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

Given the sets $A\; =\; \backslash \{-1,\; 2,\; 3\backslash \}$ and $B\; =\; \backslash \{1,\; 3\backslash \}$.

1. $A\; \backslash times\; B$: To find the Cartesian product $A\; \backslash times\; B$, we pair each element of set A with each element of set B.

   $A\; \backslash times\; B\; =\; \backslash \{(-1,\; 1),\; (-1,\; 3),\; (2,\; 1),\; (2,\; 3),\; (3,\; 1),\; (3,\; 3)\backslash \}$

2. $B\; \backslash times\; A$: To find the Cartesian product $B\; \backslash times\; A$, we pair each element of set B with each element of set A.

   $B\; \backslash times\; A\; =\; \backslash \{(1,\; -1),\; (1,\; 2),\; (1,\; 3),\; (3,\; -1),\; (3,\; 2),\; (3,\; 3)\backslash \}$

3. $B\; \backslash times\; B$: To find the Cartesian product $B\; \backslash times\; B$, we pair each element of set B with each element of set B.

   $B\; \backslash times\; B\; =\; \backslash \{(1,\; 1),\; (1,\; 3),\; (3,\; 1),\; (3,\; 3)\backslash \}$

4. $A\; \backslash times\; A$: To find the Cartesian product $A\; \backslash times\; A$, we pair each element of set A with each element of set A.

   $A\; \backslash times\; A\; =\; \backslash \{(-1,\; -1),\; (-1,\; 2),\; (-1,\; 3),\; (2,\; -1),\; (2,\; 2),\; (2,\; 3),\; (3,\; -1),\; (3,\; 2),\; (3,\; 3)\backslash \}$

First, let's determine the elements of sets P and Q.

Set P consists of natural numbers (x ∈ ℕ) that are less than 3. Natural numbers start from 1. So, P = \{1, 2\}.

Set Q consists of whole numbers (x ∈ W) that are less than or equal to 2. Whole numbers start from 0. So, Q = \{0, 1, 2\}.

Now, we find the union and intersection of P and Q:

Union (P \cup Q): This is the set of all elements that are in P, or in Q, or in both.

P \cup Q = \{0, 1, 2\}

Intersection (P \cap Q): This is the set of all elements that are common to both P and Q.

P \cap Q = \{1, 2\}

Finally, we find the Cartesian product of (P \cup Q) and (P \cap Q):

(P \cup Q) \times (P \cap Q) = \{0, 1, 2\} \times \{1, 2\}

This involves pairing each element from \{0, 1, 2\} with each element from \{1, 2\}:

= \{(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)\}

First, let's determine the elements of sets A, B, and C.

Set A consists of whole numbers (x ∈ W) that are less than 2. Whole numbers start from 0. So, $A\; =\; \backslash \{0,\; 1\backslash \}$.

Set B consists of natural numbers (x ∈ ℕ) such that 1 < x < 5. So, $B\; =\; \backslash \{2,\; 3,\; 4\backslash \}$.

Set C is given as $C\; =\; \backslash \{3,\; 5\backslash \}$.

1. $A\; \backslash times\; (B\; \backslash cap\; C)$:

   First, find the intersection of B and C (B \cap C), which is the set of elements common to both B and C.

   B \cap C = \{3\}

   Now, find the Cartesian product of A and (B \cap C):

   A \times (B \cap C) = \{0, 1\} \times \{3\}

   Pairing each element of A with the element of (B \cap C) gives:

   = \{(0, 3), (1, 3)\}

2. $A\; \backslash times\; (B\; \backslash cup\; C)$:

   First, find the union of B and C (B \cup C), which is the set of all elements in either B or C or both.

   B \cup C = \{2, 3, 4, 5\}

   Now, find the Cartesian product of A and (B \cup C):

   A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\}

   Pairing each element of A with each element of (B \cup C) gives:

   = \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\}

We use the property that two ordered pairs are equal if and only if their corresponding elements are equal.

1. Given $(2a\; +\; b,\; a\; -\; b)\; =\; (8,\; 3)$:

   Equating the corresponding elements, we get two linear equations:

   1. 2a + b = 8

   2. a - b = 3

   From equation (2), we can express b in terms of a: b = a - 3.

   Substitute this expression for b into equation (1):

   2a + (a - 3) = 8

   3a - 3 = 8

   3a = 11

   a = \frac{11}{3}

   Now, substitute the value of a back into the expression for b:

   b = a - 3 = \frac{11}{3} - 3 = \frac{11 - 9}{3} = \frac{2}{3}

   So, a = \frac{11}{3} and b = \frac{2}{3}.

2. Given $\backslash left(\backslash frac\{a\}\{4\},\; a\; -\; 2b\backslash right)\; =\; (0,\; 6\; +\; b)$:

   Equating the corresponding elements, we get two equations:

   1. \frac{a}{4} = 0

   2. a - 2b = 6 + b

   From equation (1), multiplying both sides by 4 gives:

   a = 0

   Now, substitute a = 0 into equation (2):

   0 - 2b = 6 + b

   -2b = 6 + b

   Subtract b from both sides:

   -3b = 6

   Divide by -3:

   b = -2

   So, a = 0 and b = -2.

We are given the set $A\; =\; \backslash \{1,\; 2,\; 3,\; 4,\; 5\backslash \}$ and $S\; =\; \backslash \{(x,\; y)\; :\; x\; \backslash in\; A,\; y\; \backslash in\; A\backslash \}$. This means S is the Cartesian product $A\; \backslash times\; A$, and we need to find pairs (x, y) from this set that satisfy the given conditions.

1. Condition: $x\; +\; y\; =\; 5$

   We need to find pairs (x, y) from A \times A such that their sum is exactly 5. Let's check possible pairs:

   If x = 1, y = 4 (since 1 + 4 = 5). Both 1 and 4 are in A. So, (1, 4) is a solution.

   If x = 2, y = 3 (since 2 + 3 = 5). Both 2 and 3 are in A. So, (2, 3) is a solution.

   If x = 3, y = 2 (since 3 + 2 = 5). Both 3 and 2 are in A. So, (3, 2) is a solution.

   If x = 4, y = 1 (since 4 + 1 = 5). Both 4 and 1 are in A. So, (4, 1) is a solution.

   If x = 5, y = 0 (since 5 + 0 = 5), but 0 is not in A.

   The set of ordered pairs satisfying x + y = 5 is \{(1, 4), (2, 3), (3, 2), (4, 1)\}.

2. Condition:

   We need to find pairs (x, y) from $A\; \backslash times\; A$ such that their sum is less than 5.

   • If x = 1:
     • y = 1 (1 + 1 = 2 < 5)
     • y = 2 (1 + 2 = 3 < 5)
     • y = 3 (1 + 3 = 4 < 5)
     • y = 4 (1 + 4 = 5, not < 5)
     \nSo, (1, 1), (1, 2), (1, 3) are solutions.
   • If x = 2:
     • y = 1 (2 + 1 = 3 < 5)
     • y = 2 (2 + 2 = 4 < 5)
     • y = 3 (2 + 3 = 5, not < 5)
     \nSo, (2, 1), (2, 2) are solutions.
   • If x = 3:
     • y = 1 (3 + 1 = 4 < 5)
     • y = 2 (3 + 2 = 5, not < 5)
     \nSo, (3, 1) is a solution.
   • If x = 4 or x = 5, the smallest possible sum with y=1 is 4+1=5 or 5+1=6, neither of which is less than 5.

   The set of ordered pairs satisfying is $\backslash \{(1,\; 1),\; (1,\; 2),\; (1,\; 3),\; (2,\; 1),\; (2,\; 2),\; (3,\; 1)\backslash \}$.

3. Condition: $x\; +\; y\; >\; 8$

   We need to find pairs (x, y) from $A\; \backslash times\; A$ such that their sum is greater than 8.

   • If x = 1, 2, 3, the maximum sum possible is 3 + 5 = 8, which is not greater than 8.
   • If x = 4:
     • y = 5 (4 + 5 = 9 > 8). So, (4, 5) is a solution.
     • y = 4 (4 + 4 = 8, not > 8)
   • If x = 5:
     • y = 4 (5 + 4 = 9 > 8). So, (5, 4) is a solution.
     • y = 5 (5 + 5 = 10 > 8). So, (5, 5) is a solution.

   The set of ordered pairs satisfying $x\; +\; y\; >\; 8$ is $\backslash \{(4,\; 5),\; (5,\; 4),\; (5,\; 5)\backslash \}$.