Question

Let $f(x)=\frac{x^2-2 a x+b}{b x^2-2 a x+1}$. If the equation $f(x)=m$ has two real and distinct roots $\forall m \in R$, then the range of $a$ is $(-\infty, p) \cup(q, \infty)$. Find the value of $2\left(p^2+q^2\right)$.

Answer 1

$\frac{x^2-2 a x+b}{b x^2-2 a x+1}=m$
Since $f ( x )= m$ has 2 real and distinct roots $\forall m \in R$
Which is possible only when $f(x)=0$ has no horizontal asymptotes i.e. $b$ must be equal to zero otherwise $f ( x )= m$ has two real and distinct root $\forall m \in R$ is not possible and $x ^2-2 ax + b =0$ and $bx ^2-2 ax +1=0$ must not have any common root.
Hence $\frac{x^2-2 a x}{1-2 a x}=m \Rightarrow x^2-2 a x=m-2 a x m$
$x ^2-2 ax (1- m )- m =0$
for distinct roots $D >0$
$\Rightarrow 4 a ^2(1- m )^2+4 m >0 \quad \forall m \in R \Rightarrow a ^2 m ^2- m \left(2 a ^2-1\right)+ a ^2>0 \forall m $
$D <0$
$\left(2 a^2-1\right)^2-4 a^4<0 \Rightarrow 1-4 a^2<0 \Rightarrow 4 a^2-1>0 \Rightarrow a \in\left(-\infty, \frac{-1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)$
[Note : At $a = \pm \frac{1}{2} x ^2-2 ax =0$ and $1-2 ax =0$ has a common root.]
$p =\frac{-1}{2}$ and $q =\frac{1}{2} \Rightarrow 2\left( p ^2+ q ^2\right)=2\left(\frac{1}{4}+\frac{1}{4}\right)=1$.