Discussion Forum

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

Answer:

Explanation:

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

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

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

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

 $\therefore$   f(x)  is monoatomic increasing , when x>0.

$\Rightarrow$  f(x)  > f (0)

 now, f(0) = log 1-0=0

 $\therefore$ f(x) >0

$\Rightarrow$          $\log(1+x)-\frac{x}{1+x} >0$

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

                  Also, for x >0,

 $x^{2}>0\Rightarrow x^{2}+x>x$

$\Rightarrow$   x(x+1) >x

$\Rightarrow$    $x > \frac{x}{x+1}$  ...........(ii)

 From eqs. (i) and (ii) , we get

 $\frac{x}{x+1}<\log (1+x)<x$

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