Let f : [ -2 , 2 ] R be a differentiable function such that f ( 0 ) = 1 2 . If the lim x 0 x 0 x f ( t ) ? d t e x 2 - 1 = , then 8 2 is equal to [2024]

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Solution

Q.
Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2} .$ If the $\lim_{x \to 0} \frac{x \int_{0}^{x} f (t) d t}{e^{x^{2}} - 1} = α,$ then $8 α^{2}$ is equal to [2024]

1 2

2 4

3 1

4 16

Ans.
(1)

We have, $\lim_{x \to 0} \frac{x \int_{0}^{x} f (t)}{e^{x^{2}} - 1} d t = α$

Using L'Hospital's rule, we get $\lim_{x \to 0} \frac{x f (x) + \int_{0}^{x} f (t) d t}{2 x e^{x^{2}}} = α$

Again applying L'Hospital's rule, we get

$\lim_{x \to 0} \frac{f (x) + x f' (x) + f (x)}{2 e^{x^{2}} + 4 x^{2} e^{x^{2}}} = α \Rightarrow α = \frac{2 f (0)}{2} = f (0)$

$\Rightarrow α = \frac{1}{2} [∵ f (0) = \frac{1}{2}]$

$∴ 8 α^{2} = 8 \times \frac{1}{4} = 2$

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