Question

The number of integral values of $a$ for which $y=\frac{a{x}^2-7x+5}{5x^2-7x+a}$ can take all real values, where $x\in R$ (wherever the function is defined), is

Answer 1

$y=\frac{a{x}^2-7x+5}{5x^2-7x+a}\Rightarrow (5y-a)x^2-7(y-1)x+(ay-5)=0$
As $x$ is real,
$D\ge 0\Rightarrow 49(y-1{)}^2-4(5y-a\left)\right(ay-5)\ge 0\Rightarrow (49-20a)y^2+2(1+2a^2)y+(49-20a)\ge 0,\forall y\in R$

So, $49-20a>0\Rightarrow a<\frac{49}{20}$
and $D\le 0$
$\Rightarrow 4(1+2a^2{)}^2-4(49-20a{)}^2\le 0\Rightarrow (a^2-10a+25)(a^2+10a-24)\le 0\Rightarrow (a-5{)}^2(a+12\left)\right(a-2)\le 0\Rightarrow a\in [-12,2]$

But when $a=-12$
$y=\frac{-12x^2-7x+5}{5x^2-7x-12}\Rightarrow y=-\frac{(12x-5)(x+1)}{(5x-12)(x+1)}=-\left(\frac{12x-5}{5x-12}\right)$

When $a=2$
$y=\frac{2x^2-7x+5}{5x^2-7x+2}\Rightarrow y=\frac{(2x-5)(x-1)}{(5x-2)(x-1)}=\frac{2x-5}{5x-2}$


For $a=-12,2,y\ne R$

Therefore, $a\in (-12,2)$
Hence, the number of integral values of $a$ is $13$