting AP is a , ar , ar − 2
So
Tn − Tn −1 = Tn +1 − Tn
)
(
)
(ar − 2) − 2ar + a = 0
ar − a = ar 2 − 2 − ar
2
2
⎛ 12 ⎞
⎛ 12 ⎞
a ⎜ ⎟ − 2 − 2a ⎜ ⎟ + a = 0
⎝a⎠
⎝a⎠
a 2 − 26a + 144 = 0
a = 18
or
2
r=
3
So
a =8
3
r=
2 (NA as series is convergent)
Therefore the three terms of the GP are 18, 12, 8.
(ii)
For convergent series,
S=
(iii)
r <1
a
18
18
=
= = 54
2 1
1− r
1−
3 3
The AP has first term a = 18
⎛2⎞
d = Tn − Tn −1 = ar − a = 18⎜ ⎟ − 18 = −6
⎝3⎠
and common difference
n
[2a + (n − 1)d ]
2
n
= [36 − 6(n − 1)]
2
n
= [42 − 6n]
2
= 21n − 3n 2
Sn =
5Sn > S
5( 21n − 3n 2 ) > 54
15n 2 − 105n + 54 < 0
0.559 < n < 6.441
n = 1,2,3,4,5,6
4.
(i)
Since every horizontal line cuts the graph of f(x) at most once, therefore f is oneone. Since f is one-one, hence f −1 exists.
(ii)
1
1 − e−x
1
1 − e −x =
y
1
e −x = 1 −
y
y=
⎛ 1⎞
x = − ln⎜⎜1 − ⎟⎟
y⎠
⎝
⎛ 1⎞
f −1 ( x) = − ln⎜1 − ⎟, x > 1
⎝ x⎠
(iii) Since R g = (−∞, 0) ⊄ D f = (0, ∞) , fg does not exist.
Since R f = (1, ∞) ⊂ D g = (0, ∞) , gf exists.
1
⎛ 1 ⎞
gf ( x) = g ⎜
= −1 −
, x>0
−x ⎟
1 − e −x
⎝1− e ⎠
(iv) k =1
y = 1− x
x = 1− y
g −1 ( x) = 1 − x, x ≤ 0
5.
(i) Using GC, the root α = 0.16744919 ≈ 0.16745 (to 5 decimal places)
1 3
1
(ii) Using x1 = 0, x2 = 6 (x 1 + 1) = 6 ( [0]3 + 1) = 0.166 666 666 7
1
x3 = 6 ( [0.166 666 666 7]3 + 1) = 0.167 438 271 6
1
Using GC command: 6 ( [Ans]3 + 1), we generate the rest of the xi values
x4 = 0.167 449 038
x5 = 0.167 449 189 (stabilised at 5 decimal places)
So the sequence converges to α ≈ 0.16745
(iii) Since when n → ∝ , xn → a and also xn + 1 → α
1 3
1
Hence, the recurrence relation xn + 1 = 6 (x n + 1) tends to α = 6 ( α3 + 1)
⇒
6 α = α3 + 1
3
⇒ α −6α+1=0
But this is just the equation where α is its root.
Hence the sequence can be used to estimate the root α.
6.
(a)
y=
y
(x − 4)2
x2 −4
1
0 (1,−3) 2
−2
x
4
−4
For no real roots, −3 < m < 0
(b)(i) y
= f (3x – 1)
y
(1, 3)
2
0
(ii) y2
= f (x)
x
1
3
y
(2, 3 )
2
−1
x
0
− 2
(2,− 3 )
(iii) y = f ′( x )
y
−1
7
(i)
dx
= cos t
dt
0
2
dy
= −k sin t
dt
dy
= − k tan t
dx
(ii) For tangents parallel to y-axis, t = −
π π
,
2 2
For tangents parallel to x-axis, t = − π ,0, π
π
dy
= −k
4
dx
1
Grad of normal =
k
π
1
x = k + sin = k +
4
2
π
k
y = k cos =
4
2
(iii) At t =
,
Equation of normal:
y−
k
2
=
1⎛
1 ⎞
⎜x −k −
⎟
k⎝
2⎠
Since normal has a y-intercept of -1,
k
1⎛
1 ⎞
= ⎜0 − k −
⎟
2 k⎝
2⎠
−1−
k =1
(iv) When k = 1, equation of normal:
y−
1
= x −1−
2
y = x −1
1
2
x
At point P,
cos t = 1 + sin t − 1
cos t = sin t
tan t = 1
3π
t=−
4
3π
When t = −
,
4
1
x = 1−
2
1
y=−
2
⎛
1
Intersection point P = ⎜1 −
⎝
8.
Solution: (a)(i)
2
∫ sin
,−
1 ⎞
⎟ (Ans.)
2⎠
−1 2 x dx = x sin −1 2 x −
∫
= x sin −1 2 x +
(a)(ii)
∫x
2
2x
1− 4x2
dx
1
1 − 4x 2 + c
2
1 (2 x − 2)
1
x
dx + ∫
dx = ∫
2
2 x − 2x − 3
− 2x + 3
( x − 1) 2 +
( 2)
2
dx
=
1
1
⎛ x −1⎞
tan −1 ⎜
ln x 2 − 2 x + 3 +
⎟+c
2
2
⎝ 2 ⎠
(b)
2
∫
4x 2 + 1
dx = ∫
sec 2 θ
tan 2 θ + 1
= ∫ sec θ dθ
dθ
= ln sec θ + tan θ + c
= ln 1 + 4 x 2 + 2 x + c
(c)
∫
4
1
4
a
a 2 − x 2 dx = ∫ a 2 − x 2 dx + ∫ x 2 − a 2 dx
1
a
a
4
⎡
⎤
x ⎤ ⎡ x3
= ⎢a 2 x − ⎥ + ⎢ − a 2 x ⎥
3 ⎦1 ⎣ 3
⎣
⎦a
3
4a 3
65
− 5a 2 +
=
3
3
9.
(a)
2(cos
π
π
a
2b
+ i sin ) =
+
2
2
1+ i 1− i
a (1 − i ) + 2b(1 + i )
2
0 + 2i =
a
a
0 + 2i = ( + b) + (b − )i
2
2
∴b = −
a
2
b−
and
a
=2
2
Solving a = -2, b = 1
(b)
The other root is –i
P ( z ) = ( z − 1)( z 2 − i 2 ) = ( z − 1)( z 2 + 1)
= z3 − z 2 + z −1
(c)
iz 4 = −81
z4 = −
81 i
× = 81i
i i
π
i
= 81e 2 ⋅ e 2 nπi = 81e
z = 3e
(
z1 = 3e
10.
nπ π
+ )i
2 8
(−
7π
)i
8
π
( 2 nπ + ) i
2
, n = −2, − 1 , 0, 1
, n = −2,−1,0,1
, z 2 = 3e
(−
3π
)i
8
π
, z 3 = 3e
Differentiating y = v + x 2 , with respect to x,
y
dy
= 1 + 2x − 2
dx
x
dv
v + x2
+ 2x = 1 + 2x −
dx
x2
dv
v
v
= 1− 2 −1 = − 2
dx
x
x
( )i
8
, z 4 = 3e
(
5π
)i
8
dy dv
=
+ 2x .
dx dx
Correct substitution and simplification.
(Shown)
dv
dx
=− 2
v
x
dv
dx
∫ v = −∫ x 2
x −1
1
ln v = −
+c = +c
−1
x
When x = 1, y = 2, v = 1.
1
c = ln 1 − = −1
1
1
∴ ln v = − 1
x
v=e
1
−1
x
1
y − x2 = e x
y = x +e
2
11
−1
1
−1
x
→
AQ = q − a
= λ a + (1 − λ)b − a
= (1 − λ)( b − a)
→
→ →
= (1 − λ)AB
⇒ AQ // AB and having a common point A.
Hence, the vector q is collinear with points A and B.
⎛4⎞
⎛1⎞
⎜ ⎟
⎟
⎜
Given a = ⎜ 1 ⎟ and b = ⎜−2⎟
⎝ 0⎠
⎝ 6⎠
Given also angle between OA and OQ is 60°
→
|OA | = 2
1
⎛4⎞
⎛ 4−3λ ⎞
⎝ 0⎠
⎝ 6⎠
⎝ 6−6λ⎠
⎛ ⎞
→
⎟
⎜ ⎟ ⎜
⎜ ⎟
OQ = λ a + (1 − λ)b = λ ⎜ 1 ⎟ + (1 − λ)⎜−2⎟ = ⎜ 3λ−2 ⎟
→
|OQ | =
2
2
(4−3λ) +(3λ−2)2+(6−6λ) =
2
54λ −108λ + 56
→ →
→ →
We’ve OA .OQ = |OA ||OQ | cos 60°
⎛ 1 ⎞ ⎛ 4−3λ ⎞
1
⎟
⎜ ⎟ ⎜
2
⇒ ⎜ 1 ⎟ . ⎜ 3λ−2 ⎟ = 2 54λ −108λ + 56 (2 )
⎝ 0⎠ ⎝ 6−6λ⎠
2
⇒ 4 = 108λ −216λ + 112
2
⇒ 42 = 108λ −216λ + 112
2
⇒ 0 = 108λ −216λ + 96
⇒ (3λ − 4)(3λ − 2) = 0
4
2
⇒ λ = 3 or λ = 3
12.
(a) i ( z + 2i ) = z − 2
z − (0 − 2i ) = z − (2 + 0i )
2
–2
x
–2
z − (−2) = 2 sin
12.
π
4
=
2
= 2
2
OP
|z|
(b) (i) OQ = |1/z*|
|z|
= |1/z|
= |z|2
So OP : OQ = |z|2 : 1
1
(ii) w = z + z*
1
2(cos θ − i sin θ)
1
1
cos θ + i sin θ
= 2(cos θ + i sin θ) + 2
×
(cos θ − i sin θ) cos θ + i sin θ
1 cos θ + i sin θ
= 2(cos θ + i sin θ) + 2
(cos2 θ + sin2 θ)
5
= 2 (cos θ + i sin θ)
5
5
Re(w) = 2 cos θ, Im(w) = 2 sin θ
= 2(cos θ + i sin θ) +