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