## Rajasthan Board RBSE Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise

Question 1.

If A = {a, b, c, d} and B = {p, q, r, s} then relation from A to B is

(A) {(a, p), (b, s, (c, r)}

(B) {(a, p), (b, q), (c, r), (s, d)}

(C) {{b, a), (q, b), (c, r)}

(D) {(c, s), (d, s), (r, a), (q, b)}

Solution:

A × B = {(a, p), (a, q), (a, r), (a, s), {b, p), (b, q), {b, r), (b, s), (c, p), (c, q), (c, r) (c, s) (d, p), (d, q), (d, r), (d, s)}

{(a, p), (b, r), (c, r)}

Because {(a, p), (b, r), (c, r)} ⊆ A × B.

Hence, option (A) is correct.

Question 2.

A relation R in N is defined such that xRy ⇔ x + 4y = 16, then the range of R is

(A) {1, 2, 4}

(B) {1, 3, 4}

(C) {1, 2, 3}

(D) {2, 3, 4}

Solution:

Given,set N = set of natural number = {1, 2, 3,…}

A relation R in Nto N is defined as

xRy ⇔ x + 4y = 16 ∀ x, y ∈ N

or xRy ⇔ y = \(\frac { 16-x }{ 2 }\) ∀ x, y ∈ N

when x = 4 then

So, R = {(4, 3), (8, 2), (12, 1)}

So, range of R= {3, 2, 1} and {1, 2, 3}

Hence, option (C) is correct.

Question 3.

Rule form of relation {(1, 2), (2, 5), (3, 10), (4, 17), …} in N

(A) {(x, y) : x, y ∈ N, y = 2x + 1}

(B) {(x, y) : x, y ∈ N, y = x^{2} + 1}

(C) {(x, y) : x, y ∈ N, y = 3x – 1}

(D) {(x, y) : x, y ∈ N, y = x + 3}

Solution:

{(x, y) : x, y ∈ N, y = x^{2} + 1}

when x = 1, then y = (1)^{2} + 1 = 2 ⇒ (1, 2) ∈ N

when x = 2, then y = (2)^{2} + 1 = 4 + 1 = 5 ⇒ (2, 5) ∈ N

when x = 3, then y = (3)^{2} + 1 = 9 + 1 = 10 ⇒ (3, 10) ∈ N

when x = 4, then y = (4)^{2} + 1 = 16 + 1 = 17 ⇒ (4, 17) ∈ N

So, R = {(1, 2), (2, 5), (3, 10), (4, 17), …}

= {(x, y) : x, y ∈ N, y = x^{2} + 1}

Hence, option (B) is correct.

Question 4.

If A = {2, 3, 4} and B = {3, 4, 5, 6, 7, 8} : A relation if from A to B is defined such that “x divides y” then R^{-1} is

(A) {(4, 2), (6, 2), (8, 2), (3, 3), (6,3), (4, 4), (8, 4)}

(B) {(2, 4), (2, 6), (2, 8), (3, 3), (3, 6), (4, 4), (4, 8)}

(C) {(3, 3), (4, 4), (8, 4)}

(D) {(4, 2), (6, 3), (8, 4)}

Solution:

Given sets are

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

Relation R from A to B is defined as

“x divides y” ∀ x ∈ A, y ∈ B

So, when x = 2 ∈ A

Then 2 divides the element 4, 6, 8 of B.

So, (2, 4), (2, 6), (2, 8) ∈ R

when x = 3 ∈ A, then 3 divides elements 3, 6 of B

So, (3, 3), (3, 6) ∈ R

when x = 4 ∈ A then 4 divides the elements 4, 8 of B.

So, (4, 4), (4, 8) ∈ R

R = {(2, 4), (2, 6), (2, 8), (3, 3), (3, 6), (4, 4), (4, 8)}

R^{-1} = {(4, 2), (6, 2), (8, 2), (3, 3), (6, 3), (4, 4), (8, 4)}

Hence, option (A) is correct.

Question 5.

In the set of real numbers, Relation “x is smaller than y” will be

(A) Reflexive and Transitive

(B) Symmetric and Transitive

(C) Anti-symmetric and Transitive

(D) Reflexive and Anti-symmetric

Solution:

Because this relationship is only transitive

If “x is less than or equal to y” then this relation is reflexive and anti-symmetric.

Hence, option (C) is correct.

Question 6.

A relation R inset of non zero integers is defined as xRy ⇔ xy = yx then R is

(A) reflexive and symmetric but not transitive

(B) reflexive and anti-symmetric but not transitive

(C) reflexive, anti-symmetric and transitive

(D) reflexive, symmetric and transitive

Solution:

Given, Set I_{0} = set of non-zero integers = {±1, ±2, …}

A relation in I_{0}xRy ⇔ xy = yx ∀ x, y ∈ I_{0}

⇔ y log x = x log y ∀ x, y ∈ I_{0}

To prove R is equivalence relation we have to prove that R is reflexive, symmetric and transitive relation.

(i) Reflexivity: Let a ∈ I_{0}

a ∈ I_{0}

⇒ a log a = a log a ∀ a ∈ I_{0}

R is reflexive.

(ii) Symmetricity: Let a, b ∈ I_{0}

Now, (a, b) ∈ R

(a, b) ∈ R

⇒ b log a = a log b

⇒ (b, a) ∈ R

So, (a, b) ∈ R

⇒ (b, a) ∈ R ∀ a, b ∈ I_{0}

R is a symmetric relation.

(iii) Transitivity: Let a, b, c ∈ I_{0}

Now, (a, b) ∈ R, (b, c) ∈ R

(a, b) ∈ R ⇒ b log a = a log b ……(1)

(b, c) ∈ R ⇒ c log b = b log c

⇒ b log c = c log b …(2)

By dividing equation (1) from equation (2).

R is a transitive relation.

Thus, it is proved that R is reflexive, symmetric and transitive relation.

Hence, the option (D) is correct.

Question 7.

If relation R is defined as “x is divisor of y” then from the following subet of N. Which is a total ordered set ?

(A) {36, 3, 9}

(B) {7, 77, 11}

(C) {3, 6, 9, 12, 24}

(D) {1, 2, 3, 4, …}

Solution:

Given : N = set of natural numbers = (1, 2, 3, 4, …}

A relation R is N is defined as

xRy ⇒ “x is divisor of y” ∀ x, y ∈ N

⇒ \(\frac { y }{ x }\) = k ∈ N

where N is a set of natural numbers ∀ x, y ∈ N

From option ‘A’, \(\frac { 36 }{ 9 }\), \(\frac { 36 }{ 3 }\), \(\frac { 9 }{ 3 }\) all are natural numbers

From option ‘B’, \(\frac { 11 }{ 7 }\) ∉ N, from option ‘C’, \(\frac { 9 }{ 6 }\) ∉ N

From option ‘D’, \(\frac { 3 }{ 2 }\) ∉ N, Hence, option (A) is correct.

Question 8.

From the following relations defined on set Z of integers, which of the relation is not equivalence relation

(A) aR_{1}b ⇔ (a + b) is an even integer

(B) aR_{2}b ⇔ (a – b) is an even integer

(C) aR_{3}b ⇔ a < b

(D) aR_{4}b ⇔ a = b

Solution:

Correct option (C)

aR_{3}b ⇔ a < b, R_{3} is not an equivalence relation because it is not reflexive and symmetric.

Hence, option (C) is correct.

Question 9.

A relation R is defined on set A = {1, 2, 3}, where R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, then R is

(A) reflexive but not transitive

(B) reflexive but not symmetric

(C) symmetric and transitive

(D) neither symmetric nor reflexive

Sotution:

Given : Set A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

(i) Reflexivity: (1, 1) ∈ R, (2, 2) ∈ R, (3, 3) ∈ R

(a, a) ∈ R

R is reflexive.

(ii) Symmetricity: (1, 2) = (2, 1) ∉ R

(1, 3) = (3, 1 ) ∉ R

(2, 3) = (3, 2) ∉ R

(a, b) ∈ R ⇒ (b, a) ∉ R So, R is not symmetric

Hence, option (B) is correct.

Question 10.

If A = {a, b, c}, then number of possible non-zero relations in A is

(A) 511

(B) 512

(C) 8

(D) 7

Solution:

A = {a, b, c} Number of elements in A = n(A) = 3

Then, numbers of elements in A × A = n(A × A) = 3^{2} = 9

So, the number of relations in A are = 2^{n} – 1 = 2^{9} – 1 = 512 – 1 = 511.

Hence, option (A) is correct.

Question 11.

If A = {1, 2, 3, 4} then which of the following is a function in A

(A) f_{1} = {(x, y) : y = x + 1}

(B) f_{2} = {(x, y) : x + y > 4}

(C) f_{3} = {(x, y) : y < x}

(D) f_{4} = {(x, y) : x + y = 5}

Solution:

Here, f_{1} = {(x, y) : y = x + 1} = {(1, 2), (2, 3), (3, 4)}

f_{2} = {(x, y) : x + y > 4} = (1, 4), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

f_{3} = {(x, y) : y < x} = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}

f_{4} = {(x, y) : x + y = 5} = {(x, y) : x = 5 – y} = {(1, 4), (2, 3), (3, 3), (4, 1)}

It is clear that only f_{4} is a function because eveiy element of A corresponts to unique element in B.

Hence, the option (D) is correct.

Question 12.

Function f : N → N, f(x) = 2x + 3 is

(A) One-one onto

(B) One-one into

(C) Many one-onto

(D) Many-one into

Solution:

Given: f : N → N and f(x) = 2x + 3

where N = set of natural numbers

Let x_{1}, x_{2} ∈ N is such that f(x_{1}) = f(x_{2})

f(x_{1}) = f(x_{2})

⇒ 2x_{1} + 3 = 2x_{2} + 3

⇒ 2x_{1} = 2x_{2}

⇒ x_{1} = x_{2}

f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2} ∀ x_{1}, x_{2} ∈ N

f is one-one function.

Again, let y ∈ N (co-domain) if possible than let pre image of x is in domain N then f(x) = y

f(x) = y

⇒ 2x + 3 = y

⇒ x = \(\frac { y-3 }{ 2 }\) ∈ N

At y = 1, then x = \(\frac { 1-3 }{ 2 }\) = -1 ∉ N

If this way, y has many values for which x is not exist in domain A. So, f is into function.

Hence, option (B) is correct.

Question 13.

Which one of the following is onto function define from R to R.

(A) f(x) = |x|

(B) f(x) = e^{-x}

(C) f(x) = x^{3}

(D) f(x) = sin x

Solution:

f(x) = x^{3}

Given, f : R → R and f(x) = x^{3}

Let y ∈ R (co-domain) if possible, then let pre-image of y is x in domain R, then

f(x) = y

⇒ x^{3} = y

⇒ x = y^{1/3} ∈ R ∀ y ∈ R

So, pre-image of each value ofy is exist in domain R.

So, R is onto function.

Hence, option (C) is correct.

Question 14.

Which of the following is one-one function defined from R to R

(A) f(x) = |x|

(B) f(x) = cos x

(C) f(x) = e^{x}

(D) f(x) = x^{2}

Solution:

(C) f(x) = e^{x}

Let x, y ∈ R are such that

f(x) = f(y)

⇒ e^{x} = e^{y}

⇒ x log_{e} e = y log_{e} e

⇒ x = y

f(x) = f(y)

⇒ x = y [∵ log_{e} e = 1]

f is one-one function ∀ x, y ∈ R

Question 15.

f : R → R, f(x) = x^{2} + x is:

(A) One-one one

(B) One-one into

(C) Many-one onto

(D) Many one onto

Solution:

Given: f : R → R and f(x) = x^{2} + x

where R is a set of real numbers.

Let x_{1}, x_{2} ∈ R are such that f(x_{1}) = f(x_{2})

f(x_{1}) = f(x_{2})

⇒ x_{1}^{2} + x_{1} = x_{2}^{2} + x_{2}

⇒ x_{1}^{2} – x_{2}^{2} + x_{1} – x_{2} = 0

⇒ (x_{1} – x_{2})(x_{1} + x_{2}) + 1 (x_{1} – x_{2}) = 0

⇒ (x_{1} – x_{2}) (x_{1} + x_{2} + 1} = 0

⇒ x_{1} = x_{2}, x_{1} = -(x_{2} + 1) ∀ x_{1}, x_{2} ∈ R

Here, element of set A relates to two elements of set B.

So, it is a many-one function.

Again, let y ∈ R (co-domain)

If possible then let pre-image of y is x in domain R.

then f(x) = y

⇒ x^{2} + x = y

⇒ x(x + 1) = y

⇒ x = y, x = y – 1

If y < 1, then there is no real value of x.

So, pre-image of many elements of R does not exist in domain R, so, f is an into function.

Thus, f is many-one, into function.

Hence, the option (D) is correct.

Question 16.

Which of the following is onto function—

(A) f : Z → Z, f(x) = |x|

(B) f : N → Z, f(x) = |x|

(C) f : R_{0} → R^{+}, f(x) = |x|

(D) f : C → R, f(x) = |x|

Solution:

f : R_{0} → R^{+}, f(x) = |x|

Pre-image of every positive real number is exists in domain R_{0}. So, function is onto.

Hence, option (C) is correct.

Question 17.

Domain of function

\(f(x)=\frac { 1 }{ \sqrt { \left| x \right| -x } }\)

(A) R^{+}

(B) R^{–}

(C) R_{0}

(D) R

Solution:

It is clear that function is defined is x < 0 because [x ≥ 0 then f(x) = ∞]

Required domain of f(y) = R^{–}

Hence, option (B) is correct.

Question 18.

If x is real number than the range of

Solution:

Question 19.

Range of function f(x) = \(\cos { \frac { x }{ 3 } }\) is

(A) (0, ∞)

(B) (\(\frac { -1 }{ 3 }\), \(\frac { 1 }{ 3 }\))

(C) [-1, 1]

(D) [0, 1]

Solution:

Let y = \(\cos { \frac { x }{ 3 } }\)

-1 ≤ \(\cos { \frac { x }{ 3 } }\) ≤ 1

-1 ≤ y ≤ 1

x is defined if -1 ≤ y ≤ 1

Hence, required range = [-1, 1]

Hence, option (C) is correct.

Question 20.

From ]\(\frac { -\pi }{ 2 } ,\frac { -\pi }{ 2 }\)[ which of the following is one-one onto function defined in R

(A) f(x) = tan x

(B) f(x) = sin x

(C) f(x) = cos x

(D) f(x) = e^{x} + e^{-x}

Solution:

f(x) = tan x

Hence, option (A) is correct.

Question 21.

Find the domain and range of the relaton R

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

Solution:

Given relation

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

Then, domain of relation R {(x + 1) : x ∈ {0, 1, 2, 3, 4, 5}

Domain of R = {1, 2, 3, 4, 5, 6}

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

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

Question 22.

If A = {1, 2}, then write all non-zero relations defined in A.

Solution:

All non-zero relation are:

{(1, 1)}, {(2, 2)}, {(1, 2)}, {(2, 1)}

{(1, 1), (1, 2)}, {(1, 1), (2, 1)}, {(2, 2), (1,2)}, {(2, 2), (2, 1)}, {(1, 2), (2, 1)}

{(1, 1), (2, 2), (1, 2)}, {(1, 1), (2, 2), (2, 1)}, {(1, 1), (1, 2), (2, 1)}, {(2, 2), (1, 2), (2, 1)}

{(1, 1), (1, 2), (2, 1), (2, 2)}

Question 23.

Find the domain and range of the following relations

(i) R_{1} = {(x, y) : x, y ∈ N, x + y = 10}

(ii) R_{2} = {(x, y) : y = |x – 1|, x ∈ z and |x| ≤ 3}

Solution:

(i) Given set N = {1, 2, 3,…}

A relation R_{1} in N is defined as

R_{1} = {(x, y) : x, y ∈ N, x + y = 10} ∀ x, y ∈ N

So, xR_{1}y ⇔ x + y = 10 ⇔ y = 10 – x

when x = 1, then y = 10 – 1 = 9 ∈ N then (1, 9) ∈ R_{1}

when x = 2, then y = 10 – 2 = 8 ∈ N then (2, 8) ∈ N

when x = 3, then y = 10 – 3 = 7 ∈ N then (3, 7) ∈ R_{1}

when x = 4, then y = 10 – 4 = 6 ∈ N then (4, 6) ∈ R_{1}

Similarly, (5, 5) ∈ R_{1}, (6, 4) ∈ R_{1}, (7, 3) ∈ R_{1}, (8, 2) ∈ R_{1}, (9, 1) ∈ R_{1}

R_{1} = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}

Domain of R_{1} ={1, 2, 3, 4, 5, 6, 7, 8, 9}

Range of R_{1} = {9, 8, 7, 6, 5, 4, 3, 2, 1}

(ii) Given set Z = set of integers = {0, ±1, ±2, ±3,…}

A relation R in Z is defined as

R_{2} = {(x, y), y = |x – 1|, x ∈ z and x ≤ 3}

or xRy ⇔ y = |x – 1|, |x| ≤ 3 ∀ y ∈ Z

Here |x| ≤ 3 ⇒ -3 ≤ x ≤ 3, x ∈ Z

Domain of R_{2} = {-3, -2, -1, 0, 1, 2, 3}

Range of R_{2} = {4, 3, 2, 1, 0}

Question 24.

In the set of real numbers R, two relations R_{1} and R_{2} can be defined as

(i) aR_{1}b ⇔ a – b > 0

(ii) aR_{2}b ⇔ |a| ≤ b

Also test the reflexivity, symmetricity and transitivity of R_{1} and R_{2}.

Solution:

(i) Given set R = Set of real numbers

Reflexivity R aR_{1}b ⇔ a – b > 0 ∀ a, b ∈ R

Reflexivity : Let a ∈ R

a ∈ R

⇒ a – a = 0 > 0

⇒ a – a \(\ngtr \) 0

⇒ (a, a) ∉ R_{1} a ∈ R

R is not reflexive relation.

Symmetricity : Let a, b ∈ R are such that (a, b) ∈ R_{1}

(a, b) ∈ R_{1}

⇒ a – b > 0

⇒ b – a < 0

⇒ (b, a) ∉ R_{1}

So, (a, b) ∈ R_{1}

⇒ (b, a) ∉ R_{1}

R_{1} is not symmetric.

Transitivity : Let a, b, c ∈ R are such that

(a, b) ∈ R_{1}, (b, c) ∈ R_{1}

(a, b) ∈ R_{1}

⇒ a – b > 0 …(1)

(b, c) ∈ R_{1} ⇒ b – c > 0 …(2)

and a – c = (a – b) + (b – c) > 0 [∴ From (1) and (2)]

⇒ a – c > 0

⇒ (a, c) ∈ R_{1}

So, (a, b) ∈ R_{1}, (b, c) ∈ R_{1}

⇒ (a, c) ∈ R_{1} ∀ a, b, c ∈ R

R_{1} is transitive.

From above it is clear that R_{1} is not reflexive and symmetric relation.

It is only a transitive relation.

(ii) Given: Relation R = set of real numbers

A relation R_{2} in R is defined as

aR_{2}b ⇔ |a| ≤ b ∀ a, b ∈ R

Reflexivity: Let a ∈ R

a ∈ R ⇒ |a| ≤ a is not necessary

For example a = -2 ∈ R

and |-2| ≤ -2 ⇒ (-2, -2) ∈ R_{2}

So, R_{2} is not a reflexive relation.

Symmetricity: Let (a, b) ∈ R are such that (a, b) ∈ R_{2}

(a, b) ∈ R_{2} ⇒ |a| ≤ b

Then |b| ≤ a is not necessary.

For example : a = -2, b = 3

and (-2, 3) ∉ R_{2} as |-2| ≤ 3

But (3, -2) ∉ R_{2} because |3| ≤ -2

R_{2} is not symmetric.

Transitivity : Let a, b, c ∈ R are such that

(a, b) ∈ R_{2} and (b, c) ∈ R_{2}

(a, b) ∈ R_{2} ⇒ |a| ≤ b …(1)

(b, c) ∈ R_{2} ⇒ |b| ≤ c …(2)

From equation (1) and (2),

|a| ≤ b ≤ |b| ≤ c

⇒ |a| ≤ c

⇒ (a, c) ∈ R_{2}

So, (a, b) ∈ R_{2}, (b, c) ∈ R_{2}

⇒ (a, c) ∈ R_{2} ∀ a, b, c ∈ R

R_{2} is transitive.

Hence, from above it is clear that R_{2} is not reflexive and symmetric relation.

It is only a transitive relation.

Question 25.

In a set of natural numbers, a relation R is defined as

aRb ⇔ a^{2} – 4ab + 3b^{2} = 0, (a, b ∈ N)

Prove that R is reflexive but not symmetric and transitive.

Solution:

Given set N = {1, 2, 3, 4,….}

A relation R in N is defined as

aRb ⇔ a^{2} – 4ab + 3b^{2} = 0, ∀ a, b ∈ N

Question 26.

In the set of real numbers R, two relations R_{1} and R_{2} are defined as

(i) aR_{1}b ⇔ |a| = |b|

(ii) aR_{2}b ⇔ |a| ≤ |b|

Prove that R_{1} is an equivalence relation but R_{2} is not.

Solution:

(i) Given set R = set of real numbers.

A relation R_{1} in R is defined as

aR_{1}b = |a| = |b| ∀ a, b ∈ R

For proving is equivalence relation we have to prove that R_{1} is reflexive, symmetric and transitive relation.

Question 27.

A relation R in set A = {1, 2, 3} is defined as:

R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2)}

Test the reflexivity, symmetricity and transitivity of R.

Solution:

Given : Set A = {1, 2, 3}

Relation R in A is defined as

R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1) (2, 3), (3, 2)}

(i) Reflexivity : Here

(1, 1) ∈ R

(2, 2) ∈ R

(3, 3) ∈ R

So, ∀ a ∈ A ⇒ (a, a) ∈ R

R is not reflexive.

(ii) Symmetricity :

Here

(1, 2) ∈ R ⇔ (2, 1) ∈ R

(2, 3) ∈ R ⇔ (3, 2) ∈ R

(1, 3) ∈ R ⇔ (3, 1) ∈ R

So, (a, b) ∈ R ⇒ (b, a) ∈ R

R is symmetric relation.

(iii) Transitivity:

(1, 2) ∈ R, (2, 1) ∈ R ⇔ (1, 1) ∈ R etc.,

So, by definition of transitive relation.

R is transitive if

(a, b) ∈ R, (b, c) ∈ R

⇒ (a, c) ∈ R ∀ a, b, c ∈ A

Question 28.

Find the domain of function

\(\frac { 1 }{ \sqrt { \left( x+1 \right) \left( x+2 \right) } }\)

Solution:

Let f(x) = \(\frac { 1 }{ \sqrt { \left( x+1 \right) \left( x+2 \right) } }\)

Function f(x) is defiined if (x + 1) (x + 2) > 0

x > -1, x > -2

⇒ x ∈ (-∞, -2) ∪ (-1, ∞)

Hence, required domain (-∞, -2) ∪ (-1, ∞).