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Set Theory Test 22

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Set Theory Test 22
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  • Question 1
    1 / -0
    If X and Y are two sets such that n(X)=17, n(Y)=23 and n(X $$\cup$$ Y)=38, find n(X $$\cap$$ Y).
    Solution

    $$\textbf{Step 1: Label the given information and use relevant formula}$$

                    $${\text{It is given that n(X) = 17,n(Y) = 23,n(X}} \cup {\text{Y) = 38}}$$

                    $${\text{As we know that, n(X}} \cup {\text{Y) = n(X) + n(Y) - n(X}} \cap {\text{Y)}}$$

                    $${\text{Using the formula we get}}$$

                    $${\text{n(X) + n(Y) - n(X}} \cap {\text{Y) = (X}} \cup {\text{Y)}}$$

                    $$\therefore {\text{17 + 23 - n(X}} \cap {\text{Y) = 38}}$$

                    $$ \Rightarrow {\text{n(X}} \cap {\text{Y) = 40 - 38 = 2}}$$

    $${\textbf{Hence, the value is 2.}}$$

  • Question 2
    1 / -0
    If $$X$$ and $$Y$$ are two sets, then $$X\cap \left( X\cup Y \right)$$ equals
    Solution
    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 $$X\cap X=X$$
  • Question 3
    1 / -0
    If a relation from a finite set A having m elements to a finite set B having n elements, then the number of relations from A to B is
  • Question 4
    1 / -0
    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
    Solution
    Given $$A=\left\{ x:x\in R\left| x \right| <1 \right\} $$

    If $$x$$ is positive, $$\left| x \right| =x$$
    $$\therefore x<1$$

    If $$x$$ is negative, $$\left| x \right| =-x$$
    $$\therefore -x<1$$
    $$\therefore x>-1$$

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

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

    If $$\left| x-1 \right| $$ is positive, $$\left| x-1 \right| =x-1$$
    $$\therefore x-1\ge 1$$
    $$\therefore x\ge 2$$

    If $$\left| x-1 \right| $$ 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.

    $$\therefore D:-1\le x<2$$
  • Question 5
    1 / -0
    Consider the word $$W=MISSISSIPPI$$.
    If $$N$$ denotes the number of different selections of $$5$$ letters from the word $$W = MISSISSIPPI$$ then $$N$$ belongs to the set,
    Solution
    In $$MISSISSIPPI$$
    $$M=1\quad I=4\quad S=4\quad P=2$$
    $$5$$ 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 $$\left\{20, 21, 22, 23, 24\right\}$$
  • Question 6
    1 / -0
    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 
    Solution
    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.

    $$\therefore A\cap B=\phi $$
  • Question 7
    1 / -0
    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 
    Solution
    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 $$A\cap B\neq \phi $$.
  • Question 8
    1 / -0
    If $$X$$ and $$Y$$ are two sets, then $$X\cap \left( Y\cup X \right)'$$ equals
    Solution

    $$X\cap \left( X' \cap Y'  \right) =\left( X\cap X'  \right) \cap Y' =\phi \cap Y' =\phi$$
  • Question 9
    1 / -0
    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 
    Solution

  • Question 10
    1 / -0
    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:
    Solution
    65% of families own neither a cell phone nor a scooter
    35% of families own either a phone or a scooter
    (A$$\cup$$B)=A+B-(A$$\cap$$B)
    35%x=25%x+15%x-(A$$\cap$$B)
    (A$$\cap$$B)=5%x=$$\dfrac{5}{100}x=1500$$
    number of families=30000

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