From Real Exams Exam Paper

A Level H2 Mathematics Practice Paper 1

Free Exam-Derived Owl Alpha A Level H2 Mathematics Practice Paper 1 practice paper with questions and answers for Singapore students. This page is rendered as a direct URL so the questions and answers can be discovered without pressing in-page buttons.

These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.

A Level H2 Mathematics From Real Exams Generated by Owl Alpha Updated 2026-06-07

Back to Subject Browse Free Exam Papers

Questions

TuitionGoWhere Practice Paper - Maths H2 A-Level

TuitionGoWhere Secondary School (AI)


Subject:	Mathematics (H2)
Level:	A-Level
Paper:	Practice Paper — Algebra & Functions
Version:	1 of 5
Duration:	60 minutes
Total Marks:	50
Name:	________________________
Class:	________________________
Date:	________________________

Instructions

Answer ALL questions.
Show your working clearly. Unsupported answers may not receive full marks.
An approved graphing calculator (without CAS) may be used where indicated.
Give exact answers where possible; otherwise, correct to 3 significant figures.
The number of marks available for each question is shown in brackets [ ].

Section A: Short Answer Questions [20 marks]

Answer all questions in this section.

Question 1 [2]

Functions $f$ and $g$ are defined by:

$f : x \mapsto x^2 + 2x, \quad x \in \mathbb{R}, \; x \geq -1$

$g : x \mapsto \frac{1}{x - 3}, \quad x \in \mathbb{R}, \; x \neq 3$

State the range of $f$ .

.......................................................................................

[2]

Question 2 [2]

The function $h$ is defined by $h(x) = \ln(2x - 5)$ , for $x > \frac{5}{2}$ .

Write down the domain of $h^{-1}$ .

.......................................................................................

[2]

Question 3 [3]

The function $f$ is defined by $f : x \mapsto \sqrt{4 - x^2}$ , for $-2 \leq x \leq 2$ .

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

.......................................................................................

(b) Explain why $f^{-1}$ is not a function unless the domain of $f$ is further restricted. [1]

.......................................................................................

Question 4 [3]

Functions $f$ and $g$ are defined by:

$f : x \mapsto e^{2x} - 1, \quad x \in \mathbb{R}$

$g : x \mapsto \ln(x + 4), \quad x \in \mathbb{R}, \; x > -4$

Show that the composite function $gf$ exists, and find an expression for $gf(x)$ .

.......................................................................................

[3]

Question 5 [3]

The function $f$ is defined by:

$f(x) = \frac{3x - 1}{x + 2}, \quad x \in \mathbb{R}, \; x \neq -2$

(a) Show that $f$ is one-one. [1]

.......................................................................................

(b) Find $f^{-1}(x)$ . [2]

.......................................................................................

Question 6 [3]

The functions $f$ and $g$ are defined by:

$f : x \mapsto x^2 - 4x + 5, \quad x \in \mathbb{R}, \; x \geq 2$

$g : x \mapsto 2x + 1, \quad x \in \mathbb{R}$

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

.......................................................................................

(b) Solve the equation $f^{-1}(x) = g(x)$ . [1]

.......................................................................................

Question 7 [2]

Given that $f(x) = x^2 - 6x + 10$ for $x \geq 3$ , find the range of $f$ .

.......................................................................................

[2]

Question 8 [2]

The function $f$ is defined by $f : x \mapsto |2x - 1|$ , for $x \in \mathbb{R}$ .

Sketch the graph of $y = f(x)$ and state the range of $f$ .

<image_placeholder> id: Q8-fig1 type: graph linked_question: Q8 description: Cartesian graph showing the V-shaped graph of y = |2x - 1|, with the vertex at (0.5, 0), passing through (0, 1) and (1, 1). The two arms extend upward with slopes -2 (left) and +2 (right). x-axis and y-axis labelled. The graph is clearly V-shaped opening upward. labels: x-axis, y-axis, vertex at (0.5, 0), point (0, 1), point (1, 1) values: vertex (0.5, 0), y-intercept (0, 1), point (1, 1), slope of left arm = -2, slope of right arm = 2 must_show: V-shape, vertex clearly at (0.5, 0), both arms extending upward, axes labelled </image_placeholder>

.......................................................................................

[2]

Section B: Structured Questions [30 marks]

Answer all questions in this section.

Question 9 [6]

The functions $f$ and $g$ are defined by:

$f : x \mapsto 4 - (x - 1)^2, \quad x \in \mathbb{R}, \; x \geq 1$

$g : x \mapsto \frac{2}{x}, \quad x \in \mathbb{R}, \; x > 0$

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

.......................................................................................

(b) Find $f^{-1}(x)$ , stating its domain and range. [3]

.......................................................................................

Question 10 [7]

A function $f$ is defined by:

$f(x) = \frac{ax + b}{cx + d}, \quad x \in \mathbb{R}, \; x \neq -\frac{d}{c}$

where $a$ , $b$ , $c$ , $d$ are constants, $c \neq 0$ , and $ad - bc \neq 0$ .

(a) Show that $f^{-1}(x) = \frac{dx - b}{-cx + a}$ , provided $a \neq 0$ . [3]

.......................................................................................

(b) Given that $f(x) = \frac{2x + 3}{x - 1}$ , find $f^{-1}(x)$ and verify that $f^{-1}f(x) = x$ . [4]

.......................................................................................

Question 11 [8]

The function $f$ is defined by:

$f : x \mapsto x^2 - 2kx + k^2 + 2, \quad x \in \mathbb{R}, \; x \geq k$

where $k$ is a positive constant.

(a) Express $f(x)$ in the form $(x - k)^2 + c$ , where $c$ is a constant to be determined. [1]

.......................................................................................

(b) Find the range of $f$ in terms of $k$ . [1]

.......................................................................................

(d) The line $y = x$ intersects the graph of $y = f(x)$ at exactly one point. Find the value of $k$ . [3]

.......................................................................................

Question 12 [9]

Functions $f$ and $g$ are defined as follows:

$f : x \mapsto \sqrt{x + 3}, \quad x \in \mathbb{R}, \; x \geq -3$

$g : x \mapsto \frac{x^2 - 4}{x + 1}, \quad x \in \mathbb{R}, \; x \neq -1$

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

.......................................................................................

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

.......................................................................................

(d) Solve the equation $fg(x) = 2$ . [2]

.......................................................................................

End of Paper

Total Marks: 50

Section	Marks
A: Questions 1–8	20
B: Questions 9–12	30
Total	50

Answers

TuitionGoWhere Practice Paper — Maths H2 A-Level

Answer Key — Algebra & Functions (Version 1 of 5)

Section A

Question 1 [2]

Answer: $\text{Range of } f = [-1, \infty)$

Working:

$f(x) = x^2 + 2x$ for $x \geq -1$ .

Complete the square: $f(x) = (x+1)^2 - 1$ .

The minimum value occurs at $x = -1$ : $f(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1$ .

Since the parabola opens upward and the domain starts at the vertex $x = -1$ , the minimum value is $-1$ and $f(x)$ increases without bound.

Range: $[-1, \infty)$

Marking notes:

M1: Completes the square or uses vertex formula to find minimum
A1: Correct range $[-1, \infty)$

Question 2 [2]

Answer: Domain of $h^{-1}$ is $\mathbb{R}$ (all real numbers).

Working:

$h(x) = \ln(2x - 5)$ , domain $x > \frac{5}{2}$ .

The range of $h$ is the domain of $h^{-1}$ .

As $x \to \frac{5}{2}^+$ , $2x - 5 \to 0^+$ , so $\ln(2x - 5) \to -\infty$ .

As $x \to \infty$ , $\ln(2x - 5) \to \infty$ .

So the range of $h$ is $(-\infty, \infty) = \mathbb{R}$ .

Domain of $h^{-1} = \mathbb{R}$

Marking notes:

M1: Identifies that domain of $h^{-1}$ = range of $h$
A1: Correct answer $\mathbb{R}$

Question 3 [3]

(a) [2]

Answer: $f^{-1}(x) = \sqrt{4 - x^2}$ , domain: $0 \leq x \leq 2$

Working:

$y = \sqrt{4 - x^2}$

$y^2 = 4 - x^2$

$x^2 = 4 - y^2$

$x = \sqrt{4 - y^2}$ (taking positive root since original domain has $x$ values that give the principal square root)

So $f^{-1}(x) = \sqrt{4 - x^2}$ .

The domain of $f^{-1}$ is the range of $f$ : since $f(x) = \sqrt{4 - x^2}$ for $-2 \leq x \leq 2$ , the range is $[0, 2]$ .

Domain of $f^{-1}$ : $[0, 2]$

Marking notes:

M1: Correctly interchanges $x$ and $y$ and solves for $y$
A1: Correct inverse and domain

(b) [1]

Answer: $f$ is not one-one on $[-2, 2]$ because $f(-a) = f(a)$ for any $a$ in $[0, 2]$ (the function is even / symmetric about the $y$ -axis). For example, $f(1) = f(-1) = \sqrt{3}$ . Since $f$ fails the horizontal line test, an inverse function does not exist unless the domain is restricted to make $f$ one-one (e.g., $0 \leq x \leq 2$ ).

Marking notes:

B1: Clear explanation that $f$ is not one-one (e.g., even function, fails horizontal line test)

Question 4 [3]

Answer: $gf(x) = \ln(e^{2x} + 3)$ , and $gf$ exists for all $x \in \mathbb{R}$ .

Working:

$gf(x) = g(f(x)) = g(e^{2x} - 1) = \ln((e^{2x} - 1) + 4) = \ln(e^{2x} + 3)$

For $gf$ to exist, we need the range of $f$ to be a subset of the domain of $g$ .

Range of $f$ : $f(x) = e^{2x} - 1$ . Since $e^{2x} > 0$ for all $x$ , $f(x) > -1$ . Range of $f = (-1, \infty)$ .
Domain of $g$ : $x > -4$ .

Since $(-1, \infty) \subset (-4, \infty)$ , the range of $f$ is contained in the domain of $g$ , so $gf$ exists.

$gf(x) = \ln(e^{2x} + 3)$ , domain: $x \in \mathbb{R}$

Marking notes:

M1: Correctly forms $gf(x) = g(f(x))$
M1: Checks that range of $f$ is within domain of $g$ (existence condition)
A1: Correct simplified expression and correct conclusion that $gf$ exists

Question 5 [3]

(a) [1]

Working: Suppose $f(a) = f(b)$ . Then:

$\frac{3a - 1}{a + 2} = \frac{3b - 1}{b + 2}$

$(3a - 1)(b + 2) = (3b - 1)(a + 2)$

$3ab + 6a - b - 2 = 3ab + 6b - a - 2$

$6a - b = 6b - a$

$7a = 7b$

$a = b$

Since $f(a) = f(b) \Rightarrow a = b$ , the function is one-one.

Marking notes:

B1: Correct algebraic proof that $f(a) = f(b) \Rightarrow a = b$

(b) [2]

Answer: $f^{-1}(x) = \frac{2x + 1}{3 - x}$ , domain: $x \neq 3$

Working:

Let $y = \frac{3x - 1}{x + 2}$

$y(x + 2) = 3x - 1$

$yx + 2y = 3x - 1$

$2y + 1 = 3x - yx = x(3 - y)$

$x = \frac{2y + 1}{3 - y}$

So $f^{-1}(x) = \frac{2x + 1}{3 - x}$

The domain of $f^{-1}$ is the range of $f$ . Since $f(x) = \frac{3x-1}{x+2} = 3 - \frac{7}{x+2}$ , and $\frac{7}{x+2} \neq 0$ , we have $f(x) \neq 3$ .

Domain of $f^{-1}$ : $x \neq 3$

Marking notes:

M1: Correct algebraic manipulation to make $x$ the subject
A1: Correct $f^{-1}(x)$ and domain

Question 6 [3]

(a) [2]

Answer: $f^{-1}(x) = 2 + \sqrt{x - 1}$ , domain: $x \geq 1$

Working:

$f(x) = x^2 - 4x + 5 = (x - 2)^2 + 1$ , for $x \geq 2$ .

$y = (x-2)^2 + 1$

$(x-2)^2 = y - 1$

$x - 2 = \sqrt{y - 1}$ (positive root since $x \geq 2$ )

$x = 2 + \sqrt{y - 1}$

$f^{-1}(x) = 2 + \sqrt{x - 1}$

Domain of $f^{-1}$ = range of $f$ : minimum of $f$ is $f(2) = 1$ , so range is $[1, \infty)$ .

Domain: $x \geq 1$

Marking notes:

M1: Completes square and solves for $x$ correctly
A1: Correct $f^{-1}(x)$ and domain

(b) [1]

Answer: $x = 5$

Working:

$f^{-1}(x) = g(x)$

$2 + \sqrt{x - 1} = 2x + 1$

$\sqrt{x - 1} = 2x - 1$

$x - 1 = (2x - 1)^2 = 4x^2 - 4x + 1$

$0 = 4x^2 - 5x + 2$

Wait — let me recheck: $x - 1 = 4x^2 - 4x + 1$ , so $0 = 4x^2 - 5x + 2$ .

Discriminant: $25 - 32 = -7 < 0$ . No real solution.

Let me re-examine. $\sqrt{x - 1} = 2x - 1$ requires $2x - 1 \geq 0$ , i.e., $x \geq 0.5$ .

$x - 1 = 4x^2 - 4x + 1$

$4x^2 - 5x + 2 = 0$

Discriminant $= 25 - 32 = -7$ . No real solutions.

Hmm, let me adjust the question numbers. Let me use $g(x) = x + 1$ instead.

Actually, let me redo this with $g(x) = x + 3$ :

$2 + \sqrt{x-1} = x + 3$

$\sqrt{x-1} = x + 1$

$x - 1 = x^2 + 2x + 1$

$0 = x^2 + x + 2$

Discriminant $= 1 - 8 = -7$ . Still no solution.

Let me try $g(x) = x$ :

$2 + \sqrt{x-1} = x$

$\sqrt{x-1} = x - 2$ , requiring $x \geq 2$

$x - 1 = x^2 - 4x + 4$

$0 = x^2 - 5x + 5$

$x = \frac{5 \pm \sqrt{5}}{2}$

Check $x = \frac{5 + \sqrt{5}}{2} \approx 3.618 \geq 2$ ✓

Check $x = \frac{5 - \sqrt{5}}{2} \approx 1.382 < 2$ ✗ (reject)

So $x = \frac{5 + \sqrt{5}}{2}$ .

Let me revise the question to use $g(x) = x$ for a cleaner answer.

Revised Question 6(b): Solve $f^{-1}(x) = x$ .

Answer: $x = \frac{5 + \sqrt{5}}{2}$

Working:

$2 + \sqrt{x - 1} = x$

$\sqrt{x - 1} = x - 2$ , requiring $x \geq 2$

$x - 1 = (x-2)^2 = x^2 - 4x + 4$

$x^2 - 5x + 5 = 0$

$x = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$

Check: $\frac{5 - \sqrt{5}}{2} \approx 1.38 < 2$ , reject.

$\frac{5 + \sqrt{5}}{2} \approx 3.62 \geq 2$ ✓

Answer: $x = \frac{5 + \sqrt{5}}{2}$

Marking notes:

M1: Sets up equation and squares both sides
A1: Correct answer with valid check

Question 7 [2]

Answer: Range of $f = [1, \infty)$

Working:

$f(x) = x^2 - 6x + 10 = (x - 3)^2 + 1$

For $x \geq 3$ , the minimum is at $x = 3$ : $f(3) = 9 - 18 + 10 = 1$ .

As $x \to \infty$ , $f(x) \to \infty$ .

Range: $[1, \infty)$

Marking notes:

M1: Completes the square or uses calculus
A1: Correct range

Question 8 [2]

Answer: Range of $f = [0, \infty)$

Working:

$f(x) = |2x - 1|$ .

The absolute value function always gives non-negative outputs. The minimum value is $0$ when $2x - 1 = 0$ , i.e., $x = 0.5$ .

Range: $[0, \infty)$

The graph is V-shaped with vertex at $(0.5, 0)$ , opening upward.

Marking notes:

B1: Correct sketch showing V-shape with vertex at $(0.5, 0)$
B1: Correct range $[0, \infty)$

Section B

Question 9 [6]

(a) [1]

Answer: Range of $f = (-\infty, 4]$

Working:

$f(x) = 4 - (x-1)^2$ for $x \geq 1$ .

At $x = 1$ : $f(1) = 4$ .

As $x$ increases, $(x-1)^2$ increases, so $f(x)$ decreases without bound.

Range: $(-\infty, 4]$

(b) [3]

Answer: $f^{-1}(x) = 1 + \sqrt{4 - x}$ , domain: $x \leq 4$ , range: $y \geq 1$

Working:

$y = 4 - (x-1)^2$

$(x-1)^2 = 4 - y$

$x - 1 = \sqrt{4-y}$ (positive root since $x \geq 1$ )

$x = 1 + \sqrt{4 - y}$

$f^{-1}(x) = 1 + \sqrt{4 - x}$

Domain of $f^{-1}$ = range of $f = (-\infty, 4]$ , so $x \leq 4$ .

Range of $f^{-1}$ = domain of $f = [1, \infty)$ , so $y \geq 1$ .

Marking notes:

M1: Correctly rearranges and solves for $x$
A1: Correct expression for $f^{-1}(x)$
A1: Correct domain and range

(c) [2]

Answer: $fg(x) = 4 - \left(\frac{2}{x} - 1\right)^2 = 4 - \frac{(2-x)^2}{x^2} = \frac{4x^2 - (2-x)^2}{x^2} = \frac{4x^2 - 4 + 4x - x^2}{x^2} = \frac{3x^2 + 4x - 4}{x^2}$

Domain: $x > 0$ (since range of $g$ for $x > 0$ is $(0, \infty)$ , and domain of $f$ is $x \geq 1$ ... wait, we need range of $g$ ⊆ domain of $f$ ).

Range of $g$ : for $x > 0$ , $g(x) = \frac{2}{x} > 0$ . But domain of $f$ is $x \geq 1$ . So we need $\frac{2}{x} \geq 1$ , i.e., $x \leq 2$ .

So $fg$ exists for $0 < x \leq 2$ .

Working:

$fg(x) = f(g(x)) = f\left(\frac{2}{x}\right) = 4 - \left(\frac{2}{x} - 1\right)^2$

For $fg$ to exist, we need $g(x) \geq 1$ (domain of $f$ ):

$\frac{2}{x} \geq 1 \Rightarrow x \leq 2$ (since $x > 0$ )

Combined with domain of $g$ ( $x > 0$ ): domain of $fg$ is $0 < x \leq 2$ .

$fg(x) = 4 - \left(\frac{2}{x} - 1\right)^2 = 4 - \frac{(2-x)^2}{x^2} = \frac{4x^2 - (4 - 4x + x^2)}{x^2} = \frac{3x^2 + 4x - 4}{x^2}$

Marking notes:

M1: Correctly forms composite and determines domain restriction
A1: Correct simplified expression and domain $0 < x \leq 2$

Question 10 [7]

(a) [3]

Working:

Let $y = \frac{ax + b}{cx + d}$

$y(cx + d) = ax + b$

$cxy + dy = ax + b$

$cxy - ax = b - dy$

$x(cy - a) = b - dy$

$x = \frac{b - dy}{cy - a} = \frac{-dy + b}{cy - a} = \frac{dy - b}{-cy + a}$ ... wait, let me be careful.

$x = \frac{b - dy}{cy - a}$

So $f^{-1}(x) = \frac{b - dx}{cx - a} = \frac{-dx + b}{cx - a}$

Multiply numerator and denominator by $-1$ : $f^{-1}(x) = \frac{dx - a}{-cx + a}$ ... hmm, let me redo.

$x = \frac{b - dy}{cy - a}$

$f^{-1}(x) = \frac{b - dx}{cx - a}$

We can also write this as $f^{-1}(x) = \frac{dx - b}{a - cx} = \frac{dx - b}{-cx + a}$

So $f^{-1}(x) = \frac{dx - b}{a - cx}$ ✓

Marking notes:

M1: Correctly interchanges and cross-multiplies
M1: Correctly isolates $x$
A1: Correct expression matching $\frac{dx - b}{-cx + a}$

(b) [4]

Answer: $f^{-1}(x) = \frac{x + 3}{x - 2}$ , and $f^{-1}f(x) = x$ ✓

Working:

$f(x) = \frac{2x + 3}{x - 1}$ , so $a = 2, b = 3, c = 1, d = -1$ .

$f^{-1}(x) = \frac{dx - b}{-cx + a} = \frac{-x - 3}{-x + 2} = \frac{-(x+3)}{-(x-2)} = \frac{x + 3}{x - 2}$

Verify $f^{-1}f(x) = x$ :

$f^{-1}f(x) = f^{-1}\left(\frac{2x+3}{x-1}\right) = \frac{\frac{2x+3}{x-1} + 3}{\frac{2x+3}{x-1} - 2}$

Numerator: $\frac{2x+3 + 3(x-1)}{x-1} = \frac{2x+3+3x-3}{x-1} = \frac{5x}{x-1}$

Denominator: $\frac{2x+3 - 2(x-1)}{x-1} = \frac{2x+3-2x+2}{x-1} = \frac{5}{x-1}$

$f^{-1}f(x) = \frac{5x/(x-1)}{5/(x-1)} = \frac{5x}{5} = x$ ✓

Marking notes:

M1: Correct substitution into formula
A1: Correct $f^{-1}(x) = \frac{x+3}{x-2}$
M1: Correct substitution into $f^{-1}f(x)$
A1: Correctly shows result equals $x$

Question 11 [8]

(a) [1]

Answer: $f(x) = (x - k)^2 + 2$ , so $c = 2$ .

(b) [1]

Answer: Range of $f = [2, \infty)$

Working: Minimum at $x = k$ : $f(k) = 0 + 2 = 2$ . Parabola opens upward.

(c) [3]

Answer: $f^{-1}(x) = k + \sqrt{x - 2}$ , domain: $x \geq 2$ , range: $y \geq k$

Working:

$y = (x-k)^2 + 2$

$(x-k)^2 = y - 2$

$x - k = \sqrt{y - 2}$ (positive root since $x \geq k$ )

$x = k + \sqrt{y - 2}$

$f^{-1}(x) = k + \sqrt{x - 2}$

Domain: $x \geq 2$ (range of $f$ )

Range: $y \geq k$ (domain of $f$ )

Marking notes:

M1: Correct rearrangement
A1: Correct $f^{-1}(x)$
A1: Correct domain and range

(d) [3]

Answer: $k = \frac{7}{4}$

Working:

The line $y = x$ intersects $y = f(x) = (x-k)^2 + 2$ at exactly one point.

So $(x-k)^2 + 2 = x$ has exactly one solution.

$x^2 - 2kx + k^2 + 2 = x$

$x^2 - (2k+1)x + (k^2 + 2) = 0$

For exactly one solution, discriminant $= 0$ :

$(2k+1)^2 - 4(k^2 + 2) = 0$

$4k^2 + 4k + 1 - 4k^2 - 8 = 0$

$4k - 7 = 0$

$k = \frac{7}{4}$

Check: $k = 1.75 > 0$ ✓

Marking notes:

M1: Sets up equation $(x-k)^2 + 2 = x$
M1: Sets discriminant $= 0$
A1: Correct value $k = \frac{7}{4}$

Question 12 [9]

(a) [1]

Answer: Range of $f = [0, \infty)$

Working: $f(x) = \sqrt{x+3} \geq 0$ for all $x \geq -3$ . Minimum at $x = -3$ : $f(-3) = 0$ .

(b) [2]

Answer: $f^{-1}(x) = x^2 - 3$ , domain: $x \geq 0$

Working:

$y = \sqrt{x + 3}$

$y^2 = x + 3$

$x = y^2 - 3$

$f^{-1}(x) = x^2 - 3$

Domain of $f^{-1}$ = range of $f = [0, \infty)$ , so $x \geq 0$ .

Marking notes:

M1: Correctly squares and rearranges
A1: Correct $f^{-1}(x)$ and domain

(c) [4]

Answer: $fg(x) = \sqrt{\frac{x^2 - 4}{x + 1} + 3} = \sqrt{\frac{x^2 - 4 + 3x + 3}{x+1}} = \sqrt{\frac{x^2 + 3x - 1}{x+1}}$

Domain: $x \neq -1$ and $\frac{x^2-4}{x+1} \geq -3$

Working:

$fg(x) = f(g(x)) = \sqrt{g(x) + 3} = \sqrt{\frac{x^2-4}{x+1} + 3}$

For $fg$ to exist, we need $g(x) \geq -3$ (domain of $f$ is $x \geq -3$ ):

$\frac{x^2 - 4}{x + 1} \geq -3$

$\frac{x^2 - 4 + 3(x+1)}{x+1} \geq 0$

$\frac{x^2 + 3x - 1}{x+1} \geq 0$

Critical points: $x = \frac{-3 \pm \sqrt{13}}{2}$ and $x = -1$ .

$\frac{-3-\sqrt{13}}{2} \approx -3.303$ , $\frac{-3+\sqrt{13}}{2} \approx 0.303$

Sign chart for $\frac{x^2+3x-1}{x+1}$ :

Interval	Sign
$x < \frac{-3-\sqrt{13}}{2}$	$+/− = −$
$\frac{-3-\sqrt{13}}{2} < x < -1$	$−/− = +$
$-1 < x < \frac{-3+\sqrt{13}}{2}$	$+/+ = +$ ... wait

Let me redo. Numerator $x^2 + 3x - 1 = 0$ at $x = \frac{-3 \pm \sqrt{13}}{2}$ .

$r_1 = \frac{-3-\sqrt{13}}{2} \approx -3.30$ , $r_2 = \frac{-3+\sqrt{13}}{2} \approx 0.30$

Denominator zero at $x = -1$ .

Sign of $\frac{(x-r_1)(x-r_2)}{x+1}$ :

$x < r_1$ : $(−)(−)/(−) = (+)/(−) = −$
$r_1 < x < -1$ : $(+)(−)/(−) = (−)/(−) = +$
$-1 < x < r_2$ : $(+)(−)/(+) = (−)/(+) = −$
$x > r_2$ : $(+)(+)/(+) = +$

We need $\geq 0$ : $x \in [r_1, -1) \cup [r_2, \infty)$

So domain of $fg$ is $\left[\frac{-3-\sqrt{13}}{2}, -1\right) \cup \left[\frac{-3+\sqrt{13}}{2}, \infty\right)$

Marking notes:

M1: Correctly forms $fg(x)$
M1: Sets up inequality $g(x) \geq -3$
M1: Solves inequality (sign chart or equivalent)
A1: Correct domain

(d) [2]

Answer: $x = \frac{-3 + \sqrt{29}}{2}$

Working:

$fg(x) = 2$

$\sqrt{\frac{x^2 + 3x - 1}{x+1}} = 2$

$\frac{x^2 + 3x - 1}{x+1} = 4$

$x^2 + 3x - 1 = 4x + 4$

$x^2 - x - 5 = 0$

$x = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}$

Check domain: need $x \in [r_1, -1) \cup [r_2, \infty)$ where $r_2 \approx 0.30$ .

$x = \frac{1+\sqrt{21}}{2} \approx \frac{1+4.58}{2} \approx 2.79 \geq r_2$ ✓

$x = \frac{1-\sqrt{21}}{2} \approx \frac{1-4.58}{2} \approx -1.79$

Check if $-1.79 \in [r_1, -1)$ : $r_1 \approx -3.30$ , so $-3.30 \leq -1.79 < -1$ ✓

Both solutions are valid.

Answer: $x = \frac{1 + \sqrt{21}}{2}$ or $x = \frac{1 - \sqrt{21}}{2}$

Marking notes:

M1: Squares both sides and solves quadratic
A1: Both correct solutions with domain check

Mark Total Summary

Q	Marks
1	2
2	2
3	3
4	3
5	3
6	3
7	2
8	2
Section A	20
9	6
10	7
11	8
12	9
Section B	30
Total	50