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Let g(x)=cos^2 x,f(x)=sqrtx and alpha,be...

Let g(x)=cos^2 x,f(x)=sqrtx and alpha,beta (alpha

A

` (1)/(2)(sqrt(3)-1)`

B

` (1)/(2)(sqrt(3)+1)`

C

` (1)/(2)(sqrt(3)-sqrt(2))`

D

` (1)/(2)(sqrt(2)-1)`

Text Solution

Verified by Experts

The correct Answer is:
A

We have,
` rArr 18x^(2)-9pix +pi^(2)=0 `
` rArr 18x^(2)-6pi x-3pix +pi^(2)=0 `
` (6x-pi)(3x-pi )=0 `
`rArr x=(pi)/(6),(pi)/(3) `
` " Now, " alpha lt beta " " alpha=(pi)/(6), `
` beta=(pi)/(3)`
` "Given, " g(x)=cosx^(2) " and " f(x)=sqrt(x) `
` y= gof(x) `
` therefore y=g(f(x))=cosx `
Area of region bounded by ` x= alpha, x= beta , y=0 ` and curve ` y=g(f(x)) ` is
`A=int_(pi//6)^(pi//3) cosx dx `
` A=[sinx]_(pi//6)^(pi//3) `
`A="sin"(pi)/(3)-"sin"(pi)/(6)=(sqrt(3))/(2)-(1)/(2)`
` A=((sqrt(3)-1)/(2))`
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