Question

If $a\in R$ and the equation $-3(x-[x]{)}^2+2(x-\left[x\right])+a^2=0$, (where, $\left[x\right]$ denotes the greatest integer $\le x$) has no integral solution, then all the possible values of $a$ lie in the interval :

• A. $(-2,-1)$
• B. $(-1,0)\cup (0,1)$
• C. $(-\infty ,-2)\cup (2,\infty )$
• D. $(1,2)$

Answer 1

The correct option is B $(-1,0)\cup (0,1)$

Consider $-3(x-[x]{)}^2+2(x-\left[x\right])+a^2=0$
$\Rightarrow -3\{x{\}}^2-2\{x\}-a^2=0$ $(\because x-[x]=\{x\left\}\right)$
$\Rightarrow 3\left(\right\{x{\}}^2-\frac{2}{3}\left\{x\right\})=a^2,a\ne 0$
$\Rightarrow a^2=3\left\{x\right\}\left(\right\{x\}-\frac{2}{3})$


Now, $\left\{x\right\}\in (0,1)$ and $\frac{-1}{3}\le a^2\le 1\Rightarrow 0\le a^2\le 1$
Since, $x$ is not an integer

$\therefore a\in (-1,1)-\left\{0\right\}\Rightarrow a\in (-1,0)\cup (0,1)$