Question

If $a\in R$ and the equation $-3(x-[x]{)}^2+2(x-[x])+a^2=0$ (where $,[x]$ denotes 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

Answer: (A)

Put $t=x-[x]=\{X\}$, which is a fractional part function
and lie between $0\le \{X\}<1$ and then solve it.
Given, $a\in R$ and equation is
$-3\{x-[x]{\}}^2+2\{x-[x]\}+a^2=0$
Let $t=x-[x]$, then equation is
$-3t^2+2t+a^2=0$
$\Rightarrow t=\frac{1\pm \sqrt{1+3a^2}}{3}$
$\because t=x-[x]=X$ [fractional part]
$\therefore 0\le 1\le 1$
$0\le \frac{1\pm \sqrt{1+3a^2}}{3}\le 1$
Taking positive sign, we get
$0\le \frac{1+\sqrt{1+3a^2}}{3}<1$ [ $\because (x)>0]$
$\Rightarrow \sqrt{1+3a^2}<2\Rightarrow 1+3a^2<4$
$\Rightarrow a^2-1<0$
$\Rightarrow (a+1)(a-1)<0$
$\therefore a\in (-1,1),$ for no integer solution of $a$, we consider $(-1,0)\cup (0,1)$