JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 10)
Let $$f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$$ be a differentiable function such that $$f(0)=\frac{1}{2}$$. If the $$\lim _\limits{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{\mathrm{e}^{x^2}-1}=\alpha$$, then $$8 \alpha^2$$ is equal to :
4
2
1
16
Explanation
$$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{\left(\frac{e^{x^2}-1}{x^2}\right) \times x^2} \\\\
& \lim _{x \rightarrow 0} \frac{\int_0^x f(t) d t}{x} \quad\left(\lim _{x \rightarrow 0} \frac{e^{x^2}-1}{x^2}=1\right) \\\\
& =\lim _{x \rightarrow 0} \frac{f(x)}{1} \text { (using L Hospital) } \\\\
& f(0)=\frac{1}{2} \\\\
& \alpha=\frac{1}{2} \\\\
& 8 \alpha^2=2
\end{aligned}$$
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