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 $\left(1 - a^{2}\right) x^{2} + 2 a x - 1 = 0$
Its discriminant $D=4$ and roots are $\frac{1}{a - 1} , \frac{1}{a + 1}$
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 \, \, $ $⇒ \, \frac{1}{a - 1} - 1 < 0$
$⇒ \, \, \frac{2 - a}{a - 1} < 0$ $\Rightarrow a < 1 \, or \, a > 2$
$∴ \, a > 2$ .... (1)
and $\frac{1}{a + 1} > 0 \Rightarrow a > - 1$
and $\frac{1}{a + 1} < 1 \Rightarrow - \frac{a}{a + 1} < 0$
$\Rightarrow $ $a < - 1 \, or \, a > 0$
$∴ \, $ $a>0 \, \, \, $ ... (2)
From (1) and (2), $a > 2$