The function $\mathrm{f}(\mathrm{x})=\log (1+\mathrm{x})-\frac{2 \mathrm{x}}{2+\mathrm{x}}$ is increasing on
The function $\mathrm{f}(\mathrm{x})=\log (1+\mathrm{x})-\frac{2 \mathrm{x}}{2+\mathrm{x}}$ is increasing on
- $(-\infty, \infty)$
- $(-5, \infty)$
- $(-\infty, 0)$
- $(-1, \infty)$
Solution
$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\log (1+\mathrm{x})-\frac{2 \mathrm{x}}{2+\mathrm{x}} \Rightarrow \mathrm{x} \neq-2 \\
& \begin{aligned}
\therefore \mathrm{f}^{\prime}(\mathrm{x}) & =\frac{1}{(1+\mathrm{x})}-\left[\frac{(2+\mathrm{x})(2)-(2 \mathrm{x})(1)}{(2+\mathrm{x})^2}\right] \\
& =\frac{1}{1+\mathrm{x}}-\left[\frac{4}{(2+\mathrm{x})^2}\right]=\frac{(\mathrm{x}+2)^2-4(\mathrm{x}+1)}{(\mathrm{x}+2)^2(1+\mathrm{x})}
\end{aligned}
\end{aligned}$
When $f^{\prime}(x)>0$, we write
$\frac{x^2}{(x+2)^2(1+x)}>0$
Since $x^2>0$ and $(x+2)^2>0$, we write $(1+x)>0 \Rightarrow x>-1$
Asked in: MHT CET 2021 (21 Sep Shift 2)
Practice more Applications of Derivatives questions on Aicharya