Question

The value of $a$ for which $2x^2-2(2a+1)x+a(a+1)=0$ may have one root less than $a$ and other root greater than $a$

• A. $-1<a<0$
• B. $a>0$ or $a<-1$
• C. $a\ge 0$
• D. $-\frac{1}{2}<a<0$

Answer 1

The correct option is B $a>0$ or $a<-1$

Given equation $2x^2-2(2a+1)x+a(a+1)=0$
Let $f\left(x\right)=2x^2-2(2a+1)x+a(a+1)$
The condition for $a$ to lie in between the roots is
$a.f\left(a\right)<0$
$\Rightarrow 2a^2-2a(2a+1)+a(a+1)<0(\because A=2>0,$ so $f\left(a\right)<0)$
$\Rightarrow -a^2-a<0$
$\Rightarrow a^2+a>0$
$\Rightarrow a>0$ or $a<-1$.