SET THEORY QUESTIONS FOR GRADE 11

Question 1 :

Justify the trueness of the statement:

“An element of a set can never be a subset of itself.”

Solution :

The given statement is wrong. Because every element is the subset of itself.

Let us look into the example problem

A = {1, 2, 3}

Subset of A  =  {{ } {a}, {b}, {c}, {a, b}, {b, c}, {c, a}, {a, b, c} }

Every element is the part of itself. Hence the given statement is wrong.

Question 2 :

If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B).

Solution :

P(A) means power set of A 

power set of A,  n[P (A)]  =  2^n(A) 

n(P(A)) = 1024 ==> 2¹⁰

n(A)  =  10

n(P(B)) = 32  ==>  2⁵

n(B)  =  5

n(AnB)  =  n(A) + n(B) - n(AUB)

=  10 + 5 - 15

n(A n B)  =  0

Question 3 :

If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(AΔB)).

Solution :

(AΔB)  =  (AUB) - (AnB)

n(AΔB)  =  n(AUB) - n(AnB)

n(AΔB)  =  10 - 3  ==>  7

n(P(AΔB))  =  2 ^n(AΔB)

  =  2 ⁷ 

  =  128

Hence the answer is 128.

Question 4 :

For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.

Solution :

From the given question, we come to know that set A will have 4 elements.

By combining the elements of A and A, we get the elements of A × A. From the given element, we have four different values.

So the elements of A are {0, 1, 2, 3}.

Question 5 :

Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.

Solution :

We get elements of A × B by combining the elements of set A to set B.

In (x, 1), (y, 2), (z, 1), the first terms will be the elements of set A and second terms will be the elements of set B.

Hence A = {x, y, z} and B = {1, 2}.

Question 6 :

If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (−1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.

Solution :

From the given two set of elements, we may list out the elements of A.

A = {-1, 0, 1, 2}

A x A  =  {(-1, -1) (-1, 0) (-1, 1) (-1, 2) (0, -1) (0, 0) (0, 1) (0, 2) (1, -1) (1, 0) (1, 1) (1, 2)(2, -1) (2, 0) (2, 1) (2, 2)}

Hence the remaining elements are {(-1, -1) (-1, 0) (-1, 1) (0, -1) (0, 0) (0, 2) (1, -1) (1, 0) (1, 1) (1, 2)(2, -1) (2, 0) (2, 1) (2, 2)}

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