TOPICS
Exercise - 1.2

Question-1 :-  Show that the function f : R∗ → R∗ defined by f(x) = 1/x is one-one and onto, where R∗ is the set of all non-zero real numbers. Is the result true, if the domain R∗ is replaced by N with co-domain being same as R∗

Solution :-
     Given that f : R∗ → R∗ defined by f(x) = 1/x.
  ⇒ f(x) = f(y)
  ⇒ 1/x = 1/y
  ⇒ x = y
   ∴f is one-one.
    Now, f(x) = 1/x
  ⇒ y = 1/x
  ⇒ x = 1/y 
  ⇒ f(1/y) = x
   ∴f is on-to.

    Again, g : R∗ → R∗ defined by g(x) = 1/x.
  ⇒ g(x₁) = g(x₂)
  ⇒ 1/x₁ = 1/x₂
  ⇒ x₁ = x₂ and (x₁, x₂) ∉ N
   ∴g is one-one.
    Now, g(x) = 1/x
    But every real number belonging to co-domain may not have a pre-image in N. e.g., 1/3 = 3/2 ∉ N.    
   ∴g is not on-to.
    

Question-2 :-  Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : R → R given by f(x) = x²
(iv) f : N → N given by f(x) = x³
(v) f : Z → Z given by f(x) = x³

Solution :-
(i)  f : N → N given by f(x) = x² 
  ⇒ f(x₁) = f(x₂)
  ⇒ x₁² = x₂²
  ⇒ x₁ = x₂
   ∴f is one-one.
    Now, f(x) = x²
    There are such numbers of co-domain which have no image in domain N.
    e.g., 3 ∈ co-domain N, but there is no pre-image in domain of f. 
   ∴f is not on-to.
    Hence, f is injective but not surjective.         
(ii) f : Z → Z given by f(x) = x² 
     since, Z = {0, ±1, ±2, ±3...}
     Therefore, f(-1) = f(1) = 1
  ⇒ -1  and 1 have same image.
    ∴f is not one-one.
     Now, f(x) = x²
     There are such numbers of co-domain which have no image in domain Z.
     e.g., 3 ∈ co-domain Z, but there is no √3 ∉ Z domain of f. 
    ∴f is not on-to.
     Hence, f is neither injective nor surjective.
(iii) f : R → R given by f(x) = x² 
      since, R = {0, ±1, ±2, ±3...}
      Therefore, f(-1) = f(1) = 1
  ⇒  -1  and 1 have same image.
     ∴f is not one-one.
      e.g., 3 ∈ co-domain R, but there is no √-3 ∉ R domain of f. 
     ∴f is not on-to.
      Hence, f is neither injective nor surjective.
(iv)  f : N → N given by f(x) = x³ 
  ⇒  f(x₁) = f(x₂)
  ⇒  x₁³ = x₂³
  ⇒  x₁ = x₂
     for every x ∈ N, has a unique image in its co-domain. 
    ∴f is one-one.
     There are many such members of co-domain of f which do not have pre-image in its domain. e.g., 2, 3 etc. 
    ∴f is not on-to.
     Hence, f is injective but not surjective.
        
(v)  f : Z → Z given by f(x) = x³
  ⇒ f(x₁) = f(x₂)
  ⇒ x₁³ = x₂³
  ⇒ x₁ = x₂
    for every x ∈ Z, has a unique image in its co-domain. 
   ∴f is one-one.
    There are many such members of co-domain of f which do not have pre-image in its domain.
   ∴f is not on-to. 
    Hence, f is injective but not surjective.
    

Question-3 :-  Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Solution :-
    f: → R → R is given by,
    f(x) = [x]
    It is seen that f(1.4) = [1.4] = 1, f(1.8) = [1.8] = 1.
    ∴ f(1.4) = f(1.8), but 1.4 ≠ 1.8.
    ∴ f is not one-one. 
    Now, consider 0.8 ∈ R.
    It is known that f(x) = [x] is always an integer. 
    Thus, there does not exist any element x ∈ R such that f(x) = 0.8.
    ∴ f is not on-to.
    Hence, the greatest integer function is neither one-one nor on-to.
    

Question-4 :-  Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.

Solution :-
    Modulus Function f : R → R, given by f(x) = |x|
    modulus function
    f contains (-1,1),(1,1),(-3,3),(3,3)
    Thus negative integers are not images of any element.  
   ∴f is not one-one.    
    Also second set R contains some negative numbers which are not images of any real number.
    i.e., f(x) =  |x| = -1
   ∴f is not on-to.
    Hence, f is neither one-one nor on-to.
    

Question-5 :-  Show that the Signum Function f : R → R, given by signum function is neither one-one nor onto.

Solution :-
    Signum Function f : R → R, given by answer
 ⇒  f(1) = f(2) = 1
 ⇒  f(x₁) = f(x₂) = 1 for x>1
 ⇒  x₁ ≠ x₂
   ∴f is not one-one.
    Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain, 
    there does not exist any x in domain such that f(x) = −2.
   ∴f is not onto.
    Hence, f is neither one-one nor on-to.
    

Question-6 :-  Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Solution :-
    Given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
    f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
   ∴f (1) = 4, f (2) = 5, f (3) = 6
    It is seen that the images of distinct elements of A under f are distinct.
    Hence, function f is one-one.
    

Question-7 :-  In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x²

Solution :-
(i)  f : R → R defined by f(x) = 3 – 4x 
  ⇒ f(x₁) = f(x₂)
  ⇒ 3 – 4x₁ = 3 – 4x₂
  ⇒ x₁ = x₂
   ∴f is one-one.
    Now, f(x) = 3 – 4x, Let f(x) = y
  ⇒ y = 3 – 4x
  ⇒ x = (3 - y)/4 ∈ R
  ⇒ f((3 - y)/4 ) = 3 - 4[(3 - y)/4 ] = 3 - 3 + y = y 
   ∴f is on-to.
    Hence, f is injective and surjective or f is bijective function.
(ii) f : R → R defined by f(x) = 1 + x²
  ⇒ f(x₁) = f(x₂)
  ⇒ 1 + x₁² = 1 + x₂²
  ⇒ x₁² = x₂²
  ⇒ x₁ = ± x₂
   ∴f is not one-one.
    Now, f(x) = 1 + x², Let f(x) = y
  ⇒ y = 1 + x²
  ⇒ x = ± √y - 1
  ⇒ f(√y - 1) = 1 + (1-y) = 2 - y ≠ y
   ∴f is not on-to.
    Hence, f is neither injective nor surjective so, f is not bijective function.
    

Question-8 :-  Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function

Solution :-
    f : A × B → B × A such that f(a, b) = (b, a)
  ⇒ Let (a₁, b₁) & (a₂, b₂) ∈ A x B such that f(a₁, b₁) = f(a₂, b₂)
  ⇒ (b₁, a₁) = (b₂, a₂)
  ⇒ a₁ = a₂, b₁ = b₂
  ⇒ (a₁, b₁) = (a₂, b₂) 
  ⇒ a₁ = a₂, b₁ = b₂
    So, for all (a₁, b₁),(a₂, b₂) ∈ A x B 
   ∴f is one-one.
    Now, let (b, a) ∈ B × A be any element.
    Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a).
    ∴f is on-to.
    Hence, f is injective and surjective so, f is bijective function.
    

Question-9 :-  Let f : N → N be defined by function State whether the function f is bijective. Justify your answer.

Solution :-
     Let f : N → N be defined by domain
(a)  f(1) = (n+1)/2 = (1+1)/2 = 2/2 = 1 and 
     f(2) = n/2 = 2/2 = 1
     The elements 1, 2, belonging to domain of f have the same image 1 in its co-domain. 
    ∴f is one-one.

(b)  Every number of co-domain has pre-image in its domain e.g., 1 has two pre-images 1 and 2. 
    ∴f is on-to.
     Hence, f is not injective but surjective. So, f is not bijective function.  
    

Question-10 :-  Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2)/(x-3). Is f one-one and onto? Justify your answer

Solution :-
     Let A = R – {3} and B = R – {1} and the function  f : A → B defined by f(x) = (x-2)/(x-3).
  ⇒ f(x₁) = f(x₂)
  ⇒ (x₁-2)/(x₁-3) = (x₂-2)/(x₂-3)
  ⇒ (x₁-2)(x₂-3) = (x₁-3)(x₂-2)
  ⇒ x₁x₂ - 3x₁ - 2x₂ + 6 = x₁x₂ - 2x₁ - 3x₂ + 6 
  ⇒ - 3x₁ - 2x₂ = - 2x₁ - 3x₂
  ⇒ - 3x₁ + 2x₁ = - 3x₂ + 2x₂
  ⇒ -x₁ = -x₂
  ⇒ x₁ = x₂
   ∴f is one-one. 
    Now, f(x) = (x-2)/(x-3)
    Let f(x) = y
    Then y = (x-2)/(x-3)
  ⇒ y(x-3) = x - 2
  ⇒ xy - 3y = x - 2
  ⇒ xy - x = 3y - 2
  ⇒ x(y-1) = 3y - 2
  ⇒ x = (3y-2)/(y-1)
     onto
  ⇒ f(x) = y
    ∴f is on-to.
    Hence, f is injective and surjective so, f is bijective function.
    

Question-11 :-  Let f : R → R be defined as f(x) = x⁴. Choose the correct answer.
(A) f is one-one onto,
(B) f is many-one onto,
(C) f is one-one but not onto,
(D) f is neither one-one nor onto.

Solution :-
     Let f : R → R be defined as f(x) = x⁴
  ⇒ f(x₁) = f(x₂)
  ⇒ x₁⁴ = x₂⁴
  ⇒ ± x₁ = ± x₂
   ∴f is not one-one.
    Now, f(x) = x⁴
    Let f(x) = y
    Then y = x⁴
  ⇒ x = ± (y)1/4
    f(x) =  (- y1/4)⁴
  ⇒ f(x) = y
   ∴f is not on-to.
    Hence, f is neither one-one nor on-to.
    Therefore, option (D) is correct. 
    

Question-12 :-  Let f : R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto.

Solution :-
     Let f : R → R be defined as f(x) = 3x
  ⇒ f(x₁) = f(x₂)
  ⇒ 3x₁ = 3x₂
  ⇒ x₁ = x₂
   ∴f is one-one.  
    Now, f(x) = 3x
    Let f(x) = y
    Then y = 3x
    x = y/3
    f(x) = 3 x (y/3) = y
   ∴f is not on-to.
    Hence, f is one-one and on-to.
    Therefore, option (A) is correct. 
    
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