Question

Both roots of $(a^2-1)x^2+2ax+1=0$ belong to the interval $(0,1)$ then exhaustive set of values of ${}^{\prime }{a}^{\prime }$ is :

• A. None of these
• B. $(-\infty ,-2)$
• C. $(-\infty ,-2)\cup (0,\infty )$
• D. $(-2,\frac{-1+\sqrt{5}}{2})$

Answer 1

The correct option is B

$(-\infty ,-2)$

Let $f\left(x\right)=(a^2-1)x^2+2ax+1$

Case $1$: $(a^2-1)>0$

Given both roots of $f\left(x\right)=0$ are in $(0,1)$
$\Rightarrow f\left(0\right)>0,f\left(1\right)>0$

$f\left(1\right)>0\Rightarrow a^2+2a>0$ $\Rightarrow a<-2$ or $a>0$

$\therefore a\in (-\infty ,-2)\cup (1,\infty )$

Case $2$ : $(a^2-1)<0,f\left(1\right)<0$ & $f\left(0\right)<0$ which is not possible.