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

JEE · Math · previous-year question

  1. A.$$\left( { - 2, - 1} \right)$$
  2. B.$$\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right)$$
  3. C.$$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$correct
  4. D.$$\left( {1,2} \right)$$

Answer

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

Explanation

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

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