Class 9 Mathematics Chapter 1: Relations and Functions – Complete Important Questions Bank

Chapter 1 serves as the gateway to calculus and algebra. Relations and Functions directly connect to Class 10 polynomial equations, Class 11 functions and limits, and beyond. In the 2024-25 rationalized CBSE curriculum, this chapter reduced rote learning but *increased* conceptual depth. Board examiners now ask: "Prove this relation is an equivalence relation" (requiring all three properties), "Find the inverse of a function and verify" (demanding algebraic skill + logical proof), and "Apply function composition to real-world scenarios" (HOTS-style). The new pattern tests *understanding*, not formula memorization. Questions no longer ask "Define a function"—they ask you to construct one from a word problem, verify its properties, and compute its composition. This shift means students who can reason about abstract sets and mappings score consistently above 90%. The 5-mark questions in particular reward students who explain the *logic* behind each step. Knowing the types of relations (reflexive, symmetric, transitive, equivalence) and being able to check each property systematically is non-negotiable. Similarly, function composition f∘g and inverse functions f⁻¹ require fluency in algebraic manipulation and set theory. These skills form the skeleton of Class 11 Relations and Functions (a full chapter by itself), making Chapter 1 Class 9 truly a make-or-break foundation.

**Question 1:** If A = {1, 2, 3} and R is a relation on A defined by R = {(1,1), (2,2), (3,3)}, which property does R *not* satisfy? (a) Reflexive (b) Symmetric (c) Transitive (d) Equivalence **Answer: (d)** Although R is reflexive, symmetric, and transitive (all three), the key here is that a relation must satisfy **all three** to be called an equivalence relation. R does satisfy all three, so technically (a), (b), and (c) are correct properties it possesses. However, this MCQ likely expects learners to recognize that R is equivalence. Revised: R *does* satisfy equivalence. The tricky wording tests whether students confuse "equivalence" with individual properties. **Question 2:** Let f: ℝ → ℝ be defined by f(x) = 2x + 3. What is f⁻¹(7)? (a) 2 (b) 3 (c) 17 (d) 5 **Answer: (a)** To find f⁻¹(7), we solve f(x) = 7. So 2x + 3 = 7 → 2x = 4 → x = 2. Thus f⁻¹(7) = 2. **Question 3:** A relation R on the set {a, b, c} is reflexive if: (a) (a,a), (b,b), (c,c) ∈ R (b) If (x,y) ∈ R, then (y,x) ∈ R (c) If (x,y) and (y,z) ∈ R, then (x,z) ∈ R (d) R is a function **Answer: (a)** Reflexivity requires every element related to itself. The definition states: for all x ∈ A, (x,x) ∈ R. **Question 4:** If f(x) = x² and g(x) = 2x + 1, then (f∘g)(2) equals: (a) 9 (b) 25 (c) 49 (d) 5 **Answer: (b)** (f∘g)(2) = f(g(2)). First, g(2) = 2(2) + 1 = 5. Then f(5) = 5² = 25. **Question 5:** Which of the following is a function from A = {1, 2, 3} to B = {p, q}? (a) {(1,p), (2,q), (3,p)} (b) {(1,p), (2,q)} (c) {(1,p), (1,q), (2,p)} (d) {(1,p), (2,p), (2,q), (3,p)} **Answer: (a)** A function must map *every* element of A to exactly one element in B. Option (a) assigns 1→p, 2→q, 3→p (all covered, no repetition). Option (b) misses 3. Option (c) maps 1 to two elements. Option (d) maps 2 to two elements.

**Question 1:** Define a reflexive relation and give an example on the set {1, 2, 3}. **Solution:** A relation R on a set A is reflexive if for every element x ∈ A, the pair (x,x) ∈ R. In other words, every element is related to itself. Example: R = {(1,1), (2,2), (3,3), (1,2)} on {1, 2, 3}. This relation is reflexive because all three elements have self-pairs; the pair (1,2) is extra but doesn't violate reflexivity. **Question 2:** Check whether the relation R = {(1,2), (2,1), (2,3), (3,2)} on A = {1, 2, 3} is symmetric. **Solution:** A relation is symmetric if whenever (a,b) ∈ R, then (b,a) ∈ R. Checking: (1,2) ∈ R and (2,1) ∈ R ✓; (2,3) ∈ R and (3,2) ∈ R ✓. Yes, R is symmetric because for each ordered pair, its reverse is also in R. **Question 3:** Let f: {1, 2, 3} → {a, b, c, d} be a function with f(1) = a, f(2) = c, f(3) = b. Find the range of f. **Solution:** The range of a function is the set of all output values actually mapped to by the domain elements. Here, the domain is {1, 2, 3} and the outputs are a, c, b respectively. Therefore, Range(f) = {a, b, c}. Note: The codomain is {a, b, c, d}, but the range is {a, b, c} because d is never mapped to. **Question 4:** Show that the function f(x) = 3x − 2 is one-to-one (injective). **Solution:** A function is one-to-one if different inputs give different outputs. Assume f(x₁) = f(x₂). Then 3x₁ − 2 = 3x₂ − 2 → 3x₁ = 3x₂ → x₁ = x₂. This shows that if the outputs are equal, the inputs must be equal, proving f is injective. **Question 5:** If f(x) = x + 5 and g(x) = 2x, find (g∘f)(x). **Solution:** (g∘f)(x) = g(f(x)). First evaluate f(x) = x + 5. Then apply g: g(x + 5) = 2(x + 5) = 2x + 10. Therefore, (g∘f)(x) = 2x + 10.

**Question 1:** On the set A = {1, 2, 3, 4}, the relation R is defined by R = {(a,b) : a − b is divisible by 2}. Check whether R is reflexive, symmetric, and transitive. Is it an equivalence relation? **Solution:** *Reflexive?* For all a ∈ A, is a − a = 0 divisible by 2? Yes, 0 is divisible by 2. So (a,a) ∈ R for all a. R is reflexive. ✓ *Symmetric?* If (a,b) ∈ R, is (b,a) ∈ R? If a − b is divisible by 2, then a − b = 2k for some integer k. So b − a = −2k = 2(−k), which is also divisible by 2. Thus (b,a) ∈ R. R is symmetric. ✓ *Transitive?* If (a,b) ∈ R and (b,c) ∈ R, is (a,c) ∈ R? If a − b = 2m and b − c = 2n, then a − c = (a − b) + (b − c) = 2m + 2n = 2(m + n), divisible by 2. So (a,c) ∈ R. R is transitive. ✓ **Conclusion:** R is reflexive, symmetric, and transitive, so R is an **equivalence relation**. **Question 2:** Let f: ℝ → ℝ be defined by f(x) = 2x + 3. Show that f is bijective (both one-to-one and onto), and find its inverse function f⁻¹. **Solution:** *One-to-one (Injective)?* Assume f(x₁) = f(x₂). Then 2x₁ + 3 = 2x₂ + 3 → 2x₁ = 2x₂ → x₁ = x₂. So f is injective. ✓ *Onto (Surjective)?* For any y ∈ ℝ, we need to find x such that f(x) = y. So 2x + 3 = y → x = (y − 3)/2. Since (y − 3)/2 ∈ ℝ for all y ∈ ℝ, every y has a pre-image. So f is surjective. ✓ f is bijective. To find f⁻¹: Let y = f(x) = 2x + 3. Solve for x: x = (y − 3)/2. Swap x and y: **f⁻¹(x) = (x − 3)/2**. **Verification:** f(f⁻¹(x)) = f((x − 3)/2) = 2 · (x − 3)/2 + 3 = (x − 3) + 3 = x. ✓ **Question 3:** Given f(x) = x + 1 and g(x) = 2x², find (f∘g)(x), (g∘f)(x), and verify whether f∘g = g∘f. **Solution:** (f∘g)(x) = f(g(x)) = f(2x²) = 2x² + 1. (g∘f)(x) = g(f(x)) = g(x + 1) = 2(x + 1)² = 2(x² + 2x + 1) = 2x² + 4x + 2. Compare: (f∘g)(x) = 2x² + 1 and (g∘f)(x) = 2x² + 4x + 2. These are not equal; in fact, (f∘g)(x) ≠ (g∘f)(x). Therefore, **f∘g ≠ g∘f**, showing that function composition is **not commutative**. **Question 4:** On the set A = {2, 3, 4, 5}, define a relation R by (a,b) ∈ R if gcd(a,b) = 1 (a and b are coprime). Is R symmetric? Is R transitive? Give reasons. **Solution:** *Symmetric?* If gcd(a,b) = 1, then gcd(b,a) = 1 (gcd is symmetric). So if (a,b) ∈ R, then (b,a) ∈ R. R is symmetric. ✓ *Transitive?* Check with a counterexample. (2,3) ∈ R (gcd = 1) and (3,4) ∈ R (gcd = 1), but is (2,4) ∈ R? gcd(2,4) = 2 ≠ 1, so (2,4) ∉ R. R is **not transitive**. ✗

**Question 1:** Let A = {1, 2, 3, 4} and B = {a, b, c, d, e}. Define a function f: A → B by f(1) = a, f(2) = c, f(3) = a, f(4) = e. (a) Is f one-to-one? Justify. (b) Is f onto? Why or why not? (c) Can f have an inverse? Explain. (d) Find the pre-image of c and the pre-image of b. **Solution:** (a) **Is f one-to-one (injective)?** No. A function is one-to-one if distinct domain elements map to distinct codomain elements. Here, f(1) = a and f(3) = a; both 1 and 3 map to a. Therefore, f is **not injective** because two different inputs (1 and 3) produce the same output (a). *Justification complete.* (b) **Is f onto (surjective)?** No. For f to be onto, every element in the codomain B must be the image of at least one element in A. The range of f is {a, c, e} (the actual outputs). Since b and d are in B but not in the range, f is **not surjective**. Elements b and d are never mapped to. *Justification: B has 5 elements, but f's range has only 3 elements.* (c) **Can f have an inverse?** No. For a function to have an inverse (be bijective), it must be both injective and surjective. We showed f is neither injective nor surjective. Therefore, **f cannot have an inverse function** f⁻¹. *Reason: Inverses exist only for bijective functions.* (d) **Find the pre-image of c and the pre-image of b.** - Pre-image of c: The elements in A that map to c. From the definition, f(2) = c, so the pre-image of c is {2}. - Pre-image of b: The elements in A that map to b. Looking at the function definition, no element in A maps to b. So the pre-image of b is the **empty set ∅** (or {}, no pre-image exists). --- **Question 2:** Let f: ℝ → ℝ be defined by f(x) = 3x − 5 and g: ℝ → ℝ be defined by g(x) = (x + 5)/3. (a) Find (f∘g)(x) and (g∘f)(x). (b) What is the relationship between f and g? (c) Verify that f and g are both bijective. (d) If h(x) = 2x, find (h∘f)(3). **Solution:** (a) **Find (f∘g)(x) and (g∘f)(x):** (f∘g)(x) = f(g(x)) = f((x + 5)/3) = 3 · (x + 5)/3 − 5 = (x + 5) − 5 = **x**. (g∘f)(x) = g(f(x)) = g(3x − 5) = ((3x − 5) + 5)/3 = 3x/3 = **x**. (b) **Relationship between f and g:** Since (f∘g)(x) = x and (g∘f)(x) = x, we conclude that **g is the inverse function of f**, i.e., **g = f⁻¹**. This means f and g are inverse functions of each other. (c) **Verify f and g are bijective:** *For f(x) = 3x − 5:* - Injective: If f(x₁) = f(x₂), then 3x₁ − 5 = 3x₂ − 5 → x₁ = x₂. ✓ Injective. - Surjective: For any y ∈ ℝ, solve 3x − 5 = y → x = (y + 5)/3 ∈ ℝ. Every y has a pre-image. ✓ Surjective. - f is bijective. ✓ *For g(x) = (x + 5)/3:* - Injective: If g(x₁) = g(x₂), then (x₁ + 5)/3 = (x₂ + 5)/3 → x₁ = x₂. ✓ Injective. - Surjective: For any y ∈ ℝ, solve (x + 5)/3 = y → x = 3y − 5 ∈ ℝ. Every y has a pre-image. ✓ Surjective. - g is bijective. ✓ (d) **Find (h∘f)(3):** (h∘f)(3) = h(f(3)). First, f(3) = 3(3) − 5 = 9 − 5 = 4. Then h(4) = 2(4) = 8. Therefore, **(h∘f)(3) = 8**. --- **Question 3:** On the set Z (integers), define a relation R by aRb if and only if a ≡ b (mod 3), meaning a and b leave the same remainder when divided by 3. (a) Prove R is reflexive. (b) Prove R is symmetric. (c) Prove R is transitive. (d) List the equivalence classes. (e) How does this relate to partitioning Z? **Solution:** (a) **Prove R is reflexive:** For any a ∈ Z, we must show aRa, i.e., a ≡ a (mod 3). This means a and a leave the same remainder when divided by 3. This is trivially true: a mod 3 = a mod 3. Thus, aRa for all a ∈ Z. **R is reflexive.** ✓ (b) **Prove R is symmetric:** Assume aRb, i.e., a ≡ b (mod 3). This means a − b is divisible by 3, so a − b = 3k for some integer k. Then b − a = −3k = 3(−k), also divisible by 3. Therefore, b ≡ a (mod 3), i.e., bRa. **R is symmetric.** ✓ (c) **Prove R is transitive:** Assume aRb and bRc, i.e., a ≡ b (mod 3) and b ≡ c (mod 3). Then a − b = 3m and b − c = 3n for integers m, n. Adding: (a − b) + (b − c) = 3m + 3n → a − c = 3(m + n), divisible by 3. So a ≡ c (mod 3), i.e., aRc. **R is transitive.** ✓ (d) **List the equivalence classes:** Since R is an equivalence relation (reflexive, symmetric, transitive), it partitions Z into equivalence classes. Each class contains all integers with the same remainder mod 3: - [0] = {..., −3, 0, 3, 6, 9, ...} (remainder 0) - [1] = {..., −2, 1, 4, 7, 10, ...} (remainder 1) - [2] = {..., −1, 2, 5, 8, 11, ...} (remainder 2) These three classes cover all integers without overlap. (e) **Partition of Z:** The equivalence classes [0], [1], [2] form a partition of Z. This means: - Each equivalence class is non-empty. - The classes are pairwise disjoint (no element in two classes simultaneously). - Their union is Z (every integer belongs to exactly one class). This demonstrates that every equivalence relation induces a natural partition of the underlying set.

**Question:** A school uses a digital attendance system where each student is assigned a unique ID (domain: A = {101, 102, 103, 104}) and a status code (codomain: B = {P, A, L, S}), where P = Present, A = Absent, L = Late, S = Sick Leave. The attendance officer defines a function f: A → B recording today's status: f(101) = P, f(102) = A, f(103) = L, f(104) = P. Later, the system creates a secondary relation R on B as follows: two status codes are related if they both refer to students who came to school today (either P or L). So P R P, P R L, L R P, and L R L. **(i) Examine f and determine whether it is onto. If not, which element in B has no pre-image?** **(ii) Determine whether the relation R on B is reflexive, symmetric, and transitive. Is R an equivalence relation?** **(iii) If the school then defines a new function g: B → {1, 2} where g(x) = 1 if x represents attendance (P or L) and g(x) = 2 if x represents non-attendance (A or S), verify that g is onto and find the composition (g∘f)(x). What does (g∘f) tell us in practical terms?** **(iv) The attendance officer wants to create a "reverse lookup" that maps from a student's input ID to confirm their status. Explain whether such a reverse mapping (inverse function) exists for f. What mathematical property would f need to satisfy?** --- **Detailed Step-by-Step Solution:** **(i) Is f onto?** f is onto (surjective) if every element in B = {P, A, L, S} is mapped to by at least one element in A. Range of f = {f(101), f(102), f(103), f(104)} = {P, A, L, P} = {P, A, L}. Compare range {P, A, L} with codomain {P, A, L, S}: The element **S (Sick Leave) is not in the range**. Therefore, f is **not onto**. The element S has no pre-image in A. *Practical interpretation:* No student took sick leave today. **(ii) Is R on B reflexive, symmetric, and transitive?** Recall: x R y if both x and y represent attendance (i.e., both are in {P, L}). Explicit: R = {(P,P), (P,L), (L,P), (L,L)}. *Reflexive?* For all x ∈ B, is (x,x) ∈ R? - P ∈ {P,L}? Yes, so (P,P) ∈ R. ✓ - A ∈ {P,L}? No, so (A,A) ∉ R. ✗ - L ∈ {P,L}? Yes, so (L,L) ∈ R. ✓ - S ∈ {P,L}? No, so (S,S) ∉ R. ✗ **R is not reflexive** because (A,A) and (S,S) are missing. *Symmetric?* If (x,y) ∈ R, is (y,x) ∈ R? - (P,P) ∈ R and (P,P) ∈ R. ✓ - (P,L) ∈ R and (L,P) ∈ R. ✓ - (L,P) ∈ R and (P,L) ∈ R. ✓ - (L,L) ∈ R and (L,L) ∈ R. ✓ **R is symmetric.** ✓ *Transitive?* If (x,y), (y,z) ∈ R, is (x,z) ∈ R? - (P,P), (P,P) ∈ R → (P,P) ∈ R. ✓ - (P,P), (P,L) ∈ R → (P,L) ∈ R. ✓ - (P,L), (L,P) ∈ R → (P,P) ∈ R. ✓ - (P,L), (L,L) ∈ R → (P,L) ∈ R. ✓ - (L,P), (P,P) ∈ R → (L,P) ∈ R. ✓ - (L,P), (P,L) ∈ R → (L,L) ∈ R. ✓ - (L,L), (L,P) ∈ R → (L,P) ∈ R. ✓ - (L,L), (L,L) ∈ R → (L,L) ∈ R. ✓ **R is transitive.** ✓ **Conclusion:** R is symmetric and transitive but **not reflexive**. Therefore, **R is not an equivalence relation** (all three properties are required). *Observation:* R acts like an "equivalence" on the subset {P,L} only. **(iii) Verify g is onto and find (g∘f):** Domain of g: B = {P, A, L, S}. Codomain: {1, 2}. - g(P) = 1 (attendance) - g(A) = 2 (non-attendance) - g(L) = 1 (attendance) - g(S) = 2 (non-attendance) Range of g = {1, 2}. Since range = codomain, **g is onto.** ✓ **Composition (g∘f):** For each student ID x in A, (g∘f)(x) = g(f(x)). - (g∘f)(101) = g(f(101)) = g(P) = 1 - (g∘f)(102) = g(f(102)) = g(A) = 2 - (g∘f)(103) = g(f(103)) = g(L) = 1 - (g∘f)(104) = g(f(104)) = g(P) = 1 So (g∘f): A → {1, 2} with (g∘f) = {(101,1), (102,2), (103,1), (104,1)}. **Practical interpretation:** (g∘f) is a binary flag—1 for "attended today" (present or late), 2 for "did not attend" (absent or sick). It simplifies the 4-state attendance system into a 2-state binary indicator, useful for quick attendance counting. **(iv) Does an inverse function exist for f?** An inverse function f⁻¹ exists if and only if f is bijective (both injective and surjective). *Is f injective?* Check: f(101) = P, f(104) = P. Two different students (101 and 104) have the same status P. Therefore, **f is not injective** (not one-to-one). *Is f surjective?* We showed in (i) that S has no pre-image. So **f is not surjective** (not onto). **Conclusion:** f is neither injective nor surjective, so **no inverse function exists** for f. *Mathematical requirement:* For a reverse lookup to work, each status code must correspond to exactly one student (injectivity), which fails here because two students share the same status. A bijective function (one-to-one correspondence) is necessary for an inverse to exist.

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