SPM Additional Mathematics (Model Test Paper)


SPM Additional Mathematics (Model Test Paper)

Section B
[40 marks]
Answer any four questions from this section.


Question 7
(a) Prove that ( cosec xsecx secx cosec x ) 2 =1sin2x  
[3 marks]
(b)(i) Sketch the graph of y = 1 – sin2x for 0 ≤ x ≤ 2π.

(b)(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation 2 ( cosec xsecx secx cosec x ) 2 = x π  for 0 ≤ x ≤ 2π.
State the number of soultions. 
[7 marks]

Answer and Solution:
(a)
LHS = ( cosec xsecx secx cosec x ) 2 = ( cosec x secx cosec x secx secx cosec x ) 2 = ( 1 secx 1 cosec x ) 2 = ( cosxsinx ) 2 = cos 2 x2cosxsinx+ sin 2 x =1sin2x (RHS)

(b)(i)




(b)(ii)
2 ( cosec xsecx secx cosec x ) 2 = x π 2( 1sin2x )= x π From 7(a) 2y= x π From 7(b)(i) y=2 x π (suitable straight line)

x
0
π
y
2
1
0

From the graph, there is 3 number of solutions.



Question 8
Diagram 4 shows part of a curve x = y2 + 2. The gradient of a straight line QR is –1.

Find
(a) the equation of PQ,   [2 marks]
(b) the area of shaded region, [4 marks]
(c) the volume of revolution, in terms of π, when the shaded region is rotated through 360o about the y-axis. [4 marks]

Answer and Solution:
(a)
– 2 = –1 (x – 6)
= –x + 6 + 2
y = –x + 8

(b)
= –x + 8, at y-axis, x = 0
= 8
Point R = (0, 8)
Area of shaded region
= Area under the curve + Area of triangle
= 0 2 x dy + 1 2 ( 82 )( 6 ) = 0 2 ( y 2 +2 ) dy+ 1 2 ( 6 )( 6 ) = [ y 3 3 +2y ] 0 2 +18 =[ 2 3 3 +2( 2 )0 ]+18 = 20 3 +18 =24 2 3 uni t 2

(c)
Volume of revolution
= Volume under the curve + Volume of cone
=π 0 2 x 2 dy + 1 3 π r 2 h =π 0 2 ( y 2 +2 ) 2 dy+ 1 3 π ( 6 ) 2 ( 6 ) =π 0 2 ( y 4 +4 y 2 +4 ) dy+72π =π [ y 5 5 + 4 y 3 3 +4y ] 0 2 +72π =π[ 2 5 5 + 4 ( 2 ) 3 3 +4( 2 )0 ]+72π =π[ 32 5 + 32 3 +8 ]+72π = 376 15 π+72π =97 1 15 π unit 3


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