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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 AdvancedJEE Advanced 2014Complex Numbers and Quadratic Equations
Solution:
Put $t = x - [x] = \{X\}$, which is a fractional part function
and lie between $0 \le \{X\} < 1$ and then solve it.
Given, $a \in R$ and equation is
$- 3 \{ x - [ x ] \}^2 + 2 \, \{ x - [ x ] \} + a^2 = 0 $
Let $t = x - [x]$, then equation is
$ - 3t^2 + 2t + a^2 = 0 $
$\Rightarrow t = \frac{ 1 \pm \sqrt{ 1 + 3a^2 }}{ 3} $
$\because t = x - [ x ] = { X} $ [fractional part]
$ \therefore 0 \le 1 \le 1 $
$ 0 \le \frac{ 1 \pm \sqrt{ 1 + 3a^2 }}{ 3} \le 1 $
Taking positive sign, we get
$ 0 \le \frac{ 1 + \sqrt{ 1 + 3a^2 }}{ 3 } < 1 $ [ $ \because (x) > 0 ] $
$ \Rightarrow \sqrt{ 1 + 3a^2 } < 2 \Rightarrow 1 + 3a^2 < 4 $
$\Rightarrow a^2 - 1 < 0 $
$\Rightarrow (a + 1) \, (a - 1) < 0 $
$\therefore a \, \in \, (-1, 1) , $ for no integer solution of $a$, we consider
$ (-1, 0 ) \cup (0, 1)$