5. List all the elements of the following sets :
(i) A = {x : x is an odd natural number}
(ii) B = {x : x is an integer,
(iii) C = {x : x is an integer, }
(iv) D = {x : x is a letter in the word “LOYAL”}
(v) E = {x : x is a month of a year not having 31 days}
(vi) F = {x : x is a consonant in the English alphabet which precedes k }.
5. List all the elements of the following sets :
(i) A = {x : x is an odd natural number}
(ii) B = {x : x is an integer,
(iii) C = {x : x is an integer, }
(iv) D = {x : x is a letter in the word “LOYAL”}
(v) E = {x : x is a month of a year not having 31 days}
(vi) F = {x : x is a consonant in the English alphabet which precedes k }.
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1 Answer
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5. (i) A = {1, 3, 5, 7, …}
(ii) B = {0,1,2,3,4} (as =4.5 and = –0.5)
(iii) x2 ≤ 4
x2 ≤ 22
x ≤ ± 2
So, C = {–2, –1,0,1,2}
(iv) D = {L, O, Y, A}
(v) E = {February, April, June, September, November}
(vi) F = {b, c, d, f, g, h, j}
Similar Questions for you
(|x| - 3)|x + 4| = 6

(-x - 3) (- (x + 4) = 6
(x + 3) (x + 4) = 6 ⇒ x² + 7x + 12 = 6 ⇒ x² + 7x + 6 = 0
(x + 1) (x + 6) = 0 ⇒ x = -6 (since x < -4)
Case (ii) -4 ≤ x < 0
(-x - 3) (x + 4) = 6
⇒ -x² - 7x - 12 = 6
⇒ x² + 7x + 18 = 0
The discriminant is D = 7² - 4 (1) (18) = 49 - 72 < 0, so no real solution.
Case (iii) x ≥ 0
(x - 3) (x + 4) = 6
⇒ x² + x - 12 = 6
⇒ x² + x - 18 = 0
x = [-1 ± √ (1² - 4 (1) (-18)] / 2 = [-1 ± √73] / 2
Since x ≥ 0 ⇒ x = (√73 - 1) / 2
Only two solutions.
Given n = 2x. 3y. 5z . (i)
On solving we get y = 3, z = 2
So, n = 2x. 33. 52
So that no. of odd divisor = (3 + 1) (2 + 1) = 12
Hence no. of divisors including 1 = 12
Let A = {a, b, c}, B = {1, 2, 3, 4, 5} n (A × B) = 15
x = number of one-one functions from A to B.
y = number of one-one functions for A to (A × B)
2cos
RHS 2
66. Given series is 1× 2× 3 + 2× 3 ×4 + 3× 4 ×5 + … to n term
an = (nth term of A. P. 1, 2, 3, …) ´× (nth terms of A. P. 2, 3, 4) ×
i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1
(nth term of A. P. 3, 4, 5)
i e, a = 3, d = 3 -4 = 1.
= [1 + (n -1) 1] ×[2 + (n -1):1]× [3 + (n- 1) 1]
= (1 + n -1)×(2 + n -1)×(3 + n -1)
= n (n + 1)(n + 2)
= n(n2 + 2n + n + 2)
=n3 + 2n2 + 2n.
Sn = ∑n3 + 3 ∑n2 + 2 ∑n
=
=
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