If a   ∈  R   and the equation - 3 ( x   -   [ x ] ) 2 + 2 ( x   - &#160…

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. (-2, -1)
  2. $(-\infty, -2) \cup (2, \infty)$
  3. $(-1,0) \cup (0,1)$
  4. (1,2)

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    =-2±4-4-3a2-6

              =-2±4+12a2-6

              =-1±1+3a2-3=1∓1+3a23

As we observe, one root is surely less than zero,

i.e.,   1-1+3a23

∴    For a solution to exist,

       1+1+3a23<1

∴   1+1+3a2<3

∴   1+3a2<2

∴   1+3a2<4

∴   3a2<3

∴   a2<1

∴   a∈-1,0∪0,1.     ∵   a≠0

Asked in: JEE Main 2014 (06 Apr)

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