Let $\mathrm{g}(x)=\cos x^{2}, \quad f(x)=\sqrt{x}$, and $\alpha, \beta(\alpha<\beta)$ 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 $y=\left(g_{\circ} f\right)(x)$ and the lines $x=\alpha, x=\beta$ and $y=0$, is :

(D) $\frac{1}{2}(\sqrt{3}-\sqrt{2})$

JEE Main · 2018 · 08 Apr 2018 · Offline · Area Under Curves

Let

g (x) = cos x^{2}, f (x) = x

, and

α, β (α < β)

be the roots of the quadratic equation

18 x^{2} - 9 π x + π^{2} = 0

. Then the area (in sq. units) bounded by the curve

y = (g_{\circ} f) (x)

and the lines

x = α, x = β

and

y = 0

, is :

Answer: (B)

Composing the functions,

(g \circ f) (x) = g (x) = cos ((x)^{2}) = cos x

. The quadratic

18 x^{2} - 9 π x + π^{2} = 0

has roots

x = \frac{9 π \pm 81 π ^{2} - 72 π ^{2}}{36} = \frac{9 π \pm 3 π}{36}

, namely

α = \frac{π}{6}

and

β = \frac{π}{3}

. Over this interval

cos x

stays positive, so the bounded area is

\int_{π /6}^{π /3} cos x d x = sin \frac{π}{3} - sin \frac{π}{6} = \frac{3}{2} - \frac{1}{2} = \frac{3 - 1}{2}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option