1202 QB Additional Mathematics Form 4

Book extend = 152 pg
Spine = 7.08 mm



















Titles in this series:


Subject/Form 4 5 QUESTION BANK
QUESTION BANK Tatabahasa

English
Grammar

Matematik
Sains
This series is published to help students
familiarise themselves with the types of Sejarah
questions that appear in the new SPM Ekonomi
examination format. Students can build QUESTION BANK
their confidence in answering questions Perniagaan
with various specially formulated topical
C Pendidikan Moral
practices. The questions in the book are
written with a focus on Higher Order Prinsip Perakaunan
M
Thinking Skills (HOTS) in addition to a ADDITIONAL MATHEMATICS
Y Pendidikan Seni Visual
collection of cloned SPM questions from
CM
past year examinations. Kimia Form
MY
Fizik ©PAN ASIA PUBLICATIONS
CY
CMY Biologi
K
Matematik Tambahan 4
EXTRA FEATURES: Mathematics ADDITIONAL
MUST KNOW Chemistry
Important Facts, Common Mistakes Physics KSSM

SOS TIPS Biology MATHEMATICS
Hints for questions Additional Mathematics




Dr. M. K. Wong (Textbook Author)


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MUST


KNOW Important Facts







Functions Solving Quadratic Equation
1. Function: 1. Three ways to solve the quadratic equations:
Domain = {a, b, c} (a) Factorisation
a ● ● 1 Codomain = {1, 2, 3} (Use the principle “If pq = 0, then p = 0 or q = 0”)
b ● ● 2 Objects = a, b, c (b) Completing the squares
©PAN ASIA PUBLICATIONS
c ● ● 3 Images = 1, 2, 3 2
b – 4ac
Range = {1, 3} (c) Formula x = –b + ABBBBBB
2. 4 types of relations: 2a
One-to-one Many-to-one 2. If the roots are given, then the equation can be obtained by:
x – (Sum of roots)x + (Product of roots) = 0
2
A function A function y
Solving Quadratic Inequalities y = f(x) = (x + 2)(x – 4)
One-to-many Many-to-many 1. Three ways to solve the y > 0 y > 0
quadratic inequalities: x < –2 x > 4
(a) Graph sketching –2 0 4 x
Not a Not a
function function (b) Number line
(c) Table –2< x < 4
y < 0
Important Facts (Chapter 1) 1 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 7 @ Pan Asia Publications Sdn. Bhd.


Identify a Function Forms of Quadratic Functions
1. General form
1. If f : x → y, then f (x) = y.
f (x) = ax + bx + c, a ≠ 0, b and c are constant
2
2. By using vertical line test: 2. Vertex form
If any vertical line intersects f(x) graph at not more than one f (x) = a(x + h) + k, a ≠ 0, h and k are constant
2
point, then it is a function.
3. Intercept form
y y y f (x) = a(x – p)(x – q), a ≠ 0, p and q are constant
Expansion Factorisation or formula
x x x
f (x) = a(x – h) + k f (x) = ax + bx + c f (x) = a(x – p)(x – q)
2
2
A function Not a function
The vertical line cuts the graph at Completing the square Expansion
two points


Important Facts (Chapter 1) 3 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 9 @ Pan Asia Publications Sdn. Bhd.

Composite Function and Inverse Function Type of Roots for Quadratic Equations
1. Composite function: The type of roots obtained depends on the discriminant, D = b – 4ac
2
gf Discriminant
gf (x) ≠ fg(x) a . 0 a , 0
2
f (x) = ff (x), D = b – 4ac
2
3
● f ● g ● f (x) = fff (x) D . 0 y y
x f(x) g[f(x)] = f f (x)
2
f –1 g –1 Two real and x
= ff (x)
2
distinct roots x
f g = (gf ) –1 y y
–1
–1
2. Characteristics of inverse y f(x) D = 0 x
function: y = x Two real and
(a) Only one-to-one function has f (x) equal roots x
–1
an inverse function. y y
(b) If (a, b) is a point on the D , 0 x
graph f (x), then (b, a) is its x No real roots
corresponding point on f (x). x
–1
Important Facts (Chapter 1) 5 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 11 @ Pan Asia Publications Sdn. Bhd.
00B_1202 QB AMath F4.indd 3 09/05/2022 11:30 AM

MUST


KNOW Common Mistakes







Solving Quadratic Equations Functions
Use the method of completing the squares to solve 2x + 5x + 1 = 0 . The diagram shows the f g
2
Correct Wrong function f maps set A to
2x + 5x + 1 = 0 2x + 5x + 1 = 0 set B and g maps set B x x + 2 5x + 4
2
2
to set C. Find g(x).
5
2
2
1
2
2
2 x + x + 1 = 0 1 2x + 5 + 1 = 0
2
2
31
2 x + 5 4 2 2 – 25 4 + 1 = 0 Correct A B Wrong C
16
1
2 x + 5 4 2 2 – 25 + 1 = 0 Note: 1 5 is g(x + 2) = 5x + 4 g(x + 2) = 5x + 4
2
2
When 2x +
8
2
1
2 x + 5 4 2 2 = 17 expended, it becomes Let y = x + 2 g(y) = 5y + 4
Then, x = y − 2
8
1 x + 5 4 2 2 = 17 4x + 10x + 25 . g(y) = 5(y – 2) + 4
2
4
16
= 5y – 10 + 4

x = – ± ABBB = 5y – 6
17
5
4 16 ∴ g(x) = 5x – 6
= –0.219, –2.281
Common Mistakes (Chapter 2) 8 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 2 @ Pan Asia Publications Sdn. Bhd.
Vertex Form of Quadratic Equations Composite Function
Complete the square for f (x) = 2x + 4x – 3. Hence, find the If f (x) = 2x + 1 and g(x) = x – 5, find fg(x).
2
coordinates of the minimum point. Correct Wrong
Correct Wrong fg(x) = f (x – 5) fg(x) = g(2x + 1)


= 2x + 1 – 5
= 2(x – 5) + 1
f (x) = 2x + 4x – 3 6 ©PAN ASIA PUBLICATIONS
f (x) = 2x + 4x – 3
2
2


= 2x – 10 + 1
= 2x – 4
1
2
= 2 x + 2x – 3 2 = (2x + 2) – 4 – 3 = 2x – 9
2
2
= (2x + 2) – 7
2
1
= 2 (x + 1) – 1 – 3 2 Note: Substitute g(x) first followed Do not substitute f (x) first.
2
2
2
2
= 2(x + 1) – 5 (2x + 2) ≠ 2x + 4x + 4 by f (x).
2
The coordinates of the The coordinates of the
minimum point is (–1, –5). minimum point is (1, –5).
Note:
The vertex for
f (x) = a(x – h) + k is (h, k).
2
Common Mistakes (Chapter 2) 10 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 4 @ Pan Asia Publications Sdn. Bhd.
Roots for Quadratic Equations Inverse Function
Determine the sum of roots and product of roots for the quadratic If f (x) = x – 1 , g (x) = x + 5 and fg(x) = 2x – 9, find (fg) (x).
–1
–1
–1
equation 2x + x – 6 = 0. 2
2
Correct Wrong
Correct Wrong fg(x) = 2x – 9 (fg) (x) = f g (x)
–1
–1
–1
1 Sum of roots = 1 or –1 Let y = 2x – 9 = f (x + 5)
–1
Sum of roots = –
2 Product of roots = –6 then x = y + 9 (x + 5) – 1
Product of roots = – 2 = 2
2 Note: –1 y + 9 x + 4
= –3 For ax + bx + c = 0. (fg) (y) = 2 = 2
2
Then, (fg) (x) = x + 9
–1
2
Sum of roots = – b Note:
a
Product of roots = c a (fg) (x) ≠ f g (x)
–1
–1
–1
(fg) (x) = g f (x)
–1
–1
–1
fg(x) ≠ gf(x)
Common Mistakes (Chapter 2) 12 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 6 @ Pan Asia Publications Sdn. Bhd.
00B_1202 QB AMath F4.indd 4 09/05/2022 11:30 AM

1 Functions
Chapter





NOTES


1.1 Functions 8. A function is not defined if there exists a function in
fraction form and its denominator is zero.
1. Function is a relation between two sets, domain and For example,
©PAN ASIA PUBLICATIONS
codomain.
1
2. Elements in the domain are called objects while the f : x → x if x = , then
elements in the codomain are called images. 2x – 1 2
3. The diagram shows a function that connecting set X 1 1
to set Y which represents in an arrow diagram. f (x) = 2 = 2 , no solution
1
Set X Set Y 2 1 2 – 1 0
2
a ● ● 1
9. Vertical line test:
b ● ● 2
If a vertical line parallel to the y-axis intersects a graph
c ● ● 3 more than once, then the algebraic expression is not
Function a function. If it intersects once, then it is a function.
The function is denoted by f : x → y or f (x) = y. y
Then object = {a, b, c} Vertical line test
image = {1, 2, 3} f
4. There are four types of relations:
(a) One-to-one (b) Many-to-one

x
a ● ● 1 a ● ● 1 0
b ● ● 2 b ● ● 2 10. An absolute function is a function whose value is
positive only.
c ● ● 3 c ● ● 3
x, if x > 0
(c) One-to-many (d) Many-to-many f (x) = |x| = –x, if x , 0
f(x)
a ● ● 1 a ● ● 1
f(x) = |x|
b ● ● 2 b ● ● 2
Graph f(x)
c ● ● 3 c ● ● 3 is always
positive x
5. Function is a special relation such that:
(a) Every element in the domain must map to one
element in the codomain. f(x) = x
(b) More than one element in the domain map to one 11. Discrete function is a function where the points on
element in the codomain. the graph is real, separated and not connected by a
6. A function maps onto itself if straight line or curve.
f : x → x or f (x) = x f(x)
7. Relation representations:
(a) Arrow diagram (b) Graph 3
y 2
1
p ● ● 4 8
x
6 –1 0 1 2 3
q ● ● 6
4
r ● ● 8 2 Domain = {−1, 1, 2, 3}
x Codomain = {1, 2, 3}
p q r Range = {1, 2, 3}
(c) Ordered pairs
(p, 6), (q, 4) , (r, 8)
1





01_1202 QB AMath F4.indd 1 09/05/2022 11:30 AM

12. Continuous function is a function where the points
are connected by a straight line or a curve within the f
given interval.
x ● ● y
f(x)
f –1
4
3
If f (x) = y, then f (y) = x
–1
2. Properties of inverse function:
x
0 2 4 (a) A function f that maps set A to set B has an inverse
–1
Domain of f is 0 < x < 4 function f if f is a one-to-one function.
©PAN ASIA PUBLICATIONS
Codomain of f is 0 < f (x) < 4 (b) fg(x) = x where x in the domain of g and gf (x) = x
Range of f is 0 < f (x) < 4 where x in the domain of f.
(c) If two functions, f and g are inverse to each other,
1.2 Composite Functions then
(i) domain of f = range of g,
1. Composite function: (ii) domain of g = range of f,
(iii) graph g is a reflection of graph f.
f g (d) For any real numbers, a and b, if the point (a, b)
is on the graph f, then the point (b, a) is on the
x
y = f(x) z = g(f(x)) graph g.
3. Horizontal line test:
gf
If the horizontal line intersects the graph of a function
A B C
at only one point, then this type of function is
2. If a function f maps x to y, written as f (x) = y and one-to-one function and it has an inverse function.
another function g maps y to z, written as g(y) = z, f(x)
then gf (x) maps x to z and is written as gf (x) = z. f
3. gf (x) ≠ fg(x)
4. f (x) = fff (x) = ff (x) = f f (x) Horizontal
2
2
3
line test
1.3 Inverse Functions 0 x
1. An inverse function maps an image to its object.





PAPER 1

Section A

1. The diagram shows a relation between set A and Answer:
SPM set B. (a)
CLONE
A B
f
6 ● ● 13
8 ● ● 17

p ● ● q
(b)
State
(a) the function of f (p), [2 marks]
(b) q in terms of p. [3 marks]



2 Question 1(a):
SOS TIP Write f (x) = ax + b, then find a and b. 22










01_1202 QB AMath F4.indd 2 09/05/2022 11:30 AM

Section B

11. (a) The diagram shows a graph for the function (ii)
SPM 1
|
|
CLONE f : x → 3 – x for the domain –2 < x < 9.
2
f(x)
(–2, 4)

(iii)

1
©PAN ASIA PUBLICATIONS
x
0 4 9

State
(i) the object for 4, [1 mark]
(ii) the image for 1, [1 mark]
(iii) the domain for 0 < f (x) < 1. [2 marks] 12. (a) Sketch the graph f (x) = |2x – 1| for –1 < x < 2.
(b) The diagram shows an inverse function where [3 marks]
x ≠ –q. (b) Given f : x → 2x.
(i) Find
g –1 p
3
2
x –––– (a) f , (b) f .
q + x
[2 marks]
n
3 –2 (ii) Hence, find the expression for f , in terms
of n where n = 1, 2, 3,…. [1 mark]
(iii) Find the value of x if f (x) = 16. [2 marks]
5
1 –4
–
2 Answer:
(a)
Find
(i) the value of p and q, [1 mark]
(ii) g(x) with the value of p and q from (i),
[2 marks]
(iii) the value of x if g(x) = 8. [1 mark]
Answer:
(a) (i)
(b) (i) (a)
(ii)




(b)

(iii)

(ii)




(b) (i)
(iiii)






4 Question 11(a)(iii): 5
SOS TIP 0 < f (x) < 1 is the same as 0 < |3 – x| < 1. SOS TIP
1
2
Question 12:
44
(a) Sketch the graph y = 2x – 1. Then, reflect the part of the graph that is below the x-axis on the x-axis.
(b) After finding the expression for ff (x) and fff (x), observe for the pattern formed. 5



01_1202 QB AMath F4.indd 5 09/05/2022 11:30 AM

PAPER 2


Section A

3 1 5. The diagram shows the graph f : x → 2x + 3 for
1. Given two functions defined by f : x → x + and –1 < x < 3.
4
2
5 2
g : x → – x. f(x)
4 3
(a) Is f (2) + f (3) = f (2 + 3)? Explain your answer. 10
[2 marks] 8 f(x) = 2x + 3

(b) Is g(4) − g(2) = g(4 − 2)? Show your working. 6
©PAN ASIA PUBLICATIONS
[2 marks]
(c) Find the value of k if f(2k) = 6g(k). [2 marks] 4
(d) Find the value of k if f(k) + g(k) = 5. [2 marks] 2
x
0
2
6
8
10
–2
4
2. (a) The function g is defined by g : x → x + 1 , –2
x – 2
x ≠ 2, find –1
(i) g , [2 marks] (a) Find f (x). [1 mark]
2
(ii) g . [2 marks] (b) Based on (a), find the corresponding coordinates
−1
ax + 1 for the coordinates (1, 5).
(b) The function h as defined by h : x → x , [1 mark]
–1
x ≠ 0 is given by hg (4) = 6, find the value of a. (c) On the same axes, sketch the graph f (x) and
–1
[2 marks] state its domain.
[3 marks]
(d) Hence, draw a line of symmetry for f and f .
–1
3. Given g : x → x + 5, find [2 marks]
2
(a) an expression for each of the following.
(i) g(a + 1), [1 mark]
(ii) g(a ), [2 marks] 6. (a) The functions f and g are defined by f : x → 3x – a
2
b
(iii) g(2b – 1) – g(b). [2 marks] and g : x → , x ≠ 0 where a and b are constants.
x
(b) the possible values of x if g(x) = 5x – 1. Given that f (2) = 0 and fg(2) = 16, find the
2
[2 marks] values of a and b.
[4 marks]
2
4. The diagram shows a part of the mapping for the (b) Hence, find the value of g f (x).
function f : x → ax + b where a and b are constants. [3 marks]
2
f
x ax + b 7. (a) The function g is defined by g : x → 8 – 3x.
2
Find
10 (i) the expression for g and g ,
–1
2
3
(ii) the value of x if g (x) = g (x).
–1
2
[4 marks]
–2 –10
(b) The function h is defined by h : x → ax + b,
a ≠ –1 for the domain 0 < x < 5 . Given that the
(a) Find the value of a and of b. [2 marks] graph y = h(x) passes through the point (8, 5) and
1
(b) Given the mapping starts with x = , where will the graph y = h(x) and y = h (x) intersects at the
−1
2
be the end of the arrow point? [2 marks] point whose x-coordinate is 3. Find the value of a
(c) Find another value of x so that the function f will and of b.
map to −10. [2 marks] [3 marks]
HOTS Analysing
8 Question 2(a)(ii) : 9
SOS TIP 88 To find g (x), let y = x + 1 –1 9 SOS TIP
–1
x – 2
Question 7(b) :
Find h (x) and then solve h (x) = h(x) to find the value of x which is given to be 3.
–1
01_1202 QB AMath F4.indd 9 09/05/2022 11:30 AM

1 11. The number of story books read by Amin depends
8. The function f is defined by f (x) = x + x – 12.
2
2 on his spare time and his spare time depends on the
HOTS Analysing amount of home work given by the school. Given that
(a) Find, in a similar form, x(t) = 3t – 5 where x is the number of story books,
(i) f (x + a), t is the spare time in hours and t(k) = 4 + 2k where k
f (x + a) – f (x) is the number of homework given.
(ii) a . (a) Find the number of story books he can read if he
[4 marks]
has 2 homeworks.
f (x + a) – f (x) [4 marks]
(b) Hence, find the value for a if (b) If he can read 7 story books, find the amount of
x = 0.1 when a = 2. [3 marks] spare time he has and the number of homework
given.
©PAN ASIA PUBLICATIONS
[4 marks]
9. A train moves on a straight line. Its acceleration,
a m s , is depends on time t, in seconds, and is given
−2
by a(t) = pt + q where p and q are constants. 12. The diagram shows a cylinder whose volume depends
(a) Given that when the time t = 0 s, the acceleration on its radius of the base, r m and its height, t m.
is −5 m s and when the time t = 4 s, the
−2
acceleration is 15 m s .
−2
Find the values of p and q. [4 marks]
(b) Find the time when the acceleration is 25 m s .
−2
[3 marks] V(t) = t + 1
2

10. The expenditure, RMC, for an annual dinner of a
2
3
company depends on the number of employees in the Given that the volume V(t) = (t + 1) m and the height
factory. In a certain year, the number of employees is t(r) = 1 1 r + 4 m.
2
x and the expenditure of employees per head for the 2
annual dinner is RM(x + p). (a) Express the volume, V, in terms of r. [4 marks]
(a) Express the total expenditure in that year, in (b) Find the volume and radius of the base if the
terms of p. [3 marks] height of the cylinder is 6.5 m. [4 marks]
(b) Find the value of p if the expenditure is RM2 650 HOTS Analysing
and the number of employees is 50. [4 marks]






Section B

13. A function is defined by HOTS Analysing 14. A function f is defined by f : x → x , x ≠ k.
2x + 1
f (x) = |1 – x| for x < 2 (a) State the value of k. [1 mark]
x – 4 for x . 2 (b) Find f –1 1 2 . [2 marks]
2
(a) Sketch the graph of f (x) for the domain 5
1
0 < x < 4. [4 marks] (c) Show that f (x) = x , where x ≠ – .
2
4
(b) Hence, find the corresponding range for the given 4x + 1 [3 marks]
domain of f (x). [2 marks] n x
(c) Find the values of x if f (x) = 1 for the domain (d) Hence, show that f (x) = 2nx + 1 , where
0 < x < 4. [4 marks] n = 1, 2, 3….. and x ≠ – 1 . [4 marks]
2n





10 Question 12:
SOS TIP (a) V(t) = t + 1, t(r) = r + 4. Thus Vt(r) will express V in terms of r. 1010
1
2
2
(b) Find V when t = 6.5.
Find r when t = 6.5.





01_1202 QB AMath F4.indd 10 09/05/2022 11:30 AM

6 Linear Law
Chapter





NOTES


6.1 Linear and Non-Linear Relations For example,
y = ax + b
2
©PAN ASIA PUBLICATIONS
1. A line of best fit is a line that has the following
2
characteristics: To form a linear equation, let Y = y and X = x .
(a) The line passes through as many given points as We have Y = aX + b where a is the gradient and b is
possible. the Y-intercept.
(b) The points which do not lie on the line are fairly y Y
distributed on two sides of the line.
Y = aX + b
y = ax + b
2
6.2 Linear Law and Non-Linear
Relations b b
x X
1. A non-linear relation of two variables can be 0 0
reduced to a linear form by substitution method. A non-linear graph A linear graph



PAPER 1

Section A

1. Reduce the following non-linear relations to the linear 2. The diagram shows a line of best fit obtained by
form Y = mX + c. Identify and state the quantities plotting a graph of y against x.
represented by Y, X, m and c. y
(a) yx = 3x + x [2 marks]
3
x – y 1
(b) xy = 3 [2 marks]
(c) y = 100a [2 marks] (–1, 3)
2x
Answer: (1, 1)
(a) x
0
(3, k)
Find
(a) the equation of the line of best fit, [2 marks]
(b) (b) the value of k. [2 marks]
Answer:
(a)

(c)

(b)




PB Question 2: 49
SOS TIP PBPB (a) Find the gradient of the line, m. 49 SOS TIP


Then, use (y – y ) = m(x – x ) to find the equation of the line
1
1
(b) Insert x = 3, y = k in the equation of the line to get the value of k



06_1202 QB AMath F4.indd 49 09/05/2022 11:33 AM

3. The diagram shows a straight line obtained by plotting 5. The diagram shows two straight lines with the same
y
a graph of against x. equation.
x
y y
y
— — –—
x x x 2
(2, h)
3
1
x
0 1
1
1
x —
–3 0 0 x
©PAN ASIA PUBLICATIONS
The variables x and y are related by the equation I II
y = ax + bx where a and b are constants. Find the (a) Find the equation. [3 marks]
2
values of (b) Calculate the value of h. [2 marks]
(a) a, [3 marks]
(b) b. [2 marks] Answer:
(a)
Answer:
(a)


(b)

(b)


6. The diagram shows a straight line obtained by plotting
a graph of log y against (x – 2).
4. The diagram shows a straight line obtained by plotting 10
a graph of log y against x. log y
10
10
log y
10
(6, 5)
(1, 3)

(3, 2)
x – 2
x 0
0
–1
Given the variables x and y are related by the equation
The variables x and y are related by the equation y = an x – 2 where a and n are constants. Find the
ay = b where a and b are constants. Find the values of value of
x
(a) a, [3 marks] (a) n, [3 marks]
(b) b. [2 marks] (b) a. [2 marks]
Answer: Answer:
(a) (a)





(b) (b)






50 Question 4:
SOS TIP From the graph, find the gradient of the line, m. (mx – 1) = y or 10 =10y.....…

mx
Then, log y = mx – 1 which can be written as 10
10
x
Compare  with the given equation ay = b to determine the values of a and b.
5050


06_1202 QB AMath F4.indd 50 09/05/2022 11:33 AM

Section B

11. The variables x and y are related by the equation 13. The diagram shows the graph of a quadratic function.
px y
y = qx + 2 where p and q are constants. If the graph
of y against x is plotted, a curve which passes through —
9
4
1 1
point (2, 2) is obtained. If the graph of against is
y
x
1
drawn, a straight line with a gradient of is obtained. (3, 0) x
3
Find the values of 0
©PAN ASIA PUBLICATIONS
(a) p, [4 marks] (a) Find the equation of the quadratic function.
(b) q. [4 marks] [4 marks]
Answer: (b) If a straight line is obtained by plotting a graph of
(a) Y against X such that the Y-intercept is equal to 3,
state the quantities represented by X and Y.
[4 marks]
Answer:
(a)
(b)



(b)

1
12. The diagram shows a straight line graph of xy against
SPM
CLONE 1 .
x 3 1
—– 14. The variables x and y are related by the equation
xy
y = hx where h is a constant.
3
(a) Convert the equation y = hx into the linear form
3
Y = mX + c. [3 marks]
(2, h)
(b) The diagram shows a straight line obtained by
plotting a graph of log y against log x.
10 10
(k, 1)
log y
10
1
– — (5, k)
0 x 3
Given that the variables x and y are related by
4x 2
y = , find the value of
5 – x 3
(a) k, [4 marks] 0 log x
10
(b) h. [4 marks] –1
Answer: Find the value of
(a) (i) log h, [3 marks]
10
(ii) k. [2 marks]
Answer:
(a)
(b)
(b) (i)

(ii)



52 Question 13:
SOS TIP The quadratic function in vertex form is y = – x – 3 2 2 + 9 5252
1
2
4
2
Then y = –x + 3x...........
Since the Y-intercept is 3, then  needs to be divided by x.



06_1202 QB AMath F4.indd 52 09/05/2022 11:33 AM

2
19. The variables x and y are related by the equation 20. Diagram I shows the curve y = 2x – 4. Diagram II
y = x + 5x – 1. SPM shows a straight line obtained when y = 2x – 4 is
2
2
CLONE
(a) Convert the equation y = x + 5x – 1 into the expressed in the linear form Y = –4X + c.
2
linear form Y = mX + c such that the straight line y Y
has a gradient of 5. [4 marks]
(b) Hence, state the quantities represented by X, Y
and c. [2 marks] y = 2x – 4 2
2
(c) Find the value of X in (b) if Y = 6 cm. [2 marks]
x X
Answer: 0 0
(a)
©PAN ASIA PUBLICATIONS
I II
In terms of x and/or y, express
(a) X, [3 marks]
(b) (b) Y. [2 marks]
Answer:
(a)
(c)

(b)





PAPER 2

Section B

1. The table shows the values of two variables, x and (a) Plot xy against x. Hence, draw the line of best fit.
y obtained from an experiment. A straight line is [5 marks]
y 2 1 (b) From the graph in (a), find the value of
obtained when graph x against is plotted. (i) p, (ii) q. [5 marks]
x
x 1 2 3 4 5 3. The table shows the values of two variables, V and

y 1.87 2.24 2.55 2.83 3.08 P, obtained from an experiment. It is predicted that
the variables P and V are related by the equation
(a) Based on the table given, construct another table P = a(V + 1) where a and b are constants.
b
1 y 2
for the values and x . [2 marks] V 1 2 3 4 5 6
x
y 2 1 P
(b) Plot against and draw the line of best fit. 4 13.5 32 62.5 108 171.5
x x
[3 marks] (a) Convert the equation P = a(V + 1) into the linear
b
(c) By using the graph on (b), form Y = mX + c. [2 marks]
(i) find the value of y when x = 3.5. [2 marks] (b) Plot Y against X. Hence, draw the line of best fit.
(ii) express y in terms of x. [3 marks] [3 marks]
(c) From the graph in (b), determine the value of
2. The table shows the values of two variables, x and (i) a, (ii) b. [5 marks]
SPM y, obtained from an experiment. It is known that
CLONE 4. The table shows the values of two variables, x and
the variables x and y are related by the equation y, obtained from an experiment. It is predicted that
p + qx
y = where p and q are constants. the variables x and y are related by the equation
x y = qp x + 1 where p and q are constants.
x 1 2 3 4 5 6 x 1 2 3 4 5
y 6 4 3.3 3 2.8 2.7 y 2.25 6.75 20.25 60.75 182.25

54 Question 4:
SOS TIP Given y = qp x + 1 . 10 10 10 10 5454

Take logarithm on both sides, to get log y = log q + (x + 1) log p
The suitable graph will be log y against (x + 1).






06_1202 QB AMath F4.indd 54 09/05/2022 11:33 AM

(a) Plot a suitable straight line graph that relates (a) Plot y against AB x . Hence, draw the line of
x and y. [6 marks] AB x
(b) From the graph in (a), determine the values of best fit. [5 marks]
p and q. [4 marks] (b) Use your graph in (a) to find the value of
(i) p, [2 marks]
(ii) q, [2 marks]
5. The table shows the values of two variables, x and (iii) y if x = 3.61. [1 mark]
SPM y, obtained from an experiment. It is known that
CLONE
the variables x and y are related by the equation
2
y = (q + 1)p where p and q are constants. 9. The table shows the experimental values of two
x
variables, x and y. One of the values of y is wrongly
x 1 1.5 2 2.5 3 3.5 recorded. It is known that the variables x and y are
©PAN ASIA PUBLICATIONS
y 2.25 3.74 7.6 18.9 57.7 215.4 related by the equation 4a x = (y + b) where a and b
2
2
(a) Plot log y against x . Hence, draw the line of are constants.
2
10
best fit. [5 marks] x 1 2 3 4 5 6
(b) Use the graph in (a) to estimate the value of y 2 4.1 5.7 6 8.2 9.2
(i) p, (ii) q.
[5 marks] (a) Plot y against AB x. Hence, draw the line of best fit.
[6 marks]
6. The table shows the values of two variables, x and y, (b) From the graph in (a),
SPM obtained from an experiment. The variables x and y (i) identify the incorrect value of y and
CLONE
are related by the equation y = ax b – 1 where a and b are determine its correct value, [2 marks]
constants. (ii) find the values of a and b. [2 marks]
x 1 2 3 4 5 6
y 100 25 11.1 6.3 4 2.8 10. The table shows the values of two variables, x and y,
SPM obtained from an experiment. The variables x and y
(a) Plot log y against log x. Hence, draw the line CLONE y
10
10
of best fit. [6 marks] are related by the equation a + x = 3bx where a and b
(b) Use your graph in (a) to find the value of are constants. HOTS Analysing
(i) a, (ii) b. x 1 2 3 4 5
[4 marks]
y 3 0 –9 –24 –45
7. The table shows the values of two variables, x and y, (a) Convert the equation y = 3bx into the linear
SPM obtained from an experiment. The variables x and y a + x
CLONE form Y = mX + c. Hence, draw the line of best fit.
are related by the equation y = hk 2x – 1 where h and k
are constants. [6 marks]
(b) From your graph in (a), find the values of a and b.
x 1 2 3 4 5 6 [4 marks]
y 0.24 0.35 0.5 0.72 1.03 1.49

(a) Plot log y against (2x – 1). Hence, draw the line 11. The table shows the experimental values of two
10
of best fit. variables, x and y. It is known that the variables x and
[6 marks] y are related by the equation px y = q where p and q
2
(b) Use your graph in (a) to find the value of are constants.
(i) h, (ii) k.
[4 marks] x 0.2 0.3 0.4 0.5 0.6
y 8.3 3.7 2.1 1.3 0.93
8. The table shows the values of two variables, x and (a) Reduce the equation px y = q to the linear form
2
SPM y, obtained from an experiment. It is known that
CLONE Y = mX + c. Hence, draw the line of best fit.
the variables x and y are related by the equation [6 marks]
y = pAB x + qx where p and q are constants. (b) Use your graph in (a) to find
q
x 1 2 3 4 5 6 (i) the value of , [2 marks]
p
y –1 –0.24 0.8 2 3.3 4.65 (ii) the value of q if p = 6. [2 marks]
54 Question 10: 55
SOS TIP 5454 Given a + x y = 3bx(a + x) 2 55 SOS TIP
y
= 3bx
Then
y = 3abx +3bx

Divide both sides by x and continue from there.




06_1202 QB AMath F4.indd 55 09/05/2022 11:33 AM

Form 4 Assessment






PAPER 1

Time: 2 hours
©PAN ASIA PUBLICATIONS
Section A
[64 marks]
Answer all questions.

1. (a) Diagram 1 shows a graph f (x) for the domain 2. (a) Given the function g : x → 2 – 3x and
a < x < b. h : x → x , x ≠ k. Find
–1
f(x) 2 – x
(i) the value of k,
(ii) hg(x).
[3 marks]
x
–2 0 4 (b) Diagram 2 shows a graph of f (x).
f(x)
Diagram 1
(i) State the value of a and b. x
(ii) Determine whether f (x) has an inverse 0
function for the domain given. Explain your
answer.
[3 marks] Diagram 2
(b) The function f and g are defined by f : x → 4x + 5 State whether f (x) is a function. Justify the test
and g : x → x + 8. Find the possible values of x used in determining the answer. [2 marks]
2
so that f (x) = g(x). [2 marks]
Answer:
Answer: (a) (i)
(a) (i)
(ii)
(ii)










(b)



(b)













89





11_1202 QB AMath F4.indd 89 09/05/2022 11:36 AM

Section B
[16 marks]
Answer any two questions from this section.

13. Diagram 8 shows a graph log y against x which Answer:
10
relating variables x and y in an experiment. (a)
log y
10
2
©PAN ASIA PUBLICATIONS
1.5
(b)
1
0.5

x
0 1 2 3 4 (c)
Diagram 8
Based on the graph above,
(a) find the gradient and the y-intercept,
(b) form an equation which relating x and y, (d)
(c) find the value of y when x = 0.5,
(d) find the value of x when y = 40.
[8 marks]
Answer:
→
→
(a) 15. O is the origin. OA and OB are 2i + j and 2i + 3j
~
respectively. E is the midpoint of AB. ~ ~ ~
→
(a) Find the vector OE.
→
(b) (b) OAPB is a parallelogram, find the vector of OP.
→
→
(c) Express BP – BE, in terms of i and j.
~
(d) Find ∠AOB. ~
(c) [8 marks]
Answer:
(a)
(d)



14. Diagram 9 shows a cyclic quadrilateral KLMN (b)
with centre of O. NOK is the diameter, ML = 6 cm,
LK = MN = 5 cm and ∠MNK = 65°.
L (c)
6 cm
M K

65° O
N
(d)
Diagram 9
Find
(a) the length of KM, [2 marks]
(b) ∠KML, [2 marks]
(c) the radius of the circle, [2 marks]
(d) the area of quadrilateral KLMN. [2 marks]
END OF QUESTION PAPER
93





11_1202 QB AMath F4.indd 93 09/05/2022 11:36 AM

PAPER 2

Time: 2 hours 30 minutes

Section A
[50 marks]
Answer all questions.

1
1. If the line x + 3y = 1 intercepts the curve y – 9 = xy at two points P and Q, find
2
2
©PAN ASIA PUBLICATIONS
(a) the coordinates of P and Q as the x-coordinate of P is positive,
(b) the equation of the line that passes through point Q and is perpendicular to x + 3y = 1.
[7 marks]



1
2. The roots of the equation 3x + bx + c = 0 are m and m .
2
(a) Find the value of c. [1 mark]
1 46
(b) Given m + = , find the possible values of b. [3 marks]
2
m 2 9
(c) Hence, find the possible values of m for b . 0. [3 marks]



3. Diagram 3 shows a piece of land, PQRS drawn on a Cartesian plane. Given the scale 1 unit on the Cartesian plane
represents 5 m on the real ground.

y

Q(5, 10)

R(7, y)
M


x
P(–5, 0) S

Diagram 3
P lies on the x-axis and M divides the line PQ in the ratio 2 : 3. Find
(a) the coordinates of M and R,
[2 marks]
(b) the equation of the line MR,
[3 marks]
(c) the actual area of PQRS, in m .
2
[3 marks]



4. A coil of wire with length t cm is cut into a few parts. Each part is bent into a square. Given that the length of each
side of the square follows an arithmetic progression with the common difference 1.5 cm. The smallest and the largest
squares have sides 1.5 cm and 16.5 cm respectively.
Find the number of squares can be made and the value of t.
[8 marks]




94





11_1202 QB AMath F4.indd 94 09/05/2022 11:36 AM

10. Diagram 10 shows a parallelogram ABCD.
A
B







M
©PAN ASIA PUBLICATIONS
D L
N
C
Diagram 10
→
1 → →
→ →
→
NLM is a straight line and AL is perpendicular to DC. Given DL = DC, CB = 4CM, LA = 2x and DA = 8y.
~
~
4
(a) Express, in terms of x and y,
→
(i) DL,
→
(ii) LM.
→
→
→
(b) Given ND = ky and NL = hLM, find the value of h and k.
~
(c) If |x| = 3 units and |y| = 2.5 units, find
~ →
(i) |DC|, ~
(ii) the area of the parallelogram ABCD.
[10 marks]
11. A straight line with gradient of 4 passes through A(–2, 5) and intercepts the x-axis at B. Another straight line passes
through A and intersects the x-axis at C(2, 0).
(a) Find the equation of the line AB and AC. [2 marks]
(b) Calculate the area of the triangle ABC. [2 marks]
(c) Find the coordinates of D so that ABCD is a parallelogram. [3 marks]
(d) A point P(x, y) moves such that ∠APC is always 90°. Find the equation of locus of point P. [3 marks]


Section C
[20 marks]
Answer any two questions from this section.



12. Diagram 12 shows a pyramid with a equilateral base of side 8 cm.
V




Q

L
P
M
8 cm
N
Diagram 12
V is 5 cm vertically above P where P is the midpoint of LM. Given that ∠VNQ = ∠QNM, find
(a) the length of VN, [2 marks]
(b) ∠QNM, [3 marks]
(c) the length of QM, [2 marks]
(d) the area of the inclined plane, VNM. [3 marks]
96





11_1202 QB AMath F4.indd 96 09/05/2022 11:36 AM

13. A grocery shop sells three types of can drinks. Table 13 shows the price, the price indices and the percentage of can
drinks sold.
Price per can (RM) Price index in the
Type of can drink year 2018 based on Percentage of can
2016 2018 the year 2016 sold in 2018
P 1.60 1.80 x 25
Q 1.20 1.50 150 p
R y 0.96 120 55
Table 13
©PAN ASIA PUBLICATIONS
(a) Find the value of x, y and p. [3 marks]
(b) (i) The price of can drink P increased by 5% from 2018 to 2019. Find the price index of can drink P in the year
2019 based on the year 2016. [2 marks]
(ii) The price index of can drink R in the year 2016 based on the year 2012 was 130. Calculate the price index
of can drink R in the year 2018 based on the year 2012 and its price in the year 2012. [2 marks]
(c) Calculate the composite index for the can drinks sold by the grocery shop in the year 2018 based on the
year 2016. [3 marks]

14. A carpenter wants to make a sculptor as shown in the Diagram 14.
A B



F G
C
D
5 cm

E J H
24 cm
Diagram 14
Given the height of the sculptor is 5 cm, EH = 24 cm and EJ : JH = 1 : 2. Find
(a) the length of CJ, [3 marks]
(b) the length of BC if the area of the triangle BCJ is 39 cm , [2 marks]
2
(c) the angle between JB and the plane EFGH, [3 marks]
(d) ∠ABJ. [2 marks]

15. Diagram 15 shows the front view of a tunnel in the parabolic form with the highest point of 10 m from the ground.

y



L L
A 1 2 B

10 m


x
O P
30 m
Diagram 15
Given the width of the tunnel is 30 m.
(a) Write three possible equations in the vertex form to represent the shape of the parabola. [6 marks]
(b) Two lights is hung from the roof of the tunnel such that AL = L L = L B. Find the height of the lights that hung
1 2
2
1
from the ground. [4 marks]

END OF QUESTION PAPER
97





11_1202 QB AMath F4.indd 97 09/05/2022 11:36 AM

©PAN ASIA PUBLICATIONS













Answers































































12A_1202 QB AMath F4.indd 98 09/05/2022 11:38 AM

CHAPTER 1 Let y = 2 – 4 x  = 2: –2q – 6 = –4q – 2
2q = 4
Paper 1 x = 2 – y q = 2
4 Substitute q = 2 into ,
Section A x = 8 – 4y p = –2(2) – 6
1. (a) 2(6) + 1 = 13 q(y) = 6(8 – 4y) – 3 = –10
2(8) + 1 = 17 = 45 – 24y 10
–1
Thus, the function is f (p) = 2p + 1, Thus, q(x) = 45 – 24x. (ii) g (x) = – 2 + x , x ≠ –2
(b) 2(p) + 1 = q 9. (a) f (x) = 2x – 5 10
q = 2p + 1 x + 2 Let y = – 2 + x
2. (a) y Let y = 2x – 5 2y + xy = –10
6 x + 2 xy = –10 – 2y
y(x + 2) = 2x – 5 –10 – 2y
©PAN ASIA PUBLICATIONS
yx + 2y = 2x – 5 x = y
4
yx – 2x = –2y – 5 –10 – 2x
(y – 2)x = –2y – 5 Thus, g(x) = x , x ≠ 0
2 –2y – 5 (iii) g(x) = 8
x = y – 2 –10 – 2x = 8
0 x –2y – 5 x
2 4 6 8 10 Thus, f (y) = –10 – 2x = 8x
–1
(b) Many-to-many y – 2 10x = –10
(c) 2, 4 and 8 f (x) = –2x – 5 , x ≠ 2 x = –1
–1
3. (a) Since –6 < y < 4, thus a = –6 and x – 2 12. (a) 1
b = 4. (b) f (x) = 2x – 5 x –1 0 2 1 2
(b) –3 f (k) = 3 x + 2 f (x) 3 1 0 1 3
(c) –2 < x < 5
4. (a) f (x) 2k – 5 = 3 y
–1
(b) g f (x) or f g(x) k + 2
–1
–1
5. (a) f (p) = –6 2k – 5 = 3k + 6 3
k = –11
4p – 5p = –6 Thus, the value of k is –11. 2
p = 6 1
2
(b) (i) f (–2) = (–2) – 5(–2) 10. (a) f (x – 2) = 3 – 2(x – 2)


= 3 – 2x + 4
= 4 + 10 = 7 – 2x –1 0 1 2 x
= 14
2
(ii) x – 5x = 6 (b) Let y = 3 – 2x (b) (i) (a) ff (x) = f (2x)
2x = 3 – y

x – 5x – 6 = 0 3 – y = 4x
2
(x – 6)(x + 1) = 0 x = 2 = 2 x
2
x = 6 or x = –1 3 – x (b) fff (x) = ff (2x)
–1
6. ts(x) = t[s(x)] So, f (x) = 2 . = f (4x)
= t(6 – 4x) 2f (k) = f (k – 2) = 8x
–1
= p(6 – 4x) – 3 3 – k = 2 x
3
= 6p – 4px – 3 2 1 2 2 = 7 – 2k (ii) f (x) = 2 x where n = 1, 2,
n
n
Compare with ts(x) = q – 4px: 3 – k = 7 – 2k 3,….
Thus, q = 6p – 3. k = 4 (iii) f (x) = 16
5
7. (a) gf (–2) = 5 2 (x) = 16
5
g(2(–2) – 6) = 5 Section B 32x = 16
g(–10) = 5 11. (a) (i) –2 x = 1
1
h + 10k = 5 (ii) f (x) = |3 – x| 2
2
h = 5 – 10k 13. (a) (i) If f maps set L to M, then the
1
(b) (i) k = 1 f (1) = |3 – (1)| element in set M is represented
3x 2 by f (x). Thus, the function
(ii) = x 5
x – 1 = 2 that maps M to N must be g(x)
3x = x – x because from L to N is gf (x).
2
1
|
x – 4x = 0 (iii) 3 – x| = 1 x – 1
2
x(x – 4) = 0 2 Thus, g(x) = 2 .
1
x = 0, x = 4 3 – 1 x = –1 and 3 – x = 1 (ii) gf (x) = x – x + 2
2
(iii) h(x) = 3x 2 2 f (x) – 1 2
1
x – 1 – 1 x = –4 – x = –2 2 = x – x + 2
3x 2 2
2
Let y = x – 1 x = 8 x = 4 f (x) – 1 = 2x – 2x + 4
2
yx – y = 3x Thus, the domain is 4 < x < 8. f (x) = 2x – 2x + 5
yx – 3x = y –1 p (b) (i) Let y = 3x – 4
x(y – 3) = y (b) (i) Given g (x) = q + x , x ≠ –q 3x = y + 4
y g (3) = –2 x = y + 4
–1
x = 3
y – 3 p x + 4
–1
x q + 3 = –2 Thus, g (x) = .
Hence, h (x) = , x ≠ 3. 3
–1
x – 3 p = –2q – 6 ............. 2 p p
= g g
1
1 2 1 2
3
3
8. (a) qp – 1 2 = 6 – 1 2 – 3 g –1 1 2 = –4 (ii) g 1 2 3 1 24
2
p
3 1 2 4
= –6 p = –4 = g 3 3 – 4
(b) Given qp(x) = 6x – 3 q + 1 = g(p – 4)
1
q 2 – 4 x 2 = 6x – 3 2 p = –4q – 2 .........2 = 3(p – 4) – 4
= 3p – 16
99
12A_1202 QB AMath F4.indd 99 09/05/2022 11:38 AM

p
g 2 1 2 = 20 18. (a) 12 = 3p + q ................  11k = 14
3
2
2

24 = 5p + q ................ 2
3p – 16 = 20 2 – : 12 = 2p k = 14
3p = 36 p = 6 11
p = 12 Substitute p = 6 into : (d) f (k) + g(k) = 5
2
14. (a) (i) f (–2) = 1 – 2(–2) 12 = 3(6) + q 3 3 k + 1 4 3 5 – k = 5
4
+
= 5 q = –6 4 2 4 3
(ii) f (3) = 3 – 1 Thus, p = 6 and q = –6. 7 + 1 k = 5
= 2 (b) K(4.5) = 6(4.5) – 6 4 12
1
= 21
(iii) x –3 –2 –1 0 Thus, the cost is RM21. 12 k = 13
4
f (x) 7 5 3 1 19. (a) y k = 39
©PAN ASIA PUBLICATIONS
B(3, 7) y = x x + 1
x 1 2 3 4 2. (a) g(x) = x ≠ 2
6 x – 2
f (x) –1 1 2 3 (i) gg(x) = g 3 x + 1 4
4 f B'(7, 3) x – 2
y x + 1
A'(0, 2) + 1
2 f –1 x – 2
6 = x + 1
A(2, 0) – 2
x
4 0 2 4 6 x – 2
Thus, Aʹ(0, 2) and Bʹ(7, 3). = 2x – 1 , x ≠ 5
2 5 – x
(b) Domain is 0 < x < 7.
x 20. (a) By using the vertical line test, the x + 1
–2 0 2 4 (ii) Let = y
line intersect only once. Hence, f (x) x – 2
(b) (i) (kh) (–5) = 4 is a function. x + 1 = xy – 2y
–1
kh(x) = k(y) = z (b) By using horizontal line test, the 1 + 2y = x[y – 1]
x = (kh) (z) line intersects the curve more than x = 1 + 2y
–1
Thus, (kh) (–5) = 4. once. Hence, f (x) does not have an y – 1
–1
(ii) hh (2) = 2 inverse function. ∴ g (x) = 1 + 2x , x ≠ 1
–1
–1
(iii) h k (–5) = h (2) = 4 (c) Range of f (x) is f (x) . 0 x – 1
–1 –1
–1
(kh) (–5) = 4 (b) h(x) = ax + 1
–1
Thus, h k (–5) = (kh) (–5). x
–1 –1
–1
–1
15. (a) p = 1 Paper 2 hg (4) = 6
(b) Let y = x + 2 Section A h 3 1 + 2(4) 4 = 6
1 – x 3 1 4 – 1
y – xy = x + 2 1. f (x) = x + 2 h[3] = 6
4
x + xy = y – 2 5 2 3a + 1
(1 + y)x = y – 2 g(x) = – x 3 = 6
3
4
x = y – 2 3a + 1 = 18
1 + y (a) f (2) + f (3) 3a = 17
x – 2 3 1 3 1 17
h(x) = , = 3 (2) + 4 3 (3) + 4 a =
+
1 + x 4 2 4 2 3
Thus, a = –2 , b = 1. = 2 + 11 3. (a) (i) g(a + 1) = (a + 1) + 5
2
(c) q = –1 4 = a + 2a + 6
2
(d) h (x) + h(x) = 0 = 19 (ii) g(a ) = a + 5
–1
4
2
x + 2 + x – 2 = 0 4 (iii) g(2b – 1) – g(b)
1 – x 1 + x f (2 + 3) = f (5) = (2b – 1) + 5 – [b + 5]
2
2
(x – 2)(1 + x) + (1 – x)(x – 2) 3 1
2
= 0 = (5) + = 4b – 4b + 6 – b – 5
2
(1 – x)(1 + x) 4 2 2
6x = 0 15 1 = 3b – 4b + 1
x = 0 = 4 + 2 (b) g(x) = 5x – 1
2
16. (a) When t = 2, s(2) = 2 – 5(2) + 6 = 17 x + 5 = 5x – 1
2
2
= 0 m 4 x – 5x + 6 = 0
(b) t – 5t + 6 = 30 ∴ f (2) + f (3) ≠ f (2 + 3) (x – 3)(x – 2) = 0
2
t – 5t – 24 = 0 (b) g(4) – g(2) x = 3 or x = 2
2
2
2
2
4 3
(t – 8)(t + 3) = 0 = 3 5 – (4) – 5 – (2) 4 4. (a) f (x) = ax + b
t = 8 or t = –3 (Not applicable) 4 3 4 3 f (3) = 9a + b = 10 .......... 
1
Thus, t = 8 seconds. = – 17 – – 1 2 f (–2) = 4a + b = –10 ........ 2
(c) Displacement is negative, 12 12  – 2: 5a = 20
4
s(t) , 0 = – a = 4
3
t – 5t + 6 , 0 g(4 – 2) = g(2) Substitute a = 4 into :
2
(t – 2)(t – 3) , 0 5 2 9(4) + b = 10
Thus, 2 , t , 3. = – (2) b = 10 – 36
17. (a) V(5) = 5 + 4(5) + 3 4 1 3 = –26
2
2
= 48 m 3 = – 12 (b) f (x) = 4x – 26
1
1
(b) t + 4t + 3 = 63 ∴ g(4) – g(2) ≠ g(4 – 2) f 1 2 1 2 2 – 26
2
= 4
t + 4t – 60 = 0 (c) f (2k) = 6g(k) 2 4
2
(t + 10) (t – 6) = 0 3 1 5 2 = –25
t = 6 m or t = –10 4 (2k) + = 6 3 4 – k 4
2
3
(Not applicable) 3k 1 15
Thus, the height of the tank is 6 m. + = – 4k
2 2 2
100
12A_1202 QB AMath F4.indd 100 09/05/2022 11:38 AM

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