Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions
7. Which of the following are examples of the null set
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) {x : x is a natural numbers, x < 5 and x > 7}
(iv) {y : y is a point common to any two parallel lines}
Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions
7. Which of the following are examples of the null set
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) {x : x is a natural numbers, x < 5 and x > 7}
(iv) {y : y is a point common to any two parallel lines}
1. (i) There is no odd number that can be divided by 2.
?The given set is a null set.
(ii) An even prime number is 2. Hence, the set will have 2 as element.
? The given set is not a null set.
(iii) As there is no natural number which is both less than 5 and greater than 7.
? The given set is a null set.
(
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
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)
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
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