Q104 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
If $f(x)=\frac{2 x}{4+3|x|}, x \in R$, then $f^{\prime}(0)=$
- A0
- B$\frac{1}{4}$
- Correct answerC$\frac{1}{2}$
- D$\frac{3}{4}$
Solution
We have, $f(x)=\frac{2 x}{4+3|x|}, x \in R$
$$
\begin{aligned}
& = \begin{cases}\frac{2 x}{4+3 x}, & \text { if } x \geq 0 \\
\frac{2 x}{4-3 x}, & \text { if } x < 0\end{cases} \\
\therefore \quad f^{\prime}(x) & = \begin{cases}\frac{(4+3 x) \cdot 2-2 x \cdot 3}{(4+3 x)^2}, & \text { if } x \geq 0 \\
\frac{(4-3 x) \cdot 2-2 x(-3)}{(4-3 x)^2}, & \text { if } x < 0\end{cases}
\end{aligned}
$$
$$
\begin{aligned}
& = \begin{cases}\frac{8+6 x-6 x}{(4+3 x)^2}, & \text { if } x \geq 0 \\
\frac{8-6 x+6 x}{(4-3 x)^2}, & \text { if } x < 0\end{cases} \\
& = \begin{cases}\frac{8}{(4+3 x)^2}, & \text { if } x \geq 0 \\
\frac{8}{(4-3 x)^2}, & \text { if } x < 0\end{cases}
\end{aligned}
$$
Now, $L f^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{8}{(4-3 x)^2}=\frac{8}{16}=\frac{1}{2}$
$$
\text { and } R f^{\prime}(0)=\lim _{x \rightarrow 0^{+}} \frac{8}{(4+3 x)^2}=\frac{8}{16}=\frac{1}{2}
$$
Since, $\quad L f^{\prime}(0)=R f^{\prime}(0)$
$$
\therefore \quad f^{\prime}(0)=\frac{1}{2}
$$
Hence, the correct answer is option: 'C'.