Question -
Answer -
(i) {A,P, L, E} ⇔ {x:x is a letter of word “APPLE”}
(ii) {5,-5} ⇔ {x:x2 – 25 =0}
The solution set of x2 –25 = 0 is x = ±5
(iii) {0} ⇔ {x:x+5=5, x ∈ z}
The solution set of x + 5 = 5 is x = 0.
(iv) {1,2, 5, 10} ⇔ {x: x is a natural and divisor of 10}
The natural numbers which are divisor of 10 are 1, 2, 5, 10.
(v) {A,H, J, R, S, T, N} ⇔ {x: x is a letter of the word “RAJASTHAN”}
The distinct letters of word “RAJASTHAN” are A, H, J, R, S, T, N.
(vi) {2,5} ⇔ {x:x is a prime natural number and a divisor of 10}
The prime natural numbers which are divisor of 10 are 2, 5.