MCQ
Let $f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow R$ be a differentiable function such that $f(0)=\frac{1}{2}$, If the $\lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{e^{x^2}-1}=\alpha$, then $8 \alpha^2$ is equal to :
  • A
    $16$
  • $2$
  • C
    $1$
  • D
    $4$

Answer

Correct option: B.
$2$
b
$ \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$

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