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**NCERT Class 11 Mathematics Chapter 2 Relations and Functions**

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**Solutions****NCERT Class 11 Mathematics Chapter 2 Relations and Functions****Relations and Functions**

**Relations and Functions****Chapter – 2**

Exercise 2.1 |

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

Ans:

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

Ans: It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

⇒ Number of elements in set B = 3

Number of elements in (A x B)

= (Number of elements in A) x (Number of elements in B)

= 3 x 3 = 9

Thus, the number of elements in (A x B) is 9.

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

Ans: G = {7, 8} and H = {5, 4, 2}

We know that the Cartesian product P x Q of two non-empty sets P and Q is defined as

P x Q = {(p, q): P∈ P, q ∈Q}

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

HxG = {(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)}.**

Ans: False

If P = {m, n} and Q = {n, m}, then

P x Q = {(m, m), (m, n), (n, m), (n, n)}.

**(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.**

Ans: True.

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

Ans: True.

**5. If A = {- 1, 1} find A × A × A.**

Ans: It is known that for any non-empty set A, A x A x A is defined as

A x A x A = {(a, b, c): a, b, c ∈ A}

It is given that A = {- 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.**

Ans: It is given that A x B = {(a, x), (a, y), (b, x), (b, y)}

We know that the Cartesian product of two non-empty sets P and Q is defined as P x Q = {(p. q): p ∈ P, q ∈Q}

∴ A is the set of all first elements and B is the set of all second elements.

Thus, A = {a, b} and 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).**

Ans: To verify: A x (B ∩ C) = (A x B) ∩ (A x C)

We have B ∩ C = {1, 2, 3, 4} ∩ {5,6} = ∅

∴ L.H.S.=A x (B ∩ C)= A x ∅ = ∅

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

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

∴ R .H.S.=(A x B) ∩ (A x C) = ∅

∴ L .H.S.= R.H.S

Hence, A x (B ∩ C) = (A x B) ∩ (A x C).

**(ii) A × C is a subset of B × D.**

Ans: To verify: A x C is a subset of B x D

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

B x 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)}

We can observe that all the elements of set A x C are the elements of set B x D.

Therefore, A x C is a subset of B x D.

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

Ans: A = {1, 2} and B = {3, 4}

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

⇒n(A x B) = 4

We know that if C is a set with n(C) = m, then n[P(C)] = 2ᵐ.

Therefore, the set A x B has 2⁴= 16 subsets. These are

Φ {(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)}.

**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.**

Ans: It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A x B.

We know that A = Set of first elements of the ordered pair elements of A x B

B = Set of second elements of the ordered pair elements of A x B.

∴ x, y, and zare the elements of A; and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2 , it is clear that A = {x, y, z) and 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.**

Ans: We know that if n(A) = p and n(B) = q; then n(A × B) = pq.

∴ n(A × A) = n(A) × n(A)

It is given that n(A × A) = 9

∴ n(A) × n(A) = 9

⇒n(A) = 3

The ordered pairs (-1, 0) and (0, 1) are two of the nine elements of A x A.

We know that A ⨯ A= {(a, a): a ∈A). Therefore, -1, 0, and 1 are elements of A.

Since n(A) = 3 it is clear that A = {- 1, 0, 1}

The remaining elements of set A x A are (-1,-1), (-1, 1), (0, -1), (0, 0), (1,-1), (1, 0), and (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.**

Ans: The relation R from A to A is given as

R = {(x, y): 3x – y = 0 where x, y ∈ A}

i.e., R = {(x, y): 3x = y, where x, y ∈ A}

∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

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

∴ Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the relation R.

∴ Codomain of R = A = {1, 2, 3, …, 14}

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

∴ Range of R = {3, 6, 9, 12}.

**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.**

Ans: R= {(x, y): y = x + 5 , xis a natural number less than 4, x, y ∈ N}

The natural numbers less than 4 are 1, 2, and 3.

∴ R = {(1, 6), (2, 7), (3, 8)}

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

∴ Domain of R = {1, 2, 3}

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

∴ Range of R = {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.**

Ans: A = {1, 2, 3, 5} and B = {4, 6, 9}

R = {(x, y): the difference between xand yis odd; x ∈ A, y ∈B}

∴ R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

**4. The Fig2.7 shows a relationship between the sets P and Q. Write this relation.**

**What is its domain and range?**

**(i) in set-builder form.**

Ans: R = {(x, y): y = x- 2; x ∈P} or R={(x, y): y = x – 2 for x = 5, 6, 7}.

**(ii) roster form.**

Ans: R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {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.**

Ans: R = {(1, 1), (1, 2), (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.**

Ans: Domain of R = {1, 2, 3, 4, 6}.

**(iii) Find the range of R.**

Ans: 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}}.**

Ans: R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}

∴R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

∴ Domain of R = {0, 1, 2, 3, 4, 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.**

Ans: R = {(x, x³): x is a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

∴R = {(2, 8), (3, 27), (5, 125), (7, 343)}.

**8. Let A = {x, y, z} and B = {1, 2} . Find the number of relations from A to B.**

Ans: It is given that A = {x, y, z} and B = {1, 2}

∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

Since n(A × B) = 6, the number of subsets of A × B is 2⁶.

Therefore, the number of relations from A to B is 2⁶.

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

Ans: R = {(a, b): a, b ∈ Z, a – bis an integer}

It is known that the difference between any two integers is always 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)}.**

Ans: {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 5, 8, 11, 14, 17) and range = {1}.

**(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}.**

Ans: {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14) and range = {1, 2, 3, 4, 5, 6, 7}.

**(iii) {(1, 3),(1,5),(2,5).**

Ans: {(1, 3), (1, 5), (2, 5)}

Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.

**2. Find the domain and range of the following real functions:**

**(i) f(x) = – |x|**

Ans: f(x) = – |x|

f(- 1) = – |- 1| = – 1

f(o) = – |0| = 0

f(1) = – |1| = – 1

Here, x can be any real number, but f(x) will always be negative or zero.

Therefore,

Domain of function = R (All real numbers)

Range of the function = Negative real numbers

**(ii) f(x) = √9 – x²**

Ans: f(x) = √9 – x²

9 – x² ≥ 0

x² < 9

x < ±3

x ∈ [- 3, 3]

f(- 3) = √9 – (- 3)² = 0 (Real number)

f(0) = √9 – 0 = 3 (Real number)

f(3) = √9 – (3) = 0 (Real number)

Here,

– 3 ≤ x ≤ 3

0 ≤ f(x) ≤ 3

Therefore,

Domain of function = [- 3, 3]

Range of the function = [0, 3]

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

**(i) f (0).**

Ans: f(0) = 2 × 0 – 5 = 0 – 5 = – 5.

**(ii) f (7).**

Ans: f(7) = 2 × 7 – 5 = 14 – 5 = 9.

**(iii) f (-3).**

Ans: f(- 3) = 2 × (- 3) – 5 = – 6 – 5 = – 11.

**4. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) ****Find:**

**(i) t (0).**

Ans:

**(ii) t(28).**

Ans:

**(iii) t(-10).**

Ans:

**(iv) The value of C, when t(C) = 212.**

Ans:

**5. Find the range of each of the following functions.**

**(i) f(x) = 2 – 3x, x ∈ R, x > 0.**

Ans: Given x > 0

⇒ 3x > 0

⇒ – 3x < 0

⇒ 2 – 3x < 2

⇒ f(x) < 2

∴ Range of f = (-∞, 2)

**(ii) f(x) = x² + 2, x is a real number.**

Ans: f(x) = x² + 2 x, is a real number

The values of f(x) for various values of real numbers xcan be written in the tabular form as

x | 0 | ±0.3 | ±0.8 | ±1 | ±2 | ±3 | …. |

f(x) | 2 | 2.09 | 2.64 | 3 | 6 | 11 | ……….. |

Thus, it can be clearly observed that the range of fis the set of all real numbers greater than 2.

i.e., range of f = [2, ∞)

Alter:

Let x be any real number.

Accordingly.

x² ≥ 0

⇒ x² + 2 ≥ 0 + 2

⇒ x² + 2 ≥ 2

⇒ f(x) ≥ 2

∴ Range of f = [2, ∞)

**(iii) f(x) = x, x is a real number.**

Ans: f(x) = x, x is a real number

It is clear that the range of f is the set of all real numbers.

∴ Range of f = R.

Miscellaneous Exercise on Chapter 2 |

Ans: The relation fig defined as

It is observed that for

0 ≤ x < 3, f(x) = x²

3 < x ≤ 10 f(x) = 3x

Also, at x = 3 f(x) = 3² = 9 or f(x) = 3 × 3 = 9

i.e., at x = 3 f(x) = 9

Therefore, for 0 ≤ x ≤ 10 the images of f(x) are unique.

Thus, the given relation is a function.

The relation gis defined as

It can be observed that for x = 2 g(x) = 2² = 4 and g(x) = 3 × 2 = 6

Hence, element 2 of the domain of the relation corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.

Ans:

Ans: It can be seen that function f is defined for all real numbers except at x = 6 and x = 2

Hence the domain of f is R – {2, 6}

**4. Find the domain and the range of the real function f defined by f(x) = √(x – 1) .**

Ans: The given real function is f(x) = √x – 1

It can be seen that √x – 1 is defined for (x – 1) ≥ 0

i.e., f(x) = √(x – 1) is defined for x ≥ 1

Therefore the domain of f is the set of all real numbers greater than or equal to

1 i.e.,

the domain of f = [1, ∞)

As X ≥ 1

⇒ (x – 1) ≥ 0

⇒ √x – 1 ≥ 0

f(x) ≥ 0

Therefore the range of f is the set of all real numbers greater than or equal to o i.e., the range of f = [0, ∞)

**5. Find the domain and the range of the real function f defined by f(x) = |x – 1|.**

Ans: The given real function isf(x) = |x-1|.

It is clear that |x-1| is defined for all real numbers.

∴ Domain of f= R

Also, for x ∈ R, |x-1| assumes all real numbers.

Hence, the range of fis the set of all non-negative real numbers.

Ans:

The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.

[Denominator is greater numerator]

Thus, range of f = [0, 1)

**7. Let f, g : R → R be defined, respectively by f(x) = x + 1 g(x) = 2x – 3 Find f + g, f – g and f/g.**

Ans: f, g : R → R is defined as

f(x) = x + 1

g(x) = 2x – 3

Now, (f + g)(x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2

∴ (f + g)(x) = 3x – 2

Now, (f – g)(x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4

∴ (f – g)(x) = – x + 4

**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.**

Ans: f = {(1, 1), (2, 3), (0, – 1), (- 1, – 3)}

f(x) = ax + b

(1, 1) ∈ f

⇒ f(1) = 1

⇒ a x 1 + b = 1

⇒ a + b = 1

(0, – 1) ∈ f

⇒ f(0) = – 1

⇒ a x 0 + b = – 1

⇒ b = – 1

On substituting b = – 1 in a + b = 1 we obtain a+(-1)=1 ⇒a = 1 + 1 = 2 .

Thus, the respective values of a and bare 2 and -1.

**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? **

Justify your answer in each case.

(i) (a,a) ∈ R, for all a ∈ N.

Ans: It can be seen that 2 ∈ N;however, 2 ≠ 2² = 4.

Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) (a,b) ∈ R, implies (b,a) ∈ R.

Ans: It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3².

Now, 3 ≠ 9² = 81; therefore, (3, 9) ∠”° N

Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R.

Ans: It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4² and 4 = 2².

Now, 16 ≠ 2² = 4; therefore, (16, 2)∠ “° N

Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

**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?**

**(i) f is a relation from A to B Justify your answer in each case.**

Ans: A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A x B.

It is observed that fig a subset of A x B.

Thus, fig a relation from A to B.

**(ii) f is a function from A to B.**

Ans: Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation 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.**

Ans: The relation fis defined as f= {(ab, a + b):a, b ∈ Z}

We know that a relation from a set A to a set B is said to be a function if every element of set A has unique images in set B.

Since 2, 6, -2, – 6 ∈ Z ( 2 × 6, 2 + 6), ( – 2 × -6,-2 +(-6)) ∈ f

i.e., (12, 8), (12, – 8) ∈ f

It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and -8. Thus, relation fis not a function.

**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.**

Ans: A = {9, 10, 11, 12, 13}

f: A →Nis defined as

f(n) = The highest prime factor of n

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3

Prime factor of 13 = 13

∴ f(9) = The highest prime factor of 9 = 3

f(10) = The highest prime factor of 10 = 5

f(11) = The highest prime factor of 11 = 11

f(12) = The highest prime factor of 12 = 3

f(13) = The highest prime factor of 13 = 13

The range of fis the set of all f(n) where n ∈ A.

∴Range of f = {3, 5, 11, 13}.