Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$ Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$
My working:
Let  $f(x)=\frac{ax^2+x-2}{a+x-2x^2}=y$
therefore,
$$(a+2y)x^2+(1-y)x-(2+ay)=0$$
Now, for $x$ to have real values, the discriminant of the above equation must not be less than zero.
So,
$$(1+8a)y^2+(14+4a^2)y+(1+8a)\ge 0$$
Now, somehow from this, we have to land on the condition for $a$ which I am unable to do.

So, we need $$(14+4a^2)^2\ge4(1+8a)^2$$
$$\iff(4a^2+14-2-8a)(4a^2+14+2+8a)\ge0$$
$4a^2+14+2+16a=4(a^2+4a+4)=4(a+2)^2\ge0$
So,the sufficient condition is $4a^2+14-2-16a=4(a^2-4a+3)=4(a-1)(a-3)\ge0$
$\implies$ either $a\ge$max$(1,3)$ or $x\le$min$(1,3)$
