Plus One Math's Solution Ex 2.3 Chapter2 Relations and Functions

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In this chapter, you’ll get a clear understanding of Relations and Functions. Both Relations and Functions have a different meaning in mathematics; however many get confused and use these words interchangeably. A ‘relation’ means a relationship between two elements of a set.  It is a set of inputs and outputs, denoted as ordered pairs (input, output). We can also represent a relation as a mapping diagram or a graph. A relation can either be symbolised by Roster method or Set-builder method. On the other hand, a ‘function’ is a special type of relation, in which each input is related to a unique output. So, all functions are relations, but not all relations are functions.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus One 
Subject Math's Textbook Solution
Chapter Chapter 2
Exercise Ex 2.3
Chapter Name Relations and Functions
Category Plus One Kerala


Kerala Syllabus Plus One Math's Textbook Solution Chapter  2 Relations and Functions Exercises 2.3


Chapter  2  Relations and Functions Textbook Solution



Kerala plus One maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

Chapter  2  Relations and Functions Exercise   2.3

    Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

    (i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

    (ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

    (iii) {(1, 3), (1, 5), (2, 5)}

    (i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

    Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.

    Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

    (ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

    Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.

    Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

    (iii) {(1, 3), (1, 5), (2, 5)}

    Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.

    Find the domain and range of the following real function:

    (i) f(x) = –|x| (ii) 

    (i) f(x) = –|x|, x ∈ R

    We know that |x| = 

    Since f(x) is defined for x ∈ R, the domain of f is R.

    It can be observed that the range of f(x) = –|x| is all real numbers except positive real numbers.

    ∴The range of f is (–, 0].

    (ii) 

    Sinceis defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].

    For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.

    ∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3]

    A function f is defined by f(x) = 2x – 5. Write down the values of

    (i) f(0), (ii) f(7), (iii) f(–3)

    The given function is f(x) = 2x – 5.

    Therefore,

    (i) f(0) = 2 × 0 – 5 = 0 – 5 = –5

    (ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9

    (iii) f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11

    The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by.

    Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t(C) = 212

    The given function is.

    Therefore,

    (i) 

    (ii) 

    (iii) 

    (iv) It is given that t(C) = 212

    Thus, the value of t, when t(C) = 212, is 100.

    Find the range of each of the following functions.

    (i) f(x) = 2 – 3xx ∈ Rx > 0.

    (ii) f(x) = x2 + 2, x, is a real number.

    (iii) f(x) = xx is a real number

    (i) f(x) = 2 – 3xx ∈ Rx > 0

    The values of f(x) for various values of real numbers x > 0 can be written in the tabular form as

    x

    0.01

    0.1

    0.9

    1

    2

    2.5

    4

    5

    f(x)

    1.97

    1.7

    –0.7

    –1

    –4

    –5.5

    –10

    –13

    Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2.

    i.e., range of f = (–, 2)

    Alter:

    Let x > 0

    ⇒ 3x > 0

    ⇒ 2 –3x < 2

    ⇒ f(x) < 2

    ∴Range of f = (–, 2)

    (ii) f(x) = x2 + 2, x, is a real number

    The values of f(x) for various values of real numbers x can be written in the tabular form as

    x

    0

    ±0.3

    ±0.8

    ±1

    ±2

    ±3

     

    f(x)

    2

    2.09

    2.64

    3

    6

    11

     

    …..

    Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2.

    i.e., range of f = [2,)

    Alter:

    Let x be any real number.

    Accordingly,

    x2≥ 0

    ⇒ x2 + 2 ≥ 0 + 2

    ⇒ x2 + 2 ≥ 2

    ⇒ f(x) ≥ 2

    ∴ Range of f = [2,)

    (iii) f(x) = x, x is a real number

    It is clear that the range of f is the set of all real numbers.

    ∴ Range of f = R

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