ADD MATHS P2

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Class : ………………………………………………….

ADDITIONAL MATHEMATICS
PAPER 2

FOR SPM 2019 / 2010

FORM TOPICS PAGE REMARK
4 Chapter 4 Simultaneous Equations 3 – 11
4 Chapter 2 Quadratic Equations 12 – 15
4 Chapter 3 Quadratic Functions 16 – 20
4 Chapter 1 Functions 21 – 24
5 Chapter 1 Progressions 25 – 32
4 Chapter 7 Statistics 33 – 37
4 Chapter 5 Indices And Logarithms 38 – 42
4 Chapter 6 Coordinate Geometry 43 – 46
4 Chapter 9 Differentiation 47 – 52
5 Chapter 4 Vectors 53 – 57
5 Chapter 5 Trigonometric Functions 58 – 66
4 Chapter 8 Circular Measures 67 – 78
5 Chapter 2 Linear Law 79 – 89
5 Chapter 8 Probability Distributions 90 – 97
5 Chapter 10 Linear Programming
4 Chapter 11 Index Number 98 – 108
4 Chapter 10 Solution Of Triangles 109 – 118
119 – 127

FORMULAE

ALGEBRA

1. x = 8. loga b =
2. am × an = a m + n
3. am ÷ an = a m – n 9. Tn = a + (n – 1)d
4. (am)n = amn 10. Sn = [2a + (n – 1)d]
5. loga mn = loga m + loga n 11. Tn = arn – 1
6. loga = loga m – loga n
7. loga m n = n loga m 12. Sn = = ,r1

1. y= uv, = u + v 13. S = , r < 1

CALCULUS
4. Area under a curve, =

2. y = , = =
3. = × 5. Volume of revolution =

=

BILINGUAL
DWIBAHASA

Prepared By: WongYH
SMK Merbau Miri

1

STATISTICS

1. = 6. I = × 100
2. = 7. =
8. nPr =
3.  =
9. nPr =
= 10. P(A  B) = P(A) + P(B) – P(AB)
4.  = 11. P(X = r) = nCr pr q n – r p + q = 1
12. Mean,  = np
= )c 13.  =
5. m = L + ( 14. Z =

GEOMETRY

1. Distance =

2. Midpoint, (x, y) = ( , )

3. A point dividing a segment of a line

(x, y) = ( , )

4. Area of triangle

= (x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)

5. r =

6. =

TRIGONOMETRY

1. Arc length, s = r θ

2. Area of sector, A = r2 θ

3. sin2 A + cos2 A = 1
4. sec2 A = 1 + tan2 A
5. cosec2 A = 1 + cot2 A

6. sin 2A = 2 sin A cos A
7. cos 2A = cos2 A – sin2 A

= 2 cos2 A – 1
= 1 – 2 sin2 A

8. sin (A  B) = sin A cos B  cos A sin B

9. cos (A  B) = cos A cos B sin A sin B

10. tan (A  B) =

11. tan 2A =

12. = =
13. a2 = b2 + c2 – 2 b c cos A
14. Area of triangle = a b sin C

2

FORM 4 CHAPTER 4 SIMULTANEOUS EQUATION
PAPER 2 SECTION A

1 Solve the simultaneous equations.

Selesaikan persamaan serentak berikut.

2x + y = 1 and x2 + y2 + xy = 7

Solution
Remember…..try to avoid fraction AND can be factorised as the question does not mention the answer in decimal place

Linear: 2x + y = 1

y = 1 – 2x ----------------(1)

Not linear: x2 + y2 + xy = 7 ----------------(2)

Substitute (1) into (2):

x2 + (1 – 2x)2 + x (1 – 2x) = 7 (1 – 2x)2 = 1 + 4x2 [Wrong] ← Don’t do this
(1 – 2x)2 = (1 – 2x)(1 – 2x) [Correct]
x2 + (1 – 2x)(1 – 2x) + x (1 – 2x) = 7 (1 – 2x)2 = 12 + (1)( –2)(2)x + (–2x)2

x2 + 1 – 2x – 2x + 4x2 + x – 2x2 – 7 = 0 = 1 – 4x + 4x2

3x2 – 3x – 6 = 0

x2 – x – 2 = 0

(x + 1)(x – 2) = 0

(x + 1) = 0 or (x – 2) = 0

x=–1 x=2

From (1), y = 1 – 2x
When x = – 1, y = 1 – 2(–1) When x = 2, y = 1 – 2(2)

=3 =–3

So, x = – 1 when y = 3 and x = 2 when y = – 3

2 Solve the following simultaneous equations. Give your answers correct to three decimal places.

Selesaikan persamaan serentak berikut. Berikan jawapan anda tepat kepada 3 tempat perpuluhan.
x2 + 4y2 = 37 and x – 2y = 8

Solution
Remember…..try to avoid fraction AND cannot be factorised as the question mentions the answer in decimal places

Linear: x – 2y = 8

x = 8 + 2y ----------------(1)
Not linear: x2 + 4y2 = 37 -----------------(2)

Substitute (1) into (2): (8 + 2y)2 = 64 + 4y2 [Wrong] ← Don’t do this
(8 + 2y)2 + 4y2 = 37 (8 + 2y)2 = (8 + 2y)(8 + 2y) [Correct]
(8 + 2y)2 = 82 + (8)(2)(2)y + (2y)2
(8 + 2y)(8 + 2y) + 4y2 = 37
64 + 16y + 16y + 4y2 + 4y2 = 37 = 64 + 32y + 4y2
64 + 16y + 16y + 4y2 + 4y2 – 37 = 0
←Must show this step
8y2 + 32y + 27 = 0
↑↑↑

a = 8 b = 32 c = 27

From formula, y =  b  b2  4ac
2a

 32  322  4(8)(27)
=

2(8)
= – 1.209 or – 2.791

From (1), x = 8 + 2y When y = – 1.209 , x = 8 + 2(– 2.791)
When y = – 1.209 , x = 8 + 2(– 1.209) = 2.418

= 5.582

So, x = 5.582 when y = – 1.209 and x = 2.418 when y = – 1.209

3

1 Solve the simultaneous equations EXERCISE [5 marks] [Ans: x = 2, y = – 4; x = 3, y = – 2]
Selesaikan persamaan serentak y – 2x = – 8 and x2 – 3x – y = 2

Answer space

2 Solve the simultaneous equations [5 marks] [Ans: k = 3.624, h = 0.624; k = -6.624, h = -9.624]
Selesaikan persamaan serentak
k – h = 3 and k2 + 3h = 15
Give your answers correct to three decimal places.
Berikan jawapan anda tepat kepada tiga tempat perpuluhan.

Answer space

4

3 Solve the simultaneous equations [5 marks] [Ans: x= 8/3, y =11/3; x = 3, y =2]

Selesaikan persamaan serentak
y + 5x = 17 and y2 + 5x2 = 49

Answer space

4 Solve the simultaneous equations [5 marks] [Ans: x = 1.120, y = -1.080; x = -1.786, y = 0.857]

Selesaikan persamaan serentak
2x + 3y + 1 = 0 and x2 + 6xy + 6 = 0

Give your answers correct to three decimal places.

Berikan jawapan anda tepat kepada tiga tempat perpuluhan.

Answer space

5

5 The straight line y – x = 3 intersects the curve x2 + 2xy – 3y – 2x – 5 = 0 at two points. Find the coordinates of the points.
Garis lurus y – x = 3 menyilangi lengkung x2 + 2xy – 3y – 2x – 5 = 0 pada dua titik. Cari koordinat titik-titik tersebut.

[6 marks] [Ans: (-7/3, 2/3); (2, 5)]

Answer space

6 In Diagram 1, PQRS is a piece of manila card with an area of 588 cm2. A semicircular PAS is cut out from the manila

card. The perimeter of the remaining manila card is 186 cm. find the integer value of x and y.

Dalam Rajah 1, PQRS ialah sekeping kad manila dengan luas 588 cm2. Satu semibulatan PAS dipotong keluar daripada kad itu.

Perimeter bagi kad manila yang tertinggal ialah 186 cm. Cari nilai integer x dan y. [5 marks] [Ans: x = 6, y = 1]

P 14x cm Q

A 7y cm

SR

Diagram 1 / Rajah 1

Answer space

6

INTENSIVE EXERCISE

1 SPM 2018 P2 Q4
Diagram 1 shows the plan of a rectangular garden PQRS. The garden consists of a semicircular pond PTS
and grassy area PQRST.
Rajah 1 menunjukkan pelan bagi suatu taman segiempat PQRS. Taman itu terdiri daripada semibulatan kolam PTS
dan kawasan berumput PQRST.

Diagram 1 / Rajah 1

It is given that SR = 6y metre and QR = 7x metre, x  y. The area of the rectangular garden PQRS is 168
metre2 and the perimeter of the grassy area is 60 metre. The pond with uniform depth contains 15.4 metre2 of

water. By using π = , find the depth, in metre, of the water in the pond. [7 marks / markah]

Diberi bahawa SR = 6y meter dan QR = 7x meter, x  y. Luas taman segiempat PQRS itu ialah 168 meter2 dan
perimeter kawasan berumput itu ialah 60 meter. Kolam dengan kedalaman seragam mengandungi 15.4 meter2 air.

Answer Space

2 SPM 2017 P2 Q1 [5 marks / markah] [Ans: x = 2, y = 1/3, x = -1, y = -2/3]
Solve the following simultaneous equations:

Selesaikan persamaan serentak berikut:
x – 3y = 1 , x2 + 3xy + 9y2 = 7

Answer Space

7

3 SPM 2016 P2 Q3

Adam planted vegetables on a piece of land. The shape of the land is a right angled triangle. Given the
longest side of the land is y metre. The other two sides of the land are x metre and (2x – 1) metre

respectively. He fenced the land with 40 metre of barbed wire.

Find the length, in metre, of each side of the land. [7 marks / markah] [Ans: 8, 15, 17]

Adam menanam sayur-sayuran pada sebidang tanah. Bentuk tanah ialah segitiga bersudut tegak. Diberi panjang sisi
terpanjang bagi tanah itu ialah y meter. Dua sisi tanah yang lain ialah x meter dan (2x – 1) meter masing-masing. Dia

memagari tanah itu dengan 40 meter dawai berduri.

Cari panjang, dalam meter, bagi setiap sisi tanah itu.

Answer Space

4 SPM 2015 P2 Q1 [5 marks / markah] [Ans: x = 4/5, y = -8/5, x = 4, y = 8]
Solve the following simultaneous equations:

Selesaikan persamaan serentak berikut:
3x – y – 4 = 0 , 5x2 + 2y2 – 6xy = 16

Answer Space

8

5 SPM 2014 P2 Q1

Solve the following simultaneous equations:

Selesaikan persamaan serentak berikut:
y – 2x + 1 = 0 , x2 – 2y2 – 3y + 2 = 0

Give your answer correct to three decimal places.

Beri jawapan anda betul kepada tiga tempat perpuluhan.

Answer Space [5 marks / markah][x = 0.813, y = 0.626, x = -0.527, y = -2.054]

6 SPM 2012 P2 Q1
Solve the simultaneous equations: y – 2x + 1 = 0 and 4x2 + 3y2 – 2xy = 7.

Selesaikan persamaan serentak:

Give the answers correct to three decimal places.

Beri jawapan betul kepada tiga tempat perpuluhan.

Answer Space [5 marks / markah][x = 1.129, y = 1.258, x = -0.295, y = -1.590]

9

7 P is a two digit number. When the two digits are reversed, a new number, Q, is obtained. The product of P

and Q is 574 and their difference is 27. Find the two numbers of P and Q.

P ialah satu nombor dengna dua digit. Apabila dua digit itu diterbalikkan, satu nombor baru, Q, diperolehi. Hasil

darab P dan Q ialah 574 dan perbezaannya ialah 27. Cari dua nombor P dan Q itu.

Answer Space [7 marks / markah][14, 41]

8 A piece of wire, 40 cm long, is cut into two parts which are then bent to form two squares as shown in the
diagram below.
Seutas dawai, 40 cm panjang, dipotong kepada dua bahagian di mana ianya dibengkok untuk membentuk dua
segiempat sama sisi seperti ditunjukkan dalam rajah bawah.

The total area of the two squares is 58 cm2. Find the lengths of the sides of the two squares.

Jumlah luas bagi kedua-dua segiempat ialah 58 cm2. Cari panjang setiap sisi segiempat sama sisi itu.

Answer Space [7 marks / markah][7, 3]

10

9 Find the values of p if the straight line 4x + y − p = 0 is a tangent to the curve x 2 = 3 – xy.
Cari nilai-nilai bagi p jika garis lurus 4x + y − p = 0 ialah tangen kepada lengkungan x 2 = 3 – xy.

Answer Space [6 marks / markah][Ans: p = ± 6]

10 In the diagram below, PQRS is a rectangular sheet of paper with an area of 224 cm2. QAR which is semicircular in
shape is cut out from the paper. The perimeter of the remaining paper is 72 cm.
Dalam rajah bawah, PQRS ialah satu segiempat tepat bagi sekeping kertas dengan luas 224 cm2. QAR ialah satu

semibulatan yang dipotong keluar daripada kertas itu. Perimeter bagi kertas yang tinggal ialah 72 cm.

Find the values of x and y. [Use / Guna π = ]
Cari nilai bagi x da y. [Guna π = ]
Answer Space [7 marks / markah][Ans: x = 24 , y = or x = 12, y = ]

11

FORM 4 CHAPTER 2 QUADRATIC EQUATIONS
PAPER 2 SECTION A

1 (a) The quadratic equation 3x2 + kx + h = 0, where k and h are constants, has roots 1 and 2. Find the value of k and
3

of h.

Persamaan 3x2 + kx + h = 0, dengan keadaan k dan h adalah pemalar, mempunyai punca-punca 1 and 2. Cari nilai k dan
3

nilai h.

Solution

The quadratic equation given in question must be arranged to the general form: ax2 + bx + c = 0

Recall the formula: SOR = – b POR = c
a a

From 3x2 + kx + h = 0 ←Already in general form

We know that a = 3, b = k, c = h

Given the roots are 1 and 2
3

SOR = – b POR = c
a a

1 + (2) = – k 1 × (2) = h
3 3 3 3

5 =– k 2 = h
3 3 3 3

k = 5 h=–2
3 3

k=5

1 (b) Given that 4 and k are a roots of the quadratic equation x2 – 5x + p + 7 = 0, find the value of k and of p.
Diberi bahawa 4 dan k adalah punca-punca bagi persamaan kuadratik x2 – 5x + p + 7 = 0, cari nilai k dan nilai p.

Solution

From the quadratic equation x2 – 5x + p + 7 = 0 ← is already in general form

We know that a = 1, b = – 5 and c = p + 7

Given the roots are 4 and k

SOR = – b POR = c
a a

4 + k = – 5 4×k= p7
1 1

4+k=5 4k = p + 7
k=1
4(1) = p + 7
4–7=p

p=–3

1 (c) Solve the quadratic equation 5(2x – 1) = (3x + 1)(x – 3). Give your answer correct to four significant figures.
Selesaikan persamaan kuadratik 5(2x – 1) = (3x + 1)(x – 3). Berikan jawapan anda betul kepada empat angka bererti.

Solution

Expand the 5(2x – 1) = (3x + 1)(x – 3) to general form
10x – 5 = 3x2 – 9x + x – 3

3x2 – 9x + x – 3 = 10x – 5

3x2 – 9x + x – 3 – 10x + 5 = 0
3x2 – 18x + 2 = 0

So, a = 3, b = – 18, c = 2

From formula, x =  b  b2  4ac
2a

 (18)  (18)2  4(3)(2)
x=

2(3)
x = 5.8868 or – 0.1132

12

1 (d) The quadratic equation 2x(x + 3) = 3 – 2k , where k is a constant, has two equal roots. Find the value of k.
Persamaan kuadratik 2x(x + 3) = 3 – 2k , dengan keadaan k ialah pemalar, mempunyai dua punca yang sama. Cari nilai k.

Solution

The 2x(x + 3) = 3 – 2k must be arranged to general form!
2x2 + 6x = 3 – 2k

2x2 + 6x – 3 + 2k = 0

So, a = 2, b = 6 and c = – 3 + 2k

Remember the formulae:
Two equal roots OR Intersect at one point means b2 – 4ac = 0
Two distinct roots OR Two different roots means b2 – 4ac > 0
No root OR No solution OR Does not intersect means b2 – 4ac < 0

Tangent to the curve means Solve by simultaneous equation like Chapter 4 Form 4

Then, b2 – 4ac = 0
(6)2 – 4(2)(– 3 + 2k) = 0
36 – 8(– 3 + 2k) = 0
36 – (– 24 + 16k) = 0
36 + 24 – 16k = 0
60 – 16k = 0
– 16k = – 60

16k = 60

k= 60
16

k= 15
4

1 (e) The straight line y = m – 2x does not intersect the curve y2 + xy + 8 = 0. Find the range of values of m.
Garis lurus y = m – 2x tidak bersilang dengan lengkung y2 + xy + 8 = 0. Carikan julat nilai m.
Solution

Let y = m – 2x ------------(1) and y2 + xy + 8 = 0 ------------(2)

Substitute (1) into (2): ← to general form
(m – 2x)2 + x (m – 2x) + 8 = 0

(m – 2x) (m – 2x) + x (m – 2x) + 8 = 0
m2 – 2mx – 2mx + 4x2 + mx – 2x2 + 8 = 0

2x2 – 3mx + m2 + 8 = 0
So, a = 2, b = – 3m and c = m2 + 8

Does not intersect means b2 – 4ac < 0
So, the range is – 8 < m < 8 (– 3m)2 – 4(2)(m2 + 8) < 0

9m2 – 8(m2 + 8) < 0
9m2 – (8m2 + 64) < 0

9m2 – 8m2 – 64 < 0
m2 – 64 < 0
m2 < 64
m < 8 or m < – 8

1 (f) Given α and  are the roots of the quadratic equations 2x2 + 8x + 5 = 0.
Form the quadratic equation which has the roots 2α and 2.
Diberi α dan  ialah punca persamaan kuadratik 2x2 + 8x + 5 = 0.
Bentukkan persamaan kuadratik yang mempunyai punca 2α dan 2.
Solution

From 2x2 + 8x + 5 = 0 ← Remember must in general form!
So, a = 2, b = 8 and c = 5 with the roots are α and .

Then, SOR = – b and POR = c
a a

α +  = – 8/2 and α ×  = 5/2
α+=–4 α  = 5/2

Form the quadratic equation using 2α and 2

New SOR = 2α + 2 and new POR = 2α × 2
=4α
= 2(α + ) = 4 (5/2)

= 2(– 4) =5

=–8

So, the new equation is in the form of x2 – (SOR) x + (POR) = 0  Remember this formula
x2 – (– 8) x + (5) = 0
x2 + 8x + 5 = 0

13

EXERCISE
1 The quadratic equation hx2 + kx + 3 = 0, where h and k are constants has two equal roots. Express h in term of k.

Persaman kuadratik hx2 + kx + 3 = 0, dengan keadaan h dan k adalah pemalar, mempunyai dua punca yang sama.
Ungkapkan h dalam sebutan k.

[2 marks] [Ans: h = k2/12]
Answer space

2 The quadratic equation 2x2 = px2 – 4(x + 2), where p is a constant, has no real roots. Find the range of the values of p.
Persamaan kuadratik 2x2 = px2 – 4(x + 2), dengan keadaan p ialah pemalar, tidak mempunyai punca nyata. Cari julat nilai p.
[3 marks] [Ans: p < 3/2]
Answer space

3 Find the values of n for which the curve y = n + 8x – x2 intersect the straight line y = 3 at one point. [4 marks] [Ans: -13]
Cari nilai n di mana lengkung y = n + 8x – x2 menyilang garis lurus y = 3 pada satu titik.
Answer space

14

4 Given one of the roots of a quadratic equation x2 + px + 4 = 0, where p is a constant, is four times the other root.
Diberi satu daripada punca persamaan kuadratik x2 + px + 4 = 0, dengan keadaan p ialah pemalar, adalah empat kali

punca yang satu lagi.

(a) Find the value of p if the roots are positive. [4 marks] [Ans: p = -5]

Cari nilai p jika punca-puncanya bernilai positif.

(b) Hence, form the quadratic equation which has the roots (p – 3) and 1 p. [3 marks] [Ans: 2x2 + 21x + 40 = 0]
2

Seterusnya, bentukkan persamaan kuadratik yang mempunyai punca (p – 3) dan 1 p.
2

Answer space

5 A quadratic equation x2 + 4(3x + m) = 0, where k is a constant has roots p and 2p, p  0.

Persamaan kuadratik x2 + 4(3x + m) = 0, dengan keadaan k ialah pemalar mempunyai punca-punca p and 2p, p  0.

(a) Find the value of p and of m. [4 marks] [Ans: p = – 4, m = 8]

Cari nilai p dan nilai m.

(b) Hence, form the quadratic equation which has roots m + 1 and m – 6. [3 marks] [Ans: x2 – 11x + 18 = 0]
Seterusnya, bentukkan persamaan kuadratik dengan punca-punca m + 1 dan m – 6.

Answer space

15

FORM 4 CHAPTER 3 QUADRATIC FUNCTIONS
PAPER 2 SECTION A

If given the function in the form of f (x) = a(x + b)2 + c
a > 0 → minimum point

a < 0 → maximum point

The maximum / minimum point is (–b, c) The maximum / minimum value is c
Axis of symmetry is x = – b

1 By expressing the function f (x) = 3x2 – 6x + 5 in form of f (x) = a(x – p)2 + q, find the minimum value of f (x).
Dengan mengungkapkan fungsi f (x) = 3x2 – 6x + 5 dalam bentuk f (x) = a(x – p)2 + q, cari nilai minimum bagi f (x).

Solution

For maximum or minimum point, do the completing the square f (x) = a(x + b )2 + c – ( b )2
2a 2a

From f (x) = 3x2 – 6x + 5 ← remember must be in general form

Taking out the 3:

f (x) = 3[ x2 – 2x + 5 ]
3

= 3[ x2 – 2x + ( 2 )2 + 5 – ( 2 )2 ]
2 3 2

= 3[ x2 – 2x + (– 1)2 + 5 – (– 1)2 ]
3

= 3[(x – 1)2 + 5 – (– 1)2 ]
3

= 3[(x – 1)2 + 2 ]
3

= 3(x – 1)2 + 2

So, a = 3, p = 1 and q = 2. The minimum value of f (x) is 2.

2 Find the maximum or minimum value for the function f (x) = 1 + 3x – 2x2. Thus, find the equation of the axis of
symmetry for the function.
Cari nilai maksimum atau minimum bagi fungsi f (x)= 1 + 3x – 2x2. Seterusnya, carikan persamaan paksi simetri bagi graf fungsi itu.
Solution

From f (x) = 1 + 3x – 2x2

To general form: f (x) = –2x2 + 3x + 1

Taking out – 2: f (x) = –2[x2 – 3 x – 1 ]
2 2

↑ ↑

Add in +[ – 3 ÷2] – [ – 3 ÷2]
2 2

f (x) = –2[ x2 – 3 x + ( 3 )2 – 1 – ( 3 )2 ]
2 4 2 4

-------------------- ---------------

= –2[(x – 3 )2 – 1 – ( 3 )2 ]
4 2 4

---------- ----------------

= –2[(x – 3 )2 – 17 ]
4 16

= –2(x – 3 )2 + 17 ]
4 8

The maximum value is 17 . The axis of symmetry is x = 3 . Extra info: Maximum point is ( 3 , 17 )
8 4 4 8

EXERCISE [1 mark] [Ans: (– 3, 9)]
1 Given the quadratic function y = – 2(x + 3)2 + 9. State [1 mark] [Ans: x = – 3]

Diberi fungsi kuadratik y = – 2(x + 3)2 + 9. Nyatakan
(a) the coordinates of the maximum point,

koordinat titik maksimum,
(b) the equation of the axis of symmetry.

persamaan paksi simetri.

Answer space

16

2 Diagram 2 shows the graph of the function y = – (x + 1)2 + 9, where m is a constant. The curve touches the line y = m at
point A and cut the y-axis at point B. The curve also cut the x–axis at point P.
Rajah 2 menunjukkan graf fungsi y = – (x + 1)2 + 9, dengan keadaan m ialah pemalar. Lengkung itu menyentuh garis y = m di titik A
dan menyilang paksi-y di titik B. Lengkung itu juga menyilang paksi-x di titik P.
y

A y=m

B (0, k)

P x

O [2 marks] [Ans: 9, 8]
[2 marks] [Ans: (-4,0)]
Diagram 2 / Rajah 2
(a) Determine the value of m and of k. / Tentukan nilai m dan nilai k.
(b) State the coordinates of point P. / Nyatakan koordinat bagi titik P.
Answer space

3 Diagram 3 shows the graph of a quadratic function f (x) = 3(x + p)2 + 4, where p is a constant.
Rajah 3 menunjukkan graf fungsi kuadratik f (x) = 3(x + p)2 + 4, dengan keadaan p ialah pemalar.
y

y = f (x)

(– 2, q) x
O

Diagram 3 / Rajah 3 [1 mark] [Ans: 2]
The curve y = f (x) has the minimum point (2, q), where q is a constant. [1 mark] [Ans: 4]
Lengkung y = f (x) mempunyai titik minimum (2, q), dengan keadaan q adalah pemalar. [1 mark] [Ans: x = – 2]
State / Nyatakan
(a) the value of p / nilai p
(b) the value of q / nilai q
(c) the equation of the axis of symmetry / persamaan paksi simetri.
Answer space

17

4 Diagram 4 shows the graph of a quadratic function y = f (x). The straight line y = – 9 is a tangent to the curve y = f (x).
Rajah 4 menunjukkan graf fungsi kuadratik y = f (x). Garis lurus y = – 9 ialah tangen pada lengkung y = f (x).
y

y = f (x)

01 x
7

y=–9 [1 mark] [Ans: x = 4]
[2 marks] [Ans: -4, -9]
Diagram 4 / Rajah 4
(a) Write the equation of the axis of symmetry of the curve.

Tuliskan persamaan paksi simetri bagi lengkung itu.
(b) Express f (x) in the form of (x + p)2 + q, where p and q are constants.

Ungkapkan f (x) dalam bentuk (x + p)2 + q, dengan keadaan p dan q adalah pemalar.
Answer space

5 Find the range of values of x for which (3x + 2)(x – 5) > x – 5 . [3 marks] [Ans: x < – 1/3, x > 5]
Cari julat nilai x bagi (3x + 2)(x – 5) > x – 5 .

Answer space

6 Find the range of values of x for which (3x + 1)(2x – 1) > 3 + x . [3 marks] [Ans: x < – 2/3, x >1]
Cari julat nilai x bagi (3x + 1)(2x – 1) > 3 + x .

Answer space

18

7 Given that the graph of quadratic function f (x) = 2x2 + bx + 8 always lies above the x-axis. Find the range of values of b.
Diberi graf fungsi kuadratik f (x) = 2x2 + bx + 8 sentiasa berada di atas paksi-x. Cari julat nilai b. [3 marks] [Ans: -8<x<8]
Answer space

Always lies above the x-axis → No solution and then use b2 – 4ac < 0

8 Given the quadratic function f (x) = 6x – 1 – 3x2.

Diberi fungsi kuadratik f (x) = 6x – 1 – 3x2.
(a) Express the quadratic function f (x) in the form k + m(x + n)2, where k, m and n are constants. Determine whether

f (x) has a maximum or minimum value and state the value.
Ungkapkan fungsi kuadratik f (x) dalam bentuk k + m(x + n)2, dengan keadaan k, m dan n ialah pemalar. Tentukan sama ada

fungsi f (x) mempunyai nilai maksimum atau minimum dan nyatakan nilainya. [2 marks] [Ans: f (x) = 2 – 3(x – 1)2; 2]

(b) Sketch the graph function of f (x). [3 marks]

Lakarkan graf fungsi f (x). [3 marks] [Ans: p < 2]
(c) Find the range of p such that the equation 6x – 1 – 3x2 = p has two distinct roots.

Carikan julat p supaya persamaan 6x – 1 – 3x2 = p mempunyai dua punca yang berbeza.

Answer space

19

9 The function f (x) = x2 – 4kx + 5k2 + 1 has a minimum value of r2 + 2k, where r and k are constants.
Fungsi f (x) = x2 – 4kx + 5k2 + 1 mempunyai nilai minimum r2 + 2k, dengan keadaan r dan k adalah pemalar.
(a) By using the method of completing the square, show that r = k – 1.
Dengan menggunakan kaedah melengkapkan kuasa dua, tunjukkan bahawa r = k – 1.
(b) Hence, of otherwise, find the values of r and k if the graph of the function is symmetrical about x = r2 – 1.
Seterusnya, atau dengan cara lain, cari nilai r dan nilai k jika graf bagi fungsi itu bersimetri pada x = r2 – 1.

Answer space

10 Given the quadratic function f (x) = 4 – 3x – x2. [2 marks] [Ans: (-3/2, 25/4)]
Diberi fungsi kuadratik f (x) = 4 – 3x – x2. [2 marks]

(a) Find the coordinates of the maximum point. [1 marks] [Ans: -6 < f (x) < 25/5]

Carikan koordinat bagi titik maksimumnya.
(b) Sketch the graph of f (x) for domain – 4  x  2.

Lakarkan graf f (x) itu untuk domain – 4  x  2.

(c) State the range to f(x).

Nyatakan julat yang sepadan bagi f (x).

Answer space

20

FORM 4 CHAPTER 1 FUNCTIONS
PAPER 2 SECTION A

1 Diagram 1 shows the relation between set A and set B / Rajah 1 menunjukkan hubungan antara set A dan set B.

p2
q4

6
r8

Set A Set B

Diagram 1 / Rajah 1

State / Nyatakan:

(a) Domain, codomain, object, image / Domain, kodomain, Objek, Imej

Solution Domain = {p, q, r} Codomain = {2, 4, 6, 8} Object = {p, q, r} Image = {2, 4, 6, 8}

(b) The range of the relation, / Julat hubungan itu,

Solution Range = {4, 8}

(c) Type of the relation / Jenis hubungan itu

Solution Many to one

2 Diagram 2 below shows function g where g (x) = m / Rajah 2 menunjukkan fungsi g di mana g (x) = m .
nx2 nx2

x g(x)

12

3 -2

Diagram 2 / Rajah 2

Find the value of m and n / Cari nilai bagi m dan n.

Solution

From g(x) = m
nx2

When x = 1, g(x) = 2 When x = 3, g(x) = – 2

So, 2= m So, – 2 = m
n(1)2 n(3)2

2= m –2= m
n2 3n2

2(n – 2) = m –2(3n – 2) = m
m = 2n – 4 ---------(1) m = – 6n + 4 ------------(2)

(1) = (2) : 2n – 4 = – 6n + 4
2n + 6n = 4 + 4

8n = 8
n=1

From (1), m = 2n – 4
= 2(1) – 4

m=–2

3 Given the function f (x) = 2x + 1 and g(x) = 3 – kx , find / Diberi fungsi f (x) = 2x + 1 dan g(x) = 3 – kx , cari

(a) f (2)
Solution

f (2) means substitute 2 into the function f(x)
From f (x) = 2x + 1

f (2) = 2(2) + 1
=5

(b) the value of k such that g f (2) = – 7 / nilai bagi k dengan keadaan g f (2) = – 7.
Solution

From 5(a), f (2) = 5
Given g f (2) = – 7 and g(x) = 3 – kx

So, g[ f (2)] = – 7
g [5] = – 7

3 – k (5) = – 7
3 – 5k = – 7
– 5k = – 7 – 3
– 5k = – 10

5k = 10

k=2

21

4 Given the function f (x) = 4x + 3 and g (x) = x2 – 5x + 1 / Diberi bahawa f (x) = 4x + 3 dan g(x) = x2 – 5x + 1.
Find / Cari

(a) f –1
Solution

From f (x) = 4x + 3

For inverse function, let 4x + 3 = y and find x !!
4x = y – 3

x= y3
4

So, f – 1 (x) = x  3
4

(b) f -1 (7)

Solution

From question 4(a), we know that f – 1 (x) = x  3
4

Then, f – 1 (7) = 7  3
4

f – 1 (7) = 1

(c) g f (x)

Solution

g f (x) means substitute function f (x) into the function g (x) as shown below:

f (x) = 4x + 3 and g (x) = x2 – 5x + 1

So,

g f (x) = g [f (x)]
= (4x + 3)2 – 5(4x + 3) + 1
= (4x + 3)(4x + 3) – 5(4x + 3) + 1
= 16x2 + 12x + 12x + 9 – 20x – 15 + 1
= 16x2 + 4x – 5

5 Given the function f (x) = x + 1. Find the function g if fg(x) = x2 + 3x + 5

Diberi fungsi f (x) = x + 1. Cari fungsi g jika fg(x) = x2 + 3x + 5

Solution

From fg(x) = x2 + 3x + 5

f [g(x)] = x2 + 3x + 5

g(x) + 1 = x2 + 3x + 5

g(x) = x2 + 3x + 4

6 Given the function f (x) = x + 3. Find the function g if gf (x) = x2 + 6x + 7

Diberi fungsi f (x) = x + 3. Cari fungsi g jika fg(x) = x2 + 6x + 7

Solution

From gf (x) = x2 + 3x + 5

g [f (x)] = x2 + 3x + 5

g[x + 3] = x2 + 3x + 5 ----------------(1)

Let x + 3 = y
x=y–3

From (1), g[y] = (y – 3)2 + 3(y – 3) + 5
g(y) = (y – 3)(y – 3) + 3(y – 3) + 5
g(y) = y2 – 3y – 3y + 9 + 3y – 9 + 5
g(y) = y2 – 3y + 5

So, g(x) = x2 – 3x + 5

22

EXERCISE
1 Diagram 1 shows the relation between set A and set B.

Rajah 1 menunjukkan hubungan antara set A dan set B.
Set B

8

6

4

2

Set A
01234

Diagram 5

(a) Represent the above relation in set of ordered pairs.
Wakilkan hubungan di atas dalam bentuk set hubungan bertertib.
Answer space

(b) State the images of 1.
Nyatakan imej bagi 1.
Answer space

(c) State the type of the relation.
Nyatakan jenis hubungan.
Answer space

(d) State the range of the relation.
Nyatakan julat bagi hubungan itu.
Answer space

2 Diagram 2 shows the function g : x → x  k , x  0 where k is a constant. [2 marks] [Ans: k = 1]
x

Rajah 2 menunjukkan fungsi g : x → x  k , x  0 dengan keadaan k ialah pemalar.
x

x x+k
x

3

1
2

Diagram 2 / Rajah 2

Find the value of k / Cari nilai bagi k.
Answer space

23

3 Given that the inverse function of f : x → 3x + m is f -1: x → x  7 , find the values of m and n. [3 marks][Ans:m= – 7, n= 3]
n

Diberi fungsi songsang bagi f : x → 3x + m ialah f -1: x → x  7 , cari nilai m dan n.
n

Solution space

4 Given the function f : x → 3x – 4, find / Diberi fungsi f : x → 3x – 4, cari [1 mark] [Ans: (x + 4)/3]
(a) f –1(x)
Solution space

(b) the value of p such that f –1 (2p – 1) = p / nilai p dengan keadaan f –1 (2p – 1) = p [2 marks] [Ans: p = 3/2]
Solution space

5 Given the function g : x → 3x – 2 and g f : x → 3x2 + 4, find / Diberi fungsi g : x → 3x – 2 dan g f : x → 3x2 + 4, cari
(a) g –1 (x)
[2 marks] [Ans: (x + 2)/3]

Solution space

(b) f (x) [2 marks] [Ans: x2 + 2]
Solution space

6 Given f : x → k – mx. Find / Diberi f : x → k – mx. Carikan [2 marks] [Ans: [k – x)/m]
(a) f –1(x) in terms of k and m, / f –1(x) dalam sebutan k dan m [2 marks] [Ans: m = 3, k = 2]
Solution space

(b) The values of k and m if f –1(14) = – 4 and f (5) = – 13
Nilai-nilai k dan m jika f –1(14) = – 4 dan f (5) = – 13
Solution space

24

FORM 5 CHAPTER 1 PROGRESSION

PAPER 2 SECTION A

Arithmetic Progression Geometric Progression

1) Sebutan pertama, a 1) Sebutan pertama, a ,r>1
2) Beza sepunya, d = T2 – T1 2) Nisbah sepunya, r = ,r<1
3) The nth term, Tn = a + (n – 1)d 3) The nth term, Tn = ar n – 1
4) The last term, l = a + (n – 1)d 4) The last term, l = ar n – 1

5) Sum of first n terms, Sn = [2a + (n – 1)d] 5) Sum of first n terms, Sn =

6) If given formula Sn, then Tn = Sn – Sn – 1

and T1 = S1 or Sn =

6) Sum to infinity, S =

7) If given formula Sn, then Tn = Sn – Sn – 1

and T1 = S1

1. The third and eight terms of an arithmetic progression are – 5 and 15 respectively. Find

Sebutan ketiga dan kelapan bagi suatu janjang aritmetik ialah – 5 dan 15 masing-masing. Cari

(a) the first term and the common difference / sebutan pertama dan beza sepunya

(b) the sum of the first 10 terms / hasil tambah bagi 10 sebutan pertama [(a) a = -13, d = 4 (b) 50]

Answer Space

2. The first three terms of an arithmetic progression are 2k, 3k + 3 and 5k + 1. Find [(a) k = 5 (b) 990]

Tiga sebutan pertama suatu janjang aritmetik ialah 2k, 3k + 3 dan 5k + 1. Cari

(a) the value of k, / nilai k,
(b) the sum of the first 15 terms of the progression

hasil tambah bagi 15 sebutan yang pertama untuk janjang itu
Answer Space

3. 51, 58, 65,….191 are the first n terms of an arithmetic progression. Find the value of n.

51, 58, 65,….191 ialah n sebutan pertama bagi suatu janjang aritmetik. Cari nilai n. [Ans: 21]

Answer Space

25

4. Find the sum of all the multiples of 7 between 100 and 500. [17157]

Cari hasil tambah semua gandaan bagi 7 antara 100 dan 500.

Answer Space

5. The sum of the first 6 terms of an arithmetic progression is 39 and the sum of next 6 terms is – 69.

Find

Hasil tambah bagi 6 sebutan pertama suatu janjang aritmetik ialah 39 dan jumlah bagi 6 sebutan seterusnya ialah – 69.

Cari

(a) the first term and the common difference / sebutan pertama dan beza sepunya

(b) the sum of all the terms from the 15th term to the 25th term. [(a) a = 14, d = -3 (b) -473]

hasil tambah semua sebutan dari sebutan ke-15 hingga ke-25

Answer Space

6. Given the Sn = n (2n + 1) represents the sum of the first n terms of the arithmetic progression.

2

Find

Diberi Sn = n (2n + 1) mewakili hasil tambah bagi n sebutan pertama suatu janjang aritmetik. Cari
2

(a) first term / sebutan pertama

(b) second term / sebutan kedua

(c) sum of first 8 terms / hasil tambah 8 sebutan pertama

Answer Space

26

7. Given the first 3 terms in geometric progression: 18, 9, 4 ½ …….Find

Diberi 3 sebutan pertama dalam janjang geometri: 18, 9, 4 ½ …….Cari

(a) the common ratio / nisbah sepunya (c) the sum of first 6 terms / hasil tambah 6 sebutan pertama
(b) the 7th term / sebutan ke-7 (d) the sum to infinity / hasil tambah hingga ketakterhinggaan

Answer Space

8. Find the following recurring decimal as a fraction in its simplest form.

Cari nombor perpuluhan jadi semula sebagai pecahan dalam bentuk termudah bagi setiap berikut.

(a) 0.181818…… (b) 0.83333….. (c) 1.7777…..

Answer Space

9. Given the 2x + 3, x and x – 2 are the first three consecutive terms of a geometric progression with a

common ratio, r. Find

Diberi 2x + 3, x dan x – 2 ialah tiga sebutan berturutan dalam suatu janjang geometri dengan nisbah sepunya, r. Cari

(a) the value of x, / nilai x,
(b) the first term and value of common ratio, r / sebutan pertama dan nilai bagi nisbah sepunya, r

(c) the sum to infinity for the progression / hasil tambah hingga ketakterhinggaan bagi janjang itu
Answer Space

27

10. Given that 1 = 0.16666….. / Diberi bahawa 1 = 0.16666…..

p p
= 0.1 + a + b + ……
= 0.1 + a + b + ……

Find the values of a and b. Hence, find the value of p.
Cari nilai bagi a dan b. Seterusnya, cari nilai bagi p.
Answer Space

11. Encik Ali and Encik Tan start to save money at the same time.
Encik Ali dan Encik Tan mula menyimpan duit pada masa yang sama.

(a) Encik Ali saves RMx in the first month and his savings increases constantly by RMy every
subsequent month. He saves RM200 in the 6th month and the total savings for 12 months is

RM2520. Find the values of x and y.

Encik Ali menyimpan RMx pada bulan pertama dan simpanannya bertambah secara tetap dengan RMy setiap

bulan seterusnya. Cari nilai x dan y.

(b) Encik Tan saves RM200 in the first month and his savings increases constantly by RM10 every
month. Find the value of n when both of them save the same amount of the money in nth

month.

Encik Tan menyimpan RM200 pada bulan pertama dan simpanannya bertambah secara tetap dengan RM10 setiap

bulan. Cari nilai n apabila kedua-dua mereka mempunyai simpanan duit yang sama dalam bulan ke-n.
Answer Space

28

12. Diagram shows the arrangement of equilateral triangles. The number of triangles in the lowest is 1.
For each of the other level, the number of triangles is 2 more than the level below it. The height of
each level is 3 cm. The highest level has 401 triangles.

Rajah menunjukkan susunan segitiga sama sisi. Bilangan segitiga di aras bawah ialah 1. Untuk setiap aras berikutnya,
bilangan segitiga ialah 2 lebih daripada aras bawah. Tinggi setiap aras ialah 3 cm. Aras tertinggi mempunyai 401

segitiga.

3 cm

Calculate / Hitung
(a) the height, in cm, to the highest level,

tinggi, dalam cm, ke aras tertinggi,
(b) the total number of triangles for the whole structure.

bilangan segitiga untuk seluruh struktur itu.
Answer Space

13. Diagram shows several trapeziums with areas A1, A2, A3, A4, A5 and so on. The height of the first
perpendicular edge is 2 cm and the height increases by 2 cm subsequently. The distance between the
two perpendicular edge of each trapezium is 2cm.

Rajah menunjukkan beberapa trapezium dengan luas A1, A2, A3, A4, A5 dan seterusnya. Tinggi untuk tepi serenjang yang
pertama ialah 2 cm dan tinggi itu bertambah sebanyak 2 cm seterusnya. Jarak antara dua sisi serenjang untuk setiap

trapezium ialah 2 cm.

10 cm
8 cm
6 cm
4 cm
2 cm A1 A2 A3 A4 A5
2 cm 2 cm 2 cm 2 cm

Find / Cari
(a) the height of the 25th perpendicular edge, / tinggi sisi serenjang yang ke-25,
(b) the sum of the areas for the first 10 trapeziums. / hasil tambah luas untuk 10 trapezium yang pertama
Answer Space

29

SPM DRILLING (KBAT)
1 The volume of water in a tank is 500 litres on the first day. The tank leaks at a rate of 15 litres per day. Find

the volume of water left at the end of the tenth day.
Isipadu air di dalam sebuah tangki ialah 500 liter pada hari pertama. Tangki itu bocor dan menitis dengan kadar 15 liter
sehari. Cari isipadu air yang tinggal pada akhir hari kesepuluh.
Answer Space

2 The price of an antique painting increases by 10% every year. If the price of the painting at the beginning of
the year 2000 was RM25000, what was its price at the beginning of the year 2010?
Harga sebuah lukisan antik bertambah sebanyak 10% setiap tahun. Jika harga lukisan itu pada awal tahun 2000 ialah
RM25000, berapakah harga lukisan itu pada awal tahun 2010?

30

3 Sunita has RM 250 in her piggy bank. Subsequently, she saves RM4 every day beginning from the 1st of
November. Calculate the total amount of money in her piggy bank at the end of November.
Sunita mempunyai RM250 dalam wang tabungnya. Kemudian dia menyimpan RM4 setiap hari bermula dari 1
November. Hitung jumlah wang dalam tabungnya pada akhir bulan November.
Answer Space

4 Given = 0.272727… is a recurring decimal. Find the value of k.
Diberi = 0.272727…ialah nombor perpuluhan jadi semula. Cari nilai k.
Answer Space

5 A ball is dropped from a height of 3 m. It rebounds to a height that is 4/5 of its previous height. The process
continues until the ball stops rebounding. Find the total distance travelled by the ball until it stops.
Sebiji bola dijatuhkan dari ketinggian 3 m. Bola itu melantun ke ketinggian iaitu 4/5 daripada ketinggian sebelumnya.
Proses ini berterusan sehingga bola itu berhenti. Cari jumlah jarak yang dilalui oleh bola itu sehingga ia berhenti.
Answer Space

31

6 A ball is dropped from a height of 80 m above a floor. Each time after the ball hits the floor, it will rebound
three-quarters of the distance fullen. Find
Sebiji bola dilepaskan dari ketinggian 80 m dari lantai. Setiap kali bola itu menghentam lantai, bola itu akan melantun
setinggi tiga perempat daripada jarak yang jatuh. Cari
(a) The height, in m, that the ball will reach when it rebound for the fifth time.
Tinggi, dalam m, yang dicapai oleh bola itu apabila melantun pada kali kelima.
(b) The total distance, in m, travelled when the ball hits the floor for the fifth time.
Jumlah jarak, dalam m, yang dilalui apabila bola itu menghentam lantai pada kali kelima.
(c) The total distance, in m, travelled before the ball comes to rest.
Jumlah jarak, dalam m, yang dilalui sebelum bola itu berhenti.
Answer Space

7 (a) The bacteria in a culture solution double every 5 minutes. Professor Lee begins an experiment with 20
bacteria at 10.00 am, find
Bakteria di dalam suatu larutan kultur berganda setiap 5 minit. Profesor Lee memulakan suatu eksperimen
dengan 20 bakteria pada pukul 10.00 am, cari
(i) The number of bacteria at 10.30 am
Bilangan bakteria pada pukul 10.30 am
(ii) The time when the number of bacteria is 10240
Masa apabila bilangan bakteria ialah 10240

(b) At 11.00 am, the bacteria stop growing and begin to die at a rate of 500 per minutes. Determine the
number of bacteria at 11.15 am.
Pada pukul 11.00 am, bakteria itu berhenti bertambah dan mula mati pada kadar 500 seminit. Tentukan bilangan
bakteria pada pukul 11.15 am.

Answer Space

32

FORM 4 CHAPTER 7 STATISTICS
PAPER 2 SECTION A

Remember the formulae [Refer to the formula page given]

Ungrouped data Grouped data
Mode → nombor repeated the most
Mean, x =  xf
Mean, x =  x f

N Median, M = L + ( 1 N  F ) c
2
Median → data located at the centre
fm
Variance, 2 = x2
N – ( x )2 First quartile, Q1 = L+ ( 1 N F ) c
4

Standard deviation,  = x2  (x)2 fm

N First quartile, Q3 = L+ ( 3 N F ) c
4 fm

Effect on the range, interquartile, variance and standard Variance, 2 = x2 f – ( x )2
deviation when each value in a set are changed f
New mean + , – , × , ÷
Standard deviation,  = x2 f  (x)2
New median + , – , × , ÷ f

New range × , ÷

New interquartile range × , ÷ L = Lower class boundary
New variance × , ÷ [must square] F = cumulative frequency before the class
New standard deviation × , ÷ f m =frequency of the class
c = size of class

1 A set of examination marks x1, x2, x3, x4, x5, x6, has a mean of 5 and a standard deviation of 1.5.

Satu set markah peperiksaan x1, x2 , x3, x4 , x5, x6 , mempunyai min 5 dan sisihan piawainya ialah 1.5.

(a) Cari / find

 (i) the sum of the marks, x / jumlah markah x

(ii) sum of the squares of the marks, x2

jumlah kuasa dua markah-markah tersebut, x2

.

(b) Each mark is multiplied by 2 and then 3 is added to it. Find, for this new set of marks,
Setiap markah didarab dengan 2 dan kemudian ditambah 3. Cari untuk set markah baru ini,
(i) Mean / min,
(ii) Variance / varians

Solution

(a)(i) Mean = 5 (a)(i) Standard deviation = 1.5

From x =  x From  = x2  (x)2
N
N
1.5 = x2
5= x 6  (5)2

6 1.52 = x2
– (5)2
 x = 30

6

 x2 = 163.5

(b) Each mark is multiplied by 2 and then 3 is added to it

(i) Mean = 5 × 2 + 3
= 13

(ii) Given the standard deviation = 1.5 Variance = (Standard deviation)2

New standard deviation = 1.5 × 2

=3
So, new variance = (3)2

=9

33

2 The table below shows the cumulative frequency distribution for the scores of 50 students in a competition.

Jadual di bawah menunjukkan taburan kekerapan longgokan bagi skor 50 orang murid dalam suatu pertandingan.

Score / Skor < 10 < 20 < 30 < 40 < 50

Number of students / Bilangan pelajar 5 19 37 44 50

(a) Based on the table above, copy and complete the following table.

Berdasarkan jadual di atas, salin dan lengkapkan Jadual.

Score / Skor 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49

Number of students / Bilangan pelajar

Solution

Score / Skor 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49

Number of students / Bilangan pelajar 5 14 18 7 6

(b) Without drawing an ogive, find the median and interquartile range of the distribution.
Tanpa melukis ogif, cari median dan julat antara kuartil bagi taburan itu.
Solution

Median First quartile, Q1 Third quartile, Q3 Interquartile range
= Q3 – Q1
= Observe at-( 1 × N)th = Observe at-( 1 × N)th = Observe at-( 3 × N)th = 30.2143 – 14.8571
2 4 4
= 15.3572
= Observe at-( 1 × 50)th = Observe at-( 1 × 50)th = Observe at-( 3 × 50)th
2 4 4

= Observe at-25th = Observe at-12.5th = Observe at-37.5th

= In class 20 – 29 = In class 10 – 19 = In class 30 – 39

= L + ( 1 N  F ) c = L+ ( 1 N  F ) c = L+ ( 1 N  F ) c
2 4 4

fm fm fm

= 19.5 + ( 1  50  19 )10 = 9.5 + ( 1  50  5 )10 = 29.5 + ( 3  50  37 )10
2 4 4

18 14 7

= 22.83 = 14.8571 = 30.2143

3 The mean of a set of numbers 3, x, 6, 3x – 2, 9 and 12 is 8.
Min bagi suatu set nombor 3, x, 6, 3x – 2, 9 dan 12 ialah 8.
(a) Find / Cari
(i) The value of x, / nilai x
(ii) The variance, / varians
(b) Each number in the set is multiplied by 4 and then 10 is taken away from it. For this set of numbers, find
Setiap nombor dalam set itu didarabkan dengan 4 dan kemudiannya ditolak dengan 10. Bagi set nombor ini, cari
(i) The mean, / min
(ii) The standard deviation / sisihan piawai

Solution

(a) (i) From mean, x =  x (ii) Set data: 3, x, 6, 3x – 2, 9, 12 become 3, 5, 6, 13, 9, 12

N From variance, 2 = x2
N – ( x )2
8 = 3  x  6  3x  2  9 12
= 32  52  62  132  92  122 – (8)2
6 6

8 = 4x  28 = 13.3333

6

48 = 4x + 28
48 – 28 = 4x

20 = 4x

x=5

(b) (i) New mean = Old mean × 4 – 10 (ii) From (a)(ii), variance = 13.3333
= 8 × 4 – 10 Then, standard deviation = 13.3333
= 3.6515
= 22 So, new standard deviation = Old standard deviation × 4
= 3.6515 × 4
= 14.606

34

EXERCISE

1 Table shows the marks obtained by 36 candidates in an examination.

Jadual di bawah menunjukkan markah yang diperolehi 36 orang calon dalam suatu peperiksaan.

Marks / Markah Number of candidates / Bilangan calon
40 – 49 4
50 – 59 5
60 – 69 6
70 – 79 9
80 – 89 8
90 – 99 4

(a) Without drawing an ogive, find the third quartile. [3 marks] [Ans: 83.25]
Tanpa melukis ogif, cari kuartil ketiga.
[2 marks] [Ans: 71.1667]
(b) Hitung / Calculate [3 marks] [Ans: 15.0922]
(i) the mean, / min
(ii) the standard deviation / sisihan piawai.

Answer space

2 A set of eight numbers has a mean of 11. [1 mark] [Ans: 88]
Satu set lapan nombor mempunyai min 11. [2 marks] [Ans: 2]

(a) Cari / Find  x .

(b) When a number k is added to this set, the new mean is 10. Find the value of k.
Apabila nombor k ditambah ke dalam set itu, min yang baru ialah 10. Cari nilai k.

Answer space

3 A set of 5 numbers, y1, y2 , y3 , y4 and y5 , has a mean of 6 and a variance of 4. [2 marks] [Ans: 19]
For the set of data 3y1 1, 3y2 1,3y3 1, 3y4 1 and 3y5 +1 , calculate [2 marks] [Ans: 36]

Satu set lima nombor, y1, y2 , y3 , y4 dan y5 mempunyai min 6 dan varian ialah 4.
Bagi set data 3y1 1, 3y2 1,3y3 1, 3y4 1 dan 3y5 + 1 ,hitung

(a) The mean / min
(b) Variance / varians
Answer space

35

4 The heights, in cm, of five players in a basketball team are 168, 170, 172, 175, 180. [1 mark] [Ans: 173]
Tinggi, dalam cm, lima pemain dalam pasukan bola keranjang ialah 168, 170, 172, 175, 180. [3 marks] [Ans: 4.195]

For the heights of the players, find
Untuk ketinggian pemain-pemain itu, cari

(a) mean / min
(b) the standard deviation / sisihan piawai.
Answer space

5 Diagram 5 is a histogram which represents the distribution of the marks obtained by 60 students in a test.
Rajah 5 menunjukkan histogram yang mewakili taburan markah yang diperoleh oleh 60 pelajar di dalam satu ujian.

Number of students 20
18
16 1 10.5 20.5 30.5 40.5 50.5 60.5
14
12 M ar k s
10
8
6
4
2
0

0.5

Diagram 5 / Rajah 5 [3 marks] [Ans: 25.5]
(a) Without using an ogive, calculate the median mark. [4 marks] [Ans: 13.92]

Tanpa menggunakan ogif, hitung markah median.
(b) Calculate the standard deviation of the distribution.

Hitung sisihan piawai taburan itu.
Answer space

36

6 Table shows the total time spent on doing school homework by 120 students for a period of 4 weeks.

Jadual menunjukkan jumlah masa yang diluangkan untuk membuat kerja rumah sekolah oleh 120 pelajar dalam jangka masa 4minggu.

Total Time (hours) Number of students

Jumlah masa ( jam) Bilangan pelajar
5 – 14 12
15 – 24 17
25 – 34 26
35 – 44 31
45 – 54 16
55 – 64 10
65 – 74 8

Calculate / Hitung

(a) The mean, / min [2 marks] [Ans: 36.5]

(b) The interquartile range / julat antara kuartil [4 marks] [Ans: 22.12]

Answer space

7 A set of data consists of nine numbers. The sum of the number is 171 and the sum of the squares of the numbers is 3291.

Suatu set data mengandungi sembilan nombor. Hasil tambah nombor-nombor itu ialah 171 dan hasil tambah kuasa dua nombor-

nombor itu ialah 3291.

(a) For the nine number, find

Bagi sembilan nombor itu, cari

(i) the mean / min [1 mark] [Ans: 19]

(ii) the standard deviation / sisihan piawai [2 marks] [Ans: 2.16]

(b) If 6 is added to each number in the set, find

Jika 6 ditambahkan kepada setiap nombor di dalam set, cari

(i) the new mean / min baru [1 mark] [Ans: 25]

(ii) the new standard deviation / sisihan piawai baru [1 mark] [Ans: 2.16]

Answer space

37

FORM 4 CHAPTER 5 INDICES AND LOGARITHMS [2 marks] [Ans: -5/3]
PAPER 2 SECTION A

1 Solve / Selesaikan

8x  1
32

Answer space

→ Find the smallest common factor for both 8 and 32.
→ Use the formula: am × an = a m + n OR am ÷ an = a m – n
→ Remember a0 = 1
→ Remember 1 = a – n

an

2 Solve the equation / Selesaikan persamaan [2 marks] [Ans: 13]

16x1  8x3

Answer space

3 Solve the equation / Selesaikan persamaan [3 marks] [Ans: -3/5]

273x3  1
9x3

Answer space

→ Find the smallest common factor for both 27 and 9.
→ Use the formula: am × an = a m + n OR am ÷ an = a m – n
→ Remember 1 = a – n

an

→ Remember that means power of 1

2

38

4 Sove the equation / Selesaikan persamaan [4 marks] [Ans: 3]
22x + 1 – 13(2x) – 24 = 0

Answer space

→ Indices involving addition or subtraction: Move one part to the right side or straight using substitution method!!

22x + 1 – 13(2x) – 24 = 0

22x × 21 – 13(2x) – 24 = 0

(2x )2 × 2 – 13(2x) – 24 = 0

Substitute 2x with y:

So, y2 × 2 – 13y – 24 = 0

2y2 – 13y – 24 = 0

(2y + 3)(y – 8) = 0

2y + 3 = 0 or y – 8 = 0

y=–¾ y=8

From 2x = y , When y = -3/4, When y = 8,
2x = -3/4 [rejected] 2x = 8
2x = 23
5 Sove the equation / Selesaikan persamaan
165x – 49x – 6 = 0 x=3
Answer space
[4 marks] [Ans: 3]

→ Indices involving addition or subtraction: Move one part to the right side or straight using substitution method!!

6 Given that log3 K – log9 L = 2 , express K in terms of L. [4 marks] [Ans: K = 9 L]
Diberi log3 K – log9 L = 2, ungkapkan K dalam sebutan L
Answer space

→Rule 1: Logarithms involving addition or subtraction: Change to the same base and must combine together!

→Rule 2: loga m n = loga m + loga n

→Rule 3: loga m = loga m – loga n
n

→Rule 4: log a m n = n loga m
→Rule 5: log a a = 1

→Rule 6: loga b = logc b ← only this one you will see the log divided by log
logc a

→Rule 7: If ab = c, then loga c = b

Given log3 K – log9 L = 2 ←Under Rule 1
Change base!
log3 K – log3 L =2 ←Under Rule 6: log9 L= log3 L
log3 9 log3 9

log3 K – log3 L =2
log3 32

log3 K – log3 L =2 ←Under Rule 4: log334 = 4log33
2log3 3

log3 K – log3 L =2 ←Under Rule 5: log33 = 1
2

Multiply all terms by 2: 2log3 K – log3L = 4
log3 K 2 – log3L = 4
←Under Rule 4: 2log3K = log3K2

log3 K2 =4 ←Under Rule 3: log3K2 – log3L = log3 K2
L L

K 2 = 34 ←Under Rule 7

L

K 2 = 81

L

K2 = 81L

K=9 L

39

7 (a) Given that logp3 = r and logp7 = s, express logp( 49 p ) in terms of r and s.
27

Diberi logp3 = r dan logp7 = s, ungkapkan logp( 49 p ) dalam sebutan r dan s.
27

Solution

If the log form given is in fraction, you must separate it into parts to solve!!!!

logp( 49 p ) = logp 49p – logp 27 ← Formula: loga m = loga m – loga n
n
27

= logp (7 × 7 × p) – logp (3 × 3 × 3)
= logp 7 + logp 7 + logp p – [logp 3 + logp 3 + logp 3] ← Formula: loga m × n = loga m + loga n
= s + s + 1 – [r + r + r]

= 2s – 3r + 1

(b) Given that logx 2 = p and logx 7 = q, express logx ( 56 ) in terms of p and q. [4 marks] [Ans: q + 3p – 2]
x2

Diberi logx 2 = p dan logx 7 = q, ungkapkan logx ( 56 ) dalam sebutan p dan q.
x2

Answer space

(c) Given that log5 2 = q and log5 9 = p, express log5 8.1 in terms of q and p. [4 marks] [Ans: 2p – q – 1]
Diberi log5 2 = q dan log5 9 = p, ungkapkan log5 8.1dalam sebutan q dan p
Answer space

Change 8.1 to fraction number first! 8.1 = 81/10

8 (a) Solve the equation / Selesaikan persamaan ← 3 = 3 × 1, 3 = 3 × log33, 3 = log333
3 + log3 (2x – 1) = log3x ← Formula: loga m × n = loga m + loga n

Solution
Change all to log3: log333 + log3 (2x – 1) = log3 x

log327 + log3 (2x – 1) = log3 x
log3[27 × (2x – 1)] = log3 x
log3[27(2x – 1)] = log3 x

By comparing: 27(2x – 1) = x
54x – 27 = x
54x – x = 27

53x = 27

x = 27/53

(b) Given that log3 mn = 3 + 2log3 m – log3 n, express m in terms of n. [4 marks] [Ans: m = n2/27]
Diberi log3 mn = 3 + 2log3 m – log3 n, ungkapkan m dalam sebutan n.
Answer space

40

9 (a) Given that log9 y = log3 18, find the value of y. [3 marks] [Ans: y = 324]
Diberi log9 y = log3 18, cari nilai y.
Answer space

Hint: Change to same base first!!!

(b) Given that log3 m = v and log3 n = w, express log9 ( 81m ) in terms of v and w.

n

Diberi log3 m = v dan log3 n = w, ungkapkan log9 ( 81m ) dalam sebutan v dan w.
n

Solution:

Hint: Change the base first! Then separate the fraction!!

(c) Solve the equation / Selesaikan persamaan [3 marks] [Ans: 0.1813]
3(53x + 1) = 36

Answer space

Hint: Move the 3 to right side first! Then use log10

(d) 3n2  9n3 [3 marks] [Ans: 1/3]
Simplify / Permudahkan 27 n3

Answer space

Hint: change all to base 3

(e) Solve 27t = 9(3t – 1), find the value of t / Selesaikan 27t = 9(3t – 1), cari nilai t . [3 marks] [Ans: t = ½]
Answer space

41

10 (a) Given that log8 n = 1 , find the value of n / Diberi log8 n = 1 , cari nilai n. [1 mark] [Ans: 2]
3
3

(b) Given that 2r = 3s = 6t, express t in terms of r and s. [4 marks] [Ans: t = rs/(s+r)]

Diberi 2r = 3s = 6t, ungkapkan t dalam sebutan r dan s.

(c) Given that y = kxm where k and m are constants, y = 4 when x = 2 and y = 8 when x = 5, find the values of k and m.

Diberi y = kxm dengan keadaan k dan m ialah pemalar , y = 4 apabila x = 2 dan y = 8 apabila x = 5, cari nilai k dan m.

[5 marks] [Ans: k = 2.368, m = 0.7565]

Answer space

11 The value of a house increases by 5% at the beginning of each year. If the initial value of the house is RM90 000, the
value of the house after t years, RM p, is given by p = 90000(1.05)t.

Nilai sebuah rumah meningkat sebanyak 5% pada setiap awal tahun daripada harga asal. Jika nilai rumah itu bermula dengan
RM90 000, selepas t tahun, didapati nilainya RM p, diberi oleh p = 90000(1.05)t.

Find / Cari

(a) the value of the house after 6.5 years / nilai rumah itu selepas 6.5 tahun, [1 mark] [Ans: 120608.61]

(b) minimum number of years for the value of the house to be more than RM150 000 [4 marks] [Ans: 11 years]

bilangan tahun minimum untuk nilai rumah itu melebihi RM150 000.

Answer space

12 Solve the simultaneous equations / Selesaikan persamaan serentak [7 marks] [Ans: x = 2/9, y = 4/9]
log9 y = 1 + log3 x
9x = 3y

Answer space

42

FORM 4 CHAPTER 6 COORDINATE GEOMETRY
PAPER 2 SECTION A / B

1 A straight line passes through the points A(– 2, – 5) and B(6, 7).
Suatu garis lurus melalui titik A(– 2, – 5) dan B(6, 7).

(a) Point D divides the line segment AB internally in the ratio 1 : 3. Find the coordinates of D.
Titik D membahagikan dalam tembereng garis AB dalam nisbah 1 : 3. Cari koordinat D.
Solution

Remember the formula for a point dividing a segment between two point with ratio:

The point which divide, (x, y) = ( nx1  mx2 , ny1  my2 )
mn mn

Do labeling first!!
Let A(– 2, – 5) = (x1, y1), B(6, 7) = (x2, y2), ratio 1 : 3 = m : n and the point D(x, y).

So, point D = ( (3)(6)  (1)(6) , (3)(5)  (1)(7) )
13 13

= (6, – 2)

(b) Given that the point C(h, 10) lies on the straight line AB, find the value of h.
Diberi titik C(h, 10) terletak di atas garis lurus AB, cari nilai h.
Solution

Remember the formula of gradient → m = y2  y1
x2  x1

Lie on the straight line → same gradient
Is parallel to….. → same gradient

C lies on the straight line AB → same gradient

So, gradient AC = gradient BC ←can also gradient AB = gradient AC = Gradient BC
10  (5) = 10  7
h  (2) h  6

15 = 3
h2 h6
Cross multiply: 15(h – 6) = 3(h + 2)

15h – 90 = 3h + 6
15h – 3h = 6 + 90

12h = 96

h = 96/12

h=8
(c) The straight line 8x + 4hy – 6 = 0 is perpendicular to the straight line 3x + y = 16. Find the value of h.

Garis lurus 8x + 4hy – 6 = 0 adalah berserenjang dengan garis lurus 3x + y = 16. Cari nilai h.

Solution

Line A which is perpendicular to Line B
↑↑

m1 m2

Perpendicular → m1 × m2 = –1

Gradient of straight line 8x + 4hy – 6 = 0 Gradient of straight line 3x + y = 16
4hy = – 8x + 6 y = – 3x + 16

y = 8x  6 m2 = – 3
4h

y = 2 x  3
h 2h

m1 = 2
h

For perpendicular: m1 × m2 = –1
( 2 ) × (– 3) = –1

h
6 =1
h
h=6

43

2 (a) The straight line x  y  1 has a y-intercept of 3 and is parallel to the straight line y + nx = 0. Determine the value

14 m

of m and of n.

Garis lurus x  y  1 mempunyai pintasan-y = 3 dan selari dengan garis lurus y + nx = 0. Tentukan nilai m dan n.
14 m

Solution

Remember the general equation of straight line → y = mx + c m = gradient c = y-intercept
And parallel → same gradient

Change x  y  1 to general form for straight line y = mx + c For straight line y + nx = 0.
y = – nx
14 m
Gradient is – n
Multiply with 14m: mx + 14y = 14m
14y = – mx + 14 m

y = mx 14m
14

y = m x + m

14

Given y-intercept is 3. So, the m = 3. The gradient is m = 3

14 14

Parallel means same gradient! So, 3 = – n

14

n= 3

14

(b) Given the points A(1, 8), point B(5, 4) and C(0, 1) are the vertices of a triangle.

Diberi titik-titik A(1, 8), titik B(5, 4) dan titik C(0, 1) merupakan bucu-bucu bagi suatu segitiga

(i) Given (3, b) is the midpoint of the points A and B. Find the value of b.

Diberi (3, b) ialah titik tengah bagi titik-titik A dan B. Cari nilai bagi b.

Solution

Remember the formula → Midpoint (x, y) = ( x1  x2 , y1  y2 )

22

Label the A(1, 8) = (x1, y1), point B(5, 4) = (x2, y2). Midpoint (x, y) = ( x1  x2 , y1  y2 )

22

(3, b) = ( 1  5 , 8  4 )

22

(3, b) = (3, 6)
So, b = 6
(ii) Find the distance between point A and point B / Cari jarak antara titik A dan titik B.
Solution

Remember the formula → Distance between two points = (x2  x1)2  (y2  y1)2

Label the A(1, 8) = (x1, y1) and point B(5, 4) = (x2, y2).
Distance between AB = (x2  x1)2  (y2  y1)2

= (5 1)2  (4  8)2

= 32
(iii) Point Q moves such that the distance between the QB is always 4 units. Find the locus of Q.

Titik Q bergerak dengan keadaan jarak antara QB adalah sentiasa 4 unit. Cari lokus Q.
Solution

Remember the formula to find locus → Use the formula of distance (x2  x1)2  (y2  y1)2

Let B(5, 4) = (x1, y1) and Q(x, y) = (x2, y2). So, (x2  x1)2  (y2  y1)2 = 4 ← closed point

(x  5)2  ( y  4)2 = 4
Taking square for both sides: (x – 5)2 + (y – 4)2 = 16

(x – 5)(x – 5) + (y – 4)(y – 4) = 16
x2 – 10x + 25 + y2 – 8y + 16 = 16

x2 + y2 – 10x – 8y + 25 = 0
(iv) Find the area of triangles ABC / Cari luas segitiga ABC.

Solution
Remember the formula of area of triangle → Area = 1 │ x1 x2 x3 x1 │
2 y1 y2 y3 y1

Area of triangles ABC = 1 │ 1 5 0 1 │
2 8418

= ½ │(1×4 + 5×1+0×8) – (8×5+4×0+1×1)│

= ½ │ 9 – 41│
= 16 unit2

44

1 Diagram 1 shows a kite ABCD. EXERCISE
Rajah 1 menunjukkan satu lelayang ABCD.
y
C(4, 15)

D

A(– 2, 3) B(11, 4)

O x

Diagram 1 / Rajah 1 [1 mark] [Ans: (1, 9)]
Find / Cari: [3 marks] [Ans: 2y = – x + 19]
(a) the coordinates of the midpoint of diagonal AC.

koordinat titik tengah pepenjuru AC.
(b) the equation of diagonal BD.

persamaan pepenjuru BD .
Solution space

2 In Diagram 2, ACD and BCE are straight lines. C is the midpoint of AD and BC : BE = 1 : 5.
Dalam Rajah 2, ACD dan BCE ialah garis lurus. C ialah titik tengah AD dan BC : BE = 1 : 5.
y
D (5, 6)
B (2, 9/2)

C

A (1, 2) E

Ox

Diagram 2 / Rajah 2

Find / Cari [2 marks] [Ans: (3, 4)]
(a) the coordinates of point C. [2 marks] [Ans: (7, 2)]

koordinat titik C.
(b) the coordinates of point E.

koordinat titik E.
Solution space

45

3 Diagram 3 shows straight line CD which meets straight line AB at the point E. Point C lies on the y-axis.
Rajah 3 menunjukkan garis lurus CD bertemu garis lurus AB di titik E. Titik C terletak pada paksi-y.
y

C

O x
A (0, -3) E B (12, 0)

D [1 marks] [Ans: x/12 – y/3 = 1]
[1 marks] [Ans: (4, -2)]
Diagram 3 / Rajah 3 [3 marks] [Ans: 14]
(a) Write down the equation of AB in the form of intercepts.

Tulis persamaan AB dalam bentuk pintasan.
(b) Given that 2AE = EB, find the coordinates of E.

Diberi 2AE = EB , cari koordinat E.
(c) Given that CD is perpendicular to AB, find the y-intercept of CD.

Diberi CD adalah berserenjang dengan AB, cari pintasan-y bagi CD.
Answer space

4 Diagram 4 shows a triangle AOB where O is the origin. Point C lies on the straight line AB.
Rajah 4 menunjukkan segitiga AOB di mana O adalah asalan. Titik C terletak pada garis lurus AB.

y
A (– 2, 5)

C x
O

B (5, – 1)

Diagram 4 / Rajah 4 [2 marks] [Ans: 23/2]
(a) Calculate the area, in unit2, of triangle AOB.

Hitung luas, dalam unit2, bagi segitiga AOB.

(b) Given that AC : CB = 3 : 2, find the coordinates of C. [2 marks] [Ans: (11/5, 7/5)]

Diberi AC : CB = 3 : 2, cari koordinat C.

(c) A point P moves such that its distance from point A is always twice its distance from point B.

Titik P bergerak dengan keadaan jaraknya dari titik A sentiasa dua kali jaraknya dari titik B.

(i) Find the equation of the locus of P. [3 marks] [Ans: 3x2 + 3y2 – 44x + 18y + 75]

Cari persamaan bagi lokus P.

(ii) Hence, determine whether the locus of P intersects the y-axis. [3 marks] [Ans: Tidak]

Seterusnya , tentukan sama ada lokus P memintas paksi–y.

Answer space

46

CHAPTER 9 DIFFERENTIATION

PAPER 2 SECTION A / B

1 (a) Find the lim x2 4 / Cari had x2  4 .

x2 x2 x2 x  2

Solution

Clue: If x is near to certain number → factorise and simplify first before substitution!

lim x2 4 = lim x2 22
x2
x2 x2 x2

= lim (x  2)(x  2) ← Remember the formula a2 – b2 = (a + b)(a – b)

x2 x2

= lim (x  2)

x2

= (–2 – 2)
=–4

(b) Find the 2x2 1 / Cari 2x2 1 .
x  3x2 x  3x2
lim ( ) had ( )

x x

Solution

Clue: If x is near to infinity → divide each term by xn before substitution!

2x2 1 = 2x2  1
x  3x2 x2 x2
lim ( ) lim ( )
x 3x2
x x x2  x2

2  1
x2
= lim ( )
1 
x x 3

= (20)

03

=2

3

(c) Find the lim ( 3x  2x2 / Cari 3x 2 x2 .
x x
x0 ) had ( )

x0

Solution

Clue: If x is near to zero → divide each term by xn before substitution!

lim ( 3x  2x2 ) = lim ( 3x  2x2 )
x x x
x0 x0

= lim(3  2x)

x0

= 3+2(0)

=3

(d) Find the lim(3x  5) / Cari niai bagi lim(3x  5) .
x0 x0

Solution

Clue: If it is not in fraction form → direct substitution!

lim(3x  5) = 3(0) + 5

x0

=5

2 Find the dy by using the first principle / Cari dy dengan menggunakan kaedah pertama.

dx dx

(a) y = 5
(b) y = 3x2 – 2
(c) y = 4

x

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3 Find the dy for / Cari dy untuk: (b) y = 6x1/2 – 4x3 + 5x – 1
Answer space
dx dx

(a) y = 3x2 – 8

Apply the: y = xn, then dy = nxn – 1

dx

Answer space

4 Find the dy for / Cari dy untuk: (b) y= 3 + 23 –2
x2
dx dx 5x2 3

(a) y = 2 – 2x5 + 8 – 2x Answer space

x x3

Move up the x first before differentiation!!
Answer space

5 Find the dy for / Cari dy untuk: (b) y = (3x – 1)(x + 5)
Answer space
dx dx

(a) y = 3(2x – 8) + 4x2(x + 2)

Expand first before differentiation!!

Answer space

6 Find the dy for / Cari dy untuk: (b) y = 4(x2 + 3)3 + 3(x + 2)
Answer space
dx dx

(a) y = (3x – 1)5 + 3x
Diff the power first! Then multiply with the diff inside!!

Formula: y = (ax + b)n, dy = n(ax + b)n –1(a)

dx

Answer space

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7 Find the dy for / Cari dy untuk: (b) y = (4x3 + 2x)(2x + 1)4
Answer space
dx dx

(a) y = (x + 5)(3x – 1)3

Formula: If y = uv, then dy = u dv  v du

dx dx dx

Answer space

8 Find the dy for / Cari dy untuk:

dx dx

(a) y = 3x (b) y = 6x2  5

x2 1 (2x  3)2

Formula: If y = u , then dy = v du  u dv
dx dx

v dx v2

Answer space

9 Given that f (x) = x(5 – 3x)2, find f '(2) / Diberi f (x) = x(5 – 3x)2, cari f '(2). [3 marks] [Ans: 13]
Answer space

10 The gradient of the tangent to the curve y = x2 (px  4) at x =  1 is 14. Find the value of p. [3 marks] [Ans: 2]
Kecerunan tangen kepada lengkung y = x2 (px  4) di x =  1 ialah 14. Cari nilai p.

Answer space

Remember: Gradient means dy/dx

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11 The point P lies on the curve y = (x  5)2. It is given that the equation of the normal to the curve at P is y = – 1 x + 23 .

44

Find the coordinates of P. [4 marks] [Ans: (7,4)]

Titik P terletak pada lengkung y = (x  5)2. Diberi bahawa persamaan normal kepada lengkung pada P ialah y = – 1 x + 23 . Cari
4
4

koordinat P.

Answer space m1 is the dy/dx for first equation m2 is the dy/dx for second equation
Remember: Normal means m1m2 = – 1

12 The normal to the curve y = x2 + 1 at point P is parallel to the straight line 2y + x = 8. Find the equation of the normal to

the curve at point P. [4 marks] [Ans: x + 2y = 5]

Garis normal kepada lengkung y = x2 + 1 pada titik P adalah selari dengan garis lurus 2y + x = 8. Cari persamaan garis normal

kepada lengkung itu pada titik P.

Answer space

Remember: Normal means m1m2 = – 1 , Parallel means same gradient , equation means y = mx + c

13 A cube expands in such a way that its sides change at a rate of 2cms-1. Find the rate of change of the total surface area

when its volume is 125 cm3. [4 marks] [Ans: 120]

Sebuah kubus mengembang dengan keadaan sisinya berubah pada kadar 2 cms-1. Cari kadar perubahan jumlah luas permukaan

apabila isi padunya adalah 125 cm3.

Answer space

Apply chain rule……… dy = dy × d....
dx d.... dx

14 The curve y = –3x2 – 12x + 7 has a maximum point at x = k, such that k is a constant. Find the value of k.[3 marks][Ans:-2]

Lengkung y = –3x2 – 12x + 7 mempunyai titik maksimum pada titik x = k, dengan keadaan k adalah pemalar. Cari nilai k.

Answer space
Remember: Maximum or minimum point means dy/dx OR do the completing the square method!

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