NCERT Solutions for Class 11 Maths Chapter 2

Relations and Functions Class 11

Chapter 2 Relations and Functions Exercise 2.1, 2.2, 2.3, miscellaneous Solutions

 Exercise 2.1 : Solutions of Questions on Page Number : 33                                                                                                                 

Q1 :

If  , find the values of xand y.

Answer :

It is given that .

Since the ordered pairs are equal, the corresponding elements will also be equal. Therefore, and .

∴ x= 2 and y= 1

Q2 :

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

Answer :

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.

Q3 :

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

Answer :

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

Q4 :

State whether each of the following statement 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 x Q = {(m, n), (n, m)}.

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

(iii)  If A = {1, 2}, B = {3, 4}, then A x (B ∩ Φ) = Φ.

Answer :

(i)  False

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

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

(ii)  True

(iii) 

Q5 :

If A = {-1, 1}, find A x A x A.

Answer :

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 x A x A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1),

Q6 :

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

Answer :

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}

Q7 :

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

(i)  A x (B ∩C) = (A x B) ∩(A x C)

(ii)  A x C is a subset of B x D

Answer :

(i)  To verify: A x (B ∩C) = (A x B) ∩(A x C) We have B ∩C = {1, 2, 3, 4} ∩{5, 6} = Φ

∴L.H.S. = A x (B ∩C) = A x Φ = Φ

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) = Φ

∴L.H.S. = R.H.S

Hence, A x (B ∩C) = (A x B) ∩(A x C)

(ii)  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.

Q8 :

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

Answer :

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)},

Q9 :

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 x B, find A and B, wherex, y and z are distinct elements.

Answer :

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}.

Q10 :

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

Answer :

We know that if n(A) = p and n(B) = q, then n(A x B) = pq.

∴ n(A x A) = n(A) x n(A)

It is given that n(A x A) = 9

∴ n(A) x 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 x 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),

 Exercise 2.2 : Solutions of Questions on Page Number : 35                                                                                                                 

Q1 :

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.

Answer :

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 codomainof 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.

Q2 :

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

Answer :

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.

Q3 :

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 xand yis odd; x ∈A, y ∈B}. Write R in roster form.

Answer :

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

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

Q4 :

The given figure shows a relationship between the sets P and Q. write this relation

(i) in set-builder form (ii) in roster form. What is its domain and range?

Answer :

According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}

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

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

Domain of R = {5, 6, 7}

Q5 :

Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by

{(a, b): a, b ∈A, bis exactly divisible by a}.

(i)  Write R in roster form

(ii)  Find the domain of R

(iii)  Find the range of R.

Answer :

A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈A, bis exactly divisible by a}

(i) 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)  Domain of R = {1, 2, 3, 4, 6}

(iii) 

Q6 :

Determine the domain and range of the relation R defined by R = {(x, x+ 5): x ∈{0, 1, 2, 3, 4, 5}}.

Answer :

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}

Q7 :

Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.

Answer :

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

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

Q8 :

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

Answer :

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

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

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

Q9 :

Let R be the relation on Zdefined by R = {(a, b): a, b ∈ Z, a - bis an integer}. Find the domain and range of R.

Answer :

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

 Exercise 2.3 : Solutions of Questions on Page Number : 44                                                                                                                 

Q1 :

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)}

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

(iii) {(1, 3), (1, 5), (2, 5)}

Answer :

(i) {(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)}

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)}

Q2 :

Find the domain and range of the following real function:

(i) f(x) = –|x| (ii)   Answer :

(i) f(x) = –|x|, x ∈R

| =

Since f(x) is defined for x ∈ R, the domain of fis R.

It can be observed that the range of f(x) = –|x| is all real numbers except positive real numbers.

∴The range of fis (– , 0].

(ii)

Since  is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].

For any value of xsuch that –3 ≤ x≤ 3, the value of f(x) will lie between 0 and 3.

Q3 :

A function fis defined by f(x) = 2x- 5. Write down the values of (i) f(0), (ii) f(7), (iii) f(-3)

Answer :

The given function is f(x) = 2x- 5. Therefore,

(i) f(0) = 2 x 0 - 5 = 0 - 5 = -5

(ii) f(7) = 2 x 7 - 5 = 14 - 5 = 9

Q4 :

The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by .

Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212

Answer :

The given function is . Therefore,

(i)

(ii)

(iii)

(iv) 

Thus, the value of t,when t(C) = 212, is 100.

Q5 :

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

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

(iii)  f(x) = x, xis a real number

Answer :

(i) f(x) = 2 – 3x, x ∈ R, x> 0

The values of f(x) for various values of real numbers x> 0 can be written in the tabular form as

Thus, it can be clearly observed that the range of fis the set of all real numbers less than 2. i.e., range of f= (– , 2)

Alter:

Let x > 0

⇒3x > 0

⇒ 2 –3x< 2

⇒ f(x) < 2

∴Range of f = (– , 2)

(ii)  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

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

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

 Exercise Miscellaneous : Solutions of Questions on Page Number : 46                                                                                              

Q1 :

The relation f is defined by

The relation gis defined by

Show that f is a function and g is not a function.

Answer :

The relation fis 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

Q2 :

If f(x) = x2, find .

Answer :

Q3 :

Find the domain of the function   Answer :

The given function is .

It can be seen that function fis defined for all real numbers except at x= 6 and x= 2. Hence, the domain of fis R – {2, 6}.

Q4 :

Find the domain and the range of the real function fdefined by . Answer :

The given real function is .

It can be seen that is defined for (x– 1) ≥0. i.e., is defined for x ≥1.

Therefore, the domain of fis 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 ⇒

= [0, ).

Q5 :

Find the domain and the range of the real function fdefined by f(x) = |x- 1|.

Answer :

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.

Q6 :

The range of fis 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)

Q7 :

Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f+ g, f– gand  . Answer :

f, g: R → R is defined as f(x) = x + 1, g(x) = 2x – 3 (f+ g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2

∴(f + g) (x) = 3x – 2

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

∴ (f – g) (x) = –x+ 4

Q8 :

Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a function from Zto Zdefined by f(x) = ax+ b, for some integers a, b. Determine a, b.

Answer :

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 aand bare 2 and -1.

Q9 :

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

(ii)  (a, b) ∈ R, implies (b, a) ∈ R

(iii)  (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R. Justify your answer in each case.

Answer :

R = {(a, b): a, b ∈ N and a = b2}

(i)  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)  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) 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

Q10 :

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) fis a relation from A to B (ii) fis a function from A to B. Justify your answer in each case.

Answer :

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

∴A x 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)}

It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i)  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 fis a subset of A x B.

Thus, fis a relation from A to B.

(ii) 

Q11 :

Let fbe the subset of Z x Zdefined by f = {(ab, a+ b): a, b ∈ Z}. Is fa function from Zto Z: justify your answer.

Answer :

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

We know that a relation ffrom 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 x 6, 2 + 6), (-2 x -6, -2 + (-6)) ∈ f

i.e., (12, 8), (12, -8) ∈ f

Q12 :

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

Answer :

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}