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Set Theory Test 21

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Set Theory Test 21
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  • Question 1
    1 / -0
    If X={4n3n1;nR}X=\left\{ { 4 }^{ n }-3n-1;n\in R \right\} and Y={9(n1);nN}Y=\left\{ 9\left( n-1 \right) ;n\in N \right\} , then XY=X\cap Y=
    Solution
    Let us restrict the domain of XX to natural numbers,
    4n3n1(1+3) n3n11+3n+nC232+...+3n3n19k\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 kk is some natural number,
    So,the elements of XX are multiples of 99,
    The elements of the set YY are also multiples of 99,
    But if we extend the domain of XX to the set of real numbers there would be other elements which are not present in YY as the elements of YY are only natural numbers where as the elements of XX can also be rational and irrational,
    So the elements of YY are contained in XX for all natural numbers,
    XY=Y\therefore X\cap Y=Y.
  • Question 2
    1 / -0
    The number of binary operations on the set {1,2,3}\{1, 2, 3\} is _________.
    Solution
    Let us denote this set by SS, then S=3|S| = 3.
    A binary relation defined on the elements of SS maps all elements in S×SS\times S to elements in SS by definition.
    In this case any binary relation will thus have 32=93^2 =9 inputs each of which is an ordered pair of elements from SS and only 33 number of possible outputs. 
    If all possible binary operations are considered then it is possible to assign any of the 33 outputs to any of the 99 inputs. So the number of all binary operations would exactly be 393^9.
    So option A is the correct answer.  
  • Question 3
    1 / -0
    Choose the correct answers from the alternatives given.
    Directions for questions 7171 to 7575 : Study the pie chart carefully to answer the questions that follow.
    Percentage distribution of students in different courses.
    What is the value of half of the difference between the number of students in MBA and MBBS?
    The Total number of student = 65006500

    Solution

  • Question 4
    1 / -0
    If a.N={ax: xN}a.N = \left\{ ax\thinspace :\thinspace x\in N \right\} then 3N7N=3N\cap 7N=
    Solution
    Clearly the set aNaN will include all multiples of aa.

    Hence 3N3N and 7N7N contain multiples of 3 and 7 respectively.

    Their intersection must have the multiples of their L.C.M,i.e,21

    Hence 3N7N=21N3N\cap 7N=21N
  • Question 5
    1 / -0
    The relation S={(3,3),(4,4)}S=\{(3, 3), (4, 4)\} on the set A={3,4,5}A=\{3, 4, 5\} is __________.
    Solution
    given, relation S={(3,3),(4,4)}S=\{(3,3),(4,4)\} on the set A={3,4,5}A=\{3,4,5\}

    It is not reflexive because if every element of AA is not related to itself (i.e.,) (5,5)(5,5) is absent.

    AA is symmetric because if for all aa and bb in AA that aa is related to bb if and only if bb is related to aa.

    AA is transitive because if aa is related to bb and bb is related to cc then aa is also related to cc.

    Hence, not reflexive but symmetric and transitive.
  • Question 6
    1 / -0
    A set of nn numbers has the sum ss. Each number of the set is increased by 2020, then multiplied by 55, and then decreased by 2020. The sum of the numbers in the new set thus obtained is:
    Solution
    s=a1+a2+....+ans={ a }_{ 1 }+{ a }_{ 2 }+....+{ a }_{ n }
    s=5(a1+20)20+5(a2+20)20+.....=5a1+80+5a2+80+....+5an+80s'=5({ a }_{ 1 }+20)-20+5({ a }_{ 2 }+20)-20+.....=5{ a }_{ 1 }+80+5{ a }_{ 2 }+80+....+5{ a }_{ n }+80
    =5(a1+a2+....+an)+80n=5s+80n=5\left( { a }_{ 1 }+{ a }_{ 2 }+....+{ a }_{ n } \right) +80n=5s+80n
    or i=1nai=s,i=1n[5(ai+20)20]=5i=1nai+i=1n80=5s+80n\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
  • Question 7
    1 / -0
    If A={x:xA= \{x:x is a multiple of 2},  B={x:x2\}, \,\,B= \{x:x is a multiple of 5}5\} and C={x:xC = \{x:x is a multiple of 10}\}, then A(BC) A\cap (B\cap C) is equal to 
    Solution
    A={x:xA= \{x:x is a multiple of 2}={2,4,6,8,10,12,14,....}2\}=\{2,4,6,8,10,12,14,....\}
    B={x:xB= \{x:x is a multiple of 5}={5,10,15,20,25,....}5\}=\{5,10,15,20,25,....\} and 
    C={x:xC = \{x:x is a multiple of 10}={10,20,30,40,....}10\}=\{10,20,30,40,....\}

    Here, CAC \subset A and CBC \subset B
    C=AB=A(BC)C = A \cap B = A \cap (B \cap C)
    =AC=C= A \cap C = C

  • Question 8
    1 / -0
    If M={x:x7  andxN}M = \left\{ {x:x \geqslant 7\,\,{\text{and}}\,x \in N} \right\} for universal set of natural numbers, then MM' is
    Solution
    Given M={x:x7,xN}M =\{x:x\ge 7 , x\in N\}. xx is a natural number such that x7x\ge 7. Hence x={7,8,9,10,11,12,.......}x= \{7,8,9,10,11,12,.......\infty\}
    M=NM\therefore M' = N- M
    ={1,2,3,4,5,6}=\{1,2,3,4,5,6\}
  • Question 9
    1 / -0
    In the equation (xm) 2(xn) 2=(mn) 2{ \left( x-m \right)  }^{ 2 }-{ \left( x-n \right)  }^{ 2 }={ \left( m-n \right)  }^{ 2 }, mm is a fixed positive number, and nn is a fixed negative number. The set of values xx satisfying the equation is:
    Solution

  • Question 10
    1 / -0
    Given : A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
    Find : (A×B)(B×C)(A \, \times \, B) \, \cap \, (B \, \times \, C).
    Solution
    Given A={1,2,3} A=\left\{ 1,2,3 \right\} , B={3,4} B=\left\{ 3,4 \right\}  and C={4,5,6} C=\left\{ 4,5,6 \right\} 

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

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

    (A×B)(B×C)={(1,3),(1,4),(2,3),(2,4),(3,3),(3,4) }{(3,4),(3,5),(3,6),(4,4),(4,5),(4,6) } \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\} 

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