Short Question 14 & 15


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

Solution:
5 ( 2x+3 ) n dx= 5 ( 2x+3 ) n dx = 5 ( 2x+3 ) n+1 ( n+1 )×2 +c = 5 2( 1n ) × 1 ( 2x+3 ) n1 +c = 5 2( 1n ) ( 2x+3 ) n1 +c Compare  5 2( 1n ) ( 2x+3 ) n1 with  p ( 2x+3 ) 5 n1=5 n=6 5 2( 1n ) =p 5 2( 16 ) =p 5 2( 5 ) =p p= 1 2



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+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