Question

If $t$ lies between real roots of the equation $2x^2-2(2t+1)x+t(t+1)=0,$ then $t$ cannot be

• A. $1$
• B. $-2$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $-\frac{1}{2}$

For real roots, $D>0$
$\Rightarrow 4(2t+1{)}^2-4\times 2t(t+1)>0$
$\Rightarrow 2t^2+2t+1>0,$ which is true $\forall t\in R$

Now, if $t$ lies between real roots of given quadratic, then
$f\left(t\right)<0$
$\Rightarrow 2t^2-2(2t+1)t+t(t+1)<0$
$\Rightarrow 2t^2-4t^2-2t+t^2+t<0$
$\Rightarrow -t^2-t<0\Rightarrow t^2+t>0\Rightarrow t\in (-\infty ,-1)\cup (0,\infty )$
$\therefore t$ cannot be $-\frac{1}{2}$