JEE Advanced 2016 · Paper 2, Mathematics question 41

Intersections of $y=\frac{x+9}5$ with $y=\sqrt{|x+3|}$.

• For $x\ge-3$: $x=1$ and $x=6$.
• For $x<-3$: $x=-4$.

The line lies above the curve only for $-4\le x\le1$ (between 1 and 6 the curve is above it). The bound $x\le6$ adds nothing.

Area.
$$\int_{-4}^1\frac{x+9}5\,dx-\int_{-4}^1\sqrt{|x+3|}\,dx=\frac{15}2-\left(\frac23+\frac{16}3\right)=\frac32.$$

(C)