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


Question 5 (3 marks):
Find the value of
( a )  lim x→1 ( 7− x 2 ), ( b ) f‘‘( 2 ) if f‘( x )=2 x 3 −4x+3.

Solution:
(a)
lim x→1 ( 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   =24−4   =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 =4−2t x=3+6t dx dt =6 dL dx = dL dt × dt dx dL dx =( 4−2t )× 1 6 = 4−2t 6 = 2−t 3


(b)
δL=3.4−3=0.4 δL δx ≈ dL dx δx=δL÷ δL δx δx=δL× δx δL =0.4× 3 2−t = 2 5 × 3 2−t = 6 5( 2−t ) When t=1,  δx= 6 5( 2−1 ) = 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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