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1
Paper 1
1. The relation in the given graph can be representedusing the following arrow diagram.
Based on the above arrow diagram,(a) the object of 40 is 3,(b) the type of the relation is many-to-many
relation.
2.
(a) The above relation is a many-to-one relation.(b) The function which represents the above
relation is f(x) = x2.
3. f 2 (x) = ff (x)= f (px + q)= p (px + q) + q= p2 x + pq + q
It is given that f 2 (x) = 4x + 9By comparison,p2 = 4 pq + q = 9p = – 2 –2q + q = 9
–q = 9q = –9
16
9
4
– 4
– 3
– 2
2
3
4
1
2
3
4
10
20
30
40
A B
4. (a) gf : x → x2 + 6x + 2gf (x) = x2 + 6x + 2
g(x + 4) = x2 + 6x + 2
Let x + 4 = ux = u – 4
g(u) = (u – 4)2 + 6(u – 4) + 2= u2 – 8u + 16 + 6u – 24 + 2= u2 – 2u – 6
∴ g(x) = x2 – 2x – 6
(b) fg(4) = f [42 – 2(4) – 6]= f(2)= 2 + 4= 6
5. Let g–1(x) = yg(y) = x
3y + k = x
y =
y = x –
∴ g–1(x) = x –
It is given that g–1(x) = mx –
Hence, by comparison,
m = and – = – ⇒ k = 52
56
k3
13
56
k3
13
k3
13
x – k3
SPM ZOOM–INForm 4: Chapter 1 Functions
The questionrequires p < 0.
2
Paper 2
1. (a) f : x →
f(x) =
Let f –1(x) = yf(y) = x
= x
hy = x (y – 3)hy = xy – 3x3x = xy – hy3x = y(x – h)
y =
∴ f –1 (x) =
But it is given that f –1(x) = , x ≠ 2.
Hence, by comparison, h = 2 and k = 3.
(b) gf –1(x) = g[f –1(x)]= g� �=
=
gf –1(x) = –5x
= –5x
x – 2 = –15x2
15x2 + x – 2 = 0(3x – 1)(5x + 2) = 0
x = or – 25
13
x – 23x
x – 23x
13x
x – 2)(
3xx – 2
kxx – 2
3xx – h
3xx – h
hyy – 3
hxx – 3
hxx – 3
2. (a) Let f –1(x) = yf(y) = x
– 2 = x
= x + 2
y = 2(x + 2)y = 2x + 4
∴ f –1(x) = 2x + 4∴ f –1(3) = 2(3) + 4 = 10
(b) f –1g(x) = f –1[g(x)]= f –1(3x + k)= 2(3x + k) + 4= 6x + 2k + 4
But it is given thatf –1g : x → 6x – 4f –1g (x) = 6x – 4
Hence, by comparison,2k + 4 = –4
2k = –8k = –4
(c) hf(x) : x → 9x – 3h[f(x)] = 9x – 3
h� – 2� = 9x – 3
Let – 2 = u
= u + 2
x = 2u + 4
h(u) = 9(2u + 4) – 3= 18u + 33
∴ h : x → 18x + 33
x2
x2
x2
y2
y2
3
SPM Zoom-InForm 4: Chapter 2 Quadratic Equations
Paper 1
1. 12x2 – 5x(2x – 1) = 2(3x + 2)12x2 – 10x2 + 5x = 6x + 4
12x2 – 10x2 + 5x – 6x – 4 = 02x2 – x – 4 = 0
2. Sum of roots = – + �– � = –
Product of roots = �– ��– � =
The required quadratic equation is
x2 + x + = 025
1915
25
35
23
1915
35
23
–b � b2 – 4ac2a
–(–1) � (–1)2 – 4(2)(–4)2(2)
1 � 334
x =
x =
x =
x = 1.6861 or –1.1861
4. x2 + 2x – 1 + k(2x + k) = 0x2 + 2x – 1 + 2kx + k2= 0x2 + 2x + 2kx + k2 – 1= 0x2 + (2 + 2k)x + k2 – 1 = 0
a = 1, b = 2 + 2k, c = k2 – 1
If a quadratic equation has two real and distinctroots, then b2 – 4ac > 0.
b2 – 4ac > 0(2 + 2k)2 – 4(1)(k2 – 1) > 04 + 8k + 4k2 – 4k2 + 4 > 0
8k + 8 > 08k > –8k > –1
5. 3(x2 + 4) = 2mx3x2 + 12 = 2mx
3x2 – 2mx + 12 = 0
a = 3, b = –2m, c = 12
If a quadratic equation has equal roots, then b2 – 4ac = 0.
b2 – 4ac = 0(–2m)2 – 4(3)(12) = 0
4m2 – 144 = 04m2 = 144m2 = 36m = ±6
15x2 + 19x + 6 = 03. 3x2 + 4p + 2x = 0
3x2 + 2x + 4p = 0
a = 3, b = 2, c = 4p
If a quadratic equation does not have real roots,then b2 – 4ac < 0.
b2 – 4ac < 022 – 4(3)(4p) < 0
4 – 48p < 0–48p < –4
p >
p > 112
–4–48
x2 – (sum of roots)x + (product of roots) = 0
4
6. x2 + 2x – 8 = 0a = 1, b = 2, c = –8
The roots are p and q.
Sum of roots = –
p + q = –
p + q = –2
Product of roots =
pq = –
pq = –8
The new roots are 2p and 2q.
Sum of new roots= 2p + 2q= 2(p + q)= 2(–2)= –4
Product of new roots= (2p)(2q)= 4pq= 4(–8)= –32
The quadratic equation that has the roots 2p and2q is x2 + 4x – 32 = 0.
81
ca
21
ba
7. x2 – (k + 2)x + 2k = 0a = 1, b = –(k + 2), c = 2k
If one of the roots is α, then the other root is 2α.
Sum of roots = –
α + 2α = –� �3α = k + 2
α = …
Product of roots =
2α2 =
α2 = k …
Substituting into :
� �2= k
= k
(k + 2)2 = 9kk2 + 4k + 4 = 9kk2 – 5k + 4 = 0
(k – 1)(k – 4) = 0k = 1 or 4
(k + 2)2
9
k + 23
21
2
2k1
ca
1k + 23
–(k + 2)1
ba
5
Paper 2
1. (2x – 1)(x + 3) = 2x – 3 – k2x2 + 6x – x – 3 = 2x – 3 – k
2x2 + 3x + k = 0a = 2, b = 3, c = k
The roots are –2 and p.
Sum of roots = –
–2 + p = –
–p = – + 2
p =
Product of roots =
–2p =
–2� � =
k = –2
2. 2x2 + (3 – k)x + 8m = 0a = 2, b = 3 – k, c = 8m
The roots are m and 2m.
Sum of roots = –
m + 2m = –
6m = k – 3 …
Product of roots =
m(2m) =
2m2 = 4mm2 = 2m
m2 – 2m = 0m(m – 2) = 0
m = 0 or 2m = 0 is not accepted.
∴ m = 2
8m2
ca
1
3 – k2
ba
k2
12
k2
ca
12
32
32
ba
From :When m = 2,
6(2) = k – 3k = 12 + 3k = 15
3. (a) 2x2 + px + q = 0a = 2, b = p, c = q
The roots are – and 2.
Sum of roots = –
– + 2 = –
= –
p = –1
Product of roots =
– � 2 =
q = –6
(b) 2x2 – x – 6 = k2x2 – x – 6 – k = 0a = 2, b = –1, c = –6 – k
If the quadratic equation does not have realroots, then b2 – 4ac < 0.
When b2 – 4ac < 0,(–1)2 – 4(2)(–6 – k) < 0
1 + 48 + 8k < 08k < –49
k < –
k < –6 18
498
q2
32
ca
p2
12
p2
32
ba
32
1
6
SPM ZOOM–INForm 4: Chapter 3 Quadratic Functions
Paper 1
1. f(x) = 2x2 + 8x + 6= 2(x2 + 4x + 3)
= 2[x2 + 4x + � �2
– � �2
+ 3]= 2(x2 + 4x + 22 – 22 + 3)= 2[(x + 2)2 – 1]= 2(x + 2)2 – 2
∴ a = 2, p = 2, q = –2
2. From f(x) = – (x – 4)2 + h, we can state that thecoordinates of the maximum point are (4, h). But itis given that the coordinates of the maximum pointare (k, 9). Hence, by comparison,(a) k = 4(b) h = 9(c) The equation of the tangent to the curve at its
maximum point is y = 9.
3. (a) y = (x + m)2 + nThe axis of symmetry is x = –m.But it is given that the axis of symmetry is x = 1.∴ m = –1
When m = –1, y = (x – 1)2 + nSince the y-intercept is 3, the point is (0, 3).∴ 3 = (0 – 1)2 + n
n = 2
(b) When m = –1 and n = 2,y = (x – 1)2 + 2Hence, the minimum point is (1, 2).
4. (2 + p)(6 – p) < 712 + 4p – p2 – 7 < 0
–p2 + 4p + 5 < 0p2 – 4p – 5 > 0
(p + 1)(p – 5) > 0
42
42 Hence, the required range of values of p is
p < –1 or p > 5.
5. 3x2 + hx + 27 = 0a = 3, b = h, c = 27
If a quadratic equation does not have real roots,b2 – 4ac < 0
h2 – 4(3) (27) < 0h2 – 324 < 0
(h + 18)(h – 18) < 0
Hence, the required range of values of h is –18 < h < 18.
6. g(x) = (2 – 3k)x2 + (4 – k)x + 2a = 2 – 3k, b = 4 – k, c = 2
If a quadratic curve intersects the x-axis at twodistinct points, then
b2 – 4ac > 0(4 – k)2 – 4(2 – 3k)(2) > 0
16 – 8k + k2 – 16 + 24k > 0k2 + 16k > 0
k(k + 16) > 0
Hence, the required range of values of k isk < –16 or k > 0.
k–16 0
h–18 18
p–1 5
7
Paper 2
(a) f(x) = 2x2 + 10x + k
= 2 �x2 + 5x + �= 2 �x2 + 5x + – + �= 2 [�x + �2
– + ]= 2 �x + �2
– + k
(b) (i) Minimum value = 32
– + k = 32
k =
(ii) b2 – 4ac < 0102 – 4(2)(k) < 0
100 – 8k < 0– 8k < –100
k >
k >
(c) Minimum point is �–2 , 32�.2. (a) g(x) = –2x2 + px – 12 = –2(x + q)2 – 4
–2x2 + px – 12 = –2(x2 + 2qx + q2) – 4= –2x2 – 4qx – 2q2 – 4
By comparison,p = – 4q … and –12 = –2q2 – 4
–2q2 = –8q2 = 4q = ±2
From :When q = 2, p = –4(2) = –8 (Not accepted)When q = –2, p = –4(–2) = 8 (Accepted) because p > 0 and q < 0)
1
1
12
252
–100–8
892
252
252
52
k2
254
52
k2
254
254
k2
� � 5�2
= 254
12
(b) g(x) = –2x2 + 8x – 12 = –2(x – 2)2 – 4
The maximum point is (2, –4).When x = 0, y = –12 ∴ (0, –12)The graph of the function g(x) is as shown below.
3. y = h – 2x…
y2 + xy + 8 = 0 …
Substituting into :(h – 2x)2 + x(h – 2x) + 8 = 0h2 – 4hx + 4x2 + hx – 2x2 + 8 = 02x2 – 3hx + h2 + 8 = 0
a = 2, b = –3h, c = h2 + 8
If a straight line does not meet a curve, thenb2 – 4ac < 0
(–3h)2 – 4(2) (h2 + 8) < 09h2 –8h2 – 64 < 0
h2 – 64 < 0(h + 8)(h – 8) < 0
Hence, the required range of values of h is –8 < h < 8.
h–8 8
21
2
1
y
O
–12
(2, –4) x
8
SPM ZOOM–INForm 4: Chapter 4 Simultaneous Equations
Paper 2
1. 2x – 3y = 2 …
x2 – xy + y2 = 4 …
From :
x = …
Substituting into :
� �2
– y� � + y2 – 4 = 0
2– + y2 – 4 = 0
(2 + 3y)2 – 2y(2 + 3y) + 4y2 – 16 = 04 + 12y + 9y2 – 4y – 6y2 + 4y2 – 16 = 0
7y2 + 8y – 12 = 0(7y – 6)(y + 2) = 0
y = or –2
From :
When y = –2, x = = –2
Hence, the points of intersection are �2 , �and (–2, –2).
2. 4x + y = 2 …
x2 + x – y = 2 …
From : y = 2 – 4x…
Substituting into ,x2 + x – (2 – 4x) = 2
x2 + 5x – 4 = 0
= 0.70156 or –5.70156
= x 2(1)
52 – 4(1)(–4) –5 ±
= 2
41 –5 ±
23
31
2
1
67
27
2 + 3(–2)2
)(When y = , x = =
67
67
2 + 3 1672
3
67
y(2 + 3y)2
(2 + 3y)4
2 + 3y2
2 + 3y2
23
32 + 3y2
1
2
1
From ,When x = 0.70156, y = 2 – 4(0.70156)
= –0.80624
When x = –5.70156, y = 2 – 4(–5.70156)= 24.80624
Hence, the solutions are x = 0.70156, y = –0.80624 or x = –5.70156, y = 24.80624 (correct to five decimal places).
3. (a) Since (16, m) is a point of intersection of
y = x – 2 and y2 + ky – x – 4 = 0, then
x = 16 and y = m satisfy both the equations.
Therefore,
m = (16) – 2 = 2 and
m2 + km – 16 – 4 = 022 + k(2) – 16 – 4 = 0
2k = 16k = 8
(b) When k = 8,
y = x – 2 …
y2 + 8y – x – 4 = 0 …
From :4y = x – 8x = 4y + 8 …
Substituting into ,y2 + 8y – (4y + 8) – 4 = 0
y2 + 8y – 4y – 8 – 4 = 0y2 + 4y – 12 = 0
(y – 2)(y + 6) = 0y = 2 or –6
From :When y = 2, x = 4(2) + 8 = 16When y = –6, x = 4(–6) + 8 = –16
Hence, the other point of intersection,other than (16, 2), is (–16, –6).
3
23
3
1
2
114
14
14
3
9
SPM ZOOM–INForm 4: Chapter 5 Indices and Logarithms
Paper 1
1. 2 x + 3 + 2x + 16 (2x – 1)
= 2x.23 + 2x + 16� �= 8(2x) + 2x + 8(2x)= (8 + 1 + 8)( 2x)= 17(2x)
2. 3x + 3 – 3x + 2 = 63x (33) – 3x (32) = 6
27(3x) – 9(3x) = 6(27 – 9)(3x) = 6
18(3x) = 6
3x =
3x =
3x = 3–1
x = –1
3. m = 3a n = 3b
log3 m = a log3 n = b
log3 � �= log3 m + log3 n4 – log3 27= log3 m + 4 log3 n – log3 33
= a + 4b – 3
mn4
27
13
618
2x
2
4. 5x = 32x – 1
lg 5x = lg 32x –1
x lg 5 = (2x – 1) lg 3x lg 5 = 2x lg 3 – lg 3
x lg 5 – 2x lg 3 = – lg 3x(lg 5 – 2lg 3) = –lg 3
x =
x = 1.87
5. log10 (p + 3) = 1 + log10 plog10 (p + 3) – log10 p = 1
log10 � � = 1
= 101
p + 3 = 10p9p = 3
p =
6. log2 y – log8 x = 1
log2 y – = 1
log2 y – = 1
3 log2 y – log2 x = 3log2 y3 – log2 x = 3
log2 � � = 3
= 23
y3 = 8x
x = y3
8
y3
x
y3
x
log2 x3
log2 xlog2 8
13
p + 3p
p + 3p
–lg 3lg 5 – 2 lg 3
log2 8 = log2 23 = 3
10
Paper 1
1. Let point A be (0, k).AB = 10
= 10
64 + k2 – 14k + 49 = 102
k2 – 14k + 13 = 0(k – 1)(k – 13) = 0
k = 1 or 13
Based on the diagram, k < 7.∴ k = 1∴ A(0, 1)
2. (a) x + 2y + 6 = 0x + 2y = –6
+ =
+ = 1
(b) mMN = – � � = –
Therefore, the gradient of the perpendicularline is 2.
Hence, the equation of the straight line whichpasses through the point N and isperpendicular to the straight line MN is y = 2x – 3.
3. – = 1
At point P (on the x-axis), y = 0.
– = 1 ⇒ x = 4
∴ P is point (4, 0).
At point Q (on the y-axis), x = 0.
– = 1 ⇒ y = –3
∴ Q is point (0, –3).
y3
04
03
x4
y3
x4
12
–3–6
y(–3)
x(–6)
–6–6
2y(–6)
x(–6)
(0 – 8)2 + (k – 7)2
Hence, the area of ∆PQR
=
= |–12 – (–6 + 20)|
= |–26|
= � 26
= 13 units2
4. (a) 2y = 3x – 12At point L (on the x-axis), y = 02(0) = 3x – 12
x = 4∴ L (4, 0)
At point N (on the y-axis), x = 0.2y = 3(0) – 12y = –6
∴ N (0, –6)
∴ M = � , � = (2, –3)
(b) mLN = =
∴ Gradient of perpendicular line = –
Hence, the equation of the perpendicular line is
y – y1 = m(x – x1)
y – (–3) = – (x – 2)
3(y + 3) = –2(x – 2)3y + 9 = –2x + 4
3y = –2x – 5
5. PA = PB
= (x – 1)2 + (y – 2)2 = (x – 0)2 + (y – 3)2
x2 – 2x + 1 + y2 – 4y + 4 = x2 + y2 – 6y + 9–2x – 4y + 5 = –6y + 9–2x + 2y – 4 = 0
–x + y – 2 = 0y = x + 2
(x – 0)2 + (y – 3)2 (x – 1)2 + (y – 2)2
23
23
32
–6 – 00 – 4
0 + (–6)2
4 + 02
12
12
12
12
40
0–3
25
40
SPM ZOOM–INForm 4: Chapter 6 Coordinate Geometry
Intercept form:
+ = 1 yb
xa
m = – y-interceptx-intercept
11
Paper 2
1. (a) y – 3x + 6 = 0At point B (x-axis), y = 0.0 – 3x + 6 = 0 ⇒ x = 2∴ B is point (2, 0).
y – 3x + 6 = 0At point C (y-axis), x = 0.y – 3(0) + 6 = 0 ⇒ y = –6∴ C is point (0, –6).
y = 3x – 6mBC = 3
∴mAC = –
Let A(k, 0).
∴ mAC = –
= –
–k = 18k = –18
∴ A is point (–18, 0).
(b) Let D (p, q).Midpoint of BD = Midpoint of AC
� , � = � , �� , � = (–9, –3)
Equating the x-coordinates,
= –9
p = –20
Equating the y-coordinates,
= –3
q = –6∴ D is point (–20, –6).
q2
2 + p2
q2
2 + p2
0 + (–6)2
–18 + 02
0 + q2
2 + p2
13
0 – (–6)k – 0
13
13
(c) A(–18, 0), B(2, 0), C(0, –6), D(–20, –6)Area of ABCD
=
= |–12 – (120 + 108)|
= � 240
= 120 units2
2. (a) (i) y – 3x + 6 = 0At point P (on the y-axis), x = 0.y – 3(0) + 6 = 0 ⇒ y = –6∴ P is point (0, –6).
(ii) The coordinates of point S are
� , � = (3, 3)
(b) Area of ∆QRS = 48 units2
= 48
15k + 21 – (45 + 3k) = 9612k – 24 = 96
12k = 120k = 10
(c) S(3, 3), Q(10, 0), T(x, y)TS : TQ = 2 : 3
=
3TS = 2TQ9(TS)2 = 4(TQ)2
9[(x – 3)2 + (y – 3)2] = 4[(x – 10)2 + (y – 0)2]9(x2 – 6x + 9 + y2 – 6y + 9) =
4(x2 – 20x + 100 + y2)9x2 – 54x + 81 + 9y2 – 54y + 81 =
4x2 – 80x + 400 + 4y2
5x2 + 26x + 5y2 – 54y – 238 = 0
23
TSTQ
12
k0
715
33
k0
4(–6) + 3(15)3 + 4
4(0) + 3(7)3 + 4
12
12
12
–180
20
0–6
–20–6
–180
12
SPM ZOOM–INForm 4: Chapter 7 Statistics
Paper 1
1. After the given score are arranged in ascending order, we have
6 6 6 k k 9
Since the mode is 6, then k ≠ 9.
For 7 to be the median, k = 8, as shown below.
6 6 6 8 8 9
After two new scores, 7 and 10, are added to theoriginal scores, the mean of the eight scores
=
= 7.5
2. (a) x– =
27 =
n =
n = 7
(b)
3.
∑x2 _� = —— – ( x )2
n
5278=
=
——— – 272
7
25= 5
18927
189n
∑ xn
6 + 6 + 6 + 8 + 8 + 9 + 7 + 108
Median = 7
x f fx fx2
30 3 90 2700
32 5 160 5120
34 2 68 2312
Sum 10 318 10 132
4.
∑ fx
∑ f
2
2 � =
10 132= –
–
10
= 1.96
318
10
∑ fx
∑ f
2
2
����
Number 1 k 6 k + 3 11 13
Frequency 2 2 1 3 1 1
(a) 1 < k < 6 6 < k + 3 < 11k = 2, 3, 4, 5 3 < k < 8
k = 4, 5, 6, 7
Taking into consideration both cases,k = 4 or 5
(b) 1, 1, 4, 4, 6, 7, 7, 7, 11, 13
M Q3
∴ Q3 = 7
Paper 2
1.Mass (kg) Frequency
Cumulativefrequency
1.1 – 2.0 5 5
2.1 – 3.0 9 14
3.1 – 4.0 12 26
4.1 – 5.0 8 34
5.1 – 6.0 6 40
Frequency
2
4
6
8
10
12
0 1.05 2.05 3.05 4.05 5.05 6.05Mass (kg)(Mode)3.5
(a)
Mode = 3.5 kg
13
(b) The Q1 class is given by
The Q3 class is given by
Hence, the interquartile range= Q3 – Q1 = 4.55 – 2.61 = 1.94 kg
(c) New interquartile range= Original interquartile range= 1.94 kg
2.
3 (40)4
T 30T= 4.1 – 5.0=
3Q 4.05 +
34
8(1) = 4.55 kg
(40) – 26= � �
404
T 10T= 2.1 – 3.0=
1Q 2.05 +
404
9(1) = 2.61 kg
– 5= � �
(a) Median = 46.5
(b)
L +– F
fmc = 46.5
n2� �
39.5 +– –11
k(10) = 46.5
26 + k2� �
– –11
k(10) = 7
26 + k2� �
– 11 = 0.7k26 + k2
26 + k – 22 = 1.4k
0.4k = 4
k = 10
Marks fCumulativefrequency
20 – 29 4 4
30 – 39 7 11
40 – 49 k 11 + k
50 – 59 8 19 + k
60 – 69 5 24 + k
70 – 79 2 26 + k
Marks fMid-
fx fx2
point (x)
20 – 29 4 24.5 98.0 2401.00
30 – 39 7 34.5 241.5 8331.75
40 – 49 10 44.5 445.0 19802.50
50 – 59 8 54.5 436.0 23762.00
60 – 69 5 64.5 322.5 20801.25
70 – 79 2 74.5 149.0 11100.50
36 1692 86199
(c) (i) New median = Original median + 10= 46.5 + 10= 56.5
(ii) New variance = Original variance= 185.42
∑ fx
∑ f
2
=
86 199= –
–
36
Variance
= 185.42
1692
36
∑ fx
∑ f
2
2
����
14
SPM ZOOM–INForm 4: Chapter 8 Circular Measure
Paper 1
1.
∠BOC = π – ∠AOB – ∠COD= 3.142 – 0.9 – 0.9= 1.342 rad.
Area of sector BOC
= � 82 � 1.342
= 42.94
12
O
C
0.9 rad8 cm8 cm
π – 1.8
0.9 rad
B
DA r = sθ
2. ∠BOC = 20º = �20 � � rad
3. Area of the shaded region= Area of sector OAB – Area of sector OXY
= 38.4 – 10= 28.4 cm2
= � 82 � 1.2 – � 5 � 412
12
r2θ12
� �
rs12
OB = = 44 cm20 �
15.36
3.142180 )(
3.142180
15
Paper 2
1.
(a) MO = r – 6In ∆OMB, using Pythagoras’ theorem,MO2 + MB2 = OB2
(r – 6)2 + 82 = r2
r2 – 12r + 36 + 64 – r2 = 0–12r + 100 = 0
r = 8
(b) In ∆BOM,
sin ∠BOM =
∠BOM = 1.287 rad.∴ ∠AOB = 2 � 1.287 = 2.574 rad.
(c) Area of the shaded region
= � �8 �2
�2.574 – sin 2.574r�= 70.71 cm2
2. (a) ∠BOA = = 1.160 rad.
∴ ∠BOQ = π – 1.160 = 1.982 rad.
(b) Area of the shaded region
= r 2 (θ – sin θ)
= (10)2 (1.982 – sin 1.982r)
= 53.27 cm2
12
12
π – 0.8222
13
12
2425
sin ∠BOM =8
8
13
13
O
A
C
B
r cm(r – 6) cm
8 cm8 cm
6 cm
M
(c) Perimeter of the shaded region
= 2r sin + rθ
= 2(10) sin� �r
+ 10(1.982)
= 16.73 + 19.82= 36.55 cm
3.
(a) Since ∆ADB is inscribed in a semicircle,it is a right-angled triangle.
In ∆ADB,
cos ∠ABD =
∠ABD = 0.6435 rad
(b) ∠AOD = 2 � ∠ABD= 2 � 0.6435= 1.2870 rad
∴ Length of the arc AD= 5 � 1.2870= 6.435 cm
(c) Area of ∆ODB
= � 8 � 3
= 12 cm2
Area of sector BOC
= � 52 � 0.6435
= 8.04375 cm2
Hence, the area of the shaded region= Area of ∆ODB – Area of sector BOC= 12 – 8.04375= 3.956 cm2
12
12
810
OA
D
B
C
5 cm
3 cm
4 cm
4 cm
5 cm
1.9822
θ2
MB = AB = � 16 = 8 cm12
12
The angle at the centreis twice the angle atcircumference.
16
Paper 1
1. f (x) = = (5x – k)–2
f ′(x) = –2(5x – k)–3 (5)
=
f ′(1) = 10
= 10
(5 – k)3 = –15 – k = – 1
k = 6
2. y = (x + 1) (2x – 1)2
= (x + 1) [2 (2x – 1)1 (2)] + (2x – 1)2 (1)
= (2x – 1)[4(x + 1) + (2x – 1)]= (2x – 1)(6x + 3)
3. y = 2x3 – 4x + 5
= 6x2 – 4
Gradient at the point (–1, 7)= 6 (–1)2 – 4= 2
Equation of the tangent isy – 7 = 2[x – (–1)]y – 7 = 2(x + 1)y – 7 = 2x + 2
y = 2x + 9
dydx
dydx
–10[5(1) – k]3
–10(5x – k)3
1(5x – k)2
4. z = xyz = x(30 – x)z = 30x – x2
= 30 – 2x
When z has a stationary value,
= 0
30 – 2x = 0x = 15
= –2 (negative)
Hence, the maximum value of z= 30(15) – 152
= 225
5. y = = (2x – 5)–3
= –3 (2x – 5)–4 (2) =
≈
δy ≈ � δy
= � (3.01 – 3)
= � 0.01
= – 0.06
6. A = 2πr 2 + 2πrh= 2πr 2 + 2πr(3r)= 8πr 2
= �
= 16πr � 0.1= 16π (5) � 0.1= 8π cm2 s–1
drdt
dAdr
dAdt
– 6[2(3) – 5]4
– 6(2x – 5)4
dydx
dydx
δyδx
– 6(2x – 5)4
dydx
1(2x – 5)3
d 2zdx2
dzdx
dzdx
SPM ZOOM–INForm 4: Chapter 9 Differentiation
17
Paper 2
1. y = – = x–2 – x–3
= –2x – 3 + 3x – 4 = – +
= 6x– 4 – 12x– 5 = –
x4 � + � + x 2y + 5 = 0
x4 �– + + – � +
x 2 � – � + 5 = 0
–2x + 3 + 6 – + 1 – + 5 = 0
–2x + 15 – = 0
–2x2 + 15x – 13 = 02x2 – 15x + 13 = 0
(2x – 13)(x – 1) = 0
x = or 1
2. (a) y = px3 + kx
= 3px2 + k
At (1, 1), x = 1 and m = = –5.
∴ 3px2 + k = –53p(1)2 + k = –5
3p + k = –5 …
The curve passes through point (1, 1).∴ 1 = p(1)3 + k(1)p + k = 1 …
– : 2p = –6 ⇒ p = –3
From : –3 + k = 1 ⇒ k = 42
21
2
1
dydx
dydx
132
13x
1x
12x
1x 3
1x 2
12x 5
6x 4
3x 4
2x 3
d 2ydx 2
dydx
12x5
6x4
d 2ydx2
3x4
2x3
dydx
1x 3
1x 2
(b) When p = –3 and k = 4,y = –3x3 + 4x
= –9x 2 + 4
= –18x
At turning points,
= 0
–9x2 + 4 = 0
x2 =
x = ±
When x = ,
y = –3� �3+ 4� � = 1
= – 18� � = –12 (< 0)
∴ � , 1 � is a turning point which is
a maximum.
When x = – ,
y = –3�– �3+ 4�– � = –1
= –18 �– � = 12 (> 0)
∴ �– , –1 � is a turning point which is
a minimum.
79
23
23
d 2ydx 2
79
23
23
23
79
23
23
d 2ydx 2
79
23
23
23
23
49
dydx
d 2ydx2
dydx
18
3.
(a) Volume of the cuboid = 5832 cm3
(6x)(6x)(y) = 583236x2y = 5832
x2y = 162
y =
L = Area of ABCD + 4 (Area of GBCH) + 4 (Area of VGH)
L = (6x)2 + 4(6xy) + 4 � (6x)(5x)
L = 36x2 + 24xy + 60x2
L = 96x2 + 24xy
L = 96x2 + 24x � �L = 96x2 + (shown)3888
x
162x2
12
162x2
A B
C
H
V
GF
E
D
y m
4x m5x m
3x m
6x m
6x m
(b) L = 96x2 + = 96x2 + 3888x–1
= 192x – 3888x–2 = 192x –
At stationary point,
= 0
192x – = 0
192 x =
x 3 =
x 3 = 20.25
x = 2.73
= 192 + 7776x–3 = 192 + (> 0)
∴ L is a minimum.
4. y = = h(1 + 2x)–2
= –2h(1 + 2x)–3 (2) = –
δy = � δx
– = – � c
– = – � c
– = –
h = �
h = 18
274
83
4hc27
8c3
4h[1 + 2(1)]3
8c3
4h(1 + 2x)3
8c3
dydx
4h(1 + 2x)3
dydx
h(1 + 2x)2
7776x3
d 2Ldx 2
3888192
3888x 2
3888x 2
dLdx
3888x2
dLdx
3888x
19
Paper 2
1. (a) ∠UST = 180º – 65º = 115º∠SUT = 180º – 43º – 115º = 22º
In ∆UST, using the sine rule,
=
US = � sin 43º
= 16.385 cm
(b) In ∆USR, using the cosine rule,UR2 = 72 + 16.3852 – 2(7)(16.385)cos 65ºUR2 = 220.5238UR = 14.85 cm
(c) Area of ∆RSV = 41.36 cm2
� 7 � 12 � sin ∠RSV = 41.36
sin ∠RSV = 0.98476Basic ∠ = 79.98º
∠RSV = 180º– 79.98º= 100.02º
In ∆RSV, using the cosine rule,RV 2 = 72 + 122 – 2(7)(12)cos 100.02ºRV 2 = 222.23064RV = 14.91 cm
12 cm
7 cm SR
V
100.02°
12
U
R S T
22°
65°115°
16.385 cm
43°7 cm 9 cm
9sin 22º
9sin 22º
USsin 43º
2. (a) In ∆PQS, using the sine rule,
=
sin ∠QSP = � 8
sin ∠QSP = 0.65552∠QSP = 40.96º
∴ ∠PQS = 180º – 35º – 40.96º= 104.04º
Hence, the area of ∆PQS
= � 8 � 7 � sin 104.04º
= 27.16 cm2
(b) This problem involves the ambiguous case of sine rule. The sketch of ∆QRS1 is as shown below.
In ∆QRS, using the sine rule,
=
sin ∠QSR = � 10
sin ∠QSR = 0.974283
Basic ∠ = 76.98º∴ ∠QSR = 76.98º or ∠QS1R = 103.02º
In ∆QS1R,∠RQS1 = 180º – 43º – 103.02º = 33.98º
In ∆QS1R, using the sine rule,
=
=
RS1 = � sin 33.98º
= 5.737 cm
10sin 103.02º
10sin 103.02º
RS1
sin 33.98º
10sin ∠RS1Q
RS1
sin ∠RQS1
sin 43º7
sin 43º7
sin ∠QSR10
43°
Q
S1R S
10 cm7 cm 7 cm
12
sin 35º7
sin 35º7
sin ∠QSP8
SPM ZOOM–INForm 4: Chapter 10 Solution of Triangles
20
Paper 2
1. (a) � 100 = 120
x = 480
y = � 100 = 105
� 100 = 110
z = 600
(b) –I = 115
= 115
= 115
15 200 + 130m = 16 100 + 115m15m = 900
m = 60
(c) –I 2006 (based on 2002)
= � –I 2004
= � 115
= 143.75
(d) Total yearly cost in 2006
= � 5 500 000
= RM7 906 250
2.
143.75100
125100
100 + 25100
15 200 + 130m140 + m
(120 � 20) + 130m + (105 � 80) + (110 � 40)20 + m + 80 + 40
660z
525500
x400
SPM ZOOM–INForm 4: Chapter 11 Index Numbers
I = � 100P2004
P2002
Health I2004 (based I2006 (basedWeightage
supplement on 2002) on 2004)
A 115 150 3
B 120 130 2
C 105 120 x
(a) Supplement A
I2004 (based on 2002) = 115
� 100 = 115
� 100 = 115
P2002 =
P2002 = RM60.00
(b) Supplement BI2006 (based on 2002)
= � 100
= � � 100
= � � 100
= 156
(c)–I 2004 (based on 2002) = 111
= 111
= 111
585 + 105x = 555 + 111x30 = 6x
x = 5(d)
–I 2006 (based on 2004)
=
=
= 131
Thus, � 100 = 131
� 100 = 131
P2006 =
P2006 = RM393
131 � 300100
P2006
300
P2006
P2004
131010
(150 � 3) + (130 � 2) + (120 � 5)3 + 2 + 5
585 + 105x5 + x
(115 � 3) + (120 � 2) + 105x3 + 2 + x
120100
130100
P2004
P2002
P2006
P2004
P2006
P2002
69 � 100115
69P2002
P2004
P2002
21
Paper 1
1. (a) T6 = 38a + 5d = 38
a + 5(7) = 38a = 3
(b) S9 – S3
= [2(3) + 8(7)] – [2(3) + 2(7)]
= 279 – 30= 249
2. (a) T2 – T1 = T3 – T2
2h – 1 – (h – 2) = 4h – 7 – (2h – 1)h + 1 = 2h – 6
h = 7
(b) When h = 7, the arithmetic progression is 5,13, 21, … with a = 5 and d = 8.
S8 – S3
= [2(5) + 7(8)] – [2(5) + 2(8)]
= 264 – 39= 225
3. =
=
(x= (x – 4)(9x + 4)
x2 + 4x + 4 = 9x2 – 32x – 168x2 – 36x – 20 = 0
2x2 – 9x – 5 = 0(2x + 1)(x – 5) = 0
x = – or 512
x – 4x + 2
x + 29x + 4
T3
T2
T2
T1
32
82
32
92
4. T3 – T2 = 3ar 2 – ar = 34r 2 – 4r = 3
4r 2 – 4r – 3 = 0(2r + 1)(2r – 3) = 0
r = – or
5. 0.242424 …= 0.24 + 0.0024 + 0.000024 + …
=
=
=
=
6. The numbers of bacteria form a geometricprogression 3, 6, 12, …
The number of bacteria after 50 seconds= T11 = ar10 = 3(210) = 3072
Paper 2
1. (a) The volumes of cylinders are πr 2h, πr 2 (h + 1), πr 2 (h + 2), …
T2 – T1 = πr 2 (h + 1) – πr 2h= πr 2h + πr 2 – πr 2h= πr2
T3 – T2 = πr 2 (h + 2) – πr 2 (h + 1)= πr 2h + 2πr 2 – πr 2h – πr 2
= πr2
Since T2 – T1 = T3 – T2 = πr 2, the volumes of cylinders form an arithmetic progression with a common difference of πr2.
833
2499
0.240.99
0.241 – 0.01
32
12
SPM ZOOM–INForm 5: Chapter 1 Progressions
S∞ = a
1 – r
22
(b) a = πr 2h, d = πr 2
T4 = 32πa + 3d = 32π
πr 2h + 3πr 2 = 32πr 2h + 3r 2 = 32
r 2 (h + 3) = 32 …
S4 = 104π
(2a + 3d) = 104π
4a + 6d = 104π4πr 2h + 6πr 2 = 104π
2r 2h + 3r 2 = 52r 2 (2h + 3) = 52 …
: =
=
16h + 24 = 13h + 393h = 15h = 5
From :r 2 (5 + 3) = 32
r 2 = 4r = 2
1
138
2h + 3h + 3
5232
r 2 (2h + 3)r 2 (h + 3)
2
1
2
42
1
2. (a) S2 = 150T1 + T2 = 150a + ar = 150
a (1 + r) = 150 …
T3 – T2 = 45ar 2 – ar = 45
ar (r – 1) = 45 …
: =
=
3 + 3r = 10r2 – 10r10r 2 – 13r – 3 = 0
(2r – 3)(5r + 1) = 0
r = or –
(b) For the sum to infinity to exist, – 1 < r < 1.
Thus, r = is not accepted.
Therefore, r = –
From :
a �1 – � = 150
a = 187
��a
1 – r= ∴ S∞ 156 1
4=
21187
1 – –=
51
12
15
1
15
32
15
32
103
1 + rr (r – 1)
15045
a (1 + r)ar (r – 1)
1
2
2
1
23
SPM ZOOM–INForm 5: Chapter 2 Linear Law
Paper 1
1. y = + qx
= + q
= 2 ( ) + q
2.
3. – ax =
xy2 – ax3 = bxy2 = ax3 + b
(–1, 10): 10 = a(–1) + b …
(5, –2): –2 = a(5) + b …
– : 12 = –6aa = –2
From : –2 = –2(5) + bb = 8
2
21
2
1
bx2
y2
x
= py
xk
lg = lg py
xk
lg y – lg k = lg px
lg y – x lg k = lg p
lg y = x lg k + lg p
∴ Y = lg y, X = x, m = lg k, c = lg p
1x2
(1, 5) 5 = 2(1) + qq = 3
q = 3
p = 2(3) + 3p = 9
yx
1x2
(3, p)
yx
1x2
yx
2x2
yx
2x
Paper 2
1. (a)
(b)
y = hxk
log10 y = log10 (hxk)
log10 y = log10 h + log10 xk12
log10 y = log10 h + klog10 x
log10 y = 2log10 h + 2klog10 x
log10 y = 2k log10 x + 2 log10 h
12
x3 xy2
x 1.5 2.0 2.5 3.0 3.5
y 142 338 660 1348 1995
log10 x 0.18 0.30 0.40 0.48 0.54
log10 y 2.15 2.53 2.82 3.13 3.30
log10 xO
log10 y
0.1
0.5
1.0
1.5
2.0
2.5
3.0
3.5
1.55
0.2 0.3
3.35 – 1.75= 1.6
0.55 – 0.06 = 0.49
Graph of log10 y against log10 x
0.4 0.5 0.6
The graph of log10 y against log10 x is as shown below.
24
(c) 2k = Gradient
2k =
2k = 3.2653k = 1.63
2 log10 h = Y–intercept2 log10 h = 1.55
log10 h = 0.775h = 5.96
2. (a)
3.35 – 1.750.55 – 0.06
x 0.1 0.3 0.4 0.5 0.7 0.8
y 0.78 0.60 0.54 0.50 0.44 0.42
1.64 2.78 3.43 4.00 5.17 5.671y2
xO 0.1
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
1.1
4.6
0.2 0.3
4.6 – 2 = 2.6
0.6 – 0.16 = 0.44
0.4 0.5 0.6 0.7 0.8
Graph of against xy21y2
1
(b) (i)
=
= x +
Gradient =
= 5.91
q = 0.17
Y-intercept = 1.1
= 1.1
= 1.1
p = 0.19
(ii) When x = 0.6, from the graph,
= 4.6
y2 = 0.2174y = 0.47
1y2
p0.17
pq
1q
4.6 – 20.6 – 0.16
pq
1q
1y2
x + pq
1y2
=1y
x + pq
Squaring both sides.
25
SPM ZOOM–INForm 5: Chapter 3 Integration
Paper 1
1.∫k
5(y – 5) dy = 8
[ –5y]k
5= 8
– 5k – � – 5(5)� = 8
– 5k + = 8
k2 – 10k + 25= 16
k2 – 10k + 9 = 0(k – 1)(k – 9) = 0
k = 1 or 9
2.∫2
–13g (x) dx +
∫4
23g (x) dx
= 3[ ∫2
–1g(x) dx +
∫4
2g(x) dx]
= 3[ ∫4
–1g(x) dx]
= 3(20)= 60
3. = x4 – 8x3 + 6x2
y = ∫�x4 – 8x3 + 6x2� dx
y = – 8� � + 6 � � + c
y = – 2x4 + 2x3 + c
Since the curve passes through the point
�1, – 1 �,
– = –2 + 2 + c
c = –2
Hence, the equation of the curve is
y = – 2x4 + 2x3 – 2.x5
5
15
95
45
x5
5
x3
3x4
4x5
5
dydx
252
k2
2
52
2k2
2
y2
2
4. Area of the shaded region
= ∫2
–1y dx
= ∫2
–1(x2 – 2x + 1) dx
= [ – x2 + x]2
–1
= – 4 + 2 – �– –1 – 1�= 3 units2
Paper 2
1.
Area P∫0
–1y dx
= ∫0
–1(–x3 –x) dx
= [– – ]0
–1
= 0 – [– – ]= +
=
Area Q∫2
0y dx
= ∫2
0(–x3 – x) dx
= [– – ]2
0
= – – – 0
= –4 – 2= –6
22
224
4
x2
2x4
4
34
12
14
(–1)2
2(–1)4
4
x2
2x4
4
O1 2
y
P Q
x
y = –x3 – x
13
83
x3
3
26
Hence, the total area of the shaded region= Area P + |Area Q|
= + |–6|
= 6 units2
2. (a) y = hx2 + k
= 2hx
At the point (–2, 8), the gradient of the curve is – 4.
∴ = – 4
2hx = – 42h (–2) = – 4
– 4h = – 4h = 1
The curve y = hx2 + k passes through the point (–2, 8).∴ 8 = h(–2)2 + k
8 = 4h + k8 = 4(1) + kk = 4
dydx
dydx
34
34
(b) When h = 1 and k = 4, y = x2 + 4
Volume generated, Vx
= Volume generated by the curve – Volume generated by the straight line PQ (from x = 0 to x = 2)
= π∫3
0y2 dx – πr2h
= π∫3
0(x2 + 4)2 dx – π(4)2 (2)
= π∫3
0(x4 + 8x2 + 16) dx – π
= π[ + + 16x]3
0– π
= π[ + (3)3 + 16(3) – 0] – π
= 157 π units31415
323
83
35
5
323
8x3
3x5
5
323
13
13
y
Ox
Q
P
y = x2 + 4
x = 3
2 3
4
27
Paper 1
1. (a) EA→
= DC→
= (12p_) = 9p_
(b) EQ→
= ED→
= �EA→
+ AB→
+ BC→
+ CD→�
= �9p_ – 6r_ – 9q_ – 12p_�= �– 3p_ – 6r_ – 9q_�
2. XY→
= XD→
+ DY→
= BD→
+ DC→
= �BA→
+ AD→� + AB
→
= �–6b_ + 2a_� + (6b_)
= –3b_ + a_ + 4b_= a_ + b_
3. (a) a_ + b_ + 2c_
= 7j_ + (10i_ – 5j_) + 2(–4i_ + j_)
= 7j_ + 2i_ – j_ – 8i_ + 2j_
= – 6i_ + 8j_
(b) |a_ + b_ + 2c_| =
Hence, the unit vector in the direction of
a_ + b_ + 2c_
= �–6i_ + 8j_�= – i_ + j_
45
35
110
15
(–6)2 + 82 = 1015
15
15
23
12
23
12
23
12
12
12
12
12
34
34
4. (a) If the vectors a_ and b_ are parallel, then
a_ = hb_ (h is a constant).
2i_ – 5j_ = h(ki_ – 3j_)
2i_ – 5j_ = hki_ – 3hj_
Equating the coefficients of j_–3h = –5
h =
Equating the coefficients of i_hk = 2
k = 2
k =
(b) a_ = hb_
a_ = b_
5. (a) AC→
= AB→
+ BC→
= 9i_ – 4j_ + (–6i_ + mj_)
= 3i_ + (m – 4) j_
(b) If AC→
is parallel to the x-axis, the coefficient of j_ equals zero.m – 4 = 0
m = 4
|a|
|a| : |b| = 5 : 3
= |b|53
53
65
53
53
SPM ZOOM–INForm 5: Chapter 4 Vectors
28
Paper 2
1. (a) OT→
= OA→
+ AT→
= 4x_ + AQ→
= 4x_ + (AO→
+ OQ→
)
= 4x_ + (–4x_ + 6y_)
= x_ + 2y_
(b) OS→
= OQ→
+ QS→
= 6y_ + hQP→
= 6y_ + h(QO→
+ OP→
)
= 6y_ + h(QO→
+ 4OA→
)
= 6y_ + h[–6y_ + 4(4x_)]
= (6 – 6h) y_ + 16hx_
83
13
13
13
(c) Since the points O, T and S are collinear,
then, OT→
= kOS→
, where k is a constant.
OT→
= kOS→
x_ + 2y_ = k [(6 – 6h)y_ + 16hx_]
x_ + 2y_ = k (6 – 6h)y_ + 16hkx_
x_ + 2y_ = (6k – 6hk)y_ + 16hkx_
Equating the coefficients of x_,
= 16hk
1 = 6hk
hk = …
Equating the coefficients of y_,6k – 6hk = 2 …
Substituting into :
6k – 6� � = 2
6k = 3
k =
From : hk =
h� � =
h = 13
16
12
16
2
12
16
21
2
116
83
83
83
83
29
2. (a) (i) OM→
= OB→
= (14y_) = 10y_
(ii) AK→
= AB→
= �AO→
+ OB→�
= �–2x_ + 14y_�= – x_ + y_
(b) (i) AL→
= pAM→
= p�AO→
+ OM→�
= p(–2x_ + 10y_)
= –2px + 10py_
(ii) KL→
= qKO→
= q�KA→
+ AO→�
= q� x_ – y_ – 2x_�= q�– x_ – y_�= – qx_ – qy_
72
32
72
32
72
12
72
12
14
14
14
57
57
(c) AK→
= AL→
+ LK→
– x_ + y_ = –2px_ + 10py_ + � qx_ + qy_ �– x_ + y_ = �–2p + q� x_ + �10p + q�y_–x_ + 7y_ = (–4p + 3q)x_ + (20p + 7q)y_
Equating the coefficients of x_ ,– 4p + 3q = –1 …
Equating the coefficients of y_ ,20p + 7q = 7 …
–20p + 15q = –5 … � 520p + 7q= 7 …
22q = 2
q =
From :
– 4p + 3� � = –1
– 4p = –
p = 722
1411
111
1
111
2+
1
2
1
72
32
72
12
72
32
72
12
KA→
= –AK→
= x_ – y_72
12�
30
Paper 1
1.
sin (90º – θ)= cos θ
2. – 10 tan x = 0
3 sec2 x – 10 tan x = 03(tan2 x + 1) – 10 tan x = 03 tan2 x + 3 – 10 tan x = 03 tan2 x – 10 tan x + 3 = 0
(3 tan x – 1)(tan x – 3) = 0
tan x = or tan x = 3
When tan x = ,
x = 18.43º, 198.43º
When tan x = 3,x = 71.57º, 251.57º
∴ x = 18.43º, 71.57º, 198.43º, 251.57º
13
13
3cos2 x
1 + p2
p= –
O
θ11 + p2
–p
3. 3 tan θ = 2 tan (45º – θ)
3 tan θ = 2 � �3 tan θ = 2 � �
3 tan θ + 3 tan2 θ = 2 – 2 tan θ3 tan2 θ + 5 tan θ – 2 = 0
(3 tan θ – 1)(tan θ + 2) = 0
tan θ = or tan θ = –2
When tan θ = ,
Basic ∠ = 18.43ºθ = 18.43º, 198.43º
When tan θ = –2,Basic ∠ = 63.43º
θ = 116.57º, 296.57º
∴ θ = 18.43º, 116.57º, 198.43º, 296.57º
13
13
1 – tan θ1 + tan θ
tan 45º – tan θ1 + tan 45º tan θ
SPM ZOOM–INForm 5: Chapter 5 Trigonometric Functions
WebsiteZI F505_4th pp 10/15/08 9:40 AM Page 30
31
Paper 2
1. (a) LHS =
=
=
=
= tan x= RHS
(b) (i), (ii)The graph of y = |tan x| is as shown below.
– = 0
=
|tan x| =
Number of solutions= Number of points of intersection= 4
x2π
x2π
1 – cos 2xsin 2x
x2π
1 – cos 2xsin 2x
x
y
O32
y =
(2π, 1)
π
πππ π
sinxcos x
2 sin2 x2 sin x cos x
1 – (1 – 2 sin2 x)2 sin x cos x
1 – cos 2xsin 2x
2. (a), (b)
3 sin + = 2
3 sin = 2 –
Sketch the straight line y = 2 –
Number of solution= Number of intersection point= 1
2xπ
2xπ
x2
2xπ
x2
y
xO� 2�
y = 3 sin x
232
–2
y = 2 – 2x2
Sketch thestraight line
y = .x2π
x 0 2πy 2 –2
WebsiteZI F505_4th pp 10/15/08 9:40 AM Page 31
32
Paper 1
1.
Hence, the number of 4-digit odd numbers greaterthan 2000 but less than 3000 that can be formed= 3 � 3P2
= 18
2. Each group of boys and girls is counted as oneitem.
√ √This gives 2!.
√ √ √At the same time, B1, B2, and B3 can be arrangedamong themselves in their group. This gives 3!.
√ √ √In the same way, G1, G2, and G3 can also bearranged among themselves in their group. Thisgives another 3!.
Using the multiplication principle, the totalnumber of arrangements= 2! � 3! � 3! = 72
Number of arrangements
2 1 3P2
2 3 3P2
2 5 3P2
3. Number of different committees that can beformed
= 4C1 � 7C2 � 8C3
= 4704
SPM ZOOM–INForm 5: Chapter 6 Permutations and Combinations
B1, B2 and B3G1, G2 and G3
Choosing a femalesecretary and a femaletreasurer from 7 females
Choosing 3subcommitteemembers from 8(males or females).
Choosing a malepresident from 4 males
WebsiteZI F506_4th pp 10/15/08 9:40 AM Page 32
33
Paper 1
1. P(Not a green ball) =
=
5h + 25 = 3h + 3k + 152h = 3k – 10
h =
2. P (not getting any post)
= � �
=
3. There are 3 ‘E’ and 4 ‘E–’ in the bag
(a) P(EE) = � =
(b) P(EE–) = � = 2
746
37
17
26
37
835
47
35
23
3k – 102
35
h + 5h + k + 5
35
SPM ZOOM–INForm 5: Chapter 7 Probability
34
Paper 1
1. X – Number of penalty goals scored
X ∼ B�n, �P(X = 0) =
nCo� �0
� �n
=
(1)(1)� �n
=
� �n
= � �4
∴ n = 4
2. X ∼ N(55, 122)Area of the shaded regionP (X < 37)
= P �Z < �= P (Z < –1.5)= 0.0668
37 – 5512
25
25
16625
25
16625
25
35
16625
35
3. (a) X – Mass of a crab, in gX ∼ N(175, 15)
Z =
=
= 1
(b) P(175 < X < 190)
= P � < Z < �= P (0 < Z < 1)= 0.5 – 0.1587= 0.3413
4.
P(Z > – 0.9)= 0.8159∴ k = –0.9
190 – 17515
175 – 17515
190 – 17515
X – µσ
SPM ZOOM–INForm 5: Chapter 8 Probability Distributions
–1.5
0.0668
0.1587
O 1
0.8159
–0.9
0.1841
35
Paper 2
1. (a) X – Number of blue beads drawn
X ∼ B�10, �X ∼ B�10, �(i) P(X ≥ 3)
= 1 – P(X = 0) – P(X = 1) – P(X = 2)
= 1 – 10C0 � �0 � �10– 10C1� �1 � �9
– 10C2� �2 � �8
= 0.7009
(ii) Mean = np = 10 � = 3
(b) X – Lifespan of a species of dogX ∼ N(12, σ2)
(i) P(X > 8) = 90%
P�Z > � = 0.9
P�Z > � = 0.9
– = –1.282
σ = 3.1201 years
4σ
–1.282
0.10.9
–4σ
8 – 12σ
= npqStandard deviation
= 1.49
= 10 �31 �
32
13
13
23
13
23
13
23
13
13
618
(ii) P(10 < X < 13)
= P� < Z < �= P(–0.641 < Z < 0.321)= 1 – 0.2608 – 0.3741= 0.3651
2. (a) X – Number of customers requiring a supplementary card
X ∼ B�7, �X ∼ B�7, �(i) P(X = 3)
= 7C3 � �3 � �4
= 0.2304
(ii) P(X = 3)= P(X = 0) + P(X = 1) + P(X = 2)
= 7C0 � �0 � �7 +7C1 � �1 � �6
+
7C2 � �2 � �5
= 0.1402
(b) X – Time taken to settle invoicesX ∼ N(30, 52)
(i) P(28 ≤ X ≤ 36)
= P� ≤ Z ≤ �= P(–0.4 ≤ Z ≤ 1.2)= 1– 0.3446 – 0.1151= 0.5403
–0.4 1.2
0.3446 0.1151
36 – 305
28 – 305
1125
1425
1125
1425
1125
1425
1125
1425
1425
280500
13 – 123.1201
10 – 123.1201
–0.641 0.321
0.2608 0.3741
36
(ii) P(X < 22)
= P�Z < �= P(Z < –1.6)= 0.0548
Hence, the expected number of invoices which aregiven discounts= 0.0548 � 220= 12
22 – 305
37
Paper 2
1. (a) For particle A, at maximum velocity,
= 0
1 – 2t = 0
t =
= –2 (negative)
Hence, vmax = 12 + – � �2
= 12 m s–1
(b) sB = 2t 3 – 7t 2 – 15tWhen particle B returns to O,
sB = 02t 3 – 7t 2 – 15t = 0
t(2t 2 – 7t – 15) = 0t(2t + 3)(t – 5) = 0
t = 0, – or 5
t = 0 and t = – are not accepted
∴ t = 5
sA = ∫
vA dt
sA = ∫(12 + t – t2) dt
sA = 12t + – + c
When t = 0, sA = 0. ∴ c = 0
∴ sA = 12t + –
When t = 5,sA = 12(5) + – = 30 m5
653
352
2
t 3
3 t2
2
t3
3t 2
2
32
32
14
12
12
d 2vA
dt2
12
dvA
dt
(c) When particle A reverses its direction,vA = 0
12 + t – t2 = 0t2 – t – 12 = 0
(t + 3)(t – 4) = 0t = –3 or 4t = –3 is not accepted∴ t = 4
vB =
vB = 6t2 – 14t – 15
aB =
aB =12t – 14
When t = 4,aB =12(4) – 14 = 34 m s–2
2 (a) a = 12 – 6t
v = ∫
a dt
v = ∫
(12 – 6t) dtv = 12t – 3t2 + cWhen t = 0, v = 15. Thus, c = 15∴ v = 12t – 3t2 + 15
At maximum velocity,
= 0
12 – 6t = 0t = 2
When t = 2,v = 12(2) – 3(2)2 + 15 = 27 m s–1
= –6 (< 0)
Therefore, v is a maximum.
d2vdt2
dvdt
dvB
dt
dsB
dt
SPM ZOOM–INForm 5: Chapter 9 Motion Along a Straight Line
38
(b) s = ∫
v dt
s = ∫(12t – 3t2 + 15) dt
s = 6t 2 – t 3 + 15t + cWhen t = 0, s = 0. Thus, c = 0.∴ s = 6t 2 – t 3 + 15t
At maximum displacement,
= 0
12t – 3t2 + 15 = 03t2 – 12t – 15 = 0
t2 – 4t – 5 = 0(t – 5)(t + 1) = 0
t = 5 or –1t = –1 is not accepted∴ t = 5
When t = 5,s = 6(5)2 – 53 + 15(5) = 100 m
= 12 – 6t
When t = 5, = 12 – 6(5) = –18
Therefore, s is a maximum.
d 2sdt 2
d 2sdt 2
dsdt
(c) When the particle travels to the right,v > 0
12t – 3t 2 + 15 > 03t 2 – 12t – 15 < 0
t 2 – 4t – 5 < 0(t + 1)(t – 5) < 0
–1 < t < 5
Since the values of t cannot be negative,therefore 0 ≤ t < 5.
–1 5 t
39
Paper 2
1. (a) I 180x + 90y ≤ 54002x + y ≤ 60
II 3x + 4y ≤ 120
III y ≤ 2x
(b)
xO
20
10
30
40
50
30 40
(20, 15)
60
y = 2x2x + y = 60
y = x
3x + 4y = 120
y
34
Max (24, 12)
200x + 150y = 3000
R
10 15 20
(c) (i) =
3x = 4y
y = x
The furthest point on the straight line y = x
inside the feasible region R is (20,15).∴ xmax = 20, ymax = 15
(ii) Profits = 200x + 150yDraw the straight line200x + 150y = 3000
The optimal point is (24, 12).
Hence, the maximum profit= 200(24) + 150(12)= RM6600
2. (a) Mixing:30x + 10y ≤ 15 � 60
3x + y ≤ 90
Baking:
40x + 40y ≤ 26 � 60
x + y ≤ 40
Decorating:10x + 30y ≤ 15 � 60
x + 3y ≤ 90
23
34
34
43
xy
SPM ZOOM–INForm 5: Chapter 10 Motion Along a Straight Line
x 0 30
y 60 0
x 0 40
y 30 0
x 0 30
y 0 60
200 � 150 � 0.1 = 3000
x 0 30
y 90 0
x 0 90
y 30 0
x 0 40
y 40 0
40
(b)
(c) (i) When y = 17, xmax = 23
(ii) Profits = 5x + 10y
Draw the straight line 5x + 10y = 50.
From the graph, the optimal point is (15, 25).
Hence, the maximum profit= 5(15) + 10(25)= RM325
xO
20
10
30
40
50
30 40
60
70
80
90
3x + y = 90
x + 3y = 90
y
x + y = 40
Max (15, 25)
5x + 10y = 50
R
10 23
17
5
20 50 60 70 80 90