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Secondary 4 Elementary Mathematics Algebra Functions Quiz

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Secondary 4 Elementary Mathematics From Real Exams Generated by Qwen3.6 Plus Updated 2026-07-10

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Questions

Secondary 4 Elementary Mathematics Quiz - Algebra Functions

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 40

Duration: 50 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all necessary working clearly. No marks will be given for correct answers without working.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
The use of an approved scientific calculator is expected.

Section A: Short Questions (1 mark each)

Answer questions 1 to 10.

1. Given that $f(x) = 3x - 5$ , find the value of $f(4)$ .

2. Given that $g(x) = x^2 + 2$ , find the value of $g(-3)$ .

3. If $h(x) = \frac{12}{x}$ , find the value of $x$ when $h(x) = -4$ .

4. The function $f$ is defined by $f: x \mapsto 2x + 1$ for $x \ge 0$ . State the range of $f$ .

5. Find the inverse function $f^{-1}(x)$ if $f(x) = \frac{x}{3} - 2$ .

6. Given $f(x) = 2x$ and $g(x) = x + 5$ , find the value of $fg(3)$ .

7. Given $f(x) = x^2$ and $g(x) = 3x - 1$ , find an expression for $gf(x)$ in terms of $x$ .

8. The graph of $y = f(x)$ passes through the point $(2, 5)$ . State the coordinates of the corresponding point on the graph of $y = f(x) + 3$ .

9. The graph of $y = x^2$ is transformed to the graph of $y = (x-2)^2$ . Describe this transformation geometrically.

10. The function $f(x) = ax + b$ satisfies $f(1) = 5$ and $f(2) = 8$ . Find the value of $a$ .

Section B: Structured Questions (2 marks each)

Answer questions 11 to 15.

11. Given that $f(x) = 3x^2 - 2x + 1$ . (a) Find $f(-1)$ . (b) Solve $f(x) = 6$ .

12. The function $f$ is defined by $f(x) = \frac{2x+1}{x-3}$ for $x \neq 3$ . (a) Find $f^{-1}(x)$ . (b) State the value of $x$ for which $f^{-1}(x)$ is undefined.

13. Given $f(x) = 2x - 3$ and $g(x) = x^2$ . (a) Find an expression for $fg(x)$ . (b) Hence, solve $fg(x) = 5$ .

14. The diagram shows the graph of a quadratic function $y = f(x)$ . The vertex of the graph is at $(2, -3)$ and it passes through the point $(0, 1)$ . Find the equation of the graph in the form $y = a(x-h)^2 + k$ .

15. A function is defined by $f(x) = \sqrt{x+4}$ for $x \ge -4$ . (a) Find the range of $f$ . (b) Find the value of $x$ such that $f(x) = 5$ .

Section C: Problem Solving (3 marks each)

Answer questions 16 to 20.

16. The functions $f$ and $g$ are defined by: $f(x) = 2x + 1$ $g(x) = x^2 - 3$ (a) Find an expression for $gf(x)$ in its simplest form. (b) Find the values of $x$ for which $gf(x) = 13$ .

17. The function $f$ is defined by $f(x) = \frac{3x}{x+2}$ for $x \neq -2$ . (a) Find $f^{-1}(x)$ . (b) Hence, solve the equation $f^{-1}(x) = 4$ .

18. The graph of $y = f(x)$ is shown below. It is a parabola with vertex at $(1, 4)$ and x-intercepts at $(-1, 0)$ and $(3, 0)$ . (a) Write down the equation of the axis of symmetry. (b) Find the equation of the curve in the form $y = a(x-p)(x-q)$ . (c) Hence, find the y-intercept of the curve.

19. Given that $f(x) = x^2 - 4x + 7$ . (a) Express $f(x)$ in the form $(x-a)^2 + b$ . (b) State the minimum value of $f(x)$ and the value of $x$ at which it occurs. (c) Sketch the graph of $y = f(x)$ , indicating the coordinates of the vertex and the y-intercept.

20. Two functions are defined as follows: $f(x) = 3x - 2$ $g(x) = \frac{10}{x}$ (a) Find $fg(x)$ . (b) Find $gf(x)$ . (c) Solve the equation $fg(x) = gf(x)$ .

Answers

Secondary 4 Elementary Mathematics Quiz - Algebra Functions (Answer Key)

1. $f(4) = 3(4) - 5 = 12 - 5 = 7$ Answer: 7 [1]

2. $g(-3) = (-3)^2 + 2 = 9 + 2 = 11$ Answer: 11 [1]

3. $\frac{12}{x} = -4 \Rightarrow x = \frac{12}{-4} = -3$ Answer: -3 [1]

4. Since $x \ge 0$ , $2x \ge 0$ , so $2x+1 \ge 1$ . Answer: $f(x) \ge 1$ [1]

5. Let $y = \frac{x}{3} - 2$ . $y + 2 = \frac{x}{3}$ $x = 3(y + 2) = 3y + 6$ Replace $y$ with $x$ : $f^{-1}(x) = 3x + 6$ Answer: $3x + 6$ [1]

6. $g(3) = 3 + 5 = 8$ $f(8) = 2(8) = 16$ Answer: 16 [1]

7. $gf(x) = g(x^2) = 3(x^2) - 1 = 3x^2 - 1$ Answer: $3x^2 - 1$ [1]

8. Transformation $y = f(x) + 3$ is a translation vector $\begin{pmatrix} 0 \\ 3 \end{pmatrix}$ . New y-coordinate $= 5 + 3 = 8$ . Answer: $(2, 8)$ [1]

9. Translation by vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ (or 2 units to the right). Answer: Translation 2 units right [1]

10. $f(1) = a(1) + b = 5 \Rightarrow a+b=5$ $f(2) = a(2) + b = 8 \Rightarrow 2a+b=8$ Subtracting eq1 from eq2: $a = 3$ . Answer: 3 [1]

11. (a) $f(-1) = 3(-1)^2 - 2(-1) + 1 = 3 + 2 + 1 = 6$ [1] (b) $3x^2 - 2x + 1 = 6 \Rightarrow 3x^2 - 2x - 5 = 0$ $(3x - 5)(x + 1) = 0$ $x = \frac{5}{3}$ or $x = -1$ [1]

12. (a) Let $y = \frac{2x+1}{x-3}$ . $y(x-3) = 2x + 1$ $xy - 3y = 2x + 1$ $xy - 2x = 3y + 1$ $x(y - 2) = 3y + 1$ $x = \frac{3y+1}{y-2}$ $f^{-1}(x) = \frac{3x+1}{x-2}$ [1] (b) Undefined when denominator is zero: $x - 2 = 0 \Rightarrow x = 2$ . Answer: 2 [1]

13. (a) $fg(x) = f(x^2) = 2(x^2) - 3 = 2x^2 - 3$ [1] (b) $2x^2 - 3 = 5 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4$ $x = 2$ or $x = -2$ [1]

14. Vertex form: $y = a(x-2)^2 - 3$ . Passes through $(0, 1)$ : $1 = a(0-2)^2 - 3$ $1 = 4a - 3$ $4a = 4 \Rightarrow a = 1$ Equation: $y = (x-2)^2 - 3$ Answer: $y = (x-2)^2 - 3$ [2]

15. (a) Since $\sqrt{x+4} \ge 0$ , the range is $f(x) \ge 0$ . [1] (b) $\sqrt{x+4} = 5 \Rightarrow x+4 = 25 \Rightarrow x = 21$ . [1]

16. (a) $gf(x) = g(2x+1) = (2x+1)^2 - 3$ $= 4x^2 + 4x + 1 - 3$ $= 4x^2 + 4x - 2$ [1] (b) $4x^2 + 4x - 2 = 13$ $4x^2 + 4x - 15 = 0$ $(2x + 5)(2x - 3) = 0$ $x = -\frac{5}{2}$ or $x = \frac{3}{2}$ [2]

17. (a) Let $y = \frac{3x}{x+2}$ . $y(x+2) = 3x$ $xy + 2y = 3x$ $2y = 3x - xy$ $2y = x(3-y)$ $x = \frac{2y}{3-y}$ $f^{-1}(x) = \frac{2x}{3-x}$ [1] (b) $\frac{2x}{3-x} = 4$ $2x = 4(3-x)$ $2x = 12 - 4x$ $6x = 12 \Rightarrow x = 2$ [2]

18. (a) Axis of symmetry is $x = 1$ (midpoint of roots or x-coord of vertex). [1] (b) $y = a(x+1)(x-3)$ . Using vertex $(1, 4)$ : $4 = a(1+1)(1-3)$ $4 = a(2)(-2)$ $4 = -4a \Rightarrow a = -1$ Equation: $y = -(x+1)(x-3)$ [1] (c) y-intercept when $x=0$ : $y = -(0+1)(0-3) = -(-3) = 3$ . Answer: 3 [1]

19. (a) $x^2 - 4x + 7 = (x^2 - 4x + 4) - 4 + 7 = (x-2)^2 + 3$ . [1] (b) Minimum value is 3 at $x = 2$ . [1] (c) Sketch: Parabola opening upwards. Vertex at $(2,3)$ . Y-intercept at $(0,7)$ . [1]

20. (a) $fg(x) = f(\frac{10}{x}) = 3(\frac{10}{x}) - 2 = \frac{30}{x} - 2$ . [1] (b) $gf(x) = g(3x-2) = \frac{10}{3x-2}$ . [1] (c) $\frac{30}{x} - 2 = \frac{10}{3x-2}$ Multiply by $x(3x-2)$ : $30(3x-2) - 2x(3x-2) = 10x$ $90x - 60 - 6x^2 + 4x = 10x$ $-6x^2 + 84x - 60 = 0$ Divide by -6: $x^2 - 14x + 10 = 0$ Using quadratic formula: $x = \frac{14 \pm \sqrt{196 - 40}}{2} = \frac{14 \pm \sqrt{156}}{2} = \frac{14 \pm 2\sqrt{39}}{2} = 7 \pm \sqrt{39}$ Answer: $x = 7 \pm \sqrt{39}$ [1]

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