Question

The area of the region $\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^2}\right\}$ is equal to :

• A. $\frac{5}{2} \sin ^{-1}\left(\frac{3}{5}\right)-\frac{1}{2}$
• B. $\frac{5 \pi}{4}-\frac{3}{2}$
• C. $\frac{3 \pi}{4}+\frac{3}{2}$
• D. $\frac{5 \pi}{4}-\frac{1}{2}$

Answer 1

Answer: (D)

$|x-1|< y< \sqrt{5-x^2}$
When $|x-1|=\sqrt{5-x^2}$
$ \Rightarrow(x-1)^2=5-x^2$
$\Rightarrow x^2-x-2=0 $
$ \Rightarrow x=2,-1$
Required Area $=$ Area of $\triangle ABC +$ Area of region $BCD$
$=\frac{1}{2}\begin{vmatrix}1 & 0 & 1 \\ 2 & 1 & 1 \\ -1 & 2 & 1\end{vmatrix}+\frac{\pi}{4}(\sqrt{5})^2-\frac{1}{2}(\sqrt{5})^2$
$=\frac{5 \pi}{4}-\frac{1}{2}$