JEE Main 2023 (Online) 6th April Morning ShiftQuestion 1
Let the point $(p,p+1)$ lie inside the region $E=\left\{(x,y):3-x\le y\le\sqrt{9-x^{2}},\ 0\le x\le3\right\}$. If the set of all values of $p$ is the interval $(a,b)$, then $b^{2}+b-a^{2}$ is equal to $\_\_\_\_\_\_$ .
Circle
Answer$3$
Solution
Need $x=p\in[0,3]$, $p+1>3-p\Rightarrow p>1$, and $p^2+(p+1)^2<9\Rightarrow p^2+p-4<0\Rightarrow p<\frac{-1+\sqrt{17}}2$. $a=1$, $b=\frac{\sqrt{17}-1}2$ with $b^2+b=4$. So $b^2+b-a^2=4-1=3$.