Plus Two Math's Solution Ex 1.3 Chapter1 Relations and Function

Plus Two Math's Solution Ex 1.3 Chapter1 Relations and Function

NCERT Solutions for Plus Two Maths Chapter 1 Relations and Functions Ex 1.3, NCERT Solutions for Plus Two Maths is available here. Plus Two Math's solutions are given here as a PDF file which is easily downloadable free of cost by the students of Plus Two. If you want to learn more about NCERT Solutions then visit www.keralanotes.com; this website provides complete solutions for Plus Two mathematics

Practice the NCERT Solutions PDF Ex 1.3 to solve examples related to Relations and Functions. This NCERT Solutions for Plus Two Maths Chapter helps you understand what are Relations and Functions.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus Two
Subject Math's Textbook Solution
Chapter Chapter 1
Exercise Ex 1.3
Chapter Name Relations and Function
Category Plus Two Kerala


Kerala Syllabus Plus Two Math's Textbook Solution Chapter  1 Relation and Function Exercises 1.3


Chapter  1  Relations and Function Solution



Kerala plus two 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  1  Relations and Function Exercise   1.3

    Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by = {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}. Write down gof.

    The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as

    = {(1, 2), (3, 5), (4, 1)} and = {(1, 3), (2, 3), (5, 1)}.

    Let fg, and h be functions from to R. Show that

    To prove:

    Find goand fog, if

    (i) 

    (ii) 

    (i) 

    (ii) 

    If, show that f f(x) = x, for all. What is the inverse of f?

    It is given that.

    Hence, the given function f is invertible and the inverse of f is f itself.

    State with reason whether following functions have inverse

    (i) f: {1, 2, 3, 4} → {10} with

    f = {(1, 10), (2, 10), (3, 10), (4, 10)}

    (ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with

    g = {(5, 4), (6, 3), (7, 4), (8, 2)}

    (iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} with

    h = {(2, 7), (3, 9), (4, 11), (5, 13)}

    (i) f: {1, 2, 3, 4} → {10}defined as:

    f = {(1, 10), (2, 10), (3, 10), (4, 10)}

    From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10

    is not one-one.

    Hence, function does not have an inverse.

    (ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:

    g = {(5, 4), (6, 3), (7, 4), (8, 2)}

    From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4.

    is not one-one,

    Hence, function g does not have an inverse.

    (iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:

    h = {(2, 7), (3, 9), (4, 11), (5, 13)}

    It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.

    ∴Function h is one-one.

    Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.

    Thus, h is a one-one and onto function. Hence, h has an inverse.

    Show that f: [−1, 1] → R, given byis one-one. Find the inverse of the function f: [−1, 1] → Range f.

    (Hint: For y ∈Range fy =, for some x in [−1, 1], i.e.,)

    f: [−1, 1] → R is given as

    Let f(x) = f(y).

    ∴ f is a one-one function.

    It is clear that f: [−1, 1] → Range f is onto.

    ∴ f: [−1, 1] → Range f is one-one and onto and therefore, the inverse of the function:

    f: [−1, 1] → Range exists.

    Let g: Range f → [−1, 1] be the inverse of f.

    Let y be an arbitrary element of range f.

    Since f: [−1, 1] → Range f is onto, we have:

    Now, let us define g: Range f → [−1, 1] as

    gof = and fo

    f−1 = g

    ⇒ 

    Consider fR → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

    fR → R is given by,

    f(x) = 4x + 3

    One-one:

    Let f(x) = f(y).

    ∴ f is a one-one function.

    Onto:

    For y ∈ R, let y = 4x + 3.

    Therefore, for any ∈ R, there exists  such that

    ∴ f is onto.

    Thus, f is one-one and onto, and therefore, f−1 exists.

    Let us define gR→ R by.

    Hence, f is invertible and the inverse of f is given by

    Consider fR→ [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given by, where R+ is the set of all non-negative real numbers.

    fR+ → [4, ∞) is given as f(x) = x2 + 4.

    One-one:

    Let f(x) = f(y).

    ∴ f is a one-one function.

    Onto:

    For y ∈ [4, ∞), let y = x2 + 4.

    Therefore, for any ∈ R, there exists  such that

    .

    ∴ f is onto.

    Thus, f is one-one and onto, and therefore, f−1 exists.

    Let us define g: [4, ∞) → Rby,

    Hence, f is invertible and the inverse of f is given by

    Consider fR+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with.

    fR+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.

    Let y be an arbitrary element of [−5, ∞).

    Let y = 9x2 + 6− 5.

    f is onto, thereby range f = [−5, ∞).

    Let us define g: [−5, ∞) → R+ as

    We now have:

    and

    Hence, f is invertible and the inverse of f is given by

    Let fX → Y be an invertible function. Show that f has unique inverse.

    (Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y,

    fog1(y) = IY(y) = fog2(y). Use one-one ness of f).

    Let fX → Y be an invertible function.

    Also, suppose f has two inverses (say).

    Then, for all y ∈Y, we have:

    Hence, f has a unique inverse.

    Consider f: {1, 2, 3} → {abc} given by f(1) = af(2) = b and f(3) = c. Find f−1 and show that (f−1)−1 = f.

    Function f: {1, 2, 3} → {abc} is given by,

    f(1) = af(2) = b, and f(3) = c

    If we define g: {abc} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:

    and, where X = {1, 2, 3} and Y= {abc}.

    Thus, the inverse of exists and f−1= g.

    f−1: {abc} → {1, 2, 3} is given by,

    f−1(a) = 1, f−1(b) = 2, f-1(c) = 3

    Let us now find the inverse of f−1 i.e., find the inverse of g.

    If we define h: {1, 2, 3} → {abc} as

    h(1) = ah(2) = bh(3) = c, then we have:

    , where X = {1, 2, 3} and Y = {abc}.

    Thus, the inverse of exists and g−1 = h ⇒ (f−1)−1 = h.

    It can be noted that h = f.

    Hence, (f−1)−1 = f.

    Let fX → Y be an invertible function. Show that the inverse of f−1 is f, i.e.,

    (f−1)−1 = f.

    Let fX → Y be an invertible function.

    Then, there exists a function gY → X such that gof = IXand fo= IY.

    Here, f−1 = g.

    Now, gof = IXand fo= IY

    ⇒ f−1of = IXand fof−1= IY

    Hence, f−1Y → X is invertible and f is the inverse of f−1

    i.e., (f−1)−1 = f.

    If f→ be given by, then fof(x) is

    (A)  (B) x3 (C) x (D) (3 − x3)

    fR → R is given as.

    The correct answer is C.

    Letbe a function defined as. The inverse of f is map g: Range

    (A)  (B) 

    (C)  (D) 

    It is given that

    Let y be an arbitrary element of Range f.

    Then, there exists x ∈such that 

    Let us define g: Rangeas

    Now,

    Thus, g is the inverse of f i.e., f−1 = g.

    Hence, the inverse of f is the map g: Range, which is given by

    The correct answer is B.

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