Question

If $aϵR$ and the equation $-3(x-[x]{)}^2+2(x-\left[x\right])+a^2=0$ (where, $\left[x\right]$ denotses the greatest $\ge x$) has no integral solution, then all possible value of a lie in the interval

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

Answer 1

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

$\begin{array}{cc}\text{solution}& :-3(x-[x]{)}^2+2(x-\left[x\right])+a^2=0\\ \Rightarrow & -3\{x{\}}^2+2\{x\}+a^2=0\\ \text{where}\left\{\right\}\text{denoter fractional part function}& \end{array}$

$\left\{x\right\}=\frac{-2+\sqrt{4+12a^2}}{-6}$ and $\left\{x\right\}=\frac{-2-\sqrt{4+12a^2}}{-6}$

$\left\{x\right\}=\frac{1-\sqrt{1+3a^2}}{3}$ and $\left\{x\right\}=\frac{1+\sqrt{1+3a^2}}{3}$ we know that $\left\{x\right\}\in [0,1)$

$\frac{1+\sqrt{1+3a^2}}{3}<1+\sqrt{1+3a^2}-3<0$

$\Rightarrow \sqrt{1+3a^2}-2<0\Rightarrow \sqrt{1+3a^2}<2$

Now squaring both sides in $\sqrt{1+3a^2}<2$ we get so

$\begin{array}{cc}& 1+3a^2<4\\ \Rightarrow & 3a^2<3\\ \Rightarrow & a^2<1\\ a\in (-1& 1)\end{array}$

we also need $0⩽\underset{–}{1-\sqrt{1+3a^2}}<1$

$\begin{array}{c}\frac{1-\sqrt{1+3a^2}}{3}⩾0\\ \Rightarrow 1⩾\sqrt{1+3a^2}\end{array}$

squaring both sides, wee get $1+3a^2⩽1$

$\Rightarrow 3a^2⩽0$ From this case we only get $a=0$

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

$\Rightarrow 1-\sqrt{1+3a^2}-3<0$

$\Rightarrow -\sqrt{1+3a^2}-2<0$

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

this holds true far a $\in R$

of a' from above three cases

i.e $a\in (-1,1)$

but we have to exclude $a=0$ because at $a=0$ equation gets integral solution $\Delta 0$

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

Answer: option (A)