Let the function, be differentiable for all , where a > 1, . If the area of the region enclosed by y = f(x) and the line y = –20 is then the value of is __________. [2025]

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Solution

Q.
Let the function, $f (x) = {\begin{matrix} - 3 a x^{2} - 2, & x < 1 \\ a^{2} + b x, & x \geq 1 \end{matrix}$ 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 $α + β \sqrt{3}, α, β \in Z$ then the value of $α + β$ is __________. [2025]

Ans.
(34)

Given, f(x) is continuous and differentiable at x = 1.

$f (x) = {\begin{matrix} - 3 a x^{2} - 2, & x < 1 \\ a^{2} + b x, & x \geq 1 \end{matrix}$

$\Rightarrow f' (x) = {\begin{matrix} - 6 a x, & x < 1 \\ b, & x \geq 1 \end{matrix}$

L.H.L. at $x = 1 \Rightarrow - 3 a - 2$

R.H.L. at $x = 1 \Rightarrow a^{2} + b$

$∴$ L.H.L. = R.H.L. { $∵$ f(x) is continuous}

$\Rightarrow - 3 a - 2 = a^{2} + b$ ... (i)

L.H.D. at $x = 1 \Rightarrow - 6 a$

R.H.D. at $x = 1 \Rightarrow b$

$∴$ L.H.D. = R.H.D. [ $∵$ f(x) is differentiable]

$\Rightarrow - 6 a = b$ ... (ii)

From (i) and (ii), we get

$- 3 a - 2 = a^{2} - 6 a$

$\Rightarrow a^{2} - 3 a + 2 = 0 \Rightarrow (a - 1) (a - 2) = 0$

$\Rightarrow a = 1 o r a = 2 \Rightarrow a = 2$ ( $∵$ a > 1)

From (ii), b = –12

Now, $f (x) = {\begin{matrix} - 6 x^{2} - 2, & x < 1 \\ 4 - 12 x, & x \geq 1 \end{matrix}$

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

$= {(\frac{- 6 x^{3}}{3} - 2 x + 20 x)}_{- \sqrt{3}}^{1} + {(4 x - \frac{12 x^{2}}{2} + 20 x)}_{1}^{2}$

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

$\Rightarrow 22 + 12 \sqrt{3} = α + β \sqrt{3}$

So, $α = 22 and β = 12$

$∴ α + β = 22 + 12 = 34$

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