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A Level H2 Mathematics Algebra Functions Quiz

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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.
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.
The use of an approved graphing calculator is expected. Unsupported answers from the calculator are generally acceptable unless the question specifically requires working or proof.
Clear presentation in your working is essential.

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

Focus: Domain, Range, and Basic Function Operations

1. The function $f$ is defined by $f(x) = \sqrt{4 - x^2}$ for $-2 \le x \le 2$ .
State the range of $f$ .
[1]

Answer space

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

Answer space

3. The function $h$ is defined by $h(x) = |2x - 5|$ .
Solve the inequality $h(x) \le 7$ .
[2]

Answer space

4. The function $k$ is defined by $k(x) = e^{2x} + 1$ for $x \in \mathbb{R}$ .
Find the inverse function $k^{-1}(x)$ and state its domain.
[3]

Answer space

5. Given that $f(x) = 3x + 2$ and $g(x) = x^2 - 1$ .
Find an expression for $fg(x)$ in its simplest form.
[2]

Answer space

Section B: Composite & Inverse Functions (Questions 6–12)

Focus: Existence, Domain Restrictions, and Graphical Relationships

6. The function $f$ is defined by $f(x) = \frac{1}{x}$ for $x > 0$ .
The function $g$ is defined by $g(x) = \sqrt{x - 2}$ for $x \ge 2$ .
Explain why the composite function $fg$ does not exist.
[2]

Answer space

7. The function $p$ is defined by $p(x) = (x - 1)^2 + 3$ for $x \ge 1$ .
Find the inverse function $p^{-1}(x)$ and state its domain.
[3]

Answer space

8. The function $q$ is defined by $q(x) = \ln(x + 1)$ for $x > -1$ .
The function $r$ is defined by $r(x) = e^x - 1$ for $x \in \mathbb{R}$ .
Show that $qr(x) = x$ and state the domain of the composite function $qr$ .
[3]

Answer space

9. 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 $x$ such that $f(x) = f^{-1}(x)$ .
[3]

Answer space

10. The function $g$ is defined by $g(x) = x^2 - 4x$ for $x \in \mathbb{R}$ .
Find the largest value of $k$ such that the function $g$ restricted to the domain $x \le k$ has an inverse.
[2]

Answer space

11. The function $h$ is defined by $h(x) = \frac{ax + b}{cx + d}$ where $a,b,c,d$ are constants.
Given that $h^{-1}(x) = h(x)$ , show that $a = -d$ .
[3]

Answer space

12. The function $f$ is defined by $f(x) = \sqrt{x - 1}$ for $x \ge 1$ .
The function $g$ is defined by $g(x) = x^2 + 1$ for $x \ge 0$ .
(a) Find the expression for $gf(x)$ .
(b) State the range of $gf$ .
[3]

Answer space

Section C: Graphs, Transformations & Parametrics (Questions 13–20)

Focus: Sketching, Transformations, and Parametric Equations

13. The graph of $y = f(x)$ has a vertical asymptote at $x = 2$ and a horizontal asymptote at $y = 1$ .
On the axes below, sketch the graph of $y = f(x - 1) + 2$ , clearly indicating the new asymptotes.
[2]

(Sketch space provided in exam context)

Answer space

14. The function $f$ is defined by $f(x) = |x^2 - 4|$ .
Sketch the graph of $y = f(x)$ , stating the coordinates of any points where the graph meets the axes and the coordinates of any stationary points.
[3]

Answer space

15. The curve $C$ is defined by the parametric equations $x = t^2 + 1$ , $y = 2t$ for $t \in \mathbb{R}$ .
Find the Cartesian equation of $C$ .
[2]

Answer space

16. The curve $C$ has parametric equations $x = \cos \theta$ , $y = \sin 2\theta$ for $0 \le \theta \le 2\pi$ .
Find the gradient of the curve at the point where $\theta = \frac{\pi}{6}$ .
[3]

Answer space

17. The function $f$ is defined by $f(x) = \frac{1}{x^2 - 1}$ .
Describe fully the geometrical transformation that maps the graph of $y = f(x)$ onto the graph of $y = f(2x)$ .
[2]

Answer space

18. The function $g$ is defined by $g(x) = 2^{x} - 3$ .
The graph of $y = g(x)$ is reflected in the line $y = x$ .
Find the equation of the resulting graph.
[2]

Answer space

19. The curve $C$ is defined by $y = \frac{x^2 + 1}{x - 1}$ .
Find the equations of the asymptotes of $C$ .
[3]

Answer space

20. The function $f$ is defined by $f(x) = \ln(x^2 - 4)$ for $x > 2$ .
(a) Find the range of $f$ .
(b) Explain why $f$ does not have an inverse if the domain is restricted to $x \in \mathbb{R}, |x| > 2$ .
[3]

Answer space

End of Quiz

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Answers

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

1. Range of $f$ : $[0, 2]$ or $0 \le f(x) \le 2$ .
[1]

2. $\frac{1}{x-3} = -2 \Rightarrow 1 = -2(x-3) \Rightarrow 1 = -2x + 6 \Rightarrow 2x = 5 \Rightarrow x = 2.5$ .
[1]

3. $|2x - 5| \le 7 \Rightarrow -7 \le 2x - 5 \le 7$ .
Add 5: $-2 \le 2x \le 12$ .
Divide by 2: $-1 \le x \le 6$ .
[2]

4. Let $y = e^{2x} + 1$ . Swap $x$ and $y$ : $x = e^{2y} + 1$ .
$x - 1 = e^{2y} \Rightarrow \ln(x - 1) = 2y \Rightarrow y = \frac{1}{2}\ln(x - 1)$ .
$k^{-1}(x) = \frac{1}{2}\ln(x - 1)$ .
Domain: Argument of log must be positive, so $x - 1 > 0 \Rightarrow x > 1$ .
[3] (1 for expression, 1 for domain, 1 for correctness)

5. $fg(x) = f(g(x)) = f(x^2 - 1) = 3(x^2 - 1) + 2 = 3x^2 - 3 + 2 = 3x^2 - 1$ .
[2]

6. Range of $g$ : Since $x \ge 2$ , $\sqrt{x-2} \ge 0$ . So $R_g = [0, \infty)$ .
Domain of $f$ : $x > 0$ .
For $fg$ to exist, $R_g \subseteq D_f$ .
However, $0 \in R_g$ but $0 \notin D_f$ (since $f$ is undefined at 0).
Thus, $fg$ does not exist.
[2] (1 for identifying range/domain conflict, 1 for conclusion)

7. Let $y = (x - 1)^2 + 3$ . Swap $x$ and $y$ : $x = (y - 1)^2 + 3$ .
$x - 3 = (y - 1)^2 \Rightarrow y - 1 = \pm\sqrt{x - 3}$ .
Since original domain $x \ge 1$ , the range of inverse is $y \ge 1$ . Thus we take the positive root.
$y = 1 + \sqrt{x - 3}$ .
$p^{-1}(x) = 1 + \sqrt{x - 3}$ .
Domain of $p^{-1}$ is Range of $p$ . Min value of $p$ is 3. So Domain: $x \ge 3$ .
[3]

8. $qr(x) = q(r(x)) = q(e^x - 1) = \ln((e^x - 1) + 1) = \ln(e^x) = x$ .
Domain of $r$ is $\mathbb{R}$ . Range of $r$ is $(-1, \infty)$ .
Domain of $q$ is $(-1, \infty)$ .
Since $R_r \subseteq D_q$ , the composite exists for all $x \in \mathbb{R}$ .
Domain of $qr$ is $\mathbb{R}$ .
[3]

9. For $f(x) = f^{-1}(x)$ , the solution lies on the line $y = x$ (for increasing functions) or we solve $f(x) = x$ .
$\frac{2x + 1}{x - 3} = x \Rightarrow 2x + 1 = x(x - 3) \Rightarrow 2x + 1 = x^2 - 3x$ .
$x^2 - 5x - 1 = 0$ .
$x = \frac{5 \pm \sqrt{25 - 4(1)(-1)}}{2} = \frac{5 \pm \sqrt{29}}{2}$ .
Both values are valid as they are not 3.
[3]

10. $g(x) = x^2 - 4x = (x - 2)^2 - 4$ .
This is a parabola with vertex at $x = 2$ .
For an inverse to exist, the function must be one-to-one.
Restricting to $x \le k$ , the largest interval ending at the vertex is $x \le 2$ .
So $k = 2$ .
[2]

11. $y = \frac{ax + b}{cx + d} \Rightarrow y(cx + d) = ax + b \Rightarrow cxy + dy = ax + b$ .
$x(cy - a) = b - dy \Rightarrow x = \frac{b - dy}{cy - a} = \frac{-dy + b}{cy - a}$ .
$f^{-1}(x) = \frac{-dx + b}{cx - a}$ .
Given $f^{-1}(x) = f(x) = \frac{ax + b}{cx + d}$ .
Comparing coefficients: $\frac{-dx + b}{cx - a} = \frac{ax + b}{cx + d}$ .
This implies $-d = a$ (comparing numerator $x$ coeff) and $-a = d$ (comparing denominator constant).
Thus $a = -d$ .
[3]

12. (a) $gf(x) = g(\sqrt{x - 1}) = (\sqrt{x - 1})^2 + 1 = (x - 1) + 1 = x$ .
(b) Domain of $f$ is $x \ge 1$ . Range of $f$ is $[0, \infty)$ .
Domain of $g$ is $x \ge 0$ . Since $R_f \subseteq D_g$ , composite exists.
Range of $gf$ : Since $gf(x) = x$ and domain is $x \ge 1$ , Range is $[1, \infty)$ .
[3]

13. $y = f(x - 1) + 2$ represents a translation by vector $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ .
Old VA $x = 2 \rightarrow$ New VA $x = 3$ .
Old HA $y = 1 \rightarrow$ New HA $y = 3$ .
Sketch should show hyperbola shape in appropriate quadrants relative to new asymptotes.
[2]

14. $y = |x^2 - 4|$ .
Roots of $x^2 - 4 = 0$ are $x = \pm 2$ .
For $-2 < x < 2$ , $x^2 - 4$ is negative, so graph reflects above x-axis.
Vertex of original parabola $(0, -4)$ becomes $(0, 4)$ .
Intercepts: $(\pm 2, 0)$ and $(0, 4)$ .
Stationary points: $(0, 4)$ is a local max. $(\pm 2, 0)$ are minima (cusps).
[3]

15. $y = 2t \Rightarrow t = y/2$ .
Substitute into $x$ : $x = (y/2)^2 + 1 = \frac{y^2}{4} + 1$ .
$4(x - 1) = y^2$ or $y^2 = 4x - 4$ .
[2]

16. $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ .
$x = \cos \theta \Rightarrow \frac{dx}{d\theta} = -\sin \theta$ .
$y = \sin 2\theta \Rightarrow \frac{dy}{d\theta} = 2\cos 2\theta$ .
At $\theta = \pi/6$ :
$\frac{dx}{d\theta} = -\sin(\pi/6) = -0.5$ .
$\frac{dy}{d\theta} = 2\cos(\pi/3) = 2(0.5) = 1$ .
Gradient $= \frac{1}{-0.5} = -2$ .
[3]

17. $y = f(2x)$ is a stretch parallel to the x-axis with scale factor $\frac{1}{2}$ .
[2]

18. Reflection in $y = x$ gives the inverse function.
$y = 2^x - 3 \Rightarrow y + 3 = 2^x \Rightarrow x = \log_2(y + 3)$ .
Equation: $y = \log_2(x + 3)$ .
[2]

19. $y = \frac{x^2 + 1}{x - 1}$ .
VA: Denominator zero $\Rightarrow x = 1$ .
Oblique Asymptote: Perform division.
$\frac{x^2 + 1}{x - 1} = \frac{x(x - 1) + x + 1}{x - 1} = x + \frac{x + 1}{x - 1} = x + \frac{(x - 1) + 2}{x - 1} = x + 1 + \frac{2}{x - 1}$ .
As $x \to \infty$ , $y \to x + 1$ .
OA: $y = x + 1$ .
[3]

20. (a) As $x \to 2^+$ , $x^2 - 4 \to 0^+$ , so $\ln(x^2 - 4) \to -\infty$ .
As $x \to \infty$ , $\ln(x^2 - 4) \to \infty$ .
Range is $\mathbb{R}$ (or $(-\infty, \infty)$ ).
(b) If domain is $|x| > 2$ , it includes $x < -2$ and $x > 2$ .
$f(-3) = \ln(5)$ and $f(3) = \ln(5)$ .
Since $f(-3) = f(3)$ but $-3 \neq 3$ , the function is not one-to-one.
Therefore, it does not have an inverse.
[3]