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 :

Let g(x) = cos x², f(x) = √x, and α, β (α < β) be the roots of the quadratic equation 18x² - 9π x + π² = 0. Then the area (in sq. units) bounded by the curve y = (gof)(x) and the lines x = α, x = β and y = 0, is :

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

$18x^2 - 9\pi x + \pi^2 = 0, \quad \text{roots } \alpha<\beta$

$\text{To find: area bounded by } y=(g\circ f)(x),\ x=\alpha,\ x=\beta,\ y=0$

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

$\Rightarrow y = \cos x$

$D = (-9\pi)^2 - 4(18)(\pi^2) = 81\pi^2 - 72\pi^2 = 9\pi^2$

$x = \dfrac{9\pi \pm 3\pi}{36} = \dfrac{\pi}{3},\ \dfrac{\pi}{6}$

$\alpha = \dfrac{\pi}{6}, \quad \beta = \dfrac{\pi}{3}$

$A = \int_{\pi/6}^{\pi/3} \cos x\,dx$

$dA = \cos x\,dx \quad (\text{height } \cos x,\ \text{width } dx)$

$A = \Big[\sin x\Big]_{\pi/6}^{\pi/3} = \sin\dfrac{\pi}{3} - \sin\dfrac{\pi}{6}$

$A = \dfrac{\sqrt3}{2} - \dfrac{1}{2} = \dfrac{1}{2}(\sqrt3 - 1)$

$\boxed{A = \dfrac{1}{2}(\sqrt3 - 1) \;\Rightarrow\; \text{option (4)}}$