Question

Let $g\left(x\right)=\cos x^2,f\left(x\right)=\sqrt{x}$, and $\alpha ,\beta (\alpha <\beta )$ be the roots of the quadratic equation $18x^2-9\pi x+{\pi }^2=0$. 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. $\frac{1}{2}(\sqrt{3}-\sqrt{2})$
• B. $\frac{1}{2}(\sqrt{2}-1)$
• C. $\frac{1}{2}(\sqrt{3}-1)$
• D. $\frac{1}{2}(\sqrt{3}+1)$

Answer 1

The correct option is C $\frac{1}{2}(\sqrt{3}-1)$

The given quadratic equation is $18x^2-9\pi x+{\pi }^2=0$

$⟹x=\frac{9\pi \pm \sqrt{81{\pi }^2-72{\pi }^2}}{2\left(18\right)}$

$⟹x=\frac{9\pi \pm 3\pi }{36}$

$⟹\alpha =\frac{\pi }{3}$

$⟹\beta =\frac{\pi }{6}$

Now,

$y=g\left(f\right(x\left)\right)=\cos \left(f\right(x{)}^2)=\cos x$

Area under the curve $=$ ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\cos xdx=\sin x{|}_{\pi /6}^{\pi /3}=\frac{\sqrt{3}-1}{2}$

This is the required answer.