Chapter 1 Relations and Functions

A relation R on a set A is said to be reflexive if every element of A is related to itself.

Thus, R is reflexive, iff (a, a) ∈ R, for every a ∈ A

We have,

A = {a, b, c} and R = {(a, a), (b, c), (a, b)}

Therefore by definition of reflexive relation, if a, b, c ∈ A then (a, a),(b, b) and (c, c) ∈ R.

We observe that since (b, b) and (c, c) does not belong to R so R is not reflexive. We need to add (b, b) and (c, c) in R in order to make it reflexive.

A relation R on a set A is said to be transitive iff aRb and bRc then aRc for all a,b,c ∈ A

i.e., (a, b) ∈ R and (b, c) ∈ R

⇒ (a, c) ∈ R for all a,b,c ∈ A

We observe that since (a, c) does not belong to R so R is not transitive. We need to add (a, c) in R in order to make it transitive.

Hence, minimum number of ordered pairs to be added in R to make R reflexive and transitive are (b, b), (c, c) and (a, c).

Now, R = {(a, a), (b, b), (c, c), (a, c), (b, c), (a, b)} is Reflexive and Transitive.

The set of values​​for which the function can be defined is domain of a real valued function.

We have,

In order to find domain of f(x) we need to find those real values of x which will give real value to f(x) after solving.

Now, f is a real valued function if 25-x² ≥ 0

⇒ 25-x² ≥ 0

⇒ -(x²-25) ≥ 0

On multiplying the inequality by -1 we get,

⇒ x²-25 ≤ 0

Note : The sign of inequality is reversed if it is multiplied by a negative quantity.

⇒ (x+5)(x-5) ≤ 0

⇒ x ∈ [-5,5]

Hence, Domain(f) = {x : x ∈ [-5,5] }

We have,

gof(x) = g(f(x))

⇒ gof(x) = g(2x+1)

= (2x+1)² – 2

= (4x²+1+4x) – 2

= 4x²+4x-1

∴ gof = 4x²+4x-1

Given that f(x) = 2x-3

Let f(x) = y, then 2x-3 = y

⇒

⇒ [∵ f(x) = y, ⇒ x = f^-1(y)]

Hence, f^-1 : R → R is given by

We have, f = {(a, b), (b, d), (c, a), (d, c)}

We can get f^-1 by interchanging the components of ordered pairs in f

∴ f^-1 = {(b, a), (d, b), (a, c), (c, d)}

We have,

f(x) = x² – 3x + 2

∴ f(f(x)) = f (x² – 3x + 2)

= (x² – 3x + 2)² – 3(x² – 3x + 2) + 2

= x⁴ + 9x² + 4 – 6x³ – 12x + 4x² – 3x² + 9x – 6 + 2

= x⁴ – 6x³ + 10x² – 3x

⇒ f(f(x)) = x⁴ – 6x³ + 10x² – 3x

In order to determine if g = {(1, 1), (2, 3), (3, 5), (4, 7)} represents a function or not, we need to validate if g satisfies the condition of a relation to be a function.

A relation f from a set A to a set B is said to be a function if every

element of set A has one and only one image in set B.

By definition of function we can say that no two distinct ordered pairs in a function have the same first element.

We have,

g = {(1, 1), (2, 3), (3, 5), (4, 7)}

we observe that each first element of ordered pairs is related to only one element.

Hence, g is a function.

Given,

g(x) = α x + β and g(1) = 1

⇒ α + β = 1 ……(i)

g(2) = 3, we get

⇒ α 2 + β = 3 ……(ii)

g(3) = 5, we get

⇒ α 3 + β = 5 ……(iii)

g(4) = 7, we get

⇒ α 4 + β = 7 ……(iv)

Solve any 2 equations from (i),(ii),(iii) and (iv) to find two unknowns α and β

On solving (i) and (ii), we get

α = 2 and β = -1

Hence, the function g(x) = 2x – 1

(i){(x, y): x is a person, y is the mother of x}.

Here, (x, y) is an ordered pair which associates each person (x) to his(her) mother (y).

Now, each person will have only one mother so we can say that for each value of x there is an association to a unique value of y.

Hence,(x, y) represent a function.

Also, more than one person may have same mother.

i.e., two or more distinct value of x may have same value of y.

So, (x, y) is many-one function or surjective.

(ii){(a, b): a is a person, b is an ancestor of a}.

Here, (a, b) is an ordered pair which associates each person (a) to his(her) ancestor (b).

Now, any person can have more than one ancestor so we can say that each value of a does not have a unique value of b.

Hence, (a,b) does not represent a function.

We have,

f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}

Domain of g = {2,5,1}

Now, fog(2) = f(g(2)) = f(3) = 5

fog(5) = f(g(5)) = f(1) = 2

fog(1) = f(g(1)) = f(3) = 5

∴ fog = {(2, 5), (5, 2), (1,5) }

We have,

f : C → R given by f (z) = |z|, ∀ z ∈ C

In order to prove that f is one-one, it is sufficient to prove that f(z₁)=f(z₂) ⇒ z₁=z₂∀ z₁, z₂ ∈ C .

Let z₁ = 2+3i and z₂ = 2-3i are two distinct complex numbers.

Now,

f(z₁) = |z₁| = |2+3i | = = √ 13

f(z₂) = |z₂| = |2-3i | = = √ 13

here, we observe that f(z₁) = f(z₂) but z₁ ≠ z₂

This shows that different element of C may have the same value in R.

Thus, f(z) is not one-one.

f is onto if every element of R is the f-image of some element of C.

We have, f(z) = |z|, ∀ z ∈ C and |z| ≥ 0

We observe that negative real numbers in R do not have their pre-images in C.

Thus, f is not onto.

Hence, f(z) is neither one-one nor onto.

We have,

f (x) = cosx, ∀ x ∈ R

In order to prove that f is one-one, it is sufficient to prove that f(x₁)=f(x₂) ⇒ x₁=x₂∀ x₁, x₂ ∈ A .

Let x₁ = 0 and x₂ = 2π are two different elements in R.

Now,

f(x₁) = f(0) = cos0 = 1

f(x₂) = f(2π) = cos2π = 1

we observe that f(x₁)=f(x₂) but x₁ ≠ x_2.

This shows that different element in R may have same image.

Thus, f(x) is not one-one.

We know that cosx lies between -1 and 1.

So, the range of f is [-1,1] which is not equal to its co-domain.

i.e., range of f ≠ R (co-domain)

In other words, range of f is less than co-domain, i.e there are elements in co-domain which does not have any pre-image in domain.

so, f is not onto.

Hence, f is neither one-one nor onto.

We have,

X = {1, 2, 3} and Y = {4, 5}

∴ X × Y = {(1, 4),(1, 5),(2, 4),(2, 5),(3, 4),(3, 5)}

(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)}

Now, f(1) = 4 and f(1) = 5

We observe that one element of domain maps to two distinct values.

i.e., ‘1’ has no unique image.

Thus, f is not a function.

(ii) g = {(1, 4), (2, 4), (3, 4)}

Now, g(1) = 4, g(2) = 4, g(3) = 4

We observe that each distinct element of domain has unique image.

Thus, g is a function.

(iii) h = {(1,4), (2, 5), (3, 5)}

Now, h(1) = 4, h(2) = 5, h(3) = 5

We observe that each distinct element of domain has unique image.

Thus, h is a function.

(iv) k = {(1,4), (2, 5)}.

Now, k(1) = 4 and k(2) = 5

We observe that ‘3’ does not have any image under the mapping.

Thus, k is not a function.

Given : gof = I_A

In order to prove that f is one-one, it is sufficient to prove that f(x)=f(y) ⇒ x=y ∀ x,y ∈ A .

Let x,y ∈ A such that f(x) = f(y). Then,

f(x) = f(y)

⇒ g(f(x)) = g(f(y))

⇒ gof(x) = gof(y)

⇒ I_A(x) = I_A(y) [∵ gof = I_A is given ]

⇒ x = y [by definition of Identity function, I(x) = x]

Thus, f is one-one.

Now, in order to prove that g : B → A is onto, it is sufficient to prove that each element in A has pre-image in B.

Let x ∈ A.

Also, f : A → B is a function ∴ f(x) ∈ B

Now,

Let f(x) = y

⇒ g(f(x)) = g(y)

⇒ gof(x) = g(y)

⇒ I_A(x) = g(y) [∵ gof = I_A is given ]

⇒ x = g(y) [by definition of Identity function, I(x) = x]

Thus, for every x ∈ A there exists y ∈ B such that g(y) = x.

⇒ g is onto.

We have,

Let

⇒ 2y–ycosx = 1

⇒ -ycosx = 1-2y

⇒

⇒

Now, we know that range of cosx is [-1,1].

⇒ -1 ≤ cosx ≤ 1

⇒

⇒

On multiplying the inequality by -1 we get,

⇒

Note : The sign of inequality is reversed if it is multiplied by a negative quantity.

⇒

⇒

Thus, range of

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, ∀ a, b ∈Z, aRb if and only if a – b is divisible by n.

Now,

R is Reflexive if (a,a)∈R∀a∈Z

aRa ⇒ (a-a) is divisible by n.

a-a = 0 = 0 × n [since 0 is multiple of n it is divisible by n]

⇒ a-a is divisible by n

⇒ (a,a) ∈ R

Thus, R is reflexive on Z.

R is Symmetric if (a,b)∈R⇒(b,a)∈R∀a,b∈Z

(a,b) ∈ R ⇒ (a-b) is divisible by n

⇒ (a-b) = nz for some z ∈ Z

⇒ -(b-a) = nz

⇒ b-a = n(-z) [∵ z ∈ Z ⇒ -z ∈ Z ]

⇒ (b-a) is divisible by n

⇒ (b,a) ∈ R

Thus, R is symmetric on Z.

R is Transitive if (a,b)∈R and (b,c)∈R⇒(a,c)∈R∀a,b,c∈Z

(a,b) ∈ R ⇒ (a-b) is divisible by n

⇒ a-b = nz₁ for some z₁∈ Z

(b,c) ∈ R ⇒ (b-c) is divisible by n

⇒ b-c = nz₂ for some z₂∈ Z

Now,

a-b = nz₁ and b-c = nz₂

⇒ (a-b) + (b-c) = nz₁ + nz₂

⇒ a-c = n(z₁ + z₂ ) = nz₃ where z₁ + z₂ = z₃

⇒ a-c = nz₃ [∵ z₁,z₂ ∈ Z ⇒ z₃∈ Z]

⇒ (a-c) is divisible by n.

⇒ (a, c) ∈ R

Thus, R is transitive on Z.

Since R is reflexive, symmetric and transitive it is an equivalence relation on Z.

Given that, A = {1, 2, 3, 4 }

(a) Reflexive, transitive but not symmetric

Let R be a relation defined by

R = {(1,1),(1,2),(1,4),(2,2),(2,3),(3,2),(3,3),(4,2),(4,4)} on set A.

R is reflexive ∵ (1,1),(2,2),(3,3),(4,4) ∈ R

R is transitive ∵ (1,4) ∈ R and (4,2) ∈ R ⇒ (1,2) ∈ R

R is not symmetric ∵ (1,4) ∈ R but (4,1) ∉ R

Hence, R is reflexive, transitive but not symmetric.

(b) Symmetric but neither reflexive nor transitive

Let R be a relation defined by

R = {(1,2),(2,1),(2,3),(3,2)} on set A.

R is not reflexive ∵ (1,1),(2,2),(3,3),(4,4) ∉ R

R is symmetric ∵ (1,2) ∈ R ⇒ (2,1) ∈ R and (2,3) ∈ R ⇒ (3,2) ∈ R

R is not transitive ∵ (1,2) ∈ R and (2,1) ∈ R ⇒ (1,1) ∉ R

Hence, R is symmetric but neither reflexive nor transitive.

(c) Reflexive, symmetric and transitive.

Let R be a relation defined by

R = {(1,1),(1,2),(1,4),(2,1),(2,2),(2,3),(3,2),(3,3),(4,1),(4,4)} on set A.

R is reflexive ∵ (1,1),(2,2),(3,3),(4,4) ∈ R

R is symmetric ∵ (1,2),(1,4),(2,3) ∈ R ⇒ (2,1),(4,1),(3,2) ∈ R

R is transitive ∵ (1,2) ∈ R and (2,1) ∈ R ⇒ (1,1) ∈ R

Hence, R is reflexive, symmetric and transitive.

Given that, R = {(x, y): x ∈N, y ∈N, 2x + y = 41}

Now, 2x + y = 41

⇒ y = 41 - 2x (i)

Since x ∈N, y ∈N from (i) we get the relation

R = {(1,39),(2,37),(3,35),(4,33),(5,31),(6,29),(7,27),(8,25),

(9,23),(10,21),(11,19),(12,17),(13,15),(14,13),(15,11),

(16,9),(17,7),(18,5),(19,3),(20,1)}

Domain(R) ={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}

Range(R) ={1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,

37,39}

R is not reflexive ∵ (1,1),(2,2)…(20,20) ∉ R

R is not symmetric ∵ (1,39) ∈ R but (39,1) ∉ R

R is not transitive ∵ (12,17),(17,7) ∈ R but (12,7) ∉ R

Hence, R is neither reflexive nor symmetric nor transitive.

Given that, A = {2, 3, 4}, B = {2, 5, 6, 7}

(a) an injective mapping from A to B

Let f : A → B denote a mapping f = {(x,y) : y=2x }

Now, y = 2x

When x=2 we get y = 4

Similarly, x=3 and 4 will give y=6 and 8 respectively.

∴ f = {(2,4),(3,6),(4,8)}

We observe that each element of A has unique image in B.

Thus, f is injective.

(b) a mapping from A to B which is not injective

Let g: A → B denote a mapping such that g = {(2,2),(3,5),(4,2)}

We observe that 2 and 4 ∈ A does not have unique image.

Thus, g is not injective.

(c) a mapping from B to A.

Let h : B → A denote a mapping such that

h = {(2,3),(5,2),(6,3),(7,4)}

Let f : N → N, be a function given by f(x) = 2x.

In order to prove that f is one-one, it is sufficient to prove that f(x₁)=f(x₂) ⇒ x₁=x₂∀ x₁, x₂ ∈ N

Now, let f(x₁) = f(x₂)

⇒ 2x₁= 2x₂

⇒ x₁= x₂

⇒ f is one-one.

f is not onto, as for 1 ∈ N, there does not exist any x in N such that f(x) = 2x = 1.

Thus, f : N → N, be a function given by f(x) = 2x, which is one-one but not onto.

Let f : N → N, be a function given by f(1) = f(2) = 1 and f(x) = x-1 for every x > 2.

Since, f(1) = 1 = f(2)

⇒ 1 and 2 does not have unique image.

Thus, f is not one-one.

Let f(x) = y

⇒ y = x-1

⇒ x = y+1

⇒ for each y ∈ R there exists x ∈ N such that f(x) = y.

Thus, f is onto.

Let the function f :R → R be defined by f (x) = cosx, ∀ x ∈ R.

We have,

f (x) = cosx, ∀ x ∈ R

In order to prove that f is one-one, it is sufficient to prove that f(x₁)=f(x₂) ⇒ x₁=x₂∀ x₁, x₂ ∈ A .

Let x₁ = 0 and x₂ = 2π are two different elements in R.

Now,

f(x₁) = f(0) = cos0 = 1

f(x₂) = f(2π) = cos2π = 1

we observe that f(x₁)=f(x₂) but x₁ ≠ x_2.

This shows that different element in R may have same image.

Thus, f(x) is not one-one.

We know that cosx lies between -1 and 1.

So, the range of f is [-1,1] which is not equal to its co-domain.

i.e., range of f ≠ R (co-domain)

In other words, range of f is less than co-domain, i.e there are elements in co-domain which does not have any pre-image in domain.

so, f is not onto.

Hence, f is neither one-one nor onto.

Given that,

In order to prove that f is one-one, it is sufficient to prove that f(x₁)=f(x₂) ⇒ x₁=x₂∀ x₁, x₂ ∈ A .

Let f(x₁)=f(x₂)

⇒

⇒ (x₁-2)(x₂-3) = (x₂-2)(x₁-3)

⇒ x₁x₂-3x₁-2x₂+6 = x₁x₂-3x₂-2x₁+6

⇒ -3x₁-2x₂ = -3x₂-2x₁

⇒ -3x₁+2x₁ = -3x₂+2x₂

⇒ (-3+2) x₁ = (-3+2)x₂

⇒ x₁ = x₂

∴ f is one-one.

f is onto if every element of B is the f-image of some element of A.

let f(x) = y

⇒

⇒ x-2 = y(x-3)

⇒ x-2 = xy-3y

⇒ x-xy = -3y+2

⇒ x(1-y) = -3y+2

⇒

⇒

Thus, for each y ∈ B there exists such that f(x) = y.

Hence, f is onto.

Since, f is one-one and onto therefore f is bijective.

Given that, A = [–1, 1]

Let f(x₁) = f(x₂)

⇒

⇒ 2x₁ = 2x₂

⇒ x₁ = x₂

⇒ f is one one

Now, let f(x) = y

⇒

⇒ 2y= x

⇒ for y=1 ∈ A we will have x = 2 ∉ A

⇒ f is not onto

Thus, f is one-one and not onto.

(ii) Given that, A = [–1, 1]

let g(x₁) = g(x₂)

⇒ |x₁|= |x₂|

⇒ x₁= ± x₂

⇒ x₁= x₂ and x₁= - x₂

For e.g., g(-1) = |-1| = 1 and g(1) = |1| = 1

⇒ g is not one-one.

We observe that -1 does not have any pre-image in the domain since g(x) = |x| assumes only non-negative values.

i.e. we cannot find any number in domain which will give -1 in co-domain.

⇒ g is not onto

Hence, g is neither one one nor onto.

Given that, A = [–1, 1]

let h(x₁) = h(x₂)

⇒ x₁|x₁|= x₂|x₂|

if x_1,x₂ >0

⇒ x₁² = x₂²

⇒ x₁= x₂

if x_1,x₂ < 0

⇒ x₁² = x₂²

⇒ x₁= x₂

⇒ h is one-one.

Let h(x) = y

⇒ y = x |x|

⇒ y = x²

Thus, for each y co domain there exists x in domain.

⇒ h is onto.

Hence, h is one one and onto.

So, h is bijective.

(iv) Given that, A = [–1, 1]

let k(x₁) = k(x₂)

⇒ x₁²= x₂²

⇒ x₁= ± x₂

⇒ x₁= x₂ and x₁= - x₂

For e.g., k(-1) = |-1| = 1 and k(1) = |1| = 1

⇒ k is not one-one.

We observe that -1 does not have any pre-image in the domain since k(x) = x² assumes only non-negative values.

i.e. we cannot find any number in domain which will give -1 in co-domain.

⇒ k is not onto

⇒ Hence, k is neither one one nor onto.

Let R = {(x,y): x is greater than y ∀ x,y ∈ N } be a relation defined on N.

Now,

We observe that, any element x ∈ N cannot be greater than itself.

⇒ (x,x) ∉ R ∀ x ∈ N

⇒ R is not reflexive.

Let (x,y) ∈ R ∀ x, y ∈ N

⇒ x is greater than y

But y cannot be greater than x if x is greater than y.

⇒ (y,x) ∉ R

For e.g., we observe that (5,2) ∈ R i.e 5 > 2 but 2 ≯ 5 ⇒ (2,5) ∉ R

⇒ R is not symmetric

Let (x,y) ∈ R and (y,z) ∈ R ∀ x, y,z ∈ N

⇒ x > y and y > z

⇒ x > z

⇒ (x,z) ∈ R

For e.g., we observe that

(5,4) ∈ R ⇒ 5 > 4 and (4,3) ∈ R ⇒ 4 > 3

And we know that 5 > 3 ∴ (5,3) ∈ R

⇒ R is transitive.

Thus, R is transitive but not reflexive not symmetric.

Let R = {(x,y): x + y =10, ∀ x,y ∈ N } be a relation defined on N.

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

Now,

R is not reflexive ∵ (2,2) ∉ R.

R is symmetric ∵ (3,7) ∈ R ⇒ (7,3) ∈ R

R is not transitive ∵ (1,9) ∈ R and (9,1) ∈ R but (1,1) ∉ R.

Thus, R is symmetric but not reflexive not transitive.

Let R = {(x,y): x y is square of an integer, ∀ x,y ∈ N } be a relation defined on N.

R is reflexive ∵ x² is square of an integer ∀ x ∈ N ⇒ (x,x) ∈ R

Let (x,y) ∈ R ∀ x, y ∈ N

⇒ x y is square of an integer

⇒ y x is square of an integer

⇒ (y,x) ∈ R

⇒ R is symmetric

Let (x,y) ∈ R and (y,z) ∈ R ∀ x, y,z ∈ N

⇒ x y is square of an integer and yz is square of an integer

let xy = p² and yz = q² for some p,q ∈ Z

⇒

⇒ , which is square of an integer

⇒ (x,z) ∈ R

⇒ R is transitive.

Thus, R is reflexive, symmetric and transitive.

Let R = {(x,y): x + 4y =10, ∀ x,y ∈ N } be a relation defined on N.

R = {(2,2),(6,1)}

Now,

R is not reflexive ∵ (1,1) ∉ R.

R is not symmetric ∵ (6,1) ∈ R but (1,5) ∉ R

R is not transitive

∵ (x,y) ∈ R ⇒ x+4y=10 and (y,z) ∈ R ⇒ y+4z=10

⇒ x-16z = -30

⇒ (x,z) ∉ R

Thus, R is neither symmetric nor reflexive nor transitive.

Given that, A = {1, 2, 3, ... 9} and R be the relation in A ×A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A ×A.

Let (a,b) R (a,b)

⇒ a+b = b+a

which is true since addition is commutative on N.

⇒ R is reflexive.

Let (a,b) R (c,d)

⇒ a+d = b+c

⇒ b+c = a+d

⇒ c+b = d+a [since addition is commutative on N]

⇒ (c,d) R (a,b)

⇒ R is symmetric.

Let (a,b) R (c,d) and (c,d) R (e,f)

⇒ a+d = b+c and c+f = d+e

⇒ (a+d) – (d+e) = (b+c ) – (c+f)

⇒ a-e= b-f

⇒ a+f = b+e

⇒ (a,b) R (e,f)

⇒ R is transitive.

Hence, R is an equivalence relation.

The equivalence class [(2,5)] = {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}

f : A → B is one-one if the images of distinct elements of A under f are distinct, i.e.,for every a, b ∈ A, f(a) = f(b)⇒ a = b

we suppose that f : A → B is not one-one function.

Let f(a) = x and f(b) = x

⇒ f^-1(x) = a and f^-1(x) = b

⇒ inverse function cannot be defined as we have two images ‘a’ and ‘b’ for one pre-image ‘x’.

So, f can be invertible if it is one-one.

Now, suppose that f : A → B is not onto function.

Let B = {x,y,z} and range of f = {x,y}

We can observe that ‘z’ does not have any pre-image in A.

But f^-1 has z as a pre-image which does not have any image in A.

So, f can be invertible if it is onto.

Hence, f is invertible if and only if it is both one-one and onto.

Given that, f (x) = x² + 3x + 1, g (x) = 2x – 3

(i) f o g

fog = f(g(x)) = f(2x-3)

= (2x-3)² + 3(2x-3) + 1

= (4x²-12x+9) + 6x – 9 +1

= 4x² - 6x + 1

∴ fog = 4x² - 6x + 1

(ii) g o f

gof = g(f(x)) = g(x² + 3x + 1)

= 2(x² + 3x + 1) – 3

= 2x² + 6x + 2 – 3

= 2x² + 6x – 1

∴ gof = 2x² + 6x – 1

(iii) f o f

fof = f(f(x)) = f(x² + 3x + 1)

= (x² + 3x + 1)² + 3(x² + 3x + 1) + 1

= x⁴+9x²+1+6x³+6x+2x²+3x²+9x+3+1

= x⁴+6x³+14x²+15x+5

∴ fof = x⁴+6x³+14x²+15x+5

(iv) g o g

gog = g(g(x)) = g(2x-3)

= 2(2x-3) – 3

= 4x-6-3

= 4x-9

∴ gog = 4x-9

Given that, * be the binary operation defined on Q.

A binary operation ‘*’ is commutative if a*b = b*a∀a, b∈Q

(i) a * b = a – b ∀ a, b ∈Q

a * b = a – b = -b+a = -(b-a) = -(b*a)

∴ a*b ≠ b*a

Hence, ‘*’ is not commutative on Q.

(ii) a * b = a² + b² ∀ a, b ∈Q

a * b = a² + b²

= b² + a² [∵ addition is commutative on Q ⇒ a+b = b+a ]

= b*a

∴ a*b = b*a

Hence, ‘*’ is commutative on Q.

(iii) a * b = a + ab ∀ a, b ∈ Q

a * b = a + ab and b*a = b + ab

⇒ a + ab ≠ b + ab

∴ a*b ≠ b*a

Hence, ‘*’ is not commutative on Q.

(iv) a * b = (a – b)² ∀ a, b ∈Q

a * b = (a – b)² = {-(b-a)}²

= (b-a)²

=b*a

∴ a*b = b*a

Hence, ‘*’ is commutative on Q.

Given that,

‘*’ be binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ R

A binary operation ‘*’ is commutative if a*b = b*a∀a, b∈R

Now,

a*b = 1+ab = 1+ba [∵ ab=ba since multiplication is commutative

on R]

⇒ 1+ba = b*a

∴ a*b = b*a ∀ a, b ∈ R

So, ‘*’ is commutative on R.

A binary operation ‘*’ is associative if (a*b)*c = a*(b*c)∀a, b,c∈R

Now,

(a*b)*c = (1+ab)*c = 1+(1+ab)c = 1+c+abc

a*(b*c) = a*(1+bc) = 1+a(1+bc) = 1+a+abc

∴ (a*b)*c ≠ a*(b*c)

So, ‘*’ is not associative on R.

Hence, ‘*’ is commutative but not associative on R.

Given that,

R be a relation on T defined as aRb if a is congruent to b ∀ a, b ∈ T

Now,

aRa ⇒ a is congruent to a, which is true since every triangle is congruent to itself.

⇒ (a,a) ∈ R ∀ a ∈ T

⇒ R is reflexive.

Let aRb ⇒ a is congruent to b

⇒ b is congruent to a

⇒ bRa

∴ (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a, b ∈ T

⇒ R is symmetric.

Let aRb ⇒ a is congruent to b and bRc ⇒ b is congruent to c

⇒ a is congruent to c

⇒ aRc

∴ (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a, b,c ∈ T

⇒ R is transitive.

Hence, R is an equivalence relation.

Given that, relation R is defined as aRb if a is brother of b

Now,

aRa ⇒ a is brother of a, which is not true.

⇒ (a,a) ∉ R

⇒ R is not reflexive

aRb ⇒ a is brother of b but this does not mean that b is brother of a,b can be sister of a.

Thus, (a,b) ∈ R ⇒ (b,a) ∉ R

⇒ R is not symmetric.

aRb ⇒ a is brother of b and bRc ⇒ b is brother of c

⇒ a is a brother of c.

⇒ R is transitive.

Hence, R is transitive but not symmetric.

An equivalence relation is one which is reflexive, symmetric and transitive.

Given that, A = {1, 2, 3}

We can define equivalence relation on A as follows.

R₁ = A × A = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),

(3,1),(3,2),(3,3)}

R₁ is reflexive ∵ (1,1),(2,2),(3,3) ∈ R

R₁ is symmetric ∵ (1,2),(1,3),(2,3) ∈ R ⇒ (2,1),(3,1),(3,2) ∈ R

R₁ is Transitive ∵ (1,2) ∈ R and (2,3) ∈ R ⇒ (1,3) ∈ R

Similarly,

R₂ = {(1,1),(2,2),(3,3),(1,2),(2,1)}

R₃ = {(1,1),(2,2),(3,3),(1,3),(3,1)}

R₄ = {(1,1),(2,2),(3,3),(2,3),(3,2)}

R₅ = {(1,1),(2,2),(3,3)}

∴ maximum number of equivalence relation on A is ‘5’.

Given that, R on the set {1, 2, 3} be defined by R = {(1, 2)}

Now,

R is not reflexive ∵ (1,1),(2,2),(3,3) ∉ R

R is not symmetric ∵ (1,2)∈ R but (2,1) ∉ R

R is not transitive ∵ R has only one ordered pair.

Thus, R is neither reflexive nor transitive nor symmetric.

Hence, option ‘D’ is corect.

Given that, aRb if a ≥ b

Now,

We observe that, a ≥ a since every a ∈ R is greater than or equal to itself.

⇒ a ≥ a ⇒ (a,a) ∈ R ∀ a ∈ R

⇒ R is reflexive.

Let (a,b) ∈ R

⇒ a ≥ b

But b cannot be greater than a if a is greater than b.

⇒ (b,a) ∉ R

For e.g., we observe that (5,2) ∈ R i.e 5 ≥ 2 but 2 ≱ 5 ⇒ (2,5) ∉ R

⇒ R is not symmetric

Let (a,b) ∈ R and (b,c) ∈ R

⇒ a ≥ b and b ≥ c

⇒ a ≥ c

⇒ (a,c) ∈ R

For e.g., we observe that

(5,4) ∈ R ⇒ 5 ≥ 4 and (4,3) ∈ R ⇒ 4 ≥ 3

And we know that 5 ≥ 3 ∴ (5,3) ∈ R

⇒ R is transitive.

Thus, R is reflexive, transitive but not symmetric.

Given that, A = {1, 2, 3} and

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

Now,

R is reflexive ∵ (1,1),(2,2),(3,3) ∈ R

R is not symmetric ∵ (1,2),(2,3),(1,3) ∈ R but (2,1),(3,2),(3,1) ∉ R

R is transitive ∵ (1,2) ∈ R and (2,3) ∈ R ⇒ (1,3) ∈ R

Thus, R is reflexive, transitive but not symmetric.

Hence, option ‘A’ is correct.

Given that,

Let e be the identity element for * such that

a*e = e*a = a

Now,

⇒

⇒ 2a = ae

⇒ e = 2

Number of elements in A = 5

Number of elements in B = 6

Now, for a function to be one-one all elements of domain should have unique image in co-domain.

Here, 5 elements of A can be mapped to 5 distinct element of B.

⇒ function is one-one

But range of B is 6 that means there is one element in B which does not have pre-image in A.

Thus the function cannot be onto if it is one-one.

Also, if all 6 elements of B are mapped to all 5 elements of A, then we can observe there will be atleast 2 elements of B which have same pre-image in A.

⇒ one element of A will have two image in B.

⇒ function is not one-one

Thus the function cannot be one-one if it is onto.

Hence, there cannot be any one-one and onto mapping from A to B.

Given that, A = {1, 2, 3, ...n} and B = {a, b}

Number of elements in A = n

Number of elements in B = 2

No. of possible function from A → B is n² (i.e. number of possible ways n elements of A can be mapped to 2 elements of B.

Now, not all of these functions are surjective.

we know that function f : A → B is surjective if both the elements of B are mapped.

Out of these n² functions there will be two functions where all the elements of A are mapped to first element of B and where all the elements of A are mapped to second element of B, those functions are not surjective.

⇒ number of surjective functions from A → B are n² – 2.

Given that,

If x=0 then

Thus, f is not defined.

Given that, f (x) = 3x² – 5 and

Now,

gof(x) = g(f(x))

= g(3x²-5)

=

=

=

A function is bijective iff it is one-one and onto.

Option A. f (x) = x³

Let f(x₁) = f(x₂)

⇒ x₁³= x₂³

⇒ x₁= x₂

⇒ f is one one

Let f(x) = y, y ∈ Z

⇒ y = x³

⇒ x = y^1/3 but y^1/3∉ Z

⇒ f is not onto

Thus, f is not bijective.

Option B. f (x) = x + 2

Let f(x₁) = f(x₂)

⇒ x₁+2 = x₂+2

⇒ x₁= x₂

⇒ f is one one

Let f(x) = y, y ∈ Z

⇒ y = x + 2

⇒ x = y – 2

⇒ for each y ∈ Z there exists x ∈ Z (domain) such that f(x) = y.

⇒ f is onto

Thus, f is bijective.

Option C. f (x) = 2x + 1

Let f(x₁) = f(x₂)

⇒ 2x₁+1 = 2x₂+1

⇒ x₁= x₂

⇒ f is one one

Let f(x) = y, y ∈ Z

⇒ y = 2x + 1

⇒ y - 1 = 2x

⇒

We observe that if we put y=0, then .

Thus, y = 0 ∈ Z does not have pre image in Z (domain)

⇒ f is not onto.

Thus, f is not bijective.

Option D. f (x) = x² + 1

let f(x₁) = f(x₂)

⇒ x₁² + 1 = x₂² + 1

⇒ x₁² = x₂²

⇒ x₁ = ± x₂

⇒ x₁= x₂ and x₁= - x₂

For e.g., f(-1) = |-1| = 1 and f(1) = |1| = 1

⇒ f is not one-one.

Since, f is not one one it cannot be bijective.

Given that, f(x) = x³ + 5

Let f(x) = y

⇒ y = x³ + 5

⇒ y – 5 = x³

⇒ x³ = y – 5

⇒

⇒ [∵ f(x) = y, ⇒ x = f^-1(y)]

⇒

Given that, f : A → B and g : B → C be the bijective functions.

Let A = {1,3,4}, B ={2,5,1} and C = {3,1,2}

f : A → B is bijective function.

∴ f = {(1, 2), (3, 5), (4, 1)

f^-1 = {(2,1),(5,3),(1,4)}

g : B → C is bijective function.

∴ g = {(2, 3), (5, 1), (1, 4)}

g^-1 ={(3,2),(1,5),(4,1)}

Now,

gof (1) = g(f(1)) = g(2) = 3

gof (3) = g(f(3)) = g(5) = 1

gof (4) = g(f(4)) = g(1) = 4

∴ gof = {(1,3),(3,1),(4,4)} (1)

(gof)^-1 = {(3,1),(1,3),(4,4)} (2)

fog (2) = f(g(2)) = f(3) = 5

fog (5) = f(g(5)) = f(1) = 2

fog (1) = f(g(1)) = f(4) = 1

∴ fog = {(2,5),(5,2),(1,1)} (3)

(fog)^-1 = {(5,2),(2,5),(1,1)} (4)

f^-1og^-1 (3) = f^-1(g^-1(3)) = f^-1(2) = 1

f^-1og^-1 (1) = f^-1(g^-1(1)) = f^-1(5) = 3

f^-1og^-1 (4) = f^-1(g^-1(4)) = f^-1(1) = 4

∴ f^-1og^-1 = {(3,1),(1,3),(4,4)} (5)

g^-1of^-1 (2) = g^-1(f^-1(2)) = g^-1(1) = 5

g^-1of^-1 (5) = g^-1(f^-1(5)) = g^-1(3) = 2

g^-1of^-1 (1) = g^-1(f^-1(1)) = g^-1(4) = 1

∴ g^-1of^-1 = {(2,5),(5,2),(1,1)} (6)

On comparing 1,2,3,4,5 and 6 we observe that 2 and 5 are same.

i.e (g o f)^–1 = f^-1og^-1

Given that,

Let f(x) = y

⇒

⇒ (5x-3)y = 3x+2

⇒ 5xy-3y = 3x+2

⇒ 5xy–3x = 2+3y

⇒ (5y-3)x = 2+3y

⇒

⇒ [∵ f(x) = y, ⇒ x = f^-1(y)]

⇒

⇒ f^-1(x) = f(x)

Given that, f : [0, 1] → [0, 1] be defined by

Now,

(fof)(x) =

=

=

∴ (fof)(x) = x

Given that, f (x) = x²–4x+5

Let f(x) = y

⇒ y = x²–4x+5

⇒ y = x²–4x+4+1

⇒ y = (x-2)²+1

⇒ y-1 = (x-2)²

⇒ (x-2)²= y-1

⇒

⇒

Now, if f is real valued function then

⇒ y-1 ≥ 0

⇒ y ≥ 1

∴ the range of f is [1, ∞).

Given that, and g (x) = x + 2

Now,

gof(x) = g(f(x))

⇒

⇒ 1+2 = 3

Given that,

Now,

f(-1) = 3(-1) = -3 [since -1<1 and f(x) = 3x for x≤ 1]

f(2) = 2² = 4 [since 2<3 and f(x) = x² for 1<x≤ 3]

f(4) = 2(4) = 8 [since 4>3 and f(x) = 2x for x>3]

∴ f (– 1) + f (2) + f (4) = -3+4+8 = 9

Given that, f (x) = tan x

Let f(x) = y

⇒ y = tanx

⇒ x = tan^-1y

⇒ f^-1(y) = tan^-1y [∵ f(x) = y ⇒ x = f^-1(y) ]

⇒ f^-1(x) = tan^-1x

⇒ f^-1(1) = tan^-11

⇒

⇒

Given that, 2a + 3b = 30

⇒ 3b = 30 - 2a

⇒

Since a,b ∈ N, ‘a’ must be a multiple of 3.

Substituting a = 3 in above equation we get,

⇒

∴ for a = 3, b = 8

Similarly,

For a = 6, b = 6

For a = 9, b = 4

For a = 12,b = 2

Thus, R = {(3,8),(6,6),(9,4),(12,2)}

Given that, A = {1, 2, 3, 4, 5} and R = {(a, b) : |a² – b²| < 8}

Let us check the relation for a=1,b=2

⇒ |12 – 22| = |-3|= 3 < 8

⇒ (1,2) ∈ R

Similarly, we can check for all the ordered pairs of (a,b) which satisfies the relation.

Hence, R = {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3),(3,4),(4,3),

(4,4),(5,5)}

Given that,

f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}

Now,

gof (1) = g(f(1)) = g(2) = 3

gof (3) = g(f(3)) = g(5) = 1

gof (4) = g(f(4)) = g(1) = 3

∴ gof = {(1,3),(3,1),(4,3)}

fog (2) = f(g(2)) = f(3) = 5

fog (5) = f(g(5)) = f(1) = 2

fog (1) = f(g(1)) = f(3) = 5

∴ fog = {(2,5),(5,2),(1,5)}

Given that,

Now, (fofof)(x) = f[f(f(x))]

=

=

=

=

=

=

=

=

∴ (f o f o f) (x) =

Given that, f (x) = {4 – (x–7)³}

Let f(x) = y,

⇒ y = {4 – (x–7)³}

⇒ y – 4 = – (x–7)³

⇒ 4 – y = (x–7)³

⇒ (x–7)³ = 4 – y

⇒

⇒

⇒ [∵ f(x) = y, ⇒ x = f^-1(y)]

⇒

False

Given that, R = {(3, 1), (1, 3), (3, 3)} be a relation defined on the set A = {1, 2, 3}

Now,

R is not reflexive ∵ (1,1),(2,2) ∉ R.

R is symmetric ∵ (3,1) ∈ R ⇒ (1,3) ∈ R

R is not transitive ∵ (1,3) ∈ R and (3,1) ∈ R but (1,1) ∉ R.

False

Given that, f : R → R be the function defined by

f (x) = sin (3x+2) ∀ x ∈R

f is invertible if it is bijective that is f should be one-one and onto.

Now, we know that sin x lies between -1 and 1.

So, the range of f(x) = sin (3x+2) is [-1,1] which is not equal to its co-domain.

i.e., range of f ≠ R (co-domain)

In other words, range of f is less than co-domain, i.e there are elements in co-domain which does not have any pre-image in domain.

so, f is not onto.

Hence, f is not invertible.

False

Let R = {(1, 1), (1, 2), (2, 1), (3, 3)} be a relation defined on set A= {1,2,3}

Now,

R is not reflexive ∵ (2,2) ∉ R.

R is symmetric ∵ (1,2) ∈ R ⇒ (2,1) ∈ R

R is transitive ∵ (1,2) ∈ R and (2,1) ∈ R ⇒ (1,1) ∈ R.

Thus, It is not true that every relation which is symmetric and transitive is also reflexive.

False

Let R be the relation defined on Z by mRn if m is a integral multiple of n.

Let mRm ∈ R

⇒ m is a integral multiple of m.

Which is true since m is integral multiple of itself.

Thus, R is reflexive.

Let mRn ∈ R

⇒ m is a integral multiple of n

⇒ m= zn ∀ z ∈ Z

⇒

Since,

⇒ n is not integral multiple of m.

⇒ nRm ∉ R

Thus, R is not symmetric.

Let mRn ∈ R and nRp ∈ R

⇒ m is a integral multiple of n and n is a integral multiple of p

⇒ m is a integral multiple of p

⇒ mRp ∈ R

Thus, R is transitive.

Hence, the given statement is false.

True

Given that, f :N → A defined by f (2n–1) = 0, f (2n) = 1,∀n ∈N,

We observe that for each element of A(co-domain) there exist a pre image in N(domain).

For 0 ∈ A there exist (2n-1),∀ n ∈N

For 1 ∈ A there exist (2n),∀ n ∈N

Thus, the mapping is onto.

False

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

Now,

R is not reflexive ∵ (2,2) ∉ R

False

In order to prove composition of functions is commutative we need to show

(fog)(x) = (gof)(x)

Let us suppose, f(x) = 2x, g(x) = 1 + x²

Now,

(fog)(x) = f(g(x)) = f(1+x²)

= 2(1+x²) = 2+2x² (i)

(gof)(x) = g(f(x)) = g(2x)

= 1+(2x)² = 1+4x² (ii)

From (i) and (ii), we observe that

(fog)(x) ≠ (gof)(x)

Thus, composition of functions is not commutative.

True

In order to prove composition of functions is associative we need to show

[fo(goh)](x) = [(fog)oh](x)

Let us suppose, f(x) = x, g(x) = 2x, h(x) = x + 2

Now,

[fo(goh)](x) = f(g(h(x))) = f(g(x+2)) [∵ h(x) = x + 2 ]

= f(2(x+2)) = f(2x+4)

= 2x+4 (i)

[(fog)oh](x) = (fog)oh(x) = (fog)(h(x))

= (fog)(x+2) = f(g(x+2))

= f(2(x+2)) = f(2x+4)

= 2x+4 (ii)

From (i) and (ii), we observe that

[fo(goh)](x) = [(fog)oh](x)

Thus, composition of functions is associative.

False

A function is invertible if and only if it is one-one and onto. Hence, only bijective functions are invertible.

False

We know that, ‘e’ is called an identity element for the binary operation ‘*’ on set S, such that

a * e = e * a = a ∀ a ∈ S

Let ‘+’ be a binary operation on set N.

Now, we need to find e ∈ N such that

n + e = e + n = n ∀ n ∈ N

n + 0 = 0 + n = n ∀ n ∈ N

We observe that addition of any natural number with 0 gives the desired result but 0 ∉ N, hence 0 is not the identity element for the addition operation on N.

∴ The binary operation ‘+’ on N does not have any identity element.