Question

The value of $a$ for which the equation $(1-a^2)x^2+2ax-1=0$ has roots belonging to $(0,1)$ is

• A. $a>\frac{1+\sqrt{5}}{2}$
• B. $a>2$
• C. $\frac{1+\sqrt{5}}{2}<a<2$
• D. $a>\sqrt{2}$

Answer 1

The correct option is B $a>2$

Let $f\left(x\right)=(1-a^2)x^2+2ax-1$. then $f\left(x\right)=0$ has roots between $0$ and $1$ if
(1) $D\ge 0$
(2) $(1-a^2)f\left(0\right)>0$ and $(1-a^2)f\left(1\right)>0$
Now, $D\ge 0\Rightarrow 4a^2+4(1-a^2)>0$ which is always true.
$(1-a^2)f\left(0\right)>0\Rightarrow -(1-a^2)>0\Rightarrow a^2-1>0\Rightarrow a<-1$ and $a>1$ ...(i)
and $(1-a^2)f\left(1\right)>0\Rightarrow (1-a^2)(2a-a^2)>0$
$\Rightarrow a(a-1)(a+1)(a-2)>0\Rightarrow a<-1$ or $a>2$ or $0<a<1$ ...(ii)
From (i) and (ii) we get
$a<-1$ and $a>2$