If $$A=\{1,2,3\}, B=\{3,4\}$$ and $$C=\{1,3,5\}$$, then $$A \times (B - C) =$$

$$(A \times B ) - (B \times C )$$

$$(A \times B ) - (C \times A )$$

MCQ Questions for Class 11 Engineering Maths Sets Quiz 9

A set of

$n$ numbers has the sum

$s$ . Each number of the set is increased by

$20$ , then multiplied by

$5$ , and then decreased by

$20$ . The sum of the numbers in the new set thus obtained is:

0%
$s+20n$
0%
$5s+80n$
0%
$s$
0%
$5s$

Explanation

$s={ a }_{ 1 }+{ a }_{ 2 }+....+{ a }_{ n }$

$s'=5({ a }_{ 1 }+20)-20+5({ a }_{ 2 }+20)-20+.....=5{ a }_{ 1 }+80+5{ a }_{ 2 }+80+....+5{ a }_{ n }+80$

$=5\left( { a }_{ 1 }+{ a }_{ 2 }+....+{ a }_{ n } \right) +80n=5s+80n$
or

$\sum _{ i=1 }^{ n }{ { a }_{ i } } =s,\sum _{ i=1 }^{ n }{ { [5(a }_{ i }+20)-20] } =5\sum _{ i=1 }^{ n }{ { a }_{ i } } +\sum _{ i=1 }^{ n }{ 80 } =5s+80n\quad \quad$

$A= \{x:x$ is a multiple of

$2\}, \,\,B= \{x:x$ is a multiple of

$5\}$ and

$C = \{x:x$ is a multiple of 10

$\}$ , then

$A\cap (B\cap C)$ is equal to

0%
$A$
0%
$B$
0%
$C$
0%
$\{x:x$ is a multiple of $100\}$

Explanation

$A= \{x:x$ is a multiple of

$2\}=\{2,4,6,8,10,12,14,....\}$

$B= \{x:x$ is a multiple of

$5\}=\{5,10,15,20,25,....\}$ and

$C = \{x:x$ is a multiple of

$10\}=\{10,20,30,40,....\}$

Here,

$C \subset A$ and

$C \subset B$

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

$= A \cap C = C$

$M = \left\{ {x:x \geqslant 7\,\,{\text{and}}\,x \in N} \right\}$ for universal set of natural numbers, then

$M'$ is

0%
$\left\{ {1,2,3,4,5} \right\}$
0%
$\left\{ {1,2,3,4,5,6,7} \right\}$
0%
$\left\{ {1,2,3,4,5,6} \right\}$
0%
$\left\{ {0,1,2,3,4,5,6} \right\}$

Explanation

Given

$M =\{x:x\ge 7 , x\in N\}$ .

$x$ is a natural number such that

$x\ge 7$ . Hence

$x= \{7,8,9,10,11,12,.......\infty\}$

$\therefore M' = N- M$

$=\{1,2,3,4,5,6\}$

$X=\left\{ { 4 }^{ n }-3n-1;n\in R \right\}$ and

$Y=\left\{ 9\left( n-1 \right) ;n\in N \right\}$ , then

$X\cap Y=$

0%
$X$
0%
$Y$
0%
$\phi$
0%
$\left\{ 0 \right\}$

Explanation

Let us restrict the domain of

$X$ to natural numbers,

$\Longrightarrow { 4 }^{ n }-3n-1\\ \Longrightarrow { \left( 1+3 \right) }^{ n }-3n-1\\ \Longrightarrow 1+3n+{ n }_{ { C }_{ 2 } }{ 3 }^{ 2 }+...+{ 3 }^{ n }-3n-1\\ \Longrightarrow 9k$

Where

$k$ is some natural number,

So,the elements of

$X$ are multiples of

$9$ ,

The elements of the set

$Y$ are also multiples of

$9$ ,

But if we extend the domain of

$X$ to the set of real numbers there would be other elements which are not present in

$Y$ as the elements of

$Y$ are only natural numbers where as the elements of

$X$ can also be rational and irrational,

So the elements of

$Y$ are contained in

$X$ for all natural numbers,

$\therefore X\cap Y=Y$ .

In the equation

${ \left( x-m \right) }^{ 2 }-{ \left( x-n \right) }^{ 2 }={ \left( m-n \right) }^{ 2 }$ ,

$m$ is a fixed positive number, and

$n$ is a fixed negative number. The set of values

$x$ satisfying the equation is:

0%
$x\ge 0$
0%
$x\le n$
0%
$x=0$
0%
the set of all real numbers
0%
none of these

The number of binary operations on the set

$\{1, 2, 3\}$ is _________.

0%
$3^9$
0%
$9^3$
0%
$27$
0%
$3!$

Explanation

Let us denote this set by

$S$ , then

$|S| = 3$ .

A binary relation defined on the elements of

$S$ maps all elements in

$S\times S$ to elements in

$S$ by definition.

In this case any binary relation will thus have

$3^2 =9$ inputs each of which is an ordered pair of elements from

$S$ and only

$3$ number of possible outputs.

If all possible binary operations are considered then it is possible to assign any of the

$3$ outputs to any of the

$9$ inputs. So the number of all binary operations would exactly be

$3^9$ .

So option A is the correct answer.

The relation

$S=\{(3, 3), (4, 4)\}$ on the set

$A=\{3, 4, 5\}$ is __________.

0%
Not reflexive but symmetric and transitive
0%
Reflexive only
0%
Symmetric only
0%
An equivalence relation

Explanation

given, relation

$S=\{(3,3),(4,4)\}$ on the set

$A=\{3,4,5\}$

It is not reflexive because if every element of

$A$ is not related to itself (i.e.,)

$(5,5)$ is absent.

$A$ is symmetric because if for all

$a$ and

$b$ in

$A$ that

$a$ is related to

$b$ if and only if

$b$ is related to

$a$ .

$A$ is transitive because if

$a$ is related to

$b$ and

$b$ is related to

$c$ then

$a$ is also related to

$c$ .

Hence, not reflexive but symmetric and transitive.

In order to draw a graph of

$f(x) = ax^{2} + bx + c$ , a table of values was constructed. These values of the function for a set of equally spaced increasing values of

$x$ were

$3844, 4096, 4227, 4356, 4489, 4624$ , and

$4761$ . The one which is incorrect is

0%
$4096$
0%
$4356$
0%
$4489$
0%
$4761$
0%
None of these

Explanation

We are told that the values of

$f(x)$ listed correspond to
7

$f(x), f(x + h), f(x + 2h), ...., f(x + 7h)$ .
Observe that the difference between successive values is given by

$f(x + h) - f(x) = a(x + h)^{2} + b(x + h) + c - (ax^{2} + bx + c)$

$= 2ahx + ah^{2} + bh$ .
Since the difference is a linear function of

$x$ , it must change by the same amount whenever

$x$ is increased by

$h$ . But the successive differences of the listed values

$3844\ 3969\ 4096\ 4227\ 4356\ 4489\ 4624\ 4761$
are

$125\ 127\ 131\ 129\ 133\ 135\ 137$
so that, if only one value is incorrect;

$4227$ is the value.

State true or false.

Let

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

$n(A - B) = 2$

0%
True
0%
False

Explanation

Given

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

$B=\left\{ 2,4,6,8 \right\}$ and

$C=\left\{ 3,4,5,6 \right\}$

$\therefore A-B=\left\{ x:x\in A\quad and\quad x\notin B \right\}$

$\therefore A-B=\left\{ 1,3 \right\}$

$\therefore n\left( A-B \right) =2$

$X$ and

$Y$ are two sets, then

$X\cap \left( X\cup Y \right)$ equals

0%
$X$
0%
$Y$
0%
$\phi$
0%
$None\ of\ these$

Explanation

Ans.

$(a).$
If

$X\subset Y$ , then

$X\cup Y=Y$ and

$X\cap Y=X$
If

$Y\subset X$ , then

$X\cup Y=X$ and

$X\cap X=X$

If X = {

$4^n \,-\,3n\,-\,1\,:\,\epsilon N$ } and
Y = {

$9(n\,-\,1) \,:\,\epsilon N$ }
then

$X\subset Y$

0%
True
0%
False

Explanation

Given

$X=\left\{ { 4 }^{ n }-3n-1;n\in N \right\}$

For

$n=0$ ,

$X={ 4 }^{ 0 }-0-1$

$\therefore X=0$

For

$n=1$ ,

$X={ 4 }^{ 1 }-3\left( 1 \right) -1$

$\therefore X=0$

For

$n=2$ ,

$X={ 4 }^{ 2 }-3\left( 2 \right) -1$

$\therefore X=16-6-1=9$

For

$n=3$ ,

$X={ 4 }^{ 3 }-3\left( 3 \right) -1$

$\therefore X=64-9-1=54$

For

$n=4$ ,

$X={ 4 }^{ 4 }-3\left( 4 \right) -1$

$\therefore X=256-12-1=243$

$\therefore X=\left\{ 0,0,9,54,243....... \right\}$ (1)

Consider second set

$Y=\left\{ 9\left( n-1 \right) ;n\in N \right\}$

For

$n=0$ ,

$Y=9\left( 0-1 \right)$

$\therefore Y=-9$

For

$n=1$ ,

$Y=9\left( 1-1 \right)$

$\therefore Y=0$

For

$n=2$ ,

$Y=9\left( 2-1 \right)$

$\therefore Y=9$

For

$n=3$ ,

$Y=9\left( 3-1 \right)$

$\therefore Y=18$

For

$n=4$ ,

$Y=9\left( 4-1 \right)$

$\therefore Y=27$

$\therefore Y=\left\{ -9,0,9,18,27,36,45,54,63,72......... \right\}$

$\therefore Y=\left\{ y:y\quad is\quad multiple\quad of\quad 9\quad and\quad y\ge -9 \right\}$ (2)

From (1) and (2),

$X\subset Y$

Given : A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
Find :

$(A \, \times \, B) \, \cap \, (B \, \times \, C)$ .

0%
{4,4}
0%
{3,4}
0%
{3,4}, {3,3}
0%
{3,3}

Explanation

Given

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

$B=\left\{ 3,4 \right\}$ and

$C=\left\{ 4,5,6 \right\}$

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

$\therefore A\times B=\left\{ \left( 1,3 \right) ,\left( 1,4 \right) ,\left( 2,3 \right) ,\left( 2,4 \right) ,\left( 3,3 \right) ,\left( 3,4 \right) \right\}$

$B\times C=\left\{ 3,4 \right\}\times \left\{ 4,5,6 \right\}$

$\therefore B\times C=\left\{ \left( 3,4 \right) ,\left( 3,5 \right) ,\left( 3,6 \right) ,\left( 4,4 \right) ,\left( 4,5 \right) ,\left( 4,6 \right) \right\}$

$\therefore \left( A\times B \right) \cap \left( B\times C \right) =\left\{ \left( 1,3 \right) ,\left( 1,4 \right) ,\left( 2,3 \right) ,\left( 2,4 \right) ,\left( 3,3 \right) ,\left( 3,4 \right) \right\} \cap \left\{ \left( 3,4 \right) ,\left( 3,5 \right) ,\left( 3,6 \right) ,\left( 4,4 \right) ,\left( 4,5 \right) ,\left( 4,6 \right) \right\}$

$\therefore \left( A\times B \right) \cap \left( B\times C \right) =\left\{ \left( 3,4 \right) \right\}$

Let

$A=\left\{ x:x\ \in\ R,\left| x \right| <1 \right\}$

$B=\left\{ x:x\ \in\ R,\left| x-1 \right| \ge 1 \right\}$
and

$A\cup B=R-D$ , then set

$D$ is

0%
$\left\{ x:1 < x \le 2 \right\}$
0%
$\left\{ x:1\le x<2 \right\}$
0%
$\left\{ x:1\le x\le 2 \right\}$
0%
None of these

Explanation

Given

$A=\left\{ x:x\in R\left| x \right| <1 \right\}$

$x$ is positive,

$\left| x \right| =x$

$\therefore x<1$

$x$ is negative,

$\left| x \right| =-x$

$\therefore -x<1$

$\therefore x>-1$

$\therefore A=\left\{ -1<x<1 \right\}$ (1)

Given

$B=\left\{ x:x\in R,\left| x-1 \right| \ge 1 \right\}$

$\left| x-1 \right|$ is positive,

$\left| x-1 \right| =x-1$

$\therefore x-1\ge 1$

$\therefore x\ge 2$

$\left| x-1 \right|$ is negative,

$\left| x-1 \right| =-\left( x-1 \right)$

$\therefore -\left( x-1 \right) \ge 1$

$\therefore \left( x-1 \right) \le -1$

$\therefore x\le 0$

$\therefore B=\left\{ x\le 0\quad and\quad x\ge 2 \right\}$ (2)

Thus,

$A\cup B=\left\{ -1<x<1 \right\} \cup \left\{ x\le 0\quad and\quad x\ge 2 \right\}$

The region which is not included in this domain has to be subtracted from remaining domain of real numbers.

$\therefore A\cup B=R-D$

Thus, set D is set of excluded region which is the region between 1 and 2 including 1.

$\therefore D:-1\le x<2$

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.

$\therefore A\cap B=\phi$

Let

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

$B=\left\{ \left( x,y \right) :y={ e }^{ -x },x\in R \right\}$ . Then

0%
$A\cap B=\phi$
0%
$A\cap B\neq \phi$
0%
$A\cup B={ R }^{ 2 }$
0%
$none\ of\ these$

Explanation

Ans.

$(b)$ .

$y={e}^{x},\ y={e}^{-x},$ meet where

${ e }^{ x }=\ { e }^{ -x }$ or

$e^{2x}=1$

$\therefore x=0$ and hence

$y=1$ . These curve meet at

$(0,1)$ so that

$A\cap B=(0,1)$ .
In other words

$A\cap B\neq \phi$ .

$X$ and

$Y$ are two sets, then

$X\cap \left( Y\cup X \right)'$ equals

0%
$X$
0%
$Y$
0%
$\phi$
0%
$None\ of\ these$

Explanation

$X\cap \left( X' \cap Y' \right) =\left( X\cap X' \right) \cap Y' =\phi \cap Y' =\phi$

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

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

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

$C=\{1,3,5\}$ , then

$A \times (B - C) =$

0%
$(A \times B ) - (A \times C )$
0%
$(A \times B ) + (A \times C )$
0%
$(A \times B ) - (B \times C )$
0%
$(A \times B ) - (C \times A )$

Explanation

$(B-C)$ is set of all elements present in

$B$ and not in

$C$

$\therefore (B-C)=\{4 \}$

$A\times(B-C)=\{1,2,3\}\times\{4\}=\{(1,4),(2,4),(3,4)\}$

Lets find

$(A\times B)$ and

$(A\times C)$

$(A\times B)=\{1,2,3\}\times \{3,4\}$

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

$(A\times C)=\{1,2,3\}\times \{1,3,5\}$

$=\{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5)\}$

$(A\times B)-(A\times C)=\{(1,3),(2,3),(3,3),(1,4),(2,4),(3,4)\}-\{(1,1),(1,3),(1,5),(2,1),(2,3),$

$(2,5),(3,1),(3,3),(3,5)\}$

$(A\times B)-(A\times C)=\{(1,4),(2,4),(3,4)\}$

$\therefore A\times(B-C)=(A\times B)-(A\times C)$

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

Which is the simplified representation of

$(A' \cap B'\cap C)\cup(B\cap C) \cup (A\cap C)$ where A,B,C are subsets of set X

0%
$A$
0%
$B$
0%
$C$
0%
$X\cap(A \cup B\cup C)$

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

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

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

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

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

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

$= {99^2}$

$S_1:(p\Rightarrow q) V ( q \Rightarrow p )$ 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

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

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)]

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

CBSE Questions for Class 11 Engineering Maths Sets Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sets Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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