If A = x ∈ R : |x - 2| > 1,
B = x ∈ R : √(x² - 3) > 1,
C = x ∈ R : |x - 4| ≥ 2 and Z is the set of all integers, then the number of subsets of the
set (A ∩ B ∩ C)ᶜ ∩ Z is ________________.

Solution

Maths sir · 4 steps

Write each set as intervals, intersect them, then flip to the complement and count the integers.

1. Solve each set. |x-2| > 1 gives A = (-∞, 1) ∪ (3, ∞). √(x²-3) > 1 means x² - 3 > 1, so x² > 4 and B = (-∞, -2) ∪ (2, ∞). |x-4| ≥ 2 gives C = (-∞, 2] ∪ [6, ∞).
2. Intersect. A ∩ B = (-∞, -2) ∪ (3, ∞). Intersecting with C keeps only x < -2 or x ≥ 6, so A ∩ B ∩ C = (-∞, -2) ∪ [6, ∞).
3. Take the complement. The complement is everything left over: [-2, 6). Here -2 is included because it is not in B, since √1 = 1 is not greater than 1. And 6 is excluded because it lies in A, B and C, so it is in the intersection.

   (A ∩ B ∩ C)ᶜ ∩ ℤ = {-2, -1, 0, 1, 2, 3, 4, 5}

4. So. This set has 8 elements, so it has 2⁸ = 256 subsets.

The common slip: 512. Counting 6 in the complement gives 9 integers and 2⁹ = 512. But |6-4| = 2 satisfies ≥ 2, so 6 is in C and in the intersection.

The idea to keep. For set questions built on inequalities, draw every set on one number line and watch each endpoint: whether it is open or closed decides whether it lands in the complement.

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