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. $\frac{1}{2}(\sqrt{3}+1)$
• B. $\frac{1}{2}(\sqrt{3}-1)$
• C. $\frac{1}{2}(\sqrt{2}-1)$
• D. $\frac{1}{2}(\sqrt{3}-\sqrt{2})$

Answer 1

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

Given: $18x^2-9\pi x+{\pi }^2=0$
$\Rightarrow 18x^2-6\pi x-3\pi x+{\pi }^2=0$
$\Rightarrow 6x(3x-\pi )-\pi (3x-\pi )=0$
$\Rightarrow (6x-\pi )(3x-\pi )=0$
$\therefore \alpha =\frac{\pi }{6},\beta =\frac{\pi }{3}$
Also, $y=\left(gof\right)\left(x\right)=g\left(f\right(x\left)\right)=g\left(\sqrt{x}\right)$
$\therefore y=\cos x$

Figure:

$\therefore$ Required area
$=\underset{\pi /6}{\overset{\pi /3}{\int }}\left(\cos x\right)dx={\left[\sin x\right]}_{\pi /6}^{\pi /3}$
$=\frac{1}{2}(\sqrt{3}-1)$ sq. units