4. Write the following sets in the set-builder form:
(i) (3, 6, 9, 12}
(ii) {2,4,8,16,32}
(iii) {5, 25, 125, 625}
(iv) {2, 4, 6, . . .}
(v) {1,4,9, . . .,100}
4. Write the following sets in the set-builder form:
(i) (3, 6, 9, 12}
(ii) {2,4,8,16,32}
(iii) {5, 25, 125, 625}
(iv) {2, 4, 6, . . .}
(v) {1,4,9, . . .,100}
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1 Answer
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4. (i) {3,6,9,12}= {3 × 1, 3 × 2, 3 × 3, 3 × 4}
= {x : x = 3n, n is natural number and 1≤ n ≤ 4}
(ii) {2,4,8,16,32}= {21, 22, 23, 24, 25}
= {x : x = 2n, n is natural number and 1 ≤ n ≤ 5}
(iii) {5,25,125,625}= {51, 52, 53, 54}
= {x : x = 5n, n is natural number and 1 ≤ n ≤ 4}
(iv) {2,4,6, }= {2 × 1, 2 × 2, 2 × 3, …}
= {x : x = 2n, n is a natural number}
(v) {1,4,9, …, 100}= {12, 22, 32, …, 102}
= {x : x = n2, x is a natural number and 1 ≤ n ≤ 10}
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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