Question

If the roots of $x^2-2x-a^2+1=0$ lie between the roots of $x^2-2(a+1)x+a(a-1)=0$, then the range of $a$ is

• A. $(1,\infty )$
• B. $(-\frac{1}{4},1)$
• C. $(-\infty ,0)$
• D. $(-\infty ,-\frac{1}{4})$

Answer 1

The correct option is B $(-\frac{1}{4},1)$

$x^2-2x-a^2+1=0\cdots \left(i\right)$
Let roots be $\alpha ,\beta$
$f\left(x\right)=x^2-2(a+1)x+a(a-1)\cdots \left(ii\right)$
Let the roots of $f\left(x\right)=0$ be $t_1,t_2$


From Equation $\left(i\right)$,
$x^2-2x+1=a^2$
$\Rightarrow (x-1{)}^2=a^2$
$\Rightarrow x=1\pm a$
Assuming $\alpha =1-a,\beta =1+a$
Now $\alpha ,\beta$ lies in between the roots of the equation $\left(ii\right)$,
Required conditions are
$\left(i\right)f\left(\alpha \right)<0\Rightarrow f(1-a)<0$
$\Rightarrow (1-a{)}^2-2(1-a^2)+a^2-a<0$
$\Rightarrow 4a^2-3a-1<0$
$\Rightarrow (4a+1)(a-1)<0\Rightarrow a\in (-\frac{1}{4},1)$
$\left(ii\right)f\left(\beta \right)<0\Rightarrow f(1+a)<0$
$\Rightarrow -3a-1<0$
$\Rightarrow a>-\frac{1}{3}$
$\therefore a\in (-\frac{1}{4},1)$