JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Linear Inequalities

The point $(p,p+1)$ lies on the line $y=x+1$. To lie inside $E$ it must be above $y=3-x$, giving $p+1>3-p$, i.e. $p>1$, and below the semicircle $y=\sqrt{9-x^2}$, giving $p+1<\sqrt{9-p^2}$. Squaring the second, $(p+1{)}^2<9-p^2$, i.e. $2p^2+2p-8<0$, so $p^2+p-4<0$, which holds for $p<\frac{-1+\sqrt{17}}{2}$. Thus $a=1$ and $b=\frac{-1+\sqrt{17}}{2}$, where $b$ satisfies $b^2+b-4=0$, so $b^2+b=4$. Therefore $b^2+b-a^2=4-1=3$.