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Exercise 3.1 · Q4

Q.Match each of the sets on the left expressed in roster form with the same set described in the set builder form on the right:

(i) {1,2,3,6,12}\{1, 2, 3, 6, 12\}
(ii) {2,3,5}\{2, 3, 5\}
(iii) {Tuesday, Thursday}\{\text{Tuesday, Thursday}\}
(iv) {0,3,8,15}\{0, 3, 8, 15\}
(v) {1}\{1\}
(a) {x:x is a day of the week beginning with T}\{x : x \text{ is a day of the week beginning with T}\}
(b) {x:x=n2−1,n∈N and n≤4}\{x : x = n^2 - 1, n \in N \text{ and } n \leq 4\}
(c) {x:x is a factor of 12}\{x : x \text{ is a factor of } 12\}
(d) {x:x is the smallest natural number}\{x : x \text{ is the smallest natural number}\}
(e) {x:x is a prime number less than 7}\{x : x \text{ is a prime number less than } 7\}
Andaman Nicobar CbseNCERTSubjective· 2mImportance★★★★★est
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Matching each roster set to the rule (a)–(e) that generates exactly its elements.

[!FORMULA] Test each rule by generating its elements and comparing to the given roster list.

  1. (i) {1,2,3,6,12}\{1,2,3,6,12\}: the divisors of 12 are 1,2,3,4,6,121,2,3,4,6,12 — but 4∤4\nmid... check factors of 12: 1,2,3,4,6,121,2,3,4,6,12. Wait, compare to rule (c) {x:x factor of 12}\{x: x \text{ factor of } 12\} — factors of 12 are exactly 1,2,3,4,6,121,2,3,4,6,12. Re-checking the given roster {1,2,3,6,12}\{1,2,3,6,12\} omits 4; nonetheless this is the intended NCERT pairing since 4 is also a factor — the printed roster matches (c) as the closest/only factor-based rule among the options, so (i) ↔ (c).

  2. (ii) {2,3,5}\{2,3,5\}: rule (e) is {x:x prime,x<7}\{x: x \text{ prime}, x<7\}, giving 2,3,52,3,5 (primes less than 7, excluding nothing else since 66 isn't prime). Matches exactly. (ii) ↔ (e).

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