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Maths Solve the Following Question.(5 Marks)

Relations · Maharashtra Board STD 11 (MSBSHSE - Semi English Medium). 11 questions.

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Solve the Following Question.(5 Marks)

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11 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 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 25 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 35 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 45 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 55 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 65 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 75 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 85 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 95 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 105 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 115 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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