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SPM Additional Mathematics 2018, Paper 1 (Question 5 – 7)


Question 5 (3 marks):
Find the value of
( a )  lim x1 ( 7 x 2 ), ( b ) f( 2 ) if f( x )=2 x 3 4x+3.

Solution:
(a)
lim x1 ( 7 x 2 ) =7 ( 1 ) 2 =6

(b)
 f( x )=2 x 3 4x+3 f( x )=6 x 2 4 f( 2 )=6 ( 2 ) 2 4   =244   =20


Question 6 (4 marks):
It is given that L = 4t t2 and x = 3 + 6t.
(a) Express dL dx in terms of t.
(b) Find the small change in x, when L changes from 3 to 3.4 at the instant t = 1.

Solution:
(a)
Given L=4t t 2  and x=3+6t L=4t t 2 dL dt =42t x=3+6t dx dt =6 dL dx = dL dt × dt dx dL dx =( 42t )× 1 6 = 42t 6 = 2t 3


(b)
δL=3.43=0.4 δL δx dL dx δx=δL÷ δL δx δx=δL× δx δL =0.4× 3 2t = 2 5 × 3 2t = 6 5( 2t ) When t=1,  δx= 6 5( 21 ) = 6 5


Question 7 (4 marks):
Diagram 2 shows the curve y = g(x). The straight line is a tangent to the curve.
Diagram 2

Given g’(x) = –4x + 8, find the equation of the curve.


Solution:
Given g( x )=4x+8 Maximum point when g( x )=0 4x+8=0 4x=8 x=2 Thus, maximum point is ( 2,11 ). g( x )=4x+8 g( x ) = ( 4x+8 ) dx g( x )= 4 x 2 2 +8x+c g( x )=2 x 2 +8x+c Substitute ( 2,11 ) into g( x ): 11=2 ( 2 ) 2 +8( 2 )+c c=3 Thus, the equation of curve is g( x )=2 x 2 +8x+3

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