Q36 / 90 · JEE Main 2018 (08 Apr) · Mathematics

Let $g (x) = \cos x^{2}, f (x) = \sqrt{x},$ 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 $y = (g o f) (x)$ and the lines $x = α, x = β$ and $y = 0,$ is

A
$\frac{1}{2} (\sqrt{2} - 1)$
Correct answer
B
$\frac{1}{2} (\sqrt{3} - 1)$
C
$\frac{1}{2} (\sqrt{3} + 1)$
D
$\frac{1}{2} (\sqrt{3} - \sqrt{2})$

Solution

Given, $g (x) = \cos x^{2}, f (x) = \sqrt{x}$

Given quadratic is $18 x^{2} - 9 π x + π^{2} = 0$

The roots are $x = \frac{9 π \pm \sqrt{81 π^{2} - 72 π^{2}}}{2 \times 18}$

$\Rightarrow x = \frac{9 π \pm \sqrt{9 π^{2}}}{36}$

$\Rightarrow x = \frac{(9 \pm 3) π}{36}$

$\Rightarrow x = \frac{π}{3}, \frac{π}{6}$

$y = g o f (x) = \cos {(\sqrt{x})}^{2} = \cos x$

Required Area $= \int_{\frac{π}{6}}^{\frac{π}{3}} \cos x d x$ $= {[\sin x]}_{\frac{π}{6}}^{\frac{π}{3}}$
$\Rightarrow \frac{\sqrt{3}}{2} - \frac{1}{2}$

Hence, the correct answer is option: 'B'.

Let gx=cosx2, fx=x, and α,βα<β be the roots of the quadratic equation 18x2-9πx+π2=0. Then the area (in sq. units) bounded by the curve y=gofx and the lines x=α,x=β and y=0, is

Solution

Relevant Questions

Let $g (x) = \cos x^{2}, f (x) = \sqrt{x},$ 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 $y = (g o f) (x)$ and the lines $x = α, x = β$ and $y = 0,$ is