Discussion Forum

If $f(x)=\frac{x}{x^{2}+1}$   is increasing function , then the value of x lies in 

Answer:

Explanation:

 Given, $f(x)=\frac{x}{x^{2}+1}$

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

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

since, f(x)  is increasing function,

$\therefore$     f '(x)  >0

$\Rightarrow$    $\frac{1-x^{2}}{(x^{2}+1)^{2}}>0$

$\Rightarrow$  $1- x^{2} >0$

$\Rightarrow$    $x^{2} <1$

$\Rightarrow$   -1 < x< 1

 $\therefore$  x ε  (-1,1)

  Hence, option (d)  is correct.