Question

The values of $a$ for which the equation $2x^2-2(2a+1)x+a(a+1)=0$ may have one root less than $a$ and other root greater than $a$ are given by

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

Answer 1

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

For $f\left(x\right)=2x^2-2(2a+1)x+a(a+1)=0$ to have one root less than $a$ and other root greater than $a$, it must have
1.Discriminant $>0$
$b^2-4ac>0$
$\Rightarrow 4(2a+1{)}^{{}^2}-8a(a+1)>0$
$\Rightarrow 4a^2+4a+1-2a^2-2a>0$
$\Rightarrow 2a^2+2a+1>0$
$\Rightarrow a^2+(a+1{)}^2>0$ which is always true.
2. $f\left(a\right)<0$
$2a^2-2a(2a+1)+a(a+1)<0$
$\Rightarrow 2a^2-4a^2-2a+a^2+a<0$
$\Rightarrow -a^2-a<0$
$\Rightarrow a^2+a>0$ $\Rightarrow a(a+1)>0$
$\Rightarrow a<-1$ or $a>0$.
Hence, option 'D' is correct.