Class 12 RD Sharma Mathematics Solutions - Chapter 2 Functions - Exercise 2.1 | Set 1

Chapter 2 of RD Sharma's Class 12 Mathematics textbook delves into the concept of functions which is foundational in the higher mathematics. Exercise 2.1 | Set 1 introduces students to the basic concepts and types of functions laying the groundwork for the more advanced topics. Understanding functions is crucial for the solving the various mathematical problems in the calculus, algebra and beyond.

Functions

A function is a special relationship between the two sets where each element of the first set known as the domain is related to the exactly one element of the second set called the range. The Functions are often represented as f(x) where x is an element of the domain and f(x) is the corresponding element in the range. The Functions can be classified into different types such as the linear, quadratic, polynomial and trigonometric based on their behavior and form.

Class 12 RD Sharma Mathematics Solutions -Functions - Exercise 2.1 | Set 1

Question 1. Give an example of a function

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

Solution:

Let f: R → 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)

⇒ 3x + 2 =3y + 2

⇒ 3x = 3y

⇒ x = y

⇒ f(x) = f(y)

⇒ 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

⇒ 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) Which is not one-one but onto.

Solution:

Example for the 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 the domain (Z),

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

⇒ |x| = |y|

⇒ x = ± 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

⇒ |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) Which is neither one-one nor onto.

Solution:

Example for the 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 the 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

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

Thus, f is not onto.

Question 2. Which of the following functions from A to B are one-one and 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

⇒ 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) 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

⇒ 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) 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

⇒ 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 → N, defined by f(x) = x² + x + 1, is one-one but not onto

Solution:

Given f: N → N, defined by f(x) = x² + 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).

⇒ x² + x + 1 = y² + y + 1

⇒ (x² – y²) + (x - y) = 0 `

⇒ (x + y) (x- y ) + (x - y ) = 0

⇒ (x - y) (x + y + 1) = 0

⇒ x - y = 0 [x + y + 1 cannot be zero because x and y are natural numbers

⇒ x = y

So, f is one-one.

Surjectivity:

When x = 1

x² + x + 1 = 1 + 1 + 1 = 3

⇒ x² + x +1 ≥ 3, for every x in N.

⇒ f(x) will not assume the values 1 and 2.

So, f is not onto.

Question 4. Let A = {−1, 0, 1} and f = {(x, x²) : x ∈ A}. Show that f : A → A is neither one-one nor onto.

Solution:

Given A = {−1, 0, 1} and f = {(x, x²): x ∈ A} Also given that, f(x) = x²

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

Injectivity: Let x = 1

Therefore f(1) = 1²=1 and f(-1)=(-1)²=1

⇒ 1 and -1 have the same images. So, f is not one-one.

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

f(1) = 1² = 1, f(-1) = (-1)² = 1 and f(0) = 0 ⇒ Range of f = {0, 1} So, both are not same. Hence, f is not onto

Question 5. Classify the following function as injection, surjection or bijection:

(i) f: N → N given by f(x) = x²

Solution:

Given f: N → N, given by f(x) = x²

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)

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) f: Z → Z given by f(x) = x²

Solution:

Given f: Z → Z, given by f(x) = x²

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)

x² = y²

x = ±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

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) f: N → N given by f(x) = x³

Solution:

Given f: N → N given by f(x) = x³

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)

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 = ∛y which may not be in N.

For example, if y = 3,

X = ∛3 is not in N.

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

(iv) f: Z → Z given by f(x) = x³

Solution:

Given f: Z → Z given by f(x) = x³

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)

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 = ∛y which may not be in Z.

For example, if y = 3,

x = ∛3 is not in Z.

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

(v) f: R → R, defined by f(x) = |x|

Solution:

Given f: R → 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 = ±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 = ± y ∈ Z

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

(vi) f: Z → Z, defined by f(x) = x² + x

Solution:

Given f: Z → Z, defined by f(x) = x² + 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)

x²+ x = y² + y

Here, we cannot say that x = y.

For example, x = 2 and y = - 3

Then,

x² + x = 2² + 2 = 6

y2 + y = (−3)² – 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

x² + x = y

Here, we cannot say x ∈ Z.

For example, y = - 4.

x² + x = − 4

x² + x + 4 = 0

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

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

(vii) f: Z → Z, defined by f(x) = x − 5

Solution:

Given f: Z → Z, defined by f(x) = x – 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 → R, defined by f(x) = sin x

Solution:

Given f: R → 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 π.

So, 0 and π 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 → R, defined by f(x) = x³ + 1

Solution:

Given f: R → R, defined by f(x) = x³ + 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)

x³+1 = y³+ 1

x³= y³

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

x³+1=y

x = ∛ (y - 1) ∈ R

So, f is a surjection.

So, f is a bijection.

(x) f: R → R, defined by f(x) = x³ − x

Solution:

Given f: R → R, defined by f(x) = x³ − 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)

x³ – x = y³ − y

Here, we cannot say x = y.

For example, x = 1 and y = -1

x³ − x = 1 − 1 = 0

y³ – y = (−1)³− (−1) – 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

x³ − x = y

By observation we can say that there exist some x in R, such that x³ - x = y.

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

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

Solution:

Given f: R → 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 − {3} → Q, defined by f (x) = (2x +3)/(x-3)

Solution:

Given f: Q − {3} → 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 − {3}), such that f(x) = f(y).

f(x) = f(y)

(2x + 3)/(x - 3) = (2y + 3)/(y - 3)

(2x + 3) (y − 3) = (2y + 3) (x − 3)

2xy − 6x + 3y − 9 = 2xy − 6y + 3x − 9

9x = 9y

x = y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q − {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 − 3y

2x – x y = −3y − 3

x (2−y) = −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 → Q, defined by f(x) = x³ + 1

Solution:

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

x³ + 1 = y³ + 1

x³ = y³

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

x³+ 1 = y x = ∛(y-1), which may not be in Q.

For example, if y= 8,

x³+ 1 = 8

x³= 7 x = ∛7, which is not in Q.

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

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

Solution:

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

5x³ + 4 = 5y³ + 4

5x³= 5y³

x³ = y³

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

5x³+ 4 = y

x³ = (y - 4)/5 ∈ R

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

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

Solution:

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

5x³ + 4 = 5y³ + 4

5x³ = 5y³

x³ = y³

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

5x³ + 4 = y

x³ = (y - 4)/5 ∈ R

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

(xvi) f: R → R, defined by f(x) = 1 + x²

Solution:

Given f: R → R, defined by f(x) = 1 + x²

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 + x² = 1 + y²

x² = y²

x = ± 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 + x² = y

x² = y − 1

x = ± √-1 = ± i` is not in R.

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

(xvii) f: R → R, defined by f(x) = x/(x² + 1)

Solution:

Given f: R → R, defined by f(x) = x/(x² + 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 /(x² + 1) = y /(y² + 1)

x y²+ x = x²y + y

xy² − x²y + x − y = 0

−x y (−y + x) + 1 (x − y) = 0

(x − y) (1 – 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 /(x² + 1) = y

y x² – x + y = 0

x = (-(-1) ± √ (1-4y²))/(2y) if y ≠ 0

= (1 ± √ (1-4y²))/ (2y), which may not be in R

For example, if y=1, then (1 ± √ (1-4)) / (2y) = (1 ± i √3)/2, which is not in R

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

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

Solution:

Given f: A → 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 − {3} → R − {2} given by f(x) = (x-2)/(x-3) is a bijection.

Solution:

Given that f: R − {3} → R − {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 − {3}), such that f(x) = f(y).

f(x) = f(y)

⇒ (x - 2) /(x - 3) = (y - 2) /(y - 3)

⇒ (x - 2) (y - 3) = (y - 2) (x - 3)

⇒ x y – 3 x – 2 y + 6 = x y - 3y - 2x + 6

⇒ x = y

So, f is one-one.

Surjectivity:

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

f(x) = y

⇒ (x - 2) /(x - 3) = y

⇒ x - 2 = x y - 3y

⇒ x y - x = 3y - 2

⇒ x ( y - 1 ) = 3y - 2

⇒ 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.

⇒ f is onto.

Since, f is both one-one and onto, 

it is a bijection.

Question 8. Let A = [-1, 1]. Then, discuss whether the following function from A to itself is one-one, onto or bijective:

(i) f (x) = x/2

Solution:

Given f: A → 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 → 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 = ± 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) = x²

Solution:

Given h: A → A, given by h (x) = x²

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)

 x² = y²

x = ±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.

Question 9. Are the following set of ordered pair of a function? If so, examine whether the mapping is injective or surjective:

(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 → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.

Solution:

Given f: R → R is a function defined by f(x) = 4x3 + 7 Injectivity:

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

⇒ 4x3 + 7 = 4y3 + 7

⇒ 4x3 = 4y3

⇒ x3 = y3

⇒ 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

⇒ 4x3 + 7 = y

⇒ 4x3 = y − 7

⇒ x3 = (y - 7)/4

⇒ x = ∛(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.

Read more,

• What is a Function in Maths?
• Types of Functions
• Class 12 RD Sharma Mathematics Solutions - Chapter 2 Functions - Exercise 2.2
• Class 12 RD Sharma Mathematics Solutions - Chapter 2 Functions - Exercise 2.3

Conclusion

Exercise 2.1, Set 1, of Chapter 2 in RD Sharma's Class 12 Mathematics book provides a comprehensive introduction to the concept of functions. Mastering these foundational topics is crucial for success in calculus and other advanced mathematical studies. This exercise focuses on the basics of functions, including:

• Definition of a function
• Domain and codomain of functions
• Types of functions (one-to-one, onto, bijective)
• Representation of functions (set notation, arrow diagrams, graphs)
• Composition of functions
• Inverse functions

The problems in this set typically involve identifying functions, determining their properties, and working with different function representations.