Q46 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics
The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on
- Correct answerA$(0, \infty)$
- B$(-\infty, 0)$
- C$(-\infty, \infty)$
- D
None of these
Solution
Given $f(x)=\log (1+x)-\frac{2 x}{2+x}$
$\begin{aligned}<br />& f^{\prime}(x)=\frac{1}{1+x}-\frac{(2+x)(2)-2 x}{(2+x)^2} \\<br />& =\frac{1}{1+x}-\frac{4}{(2+x)^2} \\<br />& =\frac{(2+x)^2-4-4 x}{(1+x)(2+x)^2}=\frac{x^2}{(1+x)(2+x)^2}>0 \text { for all } \\<br />& x \in(0, \infty)<br />\end{aligned}$
Thus, given function $\mathrm{f}(\mathrm{x})$ is increasing on $(0$, $\infty)$.
$\begin{aligned}<br />& f^{\prime}(x)=\frac{1}{1+x}-\frac{(2+x)(2)-2 x}{(2+x)^2} \\<br />& =\frac{1}{1+x}-\frac{4}{(2+x)^2} \\<br />& =\frac{(2+x)^2-4-4 x}{(1+x)(2+x)^2}=\frac{x^2}{(1+x)(2+x)^2}>0 \text { for all } \\<br />& x \in(0, \infty)<br />\end{aligned}$
Thus, given function $\mathrm{f}(\mathrm{x})$ is increasing on $(0$, $\infty)$.
Hence, the correct answer is option: 'A'.