JEE Main 2025 Jan 22 Shift 1, Mathematics Q21: Area as Definite Integral

JEE Main2025Jan 22, Shift 1Mathematics

Let the function $f (x) = {- 3 a x^{2} - 2, x < 1 a^{2} + b x, x \geq 1$ 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 ____.

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Solution

Continuity at $x = 1$ : $- 3 a - 2 = a^{2} + b$ .

Differentiability at $x = 1$ : $- 6 a = b$ .

Substituting $b = - 6 a$ : $- 3 a - 2 = a^{2} - 6 a \Rightarrow a^{2} - 3 a + 2 = 0 \Rightarrow a = 1$ or $a = 2$ . Since $a > 1$ , $a = 2$ and $b = - 12$ .

So $f (x) = - 6 x^{2} - 2$ for $x < 1$ , and $f (x) = 4 - 12 x$ for $x \geq 1$ .

For $x < 1$ : $f (x) = - 20$ gives $- 6 x^{2} = - 18 \Rightarrow x = - 3$ (taking the relevant root). At $x = 1$ , $f (1^{-}) = - 8$ .

For $x \geq 1$ : $f (x) = - 20$ gives $4 - 12 x = - 20 \Rightarrow x = 2$ .

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

First integral $= \int_{- 3}^{1} (- 6 x^{2} + 18) d x = [- 2 x^{3} + 18 x]_{- 3}^{1} = (- 2 + 18) - (2 \cdot 33 - 183) = 16 + 123$ .

Second integral $= \int_{1}^{2} (24 - 12 x) d x = [24 x - 6 x^{2}]_{1}^{2} = (48 - 24) - (24 - 6) = 24 - 18 = 6$ .

Total area $= 22 + 123$ , so $α = 22$ , $β = 12$ , and $α + β = 34$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q21: Area as Definite Integral

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