MCQ Questions for Class 11 Engineering Maths Sets Quiz 10

If n(U) = 50, n(A) = 20,

$n((A \cup B)')$ = 18 then n(B - A) is

0%
$14$
0%
$12$
0%
$16$
0%
$20$

Explanation

$n(U)=50, n(A)=20$

$n(A\cup B)=18$

$n(B-A)=?$

So,

$n(A\cup B)=n(0)-n(A\cup B)$

So.

$n(A\cup B)=32$

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

$=32-20$

$n(B-A)=12$

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 number of elements in the set

$\left\{ \left( a,b \right) /2{ a }^{ 2 }+3{ b }^{ 2 }=35,a,b\in z \right\}$ when

$z$ is the set of all integers is

0%
$2$
0%
$4$
0%
$8$
0%
$12$

Explanation

Let

$2x+3y=35,\quad x, y > 0$

Integral solution

$\Rightarrow \quad (a^2=x, y^2=b^2)$

$\therefore \ (a^2 , b^2)=(4, a)$ and

$(16, 1)$

$\therefore \ (a, b)\cong (2, 3), (-2, -3), (2, -3), (-2, 3), (4, 1), (-4, -1), (-4, 1), (4, -1)$

$\therefore \ 8$ elements

$\Rightarrow (C)$

1406567_1098888_ans_560b3577bcd6450a9b2a06db24e2c711.png

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,

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

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

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

$\Rightarrow A=B$

$A\ and\ B$ 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

$A - ( A - B )$ is equivalent to which expression

0%
$B$
0%
$A \cup B$
0%
$A \cap B$
0%
$B-A$

Explanation

$A-\left(A-B\right)$

$=A-\left(A\cap{B}^{C}\right)$

$=A\cap{\left(A\cap{B}^{C}\right)}^{c}$

$=A\cap\left[{A}^{c}\cup{\left({B}^{c}\right)}^{c}\right]$

$=A\cap\left[{A}^{c}\cup\,B\right]$ since

${\left({B}^{c}\right)}^{c}=B$

$=\left(A\cap{A}^{c}\right)\cup\,\left(A\cap\,B\right)$

$=\phi\,\cup\,\left(A\cap\,B\right)$ since

$A\cap{A}^{c}=\phi$

$=A\cap\,B$

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%
$(-1,0)$

In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

0%
10
0%
11
0%
12
0%
13

Explanation

Let

$A$ ,

$B$ , and

$C$ be the set of people who like

$product\ A$ ,

$product\ B,\ and\ product\ C$ respectively.

Accordingly,

$n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12$ ,

$n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8$

The Venn diagram for the given problem can be drawn as above.

It can be seen that number of people who like product C only is

$={29 – (4 + 8 + 6)} = 11$

1460243_1199498_ans_24387ca5a0fc4a569288365f3a4fedcf.PNG

The number of subsets

$R$ of

$P=(1,2,3,....,9)$ which satisfies the property "There exit integers a<b<c with a

$\in$ R, b

$\in$ R,c

$\in$ R" is

0%
$512$
0%
$466$
0%
$467$
0%
None of these

Explanation

Every subset of P containing at least 3 elements from P will always satisfy the required property since each number is distinct in the set P.

So, Total no. of ways =

$^{9}C_3 + ^{9}C_4 + .... + ^{9}C_9$

$= (^{9}C_0 + ^{9}C_1 + ..... + ^{9}C_9) - ^{9}C_0 - ^{9}C_1 - ^{9}C_2$

$=2^9 - 46$

$=512 -46$

$=466$

In a universal set x,

$n\left( x \right) = 50$ ,

$n\left( A \right) = 35$ ,

$n\left( B \right) = 20$ ,

$n\left( {A' \cap B'} \right) = 5$ ,then

$n\left( {A \cup B} \right),n\left( {A \cap B} \right)$ are repsectively

0%
$45,10$
0%
$10,45$
0%
$25,30$
0%
$15,25$

Explanation

Given

$n\left(X\right) = 50$ ,

$n\left(A\right) = 35$ ,

$n\left(B\right) = 20$

$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = 5$

$(i)\,n\left(A \cup B\right)$

$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = n\left(X\right) - n\left(A\cup\,B\right)$

$n\left(A\cup\, B\right) = n\left(X\right) - n\left({A}^{\prime}\cap\,{B}^{\prime}\right)$

$n\left(A\cup\, B\right) = 50 - 5= 45$

$(ii) n\left(A \cap B\right)$

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

$n\left(A\cap\, B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cup\, B\right)$

$n\left(A\cap\, B\right) = 35 + 20 - 45= 10$

$A$ and

$B$ are two non empty sets then

$( A \cup B ) ^ { C } = ?$

0%
$A ^ { C } \cup B ^ { C }$
0%
$A ^ { C } \cap B ^ { C }$
0%
$A \cup B ^ { C }$
0%
$A ^ { C } \cap B$

Explanation

If A and B are two no empty sets then

${ \left( A\cup B \right) }^{ C }={ A }^{ C }\cap { B }^{ C }$

Let

$S={1,2,3,.....10}$ and

$P={1,2,3,4,5}$ The number of subsets

$'Q'$ of

$S$ such that

$p \cup Q=S$ , are.....

0%
$128$
0%
$256$
0%
$32$
0%
$64$

Let

$Q$ be a non empty subset of

$N$ and

$q$ is a statement as given below :

$q:$ There exists an even number

$a \in Q$ Negation of the statement

$q$ will be :

0%
There is no even number in the set $Q$
0%
Every $a \in Q$ is an odd number.
0%
$(a)$ and $(b)$ both
0%
None of these

Explanation

Let

$Q$ be a non empty subset of

$N$ and

$q.$ There exists even number

$a\in Q$ negation of statement

$q.$ There does not exist a even number in set Q.

Hence, the answer is there is no even number in the set

$Q.$

If the number of

$5$ elements subsets of the set

$A\left\{\ a_{1},a_{2}.....a_{20}\right\}$ of

$20$ distinct elements is

$k$ times the number of

$5$ elements subsets containing

$a_{4}$ , then

$k$ is

0%
$5$
0%
$\dfrac{20}{7}$
0%
$4$
0%
$\dfrac{10}{3}$

Explanation

No of

$5$ elements subset

$=^{20}C_{5}=\dfrac {20!}{15! \, 15!}=\dfrac {20 \times 19\times 18\times 17\times 16\times 15!}{15! \, 5\times 4\times 3\times 2\times 1}=15504$

No of

$5$ elements subset containing

$a_{4}$

$=^{19}C_{4}=\dfrac {19!}{4! \, 15!}$

$\Rightarrow \dfrac {20!}{15! \, 15!}=k\dfrac {19!}{4!\, 15!}$

$\Rightarrow \dfrac {20\times 19!}{5\times 4!}=k\dfrac {19!}{4!}$

$\Rightarrow k=4$

In a battle

$70$ % of the combatants lost one eye,

$80$ % an ear,

$75$ % an arm,

$85$ % a leg,

$x$ % lost all the four limbs the minimum value of

$x$ is

0%
$10$
0%
$12$
0%
$15$
0%
$5$

Explanation

$70\%$ of the combatants lost one eye,

$80\%$ an ear,

$75\%$ an arm and

$85\%$ a leg.

Now,

$\Rightarrow$ The combatants who lost one eye and one ear

$=(70+80-100)\%$

$=50\%$

$\Rightarrow$ The combatants who lost one eye, one ear and one erm

$=(50+75-100)\%$

$=25\%$

$\Rightarrow$ The combatants who lost one eye, one ear one arm and one leg

$=(25-85-100)\%$

$=10\%$

$\therefore$ Combatant who lost all the four limbs

$=10\%$

$\therefore$

$x=10\%$

The value of set

$(A\cup B\cup C)\cap(A\cap B^1\cap C^1)^1\cap C^1$ is equal to

0%
$B\cap C^1$
0%
$A\cap C$
0%
$B\cap C^1$
0%
$A\cap C^1$

Two sets A and B are defined as follows

$A=\left\{ \left( x,y \right) :y={ e }^{ 2x },x\in R \right\}$ and

$B=\left\{ \left( x,y \right) :y={ x }^{ 2 },x\in R \right\}$ , then

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

Explanation

$\begin{array}{l} A=\left\{ { \left( { x,\, y } \right) \, :\, y={ e^{ 2x } },\, x\in R } \right\} \, and \\ B=\left\{ { \left( { x,\, y } \right) \, :\, y={ x^{ 2 } },\, x\in R } \right\} \\ then, \\ A=\left( { { e^{ \circ } },\, { e^{ 1 } },\, { e^{ 2 } },....,\, { e^{ \infty } } } \right) \\ =\left( { 1,\, { e^{ 1 } },\, { e^{ 2 } },....... } \right) \\ and, \\ B=\left( { 0,\, 1,\, 4,\, ..... } \right) \\ \therefore A\subset B \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$

If two sets

$P$ and

$Q,n\left(P\right)=5,n\left(Q\right)=4$ then

$n\left(P\times Q\right)=$

0%
$20$
0%
$9$
0%
$25$
0%
$5/4$

Explanation

$\begin{aligned} n(P \times Q) &=n(P) \times n(Q) \\ &=5 \times 4 \\ &=20 \end{aligned}$

$\therefore$ option

$A$ is correct.

If A= {1, 2, 5} and B= {3, 4, 5, 9}, then

$A \bigcup B$ is equal to :

0%
$\{1, 2, 5, 9\}$
0%
$\{1, 2, 3, 4, 9\}$
0%
$\{1, 2, 3, 4, 5, 9\}$
0%
None of these

Explanation

$A=\left\{ 1,2,5 \right\} \quad B=\left\{ 3,4,5,9 \right\}$

$A\cup B=\left\{ 1,2,3,4,5,9 \right\}$

Option C.

If P(S) denotes the set of all subsets of a given set S, then the number of one to one function from the set s={1,2,3} to the set of P(S) is

0%
336
0%
8
0%
36
0%
320

Explanation

$\begin{array}{l} Given \\ P\left( s \right) =\left\{ { 1,\, 2,\, 3 } \right\} \\ P\left( s \right) =all\, subset\, of\, 5 \\ =\left\{ { \phi ,\left\{ 1 \right\} ,\left\{ 3 \right\} ,\left\{ { 1,2 } \right\} ,\left\{ { 2,3 } \right\} ,\left\{ { 3,1 } \right\} ,\left\{ { 1,2,3 } \right\} } \right\} \\ No.\, of\, element\, of\, 5=0\left( 5 \right) =3 \\ similarly \\ O\left( { p\left( s \right) } \right) =8 \\ Total\, no.\, of\, one-one\, function \\ =\frac { { n! } }{ { \left( { n-m } \right) ! } } \, ,\, where\, n=o\left( { p\left( s \right) } \right) \, and\, m=o\left( 5 \right) \\ =\frac { { 8! } }{ { \left( { 8-3 } \right) ! } } =\frac { { 8! } }{ { 5! } } \\ =336\, \, \\ No.\, of\, one-one\, function\, is\, 336 \\ Hence,\, \, option\, \, A\, \, is\, correct, \end{array}$
1217888_1309290_ans_148476409ff94e659de01d6734148e9a.PNG

1217888_1309290_ans_148476409ff94e659de01d6734148e9a.PNG

Consider following expressions
P =

$\prod_{\theta = 1}^{100}cos\theta$ ; Q =

$\prod_{\phi = 1}^{10}cos\phi$ ; R = log cosec 0.8

$\pi$
Then number of non-positive elements in the set {P, Q, R} is

0%
0
0%
1
0%
2
0%
3

$A=\left\{1,2,3,4\right\}; B=\left\{2,4,6,8\right\}; C=\left\{3,4,5,8\right\}$ then

$A\cap B\cap C=$

0%
$\phi$
0%
${4}$
0%
$\mu$
0%
${2,4}$

Explanation

$\begin{matrix} A=\left\{ { 1,2,3,4 } \right\} \\ B=\left\{ { 2,4,6,8 } \right\} \\ C=\left\{ { 3,4,5,8 } \right\} \\ A\cap B\cap C=\left\{ { 4 } \right\} . \\ \end{matrix}$

If n(A)=3, n(B)=4, then

$n(A\times B\times C)=36 find \,n(C)$ is equal to :

0%
3
0%
12
0%
108
0%
none of these

Explanation

$n(A)=3$

$n(B)=4$

$n(A\times B\times C)=36$

$n(3\times 4\times C)=36$

$n(12\times C)=36$

$\therefore n(C)=3$

So,

$n(A\times B\times C)=36$

$n(3\times 4\times 3)=36$ . H.P.

1220763_1314665_ans_877977962f05467983fd416944d5aec5.jpg

The function

$f(x)$ satisfies the condition

$(x-2)f(x)+2f\left(\dfrac{1}{x}\right)=2$ for all

$x\neq 0$ . Then the value of

$f(2)$ is

0%
$\dfrac{1}{2}$
0%
$1$
0%
$\dfrac{7}{4}$
0%
$\dfrac{-3}{2}$

Explanation

Putting

$x=2$ in

$\left( x-2 \right) f\left( x \right) +2f\left( \dfrac { 1 }{ x } \right) =2$

$\left( 2-2 \right) f\left( 2 \right) +2f\left( \dfrac { 1 }{ 2 } \right) =2$

$2f\left( \dfrac { 1 }{ 2 } \right) =2$

$f\left( \dfrac { 1 }{ 2 } \right) =1$

Let A,B are two sets such that n(A)=4 and n(B)=Then the least possible number of elements in the power set of

$(A\cup B)$ is

0%
16
0%
64
0%
256
0%
1024

Explanation

Given,

$n(A)=4,n(B)=6$

Then the least number of possible elements in

$n(A\cup B)=2^{n(A)}.2^{n(B)}=2^4.2^6=2^{10}=1024$

CBSE Questions for Class 11 Engineering Maths Sets Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sets Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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