Question

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

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

Answer 1

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

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



Now,$\left\{x\right\}ϵ(0,1)and\frac{-2}{3}\le a^2<1$ (by graph)
Since, $x$ is not an integer
$\therefore aϵ(-1,1)-0\Rightarrow aϵ(-1,0)\cup (0,1)$