This MCQ module is based on: Relations, Functions, Domain, Range, and Graphs
TOPIC 6 OF 45
Relations, Functions, Domain, Range, and Graphs
🎓 Class 11
Mathematics
CBSE
Theory
Ch 2 — Relations and Functions
⏱ ~25 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Relations, Functions, Domain, Range, and Graphs
Targeting Class 11 level in Functions, with Advanced difficulty.
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2.3 Relations
Consider two sets \(P = \{a, b, c\}\) and \(Q = \{\text{Ali, Bhanu, Binoy, Chandra, Divya}\}\). The Cartesian product of P and Q contains 15 ordered pairs. We can now obtain a subset of \(P \times Q\) by introducing a relation? R between the first element \(x\) and the second element \(y\) of each ordered pair \((x, y)\) as:
R = \(\{(x, y) : x \text{ is the first letter of the name } y,\; x \in P,\; y \in Q\}\)
Then R = \(\{(a, \text{Ali}), (b, \text{Bhanu}), (b, \text{Binoy}), (c, \text{Chandra})\}\).
Fig 2.4 -- Arrow diagram showing the relation R (first letter of name)
Definition 2 -- Relation
A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product \(A \times B\). The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in \(A \times B\). The second element is called the image of the first element.
Definition 3 -- Domain
The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain? of the relation R.
Definition 4 -- Range and Codomain
The set of all second elements in a relation R from set A to set B is called the range? of the relation R. The whole set B is called the codomain? of the relation R. Note that range \(\subseteq\) codomain.
Remarks
(i) A relation may be represented algebraically either by the Roster method or by the Set-builder method.(ii) An arrow diagram is a visual representation of a relation.
(iii) The total number of relations that can be defined from a set A to a set B is the number of possible subsets of \(A \times B\). If \(n(A) = p\) and \(n(B) = q\), then \(n(A \times B) = pq\) and the total number of relations is \(2^{pq}\).
Worked Examples -- Relations
Example 7
Let \(A = \{1, 2, 3, 4, 5, 6\}\). Define a relation R from A to A by \(R = \{(x, y) : y = x + 1\}\).
(i) Depict this relation using an arrow diagram. (ii) Write down the domain, codomain and range of R.
Solution
(i) By the definition of the relation, R = \(\{(1,2), (2,3), (3,4), (4,5), (5,6)\}\).
(ii) We can see that the domain = \(\{1, 2, 3, 4, 5\}\).
Similarly, the range = \(\{2, 3, 4, 5, 6\}\) and the codomain = \(\{1, 2, 3, 4, 5, 6\}\).
(ii) We can see that the domain = \(\{1, 2, 3, 4, 5\}\).
Similarly, the range = \(\{2, 3, 4, 5, 6\}\) and the codomain = \(\{1, 2, 3, 4, 5, 6\}\).
Fig 2.5 -- Arrow diagram: R = {(x, y) : y = x + 1} on set A
Example 8
The figure (Fig 2.6) shows a relation between the sets P and Q. Write this relation (i) in set-builder form, (ii) in roster form. What is its domain and range?
Fig 2.6 -- Relation: "x is the square of y"
Solution
It is obvious that the relation R is "x is the square of y".
(i) In set-builder form, R = \(\{(x, y) : x \text{ is the square of } y,\; x \in P,\; y \in Q\}\).
(ii) In roster form, R = \(\{(9, 3), (9, -3), (4, 2), (4, -2), (25, 5), (25, -5)\}\).
The domain of this relation is \(\{4, 9, 25\}\).
The range of this relation is \(\{-2, 2, -3, 3, -5, 5\}\).
Note that the element 1 is not related to any element in set P.
The set Q is the codomain of this relation.
(i) In set-builder form, R = \(\{(x, y) : x \text{ is the square of } y,\; x \in P,\; y \in Q\}\).
(ii) In roster form, R = \(\{(9, 3), (9, -3), (4, 2), (4, -2), (25, 5), (25, -5)\}\).
The domain of this relation is \(\{4, 9, 25\}\).
The range of this relation is \(\{-2, 2, -3, 3, -5, 5\}\).
Note that the element 1 is not related to any element in set P.
The set Q is the codomain of this relation.
Important Note
The total number of relations that can be defined from a set A to a set B is the number of possible subsets of \(A \times B\). If \(n(A) = p\) and \(n(B) = q\), then \(n(A \times B) = pq\) and the total number of relations is \(2^{pq}\).
Example 9
Let \(A = \{1, 2\}\) and \(B = \{3, 4\}\). Find the number of relations from A into B.
Solution
We have \(A \times B = \{(1,3), (1,4), (2,3), (2,4)\}\).
Since \(n(A \times B) = 4\), the number of subsets of \(A \times B\) is \(2^4\). Therefore, the number of relations from A into B will be \(2^4 = 16\).
Since \(n(A \times B) = 4\), the number of subsets of \(A \times B\) is \(2^4\). Therefore, the number of relations from A into B will be \(2^4 = 16\).
Remark
A relation R from A to A is also stated as a relation on A.
Exercise 2.2
Q1. Let \(A = \{1, 2, 3, \ldots, 14\}\). Define a relation R from A to A by \(R = \{(x, y) : 3x - y = 0,\; x, y \in A\}\). Write down its domain, codomain and range.
The relation gives \(y = 3x\). For \(x, y \in A\):
\(x = 1 \Rightarrow y = 3\), \(x = 2 \Rightarrow y = 6\), \(x = 3 \Rightarrow y = 9\), \(x = 4 \Rightarrow y = 12\).
For \(x = 5\), \(y = 15 \notin A\). So we stop.
R = \(\{(1,3), (2,6), (3,9), (4,12)\}\).
Domain = \(\{1, 2, 3, 4\}\)
Codomain = \(\{1, 2, 3, \ldots, 14\}\) (the whole set A)
Range = \(\{3, 6, 9, 12\}\)
\(x = 1 \Rightarrow y = 3\), \(x = 2 \Rightarrow y = 6\), \(x = 3 \Rightarrow y = 9\), \(x = 4 \Rightarrow y = 12\).
For \(x = 5\), \(y = 15 \notin A\). So we stop.
R = \(\{(1,3), (2,6), (3,9), (4,12)\}\).
Domain = \(\{1, 2, 3, 4\}\)
Codomain = \(\{1, 2, 3, \ldots, 14\}\) (the whole set A)
Range = \(\{3, 6, 9, 12\}\)
Q2. Define a relation R on the set N of natural numbers by \(R = \{(x, y) : y = x + 5,\; x\) is a natural number less than \(4;\; x, y \in \mathbf{N}\}\). Depict this relationship using roster form. Write down the domain and the range.
\(x\) is a natural number less than 4, so \(x \in \{1, 2, 3\}\).
\(x = 1 \Rightarrow y = 6\), \(x = 2 \Rightarrow y = 7\), \(x = 3 \Rightarrow y = 8\).
R = \(\{(1, 6), (2, 7), (3, 8)\}\).
Domain = \(\{1, 2, 3\}\), Range = \(\{6, 7, 8\}\).
\(x = 1 \Rightarrow y = 6\), \(x = 2 \Rightarrow y = 7\), \(x = 3 \Rightarrow y = 8\).
R = \(\{(1, 6), (2, 7), (3, 8)\}\).
Domain = \(\{1, 2, 3\}\), Range = \(\{6, 7, 8\}\).
Q3. \(A = \{1, 2, 3, 5\}\) and \(B = \{4, 6, 9\}\). Define a relation R from A to B by \(R = \{(x, y) :\) the difference between \(x\) and \(y\) is odd; \(x \in A, y \in B\}\). Write R in roster form.
We check all pairs where \(|x - y|\) is odd:
For \(x = 1\): \(|1-4| = 3\) (odd), \(|1-6| = 5\) (odd), \(|1-9| = 8\) (even).
For \(x = 2\): \(|2-4| = 2\) (even), \(|2-6| = 4\) (even), \(|2-9| = 7\) (odd).
For \(x = 3\): \(|3-4| = 1\) (odd), \(|3-6| = 3\) (odd), \(|3-9| = 6\) (even).
For \(x = 5\): \(|5-4| = 1\) (odd), \(|5-6| = 1\) (odd), \(|5-9| = 4\) (even).
R = \(\{(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)\}\).
For \(x = 1\): \(|1-4| = 3\) (odd), \(|1-6| = 5\) (odd), \(|1-9| = 8\) (even).
For \(x = 2\): \(|2-4| = 2\) (even), \(|2-6| = 4\) (even), \(|2-9| = 7\) (odd).
For \(x = 3\): \(|3-4| = 1\) (odd), \(|3-6| = 3\) (odd), \(|3-9| = 6\) (even).
For \(x = 5\): \(|5-4| = 1\) (odd), \(|5-6| = 1\) (odd), \(|5-9| = 4\) (even).
R = \(\{(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)\}\).
Q4. The Fig 2.7 shows a relationship between the sets P and Q. Write this relation (i) in set-builder form (ii) roster form. What is its domain and range?
(i) In set-builder form: R = \(\{(x, y) : y = x - 2,\; x \in P\}\).
(ii) In roster form: R = \(\{(5, 3), (6, 4), (7, 5)\}\).
Domain = \(\{5, 6, 7\}\), Range = \(\{3, 4, 5\}\).
(ii) In roster form: R = \(\{(5, 3), (6, 4), (7, 5)\}\).
Domain = \(\{5, 6, 7\}\), Range = \(\{3, 4, 5\}\).
Q5. Let \(A = \{1, 2, 3, 4, 6\}\). Let R be the relation on A defined by \(\{(a, b) : a, b \in A,\; b \text{ is exactly divisible by } a\}\).
(i) Write R in roster form. (ii) Find the domain of R. (iii) Find the range of R.
(i) Write R in roster form. (ii) Find the domain of R. (iii) Find the range of R.
(i) R = \(\{(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)\}\).
(ii) Domain of R = \(\{1, 2, 3, 4, 6\}\).
(iii) Range of R = \(\{1, 2, 3, 4, 6\}\).
(ii) Domain of R = \(\{1, 2, 3, 4, 6\}\).
(iii) Range of R = \(\{1, 2, 3, 4, 6\}\).
Q6. Determine the domain and range of the relation R defined by \(R = \{(x, x + 5) : x \in \{0, 1, 2, 3, 4, 5\}\}\).
R = \(\{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\}\).
Domain = \(\{0, 1, 2, 3, 4, 5\}\).
Range = \(\{5, 6, 7, 8, 9, 10\}\).
Domain = \(\{0, 1, 2, 3, 4, 5\}\).
Range = \(\{5, 6, 7, 8, 9, 10\}\).
Q7. Write the relation \(R = \{(x, x^3) : x \text{ is a prime number less than 10}\}\) in roster form.
Prime numbers less than 10 are \(\{2, 3, 5, 7\}\).
R = \(\{(2, 8), (3, 27), (5, 125), (7, 343)\}\).
R = \(\{(2, 8), (3, 27), (5, 125), (7, 343)\}\).
Q8. Let \(A = \{x, y, z\}\) and \(B = \{1, 2\}\). Find the number of relations from A to B.
\(n(A \times B) = 3 \times 2 = 6\).
Number of relations = \(2^6 = 64\).
Number of relations = \(2^6 = 64\).
Q9. Let R be the relation on \(\mathbf{Z}\) defined by \(R = \{(a, b) : a, b \in \mathbf{Z},\; a - b \text{ is an integer}\}\). Find the domain and range of R.
Since the difference of any two integers is always an integer, every pair \((a, b)\) where \(a, b \in \mathbf{Z}\) belongs to R.
Therefore R = \(\mathbf{Z} \times \mathbf{Z}\).
Domain = \(\mathbf{Z}\) and Range = \(\mathbf{Z}\).
Therefore R = \(\mathbf{Z} \times \mathbf{Z}\).
Domain = \(\mathbf{Z}\) and Range = \(\mathbf{Z}\).
2.4 Functions
In this section, we study a special type of relation called a function?. It is one of the most important concepts in mathematics. We can visualise a function as a rule that produces new elements out of some given elements. There are many terms such as "map" or "mapping" used to denote a function.
Definition 5 -- Function
A relation \(f\) from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.In other words, a function \(f\) is a relation from a non-empty set A to a non-empty set B such that the domain of \(f\) is A and no two distinct ordered pairs in \(f\) have the same first element.
If \(f\) is a function from A to B and \((a, b) \in f\), then \(f(a) = b\), where \(b\) is called the image of \(a\) under \(f\) and \(a\) is called the preimage of \(b\) under \(f\).
The function \(f\) from A to B is denoted by \(f: A \rightarrow B\).
Looking at the previous examples, we can easily see that the relation in Example 7 is a function because every element in the domain has a unique image. However, the relation in Example 8 is not a function because elements in the domain are connected to more than one image. Similarly, the relation in Example 9 is also not a function (why?).
Worked Examples -- Functions
Example 10
Let N be the set of natural numbers and the relation R be defined on N such that \(R = \{(x, y) : y = 2x,\; x, y \in \mathbf{N}\}\). What is the domain, codomain and range of R? Is this relation a function?
Solution
The domain of R is the set of natural numbers N. The codomain is also N.
The range is the set of even natural numbers.
Since every natural number \(n\) has one and only one image \(2n\), this relation is a function.
The range is the set of even natural numbers.
Since every natural number \(n\) has one and only one image \(2n\), this relation is a function.
Example 11
Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not.
(i) R = \(\{(2,1),(3,1),(4,2)\}\) (ii) R = \(\{(2,2),(2,4),(3,3),(4,4)\}\) (iii) R = \(\{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)\}\)
Solution
(i) Since 2, 3, 4 are the elements of the domain of R having their unique images, this relation R is a function.
(ii) Since the same first element 2 corresponds to two different images 2 and 4, this relation is not a function.
(iii) Since every element has one and only one image, this relation is a function.
(ii) Since the same first element 2 corresponds to two different images 2 and 4, this relation is not a function.
(iii) Since every element has one and only one image, this relation is a function.
Definition 6 -- Real Valued Function
A function which has either R or one of its subsets as its range is called a real valued function. Further, if its domain is also either R or a subset of R, it is called a real function.
Example 12
Let N be the set of natural numbers. Define a real valued function \(f: \mathbf{N} \rightarrow \mathbf{N}\) by \(f(x) = 2x + 1\). Using this definition, complete the table below.
| \(x\) | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| \(y = f(x)\) | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
Interactive: Function Evaluator
Enter a function rule and an input value to compute the output
f(3) = 7
2.4.1 Some Functions and Their Graphs
(i) Identity Function
Let \(\mathbf{R}\) be the set of real numbers. Define the real valued function \(f: \mathbf{R} \rightarrow \mathbf{R}\) by \(f(x) = x\) for each \(x \in \mathbf{R}\). Such a function is called the identity function?. Here the domain and range of \(f\) are \(\mathbf{R}\). The graph is a straight line passing through the origin (Fig 2.8).
Fig 2.8 -- Graph of the identity function f(x) = x
(ii) Constant Function
Define the function \(f: \mathbf{R} \rightarrow \mathbf{R}\) by \(y = f(x) = c\), \(x \in \mathbf{R}\), where \(c\) is a constant and each \(x \in \mathbf{R}\). Here the domain of \(f\) is \(\mathbf{R}\) and its range is \(\{c\}\) (Fig 2.9).
Fig 2.9 -- Graph of a constant function f(x) = 5
(iii) Polynomial Function
A function \(f: \mathbf{R} \rightarrow \mathbf{R}\) is said to be a polynomial function if for each \(x \in \mathbf{R}\), \(f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n\), where \(n\) is a non-negative integer and \(a_0, a_1, a_2, \ldots, a_n \in \mathbf{R}\).
The functions defined by \(f(x) = x^3 - x^2 + 2\), and \(g(x) = x^7 + \sqrt{2}\) are some examples of polynomial functions, whereas the function \(h(x) = x^3 + \sqrt[3]{x} - 2x\) is not a polynomial function (why?).
Example 13
Define the function \(f: \mathbf{R} \rightarrow \mathbf{R}\) by \(f(x) = x^2\), \(x \in \mathbf{R}\). Complete the table given below. Draw the graph of \(f\).
| \(x\) | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| \(y = f(x) = x^2\) | 16 | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
Fig 2.10 -- Graph of \(f(x) = x^2\) (parabola)
(iv) Rational Function
Rational functions are functions of the type \(\frac{f(x)}{g(x)}\), where \(f(x)\) and \(g(x)\) are polynomial functions of \(x\) defined in a domain, where \(g(x) \neq 0\).
Example 15
Define the real valued function \(f: \mathbf{R} - \{0\} \rightarrow \mathbf{R}\) defined by \(f(x) = \frac{1}{x}\), \(x \in \mathbf{R} - \{0\}\). Complete the table and draw the graph.
| \(x\) | -4 | -2 | -1 | -0.5 | 0.5 | 1 | 2 | 4 |
|---|---|---|---|---|---|---|---|---|
| \(f(x) = \frac{1}{x}\) | -0.25 | -0.5 | -1 | -2 | 2 | 1 | 0.5 | 0.25 |
The domain is all real numbers except 0, and the range is also all real numbers except 0 (Fig 2.12).
(v) The Modulus Function
The function \(f: \mathbf{R} \rightarrow \mathbf{R}\) defined by \(f(x) = |x|\) for each \(x \in \mathbf{R}\) is called the modulus function?. For each non-negative value of \(x\), \(f(x)\) equals \(x\). For negative values of \(x\), the value of \(f(x)\) is the negation of \(x\), i.e.:
\[f(x) = |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}\]Fig 2.13 -- Graph of the modulus function f(x) = |x|
(vi) Signum Function
The function \(f: \mathbf{R} \rightarrow \mathbf{R}\) defined by:
\[f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}\]is called the signum function?. The domain is \(\mathbf{R}\) and the range is \(\{-1, 0, 1\}\) (Fig 2.14).
Fig 2.14 -- Graph of the signum function
(vii) Greatest Integer Function
The function \(f: \mathbf{R} \rightarrow \mathbf{R}\) defined by \(f(x) = [x]\), \(x \in \mathbf{R}\), assumes the value of the greatest integer less than or equal to \(x\). Such a function is called the greatest integer function? (Fig 2.15).
For example: \([2.3] = 2\), \([-1.7] = -2\), \([0.5] = 0\), \([4] = 4\).
Activity: Is It a Function?
L4 Analyse
Materials: Paper, pencil
Predict: Given the arrow diagrams below, which ones represent functions and which do not?
- Diagram 1: A = {1,2,3} to B = {a,b}. Arrows: 1 to a, 2 to a, 3 to b. Is this a function?
- Diagram 2: A = {1,2,3} to B = {a,b}. Arrows: 1 to a, 1 to b, 2 to a, 3 to b. Is this a function?
- Diagram 3: A = {1,2,3} to B = {a,b}. Arrows: 1 to a, 3 to b. Is this a function?
- For each case, state the rule violated (if any).
Observe:
- Diagram 1: Yes, it is a function. Every element of A has exactly one image (multiple elements can share the same image).
- Diagram 2: Not a function. Element 1 has two images (a and b). A function requires each element to have exactly one image.
- Diagram 3: Not a function. Element 2 has no image. Every element of the domain must have an image.
Competency-Based Questions
Scenario: A city bus service assigns routes to drivers. Let D = {Driver A, Driver B, Driver C} be the set of drivers and R = {Route 1, Route 2, Route 3, Route 4} be the set of routes. The assignment is: Driver A handles Route 1 and Route 3, Driver B handles Route 2, and Driver C handles Route 4.
Q1. Write the assignment as a relation from D to R in roster form.
L1 RememberAnswer: R = {(Driver A, Route 1), (Driver A, Route 3), (Driver B, Route 2), (Driver C, Route 4)}.
Q2. Is this assignment a function from D to R? Justify using the definition of a function.
L2 UnderstandAnswer: No, this is not a function from D to R. Driver A is assigned to two routes (Route 1 and Route 3). For a valid function, each element of the domain must have exactly one image. Since Driver A has two images, the assignment fails the definition.
Q3. Is the reverse assignment (from Routes to Drivers) a function from R to D? Analyse both the original and reverse mappings.
L4 AnalyseAnswer: The reverse assignment is {(Route 1, Driver A), (Route 2, Driver B), (Route 3, Driver A), (Route 4, Driver C)}. Each route maps to exactly one driver. So yes, the reverse assignment is a function from R to D. Note that multiple routes can map to the same driver (that is allowed); what matters is that each route has exactly one driver.
Q4. Redesign the assignment so that it becomes a valid function from D to R where every driver handles exactly one route and no route is left unassigned. Is this possible? If not, explain what constraint must be relaxed.
L6 CreateAnswer: A function from D to R with n(D) = 3 can assign at most 3 routes (one per driver). Since n(R) = 4, at least one route will remain unassigned. We cannot cover all 4 routes. Example valid function: {(Driver A, Route 1), (Driver B, Route 2), (Driver C, Route 4)}. Route 3 is unassigned, but that is fine -- the range need not equal the codomain. For all routes to be covered, we would need at least 4 drivers.
Assertion--Reason Questions
Assertion (A): The relation \(R = \{(1,2), (1,3), (2,4)\}\) is not a function from \(\{1,2\}\) to \(\{2,3,4\}\).
Reason (R): In a function, every element of the domain must have exactly one image in the codomain.
Reason (R): In a function, every element of the domain must have exactly one image in the codomain.
Answer: (a) -- A is true because element 1 maps to both 2 and 3, violating the function definition. R correctly states the definition and explains why A is true.
Assertion (A): The range of a function is always a subset of its codomain.
Reason (R): The codomain is the set from which images are drawn, and the range consists of only those elements that are actually images.
Reason (R): The codomain is the set from which images are drawn, and the range consists of only those elements that are actually images.
Answer: (a) -- Both true. The range consists of actual images, which are elements of the codomain. Hence range is always a subset of codomain.
Assertion (A): The modulus function \(f(x) = |x|\) satisfies \(f(-3) = f(3) = 3\).
Reason (R): The modulus function always returns the non-negative value of any real number input.
Reason (R): The modulus function always returns the non-negative value of any real number input.
Answer: (a) -- \(|-3| = 3\) and \(|3| = 3\). Both are true. R correctly explains why the modulus function maps both \(-3\) and \(3\) to the same value.
Frequently Asked Questions
What is a relation in mathematics?
A relation R from set A to set B is a subset of the Cartesian product A x B. The set of first elements is the domain and the set of second elements is the range.
What is the difference between a relation and a function?
A function is a special relation where every element in the domain maps to exactly one element in the codomain. In a relation, one input can map to multiple outputs.
What are domain, codomain, and range?
Domain is the set of all inputs, codomain is the target set, and range is the set of actual outputs. Range is always a subset of the codomain.
What are the main types of functions in Class 11?
NCERT Class 11 covers identity function, constant function, polynomial functions, rational functions, modulus function, signum function, and greatest integer function.
How do you determine if a relation is a function?
Check that no two ordered pairs have the same first element with different second elements. Graphically, use the vertical line test.
Frequently Asked Questions — Relations and Functions
What is Relations, Functions, Domain, Range, and Graphs in NCERT Class 11 Mathematics?
Relations, Functions, Domain, Range, and Graphs is a key concept covered in NCERT Class 11 Mathematics, Chapter 2: Relations and Functions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Relations, Functions, Domain, Range, and Graphs step by step?
To solve problems on Relations, Functions, Domain, Range, and Graphs, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 11 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 2: Relations and Functions?
The essential formulas of Chapter 2 (Relations and Functions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Relations, Functions, Domain, Range, and Graphs important for the Class 11 board exam?
Relations, Functions, Domain, Range, and Graphs is part of the NCERT Class 11 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Relations, Functions, Domain, Range, and Graphs?
Common mistakes in Relations, Functions, Domain, Range, and Graphs include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Relations, Functions, Domain, Range, and Graphs?
End-of-chapter NCERT exercises for Relations, Functions, Domain, Range, and Graphs cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.
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