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O Level Elementary Mathematics Algebra Functions Quiz

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O Level Elementary Mathematics From Real Exams Generated by Qwen3.6 Plus Updated 2026-06-03

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Questions

O-Level Elementary Mathematics Quiz - Algebra Functions

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 45

Duration: 50 minutes
Total Marks: 45

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless otherwise specified.
The use of an approved calculator is expected where appropriate.

Section A: Basic Concepts and Evaluation (10 Marks)

1. Given the function $f(x) = 3x^2 - 5x + 2$ , find the value of $f(-2)$ .
[1]

2. The function $g$ is defined by $g(x) = \frac{4}{x-3}$ , for $x \neq 3$ .
Find the value of $x$ for which $g(x) = -2$ .
[2]

3. Given $h(x) = 2x + 1$ and $k(x) = x^2 - 3$ , find the value of $h(k(2))$ .
[2]

4. The function $p$ is defined by $p(x) = \sqrt{x+5}$ .
State the smallest possible value of $x$ for which $p(x)$ is defined.
[1]

5. Given $f(x) = 5 - 2x$ , find an expression for $f^{-1}(x)$ in terms of $x$ .
[2]

Section B: Graphs and Properties (15 Marks)

6. The function $q$ is defined by $q(x) = x^2 + 4x - 1$ .
Express $q(x)$ in the form $(x+a)^2 + b$ , where $a$ and $b$ are constants.
[2]

7. The diagram shows the graph of $y = f(x)$ for $-3 \le x \le 3$ .
(Note: Imagine a standard parabola opening upwards with vertex at $(1, -4)$ and passing through $(3,0)$ and $(-1,0)$ ).

(a) Write down the coordinates of the minimum point of the graph.
[1]

(b) Write down the equation of the axis of symmetry.
[1]

8. A quadratic function is given by $y = -x^2 + 6x - 5$ .

(a) Find the coordinates of the turning point of the graph.
[3]

(b) Sketch the graph of $y = -x^2 + 6x - 5$ , showing clearly the coordinates of the turning point and the points where the graph crosses the axes.
[3]

9. The function $f$ is defined by $f(x) = \frac{2x+1}{x-2}$ , for $x \neq 2$ .

(a) Find the equation of the vertical asymptote.
[1]

(b) Find the equation of the horizontal asymptote.
[1]

10. The graph of $y = g(x)$ is obtained by translating the graph of $y = x^2$ by the vector $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ .

(a) Write down the expression for $g(x)$ .
[2]

(b) State the coordinates of the minimum point of $y = g(x)$ .
[1]

Section C: Composite and Inverse Functions (10 Marks)

11. Given $f(x) = 3x - 1$ and $g(x) = \frac{x}{2} + 1$ .

(a) Find an expression for $fg(x)$ in its simplest form.
[2]

(b) Solve the equation $fg(x) = 10$ .
[2]

12. The function $h$ is defined by $h(x) = \frac{2x-3}{x+1}$ , for $x \neq -1$ .

(a) Find $h^{-1}(x)$ .
[3]

(b) Hence, or otherwise, solve the equation $h(x) = h^{-1}(x)$ .
[3]

Section D: Application and Reasoning (10 Marks)

13. A rectangle has length $(2x + 3)$ cm and width $(x - 1)$ cm.
The area of the rectangle is $A$ cm $^2$ .

(a) Show that $A = 2x^2 + x - 3$ .
[1]

(b) Given that the area of the rectangle is 25 cm $^2$ , form a quadratic equation in $x$ and solve it to find the value of $x$ .
[4]

14. The height $h$ metres of a ball above the ground $t$ seconds after it is thrown is modelled by the function $h(t) = -5t^2 + 20t + 1.5$ .

(a) Calculate the initial height of the ball.
[1]

(b) Find the maximum height reached by the ball.
[2]

15. Given the function $f(x) = 2x + 3$ defined for $x \ge 0$ .

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

(b) Explain why $f(x)$ has an inverse function.
[1]

16. Consider the function $k(x) = \frac{1}{x-4}$ .

(a) State the value of $x$ for which $k(x)$ is undefined.
[1]

(b) Find the value of $k(6) - k(2)$ .
[2]

17. Let $f(x) = x^2 - 4x$ .

(a) Find the values of $x$ for which $f(x) = 5$ .
[2]

(b) Hence, write down the solutions to the equation $x^2 - 4x - 5 = 0$ .
[1]

18. The function $m(x)$ is defined as $m(x) = |2x - 6|$ .

(a) Calculate the value of $m(1)$ .
[1]

(b) Solve the equation $m(x) = 4$ .
[2]

19. A function is defined by $y = \frac{3}{x} + 2$ .

(a) Write down the equation of the horizontal asymptote.
[1]

(b) Find the x-coordinate of the point where the graph crosses the x-axis.
[2]

20. Given $f(x) = 4x - 1$ and $g(x) = x + 5$ .

(a) Find an expression for $gf(x)$ .
[2]

(b) Find the value of $x$ such that $gf(x) = f(g(x))$ .
[2]

End of Quiz

Answers

O-Level Elementary Mathematics Quiz - Algebra Functions (Answer Key)

1. $f(-2) = 3(-2)^2 - 5(-2) + 2$
$= 3(4) + 10 + 2$
$= 12 + 10 + 2$
$= 24$
[1]

2. $\frac{4}{x-3} = -2$
$4 = -2(x-3)$
$4 = -2x + 6$
$2x = 2$
$x = 1$
[2]

3. $k(2) = 2^2 - 3 = 4 - 3 = 1$
$h(1) = 2(1) + 1 = 3$
[2]

4. $x + 5 \ge 0$
$x \ge -5$
Smallest value is $-5$ .
[1]

5. Let $y = 5 - 2x$
$2x = 5 - y$
$x = \frac{5-y}{2}$
$f^{-1}(x) = \frac{5-x}{2}$
[2]

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

7.
(a) $(1, -4)$ [1]
(b) $x = 1$ [1]
(c) $-1 < x < 3$ [2] (Accept $-1 < x < 3$ )

8.
(a) $y = -(x^2 - 6x) - 5$
$= -((x-3)^2 - 9) - 5$
$= -(x-3)^2 + 9 - 5$
$= -(x-3)^2 + 4$
Turning point at $(3, 4)$ [3]

(b) Sketch:

Parabola opening downwards (n-shape).
Vertex at $(3, 4)$ .
y-intercept: Let $x=0, y=-5$ . Point $(0, -5)$ .
x-intercepts: $-x^2+6x-5=0 \Rightarrow x^2-6x+5=0 \Rightarrow (x-1)(x-5)=0$ . Points $(1,0)$ and $(5,0)$ .
Correct shape and labels. [3]

9.
(a) $x = 2$ [1]
(b) $y = 2$ (Coefficient of $x$ in numerator / coefficient of $x$ in denominator) [1]
(c) Let $x=0$ , $y = \frac{1}{-2} = -0.5$ . Point $(0, -0.5)$ [1]

10.
(a) Translation $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ means replace $x$ with $(x-3)$ and subtract 2 from the function.
$g(x) = (x-3)^2 - 2$ [2]
(b) Minimum point at $(3, -2)$ [1]

11.
(a) $fg(x) = f(\frac{x}{2} + 1) = 3(\frac{x}{2} + 1) - 1$
$= \frac{3x}{2} + 3 - 1$
$= \frac{3x}{2} + 2$ [2]

(b) $\frac{3x}{2} + 2 = 10$
$\frac{3x}{2} = 8$
$3x = 16$
$x = \frac{16}{3}$ or $5\frac{1}{3}$ [2]

12.
(a) Let $y = \frac{2x-3}{x+1}$
$y(x+1) = 2x - 3$
$xy + y = 2x - 3$
$xy - 2x = -y - 3$
$x(y-2) = -(y+3)$
$x = \frac{-(y+3)}{y-2} = \frac{y+3}{2-y}$
$h^{-1}(x) = \frac{x+3}{2-x}$ [3]

(b) $h(x) = h^{-1}(x)$ implies $x$ lies on the line $y=x$ for self-inverse symmetry, or solve algebraically:
$\frac{2x-3}{x+1} = \frac{x+3}{2-x}$
$(2x-3)(2-x) = (x+1)(x+3)$
$4x - 2x^2 - 6 + 3x = x^2 + 3x + x + 3$
$-2x^2 + 7x - 6 = x^2 + 4x + 3$
$3x^2 - 3x + 9 = 0$
Divide by 3: $x^2 - x + 3 = 0$
Discriminant $\Delta = (-1)^2 - 4(1)(3) = 1 - 12 = -11$ .
Since $\Delta < 0$ , there are no real solutions.
[3] (Award marks for correct algebraic setup and conclusion of no real roots).

13.
(a) Area $= (2x+3)(x-1) = 2x^2 - 2x + 3x - 3 = 2x^2 + x - 3$ . Shown. [1]

(b) $2x^2 + x - 3 = 25$
$2x^2 + x - 28 = 0$
Using quadratic formula: $x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-28)}}{2(2)}$
$x = \frac{-1 \pm \sqrt{1 + 224}}{4} = \frac{-1 \pm \sqrt{225}}{4}$
$x = \frac{-1 \pm 15}{4}$
$x_1 = \frac{14}{4} = 3.5$
$x_2 = \frac{-16}{4} = -4$
Since width $x-1$ must be positive ( $x>1$ ), reject $x=-4$ .
$x = 3.5$ [4]

14.
(a) Initial height at $t=0$ : $h(0) = 1.5$ m. [1]

(b) $h(t) = -5(t^2 - 4t) + 1.5$
$= -5((t-2)^2 - 4) + 1.5$
$= -5(t-2)^2 + 20 + 1.5$
$= -5(t-2)^2 + 21.5$
Max height is 21.5 m. [2]

(c) Hit ground when $h(t) = 0$ .
$-5t^2 + 20t + 1.5 = 0$
$5t^2 - 20t - 1.5 = 0$
$t = \frac{20 \pm \sqrt{400 - 4(5)(-1.5)}}{10}$
$t = \frac{20 \pm \sqrt{400 + 30}}{10} = \frac{20 \pm \sqrt{430}}{10}$
$t \approx \frac{20 \pm 20.736}{10}$
$t_1 \approx 4.07$ , $t_2 \approx -0.07$ (reject negative time).
Time taken $\approx 4.07$ s. [2]

15.
(a) Since $x \ge 0$ , $2x \ge 0$ , so $2x+3 \ge 3$ . Range is $f(x) \ge 3$ (or $[3, \infty)$ ). [1]
(b) $f(x)$ is a strictly increasing linear function (one-to-one), so it passes the horizontal line test. [1]

16.
(a) $x = 4$ [1]
(b) $k(6) = \frac{1}{6-4} = \frac{1}{2} = 0.5$
$k(2) = \frac{1}{2-4} = \frac{1}{-2} = -0.5$
$k(6) - k(2) = 0.5 - (-0.5) = 1$ [2]

17.
(a) $x^2 - 4x = 5 \Rightarrow x^2 - 4x - 5 = 0$
$(x-5)(x+1) = 0$
$x = 5$ or $x = -1$ [2]
(b) $x = 5, x = -1$ [1]

18.
(a) $m(1) = |2(1) - 6| = |-4| = 4$ [1]
(b) $|2x - 6| = 4$
Case 1: $2x - 6 = 4 \Rightarrow 2x = 10 \Rightarrow x = 5$
Case 2: $2x - 6 = -4 \Rightarrow 2x = 2 \Rightarrow x = 1$
$x = 1, 5$ [2]

19.
(a) As $x \to \infty$ , $\frac{3}{x} \to 0$ , so $y \to 2$ . Horizontal asymptote: $y = 2$ . [1]
(b) Crosses x-axis when $y=0$ :
$0 = \frac{3}{x} + 2$
$-2 = \frac{3}{x}$
$x = -\frac{3}{2}$ or $-1.5$ [2]

20.
(a) $gf(x) = g(4x - 1) = (4x - 1) + 5 = 4x + 4$ [2]
(b) $f(g(x)) = f(x+5) = 4(x+5) - 1 = 4x + 20 - 1 = 4x + 19$
Set $gf(x) = f(g(x))$ :
$4x + 4 = 4x + 19$
$4 = 19$ (False)
There is no solution. [2]