TOPICS

Unit-2(Examples)

Relations and Functions

**Example-1 :-** If (x + 1, y – 2) = (3,1), find the values of x and y.

In the Ordered Pairs, the corresponding values are equal. So, x + 1 = 3 x = 3 - 1 x = 2 and y - 2 = 1 y = 1 + 2 y = 3

**Example-2 :-** If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P. Are these two products equal?

Given that P = {a, b, c} and Q = {r}. P x Q = {(a, r), (b, r), (c, r)} Q x P = {(r, a), (r, b), (r, c)} By equality rule, (a, r) is not equal to (r, a). So, P x Q ≠ Q x P.

**Example-3 :-** Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find

(i) A × (B ∩ C)

(ii) (A × B) ∩ (A × C)

(iii) A × (B ∪ C)

(iv) (A × B) ∪ (A × C)

Given that A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. (i) A × (B ∩ C) B ∩ C = {3, 4} ∩ {4, 5, 6} = {4} A x (B ∩ C) = {1, 2, 3} x {4} = {(1, 4), (2, 4), (3, 4)}

(ii) (A × B) ∩ (A × C) A x B = {1, 2, 3} x {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} A x C = {1, 2, 3} x {4, 5, 6} = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}

(iii) A × (B ∪ C) B ∪ C = {3, 4} ∪ {4, 5, 6} = {3, 4, 5, 6} A × (B ∪ C) = {1, 2, 3} x {3, 4, 5, 6} = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

(iv)(A × B) ∪ (A × C) A x B = {1, 2, 3} x {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} A x C = {1, 2, 3} x {4, 5, 6} = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} (A × B) ∪ (A × C) = {(1, 3), (2, 3), (3, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

**Example-4 :-** If P = {1, 2}, form the set P × P × P.

p = {1, 2} P x P = {1, 2} x {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} P x P x P = {(1, 1), (1, 2), (2, 1), (2, 2)} x {1, 2} = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

**Example-5 :-** If R is the set of all real numbers, what do the cartesian products R × R and R × R × R represent?

The Cartesian product R × R represents the set R × R = {(x, y) : x, y ∈ R} which represents the coordinates of all the points in two dimensional space and the cartesian product R × R × R represents the set R × R × R ={(x, y, z) : x, y, z ∈ R} which represents the coordinates of all the points in three-dimensional space.

**Example-6 :-** If A × B = {(p, q),(p, r), (m, q), (m, r)}, find A and B.

Given that A × B = {(p, q),(p, r), (m, q), (m, r)}. A = set of first elements = {p, m} B = set of second elements = {q, r}.

**Example-7 :-** Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x, y) : y = x + 1 }

(i) Depict this relation using an arrow diagram.

(ii) Write down the domain, codomain and range of R.

Given that A = {1, 2, 3, 4, 5, 6} and R = {(x, y) : y = x + 1 }. (i) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} Now Arrow Diagram is given below :

(ii) Domain of R = {1, 2, 3, 4, 5} Range of R = {2, 3, 4, 5, 6} Co-domain of R = {1, 2, 3, 4, 5, 6}

**Example-8 :-** The Figure shows a relation between the sets P and Q. Write this relation

(i) in set-builder form,

(ii) in roster form. What is its domain and range?

(i) In the Set-builder form of R = {(x, y): x is the square of y, x ∈ P, y ∈ Q}. (ii) In the Set-builder form of R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)} Domain of R = {9, 4, 25} Range of R = {3, -3, 2, -2, 5, -5} Co-domain of R = {5, 3, 2, 1, -1, -2, -3, -5} or set of Q.

**Example-9 :-** Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.

Given that A = {1, 2} and B = {3, 4}. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. Now n(A × B) = 4, the number of subsets of A × B is 2⁴. Therefore, the number of relations from A into B will be 2⁴.

**Example-10 :-** Let N be the set of natural numbers and the relation R be defined on N such that R = {(x, y) : y = 2x, x, y ∈ N}.

What is the domain, codomain and range of R? Is this relation a function?

The domain of R is the set of natural numbers N. The codomain is also N. The range is the set of even natural numbers. Since every natural number n has one and only one image, this relation is a function.

**Example-11 :-** Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?

(i) R = {(2, 1), (3, 1), (4, 2)},

(ii) R = {(2, 2), (2, 4), (3, 3), (4, 4)}

(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}.

(i) Since 2, 3, 4 are the elements of domain of R having their unique images, this relation R is a function. (ii) Since the same first element 2 corresponds to two different images 2 and 4, this relation is not a function. (iii) Since every element has one and only one image, this relation is a function.

**Example-12 :-** Let N be the set of natural numbers. Define a real valued function f : N → à N by f (x) = 2x + 1. Using this definition, complete the table given below.

**Example-13 :-** Define the function f: R → R by y = f(x) = x², x ∈ R.
Complete the Table given below by using this definition. What is the domain and range of this function? Draw the graph of f.

Domain of f = {x : x ∈ R}. Range of f = {x²: x ∈ R}. The graph of f is

**Example-14 :-** Draw the graph of the function f :R → R defined by f (x) = x³, x ∈ R.

We have f(0) = 0, f(1) = 1, f(–1) = –1, f(2) = 8, f(–2) = –8, f(3) = 27; f(–3) = –27, etc. Therefore, f = {(x, x³): x ∈ R}. The graph of f is

**Example-15 :-** Define the real valued function f : R – {0} → R defined by f(x) = 1/x,
x ∈ R –{0}. Complete the Table given below using this definition. What is the domain and range of this function?

The domain is all real numbers except 0 and its range is also all real numbers except 0. The graph of f is

**Example-16 :-** Let f(x) = x² and g(x) = 2x + 1 be two real functions.Find (f + g)(x), (f – g)(x), (fg)(x), (f/g)(x).

(f + g)(x) = f(x) + g(x) = x² + 2x + 1 (f - g)(x) = f(x) - g(x) = x² - 2x + 1 (f.g)(x) = f(x) . g(x) = (x²)(2x + 1) = 2x³ + x² (f/g)(x) = f(x) ÷ g(x) = x²/(2x + 1)

**Example-17 :-** Let f(x) = √x and g(x) = x be two functions defined over the set of non negative real numbers. Find Find (f + g)(x), (f – g)(x), (fg)(x), (f/g)(x).

(f + g)(x) = f(x) + g(x) = √x + x (f - g)(x) = f(x) - g(x) = √x - x (f.g)(x) = f(x) . g(x) = √x x x = x√x = x^{3/2}(f/g)(x) = f(x) ÷ g(x) = √x/x = x^{-1/2}

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