Question

Q8. If a ∈ R and the equation -3(x - [x])² + 2(x - [x]) + a² = 0 (where, [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval

(A) (-∞, - 2) ∪ (2, ∞)
(B) (-1, 0) ∪ (0, 1)
(C) (1, 2)
(D) (-2, -1)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: - \: 3(x - [x]) + 2(x - [x]) + {a}^{2} = 0$

Now, this above equation can be rewritten as

$\rm :\longmapsto\: \: 3(x - [x]) - 2(x - [x]) - {a}^{2} = 0$

Let assume that,

$\rm :\longmapsto\:x - [x] = y - - - (1)$

So, above equation reduces to

$\rm :\longmapsto\: {3y}^{2} - 2y - {a}^{2} = 0$

Now, its a quadratic equation in y.

We know the solution of quadratic equation is given by Quadratic formula.

So,

$\rm :\longmapsto\:y = \dfrac{ - ( - 2) \: \pm \: \sqrt{ {( - 2)}^{2} - 4(3)( - {a}^{2} )} }{2 \times 3}$

$\rm :\longmapsto\:y = \dfrac{2 \: \pm \: \sqrt{ 4 + 12 {a}^{2} } }{6}$

$\rm :\longmapsto\:y = \dfrac{2 \: \pm \: \sqrt{ 4(1 + 3{a}^{2} )} }{6}$

$\rm :\longmapsto\:y = \dfrac{2 \: \pm \: 2\sqrt{1 + 3{a}^{2}} }{6}$

$\rm :\longmapsto\:y = \dfrac{1 \: \pm \: \sqrt{1 + 3{a}^{2}} }{3}$

Now,

$\rm :\longmapsto\:y = x - [x]$

$\bf\implies \:y = \{x \}$

On substituting the value of y, we get

$\rm :\longmapsto\: \{x \} = \dfrac{1 \: \pm \: \sqrt{1 + 3{a}^{2}} }{3}$

Now, we know,

$\rm :\longmapsto\:0 \leqslant \{x \} < 1$

$\rm :\longmapsto\:0 \leqslant \dfrac{1 \: \pm \: \sqrt{1 + 3{a}^{2}} }{3} < 1$

Now,

Consider,

$\rm :\longmapsto\: \dfrac{1 \: \pm \: \sqrt{1 + 3{a}^{2}} }{3} < 1$

$\rm :\longmapsto\: 1 \: \pm \: \sqrt{1 + 3{a}^{2}} < 3$

$\rm :\longmapsto\: \: \pm \: \sqrt{1 + 3{a}^{2}} < 3 - 1$

$\rm :\longmapsto\: \: \pm \: \sqrt{1 + 3{a}^{2}} < 2$

On squaring both sides, we get

$\rm :\longmapsto\: \: 1 + 3{a}^{2} < 4$

$\rm :\longmapsto\: \: 3{a}^{2} < 4 - 1$

$\rm :\longmapsto\: \: 3{a}^{2} < 3$

$\rm :\longmapsto\: \: {a}^{2} < 1$

$\rm :\longmapsto\: \: {a}^{2} - 1< 0$

$\rm :\longmapsto\:(a + 1)(a - 1) < 0$

$\bf\implies \: - 1 < a < 1$

$\bf\implies \:a \: \in \: ( - 1, \: 1)$

So, for non integral solutions,

$\bf\implies \:a \: \in \: ( - 1, \: 1) - \{ 0\}$

or

$\bf\implies \:a \: \in \: ( - 1, \: 0) \cup \: (0, \: 1)$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf \: Option \: (B) \: is \: correct}}$