Question

If the roots of the equation $x^2-4x+4-a^2=0$ lie between the roots of the equation $x^2-2(a+2)x+a(a-2)=0$ such that $a$ is real then ${}^{\prime }{a}^{\prime }$ belongs to the interval

• A. $(2,4)$
• B. $[\frac{-2}{3},2)$
• C. $(\frac{-1}{2},2)$
• D. $(\frac{-2}{3},2)$

Answer 1

The correct option is C $(\frac{-1}{2},2)$

Roots of the equation $x^2-4x+4-a^2=0$ would be -
$x=\frac{4\pm \sqrt{16+4(a^2-4)}}{2}=2\pm a$

Let $f_1\left(x\right)=x^2-4x+4-a^2$ and $f_2\left(x\right)=x^2-2(a+2)x+a(a-2)$

For $f_1\left(x\right)$ to have real roots
$D\ge 0$
$\Rightarrow 16+4(a^2-4)=a^2\ge 0$ (always true)
For $f_2\left(x\right)$ to have real roots
$D\ge 0$
$\Rightarrow 4(a+2{)}^2-4a(a-2)\ge 0\Rightarrow a\ge \frac{-2}{3}$

Roots of $f_1\left(x\right)$ lie between the roots of the equation $x^2-2(a+2)x+a(a-2)=0$
$\Rightarrow f_2(2+a)<0$ and $f_2(2-a))<0$
$\Rightarrow (2+a{)}^2-2(a+2\left)\right(a+2)+a(a-2)<0\Rightarrow a>-\frac{2}{3}\cdots (1)$
and
$(2-a{)}^2-(a+2\left)\right(a-2)+a(a-2)<0$
$\Rightarrow 4a^2-6a-4<0(2a+1)(a-2)\Rightarrow a\in (\frac{-1}{2},2)\cdots \left(2\right)$
From $\left(1\right)\&\left(2\right)$ , we get
$a\in (\frac{-1}{2},2)$