Let the function,
$f(x)= \begin{cases}-3 a x^2-2, & x \lt 1 \\ a^2+b x, & x \geqslant 1\end{cases}$
be differentiable for all $x \in \mathbf{R}$, where $\mathbf{a} \gt 1, \mathbf{b} \in \mathbf{R}$. 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 ________

Question

Aicharya · Accepted Answer

$f(x)$ is continuous and differentiable
$\begin{aligned}
& \text { at } x=1, \mathrm{LHL}=\mathrm{RHL}, \mathrm{LHD}=\mathrm{RHD} \\ & -3 a-2=a^2+b,-6 a=b \\ & a=2 ; b=-12 \\ & f(x)=\left\{\begin{array}{cc}
-6 x^2-2, & x < 1 \\ 4-12 x, & x \geq 1
\end{array}\right.
\end{aligned}$

$\begin{aligned} & \left.\text { Area }=\int_{-\sqrt{3}}^1\left(-6 x^2-2+20\right) d x+\int_1^2(4-12 x+20) d x\right] \\ & =16+12 \sqrt{3}+6=22+12 \sqrt{3} \\ & \therefore \quad \alpha+\beta=34\end{aligned}$

Let the function, $f(x)= \begin{cases}-3 a x^2-2, & x \lt 1 \\ a^2+b x, & x \geqslant 1\end{cases}$ be…

Solution