Question

If the roots of \(x^2 - 2x -a^2 + 1 = 0\) lie between the roots of \(x^2 -2(a+1)x + a(a-1) = 0\), then the range of $a$ is

Answer 1

\(x^2 - 2x - a^2 + 1 = 0 \cdots(i)\)
Let roots be \(\alpha, \beta\)
\(f(x) = x^2 -2(a+1)x + a(a -1)\cdots(ii)\)
Let the roots of $f(x) = 0$ be \(t_1, t_2 \)


From Equation $(i)$,
\(x^2 - 2x + 1 = a^2\)
\(\Rightarrow (x-1)^2 = a^2 \)
\(\Rightarrow x = 1 \pm a\)
Assuming $\alpha = 1 - a,~~\beta = 1 + a$
Now \(\alpha, \beta\) lies in between the roots of the equation $(ii)$,
Required conditions are
\((i)~ f(\alpha) < 0
\Rightarrow f(1 -a) < 0\)
$\Rightarrow (1-a)^2 -2(1-a^2) + a^2 -a < 0$
\(\Rightarrow 4a^2 - 3a - 1< 0 \)
\(\Rightarrow (4a + 1) (a-1) < 0\\
\Rightarrow a\in \left ( - \dfrac{1}{4},1 \right )\)
$(ii)~f(\beta) < 0
\Rightarrow f(1 + a) < 0$
$\Rightarrow -3a-1 < 0$
$\Rightarrow a > -\dfrac{1}{3} $
\(\therefore a \in \left ( - \dfrac{1}{4}, 1 \right )\)