Question

If $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{(\mathrm{x}+1)}-\log (1+\mathrm{x}),\mathrm{x}>0$then $\mathrm{f}$ is

• A. An increasing function
• B. A decreasing function
• C. Both increasing and decreasing function
• D. None of the above

Answer 1

The correct option is B

A decreasing function

Step 1: Given information

$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{(\mathrm{x}+1)}-\log (1+\mathrm{x})$

Step 2: Differentiation of given equation

$$\begin{gathered} \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{-1}{{(\mathrm{x}+1)}^2}-\frac{1}{1+\mathrm{x}} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=-\left(\frac{1}{{(\mathrm{x}+1)}^2}+\frac{1}{1+\mathrm{x}}\right) \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)<0 \end{gathered}$$

when $\mathrm{x}>0\mathrm{or}\mathrm{f}\text{'}\left(\mathrm{x}\right)<0,\mathrm{for}\mathrm{all}\mathrm{x}>0$

therefore, $f\left(x\right)$is decreasing function

Hence, the correct option is an option (B) A decreasing function.