Question

If the equation $(a-2)(x-[x]{)}^2+2(x-\left[x\right])+a^2=0,a\in R$ has no integral solution and has exactly one solution in $[2,3)$, then $a$ lies in the interval
(where $\left[x\right]$ denotes the greatest integer function)

• A. $(-1,2)$
• B. $(0,1)$
• C. $(-1,0)$
• D. $(2,3)$

Answer 1

The correct option is C $(-1,0)$

$f\left(x\right)=(a-2)(x-[x]{)}^2+2(x-\left[x\right])+a^2=0$
Let $t=x-\left[x\right]=\left\{x\right\}$
$\Rightarrow t\in [0,1)$
$(a-2)t^2+2t+a^2=0$

It is given that the given equation does not have any integer root.
So $t\ne 0(\because \forall x\in Z,x=[x]\Rightarrow t=0)$
$\therefore t\in (0,1)$

Case $1:$ $a-2\ne 0$
$f\left(t\right)=(a-2)t^2+2t+a^2=0$
For $f\left(x\right)$ to have exactly one root in $[2,3)$, $f\left(t\right)$ should have exactly one root in the interval $(0,1)$.
$\therefore f\left(0\right)f\left(1\right)<0$
$\Rightarrow a^2(a^2+a)<0$
$\Rightarrow a\in (-1,0)$

Case $2:$ $a-2=0$
Then $2t+a^2=0$ which is not possible as $\text{LHS}>0$
$\therefore a\in (-1,0)$