Let

Then which one of the following is true?

  • Option 1)

    f\: is \: di\! f\! \! ferentiable\: at\: x= 1\: but \: not\: at\: x= 0

  • Option 2)

    f\: is\: neither \: di\! f\! \! ferentiable\: at\: x= 0\: nor\: at\: x= 1

  • Option 3)

    f\: is \: di\! f\! \! ferentiable\: at\: x= 0\: \: and\: at\: x= 1

  • Option 4)

    f\: is \: di\! f\! \! ferentiable\: at\: x= 0\: \:but\:not\: at\: x= 1

 

Answers (1)

As we learnt in 

Differentiability -

Let  f(x) be a real valued function defined on an open interval (a, b) and  x\epsilon (a, b).Then  the function  f(x) is said to be differentiable at   x_{\circ }   if

\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}


or\:\:\:\lim_{h\rightarrow 0}\:\frac{f(x)-f(x_{0})}{x-x_{0}}

-

 

f'(1)=\lim_{h \to 0 }\frac{f(1+h)-f(1)}{h}

=\lim_{h \to 0 }\frac{f(1+h-1)sin\frac{1}{(1+h-1)}-0}{h}

=\lim_{h \to 0 }\frac{h}{h}sin\left(\frac{1}{h} \right )=\lim_{h\rightarrow 0}sin(\frac{1}{h})

There fore f is not differentiable at x = 1

similarly  f'(0)=\lim_{h \to 0}\frac{f(h)-f(0)}{h}

=\lim_{h \to 0}\frac{(h-1)sin(\frac{1}{h-1})sin1}{h}

Hence f is also not differentiable at x= 0 


Option 1)

f\: is \: di\! f\! \! ferentiable\: at\: x= 1\: but \: not\: at\: x= 0

this is incorrect

Option 2)

f\: is\: neither \: di\! f\! \! ferentiable\: at\: x= 0\: nor\: at\: x= 1

this is correct

Option 3)

f\: is \: di\! f\! \! ferentiable\: at\: x= 0\: \: and\: at\: x= 1

this is incorrect

Option 4)

f\: is \: di\! f\! \! ferentiable\: at\: x= 0\: \:but\:not\: at\: x= 1

this is incorrect

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