CBSE Class 11 Mathematics Chapter 2: Relations and Functions NCERT Solutions

We are given two ordered pairs that are equal: $\backslash left(\backslash frac\{x\}\{3\}+1,\; y-\backslash frac\{2\}\{3\}\backslash right)=\backslash left(\backslash frac\{5\}\{3\},\; \backslash frac\{1\}\{3\}\backslash right)$. For two ordered pairs to be equal, their corresponding elements must be equal. Therefore, we can set up two separate equations:

1. Equating the first elements: $\backslash frac\{x\}\{3\}\; +\; 1\; =\; \backslash frac\{5\}\{3\}$
2. Equating the second elements: $y\; -\; \backslash frac\{2\}\{3\}\; =\; \backslash frac\{1\}\{3\}$

Now, we solve each equation for its respective variable:

For the first equation (x):

\frac{x}{3} = \frac{5}{3} - 1

\frac{x}{3} = \frac{5}{3} - \frac{3}{3}

\frac{x}{3} = \frac{2}{3}

Multiplying both sides by 3, we get:

x = 2

For the second equation (y):

y = \frac{1}{3} + \frac{2}{3}

y = \frac{1+2}{3}

y = \frac{3}{3}

y = 1

Thus, the values are x=2 and y=1.

We are given that set A has 3 elements. Let the number of elements in set A be denoted as |A|, so |A| = 3.

We are also given set B as B = \{3, 4, 5\}. The number of elements in set B, denoted as |B|, is 3.

The Cartesian product of two sets A and B, denoted by A \times B, is the set of all possible ordered pairs (a, b) where a \in A and b \in B. The number of elements in the Cartesian product A \times B is calculated by multiplying the number of elements in set A by the number of elements in set B.

Number of elements in A \times B = |A| \times |B|

Substituting the given values:

|A \times B| = 3 \times 3

|A \times B| = 9

Therefore, the number of elements in A \times B is 9.

We are given two sets: $G\; =\; \backslash \{7,\; 8\backslash \}$ and $H\; =\; \backslash \{5,\; 4,\; 2\backslash \}$.

The Cartesian product $G\; \backslash times\; H$ is the set of all ordered pairs $(g,\; h)$ such that $g\; \backslash in\; G$ and $h\; \backslash in\; H$. To find $G\; \backslash times\; H$, we pair each element of G with each element of H:

$G\; \backslash times\; H\; =\; \backslash \{(7,\; 5),\; (7,\; 4),\; (7,\; 2),\; (8,\; 5),\; (8,\; 4),\; (8,\; 2)\backslash \}$

Similarly, the Cartesian product $H\; \backslash times\; G$ is the set of all ordered pairs $(h,\; g)$ such that $h\; \backslash in\; H$ and $g\; \backslash in\; G$. To find $H\; \backslash times\; G$, we pair each element of H with each element of G:

$H\; \backslash times\; G\; =\; \backslash \{(5,\; 7),\; (5,\; 8),\; (4,\; 7),\; (4,\; 8),\; (2,\; 7),\; (2,\; 8)\backslash \}$

Note that $G\; \backslash times\; H\; \backslash neq\; H\; \backslash times\; G$ because the order of elements in the ordered pairs is different.

1. False. To find $P\; \backslash times\; Q$, we pair each element of P with each element of Q. Given $P\; =\; \backslash \{m,\; n\backslash \}$ and $Q\; =\; \backslash \{n,\; m\backslash \}$. $P\; \backslash times\; Q\; =\; \backslash \{(m,\; n),\; (m,\; m),\; (n,\; n),\; (n,\; m)\backslash \}$. The correct statement is: If $P\; =\; \backslash \{m,\; n\backslash \}$ and $Q\; =\; \backslash \{n,\; m\backslash \}$, then $P\; \backslash times\; Q\; =\; \backslash \{(m,\; n),\; (m,\; m),\; (n,\; n),\; (n,\; m)\backslash \}$.
2. True. By the definition of the Cartesian product, if A and B are non-empty sets, then $A\; \backslash times\; B$ is the set of all ordered pairs $(x,\; y)$ where $x\; \backslash in\; A$ and $y\; \backslash in\; B$. Since A and B are non-empty, there will be at least one such ordered pair, making $A\; \backslash times\; B$ a non-empty set.
3. True. The intersection of any set with the empty set ($\backslash Phi$) is always the empty set. So, $B\; \backslash cap\; \backslash Phi\; =\; \backslash Phi$. The Cartesian product of any set with the empty set is also the empty set. Therefore, $A\; \backslash times\; (B\; \backslash cap\; \backslash Phi)\; =\; A\; \backslash times\; \backslash Phi\; =\; \backslash Phi$.