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Miscellaneous

Question-1 :-  Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = IR.

Solution :-
    We have given that : f : R → R be defined as f(x) = 10x + 7
    Now, gof = g(f(x)) and fog = f(g(x)) = 10g(x) + 7
       ⇒ 10g(x) + 7 = IR(X) = x
       ⇒ g(x) = (x-7)/10  
    

Question-2 :-  Let f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Solution :-
    f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1. i.e.,
    answer
    For Injective : 
    Let n and m be any two odd real numberrs, then f(n) = f(m)
     so, n - 1 = m - 1 ⇒ n = m
    Let n and m be any two odd real numberrs, then f(n) = f(m)
     so, n + 1 = m + 1 ⇒ n = m
    and now n is even and m is odd, then n ≠ m.
    Also, if f(n) odd and f(m) is even, then f(n) ≠ f(m)
    Hence, n ≠ m ⇒ f(n) ≠ f(m)
    ∴ f is an injective mapping.

    For Surjective :
    Let n be an arbitrary whole number.
    If n is an odd no., then there exists an even whole no. (n + 1) such that
    f(n + 1) = n + 1 - 1 = n.
    If n is an even no., then there exists an even whole no. (n - 1) such that
    f(n - 1) = n - 1 + 1 = n.
    Therefore, every n ∈ W has its pre-image in W.
    So, f : W → W is a surjective. 
    Thus f is invertible and f-1 exists.

    For f-1 :
    y = n - 1 ⇒ n = y + 1 and y = n + 1 ⇒ n = y - 1
    Hence, f-1(y) = y.  
    

Question-3 :-  If f : R → R is defined by f(x) = x² – 3x + 2, find f (f(x)).

Solution :-
    Given that f : R → R is defined by f(x) = x² – 3x + 2.
  ⇒ f(f(x)) = f(x² – 3x + 2)
  ⇒ (x² – 3x + 2)² - 3(x² – 3x + 2) + 2
  ⇒ (x² – 3x + 2)² - 3x² + 9x - 6 + 2
  ⇒ x⁴ - 6x³ + 10x² - 3x
    

Question-4 :-  Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = x/(1+|x|) x ∈ R is one one and onto function.

Solution :-
    Given that f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = x/(1+|x|) x ∈ R
    For f is one-one :
    For any x, y ∈ R – {+1}, we have f(x) = f(y)
  ⇒ x/(1+|x|)= y/(1+|y|)
  ⇒ xy + x = xy + y
  ⇒ x = y
    Hence, f is one-one.

    For on-to :
    If f is one-one, let y = R - {1}, then f(x) = y
  ⇒ x/(x+1) = y
  ⇒ x = xy + y
  ⇒ x(1-y) = y
  ⇒ x = y/(1-y) 
    Now, f(x) = f(y/(1-y))
    
    Therefore, f is on-to function.
    

Question-5 :-  Show that the function f : R → R given by f(x) = x³ is injective.

Solution :-
    We have f : R → R given by f(x) = x³
    For Injective : 
  ⇒ Let f(x₁) = f(x₂)
  ⇒ x₁³ = x₂³
  ⇒ x₁ = x₂
    Hence, f is injective function.
    

Question-6 :-  Give Questions of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective. (Hint : Consider f(x) = x and g(x) = |x|).

Solution :-
    Given that f : N → Z and g : Z → Z
    Let f(x) = x and g(x) = x
    Here, gof(x) = f(f(x)) = g(x)
    Therefore, gof is an injective function but g is not injective.
    

Question-7 :-  Give Questions of two functions f : N → N and g : N → N such that g o f is onto but f is not onto. (Hint : Consider f(x) = x + 1 and question

Solution :-
    Let f(x) = x + 1
        g(x) = x - 1 if x > 1 and 1 if x = 1  
    These are two Questions in which gof is on-to but f is on-to.
    

Question-8 :-  Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.

Solution :-
(i)  A ⊂ A                      Therefore R is Reflexive.
(ii) A ⊂ B ≠ B ⊂ A              Therefore R is not Commutative.
(ii) A ⊂ B, B ⊂ C, then A ⊂ C   Therefore R is transitive.
    Therefore, R is not equivalent relation.
    

Question-9 :-  Given a non-empty set X, consider the binary operation ∗ : P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

Solution :-
    Let S be a non-empty set and P(S) be its power set.
    Let any two subsets A and B of S.  
 ⇒ A ∪ B ⊂ S 
 ⇒ A ∪ B ∈ P(S)
    Therefore, '∪' is an binary operation on P(S).
    Similarly, if A, B ∈ P(S) and A - B ∈ P(S), then the intersection of sets ∩ and difference of sets 
    are also binary operation on P(S) and A ∩ S = A = S ∩ A for every subset A of sets 
 ⇒ A ∩ S = A = S ∩ A for all A ∈ P(S).
 ⇒ S is the identity element for intersection (∩) on P(S).
    

Question-10 :-  Find the number of all onto functions from the set {1, 2, 3, ... , n} to itself.

Solution :-
    The number of onto functions that can be defined from a finite set A containing n elements 
    onto a finite set B containing n elements = 2n - n.
    

Question-11 :-  Let S = {a, b, c} and T = {1, 2, 3}. Find F-1 of the following functions F from S to T, if it exists.
(i) F = {(a, 3), (b, 2), (c, 1)}
(ii) F = {(a, 2), (b, 1), (c, 1)}

Solution :-
    Let S = {a, b, c} and T = {1, 2, 3}.
(i)  F = {(a, 3), (b, 2), (c, 1)} 
  ⇒ F(a) = 3, F(b) = 2, F(c) = 1
  ⇒ F-1(3) = a, F-1(2) = b, F-1(1) = c
  ⇒ F-1 = {(3, a), (2, b), (1, c)} 
        
(ii) F = {(a, 2), (b, 1), (c, 1)}
     F is not one-one function, since element b and c have the same image 1. 
     Therefore F is not one-one function. 
    

Question-12 :-  Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗ b = |a – b| and a o b = a, ∀a, b ∈ R. Show that ∗ is commutative but not associative, o is associative but not commutative. Further, show that ∀a, b, c ∈ R, a ∗ (b o c) = (a ∗ b) o (a ∗ c). [If it is so, we say that the operation ∗ distributes over the operation o]. Does o distribute over ∗? Justify your answer.

Solution :-
(i)  For Commutative :
     a * b = |a-b| also b * a = |b-a| = (a - b)
     Hence, operation * is commutative.

     For Associative :
     Now, a * (b * c) = a * |b-c| = |a-(b-c)| = |a-b+c|
     And (a * b) * c  = |a-b| * c = |a-b-c|
     Here, a * (b * c) ≠  a * (b * c)
     Operation is * not associative. 
(ii) For Coomutative :
     aob = a∀a, b ∈ R
     boa = b ⇒ aob ≠ boa
     Here, operation o is not commutative.

     For Associative :
     Now, boa = aob = a and (aob)oc = aoc = a
     Here, ao(boc) = (aob)oc
     Here, operation o is associative.
(iii) L.H.S a * (boc) = a * b = |a-b|
     R.H.S (a * b)o(a * b) = (a - b)o(a - c) = |a-b| = L.H.S (Hence Proved)
     Now, another distribution law :
     ao(b * c) = (aob)*(aob)
     L.H.S ao(b * c) = ao(|b-c|) = a
     R.H.S (aob) * (aob) = a * a = |a-a| = 0
     As L.H.S ≠ R.H.S
     Therefore, the operation o is does not distribute over. 
    

Question-13 :-  Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

Solution :-
    For every A ∈ P(X), we have
    φ * A = (φ - A) ∪ (A - φ) = φ ∪ A = A and
    A * φ = (A - φ) ∪ (φ - A) = A ∪ φ = A
    φ is the identity element for the operation * on P(X).
    Also, A * A = (A - A) ∪ (A - A) = φ ∪ φ = ∪
    Every Element A of P(X) is invertible with A-1 = A.  
    

Question-14 :-  Define a binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as binary Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.

Solution :-
    A binary operation (or composition) * on a (non-empty) set is a function * : A x A → A.
    We denote *(a, b) by a * b for every ordered pair (a, b) ∈ A x A.
    binary operation on a no-empty set A is a rule that associates with every ordered pair of elements a, b
    (distinct or equal) of A some unique element a * b of A.
    binary operation
    For all a ∉ A, we have 0 * a(mod 6) = 0
    And a * 1 = a(mod 6) = a and a * 1 = a(mod 6)
    0 is the identity element for the operation.
    Also on 0 = 0 - 0 = 0*
           2 * 1 = 3 = 1 * 2
           0-1 = 0
           0-1 = 5
    

Question-15 :-  Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined by f(x) = x² – x, x ∈ A and g(x) = 2|x - 1/2|-1 x ∈ A. Are f and g equal? Justify your answer. (Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g(a) ∀a ∈ A, are called equal functions).

Solution :-
    When x = -1 then f(x) = 1² + 1 = 2 and g(x) = 2|-1 - 1/2|-1 = 2
    At x = 0,  f(0) = 0 and g(0) = 2|-1/2|-1 = 0
    At x = 1,  f(1) = 1² - 1 = 0 and g(1) = 2|1 - 1/2|-1 = 0
    At x = 2,  f(2) = 2² - 2 = 2 and g(2) = 2|2 - 1/2|-1 = 2
    Thus for each a ∈ A, f(a) = g(a)
    Therefore, f and g are equal function.
    

Question-16 :-  Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1     (B) 2    (C) 3    (D) 4

Solution :-
     Its clear that 1 is reflexive and symmetric but not transitive.
     Therefore, option (A) is correct.
    

Question-17 :-  Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1     (B) 2    (C) 3    (D) 4

Solution :-
    Its clear that 2 is an equivalence relation.
    Therefore, option (B) is correct. 
    

Question-18 :-  Let f : R → R be the Signum Function defined as function and g : R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?

Solution :-
    Given that f : R → R. Now, gof : R → R and fog : R → R
    Consider x = 1/2 which lie on (0, 1)
    Now, gof(1/2) = g(f(1/2)) = g(1) = [1] = 1
    And fog(1/2) = f(g(1/2)) = f(1/2) = f(0) = 0
    gof fog in (0, 1].
    Therefore, option (B) is correct.
    

Question-19 :-  Number of binary operations on the set {a, b} are
(A) 10     (B) 16    (C) 20    (D) 8

Solution :-
    A = {a, b}
    A x A = {(a, a), (a, b), (b, b), (b, a)}
    n(A x A) = 4
    Number of subsets = 24 = 16
    Hence number of binary operation is 16.
    Therefore, option (B) is Correct.  
    
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