Class 9 Mathematics Chapter 1: Relations and Functions MCQ Quiz with Answers

The 2024–25 CBSE restructure has shifted focus from lengthy descriptive answers to multiple-choice and short-answer formats. For Class 9 Mathematics, Chapter 1 (Relations and Functions) appears in 15–20% of total marks, with MCQs accounting for 40–50% of chapter weightage. Why? MCQs test conceptual depth faster than traditional problems. A single MCQ on equivalence relations forces you to verify reflexivity, symmetry, and transitivity simultaneously—skills that unlock complex proofs later. Moreover, CBSE now rewards speed: students who solve 10 MCQs in 12 minutes gain a cognitive advantage over those who spend 30 minutes on one long-form answer. The new pattern also emphasizes reasoning: modern CBSE MCQs pair assertions with reasons, requiring you to justify your choice. This mirrors real-world problem-solving. By practising these 30 questions, you'll internalize the exact phrasing, trap options, and logical shortcuts that examiners use—critical for achieving 95%+ in this chapter.

**Q1.** Let A = {1, 2, 3} and R = {(1,1), (2,2), (3,3)}. Which property does R satisfy? (A) Reflexive only (B) Symmetric only (C) Transitive only (D) All three **Answer:** (D) All three **Reason:** R is the identity relation—every element pairs with itself (reflexive), (a,b) ∈ R implies (b,a) ∈ R (symmetric), and (a,b), (b,c) ∈ R implies (a,c) ∈ R (transitive). --- **Q2.** Which of the following is a function from {1, 2} to {a, b, c}? (A) {(1,a), (1,b), (2,c)} (B) {(1,a), (2,b)} (C) {(1,a)} (D) {(2,a), (2,b)} **Answer:** (B) {(1,a), (2,b)} **Reason:** A function requires every element of the domain to map to exactly one element of the codomain; only option (B) satisfies this. --- **Q3.** If f(x) = 2x + 1, what is f(3)? (A) 5 (B) 6 (C) 7 (D) 8 **Answer:** (C) 7 **Reason:** f(3) = 2(3) + 1 = 6 + 1 = 7. --- **Q4.** A relation R on {1, 2, 3} is reflexive if: (A) (1,2) ∈ R (B) (1,1), (2,2), (3,3) ∈ R (C) (1,2) ∈ R implies (2,1) ∈ R (D) (1,2), (2,3) ∈ R implies (1,3) ∈ R **Answer:** (B) (1,1), (2,2), (3,3) ∈ R **Reason:** Reflexivity means every element relates to itself; this is the defining property. --- **Q5.** Which relation is symmetric? (A) {(1,2), (2,3)} (B) {(1,2), (2,1), (2,3)} (C) {(1,2), (2,1)} (D) {(1,1), (1,2)} **Answer:** (C) {(1,2), (2,1)} **Reason:** Symmetry requires (a,b) ∈ R ⟹ (b,a) ∈ R; only option (C) satisfies this completely. --- **Q6.** The range of f(x) = x² for domain {−2, −1, 0, 1, 2} is: (A) {−2, −1, 0, 1, 2} (B) {0, 1, 4} (C) {0, 1, 2, 4} (D) {−4, −1, 0, 1, 4} **Answer:** (B) {0, 1, 4} **Reason:** f(−2) = 4, f(−1) = 1, f(0) = 0, f(1) = 1, f(2) = 4; distinct output values are {0, 1, 4}. --- **Q7.** Is the 'greater than' relation on ℝ transitive? (A) Yes, always (B) No, never (C) Only for integers (D) Only for positive reals **Answer:** (A) Yes, always **Reason:** If a > b and b > c, then a > c—this holds for all real numbers. --- **Q8.** The composition g ∘ f means: (A) f applied first, then g (B) g applied first, then f (C) f and g applied simultaneously (D) Neither f nor g applied **Answer:** (A) f applied first, then g **Reason:** (g ∘ f)(x) = g(f(x)); f is the inner function, executed first. --- **Q9.** If f(x) = x + 2 and g(x) = 3x, what is (f ∘ g)(2)? (A) 8 (B) 12 (C) 6 (D) 10 **Answer:** (A) 8 **Reason:** (f ∘ g)(2) = f(g(2)) = f(6) = 6 + 2 = 8. --- **Q10.** A function has an inverse if and only if it is: (A) Surjective (B) Injective (C) Bijective (D) None of these **Answer:** (C) Bijective **Reason:** A function must be both one-to-one (injective) and onto (surjective) to possess an inverse function.

**Q11.** Let R be the 'congruence modulo 3' relation on ℤ. Is R an equivalence relation? (A) Yes, it is reflexive but not symmetric (B) Yes, it is symmetric but not transitive (C) Yes, it satisfies reflexivity, symmetry, and transitivity (D) No, it violates at least one property **Answer:** (C) Yes, it satisfies reflexivity, symmetry, and transitivity **Reason:** a ≡ a (mod 3) [reflexive], a ≡ b (mod 3) ⟹ b ≡ a (mod 3) [symmetric], and a ≡ b, b ≡ c (mod 3) ⟹ a ≡ c (mod 3) [transitive]. --- **Q12.** The domain of the relation {(x, y) : y = √(x − 2)} in ℝ is: (A) ℝ (B) [2, ∞) (C) (2, ∞) (D) [0, ∞) **Answer:** (B) [2, ∞) **Reason:** For real square roots, x − 2 ≥ 0, so x ≥ 2; the domain includes the boundary point x = 2. --- **Q13.** If f: ℝ → ℝ is defined by f(x) = |x|, is f injective? (A) Yes (B) No (C) Only for x > 0 (D) Only for x < 0 **Answer:** (B) No **Reason:** f(2) = f(−2) = 2, so two different inputs map to the same output; f is not one-to-one. --- **Q14.** Let A = {1, 2, 3, 4} and R = {(1,2), (2,1), (2,3), (3,2)}. Is R transitive? (A) Yes (B) No, because (1,2), (2,3) ∈ R but (1,3) ∉ R (C) Yes, because (1,2), (2,1) are present (D) Cannot be determined without more pairs **Answer:** (B) No, because (1,2), (2,3) ∈ R but (1,3) ∉ R **Reason:** For transitivity, whenever (a,b) and (b,c) are in R, (a,c) must also be in R; this fails here. --- **Q15.** If f(x) = 2x − 3 and g(x) = x², what is (f ∘ g)(x)? (A) 4x² − 3 (B) 2x² − 3 (C) (2x − 3)² (D) 2x² − 6x + 9 **Answer:** (B) 2x² − 3 **Reason:** (f ∘ g)(x) = f(g(x)) = f(x²) = 2(x²) − 3 = 2x² − 3. --- **Q16.** The function f(x) = 1/(x − 1) has domain: (A) ℝ (B) ℝ − {1} (C) (0, ∞) (D) [1, ∞) **Answer:** (B) ℝ − {1} **Reason:** Division by zero is undefined; x − 1 ≠ 0, so x ≠ 1. --- **Q17.** A relation R on a finite set is symmetric. If (a, b) ∈ R, how many times must (b, a) appear? (A) Zero times (B) Exactly once (C) At least once (D) As many times as (a, b) **Answer:** (B) Exactly once **Reason:** By definition of symmetry, (b, a) must appear exactly once if (a, b) is in R; relations have no repeated pairs. --- **Q18.** If f and g are inverse functions, then (f ∘ g)(x) equals: (A) x (B) 1 (C) f(x) · g(x) (D) f(g(x)) · g(f(x)) **Answer:** (A) x **Reason:** By definition of inverse functions, f(g(x)) = x and g(f(x)) = x for all x in the respective domains. --- **Q19.** Let f: {1, 2, 3} → {a, b, c, d} be defined by f(1) = a, f(2) = b, f(3) = c. Is f surjective? (A) Yes (B) No, because d is not in the range (C) Yes, because all elements of the domain are used (D) Cannot be determined **Answer:** (B) No, because d is not in the range **Reason:** For surjectivity, every element of the codomain must be mapped to; d has no preimage. --- **Q20.** The inverse of f(x) = 3x + 2 is: (A) f⁻¹(x) = 3x − 2 (B) f⁻¹(x) = (x − 2)/3 (C) f⁻¹(x) = (x + 2)/3 (D) f⁻¹(x) = 1/(3x + 2) **Answer:** (B) f⁻¹(x) = (x − 2)/3 **Reason:** Set y = 3x + 2, solve for x: x = (y − 2)/3; swap variables to get f⁻¹(x) = (x − 2)/3.

**Q21.** **Assertion (A):** The relation 'x is a friend of y' on a set of people is an equivalence relation. **Reason (R):** Friendship is reflexive (everyone is their own friend), symmetric (if x is a friend of y, then y is a friend of x), and transitive (if x is a friend of y and y is a friend of z, then x is a friend of z). (A) Both A and R are true, and R is the correct explanation of A (B) Both A and R are true, but R is not the correct explanation of A (C) A is true, but R is false (D) A is false, but R is true **Answer:** (D) A is false, but R is true **Reason:** Friendship typically fails reflexivity (not everyone considers themselves a friend) and transitivity (friend-of-a-friend is not always a friend), so it's not an equivalence relation; R correctly describes what an equivalence relation needs, but A misapplies it. --- **Q22.** **Assertion (A):** If f: ℝ → ℝ is defined by f(x) = x³, then f is bijective. **Reason (R):** For every y ∈ ℝ, there exists a unique x ∈ ℝ such that f(x) = y, and f(x₁) = f(x₂) implies x₁ = x₂. (A) Both A and R are true, and R is the correct explanation of A (B) Both A and R are true, but R is not the correct explanation of A (C) A is true, but R is false (D) A is false, and R is false **Answer:** (A) Both A and R are true, and R is the correct explanation of A **Reason:** f(x) = x³ is injective (x₁³ = x₂³ ⟹ x₁ = x₂) and surjective (every real y has a preimage x = ∛y), making it bijective; R correctly explains why. --- **Q23.** **Assertion (A):** The composition of two functions is always commutative, i.e., f ∘ g = g ∘ f. **Reason (R):** Function composition follows the associative property, which implies commutativity. (A) Both A and R are true, and R is the correct explanation of A (B) Both A and R are true, but R is not the correct explanation of A (C) A is false, and R is false (D) A is true, but R is false **Answer:** (C) A is false, and R is false **Reason:** Function composition is not commutative in general—(f ∘ g)(x) = f(g(x)) ≠ g(f(x)) = (g ∘ f)(x) typically; R confuses associativity with commutativity. --- **Q24.** **Assertion (A):** The relation R = {(x, y) : x² + y² = 1} is a function from ℝ to ℝ. **Reason (R):** A relation is a function if and only if for every x in the domain, there is at most one corresponding y. (A) A is false, and R is true (B) A is true, and R is false (C) A is true, but R is incomplete (D) A is false, and R is false **Answer:** (A) A is false, and R is true **Reason:** For x = 0, we have y = ±1 (two values), violating the function definition; R correctly states the definition (though 'at most one' should be 'exactly one' for functions), so A fails the test. --- **Q25.** **Assertion (A):** If f: A → B is bijective, then f⁻¹: B → A exists and is also bijective. **Reason (R):** A bijective function is both injective and surjective, and the inverse of a bijective function is unique. (A) Both A and R are true, and R is the correct explanation of A (B) Both A and R are true, but R is not the correct explanation of A (C) A is true, but R is incomplete (D) A is false, because inverses only exist for injective functions **Answer:** (A) Both A and R are true, and R is the correct explanation of A **Reason:** Bijectivity guarantees the existence and uniqueness of an inverse; the inverse is also bijective (f⁻¹ is injective and surjective). --- **Q26.** **Assertion (A):** The number of reflexive relations on a set A with n elements is 2^(n² − n). **Reason (R):** A reflexive relation must include all n diagonal pairs (a, a), and we can freely choose to include or exclude each of the remaining n² − n pairs. (A) Both A and R are true, and R is the correct explanation of A (B) Both A and R are true, but R is not the correct explanation of A (C) A is false, and R is true (D) A is false, and R is false **Answer:** (A) Both A and R are true, and R is the correct explanation of A **Reason:** Reflexive relations fix the n diagonal pairs and allow 2 choices (include or exclude) for each of the n(n−1) off-diagonal pairs, yielding 2^(n(n−1)) = 2^(n² − n) total. --- **Q27.** **Assertion (A):** If f(x) = √x and g(x) = x − 1, then (f ∘ g)(x) = √(x − 1) has domain [0, ∞). **Reason (R):** The domain of f ∘ g is restricted by the requirement that the argument of the square root must be non-negative. (A) A is true, but R is incomplete (B) A is false; the domain is [1, ∞) (C) A is true, and R is correct (D) A is false, and R is false **Answer:** (B) A is false; the domain is [1, ∞) **Reason:** For (f ∘ g)(x) = √(x − 1) to be defined, we need x − 1 ≥ 0, so x ≥ 1; the domain is [1, ∞), not [0, ∞). --- **Q28.** **Assertion (A):** A relation R on ℤ defined by aRb if |a − b| ≤ 2 is an equivalence relation. **Reason (R):** The relation is reflexive and symmetric, and these two properties always imply transitivity. (A) Both A and R are true, and R is the correct explanation of A (B) A is true, but R is false (C) A is false, and R is false (D) A is true, and R is partially correct **Answer:** (C) A is false, and R is false **Reason:** While aRa (reflexive: |a−a| = 0 ≤ 2) and aRb ⟹ bRa (symmetric), transitivity fails: 1R3 (|1−3| = 2 ≤ 2) and 3R5 (|3−5| = 2 ≤ 2), but 1∤5 (|1−5| = 4 > 2). Reflexivity + symmetry ≠ transitivity. --- **Q29.** **Assertion (A):** For f(x) = (x + 1)/(x − 1) defined on ℝ − {1}, the inverse function f⁻¹(x) = (x + 1)/(x − 1). **Reason (R):** The function f is its own inverse (f ∘ f = identity), making f⁻¹ = f. (A) Both A and R are true, and R is the correct explanation of A (B) A is true, but R is false (C) A is false; R requires verification (D) Both A and R are false **Answer:** (A) Both A and R are true, and R is the correct explanation of A **Reason:** Solve y = (x+1)/(x−1) for x: y(x−1) = x+1 ⟹ x = (y+1)/(y−1), so f⁻¹(x) = (x+1)/(x−1) = f(x); f is an involution. --- **Q30.** **Assertion (A):** If f: A → B and g: B → C are bijective, then (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹. **Reason (R):** The inverse of a composition reverses the order of the original functions in the composition. (A) Both A and R are true, and R is the correct explanation of A (B) Both A and R are true, but R is not the correct explanation of A (C) A is true, but R is false (D) A is false, and R is false **Answer:** (A) Both A and R are true, and R is the correct explanation of A **Reason:** (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹ is the reversal rule for inverse compositions; verify: (f⁻¹ ∘ g⁻¹) ∘ (g ∘ f) = identity confirms R explains A correctly.

CBSE MCQ designers deliberately embed plausible wrong answers to test depth, not just recall. Here are the most common traps in Relations and Functions questions: **Trap 1: Confusing Function Definition with Relation** Wrong: "A relation with some repeated inputs is a function." Correct: A function requires *every* input to have *exactly* one output. Trap options often list relations that have multiple outputs for a single input (e.g., {(1,a), (1,b), (2,c)}) and label them as functions. Always verify: one input, one output only. **Trap 2: Reflexivity ≠ Identity Relation** Wrong: "Every reflexive relation includes (a,a) and nothing else." Correct: A reflexive relation must include (a,a) for all elements, but may include other pairs too. Many students choose a reflexive option when the correct answer also requires symmetry or transitivity. Read what the question asks: is it reflexive *alone*, or does it need to be an equivalence relation? **Trap 3: Domain vs. Codomain Confusion** Wrong: "The domain of y = √(x−2) is {0, 1, 2, ...}." or "The range of f(x) = x² is ℝ." Correct: Domain is the set of valid *inputs* (x ≥ 2); range is the set of actual *outputs* (for x² on ℝ, that's [0, ∞), not all of ℝ). Trap options swap these or omit boundary conditions. Always solve the constraint equation (e.g., x − 2 ≥ 0) carefully. **Trap 4: Composition Order Reversal** Wrong: (f ∘ g)(x) = g(f(x)). Correct: (f ∘ g)(x) = f(g(x)); f is applied *after* g. A major exam trap: examiners pair "(f ∘ g)(x) = ..." with both f(g(x)) and g(f(x)) options. Memorize the notation: ∘ order = evaluation order *reversed*. **Trap 5: Symmetry ≠ Transitivity** Wrong: "If R is symmetric, it is transitive." Correct: A symmetric relation (aRb ⟹ bRa) says nothing about transitivity (aRb ∧ bRc ⟹ aRc). The 'friend' relation is symmetric but not transitive. When a question says "complete the properties," don't assume one implies the other. **Trap 6: Injective ≠ Surjective ≠ Bijective** Wrong: "A function that maps onto its entire codomain is always injective." Correct: Surjectivity (onto) and injectivity (one-to-one) are independent. f(x) = x² is not injective on ℝ (f(2) = f(−2)), and f: ℝ → ℝ, f(x) = x² is not surjective (no preimage for −1). Always test both properties separately. **Trap 7: Inverses Exist Only for Bijections** Wrong: "Every injective function has an inverse." Correct: A function must be *bijective* (both injective and surjective) to have an inverse. Partial inverses exist for injective functions on restricted codomains, but the question asks for f⁻¹: B → A—only possible if f is bijective. Exam trap: offer options like "injective only" as distractors. **Trap 8: Equivalence Relation Requires All Three Properties** Wrong: Choosing a relation that is reflexive and symmetric but not transitive when asked if it's an equivalence relation. Correct: An equivalence relation *must* satisfy reflexivity, symmetry, *and* transitivity simultaneously. If any one fails, it's not an equivalence relation. Trap: "incomplete" answer options that omit transitivity verification. **Trap 9: Assumed Domains and Codomains** Wrong: "f: ℝ → ℝ defined by f(x) = √x is surjective." Correct: With codomain ℝ, f is not surjective (no preimage for −1). But if codomain is restricted to [0, ∞), then f *is* surjective. Always read the function signature (f: A → B) carefully; examiners change it subtly to create traps. **Trap 10: Composition Domain Restrictions** Wrong: "If f: ℝ → ℝ and g: ℝ → ℝ, then (f ∘ g): ℝ → ℝ." Correct: (f ∘ g) has domain = {x ∈ domain(g) : g(x) ∈ domain(f)}. If g(x) = √x (domain [0,∞)) and f(x) = 1/(x−1), then (f ∘ g) excludes x=0 and requires g(x) ≠ 1. Trap options ignore this intersection. **Quick Checklist Before Submitting an Answer:** - ✓ Is the question asking for one property or all three (equivalence)? - ✓ Did I apply composition in the correct order? - ✓ Did I verify domain and range separately? - ✓ Does my answer satisfy the explicit function/relation definition given? - ✓ Did I re-read the codomain to check surjectivity?

Relations and Functions typically appears as 8–12 MCQs in a Class 9 CBSE exam, with 20–25 total multiple-choice questions across all chapters. You have 45 minutes for the MCQ block—that's ~2–3 minutes per question. Mismanaging time here can cost 10–15 marks. Here's a battle-tested strategy: **Phase 1: Scan & Classify (First 2 minutes)** Before answering, scan all 12 R&F questions and mentally sort them: (a) Easy (definition-based, direct substitution), (b) Medium (1–2 steps, property verification), (c) Hard (composition, inverse, assertion-reason). Allocate 1.5 min to easy, 2.5 min to medium, 3–4 min to hard. This ensures you don't waste 5 minutes on a hard assertion-reason when you could solve three easy ones. **Phase 2: Easy First (5–7 minutes for ~4 questions)** Attack all straightforward MCQs: "What is f(3) if f(x) = 2x+1?" or "Which set is the range of...?" These build confidence and give you quick wins. If stuck for >45 seconds, skip; don't spiral. Mark these answers confidently without second-guessing. **Phase 3: Medium Next (8–10 minutes for ~4–5 questions)** Work through property-checking and 1–2-step compositions. For "Is R transitive?", run a mental check: pick two pairs in R that might chain (aRb, bRc), verify if aRc is also in R. Write down the pair counterexample if it fails (e.g., "(1,2), (2,3) ∈ R, but (1,3) ∉ R, so NOT transitive"). This visible reasoning prevents careless errors. **Phase 4: Hard & Assertion-Reason (5–7 minutes for ~3 questions)** Assertion-reason MCQs have 4 standard outcomes: (A) Both true + R explains A, (B) Both true + R doesn't explain, (C) A true + R false, (D) Both false. Approach: (1) Verify A independently; (2) Verify R independently; (3) Check if R logically explains A. For function composition and inverse questions, draw a small diagram (domain → codomain) to visualize whether f is injective, surjective, or bijective. **Example Timing Breakdown (12-question R&F block):** - Questions 1–4 (Easy): 1.5 min each = 6 min total - Questions 5–9 (Medium): 2.5 min each = 12.5 min total - Questions 10–12 (Hard/Assertion): 3 min each = 9 min total - **Buffer/Review:** 2 min - **Total:** 29.5 min (leaving 15+ min for other chapters) **Red Flags & Time-Savers:** - If a question involves "is X an equivalence relation?", write down: reflexive? symmetric? transitive? Check all three; don't assume one implies another. If any fails, answer is "No." (~30 seconds save). - For domain/range questions, always solve the constraint inequality first (e.g., √(x−2) ⟹ x≥2) rather than guessing. (~45 sec). - Composition questions: substitute g(x) into f first, *then* simplify. Avoid mental shortcuts that lead to (f ∘ g) = (g ∘ f) errors. (~1 min). - If you hit 2.5 minutes on a medium question with no progress, skip and mark it for review in the buffer phase. **Final 2 Minutes (Review & Corrections):** If time permits, do a final scan: (1) Have you selected exactly one option per question? (2) Are your answers reasonable (domain subset of ℝ, range ≠ entire codomain unless proven surjective)? (3) For assertion-reason, is your logic sound? Don't over-think or flip answers at the last second—your first instinct is usually correct. **Practice Hack:** Solve these 30 questions in a simulated exam setting: 45 minutes, pen & paper, no calculator. Track which section (easy/medium/hard) you struggled with. If you consistently overrun on medium questions, practice those more and improve your mental arithmetic. If assertion-reason trips you up, spend extra time on those patterns. Start a 3-day free trial at cbsetutor.ai to access video solutions for each of these 30 questions, detailed step-by-step breakdowns, and adaptive drills that adjust difficulty based on your performance.

**Relation Properties (Essential Definitions):** - **Reflexive:** aRa for all a ∈ A (every element relates to itself). - **Symmetric:** aRb ⟹ bRa for all a, b ∈ A (relation is bidirectional). - **Transitive:** aRb ∧ bRc ⟹ aRc for all a, b, c ∈ A (relation chains). - **Equivalence Relation:** A relation that is reflexive, symmetric, *and* transitive. **Function Definitions:** - **Function f: A → B:** For every a ∈ A, exactly one b ∈ B such that (a, b) ∈ f. - **Domain:** Set A (all possible inputs). - **Codomain:** Set B (the target set). - **Range:** Actual set of outputs {b : ∃a ∈ A, f(a) = b} ⊆ B. - **Injective (One-to-One):** f(a₁) = f(a₂) ⟹ a₁ = a₂ (no two inputs map to same output). - **Surjective (Onto):** For every b ∈ B, ∃a ∈ A such that f(a) = b (every element of codomain is in the range). - **Bijective:** Both injective and surjective. **Function Composition:** - (f ∘ g)(x) = f(g(x)) — apply g *first*, then f. - (f ∘ g) ≠ (g ∘ f) in general (composition is not commutative). - (f ∘ g) ∘ h = f ∘ (g ∘ h) (composition is associative). **Inverse Function:** - f⁻¹: B → A exists if and only if f: A → B is bijective. - f⁻¹(f(a)) = a for all a ∈ A (left inverse property). - f(f⁻¹(b)) = b for all b ∈ B (right inverse property). - (f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹ (inverse of composition reverses order). **Counting Formulas:** - Number of reflexive relations on an n-element set: 2^(n² − n) = 2^(n(n−1)). - Number of symmetric relations on an n-element set: 2^(n(n+1)/2). - Number of equivalence relations: Depends on partitions (more complex; use Bell numbers). - Number of functions f: A → B (where |A| = m, |B| = n): n^m. - Number of injective functions (one-to-one): P(n, m) = n!/(n−m)! (if m ≤ n). - Number of bijections (if |A| = |B| = n): n! (factorial). **Domain Restriction (Common Cases):** - √(expression): expression ≥ 0. - 1/(expression): expression ≠ 0. - log(expression): expression > 0. - |expression|: all real numbers. Use these as a quick mental checklist during the exam to avoid careless errors.