Short Question 14 & 15


Question 14 (3 marks):
It is given that 5(2x+3)ndx=p(2x+3)5+c , where c, n and p are constants.
Find the value of n and of p.

Solution:
5(2x+3)ndx=5(2x+3)ndx=5(2x+3)n+1(n+1)×2+c=52(1n)×1(2x+3)n1+c=52(1n)(2x+3)n1+cCompare 52(1n)(2x+3)n1with p(2x+3)5n1=5n=652(1n)=p52(16)=p52(5)=pp=12



Question 15 (4 marks):
Diagram shows the curve y = g(x). The straight line is a tangent to the curve.
Diagram

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