If a R and the equation - 3 ( x - [ x ] ) 2 + 2 ( x - [ x ] ) + a 2 = 0 (where [ x ] denotes the greatest integer x ) has no integral solution, then all possible values of a lie in the interval: [2014]

Question

If $a \in R$ and the equation $- 3 {(x - [x])}^{2} + 2 (x - [x]) + a^{2} = 0$ (where $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of $a$ lie in the interval: [2014]

Smart Achievers · Accepted Answer

(3)

$Consider - 3 {(x - [x])}^{2} + 2 [x - [x]] + a^{2} = 0$

$\Rightarrow 3 {x}^{2} - 2 {x} - a^{2} = 0 (∵ x - [x] = {x})$

$\Rightarrow 3 ({x}^{2} - \frac{2}{3} {x}) = a^{2}, a \neq 0$

$\Rightarrow a^{2} = 3 {({x} - \frac{1}{3})}^{2} - \frac{1}{3} (∵ 0 \leq {x} < 1)$

$- \frac{1}{3} \leq {x} - \frac{1}{3} < \frac{2}{3} \Rightarrow 0 \leq 3 ({x} \frac{- 1}{3}) < \frac{4}{3}$

$- \frac{1}{3} \leq 3 ({x} \frac{- 1}{3}) - \frac{1}{3} < 1$

$For non-integral solution 0 < a^{2} < 1$

$\Rightarrow a \in (- 1, 0) \cup (0, 1)$

Solution

Q.
If $a \in R$ and the equation $- 3 {(x - [x])}^{2} + 2 (x - [x]) + a^{2} = 0$ (where $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of $a$ lie in the interval: [2014]

1 $(- 2, - 1)$

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

3 $(- 1, 0) \cup (0, 1)$

4 $(1, 2)$

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Solution

Q. 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: [2014]

1 (-2,-1)

2 (-∞,-2)∪(2,∞)

3 (-1,0)∪(0,1)

4 (1,2)