CBSE Class 12 Mathematics Relations And Functions Assignment Set G

I. Multiple Choice Questions (MCQs)

Question. A function f : N → N defined as y = x² is
(a) one-one
(b) onto
(c) bijective
(d) None of these.

Answer: D

Question. Domain of the function: f : R → R defined as y = x + 4 is
(a) Real numbers
(b) Natural numbers
(c) Negative real numbers
(d) None of these.

Answer: A

Question. Domain of a real valued function f (x) = √36 − x² is:
(a) [–6, 6]
(b) (–6, 6)
(c) (–6, 6]
(d) [–6, 6)

Answer: A

Question. Number of function from set {1, 2} into set {a, b, c} are
(a) (2)³
(b) (3)²
(c) (2)²
(d) (3)³

Answer: B

Question. Number of one-one functions that can be defined from a set {3, 4, 5} into a set {1, 2} are
(a) ²P₃
(b) 0
(c) ²P₂
(d) ³P₃

Answer: B

Question. The function f : R → [–1, 1] defined by f (x) = cos x is
(a) both one-one and onto
(b) not one-one, but onto
(c) one-one, but not onto
(d) neither one-one, nor onto

Answer: B

Question. Every function is a
(a) relation.
(b) onto function
(c) injective function
(d) bijective function

Answer: A

Question. A function which is injective and surjective is called
(a) onto function only
(b) bijective function
(c) one-one function
(d) None of these.

Answer: B

Question. An onto function from set A to set B means every element of set B has a pre-image in
(a) Set B × A
(b) Set A
(c) Set A × B
(d) None of these.

Answer: B

Question. A function f : R → R is defined as f(x) = x³ + 1. Then the function has
(a) no minimum value
(b) no maximum value
(c) both maximum and minimum value
(d) neither maximum value nor minimum value

Answer: B

II. Short Answer Type Questions-I

Question. A function f : N → N is given by f (x) = 2x + 3. Show that f (x) is one-one but not onto.]
Answer: Hint: (i) One-one: Take x₁, x₂ ∈ N such that:f (x₁) = f (x₂)
⇒ 2x₁ + 3 = 2x₂ + 3 ⇒ x₁ = x₂
⇒ f (x) is one-one.
(ii) Onto: Take y = 2x + 3          ⇒ x = (y − 3 / 2)
For every natural number y, (y − 3/2) is not a natural number.
∴ f (x) is not onto. 

Question. A function f : Q → Q is defined as f (x) = 2x + 3. Show that f (x) is one-one and onto.
Answer: Hint: (i) One-one: Take x₁, x₂ ∈ Q such that
f (x₁) = f (x₂)
⇒ 2x₁ + 3 = 2x₂ + 3 ⇒ x₁ = x₂
⇒ f (x) is one-one.
(ii) Onto: Take y = 2x + 3 ⇒ x y =
Now, for every rational number y, y − 3/2) is a rational number.
∴ f (x) is onto.

Question. A function f : N → N is defined as f (x)= 2x² + 3. Show that f (x) is one-one but not onto.
Answer: Hint: (i) One-one: Take x₁, x₂ ∈ N such that
f (x₁) = f (x₂)
⇒ 2x₁² + 3 = 2x 2/2 + 3  ⇒ x₁² = x²₂
⇒ x₁ = x₂          [∴ x₁ and x₂ are natural numbers]
⇒ f (x) is one-one.
(ii) Onto: Take y = 2x² + 3 ⇒ x = √y − 3 / 2
For every natural number y, √y − 3/2 is not a natural number.
∴ f (x) is not onto.

Question. Write Domain and Range of signum function.
Answer: (i) Domain of signum function = real numbers.
(ii) Range of signum function = {–1, 0, 1}. 

III. Short Answer Type Questions-II

Question. Let A be the set of all 50 students of Class X in a school. Let f : A → N be a function defined by f (x) = roll numbers of the student x. Show that f is one-one, but not onto.
Answer: Hint: One-One: Every student has a different roll number from 1 to 50. No two or more students in the class can have same roll number. Therefore, f is one-one. Onto: Roll numbers of the students are from 1 to 50. There is no student whose roll number is 51, 52 or 53. So, every number in the codomain of natural numbers is not a roll number. Therefore, f is not onto.

Question. Let A and B be two sets. Show that f : A × B → B × A such that f (a, b) = (b, a) is bijective function.
Answer: Hint: A and B are two sets: f : A × B → B × A such that f (a, b) = (b, a) is bijective.
One-one: Take two ordered pairs (a₁, b₁) and (a₂, b₂) such that a₁, b₁ ∈ A and a₂, b₂ ∈ B. Now assume that f (a₁, b₁) = f (a₂, b₂)
⇒ (b₁, a₁) = (b₂, a₂) ⇒ b₁ = b₂ and a₁ = a₂
⇒ f is one-one.
Onto: For every element (b, a) from co-domain, there exists a corresponding element (a, b) in domain of f. ∴ f is onto.

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