Relations and Functions Class 11 Notes AP Inter 1st Year Maths Chapter 2

Students can go through AP Inter 1st Year Maths Notes 2nd Lesson Relations and Functions will help students in revising the entire concepts quickly.

Relations and Functions Class 11 Notes AP Inter 1st Year Maths 2nd Lesson

→ Ordered pair: A pair of elements grouped in a particular order.

→ Cartesian product: A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B}

→ In particular R × R = {(x, y): x, y ∈ R} and R × R × R = {(x, y, z): x, y, z ∈ R}

→ If (a, b) = (x, y), then a = x and b = y.

→ If n(A) = p and n(B) = q, then n(A × B) = pq.

→ A × φ = φ

→ In general, A × B ≠ B × A.

→ Relation: A relation R from a set A to a set B is a subset of the Cartesian product A × B obtained by describing a relationship between the first element x and the second element y of the ordered pairs in A × B.

→ The image of an element x under a relation R is given by y, where (x, y) ∈ R.

→ The domain of R is the set of all first elements of the ordered pairs in a relation R.

→ The range of the relation R is the set of all second elements of the ordered pairs in a relation R.

→ Function: A function f from a set A to a set B is a specific type of relation for which every element x of set A has one and only one image y in set B.
We write f: A → B, and f(x) = y. A is the domain, and B is the codomain of f.

→ The range of the function is the set of images.

→ A real function has the set of real numbers or one of its subsets both as its domain and as its range.

→ Algebra of functions: For functions, f: X → R and g: X → R, we have
(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
(kf)(x) = k(f(x)), x ∈ X, where k is a real number.
\(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}\), x ∈ X, g(x) ≠ 0

Example Problems

Question 1.
If (x + 1, y – 2) = (3, 1), find the values of x and y.
Solution:
Since the ordered pairs are equal, the corresponding elements are equal.
Therefore, x + 1 = 3 and y – 2 = 1.
Solving, we get x = 2 and y = 3.

Question 2.
If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P. Are these two products equal?
Solution:
By the definition of the Cartesian product,
P × Q = {(a, r), (b, r), (c, r)} and Q × P = {(r, a), (r, b), (r, c)}
Since, by the definition of equality of ordered pairs, the pair (a, r) is not equal to the pair (r, a),
We conclude that P × Q ≠ Q × P.
However, the number of elements in each set will be the same.

Question 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)
Solution:
(i) By the definition of the intersection of two sets,
(B ∩ C) = {4}.
Therefore, A × (B ∩ C) = {(1, 4), (2, 4), (3, 4)}
(ii) Now (A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
and (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Therefore, (A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}
(iii) Since, (B ∪ C) = {3, 4, 5, 6},
we have A × (B ∪ C) = {(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) Using the sets A × B and A × C from part (ii) above, we obtain
(A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

Question 4.
If P = {1, 2}, form the set P × P × P.
Solution:
We have, P × P × P = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

Question 5.
If R is the set of all real numbers, what do the cartesian products R × R and R × R × R represent?
Solution:
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.

Question 6.
If A × B = {(p, q), (p, r), (m, q), (m, r)}, find A and B.
Solution:
A = set of first elements = {p, m}
B = set of second elements = {q, r}

Question 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.
Solution:
(i) By the definition of the relation,
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
The corresponding arrow diagram is shown in Figure.
(ii) We can see that the domain = {1, 2, 3, 4, 5}
Similarly, the range = {2, 3, 4, 5, 6} and the codomain = {1, 2, 3, 4, 5, 6}.

Question 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?
Solution:
It is obvious that the relation R is “x is the square of y”.
(i) In set-builder form,
R = {(x, y): x is the square of y, x ∈ P, y ∈ Q}
(ii) In roster form,
R = {(9, 3),(9, -3), (4, 2), (4, -2), (25, 5), (25, -5)}
The domain of this relation is {4, 9, 25}.
The range of this relation is {-2, 2, -3, 3, -5, 5}.
Note that element 1 is not related to any element in set P.
The set Q is the codomain of this relation.

Question 9.
Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.
Solution:
We have A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Since 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.

Question 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?
Solution:
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.

Question 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)}
Solution:
(i) Since 2, 3, 4 are the Elements of the 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.

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

Solution:
The completed table is given by

Question 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.

Solution:
The completed Table is given below:

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

Question 14.
Draw the graph of the function f: R → R defined by f(x) = x³, x ∈ R.
Solution:
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 given in the Figure.

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

Solution:
The completed table is given by

The domain is all real numbers except 0, and its range is also all real numbers except 0.
The graph of f is given in the  Figure.

Question 16.
Let f(x) = x² and g(x) = 2x + 1 be two real functions. Find (f + g)(x), (f – g)(x), (fg)(x) and (\(\frac {f}{g}\))(x).
Solution:
We have, (f + g)(x) = x² + 2x + 1
(f-g)(x) = x² – 2x – 1
(fg)(x) = x²(2x + 1) = 2x³ + x²
(\(\frac {f}{g}\))(x) = \(\frac{x^2}{2 x+1}\), x ≠ \(-\frac {1}{2}\)

Question 17.
Let f(x) = √x and g(x) = x be two functions defined over the set of non-negative real numbers.
Find (f + g) (x), (f – g) (x), (fg) (x) and (\(\frac {f}{g}\))(x).
Solution:
We have, (f + g) (x) = √x + x
(f-g) (x) = √x – x
(fg)(x) = √x . (x) = x^3/2
\(\left(\frac{f}{g}\right)(x)=\frac{\sqrt{x}}{x}=x^{-1 / 2}\), x ≠ 0

Question 18.
Let R be the set of real numbers. Define the real function f: R → R by f(x) = x + 10 and sketch the graph of this function.
Solution:
Here f(0) = 10, f(1) = 11, f(2) = 12, …, f(10) = 20, etc.,
and f(-1) = 9, f(-2) = 8,…, f(-10) = 0 and so on.
Therefore, the shape of the graph of the given function assumes the form shown in the Figure.

Question 19.
Let R be a relation from Q to Q defined by R = {(a, b): a, b ∈ Q and a – b ∈ Z}. Show that
(i) (a, a) ∈ R for all a ∈ Q
(ii) (a, b) ∈ R implies that (b, a) ∈ R
(iii) (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R
Solution:
(i) Since, a – a = 0 ∈ Z, if follows that (a, a) ∈ R.
(ii) (a, b) ∈ R implies that a – b ∈ Z.
So, b – a ∈ Z.
Therefore, (b, a) ∈ R
(iii) (a, b) and (b, c) ∈ R implies that a-b ∈ Z, b-c ∈ Z.
So, a – c = (a – b) + (b – c) ∈ Z.
Therefore, (a, c) ∈ R

Question 20.
Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a linear ftmction from Z into Z. Find f(x).
Solution:
Since f is a linear function, f(x) = mx + c.
Also, since (1, 1), (0, -1) ∈ R,
f(1) = m + c = 1 and f(0) = c = -1.
This gives m = 2 and f(x) = 2x – 1.

Question 21.
Find the domain of the function f(x) = \(\frac{x^2+3 x+5}{x^2-5 x+4}\)
Solution:
Since x² – 5x + 4 = (x – 4) (x – 1),
the function f is defined for all real numbers except at x = 4 and x = 1.
Hence, the domain of f is R – {1, 4}.

Question 22.
The function f is defined by f(x) = \(\begin{cases}1-x, & x<0 \\ 1, & x=0 \\ x+1, & x>0\end{cases}\)
Draw the graph of f(x).
Solution:
Here, f(x) = 1 – x, x < 0, this gives
f(-4) = 1 – (-4) = 5;
f(-3) = 1 – (-3) = 4;
f(-2) = 1 – (-2) = 3;
f(-1) = 1 – (-1) = 2; etc,
and f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5 and so on for f(x) = x + 1, x > 0.
Thus, the graph of f is as shown in the Figure.

Multiple Choice Questions

Question 1.
If (2x + 1, \(\frac {y}{2}\)) = (3, 3) then the values of x & y are
(1) 1, 6
(2) 2, 2
(3) 3, 3
(4) 7, \(\frac {3}{2}\)
Answer:
(1) 1, 6

Question 2.
If A = {1, 2, 3}, B = {a, b} then A × B = ____________
(1) {(1, a) (2, b) (3, a)}
(2) {(a, 1) (b, 2) (a, 3) (b, 3)}
(3) {(1, a) (1, b) (2, a) (2, b) (3, a) (3, b)}
(4) {(1, a) (1, b) (2, b) (3, b)}
Answer:
(3) {(1, a) (1, b) (2, a) (2, b) (3, a) (3, b)}

Question 3.
A relation can be represented as
(1) Roster method
(2) Set-builder method
(3) An arrow diagram
(4) In the above three forms
Answer:
(4) In the above three forms

Question 4.
If A = {1, 2, 3, 4} and a relation R from A to A is defined by R = {(x, y): y = 2x}, then the co-domain of R is
(1) {2, 4}
(2) {1, 2, 3, 4}
(3) {1, 3}
(4){1, 4}
Answer:
(2) {1, 2, 3, 4}

Question 5.
If f(x) = \(\frac{|\mathbf{x}|}{\mathbf{x}}\), then f(0) = ____________
(1) 1
(2) -1
(3) undefined
(4) 0
Answer:
(3) undefined

Question 6.
The domain of the function \(\frac{1}{\sqrt{x^2-25}}\) is
(1) (-∞, -5) ∪ (5, ∞)
(2) (-∞, -5] ∪ [5, ∞)
(3) (-∞, -5] ∪ (5, ∞)
(4) (-∞, -5) ∪ [5, ∞)
Answer:
(1) (-∞, -5) ∪ (5, ∞)

Question 7.
Range of the function f(x) = x², x ∈ R is
(1) [0, ∞)
(2) (-∞, ∞)
(3) (-∞, 0)
(4) (0, ∞)
Answer:
(1) [0, ∞)

Question 8.
A function f(x) is defined by f(x) = x² + 2x – 7, then the value of f(3) = ____________
(1) 2
(2) -7
(3) 8
(4) 9
Answer:
(3) 8

Question 9.
Let f = {(0, 1), (1, 3), (2, 5)} be a linear function from {0, 1, 2} to N then f(x) = ____________
(1) 2x – 1
(2) 2x + 1
(3) x² – 1
(4) x² + 1
Answer:
(2) 2x + 1

Question 10.
The domain of \(\sqrt{|\mathbf{x}|-\mathbf{x}}\) is
(1) R
(2) [0, ∞)
(3) (-∞, 0]
(4) Z
Answer:
(1) R