Page 18 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

Form
4
Chapter 6 Linear Law Additional Mathematics

3. The diagram below shows a part of graph y Paper 2
–x
against x for equation y = pq , where p and q 4. The table below shows the value of variables,
are constant. C4 x and y, obtain from an experiments. The
variables x and y are related by equation
y hy – k
x = where h and k are constants. C4
x
x 0.5 1.0 1.5 2.0 2.5
y = pq –x
y 14 24 32 48 69
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(a) Plot a graph of y against x using a scale
2
of 2 cm to 10 units on y-axis and 2 cm to
1 unit on x -axis. Hence, draw the line of
2
x best fit.
O
(b) From the graph from (b), find the values of
(i) h and k,
(a) Find the possible equation of the line of (ii) x when y = 43.
best fit for the non-linear graph.
(b) Based on your answer in (a), find the value Examiner's Comment:
of p and q, if y-intercept is 2 and gradient (a) Build a table of x .
2
is –3.
x 0.5 1.0 1.5 2.0 2.5 CHAP
Examiner's Comment: x 2 0.25 1.0 2.25 4.0 6.25 6

(a) Convert the equation y = pq into the y 14 21 32 48 69
–x

linear form, Y = mX + c.
Graph y against x 2
y = pq –x Taking logarithm to the y
base of 10 to both side.
log y = log p + log q –x 70
10 10 10
log y = log p – x log q 60
10
10
10
log y = –log q(x) + log p
10 10 10 50
Thus, the possible equation of line of 40
best fit is
log y = –log q(x) + log p. 30
10 10 10
20
(b) Compare to Y = mX + c.
10
log y = –log q(x) + log p
10 10 10 x 2
Hence, Y = log y, X = x, m = – log q 0 1 2 3 4 5 6
10
10
and c = log p.
10
When y-intercept = 2 (b) Convert the equation of x = hy – k into
log p = 2 x
10 linear form, Y = mX + c.
p = 10 2 hy – k
= 100 x = x
2
When gradient, m = –3 x = hy – k
2
– log q = –3 hy = x + k
10
log q = 3
10 1 2 k
q = 10 3 y =   x + h
h
= 1 000
Compare to Y = mX + c.
Thus, the values of p and q are 100 and
k
1
1 000 respectively. Hence, Y = y, X = x , m = and c = .
2
h h
119