Tag: mathematics and statistics

Questions Related to mathematics and statistics

If two sets $A$ and $B$ are having $39$ elements in common, then the number of elements common to each of the sets $A\times B$ and $B\times A$ are

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. ${ 2 }^{ 39 }$

  2. ${ 39 }^{ 2 }$

  3. $78$

  4. $351$

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Correct Option: B
Explanation:

If set $A$ and set $B$ have $39$ common elements, then the number of common elements in set $A\times B$ and set $B\times A\,=39^2$

If ${ y }^{ 2 }={ x }^{ 2 }-x+1$ and $\quad { I } _{ n }=\int { \cfrac { { x }^{ n } }{ y }  } dx$ and $A{ I } _{ 3 }+B{ I } _{ 2 }+C{ I } _{ 1 }={ x }^{ 2 }y$ then ordered triplet $A,B,C$ is

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right) $

  2. $\left( 3,1,0 \right) $

  3. $\left( 1,-1,2 \right) $

  4. $\left( 3,-\cfrac { 5 }{ 2 } ,2 \right) $

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Correct Option: A
Explanation:

We have given


${y^2} = {x^2} - x + 1$

${I _n} =\displaystyle \int {\dfrac{{{x^n}}}{y}dx} $
And,

$A{I _3} + B{I _2} + C{I _1} = {x^2}y$

Order triplet $A,\, B,\,C$ $ = \left( {\dfrac{1}{2},\,\dfrac{{ - 1}}{2},1} \right)$

Hence, the option $(A)$ is correct.

Suppose $S=\{1,2\}$  and $T=\{a,b\}$  then $T \times S$
mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $(a,1),(a,2),(b,1),(b,2)$
    B×A={(a,1),(a,2),(b,1),(b,2)}

  2. $(1,a),(2,b),(b,1),(b,2)$

  3. $(a,1),(a,2),(1,b),(2,b)$

  4. None of the above

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Correct Option: A
Explanation:
Given : $S=\{ 1,2\} ,T=\{ a,b\} $
$T\times S$ is the set of ordered pair of elements of $T$ and $S$ respectively
Then, $T\times S=\{a,b\}\times \{1,2\}$
$T\times S=\{ (a,1),(a,2),(b,1),(b,2)\} $

Given $A={b,c,d}$ and  $B={x,y}$ : find element of  $A\times B$ .

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. ${b,x}$

  2. ${b,y}$

  3. ${c,x}$

  4. All of the above

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Correct Option: D
Explanation:

Given $A={b,c,d}$ and  $B={x,y}$, then

$A\times B={(b,x),(c,x),(d,x),(b,y),(c,y),(d,y)}$
Hence, all the elements given in options are elements of $A\times B$.

$M={0,1,2}$ and $N={1,2,3}$: find (N-M) $\times$(N $\cap$M)

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. ${3,1}$

  2. ${3,2}$

  3. ${3,3}$

  4. None of the above

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Correct Option: A,B
Explanation:

For two sets, A and B the set difference of set B from set A is the set of all element in A but not in B.
Example:
$A={a,b,c,d}$
$B={c,d,e}$
$A-B={a,b}$

$M={0,1,2}$
$N={1,2,3}$
$N-M={3}$
$N \cap M={1,2}$
$(N-M)\times (N\cap M)$ $={3,1},{3,2}$

Given M={0,1,2} and N={1,2,3}, then (M $\cup$ N) $\times$(M-N) contains

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. ${0,0}$

  2. ${1,0}$

  3. ${2,0}$

  4. ${3,0}$

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Correct Option: A,B,C,D
Explanation:
$M=\{0,1,2\}$
$N=\{1,2,3\}$
$M\cup N=\{0,1,2,3\}$
$M-N=\{0\}$
$(M\cup N)\times (M-N)$ $=\{0,0\},\{1,0\},\{2,0\},\{3,0\}$

If $A={b,c,d}$ and $B={x,y}$. Find which of the following are elements of $A \times A$.

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. ${b,b}$

  2. ${b,c}$

  3. ${b,d}$

  4. All of the above

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Correct Option: D
Explanation:

$A\times A \Rightarrow$ the first element will be from $A$ and the second element will also be from $A$.

$A \times A = \left{{b,b}, {b,c}, {b,d},{c,b},{c,c},{c,d},{d,b},{d,c},{d,d}\right}$
Thus, all the options $A,B$ and $C$ are the elements of $A \times A$

$n(A)=4 $ and  $n(B) =5$: $n(A \times B)=$

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. $20$

  2. $10$

  3. $30$

  4. None of the above

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Correct Option: A
Explanation:

If $n(A)=m$,and $n(B)=n$,then $n(A\times B)=mn$

so$n(A\times B)=5.4=20$

n(A)=m and n(B)=n ; then

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. n(A)+n(B)=n(A+B)

  2. n(A)-n(B)=n(A+B)

  3. A$\times$B=mn

  4. n(A) $\times$n(B =n(A $\times$B)

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Correct Option: D
Explanation:

C is not even logical while in the first 2 cases:


If there are elements common in A and B L.H.S of the first statement will be greater than R.H.S.

Clearly option B is also untrue as the combination of A and B cannot have lesser number of elements than A alone.

The last option has to be true as cartesian product of 2 sets gives a matrix having n(A) and n(B) as columns and rows.whose product will give the number of elements in the matrix.

n (A $\times$ B) =

mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

  1. n(A) x n(B)

  2. n(A $\bigcap$ B)

  3. n(A $\bigcup$ B)

  4. all of these

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Correct Option: A