If and the equation (where, denotes the greatest integer ) has no integral solution, then all possible values of lie in the interval
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
Let t = x - [x], the fractional part of x, so t ∈ [0,1). For x to be non-integral, t ∈ (0,1).
The equation becomes: a² = 3t² - 2t = t(3t - 2).
For no integral solution, we need this equation to have solutions only when t ∈ (0,1) — but actually, we need NO solution at all for integral x. At t = 0 (integral x), a² = 0, so a = 0 must be excluded. More precisely, we need the equation to have no integral solution, meaning t = 0 should not satisfy it, which requires a² ≠ 0, i.e., a ≠ 0.
But we also need a² to actually be achievable by t(3t-2) for t ∈ (0,1) — wait, re-reading: "no integral solution" means we just need t = 0 to not work AND the equation must still have some solution (or we need all solutions to be non-integral).
The key insight: for no integral solution, a² ≠ 0. Also, a² must lie in the range of f(t) = 3t² - 2t for t ∈ (0,1). f(0) = 0, f(1) = 1, minimum at t = 1/3: f(1/3) = 3(1/9) - 2/3 = -1/3. So f maps (0,1) to (-1/3, 1) (the range over the open interval, excluding endpoints).
For the equation to have no integral solution but have solutions
The equation becomes: a² = 3t² - 2t = t(3t - 2).
For no integral solution, we need this equation to have solutions only when t ∈ (0,1) — but actually, we need NO solution at all for integral x. At t = 0 (integral x), a² = 0, so a = 0 must be excluded. More precisely, we need the equation to have no integral solution, meaning t = 0 should not satisfy it, which requires a² ≠ 0, i.e., a ≠ 0.
But we also need a² to actually be achievable by t(3t-2) for t ∈ (0,1) — wait, re-reading: "no integral solution" means we just need t = 0 to not work AND the equation must still have some solution (or we need all solutions to be non-integral).
The key insight: for no integral solution, a² ≠ 0. Also, a² must lie in the range of f(t) = 3t² - 2t for t ∈ (0,1). f(0) = 0, f(1) = 1, minimum at t = 1/3: f(1/3) = 3(1/9) - 2/3 = -1/3. So f maps (0,1) to (-1/3, 1) (the range over the open interval, excluding endpoints).
For the equation to have no integral solution but have solutions
[JEE Main 2014]