QJEEM14Ma89a4: JEE Main 2014 Mathematics - Class 11 JEEM14Ma89a4 Solution: If and the equation (where [ ] denotes the greater integer )…

Given, $- 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$
As we know, $\left[ x \right] + \left\{ x \right\} = x$
where $\left[ x \right]$ is integral part and $\left\{ x \right\}$ is fractional part.
$\therefore$ $\left\{ x \right\} = x - \left[ x \right]$
Now put $\left\{ x \right\}$ inplace of $x - \left[ x \right]$ in the equation.
The new equation is $- 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0$

[Note : Question says this equation has no integral solution, it means $\left\{ x \right\} \ne$ 0. So, $x$ is not a integer.]

$\therefore$ $\left\{ x \right\}$ = ${{ - 2 \pm \sqrt {4 - 4 \times \left( { - 3} \right){a^2}} } \over { - 6}}$
             = ${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$
As $\left\{ x \right\}$ is fractional part so it is lies between 0 to 1( 0 \le \left\{ x \right\} < 1 ).

By considering positive sign, we get

0 \le {{ - 2 + \sqrt {4 + 12{a^2}} } \over { - 6}} < 1

$\Rightarrow$ 0 \ge - 2 + \sqrt {4 + 12{a^2}} > - 6

$\Rightarrow$ 2 \ge + \sqrt {4 + 12{a^2}} > - 4

$\because$ $+ \sqrt {4 + 12{a^2}}$ is always positive which is greater than any negative no. So can ignore the inequality + \sqrt {4 + 12{a^2}} > - 4

Consider this inequality,
$2 \ge + \sqrt {4 + 12{a^2}}$
$\Rightarrow$ $4 \ge 4 + 12{a^2}$
$\Rightarrow$ $12{a^2} \le 0$
$\Rightarrow$ ${a^2} \le 0$
$\Rightarrow$ ${a^2} = 0$
$\Rightarrow$ ${a} = 0$

If $a$ = 0 then $- 3{\left\{ x \right\}^2} + 2\left\{ x \right\} = 0$ so $\left\{ x \right\}$ becomes 0 but question says $\left\{ x \right\}$ $\ne$ 0.
So $a$ can't be 0.

Now by considering negative sign, we get


0 \le {{ - 2 - \sqrt {4 + 12{a^2}} } \over { - 6}} < 1

$\Rightarrow$ 0 \ge - 2 - \sqrt {4 + 12{a^2}} > - 6

$\Rightarrow$ 2 \ge - \sqrt {4 + 12{a^2}} > - 4

As 2 is always greater than ${ - \sqrt {4 + 12{a^2}} }$ . Ignore this inequality.

Now consider this inequality,

- \sqrt {4 + 12{a^2}} > - 4

$\Rightarrow$ \sqrt {4 + 12{a^2}} < 4

$\Rightarrow$ 4 + 12{a^2} < 16

$\Rightarrow$ 12{a^2} < 12

$\Rightarrow$ {a^2} < 1

$\Rightarrow$ \left( {{a^2} - 1} \right) < 0

$\Rightarrow$ \left( {a + 1} \right)\left( {a - 1} \right) < 0

$\Rightarrow$ - 1 < a < 1

But earlier we found that $a$ $\ne$ 0.

So, the range of $a$ is = $\left( { - 1,0} \right) \cup \left( {0,1} \right)$