Question

If $a\ne R$ and the equation $-3(x-[x]{)}^2+2(x-\left[x\right])+a^2=0$ ( where $\left[x\right]$ denotes the greatest integer $\le x$) has no integral solution, then all possible values of a lie in the interval :

• A. $(-2,-1)$
• B. $(-\infty ,-2)\cup (2,\infty )$
• C. $(-1,0)\cup (0,1)$
• D. $(1,2)$

Answer 1

The correct option is C $(-1,0)\cup (0,1)$

REF.Image

As $x-\left[x\right]=\left\{x\right\}$, let $\left\{x\right\}=y$

As x is non-integral, $0<y<1$ $(y\ne 0)$

equation becomes, $-3y^2+2y+a^2=0$

$\Rightarrow a^2=3y^2-2y=y(3y-2)$

Drawing graph of $3y^2-2y=0$ (parabola)

Minima $\Rightarrow \frac{d}{dy}(3y^2-2y)=0\Rightarrow 6y-2=0\Rightarrow y=1/3$

Taking positive solutions $(a^2\ge 0)$

we get $0<a^2<1$

$\Rightarrow aϵ(-1,0)\cup (0,1)$

Option (C)