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


Question 5 (3 marks):
Find the value of
(a) limx1(7x2),(b) f(2) if f(x)=2x34x+3.

Solution:
(a)
limx1(7x2)=7(1)2=6

(b)
 f(x)=2x34x+3f(x)=6x24f(2)=6(2)24  =244  =20


Question 6 (4 marks):
It is given that L = 4t t2 and x = 3 + 6t.
(a) Express dLdx 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=4tt2 and x=3+6tL=4tt2dLdt=42tx=3+6tdxdt=6dLdx=dLdt×dtdxdLdx=(42t)×16=42t6=2t3


(b)
δL=3.43=0.4δLδxdLdxδx=δL÷δLδxδx=δL×δxδL=0.4×32t=25×32t=65(2t)When t=1, δx=65(21)=65


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+8Maximum point when g(x)=04x+8=04x=8x=2Thus, maximum point is (2,11).g(x)=4x+8g(x)=(4x+8)dxg(x)=4x22+8x+cg(x)=2x2+8x+cSubstitute (2,11) into g(x):11=2(2)2+8(2)+cc=3Thus, the equation of curve isg(x)=2x2+8x+3

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