JEE Main | 2014Math miscellaneousHard
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
-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
Options
A.(-2, -1)
B.(-∞, - 2) ∪ (2, ∞)
C.(-1, 0) ∪ (0, 1)
D.(1, 2)
Solution
-3(x- [x])2 + 2[x - [x]) + a2 = 0
3 {x}2 - 2{x} - a2 = 0
a ≠ 0, 3
= a2
a2 = 3
0 ≤ {x} < 1 and -
≤ {x} - 
0 ≤ 3
For non-integral solution
0 < a2 < 1 and a ∈ (-1, 0) υ (0, 1)
Alternative
-3{x}2 + 2{x} + a2 = 0
Now, -3{x}2 + 2{x}
to have no integral roots 0 < a2 < 1
∴ a ∈ (-1, 0) υ(0, 1)
3 {x}2 - 2{x} - a2 = 0
a ≠ 0, 3
= a2a2 = 3
0 ≤ {x} < 1 and -
≤ {x} - 0 ≤ 3
For non-integral solution
0 < a2 < 1 and a ∈ (-1, 0) υ (0, 1)
Alternative
-3{x}2 + 2{x} + a2 = 0
Now, -3{x}2 + 2{x}
to have no integral roots 0 < a2 < 1
∴ a ∈ (-1, 0) υ(0, 1)
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