The area (in square units) bounded by the curves $|x|=2,|y|=2$ and $x y \leq \frac{1}{2}$ is

• Correct answer
  A

  $9+3 \log 2$
• B

  $3+\log 2$
• C

  $2+\log 3$
• D

  $8+2 \log 3$

Solution

We have, $|x|=2,|y|=2, x y \leq \frac{1}{2}$
$\begin{aligned} \text { Required area } & =2[2 \times 2]+2\left[\int_0^{1 / 4} 2 d x+\int_{1 / 4}^2 \frac{1}{2 x} d x\right] \\ & =8+2\left[2[x]_0^{1 / 4}+\frac{1}{2}[\log x]_{1 / 4}^2\right] \\ & =8+4\left[\frac{1}{4}-0\right]+\left[\log 2-\log \frac{1}{4}\right] \\ & =8+1+\log 2-\log \left(2^{-2}\right) \\ & =9+\log 2+2 \log 2=9+3 \log 2\end{aligned}$

Hence, the correct answer is option: 'A'.