78
1 Paper 1 1 Based on the given ordered pairs {(2, 1), (4, 3), (6, 5), (6, 7)}, an arrow diagram can be drawn as shown below. (a) The image of 2 is 1. (b) The object of 7 is 6. 2 (a) Let g 1 (7) = y Thus, g(y) = 7 4y 1 = 7 4y = 8 y = 2 g 1 (7) = 2 (b) hg(x) = h(4x 1) = (4x 1) 2 3(4x 1) + 5 = 16x 2 8x + 1 12x + 3 + 5 = 16x 2 20x + 9 3 (a) The range is {3, 7}. (b) The above relation is a many-to-one relation. Form 4: Chapter 1 (Functions) SPM Flashback Fully-Worked Solutions '3' and '7' are linked to object(s) but '5' and '11' are not linked to any object(s). Therefore, the range is {3, 7}. Element '7' in the codomain is linked to two elements, i.e. '28' and '49' in the domain. Therefore, it is a many-to-one relation.

-Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

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Page 1: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

1

Paper 1 1 Based on the given ordered pairs

{(2, 1), (4, 3), (6, 5), (6, 7)}, an arrow diagram can be drawn as shown below.

(a) The image of 2 is 1. (b) The object of 7 is 6. 2 (a) Let g−1(7) = y

Thus, g(y) = 7 4y − 1 = 7 4y = 8 y = 2 ∴ g−1(7) = 2

(b) hg(x)

= h(4x − 1) = (4x − 1)2 − 3(4x − 1) + 5 = 16x2 − 8x + 1 − 12x + 3 + 5 = 16x2 − 20x + 9

3

(a) The range is {3, 7}. (b) The above relation is a many-to-one

relation.

Form 4: Chapter 1 (Functions) SPM Flashback

Fully-Worked Solutions

'3' and '7' are linked to object(s) but '5' and '11' are not linked to any object(s). Therefore, the range is {3, 7}.

Element '7' in the codomain is linked to two elements, i.e. '28' and '49' in the domain. Therefore, it is a many-to-one relation.

Page 2: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

2

4 h : x → 2x + m h(x) = 2x + m Let h−1(x) = y h(y) = x 2y + m = x

y =

x − m2

∴ h−1(x) =

x − m2

=x2

−m2

But it is given that h−1(x) = 3kx +

32

.

Hence, by comparison,

3k =

12

⇒ k =16

and

m2

=32

⇒ m = −3 .

5 (a) hg(x) = 6x − 2

h[g(x)] = 6x − 2 3g(x) + 1 = 6x − 2 3g(x) = 6x − 3 g(x) = 2x − 1

(b) gh(x) = g(3x + 1) = 2(3x +1) −1 = 6x +1

When gh(x) =13

,

6x +1 =13

18x + 3 = 118x = −2

x = −19

6 (a) From the given arrow diagram,

f (−2) = −5. Hence, f−1(−5) = −2 .

(b) gf (−2) = 3

7 (a) Let w−1(x) = z

w(z) = x6

3− 2z= x

6 = x(3− 2z)6 = 3x − 2xz

2xz = 3x − 6

z =3x − 6

2x

∴ w−1( x) =3x − 6

2x, x ≠ 0

(b) w−1h −

52

⎝ ⎜

⎠ ⎟

= w−1 2 −52

⎝ ⎜

⎠ ⎟ + 3

⎣ ⎢ ⎤

⎦ ⎥

= w−1(−2)

=3(−2) − 6

2(−2)= 3

8 Let

n−1(x) = yn( y) = x

4 y − 1 = x4 y = x +1

y =x +1

4

n−1(x) =x +1

4

mn−1(x) = m x +14

⎝ ⎜

⎠ ⎟

=3

8 x + 14

⎝ ⎜

⎠ ⎟ − 5

=3

2x − 3, x ≠

32

It is given that h(x) = 3x + 1. Hence, h[g(x)] = 3g(x) + 1.

This is a composite function gf(x) which maps x directly onto z.

Change the subject of the formula to z.

Page 3: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

3

9 (a) The relation between set P and set Q is a many-to-one relation.

(b) The relation can be represented by

f ( x) = x4 .

10

m(2) = 72 − h

h= 7

2 − h = 7h8h = 2

h =14

Function notation

Page 4: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

4

Paper 2 1 (a)

f : x → 2x − 3f (x) = 2x − 3

Let f −1(x) = yf ( y) = x

2y − 3 = x

y =x + 3

2

∴ f −1(x) =x + 3

2

f −1g(x)

= f −1 x2

+ 2⎛

⎝ ⎜

⎠ ⎟

=

x2

+ 2⎛

⎝ ⎜

⎠ ⎟ + 3

2

=

x + 4 + 622

=x +10

4

∴ f −1g : x →x + 10

4

(b) hg : x → 2x + 4

hg(x) = 2x + 4

h x2

+ 2⎛

⎝ ⎜

⎠ ⎟ = 2x + 4

Let x2

+ 2 = u

x2

= u − 2

x = 2u − 4

h(u) = 2(2u − 4) + 4= 4u − 8 + 4= 4u − 4

∴ h : x → 4x − 4

Page 5: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

5

Paper 1 1

2x(x − 3) = (2 − x)(x +1)2x2 − 6x = 2x + 2 − x2 − x

3x2 − 7x − 2 = 0

x =−(−7) ± (−7)2 − 4(3)(−2)

2(3)

x =7 ± 49 + 24

6

x =7 ± 8.5440

6x = 2.591 or −0.2573

2 The roots are 23

and −

52

.

Sum of roots =

23

+ −52

⎝ ⎜

⎠ ⎟ = −

116

Product of roots =

23

× −52

⎝ ⎜

⎠ ⎟ = −

53

The quadratic equation is

x2 +

116

x −53

= 0

6x 2 + 11x − 10 = 0

3 y = 5x − 2 …

y = 3x2 + 3x + k … Substituting into :

3x2 + 3x + k = 5x − 23x2 − 2x + k + 2 = 0

a = 3, b = −2, c = k + 2 In the case where a curve does not meet a straight line, b2 − 4ac < 0 is applied.

b2 − 4ac < 0(−2)2 − 4(3)(k + 2) < 0

4 −12k − 24 < 0−12k − 20 < 0

−12k < 20

k >20

−12

k > −1 23

4

7 − 2(1+ x)2 = x(x + 5)7 − 2(1+ 2x + x2 ) = x2 + 5x

7 − 2 − 4x − 2x2 = x2 + 5x3x2 + 9x − 5 = 0

x =−b ± b2 − 4ac

2a

x =−9 ± 92 − 4(3)(−5)

2(3)

x =−9 ± 141

6x = 0.4791 or −3.479

Form 4: Chapter 2 (Quadratic Equations) SPM Flashback

Fully-Worked Solutions

x2 − (sum of roots)x + (product of roots) = 0

Page 6: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

6

5

9x2 + qx + 1 = 4x9x2 + qx − 4x + 1 = 09x2 + (q − 4)x + 1 = 0

a = 9, b = q − 4, c = 1 If the equation has equal roots, then

b2 − 4ac = 0(q − 4)2 − 4(9)(1) = 0q2 − 8q +16 − 36 = 0

q2 − 8q − 20 = 0(q + 2)(q −10) = 0

q = −2 or 10

Page 7: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

7

Paper 1 1

x(x + 2) = 3px − 4x2 + 2x = 3px − 4

x2 + 2x − 3px + 4 = 0x2 + (2 − 3p)x + 4 = 0

a = 1, b = 2 − 3p, c = 4 For the case of two distinct roots, b2 − 4ac > 0 is applied.

(2 − 3p)2 − 4(1)(4) > 04 − 12p + 9p2 − 16 > 0

9p2 −12p − 12 > 03p2 − 4p − 4 > 0

(3p + 2)( p − 2) > 0

Hence, the ranges of values of p are

p < −

23

or p > 2 .

2

x(x − 1) > 12x2 − x > 12

x2 − x −12 > 0(x + 3)(x − 4) > 0

The ranges of values of x are x < −3 or x > 4 .

3

(a) The curve passes through the point (0, −4).

y = −(x − k )2 − 3−4 = −(0 − k )2 − 3−1 = −k 2

k 2 = 1k = 1

(b) The equation of the curve is

y = −(x − 1)2 − 3. Hence, the equation of the axis of symmetry is x −1 = 0 ⇒ x = 1.

(c) The coordinates of the maximum point are

(1, −3). 4 f (x) = −2(x + p)2 − 2

The maximum point of the graph of f (x) is

(−p, −2) . But it is given that the maximum point of the graph of f (x) is (−3, q) . Hence, by comparison,

(a) −p = −3 ⇒ p = 3 (b) q = −2 (c) the equation of the axis of symmetry of the

curve is x = −3.

Form 4: Chapter 3 (Quadratic Functions) SPM Flashback

Fully-Worked Solutions

Page 8: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

8

5 (a) The equation of the axis of symmetry is

x =−6 + (−2)

2x = −4

(b) The minimum point is (−4, −5). Hence,

f (x) = ( x + 4)2 − 5 .

6

(1− 2x)(3 + x) > x + 33+ x − 6x − 2x2 > x + 3

−2x2 − 6x > 02x2 + 6x < 02x(x + 3) < 0

Hence, the required range of values of x is −3 < x < 0 .

Page 9: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

9

Paper 2 1 (a)

f (x) = −x2 + 4kx − 5k 2 − 1= −(x2 − 4kx + 5k 2 +1)= −(x2 − 4kx + 4k 2 − 4k 2 + 5k 2 +1)

= −[(x − 2k )2 + k 2 + 1]= −(x − 2k )2 − k 2 −1

Maximum value = −k 2 −1 But it is given that the maximum value = −r2 − 2k . By comparison,

−r2 − 2k = −k 2 − 1r2 = k 2 − 2k +1r2 = (k −1)2

r = k − 1 (shown)

(b) From f (x) = −(x − 2k )2 − k 2 − 1, the axis

of symmetry is x = 2k . But it is given that the axis of symmetry is x = r2 − 1. By comparison, r2 − 1 = 2k Solve the following simultaneous equations: r = k − 1 … r2 − 1 = 2k … Substitute into :

(k −1)2 −1 = 2kk 2 − 2k +1− 1− 2k = 0

k 2 − 4k = 0k(k − 4) = 0

k = 0 or k = 4k = 0 is not accepted.∴ k = 4When k = 4, r = 4 − 1 = 3

Add and subtract

12

× (−4k )⎡

⎣ ⎢ ⎤

⎦ ⎥ 2

= 4k 2

Page 10: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

10

Paper 2 1 4x + y = x2 + x − y = −3

4x + y = −3 …

x2 + x − y = −3 …

From : y = −3− 4x …

Substitute into :

x2 + x − (−3− 4x) = −3x2 + x + 3+ 4x + 3 = 0

x2 + 5x + 6 = 0(x + 2)(x + 3) = 0

x = −2 or −3

From : When x = −2, y = −3 − 4(−2) = 5 When x = −3, y = −3− 4(−3) = 9

Hence, the solutions are x = −2, y = 5

or x = −3, y = 9 . 2 x − y = 1 …

x2 + 3y = 6 …

From : x = 1+ y …

Substituting into , we have:

(1+ y)2 + 3y = 61+ 2y + y2 + 3y − 6 = 0

y2 + 5y − 5 = 0

y =−b ± b2 − 4ac

2a

y =−5 ± 52 − 4(1)(−5)

2(1)

y =−5 ± 45

2(1)y = 0.854 or −5.854

From : When y = 0.854, x = 1+ 0.854 = 1.854 When y = −5.854, x = 1+ (−5.854) = −4.854

Hence, the solutions are x = 1.854, y = 0.854 or x = −4.854, y = −5.854 (correct to 3 decimal places).

3 x +

y2

= 1 …

3xy − 7 y = 2 … From :

2x + y = 2 y = 2 − 2x … Substituting into :

3x(2 − 2x) − 7(2 − 2x) − 2 = 06x − 6x2 − 14 + 14x − 2 = 0

−6x2 + 20x −16 = 03x2 −10x + 8 = 0

(3x − 4)(x − 2) = 0

x =43

or 2

From :

When x =

43

, y = 2 − 2 4

3⎛

⎝ ⎜

⎠ ⎟ = −

23

When x = 2, y = 2 − 2(2) = −2

Hence, the solutions are x = 1 1

3, y = −

23

or

x = 2, y = −2 .

Form 4: Chapter 4 (Simultaneous Equations) SPM Flashback

Fully-Worked Solutions

Page 11: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

11

4 3x + y = 2 …

x2 + 2 y2 + xy = 4 …

From : y = 2 − 3x … Substituting into :

x2 + 2(2 − 3x)2 + x(2 − 3x) = 4x2 + 2(4 −12x + 9x2 ) + 2x − 3x2 − 4 = 0

x2 + 8 − 24x + 18x2 + 2x − 3x2 − 4 = 016x2 − 22x + 4 = 0

8x2 − 11x + 2 = 0

x =−(−11) ± (−11)2 − 4(8)(2)

2(8)

x =11± 57

16x = 1.159 or 0.216

From : When x = 1.159, y = 2 − 3(1.159) = −1.477 When x = 0.216, y = 2 − 3(0.216) = 1.352 Hence, the solutions are x = 1.159, y = −1.477 or x = 0.216, y = 1.352 (correct to 3 decimal places).

Page 12: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

12

Paper 1 1 log2 R − log4 Q = 2

log2 R −log2 Qlog2 4

= 2

log2 R −log2 Qlog2 22

= 2

log2 R −log2 Q

2= 2

2log2 R − log2 Q = 4log2 R2 − log2 Q = 4

log2R2

Q= 4

R2

Q= 24

R2 = 24QR = 24QR = 22 QR = 4 Q

2 42x− 3 = 5x

log 42x− 3 = log 5x

(2x − 3) log 4 = x log 52x log 4 − 3log 4 = x log 52x log 4 − x log 5 = 3log 4x(2log 4 − log 5) = 3log 4

x =3log 4

2log 4 − log 5x = 3.576

3

2434x = 98 x+6

(35 )4x = (32 )8x+6

320x = 32(8x+6)

Equating the powers, we have:

20x = 2(8x + 6)20x = 16x +12

4x = 12x = 3

4 log5 2.7

= log5 2 710

= log52710

= log533

2 × 5⎛

⎝ ⎜

⎠ ⎟

= log5 33 − (log5 2 + log5 5)= 3log5 3 − log5 2 − log5 5= 3 p − m − 1

5 3s + 2 − 3s + 1 =

23

3s(32) − 3s(31) =23

9(3s) − 3(3s) =23

(9 − 3)(3s) =23

6(3s) =23

3s =23

×16

3s =19

3s = 3−2

∴ s = −2

6 log5 (2y − 1) = 1 + log5 ( y − 8)

log5 (2 y − 1) − log5 ( y − 8) = 1

log52 y − 1y − 8

⎝ ⎜

⎠ ⎟ = 1

2y − 1y − 8

= 51

2y − 1 = 5( y − 8)2y − 1 = 5y − 40

3y = 39y = 13

Form 4: Chapter 5 (Indices and Logarithms) SPM Flashback

Fully-Worked Solutions

Page 13: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

13

7

2 = ms 3 = mt

logm 2 = s logm 3 = t

logmm6

⎝ ⎜

⎠ ⎟

= logmm

12

2 × 3

⎜ ⎜ ⎜

⎟ ⎟ ⎟

= logm m12 − logm 2 − logm 3

=12

− s − t

8 272x−5 =

19x+1

33(2x−5) =1

32( x+1)

33(2x−5) =1

[32( x+1) ]12

33(2x−5) =1

3( x+1)

33(2x−5) = 3−( x+1)

Equating the powers, 3(2x − 5) = −(x +1)

6x −15 = −x −17x = 14

x = 2

9 log3

x4

y⎛

⎝ ⎜

⎠ ⎟ = 2 + 2 log3 x + log3 y

log3x4

y⎛

⎝ ⎜

⎠ ⎟ − 2 log3 x − log3 y = 2

log3x4

y⎛

⎝ ⎜

⎠ ⎟ − log3 x2 − log3 y = 2

log3

x4

yx2 y

⎜ ⎜ ⎜ ⎜

⎟ ⎟ ⎟ ⎟

= 2

log3x4

y⎛

⎝ ⎜

⎠ ⎟

1x2 y

⎝ ⎜

⎠ ⎟ = 2

log3x2

y2

⎝ ⎜

⎠ ⎟ = 2

x2

y2= 32

xy

⎝ ⎜

⎠ ⎟

2

= 32

xy

= 3

y =x3

10 2 + 2log4 ( p − 6) = log2 p

2 log4 ( p − 6) − log2 p = −2

2 log2 ( p − 6)log2 4

⎣ ⎢ ⎤

⎦ ⎥ − log2 p = −2

2 log2 ( p − 6)2

⎣ ⎢ ⎤

⎦ ⎥ − log2 p = −2

log2 ( p − 6) − log2 p = −2

log2p − 6

p⎛

⎝ ⎜

⎠ ⎟ = −2

p − 6p

= 2−2

p − 6p

=14

4p − 24 = p3p = 24

p = 8

log2 4 = log2 22 = 2log2 2 = 2(1) = 2

Page 14: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

14

Paper 1 1

B = (b, c)4(2a) + 3(2b)

3 + 4, 4a + 3(3c)

3 + 4⎛

⎝ ⎜

⎠ ⎟ = (b, c)

8a + 6b7

, 4a + 9c7

⎝ ⎜

⎠ ⎟ = (b, c)

Equating the x-coordinates:

8a + 6b7

= b

8a + 6b = 7b8a = b

a =

b8

Equating the y-coordinates:

4a + 9c7

= c

4a + 9c = 7c4a + 2c = 0

2a + c = 0 … Substitute into :

2 b8

⎝ ⎜

⎠ ⎟ + c = 0

b4

+ c = 0

b + 4c = 0b = −4c

2 yk

+x3

= 1

3y + kx3k

= 1

3y + kx = 3k3y = −kx + 3k

y = −k3

x + k

∴ m1 = −

k3

5y = 3x + 25

y =35

x + 5

∴ m2 =

35

m1m2 = −1

−k3

⎝ ⎜

⎠ ⎟

35

⎝ ⎜

⎠ ⎟ = −1

−k5

= −1

k = 5

Form 4: Chapter 6 (Coordinate Geometry) SPM Flashback

Fully-Worked Solutions

Page 15: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

15

3 x3

+y4

= 1 is an equation in the intercept form

where the x-intercept is 3 and the y-intercept is 4.

∴ Q = (0, 4)∴ P = (3, 0)

mQP =

4 − 00 − 3

= −43

Gradient of the perpendicular line

= −1

−43

⎝ ⎜

⎠ ⎟

=34

Hence, the equation of the straight line that is perpendicular to PQ and passes through the point Q(0, 4) is

y − y1 = m(x − x1 )

y − 4 =34

(x − 0)

4( y − 4) = 3x4y − 16 = 3x

4y = 3x + 16

4 A = (−2, −2), B = (3, 5)

Let P = (x, y)

PA : PB = 2 : 3PAPB

=23

3PA = 2PB3 [x − (−2)]2 + [ y − (−2)]2 = 2 (x − 3)2 + ( y − 5)2

Squaring both sides, we have:

32[(x + 2)2 + ( y + 2)2 ] =22[(x − 3)2 + ( y − 5)2 ]

9(x2 + 4x + 4 + y2 + 4y + 4) =4(x2 − 6x + 9 + y2 − 10y + 25)

9x2 + 36x + 36 + 9y2 + 36y + 36 =4x2 − 24x + 36 + 4y2 − 40y + 100

5x2 + 5y2 + 60x + 76y − 64 = 0 Hence, the equation of the locus of P is 5x2 + 5y 2 + 60x + 76y − 64 = 0 .

5 AB : y =

h2

x +k2

PQ : y =k + 1

3x +

h3

mAB =h2

mPQ =k + 1

3

Since the straight lines AB and PQ are perpendicular to each other,

(mAB ) (mPQ ) = −1h2

k + 13

⎝ ⎜

⎠ ⎟ = −1

h =−6

k + 1

6 The equation of PQ is

2y = −x + 4

y = −12

x + 2

∴ mPQ = −12

∴ mQR = −1

mPQ= 2

Hence, the equation of QR is y = 2x − 3. Equation of PQ: 2y = −x + 4 … Equation of QR: y = 2x − 3 …

× 2 + : 5y = 5 y = 1 From : 2(1) = −x + 4 x = 2 Hence, the coordinates of point Q are (2, 1) .

Page 16: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

16

Paper 2 1

(a) (i) Radius of the circle

= MA= (−3 − 1)2 + (0 − 3)2

= 16 + 9= 25= 5

MR = MA(x − 1)2 + ( y − 3)2 = 5(x − 1)2 + ( y − 3)2 = 52

x2 − 2x +1+ y2 − 6y + 9 = 25x2 − 2x + y2 − 6y −15 = 0

Hence, the equation of the locus of the point R(x, y) is x

2 − 2x + y2 − 6y − 15 = 0 .

(ii) Point B(4, k) lies on the circumference of the circle.

42 − 2(4) + k 2 − 6k − 15 = 0k 2 − 6k − 7 = 0

(k − 7)(k + 1) = 0k = 7 or −1

(b) mMA =

0 − 3−3− 1

=34

∴ mAC = −1

mMA= −

134

= −43

Equation of the straight line AC is

y − 0 = −43

[x − (−3)]

3y = −4(x + 3)3y = −4x − 12

At point C (y-axis), 0=x .

3y = −4(0) −12 ⇒ y = −4∴ C = (0, − 4)

Area of ∆OAC

=12

0 −3 0 00 0 −4 0

=12

12

= 6 units2

Page 17: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

17

2 (a)

The equation of AB in the intercept form is

x6

+y

(−3)= 1 .

(b)

2AD = DBADDB

=12

D =

2(0) + 1(6)1+ 2

, 2(−3) + 1(0)1+ 2

⎝ ⎜

⎠ ⎟ = (2, −2)

(c)

mAB =0 − (−3)

6 − 0=

12

mCD = −1

mAB= −

112

⎝ ⎜

⎠ ⎟

= −2

Hence, the equation of CD is

y − y1 = m(x − x1 )y − (−2) = −2(x − 2)

y + 2 = −2x + 4y = −2x + 2

∴ y-intercept = 2

Page 18: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

18

3 (a) (i) The gradient of the straight line 2x − y − 5 = 0 ⇒ y = 2x − 5 is 2.

∴ mBC = 2

∴ mAB = −1

mBC= −

12

Hence, the equation of the straight line AB is:

y − y1 = m(x − x1)

y − (−3) = −12

[x − (−9)]

2( y + 3) = −(x + 9)2y + 6 = −x − 9

x + 2y + 15 = 0 (general form)

(ii) Equation of BC: 2x − y − 5 = 0 …

Equation of AB: x + 2y +15 = 0 …

+

4x − 2y −10 = 0x + 2y + 15 = 0

5x + 5 = 0∴ x = −1

K × 2K

From :

2(−1) − y − 5 = 0∴ y = −7

∴ B is point (−1, −7) .

(b)

B = (−1, −7)3(−9) + 2h

2 + 3, 3(−3) + 2k

2 + 3⎛

⎝ ⎜

⎠ ⎟ = (−1, −7)

−27 + 2h5

, −9 + 2k5

⎝ ⎜

⎠ ⎟ = (−1, −7)

Equating the x-coordinates:

−27 + 2h5

= −1

2h = 22h = 11

Equating the y-coordinates:

−9 + 2k5

= −7

2k = −26k = −13

∴D is point (11, −13). Area of ∆ADO

=12

−9 11 0 −9−3 −13 0 −3

=12

117 − (−33)

=12

150

= 75 units2

(c) Let P be point (x, y).

PA = 2[x − (−9)]2 + [y − (−3)]2 = 2

(x + 9)2 + ( y + 3)2 = 22

x2 + 18x + 81+ y2 + 6y + 9 = 4x 2 + 18x + y2 + 6y + 86 = 0

At the point A(−9, −3), x1 = −9, y1 = −3.

Page 19: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

19

4 (a) (i) Area of ∆OAB

=12

0 −3 6 00 −5 1 0

=12

−3− (−30)

=12

27

= 13.5 units2

(ii) AB = [6 − (−3)]2 + [1− (−5)]2

= 92 + 62

= 117= 10.82 units

Area of ∆OAB = 13.512

× AB × h = 13.5

12

×10.82 × h = 13.5

h = 2.50

Hence, the perpendicular distance from O to AB is 2.50 units.

(b) (i) Let P = (x, y)

PA = 2PB(PA)2 = (2PB)2

PA2 = 4PB2

[x − (−3)]2 + [y − (−5)]2 =4[(x − 6)2 + ( y − 1)2 ]

(x + 3)2 + ( y + 5)2 =4[(x − 6)2 + ( y − 1)2 ]

x2 + 6x + 9 + y2 +10y + 25 =4(x2 −12x + 36 + y2 − 2y + 1)

x2 + 6x + 9 + y2 +10y + 25 =4x2 − 48x +144 + 4y2 − 8y + 4

0 = 3x2 − 54x + 3y2 −18y +114x 2 + y2 − 18x − 6y + 38 = 0

(ii) At the y-axis, x = 0.

∴02 − 18(0) + y2 − 6y + 38 = 0y2 − 6y + 38 = 0

b2 − 4ac= (−6)2 − 4(1)(38)= 36 − 152= −116

Since b2 − 4ac < 0 , there are no real roots and hence, the locus will not cut the y-axis.

Page 20: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

20

Paper 1 1 σ = k

2255 km −=

−=

=−

=⎟⎟

⎜⎜

⎛−

=⎟⎟

⎜⎜

⎛−

=⎟⎟

⎜⎜

⎛−

∑∑

∑∑

x

kmx

xkm

kxm

kxm

kn

xn

x

22

2

2

2

2

2

22

255

255

55

55

)(

)(

2 Variance = 14

x2∑n

−x∑

n

⎜ ⎜

⎟ ⎟

2

= 14

12 + 42 + k 2

3−

1 + 4 + k3

⎝ ⎜

⎠ ⎟

2

= 14

17 + k 2

3−

5 + k3

⎝ ⎜

⎠ ⎟

2

= 14

17 + k 2

3−

(5 + k )2

9= 14

3(17 + k 2 ) − (5 + k )2 = 12651 + 3k 2 − (25 + 10k + k 2 ) = 126

51 + 3k 2 − 25 − 10k − k 2 − 126 = 02k 2 − 10k − 100 = 0

k 2 − 5k − 50 = 0(k + 5)(k − 10) = 0

k = −5 or 10k = −5 is not accepted.

∴ k = 10

Form 4: Chapter 7 (Statistics) SPM Flashback

Fully-Worked Solutions

Page 21: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

21

Paper 2 1 (a) (i) Given that x = 10,

x∑6

= 10

x = 60∑

(ii) Given that σ = 3,

=

=−

=−

=

6542

22

22

2

9106

9)(

9

x

x

xn

x

σ

(b) If each mark is multiplied by 4 and then 5

is added to it: (i) New mean

= (4 × original mean) + 5= (4 ×10) + 5= 45

(ii) New standard deviation

= (4 × original standard deviation)= 4 × 3= 12

Hence, the new variance

= 122

= 144

2 (a) Mean, x =

x∑n

=20010

= 20

Variance, σ 2 =x2∑

n− (x)2

=440010

− 202

= 40

(b) Assume that a number k is added to the

first set of data. (i)

New mean = 20 + 2

x∑ + k

11= 22

200 + k11

= 22

k = 42

(ii) New standard deviation

=x2∑ + 422

11− 222

=4400 + 422

11− 222

= 8.739

Page 22: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

22

3 (a) The frequency table which represents the given histogram is as follows.

Marks Frequency Cumulative frequency

10 − 24 2 2 25 − 39 6 8 40 − 54 8 16 55 − 69 13 29 70 − 84 5 34 85 − 99 2 36

The median class is given by

Tn

2

= T362

= T18 .

Thus, the median class is 55 − 69. Median

= L +

n2

− F

fm

⎜ ⎜ ⎜

⎟ ⎟ ⎟ c

= 54.5 +

362

− 16

13

⎜ ⎜ ⎜

⎟ ⎟ ⎟ (15)

= 56.81

(b) Marks f Midpoint

(x) fx fx2

10 − 24 2 17 34 578 25 − 39 6 32 192 6144 40 − 54 8 47 376 17 672 55 − 69 13 62 806 49 972 70 − 84 5 77 385 29 645 85 − 99 2 92 184 16 928

f∑= 36

fx∑

= 1977

fx2∑

= 120 939 Standard deviation, σ

=fx2∑f∑

−fx∑f∑

⎜ ⎜

⎟ ⎟

2

=120 939

36−

197736

⎝ ⎜

⎠ ⎟

2

= 18.54

Median class

Page 23: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

23

4 (a)

Marks Frequency Cumulative frequency

20 − 29 2 2 30 − 39 4 6 40 − 49 6 (12) 50 − 59 (12) 24 60 − 69 k 24 + k 70 − 79 2 26 + k

f = 26 + k∑

If the median is 52, then the median class is 50 − 59.

Median = 52

L +

n2

− F

fm

⎜ ⎜ ⎜

⎟ ⎟ ⎟ c = 52

49.5 +

26 + k2

− 12

12

⎜ ⎜ ⎜

⎟ ⎟ ⎟ (10) = 52

26 + k − 242

12

⎜ ⎜ ⎜

⎟ ⎟ ⎟ (10) = 52 − 49.5

2 + k2

12

⎜ ⎜ ⎜

⎟ ⎟ ⎟ (10) = 2.5

2 + k24

(10) = 2.5

2 + k = 6k = 4

(b)

Modal mark = 53.75 (c) New mode = Original mode − 3

= 53.75 − 3= 50.75

Median class

Page 24: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

24

Paper 1 1 Let OR = r cm

Perimeter of the sector ROS = 40 cm

r + r + 8 = 402r = 32r = 16

θ =

sr

=8

16= 0.5 rad.

2

Length of the major arc AB = 46.64 cmr(2π − 0.454) = 46.64

r[(2 × 3.142) − 0.454] = 46.64r(5.83) = 46.64

r = 8∴ Radius = 8 cm

3 (a) Area of sector = 20 cm2

12

r2(0.4) = 20

r2 = 100r = 10

Length of the arc AB

= 10 × 0.4= 4 cm

(b) Reflex angle AOB

= 2π − 0.4= (2 × 3.142) − 0.4= 5.884 rad.

= 5.884 ×180

3.142= 337° ′ 5

4 ∠AOB =

812

=23

rad.

Area of the shaded region

= Area of sector OAB − Area of sector AMN

=12

(12)2 23

⎝ ⎜

⎠ ⎟ −

12

(6)2 (0.7)

= 48 −12.6= 35.4 cm2

Form 4: Chapter 8 (Circular Measures) SPM Flashback

Fully-Worked Solutions

Page 25: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

25

Paper 2 1

(a) From ∆OPB,

cos θ =9

15=

35

θ = 0.9273 rad.

(b) Area of the shaded region

= Area of sector OAB − Area of ∆OPB

=12

(15)2(0.9273) −12

× 9 ×12

= 104.32125 − 54= 50.32 cm2

OP =35

×15

= 9 cm

Make sure that your calculator is in the radian mode.

Page 26: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

26

2 (a)

Insert a line OQ in the diagram. • Since OPQR is a rhombus,

OP = PQ = QR = RO . • Since the radius of a circle is a constant,

OP = OQ = OR.

It can now be concluded that ∆OPQ is an equilateral triangle because OP = PQ = OQ .

It can also be concluded that ∆ORQ is an equilateral triangle because OR = RQ = OQ . Therefore, ∠POQ = ∠ROQ = 60°.

Hence, ∠POR = α= 120°

= 120 ×π

180

=23

π rad.

(b) Based on ∆JOQ ,

cos ∠JOQ =OQOJ

cos 60° =8

OJ

OJ =8

cos 60°OJ = 16 cm

Length of the arc JLK

= OJ × ∠JOK

= 16 ×23

π

= 33.51 cm

(c) Area of the shaded segment

=12

r2 (α − sin α)

=12

(16)2 23

π − sin 120°⎛

⎝ ⎜

⎠ ⎟

= 157.23 cm2

Page 27: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

27

3 (a) In ∆OQR ,

cos π3

⎝ ⎜

⎠ ⎟

r

=7

OR

cos 60° =7

OR

OR =7

cos 60°OR = 14 cm

In ∆OQR ,

tan 60° =QR7

QR = 7 × tan 60°QR = 12.1244 cm

PR = OR − OP= 14 − 7= 7 cm

Length of arc PQ

= 7 ×π3

= 7 ×3.142

3= 7.3313 cm

Hence, the perimeter of the shaded region A = PR + QR + Length of arc PQ = 7 + 12.1244 + 7.3313 = 26.46 cm

(b) Area of ∆OQR

=12

× 7 ×12.1244

= 42.4354 cm2

Area of the sector ORS

=12

×142 ×3.142

3= 102.6387 cm2

Hence, the area of the shaded region B = Area of sector ORS − Area of ∆OQR = 102.6387 − 42.4354 = 60.20 cm2

Page 28: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

28

4 (a)

∆CAO is an isosceles triangle.

∴∠CAO = ∠COA= π −1.982= 3.142 −1.982= 1.16 rad.

(b)

Perimeter of the shaded region

= CR + Length of arc CQ + Length of arc RQ

= 8 + (10 ×1.982) + (18 ×1.16)= 48.7 cm

(c) MC = CO2 − MO2

= 102 − 42

= 84 cm

Area of shaded region

= Area of sector ARQ − Area of ∆ACO− Area of sector COQ

=12

(18)2(1.16) −12

× 8 × 84 −12

(10)2(1.982)

= 52.16 cm2

Page 29: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

29

Paper 1 1 (a) y = 15x(3 − x) = 45x − 15x2

dydx

= 45 − 30x

When dydx

= 0,

45 − 30x = 0,x = 1.5

(b) When x = 1.5, y = 15(1.5)(3−1.5) = 33.75

d 2 ydx2

= −30 (negative)

Therefore, y is a maximum.

2 δyδx

≈dydx

δy ≈dydx

× δx

= (2x + 3) × (2.01− 2)= (2 × 2 + 3) × 0.01= 0.07

3 y = 3x3(2x −1)3

Let u = 3x3 and v = (2x −1)3

dudx

= 9x2 dvdx

= 3(2x − 1)2(2)

= 6(2x −1)2

dydx

= u dvdx

+ v dudx

= 3x3[6(2x −1)2 ]+ (2x − 1)3(9x2 )= 18x3(2x −1)2 + 9x2 (2x − 1)3

= 9x2 (2x −1)2[2x + (2x −1)]= 9x 2(2x − 1)2(4x − 1)

4 y = 4x +

4x

= 4x + 4x−1

dydx

= 4 − 4x−2 = 4 −4x2

∴dxdy

=1dydx

=1

4 − 4x2

⎝ ⎜

⎠ ⎟

Rate of change of x:

dxdt

=dxdy

×dydt

=1

4 −4x2

⎝ ⎜

⎠ ⎟

× 2

=1

4 −432

⎝ ⎜

⎠ ⎟

× 2

=9

16units s−1

5 g(x) =

16(3x − 4)2

=

16

(3x − 4)−2

′ g (x) =−26

(3x − 4)−3(3)

= −(3x − 4)−3

′ ′ g (x) = 3(3x − 4)−4 (3)= 9(3x − 4)−4

=9

(3x − 4)4

∴ ′ ′ g 13

⎝ ⎜

⎠ ⎟ =

9

3× 13

− 4⎛

⎝ ⎜

⎠ ⎟

4=

19

Form 4: Chapter 9 (Differentiation) SPM Flashback

Fully-Worked Solutions

Rate of change of y = 2 units

Page 30: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

30

6 V =

13

πh2 (21− h)

V = 7πh2 −13

πh3

dVdh

= 14πh − πh2

dhdV

=1

14πh − πh2

Rate of change of depth of water

=dhdt

=dhdV

×dVdt

=1

14πh − πh2× 9

=1

14π (3) − π (3)2× 9

=3

11πcm s−1

7 y = (x + 3)2 = x2 + 6x + 9

dydx

= 2x + 6

If the gradient of the normal is −

16

, then the

gradient of the tangent is 6.

∴dydx

= 6

2x + 6 = 6x = 0

When x = 0, y = (0 + 3)2 = 9 Hence, the coordinates of point Q are (0, 9) .

8 y =

13

u6

y =13

(3x − 6)6

dydx

=63

(3x − 6)5(3)

= 6(3x − 6)5

9 (a) y = 3 + 14x − 2x3

dydx

= 14 − 6x2

When x = 2, dy

dx= 14 − 6(2)2 = −10

(b) δy ≈

dydx

× δx

= (−10) × [(2 + k ) − 2]= −10k

Page 31: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

31

Paper 2 1

Using the concept of similar triangles,

r0.5

=h

0.7

r =h

0.7× 0.5

r =57

h

V =13

π r2h

V =13

π 57

h⎛

⎝ ⎜

⎠ ⎟

2

h

V =25

147πh3

dVdh

=25

147π (3h2 )

dVdh

=2549

π h2

dhdV

=49

25πh2

Rate of change of the height of the water level:

dhdt

=dhdV

×dVdt

dhdt

=49

25πh2× 0.1

dhdt

=49

25π(0.3)2× 0.1

dhdt

=49

25(3.142)(0.3)2× 0.1

= 0.6931 m s−1

2 y = 2x3 − 3x2 − 12x + 11

dydx

= 6x2 − 6x − 12

d 2 ydx2

= 12x − 6

(a) At turning point,

dydx

= 0

6x2 − 6x −12 = 0x2 − x − 2 = 0

(x + 1)(x − 2) = 0x = −1 or 2

When x = −1,

y = 2(−1)3 − 3(−1)2 −12(−1) + 11 = 18 ∴ (−1, 18) is a turning point.

When x = −1,

d 2 ydx2

= 12(−1) − 6 = −18 (negative)

∴ (−1, 18) is a maximum point.

When x = 2,

y = 2(2)3 − 3(2)2 − 12(2) + 11 = −9 ∴ (2, −9) is a turning point.

When x = 2,

d 2 ydx2

= 12(2) − 6 = 18 (positive)

∴ (2, −9) is a minimum point. (b) At point (3, 2),

dydx

= 6(3)2 − 6(3) − 12 = 24

mgradient = 24

∴ mnormal = −124

The equation of normal is

y − 2 = −124

(x − 3)

24( y − 2) = −(x − 3)24y − 48 = −x + 3

24y = −x + 51

At the x-axis, y = 0.

24(0) = −x + 51x = 51

∴ P is point (51, 0).

Rate of increase of the volume of water:

dVdh

= 0.1 m3 s−1

Page 32: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

32

Paper 2 1 (a)

Area of ∆PRQ = 4 m2

12

(3)(3)sin∠RPQ = 4

sin∠RPQ =89

∠RPQ = 62.73°

Using the cosine rule,

RQ2 = 32 + 32 − 2(3)(3) cos 62.73°RQ2 = 9.75268RQ = 3.123 m (correct to 4

significant figures)

(b) Step 1 (Find PM where M is the midpoint

of QR)

Area of ∆PRQ = 4 m2

12

× RQ × PM = 4

12

× 3.123× PM = 4

PM = 2.5616 m

Step 2 (Determine ∠VMP )

The angle between the inclined plane VQR and the base PQR is 55°. Therefore, ∠VMP = 55°. Step 3 (Find VM) From ∆VRM , using the Pythagoras' Theorem, VM = 22 −1.56152 = 1.2497 m Step 4 (Calculate PV)

Using the cosine rule,

VP2 = 2.56162 + 1.24972 −2(2.5616)(1.2497) cos 55°

VP = 2.110 m (correct to 4significant figures)

(c) Step 1 (Find ∠VQP)

cos ∠VQP =22 + 32 − 2.1102

2(2)(3)cos ∠VQP = 0.712325cos ∠VQP = 44.58°

Step 2 (Find the area of ∆VPQ ) Area of ∆VPQ

=12

× 2 × 3× sin 44.58°

= 2.106 m2

Form 4: Chapter 10 (Solution of Triangles) SPM Flashback

Fully-Worked Solutions

Page 33: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

33

2 (a)

(i) Based on ∆ABC , using the sine rule,

sin ∠ABC15

=sin 30°

9

sin ∠ABC =sin 30°

9×15

sin ∠ABC = 0.83333∠ABC = 56.44°

(ii) Based on ∆ADC , using the cosine

rule,

152 = 102 + 82 − 2(10)(8)cos ∠ADC

cos ∠ADC =102 + 82 −152

2(10)(8)cos ∠ADC = −0.38125

∠ADC = 112.41°

(iii) Based on ∆ABC ,

∠ACB = 180° − 30° − 56.44°= 93.56°

Area of the quadrilateral ABCD

= Area of ∆ADC + Area of ∆ABC

=12

(10)(8) sin 112.41° +

12

(15)(9) sin 93.56°

= 36.9792 + 67.3697= 104.35 cm2

(b) (i) Based on ∆ABC , since the length of BC is shorter than the length of AC and ∠BAC is an acute non-included angle, the ambiguous case will occur. Another triangle (∆A ′ B C) that can be drawn is as shown below.

(ii)

∠A ′ B C = 180° − 56.44°

= 123.56°

3 (a) In ∆PQR , using the cosine rule,

PR2 = 102 + 72 − 2(10)(7) cos 75°= 112.76533

PR = 10.62 cm

(b)

In ∆PSR , using the sine rule,

sin ∠PSR10.62

=sin 40°

8sin ∠PSR = 0.85330

∠PSR = 58.57° or 121.43°

∠PS2 R ∠PS1R

Page 34: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

34

(c) (i) The obtuse ∠PSR is represented by

∠PS1R . In ∆PS1R ,

∠PRS1 = 180° − 40° − 121.43°

= 18.57°

In ∆PS1R , using the sine rule,

PS1

sin 18.57°=

10.62sin 121.43°

PS1 =10.62

sin 121.43°× sin 18.57°

PS1 = 3.964 cm

(ii) Area of quadrilateral PQRS

= Area of ∆PQR + Area of ∆PS1R

=12

×10 × 7 × sin 75° +

12

× 3.964 ×10.62 × sin 40°

= 47.34 cm2

4 (a) Area of ∆DBC = 29 cm2 12

× 6 ×10 × sin ∠DCB = 29

sin ∠DCB = 0.96667Basic ∠ = 75.16°∠DCB = 180° − 75.16°∠DCB = 104.84° (obtuse)

(b)

BD2 = 62 +102 − 2(6)(10) cos 104.84°BD2 = 166.73448BD = 12.91 cm

(c)

In ∆ABD,

sin ∠A12.91

=sin 45°

9.5

sin ∠A =sin 45°

9.5×12.91

sin ∠A = 0.96092∠A = 73.93°

∴∠DAB = 73.93°

and ∠D ′ A B = 180° − 73.93°

= 106.07°

Thus, ∠ABD = 180° − 45° − ∠DAB= 180° − 45° − 73.93°= 61.07°

Thus, ∠ ′ A BD = 180° − 45° − ∠D ′ A B= 180° − 45° −106.07°= 28.93°

First quadrant

Second quadrant

Page 35: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

35

Paper 2 1

Item Price in 2000

Price in 2002

Price index

for 2002

based on 2000 (I)

Weekly expenses(weigh-tage, w)

Iw

P RMp RM1.75 140 12 1680 Q RM2.00 RM2.30 115 28 3220 R RM4.00 RM4.80 q = 120 20 2400 S RM6.00 RM7.50 125 30 3750 T RM2.50 RMr 110 10 1100

w =∑

100

Iw =∑12 150

(a) (i) For item P,

I2002 =P2002

P2000×100

140 =1.75

p×100

p = 1.25

(ii) For item R,

I2002 =P2002

P2000×100

q =4.804.00

×100

q = 120

(iii) For item T,

I2002 =P2002

P2000×100

110 =r

2.50×100

r = 2.75

(b)

I =Iw∑w∑

=12 150

100= 121.5

(c)

I =P2002

P2000×100

121.5 =P2002

RM500×100

P2002 = RM607.50

Hence, the corresponding total monthly expenses in the year 2002 was RM607.50.

(d)

Hence, I 2004 (based on the year 2000)

=100 + 20

100× I 2002

=120100

×121.5

= 145.8

Form 4: Chapter 11 (Index Numbers) SPM Flashback

Fully-Worked Solutions

+ 21.5% + 20% Year 2000 Year 2002 Year 2004

Page 36: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

36

2 (a) (i) For item S,

I2004 =P2004

P2002×100

110 =1.50P2002

×100

P2002 =1.50110

×100

P2002 = RM1.36

Hence, the price of item S in the year 2002 was RM1.30.

(ii) For item P,

It is given that:

I2002 (based on 2000) = 105P2002

P2000×100 = 105

∴ P2000 =

P2002 ×100105

From the table, we can see that:

I2004 (based on 2002) = 115P2004

P2002×100 = 115

∴ P2004 =

115 × P2002

100 …

I2004 (based on 2000)

=P2004

P2000×100

=

115 × P2002

100P2002 ×100

105

×100

=115100

×105100

×100

= 120.75

Hence, the price index of item P for the year 2004 based on the year 2000 is 120.75.

(b) (i) I = 110

Iw∑w∑

= 110

(115 × 20) + (10x) + (105 × 40) + (110 × 30)

20 + 10 + 40 + 30= 110

9800 + 10x100

= 110

9800 + 10x = 11 00010x = 1200

x = 120

(ii) I 2004 =

P2004

P2002× 100

110 =22

P2002×100

P2002 =22

110×100

P2002 = 20.00

Hence, the price of a box of icecream in the year 2002 was RM20.00.

From

From

Page 37: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

37

3 (a) I2004 (based on 2002)

=

P2004

P2002×100

For material K,

p =

1.751.40

×100 = 125

For material M,

q2

×100 = 140

q =140 × 2

100q = 2.80

For material N,

2.40r

×100 = 80

r =2.40 ×100

80r = 3.00

(b) (i)

Material I2004 (based on the year

2002)

Angle of pie chart (degrees)

w

K 125 75 15 L 150 40 8 M 140 155 31 N 80 90 18

I 2004 (based on 2002)

=Iw∑w∑

=(125 × 15) + (150 × 8) + (140 × 31) + (80 × 18)

15 + 8 + 31 + 18

=885572

= 122.99

(ii) I 2004 (based on 2002) = 122.99

P2004

P2002×100 = 122.99

RM9P2002

×100 = 122.99

P2002 =RM9 ×100

122.99= RM7.32

(c)

I 2002+ 22.99%⎯ → ⎯ ⎯ ⎯ I 2004

+ 20%⎯ → ⎯ ⎯ I 2006

(100) (122.99) (?)

I 2006 (based on 2002)

=100 + 20

100×122.99

= 147.59

Page 38: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

38

4 (a) For component U,

I2006 = 120P2006

P2004×100 = 120

h50

×100 = 120

h = 60

(b) For component S,

I2006 = 125P2006

P2004×100 = 125

mk

×100 = 125

100m = 125k

4m = 5k … P2006 = 20 + P2004

m = 20 + k … Substituting into :

4(20 + k ) = 5k80 + 4k = 5k

k = 80

From : m = 20 + 80 = 100

(c) (i) I = 132

P2006

P2004×100 = 132

1716P2004

×100 = 132

P2004 = RM1300

(ii)

Component I2006 w U 120 1

R 4025

×100 = 160 3

S 125 4

T 4440

×100 = 110 p

I = 132(120 × 1) + (160 × 3) + (125 × 4) + 110p

1 + 3 + 4 + p= 132

1100 + 110p8 + p

= 132

1100 + 110p = 1056 + 132p44 = 22pp = 2

Page 39: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

39

Paper 1 1 (a) k + 1, k + 5, 2k + 6 , … arithmetic progression

d = T2 − T1 = k + 5 − (k + 1) = 4d = T3 − T2 = 2k + 6 − (k + 5) = k + 1

Since common difference is always the same, k +1 = 4 ⇒ k = 3

(b) When k = 3, we have 4, 8, 12, …

S8 =

82

[2(4) + 7(4)] = 144

2 (a) T4 = 24

ar3 = 2481r3 = 24

r3 =2481

=827

r =23

(b)

S∞ =a

1− r=

81

1− 23

= 243

3 Since k, 3, 9

k, m are four consecutive terms

of a geometric progression,

3k

=m9k

3k

=mk9

mk 2 = 27

m =27k 2

4 The arithmetic progression 5, 9, 13, … has a common difference of 4.

S3 = 5732

[2a + (3 − 1)(4)] = 57

32

(2a + 8) = 57

3a +12 = 57a = 15

Hence, the three consecutive terms which sum up to 57 are 15, 19 and 23.

+4 +4 5 Volumes of water in litres:

410, 425, 440, … Volume of water at the end of the 8th day

= T8

= a + 7d= 410 + 7(15)= 515 litres

6 0.848484K

= 0.84 + 0.0084 + 0.000084 +K

=0.84

1− 0.01

=0.840.99

=8499

=2833

Form 5: Chapter 12 (Progressions) SPM Flashback

Fully-Worked Solutions

S∞ =

a1− r

These three terms are not the first three terms but any three consecutive terms with a common difference of 4. Therefore, a new value of a (first term) has to be determined.

Page 40: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

40

7 (a) If 3, k, 48 are in an arithmetic progression, then

k − 3 = 48 − k2k = 51

k = 25.5

(b) If 3, k, 48 are in a geometric progression,

then

k3

=48k

k 2 = 144k = 12

8 (a) The common difference of the arithmetic

progression 2, 5, 8, … is 5 − 2 = 3 . (b) The sum of all the terms from the 4th term

to the 23rd term

= S23 − S3

=232

[2(2) + (23− 1)(3)]−32

[2(2) + (3− 1)(3)]

= 805 −15= 790

9 (a) The common ratio of the geometric

progression 2, 6, 18, K is 62

= 3 .

(b) Sn = 6560

a(r n −1)r − 1

= 6560

2(3n −1)3 − 1

= 6560

3n −1 = 65603n = 65613n = 38

∴ n = 8

10 138 += kT

a + 7d = 3k + 1a + 7(4) = 3k + 1

a = 3k − 27 …

S8 = 13k + 682

(2a + 7d ) = 13k + 6

8a + 28d = 13k + 68a + 28(4) = 13k + 6

8a = 13k −106 … Substituting into ,

8(3k − 27) = 13k − 10624k − 216 = 13k − 106

11k = 110k = 10

11 (a) T3 = 10

ar2 = 10 …

T3 + T4 = 1510 + T4 = 15

T4 = 5

ar3 = 5 …

: ar3

ar2=

510

r =12

From :

a 12

⎝ ⎜

⎠ ⎟

2

= 10

a 14

⎝ ⎜

⎠ ⎟ = 10

a = 40

(b)

S∞ =40

1− 12

= 80

Page 41: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

41

Paper 2 1 (a)

T1 = Area of the first triangle

=12

bh

T2 = Area of the second triangle

=12

b2

⎝ ⎜

⎠ ⎟

h2

⎝ ⎜

⎠ ⎟ =

18

bh

T3 = Area of the third triangle

=12

b4

⎝ ⎜

⎠ ⎟

h4

⎝ ⎜

⎠ ⎟ =

132

bh

T2

T1=

18

bh

12

bh=

14

T3

T2=

132

bh

18

bh=

14

Since

T2

T1=

T3

T2=

14

(a constant), the

areas of the triangles form a geometric

progression with a common ratio of 14

.

(b) (i)

T1 =

12

(160)(80) = 6400

T2 =18

(160)(80) = 1600

T3 =1

32(160)(80) = 400

Tn = 25ar n−1 = 25

6400 14

⎝ ⎜

⎠ ⎟

n−1

= 25

14

⎝ ⎜

⎠ ⎟

n−1

=1

256

14

⎝ ⎜

⎠ ⎟

n−1

=14

⎝ ⎜

⎠ ⎟

4

n − 1 = 4n = 5

Hence, the 5th triangle has an area of 25 cm2.

(ii) Sum to infinity,

S∞ =a

1− r

=6400

1− 14

= 8533 13

cm2

Page 42: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

42

2 (a) The number of cubes in each storey forms a geometric progression 1, 4, 16, 64, …, where a = 1 and r = 4.

Tn = 4096ar n−1 = 4096

1(4)n−1 = 40964n−1 = 46

n − 1 = 6n = 7

Hence, the height of the model is 7 × 4 = 28 cm .

(b)

S7 =1(47 −1)

4 − 1= 5461

Hence, the total price of cubes used

= 5461× 80= 436 880 sen= RM4368.80

3 (a) T5 = 320 a + 4d = 320

h + 4k = 320 …

S8 = 244082

(2a + 7d ) = 2440

4(2h + 7k) = 2440

2h + 7k = 610 …

× 2 − : k = 30 From : h + 4(30) = 320 h = 200

(b) Tn (of Epsilon) = Tn (of Sigma)

200 + (n − 1)(30) = 160 + (n − 1)(35)200 + 30n − 30 = 160 + 35n − 35

45 = 5nn = 9

Page 43: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

43

Paper 1 1

y = px2 + qxyx

= px + q

For the point (2, 6), x = 2 and yx

= 6.

∴ 6 = p(2) + q …

For the point (10, 2), x = 10 and yx

= 2 .

∴ 2 = p(10) + q …

− : −4 = 8 p ⇒ p = −

12

From : 6 = −

12

⎝ ⎜

⎠ ⎟ (2) + q ⇒ q = 7

2

y = 7x2 − x3

yx2

= 7 − x

The straight line passes through the

point (2, h). Thus, x = 2 and y

x2= h .

yx2

= 7 − x

h = 7 − 2h = 5

The straight line passes through the

point (k, 3). Thus, x = k and y

x2= 3.

yx2

= 7 − x

3 = 7 − kk = 4

3 (a) 2mxy =

log10 y = log10 mx2

log10 y = log10 m + log10 x2

log10 y = log10 m + 2 log10 xlog10 y = 2 log10 x + log10 m

(b) (i) Y-intercept = −1

log10 m = −1m = 10−1

m =1

10

(ii) Gradient = 2

k − (−1)2 − 0

= 2

k +1 = 4k = 3

4 y = −3x3 + 4

yx3

= −3 +4x3

yx3

= 4 1x3

⎝ ⎜

⎠ ⎟ − 3

(Y = 4X + c)

By comparison, Y =

yx 3

and X =

1x 3

.

Form 5: Chapter 13 (Linear Law) SPM Flashback

Fully-Worked Solutions

Gradient Y-intercept

Divide throughout by x3.

Rearrange

Page 44: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

44

Paper 2 1 (a)

x 1 2 3 4 5 y 1.32 1.76 2.83 5.51 13.00x2 1 4 9 16 25

log10 y 0.121 0.246 0.452 0.741 1.114 The graph of log10 y against x2 is as shown below.

(b)

y = abx2

log10 y = log10 a + x2 log10 b

(i) Y-intercept = 0.08

log10 a = 0.08a = antilog 0.08a = 1.2

(ii) Gradient =

0.74 − 0.1216 − 1

log10 b =0.6215

= 0.04133

b = antilog 0.04133b = 1.1

→↑

Non-linear

Linear

Page 45: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

45

2 (a) x 2 4 6 8 10 12 y 5.18 11.64 26.20 58.95 132.63 298.42

log10 y 0.71 1.07 1.42 1.77 2.12 2.47

y = pk x

log10 y = log10 p + x log10 k

The graph of log10 y against x is a straight-line graph, as shown below:

(b) (i) log10 p = Y-intercept

log10 p = 0.36p = 2.29

(ii) log10 k = gradient

log10 k =2.12 − 1.42

10 − 6

log10 k =0.74

log10 k = 0.175k = 1.5

Page 46: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

46

3 (a) x 2.5 3.0 3.5 4.0 4.5 5.0 y 7.0 7.7 8.4 9.9 10.1 11.0

xy 17.5 23.1 29.4 39.6 45.5 55.0x2 6.3 9.0 12.3 16.0 20.3 25.0

The graph of xy against x2 is as shown below.

(b) (i) From the graph, the value of y which is incorrectly recorded is 9.9. The actual value of y is given by:

xyactual = 374(yactual) = 37

yactual = 9.25

(ii)

y = qx +p

qx

xy = qx2 +pq

q = Gradient

q =55 − 525 − 0

q = 2

pq

= Y-intercept

pq

= 5

p2

= 5

p = 10

→↑

Page 47: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

47

4 (a) x 1 3 5 7 9 11 y 5 20 80 318 1270 5050

x + 1 2 4 6 8 10 12 log10 y 0.70 1.30 1.90 2.50 3.10 3.70

The graph of log10 y against (x + 1) is as shown below.

(b) y = hq x +1

log10 y = log10 h + (x +1) log10 qlog10 y = (x +1) log10 q + log10 h

Gradient Y-intercept

Y-intercept = 0.1log10 h = 0.1

h = 1.26

Gradient =3.7 − 0.712 − 2

log10 q = 0.3q = 2

→↑

Page 48: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

48

Paper 1

1

3(1+ 2x)4

dx∫

= 3(1+ 2x)−4 dx∫= 3 (1+ 2x)−3

−3(2)⎛

⎝ ⎜

⎠ ⎟ + c

= −12

(1+ 2x)−3 + c

But it is given that

3(1+ 2x)4

dx∫ = k(1+ 2x)n + c , thus by

comparison, k = −

12

and n = −3.

2 Area of the shaded region = 54 units2

y dx0

k

∫ = 54

6x2 dx0

k

∫ = 54

6 x3

3⎛

⎝ ⎜

⎠ ⎟

⎣ ⎢ ⎤

⎦ ⎥ 0

k

= 54

[2x3 ]0k = 54

2(k 3 − 03 ) = 54k 3 = 27k = 3

3

(4x −1) dx−1

k

∫ = 3

4x2

2− x

⎣ ⎢ ⎤

⎦ ⎥ −1

k

= 3

[2x2 − x]−1k = 3

2k 2 − k − [2(−1)2 − (−1)] = 32k 2 − k − 3 = 32k 2 − k − 6 = 0

(2k + 3)(k − 2) = 0

k = −32

or 2

k = −32

is not accepted

∴ k = 2

4

f (x) dx1

2

∫ + [ f (x) + cx] dx2

3

∫ = 30

f (x) dx1

2

∫ + f (x) dx2

3

∫ + cx dx2

3

∫ = 30

f (x) dx1

3

∫ + c x2

2⎡

⎣ ⎢ ⎤

⎦ ⎥ 2

3

= 30

5 + c 32 − 22

2⎛

⎝ ⎜

⎠ ⎟ = 30

52

c = 25

c = 10

5

12

f (x) dxa

b

∫ =12

f (x) dxa

b

∫ =12

(−6) = −3

Form 5: Chapter 14 (Integration) SPM Flashback

Fully-Worked Solutions

The area of the region below the x-axis has a negative sign.

Page 49: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

49

6

[nx − g(x)] dx = 91

4

nx dx1

4

∫ − g(x) dx1

4

∫ = 9

n x2

2⎡

⎣ ⎢ ⎤

⎦ ⎥ 1

4

+ g(x) dx4

1

∫ = 9

n 42

2−

12

2⎛

⎝ ⎜

⎠ ⎟ − 6 = 9

152

n − 6 = 9

152

n = 15

n = 2

g(x) dxa

b

∫ = − g(x) dxb

a

Page 50: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

50

Paper 2

1 (a) dydx

= 2x + 4

y = (2x + 4) dx∫y =

2x2

2+ 4x + c

y = x2 + 4x + c

y = 7 when x = 1 :7 = 12 + 4(1) + cc = 2

∴ y = x 2 + 4x + 2

(b) x2 d 2 y

dx2+ (x − 1) dy

dx+ y + 3 = 0

x2 (2) + (x −1)(2x + 4) + (x2 + 4x + 2) + 3 = 02x2 + 2x2 + 2x − 4 + x2 + 4x + 2 + 3 = 0

5x2 + 6x +1 = 0(5x + 1)(x +1) = 0

x = −15

or −1

2 8y = 3x

x =

83

y …

x = y2 −1 … Substitute into :

83

y = y2 − 1

8y = 3y2 − 33y2 − 8y − 3 = 0

(3y +1)( y − 3) = 0

y = −13

or 3

It is obvious that for point A, y = 3. From , when y = 3,

x =83

y =83

(3) = 8

∴ A = (8, 3)

Volume of the solid generated when the shaded region is revolved through 360° about the y-axis = Volume of the cone generated by the line

segment OA − Volume of the solid generated by the curve x = y2 − 1 from y = 1 to y = 3

=13

πr2h − π x2 dy1

3

∫=

13

π (8)2 (3) − π ( y2 − 1)2 dy1

3

∫= 64π − π ( y4 − 2y2 +1) dy

1

3

∫= 64π − π y5

5−

2 y3

3+ y

⎣ ⎢ ⎤

⎦ ⎥ 1

3

= 64π − π 35

5−

2(3)3

3+ 3−

15

−23

+ 1⎛

⎝ ⎜

⎠ ⎟

⎣ ⎢ ⎤

⎦ ⎥

= 64π − π 2435

−18 + 3−15

+23

− 1⎛

⎝ ⎜

⎠ ⎟

= 64π −49615

π

= 30 1415

π units3

Page 51: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

51

3 (a) dydx

= 3x2 − 4x

y = (3x2 − 4x) dx∫y =

3x3

3−

4x2

2+ c

y = x3 − 2x2 + c

The curve passes through the point A(1, −9).

−9 = 13 − 2(1)2 + cc = −8

Hence, the equation of the curve is y = x3 − 2x2 − 8.

(b) At turning points, dydx

= 0.

dydx

= 0

3x2 − 4x = 0x(3x − 4) = 0

x = 0 or 1 13

d 2 ydx2

= 6x − 4

When x = 0, y = (0)3 − 2(0)2 − 8 = −8 ∴ (0, −8) is a turning point.

d 2 ydx2

= 6(0) − 4 = −4 (negative)

Hence, (0, −8) is a maximum point.

When x = 1 1

3,

y =

43

⎝ ⎜

⎠ ⎟

3

− 2 43

⎝ ⎜

⎠ ⎟

2

− 8 = −9 527

∴ 1 1

3, −9 5

27⎛

⎝ ⎜

⎠ ⎟ is a turning point.

d 2 ydx2

= 6 43

⎝ ⎜

⎠ ⎟ − 4 = 4 (positive)

Hence, 1 1

3, −9 5

27⎛

⎝ ⎜

⎠ ⎟ is a minimum point.

Page 52: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

52

4 (a) y =

4(2x − 1)2

= 4(2x −1)−2

dydx

= −8(2x −1)−3(2)

= −16

(2x −1)3

At the point A(1, 4),

m =dydx

= −16

[2(1) − 1]3= −16

Hence, the equation of the tangent at the point A(1, 4) is

y − y1 = m(x − x1 )y − 4 = −16(x − 1)y − 4 = −16x + 16

y = −16x + 20

(b)

(i) Area of the shaded region

= y dx2

3

∫=

4(2x − 1)22

3

∫ dx

= 4(2x −1)−22

3

∫ dx

=4(2x −1)−1

−1(2)⎡

⎣ ⎢ ⎤

⎦ ⎥ 2

3

= −2

2x − 1⎡

⎣ ⎢ ⎤

⎦ ⎥ 2

3

= −2

2(3) − 1− −

22(2) −1

⎝ ⎜

⎠ ⎟

= −25

− −23

⎝ ⎜

⎠ ⎟

=4

15 units2

(ii) Volume generated, Vx

= π y2 dx2

3

∫= π 4

(2x − 1)2

⎝ ⎜

⎠ ⎟

2

dx2

3

∫= π 16

(2x − 1)4dx

2

3

∫= π 16(2x − 1)−4 dx

2

3

∫= π 16(2x −1)−3

−3(2)⎡

⎣ ⎢ ⎤

⎦ ⎥ 2

3

= π 8−3(2x − 1)3

⎣ ⎢ ⎤

⎦ ⎥ 2

3

= −83

π 1(2x − 1)3

⎣ ⎢ ⎤

⎦ ⎥ 2

3

= −83

π 1[2(3) − 1]3

−1

[2(2) − 1]3

⎝ ⎜

⎠ ⎟

= −83

π 1125

−127

⎝ ⎜

⎠ ⎟

=784

10125π units3

Page 53: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

53

5 (a) The gradient of the straight line x + y − 8 = 0 ⇒ y = −x + 8 is −1.

The gradient function is dydx

= px3 + x2.

dydx

= −1

px3 + x2 = −1p(−1)3 + (−1)2 = −1

−p +1 = −1p = 2

(b) dydx

= px3 + x2 = 2x3 + x2

y = (2x3 + x2 ) dx∫y =

2x4

4+

x3

3+ c

y =x4

2+

x3

3+ c

At the point (−1, 0), x = −1, y = 0

∴ 0 =(−1)4

2+

(−1)3

3+ c

0 =12

−13

+ c

c = −16

Hence, the equation of the curve is

y =

x4

2+

x 3

3−

16

6 (a) Equation of RAQ: 2y = −x + 10 At point Q (on the x-axis), y = 0.

2(0) = −x + 10x = 10

∴ Q = (10, 0)

Area of the shaded region = Area under the curve from x = 0 to

x = 2 + Area under the straight line AQ

=x2

2+ 2

⎝ ⎜

⎠ ⎟ dx

0

2

∫ +12

× 8 × 4⎛

⎝ ⎜

⎠ ⎟

=x3

6+ 2x

⎣ ⎢ ⎤

⎦ ⎥ 0

2

+ 16

=23

6+ 2(2) − 0 +16

= 21 13

units2

(b)

Volume generated = Volume generated by the curve from

y = 2 to y = 4

= π (2y − 4) dy2

4

∫= π[y2 − 4y]2

4

= π[42 − 4(4) − (22 − 4(2))]= 4π units3

x2

2= y − 2

x2 = 2y − 4

At the point (−1, 0), x = −1.

Page 54: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

54

7 (a) y = x − 4 …

x = ( y − 2)2 … Substituting into :

y = ( y − 2)2 − 4y = y2 − 4y + 4 − 4y = y2 − 4y0 = y2 − 5y0 = y( y − 5)y = 0 or 5

From : When y = 0, x = (0 − 2)2 = 4 ∴ B = (4, 0) When y = 5, x = (5 − 2)2 = 9 ∴ A = (9, 5)

(b)

Area of the shaded region P = Area of trapezium − Area under the

curve

=12

(4 + 9)(5) − x dy0

5

∫= 32.5 − ( y − 2)2 dx

0

5

∫= 32.5 −

( y − 2)3

3(1)⎡

⎣ ⎢ ⎤

⎦ ⎥ 0

5

= 32.5 −(5 − 2)3

3−

(0 − 2)3

3⎛

⎝ ⎜

⎠ ⎟

= 20 56

units2

(c) Vy = π x2 dy

0

2

= π ( y − 2)4 dy0

2

∫= π ( y − 2)5

5(1)⎡

⎣ ⎢ ⎤

⎦ ⎥ 0

2

= π (2 − 2)5

5−

(0 − 2)5

5⎛

⎝ ⎜

⎠ ⎟

= π 0 − −325

⎝ ⎜

⎠ ⎟

⎣ ⎢ ⎤

⎦ ⎥

= 6 25

π units3

Page 55: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

55

Paper 1

1 (a) Since P = (4, 2), OP = 4

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

(b) Since Q = (−6, 3), OQ = −6i + 3 j .

Hence, QO = −OQ = 6i − 3 j . 2 r = 2p − 3q

r = 2(3a + 2b) − 3(3a − b)r = 6a + 4b − 9a + 3br = −3a + 7b

But it is given that r = ha + (h + k )b . By comparison,

h = −3 and h + k = 7−3+ k = 7

k = 10

3

(a)

BD = BA + AD= −12 p + 6q

(b) EC = EB + BC

=13

DB + BC

=13

(−BD) + BC

=13

(12 p − 6q) + 6q

= 4p − 2q + 6q

= 4 p + 4q

4 (a) AB = OB − OA

= 3i + 17 j − (−2i + 5 j)

= 5i + 12 j

(b)

AB = 52 +122 = 169 = 13

Unit vector in the direction of AB

=1

AB( AB)

=1

13(5i +12 j)

=5

13i +

1213

j

5 AB + 2BC = 11i −13 j

(OB − OA) + 2(OC − OB) = 11i − 13 j

53

⎛ ⎝ ⎜

⎞ ⎠ ⎟ − −1

7⎛ ⎝ ⎜

⎞ ⎠ ⎟ + 2 m

p⎛ ⎝ ⎜

⎞ ⎠ ⎟ − 5

3⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎡ ⎣ ⎢

⎤ ⎦ ⎥ =

11−13

⎛ ⎝ ⎜

⎞ ⎠ ⎟

6−4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + 2 m − 5

p − 3⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 11

−13⎛ ⎝ ⎜

⎞ ⎠ ⎟

6 + 2m − 10−4 + 2 p − 6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 11

−13⎛ ⎝ ⎜

⎞ ⎠ ⎟

2m − 42 p −10

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 11

−13⎛ ⎝ ⎜

⎞ ⎠ ⎟

Equating the x-components:

2m − 4 = 11

m = 7 12

Equating the y-components:

2p − 10 = −13

p = −1 12

Form 5: Chapter 15 (Vectors) SPM Flashback

Fully-Worked Solutions

Page 56: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

56

6 (a) OQ = 9

−12⎛ ⎝ ⎜

⎞ ⎠ ⎟

(b)

OQ = 92 + (−12)2 = 15

Hence, the unit vector in the direction of

OQ =1

159

−12⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

35

− 45

⎜ ⎜ ⎜

⎟ ⎟ ⎟

.

7 BM = BA + AM

= CO +12

AO

= (4i − 2 j) +12

(−2i − 6 j)

= 4i − 2 j − i − 3 j

= 3i − 5 j

8 (a) A(6, 3) ⇒ OA = 6

3⎛ ⎝ ⎜

⎞ ⎠ ⎟

B(0, −5) ⇒ OB = 0

−5⎛ ⎝ ⎜

⎞ ⎠ ⎟

AB = OB − OA

= 0−5

⎛ ⎝ ⎜

⎞ ⎠ ⎟ − 6

3⎛ ⎝ ⎜

⎞ ⎠ ⎟

= −6−8

⎛ ⎝ ⎜

⎞ ⎠ ⎟

(b)

AB = (−6)2 + (−8)2 = 100 = 10

Unit vector in the direction of AB

=1

AB( AB)

=1

10−6−8

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=−

35

−45

⎜ ⎜ ⎜

⎟ ⎟ ⎟

9 (a) Since the point X, Y and Z are collinear,

XY = kYZ2a + 4b = k(3a + mb)2a + 4b = 3ka + kmb

Equating the coefficients of a ,

3k = 2

k =23

Equating the coefficients of b ,

km = 423

m = 4

m = 6

(b) XY =

23

YZ

XY =23

YZ

XY

YZ=

23

XY : YZ = 2 : 3

k =

23

Page 57: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

57

Paper 2 1 (a) Using the concept of position vectors:

AB = 1014

⎛ ⎝ ⎜

⎞ ⎠ ⎟

OB − OA = 1014

⎛ ⎝ ⎜

⎞ ⎠ ⎟

46

⎛ ⎝ ⎜

⎞ ⎠ ⎟ − OA = 10

14⎛ ⎝ ⎜

⎞ ⎠ ⎟

OA = 46

⎛ ⎝ ⎜

⎞ ⎠ ⎟ − 10

14⎛ ⎝ ⎜

⎞ ⎠ ⎟

OA = −6−8

⎛ ⎝ ⎜

⎞ ⎠ ⎟

∴ A = (−6, − 8)

(b)

OA = (−6)2 + (−8)2 = 100 = 10

Unit vector in the direction of OA

=1

OA(OA)

=1

10−6−8

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=− 3

5− 4

5

⎝ ⎜

⎠ ⎟

(c) Since CD is parallel to AB, CD = k AB (k is a constant).

m7

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = k 10

14⎛ ⎝ ⎜

⎞ ⎠ ⎟

m7

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 10k

14k⎛ ⎝ ⎜

⎞ ⎠ ⎟

Equating the x-components: m = 10k … Equating the y-components:

7 = 14k

k =12

From , when k =

12

,

m = 10 1

2⎛

⎝ ⎜

⎠ ⎟ = 5

Page 58: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

58

2

(a) (i)

AP = AO + OP

AP = −4a + 4 p

(ii) OQ = OA + AQ

OQ = OA +14

AB

OQ = OA +14

( AO + OB)

OQ = OA +14

( AO + 3OP)

OQ = 4a +14

[−4a + 3(4 p)]

OQ = 4a − a + 3p

OQ = 3a + 3 p

(b) (i)

AR = hAP = h(−4a + 4p)

AR = −4ha + 4h p

(ii) RQ = kOQ = k(3a + 3p)

RQ = 3ka + 3k p

(c) AQ = AR + RQ

−a + 3p = −4ha + 4hp + 3ka + 3k p

−a + 3p = (−4h + 3k )a + (4h + 3k )p

Equating the coefficients of a and of b respectively, we have:

+

−4h + 3k = −14h + 3k = 3

6k = 2

k = 13

KK

From : −4h + 3 1

3⎛

⎝ ⎜

⎠ ⎟ = −1

−4h = −2

h =12

Given that

OP =

13

OB ,

∴ OB = 3OP .

Given AQ =

14

AB

From (a) (ii),

AQ = −a + 3p

Page 59: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

59

3 (a)

(i)

AC = AO + OC= −6a + 8b

(ii)

BD = BC + CD= 5b + (8a − 5b)= 8a

(b) If the points A, E and C are collinear, then

AC = k AE

−6a + 8b = k( AO + OB + BE)

−6a + 8b = k(−6a + 3b + hBD)−6a + 8b = k[−6a + 3b + h(8a)]−6a + 8b = −6ka + 3kb + 8hka−6a + 8b = (−6k + 8hk )a + 3kb

Equating the coefficients of b :

3k = 8

k =83

Equating the coefficients of a :

−6 = −6k + 8hk

−6 = −6 83

⎝ ⎜

⎠ ⎟ + 8h 8

3⎛

⎝ ⎜

⎠ ⎟

−18 = −48 + 64h30 = 64h

h =1532

(c)

OC = 8b = 8b = 8(2) = 16

OA = 6a = 6a = 6(2) = 12

AC = OC2

+ OA2

= 162 + 122

= 20

Given that OB =

38

OC , thus OC =

83

OB .

Therefore, OC =

83

OB =83

(3b) = 8b .

This gives BC = 8b − 3b = 5b.

Page 60: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

60

4

(a) PR = PQ + QR

= 4u +32

PS

= 4u +32

(12v)

= 4u + 18v

(b) (i) TX = mPQ

= m(4u)= 4mu

(ii) P, X and R are collinear.

PX = k PR

PT + TX = k(4u +18v)34

PS + 4mu = 4ku +18kv

34

(12v) + 4mu = 4ku +18kv

9v + 4mu = 4ku +18kv

Equating the coefficients of v ,

9 = 18k

k =12

Equating the coefficients of u ,

4m = 4k

4m = 4 12

⎝ ⎜

⎠ ⎟

m =12

Given that

PS =

23

QR ,

thus

QR =

32

PS .

Given From (b) (i)

Page 61: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

61

Paper 1 1

(a) cot (−θ )

=1

tan (−θ )

=1

−tan θ

= −1t

(b) cos (90° − θ )

= sin θ

=t

1+ t 2

2 6 sec2 θ − 20 tan θ = 0

6(1+ tan2 θ ) − 20 tan θ = 06 + 6 tan2 θ − 20 tan θ = 06 tan2 θ − 20 tan θ + 6 = 03 tan2 θ −10 tan θ + 3 = 0(3 tan θ −1)(tan θ − 3) = 0

tan θ =13

or 3

When tan θ =13

,

θ = 18.43°, 198.43°

When tan θ = 3,θ = 71.57°, 251.57°

∴θ = 18.43°, 71.57°, 198.43°, 251.57°

3 2 sin2 x + cos x = 1

2(1− cos2 x) + cos x −1 = 02 − 2 cos2 x + cos x −1 = 0−2 cos2 x + cos x +1 = 0

2 cos2 x − cos x −1 = 0(2 cos x +1)(cos x −1) = 0

cos x = −

12

or cos x = 1

When cos x = −12

,

x = 120°, 240°

When cos x = 1,x = 0°, 360°

∴ x = 0°, 120°, 240°, 360°

4 cos 2θ − 3 sin θ = 2

1− 2 sin2 θ − 3 sin θ − 2 = 0−2 sin2 θ − 3 sin θ −1 = 0

2 sin2 θ + 3 sin θ +1 = 0(2 sin θ +1)(sin θ +1) = 0

sin θ = −

12

or sin θ = −1

When sin θ = −12

,

Basic ∠ = 30°θ = 210°, 330°

When sin θ = −1,θ = 270°

∴θ = 210°, 270°, 330°

Form 5: Chapter 16 (Trigonometric Functions) SPM Flashback

Fully-Worked Solutions

tan θ = t

Page 62: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

62

5 15 cos2 x − 7 cos x = 4 cos 60°

15 cos2 x − 7 cos x = 4(0.5)15 cos2 x − 7 cos x = 2

15 cos2 x − 7 cos x − 2 = 0(3 cos x − 2)(5 cos x +1) = 0

cos x =

23

or cos x = −15

When cos x =23

Basic ∠ = 48.19°∴ x = 48.19°, 311.81°

When cos x = −15

Basic ∠ = 78.46°∴ x = 101.54°, 258.46°

∴ x = 48.19°, 101.54°, 258.46°, 311.81°

Page 63: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

63

Paper 2 1 (a) LHS

= tan x2+ cot x

2

=sin x

2

cos x2

+cos x

2

sin x2

=sin2 x

2+ cos2 x

2

sin x2

cos x2

=1

sin x2

cos x2

=2

2 sin x2

cos x2

=2

sin x= 2 cosec x= RHS

(b) (i)

(ii) sin 3

2x =

34π

x −1

2 sin 3

2x =

32π

x − 2

The solutions to the equation

2 sin 3

2x =

32π

x − 2 are given by the

x-coordinates of the intersection

points of the graphs of y = 2 sin 3

2x

and y =

32π

x − 2.

Hence, the equation of the straight line for solving the equation

sin 3

2x =

34π

x −1 is y =

32π

x − 2 .

Number of solutions = Number of intersection points = 3

This is a y = sin θ graph

with 1 1

2 cycles

because θ = 1 1

2x.

x 0 2π y −2 1

Page 64: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

64

2 (a) The sketch of the graph of y = cos 2x for 0 ≤ x ≤ π is as shown below:

(b) 2 sin2 x = 2 − x

π

1− cos 2x = 2 − xπ

−cos 2x = 1− xπ

cos 2x =xπ−1

The straight line that has to be drawn is

y = x

π− 1.

x 0 π y −1 0

Hence, the number of solutions to the

equation 2 sin2 x = 2 − x

π for 0 ≤ x ≤ π

= Number of intersection points = 2

3 (a) LHS = −2 cos2 x + cosec2 x − cot2 x

= −2 cos2 x +1

= −(2 cos2 x −1)= −cos 2x= RHS

(b) (i) The sketch of the graph of

y = −cos 2x is as shown below.

(ii) 2(−2 cos2 x + cosec2 x − cot2 x) = x

π−1

2(−cos 2x) = xπ−1

−cos 2x =x

2π−

12

The straight line that has to be

sketched is y =

x2π

−12

.

x 0 2π

y −

12

12

Number of solutions = Number of intersection points = 4

If cot2 x +1 = cosec2 x , then cosec2 x − cot2 x = 1.

cos 2x = 1− 2 sin2 x∴ 2 sin2 x = 1− cos 2x

Page 65: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

65

4 (a), (b)

sin 2x + 2 cos x = 0

sin 2x = −2 cos x

Number of solutions = Number of intersection points = 2

Page 66: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

66

Paper 1 1 Number of codes that can be formed

= 6P3 × 4P2 = 120×12 = 1440. 2 (a) Number of teams that can be formed if

each team consists of 3 boys (and 5 girls)

= 7C3 × 6C5

= 35× 6= 210

(b) Available: 7 boys and 6 girls

Needed: 7 boys and 1 girl or 8 boys and 0 girls Hence, the number of teams that can be formed if each team consists of not more than 1 girl = 7C7 × 6C1 = 6

3 (a) Number of arrangements = 5! = 120 (b) Step 1

If the letter 'O ' and 'E ' have to be side by side, they will be counted as 1 item. Together with the letters 'P ', 'W ' and 'R ', there are altogether 4 items, as illustrated below:

Number of arrangements = 4! Step 2 But the letters 'O ' and 'E ' can also be arranged among themselves in their group.

Number of arrangements = 2!

Step 3 Hence, using the rs multiplication principle, the number of arrangements of all the letters of the word 'POWER ' in which the letters O and E have to be side by side = 4!× 2!= 24 × 2 = 48

4 (a) If there is no restriction, the number of

ways the dancing groups can be formed

= 16C6

= 8008

(b) If the group of dancers must consist of 1

Form Five student and exactly 2 Form Six students, the number of Form Four students required is 3. Hence, the number of ways the dancing groups can be formed

= 7C3 × 5C1 × 4C2

= 1050

5 Number of arrangements

N 7P3

T 7P3

G 7P3

R 7P3

L 7P3

Hence, the total number of arrangements = 7P3 × 5 = 1050

Form 5: Chapter 17 (Permutations and Combinations) SPM Flashback

Fully-Worked Solutions

7C7 × 6C1 Impossible

7 Form Four + 5 Form Five + 4 Form Six students

Page 67: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

67

Paper 1

1 P(Blue) =

25

k6 + k

=25

5k = 12 + 2k3k = 12k = 4

2 P(both students participate in the same game)

= P(TT or BB or HH )= P(TT ) + P(BB) + P( HH )

=5

14×

413

⎝ ⎜

⎠ ⎟ +

614

×5

13⎛

⎝ ⎜

⎠ ⎟ +

314

×2

13⎛

⎝ ⎜

⎠ ⎟

=4

13

3 Let A − Azean D − Dalilah N − Nurur P(only one of them will qualify)

= P( A D N ) + P( A D N ) + P( A D N )

=13

×25

×47

⎝ ⎜

⎠ ⎟ +

23

×35

×47

⎝ ⎜

⎠ ⎟ +

23

×25

×37

⎝ ⎜

⎠ ⎟

=8

105+

835

+435

=44

105

Form 5: Chapter 18 (Probability) SPM Flashback

Fully-Worked Solutions

Initially, there are 5 students playing table tennis out of 14 students.

After 1 student is chosen, it is left with 4 students playing table tennis out of 13 students.

Page 68: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

68

Paper 1 1

P(Z > k ) = 0.5 − 0.2580 = 0.2420

2 X − Number of students who passed

X ~ B(7, 0.6)

P( X = 5)= 7C5(0.6)5 (0.4)2

= 0.2613

3 (a) X ~ N (4.8, 1.22 )

Z =X − µ

σ

Z =4.2 − 4.8

1.2Z = −0.5

(b) P(4.8 < X < 5.2)

= P 4.8 − 4.81.2

< Z <5.2 − 4.8

1.2⎛

⎝ ⎜

⎠ ⎟

= P(0 < Z < 0.333)= 0.5 − 0.3696= 0.1304

4 X − Volume, in ml X ~ N(650, 252)

(a) Z =

25

X − 65025

=25

X − 650 = 10X = 660

Hence, the volume which is equivalent to

the standard score of 25

is 660 ml .

(b) P( X > 620)

= P Z >620 − 650

25⎛

⎝ ⎜

⎠ ⎟

= P(Z > −1.2)= 1− 0.1151= 0.8849

Hence, the percentage of bottles of soy sauce that have volumes of more than 620 ml

= (0.8849 ×100)%= 88.49%

5 P(Z > m) = 0.5 − 0.1985

P(Z > m) = 0.3015

∴ m = 0.52

Form 5: Chapter 19 (Probability Distributions) SPM Flashback

Fully-Worked Solutions

Page 69: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

69

Paper 2 1 (a) X − Number of children

X ~ B(8, 0.4)

(i) P( X ≥ 2)

= 1− P( X = 0) − P( X = 1)= 1− 8C0 (0.4)0 (0.6)8 − 8C1(0.4)1 (0.6)7

= 1− 0.01680 − 0.08958= 0.8936

(ii) Variance = 192

npq = 192n(0.4)(0.6) = 192

n = 800

Hence, the population of the housing estate is 800.

(b) X − Mass of a worker

X ~ N(65, 62)

P(58 < X < 70)

= P 58 − 656

< Z <70 − 65

6⎛

⎝ ⎜

⎠ ⎟

= P(−1.167 < Z < 0.833)

= 1− Q(1.167) − Q(0.833)= 1− 0.1216 − 0.2025= 0.6759

Number of workers whose masses are between 58 kg and 70 kg = 142 P(−1.167 < Z < 0.833) × N = 142

0.6759 × N = 142

N =142

0.6759N = 210

Hence, the total number of workers = 210.

2 (a) X − Number of goals

X − B(8, p)

(i) Mean = 6.4

np = 6.48p = 6.4

p = 0.8

(ii) P( X ≥ 3)

= 1− P( X = 0) − P( X = 1) − P( X = 2)= 1− 8C0 (0.8)0 (0.2)8 − 8C1(0.8)1 (0.2)7

− 8C2(0.8)2 (0.2)6

= 0.9988

(b) X − Body-mass of a student

X ~ N(52, 122)

(i) P( X < 40)

= P Z <40 − 52

12⎛

⎝ ⎜

⎠ ⎟

= P(Z < −1)= 0.1587

(ii) P( X > m) = 8%

P Z >

m − 5212

⎝ ⎜

⎠ ⎟ = 0.08

m − 5212

= 1.406

m = 68.872

Page 70: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

70

3 (a) X − Number of students who scored a distinction in Mathematics X ~ B(n, p) X ~ B(8, 0.7)

(i)

P( X = 3) = 8C3(0.7)3 (0.3)5

= 0.04668

(ii) P( X < 3)

= P( X = 0) + P( X = 1) + P( X = 2)= 8C0 (0.7)0 (0.3)8 + 8C1(0.7)1 (0.3)7

+ 8C2 (0.7)2 (0.3)6

= 0.0113

(b) X − Volume of milk, in ml

X ~ N(µ, σ 2) X ~ N(1000, 202)

(i) P( X < 1050)

= P Z <1050 − 1000

20⎛

⎝ ⎜

⎠ ⎟

= P(Z < 2.5)= 1− 0.00621= 0.9938

(ii) P( X > v) = 0.7

P Z >

v − 100020

⎝ ⎜

⎠ ⎟ = 0.7

v − 100020

= −0.524

v = 989.52 ml

v = 0.9895 l

4 X − Mass of a pineapple, in kg X ~ N(1.3, 0.22)

(a) P(grade A)

= P( X > 1.4)

= P Z >1.4 − 1.3

0.2⎛

⎝ ⎜

⎠ ⎟

= P(Z > 0.5)= 0.3085

(b) P(grade B)

= P(1.2 < x < 1.4)

= P 1.2 − 1.30.2

< Z <1.4 − 1.3

0.2⎛

⎝ ⎜

⎠ ⎟

= P(−0.5 < Z < 0.5)= 1− 0.3085 − 0.3085= 0.383

Hence, the number of grade B pineapples

= 0.383 ×1000= 383

(c) P( X > m) = 93.32%

P Z >

m − 1.30.2

⎝ ⎜

⎠ ⎟ = 0.9332

m − 1.30.2

= −1.5

m = 1.0

Page 71: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

71

Paper 2 1 (a) (i)

v = a dt∫

v = (8 − 2t ) dt∫v = 8t −

2t 2

2+ c

v = 8t − t 2 + c

When t = 0, v = 20 m s−1. Thus, c = 20. ∴ v = 8t − t2 + 20 At the maximum velocity,

dvdt

= 0

8 − 2t = 0t = 4

d 2vdt 2

= −2 (negative)

Therefore, v is a maximum. When t = 4, v(max.) = 8(4) − 42 + 20 = 36 m s−1

(ii) When v = 0,

8t − t 2 + 20 = 0,t 2 − 8t − 20 = 0

(t + 2)(t − 10) = 0t = −2 or 10t = −2 is not accepted

∴ t = 10∴ n = 10

(b)

Total distance travelled during the period of 0 ≤ t ≤ 10

= Area below the v − t graph

= v dt0

10

∫= (8t − t2 + 20) dt

0

10

∫=

8t2

2−

t3

3+ 20t

⎣ ⎢ ⎤

⎦ ⎥ 0

10

= 4t 2 −t3

3+ 20t

⎣ ⎢ ⎤

⎦ ⎥ 0

10

= 4(10)2 −103

3+ 20(10) − 0

= 266 23

m

Form 5: Chapter 20 (Motion Along a Straight Line) SPM Flashback

Fully-Worked Solutions

Page 72: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

72

2 (a) v = 3t(4 − t ) = 12t − 3t2

dvdt

= 12 − 6t

At maximum velocity,

dvdt

= 0

12 − 6t = 0t = 2

d 2vdt 2

= −6 (negative)

Hence, v is a maximum. ∴ vmax = 12(2) − 3(2)2 = 12 m s−1

(b)

s = v dt∫

s = (12t − 3t 2 ) dt∫s =

12t 2

2−

3t 3

3+ c

s = 6t2 − t3 + c

When t = 0, s = 0. Thus, c = 0. ∴ s = 6t2 − t 3 Distance travelled during the 3rd second

= s3 − s2

= [6(3)2 − 33]− [6(2)2 − 23]

= 27 − 16

= 11

= 11 m

(c) When the particle passes through point O again, s = 0.

When s = 0,6t 2 − t 3 = 0

t 2 (6 − t) = 0t = 0 or 6t = 0 is not accepted

∴ t = 6

(d) When a particle reverses its direction, it is

at instantaneous rest, i.e. v = 0.

When v = 0,3t(4 − t ) = 0

t = 0 or 4t = 0 is not accepted

∴ t = 4

Page 73: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

73

3 (a) vP = 4 + 2t − 2t2 When object P travels at a maximum

velocity, dvP

dt= 0 .

When dvP

dt= 0,

2 − 4t = 0

t =12

d 2vP

dt 2= −4 (negative)

Hence, the velocity of object P is a maximum.

∴ vP (max) = 4 + 2 12

⎝ ⎜

⎠ ⎟ − 2 1

2⎛

⎝ ⎜

⎠ ⎟

2

= 4 12

m s−1

(b) At point C, vP = 0.

4 + 2t − 2t 2 = 02 + t − t 2 = 0t 2 − t − 2 = 0

(t − 2)(t +1) = 0

t = 2 or −1 t = −1 is not accepted. ∴ t = 2 For object P,

sP = v P dt∫sP = (4 + 2t − 2t 2 ) dt∫sP = 4t + t2 −

2t 3

3+ c

When t = 0, sP = 0. Thus, c = 0.

∴ sP = 4t + t2 −

2t 3

3

When t = 2,

sP = 4(2) + 22 −2(2)3

3

sP = 6 23

m

For object Q, sQ = (vQ × t) + 24 sQ = −7t + 24

When t = 2,sQ = −7(2) + 24

= 10 m

Hence, the distance between object P and

object Q is 10 − 6 2

3= 3 1

3 m

(c) When object P and object Q meet,

sP = sQ

4t + t2 −2t3

3= −7t + 24

12t + 3t2 − 2t3 = −21t + 722t 3 − 3t 2 − 33t + 72 = 0 (shown)

This displacement is measured from point A.

Displacement = Uniform velocity × Time

We have to "plus 24" here because at the beginning of the motion, object Q is at 24 m to the right of point A.

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74

4 (a) (i) v = t 2 − 6t + k

When t = 0, v = 8

8 = 02 − 6(0) + kk = 8

(ii) When v < 0,

t2 − 6t + 8 < 0(t − 2)(t − 4) < 0

Hence, the range of values of t when the particle moves to the left is 2 < t < 4 .

(iii) a =

dvdt

= 2t − 6

When a < 0,2t − 6 < 0

2t < 6t < 3

Hence, the range of values of t when the particle decelerates is 0 ≤ t < 3 .

(b) (i) t (s) 0 1 2 3 4

v (m s−1) 8 3 0 −1 0

(ii) Total distance travelled in the first 4 s

= Area under the v − t graph

= v dt0

2∫ + v dt2

4∫= (t 2 − 6t + 8) dt

0

2∫ + (t 2 − 6t + 8) dt2

4∫

=t 3

3− 3t 2 + 8t

⎣ ⎢ ⎤

⎦ ⎥ 0

2

+t 3

3− 3t 2 + 8t

⎣ ⎢ ⎤

⎦ ⎥ 2

4

=23

3− 3(2)2 + 8(2) − 0 +

43

3− 3(4)2 + 8(4) −

23

3− 3(2)2 + 8(2)

⎝ ⎜

⎠ ⎟

=203

+163

−203

=203

+ −43

=203

+43

= 8 m

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75

Paper 2 1 (a) The total mass of seafood is not less than

20 kg. The inequality is x + y ≥ 20.

x 0 20 y 20 0

The mass of prawns is at most three times that of squids. The inequality is x ≤ 3 y .

x 0 30 y 0 10

The allocation is RM250. The inequality is 10x + 5y ≤ 250 ⇒ 2x + y ≤ 50

x 0 25 y 50 0

(b)

(c) If the restaurant buys 15 kg of squids,

y = 15. From the graph, when y = 15, the minimum value of x = 5. Therefore, if the restaurant buys 15 kg of squids, the minimum mass of prawns it can buy is 5 kg. Total expenditure = 10x + 5y When x = 5 and y = 15, the total expenditure = 10(5) + 5(15) = RM125 Hence, the maximum amount of money that could remain from its allocation is RM250 − RM125 = RM125.

Form 5: Chapter 21 (Linear Programming) SPM Flashback

Fully-Worked Solutions

Page 76: -Nota 2 Matematik Tambahan Tingkatan 4 dan 5 SPM(2)

76

2 (a) The total number of participants is at least 30. The inequality is x + y ≥ 30 .

x 0 30 y 30 0

The number of Mathematics participants is at most twice that of Science. The inequality is y ≤ 2x .

x 0 30 y 0 60

The expenditure for a Science participant and a Mathematics participant are RM80 and RM60 respectively. The maximum allocation is RM3600. The inequality is 80x + 60y ≤ 3600 ⇒

4x + 3y ≤ 180.

x 0 45 y 60 0

(b)

(c) (i) When y = 12, from the graph,

xmin = 18 and xmax = 36 . Hence, when the number of Mathematics participants is 12, the minimum and maximum number of Science participants are 18 and 36 respectively.

(ii) Cost = 80x + 60y

Draw the straight line 80x + 60y = 480

x 0 6 y 8 0

For minimum cost, from the graph, xmin = 10 and ymin = 20. Hence, the minimum cost

= 80x + 60y= 80(10) + 60(20)= RM2000

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77

3 (a) For the constraint "the total number of balls bought should not be more than 80", the inequality is x + y ≤ 80 .

x 0 80 y 80 0

For the constraint "the number of footballs bought should not be more than 4 times the number of volleyballs bought", the inequality is y ≤ 4x .

x 0 20 y 0 80

For the constraint "the number of footballs bought should exceed the number of volleyballs bought by at least 15", the inequality is y − x ≥ 15.

x 0 40 y 15 55

(b)

(c) (i) From the graph, if the number of volleyballs bought is 25 (x = 25), the range of the number of footballs bought is 40 ≤ y ≤ 55 .

(ii) Cost = 60x + 80y

Draw the straight line 60x + 80y = 4800 .

x 0 80 y 60 0

From the graph, the optimal point is (16, 64). Hence the maximum cost = 60(16) + 80(64) = RM6080.

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78

4 (a) The maximum time for the use of machine P is 12 hours:

60x + 20y ≤ 12 × 603x + y ≤ 36

x 0 12 6 y 36 0 18

The use of machine Q is at least 8 hours:

30x + 40y ≥ 8 × 603x + 4y ≥ 48

x 0 16 y 12 0

The ratio of the number of 'Premier' pewter plates produced to the number of 'Royal' pewter plates produced is at least 1 : 3:

xy

≥13

3x ≥ yy ≤ 3x

x 0 6 y 0 18

(b)

(c) (i) From the graph, if y = 12, xmax = 8

(ii) Profits = 100x + 140y Draw the straight line 100x +140y = 1400

x 0 14 y 10 0

From the graph, the optimal point = (6, 18) Hence, the maximum profit = 100(6) +140(18) = RM3120

(0, 36) is out of the graph paper. So, another point (6, 18) has to be determined.