Let $f:\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\rightarrow\mathrm{R}$ be a differentiable function such that $f(0)=\frac{1}{2}$. If $\lim_{x\rightarrow 0}\frac{x\int_{0}^{x}f(t)dt}{e^{x^{2}}-1}=\alpha$, then $8\alpha^{2}$ is equal to:

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Limits

Let

f : [- \frac{π}{2}, \frac{π}{2}] \to R

be a differentiable function such that

f (0) = \frac{1}{2}

. If

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:

Answer: (B)

x \to 0 lim \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = x \to 0 lim \frac{\int _{0}^{x} f ( t ) d t}{x} \cdot \frac{x ^{2}}{e ^{x^{2}} - 1} = x \to 0 lim \frac{\int _{0}^{x} f ( t ) d t}{x}

since

\frac{x ^{2}}{e ^{x^{2}} - 1} \to 1

. By L'Hopital's rule this equals

f (0) = \frac{1}{2} = α

, so

8 α^{2} = 8 \cdot \frac{1}{4} = 2

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option

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