Question

If the range of the function $f\left(x\right)=\frac{-x|x|}{1+x^2}$ is $(-a,a),$ then the value of $\frac{3a+4}{4a-3}$ is equal to

Answer 1

$f\left(x\right)=\{\begin{array}{cc}\frac{-x^2}{1+x^2},& \text{if}x\ge 0\\ \frac{x^2}{1+x^2},& \text{if}x<0\end{array}$

Case $1:$ $x\ge 0$
$y=\frac{-x^2}{1+x^2}$
$\Rightarrow x^2=\frac{-y}{y+1}\ge 0$
$\Rightarrow \frac{y}{y+1}\le 0,y\ne -1$
$\Rightarrow y\in (-1,0]$

Case $2:$ $x<0$
$y=\frac{x^2}{1+x^2}$
$\Rightarrow x^2=\frac{y}{1-y}>0$
$\Rightarrow \frac{y}{y-1}<0,y\ne 1$
$\Rightarrow y\in (0,1)$

Hence, Range $=(-1,1)$
$\Rightarrow a=1$
$\therefore \frac{3a+4}{4a-3}=7$