IIT-JEE 1999Question 1
Let $f(x)$ be a continuous function given by $f(x)=\begin{cases}2x, & |x|\le1\\ x^{2}+ax+b, & |x|>1\end{cases}$ Find the area of the region in the third quadrant bounded by the curves $x=-2y^{2}$ and $y=f(x)$ lying on the left of the line $8x+1=0$.
Application of Integrals
Answer$\dfrac{257}{192}$ sq. units
Solution
Continuity at $x=\pm1$ gives $a=2$, $b=-1$, so $f=x^2+2x-1$ for $x<-1$. The curve $y=-\sqrt{-x/2}$ meets $y=2x$ at $x=-\frac18$ and $y=x^2+2x-1$ at $x=-2$. Area $=\int_{-2}^{-1}\left(-\sqrt{-x/2}-x^2-2x+1\right)dx+\int_{-1}^{-1/8}\left(-\sqrt{-x/2}-2x\right)dx=\frac{257}{192}$.