Page 17 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

Form
4 Additional Mathematics Chapter 6 Linear Law

SPM Simulation HOTS Questions

Paper 1
1. The variable x and y are related by equation y = mx − nx Divide both
2
y = mx – nx where m and n are constant. x side with x .
2
2
x y – y n
y x 2 x 3 = m −
x
Compare to Y = mX + c.
x k + 2
O y 1
©PAN ASIA PUBLICATIONS
Hence Y = , X = , m = –n and c = m.
x 3 x
From the graph, c = k + 2
x m = k + 2
–3k O
k = m − 2 …❷
Diagram (a) To express m in terms of n, eliminate
variable k.
y n
y x – 2 As ❶ = ❷, m − 2 =
3
3 m − 6 = n
x k + 2
n
O m is the m = + 6
subject 3
CHAP 2. The variable x and y are related by an equation
6 O
2
–3k x y = 10 2x + 1 . The diagram below shows a
straight line graph obtained by plotting
Diagram (b) log y against x .
2
10
Diagram (a) and diagram (b) show the straight log y
line graphs obtained by plotting the relations 10
from the equation. Express m in terms of n. (p, 9)
C4
Examiner's Comment:
q
In diagram (a) x 2
O
x is the X-intercept.
y Find the values of p and q. C3
Then, reduce = mx − nx to get x as the
2
x Examiner's Comment:
variable of X. Convert the equation y = 10 2x + 1 into the
2

y = mx − nx Divide both linear form, Y = mX + c.
2
2
x side with x. y = 10 2x + 1 Taking logarithm
y = mx − n log y = log 10 2x + 1 to the base of 10
2
to both side.
x 2 10 10
2
Compare to Y = mX + c. log y = (2x + 1) log 10
10
10
2
y log y = 2x + 1
10
Hence Y = , X = x, m = m and c = −n.
x 2 Compare to Y = mX + c.
From the graph, c = −3k Hence Y = log y, X = x , m = 2 and c = 1.
2
− n = −3k 10
n From the graph, q = y-intercept
k = …❶
3 = 1
In diagram (b) Gradient, m = 2
9 – 1
y is the y-intercept. = 2
x 3 p – 0 p = 8
2
y y
Then, reduce = mx − nx to get as p = 4
2
x x 3
variable of y. Thus, p = 4 and q = 1.
118