Let A = { x ∈ R:[x + 3] + [x + 4] ≤ 3} ,
B = { x ∈ R:3ˣ( Σᵣ₌₁^∞ 3/10ʳ )ˣ⁻³ < 3⁻³ˣ }, where [t] denotes greatest integer function. Then,

Solution

Maths sir · 5 steps

Pull the integer out of the floor in A, and turn the series in B into 1/3. Both become x < -1.

1. Simplify set A. Adding an integer passes straight through the floor, so [x+3] = [x] + 3 and [x+4] = [x] + 4. The condition becomes 2[x] + 7 ≤ 3, that is [x] ≤ -2.
2. Read off A. [x] ≤ -2 holds exactly when x < -1, because every x in [-2, -1) still has [x] = -2. So A = (-∞, -1).
3. Sum the series. Σᵣ₌₁^∞ 3/10ʳ = 0.333… = (3/10)/(1 - 1/10) = 1/3. So the bracket in B is 3⁻¹.

   3ˣ · 3⁻⁽ˣ⁻³⁾ < 3⁻³ˣ ⇒ 3³ < 3⁻³ˣ ⇒ 3 < -3x

4. Read off B. The base 3 is greater than 1, so comparing exponents keeps the inequality the same way. 3 < -3x gives x < -1, so B = (-∞, -1).
5. So. Both sets are (-∞, -1), so A = B. The answer is (c).

The idea to keep. Inside a floor, an integer added to x comes out unchanged: [x + n] = [x] + n. And [x] ≤ k means x < k + 1, not x ≤ k.

Answer from the NTA final answer key. Solution by PrepWiser.