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Relation · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 11 questions.

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Question 14 Marks
Prove that: $(\text{A}\cap\text{B})\times\text{C}=(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$
Answer
Let (a, b) bean arbitrary element of $(\text{A}\cap\text{B})\times\text{C}.$ Then,$(\text{a},\text{b})\in(\text{A}\cap\text{B})\times\text{C}$
$\Rightarrow\text{a}\in\text{A}\cap\text{B}$ and $\text{b}\in\text{C}$
$\Rightarrow(\text{a}\in\text{A}\text{ or a}\in\text{B})$ and $\text{b}\in\text{C}$ [By defination]
$\Rightarrow(\text{a}\in\text{A}\text{ and b}\in\text{C})\text{ or }(\text{a}\in\text{B}\text{ and b}\in\text{C})$
$\Rightarrow(\text{a},\text{b})\in\text{A}\times\text{C and (a, b)}\in\text{B}\times\text{C}$
$\Rightarrow(\text{a},\text{b})\in(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$
$\Rightarrow(\text{a, b})\in(\text{A}\cap\text{B})\times\text{C}$
$\Rightarrow(\text{a, b})\in(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$
$\Rightarrow(\text{A}\cap\text{B})\times\text{C}\subseteq\text{A}\times(\text{B}\times\text{C})\ ...(\text{i})$
Let (x, y) be an arbitrary element of $(\text{A}\times\text{C})\cap(\text{B}\times\text{C}).$ Then,
$(\text{x, y})\in(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$
$\Rightarrow(\text{x, y)}\in\text{A}\times\text{C}\text{ and (x, y)}\in\text{B}\times\text{C}$ [By defination]
$\Rightarrow(\text{x}\in\text{A and y}\in\text{C) and (x}\in\text{B and y}\in\text{C})$
$\Rightarrow(\text{x}\in\text{A or x}\in\text{B})\text{ and y}\in\text{C}$
$\Rightarrow\text{x}\in\text{A}\cap\text{B and y}\in\text{C}$
$\Rightarrow(\text{x},\text{y})\in(\text{A}\cap\text{B})\times\text{C}$
$\Rightarrow(\text{x},\text{y})\in(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$
$\Rightarrow(\text{x, y})\in(\text{A}\cap\text{B})\times\text{C}$
$\Rightarrow(\text{A}\times\text{C})\cap(\text{B}\times\text{C})\subseteq(\text{A}\cap\text{B})\times\text{C}\ ...(\text{ii})$
Using equation (i) and equation (ii), we get
$(\text{A}\cap\text{B})\times\text{C}=(\text{A}\times\text{C})\cap(\text{B}\times\text{C})$
Hence proved.
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Question 24 Marks
Prove that: $(\text{A}\cup\text{B})\times\text{C}=(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$
Answer
Let (a, b) bean arbitrary element of $(\text{A}\cup\text{B})\times\text{C}.$ Then, $(\text{a},\text{b})\in(\text{A}\cup\text{B})\times\text{C}$ $\Rightarrow\text{a}\in\text{A}\cup\text{B}$ and $\text{b}\in\text{C}$ [By defination] $\Rightarrow(\text{a}\in\text{A}\text{ or a}\in\text{B})$ and $\text{b}\in\text{C}$ [By defination] $\Rightarrow(\text{a}\in\text{A}\text{ and b}\in\text{C})\text{ or }(\text{a}\in\text{B}\text{ and b}\in\text{C})$ $\Rightarrow(\text{a},\text{b})\in\text{A}\times\text{C or (a, b)}\in\text{B}\times\text{C}$ $\Rightarrow(\text{a},\text{b})\in(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ $\Rightarrow(\text{a, b})\in(\text{A}\cup\text{B})\times\text{C}$ $\Rightarrow(\text{a, b})\in(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ $\Rightarrow(\text{A}\cup\text{B})\times\text{C}\subseteq(\text{A}\times\text{C})\cup(\text{B}\times\text{C})\ ...(\text{i})$ Again, let (x, y) be an abitrary element of $(\text{A}\times\text{C})\cup(\text{B}\times\text{C}).$ Then, $(\text{x, y})\in(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ $\Rightarrow(\text{x, y)}\in\text{A}\times\text{C}\text{ or (x, y)}\in\text{B}\times\text{C}$ $\Rightarrow\text{x}\in\text{A and y}\in\text{C or x}\in\text{B and y}\in\text{C}$ $\Rightarrow(\text{x}\in\text{A or x}\in\text{B})\text{ and y}\in\text{C}$ $\Rightarrow\text{x}\in\text{A}\cup\text{B and y}\in\text{C}$ $\Rightarrow(\text{x},\text{y})\in(\text{A}\cup\text{B})\times\text{C}$ $\Rightarrow(\text{x},\text{y})\in(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ $\Rightarrow(\text{x, y})\in(\text{A}\cup\text{B})\times\text{C}$ $\Rightarrow(\text{A}\times\text{C})\cup(\text{B}\times\text{C})\subseteq(\text{A}\cup\text{B})\times\text{C}\ ...(\text{ii})$ Using equation (i) and equation (ii), we get $(\text{A}\cup\text{B})\times\text{C}=(\text{A}\times\text{C})\cup(\text{B}\times\text{C})$ Hence proved.
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Question 34 Marks
If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find $(\text{A}\times\text{B})\cap(\text{A}\times\text{C})$
Answer
We have, $\text{A}=\{1,2,3\},\text{ B}=\{3,4\}$ and $\text{C}=\{4,5,6\}$ $\therefore\ \text{A}\times\text{B}=\{1,2,3\}\cap\{3,4\}$ $=\{(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)\}$ And, $\text{A}\times\text{C}=\{1,2,3\}\times\{4,5,6\}$ $= \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$ $(\text{A}\times\text{B})\cap(\text{A}\times\text{C})=\{(1,4),(2,4),(3,4)\}$
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Question 44 Marks
If $\text{A}\times\text{b}\subseteq\text{C}\times\text{D and A}\times\text{B}=\phi,$ prove that $\text{A}\subseteq\text{C and B}\subseteq\text{D}$
Answer
Let (a, b) be an arbitrary element of A × B. then, $(\text{a},\text{b})\in\text{A}\times\text{B}$ $\Rightarrow\text{a}\in\text{A}\text{ and b}\in\text{B}\ ...(\text{i})$ Now, $(\text{a, b})\in\text{A}\times\text{B}$ $\Rightarrow(\text{a},\text{ b})\in\text{C}\times\text{D}$ $\big[\because\text{ A}\times\text{B}\subseteq\text{C}\times\text{D}\big]$ $\Rightarrow\text{a}\in\text{C and b}\in\text{D}\ ...(\text{ii})$ $\therefore\ \text{a}\in\text{A}$ $\Rightarrow\text{a}\in\text{C}$ [Using (i) and (ii)] $\Rightarrow\text{A}\subseteq\text{C}$ and, $\text{b}\in\text{B}$ $\Rightarrow\text{b}\in\text{D}$ $\Rightarrow\text{B}\subseteq\text{D}$ Hence, proved.
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Question 54 Marks
If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find $\text{A}\times(\text{B}\cup\text{C})$
Answer
We have, $\text{A}=\{1,2,3\},\text{ B}=\{3,4\}$ and $\text{C}=\{4,5,6\}$ $\therefore\ \text{B}\cup\text{C}=\{3,4\}\cup\{4,5,6\}=(3,4,5,6)$ $\therefore\ \text{A}\times(\text{B}\cup\text{C})=\{1,2,3\}\times\{3,4,5,6\}$ $=\{(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2 ,4), (2, 5), (2, 6), (3, 3),(3, 4), (3 ,5), (3, 6)\}$
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Question 64 Marks
If A = {1, 2, 3}, B = {4}, C = {5}, then verify that: $\text{A}\times(\text{B}-\text{C})=(\text{A}\times\text{B})-(\text{A}\times\text{C})$
Answer
We have, $\text{A}=\{1,2,3\},\text{ B}=\{4\}$ and $\text{C}=\{5\}$ $\therefore\ \text{B}-\text{C}=\{4\}$ $\therefore\ \text{A}\times(\text{B}-\text{C})=\{1,2,3\}\times\{4\}$ $\Rightarrow\text{A}\times(\text{B}-\text{C})=\{(1, 4), (2, 4), (3, 4)\}\ ...(\text{i})$ Now, $\text{A}\times\text{B}=\{1, 2, 3\}\times\{4\}$ $=\{(1, 4), (2, 4), (3, 4)\}$ and, $\text{A}\times\text{C}=\{1,2,3\}\times\{5\}$ $=\{(1, 5) , (2, 5), (3, 5)\}$ $\therefore\ (\text{A}\times\text{B})-(\text{A}\times\text{C})=\{(1, 4), (2, 4), (3, 4)\}\ ...(\text{ii})$ From equation (i) and (ii), we get $\text{A}\times (\text{B}-\text{C})=(\text{A}\times\text{B})-(\text{A}\times\text{C})$ Hence verified.
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Question 74 Marks
Determine the domain and range of the relation R defined by: $\text{R}=\{(\text{x, x,}+5):\text{x}\in\{0,1,2,3,4,5\}\}$
Answer
We have, $\text{R}=\{(\text{x, x,}+5):\text{x}\in\{0,1,2,3,4,5\}\}$ For the elements of the given sets, we find that R = {(0, 5), (1, 6), (2, 7), (3, 8),(4, 9), (5, 10)} Clearly, Domain (R) = {0, 1, 2, 3, 4, 5} and Range (R) = {5, 6, 7, 8, 9, 10}
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Question 84 Marks
If A = {-1, 1}, find A × A × A.
Answer
We have, A = {-1, 1} $\therefore$ A × A = {-1, 1} × {-1, 1} = {(-1, -1), (-1, 1), (1, -1), (1, 1)} $\therefore$ A × A × A = {-1, 1} × {(-1, -1), (-1, 1), (1, -1), (1, 1)} = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}
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Question 94 Marks
If A = {2, 3}, B = {4, 5}, C = {5, 6}, find $\text{A}\times(\text{B}\cap\text{C}),\text{ A}\times(\text{B}\cap\text{C}),(\text{A}\times\text{B})\cup(\text{A}\times\text{C}).$
Answer
We have, $\text{A}=\{2,3\},\text{ B}=\{4,5\}$ and $\text{C}=\{5,6\}$ $\therefore\ \text{B}\cup\text{C}=\{4,5\}\cup\{5,6\}=\{4,5,6\}$ $\therefore\ \text{A}\times\{\text{B}\cup\text{C}\}=\{2,3\}\times\{4,5,6\}$ $= \{(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$ Now, $\text{B}\cap\text{C}=\{4,5\}\cap\{5,6\}=\{5\}$ $\therefore\ \text{A}\times(\text{B}\cap\text{C})=\{2,3\}\times\{5\}$ $\text{A}\times(\text{B}\cap\text{C})=\{(2,5),(3,5)\}$ Now, $\text{A}\times\text{B}=\{2,3\}\times\{4,5\}$ $= \{(2,4), (2,5), (3, 4), (3, 5)\}$ and $\text{A}\times\text{C}=\{2,3\}\times\{5,6\}$ $=\{(2, 5), (2, 6), (3, 5), (3, 6)\}$ $\therefore\ (\text{A}\times\text{B})\cup(\text{A}\times\text{C})= \{(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$
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Question 104 Marks
Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that: $\text{A}\times\text{C}\subset\text{B}\times\text{D}$
Answer
We have, A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8} $\therefore$ B × D = {1, 2, 3, 4} × {5, 6, 7, 8} {(1, 5), (1, 6), (1, 7), (1, 8) , (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)} ....(i) and, A × C = (1, 2) × (5, 6) = {(1, 5), (1, 6), (2, 5), (2, 6)} ....(ii) Clearly from equation (i) and equation (ii), we get $\text{A}\times\text{C}\subset\text{B}\times\text{D}$ Hence verified.
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Question 114 Marks
If A = {1, 2, 3}, B = {4}, C = {5}, then verify that: $\text{A}\times(\text{B}\cup\text{C})=(\text{A}\times\text{B})\cup(\text{A}\times\text{C})$
Answer
We have, $\text{A}=\{1,2,3\}\times\{4\}$ $\therefore\ \text{B}\cup\text{C}=\{4\}\cup\{5\}=\{4,5\}$ $\therefore\ \text{A}\times(\text{B}\cup\text{C})=\{1,2,3\}\times\{4,5\}$ $\Rightarrow\text{A}\times(\text{B}\cup\text{C})=\{(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)\}\ ...(\text{i}) $ Now, $\text{A}\times\text{B}=\{1, 2, 3\}\times\{4\}$ $=\{(1, 4), (2, 4), (3, 4)\}$ and, $\text{A}\times\text{C}=\{1,2,3\}\times\{5\}$ $=\{(1, 5) , (2, 5), (3, 5)\}$ $\therefore\ (\text{A}\times\text{B})\cup(\text{A}\times\text{C})=\{(1, 4), (2, 4), (3, 4)\} \cup\{(1, 5), (2, 5), (3, 5)\}$ $\Rightarrow(\text{A}\times\text{B})\cup(\text{A}\times\text{C})=\{(1,4), (1,5), (2, 4), (2, 5), (3, 4), (3, 5)\}\ ...(\text{ii})$ From equation (i) and (ii), we get $\text{A}\times (\text{B}\cup\text{C})=(\text{A}\times\text{B})\cup(\text{A}\times\text{C})$ Hence verified.
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