Question

Let $$f\left( x \right) =\left\{ \begin{matrix} { x }^{ n }\sin { \frac { 1 }{ x } } ,x\neq 0 \\ \quad \quad 0,x=0 \end{matrix} \right\}$$. Then, $$f\left( x \right)$$ is continuous but not differentiable at $$x=0$$, if

A
n(0,1)
B
n[1,)
C
n(,0)
D
n=0

Solution

The correct option is A $$n\in \left( 0,1 \right)$$Since, $$f\left( x \right)$$ is continuous at $$x=0$$, therefore$$\displaystyle\lim _{ x\rightarrow 0 }{ f\left( x \right) } =f\left( 0 \right) \Rightarrow \displaystyle\lim _{ x\rightarrow 0 }{ { x }^{ n }\sin { \dfrac { 1 }{ x } } } =0,\forall n> 0$$$$f\left( x \right)$$ is differentiable at $$x=0$$, if $$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right) -f\left( 0 \right) }{ x-0 } }$$ exists finitely.$$\Rightarrow \displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right) -f\left( 0 \right) }{ x-0 } }$$ exists finitely$$\Rightarrow \displaystyle\lim _{ x\rightarrow 0 }{ { x }^{ n-1 } } \sin { \dfrac { 1 }{ x } }$$ exists finitely$$\Rightarrow n-1 > 0$$    $$\Rightarrow$$   $$n > 1$$Hence, $$f\left( x \right)$$ is continuous but not differentiable at $$x=0$$, if $$n\in \left( 0,1 \right)$$.Mathematics

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