A Level H2 Mathematics Algebra Functions Quiz

A-Level Maths H2 Quiz - Algebra Functions

Name: _________________ Class: _________________ Date: _________________

Score: _____ / 60 Duration: 45 minutes

Instructions:

• Answer ALL questions in the spaces provided
• Show all working clearly
• Calculators are allowed
• Give answers to 3 significant figures unless otherwise stated

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Section A: Functions and Transformations [20 marks]

1. The function f is defined by f(x) = 2x + 3 for x ∈ ℝ and the function g is defined by g(x) = x² - 1 for x ∈ ℝ.

(a) Find fg(x). [2]

Answer: _________________________________

(b) State the domain and range of fg. [2]

Domain: _________________________________

Range: _________________________________

2. The function h(x) = x² - 4x + 1 is defined for x ∈ ℝ.

(a) Explain why h does not have an inverse function. [1]

Answer: _________________________________

(b) Find the largest value of k such that h: [k, ∞) → ℝ has an inverse function. [2]

Answer: k = _________________________________

3. Given that f(x) = 1/(x-2), describe the sequence of transformations that maps the graph of y = 1/x to the graph of y = f(x) + 3. [2]

Answer: _________________________________

4. The curve C has parametric equations x = 2t, y = t² - 1 for t ∈ ℝ.

Find the cartesian equation of C. [3]

Answer: _________________________________

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Section B: Composite Functions and Inverses [25 marks]

5. Functions f and g are defined by: f(x) = √(x + 1) for x ≥ -1 g(x) = 2x - 3 for x ∈ ℝ

(a) Show that the composite function gf exists and find gf(x). [3]

Working:

Answer: gf(x) = _________________________________

(b) State the domain of gf. [1]

Answer: _________________________________

(c) Find the range of gf. [2]

Answer: _________________________________

6. The function p is defined by p(x) = (x + 2)/(x - 1) for x ∈ ℝ, x ≠ 1.

(a) Find p⁻¹(x). [3]

Working:

Answer: p⁻¹(x) = _________________________________

(b) State the domain and range of p⁻¹. [2]

Domain: _________________________________

Range: _________________________________

7. Given that f(x) = 3x - 2 and g(x) = x² + 1, solve the equation fg(x) = gf(x). [4]

Working:

Answer: x = _________________________________

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Section C: Advanced Function Analysis [15 marks]

8. The function f is defined by f(x) = |2x - 3| for x ∈ ℝ.

(a) Sketch the graph of y = f(x), showing clearly the coordinates of any turning points and intercepts. [3]

[GRAPH SPACE]

(b) Solve the inequality f(x) < 5. [2]

Working:

Answer: _________________________________

9. Functions h and k are defined by: h(x) = x² - 2x for x ∈ ℝ k(x) = 3 - x for x ∈ ℝ

(a) Find the values of x for which h(x) = k(x). [3]

Working:

Answer: x = _________________________________

(b) Hence sketch the graphs of y = h(x) and y = k(x) on the same axes, showing clearly the points of intersection. [2]

[GRAPH SPACE]

10. A function f is defined by f(x) = ax + b where a and b are constants. Given that f(2) = 7 and f⁻¹(3) = -1, find the values of a and b. [5]

Working:

Answer: a = _______, b = _______

A-Level Maths H2 Quiz - Algebra Functions (Answers)

Section A: Functions and Transformations [20 marks]

1. (a) fg(x) = f(g(x)) = f(x² - 1) = 2(x² - 1) + 3 = 2x² - 2 + 3 = 2x² + 1 [2]

(b) Domain: x ∈ ℝ (since g has domain ℝ and range of g is contained in domain of f) [1] Range: y ≥ 1 (since 2x² ≥ 0, so 2x² + 1 ≥ 1) [1]

2. (a) h(x) is not one-to-one / fails the horizontal line test / is a parabola opening upwards [1]

(b) Complete the square: h(x) = (x - 2)² - 3 Minimum point at x = 2, so k = 2 [2]

3. Translation 2 units to the right (or in the positive x-direction), then translation 3 units up (or in the positive y-direction) [2]

4. From x = 2t, we get t = x/2 Substituting into y = t² - 1: y = (x/2)² - 1 = x²/4 - 1 Therefore: y = x²/4 - 1 or 4y = x² - 4 [3]

Section B: Composite Functions and Inverses [25 marks]

5. (a) For gf to exist, range of f must be contained in domain of g. Range of f: y ≥ 0 (since √(x+1) ≥ 0) Domain of g: x ∈ ℝ Since [0, ∞) ⊂ ℝ, gf exists. [1] gf(x) = g(f(x)) = g(√(x+1)) = 2√(x+1) - 3 [2]

(b) Domain of gf = domain of f = x ≥ -1 [1]

(c) When x ≥ -1, √(x+1) ≥ 0, so 2√(x+1) ≥ 0, therefore 2√(x+1) - 3 ≥ -3 Range of gf: y ≥ -3 [2]

6. (a) Let y = (x + 2)/(x - 1) y(x - 1) = x + 2 yx - y = x + 2 yx - x = y + 2 x(y - 1) = y + 2 x = (y + 2)/(y - 1) Therefore p⁻¹(x) = (x + 2)/(x - 1) [3]

(b) Domain of p⁻¹: x ∈ ℝ, x ≠ 1 [1] Range of p⁻¹: y ∈ ℝ, y ≠ 1 [1]

7. fg(x) = f(x² + 1) = 3(x² + 1) - 2 = 3x² + 1 gf(x) = g(3x - 2) = (3x - 2)² + 1 = 9x² - 12x + 4 + 1 = 9x² - 12x + 5

Setting fg(x) = gf(x): 3x² + 1 = 9x² - 12x + 5 0 = 6x² - 12x + 4 0 = 3x² - 6x + 2

Using quadratic formula: x = (6 ± √(36 - 24))/6 = (6 ± √12)/6 = (6 ± 2√3)/6 = (3 ± √3)/3 [4]

Section C: Advanced Function Analysis [15 marks]

8. (a) f(x) = |2x - 3| When 2x - 3 ≥ 0 (x ≥ 3/2): f(x) = 2x - 3 When 2x - 3 < 0 (x < 3/2): f(x) = -(2x - 3) = -2x + 3

Turning point: (3/2, 0) y-intercept: f(0) = 3, so (0, 3) x-intercept: (3/2, 0) [3]

(b) |2x - 3| < 5 -5 < 2x - 3 < 5 -2 < 2x < 8 -1 < x < 4 [2]

9. (a) h(x) = k(x) x² - 2x = 3 - x x² - 2x + x - 3 = 0 x² - x - 3 = 0

Using quadratic formula: x = (1 ± √(1 + 12))/2 = (1 ± √13)/2 [3]

(b) Sketch should show:

• h(x): parabola with vertex at (1, -1)
• k(x): straight line with gradient -1, y-intercept 3
• Intersection points at x = (1 ± √13)/2 [2]

10. Given f(x) = ax + b, f(2) = 7, and f⁻¹(3) = -1

From f(2) = 7: 2a + b = 7 ... (1) From f⁻¹(3) = -1: f(-1) = 3, so -a + b = 3 ... (2)

Subtracting (2) from (1): 3a = 4, so a = 4/3 Substituting into (1): 2(4/3) + b = 7, so 8/3 + b = 7, therefore b = 7 - 8/3 = 13/3

Answer: a = 4/3, b = 13/3 [5]

Marking Notes:

• Award full marks for correct method even if arithmetic errors occur
• Partial credit available for correct setup of equations
• Graphs must show key features clearly for full marks
• Accept equivalent forms of answers (e.g., decimals where appropriate)