TOPICS
Exercise - 1.3

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) = f(f(1)) = g(2) = 3   [where f(1) = 2, g(2) = 3] 
  ⇒ gof(3) = f(f(3)) = g(5) = 1   [where f(3) = 5, g(5) = 1] 
  ⇒ gof(4) = f(f(4)) = g(1) = 3   [where f(4) = 1, g(1) = 3] 
  ⇒ gof = {(1,3), (3,1), (4,3)}
    

Question-2 :-  Let f, g and h be functions from R to R. Show that (a) (f + g)oh = foh + goh, (b) (f . g)oh = (foh) . (goh)

Solution :-
(a)  (f + g)oh = foh + goh 
   ⇒ L.H.S (Left Hand Side)
   ⇒ Consider (f + g)oh(x)
   ⇒ (f+g)(h(x))
   ⇒ f(h(x)) + g(h(x))
   ⇒ foh(x) + goh(x)
   ⇒ {(foh) + (goh)}(x)
    ∴((f+g)oh)(x) = {(foh) + (goh)}(x) for all x ∈ R
      Hence, (f + g)oh = foh + goh.
        
(b)  (f . g)oh = (foh) . (goh) 
   ⇒ L.H.S (Left Hand Side)
   ⇒ Consider (f . g)oh(x)
   ⇒ (f . g)(h(x))
   ⇒ f(h(x)) . g(h(x))
   ⇒ foh(x) . goh(x)
   ⇒ {(foh) . (goh)}(x)
    ∴((f . g)oh)(x) = {(foh) . (goh)}(x) for all x ∈ R
      Hence, (f . g)oh = (foh) . (goh) 
    

Question-3 :-  Find gof and fog, if (i) f(x) = |x| and g(x) = | 5x – 2 |,     (ii) f(x) = 8x³ and g(x) = x1/3.

Solution :-
(i)  f(x) = |x| and g(x) = | 5x – 2 |
     gof(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) = x1/3
     gof(x) = g(f(x))
            = g(8x³)
            = (8x³)1/3
            = (2x)3 x 1/3
            = 2x

    fog(x) = f(g(x)) 
           = f(x1/3)
           = 8 (x1/3)³
           = 8x
    

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 :-
     Given that f(x) = (4x+3)/(6x-4), x ≠ 2/3
    answer
     Therefore, fof(x) = x, for all x ≠ 2/3.
     fof(x) = I
     Hence, the given function f is invertible and the inverse of f is f itself.
    

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 :-
(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 on-to function. 
     Hence, h has an inverse.
    

Question-6 :-  Show that f : [–1, 1] → R, given by f(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),  
     Let f(x) = f(y).
 ⇒  x/(x+2) = y/(y+2)
 ⇒  x(y+2) = y(x+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 on-to.
    ∴ f: [−1, 1] → Range f is one-one and on-to 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 on-to, we have:
    y = f(x) for same x ∈ [-1,1]
 ⇒ y = x/(x+2)
 ⇒ y(x+2) = x
 ⇒ xy + 2y = x
 ⇒ x = 2y/(1-y), y ≠ 1
    Now, let us define g: Range f → [−1, 1] as
    g(y) = 2y/(1-y), y ≠ 1
    answer
    f-1(y) = 2y/(1-y), y ≠ 1
    

Question-7 :- Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Solution :-
     Given that  f: R → R given by f(x) = 4x + 3
     Let f(x) = f(y).
  ⇒ 4x + 3 = 4y + 3
  ⇒ 4x = 4y
  ⇒ x = y
    ∴f is one-one.
     For y ∈ R , let y = 4x + 3.
  ⇒ 4x = y - 3
  ⇒ x = (y-3)/4 ∈ R
     Therefore, for any y ∈ R, there exists x = (y-3)/4 ∈ R such that
  ⇒ f(x) = 4((y-3)/4) + 3
  ⇒ f(x) = y - 3 + 3
  ⇒ f(x) = y
   ∴f is on-to.
    Thus, f is one-one and on-to and therefore, f-1 exists.

    Let us define g: R → R by g(x) = (y-3)/4
    question
    gof = fog = I
    Hence, f is invertible and the inverse of f is given by
    f-1(y) = g(y) = (y-3)/4
    

Question-8 :-  Consider f : R+ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f-1 of f given by f-1(y) = √y - 4 , where R+ is the set of all non-negative real numbers.

Solution :-
     f : R+ → [4, ∞) given by f(x) = x² + 4.    
     Let f(x) = f(y).
 ⇒  x² + 4 = y² + 4
 ⇒  x² = y²
 ⇒  x = y
   ∴ f is a one-one function.
    f : R+ → [4, ∞), Let y = x² + 4.
 ⇒  x² = y - 4
 ⇒  x = √y - 4 > 0
 ⇒  f(√y - 4) = (√y - 4)² + 4 = y
    ∴f is on-to.
    Thus, f is one-one and on-to and therefore, f-1 exists.

    Let us define f : R+ → [4, ∞) given by g(y) = √y - 4
    answer
    gof = fog = I
    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.
    f(x) = f(y)
    9x² + 6x − 5 = 9y² + 6y − 5
 ⇒ 9x² + 6x = 9y² + 6y
 ⇒ 9x² - 9y²= 6y - 6x
 ⇒ 9(x²-y²) = -6(x-y)
 ⇒ 9(x+y)(x-y) - 6(x-y) = 0
 ⇒ (x-y)[9(x+y) - 6] = 0
 ⇒ x - y = 0
 ⇒ x = y
   ∴f is one-one.
    Now, f:  R+ → [−5, ∞) is given as f(x) = 9x² + 6x − 5
    Let f(x) = y
    y = 9x² + 6x − 5
 ⇒ 9x² + 6x - (y + 5) = 0
    By discrimination method values find of x
 ⇒ a = 9, b = 6, c = -(y + 5)
 ⇒ d = b² - 4ac
 ⇒ d = 6² - 4 x 9 x (-(y + 5))
 ⇒ d = 36 + 36y + 180
 ⇒ d = 216 + 36y
 ⇒ d = 36(6 + y)
 ⇒ d = 36(y + 6)
 ⇒ x = (-b ± √d)/2a = (-6 ± √36(y+6))/(2 x 9)
 ⇒ x = (-6 ± 6√(y+6))/18
 ⇒ x = (√y+6-1)/3
   Therefore, ∴f is on-to. answer
    

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

Solution :-
     Given f : X → Y be an invertible function.
     Also, suppose f has two inverses (g₁ and g₂)
     Then for all y ∈ Y, fog₁(y) = I(y) = fog₂(y)
 ⇒  f(g₁(y)) = f(g₂(y))
 ⇒  g₁(y) = g₂(y)
 ⇒  g₁ = g₂
     The inverse is unique and 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, then 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 = I, fog = I

    f = {(1,a),(2,b),(3,c)}
    Therefore, f-1 = {(1,a),(2,b),(3,c)}
    (f-1)-1 = {(1,a),(2,b),(3,c)} = 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-1of = 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) x1/3
(B) x³
(C) x
(D) (3 – x³).

Solution :-
     Given f: R → R be given by f(x) = (3 - x³)1/3
      answer
     The correct answer is C. 
    

Question-14 :-  Let f: R-{-4/3} → R be a function defined as f(x) = 4x/(3x + 4). The inverse of f is the 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 :-
    Given f: R-{-4/3} → R be a function defined as f(x) = 4x/(3x + 4).
    Now, Range of f → R-{-4/3}
    Let y = f(x)
    y = 4x/(3x + 4)
 ⇒  y(3x + 4) = 4x
 ⇒  3xy + 4y = 4x
 ⇒  3xy - 4x = -4y
 ⇒  x(3y-4) = -4y
 ⇒  x = 4y/(4-3y) 
    f-1(y) = g(y) = 4y/(3-4y)
    Therefore, option (B) is correct.
    
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