Answer:
Option C
Explanation:
Here, {(x,y)}ε R2:y≥√∣x+3∣,5y≤(x+9)≤15 }
∴
\Rightarrow y\geq\begin{cases}\sqrt{x+3},& when & x \geq -3\\\sqrt{-x-3}& when & x \leq-3\end{cases}
or y^{2}\geq\begin{cases}x+3 & when& x \geq -3\\-3-x & when & x \leq -3\end{cases}
shown as

Also, 5y\leq (x+9)\leq 15
\Rightarrow (x+9)\geq 5y and x\leq 6
shown as

\therefore { (x,y)\epsilon R^{2}:y\geq\sqrt{\mid x+3 \mid}, 5y\leq(x+9)\leq15}

\therefore Required area
= Area of trapezium ABCD - Area of ABE under parabola- Area of CDE under parabola
= \frac{1}{2}(1+2)(5)-\int_{-4}^{-3}\sqrt{-(x+3)}dx-\int_{-3}^{1}\sqrt{(x+3) } dx
=\frac{15}{2}-[\frac{(-3-x)^{\frac{3}{2}}}{-\frac{3}{2}}]_{-4}^{-3}-[\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}]^{1}_{-3}
=\frac{15}{2}+\frac{2}{3}[0-1]-\frac{2}{3}[8-0]
=\frac{15}{2}-\frac{2}{3}-\frac{16}{3}=\frac{15}{2}-\frac{18}{3}=\frac{3}{2}