MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 11

Number of functions from Set-A containing

$5$ elements to a set-

$B$ containing

$4$ elements is

0%
${5}^{4}$
0%
${4}^{5}$
0%
$4!$
0%
$5!$

For any set

$A$ , if

$A\subseteq \phi \Leftrightarrow A=\phi$ .

0%
True
0%
False

Explanation

True

Possible sunsets of

$\phi={\phi}$

$A\subseteq \phi$

$\rightarrow A=\phi$

$A=\left\{ 1,2,4 \right\} ,B=\left\{ 2,4,5 \right\}$ and

$C=\left\{ 2,5 \right\}$ , then

$\left( A-B \right) \times \left( B-C \right) =$

0%
$\left\{ \left( 1,2 \right) ,\left( 1,5 \right) ,\left( 2,5 \right) \right\}$
0%
$\left\{ \left\{ 1,4 \right\} \right\}$
0%
$\left( 1,4 \right)$
0%
$\left\{ \left( 1,2 \right) \right\}$

Explanation

$A=\left\{1,2,4\right\}$ and

$B=\left\{2,4,5\right\}$

$A-B=\left\{1,2,4\right\}-\left\{2,4,5\right\}=\left\{1\right\}$

$B=\left\{2,4,5\right\}$ and

$C=\left\{2,5\right\}$

$B-C=\left\{2,4,5\right\}-\left\{2,5\right\}=\left\{4\right\}$

$\left(A-B\right)\times\left(B-C\right)=\left\{1\right\}\times \left\{4\right\}=\left(1,4\right)$

$A=\left\{x:x^{2}=1\right\}$ and

$B=\left\{x:x^{4}=1\right\}$ , then

$A \Delta B$ is equal to

0%
$\left\{i,-i\right\}$
0%
$\left\{-1-1\right\}$
0%
$\left\{1-1,i,-i\right\}$
0%
$\left\{1,i\right\}$

Explanation

Given:

$A=\{x:x^2=1 \}$ and

$B=\{x:x^4=1 \}$

Consider,

$A=\{x:x^2=1 \}$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm1$

So,

$A=\{-1,1 \}$

and

$B=\{x:x^4=1 \}$

$\Rightarrow x^2=\pm1$

$\Rightarrow x^2=1$ or

$x^2=-1$

$\Rightarrow x=\pm1$ or

$x=\pm-i$

So,

$B=\{-1,1,-i,i \}$

Now

$A\Delta B=(A-B)\cup(B-A)=\phi\cup\{i,-i \}$

So,

$\text{A}$ is correct option.

{

$x \epsilon R : \dfrac{14x}{x+1} - \dfrac{9x-30}{x-4} <0$ } is equal to

0%
$(-1, 4)$
0%
$(1, 4) \cup (5, 7)$
0%
$(1, 7)$
0%
$(-1, 1) \cup (4, 6)$

Explanation

Given,

$\dfrac{14x}{x+1}-\dfrac{9x-30}{x-4}<\:0$

$\dfrac{5x^2-35x+30}{\left(x+1\right)\left(x-4\right)}<0$

$\dfrac{(x-1)(x-6)}{\left(x+1\right)\left(x-4\right)}<0$

For

$(x-1),(x-6)$ and

$(x+1),(x-4)$ we get,

$-1<x<1\quad \mathrm{or}\quad \:4<x<6$

$\begin{bmatrix}\mathrm{Solution:}\:&\:-1<x<1\quad \mathrm{or}\quad \:4<x<6\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-1,\:1\right)\cup \left(4,\:6\right)\end{bmatrix}$

$n(A \cap B)=2x$ . If

$n(A)=2(n(B))$ then

$'x'$ is

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Let A and B be two sets such that

$\\ A\times B$ has 6 elements. If three elements of

$\\ A\times B$ are

$\\ \left\{ \left( 1,4 \right) ,\left( 2,6 \right) ,\left( 3,6 \right) \right\}$ , then

0%
$A=\left\{ 1,2 \right\}$ and $B=\left\{ 3,4,6 \right\}$
0%
$A=\left\{ 4,6 \right\}$ and $B=\left\{ 1,2,3 \right\}$
0%
$A=\left\{ 1,2,3 \right\}$ and $B=\left\{ 4,6 \right\}$
0%
$A=\left\{ 1,2,4 \right\}$ and $B=\left\{ 3,6 \right\}$

$a * b = 2 ^ { a b }$ on

$N \cup \{ 0 \}$ then

$*$ is

0%
commutative
0%
associate
0%
$(a) \& (b)$ both
0%
none of these

$A = \{ 2,3,4,5,7 \} , B = \{ 7,8,9 \}$ , then find

$n ( A \cup B ).$

0%
$1$
0%
$3$
0%
$5$
0%
$7$

Explanation

$A=\left \{ 2,3,4,5,7 \right \}$

$n(A)=5$

$B=\left \{ 7,8,9 \right \}$

$n(B)= 3$

$n(A \cap B)=1$

$\therefore (A\cup B)=n(A)+n(B)-n(A\cap B)$

$(A\cup B)=5+3-1=7$

$I$ is the set of isosceles triangle and

$E$ is the equilateral triangles then _____________.

0%
$I\subset E$
0%
$E \subset I$
0%
$E=I$
0%
None of these

Explanation

Given,

$I$ is the set of isosceles triangle and

$E$ is the equilateral triangles.

We know that every equilateral triangle is an isosceles triangle but the converse is not true.

Hence

$E\subset I$ .

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

$A=\left\{x|x\in N\quad and\quad x < 6\dfrac{1}{4}\right\}$ and

$B=\left\{x|x\in N\quad and\quad x^2\leq 5\right\}$ . Then the number of subsets of set

$A\times (A\cap B)$ which contains exactly

$3$ elements is?

0%
$126$
0%
$220$
0%
$280$
0%
$144$

Let

$A, B$ and

$C$ be sets such that

$\phi = A\cap B \subseteq C$ . Then which of the following statements is not true?

0%
If $(A - C) \subseteq B$ , then $A\subseteq B$
0%
$(C \cup A)\cap (C\cup B) = C$
0%
If $(A - B)\subseteq C$ , then $A\subseteq C$
0%
$B\cap C \neq \phi$

Explanation

for

$A = C, A - C = \phi$

$\Rightarrow \phi \subseteq B$
But

$A\not {\subseteq} B$

$\Rightarrow$ option

$1$ is NOT true
Let

$x\epsilon (C \cup A)\cap (C\cup B)$

$\Rightarrow x\epsilon (C\cup A)$ and

$x \epsilon (C\cup B)$

$\Rightarrow (x \epsilon C$ or

$x \epsilon A)$ and

$(x\epsilon C$ or

$x \epsilon B)$

$\Rightarrow x \epsilon C$ or

$x \epsilon (A\cap B)$

$\Rightarrow x \epsilon C$ or

$x\epsilon C$ (as

$A\cup B\subseteq C)$

$\Rightarrow x \epsilon C$

$\Rightarrow (C \cup A)\cap (C\cup B)\subseteq C....(1)$
Now

$x \epsilon C\Rightarrow x \epsilon (C\cup A)$ and

$x \epsilon (C \cup B)$

$\Rightarrow x\epsilon (C\cup A)\cap (C\cup B)$

$\Rightarrow C\subseteq (C\cup A)\cap (C \cup B) .....(2)$

$\Rightarrow$ from (1) and (2)

$C = (C\cup A)\cap (C\cup B)$

$\Rightarrow$ option 2 is true
Let

$x \epsilon A$ and

$x \not {\epsilon} B$

$\Rightarrow x \epsilon (A - B)$

$\Rightarrow x \epsilon C$ (as

$A - B \subseteq C)$
Let

$x \epsilon A$ and

$x \epsilon B$

$\Rightarrow x \epsilon (A\cap B)$

$\Rightarrow x \epsilon C$ (as

$A\cap B\subseteq C)$
Hence

$x \epsilon A \Rightarrow x \epsilon C$

$\Rightarrow A \subseteq C$

$\Rightarrow$ Option 3 is true
as

$C\supseteq (A\cap B)$

$\Rightarrow B\cap C\supseteq (A\cap B)$
as

$A\cap B\neq \phi$

$\Rightarrow B\cap C \neq \phi$

$\Rightarrow$ Option 4 is true.
1247200_1615075_ans_005ea1b2d1a44e42b1b0826402402ee1.png

State whether the following statement is true or false. Give reason to support your answer.
Every subset of a finite set is finite.

0%
True
0%
False

$u=\{2, 3, 5, 7, 9\}$ is the universal set and

$A=\{3, 7\}, B=\{2, 5, 7, 9\}$ , then find the following statement is true/false.

$(A\cap B)'=A'\cap B'$ .

0%
True
0%
False

Explanation

The given sets are:

$u=\left\{2,3,5,7,9 \right\}\ A=\left\{3,7 \right\}\ B=\left\{2,5,7,9 \right\}$

$A\cap B=\left\{7 \right\}$

$(A\cap B)'=u-(A\cap B)=\left\{2,3,5,9 \right\}$

$A'=u-A=\left\{2,5,9 \right\}$

$B'=u-B=\left\{3 \right\}$

$A' \cap B'=\phi$

$\Rightarrow \ \boxed {(A\cap B)' \neq A' \cap B'}$

Mark the correct alternative of the following.
The number of subsets of a set containing n elements is?

0%
n
0%
$2^n-1$
0%
$n^2$
0%
$2^n$

State whether the following statement is true or false. If the statement is false, re-write the given statement correctly.
If

$A=\{1, 2\}, B=\{3, 4\}$ , then

$A\times (B\cap \phi)=\phi$ .

0%
True
0%
False

State whether the following statement is true or false. If the statement is false, re-write the given statement correctly.
If

$P=\{m, n\}$ and

$Q=\{n, m\}$ , then

$P\times Q=\{(m, n), (n, m)\}$ .

0%
True
0%
False

Examine whether the following statements are true or false:

$a\in \left\{\{a\}\, b \right\}$

0%
True
0%
False

Examine whether the following statements are true or false:

$(a,e) \subset$ (

$x : x$ is a vowel in the English alphabet)

0%
True
0%
False

Explanation

True,

$a,e$ are two vowels of the English alphabet.

Examine whether the following statements are true or false:
(

$x : x$ is an even natural number less than

$6$ )

$\subset$ (

$x: x$ is a natural number which divide

$36$ .

0%
True
0%
False

Explanation

True. (

$x : x$ is an even natural number less than

$6$ )=

$(2,4)$
(

$x : x$ is a natural number which divides

$36$ )=

$(1,2,3,4,6,9,12,18,36)$ .

Examine whether the following statements are true or false:

$(a)\subset (a,b,c)$

0%
True
0%
False

Explanation

True. Each element of

$(a)$ is also an element of

$(a, b, c)$ .

Examine whether the following statements are true or false:

$(a,b)\not{\subset}(b,c,a)$

0%
True
0%
False

Explanation

False. Each element of

$(a,b)$ is also an element of

$(b,c,a)$ .

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
If

$x A$ and

$A B$ , then

$x B$

0%
True
0%
False

Explanation

False
Let

$A = \left \{ 1,2 \right \}$ and

$B = \left \{ 1,\left \{ 1,2 \right \} ,\left \{ 3 \right \}\right \}$
Now,

$2 ∈ \left \{ 1,2 \right \}$ and

$\left \{ 1,\left \{ 1,2 \right \} ,\left \{ 3 \right \}\right \}$

$\therefore A ∈ B$
However,

$2 ∉ \left \{ \left \{ 3 \right \},1,\left \{ 1,2 \right \} \right \}$ .

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
If

$A B$ and

$B C$ , then

$A C$

0%
True
0%
False

Explanation

False
Let

$A=\left \{ 1,2 \right \},B=\left \{ 0,6,8 \right \}$ and

$C=\left \{ 0,1,2,6,9 \right \}$
Accordingly

$A ⊄ B$ and

$B ⊄ C$
However,

$A ⊂ C$ .

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
If

$A B$ and

$x B$ , then

$x A$

0%
True
0%
False

Explanation

True
Let,

$A ⊂ B$ and

$x ∉ B$
To show:

$x \notin A$ ,
If possible, suppose

$x ∈ A$ ,
Then,

$x\in B$ , which is a contradiction as

$x ∉ B$

$\therefore x ∉ A$

Choose the correct answer from the given four options
Which of the following collection doesn't form a set

0%
Collection of 5 odd prime numbers
0%
Collection of 3 most intelligent students of your class
0%
Collection of 4 vowels of the English alphabet
0%
Collection of first 6 months of a year

State whether the following statements are true(T) or false(F):
A collection of books is a set.

0%
True
0%
False

Choose the correct answer from the given four options
If A = {x | x is a positive multiple of 3 less than 20} and B = {x | x is a prime number less than 20}, then n(A) + n(B) is

0%
$6$
0%
$8$
0%
$13$
0%
$14$

Explanation

$A = \{x | x$ is a positive multiple of

$3$ less than

$20\}$

$=\left\{3,6,9,12,15,18\right\}\Rightarrow n(A)=6$

$B = \{x | x$ is a prime number less than

$20\}$

$=\left\{2,3,5,7,11,13,17,19\right\} \Rightarrow n(B)=8$

$n(A) + n(B) = 6+8=14$

State true or false for each of the following. Correct the wrong statement If

$A = \{0\}$ , then

$n (A) = 0$

0%
True
0%
False

Explanation

Given
If

$A = \{0\}$ , then

$n (A) = 0$
The statement given here is false
Correct statement: If

$A = \{0\}$ , then

$n (A) = 1$

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 11 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 11 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.