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Q.
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 :
JEE MainJEE Main 2014Complex Numbers and Quadratic Equations
Solution:
$-3(x-[x])^{2}+2[x-[x])+a^{2}=0$
$3\{x\}^{2}-2\{x\}-a^{2}=0$
$a \neq 0,3\left(\{x\}^{2}-\frac{2}{3}\{x\}\right)=a^{2}$
$a^{2}=3\left(\{x\}-\frac{1}{3}\right)^{2}-\frac{1}{3}$
$0 \leq\{x\}<\,1 $ \and $3-\frac{1}{3} \leq\{x\}-\frac{1}{3}<\,\frac{2}{3}$
$0 \leq 3\left(\{x\}-\frac{1}{3}\right)^{2}<\frac{4}{3}$
$-\frac{1}{3} \leq 3\left(\{x\}-\frac{1}{3}\right)^{2}-\frac{1}{3}<\,1$
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$
$\therefore a \in(-1,0) \cup(0,1)$
