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

Answer:

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.

Answer:

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 ?

Answer:

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 .

Answer:

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) .

Answer:

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 \}

Answer:

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.

Answer:

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 \} \}

Answer:

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?

Answer:

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.

Answer:

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) \}

Answer:

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 .

Answer:

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\} .

Answer:

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) \}

Answer:

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 .

Answer:

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.

Answer:

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}

Answer:

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 \}

Answer:

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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