Mathematics

Problem

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 = (g \circ f)(x)

and the lines

x = \alpha

x = \beta

and

y = 0

, is [JEE-MAIN-2018] (1)

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

(2)

\frac{1}{2}(\sqrt{3}+1)

(3)

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

(4)

\frac{1}{2}(\sqrt{2}-1)