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

Engineering

Mathematics

Piecewise Defined Functions

Question

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

$(- 1, 0) \cup (0, 1)$

$(- \infty, - 2) \cup (2, \infty)$

(– 2, – 1)

(1, 2)

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Let x – [x] = f

$∴$ 3f² – 2f – a² = 0 …….(1)

$f = \frac{1 + \sqrt{1 + 3 a^{2}}}{3}$

for no integral solution

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

⇒ 0 $\leq$ a² < 1

$∴$ a $\in$ (– 1, 1)

But if a = 0, then from equation (i), f = 0

But it will give integral value of x.

$∴$ a = 0 is not possible.

So, a $\in$ (– 1, 1) – {0}