Question

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

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

Answer 1

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

As we know,
$x-\left[x\right]=\left\{x\right\}$
So, $-3\{x{\}}^2+2\{x\}+a^2=0...(1)$
$\left\{x\right\}=\frac{-2\pm \sqrt{4-4(-3)\left(a^2\right)}}{2\times (-3)}$
$=\frac{-1\pm \sqrt{1+3a^2}}{-3}$
$=\frac{1\mp \sqrt{1+3a^2}}{3}$
As, Range of $\left\{x\right\}=[0,1)$

Case 1: $0\le \frac{1-\sqrt{1+3a^2}}{3}<1$
$\Rightarrow a^2=0$
From $\left(1\right)$, $\left\{x\right\}=0$ which gives integal solution or $-\frac{2}{3}$ (which is not possible)

Case 2: $0\le \frac{1+\sqrt{1+3a^2}}{3}<1$
$\Rightarrow 0\le a^2<1$
$\Rightarrow a\in (-1,1)$
At, $a=0$, equation $\left(1\right)$ has integral solution
So, interval of $a$ is $(-1,0)\cup (0,1)$