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CBSE Questions for Class 11 Engineering Maths Sets Quiz 10 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sets
Quiz 10
If n(U) = 50, n(A) = 20, $$n((A \cup B)')$$ = 18 then n(B - A) is
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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
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$$I-N=N_p$$
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$$N \cap {N_p} = \phi $$
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.
So $$I-N=N\cup N_p-N=N_p$$
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
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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 $$\Rightarrow (C)$$
The set which begins with additive identity is
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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 $$W$$.
If $$A$$ 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
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$$\cfrac{5}{12}$$
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$$\cfrac{3}{8}$$
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$$\cfrac{5}{8}$$
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$$\cfrac{1}{4}$$
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 $$A=\{4^n-3n-1:n\in N\}$$ and $$B=\{9(n-1): n\in N\}$$, then?
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$$B\subset A$$
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$$A\cup B = N$$
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$$A\subset B$$
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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 :
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0%
$$A\supset B$$
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$$A=B$$
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$$A\subset B$$
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$$A \subseteq B$$
Explanation
$$\Rightarrow A\cup B=A\cap B$$
$$\Rightarrow A=B$$
If $$A\ and\ B$$ are any two sets, then
(i) $$ A\subset A\cup B$$
(ii) $$ A\cup A\subset B$$
both relation is ?
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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
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$$6$$
0%
$$9$$
0%
$$8$$
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$$10$$
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
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$$ B\cap C'$$
0%
$$ A\cap C'$$
0%
$$ B\cap C$$
0%
$$A\cap B\cap C'$$
Explanation
$$ (A\cup B\cup C)\bigcap (A\bigcap {B}'\cap {C}'),'\cap {C}'$$
$$ (A\cup B\cup C)\cap ({A}'\cup B\cup C)\cap {C}' $$
$$ \left [ (A\cap {A}')\phi \cup (A\cap B)\cup (A\cap C)\cup(B\cap {A}' )\phi\cup (B\cap B)\phi \cup (B\cap C)\cap {C}' \right ]$$
$$ \left [ (A\cap B) \cup (A\cap C)\cup (B\cap C) \right ] \cap {C}'$$
$$ (A\cap B\cap {C}') \cup (A\cap \dfrac{C\cap {C}'}{\phi })\cup (B\cap \dfrac{C\cap {C}'}{\phi })$$
$$ (A\cap B\cap {C}') \cup (A\cap \phi ) \cup (B\cap \phi )$$
$$ (A\cap B\cap {C}') \cup \phi \cup \phi $$
$$ = A\cap B\cap {C}'$$
$$A - ( A - B ) $$ is equivalent to which expression
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$$B$$
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$$A \cup B$$
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$$A \cap B$$
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$$B-A$$
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:
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$$(log_25,log_2 7)$$
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$$[log_34,log_3 8]$$
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$$\left(\dfrac{-3}{2},1\right)$$
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$$(-1,0)$$
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.
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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
$$={29 – (4 + 8 + 6)} = 11$$
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
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$$512$$
0%
$$466$$
0%
$$467$$
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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
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$$45,10$$
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$$10,45$$
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$$25,30$$
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$$15,25$$
Explanation
Given $$n\left(X\right) = 50$$,
$$n\left(A\right) = 35$$,
$$n\left(B\right) = 20$$
$$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = 5$$
$$(i)\,n\left(A \cup B\right)$$
$$n\left({A}^{\prime}\cap\,{B}^{\prime}\right) = n\left(X\right) - n\left(A\cup\,B\right)$$
$$n\left(A\cup\, B\right) = n\left(X\right) - n\left({A}^{\prime}\cap\,{B}^{\prime}\right)$$
$$n\left(A\cup\, B\right) = 50 - 5= 45$$
$$(ii) n\left(A \cap B\right)$$
$$n\left(A\cup B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cap B\right)$$
$$n\left(A\cap\, B\right) = n\left(A\right) + n\left(B\right) - n\left(A\cup\, B\right)$$
$$n\left(A\cap\, B\right) = 35 + 20 - 45= 10$$
If $$A$$ and $$B$$ are two non empty sets then $$( A \cup B ) ^ { C } = ?$$
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$$A ^ { C } \cup B ^ { C }$$
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$$A ^ { C } \cap B ^ { C }$$
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$$A \cup B ^ { C }$$
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$$A ^ { C } \cap B$$
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.....
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$$128$$
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$$256$$
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$$32$$
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$$64$$
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 :
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There is no even number in the set $$Q$$
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Every $$a \in Q$$ is an odd number.
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$$(a)$$ and $$(b)$$ both
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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
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$$5$$
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$$\dfrac{20}{7}$$
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$$4$$
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$$\dfrac{10}{3}$$
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
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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
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$$B\cap C^1$$
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$$A\cap C$$
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$$B\cap C^1$$
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$$A\cap C^1$$
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
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$$A\subset B$$
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$$B\subset A$$
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$$A\bigcup B$$
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$$A\cap B=\phi $$
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 $$n\left(P\times Q\right)=$$
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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 :
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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
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0%
336
0%
8
0%
36
0%
320
Explanation
$$\begin{array}{l} Given \\ P\left( s \right) =\left\{ { 1,\, 2,\, 3 } \right\} \\ P\left( s \right) =all\, subset\, of\, 5 \\ =\left\{ { \phi ,\left\{ 1 \right\} ,\left\{ 3 \right\} ,\left\{ { 1,2 } \right\} ,\left\{ { 2,3 } \right\} ,\left\{ { 3,1 } \right\} ,\left\{ { 1,2,3 } \right\} } \right\} \\ No.\, of\, element\, of\, 5=0\left( 5 \right) =3 \\ similarly \\ O\left( { p\left( s \right) } \right) =8 \\ Total\, no.\, of\, one-one\, function \\ =\frac { { n! } }{ { \left( { n-m } \right) ! } } \, ,\, where\, n=o\left( { p\left( s \right) } \right) \, and\, m=o\left( 5 \right) \\ =\frac { { 8! } }{ { \left( { 8-3 } \right) ! } } =\frac { { 8! } }{ { 5! } } \\ =336\, \, \\ No.\, of\, one-one\, function\, is\, 336 \\ Hence,\, \, option\, \, A\, \, is\, correct, \end{array}$$
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
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0%
0
0%
1
0%
2
0%
3
If $$A=\left\{1,2,3,4\right\}; B=\left\{2,4,6,8\right\}; C=\left\{3,4,5,8\right\}$$ then $$A\cap B\cap C=$$
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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 :
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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.
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
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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
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0%
16
0%
64
0%
256
0%
1024
Explanation
Given,
$$n(A)=4,n(B)=6$$
Then the least number of possible elements in
$$n(A\cup B)=2^{n(A)}.2^{n(B)}=2^4.2^6=2^{10}=1024$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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