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 :

Question

Accepted Answer

Answer: B. 2 \begin{aligned} & \lim _{x ightarrow 0} \frac{x \int_0^x f(t) d t}{\left(\frac{e^{x^2}-1}{x^2}ight) 	imes x^2} \\ & \lim _{x ightarrow 0} \frac{\int_0^x f(t) d t}{x} \quad\left(\lim _{x ightarrow 0} \frac{e^{x^2}-1}{x^2}=1ight) \\ & =\lim _{x ightarrow 0} \frac{f(x)}{1} 	ext { (using L Hospital) } \\ & f(0)=\frac{1}{2} \\ & \alpha=\frac{1}{2} \\ & 8 \alpha^2=2 \end{aligned}

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 :

Answer

Explanation

More JEE Math questions