Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

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

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

  1. ${ 2 }^{ 39 }$
  2. ${ 39 }^{ 2 }$
  3. $78$
  4. $351$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

  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) $
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product
Suppose $S=\{1,2\}$  and $T=\{a,b\}$  then $T \times S$
  1. <span class="math">$(a,1),(a,2),(b,1),(b,2)$
    <span class="MJX_Assistive_MathML">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
Reveal answer Fill a bubble to check yourself
A Correct answer
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)\} $
Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

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

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

  4. None of the above
Reveal answer Fill a bubble to check yourself
A,B Correct answer
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:
$A={a,b,c,d}$
$B={c,d,e}$
$A-B={a,b}$

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

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

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

  1. $\{b,b\}$
  2. $\{b,c\}$
  3. $\{b,d\}$

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $A \times A$

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

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

  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)
Reveal answer Fill a bubble to check yourself
D Correct answer
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.