Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$

Question

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.

Answer 1

The minimum of the polynomial $ax^2+bx+c$ , where $a$ is positive, is $\frac{4ac - b^2}{4a}$. To see this, note that $ax^2 + bx + c = \frac{(4a^2x^2 + 4abx + b^2)}{4a} -\frac{b^2}{4a} + c = \frac 1{4a} (2ax + b)^2 + \frac{4ac - b^2}{4a}$, so the smallest value is when the square is zero, then the value is $\frac{4ac - b^2}{4a}$.

In other words, the polynomial is positive everywhere if and only if $4ac > b^2$.

Note that if $1 \leq a \leq 3$, then $(1+8a) > 0$, so you can apply the given criterion to conclude that it is enough to check that $4 \times (1+8a) \times (1+8a) > (14 + 4a^2)^2$ whenever $1 \leq a \leq 3$.

Can you check this?

Answer 2

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$(1,3)$