Question

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

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

Answer 1

Answer: (B)

A decreasing function

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

$f^{\prime }\left(x\right)=\frac{-1}{(x+1{)}^2}-\frac{1}{1+x}$

$f^{\prime }\left(x\right)=-(\frac{1}{(x+1{)}^2}+\frac{1}{1+x})$

$\because x>0,(\frac{1}{(x+1{)}^2}+\frac{1}{(1+x)})$

$\therefore f^{\prime }\left(x\right)<0$

$\Rightarrow f\left(x\right)$ is a decreasing function