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Let \mathrm{f_{\mathrm{a}}} be a function defined by \mathrm{f_a{(x)}\left\{\begin{array}{cc} x^\alpha \sin 1 / x & , x \neq 0 \\ 0 & , x=0 \end{array}\right.}

A value of \alpha so that f_{\alpha} is differentiable on \mathbf{R} and f_{\alpha}^{\prime} is differentiable on \mathbf{R} is

Option: 1

2


Option: 2

5/2


Option: 3

7/2


Option: 4

1


Answers (1)

best_answer

For \mathrm{x \neq 0}
\begin{aligned} f^{\prime}{ }_{\alpha}(x)=x^{\alpha-2} \sin 1 / x+\alpha x^{\alpha-1} \sin 1 / x \\ \\ \frac{f_{\alpha}(x)-f_{\alpha}(0)}{x-0}=x^{\alpha-1} \sin 1 / x \\ \lim _{x \rightarrow 0} \frac{f_{\alpha}(x)-f_{\alpha}(0)}{x-0} \text { exists and is equal } 0 \text { if } \alpha>1 \end{aligned}

For \mathrm{x \neq 0,=f_{\alpha}^{\prime \prime}(x)(\alpha-2) x^{\alpha-3} \sin 1 / x-x^{\alpha-4} \cos 1 / x+\alpha(\alpha-1) x^{\alpha-2} \sin 1 / x-\alpha x^{\alpha-3} \cos 1 / x}

\mathrm{\lim _{x \rightarrow 0} \frac{f_{\alpha}^{\prime}(x)-f_{\alpha}^{\prime}(0)}{x-0}=\lim _{x \rightarrow 0}\left(x^{\alpha-3} \sin 1 / x+\alpha x^{\alpha-2} \sin 1 / x\right)}

The last limit exists if \alpha>3 So \alpha=7 / 2.

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Riya

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