Question

For the equation $x^2-2ax+a^2-1=0$, the values of ${}^{\prime }{a}^{\prime }$ for which $3$ lies in between the roots of given equation is

• A. $(-\infty ,2)\cup (4,\infty )$
• B. $(-\infty ,3)\cup (4,\infty )$
• C. $(2,4)$
• D. $[3,4]$

Answer 1

The correct option is C

$(2,4)$

For roots to be real,

$b^2-4ac>0$

$\Rightarrow 4a^2-4(a^2-1)>0$

$1>0$ [Always true]

Here, $a>0$.

So, the given quadratic expression will have minimum value at $x=\frac{-b}{2a}=a$.

For $3$ to lie between the roots, $f\left(3\right)<0$.

Hence, $f\left(3\right)=9-6a+a^2-1<0$

$\Rightarrow a^2-6a+8<0$

$\Rightarrow a^2-4a-2a+8<0$

$\Rightarrow a(a-4)-2(a-4)<0$

$\Rightarrow 2<a<4$ . . . $\left(1\right)$

Hence, $a\in (2,4)$.