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  • <img alt="\mathrm{f(x)=\left\{\begin{array}{cc} x e^{-\left(\frac{1}{x}+\frac{1}{|x|}\right)}, & x \neq 0 \\ a, & , x=0 \end{array}\right.}" src="https://entrancecorner.oncodecogs.com/gif.l

\mathrm{f(x)=\left\{\begin{array}{cc} x e^{-\left(\frac{1}{x}+\frac{1}{|x|}\right)}, & x \neq 0 \\ a, & , x=0 \end{array}\right.} .  Value of ‘a’ such that \mathrm{f(x)}  is differentiable at x = 0, is equal to 

 

Option: 1

1


Option: 2

-1


Option: 3

0


Option: 4

none of these


Answers (1)

best_answer

\mathrm{\begin{aligned} & f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-a}{h} \\ & f^{\prime}(0+0)=\lim _{h \rightarrow 0} \frac{h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-a}{h}=\lim _{h \rightarrow 0} \frac{h \cdot e^{-2 / h}-a}{h}=0-\lim _{h \rightarrow 0} \frac{a}{h}=0 \end{aligned}}

provided a = 0

Also, \mathrm{f^{\prime}(0-0)=\lim _{h \rightarrow 0} \frac{-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-a}{-h}=1+\lim _{h \rightarrow 0} \frac{a}{h}=1, \text { provided } a=0}

Thus \mathrm{f(x)} 

can’t be made differentiable at x = 0.

 

Posted by

Divya Prakash Singh

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