40
1 Paper 1 1. The relation in the given graph can be represented using 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) = x 2 . 3. f 2 (x)= ff (x) = f (px + q) = p (px + q) + q = p 2 x + pq + q It is given that f 2 (x) = 4x + 9 By comparison, p 2 = 4 pq + q = 9 p = – 2 –2q + q = 9 q = 9 q = –9 16 9 4 – 4 – 3 – 2 2 3 4 1 2 3 4 10 20 30 40 A B 4. (a) gf : x x 2 + 6x + 2 gf (x)= x 2 + 6x + 2 g(x + 4) = x 2 + 6x + 2 Let x + 4 = u x = u – 4 g(u) = (u – 4) 2 + 6(u – 4) + 2 = u 2 – 8u + 16 + 6u – 24 + 2 = u 2 – 2u – 6 g(x)= x 2 – 2x – 6 (b) fg(4) = f [ 4 2 – 2(4) – 6 ] = f(2) = 2 + 4 = 6 5. Let g –1 (x)= y g(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 = 5 2 5 6 k 3 1 3 5 6 k 3 1 3 k 3 1 3 x k 3 SPM ZOOM–IN Form 4: Chapter 1 Functions The question requires p < 0.

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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 = �– ��– � =

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

π – 1.8

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.

cos ∠ABD =

(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

– 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

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 +

= 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→

= �–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

–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