Question

Let $g(x) = \cos \, x^2, f(x) = \sqrt{x}$ , 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 $y = (gof)(x)$ 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. $\frac{1}{2} ( \sqrt{3} - \sqrt{2})$
• D. $\frac{1}{2} ( \sqrt{2} - 1)$

Answer 1

Answer: (A)

$18 x^{2}-9 \pi x+\pi^{2}=0$
$(6 x-\pi)(3 x-\pi)=0$
$\therefore x=\frac{\pi}{6}, \frac{\pi}{3}$
$\alpha=\frac{\pi}{6}, \,\,\, \beta=\frac{\pi}{3}$
$y=($ gof $)(x)=\cos x$
Area $=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos \,x \,d x=(\sin x)_{\frac{\pi}{6}}^{\frac{\pi}{3}}$
$=\frac{\sqrt{3}}{2}-\frac{1}{2}$
$=\frac{1}{2}(\sqrt{3}-1)$ sq. units