Question

Which of the following inequality is true for $x>0$?

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

Answer 1

Answer: (D)

$\frac{x}{1+x}<\log (1+x)<x$

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

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

$=\frac{1}{1+x}-\frac{1}{(1+x{)}^2}=\frac{x}{(1+x{)}^2}$

which is positive. $[\because x>0]$

$\therefore f\left(x\right)$ is monotonic increasing, when $x>0$.

$⟹f\left(x\right)>f\left(0\right)$

Now, $f\left(0\right)=\log 1-0=0$

$\therefore f\left(x\right)>0$

$⟹\log (1+x)-\frac{x}{1+x}>0$

$⟹\frac{x}{1+x}<\log (1+x)....\left(i\right)$

Also, for $x>0$

$x^2>0\Rightarrow x^2+x>x$

$⟹x(x+1)>x$

$⟹x>\frac{x}{x+1}...\left(ii\right)$

From eqs. (i) and (ii), we get

$\frac{x}{x+1}<\log (1+x)<x$

$[\because \log (1+x)<x$ for $x>0]$

Hence, option D is correct.