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Let A = {z1: z1³⁰ = 1}, B = {z2: z2⁴² = 1}, and C = {z3: 1 + z3 + z3² +... + z3⁶⁹ = 0}. Match the following: List-I: (P) n(A ∩ B), (Q) n(B ∩ C), (R) n(C ∩ A), (S) n(A ∩ B ∩ C) List-II: (1) 1, (2) 6, (3) 9, (4) 13, (5) 15

1. P -> 2; Q -> 4; R -> 3; S -> 1
2. P -> 1; Q -> 3; R -> 4; S -> 2
3. P -> 4; Q -> 5; R -> 2; S -> 3
4. P -> 5; Q -> 3; R -> 1; S -> 4

Correct answer: P -> 2; Q -> 4; R -> 3; S -> 1

Solution

A = 30th roots of unity (|A|=30). B = 42nd roots of unity (|B|=42). C: 1+z+z²+...+z⁶⁹ = 0 => (z⁷⁰-1)/(z-1) = 0 => z⁷⁰=1 and z≠1. So C = {70th roots of unity} {1}, |C|=69. P: n(A∩B) = gcd(30,42) = 6. Answer: 6 => List-II value 2 (which is 6). Q: n(B∩C) = 42nd roots ∩ 70th roots that are not 1 = gcd(42,70)th roots minus {1}. gcd(42,70)=14. So B∩C = 14th roots of unity {1}, |B∩C| = 13. Answer: 13 => List-II value 4. R: n(C∩A) = 30th roots ∩ 70th roots, excluding 1. gcd(30,70)=10. C∩A = 10th roots of unity {1}, |C∩A| = 9. Answer: 9 => List-II value 3. S: n(A∩B∩C) = gcd(30,42,70)th roots {1}. gcd(30,42)=6, gcd(6,70)=2. So A∩B∩C = 2nd roots of unity {1} = {-1}, |A∩B∩C| = 1. Answer: 1 => List-II value 1. Match: P->2(6), Q->4(13), R->3(9), S->1(1).

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