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…
JEE Main 2014 — Mathematics Algebra
2014mcqhard
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
Official previous-year question
Held on 6 Apr 2014 · Verified 6 Jul 2026.
Options
A
(−2,−1)
B
(−∞,−2)∪(2,∞)
C
(−1, 0)∪(0, 1)
D
(1,2)
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Solution
Let x−[x]=t
∴−3t2+2t+a2=0
As x is not an integer, t=0∴a=0
For a root of x to exist at least one of the roots of t should be between 0 and 1.
Roots =−6−2±4−(4)(−3)a2
=−6−2±4+12a2
=−3−1±1+3a2=31∓1+3a2
As we observe, one root is surely less than zero,
i.e., 31−1+3a2
∴ For a solution to exist,
31+1+3a2<1
∴1+1+3a2<3
∴1+3a2<2
∴1+3a2<4
∴3a2<3
∴a2<1
∴a∈(−1,0)∪(0,1). [∵a=0]
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