Important Questions For Class 11 Maths Chapter 2 Relations and Functions are given here based on the new weightage prescribed by CBSE**. **These will help the students with their examination preparation for the exams of the 2020-2021 session. Students can go through with these important questions of Relations and Functions with solutions to score better in the exam.

In this chapter, two important concepts are given namely relations and functions. These concepts have a large scope in mathematics. NCERT book contains all the essential topics related to these concepts. You can refer to the detailed explanations of all the important questions of this chapter here along with the practice questions at the bottom. Students can also access important questions for Class 11 Maths for all the chapters at BYJU’S.

**Read more:**

## Important Questions & Answers For Class 11 Maths Chapter 2 Relations and Functions

**Q.1: Write the range of a Signum function.**

**Solution:**

The real function f: R → R defined by

is called the signum function. Domain of f = R, Range of f = {1, 0, – 1}

**Q.2: The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.**

**Solution:**

We know that,

If n(A) = p and n(B) = q, then n(A × B) = pq

From the given,

n(A × A) = 9

n(A) × n(A) = 9,

n(A) = 3 ……(i)

The ordered pairs (-1, 0) and (0, 1) are two of the nine elements of A × A.

Therefore, A × A = {(a, a) : a ∈ A}

Hence, -1, 0, 1 are the elemets of A. …..(ii)

From (i) and (ii),

A = {-1, 0, 1}

The remaining elements of set A × A are (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0) and (1, 1).

**Q.3: Express the function f: A—R. f(x) = x ^{2} – 1. where A = { -4, 0, 1, 4) as a set of ordered pairs.**

**Solution:**

Given,

A = {-4, 0, 1, 4}

f(x) = x^{2} – 1

f(-4) = (-4)^{2} – 1 = 16 – 1=15

f(0) = (0)^{2} – 1 = -1

f(1) = (1)^{2} – 1 = 0

f(4) = (4)^{2} – 1 = 16 – 1 =15

Therefore, the set of ordered pairs = {(-4, 15), (0, -1), (1, 0), (4, 15)}

**Q.4: Assume that A = {1, 2, 3,…,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, such that x, y ∈ A}. Determine and write down its range, domain, and codomain.**

**Solution:**

It is given that the relation R from A to A is given by R = {(x, y): 3x – y = 0, where x, y ∈ A}.

It means that R = {(x, y) : 3x = y, where x, y ∈ A}

Hence, R = {(1, 3), (2, 6), (3, 9), (4, 12)}

We know that the domain of R is defined as the set of all first elements of the ordered pairs in the given relation.

Hence, the domain of R = {1, 2, 3, 4}

To determine the codomain, we know that the entire set A is the codomain of the relation R.

Therefore, the codomain of R = A = {1, 2, 3,…,14}

As it is known that, the range of R is defined as the set of all second elements in the relation ordered pair.

Hence, the Range of R is given by = {3, 6, 9, 12}

**Q.5: Let f(x) = x ^{2} and g(x) = 2x + 1 be two real functions. Find**

**(f + g) (x), (f –g) (x), (fg) (x), (f/g ) (x)**

**Solution:**

Given,

f(x) = x^{2} and g(x) = 2x + 1

(f + g) (x) = x^{2} + 2x + 1

(f – g) (x) = x^{2} -(2x + 1) = x^{2} – 2x – 1

(fg) (x) = x^{2}(2x + 1) = 2x^{3} + x^{2}

(f/g) (x) = x^{2}/(2x + 1), x ≠ -1/2

**Q.6: Redefine the function: f(x) = |x – 1| – |x + 6|. Write its domain also.**

**Solution:**

Given function is f(x) = |x – 1| – |x + 6|

Redefine of the function is:

The domain of this function is R.

**Q.7: The function f is defined by**

**Draw the graph of f(x).**

**Solution:**

f(x) = 1 – x, x < 0, this gives

f(– 4) = 1 – (– 4)= 5;

f(– 3) =1 – (– 3) = 4,

f(– 2) = 1 – (– 2)= 3

f(–1) = 1 – (–1) = 2; etc,

Also, f(1) = 2, f (2) = 3, f (3) = 4, f(4) = 5 and so on for f(x) = x + 1, x > 0.

Thus, the graph of f is as shown in the below figure.

**Q.8: Find the domain and range of the real function f(x) = x/1+x ^{2}.**

**Solution:**

Given real function is f(x) = x/1+x^{2}.

1 + x^{2} ≠ 0

x^{2} ≠ -1

Domain : x ∈ R

Let f(x) = y

y = x/1+x^{2}

⇒ x = y(1 + x^{2})

⇒ yx^{2} – x + y = 0

This is quadratic equation with real roots.

(-1)^{2} – 4(y)(y) ≥ 0

1 – 4y^{2} ≥ 0

⇒ 4y^{2} ≤ 1

⇒ y^{2} ≤1/4

⇒ -½ ≤ y ≤ ½

⇒ -1/2 ≤ f(x) ≤ ½

Range = [-½, ½]

### Practice Questions For Class 11 Maths Chapter 2 Relations and Functions

- Let A = {1, 2, 3}, B = {4} and C = {5}
- (i) Verify that: A x (B – C) = (A x B) – (A x C)
- (ii) Find (A x B) ∩ (A x C).

- Find x and y if: (i) (4x + 3, y) = (3x + 5, – 2) (ii) (x – y, x + y) = (6, 10)
- Find the domain for which the functions f (x) = 2x
^{2}– 1 and g (x) = 1 – 3x and check whether they are equal. - Find the domain and range of the real function f(x) = 1/(1 – x
^{2}). - A relation R is defined from a set A = {2, 3, 4, 7} to a set B = { 3, 6, 9, 0} as follows R = ((x,y) ∈ R : x is relatively prime to y; x ∈ A, y ∈ B). Express R as a set of ordered pairs and determine the domain and range.
- Draw the graph of the function f: R → R defined by f (x) = x
^{3}, x ∈ R - If R
_{3}= {(x, x) | x is a real number} is a relation, then find the domain and range of R_{3}. - Redefine the function f (x) = |x − 2| + |2 + x| , – 3 ≤ x ≤ 3.
- In each of the following cases, find a and b.

(i)(2a + b, a – b) = (8, 3)

(ii) {a/4, a – 2b) = (0, 6 + b) - If R1 = {(x, y)| y = 2x + 7, where x∈ R and -5 ≤ x ≤ 5} is a relation. Then find the domain and range of R1.
- Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7.

(i) For what real numbers x, f(x) = g(x)?

(ii) For what real numbers x, f (x) < g(x)? - The ordered pair (5, 2) belongs to the relation R ={(x, y): y = x – 5, x, y∈Z}

These Important Questions For Class 11 Maths Chapter 2 Relations and Functions must have helped students in their exam preparation. They can access more study material for CBSE/ICSE/State Board/Competitive exam prepration by visiting BYJU’S website.

Questions are tough but very helpful.

I thank to buyjus for this🙋♂🙋♂

it was really very much helpful