# Class 11 Maths MCQ – Functions

This set of Class 11 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Functions”.

1. In a function from set A to set B, every element of set A has___________ image in set B.
a) one and only one
b) different
c) same
d) many

Explanation: A relation from a set A to a set B is said to be a function if every element of set A has one and one image in set B.

2. In a function from set A to set B, image can have more than one preimage.
a) True
b) False

Explanation: A relation from a set A to a set B is said to be a function if every element of set A has one and one image in set B. A preimage must have one image, an image can have more than one preimage.

3. Let R be a relation defined on set of natural numbers {(x, y): y=2x}. Is this relation can be called a function?
a) True
b) False

Explanation: Since every natural number has one and only image so this relation can be called a function.

4. Which of the following is not a function?
a) {(1,2), (2,4), (3,6)}
b) {(-1,1), (-2,4), (2,4)}
c) {(1,2), (1,4), (2,5), (3,8)}
d) {(1,1), (2,2), (3,3)}

Explanation: A relation from a set A to a set B is said to be a function if every element of set A has one and one image in set B.
In {(1,2), (1,4), (2,5), (3,8)}, since element 1 has two images 2 and 4 which is not possible in a function so, it is not a function. Rest all have one and only one image so they can be called a function.

5. Which function is shown in graph?

a) Constant
b) Modulus
c) Identity
d) Signum function

Explanation: {(1,1), (2,2), (3,3), …..}. This function involves relation {(x, y): y=x} which is involved in identity function. So, above function is identity function.
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6. Which function is shown in graph?

a) Constant
b) Modulus
c) Identity
d) Signum function

Explanation: {(1,5), (2,5), (3,5),……}. This function involves relation {(x, y): y=constant} which is involved in constant function. So, above function is constant function.

7. Which function is shown in graph?

a) Constant
b) Modulus
c) Identity
d) Signum function

Explanation: {(-1,1), (1,1), (-2,2), (2,2), (-3,3), (3,3), ……}. This function involves relation {(x, y): y = |x|} which is involved in modulus function. So, above function is modulus function.

8. f(x) = {$$\frac{|x|}{x}$$ for x≠0 and 0 for x=0}. Which function is this?
a) Constant
b) Modulus
c) Identity
d) Signum function

Explanation: f(x) = {$$\frac{|x|}{x}$$ for x≠0 and 0 for x=0}. Function is {(-3, -1), (-2, -1), (-1,1), (0,0), (1,1), (2,1), (3,1), …….}. This is signum function.

9. Find domain of function |x|.
a) Set of real numbers
b) Set of positive real numbers
c) Set of integers
d) Set of natural numbers

Explanation: Since the above function can have all real values of x. So, domain is set of real numbers.

10. Find range of function |x|.
a) Set of real numbers
b) Set of positive real numbers
c) Set of integers
d) Set of natural numbers

Explanation: Since the above function can have positive real value of y for all real values of x. So, range is set of positive real numbers.

11. f(x) = $$\sqrt{9-x^2}$$. Find the domain of the function.
a) (0,3)
b) [0,3]
c) [-3,3]
d) (-3,3)

Explanation: We know radical cannot be negative. So, 9-x,2 ≥ 0
(3-x) (3+x) ≥ 0 => (x-3) (x+3) ≤ 0 => x∈[-3,3].

12. f(x) = $$\sqrt{9-x^2}$$. Find the range of the function.
a) R
b) R+
c) [-3,3]
d) [0,3]

Explanation: We know, square root is always non-negative. So, $$\sqrt{9-x^2}$$ ≥ 0. So, the range of the function is set of positive real numbers from 0 to 3.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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