Question

If $f\left(x\right)=\frac{x}{x^2+1}$ and $f\left(A\right)=\{y:-\frac{1}{2}\le y<0\}$ then set $A$ is$[-a,0)or(-\infty ,-b]or(-\infty ,0)$ Find $a+b$

Answer 1

$\frac{-1}{2}\le f\left(x\right)\le 0$

$\frac{-1}{2}\le \frac{x}{x^2+1}\le 0$

Thus $\frac{x}{x^2+1}\le 0$ and $\frac{-1}{2}\le \frac{x}{x^2+1}$

Considering the first one

$\frac{x}{x^2+1}\le 0$

$xϵ(-\infty ,0]$ ...(i)

Considering the second one

$\frac{-1}{2}\le \frac{x}{x^2+1}$

$-x^2-1\le 2x$

$x^2+2x+1\ge 0$

$(x+1{)}^2\ge 0$

Now $(x+1{)}^2$ is always greater than 0.

Hence, $(x+1{)}^2=0$ implies

$x=-1$...(ii)

Thus domain of $f\left(x\right)$ is $(-\infty ,-1]\cup [-1,0)$

Therefore $-b=-1$ and $-a=-1$

Hence $a=b=1$

Therefore $a+b=2$.