JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
Let [t] denote the greatest integer ≤ t. Then the equation in x,
[x]² + 2[x+2] - 7 = 0 has :
- (a) no integral solution.
- (b) exactly two solutions.
- (c) exactly four integral solutions.
- (d) infinitely many solutions.
Answer and solution
Answer (d) infinitely many solutions.
Solution
Maths sir · 4 steps
Pull the 2 out of [x+2] first, then it is just a quadratic in [x].
- Shift the integer out. Adding an integer to x adds the same integer to its greatest integer, so [x+2] = [x] + 2. Put n = [x], where n must be an integer.
n² + 2(n+2) - 7 = 0 ⇒ n² + 2n - 3 = 0 ⇒ (n+3)(n-1) = 0
- Values of [x]. So [x] = 1 or [x] = -3. Both are integers, so both are allowed.
- Back to x. [x] = 1 means 1 ≤ x < 2, and [x] = -3 means -3 ≤ x < -2. Every real number in these two intervals satisfies the equation.
- So. The solution set is [-3, -2) ∪ [1, 2), which holds infinitely many values of x. The answer is (d).
The idea to keep. When you solve for [x], each integer value you get is a whole interval of x, not a single point. Always convert [x] = n back to n ≤ x < n+1.
Answer from the NTA final answer key. Solution by PrepWiser.