Question

If the range of values of a for which the roots of the equation $x^2-2x-a^2+1=0$ lie between the roots of the equation $x^2-2(a+1)x+a(a-1)=0$ is $(p,q)$, find the value of $\left(q+\frac{1}{p^2}\right)$.

Answer 1

Answer: 17

$\therefore \alpha ,\beta =\frac{2\pm \sqrt{4+4\left(a^2-1\right)}}{2}=1\pm a$

Let $f(x)=x^2-2(a+1)x+a(a-1)$
Now, $f(\alpha )<0$ and $f(\beta )<0$ must hold simultaneously.
So,
$f(\alpha )<0\Rightarrow a>\frac{-1}{3}$.......(1)
and
$f(\beta )<0\Rightarrow \frac{-1}{4}<a<1$.......(2)
$\therefore$From (1) and (2), we get
$a\in \left(\frac{-1}{4},1\right)\Rightarrow p=\frac{-1}{4}\text{ and }q=1\Rightarrow \left(q+\frac{1}{p^2}\right)=1+16=17$