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The value of $m$ for which the function $f(x)=\left\{\begin{array}{r}m x^2, x \leq 1 \\ 2 x, x\gt1\end{array}\right.$ is differentiable at $x=1$,is
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Does not exist
$L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$
$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{m(1-h)^2-m}{-h}=\lim _{h \rightarrow 0} \frac{m\left[1+h^2-2 h-1\right]}{-h} \\ & =\lim _{h \rightarrow 0} m(2-h)=2 m \text { and } R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ & =\lim _{h \rightarrow 0} \frac{2(1+h)-m}{h} . \text { For differentiability, } L f^{\prime}(1)=R f^{\prime}(1)\end{aligned}$
But for any value of $m, R f^{\prime}(1)=L f^{\prime}(1)$ not possible.
$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{m(1-h)^2-m}{-h}=\lim _{h \rightarrow 0} \frac{m\left[1+h^2-2 h-1\right]}{-h} \\ & =\lim _{h \rightarrow 0} m(2-h)=2 m \text { and } R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ & =\lim _{h \rightarrow 0} \frac{2(1+h)-m}{h} . \text { For differentiability, } L f^{\prime}(1)=R f^{\prime}(1)\end{aligned}$
But for any value of $m, R f^{\prime}(1)=L f^{\prime}(1)$ not possible.
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