Sets Test - 12

Union of the closed sets [ 2 + (1/n), 10 - (1/n)], n = 1, 2,..... is  (2,10)

A and B are two sets satisfying A − B = B − A,

which is possible only, if A = B

If A and B are finite sets, then  n(A - B) = n(A) - n(A ∩ B)

Number of elements in \( A_1 \cup A_2 \cup A_3 … \cup A_{30}\) = 30×5 = 150 (when repetition is not allowed)

But each element is repeated 10 times

\(\therefore n(S) = \frac{30 \times 5}{10} = \frac{150}{10}\)

= 15 .....(i)

Number of elements in \(B_1 \cup B_2 \cup B_3 … \cup B_n \) = 3n (when repetition is not allowed)

But each element is repeated 09 times

\(\therefore(S) = \frac{3n}{9} = \frac{n}{3}\) ......(ii)

From(i) and (ii) we get

\(\frac{n}{3} = 15\)

⇒ 15 x 3 = 45

We have the number of subsets of set containing n elements = \(2^n\)

So, according to the question, we have

\(2^m - 2^n = 112 \)

\(\implies 2^n.(2^{m-n} - 1) = 2^4.7\)

\(\therefore \,2^n = 2^4 \,and \,2^{m-n} - 1 = 7\)

⇒ n = 4 and \(2^{m-n} = 1 + 7 = 8 = 2^3\)

⇒ n = 4 and m – n = 3 ⇒ m – 4 = 3 ⇒ m = 7

We know that: \((A \cap B)' = A' \cup B'\) [De Morgan's law]

\(\therefore (A \cap B')' \cup (B \cap C)\\ = [A' \cup (B')'] \cup (B \cap C)\)

= \((A' \cup B) \cup (B \cap C) [\because (B')' = B]\)

= \(A' \cup B\)

We know that rectangles, rhombus and square in a plane is a parallelogram but trapezium is not a parallelogram.

\(\therefore F_1 = F_2 \cup F_3 \cup F_4 \cup F_1\)

Total number of students = 60 ⇒ n(U) = 60

Number of students who play cricket = 25 ⇒ n(C) = 25

Number of students who play tennis = 20 ⇒ n(T) = 20

Number of students who play cricket and tennis both = 10

\(\implies n(C \cap T) = 10 \\ \because n(C \cup T) = n(C) + n(T) - n(C \cap T) \)

= 25 + 20 – 10 = 45 – 10 = 35

\(\therefore n(C' \cap T') = n(U) - n(C \cup T)\)

= 60 – 35 = 25

Total number of persons in a town = 840 ⇒ n(U) = 840

Number of persons who read Hindi = 450 ⇒ n(H) = 450

Number of persons who read English = 300 ⇒ n(E) = 300

Number of persons who read both = 200 ⇒ n(H ∩ E) = 200

\(\because\) n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

= 450 + 300 – 200 = 550

n(H'∩ E') = n(U) – n(H ∪E)

= 840 – 550 = 290

Given that: X = {\(8^n\) – 7n – 1 |n ∈ N} = {0, 49, 490, …}

And Y = {49n – 49 | n ∈ N} = {0, 49, 98, …}

Here, it is clear that every element belonging to X is also present in Y.

\(\therefore\) X ⊂ Y

Let p% of the people watch a channel and q% of the people watch another channel

\(\because\) n(p \(\cap\) q) = x% and n(p \(\cup\) q) ≤ 100

So, n(p \(\cup\) q) = n(p) + n(q) – n(p \(\cap\) q)

100 = 63 + 76 – x

100 = 139 – x

⇒ x = 139 – 100

⇒ x = 39

Now n(p) = 63

\(\therefore\) n(p \(\cap\) q) ≤ n(p)

⇒ x ≤ 63

So 39 ≤ x ≤ 63.

Given that:

\(A = \{(x, y | y = \frac{1}{x}, 0 \neq x \in R \}\)

and \(B = \{(x, y |y = x, x \in R \}\)

It is very clear that \(y = \frac{1}{x} \) and y = -x

\(\because \frac{1}{x} \neq -x\)

\(\therefore A \cap B \) = φ

Given that: \(A \cap (A \cap B)\)

Let x ∈ A \(\cap\) (A \(\cup\) B)

⇒ x ∈ A and x ∈ (A \(\cup\) B)

⇒ x ∈ A and (x ∈ A or x ∈ B)

⇒ (x ∈ A and x ∈ A) or (x ∈ A and x ∈ B)

⇒ x ∈ A or x ∈ (A  \(\cap\) B)

⇒ x ∈ A