Case study (4 Marks)

Question 14 Marks

Ordered Pairs The ordered pair of two elements a and 3 is denoted by (a, b) : a is first element (or first component) and d is second element (or second component). Two ordered pairs are equal if their corresponding elements are equal. ie. (a, b) = (c, d)

⇒ a = c and b = d

Cartesian Product of Two Sets For two non-empty sets A and B, the cartesian product A . B is the set of all ordered pairs of elements from sets Aand B. In symbolic form, it can be written as

$\text{A}\cdot\text{B}=\{(\text{a},\text{b}):\text{a}\in\text{A},\text{b}\in\text{B}\}$

Based on the above topics, answer the following questions.

If (a - 3, 6 + 7) = (3, 7), then the value of aand d are:

6, 0

3, 7

7, 0

3, -7

If (x + 6, y - 2) = (0, 6), then the value of x and y are:

6, 8

-6, -8

-6, 8

6, -8

If (x + 2, 4) = (5, 2x + y), then the value of x and y are:

-3, 2

3, 2

-3, -2

Let A and B be two sets such that A . B consists of 6 elements. If three elements of A . B are (1, 4), (2, 6) and (3, 6), then

(A . B) = (B . A)

$(\text{A}\cdot\text{B})\neq(\text{B}\cdot\text{A})$

A . B = {(1, 4), (1, 6), (2, 4)}

None of the above

If m(A . B) = 45, then n(A) cannot be

Answer

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Question 24 Marks

Method to Find the Sets When Cartesian Product is Given For finding these two sets, we write first element of each ordered pair in first set say $A$ and corresponding second element in second set $B ($say$).$ Number of Elements in Cartesian Product of Two Sets If there are p elements in set $A$ and $g$ elements in set $B,$ then there will be $pq$ elements in $A . B$ i.e. if $n(A) = p$ and $n(B) = q,$ then $n(A . B) = pq.$
Based on the above two topic, answer the following questions.

If $A . B = \{(a, 1), (b, 3), (a, 3), (b, 1), (a, 2), (b, 2)\}.$ Then, $A$ and $B$ are:

$\{1, 3, 2\}, \{a, b\}$
$\{a, b\}, \{1, 3\}$
$\{a, b\}, \{1, 3, 2\}$
None of these

If the set $A$ has $3$ elements and set $B$ has $4$ elements, then the number of elements in $A . B$ is:

$3$
$4$
$7$
$12$

$A$ and $B$ are two sets given in such a way that $A . B$ contains $6$ elements. If three elements of $A . B$ are $(1, 3), (2, 5)$ and $(3, 3)$, then $A, B$ are:

$\{1, 2, 3\}, \{3, 5\}$
$\{3, 5,\}, \{1, 2, 3\}$
$\{1, 2\}, \{3, 5\}$
$\{1, 2, 3\}, \{5\}$

The remaining elements of $A . B$ in $(iii)$ is:

$(5, 1), (3, 2), (3, 5)$
$(1, 5), (2, 3), (3, 5)$
$(1, 5), (3, 2), (5, 3)$
None of the above

The cartesian product $P . P$ has $16$ elements among which are found $(a, 1)$ and $(b, 2).$ Then, the set $P$ is:

$\{a, b\}$
$\{1, 2\}$
$\{a, b,1, 2\}$
$\{0, b, 1, 2, 4\}$

Answer

$(c)\ \{a, b\}, \{1, 3, 2\}$
Solution:
Here, first element of each ordered pair of $A . B$ gives the elements of set A and corresponding second element gives the elements of set $B.$
$\therefore A = \{a, b\}$ and $B = \{1, 3, 2\}$
Note We write each element only one time in set, if it occurs more than one time.

$(d) 12$

Solution:
Given, $n (A) = 3$ and $n(B) = 4.$
$\therefore$ The number of elements in $A . B$ is:
$n(A . B) = n(A) . n(B) = 3 . 4 = 12$

$(a) \{1, 2, 3\}, \{3, 5\}$

Solution:
It is given that $(1, 3), (2, 5)$ and $(3, 3)$ are in $A . B.$ It follows that $1, 2, 3$ are elements of $A$ and $3, 5$ are elements of $B.$
$\therefore A = \{1, 2, 3\}$ and $B = \{3, 5\}$

$(b) (1, 5), (2, 3), (3, 5)$

Solution:
$\because A = \{1, 2, 3\}$ and $B = \{3, 5\}$
$\therefore A = \{1, 2, 3\}$ and $B = \{3, 5\}$
$= \{(1, 3), (1, 5), (2 3), (2, 5), (3, 3), (3, 5)\}$
Hence, the remaining elements of $(A . B)$ are $(1, 5), (2, 3), (3, 5).$

$\{a, b,1, 2\}$

Solution:
Given, $n(P . P) = 16$
$\Rightarrow n(P) . n(P) = 16$
$\Rightarrow n(P) = 4 ....(i)$
Now, as $(\text{a},1)\in\text{P}\cdot\text{P}$
$\therefore\text{a}\in\text{P}$ and $1\in\text{P}$
Again, $(\text{b},2)\in\text{P}\cdot\text{P}$
$\therefore\text{b}\in\text{P}$ and $2\in\text{P}$
$\Rightarrow\text{a},\text{b},1,2\in\text{P}$
From Eq. $(i),$ it is clear that $P$ has exactly four elements.

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MATHS Case study (4 Marks)

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