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 Let $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\mathrm{x}^{\mathrm{n}}\sin\frac{1}{\mathrm{x}},\quad  \mathrm{x}\neq 0\\0, \quad \mathrm{x}=0\end{array}\right.$$ , then f(x) 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 $$\mathrm{n}\in(0, 1 ] $$
Since, $$f\left( x \right)$$ is continuous at $$x=0$$, therefore, $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f\left( 0 \right) =0$$

$$\displaystyle\Rightarrow \lim _{ x\rightarrow 0 }{ { x }^{ n } } \sin { \left( \frac { 1 }{ x }  \right)  } =0\Rightarrow n>0$$

$$f(x)$$ is differential at $$x=0$$ if $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { f\left( x \right)-f\left( 0 \right)  }{ x-0 }  } $$ exists finitely

$$\displaystyle \Rightarrow \lim _{ x\rightarrow 0 }{ \frac { { x }^{ n }\sin { \left( \frac { 1 }{ x }  \right)  } -0 }{ x }  } $$ exists finitely

$$\displaystyle \Rightarrow { x }^{ n-1 }\sin { \left( \frac { 1 }{ x }  \right)  } $$ exists finitely.

$$\Rightarrow n-1>0 \Rightarrow  n>1$$

If $$n\le1$$, then $$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ n-1 }\sin { \left( \frac { 1 }{ x }  \right)  }  } $$ does not exists and hence, $$f(x)$$ is not differentiable at $$x=0$$

Hence, $$f(x)$$ is continuous but not differentiable at $$x=0$$ for $$0<n\le1$$

i.e., $$n\in (0,1]$$

Mathematics

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