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  • Let <img alt="f(x)=\left\{\begin{array}{r} (x-1) \sin \left(\frac{1}{x-1}\right), \text { if } x \neq 1 \\ 0, \text { if } x=1 \end{array}\right." src="https://entrancecorner.oncodecogs.com/gi

Let f(x)=\left\{\begin{array}{r} (x-1) \sin \left(\frac{1}{x-1}\right), \text { if } x \neq 1 \\ 0, \text { if } x=1 \end{array}\right. . Then which one of the following is true ?

Option: 1

f is differentiable at x=1 but not at x=0


Option: 2

f is neither differntiable at x=0 nor at x=1.


Option: 3

f is differentiable at x=0 and at x=1.


Option: 4

f is differentiable at x=0 but not at x=1


Answers (1)

best_answer

f(x)=\left\{\begin{array}{r} (x-1) \sin \left(\frac{1}{x-1}\right), \text { if } x \neq 1 \\ 0, \text { if } x=1 \end{array}\right.
\begin{aligned} f^{\prime}(1) & =\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{(-h) \sin \left(-\frac{1}{h}\right)-0}{-h} \\ & =\lim _{h \rightarrow 0}-\sin \left(\frac{1}{h}\right) \end{aligned} \\ \\ f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{g(1+h)-f(1)}{h}=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}}{h}=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right) \\\\ \Rightarrow Thus,\ f\ is\ not\ differentiable\ at\ x=1 \\ f(0)=\sin 1
\begin{aligned} f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{(h-1) \sin \left(\frac{1}{h-1}\right)-\sin 1}{h} \\\ & =\lim _{h \rightarrow 0} \frac{h \sin \left(\frac{1}{h-1}\right)-\left[\sin \left(\frac{1}{h-1}\right)+\sin 1\right]}{h} \\\ & =\lim _{h \rightarrow 0} \sin \left(\frac{1}{h-1}\right)-\frac{2}{h} \sin \frac{1}{2}\left(\frac{1}{h-1}+1\right) \cos \frac{1}{2}\left(\frac{1}{h-1}-1\right) \\ & =\sin (-1)-\frac{2}{\frac{h}{2(h-1)} 2(h-1)} \sin \left(\frac{h}{2(h-1)}\right) \cos \left(\frac{2-h}{2(h-1)}\right) \end{aligned}
=-sin\ 1-cos\ 1
Similarly , we can show 
f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\cos 1-\sin 1
Thus, f\left ( x \right ) is differentiable at x=0.

Posted by

himanshu.meshram

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