Question

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

Answer 1

Answer: 2

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