Let the function, f(x) = begin cases -3ax 2 - 2, & x < 1 \\ a 2 + bx, & x ge 1 end cases Be differentiable for all x in R , where a 1, b in R . If the area of the region enclosed by y=f(x) and the line y=-20 is α+β √(3), α, β, in Z , then the value of α+β is \ \ \ \ \

Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 22 January, Morning Shift — Question 21

Let the function,

f(x) = \begin{cases} -3ax^2 - 2, & x < 1 \\ a^2 + bx, & x \ge 1 \end{cases}

Be differentiable for all $\mathrm{x} \in \mathbf{R}$ , 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 $\_\_\_\_\_$

Numerical answer — enter this value.

$f(x)$ is continuous and differentiable $\text{at } x = 1;$

\qquad \text{LHL} = \text{RHL}, \text{ LHD} = \text{RHD}

-3a - 2 = a^2 + b, \qquad

-6a = b

a = 2, 1; \qquad b = -12

f(x) = \begin{cases} -6x^2 - 2; & x < 1 \\ 4 - 12x; & x \ge 1 \end{cases}

Area $=\int_{-\sqrt{3}}^{1}\left(-6 x^{2}-2+20\right) d x+\int_{1}^{2}(4-12 x+20) d x$

$16+12 \sqrt{3}+6=22+12 \sqrt{3}$

$\alpha+\beta=22+12=34$

Answer key and solution verified before publishing.

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