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

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

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

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

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

MCQ Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 9

If sets

$A$ and

$B$ are define as

$A=\left\{ \left( x,y \right) :y={ e }^{ x },x\in R \right\}$

$B=\left\{ \left( x,y \right) :y=x,x\in R \right\}$ , then

0%
$B\subset A$
0%
$A\subset B$
0%
$A\cap B=\phi$
0%
$A\cup B=A$

Explanation

Given

$A=\left\{ \left( x,y \right) :y={ e }^{ x },x\in R \right\}$

Thus, in roaster form, set A can be written as,

$A=\left\{ .............\left( -4,\frac { 1 }{ { e }^{ 4 } } \right) ,\left( -3,\frac { 1 }{ { e }^{ 3 } } \right) ,\left( -2,\frac { 1 }{ { e }^{ 2 } } \right) ,\left( -1,\frac { 1 }{ { e }^{ 1 } } \right) ,\left( 0,1 \right) ,\left( 1,{ e }^{ 1 } \right) ,\left( 2,{ e }^{ 2 } \right) ,\left( 3,{ e }^{ 3 } \right) ,\left( 4,{ e }^{ 4 } \right) ........... \right\}$ (1)

Given

$B=\left\{ \left( x,y \right) :y=x,x\in R \right\}$

Thus, in roaster form, set B can be written as,

$B=\left\{ ...........\left( -4,-4 \right) ,\left( -3,-3 \right) ,\left( -2,-2 \right) ,\left( -1,-1 \right) ,\left( 0,0 \right) ,\left( 1,1 \right) ,\left( 2,2 \right) ,\left( 3,3 \right) ,\left( 4,4 \right) ......... \right\}$ (2)

From equations (1) and (2), it is clear that no element is common in both the sets.

$\therefore A\cap B=\phi$

Range of the function f(x) = cos (K sinx) is [-1, 1], then the positive integral value of K can be?

0%
1
0%
2
0%
5
0%
4

Suppose

${A_1},{A_2},...,{A_{30}}$ are thirty sets each having 5 elements and

${B_1},{B_2},...,{B_n}$ are

$n$ sets each with 3 elements, let

$\bigcup\limits_{i = 1}^{30} {{A_i}} = \bigcup\limits_{i = 1}^n {Bj} = S$ and each element of

$S$ belongs to exactly 10 of the

${A_i}'s$ and exactly 9 of the

$B,'s$ . Then

$n$ is equal to

0%
$15$
0%
$3$
0%
$45$
0%
$35$

Consider the word

$W=MISSISSIPPI$ .

$N$ denotes the number of different selections of

$5$ letters from the word

$W = MISSISSIPPI$ then

$N$ belongs to the set,

0%
$\{ 15,\ 16,\ 17,\ 18,\ 19\}$
0%
$\{ 20,\ 21,\ 22,\ 23,\ 24\}$
0%
$\{ 25,\ 26,\ 27,\ 28,\ 29\}$
0%
$\{ 30,\ 31,\ 32,\ 33,\ 34\}$

Explanation

$MISSISSIPPI$

$M=1\quad I=4\quad S=4\quad P=2$

$5$ letters word can be made as:-

$^{2}C_{1} \ ^{3}C_{1}+ ^{2}C_{1} + ^{2}C_{1} ^{3}C_{2} + ^{3}C_{1} \times 1$

$=6+4+6+6+3$

$=23$

the set is

$\left\{20, 21, 22, 23, 24\right\}$

$X =\{1,2,3,4,5,6,7,8,9, 10\}$ 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)

$(A')' =A$

0%
Only a is true
0%
Only b and c are true
0%
Only a and b are true
0%
All three a,b and c are true.

Explanation

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

In certain town,

$25\%$ families own a cell phone,

$15\%$ families own a scooter and

$65\%$ families own neither a cell phone nor a scooter. If

$1500$ families own both a cell phone and a scooter, then the total number of families in the town is:

0%
$10000$
0%
$20000$
0%
$30000$
0%
$50000$

Explanation

65% of families own neither a cell phone nor a scooter

35% of families own either a phone or a scooter

$\cup$ B)=A+B-(A

$\cap$ B)

35%x=25%x+15%x-(A

$\cap$ B)

$\cap$ B)=5%x=

$\dfrac{5}{100}x=1500$

number of families=30000

$A \subset B$ , then

0%
$C - B \subset C - A$ .
0%
$A \cup B = A$
0%
$A \cap B = B$
0%
None

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

0%
$2^{99}$
0%
$99^{2}$
0%
$100$
0%
$18$

Explanation

Let us the

$n\left( {\left( {A \times B} \right) \cap \left( {B \times A} \right)} \right)$

$= n\left( {\left( {A \cap B} \right) \times \left( {B \cap A} \right)} \right)$

$= n\left( {A \cap B} \right).\,n\left( {B \cap A} \right)$

$= n\left( {A \cap B} \right).\,n\,\left( {A \cap B} \right)$

$= \left( {99} \right)\left( {99} \right)$

$= {99^2}$

The set of all

$x$ for which

$1 + \log x < x$ is

0%
$(1, \infty)$
0%
$(0, 1)$
0%
$(0, \infty)$
0%
None of these

Explanation

$1+\log{x}<x$

$\log{x}<x-1$

$x<{e}^{x-1}$

$\therefore x\in (1,\infty).$

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

0%
$(\alpha \times \beta)\cup (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$
0%
$(\alpha \times \beta)\cap (\xi \times \eta)=(\alpha \times \beta)\cap (\xi \times \eta)$
0%
$(\alpha \cap \beta)\cup (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$
0%
$(\alpha \cap \beta)= (\xi \cap \eta)=(\alpha \times \beta)\cup (\xi \times \eta)$

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

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

0%
$S_1$ is true, $S_2$ is false
0%
$S_1$ false
0%
$S_1$ is false, $S_2$ is false
0%
$S_1$ is true

$If\,A = \left\{ {\left( {x,y} \right)\,\left| {{x^2} + {y^2} \le \left. 4 \right\}\,and} \right.} \right.$

$B = \left\{ {\left( {x,y} \right)\,\left| {{{(x - 3)}^2} + {y^2}} \right. \le \left. 4 \right\}} \right.\,and\,the$

$po{\mathop{\rm int}} \,P\left( {a,\frac{1}{2}} \right)\,belongs\,to\,the\,set\,B - A$ then the set of possible real values of

$a$ is

0%
$\left( {\frac{{1 + \sqrt {3} }}{4},\frac{{7 + \sqrt 7 }}{4}} \right)$
0%
$\left( {\frac{{7 - \sqrt 7 }}{4},\frac{{1 + \sqrt 7 }}{4}} \right)$
0%
$\left( {\frac{{1 - \sqrt {31} }}{4},\frac{{7 - \sqrt 7 }}{4}} \right)$
0%
none of these

Explanation

${x^2} + {y^2} > 4$

${a^2} + {\left( {a - \frac{1}{2}} \right)^2} > 4$

$2{a^2} - a - \frac{5}{4} > 0$

$a = \frac{{1 \pm \sqrt {31} }}{4}$

$a \in \left( { - \infty ,\frac{{1 - \sqrt {31} }}{4}} \right)\,V\left( {\frac{{1 + \sqrt {31} }}{4},\infty } \right)$

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

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:

0%
$6$
0%
$10$
0%
$20$
0%
$30$

Explanation

$P(GD and PI)\\=P(GD)\times P(PI)\\=(\frac{20}{100})\times(\frac{30}{100})\\=(\frac{6}{100})\\\therefore no. of candidates expected\\=P(GD and PI)\times100\\=6$

So, the correct option is '6'.

Let

$A\subset B$ then

$A'\cap B'=$

0%
$A'$
0%
$B'$
0%
$B$
0%
none of these

Explanation

Now, given

$A\subset B$ .

This gives

$A\cup B=B$ ......(1).

Now,

$A'\cap B'$

$=(A\cup B)'$ [ Using De Morgan's law]

$=B'$ . [ Using (1)]

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

0%
$S={ 1 }$
0%
$S=Z$
0%
$S=N$
0%
none of these

Let

$S=\{1,2,3,4,5,6,7\}$ and let

$A=\{2,5,7\}$ then

$A'$ is

0%
$\{1,3,6\}$
0%
$\{1,3,4,6\}$
0%
$\{1,4,6\}$
0%
none of these

Explanation

Given,

$S=\{1,2,3,4,5,6,7\}$ and let

$A=\{2,5,7\}$ .

Now,

$A'=S-A=\{1,3,4,6\}$ .

If AandB are subsects of the universal set X and n(X)=

$50,$ n(A)=

$35$ ,n(B)=20 Find

$n(A\bigcup {B)}$

$n(A\bigcap {B)}$

$n(A`\bigcap {B)}$

$n(A\bigcap {B`} )$

0%
$45,10,10,25$
0%
$45,10,10,20$
0%
$40,10,10,25$
0%
$45,10,15,25$

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

0%
A -B
0%
B - A
0%
A - A'
0%
A - B'

Explanation

$A' -B' = B-A$

since

$-X' =X \ and\ X'=-X$

Let

$A=\{x:x\in R\ \&\ x^2+1=0\}$ then

$A$ is a null set.

0%
True
0%
False

Explanation

For the equation,

$x^2+1=0$ .......(1).

There is no solution in real numbers for the above equation .

So the set which contains the solutions of the equation (1) is null set.

$|x|$ represent number of elements in region X. Now the following conditions are given

$|U|=14$ ,

$|(A-B)^C|=12$ ,

$|A\cup B|=9$ and

$|A\Delta B|=7$ , where A and B are two subsets of the universal set U and

$A^C$ represents complement of set A, then?

0%
$|A|=2$
0%
$|B|=5$
0%
$|A|=4$
0%
$|B|=7$

$A=\{4^n-3n-1:n\in N\}$ and

$B=\{9(n-1): n\in N\}$ , then?

0%
$B\subset A$
0%
$A\cup B = N$
0%
$A\subset B$
0%
None of these

Explanation

${ 4^{ n } }-3n-1,n\in N \\ { \left( { 3+1 } \right) ^{ n } }-3n-1 \\ by\, u\sin g\, binomal\, ension \\ \Rightarrow \left[ { 1+3n{ +^{ n } }{ C_{ 2 } }\cdot { 3^{ 2 } }{ +^{ n } }{ C_{ 3 } }\cdot { 3^{ 3 } }+{ { ...... }^{ n } }{ C_{ n } }\cdot { 3^{ n } } } \right] -3n-1 \\ \Rightarrow 1+3n+{ 9^{ n } }{ C_{ 2 } }+{ 27^{ n } }{ C_{ 3 } }+{ ....3^{ n } }-3n-1 \\ \Rightarrow 9\left( { ^{ n }{ C_{ 2 } }+{ 3^{ n } }{ C_{ 3 } }+{ { ..... }^{ n } }{ C_{ n } }{ 3^{ \left( { n-2 } \right) } } } \right) \\ Which\, is\, divisible\, by\, \left( q \right) , \\ All\, element\, of\left( B \right) is\, including\, in\, \left( A \right) \\ B\subset A \\$

$A \cup B= A \cap B$ if :

0%
$A\supset B$
0%
$A=B$
0%
$A\subset B$
0%
$A \subseteq B$

Explanation

$\Rightarrow A\cup B=A\cap B$

$\Rightarrow A=B$

Let

$N$ be the set of non-negative integers,

$I$ the set of integers,

$N_p$ the set of non-positive integers,

$E$ the set of even integers and

$P$ the set of prime numbers. Then

0%
$I-N=N_p$
0%
$N \cap {N_p} = \phi$
0%
$E \cap P = \phi$
0%
$N\Delta {N_p} = 1 - \{ 0\}$

Explanation

By option verification,

$I-N=N_p$

$I$ is the set of integers,

$\implies N\cup N_p$

$N$ is set of non-negative integers,

$N_p$ is non-positive integers.

$I-N=N\cup N_p-N=N_p$

The set which begins with additive identity is

0%
W
0%
N
0%
Q
0%
Z

Explanation

Additive identity is

$0$ .

The set that begins with additive identity is whole numbers which us denoted by

$W$ .

$A$ and

$B$ are events such that

$P(A\cup B)=\cfrac{3}{4},P(A\cap B)=\cfrac{1}{4},P(\overline { A } )=\cfrac{2}{3}$ , then

$P(\overline { A } \cap B)$ is

0%
$\cfrac{5}{12}$
0%
$\cfrac{3}{8}$
0%
$\cfrac{5}{8}$
0%
$\cfrac{1}{4}$

Explanation

Given,

$P(A \cup B)=\dfrac {3}{4}, P(A \cap B)=\dfrac {1}{4}, P(\bar A)=\dfrac {2}{3}$

We need to find

$P(\bar A\cap B)$ .

Therefore,

$P(B)=\dfrac {2}{3}$

We know

$P(B)=P(A\cup B)+P(A \cap B)-P(A)$

Therefore,

$P(\bar A\cap B)=\dfrac {2}{3}-\dfrac {1}{4}=\dfrac {5}{12}$

If set 's' contains all the real values of x for which

$log_ {(2x+3)^{x^2}}<1$ is true, then set 'S' contain:

0%
$(log_25,log_2 7)$
0%
$[log_34,log_3 8]$
0%
$\left(\dfrac{-3}{2},1\right)$
0%
$(-1,0)$

Events

$A$ and

$C$ are independent. If the probabilities relating

$A,B$ and

$C$ are

$P\left( A \right) = \dfrac{1}{5},\,\,P\left( B \right) = \dfrac{1}{6},$

$P\left( {A \cap C} \right) = \dfrac{1}{20},\,P\left( {B \cup C} \right) = \dfrac{3}{8}$ then

0%
events $B$ and $C$ are independent
0%
events $B$ and $C$ are mutually exclusive
0%
events $B$ and $C$ are neither independent nor mutually exclusive
0%
events $B$ and $C$ are equiprobable

Explanation

Solution -

$P(A) = \frac{1}{5}$

$P(B) = \frac{1}{6}$

$P(A\cap C) = \frac{1}{20}$

$P(B\cup C) = \frac{3}{8}$

Events A and C are independent

then,

$P(\frac{A}{C}) = \frac{P(A\cap C)}{P(C)} = P(A)(P(C))$

$\frac{1/20}{P(C)} = \frac{1}{5} P(C)$

$P(C) = \frac{1}{2}$

$P(B\cap C) = P(B)+P(C)-P(B\cup C)$

$= \frac{1}{6}+\frac{1}{2}-\frac{3}{8} = \frac{7}{24}$

$P(\frac{B}{C}) = \frac{P(B\cap C)}{P(C)} = \frac{7/24}{1/2} = \frac{7}{12}$

$P(B).P(C) = \frac{1}{6}\times\frac{1}{2} = \frac{1}{2}$

Events B & C are neither independent nor mutually exclusive

C is correct

1100667_1187864_ans_6c5ff5a62ec04f1896a5c3dac8cd2aca.jpg

$A\ and\ B$ are any two sets, then
(i)

$A\subset A\cup B$
(ii)

$A\cup A\subset B$

both relation is ?

0%
True
0%
False

Let

$A=\left\{ 1,2,3,4 \right\} ,B=\left\{ 2,4,6 \right\}$ . Then the number of sets

$C$ such that

$A\cap B\subseteq C\subseteq A\cup B$ is

0%
$6$
0%
$9$
0%
$8$
0%
$10$

Explanation

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

$A\cap B=\left\{2,4\right\}\ \ \ A\cup B=\left\{1,2,3,4,6\right\}$

$A\cap B\subseteq C\subseteq A\cup B\Rightarrow C=\left\{2,4,D\right\}$ where

$D$ is subsets of

$\left\{1,3,6\right\}$

$\Rightarrow$ No.of sets

$C$ = no.of subsets of

$\left\{1,3,6\right\}=8$

The set

$\left( A\cup B\cup C \right) \cap \left( A\cap B'\cap C' \right) '\cap C'$ is equal to

0%
$B\cap C'$
0%
$A\cap C'$
0%
$B\cap C$
0%
$A\cap B\cap C'$

Explanation

$(A\cup B\cup C)\bigcap (A\bigcap {B}'\cap {C}'),'\cap {C}'$

$(A\cup B\cup C)\cap ({A}'\cup B\cup C)\cap {C}'$

$\left [ (A\cap {A}')\phi \cup (A\cap B)\cup (A\cap C)\cup(B\cap {A}' )\phi\cup (B\cap B)\phi \cup (B\cap C)\cap {C}' \right ]$

$\left [ (A\cap B) \cup (A\cap C)\cup (B\cap C) \right ] \cap {C}'$

$(A\cap B\cap {C}') \cup (A\cap \dfrac{C\cap {C}'}{\phi })\cup (B\cap \dfrac{C\cap {C}'}{\phi })$

$(A\cap B\cap {C}') \cup (A\cap \phi ) \cup (B\cap \phi )$

$(A\cap B\cap {C}') \cup \phi \cup \phi$

$= A\cap B\cap {C}'$

1171693_1144603_ans_ead037711eae4b33a6317b9d77fe577c.jpg

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 9 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Set Theory Quiz 9 - MCQExams.com

0:0:3

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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