JEE Main · 2014 · 06 Apr 2014 · Offline · Sets Relations and Functions

Let $t=x-[x]$ be the fractional part, so $0\le t<1$, and the equation becomes $a^2=3t^2-2t$. Asking for no integral solution means $x$ is not an integer, so $t$ runs over $0<t<1$. On this range $f(t)=3t^2-2t$ dips to its least value $-\frac{1}{3}$ at $t=\frac{1}{3}$ and climbs towards $f(1)=1$, while $f(0)=0$. Since $a^2\ge 0$, a fractional solution exists only for $0\le a^2<1$, and discarding $a=0$ leaves $a\in (-1,0)\cup (0,1)$.