Question

The values 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 $a$ 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

The given condition suggest that a lies between the roots.

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

For ‘$a$’ to lie between the roots we must have Discriminant $\ge$ 0 and $f\left(a\right)<0$

Now, Discriminant $\ge$ 0

$\Rightarrow 4(2a+1{)}^2–8a(a+1)\ge 0$

$\Rightarrow 8(a^2+a+\frac{1}{2})\ge 0$ which is always true

Also $f\left(a\right)<0\Rightarrow 2a^2–2a(2a+1)+a(a+1)<0$

$\Rightarrow -a^2–a<0\Rightarrow a^2+a>0\Rightarrow a(1+a)>0$

$\Rightarrow$ $a$ > 0 or $a$ < –1