Set Theory Test 23

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Set Theory Test 23

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Weekly Quiz Competition

Question 1
1 / -0

If two sets $$A$$ and $$B$$ are having $$99$$ elements in common, then the number of ordered pairs common to each of the sets $$AxB$$ and $$BxA$$ are

A
$$2^{99}$$

B
$$99^{2}$$

C
$$100$$

D
$$18$$

Solution

Let us the $$n\left( {\left( {A \times B} \right) \cap \left( {B \times A} \right)} \right)$$
$$ = n\left( {\left( {A \cap B} \right) \times \left( {B \cap A} \right)} \right)$$
$$ = n\left( {A \cap B} \right).\,n\left( {B \cap A} \right)$$
$$ = n\left( {A \cap B} \right).\,n\,\left( {A \cap B} \right)$$
$$ = \left( {99} \right)\left( {99} \right)$$
$$ = {99^2}$$
Question 2
1 / -0

If $$X =\{1,2,3,4,5,6,7,8,9, 10\}$$ is the universal set and$$ A= \{1, 2, 3,4\}, B= \{2,4,6,8\}, C= \{3,4,5,6\}$$ verify the following.
(a) $$A \cup (B\cup C) = (A \cup B) \cup C$$
(b)$$A \cap (B\cup C) = (A \cap B) \cup (A \cap C)$$
(c) $$(A')' =A$$

A
Only a is true

B
Only b and c are true

C
Only a and b are true

D
All three a,b and c are true.

Solution

given
$$X=$${$$1,2,3,4,5,6,7,8,9,10$$},$$A=\{1,2,3,4\},B=\{2,4,6,8\},C=\{3,4,5,6\}$$

now from the given data

$$A\cup B=$$$$\{1,2,3,4,6,8\}$$,$$B\cup C=$$$$\{2,3,4,5,6,8\}$$

Now $$A\cup (B\cup C)=\{1,2,3,4,5,6,8\}\quad and\quad(A\cup B)\cup C=$$$$\{1,2,3,4,5,6,8\}$$

$$\therefore A\cup (B\cup C)=(A\cup B)\cup C=\{1,2,3,4,5,6,8\}$$

$$A\cap B=$$$$\{2,4\}$$,$$A\cap C=$$$$\{3,4\}$$,$$B\cup C=$$$$\{2,3,4,5,6,8\}$$

Now $$A\cap(B\cup C)=\{2,3,4\}\quad and \quad(A\cap B)\cup (A\cap C)=$$$$\{2,3,4\}$$

$$\therefore A\cap(B\cup C)=(A\cap B)\cup (A\cap C)=$$$$\{2,3,4\}$$

$$A'=$$$$\{5,6,7,8,9,10\}$$

therefore $$(A')'=$$$$\{1,2,3,4\}$$$$=A$$

therefore all three a,b,c are true
Question 3
1 / -0

$$S_1:(p\Rightarrow q) V ( q \Rightarrow p )$$ is a tautology.
$$S_2: ((p\Rightarrow q) V ( q \Rightarrow p))$$ is a fallacy

A
$$S_1$$ is true, $$S_2$$ is false

B
$$S_1$$ false

C
$$S_1$$ is false, $$S_2$$ is false

D
$$S_1$$ is true

Solution
Question 4
1 / -0

$$If\,A = \left\{ {\left( {x,y} \right)\,\left| {{x^2} + {y^2} \le \left. 4 \right\}\,and} \right.} \right.$$
$$B = \left\{ {\left( {x,y} \right)\,\left| {{{(x - 3)}^2} + {y^2}} \right. \le \left. 4 \right\}} \right.\,and\,the$$
$$po{\mathop{\rm int}} \,P\left( {a,\frac{1}{2}} \right)\,belongs\,to\,the\,set\,B - A$$ then the set of possible real values of $$a$$ is

A
$$\left( {\frac{{1 + \sqrt {3} }}{4},\frac{{7 + \sqrt 7 }}{4}} \right)$$

B
$$\left( {\frac{{7 - \sqrt 7 }}{4},\frac{{1 + \sqrt 7 }}{4}} \right)$$

C
$$\left( {\frac{{1 - \sqrt {31} }}{4},\frac{{7 - \sqrt 7 }}{4}} \right)$$

D
none of these

Solution

$${x^2} + {y^2} > 4$$
$${a^2} + {\left( {a - \frac{1}{2}} \right)^2} > 4$$
$$2{a^2} - a - \frac{5}{4} > 0$$
$$a = \frac{{1 \pm \sqrt {31} }}{4}$$
$$a \in \left( { - \infty ,\frac{{1 - \sqrt {31} }}{4}} \right)\,V\left( {\frac{{1 + \sqrt {31} }}{4},\infty } \right)$$
$$a \in \left( {\frac{{1 + \sqrt 3 }}{4},\frac{{7 + \sqrt 7 }}{4}} \right)$$
Question 5
1 / -0

In a selection process, a hundred candidate participate in Group Discussion sessions (GD) and Personal Interview (PI). The possibilities of a candidate's good performance in GD and in PI are independent of each other. It was found that $$20$$ candidates were good in GD and $$30$$ were good in PI. The number of candidates good in both GD and PI is expected to be about:

A
$$6$$

B
$$10$$

C
$$20$$

D
$$30$$

Solution

$$P(GD and PI)\\=P(GD)\times P(PI)\\=(\frac{20}{100})\times(\frac{30}{100})\\=(\frac{6}{100})\\\therefore no. of candidates expected\\=P(GD and PI)\times100\\=6$$
So, the correct option is '6'.
Question 6
1 / -0

If $$A \subset B$$, then

A
$$C - B \subset C - A$$.

B
$$A \cup B = A$$

C
$$A \cap B = B$$

D
None

Solution
Question 7
1 / -0

The set of all $$x$$ for which $$1 + \log x < x$$ is

A
$$(1, \infty)$$

B
$$(0, 1)$$

C
$$(0, \infty)$$

D
None of these

Solution

$$1+\log{x}<x$$
$$\log{x}<x-1$$
$$x<{e}^{x-1}$$
$$\therefore x\in (1,\infty).$$
Question 8
1 / -0

$$s = \{ x \in N:2 + {\log _2}\sqrt {x + 1} > 1 - {\log _{1/2}}\sqrt {4 - {x^2}} \} $$ , then

A
$$S={ 1 }$$

B
$$S=Z$$

C
$$S=N$$

D
none of these

Solution
Question 9
1 / -0

If $$\alpha,\xi,\eta$$ are non-empty sets then:

A
$$(\alpha \times \beta)\cup (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$$

B
$$(\alpha \times \beta)\cap (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$$

C
$$(\alpha \cap \beta)\cup (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$$

D
$$(\alpha \cap \beta)= (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$$

Solution
Question 10
1 / -0

For any two sets A and B, A' - B' is equal to

A
A -B

B
B - A

C
A - A'

D
A - B'

Solution

$$A' -B' = B-A$$
since $$ -X' =X \ and\ X'=-X$$

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Re-Attempt Weekly Quiz Competition

Set Theory Test 23

Result

Set Theory Test 23

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SHARING IS CARING

Question 1

$$2^{99}$$

$$99^{2}$$

$$100$$

$$18$$

Solution

Question 2

Only a is true

Only b and c are true

Only a and b are true

All three a,b and c are true.

Solution

Question 3

$$S_1$$ is true, $$S_2$$ is false

$$S_1$$ false

$$S_1$$ is false, $$S_2$$ is false

$$S_1$$ is true

Solution

Question 4

$$\left( {\frac{{1 + \sqrt {3} }}{4},\frac{{7 + \sqrt 7 }}{4}} \right)$$

$$\left( {\frac{{7 - \sqrt 7 }}{4},\frac{{1 + \sqrt 7 }}{4}} \right)$$

$$\left( {\frac{{1 - \sqrt {31} }}{4},\frac{{7 - \sqrt 7 }}{4}} \right)$$

none of these

Solution

Question 5

$$6$$

$$10$$

$$20$$

$$30$$

Solution

Question 6

$$C - B \subset C - A$$.

$$A \cup B = A$$

$$A \cap B = B$$

None

Solution

Question 7

$$(1, \infty)$$

$$(0, 1)$$

$$(0, \infty)$$

None of these

Solution

Question 8

$$S={ 1 }$$

$$S=Z$$

$$S=N$$

none of these

Solution

Question 9

$$(\alpha \times \beta)\cup (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$$

$$(\alpha \times \beta)\cap (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$$

$$(\alpha \cap \beta)\cup (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$$

$$(\alpha \cap \beta)= (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$$

Solution

Question 10

A -B

B - A

A - A'

A - B'

Solution

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