Question

For $x>0,$ which of the following is correct

• A. $\log (1+x)\le \frac{x}{1+x}$
• B. $\log (1+x)<\frac{x}{1+x}$
• C. $\log (1+x)\ge \frac{x}{1+x}$
• D. $\log (1+x)>\frac{x}{1+x}$

Answer 1

The correct option is D $\log (1+x)>\frac{x}{1+x}$

Let $f\left(x\right)=\log (1+x)-\frac{x}{1+x}$
Differentiating the above w.r.t. $x$
$f^{\prime }\left(x\right)=\frac{1}{1+x}-\frac{(1+x)\cdot 1-x\cdot (0+1)}{(1+x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{1+x}-\frac{1}{(1+x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x}{(1+x{)}^2}>0\forall x>0$
$\therefore f\left(x\right)$ is increasing.
and $f\left(0\right)=\log (1+0)-\frac{0}{1+0}=0$
Since $x>0\Rightarrow f\left(x\right)>f\left(0\right)$
$\Rightarrow f\left(x\right)>0$
$\Rightarrow \log (1+x)-\frac{x}{1+x}>0$
$\therefore \log (1+x)>\frac{x}{1+x}$