CBSE Important Questions Class 11 Maths Chapter 2 – Relations and Functions

Class 11 Maths Chapter 2 Relations and Functions is part of the CBSE Class 11 Maths syllabus. It explains ordered pairs, the cartesian product of two sets, and what a relation is. Students then learn when a relation becomes a function, how to write the domain, codomain and range, and the graphs of standard real functions such as the identity, constant, modulus and polynomial functions.

These CBSE important questions for Class 11 Maths Chapter 2 cover identifying from a diagram whether a correspondence is a function, the condition for a relation to be a function, writing a relation in roster and set-builder form, finding the domain and range, the range of f(x) = x² + 2x + 3, and the domain of √(9 − x²). Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. State whether or not each of the following diagrams defines a function of A = {a, b, c} into B = {x, y, z}.

Answer: (i) Not a function. The single element c ∈ A is associated with two distinct elements x and y of B. A function must give exactly one image.
(ii) This is a function. Each element of A is associated with one and only one element of B.
(iii) Not a function. The element b ∈ A is not associated with any element of B, and in a function every element of the domain must have an image.

Q2. Under which condition is a relation f from A to B said to be a function?

Answer: A relation f from A to B is a function if every element of A has one and only one image in B.
In other words, two conditions must hold:
1. No element of A is left without an image.
2. No element of A has more than one image.

Q3. The given figure shows a relation between the sets P and Q. Write this relation (i) in roster form and (ii) in set-builder form.

Answer: (i) Roster form: R = {(4, 2), (9, 3), (25, 5)}
(ii) Set-builder form: R = {(x, y) : x = y², x ∈ P and y ∈ Q}
Each first element is the square of the second element, which is the rule connecting the two sets.

Q4. Which of the following relations are functions?
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) {(1, 3), (1, 5), (2, 5)}
(a) (i), (iii)
(b) (i) only
(c) (i), (ii)
(d) (i), (ii), (iii)

Answer: (c) (i), (ii)
In (i) every first element appears only once, so it is a function (a constant function).
In (ii) every first element appears only once, so it is also a function.
In (iii) the element 1 is paired with both 3 and 5, so one input has two outputs. That is not allowed, so (iii) is not a function.

Q5. In the given figure, write the relation in (i) set-builder form and (ii) roster form. Also write the domain and the range.

Answer: (i) Set-builder form: R = {(x, y) : y = 2x + 1, x ∈ A}
(ii) Roster form: R = {(1, 3), (2, 5), (3, 7)}
Domain = the set of all first elements = {1, 2, 3}
Range = the set of all second elements = {3, 5, 7}

Q6. Define a relation.

Answer: Let A and B be two sets. A relation R from A to B is a subset of the cartesian product A × B, that is R ⊆ A × B.
In simple words, a relation is a set of ordered pairs (a, b) where a comes from A and b comes from B, and the pairs follow some rule.
The set of all first elements is the domain and the set of all second elements is the range of the relation.

Q7. Let A = {−1, −2, 0, 1, 2} and f : A → Z be defined by f(x) = x² + 2x + 3. Find the range of f.

Answer: Substitute each element of A into the function:
f(−1) = 1 − 2 + 3 = 2
f(−2) = 4 − 4 + 3 = 3
f(0) = 0 + 0 + 3 = 3
f(1) = 1 + 2 + 3 = 6
f(2) = 4 + 4 + 3 = 11
Collecting the distinct values, the range is {2, 3, 6, 11}.

Q8. Let A = {−1, −2, 0, 1, 2} and f : A → Z given by f(x) = x² + 2x + 3. Then the range of f is
(a) {−2, −3, 6, 11}
(b) {2, 3, 6, 11}
(c) {1, 2, −6, −11}
(d) {−2, −3, −6, −11}

Answer: (b) {2, 3, 6, 11}
As worked out above, the images are 2, 3, 3, 6 and 11. Since a set never repeats an element, the range is {2, 3, 6, 11}.

Q9. Find the domain of the function f(x) = √(9 − x²).

Answer: A square root gives a real value only when the expression inside it is not negative.
9 − x² ≥ 0
x² − 9 ≤ 0
(x − 3)(x + 3) ≤ 0
This product is negative or zero when x lies between −3 and 3, including both.
So the domain of f(x) is [−3, 3].

Q10. If R = {(d, 1), (m, 2), (r, 3)}, then the domain of R is
(a) {1, m, r}
(b) {d, x, r}
(c) {d, m, r}
(d) {d, m, 2}

Answer: (c) {d, m, r}
The domain of a relation is the set of all first elements of its ordered pairs.
Here the first elements are d, m and r, so the domain is {d, m, r}.
(The range, which is the set of second elements, would be {1, 2, 3}.)

Q11. If A = {1, 2} and B = {3, 4}, find A × B and write the number of relations from A to B.

Answer: The cartesian product A × B is the set of all ordered pairs with the first element from A and the second from B.
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
So n(A × B) = 2 × 2 = 4.
A relation from A to B is any subset of A × B, and a set with 4 elements has 2⁴ subsets.
So the number of relations from A to B is 16.

Q12. Find the domain and range of the function f(x) = 1 / (x − 5).

Answer: Domain: The denominator must not be zero, so x − 5 ≠ 0, which gives x ≠ 5.
Domain = R − {5}, that is all real numbers except 5.
Range: Let y = 1/(x − 5). Then x − 5 = 1/y, so x = 5 + 1/y.
For x to exist, y must not be zero.
Range = R − {0}, that is all real numbers except 0.

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