Question

$y=f\left(x\right)$ be a function such that $f\left(x\right)=$ min $\{\sqrt{x(2-x)},(2-x\left)\right\}$, then the area bounded by $y$ $=f\left(x\right)$ and $x$-axis is equal to:

• A. $\frac{\pi }{2}+\frac{1}{4}$
• B. $\frac{\pi }{4}+\frac{1}{4}$
• C. $\frac{\pi }{4}+\frac{1}{2}$
• D. $\frac{\pi }{2}+\frac{1}{2}$

Answer 1

The correct option is C $\frac{\pi }{4}+\frac{1}{2}$

$x(2-x)\ge 0$

$\Rightarrow x(x-2)\le 0$

$\Rightarrow 0\le \times \le 2$
$y=\sqrt{x(2-x)}$

$\Rightarrow \frac{dy}{dx}=\frac{2-2x}{2\sqrt{x(2-x)}}=\frac{1-x}{\sqrt{x(2-x)}}$

Required area $={\int }_0^1\sqrt{1-(x-1{)}^{2}dx+\frac{1}{2}}$

$={|\frac{x-1}{2}\sqrt{2x-x^2}+\frac{1}{2}{\sin }^{-1}(x-1\left)\right|}_0^1+\frac{1}{2}=\frac{\pi }{4}+\frac{1}{2}$