JEE-Mains 2014 (3/30)

Let $u = x - \lfloor x \rfloor$ , which gives the quadratic equation:

$-3u^2 + 2u + a^2 = 0 \Rightarrow x - \lfloor x \rfloor = \frac{-2 \pm \sqrt{4 - 4(-3)(a^2)}}{-6} = \frac{1 \pm \sqrt{3a^2 + 1}}{3}$ (i).

The quantity $x - \lfloor x \rfloor$ must lie within $(0,1)$ for all $x \notin \mathbb{Z}$ . This implies for (i):

$0 < \frac{1 \pm \sqrt{3a^2 + 1}}{3} < 1$

For the RHS inequality, we have:

$\pm \sqrt{3a^2 + 1} < 2;$

or $3a^2 + 1 < 4$ ;

or $a^2 < 1$ ;

or $|a| < 1$ ;

or $-1 < a < 1$ (ii).

For the LHS inequality:

or $0 < 1 \pm \sqrt{3a^2 + 1};$

or $1 < 3a^2 + 1$ ;

or $0 < a^2$ ;

or $0 \neq a$ (iii).

Combining (ii) and (iii) gives the final result of $\boxed{ a \in (-1,0) \cup (0,1)}.$

-3p^2 + 2p+a^2=0 Where p is fractional part of x So p must lie between 1 and -1 For no integer solution p(1) and p(-1) are both less than 0 Putting p=1, a^2<1 Putting p=-1,a^2<5 Intersecting the solutions a^2<1 So -1<a<1