Question

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

$(-1,0)\cup (0,1)$

Explanation for the correct option:

Step 1. Find the Domain of the given function:

Given, $-3{(x-[x\left]\right)}^2+2(x-[x\left]\right)+a^2=0$

Put $x–\left[x\right]=t$

$\Rightarrow$$0\le t<1$

Now, $–3t^2+2t+a^2=0$

$t=\frac{-2\pm \sqrt{4+12a^2}}{–6}$

$\Rightarrow$$0\le \frac{-2\pm \sqrt{4+12a^2}}{–6}<1$

$\Rightarrow$ $0\le \frac{1\pm \sqrt{1+3a^2}}{3}<1$

Step 2. Take positive sign of $0\le \frac{1\pm \sqrt{1+3a^2}}{3}<1$

$1+\sqrt{1+3a^2}<3$

$\Rightarrow$ $\sqrt{1+3a^2}<2$

$\Rightarrow$ $1+3a^2<4$

$\Rightarrow$ $3a^2<3$

$\Rightarrow$ $a^2<1$

$\therefore$$a$ belongs to $(-1,1)$

Thus, For all integral values of $a$ the integral is belongs in the interval $\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{)}\mathbf{}\mathbf{\cup }\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{)}$

Hence, Option ‘A’ is Correct.