First Form 4 Test Additional Mathematics

1. SMK CONVENT BUKIT NANAS, KUALA LUMPUR.
YEAR 2012
TEST 1:
ADDITIONAL MATHEMATICS FORM 4
Time: 1 ½ HOUR
NAME: ..................................................................................
FORM : 4 ...................
Instruction to candidates:
This paper consists of 18 questions. Answer all questions. Write your answer clearly in the spaces provided
in the question paper. Show your working. It may help you to get marks. If you wish to change your answer,
erase the answer that you have done. Then write down the new answer. The diagrams in the questions
provided are not drawn to scale unless stated. The marks allocated for each question are shown in brackets.
This
question
paper
must
be
handed
in
at
the
end
of
examination.
1. Given that 7m + 2n= 9 and 6m + n = 7, find the value of m andn.
Answer:
[4 marks]
FF Diagram 1shows the relation between set Aand set B in the arrow diagram.
2.
Diagram 1
a) State the type of the relation. [2 marks]
b) Object(s) of p.
Answer:

2. 3.
Diagram 2 shows the relation between set S and set Tin the arrow diagram.
Diagram 2
State the
(a) image of 12
(b) object of 7
(c) domain of this relation
(d) range of this relation
[4 marks]
Answer:
4.
Diagram 3 shows the relation between set X and set Y in the graph form.
Diagram 3
State the relation in the form of ordered pairs.
[2 marks]
Answer:

3. 5. Given function f : x → 3x + 6, find the
a) image of -7
b) object when image is 15
Answer:[3 marks]
6
Given the function f :x → |2x + 2|.
(a) Find the images of −2, 0 and 2.
(b) Sketch the graph of f(x) for the domain −2 ≤ x ≤ 2.
[4 marks]
Answer:

4. 7
x
The functions of f and g are defined as f :x → x − 4 and g : x → 4 . Find the composite
function of gf and the value of gf(2).
[3 marks]
Answer:
8
Given function f : x → cosx, 0° ≤ x ≤ 180°, find the value of
(a) f(120°)
(b) x when f(x) = 0.5
[3 marks]
Answer:
9
The functions of u and v are defined as u: x → x − 3 and v :x → x2 − 6. Find the composite
function of uvand vu.
[4 marks]
Answer:

5. 10 The functions of f and g are defined as f : x → x + 4 and fg : x → 2x − 7.
Find the function g.
[2 marks]
Answer:
11 The functions of f and g are defined as f :x → x + 6 and gf : x → 3x + 8.
Find the function g.
[3 marks]
Answer:
x−7
12. The functions of h and k are defined as h: x → x + 9 and k : x → x − 2, x ≠ a. Find
(a) the value of a
(b) the composite function h2
(c) the composite function kh
[4 marks]
Answer:

6. 13 The function f is defined as f :x → 4x + 3. Find
(a) f−1(x)
(b) f−1(7)
[3 marks]
Answer:
14 The functions f and g are defined asf : x → 4 − 2x and g : x → 5x + 3.
Find gf−1(x).
[3 marks]
Answer:
15 Given the function f : x → 3x– 4 and the composite function fg : x → 5 – 6x. Find
(a) g(x)
(b) the value of x whenf2(x) = 2.
[4 marks]
Answer:

7. 16 Diagram 4represents a functionf: x → x2 + bx + c.
Diagram 4
Find the
(a) values of b and c,
(b) theimage of −9under the function.
[5 marks]
Answer:
17
Given h−1(x) →
Answer:
2
findh(x). Hence evaluate h(2).
3x 5
[4 marks]

8. 18
Given g(x) →
3 2x
find
4
a) g-1(x) ,
b) the value of x if g(x) maps to itself.
Answer:
[3 marks]
END OF THE QUESTION PAPER
Prepared by:
Checked by:
..............................................
.............................................
Miss. SazlinAinabtAbd. Ghani
Coordinator ofForm 4 Add. Maths.
Ms.GanFei Ting
Head of Panel of Add. Maths.

9. Marking Scheme and Answers for ADDITIONAL MATHEMATICS FORM 4 TEST 1
1
7m + 2n = 9  EQ 1
12m +2n = 14  EQ 2
EQ 1 – EQ 2,
-5m = -5
m=1
use EQ 1 to find t
7(1) + 2n = 9
2n = 2
n=1
2
Many-to-one relation
3
(a)
(b)
(c)
(d)
4
{(2, 8), (2, 14), (2, 16), (2, 20), (5, 20), (7, 14)}
5
(a) f(−7) = 3(−7) + 6
= −15
(b) f(x) = 15
3x + 6 = 15
3x = 9
x=3
6
(a) f(x) = |2x + 2|
f(−2) = |2(−2) + 2|
= |−4 + 2|
= |−2|
=2
f(0) = |2(0) + 2|
= |2|
=2
f(2) = |2(2) + 2|
= |4 + 2|
= |6|
=6
(b) f(x) = 0
|2x + 2|= 0
2x + 2 = 0
x = −1
2
14
{12, 14}
{2, 7}

10. 7
x
Given f(x) = x − 4 and g(x) = 4 .
gf(x) = g(f(x))
= g(x − 4)
x−4
= 4
(2) − 4
gf(2) = 4
1
= −2
8
(a) f : x → cosx
f(120°) = cos 120°
= −0.5
(b) f(x) = 0.5
cosx = 0.5
x = 60°
9
Given u(x) = x − 3 and v(x) = x2 − 6.
uv(x) = u(v(x))
= u(x2 − 6)
= (x2 − 6) − 3
= x2 − 9
vu(x) = v(u(x))
= v(x − 3)
= (x − 3)2 − 6
= (x2 − 6x + 9) − 6
= x2 − 6x + 3
10 Given f(x) = x + 4 and fg(x) = 2x − 7.
fg(x) = f(g(x))
= g(x) + 4
g(x) + 4 = 2x − 7
g(x) = 2x − 11
∴g : x → 2x − 11

11. 11 Given f(x) = x + 6 and gf(x) = 3x + 8.
gf(x) = g(f(x))
= g(x + 6)
g(x + 6) = 3x + 8
Let y = x + 6
So, x = y − 6
g(y) = 3(y − 6) + 8
= 3y − 10
Therefore, g : x → 3x − 10
12
(a) x − 2 = 0
x=2
∴a = 2
(b) h2 = h(h(x))
= h(x + 9)
= (x + 9) + 9
= x + 18
(c) kh(x) = k(h(x))
= g(x + 9)
(x + 9) − 7
= (x + 9) − 2
x+2
= x + 7, x ≠ −7
13 (a) Let f−1(7) = k
So f(k) = 7
4k + 3 = 7
k=1
Therefore, f−1(7) = 1
(b) Let f−1(x) = y
So f(y) = x
4y + 3 = x
4y = x − 3
x−3
y= 4
x−3
Therefore, f−1(x) = 4
14 Let f−1(x) = y
So f(y) = x
4 − 2y = x
4−x
y= 2
Therefore f−1(x) =
gf−1(x) = g(f−1(x))
4−x
2

12. 4−x
= g( 2 )
4−x
= 5( 2 ) + 3
26 − 5x
= 2
15 (a) Given f(x) = 3x - 4 and fg(x) = 5 - 6x .
fg(x) = f(g(x))
5-6x = 3g(x) – 4
Therefore, g : x → 3-2x
(b) ff(x) = 3(3x-4)-4 = 2
9x-12-4 = 2
x=2
16
f(x) = x2 + bx + c
f(2) = 13
(2)2 + b(2) + c = 13
4 + 2b + c = 13
2b + c = 9 −−−− E1
f(3) = 26
(3)2 + b(2) + c = 26
9 + 3b + c = 26
3b + c = 17 −−−− E2
E1 − E2:
−b = −8
b=8
Substitute b = 8 into E1.
2(8) + c = 9
c = −7
f(x) = x2 + 8x − 7
f(−9) = (−9)2 + 8(−9) − 7
= 81 − 72 − 7
=2
17 Let h−1(x) = y
So h(y) = x
2
=y
3x 5
2 5y
x
3y

13. Therefore, h(x) =
2 5x
;x
3x
18 Let g−1(x) = y
So g(y) = x
3 2y
x
4
3 4x
y
2
Therefore, g-1(x) =
When g(x) = x,
3 2x
4
x=
1
2
x
3 4x
2
0