Students can cross-reference their work with OP Malhotra Class 11 Solutions Chapter 2 Relations and Functions Ex 2(d) to ensure accuracy.
S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(d)
Question 1.
A function f given as f : {(2, 7), (3, 4), (7, 9), (- 1, 6), (0, 2), (5, 3)}.
In this function one-one onto ?
Interchange the order of the elements in the ordered pairs and form the new relation. Is this relation a function ? If it is a function, is it one-one onto.
Solution:
Given function f = {(2, 7), (3, 4), (7, 9), (- 1, 6), (0, 2), (5, 3)}
Since different elements have different images.
∴ function f is one-one.
Also every element in codomain of f have atleast one pre-image in domain of f ∴ f is onto.
Thus f is one-one and onto.
Let g = {(7, 2), (4, 3), (9, 7), (6, – 1), (2, 0), (3, 5)}
Further different elements in domain of g have different images. g is one-one. Corresponding to every element in codomain of g 3 atleast one element in domain of g
∴ g is onto. Thus g is one-one onto.
Question 2.
Determine if each function is one-one.
(i) To each person on the earth assign the number which corresponds to his age.
(ii) To each country in the world assign the latitude and longitude of its capital.
(iii) To each book written by only one author assign the author.
(iv) To each country in the world which has a prime minister assign its prime minister.
Solution:
(i) Since there are many persons on earth which have same age and thus different elements have same image.
∴ given function is many one not one-one.
(ii) Since each country in the world have unique latitude and longitude of its capital.
∴ function is one-one.
(iii) Since there are different books written by the same author. So different elements are having same image.
∴ given function is not one-one.
(iv) Since different countries in the world are having different prime ministers.
Thus f is one-one.
Question 3.
Let f : A → B. Find f (A), i.e., the range of f , if f is an onto function.
Solution:
Given f : A → B if range of f = codomain of f. Then f is onto.
∴ f is onto if f (A) = B
Question 4.
Show that the function f : R → R given by f (x) = cos x for all x ∈ R, is neither one-one nor onto.
Solution:
Given f : R → R defined by f (x) = cos x ∀ x ∈ R
One-one : Since f (0) = cos 0 = 1 ; f (2π) = cos 2π = 1
Thus the elements 0 and 2π in R (domain of f ) have same image 1 in R (codomain of f )
∴ f is many one.
∴ x, y ∈ R s.t f (x) = f (y) ⇒ cos x = cos y ⇒ x = 2nπ ± y ⇒ x ≠ y
∴ f is not one-one.
Since – 1 ≤ cos x ≤ 1
⇒ – 1 ≤ f (x) ≤ 1
∴ Rf = [- 1, 1] ≠ R = codomain of f
Thus f is not onto.
Hence f is neither one-one nor onto.
Question 5.
Let A = {- 1, 1}. Let functions f g and A of A onto A be defined by :
(i) f(x) = x, (ii) g (x) = x³, (iii) h (x) = sin x
Which function, if any, is onto ?
Solution:
Given A = {- 1, 1}
(i) Given f : A → A defined by f(x) = x ∀ x ∈ A
Let y ∈ A be any arbitrary element s.t f (x) = y ⇒ x = y ⇒ x ∈ A.
Thus for every y ∈ A ∃ x ∈ A s.t f (x) = y
∴ f is onto.
(ii) Given g : A → A defined by g (x) = x³ ∀ x ∈ A
g(- 1) = (- 1)³ = – 1 ; g(1) = 1³ = 1
∴ g = {(- 1, – 1), (1, 1)}
Thus y = – 1,1 ∈ A ∃ x = – 1, 1 ∈ A s.t g (x) = y
∴ g is onto.
(iii) Given h : A A is defined by h (x) = sin x
For y = sin x = 1 ⇒ x = \(\frac { π }{ 2 }\) ∉ A
Hence 1 is not the image of any element in the domain i.e. 1 in (codomain of h) has no pre-image in (domain of h).
∴ h is not onto.
Question 6.
Given A = {2, 3, 4}, B = {2, 5, 6, 7}, construct an example of each of the following.
(i) A one-one mapping from A to B
(ii) A mapping from A to B which is not one-one
(iii) A mapping from B to A.
Solution:
Given A = {2, 3, 4} and B = {2, 5, 6, 7}
(i) Let f = {(2, 5), (3, 6), (4, 7)}
Here f different element of f (in domain) have different image in (codomain of f ).
∴ f is one-one mapping from A to B.
(ii) Let g = {(2, 5), (3, 5), (4, 7)}
Here elements 2 and 3 in domain of f have same image 5 in codomain of f
Thus, f is many one.
(iii) Let h = {(2, 2), (5, 3), (6, 4), (7, 4)}
Clearly every element in domain of f has unique image i.e. no two ordered pairs are having same first component.
∴ h be a mapping from B to A.
Question 7.
Are the following sets of ordered pairs functions ? If so, examine whether the mapping is onto or one-one.
(i) {(x, y) : x is a person, y is the mother of
(ii) {(a, b) : a is a person, b is an ancestor of a).
Solution:
(i) Let f = {(x, y): x is a person, y is the mother of x}
Since every person has one and only one mother.
Thus the given set of ordered pair forms a function. Clearly the function f is onto.
Since different persons may have same one mother.
Hence different elements have same image.
Thus f is not one-one.
(ii) Let g = {(a, b): a is a person, b is an ancestor of a}
Since a person has many ancestors (e.g. father, mother, grandmother etc.). Thus every element of g does not have unique image.
∴ The given set of ordered pair does not form a function.
Question 8.
Is the function f : N → N (N is the set of natural numbers) defined by
f (n) = 2n + 3 for all n ∈ N onto ?
Solution:
Given f : N → N defined by f (n) = 2n + 3 ∀ n ∈ N
Let 1 ∈ N (codomain of f ) s.t f(n) = 1 ⇒ 2n + 3 = 1 ⇒ 2n = – 2 ⇒ n = – 1 ∉ N
Thus 1 is not the image of any element in the domain.
Hence the element 1 ∈ N (codomain of f ) has no pre-image in N (domain of f ) ∴ f is not onto.
Question 9.
Let A = {x : 0 ≤ x ≤ 2} and B = {1}. Give an example of a function from A to B. Can you define a function from B and A which is onto ? Give reasons for your answer.
Solution:
Given A = {x : 0 ≤ x ≤ 2} and B = {1}
Let us define a function f : A → B by f(x) = 1 ∀x ∈ A
Clearly each element of A has unique image 1.
∴ f is a function from A to B.
Further set B contains only one element 1 and set A contains infinite number of elements lying between 0 and 2 including 0 and 2.
So it is not possible to define a function from B to A which is onto.
Question 10.
Prove that the function f : R → R, f (x) = x² + x is a many-one into function.
Solution:
Given f : R → R defined by f (x) = x² + x ∀ x ∈ R
Here f (1) = 1² + 1 = 2
f (- 2) = (- 2)² – 2 = 2
Thus elements 1, – 2 ∈ R (domain of f ) have same image 2 (codomain of f ) ∴ f is many one. Since different elements have same image.
∀x, y ∈ R s.t f (x) = f (y) ⇒ x² + x = y² + y
⇒ x² – y² + x – y = 0 ⇒ (x – y)(x + y) + (x – y) = 0
⇒ (x – y) (x + y + 1) = 0
⇒ x = y or x = – y – 1
∴ x ≠ y
Thus f is not one-one.
Let – 1 ∈ R (codomain of f ) s.t f(x) = – 1
⇒ x² + x = – 1 ⇒ x² + x + 1 ⇒ x = \(\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm \sqrt{3} i}{2} \notin \mathrm{R}\)
Thus, – 1 is not the image of any element of the domain.
∴ f is not onto.
Thus f is many one and into.
Question 11.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one but not onto.
Solution:
Given A = {1,2,3} and B = {4, 5, 6, 7}
and f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.
Df = {1, 2, 3} and Rf = {4, 5, 6}
Clearly different elements in Df have different images.
∴ f is one-one.
Further Rf ≠ codomain of f
∴ f is not onto.
Hence f is one-one and into.
Question 12.
Show that the function f : R → R by f(x) = 3 – 4x is one-one onto and hence bijective.
Solution:
Given function f : R → R defined by f (x) = 3 – 4x ∀ x ∈ R
One-one : ∀ x, y ∈R s.t f (x) = f (y) ⇒ 3 – 4x = 3 – 4y ⇒ 4x = 4y ⇒ x = y
∴ f is one-one.
Onto : Let y ∈ R (codomain of f) be any arbitrary element
Then f(x) = y ⇒ 3 – 4x = y ⇒ 4x = 3 – y ⇒ x = \(\frac{3-y}{4}\)
since y ∈ R ⇒ \(\frac{3-y}{4}\) ∈ R ⇒ x ∈ R
Thus ∀ y ∈ R (codomain of f ) ∃ x ∈ R (domain of f )
s.t f(x) = f(\(\frac{3-y}{4}\)) = 3 – 4(\(\frac{3-y}{4}\)) = 3 – 3 + y = y
∴ f is onto.
Hence f is one-one and onto.
∴ f is bijective.