NCERT EXERCISE 1.3

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    Exercise 1.3

    The Exercise 1.3 deals with functions and various types of sets and several properties of sets.

    The exercise 1.3 also include the examples based on various ranges as R, Z and each type of function. The Exercise 1.3 provides a lot of working problems to make you more familiar and through with the concepts..

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    Question 1:
    Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
    Solution :
    The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as
    f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.
    gof (1) = g[f(1)] = g(2) = 3            [as f(1) = 2 and g(2) = 3]
    gof (3) = g[f(3)] = g(5) = 1            [as f(3) = 5 and g(5) = 1]
    gof (4) = g[f(4)] = g(1) = 3            [as f(4) = 1 and g(1) = 3]
    ∴ gof = {(1, 3), (3, 1), (4, 3)}
    EXERCISE 1.3                                                                                                    Question 2:
    Let f, g and h be functions from R to R. Show that
    (f + g)oh = foh + goh
    (f.g)oh = (foh).(goh)
    Solution :
    To prove: (f + g)oh = foh + goh
    LHS = [(f + g)oh](x)
    = (f + g)[h(x)] = f [h(x)] + g[h(x)]
    = (foh)(x) + (goh)(x)
    = {(foh)(x) + (goh)}(x) = RHS
    ∴ {(f + g)oh}(x) = {(foh)(x) + (goh)}(x)                       for all x ∈R
    Hence, (f+g)oh = foh + goh
    To Prove: (f.g)oh = (foh).(goh)
    LHS = [(f.g)oh](x)
    = (f.g)[h(x)] = f[h(x)] . g[h(x)]
    = (foh)(x) . (goh)(x)
    = {(foh).(goh)}(x) = RHS
    ∴[(f.g)oh](x) = {(foh).(goh)}(x)                                      for all x ∈R
    Hence, (f.g)oh = (foh).(goh)
    EXERCISE 1.3                                                                                                     Question 3:
    Find 𝑔𝑜𝑓 and 𝑓𝑜𝑔, if
    (i) 𝑓(𝑥) = |𝑥| and 𝑔(𝑥) = |5𝑥 − 2|
    (ii) 𝑓(𝑥) = 8𝑥³ and 𝑔(𝑥) = 𝑥ˆ1/3
    Solution:
    (i) 𝑓(x) = |x| and g(x) = |5x-2|
    ∴ go f(x) = g(f(x)) = g(|x|) = |5|x|-2|
    fog(x) = f(g(x)) = f(|5x-2|) = ||5x-2|| = |5x-2|
    (ii) f(x) = 8x³ and g(x) = 𝑥^1/3
    ∴gof(x) = g(f(x)) = g(8x³ ) = (8𝑋³)^1/3 = 2x
    fog(x) = f(g(x)) = f(𝑥^1/3)³ = 8(𝑥^1/3)³ = 8x
    EXERCISE 1.3                                                                                                Question 4:
    If 𝑓(𝑥) = (4x+3) /(6x−4) , 𝑥 ≠ 2/3 , show that fof (x) = x, for all 𝑥 ≠ 2/3 . What is the inverse of f ?
    Solution :
    It is given that f(x) = (4x+3) /(6x−4) , 𝑥 ≠ 2/3
    (fof)(x) = f(f(x)) = f ((4x+3) /(6x−4) = 4(4𝑥+3 6𝑥−4)+3/6(4𝑥+3 6𝑥−4)− 4
    =16𝑥+12+18𝑥−12 /(24𝑥+18−24𝑠+16)=34𝑥/34
    ∴fof(x) = x, for all x ≠ 2/3 .
    ⇒ fof = Ix
    Hence, the given function f is invertible and the inverse of f is f itself.
    EXERCISE 1.3                                                                                                 Question 5:
    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)}
    Solution 5:
    (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
    ∴f is not one – one.
    Hence, function f 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.
    ∴ g 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.
    EXERCISE 1.3                                                                                              Question 6:
    Show that f: [−1, 1] → R, given by𝑓(𝑥) = 𝑋/(𝑋+2) is one – one. Find the inverse of the function f: [−1, 1] → Range f.
    (Hint: For y ∈ Range f, y = 𝑓(𝑥) = 𝑋/(𝑋+2)) , for some x in [−1, 1], i.e., 𝑥 = 2𝑦/ (1−𝑦))
    Solution :
    f: [−1, 1] → R is given as 𝑓(𝑥) = 𝑋/ (𝑋+2)
    For one – one
    Let f(x) = f(y)
    ⇒ 𝑋/ (𝑋+2) = 𝑌/ (𝑌+2)
    ⇒xy +2x = xy +2y
    ⇒2x = 2y
    ⇒x = 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 f 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                                                                  y = f(x) for some x ∈ [−1, 1]
    ⇒ y = 𝑋/ (𝑋+2)
    ⇒ xy + 2y = x
    ⇒ x(1-y) = 2y
    ⇒ x = 2𝑦/1−𝑦,                       y ≠ 1
    Now, let us define g: Range f → [−1, 1] as
    g(y) = 2𝑦1/ −𝑦 ,                   y ≠ 1
    Now,
    (gof)(x) = g(f(x)) = g (𝑋/ (𝑋+2)) = 2 (2(x/ x+2)/(1−(x/x+2))=2x/(x+2−x) =2x/ 2 =x
    and
    (fog)(y) = f(g(y)) = f (2𝑦/ 1−𝑦) = 2𝑦/ (1−y)/ (2𝑦/ (1−y) +2)=2y/ (2y+2−2y)              =  2y /2 = y
    ∴ gof = x = I[−1,1] and fog = y = I Range f
    ∴ 𝑓¯1 = g
    ⇒𝑓¯1 (y) = 2𝑦/ 1−𝑦 , y ≠ 1
    EXERCISE 1.3                                                                                                       Question 7:
    Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
    Solution:
    f: R → R is given by, f(x) = 4x + 3
    For one – one
    Let f(x) = f(y)
    ⇒ 4x + 3 = 4y + 3
    ⇒ 4x = 4y
    ⇒ x = y
    ∴ f is a one – one function
    For onto
    For y ∈ R, let y = 4x + 3.
    ⇒ x = 𝑌−3/ 4 ∈ R
    Therefore, for any y ∈ R, there exists x = 𝑌−34 ∈ R, such that
    f(x) = f (𝑌−3/ 4) = 4 (𝑌−3 /4) + 3 = y.
    ∴ f is onto.
    Thus, f is one – one and onto and therefore, f−1 exists.                                                Let us define g: R → R by g(x) = 𝑌−34
    Now,
    (gof)(x) = g(f(x)) = g(4x + 3) = (4x+3)−3/ 4 = 4x/ 4 = x
    and
    (fog)(y) = f(g(y)) = f (𝑌−3 /4) = 4 (𝑌−3 /4) + 3 = y-3 + 3 = y
    ∴ gof = fog = IR
    Hence, f is invertible and the inverse of f is given by f ¯1 (y) = g(y) = 𝑌−3 /4 .
    EXERCISE 1.3                                                                                                   Question 8:
    Consider f: 𝑹+ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse 𝑓 ¯1 of given f by 𝑓 ¯1 (𝑦) = √(Y−4), where 𝑹+ is the set of all non-negative real numbers.
    Solution :
    f: 𝑹+ → [4, ∞) is given as f(x) = x² + 4.
    For one – one
    Let f(x) = f(y)
    ⇒ x² + 4 = y² + 4
    ⇒ x² = y²
    ⇒ x = y                                                 [𝑎𝑠 𝑥 = 𝑦 ∈ 𝑹+]
    ∴ f is a one – one function.
    For onto
    For y ∈ [4, ∞), let y = x²+ 4
    ⇒ 𝑥² = 𝑦 − 4 ≥ 0 [as 𝑦 ≥ 4]
    ⇒ x =√(Y−4) ≥ 0
    Therefore, for any y ∈ [4, ∞), there exists x = √(Y−4) ∈ R+ , such that
    f(x) = f(√Y−4) = (√Y−4)² + 4 = y – 4 + 4 = y
    ∴ f is onto.
    Thus, f is one – one and onto and therefore, f ¯1 exists.
    Let us define g: [4, ∞) → R+ by g(y) = √(Y−4)
    Now,
    (gof)(x) = g(f(x)) = g(x² + 4) = √(x²+4)−4 = √X² =x
    and
    (fog)(y) = f(g(y)) = f(√Y−4) = (√Y−4)² + 4 = y – 4 + 4 = y
    ∴ gof = fog = I_ R
    Hence, f is invertible and the inverse of f is given by f ¯1 (y) = g(y) = √(Y−4).

    Question 9:
    Consider f: 𝑹+ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with   𝑓 ¯1 (𝑦) =
    ((√𝑦+6)−1/ 3)
    Solution:
    f: R+ → [−5, ∞) is given as f(x) = 9x² + 6x − 5.
    Let y be an arbitrary element of [−5, ∞).
    Let y = 9x² + 6x – 5
    ⇒ y = (3x + 1)² -1-5 = (3x + 1)² -6
    ⇒ y + 6 = (3x + 1)²
    ⇒ 3𝑥 + 1 = √(Y+6)                                        [as 𝑦 ≥ −5 ⟹ 𝑦 + 6 > 0]
    ⇒ x = ((√𝑦+6)− 1 /3)
    ∴ f is onto, thereby range f = [−5, ∞).
    Let us define g: [−5, ∞) → R+ as g(y) = (√(𝑦+6)−1 /3)
    Now, (gof)(x) = g(f(x)) = g(9x² + 6x-5) = g((3x + 1)²-6)
    = √√((3x+1)²−6+6 −1)
    = 3x+1−1 / 3=  3x /3=  x
    and
    (fog)(y) = f(g(y)) = 𝑓(√𝑌+6−1 / 3)=[3(√y+6−1 / 3)+1 ]²−6
    = (√Y+6)²−6  =  𝑦+6 −6  =  𝑦
    ∴ gof = x = IR and fog = y = 𝑙𝑅𝑎𝑛𝑔𝑒 𝑓
    Hence, f is invertible and the inverse of f is given by
    𝑓 ¯1 (𝑦) = 𝑔(𝑦) = ((√𝑌+6−1 /3)

    EXERCISE 1.3                                                                                                 Question 10:
    Let f: X → 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 ).
    Solution:
    Let f: X → Y be an invertible function.
    Also, suppose f has two inverses (say g1 and g2)
    Then, for all y ∈Y, we have
    fog1(y) = IY(y) = fog2(y)                                                                                           ⇒ f(g1 (y)) = f(g2 (y))
    ⇒ g1 (y) = g2 (y)                                   [as f is invertible ⇒ f is one – one]
    ⇒ 𝑔1 = 𝑔2                                              [as 𝑔 is one – one]
    Hence, f has a unique inverse.

    Question 11:
    Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f−1 and show that
    (f ¯1 )¯1 = f.
    Solution:
    Function f: {1, 2, 3} → {a, b, c} is given by f(1) = a, f(2) = b, and f(3) = c
    If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3.
    We have
    (fog)(a) = f(g(a)) = f(1) = a
    (fog)(b) = f(g(b)) = f(2) = b
    (fog)(c) = f(g(c)) = f(3) = c
    and
    (gof)(1) = g(f(1)) = f(a) = 1
    (gof)(2) = g(f(2)) = f(b) = 2
    (gof)(3) = g(f(3)) = f(c) = 3
    ∴ gof = IX and fog = IY, where X = {1, 2, 3} and Y= {a, b, c}.
    Thus, the inverse of f exists and f¯1 = g.
    ∴ f¯1 :{a, b, c} → {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} → {a, b, c} as h(1) = a, h(2) = b, h(3) = c
    We have
    (goh)(1) = g(h(1)) = g(a) = 1
    (goh)(2) = g(h(2)) = g(b) = 2
    (goh)(3) = g(h(3)) = g(c) = 3
    and
    (hog)(a) = h(g(a)) = h(1) = a
    (hog)(b) = h(g(b)) = h(2) = b
    (hog)(c) = h(g(c)) = h(3) = c
    ∴ goh = IX and hog = IY, where X = {1, 2, 3} and Y = {a, b, c}.
    Thus, the inverse of g exists and g¯1 = h ⇒ (f¯1 )¯1 = h.
    It can be noted that h = f.
    Hence, (f¯1 )¯1 = f .
    Question 12:
    Let f: X → Y be an invertible function. Show that the inverse of f¯1 is f,
    i.e., (f¯1 )¯1 = f.
    Solution:
    Let f: X → Y be an invertible function.
    Then, there exists a function g: Y → X such that gof = IX and fog = IY.
    Here, f¯1 = g.
    Now,
    gof = IX and fog = IY
    ⇒ f¯1 of = IX and fof¯1 = IY
    Hence, f¯1 : Y → X is invertible and f is the inverse of f¯1 i.e., (f¯1 )¯1 = f
    Question 13:
    If f: R → R be given by 𝑓(𝑥) = (3−𝑥³) ^1/3, then fof(x) is
    (A) 1/ x³
    (B) x³
    (C) X
    (D) (3 − x³)
    Solution:
    f: R → R be given as f(x) = (3−𝑥³) ^1/3
    ∴ fof(x) = f(f(x)) = 𝑓(3−𝑥³) ^1/ 3 = [3−((3−𝑥³)^ 1/3)³] ^1/3
    = [3−(3−𝑥³)]^1/3=(x³)^ 1/3
    ∴ fof(x) = x
    The correct answer is C.
     Question 14:
    : Let f: R- {- 4 /3} → R be a function as f(x) = 4x/ 3x+4 . The inverse of f is map g: Range f → R- {- 4/ 3} given by
    (A) g(y) = 3y /3−4y
    (B) g(y) = 4y / 4−3y
    (C) g(y) = 4y / 3−4y                                                                                                  (D) g(y) = 3y / 4−3y
    Solution :
    It is given that 𝑓:𝑹 − {- 4 /3} → 𝑹 be a function as 𝑓(𝑥) = 4x / 3x+4
    Let y be an arbitrary element of Range f.
    Then, there exists x ∈ R- {- 4 /3} such that y = f(x)
    ⇒ y = 4x / 3x+4
    ⇒ 3xy + 4y = 4x
    ⇒ x(4-3y) = 4y
    ⇒ x = 4y / 4−3y
    Let us define g: Range f → R− {- 4/ 3} as g(y) = 4y /4−3y
    Now,
    𝑔𝑜𝑓(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔 (4x / 3x+4) = 4(4x /3x+4 ) / (4−3(4x / 3x+4)) =                        16x / (12x+16−12x)=                16x /16 =x
    and
    𝑓o𝑔(𝑦) = 𝑓(𝑔(𝑦)) = 𝑓 (4y /4−3y) = 4(4y /4−3y )  / 3(4y /4−3y)+4 =16y / 12y+16−12y=16y / 16 =y
    ∴ gof = IR−{4 /−3} and fog = IRange f
    Thus, g is the inverse of f i.e., f¯1 = g.
    Hence, the inverse of f is the map g: Range f → R- {-4/ 3}, which is given by                 g(y) = 4y /4−3y .
    The correct answer is B.
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