JEE MAIN - Mathematics (2025 - 22nd January Morning Shift - No. 21)
Let the function,
$$f(x)= \begin{cases}-3 \mathrm{ax}^2-2, & x<1 \\ \mathrm{a}^2+\mathrm{b} x, & x \geqslant 1\end{cases}$$
be differentiable for all $x \in \mathbf{R}$, where $\mathrm{a}>1, \mathrm{~b} \in \mathbf{R}$. If the area of the region enclosed by $y=f(x)$ and the line $y=-20$ is $\alpha+\beta \sqrt{3}, \alpha, \beta \in Z$, then the value of $\alpha+\beta$ is ___________ .
Explanation
$\mathrm{f}(\mathrm{x})$ is continuous and differentiable
$$ \begin{array}{ll} \text { at } x=1 ; & \text { LHL }=\text { RHL, LHD }=\text { RHD } \\ & -3 a-2=a^2+b,-6 a=b \\ & a=2,1 ; b=-12 \end{array} $$
$f(x)=\left\{\begin{array}{cc}-6 x^2-2, & x<1 \\ 4-12 x, & x \geq 1\end{array}\right.$
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$\begin{aligned} & \left.\text { Area }=\int_{-\sqrt{3}}^1\left(-6 x^2-2+20\right) d x+\int_1^2(4-12 x+20) d x\right] \\\\ & =16+12 \sqrt{3}+6=22+12 \sqrt{3} \\\\ & \therefore \quad \alpha+\beta=34\end{aligned}$
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