# Class 12 RD Sharma Solutions – Chapter 2 Functions – Exercise 2.1 | Set 1

### (i) Which is one-one but not onto.

Solution:

Let f: R â†’ R given by f(x) = 3x + 2

Let us check one-one condition on f(x) = 3x + 2

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

f (x) = f(y)

â‡’ 3x + 2 =3y + 2

â‡’ 3x = 3y

â‡’ x = y

â‡’ f(x) = f(y)

â‡’ x = y

So, 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

â‡’ 3x + 2 = y

â‡’ 3x = y – 2

â‡’ 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 âˆ‰ 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) Which is not one-one but onto.

Solution:

Example for the function which is not one-one but onto

Let f: Z â†’ N âˆª {0} given by f(x) = |x|

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

Such that f(x) = f(y).

â‡’ |x| = |y|

â‡’ x = Â± y

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

So, f is not 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).

f(x) = y

â‡’ |x| = y

â‡’ x = Â± 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) Which is neither one-one nor onto.

Solution:

Example for the function which is neither one-one nor onto.

Let f: Z â†’ Z given by f(x) = 2xÂ² + 1

Injectivity:

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

f(x) = f(y)

â‡’ 2xÂ²+1 = 2yÂ²+1

â‡’ 2xÂ² = 2yÂ²

â‡’ xÂ² = yÂ²

â‡’ x = Â± 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 the co-domain (Z), such that f(x) = y for some element x in Z

(domain).

f (x) = y

â‡’ 2xÂ²+1=y

â‡’ 2xÂ²= y âˆ’ 1

â‡’ xÂ² = (y-1)/2

â‡’ x = âˆš ((y-1)/2) âˆ‰ Z always.

For example, if we take, y = 4,

x = Â± âˆš ((y-1)/2)

= Â± âˆš ((4-1)/2)

= Â± âˆš (3/2) âˆ‰ Z

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

Thus, f is not onto.

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

Solution:

(i) Consider 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

â‡’ 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}

â‡’ Co-domain = range So, f1 is onto.

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

Solution:

(ii) Consider 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

â‡’ 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}

â‡’ Co-domain = range

So, f2 is onto.

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

Solution:

(iii) Consider 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

â‡’ a and b have the same image x.

Also c and d have the same image z So, f3 is not one-one.

Surjectivity: Co-domain of f3 ={x, y, z}

Range of f3 =set of images = {x, z}

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

So, f3 is not onto.

### Question 3. Prove that the function f: N â†’ N, defined by f(x) = xÂ² + x + 1, is one-one but not onto

Solution:

Given f: N â†’ N, defined by f(x) = xÂ² + x + 1

Now we have to prove that given function is one-one

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

â‡’ xÂ² + x + 1 = yÂ² + y + 1

â‡’ (xÂ² â€“ yÂ²) + (x – y) = 0 `

â‡’ (x + y) (x- y ) + (x – y ) = 0

â‡’ (x – y) (x + y + 1) = 0

â‡’ x – y = 0 [x + y + 1 cannot be zero because x and y are natural numbers

â‡’ x = y

So, f is one-one.

Surjectivity:

When x = 1

xÂ² + x + 1 = 1 + 1 + 1 = 3

â‡’ xÂ² + x +1 â‰¥ 3, for every x in N.

â‡’ f(x) will not assume the values 1 and 2.

So, f is not onto.

### Question 4. Let A = {âˆ’1, 0, 1} and f = {(x, xÂ²) : x âˆˆ A}. Show that f : A â†’ A is neither one-one nor onto.

Solution:

Given A = {âˆ’1, 0, 1} and f = {(x, xÂ²): x âˆˆ A} Also given that, f(x) = xÂ²

Now we have to prove that given function neither one-one or nor onto.

Injectivity: Let x = 1

Therefore f(1) = 1Â²=1 and f(-1)=(-1)Â²=1

â‡’ 1 and -1 have the same images. So, f is not one-one.

Surjectivity: Co-domain of f = {-1, 0, 1}

f(1) = 1Â² = 1, f(-1) = (-1)Â² = 1 and f(0) = 0 â‡’ Range of f = {0, 1} So, both are not same. Hence, f is not onto

### (i) f: N â†’ N given by f(x) = xÂ²

Solution:

Given f: N â†’ N, given by f(x) = xÂ²

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

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

f(x) = f(y)

xÂ² = yÂ²

x = y (We do not get Â± 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

xÂ²= y

x = âˆšy, which may not be in N.

For example, if y = 3,

x = âˆš3 is not in N.

So, f is not a surjection.

Also f is not a bijection.

### (ii) f: Z â†’ Z given by f(x) = xÂ²

Solution:

Given f: Z â†’ Z, given by f(x) = xÂ²

Now we have to 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)

xÂ² = yÂ²

x = Â±y

So, 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

xÂ² = y

x = Â± âˆšy which may not be in Z.

For example, if y = 3, x = Â± âˆš 3 is not in Z.

So, f is not a surjection.

Also f is not bijection.

### (iii) f: N â†’ N given by f(x) = xÂ³

Solution:

Given f: N â†’ N given by f(x) = xÂ³

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

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

f(x) = f(y)

xÂ³ = yÂ³

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

xÂ³= y

x = âˆ›y which may not be in N.

For example, if y = 3,

X = âˆ›3 is not in N.

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

### (iv) f: Z â†’ Z given by f(x) = xÂ³

Solution:

Given f: Z â†’ Z given by f(x) = xÂ³

Now we have to 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)

xÂ³ = yÂ³

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

xÂ³ = y

x = âˆ›y which may not be in Z.

For example, if y = 3,

x = âˆ›3 is not in Z.

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

### (v) f: R â†’ R, defined by f(x) = |x|

Solution:

Given f: R â†’ R, defined by f(x) = |x|

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

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

f(x) = f(y)

|x|=|y|

x = Â±y

So, f is not an injection. Surjection test:

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

f(x) = y

|x|=y

x = Â± y âˆˆ Z

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

### (vi) f: Z â†’ Z, defined by f(x) = xÂ² + x

Solution:

Given f: Z â†’ Z, defined by f(x) = xÂ² + x

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

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

f(x) = f(y)

xÂ²+ x = yÂ² + y

Here, we cannot say that x = y.

For example, x = 2 and y = – 3

Then,

xÂ² + x = 2Â² + 2 = 6

y2 + y = (âˆ’3)Â² â€“ 3 = 6

So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.

So, 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

xÂ² + x = y

Here, we cannot say x âˆˆ Z.

For example, y = – 4.

xÂ² + x = âˆ’ 4

xÂ² + x + 4 = 0

x = (-1 Â± âˆš-5)/2 = (-1 Â± i âˆš5)/2 which is not in Z.

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

### (vii) f: Z â†’ Z, defined by f(x) = x âˆ’ 5

Solution:

Given f: Z â†’ Z, defined by f(x) = x â€“ 5

Now we have to check for the given function is injection, surjection and bijection condition.

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

f(x) = f(y)

x – 5 = y – 5

x = y So, f is 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 x – 5 = y

x = y + 5, which is in Z.

So, f is a surjection and f is a bijection.

### (viii) f: R â†’ R, defined by f(x) = sin x

Solution:

Given f: R â†’ R, defined by f(x) = sin x

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y)

Sin x = sin y

Here, x may not be equal to y because sin 0 = sin Ï€.

So, 0 and Ï€ have the same image 0.

So, f is not an injection. Surjection test:

Range of f = [-1, 1]

Co-domain of f = R Both are not same. So, f is not a surjection and f is not a bijection.

### (ix) f: R â†’ R, defined by f(x) = xÂ³ + 1

Solution:

Given f: R â†’ R, defined by f(x) = xÂ³ + 1

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

xÂ³+1 = yÂ³+ 1

xÂ³= yÂ³

x = y

So, f is an injection.

Surjection test:

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

f(x) = y

xÂ³+1=y

x = âˆ› (y – 1) âˆˆ R

So, f is a surjection.

So, f is a bijection.

### (x) f: R â†’ R, defined by f(x) = xÂ³ âˆ’ x

Solution:

Given f: R â†’ R, defined by f(x) = xÂ³ âˆ’ x

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test:

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

xÂ³ â€“ x = yÂ³ âˆ’ y

Here, we cannot say x = y.

For example, x = 1 and y = -1

xÂ³ âˆ’ x = 1 âˆ’ 1 = 0

yÂ³ â€“ y = (âˆ’1)Â³âˆ’ (âˆ’1) â€“ 1 + 1 = 0

So, 1 and -1 have the same image 0.

So, f is not an injection. Surjection test:

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

f(x) = y

xÂ³ âˆ’ x = y

By observation we can say that there exist some x in R, such that xÂ³ – x = y.

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

### (xi) f: R â†’ R, defined by f(x) = sin2x + cos2x

Solution:

Given f: R â†’ R, defined by f(x) = sin2x + cos2x

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

f(x) = sin2x + cos2x

We know that sin2x + cos2x = 1

So, f(x) = 1 for every x in R.

So, for all elements in the domain, the image is 1.

So, f is not an injection. Surjection condition: Range of f = {1} Co-domain of f = R Both are not same. So, f is not a surjection and f is not a bijection

### (xii) f: Q âˆ’ {3} â†’ Q, defined by f (x) = (2x +3)/(x-3)

Solution:

Given f: Q âˆ’ {3} â†’ Q, defined by f (x) = (2x +3)/(x-3)

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test: Let x and y be any two elements in the domain (Q âˆ’ {3}), such that f(x) = f(y).

f(x) = f(y)

(2x + 3)/(x – 3) = (2y + 3)/(y – 3)

(2x + 3) (y âˆ’ 3) = (2y + 3) (x âˆ’ 3)

2xy âˆ’ 6x + 3y âˆ’ 9 = 2xy âˆ’ 6y + 3x âˆ’ 9

9x = 9y

x = y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q âˆ’ {3}), such that f(x) = y for some element x in Q (domain).

f(x) = y

(2x + 3)/(x – 3) = y

2x + 3 = x y âˆ’ 3y

2x â€“ x y = âˆ’3y âˆ’ 3

x (2âˆ’y) = âˆ’3 (y + 1)

x = -3(y + 1)/(2 – y) which is not defined at y = 2.

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

### (xiii) f: Q â†’ Q, defined by f(x) = xÂ³ + 1

Solution:

Given f: Q â†’ Q, defined by f(x) = xÂ³ + 1

Now we have to check for the given function is injection, surjection and bijection condition.

Injection test: Let x and y be any two elements in the domain (Q), such that f(x) = f(y).

f(x) = f(y)

xÂ³ + 1 = yÂ³ + 1

xÂ³ = yÂ³

x = y

So, f is an injection. Surjection test:

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

f(x) = y

xÂ³+ 1 = y x = âˆ›(y-1), which may not be in Q.

For example, if y= 8,

xÂ³+ 1 = 8

xÂ³= 7 x = âˆ›7, which is not in Q.

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

### (xiv) f: R â†’ R, defined by f(x) = 5xÂ³ + 4

Solution:

Given f: R â†’ R, defined by f(x) = 5xÂ³ + 4

Now we have to check for the given function is injection, surjection and bijection

condition.

Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x) = f(y)

5xÂ³ + 4 = 5yÂ³ + 4

5xÂ³= 5yÂ³

xÂ³ = yÂ³

x = y

So, f is an injection.

Surjection test:

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

f(x) = y

5xÂ³+ 4 = y

xÂ³ = (y – 4)/5 âˆˆ R

So, f is a surjection and f is a bijection.

### (xv) f: R â†’ R, defined by f(x) = 5xÂ³ + 4

Solution:

Given f: R â†’ R, defined by f(x) = 5xÂ³ + 4

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

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

f(x) = f(y)

5xÂ³ + 4 = 5yÂ³ + 4

5xÂ³ = 5yÂ³

xÂ³ = yÂ³

x = y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (R),

such that f(x) = y for some element x in R (domain).

f(x) = y

5xÂ³ + 4 = y

xÂ³ = (y – 4)/5 âˆˆ R

So, f is a surjection and f is a bijection.

### (xvi) f: R â†’ R, defined by f(x) = 1 + xÂ²

Solution:

Given f: R â†’ R, defined by f(x) = 1 + xÂ²

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

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

f(x) = f(y)

1 + xÂ² = 1 + yÂ²

xÂ² = yÂ²

x = Â± y

So, f is not an injection. Surjection test:

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

f(x) = y

1 + xÂ² = y

xÂ² = y âˆ’ 1

x = Â± âˆš-1 = Â± i` is not in R.

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

### (xvii) f: R â†’ R, defined by f(x) = x/(xÂ² + 1)

Solution:

Given f: R â†’ R, defined by f(x) = x/(xÂ² + 1)

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

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

f(x) = f(y)

x /(xÂ² + 1) = y /(yÂ² + 1)

x yÂ²+ x = xÂ²y + y

xyÂ² âˆ’ xÂ²y + x âˆ’ y = 0

âˆ’x y (âˆ’y + x) + 1 (x âˆ’ y) = 0

(x âˆ’ y) (1 â€“ x y) = 0

x = y or x = 1/y

So, f is not an injection. Surjection test:

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

f(x) = y

x /(xÂ² + 1) = y

y xÂ² â€“ x + y = 0

x = (-(-1) Â± âˆš (1-4yÂ²))/(2y) if y â‰  0

= (1 Â± âˆš (1-4yÂ²))/ (2y), which may not be in R

For example, if y=1, then (1 Â± âˆš (1-4)) / (2y) = (1 Â± i âˆš3)/2, which is not in R

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

### Question 6. If f: A â†’ B is an injection, such that range of f = {a}, determine the number of elements in A.

Solution:

Given f: A â†’ B is an injection

And also given that range of f = {a}

So, the number of images of f = 1 Since, f is an injection,

there will be exactly one image for each element of f .

So, number of elements in A = 1.

### Question 7. Show that the function f: R âˆ’ {3} â†’ R âˆ’ {2} given by f(x) = (x-2)/(x-3) is a bijection.

Solution:

Given that f: R âˆ’ {3} â†’ R âˆ’ {2} given by f (x) = (x-2)/(x-3)

Now we have to show that the given function is one-one and on-to

Injectivity: Let x and y be any two elements in the domain (R âˆ’ {3}), such that f(x) = f(y).

f(x) = f(y)

â‡’ (x – 2) /(x – 3) = (y – 2) /(y – 3)

â‡’ (x – 2) (y – 3) = (y – 2) (x – 3)

â‡’ x y â€“ 3 x â€“ 2 y + 6 = x y – 3y – 2x + 6

â‡’ x = y

So, f is one-one.

Surjectivity:

Let y be any element in the co-domain (R âˆ’ {2}), such that f(x) = y for some element x in R âˆ’ {3} (domain).

f(x) = y

â‡’ (x – 2) /(x – 3) = y

â‡’ x – 2 = x y – 3y

â‡’ x y – x = 3y – 2

â‡’ x ( y – 1 ) = 3y – 2

â‡’ x = (3y – 2)/ (y – 1), which is in R – {3}

So, for every element in the co-domain, there exists some pre-image in the domain.

â‡’ f is onto.

Since, f is both one-one and onto,

it is a bijection.

### (i) f (x) = x/2

Solution:

Given f: A â†’ A, given by f (x) = x/2

Now we have to show that the given function is one-one and on-to

Injection test:

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

f(x) = f(y)

x/2 = y/2

x = y

So, f is one-one. Surjection test:

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

f(x) = y

x/2 = y

x = 2y, which may not be in A.

For example, if y = 1, then x = 2, which is not in A.

So, f is not onto.

So, f is not bijective.

### (ii) g (x) = |x|

Solution:

Given g: A â†’ A, given by g (x) = |x|

Now we have to show that the given function is one-one and on-to

Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).

g(x) = g(y)

|x| = |y|

x = Â± y

So, f is not one-one. Surjection test:

For y = -1, there is no value of x in A.

So, g is not onto.

So, g is not bijective.

### (iii) h (x) = xÂ²

Solution:

Given h: A â†’ A, given by h (x) = xÂ²

Now we have to show that the given function is one-one and on-to

Injection test:

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

h(x) = h(y)

xÂ² = yÂ²

x = Â±y

So, f is not one-one. Surjection test:

For y = – 1, there is no value of x in A.

So, h is not onto.

So, h is not bijective.

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

Solution:

Let f = {(x, y): x is a person, y is the mother of x}

As, for each element x in domain set, there is a unique related element y in co-domain set.

So, f is the function.

Injection test: As, y can be mother of two or more persons So, f is not injective.

Surjection test:

For every mother y defined by (x, y), there exists a person x for whom y is mother.

So, f is surjective.

Therefore, f is surjective function.

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

Solution:

Let g = {(a, b): a is a person, b is an ancestor of a}

Since, the ordered map (a, b) does not map ‘a’ – a person to a living person.

So, g is not a function.

### Question 10. Let A = {1, 2, 3}. Write all one-one from A to itself.

Solution:

Given A = {1, 2, 3}

Number of elements in A = 3

Number of one-one functions = number of ways of arranging 3 elements = 3! = 6

(i) {(1, 1), (2, 2), (3, 3)}

(ii) {(1, 1), (2, 3), (3, 2)}

(iii) {(1, 2 ), (2, 2), (3, 3 )}

(iv) {(1, 2), (2, 1), (3, 3)}

(v) {(1, 3), (2, 2), (3, 1)}

(vi) {(1, 3), (2, 1), (3,2 )}

### Question 11. If f: R â†’ R be the function defined by f(x) = 4×3 + 7, show that f is a bijection.

Solution:

Given f: R â†’ R is a function defined by f(x) = 4×3 + 7 Injectivity:

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

â‡’ 4×3 + 7 = 4y3 + 7

â‡’ 4×3 = 4y3

â‡’ x3 = y3

â‡’ x = y

So, f is one-one.

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

f(x) = y

â‡’ 4×3 + 7 = y

â‡’ 4×3 = y âˆ’ 7

â‡’ x3 = (y – 7)/4

â‡’ x = âˆ›(y-7)/4 in R

So, for every element in the co-domain, there exists some pre-image in the domain. f is onto.

Since, f is both one-to-one and onto,

it is a bijection.

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