A Level H2 Mathematics Practice Paper 3

TuitionGoWhere Practice Paper – Maths H2 A-Level

Answer Key & Marking Scheme – Version 3

Paper: Practice Paper 3 – Algebra & Functions Total Marks: 60

Section A: Functions, Domain & Range (Questions 1–5)

1. (a) Range of f = ℝ (or (–∞, ∞)). ✓ [1]

Note: ln(2x – 1) can take all real values as x → ½⁺ gives ln(0⁺) → –∞, and as x → ∞, ln(2x – 1) → ∞.

1. (b)

• Domain of g = ℝ, range of g = (3, ∞) [since eˣ > 0 for all x, so eˣ + 3 > 3]. ✓
• Domain of f = (½, ∞).
• Since range of g = (3, ∞) ⊆ (½, ∞) = domain of f, the composite fg exists. ✓
• fg(x) = f(g(x)) = ln(2(eˣ + 3) – 1) = ln(2eˣ + 6 – 1) = ln(2eˣ + 5). ✓ [3]

2. (a)

• Range of h: h(x) = (x – 2)² + 3, for x ≥ 2. Minimum at x = 2 gives h(2) = 3. Range of h = [3, ∞). ✓
• Domain of k = [3, ∞).
• For kh to exist, range of h ⊆ domain of k.
• Range of h = [3, ∞) ⊆ [3, ∞) = domain of k. ✓
• Wait — this means kh does exist. Let's re-examine.

Correction: h(x) = x² – 4x + 7 = (x – 2)² + 3. For x ≥ 2, minimum is 3, so range of h = [3, ∞). Domain of k = [3, ∞). Since [3, ∞) ⊆ [3, ∞), the composite kh exists. The question asks to explain why it does NOT exist — this is a trick: it actually does exist. Students should recognize that range of h = [3, ∞) and domain of k = [3, ∞), so the condition is satisfied.

Alternative interpretation: If the question intended h(x) = x² – 4x + 7 with domain x ≥ 2, then h(x) ≥ 3. Domain of k is x ≥ 3. Range of h is [3, ∞) which is a subset of domain of k = [3, ∞). So kh exists. The question as written has no valid "does not exist" answer. Award marks for correct reasoning showing existence.

Marking: Award [1] for finding range of h, [1] for correct conclusion that kh exists with justification. If student incorrectly claims non-existence, award [0].

2. (b) For kh to exist, we need range of h ⊆ domain of k = [3, ∞).

h(x) = (x – 2)² + 3. For h(x) ≥ 3, we need (x – 2)² + 3 ≥ 3 ⇒ (x – 2)² ≥ 0, which is true for all x. So the condition is already satisfied for all x ≥ 2.

If the domain of h were restricted differently: h(x) ≥ 3 ⇒ (x – 2)² ≥ 0, always true. So any domain with x ≥ 2 works. The largest possible domain is x ≥ 2. [2]

Note: This question is problematic given part (a). Award marks for logical consistency.

3. (a)

• Let y = 3 – 2e⁻ˣ.
• 2e⁻ˣ = 3 – y ⇒ e⁻ˣ = (3 – y)/2. ✓
• –x = ln((3 – y)/2) ⇒ x = –ln((3 – y)/2) = ln(2/(3 – y)). ✓
• Therefore, f⁻¹(x) = ln(2/(3 – x)). ✓
• Domain of f⁻¹ = range of f.
• As x → ∞, e⁻ˣ → 0, so f(x) → 3⁻. As x → –∞, e⁻ˣ → ∞, so f(x) → –∞.
• Range of f = (–∞, 3). So domain of f⁻¹ = (–∞, 3). ✓ [3]

3. (b)

• Graph of y = f(x): decreasing, asymptotic to y = 3 as x → ∞, passes through (0, 1) since f(0) = 3 – 2 = 1.
• Graph of y = f⁻¹(x): reflection of y = f(x) in the line y = x. Domain (–∞, 3), range ℝ.
• Both graphs shown on same axes with line y = x indicated. [1]

4.

• For 0 ≤ x < 2: f(x) = x². Minimum at x = 0 gives 0, maximum as x → 2⁻ gives 4⁻. Range on this interval: [0, 4).
• For 2 ≤ x ≤ 4: f(x) = 8 – 2x. At x = 2, f(2) = 4. At x = 4, f(4) = 0. Linear decreasing: range [0, 4]. ✓
• Combining: overall range = [0, 4]. ✓ [2]

5. (a)

• g(x) = |2x – 5| + 1. The minimum of |2x – 5| is 0 (when x = 5/2).
• Therefore, minimum value of g(x) = 0 + 1 = 1. ✓ [1]

5. (b)

• g(x) = 2x ⇒ |2x – 5| + 1 = 2x ⇒ |2x – 5| = 2x – 1.
• Case 1: 2x – 5 ≥ 0, i.e., x ≥ 5/2.
  • Then |2x – 5| = 2x – 5.
  • Equation: 2x – 5 = 2x – 1 ⇒ –5 = –1 (contradiction). No solution. ✓
• Case 2: 2x – 5 < 0, i.e., x < 5/2.
  • Then |2x – 5| = –(2x – 5) = 5 – 2x.
  • Equation: 5 – 2x = 2x – 1 ⇒ 6 = 4x ⇒ x = 3/2. ✓
  • Check: x = 3/2 < 5/2, valid.
• Also check RHS: 2x – 1 must be ≥ 0 for equality since LHS = |2x – 5| ≥ 0. At x = 3/2, RHS = 2, valid. ✓
• Solution: x = 3/2. [3]

Section B: Graphs, Transformations & Parametric Equations (Questions 6–10)

6. (a)

• x = 2cos θ ⇒ cos θ = x/2.
• y = 3sin θ ⇒ sin θ = y/3.
• Using cos²θ + sin²θ = 1: (x/2)² + (y/3)² = 1. ✓
• Cartesian equation: x²/4 + y²/9 = 1. ✓ [2]

6. (b)

• Ellipse centred at origin, semi-major axis 3 along y-axis, semi-minor axis 2 along x-axis.
• Meets x-axis when y = 0: x²/4 = 1 ⇒ x = ±2. Points: (2, 0) and (–2, 0). ✓
• Meets y-axis when x = 0: y²/9 = 1 ⇒ y = ±3. Points: (0, 3) and (0, –3). ✓
• Sketch: ellipse with intercepts labelled. [2]

7. (a) y = f(x + 1)

• Translation by vector (–1, 0): shift left by 1 unit.
• Maximum point moves from (2, 5) to (1, 5).
• x-intercepts move from (0, 0) and (4, 0) to (–1, 0) and (3, 0).
• Sketch showing shifted curve with points labelled. [2]

7. (b) y = |f(x)|

• Parts of f(x) below x-axis are reflected above x-axis.
• Original f(x): maximum at (2, 5), passes through (0, 0) and (4, 0). Assume f(x) ≥ 0 for 0 ≤ x ≤ 4 (since it's a "bump" from (0,0) up to (2,5) down to (4,0)).
• If f(x) is non-negative on [0, 4], then |f(x)| = f(x) on this interval.
• If f(x) < 0 outside [0, 4], those parts are reflected.
• Sketch showing original curve for f(x) ≥ 0, reflected parts for f(x) < 0. [2]

7. (c) y = f(|x|)

• For x ≥ 0, graph is y = f(x).
• For x < 0, graph is reflection of the x ≥ 0 part in the y-axis (even function).
• Original f(x) for x ≥ 0: from (0, 0) up to (2, 5) down to (4, 0).
• For x < 0: mirror image, so curve from (0, 0) up to (–2, 5) down to (–4, 0).
• Sketch showing symmetric curve about y-axis. [2]

8. (a)

• y = (x² + 1)/(x – 2).
• Vertical asymptote: denominator = 0 ⇒ x = 2. ✓
• As x → ∞: y = (x² + 1)/(x – 2) = x + 2 + 5/(x – 2) (by polynomial division).
• Oblique asymptote: y = x + 2. ✓ [2]

8. (b)

• y = x + 2 + 5/(x – 2).
• dy/dx = 1 – 5/(x – 2)².
• Stationary points: dy/dx = 0 ⇒ 1 = 5/(x – 2)² ⇒ (x – 2)² = 5 ⇒ x = 2 ± √5.
• At x = 2 + √5: y = (2 + √5) + 2 + 5/√5 = 4 + √5 + √5 = 4 + 2√5.
• At x = 2 – √5: y = (2 – √5) + 2 + 5/(–√5) = 4 – √5 – √5 = 4 – 2√5.
• Second derivative or first derivative test to determine nature:
  • d²y/dx² = 10/(x – 2)³.
  • At x = 2 + √5: (x – 2)³ = (√5)³ > 0, so d²y/dx² > 0 ⇒ minimum.
  • At x = 2 – √5: (x – 2)³ = (–√5)³ < 0, so d²y/dx² < 0 ⇒ maximum.
• Sketch: vertical asymptote x = 2, oblique asymptote y = x + 2, local max at (2 – √5, 4 – 2√5), local min at (2 + √5, 4 + 2√5). [2]

9.

• y = f(x) → y = 3f(2x – 1) + 4.
• Step 1: f(x) → f(2x). Horizontal stretch by scale factor ½ (compression). ✓
• Step 2: f(2x) → f(2(x – ½)) = f(2x – 1). Translation by vector (½, 0). ✓
• Step 3: f(2x – 1) → 3f(2x – 1). Vertical stretch by scale factor 3. ✓
• Step 4: 3f(2x – 1) → 3f(2x – 1) + 4. Translation by vector (0, 4). ✓
• Alternative order possible if correctly described. [3]

10. (a)

• y = 1/(x² + 1). Since x² + 1 ≥ 1 for all real x, 0 < 1/(x² + 1) ≤ 1.
• Range: (0, 1]. ✓ [1]

10. (b)

• y = (x² + 1)⁻¹.
• dy/dx = –(x² + 1)⁻² · 2x = –2x/(x² + 1)². ✓
• Stationary point: dy/dx = 0 ⇒ –2x = 0 ⇒ x = 0. ✓
• At x = 0, y = 1/(0 + 1) = 1. Point: (0, 1).
• Nature: For x < 0, dy/dx > 0 (increasing). For x > 0, dy/dx < 0 (decreasing). Therefore, (0, 1) is a maximum point. ✓ [3]

Section C: Equations, Inequalities & Modulus (Questions 11–15)

11.

• (2x – 1)/(x + 3) > 1.
• (2x – 1)/(x + 3) – 1 > 0 ⇒ (2x – 1 – (x + 3))/(x + 3) > 0 ⇒ (x – 4)/(x + 3) > 0. ✓
• Critical values: x = –3 (undefined), x = 4 (numerator zero). ✓
• Sign analysis:
  • x < –3: (x – 4) < 0, (x + 3) < 0 ⇒ positive. ✓
  • –3 < x < 4: (x – 4) < 0, (x + 3) > 0 ⇒ negative.
  • x > 4: (x – 4) > 0, (x + 3) > 0 ⇒ positive. ✓
• Solution: x < –3 or x > 4. ✓ [4]

12.

• |3x – 2| = |x + 4|.
• Square both sides: (3x – 2)² = (x + 4)². ✓
• 9x² – 12x + 4 = x² + 8x + 16.
• 8x² – 20x – 12 = 0 ⇒ 2x² – 5x – 3 = 0. ✓
• (2x + 1)(x – 3) = 0 ⇒ x = –½ or x = 3. ✓
• Check both solutions satisfy original equation:
  • x = –½: LHS = |–1.5 – 2| = |–3.5| = 3.5; RHS = |–0.5 + 4| = |3.5| = 3.5. ✓
  • x = 3: LHS = |9 – 2| = 7; RHS = |3 + 4| = 7. ✓
• Solutions: x = –½, x = 3. ✓ [4]

13.

• |2x + 1| ≤ 5.
• –5 ≤ 2x + 1 ≤ 5. ✓
• –6 ≤ 2x ≤ 4. ✓
• –3 ≤ x ≤ 2. ✓
• Solution: –3 ≤ x ≤ 2. [3]

14. (a)

• fg(x) = f(g(x)) = f(2x + 1) = (2x + 1)² – 3 = 4x² + 4x + 1 – 3 = 4x² + 4x – 2. ✓
• gf(x) = g(f(x)) = g(x² – 3) = 2(x² – 3) + 1 = 2x² – 6 + 1 = 2x² – 5. ✓
• fg(x) = gf(x) ⇒ 4x² + 4x – 2 = 2x² – 5.
• 2x² + 4x + 3 = 0. ✓
• Discriminant: Δ = 16 – 24 = –8 < 0. No real solutions. ✓ [3]

14. (b)

• fg(x) > gf(x) ⇒ 4x² + 4x – 2 > 2x² – 5.
• 2x² + 4x + 3 > 0. ✓
• Discriminant = 16 – 24 = –8 < 0, and coefficient of x² is positive (2 > 0).
• Therefore, 2x² + 4x + 3 > 0 for all real x. ✓
• Solution: x ∈ ℝ. [2]

15.

• f(1) = a + b + c = 2 ... (1)
• f(2) = 4a + 2b + c = 7 ... (2)
• f(3) = 9a + 3b + c = 14 ... (3)
• (2) – (1): 3a + b = 5 ... (4) ✓
• (3) – (2): 5a + b = 7 ... (5) ✓
• (5) – (4): 2a = 2 ⇒ a = 1. ✓
• From (4): 3(1) + b = 5 ⇒ b = 2. ✓
• From (1): 1 + 2 + c = 2 ⇒ c = –1. ✓
• Therefore, a = 1, b = 2, c = –1. [4]

Section D: Applications & Problem Solving (Questions 16–20)

16. (a)

• y = (x² – 4)/(x – 1).
• Polynomial division: x² – 4 ÷ (x – 1).
  • x² ÷ x = x. Multiply: x(x – 1) = x² – x. Subtract: (x² – 4) – (x² – x) = x – 4.
  • x ÷ x = 1. Multiply: 1(x – 1) = x – 1. Subtract: (x – 4) – (x – 1) = –3. ✓
• Therefore, y = x + 1 – 3/(x – 1). ✓
• So A = 1, B = 1, C = –3. ✓ [3]

16. (b)

• From y = x + 1 – 3/(x – 1):
• Vertical asymptote: x = 1 (denominator zero). ✓
• Oblique asymptote: as x → ±∞, 3/(x – 1) → 0, so y → x + 1. Equation: y = x + 1. ✓ [2]

17. (a)

• f(x) = e²ˣ – 4eˣ + 3.
• Let y = eˣ. Then e²ˣ = (eˣ)² = y². ✓
• f(x) = y² – 4y + 3. [1]

17. (b)

• f(x) = 0 ⇒ y² – 4y + 3 = 0.
• (y – 1)(y – 3) = 0 ⇒ y = 1 or y = 3. ✓
• Since y = eˣ:
  • eˣ = 1 ⇒ x = ln 1 = 0. ✓
  • eˣ = 3 ⇒ x = ln 3. ✓
• Solutions: x = 0, x = ln 3. ✓ [3]

18. (a)

• f(x) = ln(x + 2), for x > –2.
• As x → –2⁺, x + 2 → 0⁺, ln(x + 2) → –∞.
• As x → ∞, ln(x + 2) → ∞.
• Range of f = ℝ (or (–∞, ∞)). ✓ [1]

18. (b)

• gf(x) = g(f(x)) = g(ln(x + 2)) = (ln(x + 2))² – 1. ✓
• Domain of gf = domain of f = (–2, ∞).
• Let u = ln(x + 2). As x → –2⁺, u → –∞. As x → ∞, u → ∞.
• So u ∈ ℝ. ✓
• gf(x) = u² – 1, where u ∈ ℝ.
• Minimum of u² – 1 is –1 (when u = 0, i.e., ln(x + 2) = 0 ⇒ x = –1).
• As u → ±∞, u² – 1 → ∞.
• Range of gf = [–1, ∞). ✓ [3]

19.

• f(x) = (ax + b)/(cx + d).
• Vertical asymptote at x = 2 ⇒ denominator = 0 when x = 2: 2c + d = 0 ⇒ d = –2c. ✓
• Horizontal asymptote at y = 3 ⇒ lim(x→∞) f(x) = a/c = 3 ⇒ a = 3c. ✓
• Passes through (0, 1): f(0) = b/d = 1 ⇒ b = d. ✓
• From d = –2c and b = d, we have b = –2c.
• We need integer values. Choose c = 1 (simplest).
  • Then a = 3, d = –2, b = –2. ✓
• Check ad ≠ bc: (3)(–2) – (–2)(1) = –6 + 2 = –4 ≠ 0. ✓
• Therefore, a = 3, b = –2, c = 1, d = –2.
• f(x) = (3x – 2)/(x – 2). ✓ [5]

20. (a)

• x = t² + 2t, y = t² – 2t.
• x + y = 2t² ⇒ t² = (x + y)/2. ✓
• x – y = 4t ⇒ t = (x – y)/4. ✓
• Substitute: ((x – y)/4)² = (x + y)/2.
• (x – y)²/16 = (x + y)/2. ✓
• (x – y)² = 8(x + y). ✓
• Therefore, k = 8. [4]

20. (b)

• Equation: (x – y)² = 8(x + y).
• This is a parabola. Let u = x + y, v = x – y. Then v² = 8u.
• The curve is a parabola opening to the right in the (u, v) plane.
• Intersection with axes:
  • x-axis (y = 0): x² = 8x ⇒ x(x – 8) = 0 ⇒ x = 0 or x = 8. Points: (0, 0) and (8, 0). ✓
  • y-axis (x = 0): (–y)² = 8y ⇒ y² = 8y ⇒ y(y – 8) = 0 ⇒ y = 0 or y = 8. Points: (0, 0) and (0, 8). ✓
• Sketch: parabola passing through (0, 0), (8, 0), (0, 8), symmetric about line y = x, vertex at origin. [2]

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