Question

Let $\mathrm{g}\left(\mathrm{x}\right)=\cos {\mathrm{x}}^2$,$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}$ and $\mathrm{\alpha },\mathrm{\beta }(\mathrm{\alpha }<\mathrm{\beta })$ be the roots of the quadratic equation $18{\mathrm{x}}^2-9\mathrm{\pi x}+{\mathrm{\pi }}^2=0$. Then the area in sq. units bounded by the curve $\mathrm{y}=\left(\mathrm{gof}\right)\left(\mathrm{x}\right)$ and the lines $\mathrm{x}=\mathrm{\alpha }$ and $\mathrm{x}=\mathrm{\beta }$ and$\mathrm{y}=0$ is:

• A. $\frac{\sqrt{3}-\sqrt{2}}{2}$
• B. $\frac{\sqrt{2}-1}{2}$
• C. $\frac{\sqrt{3}-1}{2}$
• D. $\frac{\sqrt{3}+1}{2}$

Answer 1

The correct option is C

$\frac{\sqrt{3}-1}{2}$

Step 1: Determine the value of the range.

We are given, a quadratic equation $18{\mathrm{x}}^2-9\mathrm{\pi x}+{\mathrm{\pi }}^2=0$

Here, $\mathrm{a}=18,\mathrm{b}=-9\mathrm{\pi },\mathrm{c}={\mathrm{\pi }}^2$. Now finding its roots,

$$\begin{gathered} \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ =\frac{9\mathrm{\pi }\pm \sqrt{81{\mathrm{\pi }}^2-72{\mathrm{\pi }}^2}}{36} \\ =\frac{9\mathrm{\pi }\pm 3\mathrm{\pi }}{36} \\ \mathrm{x}=\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{6} \end{gathered}$$

Hence, $\mathrm{\alpha }=\frac{\mathrm{\pi }}{6},\mathrm{\beta }=\frac{\mathrm{\pi }}{3}$

Step 2: Determine the area under the curve

we have, $\mathrm{y}=\mathrm{g}\left(\mathrm{f}\right(\mathrm{x}\left)\right)$

$$\begin{gathered} =\cos \left(\mathrm{f}{\left(\mathrm{x}\right)}^2\right) \\ =\cos \left({\left(\sqrt{\mathrm{x}}\right)}^2\right) \\ y=\cos \left(\mathrm{x}\right) \end{gathered}$$

Now the area under the curve is,

$$\begin{gathered} \Rightarrow \mathrm{A}={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\cos \left(\mathrm{x}\right)d\mathrm{x} \\ \Rightarrow \mathrm{A}=\sin \left(\frac{\mathrm{\pi }}{3}\right)-\sin \left(\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow \mathrm{A}=\frac{\sqrt{3}-1}{2} \end{gathered}$$

Therefore, area under the curve is $\frac{\sqrt{3}-1}{2}$ sq. units

Hence, option C is the correct answer.