Q102 / 160 · AP EAMCET 2023 (17 May Shift 1) · Mathematics
$\mathrm{f}(\mathrm{x})$ is differentiable on $\mathbb{R}$ and $\mathrm{f}^{\prime}(\mathrm{m}) \neq 0, \mathrm{~m} \in \mathbb{R}$.
If $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$, then $m=$
- A$0$
- B$-1$
- Correct answerC$1$
- D$2$
Solution
Let $f(x)=\frac{1}{3} x^3+2 x$
$\Rightarrow f^{\prime}(x)=x^2+2 \neq 0 \forall x \in R$
Now, $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$
$\begin{aligned} \Rightarrow & \lim _{x \rightarrow m} \frac{x\left(\frac{1}{3} m^3+2 m\right)-m\left(\frac{1}{3} x^3+2 x\right)}{x-m}+m^2+2 \\ & =\frac{1}{3} m^3+2 m \\ \Rightarrow & \lim _{x \rightarrow m} \frac{-\frac{1}{3} m x(m+x)(m-x)}{(m-x)}+m^2+2=\frac{1}{3} m^3+2 m \\ \Rightarrow & -\frac{2 m^3}{3}-m^2+2=\frac{1}{3} m^3+2 m \\ \Rightarrow & m^3-m^2+2 m-2=0 \\ \Rightarrow & (m-1)\left(m^2+2\right)=0 \Rightarrow m=1\end{aligned}$
Hence, the correct answer is option: 'C'.