Question
Let $$g\left( x \right) = \cos \,{x^2},\,f\left( x \right) = \sqrt x ,$$ and $$\alpha ,\,\beta \left( {\alpha < \beta } \right)$$ be the roots of the quadratic equation $$18{x^2} - 9\pi x + {\pi ^2} = 0.$$ Then the area (in sq. units) bounded by the curve $$y = \left( {gof} \right)\left( x \right)$$ and the lines $$x = \alpha ,\,x = \beta $$ and $$y=0,$$ is :
A.
$$\frac{1}{2}\left( {\sqrt 3 + 1} \right)$$
B.
$$\frac{1}{2}\left( {\sqrt 3 - \sqrt 2 } \right)$$
C.
$$\frac{1}{2}\left( {\sqrt 2 - 1} \right)$$
D.
$$\frac{1}{2}\left( {\sqrt 3 - 1} \right)$$
Answer :
$$\frac{1}{2}\left( {\sqrt 3 - 1} \right)$$
Solution :
$$\eqalign{
& {\text{Here}},\,\,\,18{x^2} - 9\pi x + {\pi ^2} = 0 \cr
& \Rightarrow \left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0 \cr
& \Rightarrow \alpha = \frac{\pi }{6},\,\,\beta = \frac{\pi }{3} \cr
& {\text{Also, }}\,gof\left( x \right) = \cos \,x \cr
& \therefore {\text{Required area :}} \cr
& = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\cos \,x\,dx} = \frac{{\sqrt 3 - 1}}{2} \cr} $$