If `g(x)=int_(sinx)^("sin"(2x))sin^(-1)(t)dt ,t h e n :`
(a) `g^(prime)(pi/2)=-2pi` (b) `g^(prime)(-pi/2)=-2pi` (c) `g^(prime)(-pi/2)=2pi` (d) `g^(prime)(pi/2)=2pi`

Options

A. `g'((pi)/(2))=-2pi`
B. `g'(-(pi)/(2))=2pi`
C. `g'((pi)/(2))=0`
D. `g'(-(pi)/(2))=0`

Answer

C, D

Explanation

We have, g(x)`=overset(sin 2x)underset(sinx)int sin^(-1)(t)` dt ,
Using Leibnitz's rule, we obtain
`g'(x)=0+(d)/(dx)(sin2x)sin^(-1)(sin2x)-(d)/(dx)(sinx)sin^(-1)(sinx)`
`rArrg'(x)=2cos 2x sin^(-1)(sin2x)-cosxsin^(-1)(sinx)`
We know that
`sin^(-1)(sinx)={:{(-pi-x","-(3pi)/(2)le x le-(pi)/(2)),(x","-(pi)/(2)le x le (pi)/(2)),(pi-x","(pi)/(2)le x le (3pi)/(2)):}`
`:. g((pi)/(2))=2cospisin^(-1)(sin-pi)-"cos"(pi)/(2)sin^(-1)("sin"(pi)/(2))`
`=-2(pi-pi)-0xx(pi)/(2)=0`
and
`g'(-(pi)/(2))=2 cos(-pi)sin^(-1)(sin(-pi))-cos(-(pi)/(2))sin^(-1)("sin"(-(pi)/(2)))`
`=-2xx(-pi+pi)-0xx(-(pi)/(2))=0`