1 Marks Question

Question 11 Mark

If A = {1, 2, 3}, B = {4, 5, 6}, the given following are relations from A to B? Give reason in support of your answer.
{(1, 6), (3,4), (5, 2)}

Answer

We have,
A = {1, 2, 3} and B = {4, 5, 6}
{(1, 6), (3, 4), (5, 2)} is not a relation from A to B as it is not a subset of A × B.

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Question 21 Mark

If $\text{R}=\{(\text{x, y}):\text{x},\text{ y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$ is a relation defined on the set Z of integers, then write domain of R.

Answer

We have,
$\text{R}=\{(\text{x, y}):\text{x},\text{ y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$
Now,
$\text{x}^2+\text{y}^2\leq4$
$\Rightarrow\text{y}^2\leq4-\text{x}^2$
$\Rightarrow\text{y}\leq\sqrt{4-\text{x}^2}$
Putting x = -2, -1, 0, 1, 2, we get $\text{y}\in\text{z}$
For other values of x, we get $\text{y}\notin\text{z}$
$\therefore\ \text{Domain(R)}=\{-2,-1,0,1,2\}$

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Question 31 Mark

If A = {1, 3, 5} and B = {2, 4}, list of elements of R, if $\text{R}=\{(\text{x, y}):\text{x, y}\in\text{A}\times\text{B and x}>\text{y}\}.$

Answer

We have,
A = {1, 3, 5} and B = {2, 4}
$\therefore$ A × B = {1, 3, 5} × {2, 4}
= {(1, 2), (1, 4), (3, 2), (3, 4), (5, 2), (5, 4)}
Clearly, $(3,2)\in\text{R},( 5,2)\in\text{R}$ and $(5, 4)\in\text{R}$
⇒ R = {(3, 2), (5, 2), (5, 4)}

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Question 41 Mark

Let A = {1, 2, 3} and $\text{R}=\{(\text{a, b}):|\text{a}^2-\text{b}^2|\leq5,\text{a, b}\in\text{A}\}.$ Then write R as set of ordered pairs.

Answer

We have,
A = {1, 2, 3} and $\text{R}=\{(\text{a, b}):|\text{a}^2-\text{b}^2|\leq5,\text{a, b}\in\text{A}\}$
For the elements of the given set A, we find that
$|1^2-1^2|\leq5,|2^2-2^2|\leq,5|3^2-3^2|\leq5,\\|1^2-2^2|\leq5,|2^2-1^2|\leq5,|2^2-3^2|\leq5,$ and $|3^2-2^2|\leq5$
$\Rightarrow\text{R}=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}$

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Question 51 Mark

If n(A) = 3, n(B) = 4, then write n(A × A × B).

Answer

We have,
n(A) = 3, n(B) = 4
$\therefore$ n(A × A × B) = n(A) × n(A × B) $\big[\because\ \text{n(A}\times\text{B})=\text{n(A)}\times\text{n}(\text{B})\big]$
= n(A) × n(A) × n(B)
= 3 × 3 × 4
= 36
⇒ n(A × A × B) = 36

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Question 61 Mark

If $\text{R}=\{(\text{x, y}):\text{x, y}\in\text{W},2\text{x}+\text{y}=8\},$ then write the domain and range of R.

Answer

We have,
$\text{R}=\{(\text{x, y}):\text{x, y}\in\text{W},2\text{x}+\text{y}=8\}$
Now,
2x + y = 8
⇒ y = 8 - 2x
Putting x = 0, 1, 2, 3, 4 we get, y = 8, 6, 4, 2, 0 respectively
For x > 4, $\text{y}\notin\text{w}$
$\therefore$ Domain(R) = {0, 1, 2, 3, 4} and
Range(R) = {8, 6, 4, 2, 0}
= {0, 2, 4, 6, 8}

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Question 71 Mark

If R = {(2, 1), (4, 7), (1, -2), ...}, then write the linear relation between the components of the ordered pairs of the relation R.

Answer

We have,
R = {(2, 1), (4, 7), (1, -2), ...}
Now,
1 = 3 × 2 - 5,
7 = 3 × 4 -5,
-2 = 3 × 1 - 5,
................
................
$\therefore$ y = 3x - 5

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Question 81 Mark

If A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}, write (A - C) × (B - C).

Answer

We have,
A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}
$\therefore$ A - C = {1, 2, 4} - {2, 5}
A - C = {1, 4}
and, B - C = {2, 4, 5} - {2, 5}
B - C = {4}
Now,
(A - C) × (B - C) = {1, 4} × {4}
(A - C) × (B - C) = {(1, 4), (4, 4)}

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Question 91 Mark

If A = {1, 2, 3}, B = {4, 5, 6}, the given following are relations from A to B? Give reason in support of your answer.
{(1, 5), (2, 6), (3, 4), (3, 6)}

Answer

We have,
A = {1, 2, 3} and B = {4, 5, 6}
{(1, 5), (2, 6), (3, 4), (3, 6)} is a subset of A × B, so it is a relation from A to B.

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Question 101 Mark

Let A = {1, 2, 3, 5}, B = {4, 6, 9} and R be a relation from A to B defined by R = {(x, y) : x - y is odd}. Write R in roster form.

Answer

We have,
A = {1, 2, 3, 5} and B = {4, 6, 9}
For the elements of the given sets A and B, we find
4 - 1 = 3, 6 - 1 = 5, 9 - 2 = 7, 4 - 3 = 1, 6 - 3 = 3, 4 - 5 = -1 and 6 - 5 = -1
$\therefore$ R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}

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Question 111 Mark

Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, write A and B.

Answer

Since (x, 1), (y, 2), (z, 1) are elements of A × B. Therefore,
$\text{x, y, z}\in\text{A}$ and $1, 2\in\text{B}$
It is given that n(A) = 3 and n(B) = 2
$\therefore\ \text{x, y, z}\in\text{A and n(A)}=3$
$\Rightarrow\text{A}=\{\text{x, y, z}\}$
$1,2\in\text{B and n(B)}=2$
$\Rightarrow\text{B}=\{1,2\}$

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Question 121 Mark

Let $\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{y}=2\text{x}-4\}.$ If (a, -2) and $(4,\text{b}^2)\in\text{R},$ then write the values of a and b.

Answer

We have,
$\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{y}=2\text{x}-4\}$
Now,
y = 2x - 4
Putting y = -2 and x = a, we get
-2 = 2a - 4
⇒ 4 - 2 = 2a
⇒ 2 = 2a
⇒ 2a = 2
$\Rightarrow\text{a}=\frac{2}{2}=1$
Putting y = b² and x = 4, we get
b²= 2 × 4 - 4
⇒ b² = 8 - 4
⇒ b² = 4
$\Rightarrow\text{b}=\pm2$
Hence, $\text{a}=1,\text{b}=\pm2$

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Question 131 Mark

A relation R is defined from a set A = {2, 3, 4, 5} to a set B = {3, 6, 7, 10} as follows:
$(\text{x, y})\in\text{R}\Leftrightarrow\text{x}$ is relatively prime to y.
Express R as a set of ordered pairs and determine its domain and range.

Answer

We have,
A = {2, 3, 4, 5} and B = {3, 6, 7, 10}
It is given that $(\text{x, y})\in\text{R}\Leftrightarrow\text{x}$ is relatively prime to y.
$\therefore\ (2,3)\in\text{R},(2,7)\in\text{R},(3,7)\in\text{R},(3,10)\in\text{R},\\\ \ \ \ \ (4,3)\in\text{R},(4,7)\in\text{R},(5,3)\in\text{R},\text{ and }(5,7)\in\text{R}$
Thus,
$\text{R}=\{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3),(5, 7)\}$
Clearly, Domain (R) = {2, 3, 4, 5} and Range = {3, 7, 10}

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Question 141 Mark

If A = {1, 2, 3}, B = {4, 5, 6}, the given following are relations from A to B? Give reason in support of your answer.

{(4, 2), (4, 3), (5, 1)}

Answer

We have,
A = {1, 2, 3} and B = {4, 5, 6}
{(4, 2), (4, 3), (5, 1)} is not a relation from A to B as it is not a subset of A × B.

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Question 151 Mark

If R is a relation from set A = {11, 12, 13} to set B = {8, 10, 12} defined by y = x - 3, then write R^-1

Answer

We have,
R is a relation from set A = {11, 12, 13} to be set B = {8, 10, 12} defined by y = x - 3
Putting x = 11, 12, 13, we get y = 8, 9, 10 respectively.
$\therefore\ (11, 8)\in\text{R},(12, 9)\notin\text{R}$ and $(13, 10)\in\text{R}$
$\therefore\ \text{R}=\{(11, 8), (13, 10)\}$
$\Rightarrow\text{R}^{-1}=\{(8,11),(10,13)\}$

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Question 161 Mark

If R is a relation defined on the set Z of integers by the rule $(\text{x, y})\in\text{R}\Leftrightarrow\text{x}^2+\text{y}^2=9,$ then write domain of R.

Answer

We have,
$(\text{x, y})\in\text{R}\Leftrightarrow\text{x}^2+\text{y}^2=9$
$\Rightarrow\text{y}^2=9-\text{x}^2$
$\Rightarrow\text{y}=\sqrt{9-\text{x}^2}$
Putting x = -3, 0, 3, we get $\text{y}=0,\pm3,0$ respectively.
For all other values of x, we get $\text{y}\notin\text{z}$
$\therefore\ \text{Domain(R)}=\{-3,0,3\}$

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