Question

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

• A. $(1,\infty )$
• B. $(-\frac{1}{2},1)$
• C. $(-\infty ,0)$
• D. $(-\infty ,-\frac{1}{4})$

Answer 1

Answer: (B)

$(-\frac{1}{2},1)$

We are given,

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

here the co-efficient of $x^2$ is positive & $g\left(x\right)=x^2-(a+1)x+(a-1)=0\to \left(2\right)$ & also here the co-efficient of $x^2$ is positive.

$\Rightarrow$ Let ${\alpha }_1,{\beta }_1$ are the roots of $f\left(x\right)$ & ${\alpha }_2,{\beta }_2$ are the roots of $g\left(x\right)$.

Now, here we can absorbed that at $pt$ ${\alpha }_1$ & ${\beta }_{1}g\left(x\right)$ will generated negative values.

So, $g\left({\alpha }_1\right),g\left({\beta }_2\right)<0$

From equati0on $\left(1\right)$

${\alpha }_1$ or ${\beta }_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-(-2)\pm \sqrt{(-1{)}^2-4(1\left)\right(a-1)}}{2\left(1\right)}$

$=\frac{2\pm \sqrt{4a^2}}{2}=1\pm a$

So, ${\alpha }_1=1+a$ & ${\beta }_1=1-a$

as $a\in R$

Now, $g\left({\alpha }_1\right)=(1+a{)}^2-(a+1\left)\right(1+a)+1(a-1)<0$

$1+2a+a^2-a^2-2a-1+a-1<0$

$a<1\to \left(1\right)$

& $g\left({\beta }_1\right)=(1-a{)}^2-(a+1\left)\right(1-a)+1(a-1)<0$

$1-2a+a^2-(a-a^2+1-a)+a-1<0$

$1-2a+a^2+a^2-1+a-2<0$

$2a^2-a-1<$

$2a^2-2a+a-1<0$

$2a(a-1)+1(a-1)<0$

$(2a+1)(a-1)<0$

$\to$ at $a>1$ quantity $>0$, so condition False

$\to$ now at $a<-1/2$ quantity $>0$, so condition again False

$\to$ now at a $\in (-1/2,1)$ quantity $<0$

So, $a\in (-1/2,1)\to \left(\right(1)$

From $\left(i\right)$ & $\left(ii\right)$

$a\in (\frac{-1}{2},1)$