JEE Main 2025 (Online) 22nd January Morning ShiftQuestion 1
Let the function,
$f(x)=\begin{cases}-3ax^{2}-2, & x<1\\ a^{2}+bx, & x\ge1\end{cases}$
be differentiable for all $x\in\mathbf{R}$, where $a>1$, $b\in\mathbf{R}$. If the area of the region enclosed by $y=f(x)$ and the line $y=-20$ is $\alpha+\beta\sqrt{3}$, $\alpha,\beta\in\mathbb{Z}$, then the value of $\alpha+\beta$ is $\_\_\_\_\_\_\_\_$ .
Application of Integrals
Answer$34$
Solution
Continuity and differentiability at $1$: $-3a-2=a^2+b$, $-6a=b$, so $a=2$, $b=-12$. Area above $y=-20$: $\int_{-\sqrt3}^1(18-6x^2)dx+\int_1^2(24-12x)dx=(16+12\sqrt3)+6=22+12\sqrt3$, so $\alpha+\beta=34$.