Tag: relations

Questions Related to relations

If A $=$ {1, 2}, B $=$ {3, 4}, then A$\times$B $=$

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

  1. {(1, 3), (1, 4), (2, 3), (2, 4)}

  2. {(1, 1), (2, 2), (3, 3), (4, 4)}

  3. {(4, 1), (3, 1), (4, 2), (3, 2)}

  4. All the above

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Correct Option: A
Explanation:

Given, $A={1,2}$ and $B={3,4}$.

Then,
$A\times B$
$={(1,3),(1,4),(2,3),(2,4)}$. [ Using (1) and (2)]

If ,$(x-1, y+2)= (7, 5)$ then values of $x$ and $y$ are

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

  1. $5$,$8$

  2. $8$,$3$

  3. $-1$,$5$

  4. $7$,$1$

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Correct Option: B
Explanation:

as the given ordered pairs are equal,

$x-1=7$  and  $y+2=5$
$\therefore x=8$ and $y=3$

Ordered pairs (a, 3) and (5, x) are equal ,the values of $a$ and $x$ are

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

  1. $2$ and $4$

  2. $3$ and $6$

  3. $5$ and $3$

  4. $1$ and $-1$

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Correct Option: C
Explanation:

If two ordered pairs are equal, then both x - coordinates are equal and both y-coordinates are equal.

Since  $ (a, 3)  =  (5,x)  $ then ,$ a = 5 , x = 6 $

Determine all ordered pairs that satisfy $(x - y)^{2} + x^{2} = 25$, where $x$ and $y$ are integers and $x \geq 0$. Find the number of different values of $y$ that occur

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

  1. $3$

  2. $4$

  3. $5$

  4. $6$

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Correct Option: B
Explanation:

${ \left( x-y \right)  }^{ 2 }+{ x }^{ 2 }=25$
Now, ${ 3 }^{ 2 }+{ 4 }^{ 2 }={ 5 }^{ 2 }$
$9+16=25$
$\therefore $There are 2 possibilities:
$I.{ \left( x-y \right)  }^{ 2 }=9$ and ${ x }^{ 2 }=16$
$\therefore x=\pm 4$ and $x-y=\pm 3$
$\left( i \right) .x-y=3\Rightarrow \left( 4,1 \right) $ and $\left( -4,-7 \right) $
$\left( ii \right) .x-y=-3\Rightarrow \left( 4,7 \right) $ and $\left( -4,-1 \right) $
$II.{ \left( x-y \right)  }^{ 2 }=16$ and ${ x }^{ 2 }=9$
$\therefore x=\pm 3$ and $x-y=\pm 4$
$\left( i \right) .x-y=4\Rightarrow \left( 3,-1 \right) $ and $\left( -3,-7 \right) $
$\left( ii \right) .x-y=-4\Rightarrow \left( 3,7 \right) $ and $\left( -3,-1 \right) $
$\therefore $ Different values of y are $1,-1,7,-7$
$\therefore 4$ different values of y occur.