Which of the following set s is null set? A = x : x ∈ R , x < 5 and x = 25 B = x : x ∈ R , x 2 < 0 C = x : x = 2 x - 3 , x ∈ R

Given, x = 25 & x < 5 but 25 is not less than 5 , hence, A = ϕ Given, x 2 < 0 but we know that the square of any number is always positive and greater than or equal to zero, hence, B = ϕ Given, C = x : x = 2 x - 3 , x ∈ R . Hence, it will be an infinite set by putting the value of x ∈ R . So, A and B are null set.

If A = 1 , 2 , 3 , 4 , 5 , B = 1 , 3 , 5 , 7 , 9 and C = 2 , 4 , 8 , 10 then verify that A ∩ B ∪ C = A ∩ B ∪ A ∩ C .

A = 1 , 2 , 3 , 4 , 5 , B = 1 , 3 , 5 , 7 , 9 , C = 2 , 4 , 8 , 10 L . H . S = A ∩ B ∪ C = A ∩ 1 , 2 , 3 , 4 , 5 , 7 , 8 , 9 , 10 = 1 , 2 , 3 , 4 , 5 = A R . H . S . = A ∩ B ∪ A ∩ C = 1 , 3 , 5 ∪ 2 , 4 = 1 , 2 , 3 , 4 , 5 = A = L . H . S . Hence verified.

Draw the Venn diagram of A ∩ B ' .

A ∩ B ' is read as the compliment of A ∩ B i.e. the elements except A ∩ B The shaded area represents A ∩ B ' .

If n A ∪ B = 30 , n A = 20 , n B = 12 , then find n A ∩ B .

We know that, n A ∪ B = n A + n B - n A ∩ B ⇒ 30 = 20 + 12 - n A ∩ B ⇒ n A ∩ B = 32 - 30 ∴ n A ∩ B = 2

Write the following intervals in set builder form. (i) - ∞ , 6 (ii) - 9 , 4

(i) x is such that x is always less than six, hence, x : x ∈ R , x < 6 (ii) x is such that x lies between - 9 & 4 , including - 9 , hence, x : x ∈ R , - 9 ≤ x < 4

If n U = 35 , n A ' = 10 , n B ' = 15 and n A ∩ B = 15 then find n A ∪ B

n U = 35 , n A ' = 10 n B ' = 15 , n A ∩ B = 15 n A ' = n U - n A ⇒ 10 = 35 - n A ∴ n A = 25 n B ' = n U − n B ⇒ 15 = 35 − n B ∴ n B = 20 So, n A ∪ B = n A + n B - n A ∩ B ⇒ n A ∪ B = 25 + 20 - 15 ∴ n A ∪ B = 30