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(i) Given as f(x) = e^x, g(x) = log_e x

Let f: R → (0, ∞); and g: (0, ∞) → R

Let us calculate fog,

Clearly, the range of g is a subset of the domain of f.

fog: (0, ∞) → R

(fog)(x) = f(g(x))

= f(log_e x)

= log_e e^x

= x 

Let us calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→ R

(gof)(x) = g(f(x))

= g(e^x)

= log_e e^x

= x

(ii) f(x) = x², g(x) = cos x

f: R→ [0, ∞); g: R → [−1, 1]

Let us calculate fog,

Clearly, the range of g is not a subset of the domain of f.

⇒ Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}

⇒ Domain (fog) = x: x ∈ R and cos x ∈ R}

⇒ Domain of (fog) = R

(fog): R→ R

(fog)(x) = f(g(x))

= f(cos x)

= cos² x

Let us calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog: R→R

(gof)(x) = g(f(x))

= g(x²)

= cos x²

(iii) Given f(x) = |x|, g(x) = sin x

f: R → (0, ∞) ; g : R → [−1, 1]

Let us calculate fog,

Clearly, the range of g is a subset of the domain of f.

⇒ fog: R → R

(fog)(x) = f(g(x))

= f(sin x)

= |sin x|

Let us calculate gof,

Clearly, the range of f is a subset of the domain of g.

⇒ fog : R→ R

(gof)(x) = g(f(x))

= g(|x|)

= sin |x|

(iv) Given f(x) = x + 1, g(x) = e^x

f: R → R ; g: R → [ 1, ∞)

Let us calculate fog:

Clearly, range of g is a subset of domain of f.

⇒ fog: R → R

(fog)(x) = f(g(x))

= f(e^x)

= e^x + 1

Let us compute gof,

Clearly, range of f is a subset of domain of g.

⇒ fog: R → R

(gof)(x) = g(f(x))

= g(x + 1)

= e^x+1

(v) Given f(x) = sin⁻¹ x, g(x) = x²

f: [−1,1] → [(-π)/2,π/2]; g : R → [0, ∞) 

Let us compute the fog:

Clearly, the range of g is not a subset of the domain of f.

Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}

Domain (fog) = {x: x ∈ R and x² ∈ [−1, 1]}

Domain (fog) = {x: x ∈ R and x ∈ [−1, 1]}

Domain of (fog) = [−1, 1]

fog: [−1,1] → R 

(fog)(x) = f(g(x))

= f(x²)

= sin⁻¹(x²)

Let us compute the gof:

Clearly, the range of f is a subset of the domain of g.

fog: [−1, 1] → R

(gof)(x) = g(f(x))

= g(sin⁻¹ x)

= (sin⁻¹ x)²

Two header files are

1. iostream.h 

2. iomanip.h

The two functions of iostream.h files are

• cin 
• cout

getline() – It is a stream function where as the others are console functions.

The iostream.h header file contains C++ streams and i/o routine functions. It is a header file which we include in our programs to perform basic input output operations. We use ‘cin’ in programs to get input from the keyboard. To use ‘cin’ and ‘cout’, we must add iostream. h header file

Proto type error. To use cin and cout the header file iostream is a must.

Given that f(x) = |x|,

Let us prove that fof = f.

Consider that (fof)(x) = f(f(x)) 

= f(|x|) 

= ||x|| 

= |x| 

= f(x)

Therefore, (fof)(x) = f(x), ∀ x ∈ R

Hence, fof = f

Given f(x) = \({\frac{4x + 3}{6x - 4}}\), x ≠ \({\frac{2}{3}}\)

Let us show fof(x) = x

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

= f((4x + 3)/(6x – 4)) 

=  (4((4x + 3)/(6x -4)) + 3)/(6((4x +3)/(6x – 4)) – 4)

= \(​​{\frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16}}\)

= (34x)/(34)

= x

So, fof(x) = x for all x ≠ 2/3

=> fof = 1

So, the given function f is invertible and the inverse of f is f itself.

f(x) = √(9-x²)

We know, 

The square of a real number is never negative. 

Clearly, 

f(x) takes real values only when 9 – x² ≥ 0 

⇒ 9 ≥ x² 

⇒ x² ≤ 9 

⇒ x² – 9 ≤ 0 

⇒ x² – 3² ≤ 0 

⇒ (x + 3)(x – 3) ≤ 0 

⇒ x ≥ –3 and x ≤ 3 

∴ x ∈ [–3, 3] 

Thus, 

Domain of f = [–3, 3]

f(x) = √(9-x²)

We know,

The square of a real number is never negative. 

Clearly, 

f(x) takes real values only when 9 – x² ≥ 0 

⇒ 9 ≥ x² 

⇒ x² ≤ 9 

⇒ x² – 9 ≤ 0 

⇒ x² – 3² ≤ 0 

⇒ (x + 3)(x – 3) ≤ 0 

⇒ x ≥ –3 and x ≤ 3 

∴ x ∈ [–3, 3] 

Thus,

Domain of f = [–3, 3] 

When x ∈ [–3, 3], we have 

0 ≤ 9 – x² ≤ 9 

Hence,

0 ≤ \(\sqrt{9-x^2}\) ≤ 3

⇒ 0 ≤ f(x) ≤ 3

∴ f(x) ∈ [0, 3] 

Thus, 

Range of f = [0, 3]

f(x) = √(x-3)

We know,

The square of a real number is never negative. 

Clearly,

f(x) takes real values only when x – 3 ≥ 0 

⇒ x ≥ 3 

∴ x ∈ [3, ∞) 

Thus, 

Domain of f = [3, ∞) 

When x ≥ 3, 

We have x – 3 ≥ 0 

Hence,

\(\sqrt{x-3}\)  ≥ 0

⇒ f(x) ≥ 0

∴ f(x) ∈ [0, ∞) 

Thus, 

Range of f = [0, ∞)

f(x) = √(x-1)

We know,

The square of a real number is never negative. 

Clearly, 

f(x) takes real values only 

When x – 1 ≥ 0

⇒ x ≥ 1 

∴ x ∈ [1, ∞) 

Thus, 

Domain of f = [1, ∞) 

When x ≥ 1, 

We have,

x – 1 ≥ 0 

Hence,

\(\sqrt{x-1}\) ≥ 0

⇒ f(x) ≥ 0

∴ f(x) ∈ [0, ∞) 

Thus,

Range of f = [0, ∞)

f(x) = √(x²-16)

We know,

The square of a real number is never negative. 

Clearly, 

f(x) takes real values only when x² – 16 ≥ 0 

⇒ x² – 4² ≥ 0 

⇒ (x + 4)(x – 4) ≥ 0 

⇒ x ≤ –4 or x ≥ 4 

∴ x ∈ (–∞, –4] ∪ [4, ∞) 

Thus,

Domain of f = (–∞, –4] ∪ [4, ∞) 

When x ∈ (–∞, –4] ∪ [4, ∞), 

We have,

x² – 16 ≥ 0

Hence,

\(\sqrt{x^2-16}\) ≥ 0

⇒ f(x) ≥ 0

∴ f(x) ∈ [0, ∞) 

Thus, 

Range of f = [0, ∞)

Given that f: R+→ R such that f(x) = log_ex 

To find: (i) Range of f 

Here, f(x) = log_ex 

We know that the range of a function is the set of images of elements in the domain. 

∴ The image set of the domain of f = R 

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

To find: (ii) {x : x ϵ R⁺ and f(x) = -2} 

We have, f(x) = -2 …(a) 

And f(x) = log_ex …(b) 

From eq. (a) and (b), we get 

log_ex = -2 

Taking exponential both the sides, we get

⇒ e^loge^x = e ^-2

[ ∵ Inverse property I . e b^logbx = x]

⇒ x = e-2 ∴{x : x ϵ R⁺ and f(x) = -2} = {e^-2 } 

To find: (iii) f(xy) = f(x) + f(y) for all x, y ϵ R 

We have, 

f(xy) = loge(xy) 

= loge(x) + loge(y) 

[Product Rule for Logarithms] 

= f(x) + f(y) [∵f(x) = logex] 

∴ f(xy) = f(x) + f(y) holds.

i. {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}} 

When x = 1, we have 

y = 3(1) = 3 

When x = 2, we have 

y = 3(2) = 6 

When x = 3, we have 

y = 3(3) = 9 

Thus, 

R = {(1, 3), (2, 6), (3, 9)} 

Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first component but different second components. 

Hence, 

The given relation R is a function.

ii. {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6} 

When x = 1, we have 

y > 1 + 1 or y > 2

⇒ y = {4, 6} 

When x = 2, we have

y > 2 + 1 or y > 3

⇒ y = {4, 6} 

Thus, 

R = {(1, 4), (1, 6), (2, 4), (2, 6)} 

Every element of set x has an ordered pair in the relation. However, two ordered pairs (1, 4) and (1, 6) have the same first component but different second components. 

Hence, 

The given relation R is not a function.

iii. {(x, y): x + y = 3, x, y∈ {0, 1, 2, 3}} 

When x = 0, we have 

0 + y = 3

⇒ y = 3 

When x = 1, we have 

1 + y = 3

⇒ y = 2 

When x = 2, we have 

2 + y = 3

⇒ y = 1 

When x = 3, we have 

3 + y = 3

⇒ y = 0 

Thus, 

R = {(0, 3), (1, 2), (2, 1), (3, 0)} 

Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first component but different second components. 

Hence, 

The given relation R is a function.

Given,

f : R⁺→ R and f(x) = log_ex. 

i. the image set of the domain of f : 

Domain of f = R⁺ 

(set of positive real numbers) 

We know the value of logarithm to the base e (natural logarithm) can take all possible real values. 

Hence, 

The image set of f is the set of real numbers. 

Thus,

The image set of f = R 

ii. {x: f(x) = –2} 

Given,

f(x) = –2 

⇒ log_ex = –2 

∴ x = e^-2 

[∵ log_ba = c ⇒ a = b^c] 

Thus, 

{x: f(x) = –2} = {e –2} 

iii. whether f(xy) = f(x) + f(y) holds. 

We have, 

f(x) = log_ex 

⇒ f(y) = log_ey 

Now, 

let us consider f(xy). 

f(xy) = log_e(xy) 

⇒ f(xy) = log_e(x × y) 

[∵ log_b(a×c) = log_ba + log_bc] 

⇒ f(xy) = log_ex + log_ey 

∴ f(xy) = f(x) + f(y) 

Hence, 

The equation f(xy) = f(x) + f(y) holds.

Given,

X = {1, 2, 3, 4} and 

Y = {1, 5, 9, 11, 15, 16} 

i. f₁ = {(1, 1), (2, 11), (3, 1), (4, 15)} 

Every element of set X has an ordered pair in the relation f₁ and no two ordered pairs have the same first component but different second components. 

Hence, 

The given relation f₁ is a function. 

ii. f₂ = {(1, 1), (2, 7), (3, 5)} 

In the relation f₂, 

The element 2 of set X does not have any image in set Y. 

However,

For a relation to be a function, 

Every element of the domain should have an image. 

Hence, 

The given relation f₂ is not a function. 

iii. f₃ = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)} 

Every element of set X has an ordered pair in the relation f₃. 

However, 

Two ordered pairs (2, 9) and (2, 11) have the same first component but different second components. 

Hence, 

The given relation f₃ is not a function.

Option : (D)

Given,

f(x) = √(4x-x²)

Here, 

4x-x²≥0 

x²-4x≤0 

x(x-4)≤ 0

So,

x ∈ [0,4]

Option : (D)

For f(x) = log |x|; 

It is defined at all positive values of x except 0.. 

But since we have |x|; 

So, 

|x|>0; 

x ∈ R-{0}

Option : (A)

Given,

f(x) = √((x+1)(x-3)/(x-2)) 

Here,

\({\frac{(x+1)(x-3)}{x-2}}\) ≥  0

But,

x ≠ 2

So,

x ∈  [-1,2) ∪ [3,∞)

Option : (C)

Given,

f(x) = √((x-2)/(x+2))+√((1-x)/(1+x))

For function to be defined,

\({\frac{x-2}{x+2}}\) ≥ 0, x ≠ - 2 

x∈(-∞,-2)∪[2, ∞) …(1)

And,

\({\frac{1-x}{1+x}}\) ≤ 0

So,

Taking common of both the solutions, we get

x ∈ ϕ.

Option : (C)

f(|x|) = |x|-1 

f(|x|) = x 

We have, 

|x| = x ;x≥0 

And |x| = -x ; x≤0 

So, 

-x-1 = x 

2x = -1

x = \(-\frac{1}{2}\)

Answer : (b) 1 

 f (– 1.75) + f (0.5) + f (1.5) 

= (– 1.75 + 2) + (0.5)² + (2 – 1.5) 

= 0.25 + 0.25 + 0.5

= 1.

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

So, f:A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f:A → B 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, f:A → B is Surjection iff for each b ∈ B, there exists a ∈ B such that f(a) = b

Now, Let, f:N → N given by f(x) = x²

Check for Injectivity:

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

So, from definition

⇒ f(x) = f(y)

⇒ x² = y²

⇒ x² – y² = 0

⇒ (x – y)(x + y) = 0

As x, y ∈ N 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 y ∈ N be arbitrary, then

⇒ f(x) = y

⇒ x² = y

⇒ x = √y

⇒ √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:N → N given by f(x) = x² is One – One but not onto.

Answer: (b) = one - one

f (x) = x + 2 and f : A → B 

∴  f (1) = 1 + 2 = 3, f (2) = 2 + 2 = 4 

f (3) = 3 + 2 = 5, f (4) = 4 + 2 = 6 

Thus each element in A has a unique image in B and no two elements in B have the same pre-image in A. 

So f is one-one. 

But not all the elements of B, i.e., 1 and 2 have a pre-image in A. 

So, the function f is not onto

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

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B 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, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B such that f(a) = b

Now, Let, f: R → R given by f(x) = x³ – x

Check for Injectivity:

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

Hence f is not One – One function

Check for Surjectivity:

Let y be element belongs to R i.e y ∈ R 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 x = α be that root

⇒ α³ - α = y

f(α) = y

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

Therefore f is onto

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

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

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B 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, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B 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.

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

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B 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, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B 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.

To Show: that f is one-one and onto

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ↔ x₁= x₂ or x₁ ≠ x₂ ↔ f(x₁) ≠ f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ Q and f(x) = \(\frac{x-1}{x-2}\).So f(x₁) = f(x₂) → \(\frac{x_1-1}{x_1-2}\) = \(\frac{(x_2-1)}{x_2-2}\), on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x₂, f(x) is one-one

Given co-domain of f(x) is R – {1}

Let y = f(x) = \(\frac{x-1}{x-2}\), So x = \(\frac{2y-1}{y-1}\) [Range of f(x) = Domain of y]

So Domain of y = Range of f(x) = R – {1}

Hence, Range of f(x) = co-domain of f(x) = R – {1}.

So, f(x) is onto function

As it is a bijective function. So it is invertible

Invers of f(x) is f^-1(y) = \(\frac{2y-1}{y-1}\)

One-one f:

Consider x₁ and x₂ ∈ dom (f)

We know that

f(x₁) = f(x₂)

It can be written as

(4x₁ + 3)/ (6x₁ – 4) = (4x₂ + 3)/ (6x₂ – 4)

So we get

(4x₁ + 3) (6x₂ – 4) = (4x₂ + 3) (6x₁ – 4)

On further calculation

24 x₁ x₂ – 16x₁ + 18 x₂ – 12 = 24 x₁ x₂ – 16x₂ + 18x₁ – 12

We get

34 x₁ = 34 x₂ where x₁ = x₂

f is one-one

Onto f:

Consider y ∈ co domain (f)

We know that y = f(x)

y = (4x + 3)/ (6x – 4)

On further calculation

6xy – 4y = 4x + 3

So we get

6xy – 4x = 3 + 4y

It can be written as

x (6y – 4) = 3 + 4y

So x = (3 + 4y)/ (6y – 4) ∈ domain Ɐ y ∈ co-domain

f is an onto function

Here, x = (3 + 4y)/ (6y – 4) where y ≠ 2/3

We get

f ^-1 (y) = (3 + 4y)/ (6y – 4) where y ≠ 2/3

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

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B 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, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B 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.

To Show: that f is one-one and onto

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ↔ x₁= x₂ or x₁ ≠ x₂ ↔ f(x₁) ≠ f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ ∈ Q and f(x) = \(\frac{(4x+3)}{(6x-4)}\).So f(x₁) = f(x₂) → \(\frac{(4x_1+3)}{(6x_1-4)}=\frac{(4x_2+3)}{(6x_2-4)}\) → = on solving we get x₁=x₂

So f(x1) = f(x2) ↔ x1= x2, f(x) is one-one

Given co-domain of f(x) is R except 3x-2=0.

Let y = f(x) = \(\frac{(4x+3)}{(6x-4)}\) So x = \(\frac{4x+3}{6x-4}\) [Range of f(x) = Domain of y]

So Domain of y is R (except 3x-2=0) = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R except 3x-2=0

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) = \(\frac{4x+3}{6x-4}\)

Since g is surjective (onto),

there exists y ∈ B for every z ∈ C such that

g(y) = z …….(i)

Since f is surjective,

there exists x ∈ A for every y ∈ B such that

f(x) = y …….(ii)

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

= g(y) ……[From (ii)]

= z …..[From(i)]

i.e., for every z ∈ C, there is x in A such that

(gof) x = z

∴ gof is surjective (onto).

(i) Given f₁ = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}

Injectivity:

f₁(1) = 3

f₁(2) = 5

f₁(3) = 7

⇒ Every element of A has different images in B.

So, f₁ is one-one.

Surjectivity:

Co-domain of f₁ = {3, 5, 7}

Range of f₁ = set of images  =  {3, 5, 7}

⇒ Co-domain = range

So, f₁ is onto.

(ii) Given f₂ = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}

f₂ = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}

Injectivity:

f₂(2) = a

f₂(3) = b

f₂(4) = c

⇒ Every element of A has different images in B.

So, f₂ is one-one.

Surjectivity:

Co-domain of f₂ = {a, b, c}

Range of f₂ = set of images = {a, b, c}

⇒ Co-domain = range

So, f₂ is onto.

(iii) Given f₃ = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Injectivity:

f₃(a) = x

f₃(b) = x

f₃(c) = z

f₃(d) = z

⇒ a and b have the same image x.

Also c and d have the same image z

Therefore, f₃ is not one-one.

Surjectivity:

Co-domain of f₁ = {x, y, z} 

Range of f₁ = set of images = {x, z}

So, the co-domain  is not same as the range.

Hence, f₃ is not onto.

SINCE, f(a, b) = (b, a).There is no same value of y at different values of x

∴function is one one

∴Range(A×B)≠Codomain(B × A)

⇒function is into

(i) Let f: Z → Z given by f(x) = 3x + 2

We have to check one-one condition on f(x) = 3x + 2

Injectivity:

Let x and y be any two elements in domain (Z), such that f(x) = f(y).

 f(x) = f(y)

⇒ 3x + 2 = 3y + 2

⇒ 3x = 3y

⇒ x = y

⇒ f(x) = f(y) 

⇒ x = y

Therefore, f is one-one.

Surjectivity:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z(domain).

Let f(x) = y

⇒ 3x + 2 = y

⇒ 3x = y – 2

⇒ x = (y – 2)/3. It may not be in the domain (Z)

Because if we take y = 3,

x = (y – 2)/3 = (3 - 2)/3 = 1/3 ∉ domain Z.

So, for every element in the co domain there need not be any element in the domain such that f(x) = y.

Thus, f is not onto.

(ii) Example for function which is not one-one but onto

Let f: Z → N ∪ {0} given by f(x) = |x|

Injectivity:

Let x and y be any two elements in domain (Z),

Such that f(x) = f(y).

⇒ |x| = |y|

⇒ x = ± y

So, different elements of domain f may give the same image.

Therefore, f is not one-one.

Surjectivity:

Let y be any element in co domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

⇒ |x| = y

⇒ x = ± y

Which is an element in Z (domain). So, for every element in the co-domain, there exists a pre-image in the domain.

Thus, f is onto.

(iii) Example for function which is neither one-one nor onto.

Let f: Z → Z given by f(x) = 2x² + 1

Injectivity:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

⇒ 2x² +1 = 2y² +1

⇒ 2x² = 2y²

⇒ x² = y²

⇒ x = ± y

So, different elements of domain f may give the same image.

Thus, f is not one-one.

Surjectivity:

Let y be any element in co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

⇒ 2x² + 1 = y

⇒ 2x² = y − 1

⇒ x² = (y - 1)/2

⇒ x = √((y - 1)/2) ∉ Z always.

For example, if we take, y = 4,

x = ± √((y - 1)/2)

= ± √((4 - 1)/2)

= ± √(3/2) ∉ Z

Therefore, x may not be in Z (domain).

Hence, f is not onto.

(i) Given as f: N → N, given by f(x) = x²

Let us check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in domain (N), such that f(x) = f(y).

f(x) = f(y)

x² = y²

x = y (We do not get ± because x and y are in N that is natural numbers)

So, f is an injection.

Surjection condition:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x²= y

x = √y, which may not be in N.

For example, if y = 3,

x = √3 is not in N.

So, f is not a surjection.

Also f is not a bijection.

(ii) Given f: Z → Z, given by f(x) = x²

Let us check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in domain (Z), such that f(x) = f(y).

f(x) = f(y)

x² = y²

x = ±y

Therefore, f is not an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x² = y

x = ± √y which may not be in Z.

For example, if y = 3,

x = ± √3 is not in Z.

So, f is not a surjection.

Also f is not bijection.

(iii) Given f: N → N given by f(x) = x³

Let us check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in domain (N), such that f(x) = f(y).

f(x) = f(y)

x³ = y³

x = y

So, f is an injection 

Surjection condition:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x³= y

x = ³√y which may not be in N.

For example, if y = 3,

X = ³√3 is not in N.

So, f is not a surjection and f is not a bijection.

(iv) Given f: Z → Z given by f(x) = x³

Let us check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y)

f(x) = f(y)

x³ = y³

x = y

So, f is an injection.

Surjection condition:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x³ = y

x = ³√y which may not be in Z.

For example, if y = 3,

x = ³√3 is not in Z.

So, f is not a surjection and f is not a bijection.

(v) Given f: R → R, defined by f(x) = |x|

Let us check for the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in domain (R), such that f(x) = f(y)

f(x) = f(y)

|x| = |y|

x = ± y

Therefore, f is not an injection.

Surjection test:

Let y be any element in co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

|x| = y

x = ± y ∈ Z

Therefore, f is a surjection and f is not a bijection.

Given that f: A → B is an injection

And also given the range of f = {a}

Therefore, the number of images of  f = 1

Since, f  is an injection, there will be exactly one image for each element of f .

Therefore, number of elements in A = 1.

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

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

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

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

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇒ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Here, Range {f} = {a}

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

Thus A has only one element