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A Level H2 Mathematics Practice Paper 4

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A Level H2 Mathematics AI Generated Generated by Qwen3.6 Plus Updated 2026-06-03

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Questions

A-Level Maths H2 Quiz - Algebra Functions

Name: _________________________
Class: _________________________
Date: _________________________
Score: ________ / 50

Duration: 60 Minutes
Total Marks: 50

Instructions:

Answer all 20 questions.
Write your answers in the spaces provided.
You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless otherwise stated.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.

Section A: Basic Concepts and Manipulation (Questions 1–5)

[10 Marks]

1. The function $f$ is defined by $f(x) = \frac{2x-1}{x+3}$ for $x \in \mathbb{R}, x \neq -3$ .
Find the value of $f^{-1}(2)$ .
[1]

2. The functions $f$ and $g$ are defined by $f(x) = x^2 + 1$ for $x \ge 0$ and $g(x) = \sqrt{x-1}$ for $x \ge 1$ .
Find the expression for $fg(x)$ in its simplest form.
[1]

3. Given that $h(x) = |2x - 4|$ , solve the inequality $h(x) \le 6$ .
[2]

4. The function $k$ is defined by $k(x) = e^{2x} - 3$ .
State the range of $k$ .
[1]

5. Find the exact set of values of $x$ for which $\frac{x-1}{x+2} > 0$ .
[2]

Section B: Composite and Inverse Functions (Questions 6–10)

[15 Marks]

6. The functions $f$ and $g$ are defined by: $f(x) = \frac{1}{x-2}, \quad x \in \mathbb{R}, x \neq 2$ $g(x) = x^2 + 1, \quad x \in \mathbb{R}$

(a) Explain why the composite function $fg$ does not exist.
[1]

(b) Find the largest possible domain of $g$ such that the composite function $fg$ exists.
[2]

7. The function $f$ is defined by $f(x) = \frac{3x+1}{x-2}$ for $x > 2$ .

(a) Find an expression for $f^{-1}(x)$ .
[2]

(b) State the domain of $f^{-1}$ .
[1]

8. The function $g$ is defined by $g(x) = (x-1)^2 + 3$ for $x \ge 1$ .

(a) Sketch the graph of $y = g(x)$ , stating the coordinates of the vertex and the $y$ -intercept.
[2]

(b) Find $g^{-1}(x)$ and state its domain.
[2]

9. Let $f(x) = \ln(x+1)$ for $x > -1$ and $g(x) = e^x - 1$ for $x \in \mathbb{R}$ .

(a) Show that $gf(x) = x$ .
[1]

(b) Hence, or otherwise, find the exact solution to the equation $f(g(x)) = 2$ .
[2]

10. The function $h$ is defined by $h(x) = \frac{ax+b}{cx+d}$ where $a,b,c,d$ are constants.
Given that $h^{-1}(x) = h(x)$ for all $x$ in the domain, show that $a + d = 0$ .
[2]

Section C: Graphs and Transformations (Questions 11–15)

[15 Marks]

11. The diagram shows the graph of $y = f(x)$ which passes through the points $A(-2, 0)$ , $B(0, 3)$ , and $C(2, 0)$ . The line $x=1$ is a vertical asymptote.

On separate diagrams, sketch the graphs of:

(a) $y = |f(x)|$
[2]

(b) $y = f(|x|)$
[2]

12. The graph of $y = \frac{1}{x}$ is transformed to the graph of $y = \frac{2}{x-1} + 3$ .
Describe the sequence of transformations geometrically.
[2]

13. Sketch the graph of $y = |2x - 1|$ . Indicate the coordinates of any points where the graph meets the axes.
[2]

14. The function $f$ is defined by $f(x) = \frac{x^2-4}{x-2}$ for $x \neq 2$ .

(a) Simplify $f(x)$ .
[1]

(b) Sketch the graph of $y = f(x)$ , indicating any holes or asymptotes.
[2]

15. Given that $f(x) = x^2 - 4x + 5$ for $x \in \mathbb{R}$ .

(a) Express $f(x)$ in the form $(x-a)^2 + b$ .
[1]

(b) Hence, sketch the graph of $y = |f(x)|$ .
[2]

Section D: Applications and Synthesis (Questions 16–20)

[10 Marks]

16. A function $f$ is defined by $f(x) = \sqrt{4-x^2}$ for $-2 \le x \le 2$ .

(a) State the range of $f$ .
[1]

(b) Explain why $f$ does not have an inverse function over its entire domain.
[1]

17. The variables $x$ and $y$ are related by the equation $y = \frac{A}{x} + B$ , where $A$ and $B$ are constants.

(a) State what graph should be plotted to obtain a straight line.
[1]

(b) If the straight line graph of $Y$ against $X$ has a gradient of $-2$ and a $Y$ -intercept of $5$ , find the values of $A$ and $B$ .
[2]

18. Solve the inequality $|x^2 - 5| < 4$ .
[2]

19. The function $f$ is defined by $f(x) = \frac{1}{x}$ for $x \neq 0$ .
Find the set of values of $x$ such that $f(f(x)) = x$ .
[1]

20. Given $f(x) = 2x + 1$ and $g(x) = x^2$ .
Find the value of $x$ for which $fg(x) = gf(x)$ .
[2]

*** End of Quiz ***

Answers

A-Level Maths H2 Quiz - Algebra Functions (Answer Key)

1.
$f(x) = 2 \implies \frac{2x-1}{x+3} = 2$
$2x - 1 = 2(x+3)$
$2x - 1 = 2x + 6$
$-1 = 6$ (Contradiction)
Alternatively, find $f^{-1}(x)$ :
$y = \frac{2x-1}{x+3} \implies y(x+3) = 2x-1 \implies xy+3y = 2x-1 \implies x(y-2) = -1-3y \implies x = \frac{3y+1}{2-y}$ .
$f^{-1}(2)$ involves division by zero.
Answer: Undefined / Does not exist.
(Note: If the question implies finding $x$ such that $f(x)=2$ , there is no solution. If asking for $f^{-1}(2)$ , it is undefined as 2 is the horizontal asymptote.)
[1]

2.
$fg(x) = f(g(x)) = f(\sqrt{x-1}) = (\sqrt{x-1})^2 + 1 = x - 1 + 1 = x$ .
Answer: $x$
[1]

3.
$|2x - 4| \le 6$
$-6 \le 2x - 4 \le 6$
$-2 \le 2x \le 10$
$-1 \le x \le 5$
Answer: $-1 \le x \le 5$
[2]

4.
$e^{2x} > 0$ for all real $x$ .
$e^{2x} - 3 > -3$ .
Answer: $y > -3$ or $(-3, \infty)$
[1]

5.
Critical values: $x=1, x=-2$ .
Test intervals:
$x > 1$ : $(+) / (+) > 0$ (True)
$-2 < x < 1$ : $(-) / (+) < 0$ (False)
$x < -2$ : $(-) / (-) > 0$ (True)
Answer: $x < -2$ or $x > 1$
[2]

6.
(a) Range of $g(x) = x^2+1$ is $[1, \infty)$ . Domain of $f(x)$ is $\mathbb{R} \setminus \{2\}$ .
Since $2 \in [1, \infty)$ , the range of $g$ is not a subset of the domain of $f$ . Specifically, $g(x)=2$ when $x=1$ , and $f(2)$ is undefined.
Answer: Range of $g$ is not contained in Domain of $f$ (or $g(1)=2$ which is not in domain of $f$ ).
[1]

(b) We must exclude $x$ such that $g(x) = 2$ .
$x^2 + 1 = 2 \implies x^2 = 1 \implies x = \pm 1$ .
Largest domain is $\mathbb{R} \setminus \{-1, 1\}$ .
Answer: $\{x \in \mathbb{R} : x \neq 1, x \neq -1\}$
[2]

7.
(a) $y = \frac{3x+1}{x-2} \implies y(x-2) = 3x+1 \implies xy - 2y = 3x + 1 \implies xy - 3x = 2y + 1 \implies x(y-3) = 2y+1 \implies x = \frac{2y+1}{y-3}$ .
$f^{-1}(x) = \frac{2x+1}{x-3}$ .
[2]

(b) Domain of $f^{-1}$ is Range of $f$ .
As $x > 2$ , $f(x)$ has vertical asymptote $x=2$ and horizontal asymptote $y=3$ .
Check monotonicity: $f'(x) = \frac{3(x-2) - (3x+1)(1)}{(x-2)^2} = \frac{-7}{(x-2)^2} < 0$ . Decreasing.
Limit $x \to 2^+ \implies f(x) \to \infty$ .
Limit $x \to \infty \implies f(x) \to 3$ .
Range is $(3, \infty)$ .
Answer: $x > 3$
[1]

8.
(a) Vertex at $(1, 3)$ . $y$ -intercept: $g(0)$ is undefined in domain $x \ge 1$ , but if extended, $(0-1)^2+3=4$ . However, domain is $x \ge 1$ , so starting point is $(1,3)$ . Graph is half-parabola opening upwards.
[2]

(b) $y = (x-1)^2 + 3 \implies y-3 = (x-1)^2 \implies x-1 = \sqrt{y-3}$ (since $x \ge 1$ ).
$x = 1 + \sqrt{y-3}$ .
$g^{-1}(x) = 1 + \sqrt{x-3}$ .
Domain: $x \ge 3$ .
[2]

9.
(a) $gf(x) = g(\ln(x+1)) = e^{\ln(x+1)} - 1 = (x+1) - 1 = x$ .
[1]

(b) $f(g(x)) = 2$ . Since $f$ and $g$ are inverses (shown in a), $f(g(x)) = x$ .
Therefore, $x = 2$ .
Check validity: $g(2) = e^2-1 > -1$ , so valid for $f$ .
Answer: $x = 2$
[2]

10.
$y = \frac{ax+b}{cx+d} \implies y(cx+d) = ax+b \implies cxy + dy = ax+b \implies x(cy-a) = b-dy \implies x = \frac{-dy+b}{cy-a}$ .
$f^{-1}(x) = \frac{-dx+b}{cx-a}$ .
Given $f^{-1}(x) = f(x) = \frac{ax+b}{cx+d}$ .
Comparing coefficients: $\frac{-d}{c} = \frac{a}{c} \implies -d = a \implies a+d=0$ .
[2]

11.
(a) Reflect negative parts of $f(x)$ in x-axis. Points A and C remain on axis. B(0,3) remains. Asymptote $x=1$ remains. Curve stays above x-axis.
[2]

(b) $y=f(|x|)$ is even symmetry. Keep right side ( $x \ge 0$ ) and reflect it to left side.
Right side has asymptote $x=1$ . Left side will have asymptote $x=-1$ .
Passes through $(2,0)$ and $(-2,0)$ . Y-intercept $(0,3)$ .
[2]

12.

Translation by vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (1 unit right).
Stretch parallel to y-axis, scale factor 2.
Translation by vector $\begin{pmatrix} 0 \\ 3 \end{pmatrix}$ (3 units up).
(Order of stretch and translations can vary if specified correctly, e.g., stretch then translate).
[2]

13.
V-shape graph. Vertex at $(0.5, 0)$ .
y-intercept: $|2(0)-1| = 1 \implies (0,1)$ .
x-intercept: $(0.5, 0)$ .
[2]

14.
(a) $f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$ for $x \neq 2$ .
[1]

(b) Straight line $y=x+2$ with a "hole" (open circle) at $x=2$ .
Coordinate of hole: $(2, 4)$ .
[2]

15.
(a) $x^2 - 4x + 5 = (x-2)^2 - 4 + 5 = (x-2)^2 + 1$ .
[1]

(b) Since $(x-2)^2 + 1 \ge 1$ , the function is always positive.
$|f(x)| = f(x)$ . Graph is standard parabola vertex $(2,1)$ .
[2]

16.
(a) Max value $\sqrt{4-0}=2$ . Min value $\sqrt{4-4}=0$ .
Answer: $0 \le y \le 2$
[1]

(b) The function is not one-to-one (fails horizontal line test, e.g., $f(1)=f(-1)$ ).
[1]

17.
(a) Plot $y$ against $\frac{1}{x}$ .
[1]

(b) Equation: $y = A(\frac{1}{x}) + B$ .
Gradient $A = -2$ .
Intercept $B = 5$ .
Answer: $A = -2, B = 5$
[2]

18.
$|x^2 - 5| < 4 \implies -4 < x^2 - 5 < 4$ .
Add 5: $1 < x^2 < 9$ .
$x^2 > 1 \implies x > 1$ or $x < -1$ .
$x^2 < 9 \implies -3 < x < 3$ .
Intersection: $(-3, -1) \cup (1, 3)$ .
Answer: $-3 < x < -1$ or $1 < x < 3$
[2]

19.
$f(f(x)) = f(1/x) = \frac{1}{1/x} = x$ .
This holds for all $x$ in the domain of $f(f(x))$ .
Domain of $f$ : $x \neq 0$ .
Domain of $f(f(x))$ : $x \neq 0$ and $f(x) \neq 0$ .
$f(x) = 1/x$ is never 0.
So valid for all $x \neq 0$ .
Answer: $\{x \in \mathbb{R} : x \neq 0\}$
[1]

20.
$fg(x) = f(x^2) = 2x^2 + 1$ .
$gf(x) = g(2x+1) = (2x+1)^2 = 4x^2 + 4x + 1$ .
$2x^2 + 1 = 4x^2 + 4x + 1$ .
$2x^2 + 4x = 0$ .
$2x(x+2) = 0$ .
$x = 0$ or $x = -2$ .
Answer: $x = 0, -2$
[2]