NCERT Solutions Class 11 Maths Chapter 2 Relations and Functions

Given,

As the ordered pairs are equal, the corresponding elements should also be equal.

Thus,

x/3 + 1 = 5/3 and y – 2/3 = 1/3

Solving, we get

x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M. and adding]

x = 2 and 3y = 3

Therefore,

x = 2 and y = 1

 

Given, set A has 3 elements, and the elements of set B are {3, 4, and 5}.

So, the number of elements in set B = 3

Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)

= 3 × 3 = 9

Therefore, the number of elements in (A × B) will be 9.

 

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

We know that,

The Cartesian product of two non-empty sets P and Q is given as

P × Q = {(p, q): p ∈ P, q ∈ Q}

So,

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

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

 

(i) The statement is false. The correct statement is

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

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

(ii) True

(iii) True

 

 

The A × A × A for a non-empty set A is given by

A × A × A = {(a, b, c): a, b, c ∈ A}

Here, it is given A = {–1, 1}

So,

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)}

 

Given,

A × B = {(a, x), (a, y), (b, x), (b, y)}

We know that the Cartesian product of two non-empty sets, P and Q is given by:

P × Q = {(p, q): p ∈ P, q ∈ Q}

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

Therefore, A = {a, b} and B = {x, y}

 

Given,

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

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

Thus,

L.H.S. = A × (B ∩ C) = A × Φ = Φ

Next,

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

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

Thus,

R.H.S. = (A × B) ∩ (A × C) = Φ

Therefore, L.H.S. = R.H.S.

Hence verified

(ii) To verify: A × C is a subset of B × D

First,

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

And,

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)}

Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.

Thus, A × C is a subset of B × D.

Hence verified

 

Given,

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

So,

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

Number of elements in A × B is n(A × B) = 4

We know that,

If C is a set with n(C) = m, then n[P(C)] = 2m.

Thus, the set A × B has 24 = 16 subsets.

And these subsets are as given below:

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)}

 

Given,

n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that,

A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B

So, clearly, x, y, and z are the elements of A; and

1 and 2 are the elements of B.

As n(A) = 3 and n(B) = 2, it is clear that set A = {x, y, z} and set B = {1, 2}

 

We know that,

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

Also, n(A × A) = n(A) × n(A)

Given,

n(A × A) = 9

So, n(A) × n(A) = 9

Thus, n(A) = 3

Also, given that the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

And, we know in A × A = {(a, a): a ∈ A}

Thus, –1, 0, and 1 have to be the elements of A.

As n(A) = 3, clearly A = {–1, 0, 1}

Hence, the remaining elements of set A × A are as follows:

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)

 

Given, A = {x, y, z} and B = {1, 2}

Now,

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

As n(A × B) = 6, the number of subsets of A × B will be 26.

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

 

Given,

Relation R = {(a, b): a, b ∈ Z, a – b is an integer}

We know that the difference between any two integers is always an integer.

Therefore,

Domain of R = Z and Range of R = Z

 

The relation R from A to A is given as:

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

= {(x, y): 3x = y, where x, y ∈ A}

So,

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

Now,

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

Hence, Domain of R = {1, 2, 3, 4}

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

Hence, 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.

Hence, Range of R = {3, 6, 9, 12}

 

The relation R is given by:

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

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

So,

R = {(1, 6), (2, 7), (3, 8)}

Now,

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

Hence, Domain of R = {1, 2, 3}

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

Hence, Range of R = {6, 7, 8}

 

Given,

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

The relation from A to B is given as

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

Thus,

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

 

From the given figure, it’s seen that

P = {5, 6, 7}, Q = {3, 4, 5}

The relation between P and Q:

Set-builder form

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

Roster form

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

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

 

Given,

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

Hence,

(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) Range of R = {1, 2, 3, 4, 6}

 

Given,

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

Thus,

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

So,

Domain of R = {0, 1, 2, 3, 4, 5} and,

Range of R = {5, 6, 7, 8, 9, 10}

 

Given,

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

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

Therefore,

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

 

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

As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called 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)}

As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called 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)}

It’s seen that the same first element, i.e., 1, corresponds to two different images, i.e., 3 and 5; this relation cannot be called a function.

 

(i) Given,

f(x) = –|x|, x ∈ R

We know that,

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

It is also seen that the range of f(x) = –|x| is all real numbers except positive real numbers.

Therefore, the range of f is given by (–∞, 0].

(ii) f(x) = √(9 – x2)

As √(9 – x2) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x2 ≥ 0.

So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3].

Now,

For any value of x in the range [–3, 3], the value of f(x) will lie between 0 and 3.

Therefore, the range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

 

Given,

Function, f(x) = 2x – 5

Therefore,

(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9

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

 

 

 

(i) Given,

f(x) = 2 – 3x, x ∈ R, x > 0

We have,

x > 0

So,

3x > 0

-3x < 0 [Multiplying by -1 on both sides, the inequality sign changes]

2 – 3x < 2

Therefore, the value of 2 – 3x is less than 2.

Hence, Range = (–∞, 2)

(ii) Given,

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

We know that,

x2 ≥ 0

So,

x2 + 2 ≥ 2 [Adding 2 on both sides]

Therefore, the value of x2 + 2 is always greater or equal to 2, for x is a real number.

Hence, Range = [2, ∞)

(iii) Given,

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

Clearly, the range of f is the set of all real numbers.

Thus,

Range of f = R

 

Given the functions 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

Thus, (f + g) (x) = 3x – 2

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

Thus, (f – g) (x) = –x + 4

f/g(x) = f(x)/g(x), g(x) ≠ 0, x ∈ R

f/g(x) = x + 1/ 2x – 3, 2x – 3 ≠ 0

Thus, f/g(x) = x + 1/ 2x – 3, x ≠ 3/2

 

The given relation f is defined as:

It is seen that for 0 ≤ x < 3,

f(x) = x2 and for 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 [Single image]

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

Therefore, the given relation is a function.

Now,

In the given relation, g is defined as

It is seen that, for x = 2

g(x) = 22 = 4 and g(x) = 3 × 2 = 6

Thus, element 2 of the domain of the relation g corresponds to two different images, i.e., 4 and 6.

Therefore, this relation is not a function.

 

Given,

f(x) = x2

Hence,

 

Given function,

.

It’s clearly seen that the function f is defined for all real numbers except at x = 6 and x = 2, as the denominator becomes zero otherwise.

Therefore, the domain of f is R – {2, 6}.

 

Given real function,

f(x) = √(x – 1)

Clearly, √(x – 1) is defined for (x – 1) ≥ 0

So, the function f(x) = √(x – 1) is defined for x ≥ 1

Thus, the domain of f is the set of all real numbers greater than or equal to 1.

Domain of f = [1, ∞)

Now,

As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √(x – 1) ≥ 0

Thus, the range of f is the set of all real numbers greater than or equal to 0.

Range of f = [0, ∞)

 

Given a real function,

f (x) = |x – 1|

Clearly, the function |x – 1| is defined for all real numbers.

Hence,

Domain of f = R

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

Therefore, the range of f is the set of all non-negative real numbers.

 

Given function,

Substituting values and determining the images, we have

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.

[As the denominator is greater than the numerator.]

Or,

We know that, for x ∈ R,

x2 ≥ 0

Then,

x2 + 1 ≥ x2

1 ≥ x2 / (x2 + 1)

Therefore, the range of f = [0, 1)

 

Given, f = {(1, 1), (2, 3), (0, –1), (–1, –3)}

And the function defined as,  f(x) = ax + b

For (1, 1) ∈ f

We have,  f(1) = 1

So, a × 1 + b = 1

a + b = 1 …. (i)

And for (0, –1) ∈ f

We have f(0) = –1

a × 0 + b = –1

b = –1

On substituting b = –1 in (i), we get

a + (–1) = 1 ⇒ a = 1 + 1 = 2.

Therefore, the values of a and b are 2 and –1, respectively.

 

Given relation R = {(a, b): a, b ∈ N and a = b2}

(i) It can be seen that 2 ∈ N; however, 2 ≠ 22 = 4.

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

(ii) Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.

Now, 3 ≠ 92 = 81; therefore, (3, 9) ∉ N

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

(iii) It’s clearly 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

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

 

Given,

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

So,

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)}

Also, 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 × B.

It’s clearly seen that f is a subset of A × B.

Therefore, f is a relation from A to B.

(ii) As the same first element, i.e., 2 corresponds to two different images (9 and 11), relation f is not a function.

 

Given relation, f is defined as

f = {(ab, a + b): a, b ∈ Z}

We know that a relation f 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.

As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f

i.e., (12, 8), (12, –8) ∈ f

It’s clearly seen that the same first element, 12, corresponds to two different images (8 and –8).

Therefore, the relation f is not a function.

 

Prime factor of 13 = 13

Thus, it can be expressed as

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 f is the set of all f(n), where n ∈ A.

Therefore,

Range of f = {3, 5, 11, 13}Given,

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

Now, f: A → N is defined as

f(n) = The highest prime factor of n

So,

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3