Verified Solution • 2025–26 Session moderate

QJEEM25Ma06d6: JEE Main 2025 Mathematics - Class 12 JEEM25Ma06d6 Solution: Let the function, . If the area of the region enclosed by and…

Let the function,

f(x)={−3ax2−2,amp;xlt;1a2+bx,amp;x⩾1f(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∈Rx \in \mathbf{R} , where \mathrm{a}>1, \mathrm{~b} \in \mathbf{R} . If the area of the region enclosed by y=f(x)y=f(x) and the line y=−20y=-20 is α+β3,α,β∈Z\alpha+\beta \sqrt{3}, \alpha, \beta \in Z , then the value of α+β\alpha+\beta is ___________ .

Step-by-Step Verified Solution

f(x)\mathrm{f}(\mathrm{x}) is continuous and differentiable

 at x=1;amp; LHL = RHL, LHD = RHD amp;−3a−2=a2+b,−6a=bamp;a=2,1;b=−12\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)={−6x2−2,amp;xlt;14−12x,amp;x≥1f(x)=\left\{\begin{array}{cc}-6 x^2-2, & x<1 \\ 4-12 x, & x \geq 1\end{array}\right.

JEE Main 2025 (Online) 22nd January Morning Shift Mathematics - Limits, Continuity and Differentiability Question 10 English Explanation

amp; Area =∫−31(−6x2−2+20)dx+∫12(4−12x+20)dx]amp;=16+123+6=22+123amp;∴α+β=34\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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