Ans: Given:
${{f = \{ (2, 4), (5, 6), (8, - 1), (10, - 3)\} ,}}$
${{g = \{ (2, 5), (7, 1), (8, 4), (10, 13), (11, - 5)\} }}{{.}}$
First, find the domain of \[{{f \text{ and } g}}{{.}}\] Then find the domain of \[{{f + g}}\].
Domain of ${{f = \{ 2, 5, 8, 10\} ,}}$
Domain of ${{g = \{ 2, 7, 8, 10, 11\} }}$
Domain of $f + g$ = (domain of $f$ $\cap$ domain of $g$ )
$\therefore$ Domain of ${{f + g = \{ 2, 8, 10\} }}{{.}}$
EXERCISE QUESTION:
1.If \[A = \{ - 1, 2, 3\} \text{ and } B = \{ 1, 3\}\], then determine,
Ans: Given: sets \[{{A = \{ - 1, 2, 3\} \text{ and } B = \{ 1, 3\} }}{{.}}\]
The terms \[{{A }} \times {{ B}}\] represents the Cartesian product and represents the relation of elements in A with B
\[{{A }} \times {{ B = \{ ( - 1, 1), ( - 1, 3), (2, 1), (2, 3), (3, 1), (3, 3)\} }}\]
Ans: Given: sets \[{{A = \{ - 1, 2, 3\} \text{ and } B = \{ 1, 3\} }}{{.}}\]
The terms \[{{A }} \times {{ B}}\] represents the Cartesian product and represents the relation of elements in A with B
\[{{B }} \times {{ A = \{ (1, - 1), (1, 2), (1, 3), (3, - 1), (3, 2), (3, 3)\} }}\]
Ans: Given: sets \[{{A = \{ - 1, 2, 3\} \text{ and } B = \{ 1, 3\} }}{{.}}\]
The terms \[{{A }} \times {{ B}}\] represents the Cartesian product and represents the relation of elements in A with B
\[{{B }} \times {{ B = \{ (1, 1), (1, 3), (3, 1), (3, 3)\} }}\]
Ans: Given: sets \[{{A = \{ - 1, 2, 3\} \text{ and } B = \{ 1, 3\} }}{{.}}\]
The terms \[{{A }} \times {{ B}}\] represents the Cartesian product and represents the relation of elements in A with B
\[{{A }} \times {{ A = \{ ( - 1, - 1), ( - 1, 2), ( - 1, 3), (2, - 1), (2, 2), (2, 3), (3, - 1), (3, 2), (3, 3)\} }}{{.}}\]
2.If \[P = \{ x : x < 3, x \in N\} , Q = \{ x : x \le \ 2, x \in W\} ,\] then find $(P \cup Q) \times(P \cap Q)$ where \[{{W}}\] is the set of whole numbers.
Ans: Given:
${{P = \{ x : x < 3, x }} \in {{ N\} }}$
${{Q = \{ x : x }} \leqslant {{ 2, x }} \in {{ W\} }}$
Use basic operations of sets like union and intersection of sets and then apply cartesian products.From the sets, we get
$\Rightarrow {{ P = \{ 1, 2\} }}$
$\Rightarrow {{ Q = \{ 0, 1, 2\} }}$
$\Rightarrow {{ P }} \cup {{ Q = \{ 0, 1, 2\} \text{ and } P }} \cap {{ Q = \{ 1, 2\} }}$
$\Rightarrow {{ (P }} \cup {{ Q) }} \times {{ (P }} \cap {{ Q) = \{ 0, 1, 2\} }} \times {{ \{ 1, 2\} }}$
$\Rightarrow {{ (P }} \cup {{ Q) }} \times {{ (P }} \cap {{ Q) = \{ (0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)\} }}$
3.If \[A = \{ x : x \in W, x < 2\} , B = \{ x : x \in N, 1 < x < 5\} \text{ and } C = \{ 3, 5\} \] then find
\[(i). A \times (B \cup C)\]
Ans: Given: The sets \[{{A = \{ x : x }} \in {{ W, x < 2\} , B = \{ x : x }} \in {{ N, 1 < x < 5\} \text{ and } C = \{ 3, 5\} }}\].
First convert the sets in roster form then use basic set operations and Cartesian products.
The sets in roster form will be,
\[{{A = \{ 0, 1\} , B = \{ 2, 3, 4\} \text{ and } C = \{ 3, 5\} }}{{.}}\]
\[{{A }} \times {{ (B }} \cap {{ C)}}\]
$\Rightarrow {{ A }} \times {{ (B }} \cap {{ C) = \{ 0, 1\} }} \times {{ \{ 3\} }}$
$\Rightarrow {{ A }} \times {{ (B }} \cap {{ C) = \{ (0, 3), (1, 3)\} }}$
\[(ii).A \times (B \cup C)\]
Ans: Given: The sets \[{{A = \{ x : x }} \in {{ W, x < 2\} , B = \{ x : x }} \in {{ N, 1 < x < 5\} \text{ and } C = \{ 3, 5\} }}\].
First convert the sets in roster form then use basic set operations and Cartesian products.
The sets in roster form will be,
\[{{A = \{ 0, 1\} , B = \{ 2, 3, 4\} \text{ and } C = \{ 3, 5\} }}{{.}}\]
\[{{A }} \times {{ (B }} \cup {{ C)}}\]
$\Rightarrow {{ A }} \times {{ (B }} \cup {{ C) = \{ 0, 1\} }} \times {{ \{ 2, 3, 4, 5\} }}$
$\Rightarrow {{ A }} \times {{ (B }} \cup {{ C) = \{ (0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}}{{.}}$
4.In each of the following cases, find a and b.
\[(i).(2a + b, a - b) = (8, 3)\]
Ans: Since, two ordered pairs are equal, if their corresponding first and second elements are equal.
\[(2{{a + b, a - b) = (8, 3)}}\]
\[\Rightarrow \;2{{a + b = 8 \text{ and } a - b = 3}}\]
On substituting \[{{b = a - 3 \text{ in } 2a + b = 8, we get}}\]
$\Rightarrow 2{{a + a - 3 = 8}}$
$\Rightarrow \;3{{a - 3 = 8}}$
$\Rightarrow {{ a = }}\dfrac{{11}}{3}{{ }}$
Substituting \[{{a = }}\dfrac{{11}}{3}{{ \text{ in } b = a - 3, we get}}\]
$\Rightarrow {{ b = }}\dfrac{{11}}{3}{{ - 3}}$
$\Rightarrow {{ b = }}\dfrac{2}{3}$
$\therefore {{ a = }}\dfrac{{11}}{3}{{ \text{ and } b = }}\dfrac{2}{3}$
\[(ii).\left( {\dfrac{a}{4}, a - 2b} \right) = (0, 6 + b)\]
Ans: Since, two ordered pairs are equal, if their corresponding first and second elements are equal.
\[\left( {\dfrac{{{a}}}{4}{{, a - 2b}}} \right){{ = (0, 6 + b)}}\]
$\Rightarrow \;\dfrac{{{a}}}{4}{{ = 0 \text {and } a - 2b = 6 + b}}$
$\Rightarrow {{ a = 0 \text{ and } 0 - 2b = 6 + b}}$
$\Rightarrow {{ a = 0 \text{ and } b = - 2}}$
5. \[A = \{ 1, 2, 3, 4, 5\} , S = \{ (x, y) : x \in A, y \in A\} ,\] then find the orders which satisfy the conditions given below.
\[(i). x + y = 5\]
Ans: Given: The sets \[{{A = \{ 1, 2, 3, 4, 5\} \text{ and } S = \{ (x, y) : x }} \in {{ A, y }} \in {{ A\} }}{{.}}\]
Use the condition for all the elements in sets till the condition is satisfied.
The set of ordered pairs satisfying \[{{x + y = 5 \text{ is }, \{ (1, 4), (2, 3), (3, 2), (4, 1)\} }}{{.}}\]
\[(ii). x + y < 5\]
Ans: Given: The sets \[{{A = \{ 1, 2, 3, 4, 5\} \text{ and } S = \{ (x, y) : x }} \in {{ A, y }} \in {{ A\} }}{{.}}\]
Use the condition for all the elements in sets till the condition is satisfied.
The set of ordered pairs satisfying \[{{x + y < 5 \text{ is }, \{ (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)\} }}{{.}}\]
\[(iii). x + y > 8\]
Ans: Given: The sets \[{{A = \{ 1, 2, 3, 4, 5\} \text{ and } S = \{ (x, y) : x }} \in {{ A, y }} \in {{ A\} }}{{.}}\]
Use the condition for all the elements in sets till the condition is satisfied.
The set of ordered pairs satisfying \[{{x + y > 8 \text{ is }, \{ (4, 5), (5, 4), (5, 5)\} }}{{.}}\]
6.If \[R = \{ (x, y) : x, y \in W, {x^2} + {y^2} = 25\} ,\]then find the domain and range of \[{{R}}{{.}}\]
Ans: Given: Relation \[{{R = \{ (x, y) : x, y }} \in {{ W, }}{{{x}}^2}{{ + }}{{{y}}^2}{{ = 25\} }}{{.}}\]
First, write the relation in roster form, then find the domain and range of \[{{R}}{{.}}\]\[{{R = \{ (0, 5), (3, 4), (4, 3), (5, 0)\} }}{{.}}\]
Domain of \[{{R}}\] is set of first element of ordered pairs in \[{{R = \{ 0, 3, 4, 5\} }}{{.}}\]
Range of \[{{R}}\] is set of second element of ordered pairs in \[{{R = \{ 5, 4, 3, 0\} }}{{.}}\]
7.If \[{R_1} = \{ (x, y) | y = 2x + 7, \text{ where } x \in R \text{ and } - 5 \le x \le 5\} \] is a relation. Then find the domain and range of \[{R_1}.\]
Ans: Given: A relation \[{{{R}}_1}{{ = \{ (x, y) | y = 2x + 7, \text{ where } x }} \in {{ R \text{ and } - 5 }} \leqslant {{ x }} \leqslant {{ 5\} }}{{.}}\]
The value of \[{{x}}\] represents the domain and the values of \[{{y for all x}}{{.}}\]
Domain of \[{{{R}}_1}{{ = \{ - 5 }} \leqslant {{ x }} \leqslant {{ 5, x }} \in {{ R\} ,}}\]
\[\Rightarrow {{ }}{{{R}}_1}{{ = [ - 5, 5]}}\].
Range of \[{{{R}}_1},\]
${{\text{when } x = - 5, \text{ then } y = 2( - 5) + 7 = - 3}}$
${{\text{when } x = 5, \text{ then } y = 2(5) + 7 = 17}}$
Range of \[{{{R}}_1},\]
$\Rightarrow {{ }}{{{R}}_1} = {{ \{ - 3 }} \leqslant {{ y }} \leqslant {{ 17, y }} \in {{ R\} }}$
$\Rightarrow {{ }}{{{R}}_1}{{ = [ - 3, 17]}}$
8.If \[{R_2} = \{ (x, y) | x \text{ and } y \text{ are integers and } {x^2} + {y^2} = 64\} \] is a relation, then find the value of \[{R_{2.}}\]
Ans: Given: A relation \[{{{R}}_2}{{ = \{ (x, y) | |\text{ x and y are integers and }}}{{{x}}^2}{{ + }}{{{y}}^2}{{ = 64\} }}\].
Use the given condition in a relation and then write the set in roster form.
Since, \[64\] is the sum of square of \[0{{ \text{ and } }} \pm {{ 8}}{{.}}\]
$\Rightarrow {{ \text{ when } x = 0, \text{ then } }}{{{y}}^2}{{ = 64,}}$
$\Rightarrow {{ y = }} \pm {{ 8}}$
$\Rightarrow {{ x = 8, \text{ then } }}{{{y}}^2}{{ = 64 - (8}}{{{)}}^2}{{ = 0}}$
$\Rightarrow {{ x = - 8, \text{ then } }}{{{y}}^2}{{ = 64 - ( - 8}}{{{)}}^2}{{ = 0}}$
$\therefore {{ }}{{{R}}_2}{{ = \{ (0, 8), (0, - 8), (8, 0), ( - 8, 0)\} }}$
9.If \[{{{R}}_3}{{ = \{ (x, | x |) | x\} }}\] is a real number is a relation, then find domain and range of \[{R_3}.\]
Ans: Given: A relation \[{{{R}}_3}{{ = \{ (x, | x |) | x\} }}\] is a real number.
The value of \[{{x}}\] represents the domain and the values of \[{\text{y for all x}}{{.}}\]
Domain of \[{{{R}}_3}{{ = \text{ real number }}}{{.}}\]
Since, the image of any real number under \[{{{R}}_3}\] is a positive real number or zero.
Range of \[{{{R}}_3}{{ = }}{{{R}}^ + }\; \cup {{ \{ 0\} or (0, }}\infty {{)}}{{.}}\]
10.Is the given relation a function? Give reason for your answer.
\[(i). h = \{ (4, 6), (3, 9), ( - 11, 6), (3, 11)\} \]
Ans: Given: Relations are given in set-builder and roster form.
A function is said to be a relation, only and only if every element in the domain has a unique image.
\[{{h = \{ (4, 6), (3, 9), ( - 11, 6), (3, 11)\} }}\]
Since, \[3\] has two images \[9{{ \text{ and } 11}}{{.}}\] So, it is not a function.
\[(ii). f = \{ (x, x) | x\} \] is a real number
Ans: Given: Relations are given in set-builder and roster form.
A function is said to be a relation, only and only if every element in the domain has a unique image.
\[{{f = \{ (x, x) | x\} }}\] is a real number
We observe that every element in the domain has a unique image. So, it is a function.
\[(iii). g = \left\{ {\left( {x, \dfrac{1}{x}} \right) | x} \right\} \] is a positive integer
Ans: Given: Relations are given in set-builder and roster form.
A function is said to be a relation, only and only if every element in the domain has a unique image.
\[{{g = }}\left\{ {\left( {{{x, }}\dfrac{1}{{{x}}}} \right){{ | \text{ x is a positive integer }}}} \right\}{{ }}\]
For every \[{{x}}\], it is a positive integer and \[\dfrac{1}{{{x}}}\] is unique and distinct. Therefore, every element in the domain has a unique image. So, it is a function.
\[(iv). s = \{ (x, {x^2})| {x}\} \] is a positive integer
Ans: Given: Relations are given in set-builder and roster form.
A function is said to be a relation, only and only if every element in the domain has a unique image.
\[{{s = \{ (x, }}{{{x}}^2})|{{ x\} }}\]
is a positive integer
Since, the square of any positive integer is unique. So, every element in the domain has a unique image. Hence, it is a function.
\[(v). t = \{ (x, 3) | x\} \] is a real number
Ans: Given: Relations are given in set-builder and roster form.
A function is said to be a relation, only and only if every element in the domain has a unique image.
\[{{t = \{ (x, 3) | x\} }}\] is a real number
Since, every element in the domain has the image \[3\]. So, it is a constant function..
11.If f and g are real functions defined by \[f(x) = {x^2} + 7 \text{ and } g(x) = 3x + 5.\] Then find each of the following.
\[(i). f(3) + g( - 5)\]
Ans: Given: \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
Substitute the value of \[{{x}}\] in the functions \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
We have,
${{f(3) = (3}}{{{)}}^2}{{ + 7 = 16 \text{ and } g( - 5) = 3( - 5) + 5 = - 10}}$
$\Rightarrow {{ f(3) + g( - 5) = 16 - 10}}$
$\therefore {{ f(3) + g( - 5) = 6}}$
\[(ii). f\left( {\dfrac{1}{2}} \right) \times g(14)\]
Ans: Given: \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
Substitute the value of \[{{x}}\] in the functions \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
We have,
${{f}}\left( {\dfrac{1}{2}} \right){{ = }}{\left( {\dfrac{1}{2}} \right)^2}{{ + 7 = }}\dfrac{{29}}{4}$
${{g(14) = 3(14) + 5 = 47}}$
$\therefore {{ f}}\left( {\dfrac{1}{2}} \right){{ }} \times {{ g(14) = }}\dfrac{{29}}{{14}}{{ }} \times {{ 47 = }}\dfrac{{1363}}{4}$
\[(iii). f( - 2) + g( - 1)\]
Ans: Given: \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
Substitute the value of \[{{x}}\] in the functions \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
We have,
${{f( - 2) = ( - 2}}{{{)}}^2}{{ + 7 = 11,}}$
${{g( - 1) = 3( - 1) + 5 = 2}}$
$\therefore {{ f( - 2) + g( - 1) = 11 + 2 = 13}}$
\[(iv). f(t) - f( - 2)\]
Ans: Given: \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
Substitute the value of \[{{x}}\] in the functions \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
We have,
${{f(t) = }}{{{t}}^2}{{ + 7,}}$
${{f( - 2) = ( - 2}}{{{)}}^2}{{ + 7 = 11}}$
$\therefore {{ f(t) - f( - 2) = }}{{{t}}^2}{{ - 4}}$
Ans: Given: \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
Substitute the value of \[{{x}}\] in the functions \[{{f(x) = }}{{{x}}^2}{{ + 7 \text{ and } g(x) = 3x + 5}}{{.}}\]
We have,
${{f(t) = }}{{{t}}^2}{{ + 7}}$
${{f(5) = (5}}{{{)}}^2}{{ + 7 = 32}}$
$\therefore {{ }}\dfrac{{{{f(t) - f(5)}}}}{{{{t - 5}}}},{{ if t }} \ne {{ 5}}$
${{ = }}\dfrac{{{{{t}}^2}{{ + 7 - 32}}}}{{{{t - 5}}}}$
$= {{ }}\dfrac{{{{{t}}^2}{{ - 25}}}}{{{{t - 5}}}}$
$= {{ }}\dfrac{{({{t - 5)(t + 5)}}}}{{({{t - 5)}}}}$
$= {{ t + 5}}$
12.Let \[{{f \text{ and } g}}\] be real functions defined by \[f(x) = 2x + 1 \text{ and } g(x) = 4x - 7.\]
Ans: Given: \[{{f(x) = 2x + 1 \text{ and } g(x) = 4x - 7}}{{.}}\]
Substitute the real functions according to given condition and obtain the real value of \[{{x}}{{.}}\]
We have,
$\Rightarrow {{ f(x) = g(x)}}$
$\Rightarrow {{ 2x + 1 = 4x - 7}}$
$\Rightarrow {{ x = 4}}$
Ans: Given: \[{{f(x) = 2x + 1 \text{ and } g(x) = 4x - 7}}{{.}}\]
Substitute the real functions according to given condition and obtain the real value of \[{{x}}{{.}}\]
We have,
$\Rightarrow {{ f(x) < g(x)}}$
$\Rightarrow {{ 2x + 1 < 4x - 7}}$
$\Rightarrow {{ - 2x + 1 < - 7}}$
$\Rightarrow {{ - 2x < - 8}}$
$\Rightarrow {{ }}\dfrac{{{{ - 2 x}}}}{{{{ - 2}}}}{{ > }}\dfrac{{{{ - 8}}}}{{{{ - 2}}}}$
\[\Rightarrow {{ x > 4}}\]
13.If \[{{f \text{ and } g}}\] are two real valued functions defined as \[f(x) = 2x + 1 \text{ and } g(x) = 2{x^2} + 1,\] then find
\[(i). f + g\]
Ans: Given: \[{{f(x) = 2x + 1 \text{ and } g(x) = 2}}{{{x}}^2}{{ + 1}}{{.}}\]
Directly perform the operation according to questions like addition, subtraction, multiplication and division.
We have,
$\Rightarrow {{ (f + g)(x) = f(x) + g(x)}}$
$\Rightarrow {{ (f + g)(x) = 2x + 1 + }}{{{x}}^2}{{ + 1}}$
$\Rightarrow {{ (f + g)(x) = }}{{{x}}^2}{{ + 2x + 2}}$
\[(ii). f - g\]
Ans: Given: \[{{f(x) = 2x + 1 \text{ and } g(x) = 2}}{{{x}}^2}{{ + 1}}{{.}}\]
Directly perform the operation according to questions like addition, subtraction, multiplication and division.
We have,
$\Rightarrow {{ (f - g)(x) = f(x) - g(x)}}$
$\Rightarrow {{ (f - g)(x) = (2x + 1) - (}}{{{x}}^2}{{ + 1)}}$
$\Rightarrow {{ (f - g)(x) = x(2 - x)}}$
\[(iii). fg\]
Ans: Given: \[{{f(x) = 2x + 1 \text{ and } g(x) = 2}}{{{x}}^2}{{ + 1}}{{.}}\]
Directly perform the operation according to questions like addition, subtraction, multiplication and division.
We have,
$\Rightarrow {{ (fg)(x) = f(x) }}{{. g(x)}}$
$\Rightarrow {{ (fg)(x) = (2x + 1)(}}{{{x}}^2}{{ + 1)}}$
$\Rightarrow {{ (fg)(x) = 2}}{{{x}}^3}{{ + }}{{{x}}^2}{{ + 2x + 1}}$
\[(iv).\dfrac{{{f}}}{{{g}}}\]
Ans: Given: \[{{f(x) = 2x + 1 \text{ and } g(x) = 2}}{{{x}}^2}{{ + 1}}{{.}}\]
Directly perform the operation according to questions like addition, subtraction, multiplication and division.
We have,
$\Rightarrow {{ }}\left( {\dfrac{{{f}}}{g}} \right){{(x) = }}\dfrac{{{{f(x)}}}}{{{{g(x)}}}}$
$\Rightarrow {{ }}\left( {\dfrac{{{f}}}{g}} \right){{(x) = }}\dfrac{{2{{x + 1}}}}{{{{{x}}^2}{{ + 1}}}}$
14.Express the following functions as a set of ordered pairs and determine their range.
Ans: Given: A function \[{{f : x }} \to {{ R, f(x) = }}{{{x}}^3}{{ + 1, \text{ where } x = \{ - 1, 0, 3, 9, 7\} }}\].
Substitute the value of \[{{x}}\] in the function to obtain values of \[{{y}}\] as range.
We have,
${{f : X }} \to {{ R, f(x) = }}{{{x}}^3}{{ + 1, \text{ where } x = \{ - 1, 0, 3, 9, 7\} }}$
${{x = - 1, \text{ then } f( - 1) = ( - 1}}{{{)}}^3}{{ + 1}}$
$\Rightarrow \;{{f( - 1) = 0}}$
${{x = 0, \text{ then } f(0) = (0}}{{{)}}^3}{{ + 1}}$
$\Rightarrow \;{{f(0) = 1}}$
$\Rightarrow \;{{f(3) = 28}}$
${{x = 9, \text{ then } f(9) = (9}}{{{)}}^3}{{ + 1}}$
$\Rightarrow {{ f(9) = 730}}$
${{x = 3, \text{ then } f(3) = (3}}{{{)}}^3}{{ + 1}}$
${{x = 7, \text{ then } f(7) = (7}}{{{)}}^3}{{ + 1}}$
$\Rightarrow {{ f(7) = 344}}$
$\Rightarrow {{ f = \{ ( - 1, 0), (0, 1), (3, 28), (9, 730), (7, 344)\} }}$
$\therefore \;{{\text{ Range of }f = \{ 0, 1, 28, 730, 344\} }}$