A Level H2 Mathematics Practice Paper 2

TuitionGoWhere Practice Paper – Maths H2 A-Level

Answer Key and Marking Scheme

Paper: Practice Paper – Algebra & Functions
Version: 2 of 5
Total Marks: 60

Section A: Functions – Domain, Range, and Composites (15 marks)

Question 1

(a) Range of f = (−∞, ∞) or ℝ. [1 mark]
Since ln(x + 2) can take any real value as x > −2.

(b) For fg to exist, the range of g must be a subset of the domain of f.
Domain of f: x > −2.
g(x) = x² − 1, x ≥ 0. Range of g = [−1, ∞).
Since [−1, ∞) ⊆ (−2, ∞), fg exists. [2 marks]
1 mark for checking condition; 1 mark for correct conclusion with justification.

(c) fg(x) = f(g(x)) = ln((x² − 1) + 2) = ln(x² + 1).
Domain of fg: {x ∈ domain of g : g(x) ∈ domain of f}.
Domain of g: x ≥ 0. g(x) = x² − 1 > −2 for all x ≥ 0 (since x² − 1 ≥ −1 > −2).
Thus, domain of fg = {x ∈ ℝ : x ≥ 0}. [3 marks]
1 mark for correct expression; 1 mark for correct domain; 1 mark for clear reasoning.

Question 2

(a) h(x) = √(4 − x²), −2 ≤ x ≤ 0.
When x = −2, h = 0; when x = 0, h = 2. As x increases from −2 to 0, 4 − x² decreases from 0 to 4, so h increases from 0 to 2.
Range of h = [0, 2]. [1 mark]

(b) h is strictly increasing on [−2, 0] (since derivative is positive), so it is one-to-one and therefore has an inverse. [1 mark]

(c) Let y = √(4 − x²). Then y² = 4 − x², so x² = 4 − y².
Since x ≤ 0, x = −√(4 − y²).
Thus, h⁻¹(x) = −√(4 − x²).
Domain of h⁻¹ = range of h = [0, 2]. [3 marks]
1 mark for correct rearrangement; 1 mark for correct sign choice; 1 mark for domain.

Question 3

(a) Let y = (2x + 1)/(x − 3).
y(x − 3) = 2x + 1
yx − 3y = 2x + 1
yx − 2x = 3y + 1
x(y − 2) = 3y + 1
x = (3y + 1)/(y − 2), y ≠ 2.
Thus, k⁻¹(x) = (3x + 1)/(x − 2), x ≠ 2.
Domain of k⁻¹ = range of k = ℝ \ {2}. [3 marks]
1 mark for algebraic manipulation; 1 mark for correct expression; 1 mark for domain.

(b) k(k⁻¹(x)) = k((3x + 1)/(x − 2)) = [2((3x + 1)/(x − 2)) + 1] / [((3x + 1)/(x − 2)) − 3]
= [(6x + 2)/(x − 2) + (x − 2)/(x − 2)] / [(3x + 1)/(x − 2) − (3x − 6)/(x − 2)]
= [(7x)/(x − 2)] / [7/(x − 2)] = x. [1 mark]

Section B: Graphs, Transformations, and Parametric Equations (15 marks)

Question 4

(a) x = 2 cos θ, y = 3 sin θ.
cos θ = x/2, sin θ = y/3.
Using cos²θ + sin²θ = 1: (x/2)² + (y/3)² = 1, i.e., x²/4 + y²/9 = 1.
Since 0 ≤ θ ≤ π, cos θ ranges from 1 to −1, so x ∈ [−2, 2]; sin θ ranges from 0 to 1 (and back to 0), so y ∈ [0, 3].
Cartesian equation: x²/4 + y²/9 = 1, with y ≥ 0 (upper half of ellipse). [2 marks]

(b) Sketch: Upper half of ellipse centred at origin, x-intercepts at (−2, 0) and (2, 0), y-intercept at (0, 3).
Coordinates: Meets x-axis at (−2, 0) and (2, 0); meets y-axis at (0, 3). [3 marks]
1 mark for correct shape; 1 mark for correct intercepts; 1 mark for correct restricted domain/range.

Question 5

Note: Since the original graph of y = f(x) is not provided in this practice paper, the following describes the expected transformations. In an actual exam, a specific graph would be given.

(a) y = |f(x)|: Reflect any part of f(x) below the x-axis in the x-axis. Parts above the x-axis remain unchanged. [2 marks]

(b) y = f(|x|): Replace the part of the graph for x < 0 with a reflection of the part for x > 0 in the y-axis. The part for x ≥ 0 remains unchanged. [2 marks]

(c) y = 1/f(x): Where f(x) = 0, vertical asymptotes occur. Where f(x) → ±∞, 1/f(x) → 0. Where f(x) = 1 or −1, the graph is unchanged. The graph approaches 0 as |f(x)| → ∞. [2 marks]

Question 6

(a) y = (x² + 1)/(x − 1).
Vertical asymptote: denominator = 0 ⇒ x = 1.
As x → ∞, y = (x² + 1)/(x − 1) = x + 1 + 2/(x − 1) → ∞ (no horizontal asymptote).
Oblique asymptote: perform division: (x² + 1) ÷ (x − 1) = x + 1 + 2/(x − 1).
As x → ±∞, 2/(x − 1) → 0, so oblique asymptote is y = x + 1. [2 marks]
1 mark for vertical asymptote; 1 mark for oblique asymptote.

(b) y = (x² + 1)/(x − 1).
dy/dx = [(2x)(x − 1) − (x² + 1)(1)] / (x − 1)² = (2x² − 2x − x² − 1)/(x − 1)² = (x² − 2x − 1)/(x − 1)².
Stationary points when dy/dx = 0: x² − 2x − 1 = 0 ⇒ x = 1 ± √2.
When x = 1 + √2: y = ((1 + √2)² + 1)/(√2) = (1 + 2√2 + 2 + 1)/√2 = (4 + 2√2)/√2 = 2√2 + 2.
When x = 1 − √2: y = ((1 − √2)² + 1)/(−√2) = (1 − 2√2 + 2 + 1)/(−√2) = (4 − 2√2)/(−√2) = −2√2 + 2.
Coordinates: (1 + √2, 2 + 2√2) and (1 − √2, 2 − 2√2). [2 marks]

Section C: Equations, Inequalities, and Modulus (15 marks)

Question 7

(x + 2)/(x − 1) > 0.
Critical points: x = −2 (numerator = 0), x = 1 (denominator = 0).
Sign analysis:
x < −2: (−)/(−) = +
−2 < x < 1: (+)/(−) = −
x > 1: (+)/(+) = +
Solution: x < −2 or x > 1. [3 marks]
1 mark for critical points; 1 mark for sign analysis; 1 mark for correct solution.

Question 8

|2x − 3| < 5
⇔ −5 < 2x − 3 < 5
⇔ −2 < 2x < 8
⇔ −1 < x < 4. [2 marks]

Question 9

(a) y = |x − 2|: V-shaped graph with vertex at (2, 0).
y = 3 − x: straight line with y-intercept 3, slope −1.
Sketch both on same axes, showing intersection. [2 marks]

(b) |x − 2| < 3 − x.
From the graph, the line y = 3 − x is above y = |x − 2| for x < 2.5 (intersection point).
Algebraic solution:
Case 1: x ≥ 2: x − 2 < 3 − x ⇒ 2x < 5 ⇒ x < 2.5. So 2 ≤ x < 2.5.
Case 2: x < 2: −(x − 2) < 3 − x ⇒ −x + 2 < 3 − x ⇒ 2 < 3, always true. So x < 2.
Combined: x < 2.5. [3 marks]
1 mark for case analysis; 1 mark for correct algebraic solution; 1 mark for final answer.

Question 10

(a) p(x) = |x² − 4|.
x² − 4 ≥ 0 when x ≤ −2 or x ≥ 2; x² − 4 < 0 when −2 < x < 2.
Thus:
p(x) = x² − 4  for x ≤ −2 or x ≥ 2
p(x) = 4 − x²  for −2 < x < 2. [2 marks]

(b) p(x) = 5.
Case 1: x² − 4 = 5 ⇒ x² = 9 ⇒ x = ±3. Both satisfy x ≤ −2 or x ≥ 2.
Case 2: 4 − x² = 5 ⇒ x² = −1, no real solutions.
Solutions: x = −3, x = 3. [3 marks]
1 mark for each case; 1 mark for final answer.

Section D: Functions in Context and Advanced Applications (15 marks)

Question 11

(a) f(x) = e^(2x) − 3.
Domain: ℝ.
Range: e^(2x) > 0, so e^(2x) − 3 > −3. Range = (−3, ∞). [1 mark]

(b) Let y = e^(2x) − 3.
y + 3 = e^(2x)
ln(y + 3) = 2x
x = (1/2) ln(y + 3).
Thus, f⁻¹(x) = (1/2) ln(x + 3).
Domain of f⁻¹ = range of f = (−3, ∞). [2 marks]

(c) Graphs: y = f(x) is exponential, passing through (0, −2), asymptote y = −3.
y = f⁻¹(x) is logarithmic, passing through (−2, 0), asymptote x = −3.
The graphs are reflections of each other in the line y = x. [2 marks]

Question 12

(a) fg(x) = f(g(x)) = f(x²) = 2x² + 1.
gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)² = 4x² + 4x + 1. [2 marks]

(b) fg ≠ gf since 2x² + 1 ≠ 4x² + 4x + 1 in general. [1 mark]

(c) gf(x) = (2x + 1)², x ∈ ℝ.
Since 2x + 1 can take any real value, (2x + 1)² ≥ 0.
Range of gf = [0, ∞). [2 marks]

Question 13

(a) f(x) = (ax + b)/(cx − a).
f(f(x)) = [a((ax + b)/(cx − a)) + b] / [c((ax + b)/(cx − a)) − a]
= [(a(ax + b) + b(cx − a)) / (cx − a)] / [(c(ax + b) − a(cx − a)) / (cx − a)]
= [a²x + ab + bcx − ab] / [acx + bc − acx + a²]
= [(a² + bc)x] / [a² + bc]
= x, provided a² + bc ≠ 0.
Thus, f is self-inverse. [3 marks]
1 mark for correct substitution; 1 mark for algebraic simplification; 1 mark for conclusion.

(b) g(x) = (kx + 4)/(x − k). For self-inverse, comparing with form in (a): a = k, b = 4, c = 1, −a = −k.
Condition: a² + bc = k² + 4 ≠ 0 (always true for real k).
Thus, any real k works? Wait—check: g(g(x)) = x.
g(g(x)) = [k((kx + 4)/(x − k)) + 4] / [((kx + 4)/(x − k)) − k]
= [(k²x + 4k + 4x − 4k) / (x − k)] / [(kx + 4 − kx + k²) / (x − k)]
= [(k² + 4)x] / [k² + 4] = x.
Thus, g is self-inverse for all real k (provided denominator ≠ 0, i.e., x ≠ k).
However, the standard form requires a² + bc ≠ 0, which is k² + 4 ≠ 0, always true.
So k can be any real number. [2 marks]
1 mark for setting up; 1 mark for correct conclusion.

END OF ANSWER KEY