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

Let `g(x)=cos^2 x`,`f(x)=sqrtx` and `alpha`,`beta` (`alpha` < `beta`) be the roots of the quadratic equation `18x^2-9pix+pi^2-0`. Then the area (in sq. units) bounded by the curve `y=(gof)(x)` and the lines `x=alpha,a=beta and y=0` is

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
B
`18x^(2)-9pix+pi^(2)=0`
`rArr" "(6x-pi)(3x-pi)=0`
`therefore" "alpha=(pi)/(6),beta=(pi)/(3)`
`y=g(f(x))=g(sqrt(x))=cos x`
Area bounded by curves `y=cos x, x=(pi)/(6),x=(pi)/(3) and y=0` is
`overset(pi//3)underset(pi//6)intcos x dx =[sin x]_(pi//6)^(pi//3)=sin""(pi)/(3)-sin""(pi)/(3)=(1)/(2)(sqrt(3)-1)`
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