Question

If $a\in R$ and the roots of $x^2-2x-a^2+1=0$ lies between the roots of $x^2-2(a+1)x+a(a-1)=0$, then $a$ belongs to

• A. $(1,\infty )$
• B. $(-\frac{1}{3},1)$
• C. $(-\infty ,0)$
• D. $(-\infty ,-\frac{1}{4})$

Answer 1

The correct option is B $(-\frac{1}{3},1)$

$f\left(x\right)=x^2-2x-a^2+1=0$

Roots are $1+a,1-a$

$g\left(x\right)=x^2-2(a+1)x+a(a-1)$ roots are

$a+1+\sqrt{3a+1},a+1-\sqrt{3a+1}ifa>\frac{-1}{3}$

say $a>0$

Then $1+a>1-a$ and $a+1+\sqrt{3a+1}>a+1-\sqrt{3a+1}$

For $f\left(x\right)$ to lie between $g\left(x\right)$, Bigger root of

$g\left(x\right)>$ bigger root of $f\left(x\right)$ and smaller root of

$g\left(x\right)<$ smaller root of $f\left(x\right)$.

$\Rightarrow 1+a<a+1+\sqrt{3a+1}$

$\Rightarrow a>\frac{-1}{3}$

$1-a>a+1-\sqrt{3a+1}$

$2a<\sqrt{3a+1}$

$4a^2<3a+1$

$(a+\frac{1}{4})(a-1)<0$

$\frac{-1}{4}<a<1$

$\Rightarrow 0<a<1$

Say $\frac{1}{3}<a<0$

$1+a<1-a$ and $a+1+\sqrt{3a+1}>a+1-\sqrt{3a+1}$

$(a+1)+\sqrt{3a+1}>1-a$

$2a+\sqrt{3a+1}>0$

And also

$(a+1)-\sqrt{3a+1}<1+a$

$3a+1>0$

$\therefore a\in (\frac{-1}{3},1)$