Question 1: Define a function as a set of ordered pairs.

A function is a set of ordered pairs in which no two different ordered pairs have the same $x -$ coordinate. An equation that produces such a set of ordered pairs defines a function.

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Question 2: Define a function as a correspondence between two sets.

A function is a special type of relation where every input has a unique output. Definition: A function is a correspondence between two sets (called the domain and the range) such that to each element of the domain, there is assigned exactly one element of the range.

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Question 3: What is the fundamental difference between a relation and a function ? Is every relation a function ?

Relation- Two or more sets can be related to each other by any means. Consider for an example two sets $A$ and $B$ having $m$ and $n$ elements respectively, we can have a relation with any ordered pair which shows a relation between the two sets.

Functions- A functions can have the same Range mapped as that of in Relation, such that a set of inputs is related with exactly one output.

Consider for an example Set $A$ and Set $B$ are related in a manner that all the elements of Set $A$ are related to exactly one element of Set $B$ or many elements of set $A$ are related to one element of Set $B$. Thus this type of relation is said to be a function.

It is to be noted that a function cannot have One to Many Relation between the set $A$ and $B$.

Relations and Functions Differences:

 Differentiating Parameter Relations Functions Definition A relation is a relationship between sets of values. Or, it is a subset of the Cartesian product A function is a relation in which there is only one output for each input. Denotation A relation is denoted by $R$ A function is denoted by $F$ or $f$. Example $R = \{ (2, x), (9, y), (2, z) \}$ ** It is not function as $2$ is input for both $x$ and $z$. $F = \{ (2, x), (9, y), (5, x) \}$ Note: Every relation is not a function. Every function is a relation.

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Question 4: Let $A = \{-2, -1, 0, 1, 2 \}$ and and $f \colon A \rightarrow Z$ be a function defined by $f(x) = x^2 - 2x - 3$. Find:   (i) range of $f$  i.e.  $f (A)$    (ii) pre-images of $6,-3$ and $5$.

Given $A = \{-2, -1, 0, 1, 2 \}$ and $f(x) = x^2 - 2x - 3$

Therefore  $f(-2) = (-2)^2 - 2(-2) - 3 = 5$

$f(-1) = (-1)^2 - 2(-1) - 3 = 0$

$f(0) = (0)^2 - 2(0) - 3 = -3$

$f(1) = (1)^2 - 2(1) - 3 = -4$

$f(2) = (2)^2 - 2(2) - 3 = -3$

i)Range $(f) = \{ -4, -3, 0, 5 \}$

ii) Let $x$ be pre-image of $6$

$\therefore f(6) = x^2 - 2x - 3 = 6$  $\Rightarrow x = 1 \pm \sqrt{10}$

Since $x = 1 \pm \sqrt{10} \notin A$, there is no pre-image of $6$.

Similarly, let $x$ be the pre-image of $-3$

$\therefore f(-3) = x^2 - 2x - 3 = -3 \Rightarrow x = 0, 2$

Since $0, 2 \in A \Rightarrow 0$ and $2$ are pre-image of $-3$

Similarly, let $x$ be the pre-image of $5$

$\therefore f(5) = x^2 - 2x - 3 = 5 \Rightarrow x = 4, -2$

Since $-2 \in A, -2$ is pre-image of $5$

Therefore pre-images of $6 , -3$, and $5$ is $\phi, \{ 0, 2 \}, -2$ respectively.

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Question 5:If a function $f \colon R \rightarrow R$ be defined by

$f(x) = \Bigg\{$ $\begin{array}{ccc} 3x-2, \hspace*{0.5cm} x <0 \\ 1, \hspace*{1.5cm} x = 0 \\ 4x+1, \hspace*{0.5cm} x > 0 \end{array}$

Find: $f (1),f (-1),f (0),f (2)$.

Given $f(x) = \Bigg\{$ $\begin{array}{ccc} 3x-2, \hspace*{0.5cm} x <0 \\ 1, \hspace*{1.5cm} x = 0 \\ 4x+1, \hspace*{0.5cm} x > 0 \end{array}$

$f(1) = 4 \times 1 + 1 = 5$

$f(-1) = 3 \times (-1) - 2 = -5$

$f(0) = 1$

$f(2) = 4(2) +1 = 9$

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Question 6: A function $f \colon R \rightarrow R$ is defined by $f (x) = x^2$. Determine:   (i) range of $f$   (ii) $\{x \colon f (x) = 4 \}$   (iii) $\{y \colon f(y) = - 1 \}$

Given $f (x) = x^2$   … … … … … i)

i) Range $(f) = R^+$ (set of all real numbers greater than or equal to $0$)

ii) We have $\{ x : f(x) = 4 \} \Rightarrow f(x) = 4$   … … … … … ii)

From i) and ii) we get $x^2 = 4 \Rightarrow x = \pm 2$

$\therefore \{ x : f(x) = 4 \} = \{ -2, 2 \}$

iii) $\{ y : f(y) = -1 \} \Rightarrow f(y) = -1$ … … … … … iii)

$\therefore x^2 = -1 \ or \ x^2 \geq 0$

$\Rightarrow f(y) \neq -1$

$\therefore \{ y : f(y) = -1 \} = \phi$

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Question 7: Let $f : R^+ \rightarrow R$, where $R^+$ is the set of all positive real numbers, be such that $f (x) = \log_e x$ Determine:     (i) the image set of the domain of $f$     (ii) $\{ x \colon f (x)=-2 \}$     (iii) whether $f (xy) = f (x) + f (y)$ holds.

Given $f : R^+ \rightarrow R$ and $f (x) = \log_e x$   … … … … … i)

i) Since $f : R^+ \rightarrow R$, Therefore the image set of the domain(f) = R

ii) Since $\{ x: f(x) = -2 \}\Rightarrow f(x) = -2$   … … … … … ii)

From i) and ii)  we get $\log_e x = -2 \Rightarrow x = e^{-2}$

$\therefore \{ x : f(x) = -2 \} = \{ e^{-2} \}$

iii) $f(xy) = \log_e x + \log_e y = f(x) + f(y)$

$\therefore f(xy) = f(x) + f(y)$

Hence proved.

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Question 8: Write the following relations as sets of ordered pairs and find which of them are functions:
(i) $\{ (x, y) \colon y = 3x, x \in \{ 1, 2, 3 \}, y \in \{3, 6, 9, 12 \} \}$

(ii) $\{ (x, y) \colon y >x+1, x=1, 2 \ and \ y = 2,4,6 \}$

(iii) $\{ (x, y) \colon x+y =3, x , y \in \{ 0,1,2, 3 \} \}$

i) We have $\{ (x, y) \colon y = 3x, x \in \{ 1, 2, 3 \}, y \in \{3, 6, 9, 12 \} \}$

Putting $x = 1, 2, 3$ in $y = 3x$ we get $y = 3, 6, 9$ respectively.

$\therefore R = \{ (1,3), (2,6), (3, 9) \}$

Yes, it is a function.

ii) We have $\{ (x, y) \colon y >x+1, x=1, 2 \ and \ y = 2,4,6 \}$

Putting $x = 1, 2$ in $y > x+1$ , we get $y > 2, y > 3$ respectively.

$\therefore R = \{(1, 4), (1, 6), (2, 4), (2, 6) \}$

It is not a function from $A$ to $B$ because two ordered paired in $R$ have same first element.

iii) We have $\{ (x, y) \colon x+y =3, x , y \in \{ 0,1,2, 3 \} \}$

Now $y = 3-x$, for $x = 0, 1, 2, 3$ we get $y = 3, 2, 1, 0$ respectively.

Yes, this is a relation.

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Question 9: Let $f \colon R \rightarrow R$ and $g \colon C \rightarrow C$ be two functions defined as $f (x) = x^2$ and $g (x) = x^2$. Are they equal functions?

We have $f \colon R \rightarrow R$ and $g \colon C \rightarrow C$

$f (x) = x^2 \Rightarrow$ Domain $(f) = R$

$g (x) = x^2 \Rightarrow$ Domain $(g) = C$

Since Domain $(f) \neq$ Domain $(g)$

$\therefore f(x)$ and $g(x)$ are not equal functions.

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Question 10: If $f, g, h$ are three functions defined from $R$ to $R$ as follows:     (i) $f(x)=x^2$     (ii)  $g(x)= \sin x$     (iii) $h(x)- x^2+1$  Find the range of each function.

i) We have $f(x)=x^2$

Range of $f(x) R^+ ($ set of all real numbers greater than or equal to $0) = \{ x \in R$ where $x \geq 0 \}$

ii) We have $g(x)= \sin x$

Range $g(x) = \{ x \in R : -1 \leq x \leq 1 \}$

iii) We have $h(x)- x^2+1$

Range $h(x) = \{ x \in R$ where $x \geq 1 \}$

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Question 11: Let $X = \{ 1, 2, 3, 4 \}$ and $Y = \{ 1,5, 9 ,11 ,15,16 \}$
Determine which of the following sets are functions from $X$ to $Y$
(i) $f_1 = \{ (1, 1), (2, 11), ( 3, 1), (4, 15) \}$ (ii) $f_2 = \{ (1, 1), (2,7),(3, 5) \}$
(iii) $f_3 = \{ (1 , 5), (2, 9), (3 , 1), (4, 5) , (2, 11) \}$

Given $X = \{ 1, 2, 3, 4 \}$ and $Y = \{ 1,5, 9 ,11 ,15,16 \}$

i) We have $f_1 = \{ (1, 1), (2, 11), ( 3, 1), (4, 15) \}$

$f_1$ is a function from $X$ to $Y$.

ii) We have $f_2 = \{ (1, 1), (2,7),(3, 5) \}$

$f_2$ is not a function from $X$ to $Y$ because there is an element $4 \in X$ which is not associated to any element of $Y$

iii) We have $f_3 = \{ (1 , 5), (2, 9), (3 , 1), (4, 5) , (2, 11) \}$

$f_3$ is not a function from $X$ to $Y$ because an element $2 \in X$ is associated to two elements $9$ and $11$ in $Y$

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Question 12: Let $A = \{ 12, 13,14, 15,16, 17 \}$ and $f \colon A \rightarrow Z$ be a function given by $f (x) =$ highest prime factor of $x$Find range of $f$.

Given $A = \{ 12, 13,14, 15,16, 17 \}$ and $f \colon A \rightarrow Z$ be defined as $f (x) =$ highest prime factor of $x$

$f(12) =$ highest prime factor of $12 = 3$

$f(13) =$ highest prime factor of $13 = 13$

$f(14) =$ highest prime factor of $14 = 7$

$f(15) =$ highest prime factor of $15 = 5$

$f(16) =$ highest prime factor of $16 = 2$

$f(17) =$ highest prime factor of $17 = 17$

$\therefore f = \{ (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17) \}$

Therefore Range $(f) = \{ 3, 13, 7, 5, 2, 17 \}$

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Question 13: If $f : R \rightarrow R$ be defined by $f (x) : x^2 +1$, then find $f^{-1} \{ 17 \}$ and $f^{-1} \{- 3\}$.

If $f: A \rightarrow B$ is such that $y \in B$, then $f^{-1} \{y \} = \{ x \in A : f(x) = y \}$

In other words, $f^{-1} \{ y \}$ is a set of pre-images of $y$

Let $f^{-1} \{ 17 \} = x$

$\Rightarrow f(x) = 17$

$\Rightarrow x^2 + 1 = 17 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$

$\therefore f^{-1} \{17 \} = \{-4, 4 \}$

Similarly, $f^{-1} \{ -3 \} = x$

$\Rightarrow f(x) = -3$

$\Rightarrow x^2 + 1 = -3 \Rightarrow x^2 = -4 \Rightarrow x = \pm \sqrt{-4}$

Clearly, there is no solution available in $R$

$\therefore f^{-1} \{-3 \} = \phi$

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Question 14: Let $A = \{p, q,r, s \}$ and $B =\{ 1,2,3 \}$. Which of the following relations from $A$ to $B$ is not a function?
(i) $R_1 = \{ (p,1),(q,2),(r,1),(s,2) \}$
(ii) $R_2 = \{ (p,1),(q,1),(r,1), (s,1) \}$              (iii) $R_3 = \{ (p, 1), (q, 2), (p, 2), (s, 3) \}$    (iv) $R_4 = \{ (p, 2), (q, 3), (r, 2), (s, 2) \}$

Given $A = \{p, q,r, s \}$ and $B =\{ 1,2,3 \}$

i) $R_1 = \{ (p,1),(q,2),(r,1),(s,2) \}$

Therefore $R_1$ is a function  [ As it has unique image in $B$ for all elements in $A$ ]

ii) $R_2 = \{ (p,1),(q,1),(r,1), (s,1) \}$

$R_2$ is a function   [ As it has unique image in $B$ for all elements in $A$ ]

iii) $R_3 = \{ (p, 1), (q, 2), (p, 2), (s, 3) \}$

$R_3$ is not a function because the elements $p \in A$ is associated with two elements $1$ and $2$ in $B$.

iv) $R_4 = \{ (p, 2), (q, 3), (r, 2), (s, 2) \}$

$R_4$ is a function   [ As it has unique image in $B$ for all elements in $A$ ]

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Question 15: Let $A = \{ 9,10,11,12,13 \}$ and let $f:A \rightarrow N$ be defined by $f (n) =$ the highest prime factor of $n$. Find the range of $f$.

Given $A = \{ 9,10,11,12,13 \}$ and $f:A \rightarrow N$ be defined by $f (n) =$ the highest prime factor of $n$

$f(9) =$ highest prime factor of $9 = 3$

$f(10) =$ highest prime factor of $10 = 5$

$f(11) =$ highest prime factor of $11 = 11$

$f(12) =$ highest prime factor of $12 = 3$

$f(13) =$ highest prime factor of $13 = 13$

$\therefore f = \{ (9, 3), (10, 5), (11, 11), (12, 3), (13, 13) \}$

Therefore Range $(f) = \{ 3, 5, 11, 13 \}$

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Question 16: The function $f$ is defined by $f(x) = \Bigg\{$ $\begin{array}{ll} x^2, \hspace*{0.5cm} 0 \leq x \leq 3 \\ 3x, \hspace*{0.5cm} 3 \leq x \leq 10 \end{array}$

The relation $g$ is defined by $g(x) = \Bigg\{$ $\begin{array}{ll} x^2, \hspace*{0.5cm} 0 \leq x \leq 2 \\ 3x, \hspace*{0.5cm} 2 \leq x \leq 10 \end{array}$

Show that $f$ is a function and $g$ is not a function.

Given $f(x) = \Bigg\{$ $\begin{array}{ll} x^2, \hspace*{0.5cm} 0 \leq x \leq 3 \\ 3x, \hspace*{0.5cm} 3 \leq x \leq 10 \end{array}$

For $x = 3, f(3) = x^2 = (3)^2 = 9$

and $f(3) = 3x = 3 \times 3 = 9$

Therefore we observe that $f(x)$ takes unique value at each point in its domain $[ 0, 10 ]$

Therefore $f(x)$ is a function.

Also $g(x) = \Bigg\{$ $\begin{array}{ll} x^2, \hspace*{0.5cm} 0 \leq x \leq 2 \\ 3x, \hspace*{0.5cm} 2 \leq x \leq 10 \end{array}$

For $x = 2, g(2) = x^2 = 2^2 = 4$

and $g(2) = 3x = 3 \times 2 = 6$

Therefore we observe that $g(x)$ does not take unique values at each point in its domain $[0, 10 ]$

Hence $g(x)$ is not a function.

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Question 17: If $f(x) = x^2$, find $\frac{f(1.1) - f(1)}{1.1 - 1}$

$f(x) = x^2$

$f(1.1) = (1.1)^2 = 1.21$

$f(1) = (1)^2 = 1$

$\therefore$ $\frac{f(1.1) - f(1)}{1.1 - 1}$ $=$ $\frac{1.21 - 1}{1.1 - 1}$ $=$ $\frac{0.21}{0.1}$ $= 2.1$

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Question 18: Express the function $f :X \rightarrow R$ given by $f(x) = x^3+1$ as set of ordered pairs, where $X = \{-1, 0, 3, 9 ,7 \}$

Given $f :X \rightarrow R$ given by $f(x) = x^3+1$

$\therefore f(-1) = (-1)^3 + 1 = -1 + 1 = 0$

$f(0) = (0)^3 + 1 = 0 + 1 = 1$

$f(3) = (3)^3 + 1 = 27 + 1 = 28$

$f(9) = (9)^3 + 1 = 729 + 1 = 730$

$f(7) = (7)^3 + 1 = 343 + 1 = 344$

$f = \{ (x, f(x) ) : x \in X \} = \{ (-1, 0), (0, 1), (3, 28), (9, 730), (7, 344) \}$

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