$\mathrm{f}(\mathrm{x})$ is differentiable on $\mathbb{R}$ and $\mathrm{f}^{\prime}(\mathrm{m})
eq 0, \mathrm{~m} \in \mathbb{R}$. If $\lim _{x ightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$, then $m=$

Question

$\mathrm{f}(\mathrm{x})$ is differentiable on $\mathbb{R}$ and $\mathrm{f}^{\prime}(\mathrm{m}) 
eq 0, \mathrm{~m} \in \mathbb{R}$. If $\lim _{x ightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$, then $m=$

Aicharya · Accepted Answer

Let $f(x)=\frac{1}{3} x^3+2 x$ $\Rightarrow f^{\prime}(x)=x^2+2 
eq 0 \forall x \in R$ Now, $\lim _{x ightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$ $\begin{aligned} \Rightarrow & \lim _{x ightarrow m} \frac{x\left(\frac{1}{3} m^3+2 might)-m\left(\frac{1}{3} x^3+2 xight)}{x-m}+m^2+2 \ & =\frac{1}{3} m^3+2 m \ \Rightarrow & \lim _{x ightarrow m} \frac{-\frac{1}{3} m x(m+x)(m-x)}{(m-x)}+m^2+2=\frac{1}{3} m^3+2 m \ \Rightarrow & -\frac{2 m^3}{3}-m^2+2=\frac{1}{3} m^3+2 m \ \Rightarrow & m^3-m^2+2 m-2=0 \ \Rightarrow & (m-1)\left(m^2+2ight)=0 \Rightarrow m=1\end{aligned}$

$\mathrm{f}(\mathrm{x})$ is differentiable on $\mathbb{R}$ and $\mathrm{f}^{\prime}(\mathrm{m}) \neq 0,…

Solution