JEE Main 2014 (Offline)Question 1
If $a\in R$ and the equation $-3(x-[x])^{2}+2(x-[x])+a^{2}=0$ (where $[x]$ denotes the greater integer $\le x$) has no integral solution, then all possible values of $a$ lie in the interval :
Functions
Solution
With $t=\{x\}\in[0,1)$: $a^2=3t^2-2t$. An integral solution means $t=0$, i.e. $a=0$. For $t\in(0,1)$, $3t^2-2t$ takes values up to (not including) $1$, so solutions exist for $0<a^2<1$, i.e. $a\in(-1,0)\cup(0,1)$.