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  • <img alt="\mathrm{\text { Let } \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl} \mathbf{1} & x>0 \\ \mathbf{0}, & x=\mathbf{0} \\ -\mathbf{1}, & x<\mathbf{O} \end{array}\right.}"

\mathrm{\text { Let } \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl} \mathbf{1} & x>0 \\ \mathbf{0}, & x=\mathbf{0} \\ -\mathbf{1}, & x<\mathbf{O} \end{array}\right.}

Option: 1

differentiable at x=0
 


Option: 2

continuous at x=0
 


Option: 3

not continuous at x=0
 


Option: 4

none of these


Answers (1)

best_answer

Let \mathrm{g(x)=f(x) \cdot \sin x=\left\{\begin{array}{cc}\sin x, & x \geq 0 \\ -\sin x, & x<0\end{array}\right.g^{\prime}(0+)=1 ; g^{\prime}(0-)=1 ; g(0+)=g(0-)=0}.

Hence g is continuous but not differentiable at \mathrm{X}=0

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chirag

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