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  • Let <img alt="f(x)=\left\{\begin{array}{ll} (x-2)^2 \sin \left(\frac{1}{x-2}\right)-|x|, & x \neq 2 \\ -2 & , \quad x=2 \end{array} \text { then the points where } f(x)\right. \text { is no

Let f(x)=\left\{\begin{array}{ll} (x-2)^2 \sin \left(\frac{1}{x-2}\right)-|x|, & x \neq 2 \\ -2 & , \quad x=2 \end{array} \text { then the points where } f(x)\right. \text { is not differentiable are }

Option: 1

x=0,2 only


Option: 2

x=0 only


Option: 3

x= 2 only


Option: 4

 none of these


Answers (1)

best_answer

\mathrm{f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{h^2 \sin \frac{1}{h}-(2+h)+2}{h}}

Since  \mathrm{(x-2)^2 \sin \left(\frac{1}{x-2}\right)} is differentiable at \mathrm{x= 0} and \mathrm{|x|} is not differentiable at \mathrm{x=0}

\mathrm{\therefore \quad f(x)} is not differentiable at \mathrm{x= 0}

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Nehul

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