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

Answer 1

Answer: (C)

$-3(x-[x]{)}^2+2[x-[x])+a^2=0$
$3\{x{\}}^2-2\{x\}-a^2=0$
$a\ne 0,3\left(\{x{\}}^2-\frac{2}{3}\{x\}\right)=a^2$
$a^2=3{\left(\{x\}-\frac{1}{3}\right)}^2-\frac{1}{3}$
$0\le \{x\}<1$ \and $3-\frac{1}{3}\le \{x\}-\frac{1}{3}<\frac{2}{3}$
$0\le 3{\left(\{x\}-\frac{1}{3}\right)}^2<\frac{4}{3}$
$-\frac{1}{3}\le 3{\left(\{x\}-\frac{1}{3}\right)}^2-\frac{1}{3}<1$
For non-integral solution
$0<a^2<1$ and $a\in (-1,0)\cup (0,1)$
Alternative
$-3\{x{\}}^2+2\{x\}+a^2=0$
Now, $-3\{x{\}}^2+2\{x\}$

to have no integral roots $0<a^2<1$
$\therefore a\in (-1,0)\cup (0,1)$