Plus One Math's Solution miscellaneous Chapter2 Relations and Functions


In this chapter, you are provided with several diagrams and examples along with their solutions for a clear understanding of Relations and Functions. To know more about 
Class 11 Maths Chapter 2 Relations and Functions, you should explore the exercises below. You can also download the Sets Class 11 NCERT Solutions PDF

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 Miscellaneous Exercise
Chapter Name Relations and Functions
Category Plus One Kerala


Kerala Syllabus Plus One Math's Textbook Solution Chapter  2 Relations and Functions Miscellaneous Exercise


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 Miscellaneous Exercise

    If f(x) = x2, find.

    The relation f is defined by 

    The relation g is defined by 

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

    The relation f is defined as

    It is observed that for

    0 ≤ x < 3, f(x) = x2

    3 < x ≤ 10, f(x) = 3x

    Also, at x = 3, f(x) = 32 = 9 or f(x) = 3 × 3 = 9

    i.e., at x = 3, f(x) = 9

    Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.

    Thus, the given relation is a function.

    The relation g is defined as

    It can be observed that for x = 2, g(x) = 22 = 4 and g(x) = 3 × 2 = 6

    Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.

    Find the domain of the function 

    The given function is.

    It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.

    Hence, the domain of f is R – {2, 6}.

    Find the domain and the range of the real function f defined by.

    The given real function is.

    It can be seen that is defined for (x – 1) ≥ 0.

    i.e., is defined for x ≥ 1.

    Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of f = [1,).

    As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ 

    Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of f = [0,).

    Find the domain and the range of the real function f defined by f (x) = |x – 1|.

    The given real function is f (x) = |x – 1|.

    It is clear that |x – 1| is defined for all real numbers.

    ∴Domain of f = R

    Also, for x ∈ R, |x – 1| assumes all real numbers.

    Hence, the range of f is the set of all non-negative real numbers.

    Letbe a function from R into R. Determine the range of f.

    The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.

    [Denominator is greater numerator]

    Thus, range of f = [0, 1)

    Let fgR → R be defined, respectively by f(x) = + 1, g(x) = 2x – 3. Find f + gf – g and.

    fgR → is defined as f(x) = + 1, g(x) = 2x – 3

    (f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2

    ∴(f + g) (x) = 3x – 2

    (f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4

    ∴ (f – g) (x) = –x + 4

    Let = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers ab. Determine ab.

    = {(1, 1), (2, 3), (0, –1), (–1, –3)}

    f(x) = ax + b

    (1, 1) ∈ f

    ⇒ f(1) = 1

    ⇒ a × 1 + b = 1

    ⇒ a + b = 1

    (0, –1) ∈ f

    ⇒ f(0) = –1

    ⇒ a × 0 + b = –1

    ⇒ b = –1

    On substituting b = –1 in a + b = 1, we obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2.

    Thus, the respective values of a and b are 2 and –1.

    Let R be a relation from N to N defined by R = {(ab): ab ∈ N and a = b2}. Are the following true?

    (i) (aa) ∈ R, for all a ∈ N

    (ii) (ab) ∈ R, implies (ba) ∈ R

    (iii) (ab) ∈ R, (bc) ∈ R implies (ac) ∈ R.

    Justify your answer in each case.

    R = {(ab): ab ∈ N and a = b2}

    (i) It can be seen that 2 ∈ N;however, 2 ≠ 22 = 4.

    Therefore, the statement “(aa) ∈ R, for all a ∈ N” is not true.

    (ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.

    Now, 3 ≠ 92 = 81; therefore, (3, 9) ∉ N

    Therefore, the statement “(ab) ∈ R, implies (ba) ∈ R” is not true.

    (iii) It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 42 and 4 = 22.

    Now, 16 ≠ 22 = 4; therefore, (16, 2) ∉ N

    Therefore, the statement “(ab) ∈ R, (bc) ∈ R implies (ac) ∈ R” is not true.

    Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?

    (i) f is a relation from A to B (ii) f is a function from A to B.

    Justify your answer in each case.

    A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

    ∴A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

    It is given that = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

    (i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

    It is observed that f is a subset of A × B.

    Thus, f is a relation from A to B.

    (ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation is not a function.

    Let f be the subset of Z × Z defined by = {(aba + b): ab ∈ Z}. Is f a function from Z to Z: justify your answer.

    When 1 value of y gets mapped from 2 values of X makes f a non function 

     

    Let easy be the way when a+ b = 0 , a = -b 

    ab = - a×a  which is multiple 

    E.g. - 4 = -2× 2     -2 + 2 = 0

             -9 = -3×3     -3 + 3 = 0 

    Hence this is many to one which is not a function

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

    A = {9, 10, 11, 12, 13}

    f: A → N is defined as

    f(n) = The highest prime factor of n

    Prime factor of 9 = 3

    Prime factors of 10 = 2, 5

    Prime factor of 11 = 11

    Prime factors of 12 = 2, 3

    Prime factor of 13 = 13

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

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

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

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

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

    The range of f is the set of all f(n), where n ∈ A.

    ∴Range of f = {3, 5, 11, 13}

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Chapter 2: Relations and Functions Miscellaneous Exercise Solution


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