Question

The value of $a$ for which both the roots of the equation $\left(1 - a^{2}\right) x^{2} + 2 a x - 1 = 0$ lie between $0$ and $1$ , will always be greater than

Answer 1

Given equation is $\left(1 - a^{2}\right)x^{2}+2ax-1=0$
$\Rightarrow x^{2}+\frac{2 a}{\left(1 - a^{2}\right)}x-\frac{1}{\left(1 - a^{2}\right)}=0$
$\Rightarrow x^{2}-\frac{2 a}{\left(a^{2} - 1\right)}x+\frac{1}{\left(a^{2} - 1\right)}=0$
$\Rightarrow x^{2}-\left[\frac{1}{\left(a - 1\right)} + \frac{1}{\left(a + 1\right)}\right]x+\frac{1}{\left(a - 1\right)}\times \frac{1}{\left(a + 1\right)}=0$
Therefore, roots are $\frac{1}{a - 1}, \frac{1}{a + 1}$
Its discriminant is $D=\left(2 a\right)^{2}-4\left(1 - a^{2}\right)\left(- 1\right)=4$
Given, $0 < \frac{1}{a - 1} < 1 , $ $0 < \frac{1}{a + 1} < 1 $
Now, $\frac{1}{a - 1} > 0 \Rightarrow a > 1 , $
$\frac{1}{a - 1} < 1$ $\Rightarrow \frac{1}{a - 1} - 1 < 0$
$\Rightarrow \frac{2 - a}{a - 1} < 0$ $\Rightarrow a < 1 or a > 2$
$\therefore a > 2$ .... (1)
and $\frac{1}{a + 1} > 0 \Rightarrow a > - 1$
and $\frac{1}{a + 1} < 1 \Rightarrow - \frac{a}{a + 1} < 0$
$\Rightarrow $ $a < - 1 or a > 0$
$\therefore $ $a>0$ ... (2)
From (1) and (2), $a > 2$