Question

Let $f(x)=\frac{x^2-2ax+b}{b{x}^2-2ax+1}$. 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

Answer: 1

$\frac{x^2-2ax+b}{b{x}^2-2ax+1}=m$
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 $x^2-2ax+b=0$ and $b{x}^2-2ax+1=0$ must not have any common root.
Hence $\frac{x^2-2ax}{1-2ax}=m\Rightarrow x^2-2ax=m-2axm$
$x^2-2ax(1-m)-m=0$
for distinct roots $D>0$
$\Rightarrow 4a^2(1-m{)}^2+4m>0\forall m\in R\Rightarrow a^2{m}^2-m\left(2a^2-1\right)+a^2>0\forall m$
$D<0$
${\left(2a^2-1\right)}^2-4a^4<0\Rightarrow 1-4a^2<0\Rightarrow 4a^2-1>0\Rightarrow a\in \left(-\infty ,\frac{-1}{2}\right)\cup \left(\frac{1}{2},\infty \right)$
[Note : At $a=\pm \frac{1}{2}{x}^2-2ax=0$ and $1-2ax=0$ has a common root.]
$p=\frac{-1}{2}$ and $q=\frac{1}{2}\Rightarrow 2\left(p^2+q^2\right)=2\left(\frac{1}{4}+\frac{1}{4}\right)=1$.