Question

If $a\in R$ and the equation $-3(x-[x]{)}^2+2(x-\left[x\right])+a^2=0$ (where, [x] denotes the greatest integer $\ge x)$ 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 $(-1,0)\cup (0,1)$

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 $-3{\{x-[x\left]\right\}}^2+2\{x-[x\left]\right\}+a^2=0$
Let t = x – [x], then equation is
$-3t^2+2t+a^2=0\Rightarrow t=\frac{1\pm \sqrt{1+3a^2}}{3}\because t=x-\left[x\right]=x\left[fractionalpart\right]\therefore 0\le t\le 1$
$0\le \frac{1+\sqrt{1+3a^2}}{3}<1[\because Fractionalpart(x)>0]\Rightarrow \sqrt{1+3a^2}<2\Rightarrow 1+3a^2<4\Rightarrow a^2-1<0\Rightarrow (a+1)(a-1)<0$
$\therefore$ a lies in (-1,1), for no integer solution we consider $(-1,0)\cup (0,1)$