Relations and Functions Chapter 2 Solutions – NCERT Class 11 Mathematics

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Exercise 2.1

1. If (x/3 + 1, y – 2/3) = (5/3, 1/3) then find values of x and y.

\begin{equation} \left(\frac{x}{3} + 1, y – \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right) \end{equation}

Comparing both sides of this equation we get

\begin{equation} \frac{x}{3} + 1 = \frac{5}{3} \\ y – \frac{2}{3} = \frac{1}{3} \\ \end{equation}

Simplifying x/3 + 1 = 5/3

\begin{equation} \frac{x}{3} + 1 = \frac{5}{3} \\ \frac{x}{3} = \frac{5}{3} – 1 \\ \frac{x}{3} = \frac{5 – 3}{3} \\ \frac{x}{3} = \frac{2}{3} \\ x = 2 \end{equation}

Simplifying y – 2/3 = 1/3

\begin{equation} y – \frac{2}{3} = \frac{1}{3} \\ y = \frac{1}{3} + \frac{2}{3} \\ y = \frac{1 + 2}{3} = \frac{3}{3} \\ y = 1 \end{equation}

Thus if (x/3 + 1, y – 2/3) = (5/3, 1/3) then value x = 2 and y = 1.


2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B)

We know that n(A × B) = n(A) × n(B)
As per question its given that n(A) = 3 and n(B) = 3
Hence n(A × B) = n(A) × n(B) = 3 × 3
n(A × B) = 3 × 3
n(A × B) = 9

Thus number of elements in A × B is 9 if set A have 3 elements and set B also have 3 elements.


3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G

G × H = {{7, 8} × {5, 4, 2}} = {(7, 5), (7, 4), (7, 2), (8,5), (8,4), (8,2)}
H × G = {{5, 4, 2} × {7, 8}} = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}


4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n),(n, m)}

False
P × Q = {{m, n} × {n, m}} = {(m, n), (m, m), (n, n), (n, m)}
Hence if P = {m, n} and Q = {n, m} then P × Q = {(m, n), (m, m), (n, n), (n, m)}

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B

True

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ

True
Here’s why it’s True
B ∩ φ = {3, 4} ∩ φ = φ
A × (B ∩ φ) = A × φ = φ


5. If A = {–1, 1} then Find A × A × A
A × A = {{-1, 1} × {-1, 1}} = {(-1, -1), (-1, 1), (1, -1), (1, 1)}
A × A × A = {(-1, -1), (-1, 1), (1, -1), (1, 1)} × {-1, 1}
A × A × A = {(-1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}


6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B
A = {a, b}
B = {x, y}


7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C)

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

A × B = {1, 2} × {1, 2, 3, 4} = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A × C = {1, 2} × {5, 6} = {(1, 5), (1, 6), (2, 5), (2, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

(A × B) ∩ (A × C) = ϕ

Hence A × (B ∩ C) = (A × B) ∩ (A × C) = ϕ

(ii) A × C is a subset of B × D
A × C = {1, 2} × {5, 6}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

B × D = {1, 2, 3, 4} × {5, 6, 7, 8}
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

As elements in set A × C are also in set B × D hence A × C is a subset of B × D


8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

A = {1, 2}
B = {3, 4}
A × B = {{1, 2} × {3, 4}} = {(1, 3), (1, 4), (2, 3), (2, 4)}
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Basic formula to find out how many subsets does a set have is 2n where n is number of elements in set.
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
=> Number of elements in A × B = 4
Hence number of subsets of A × B = 24 = 16

Subsets of A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
{(1, 3)}
{(1, 4)}
{(2, 3)}
{(2, 4)}
{(1, 3), (1, 4)}
{(1, 3), (2, 3)}
{(1, 3), (2, 4)}
{(1, 4), (2, 3)}
{(1, 4), (2, 4)}
{(2, 3), (2, 4)}
{(1, 3), (1, 4), (2, 3)}
{(1, 3), (1, 4), (2, 4)}
{(1, 3), (2, 3), (2,4)}
{(1, 4), (2, 3), (2,4)}
{(1, 3), (1, 4), (2, 3), (2, 4)}
ϕ

9. Let A and B be two sets such that n(A) = 3 and n(B) = 2.
If (x, 1), (y, 2), (z, 1) are in A × B, Find A and B, where x, y and z are distinct elements.


Given that n(A) = 3 and n(B) = 2
And (x, 1), (y, 2), (z, 1) ∈ A × B
This means that
x, y, z ∈ A
1, 2 ∈ B

As A × B = {(a, b): a ∈ A, b ∈ B}
This implies
A = {x, y, z}
B = {1, 2}


10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1).
Find the set A and the remaining elements of A × A

Number of elements in set A × A = 9
Elements in cartesian product (-1, 0) and (0, 1)
(-1, 0), (0, 1) ∈ A × A
A = {-1, 0, 1}

A × A = {{-1, 0, 1} × {-1, 0, 1}}
{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)}

Hence remaining elements in cartesian production A × A are (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)


Exercise 2.2

1. Let A = {1, 2, 3,…,14}
Define a relation R from A to A by
R = {(x, y) : 3x – y = 0, where x, y ∈ A}.
Write down its domain, codomain and range.


R = Relation from A to A = {(x, y) : 3x – y = 0 where x, y ∈ A}
On simplification R can be rewritten as
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Domain of R = {1, 2, 3, 4}
Range of R = {3, 6, 9, 12}
Co-Domain of R = {1, 2, 3, ………., 14}


2. Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}
Let’s simplify this to Roster Form
As x is a natural number less than 4
Hence
Possible values of x are
x = 1
x = 2
x = 3

Finding corresponding values of y for these values of x using y = x + 5
If x = 1, then y = 1 + 5 = 6
If x = 2, then y = 2 + 5 = 7
If x = 3, then y = 3 + 5 = 8

Hence pairs of values of x, y are (1, 6), (2, 7) and (3, 8)
Thus
Roster Form of Relation R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N} is {(1, 6), (2, 7), (3, 8)}
Domain is {1, 2, 3}
Range is {6, 7, 8}


3. A = {1, 2, 3, 5} and B = {4, 6, 9}.
Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

It’s given that
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
Possible values of x = 1, 2, 3, 5 and possible values of y = 4, 6, 9

Calculating difference between all possible values of x and y
If x = 1, y = 4 => difference = 3 (Odd)
If x = 2, y = 4 => difference = 2 (Even)
If x = 3, y = 4 => difference = 1 (Odd)
If x = 5, y = 4 => difference = 1 (Odd)
If x = 1, y = 6 => difference = 5 (Odd)
If x = 2, y = 6 => difference = 4 (Even)
If x = 3, y = 6 => difference = 3 (Odd)
If x = 5, y = 6 => difference = 1 (Odd)
If x = 1, y = 9 => difference = 8 (Even)
If x = 2, y = 9 => difference = 7 (Odd)
If x = 3, y = 9 => difference = 6 (Even)
If x = 5, y = 9 => difference = 4 (Even)

As it’s given that relation R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B} and it does contain only odd differences.
Hence relation R in Roster Form can be written as R = {(1, 4), (3, 4), (5, 4), (1, 6), (3, 6), (5, 6), (2, 9)}


4. The Fig 2.7 shows a relationship between the sets P and Q. Write this relation in
(i) Set-builder form
(ii) Roster form
What is its domain and range?

Set-builder Form of relationship between P and Q = {(x, y) : y = x – 2, x ∈ P and y ∈ Q}
Roster Form of relationship between P and Q = {(5, 3), (6, 4), (7, 5)}
Domain of Relation = {5, 6, 7}
Range of Relation = {3, 4, 5}


5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a , b ∈A, b is exactly divisible by a}.
(i) Write R in roster form
R = {(a, b): a , b ∈ A, b is exactly divisible by a}
Possible values of a = 1, 2, 3, 4, 6 and b = 1, 2, 3, 4, 6
But for pair (a, b) to be in relation R
b have to be exactly divisible by a
Hence
R = {(1, 1), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

(ii) Find the domain of R
Domain of R = {1, 2, 3, 4, 6}

(iii) Find the range of R
Range of R = {1, 2, 3, 4, 6}


6. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}
R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}
Putting
x = 0, y = 0 + 5 = 5
x = 1, y = 1 + 5 = 6
x = 2, y = 2 + 5 = 7
x = 3, y = 3 + 5 = 8
x = 4, y = 4 + 5 = 9
x = 5, y = 5 + 5 = 10

Hence Relation R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}} can be rewritten as
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
Domain of R = {0, 1, 2, 3, 5}
Range of R = {5, 6, 7, 8, 9, 10}


7. Write the relation R = {(x, x3) : x is a prime number less than 10} in Roster Form
So in relation R = {(x, x3) : x is a prime number less than 10}
Hence possible values of x = 2, 3, 5, 7
If
x = 2, x3 = 8
x = 3, x3 = 27
x = 5, x3 = 125
x = 7, x3 = 343

Thus relation R = {(x, x3) : x is a prime number less than 10} can be rewritten as {(2, 8), (3, 27), (5, 125), (7, 343)} in Roster Form.


8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B
We know that if
Number of elements in set A = n(A)
Number of elements in set B = n(B)
=> Number of possible relations from A to B = 2n(A) ✘ n(B)

As per question
A = {x, y, z} => n(A) = 3
B = {1, 2} => n(B) = 2
Hence Number of possible relations from A to B = 23 ✘ 2 = 26 = 64
So Number of relations from A to B is 64.


9. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z and a – b is an integer}
Find the domain and range of R


If R = {(a, b) : a, b Z and a – b is an integer}
Domain of R = Z
Range of R = Z


Exercise 2.3

1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

As every first element in the ordered pairs (2,1), (5,1), (8,1), (11,1), (14,1), (17,1) is mapped uniquely to other numbers that’s why relation {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)} is a Function.
Domain = Set of first components of the ordered pairs
Domain = {2, 5, 8, 11, 14, 17}

Range = Set of second components of the ordered pairs
Range = {1}

(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
As every first element in the ordered pairs (2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7) is mapped uniquely to other numbers that’s why relation {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)} is a Function.
Domain = Set of first components of the ordered pairs
Domain = {2, 4, 6, 8, 10, 12, 14}

Range = Set of second components of the ordered pairs
Range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1,3), (1,5), (2,5)}
As element 1 is mapped to both of 3, 5 thus {(1,3), (1,5), (2,5)} is not a Function.
So as its not a function hence it does not have a domain or range.


2. Find the domain and range of the following real functions:
(i) f(x) = – x

Domain of f(x) = – x is whole set of Real Numbers R
Range of f(x) = – x is (- ∞, 0]

(ii) f(x) = Square Root of 9 – x2
Domain of f(x) = Square Root of 9 – x2 is [- 3, 3]
Range of f(x) = Square Root of 9 – x2 is [0, 3]


3. A function f is defined by f(x) = 2x –5
Write down the values of

(i) f(0)
f(0) = 2 × 0 – 5 = 0 – 5 = – 5
f(0) = – 5

(ii) f(7)
f(7) = 2 × 7 – 5 = 14 – 5 = 9
f(7) = 9

(iii) f(–3)
f(- 3) = 2 × (- 3) – 5 = – 6 – 5 = – 11
f(- 3) = – 11


4. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = 9C/5 + 32
Find
(i) t(0)
t(0) = 2 × 0/5 + 32 = 0 + 32 = 32
t(0) = 32

(ii) t(28)
t(28) = 2 × 28/5 + 32 = 56/5 + 32 = 11.2 + 32 = 43.2

(iii) t(–10)
t(-10) = 2 × (-10)/5 + 32 = 2 × – 2 + 32 = – 4 + 32 = 28
t(-10) = 28

(iv) The value of C, when t(C) = 212
As per question
t(C) = 9C/5 + 32
Replacing t(C) = 212
212 = 9C/5 + 32
212 – 32 = 9C/5
18 = 9C/5
18 × 5 = 9C
(18 × 5)/9 = C
2 × 5 = C
C = 10

Thus if t(C) = 212 then value of C = 10


5. Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x > 0

Let f(x) = y = 2 – 3x where x ∈ R and x > 0
y = 2 – 3x
3x = 2 – y
x = (2 – y)/3
As per question its given that x > 0
Then
(2 – y)/3 > 0
2 – y > 0
– y > – 2
Multiplying both sides of above in-equation by -1
-1 × – y > -1 × – 2
y < 2

Thus range of f(x) = 2 – 3x, x ∈ R, x > 0 is y < 2

(ii) f(x) = x2 + 2, x is a real number
Let f(x) = y = x2 + 2 where x is a real number
y = x2 + 2
y – 2 = x2
x = ± Square Root of (y – 2)
Because x is a Real Number
y – 2 ≥ 0
=> y ≥ 2
Thus Range of f(x) = x2 + 2, x is a real number is [2, ∞)

(iii) f(x) = x, x is a real number
Let f(x) = x where x ∈ R
Hence Range of f(x) = x will also be R


Miscellaneous Exercise Solutions

Question 1

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Question 1

Question 2

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Question 2

Question 3

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Question 3

Question 4

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Question 4

5. Find the domain and the range of the real function f defined by f (x) = |x –1|
We know that absolute or modulus of a function is defined for all of Real values of x
Hence Domain of f(x) = |x – 1| is Real Numbers R

As f(x) = |x – 1| can only exist for non-negative values.
Hence
Range of f(x) = |x – 1| = R+ ∪ {0}


Question 6

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Solutions Question 6

Question 7

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Solutions Question 7

8. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b.
Determine a, b

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Solutions Question 8

9. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b2}. Are the following true?
(i) (a,a) ∈ R, for all a ∈ N

We are given that
R : N → N defined as R = {(a, b) : a, b ∈ N and a = b2}

If a = 1 then a2 = 1
If a = 2 then a2 = 4
As R = {(a, b) : a, b ∈ N and a = b2} is not True for all possibles of a, b ∈ N.
Thus R is not a relation.

(ii) (a,b) ∈ R, implies (b,a) ∈ R
We are given that
R : N → N defined as R = {(a, b) : a, b ∈ N and a = b2}
(a,b) ∈ R, implies (b,a) ∈ R
It is alos false as if a = b2
but b = a2 is not True
For example: 4 = 22 i which means (2, 4) ∈ R
But 2 ≠ 42 which means (4, 2) R
Hence R is not a relation.

(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R
We are given that
R : N → N defined as R = {(a, b) : a, b ∈ N and a = b2}
It is false as a = b2 and b = c2
Which means a = (c2)2 = c4
a = c4 which means a ≠ c2
Hence R is not a relation.


10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?
Let’s first find out what’s A ✕ B = ?
A ✕ B = {1, 2, 3, 4} ✕ {1, 5, 9, 11, 15, 16}
A ✕ B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

(i) f is a relation from A to B
f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
As f contains corresponding ordered pair where values from set A are mapped to values from set B
Hence it’s a Relation from A to B.

(ii) f is a function from A to B
f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
As order pairs (2, 9) and (2, 11) have same value 2 from set A
Hence f is not a Function


11. Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Solutions Question 11

12. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime factor of n.
Find the range of f.

NCERT Class 11 Mathematics Chapter 2 – Relations and Functions Miscellaneous Exercise Solutions 12

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