Functions

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, Let, given by f(x) = x²

Check for Injectivity:

Let x,y be elements belongs to N i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ x² = y²

⇒ x² – y² = 0

⇒ (x – y)(x + y) = 0

As therefore x + y>0

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to N i.e be arbitrary, then

⇒ f(x) = y

⇒ x² = y

⇒

⇒ not belongs to N for non–perfect square value of y.

Therefore no non – perfect square value of y has a pre image in domain N.

Hence, given by f(x) = x² is One – One but not onto.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, Let, given by f(x) = x³ – x

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ x³ – x = y³ – y

⇒ x³ – y³ – (x – y) = 0

⇒ (x – y)(x² + xy + y² – 1) = 0

As x² + xy + y² ≥ 0

⇒ therefore x² + xy + y² – 1≥ – 1

⇒ x – y≠0

⇒ x ≠ y for some

Hence f is not One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒ x³ – x = y

⇒ x³ – x – y = 0

Now, we know that for 3 degree equation has a real root

So, let be that root

⇒

Thus for clearly , there exist such that f(x) = y

Therefore f is onto

⇒ Hence, given by f(x) = x³ – x is not One – One but onto

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, Let, given by f(x) = 5

As we know

A constant function is neither one – one nor onto.

So, here f(x) = 5 is constant function

Therefore

given by f(x) = 5 is neither one – one nor onto function.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, As given,

f₁ = {(1, 3), (2, 5), (3, 7)}

A = {1, 2, 3}, B = {3, 5, 7}

Thus we can see that,

Check for Injectivity:

Every element of A has a different image from B

Hence f is a One – One function

Check for Surjectivity:

Also, each element of B is an image of some element of A

Hence f is Onto.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, As given,

f₂ = {(2, a), (3, b), (4, c)}

A = {2, 3, 4}, B = {a, b, c}

Thus we can see that

Check for Injectivity:

Every element of A has a different image from B

Hence f is a One – One function

Check for Surjectivity:

Also, each element of B is an image of some element of A

Hence f is Onto.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, As given,

f₃ = {(a, x), (b, x), (c, z), (d, z)}

A = {a, b, c, d}, B = {x, y, z}

Thus we can clearly see that

Check for Injectivity:

Every element of A does not have different image from B

Since,

f₃(a) = x = f₃(b) and f₃(c) = z = f₃(d)

Therefore f is not One – One function

Check for Surjectivity:

Also each element of B is not image of any element of A

Hence f is not Onto.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, given by f(x) = x² + x + 1

Check for Injectivity:

Let x,y be elements belongs to N i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ x² + x + 1 = y² + y + 1

⇒ x² – y² + x – y = 0

⇒ ( x – y )( x + y + 1) = 0

As therefore x + y + 1>0

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

y be element belongs to N i.e be arbitrary

Since for y > 1, we do not have any pre image in domain N.

Hence, f is not Onto function.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, We have, A = {–1, 0, 1} and f = {(x, x²) : x ∈ A}.

To Prove: – f : A → A is neither One – One nor onto function

Check for Injectivity:

We can clearly see that

f(1) = 1

and f( – 1) = 1

Therefore

f(1) = f( – 1)

⇒ Every element of A does not have different image from A

Hence f is not One – One function

Check for Surjectivity:

Since, y = – 1 be element belongs to A

i.e in co – domain does not have any pre image in domain A.

Hence, f is not Onto function.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, given by f(x) = x²

Check for Injectivity:

Let x,y be elements belongs to N i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ x² = y²

⇒ x² – y² = 0

⇒ (x – y)(x + y) = 0

As therefore x + y>0

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to N i.e be arbitrary, then

⇒ f(x) = y

⇒ x² = y

⇒

⇒ not belongs to N for non–perfect square value of y.

Therefore no non – perfect square value of y has a pre–image in domain N.

Hence, f is not Onto function.

Thus, Not Bijective also.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : Z → Z given by f(x) = x²

Check for Injectivity:

Let x₁, – x₁ be elements belongs to Z i.e such that

So, from definition

⇒ x₁ ≠ – x₁

⇒ (x₁)² = ( – x₁)²

⇒ f(x₁)² = f( – x₁)²

Hence f is not One – One function

Check for Surjectivity:

Let y be element belongs to Z i.e be arbitrary, then

⇒ f(x) = y

⇒ x² = y

⇒

⇒ not belongs to Z for non–perfect square value of y.

Therefore no non – perfect square value of y has a pre–image in domain Z.

Hence, f is not Onto function.

Thus, Not Bijective also

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : N → N given by f(x) = x³

Check for Injectivity:

Let x,y be elements belongs to N i.e such that

⇒ f(x) = f(y)

⇒ x³ = y³

⇒ x³ – y³ = 0

⇒ (x – y)(x² + y² + xy) = 0

As therefore x² + y² + xy >0

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to N i.e be arbitrary, then

⇒ f(x) = y

⇒ x³ = y

⇒

⇒ not belongs to N for non–perfect cube value of y.

Since f attain only cubic number like 1,8,27….,

Therefore no non – perfect cubic values of y in N (co – domain) has a pre–image in domain N.

Hence, f is not onto function

Thus, Not Bijective also

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : Z → Z given by f(x) = x³

Check for Injectivity:

Let x,y be elements belongs to Z i.e such that

⇒ f(x) = f(y)

⇒ x³ = y³

⇒ x³ – y³ = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to Z i.e be arbitrary, then

⇒ f(x) = y

⇒ x³ = y

⇒

⇒

Since f attain only cubic number like 1,8,27….

Therefore no non – perfect cubic values of y in Z (co – domain) have a pre–image in domain Z.

Hence, f is not onto function

Thus, Not Bijective also

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : R → R, defined by f(x) = |x|

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

Case i

⇒ x = y

⇒ |x| = |y|

Case ii

⇒ – x = y

⇒ | – x| = |y|

⇒ x = |y|

Hence from case i and case ii f is not One – One function

Check for Surjectivity:

Since f attain only positive values, for negative real numbers in R

(co – domain) there is no pre–image in domain R.

Hence, f is not onto function

Thus, Not Bijective also

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : Z → Z given by f(x) = x² + x

Check for Injectivity:

Let x,y be elements belongs to Z i.e such that

⇒ f(x) = f(y)

⇒ x² + x = y² + y

⇒ x² – y² + x – y = 0

⇒ (x – y)( x + y + 1) = 0

Either (x – y) = 0 or ( x + y + 1) = 0

Case i :

If x – y = 0

⇒ x = y

Hence f is One – One function

Case ii :

If x + y + 1 = 0

⇒ x + y = – 1

⇒ x ≠ y

Hence f is not One – One function

Thus from case i and case ii f is not One – One function

Check for Surjectivity:

As

Let x be element belongs to Z i.e be arbitrary, then

⇒ f(x) = 1

⇒ x² + x = 1

⇒ x² + x – 1 = 0

⇒

Above value of x does not belong to Z

Therefore no values of x in Z (co – domain) have a pre–image in domain Z.

Hence, f is not onto function

Thus, Not Bijective also

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : Z → Z given by f(x) = x – 5

Check for Injectivity:

Let x,y be elements belongs to Z i.e such that

⇒ f(x) = f(y)

⇒ x – 5 = y – 5

⇒ x = y

Hence, f is One – One function

Check for Surjectivity:

Let y be element belongs to Z i.e be arbitrary, then

⇒ f(x) = y

⇒ x – 5 = y

⇒ x = y + 5

Above value of x belongs to Z

Therefore for each element in Z (co – domain) there exists an element in domain Z.

Hence, f is onto function

Thus, Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : R → R, defined by f(x) = sin x

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

⇒ f(x) = f(y)

⇒ sin x = sin y

⇒

⇒ x ≠ y

Hence, f is not One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒ sin x = y

⇒

Now, for y>1 x not belongs to R (Domain)

Hence, f is not onto function

Thus, It is also not Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, Let, given by f(x) = x³ + 1

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ x³ + 1 = y³ + 1

⇒ x³ = y³

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒ x³ + 1 = y

Now, we know that for 3 degree equation has a real root

So, let be that root

⇒

Thus for clearly , there exist such that f(x) = y

Therefore f is onto

Thus, It is also Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, Let, given by f(x) = x³ + x

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ x³ – x = y³ – y

⇒ x³ – y³ – (x – y) = 0

⇒ (x – y)(x² + xy + y² – 1) = 0

Hence f is not One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒ x³ – x = y

⇒ x³ – x – y = 0

Now, we know that for 3 degree equation has a real root

So, let be that root

⇒

Thus for clearly , there exist such that f(x) = y

Therefore f is onto

Thus, It is not Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : R → R, defined by f(x) = sin²x + cos²x

Check for Injectivity and Check for Surjectivity

Let x be element belongs to R i.e such that

So, from definition

⇒ f(x) = sin²x + cos²x

⇒ f(x) = sin²x + cos²x

⇒ f(x) = 1

⇒ f(x) = constant

We know that a constant function is neither One – One function nor onto function.

Thus, It is not Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : R → R given by

Check for Injectivity:

Let x,y be elements belongs to Q i.e such that

⇒ f(x) = f(y)

⇒

⇒ (2x + 3)(y – 3) = (2y + 3)(x – 3)

⇒ 2xy – 6x + 3y – 9 = 2xy – 6y + 3x – 9

⇒ – 6x + 3y = – 6y + 3x

⇒ – 6x + 3y + 6y – 3x = 0

⇒ – 9x + 9y = 0

⇒ x = y

Thus, f is One – One function

Check for Surjectivity:

Let y be element belongs to Q i.e be arbitrary, then

⇒ f(x) = y

⇒

⇒ 2x + 3 = y (x – 3)

⇒ 2x + 3 = xy – 3y

⇒ 2x – xy = – 3(y + 1)

⇒

Above value of x belongs to Q – [3] for y = 2

Therefore for each element in Q – [3] (co – domain), there does not exist an element in domain Q.

Hence, f is not onto function

Thus, Not Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : Q → Q, defined by f(x) = x³ + 1

Check for Injectivity:

Let x,y be elements belongs to Q i.e such that

⇒ f(x) = f(y)

⇒ x³ + 1 = y³ + 1

⇒ x³ = y³

⇒ x = y

Hence, f is One – One function

Check for Surjectivity:

Let y be element belongs to Q i.e be arbitrary, then

⇒ x³ + 1 = y

⇒ x³ + 1 – y = 0

Now, we know that for 3 degree equation has a real root

So, let be that root

⇒

⇒

Thus for clearly , there exist such that f(x) = y

Therefore f is onto

Thus, It is a Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : R → R, defined by f(x) = 5x³ + 4

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

⇒ f(x) = f(y)

⇒ 5x³ + 4 = 5y³ + 4

⇒ x³ = y³

⇒ x = y

Hence, f is One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ 5x³ + 4 = y

⇒ 5x³ + 4 – y = 0

Now, we know that for 3 degree equation has a real root

So, let be that root

⇒

Thus for clearly , there exist such that f(x) = y

Therefore f is onto

Thus, It is a Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : R → R given by f(x) = 3 – 4x

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

⇒ f(x) = f(y)

⇒ 3 – 4x = 3 – 4y

⇒ x = y

Hence, f is One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒ 3 – 4x = y

⇒

Above value of x belongs to R

Therefore for each element in R (co – domain), there exists an element in domain R.

Hence, f is onto function

Thus, Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, given by f(x) = 1 + x²

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ x² + 1 = y² + 1

⇒ x² = y²

⇒ ±x = ±y

Therefore, either x = y or x = – y or x ≠ y

Hence f is not One – One function

Check for Surjectivity:

1 be element belongs to R i.e be arbitrary, then

⇒ f(x) = 1

⇒ x² + x = 1

⇒ x² + x – 1 = 0

⇒

Above value of x not belongs to R for y < 1

Therefore f is not onto

Thus, It is also not Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f: R → R given by

Check for Injectivity:

Let x,y be elements belongs to R i.e. such that

⇒ f(x) = f(y)

⇒

⇒ xy² + x = yx² + y

⇒ xy² + x – yx² – y = 0

⇒ xy (y – x) + (x – y) = 0

⇒ (x – y)(1 – xy) = 0

Case i :

⇒ x – y = 0

⇒ x = y

f is One – One function

Case ii :

⇒ 1 – xy = 0

⇒ xy = 1

Thus from case i and case ii f is One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒

⇒ x = x²y + y

⇒ x – x²y = y

Above value of x belongs to R

Therefore for each element in R (co – domain) there exists an element in domain R.

Hence, f is onto function

Thus, Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Here, Range {f} = {a}

Since it is injective map, different elements have different images.

Thus A has only one element

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f: R → R given by

To Prove: – is a bijection

Check for Injectivity:

Let x,y be elements belongs to R i.e. such that

⇒ f(x) = f(y)

⇒

⇒ (x – 2)(y – 3) = (x – 3)(y – 2)

⇒ xy – 3x – 2y + 6 = xy – 2x – 3y + 6

⇒ – 3x – 2y + 2x + 3y = 0

⇒ – x + y = 0

⇒ x = y

Hence, f is One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒

⇒ x – 2 = xy – 3y

⇒ x – xy = 2 – 3y

⇒

is a real number for all y ≠ 1.

Also, for any y

Therefore for each element in R (co – domain), there exists an element in domain R.

Hence, f is onto function

Thus, Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, here f: A → A: A = [–1, 1] given by function is

Check for Injectivity:

Let x, y be elements belongs to A i.e. such that

⇒ f(x) = f(y)

⇒

⇒ 2x = 2y

⇒ x = y

1 belongs to A then

Not element of A co – domain

Hence, f is not One – One function

Check for Surjectivity:

Let y be element belongs to A i.e be arbitrary, then

⇒ f(x) = y

⇒

⇒ x = 2y

Now,

1 belongs to A

⇒ x = 2, which not belong to A co – domain

Hence, f is not onto function

Thus, It is not Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, here f : A → A : A = [–1, 1] given by function is g(x) = |x|

Check for Injectivity:

Let x, y be elements belongs to A i.e such that

⇒ g(x) = g(y)

⇒ |x| = |y|

⇒ x = y

1 belongs to A then

⇒ g(1) = 1 = g( – 1)

Since, it has many element of A co – domain

Hence, g is not One – One function

Check for Surjectivity:

Let y be element belongs to A i.e be arbitrary, then

⇒ f(x) = y

⇒

⇒ x = 2y

Now,

1 belongs to A

⇒ x = 2, which not belong to A co – domain

Since g attain only positive values, for negative – 1 in A (co – domain) there is no pre–image in domain A.

Hence, g is not onto function

Thus, It is not Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, here f : A → A : A = [–1, 1] given by function is h(x) = x²

Check for Injectivity:

Let x, y be elements belongs to A i.e. such that

⇒ h(x) = h(y)

⇒ x² = y²

⇒ ±x = ±y

Since it has many elements of A co – domain

Hence, h is not One – One function

Check for Surjectivity:

Let y be element belongs to A i.e. be arbitrary, then

⇒ h(x) = y

⇒ x² = y

⇒ x = ±√y

Since h have no pre–image in domain A.

Hence, h is not onto function

Thus, It is not Bijective function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Here, It is given (x, y): x is a person, y is the mother of x

As we know each person “x” has only one biological mother

Thus,

Given relation is a function

Since more than one person may have the same mother

Function, not One – One (injective) but Onto (Surjective)

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Here, It is given (a, b): a is a person, b is an ancestor of a

As we know any person “a” has more than one ancestor

Thus,

Given relation is not a function

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

We have A = {1, 2, 3}

So all one – one functions from A = {1, 2, 3} to itself are obtained by re – arranging elements of A.

Thus all possible one – one functions are:

f(1) = 1, f(2) = 2, f(3) = 3

f(1) = 2, f(2) = 3, f(3) = 1

f(1) = 3, f(2) = 1, f(3) = 2

f(1) = 1, f(2) = 3, f(3) = 2

f(1) = 3, f(2) = 2, f(3) = 1

f(1) = 2, f(2) = 1, f(3) = 3

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, f : R → R, defined by f(x) = 4x³ + 7

To Prove : – f : R → R is bijective defined by f(x) = 4x³ + 7

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

⇒ f(x) = f(y)

⇒ 4x³ + 7 = 4y³ + 7

⇒ x³ = y³

⇒ x = y

Hence, f is One – One function

Check for Surjectivity:

Let y be element belongs to R i.e be arbitrary, then

⇒ f(x) = y

⇒ 4x³ + 7 = y

⇒ 4x³ + 7 – y = 0

Now, we know that for 3 degree equation has a real root

So, let be that root

⇒

⇒

Thus for clearly , there exist such that f(x) = y

Therefore f is onto

Thus, It is Bijective function

Hence Proved

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each, there exists such that f(a) = b

Now, given by f(x) = e^x

Check for Injectivity:

Let x,y be elements belongs to R i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ e^x = e^y

⇒

⇒ e^{x – y} = 1

⇒ e^{x – y} = e⁰

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Here range of f = (0,∞) ≠ R

Therefore f is not onto

Now if co – domain is replaced by R₀⁺ (set of all positive real numbers) i.e (0,∞) then f becomes an onto function.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

To Prove : – Logarithmic function f : R _{+ +} → R given by f(x) = log_a x, a > 0 is a bijection.

Now, f : R₀⁺ → R given by f(x) = log_a x, a > 0

Check for Injectivity:

Let x,y be elements belongs to R₀⁺ i.e such that

So, from definition

⇒ f(x) = f(y)

⇒ log_a x = log_a y

⇒ log_a x – log_a y = 0

⇒

⇒

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to R i.e. be arbitrary, then

⇒ f(x) = y

⇒ log_a x = y

⇒ x = a^y

Above value of x belongs to R₀⁺

Therefore, for all there exist x = a^y such that f(x) = y .

Hence, f is Onto function.

Thus, it is Bijective also

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, f: A → A where A = {1, 2, 3} and its a One – One function

To Prove: – A is Onto function

Since it is given that f is a One – One function,

Three elements of A = {1, 2, 3} must be taken to 3 different elements of co – domain A = {1, 2, 3} under f.

Thus by definition of Onto Function

f has to be Onto function.

Hence Proved

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, f : A → A where A = {1, 2, 3} and its an Onto function

To Prove: – A is a One – One function

Let's assume f is not Onto function,

Then,

There must be two elements let it be 1 and 2 in Domain A = {1, 2, 3} whose images in co–domain A = {1, 2, 3} is same.

Also, Image of 3 under f can be only one element.

Therefore,

Range set can have at most two elements in co – domain A = {1, 2, 3}

⇒ f is not an onto function

Hence it contradicts

⇒ f must be One – One function

Hence Proved

TIP: –

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, f : A → A where A = {1, 2, 3,….,n}

All onto function

It’s a permutation of n symbols 1,2,3,….n

Thus,

Total number of Onto maps from A = {1, 2, 3, …., n} to itself =

Total number of permutations of n symbols 1,2,3,….n.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

a = b for all

Let, f₁: R → R and f₂: R → R be two functions given by (Examples)

f₁(x) = x

f₁(x) = – x

From above function it is clear that both are One – One functions

Now,

⇒ (f₁ + f₂)(x) = f₁(x) + f₂(x)

⇒ (f₁ + f₂)(x) = x – x

⇒ (f₁ + f₂)(x) = 0

Therefore,

f₁ + f₂ : R → R is a function given by

(f₁ + f₂)(x) = 0

Since f₁ + f₂ is a constant function,

Hence it is not an One – One function.

TIP: –

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Let, f₁: Z → Z and f₂: Z → Z be two functions given by (Examples)

f₁(x) = x

f₁(x) = – x

From above function it is clear that both are Onto or Surjective functions

Now,

f₁ + f₂ : Z → Z

⇒ (f₁ + f₂)(x) = f₁(x) + f₂(x)

⇒ (f₁ + f₂)(x) = x – x

⇒ (f₁ + f₂)(x) = 0

Therefore,

f₁ + f₂ : Z → Z is a function given by

(f₁ + f₂)(x) = 0

Since f₁ + f₂ is a constant function,

Hence it is not an Onto/Surjective function.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

a = b for all

Let, f₁: R → R and f₂: R → R are two functions given by

f₁(x) = x

f₂(x) = x

From above function it is clear that both are One – One functions

Now, f₁×f₂ : R → Rgiven by

⇒ (f₁×f₂ )(x) = f₁(x)×f₂(x) = x²

⇒ (f₁×f₂ )(x) = x²

Also,

f(1) = 1 = f( – 1)

Therefore,

f is not One – One

⇒ f₁×f₂ : R → R is not One – One function.

Hence Proved

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

a = b for all

Let, f₁: R → R and f₂: R → R are two non – zero functions given by

f₁(x) = x³

f₁(x) = x

From above function it is clear that both are One – One functions

Now, given by

⇒

⇒

Again,

defined by

f(x) = x²

Now,

⇒ f(1) = 1 = f( – 1)

Therefore,

f is not One – One

⇒ is not One – One function.

Hence it is not necessarily to be one – one function.

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Now, f : A → B, denotes a mapping such that

⇒ f = {(x,y):y = x + 3}

It can be written as follows in roster form

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

Hence this is injective mapping

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Now, f : A → B, denotes a mapping such that

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

Hence this is not injective mapping

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Now, f : A → B, denotes a mapping such that

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

Here it is clear that every first component is from B and second component is from A

Hence this is mapping from B to A

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Now, f : A → A given by f(x) = x – [x]

To Prove: – f(x) = x – [x], is neither one – one nor onto

Check for Injectivity:

Let x be element belongs to Z i.e such that

So, from definition

⇒ f(x) = x – [x]

⇒ f(x) = 0 for

Therefore,

Range of f = [0,1] ≠ R

Hence f is not One – One function

Check for Surjectivity:

Since Range of f = [0,1] ≠ R

Hence, f is not Onto function.

Thus, it is neither One – One nor Onto function

Hence Proved

TIP: – One – One Function: – A function is said to be a one – one functions or an injection if different elements of A have different images in B.

So, is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all

⇔ f(a) = f(b)

⇒ a = b for all

Onto Function: – A function is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, is Surjection iff for each , there exists such that f(a) = b

Bijection Function: – A function is said to be a bijection function if it is one – one as well as onto function.

Now, suppose

f(n₁) = f(n₂)

If n₁ is odd and n₂ is even, then we have

⇒ n₁ + 1 = n₂ – 2

⇒ n₂ – n₁ = 2

Not possible

Suppose both n₁ even and n₂ is odd.

Then, f(n₁) = f(n₂)

⇒ n₁ – 1 = n₂ + 1

⇒ n₁ – n₂ = 2

Not possible

Therefore, both n₁ and n₂ must be either odd or even

Suppose both n₁ and n₂ are odd.

Then, f(n₁) = f(n₂)

⇒ n₁ + 1 = n₂ + 1

⇒ n₁ = n₂

Suppose both n₁ and n₂ are even.

Then, f(n₁) = f(n₂)

⇒ n₁ – 1 = n₂ – 1

⇒ n₁ = n₂

Then, f is One – One

Also, any odd number 2r + 1 in the co – domain N will have an even number as image in domain N which is

⇒ f(n) = 2r + 1

⇒ n – 1 = 2r + 1

⇒ n = 2r + 2

Any even number 2r in the co – domain N will have an odd number as image in domain N which is

⇒ f(n) = 2r

⇒ n + 1 = 2r

⇒ n = 2r – 1

Thus f is Onto function.

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

Now, f(x) = 2x + 3 and g(x) = x² + 5

gof(x) = g(2x + 3) = (2x + 3)² + 5

⇒ gof(x) = 4x² + 12x + 9 + 5 = 4x² + 12x + 14

fog (x) = f(g(x)) = f (x² + 5) = 2 (x² + 5) + 3

⇒ fog(x)= 2x² + 10 + 3 = 2x² + 13

Hence, gof(x) = 4x² + 12x + 14 and fog (x) = 2x² + 13

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f(x) = 2x + x² and g(x)=x³

Now, gof(x) = g(f (x)) =g(2x + x²)

gof (x)=(2x + x²)³ = x⁶ + 8x³ + 6x⁵ + 12x⁴

and fog(x)=f(g(x))= f(x³)

⇒ fog(x) = 2x³ + x⁶

So, gof(x) = x⁶ + 6x⁵ + 12x⁴ + 8x³ and fog(x) = 2x³ + x⁶

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f(x)=x² + 8 and g(x)=3x³ + 1

So, gof(x)= g(f(x))

gof(x)= g(x² + 8)

gof(x)= 3(x² + 8)³ + 1

⇒ gof(x)= 3(x⁶ + 512 + 24x⁴ + 192x²) + 1

⇒ gof(x)= 3x⁶ + 72x⁴ + 576x² + 1537

Similarly, fog(x)=f(g(x))

⇒ fog(x)= f(3x³ + 1)

⇒ fog(x)=(3x³ + 1)² + 8

⇒ fog(x)=(9x⁶ + 1 + 6x³) + 8

⇒ fog(x)=9x⁶ + 6x³ + 9

So, gof(x) = 3x⁶ + 72x⁴ + 576x² + 1537 and fog(x) = 9x⁶ + 6x³ + 9

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f (x) = x and g (x) = |x|

Now, gof(x)=g(f(x) =g(x)

⇒ gof(x) =|x|

and, fog(x) = f(g(x)) = f (|x|)
⇒ fog(x)=|x|

Hence, gof(x) = fog(x) = |x|

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f(x) = x² + 2x – 3 and g(x) = 3x – 4

Now, gof(x)=g(f(x))= g(x² + 2x – 3)

gof(x) = 3(x² + 2x–3) – 4

⇒ gof(x)= 3x² + 6x – 9 – 4

⇒ gof(x) = 3x² + 6x – 13

and, fog= f(g(x)) = f(3x – 4)

fog(x) = (3x – 4)² + 2(3x – 4) – 3

= 9x² + 16 – 24x + 6x – 8 – 3

∴ fog(x) = 9x² – 18x + 5

Thus, gof(x) = 3x² + 6x – 13 and fog(x) = 9x² – 18x + 5

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f(x) = 8x³ and

Now, gof(x) = g(f(x)) = g(8x³)

gof(x) = 2x

fog(x) = 8x

Thus, gof(x) = 2x and fog(x) = 8x

Let f = {(3,1), (9,3), (12,4)} and

g = {(1,3), (3,3), (4,9), (5,9)}

Now,

range of f = (1, 3, 4}

domain of f = {3, 9, 12}

range of g = {3,9}

domain of g = (1, 3, 4, 5}

since, range of f ⊂ domain of g

∴ gof is well defined.

Again, the range of g ⊆ domain of f

∴ fog in well defined.

Finally, gof = {(3,3), (9 ,3), (12,9)}

fog = {(1,1) , (3,1), (4,3), (5,3)}

We have,

f = {(1, – 1), (4, – 2) , (9, – 3), (16,4)} and

g = {(– 1, – 2), (– 2, – 4), (– 3, – 6), (4,8)}

Now,

Domain of f = {1,4,9,16}

Range of f = {– 1, – 2, – 3, 4}

Domain of g = (– 1, – 2, – 3,4}

Range of g = (– 2, – 4, – 6, 8}

Clearly range of f = domain of g

∴ gof is defined.

but, range of g ≠ domain of f
So, fog is not defined.

Now,

gof(1) = g(– 1)= – 2

gof(4) = g(– 2) = – 4

gof(9) = g (– 3) = – 6

gof(16) = g(4)= 8

So, gof = {(1, – 2), (4, – 4), (9 , – 6), (16,8)}

Given, A = {a, b, c}, B = {u, v, w} and

f = A → B and g: B → A defined by

f = {(a, v), (b, u), (c, w)} and

g = {(u, b), (v, a), (w, c)}

For both f and g, different elements of domain have different

images

∴ f and g are one – one

Again, for each element in co – domain of f and g, there is a pre – image in the domain

∴ f and g are onto

Thus, f and g are bijective.

Now,

gof = {(a, a), (b, b), (c, c^.)} and

fog = {(u, u), (v, v), (w, w)}

We have, f: R → R given by f(x) = x² + 8 and

g : R → R given by g (x) = 3x³ + 1

fog(x) = f (g(x)) = f (3x³ + 1)

= (3x³ + 1)² + 8

fog(2) = (3 × 8 + 1)^{2 +} 8 = 625 + 8 = 633

Again,

gof(x) = g(f(x)) = g(x² + 8)

= 3(x² + 8)³ + 1

gof(1) = 3(1 + 8)^{3 +} 1 = 2188

We have, f : R⁺ → R⁺ given by

f (x) = x²

g: R ⁺ → R ⁺ given by

fog (x) = f(g(x)) = f(= ()² = x

Also,

gof (x) = g(f(x)) = g(x²) = = x

Thus,

fog(x)= gof (x)

They are equal functions as their domain and range are also equal.

We have, f: R → R and g: R → R are two functions defined by

f (x)= x² and g(x) = x + 1

Now,

fog (x) = f(g(x)) = f(x + 1) =(x + 1)²

⇒ fog(x) = x² + 2x + 1 ……(i)

gof(x) = g(f(x)) = g(x²) = x² + 1 ……(ii)

from (i) & (ii)

fog ≠ gof

Let f: R → R and g: R → R are defined as

f (x) = x + 1 and g (x) = x – 1

Now,

fog(x) = f(g(x)) = f(x – 1) = x – 1 + 1

= x =I_R ……(i)

Again,

fog(x ) = f(g(x)) = g(x + 1) = x + 1 – 1

= x = I_R ……(ii)

from (i)& (ii)

fog = gof = I_R

We have, f: N → Z_o, g: Z₀ → Q and h: Q → R

Also, f(x) = 2x, and h(x) = e^x

Now, f: N → Z_o and hog: Z₀ → R

∴ (hog)of: N → R

Also, gof: N → Q and h: Q → R

∴ ho(gof): N → R

Thus, (hog)of and ho(gof) exist and are function from N to set R.

Finally. (hog)of(x) = (hog)(f(x)) = (hog)(2x)

Now, ho(gof)(x) = ho(g(2x)) = h

Hence, associativity verified.

We have,

ho(gof)(x)=h(gof(x))=h(g(f(x)))

= h(g(2x)) = h(3(2x) + 4)

= h(6x + 4) = sin(6x + 4) ∀x ∈N

((hog)of)(x) = (hog)(f(x))= (hog)(2x)

=h(g(2x))=h(3(2x) + 4)

=h(6x + 4) = sin(6x + 4) ∀x ∈N

This shows, ho(gof) = (hog)of

Define f:N → N by, f(x) = x + 1 And, g: N → N by,

We first show that f is not onto.

For this, consider element 1 in co – domain N. It is clear that this element is not an image of any of the elements in domain N.

Therefore, f is not onto.

Define f: N → Z as f(x) = x and g: N → N as g(x)=|x|.

We first show that g is not injective.

It can be observed that:

g(– 1)=| – 1| = 1

g(1) =|1| = 1

Therefore, g(– 1) = g(1), but —1 ≠ 1.

Therefore, g is not injective.

Now, gof: N → Z is defined as gof(x) = g(f(x)) =g(x)=|x|.

Let x, y ∈N such that gof(x) = gof(y).

⇒ |x|=|y|

Since x and y ∈N both are positive.

∴ |x|=|y| ⇒ x=y

Hence, gof is injective

We have, f : A → B and g : B → C are one – one functions.

Now we have to prove : gof: A → C in one – one

let x, y ∈ A such that

gof(x) = gof(y)

g(f(x))=g(f(y))

f (x) = f (y) [As, g in one – one]

x = y [As, f in one – one]

gof is one – one function

We have, f: A → B and g: B → C are onto functions.

Now, we need to prove: gof: A → C is onto.

let y ∈ C, then

gof (x) = y

g(f(x)) = y ……(i)

Since g is onto, for each element in C, there exists a preimage in B.

g(x)=y ……(ii)

From (i) & (ii)

f(x)=x

Since f is onto, for each element in B there exists a preim age in el

f(x)=x ……(iii)

From (ii)and(iii) we can conclude that for each y ∈ C, there exists a preimage in A such that gof(x) = y

∴ gof is onto.

f(x) = e^x and g (x) = log_e x

Now, fog(x) = f(g(x)) = f(log_e x) = e^log_e^x= x

⇒ fog(x) = x

gof(x) = g(f(x)) = g(e^x) = log_e e^x = x

⇒ gof(x)=x

Hence, fog(x) = x and gof(x) = x

f(x)= x, g (x) = cos x

Domain of f and Domain of g = R

Range of f = (0, ∞)

Range of g = (– 1,1)

∴ Range of f ⊂ domain of g ⇒ gof exist

Also, Range of g ⊂ domain of f ⇒ fog exist

Now,

gof (x) = g (f (x)) = g (x²) = cos x²

And

fog (x) = f(g(x)) = f (cos x) = cos² x

Hence, fog(x) = cos x² and gof(x) = cos² x

f(x)= |x| and g (x)= sin x

Range of f = (0, ∞) ⊂ Domain g (R) ⇒ gof exist

Range of g= [ – 1,1] ⊂ Domain f (R) ⇒ fog exist

Now, fog (x)= f(g(x)) = f(sin x) = |sin x| and

gof(x) = g(f(x)) = g(lxl) =sin |x|

Hence, fog(x) = |sin x| and gof(x) = sin |x|

f(x) = x + 1 and g(x) = e^x

Range of f = R ⊂ Domain of g= R ⇒ gof exist

Range of g = (0, ∞) = Domain of f = R ⇒ fog exist

Now,

gof(x) = g(f(x)) = g(x + 1) = e^{x+ 1}

And

fog (x) = f(g(x))= f(e^x) = e^x + 1

Hence, fog(x) = e^x+ 1 and gof(x) = e^x + 1

f(x) = sin ^{– 1} x and g (x) = x²

Range of f= Domain of g = R ⇒ gof exist

Range of g= (0,∞) ⊂ Domain of f = R ⇒ fog exist

Now,

fog (x) = f(g(x)) = f(x²) = sin ^{– 1}x² and

gof(x) = g(f(x)) = g (sin ^{– 1} x) = (sin ^{– 1}x)²

Hence, fog(x) = sin ^{– 1}x² and gof(x) = (sin ^{– 1}x)²

f(x) = x + 1 and g(x) = sin x

Range of f = R ⊂ Domain of g = R ⇒ gof exists

Range of g= [ – 1,1] ⊂ Domain of f ⇒ fog exists

Now,

fog(x) = f(g(x)) = f(sin x) = sin x + 1

And

gof(x) = g(f(x)) = g(x + 1) = sin(x + 1)

Hence, fog(x) = sin x + 1 and gof(x) = sin(x + 1)

f(x) = x + 1 and g(x) = 2x + 3

Range of f = R ⊂ Domain of g = R ⇒ gof exists

Range of g= R ⊂ Domain of f ⇒ fog exists

Now,

fog(x) = f(g(x) = f(2x + 3) = (2x + 3) + 1= 2x + 4 and

gof(x)= g(f(x))= g(x + 1) = 2(x + 1) + 3 = 2x + 5

So, fog(x) = 2x + 4 and gof(x) = 2x + 5

f(x) = c, c ∈ R and

g(x) = sin x²

Range of f = R ⊂ Domain of g = R ⇒ gof exists

Range of g= [ – 1,1] ⊂ Domain of f =R ⇒ fog exists

Now,

gof(x) = g(f(x)) = g(c) = sin c² and

fog(x) = f (g(x))= f(sin x²) =c

Thus, gof(x) = sin c² and fog(x) = c

f(x) = x² + 1 and

Range of f = (2,∞) ⊂ Domain of g = R ⇒ gof exists

Range of g= R – [ – 1] ⊂ Domain of f =R ⇒ fog exists

Now,

and

Hence,

We have, f (x) = x² + x + 1 and g(x) = sin x

Now,

fog(x) = f(g(x)) = f(sin x)

⇒ fog(x) = sin² x + sin x + 1

Again, gof(x) = g(f(x)) = g (x² + x + 1)

⇒ gof(x) = sin(x² + x + 1)

Clearly,

fog ≠ gof

We have, f(x) = |x|

We assume the domain of f = R and range of f = (0,∞)

Range of f ⊂ domain of f

∴ fof exists,

Now,

fof(x) = f(f(x)) = f(|x|) = ||x|| = f(x)

∴ fof = f

Hence proved.

f(x)= 2x + 5 and g(x)= x^{2 +} 1

The range of f = R and range of g = [1,∞]

The range of f ⊂ Domain of g (R) and range of g ⊂ domain of f (R)

∴ both fog and gof exist.

(i) fog(x) = f(g(x)) = f (x² + 1)

= 2(x² + 1) + 5

⇒ fog(x)=2x² + 7

Hence fog(x) = 2x² + 7

(ii) gof(x) = g(f(x)) ^– = g (2x + 5)

= (2x + 5)² + 1

gof(x)= 4x² + 20x + 26

Hence gof(x) = 4x² + 20x + 26

(iii) fof(x) = f(f(x)) = f(2x + 5)

= 2 (2x + 5) + 5

fof(x) = 4x + 15

Hence fof(x) = 4x + 15

(iv) f²(x) = [f(x)]²= (2x + 5)²

= 4x² + 20x + 25

∴ from (iii) and (iv)

fof ≠ f²

We have, f (x) = sin x and g (x) = 2x.

Domain of f and g is R

Range of f = [ – 1,1], Range of g = R

∴ Range of f ⊂ Domain g and Range of g ⊂ Domain f

fog and gof both exist.

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

⇒ gof(x) = 2sin x

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

∴ gof ≠ fog

f, g and h are real functions given by f(x) = sin x, g(x) = 2x and

h(x) = cos x

To prove: fog=go(fh)

L.H.S

fog(x) = f(g(x))

= f(2x) = sin 2x

⇒ fog(x)=2sin x cos x ……(A)

R.H .S

go(fh)(x) = go(f(x).h(x))

= g(sin x cos x) = 2sin x cos x

go(fh)(x) = 2 sin x cos x ……(B)

from A and B

fog(x) = go(fh)(x)

Hence proved

We are given that f is a real function and g is a function given by

g(x) = 2x

To prove; gof=f + f.

L.H.S

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

=f + f = R.H.S

gof=f + f

Hence proved

f(x) = , g(x) = log _e x

Domain of f and g are R.

Range of f= (– ∞, 1) Range of g = (0, e)

Range of f ⊂ Domain of g ⇒ gof exists

Range of g ⊂ Domain f ⇒ fog exists

∴ gof (x) = g(f(x)) = g

∴ gof (x) =

Again

fog (x) = f(g(x)) = f(log _e x)

fog (x) =

f: and g: [ – 1,1]R defined as f(x) = tan x and g(x) =

Range of f: let y = f(x)

⇒ y = tan x

⇒ x = tan ^{– 1} y

Since, x ϵ , y ϵ (– ∞, ∞)

As Range of f ⊂ Domain of g

∴ gof exists.

Similarly, let y = g(x)

⇒ y =

⇒ x =

∴ Range of g is [ – 1,1]

As, Range of g ⊂ Domain of f

Hence, fog also exists

Now,

fog(x) = f(g(x)) = f

⇒ fog(x) = tan

Again,

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

⇒ gof(x) =

f(x) = , g(x) = x² + 1

Now,

Domain of f = [ – 3, ∞], domain of g = (– ∞, ∞)

Range of f = [0, ∞), range of g = [1, ∞)

Then, range of f ⊂ Domain of g and range of g ⊂ Domain of f

Hence, fog and gof exists

Now,

fog(x) = f(g(x)) = f(x² + 1)

⇒ fog(x) =

Again,

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

⇒ gof(x) =

⇒ gof(x) = x + 4

We have, f(x) =

Clearly, domain of f = [2, ∞] and range of f = [0, ∞)

We observe that range of f is not a subset of domain of f

∴ Domain of (fof) = {x: x ϵ Domain of f and f(x) ϵ Domain of f}

= {x: x ϵ [2, ∞) and ϵ [2, ∞)}

= {x: x ϵ [2, ∞) and ≥ 2}

= {x: x ϵ [2, ∞) and x – 2 ≥ 4}

= {x: x ϵ [2, ∞) and x ≥ 6}

= [6, ∞)

Now,

fof(x) = f(f(x)) = f =

We have, f(x) =

Clearly, domain of f = [2, ∞] and range of f = [0, ∞)

We observe that range of f is not a subset of domain of f

∴ Domain of (fof) = {x: x ϵ Domain of f and f(x) ϵ Domain of f}

= {x: x ϵ [2, ∞) and ϵ [2, ∞)}

= {x: x ϵ [2, ∞) and ≥ 2}

= {x: x ϵ [2, ∞) and x – 2 ≥ 4}

= {x: x ϵ [2, ∞) and x ≥ 6}

= [6, ∞)

Clearly, range of f = [0, ∞) ⊄ Domain of (fof)

∴ Domain of ((fof)of) = {x: x ϵ Domain of f and f(x) ϵ Domain of (fof)}

= {x: x ϵ [2, ∞) and ϵ [6, ∞)}

= {x: x ϵ [2, ∞) and ≥ 6}

= {x: x ϵ [2, ∞) and x – 2 ≥ 36}

= {x: x ϵ [2, ∞) and x ≥ 38}

= [38, ∞)

Now,

(fof)(x) = f(f(x)) = f =

(fofof)(x) = (fof)(f(x)) = (fof) =

∴ fofof : [38, ∞) → R defined as

(fofof)(x) =

We have, f(x) =

Clearly, domain of f = [2, ∞] and range of f = [0, ∞)

We observe that range of f is not a subset of domain of f

∴ Domain of (fof) = {x: x ϵ Domain of f and f(x) ϵ Domain of f}

= {x: x ϵ [2, ∞) and ϵ [2, ∞)}

= {x: x ϵ [2, ∞) and ≥ 2}

= {x: x ϵ [2, ∞) and x – 2 ≥ 4}

= {x: x ϵ [2, ∞) and x ≥ 6}

= [6, ∞)

Clearly, range of f = [0, ∞) ⊄ Domain of (fof)

∴ Domain of ((fof)of) = {x: x ϵ Domain of f and f(x) ϵ Domain of (fof)}

= {x: x ϵ [2, ∞) and ϵ [6, ∞)}

= {x: x ϵ [2, ∞) and ≥ 6}

= {x: x ϵ [2, ∞) and x – 2 ≥ 36}

= {x: x ϵ [2, ∞) and x ≥ 38}

= [38, ∞)

Now,

(fof)(x) = f(f(x)) = f =

(fofof)(x) = (fof)(f(x)) = (fof) =

∴ fofof : [38, ∞) → R defined as

(fof)(x) = f(f(x)) = f =

(fofof)(x) = (fof)(f(x)) = (fof) =

∴ fofof : [38, ∞) → R defined as

(fofof)(x) =

(fofof)(38) =

=

We have, f(x) =

Clearly, domain of f = [2, ∞] and range of f = [0, ∞)

We observe that range of f is not a subset of domain of f

∴ Domain of (fof) = {x: x ϵ Domain of f and f(x) ϵ Domain of f}

= {x: x ϵ [2, ∞) and ϵ [2, ∞)}

= {x: x ϵ [2, ∞) and ≥ 2}

= {x: x ϵ [2, ∞) and x – 2 ≥ 4}

= {x: x ϵ [2, ∞) and x ≥ 6}

= [6, ∞)

Now,

(fof)(x) = f(f(x)) = f =

∴ fof: [6, ∞) → R defined as

(fof)(x) =

f²(x) = [f(x)]² = = x – 2

∴ f²: [2, ∞) → R defined as

f²(x) = x – 2

∴ fof ≠ f²

Range of f = [0, 3] ⊂ Domain of f

∴ fof(x) = f(f(x)) = f = f

So, fof(x) =

Domain of f(x) and g(x) is R.

Range of f(x) = [0, ∞) and range of g(x) = [0, ∞)

As, range of f ⊂ Domain of g and range of g ⊂ Domain of f

So, gof and fog exists

Now,

fog(x) = f(g(x)) = f(|x|–x)

⇒ fog(x) = ||x|–x| + |x|–x

As, range of g(x) ≥ 0 so, ||x|–x| = |x|–x

So, fog(x) = ||x|–x| + |x|–x = |x|–x + |x|–x

⇒ fog(x) = 2(|x|–x)

Also,

gof(x) = g(f(x)) = g(|x| + x) = ||x| + x| – (|x| + x)

As, range of f(x) ≥ 0 so, ||x| + x| = |x| + x

So, gof(x) = ||x| + x| – (|x| + x) = |x| + x – (|x| + x) = 0

Thus, gof(x) = 0

Now, fog(– 3) = 2(| – 3|–(– 3)) = 2(3 + 3) = 6,

fog(5) = 2(|5| – 5) = 0, gof(– 2) =0

(i) f : [1, 2, 3, 4] → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

Recall that a function is invertible only when it is both one-one and onto.

Here, we have f(1) = 10 = f(2) = f(3) = f(4)

Hence, f is not one-one.

Thus, the function f does not have an inverse.

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

Recall that a function is invertible only when it is both one-one and onto.

Here, we have g(5) = 4 = g(7)

Hence, g is not one-one.

Thus, the function g does not have an inverse.

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Recall that a function is invertible only when it is both one-one and onto.

Here, observe that distinct elements of the domain {2, 3, 4, 5} are mapped to distinct elements of the co-domain {7, 9, 11, 13}.

Hence, h is one-one.

Also, each element of the range {7, 9, 11, 13} is the image of some element of {2, 3, 4, 5}.

Hence, h is also onto.

Thus, the function h has an inverse.

(i) A = {0, –1, –3, 2}; B = {–9, –3, 0, 6} & f(x) = 3x

We have f : A → B and f(x) = 3x.

⇒ f = {(0, 3×0), (–1, 3×(-1)), (–3, 3×(-3)), (2, 3×2)}

∴ f = {(0, 0), (–1, –3), (–3, –9), (2, 6)}

Recall that a function is invertible only when it is both one-one and onto.

Here, observe that distinct elements of the domain {0, –1, –3, 2} are mapped to distinct elements of the co-domain {0, –3, –9, 6}.

Hence, f is one-one.

Also, each element of the range {–9, –3, 0, 6} is the image of some element of {0, –1, –3, 2}.

Hence, f is also onto.

Thus, the function f has an inverse.

We have f^-1 = {(0, 0), (–3, –1), (–9, –3), (6, 2)}

(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} & f(x) = x²

We have f : A → B and f(x) = x².

⇒ f = {(1, 1²), (3, 3²), (5, 5²), (7, 7²), (9, 9²)}

∴ f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}

Recall that a function is invertible only when it is both one-one and onto.

Here, observe that distinct elements of the domain {1, 3, 5, 7, 9} are mapped to distinct elements of the co-domain {1, 9, 25, 49, 81}.

Hence, f is one-one.

However, the element 0 of the range {0, 1, 9, 25, 49, 81} is not the image of any element of {1, 3, 5, 7, 9}.

Hence, f is not onto.

Thus, the function f does not have an inverse.

f : {1, 2, 3} → {a, b, c} and f(1) = a, f(2) = b, f(3) = c

⇒ f = {(1, a), (2, b), (3, c)}

Recall that a function is invertible only when it is both one-one and onto.

Here, observe that distinct elements of the domain {1, 2, 3} are mapped to distinct elements of the co-domain {a, b, c}.

Hence, f is one-one.

Also, each element of the range {a, b, c} is the image of some element of {1, 2, 3}.

Hence, f is also onto.

Thus, the function f has an inverse.

We have f^-1 = {(a, 1), (b, 2), (c, 3)}

g : {a, b, c} → {apple, ball, cat} and g(a) = apple, g(b) = ball, g(c) = cat

⇒ g = {(a, apple), (b, ball), (c, cat)}

Similar to the function f, g is also one-one and onto.

Thus, the function g has an inverse.

We have g^-1 = {(apple, a), (ball, b), (cat, c)}

We know (gof)(x) = g(f(x))

Thus, gof : {1, 2, 3} → {apple, ball, cat} and

(gof)(1) = g(f(1)) = g(a) = apple

(gof)(2) = g(f(2)) = g(b) = ball

(gof)(3) = g(f(3)) = g(c) = cat

⇒ gof = {(1, apple), (2, ball), (3, cat)}

As the functions f and g, gof is also both one-one and onto.

Thus, the function gof has an inverse.

We have (gof)^-1 = {(apple, 1), (ball, 2), (cat, 3)}

Now, let us consider f^-1og^-1.

We know (f^-1og^-1)(x) = f^-1(g^-1(x))

Thus, f^-1og^-1 : {apple, ball, cat} → {1, 2, 3} and

(f^-1og^-1)(apple) = f^-1(g^-1(apple)) = f^-1(a) = 1

(f^-1og^-1)(ball) = f^-1(g^-1(ball)) = f^-1(b) = 2

(f^-1og^-1)(cat) = f^-1(g^-1(cat)) = f^-1(c) = 3

⇒ f^-1og^-1 = {(apple, 1), (ball, 2), (cat, 3)}

Therefore, we have (gof)^-1 = f^-1og^-1.

We have f : A → B & f(x) = 2x + 1

⇒ f = {(1, 2×1 + 1), (2, 2×2 + 1), (3, 2×3 + 1), (4, 2×4 + 1)}

∴ f = {(1, 3), (2, 5), (3, 7), (4, 9)}

Function f is clearly one-one and onto.

Thus, f^-1 exists and f^-1 = {(3, 1), (5, 2), (7, 3), (9, 4)}

We have g : B → C & g(x) = x² – 2

⇒ g = {(3, 3² – 2), (5, 5² – 2), (7, 7² – 2), (9, 9² – 2)}

∴ g = {(3, 7), (5, 23), (7, 47), (9, 79)}

Function g is clearly one-one and onto.

Thus, g^-1 exists and g^-1 = {(7, 3), (23, 5), (47, 5), (79, 9)}

We know (gof)(x) = g(f(x))

Thus, gof : A → C and

(gof)(1) = g(f(1)) = g(3) = 7

(gof)(2) = g(f(2)) = g(5) = 23

(gof)(3) = g(f(3)) = g(7) = 47

(gof)(4) = g(f(4)) = g(9) = 79

⇒ gof = {(1, 7), (2, 23), (3, 47), (4, 79)}

Clearly, gof is also both one-one and onto.

Thus, the function gof has an inverse.

We have (gof)^-1 = {(7, 1), (23, 2), (47, 3), (79, 4)}

Now, let us consider f^-1og^-1.

We know (f^-1og^-1)(x) = f^-1(g^-1(x))

Thus, f^-1og^-1 : C → A and

(f^-1og^-1)(7) = f^-1(g^-1(7)) = f^-1(3) = 1

(f^-1og^-1)(23) = f^-1(g^-1(23)) = f^-1(5) = 2

(f^-1og^-1)(47) = f^-1(g^-1(47)) = f^-1(7) = 3

(f^-1og^-1)(79) = f^-1(g^-1(79)) = f^-1(9) = 4

⇒ f^-1og^-1 = {(7, 1), (23, 2), (47, 3), (79, 4)}

Therefore, we have (gof)^-1 = f^-1og^-1.

We have f : Q → Q and f(x) = 3x + 5.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ Q (domain) such that f(x₁) = f(x₂)

⇒ 3x₁ + 5 = 3x₂ + 5

⇒ 3x₁ = 3x₂

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ Q (co-domain) such that f(x) = y

⇒ 3x + 5 = y

⇒ 3x = y – 5

Clearly, for every y ϵ Q, there exists x ϵ Q (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

We have f : R → R and f(x) = 4x + 3.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ R (domain) such that f(x₁) = f(x₂)

⇒ 4x₁ + 3 = 4x₂ + 3

⇒ 4x₁ = 4x₂

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ R (co-domain) such that f(x) = y

⇒ 4x + 3 = y

⇒ 4x = y – 3

Clearly, for every y ϵ R, there exists x ϵ R (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

We have f : R⁺→ [4, ∞) and f(x) = x² + 4.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ R⁺ (domain) such that f(x₁) = f(x₂)

⇒ x₁² + 4 = x₂² + 4

⇒ x₁² = x₂²

∴ x₁ = x₂ (x₁≠–x₂as x₁, x₂ϵ R⁺)

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ [4, ∞) (co-domain) such that f(x) = y

⇒ x² + 4 = y

⇒ x² = y – 4

Clearly, for every y ϵ [4, ∞), there exists x ϵ R⁺ (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

We have

We know (fof)(x) = f(f(x))

As (fof)(x) = x = I_x (the identity function), f(x) = f^-1(x).

Thus,

We have f : R⁺→ [–5, ∞) and f(x) = 9x² + 6x – 5.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ R⁺ (domain) such that f(x₁) = f(x₂)

⇒ 9x₁² + 6x₁ – 5 = 9x₂² + 6x₂ – 5

⇒ 9x₁² + 6x₁ = 9x₂² + 6x₂

⇒ 9x₁² – 9x₂² + 6x₁ – 6x₂ = 0

⇒ 9(x₁² – x₂²) + 6(x₁ – x₂) = 0

⇒ 9(x₁ – x₂)(x₁ + x₂) + 6(x₁ – x₂) = 0

⇒ (x₁ – x₂)[9(x₁ + x₂) + 6] = 0

⇒ x₁ – x₂ = 0 (as x₁, x₂ϵ R⁺)

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ [–5, ∞) (co-domain) such that f(x) = y

⇒ 9x² + 6x – 5 = y

Adding 6 to both sides, we get

9x² + 6x – 5 + 6 = y + 6

⇒ 9x² + 6x + 1 = y + 6

⇒ (3x + 1)² = y + 6

Clearly, for every y ϵ [4, ∞), there exists x ϵ R⁺ (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

We have f : R → R and f(x) = x³ – 3.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ R (domain) such that f(x₁) = f(x₂)

⇒ x₁³ – 3 = x₂³ – 3

⇒ x₁³ = x₂³

⇒ (x₁ – x₂)(x₁² + x₁x₂ + x₂²) = 0

⇒ x₁ – x₂ = 0 (as x₁, x₂ϵ R⁺)

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ R (co-domain) such that f(x) = y

⇒ x³ – 3 = y

⇒ x³ = y + 3

Clearly, for every y ϵ R, there exists x ϵ R (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

Hence, we have

Thus, f^-1(24) = 3 and f^-1(5) = 2.

We have f : R → R and f(x) = x³ + 4.

Recall that a function is a bijection only if it is both one-one and onto.

First, we will check if f is one-one.

Let x₁, x₂ϵ R (domain) such that f(x₁) = f(x₂)

⇒ x₁³ + 4 = x₂³ + 4

⇒ x₁³ = x₂³

⇒ (x₁ – x₂)(x₁² + x₁x₂ + x₂²) = 0

As x₁, x₂ϵ R and the second factor has no real roots,

x₁ – x₂ = 0

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will check if f is onto.

Let y ϵ R (co-domain) such that f(x) = y

⇒ x³ + 4 = y

⇒ x³ = y – 4

Clearly, for every y ϵ R, there exists x ϵ R (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f is a bijection and has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

Hence, we have

Thus, f^-1(3) = –1.

We have f : Q → Q and f(x) = 2x.

Recall that a function is a bijection only if it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ Q (domain) such that f(x₁) = f(x₂)

⇒ 2x₁ = 2x₂

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ Q (co-domain) such that f(x) = y

⇒ 2x = y

Clearly, for every y ϵ Q, there exists x ϵ Q (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f is a bijection and has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus,

Now, we have g : Q → Q and g(x) = x + 2.

First, we will prove that g is one-one.

Let x₁, x₂ϵ Q (domain) such that g(x₁) = g(x₂)

⇒ x₁ + 2 = x₂ + 2

∴ x₁ = x₂

So, we have g(x₁) = g(x₂) ⇒ x₁ = x₂.

Thus, function g is one-one.

Now, we will prove that g is onto.

Let y ϵ Q (co-domain) such that g(x) = y

⇒ x + 2 = y

∴ x = y – 2

Clearly, for every y ϵ Q, there exists x ϵ Q (domain) such that g(x) = y and hence, function g is onto.

Thus, the function g is a bijection and has an inverse.

We have g(x) = y ⇒ x = g^-1(y)

But, we found g(x) = y ⇒ x = y – 2

Hence, g^-1(y) = y – 2

Thus, g^-1(x) = x – 2

We have (f^-1og^-1)(x) = f^-1(g^-1(x))

We found and g^-1(x) = x – 2

⇒ (f^-1og^-1)(x) = f^-1(x – 2)

We know (gof)(x) = g(f(x)) and gof : Q → Q

⇒ (gof)(x) = g(2x)

∴ (gof)(x) = 2x + 2

Clearly, gof is a bijection and has an inverse.

Let y ϵ Q (co-domain) such that (gof)(x) = y

⇒ 2x + 2 = y

⇒ 2x = y – 2

We have (gof)(x) = y ⇒ x = (gof)^-1(y)

But, we found (gof)(x) = y ⇒

Hence,

Thus,

So, it is verified that (gof)^-1 = f^-1og^-1.

We have f : A → B where A = R – {3} and B = R – {1}

First, we will prove that f is one-one.

Let x₁, x₂ϵ A (domain) such that f(x₁) = f(x₂)

⇒ (x₁ – 2)(x₂ – 3) = (x₁ – 3)(x₂ – 2)

⇒ x₁x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 2x₁ – 3x₂ + 6

⇒ –3x₁ – 2x₂ = –2x₁ – 3x₂

⇒ –3x₁ + 2x₁ = 2x₂ – 3x₂

⇒ –x₁ = –x₂

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ B (co-domain) such that f(x) = y

Clearly, for every y ϵ B, there exists x ϵ A (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

We have f : R⁺→ [–9, ∞) and f(x) = 5x² + 6x – 9.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ R⁺ (domain) such that f(x₁) = f(x₂)

⇒ 5x₁² + 6x₁ – 9 = 5x₂² + 6x₂ – 9

⇒ 5x₁² + 6x₁ = 5x₂² + 6x₂

⇒ 5x₁² – 5x₂² + 6x₁ – 6x₂ = 0

⇒ 5(x₁² – x₂²) + 6(x₁ – x₂) = 0

⇒ 5(x₁ – x₂)(x₁ + x₂) + 6(x₁ – x₂) = 0

⇒ (x₁ – x₂)[5(x₁ + x₂) + 6] = 0

⇒ x₁ – x₂ = 0 (as x₁, x₂ϵ R⁺)

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ [–9, ∞) (co-domain) such that f(x) = y

⇒ 5x² + 6x – 9 = y

Adding to both sides, we get

Clearly, for every y ϵ [–9, ∞), there exists x ϵ R⁺ (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

We have f : N → N and f(x) = 9x² + 6x – 5.

We need to prove f : N → S is invertible.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ N (domain) such that f(x₁) = f(x₂)

⇒ 9x₁² + 6x₁ – 5 = 9x₂² + 6x₂ – 5

⇒ 9x₁² + 6x₁ = 9x₂² + 6x₂

⇒ 9x₁² – 9x₂² + 6x₁ – 6x₂ = 0

⇒ 9(x₁² – x₂²) + 6(x₁ – x₂) = 0

⇒ 9(x₁ – x₂)(x₁ + x₂) + 6(x₁ – x₂) = 0

⇒ (x₁ – x₂)[9(x₁ + x₂) + 6] = 0

⇒ x₁ – x₂ = 0 (as x₁, x₂ϵ R⁺)

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ S (co-domain) such that f(x) = y

⇒ 9x² + 6x – 5 = y

Adding 6 to both sides, we get

9x² + 6x – 5 + 6 = y + 6

⇒ 9x² + 6x + 1 = y + 6

⇒ (3x + 1)² = y + 6

Clearly, for every y ϵ S, there exists x ϵ N (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

Hence, we have

Thus, f^-1(43) = 2 and f^-1(163) = 4.

We have f : R – → R and

We need to prove f : R – → range(f) is invertible.

First, we will prove that f is one-one.

Let x₁, x₂ϵ A (domain) such that f(x₁) = f(x₂)

⇒ (4x₁)(3x₂ + 4) = (3x₁ + 4)(4x₂)

⇒ 12x₁x₂ + 16x₁ = 12x₁x₂ + 16x₂

⇒ 16x₁ = 16x₂

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ range(f) (co-domain) such that f(x) = y

⇒ 4x = 3xy + 4y

⇒ 4x – 3xy = 4y

⇒ x(4 – 3y) = 4y

Clearly, for every y ϵ range(f), there exists x ϵ A (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

We have f: R → (–1, 1) and

Given that f^-1 exists.

Let y ϵ (–1, 1) such that f(x) = y

⇒ 10^2x – 1 = y (10^2x + 1)

⇒ 10^2x – 1 = 10^2xy + y

⇒ 10^2x – 10^2xy = 1 + y

⇒ 10^2x (1 – y) = 1 + y

Taking log₁₀ on both sides, we get

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus,

We have f: R → (0, 2) and

Given that f^-1 exists.

Let y ϵ (0, 2) such that f(x) = y

⇒ 2e^2x = y (e^2x + 1)

⇒ 2e^2x = e^2xy + y

⇒ 2e^2x – e^2xy = y

⇒ e^2x (2 – y) = y

Taking ln on both sides, we get

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus,

We have f : [–1, ∞) → [–1, ∞) and f(x) = (x + 1)² – 1

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x₁, x₂ϵ [–1, ∞) (domain) such that f(x₁) = f(x₂)

⇒ (x₁ + 1)² – 1 = (x₂ + 1)² – 1

⇒ (x₁ + 1)² = (x₂ + 1)²

⇒ x₁² + 2x₁ + 1 = x₂² + 2x₂ + 1

⇒ x₁² + 2x₁ = x₂² + 2x₂

⇒ x₁² – x₂² + 2x₁ – 2x₂ = 0

⇒ (x₁² – x₂²) + 2(x₁ – x₂) = 0

⇒ (x₁ – x₂)(x₁ + x₂) + 2(x₁ – x₂) = 0

⇒ (x₁ – x₂)[x₁ + x₂ + 2] = 0

⇒ x₁ – x₂ = 0 (as x₁, x₂ϵ R⁺)

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ [–1, ∞) (co-domain) such that f(x) = y

⇒ (x + 1)² – 1 = y

⇒ (x + 1)² = y + 1

Clearly, for every y ϵ [–1, ∞), there exists x ϵ [–1, ∞) (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

Now, we need to find the values of x for which f(x) = f^-1(x).

We have f(x) = f^-1(x)

We can write

On substituting, we get

t⁴ = t

⇒ t⁴ – t = 0

⇒ t (t³ – 1) = 0

⇒ t (t – 1)(t² + t + 1) = 0

t² + t + 1 ≠ 0 because this equation has no real root t.

⇒ t = 0 or t – 1 = 0

⇒ t = 0 or t = 1

Case – I: t = 0

⇒ x + 1 = 0

∴ x = –1

Case – II: t = 1

⇒ x + 1 = 1

∴ x = 0

Thus, S = {0, –1}

We have f : A → A where A = {x ϵ R | –1 ≤ x ≤ 1} defined by f(x) = x².

Recall that a function is invertible only when it is both one-one and onto.

First, we will check if f is one-one.

Let x₁, x₂ϵ A (domain) such that f(x₁) = f(x₂)

⇒ x₁² = x₂²

⇒ x₁² – x₂² = 0

⇒ (x₁ – x₂)(x₁ + x₂) = 0

⇒ x₁ – x₂ = 0 or x₁ + x₂ = 0

∴ x₁ = ±x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = ±x₂.

This means that two different elements of the domain are mapped to the same element by the function f.

For example, consider f(–1) and f(1).

We have f(–1) = (–1)² = 1 and f(1) = 1² = 1 = f(–1)

Thus, f is not one-one and hence f^-1 doesn’t exist.

Now, let us consider g : A → A defined by g(x) = sin

First, we will prove that g is one-one.

Let x₁, x₂ϵ A (domain) such that g(x₁) = g(x₂)

(in the given range)

∴ x₁ = x₂