Question 5 (3 marks):
Find the value of
(a) limx→1(7−x2),(b) f‘‘(2) if f‘(x)=2x3−4x+3.
Solution:
(a)
limx→1(7−x2)=7−(1)2=6
(b)
f‘(x)=2x3−4x+3f‘‘(x)=6x2−4f‘‘(2)=6(2)2−4 =24−4 =20
Find the value of
(a) limx→1(7−x2),(b) f‘‘(2) if f‘(x)=2x3−4x+3.
Solution:
(a)
limx→1(7−x2)=7−(1)2=6
(b)
f‘(x)=2x3−4x+3f‘‘(x)=6x2−4f‘‘(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 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=4t−t2 and x=3+6tL=4t−t2dLdt=4−2tx=3+6tdxdt=6dLdx=dLdt×dtdxdLdx=(4−2t)×16=4−2t6=2−t3
(b)
δL=3.4−3=0.4δLδx≈dLdxδx=δL÷δLδxδx=δL×δxδL=0.4×32−t=25×32−t=65(2−t)When t=1, δx=65(2−1)=65
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=4t−t2 and x=3+6tL=4t−t2dLdt=4−2tx=3+6tdxdt=6dLdx=dLdt×dtdxdLdx=(4−2t)×16=4−2t6=2−t3
(b)
δL=3.4−3=0.4δLδx≈dLdxδx=δL÷δLδxδx=δL×δxδL=0.4×32−t=25×32−t=65(2−t)When t=1, δx=65(2−1)=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)=0−4x+8=04x=8x=2Thus, maximum point is (2,11).g‘(x)=−4x+8∫g‘(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
Diagram 2 shows the curve y = g(x). The straight line is a tangent to the curve.

Given g’(x) = –4x + 8, find the equation of the curve.
Solution:
Given g‘(x)=−4x+8Maximum point when g‘(x)=0−4x+8=04x=8x=2Thus, maximum point is (2,11).g‘(x)=−4x+8∫g‘(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