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  • Let  <img alt="\mathrm{ f(x)=\left\{\begin{array}{r}(x-1)^2 \sin \left(\frac{1}{x-1}\right)-|x|, \text { if } x \neq 1 \\ -1, \text { if } x=1\end{array}\right.}" src="https://entrancecorner.o

Let  \mathrm{ f(x)=\left\{\begin{array}{r}(x-1)^2 \sin \left(\frac{1}{x-1}\right)-|x|, \text { if } x \neq 1 \\ -1, \text { if } x=1\end{array}\right.} be a real valued function, then the set of points where  \mathrm{ f(x) } is not differentiable.

Option: 1

\mathrm{ \{0\}}


Option: 2

\mathrm{ \{0,1\}}


Option: 3

\mathrm{ \{1\}}


Option: 4

\mathrm{ \{-1,0,1\}}


Answers (1)

best_answer

\mathrm{ \text { Given, } f(x)=\left\{\begin{array}{cc} (x-1)^2 \sin \left(\frac{1}{x-1}\right)-|x|, & \text { if } x \neq 1 \\ -1 & \text {, if } x=1 \end{array}\right.}
\mathrm{ We \, \, should \, \, check \, \, at \, \, \, x=0, x=1 \, \, \,}
\mathrm{At, x=0,(x-1)^2 \sin \left(\frac{1}{x-1}\right) is\, \, differentiable, but |x| is \, not. }
\mathrm{\begin{aligned} & \text { At, } x=1, \lim _{x \rightarrow 1^{+}}\left[(x-1)^2 \sin \left(\frac{1}{x-1}\right)-|x|\right] \\ & =\lim _{h \rightarrow 0^{+}}\left[h^2 \sin \left(\frac{1}{h}\right)-|1+h|\right]=-1 \\ & \text { And } \lim _{x \rightarrow 1^{-}}\left[(x-1)^2 \sin \left(\frac{1}{x-1}\right)-|x|\right] \\ & =\lim _{h \rightarrow 0^{+}}\left[(-h)^2 \sin \left(\frac{1}{-h}\right)-|1-h|\right]=-1 \end{aligned}}
\mathrm{Also, f^{\prime}\left(1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h}}
\mathrm{=\lim _{h \rightarrow 0^{+}} \frac{h^2 \sin \left(\frac{1}{h}\right)-|1+h|-(-1)}{h}=\lim _{h \rightarrow 0^{+}} h \sin \left(\frac{1}{h}\right)=0}
\mathrm{Similarly, f^{\prime}(-1)=0}
\mathrm{Hence, f(x) \, is\, \, non-differentiable\, \, at\, \, x=0\, \, \, only.}
 

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