CBSE Important Questions Class 12 Maths Chapter 1 – Relations and Functions

Class 12 Maths Chapter 1 Relations and Functions is the first chapter of the CBSE Class 12 Maths syllabus. It builds on what students learned in Class 11 and goes deeper. Students learn the different types of relations — reflexive, symmetric, transitive and equivalence relations — and the different types of functions — one-one (injective), onto (surjective) and bijective.

These CBSE important questions for Class 12 Maths Chapter 1 cover checking whether a set of ordered pairs is a relation or a function, proving that a relation is an equivalence relation, giving examples of relations with particular properties, finding the range of a function, and showing that a function is one-one. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}. Are the following true? Justify your answer in each case.
(i) f is a relation from A into B.
(ii) f is a function from A into B.

Answer: (i) Yes, f is a relation.
Every ordered pair in f has its first element from A and its second element from B, so f is a subset of A × B. By definition, any subset of A × B is a relation from A to B.
(ii) No, f is not a function.
Look at the pairs (2, 9) and (2, 11). The element 2 has two different images, 9 and 11.
In a function every element of the domain must have exactly one image, so f is not a function from A into B.

Q2. Is every relation a function?

Answer: No. Every relation is not a function, but every function is definitely a relation.
A relation only needs to be a set of ordered pairs. A function has two extra conditions: every element of the domain must have an image, and no element may have more than one image.
So a function is a special kind of relation.

Q3. Prove that the relation R on the set N × N defined by (a, b) R (c, d) ⇔ a + d = b + c, for all (a, b), (c, d) ∈ N × N, is an equivalence relation.

Answer: To prove R is an equivalence relation we check three properties.
Reflexive: For any (a, b) ∈ N × N, we have a + b = b + a (addition is commutative).
So (a, b) R (a, b). Hence R is reflexive.
Symmetric: Let (a, b) R (c, d), so a + d = b + c.
This can be rewritten as c + b = d + a, which means (c, d) R (a, b). Hence R is symmetric.
Transitive: Let (a, b) R (c, d) and (c, d) R (e, f).
Then a + d = b + c ...(1) and c + f = d + e ...(2)
Adding (1) and (2): a + d + c + f = b + c + d + e
Cancelling c and d from both sides: a + f = b + e
This means (a, b) R (e, f). Hence R is transitive.
Since R is reflexive, symmetric and transitive, it is an equivalence relation.

Q4. Give an example of a relation which is (i) symmetric but neither reflexive nor transitive, (ii) transitive but neither reflexive nor symmetric, (iii) reflexive and symmetric but not transitive.

Answer: (i) Let A = {1, 2, 3} and R = {(1, 2), (2, 1)}.
Not reflexive, since (1,1), (2,2), (3,3) are not in R.
Symmetric, since (1,2) and (2,1) are both in R.
Not transitive, since (1,2) and (2,1) are in R but (1,1) is not.
(ii) Let R on the set of real numbers be R = {(x, y) : x < y}.
Not reflexive, since x is never less than x.
Not symmetric, since 1 < 2 but 2 is not less than 1.
Transitive, since a < b and b < c gives a < c.
(iii) Let A = {2, 4, 6} and R = {(2,2),(4,4),(6,6),(2,4),(4,2),(4,6),(6,4)}.
Reflexive and symmetric, but not transitive because (2,4) and (4,6) are in R while (2,6) is not.

Q5. Determine whether the relation R on the set N of natural numbers, defined by R = {(x, y) : y = x + 5 and x < 4}, is reflexive, symmetric and transitive.

Answer: Listing the elements: R = {(1, 6), (2, 7), (3, 8)}
Reflexive: (1,1), (2,2) and so on are not in R, so R is not reflexive.
Symmetric: (1, 6) is in R but (6, 1) is not, so R is not symmetric.
Transitive: There is no pair in R whose first element is 6, 7 or 8, so the transitivity condition is never triggered in a way that fails... but since no chain exists at all, R is not transitive in the sense required here.

Q6. Let A = {−1, −2, 0, 1, 2} and f : A → Z be 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}
f(−1) = 1 − 2 + 3 = 2
f(−2) = 4 − 4 + 3 = 3
f(0) = 3
f(1) = 1 + 2 + 3 = 6
f(2) = 4 + 4 + 3 = 11
Collecting the distinct values, the range is {2, 3, 6, 11}.

Q7. Let X = {1, 2, 3, 4} and a function be defined from X to N as f(x) = x!/(x − 1)!. Then the range of f is
(a) {1, 2, 3, 4}
(b) {1, 4, 9, 16}
(c) {1, 2, 6, 24}
(d) {1}

Answer: (a) {1, 2, 3, 4}
Simplify first: x!/(x − 1)! = x
So f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 4.
The range is {1, 2, 3, 4}.

Q8. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Answer: A function is one-one (injective) if different elements of the domain have different images.
Looking at f:
f(1) = 4, f(2) = 5, f(3) = 6
All three images 4, 5 and 6 are different from each other.
So no two elements of A have the same image, and therefore f is one-one.
(Note that f is not onto, because 7 ∈ B has no pre-image in A.)

Q9. Define one-one, onto and bijective functions.

Answer: • One-one (injective): A function f : A → B is one-one if different elements of A have different images in B. That is, f(x₁) = f(x₂) implies x₁ = x₂.
• Onto (surjective): f is onto if every element of B is the image of at least one element of A, that is the range equals the codomain.
• Bijective: f is bijective if it is both one-one and onto. Only a bijective function has an inverse.

Q10. Show that the function f : R → R defined by f(x) = 3 − 4x is one-one and onto.

Answer: One-one: Let f(x₁) = f(x₂).
3 − 4x₁ = 3 − 4x₂
−4x₁ = −4x₂
x₁ = x₂
So f is one-one.
Onto: Let y be any real number in the codomain. Set y = 3 − 4x.
4x = 3 − y, so x = (3 − y)/4
Since y is real, x is also real, so every y has a pre-image.
So f is onto.
Being both one-one and onto, f is bijective.

Q11. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a − b| is even} is an equivalence relation.

Answer: Reflexive: For any a, |a − a| = 0, which is even. So (a, a) ∈ R.
Symmetric: If |a − b| is even, then |b − a| is the same number, so it is also even. So (b, a) ∈ R.
Transitive: Let |a − b| and |b − c| both be even. This means a and b have the same parity (both odd or both even), and b and c have the same parity. So a and c also have the same parity, which makes |a − c| even.
Since all three properties hold, R is an equivalence relation.
(The equivalence classes here are {1, 3, 5} and {2, 4}.)

Q12. If f : R → R is given by f(x) = x² and g : R → R is given by g(x) = x + 1, find (fog)(x) and (gof)(x). Are they equal?

Answer: (fog)(x) means f applied to g(x):
= f(g(x)) = f(x + 1) = (x + 1)² = x² + 2x + 1
(gof)(x) means g applied to f(x):
= g(f(x)) = g(x²) = x² + 1
Are they equal? No. (fog)(x) = x² + 2x + 1 while (gof)(x) = x² + 1.
This shows that the composition of functions is not commutative.

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