If a R and the equation -3(x-[x])^2+2(x-[x])+a^2=0 (where [x] denotes the greatest integer…

Question

If $a\in\mathbb R$ 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:

Accepted Answer

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 = $$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$

Solutions · 1