JEE Main 2023 · 24 January · evening · Single correct · +4 / −1
The set of all values of a for which lim _(x → a) ([x - 5] - [2x + 2]) = 0, where [α] denotes the greatest integer less than or equal to α is equal to
- (a) [-7.5,-6.5]
- (b) (-7.5,-6.5]
- (c) [-7.5,-6.5)
- (d) (-7.5,-6.5)
Answer and solution
Answer (d) (-7.5,-6.5)
Solution
Maths sir · 4 steps
Pull integers out of the brackets, find where it is 0, then make both sides of a stay there.
- Pull out integers. Integers come out of the bracket: [x-5] = [x]-5 and [2x+2] = [2x]+2. So the expression becomes [x]-[2x]-7, and we need [x]-[2x] = 7 for all x near a.
- Split by fraction. Write x = n + f with n an integer and 0 ≤ f < 1. If f < 1/2, then [2x] = 2n and [x]-[2x] = -n, so n = -7. If f ≥ 1/2, then [2x] = 2n+1 and [x]-[2x] = -n-1, so n = -8.
Interval of x [x] [2x] [x]-[2x]-7 [-8,-7.5) -8 -16 1 [-7.5,-7) -8 -15 0 [-7,-6.5) -7 -14 0 [-6.5,-6) -7 -13 -1 - Check the endpoints. The expression is 0 exactly on [-7.5,-6.5). At a = -7.5 the left-hand limit is 1, and at a = -6.5 the right-hand limit is -1. So the limit is not 0 at either endpoint.
- So. The limit equals 0 only when a is strictly inside, that is a ∈ (-7.5,-6.5), which is option (d).
The idea to keep. A limit needs the function to behave the same on both sides of a. Find the set where the expression has the value, then drop any endpoint where one side jumps away.
Answer from the NTA final answer key. Solution by PrepWiser.