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:

We begin with the two given functions. The inner function is $$f(x)=\sqrt{x}$$, so its domain is $$x\ge 0$$. The outer function is $$g(x)=\cos x^{2}$$. Their composition is obtained by substituting $$f(x)$$ into $$g(x)$$:

$$ (g\circ f)(x)\;=\;g\!\bigl(f(x)\bigr)\;=\;g\!\bigl(\sqrt{x}\bigr)\;=\;\cos\!\left(\bigl(\sqrt{x}\bigr)^{2}\right) \;=\;\cos x . $$

Hence the curve whose area is to be found is simply $$y=\cos x$$ for $$x\ge 0$$.

Next, we determine the limits of integration. They are the two real roots $$\alpha,\beta$$ of the quadratic equation

$$ 18x^{2}-9\pi x+\pi^{2}=0 . $$

For any quadratic $$ax^{2}+bx+c=0$$, the roots are given by the quadratic formula $$ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. $$ Here $$a=18,\; b=-9\pi,\; c=\pi^{2}$$. Substituting these values, we have

$$ x=\frac{-(-9\pi)\pm\sqrt{(-9\pi)^{2}-4\cdot18\cdot\pi^{2}}}{2\cdot18} =\frac{9\pi\pm\sqrt{81\pi^{2}-72\pi^{2}}}{36}. $$

Simplifying the discriminant, $$ 81\pi^{2}-72\pi^{2}=9\pi^{2}, $$ and using $$\sqrt{9\pi^{2}}=3\pi$$, we obtain

$$ x=\frac{9\pi\pm3\pi}{36}. $$

This gives two values:

$$ x_{1}=\frac{9\pi+3\pi}{36}=\frac{12\pi}{36}=\frac{\pi}{3},\qquad x_{2}=\frac{9\pi-3\pi}{36}=\frac{6\pi}{36}=\frac{\pi}{6}. $$

Because $$\alpha<\beta$$, we assign $$ \alpha=\frac{\pi}{6},\qquad\beta=\frac{\pi}{3}. $$

The required area is the definite integral of $$\cos x$$ from $$x=\alpha$$ to $$x=\beta$$, taken above the $$x$$-axis (the line $$y=0$$):

$$ \text{Area}= \int_{\alpha}^{\beta}\cos x\,dx = \int_{\pi/6}^{\pi/3}\cos x\,dx. $$

We recall the antiderivative formula $$ \int\cos x\,dx=\sin x+C. $$ Using this, we integrate:

$$ \text{Area}= \left[\sin x\right]_{\pi/6}^{\pi/3} = \sin\!\left(\frac{\pi}{3}\right) - \sin\!\left(\frac{\pi}{6}\right). $$

Now we substitute the standard sine values: $$ \sin\!\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2},\qquad \sin\!\left(\frac{\pi}{6}\right)=\frac{1}{2}. $$

Hence

$$ \text{Area}= \frac{\sqrt{3}}{2}-\frac{1}{2} =\frac{\sqrt{3}-1}{2}. $$

Therefore the required area, in square units, is $$\dfrac{\sqrt{3}-1}{2}$$.

Hence, the correct answer is Option B.