Secondary 2 Mathematics Algebra Functions Quiz

Secondary 2 Mathematics Quiz - Algebra Functions (Answer Key)

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Section A: Basic Skills [10 marks]

1. Solve the equation $2x + 7 = 15$. [1 mark]

Answer: $x = 4$

Working: $2x = 15 - 7 = 8$, so $x = 4$

2. If $f(x) = 3x - 4$, find $f(5)$. [1 mark]

Answer: $11$

Working: $f(5) = 3(5) - 4 = 15 - 4 = 11$

3. Factorise $6x + 9y$. [1 mark]

Answer: $3(2x + 3y)$

Working: Common factor is 3

4. Expand $(x + 4)(x - 2)$. [1 mark]

Answer: $x^2 + 2x - 8$

Working: $x^2 - 2x + 4x - 8 = x^2 + 2x - 8$

5. If $y$ is directly proportional to $x$ and $y = 12$ when $x = 4$, find the constant of proportionality. [1 mark]

Answer: $k = 3$

Working: $y = kx$, so $12 = k(4)$, therefore $k = 3$

6. Solve $\frac{x}{3} = 8$. [1 mark]

Answer: $x = 24$

Working: Multiply both sides by 3

7. If $g(x) = x^2 + 1$, find $g(-2)$. [1 mark]

Answer: $5$

Working: $g(-2) = (-2)^2 + 1 = 4 + 1 = 5$

8. Simplify $4x - 2x + 7$. [1 mark]

Answer: $2x + 7$

Working: Collect like terms

9. Factorise $x^2 - 9$. [1 mark]

Answer: $(x + 3)(x - 3)$

Working: Difference of two squares

10. If $p$ is inversely proportional to $q$, write the relationship in the form $p = \frac{k}{q^n}$. State the value of $n$. [1 mark]

Answer: $n = 1$

Working: Inverse proportion means $p = \frac{k}{q}$, so $n = 1$

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Section B: Standard Applications [20 marks]

11. $y$ is directly proportional to the square of $x$. When $x = 3$, $y = 18$. [3 marks]

(a) Find the equation connecting $y$ and $x$. [2 marks]

Answer: $y = 2x^2$

Working: $y = kx^2$. When $x = 3$, $y = 18$: $18 = k(3)^2 = 9k$, so $k = 2$

Marking: M1 for $y = kx^2$, A1 for correct value of $k$

(b) Find the value of $y$ when $x = 5$. [1 mark]

Answer: $y = 50$

Working: $y = 2(5)^2 = 2(25) = 50$

12. Solve the quadratic equation $x^2 + 5x - 14 = 0$ by factorisation. [2 marks]

Answer: $x = 2$ or $x = -7$

Working: $(x + 7)(x - 2) = 0$, so $x = -7$ or $x = 2$

Marking: M1 for correct factorisation, A1 for both solutions

13. The functions $f$ and $g$ are defined by $f(x) = 2x + 3$ and $g(x) = x^2 - 1$. [3 marks]

(a) Find $f(4)$. [1 mark]

Answer: $11$

Working: $f(4) = 2(4) + 3 = 11$

(b) Find $g(3)$. [1 mark]

Answer: $8$

Working: $g(3) = 3^2 - 1 = 9 - 1 = 8$

(c) Find $f(g(2))$. [1 mark]

Answer: $9$

Working: $g(2) = 2^2 - 1 = 3$, then $f(3) = 2(3) + 3 = 9$

14. Solve the simultaneous equations: [3 marks] $2x + y = 11$ $x - y = 4$

Answer: $x = 5$, $y = 1$

Working: Adding equations: $3x = 15$, so $x = 5$. Substituting: $y = 11 - 2(5) = 1$

Marking: M1 for correct method, M1 for finding one variable, A1 for both correct values

15. A rectangular garden has length $(x + 6)$ metres and width $(x - 2)$ metres. The area of the garden is 40 square metres. [4 marks]

(a) Write an equation for the area of the garden. [1 mark]

Answer: $(x + 6)(x - 2) = 40$

(b) Expand and simplify your equation. [1 mark]

Answer: $x^2 + 4x - 12 = 40$ or $x^2 + 4x - 52 = 0$

Working: $(x + 6)(x - 2) = x^2 + 4x - 12$

(c) Solve the equation to find the value of $x$. [2 marks]

Answer: $x = 6$

Working: $x^2 + 4x - 52 = 0$, $(x + 10)(x - 6) = 0$, so $x = 6$ (reject $x = -10$ as width would be negative)

Marking: M1 for correct factorisation, A1 for correct value considering context

16. $z$ is inversely proportional to the cube of $w$. When $w = 2$, $z = 5$. [3 marks]

(a) Find the relationship between $z$ and $w$. [2 marks]

Answer: $z = \frac{40}{w^3}$

Working: $z = \frac{k}{w^3}$. When $w = 2$, $z = 5$: $5 = \frac{k}{2^3} = \frac{k}{8}$, so $k = 40$

Marking: M1 for correct form, A1 for correct constant

(b) Find the value of $z$ when $w = 4$. [1 mark]

Answer: $z = \frac{5}{8}$ or $0.625$

Working: $z = \frac{40}{4^3} = \frac{40}{64} = \frac{5}{8}$

17. Factorise completely $2x^3 - 8x^2 + 6x$. [2 marks]

Answer: $2x(x - 1)(x - 3)$

Working: $2x(x^2 - 4x + 3) = 2x(x - 1)(x - 3)$

Marking: M1 for extracting common factor $2x$, A1 for complete factorisation

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Section C: Problem Solving [10 marks]

18. The cost $C$ dollars of hiring a car is given by $C = 50 + 0.3d$, where $d$ is the distance travelled in kilometres. [3 marks]

(a) Find the cost of hiring the car to travel 200 km. [1 mark]

Answer: C = \110$

Working: $C = 50 + 0.3(200) = 50 + 60 = 110$

(b) If the total cost was $95, find the distance travelled. [2 marks]

Answer: $d = 150$ km

Working: $95 = 50 + 0.3d$, so $45 = 0.3d$, therefore $d = 150$

Marking: M1 for correct equation setup, A1 for correct answer

19. The time $T$ hours taken to complete a job is inversely proportional to the number of workers $n$. When 6 workers are employed, the job takes 8 hours to complete. [4 marks]

(a) Find the relationship between $T$ and $n$. [2 marks]

Answer: $T = \frac{48}{n}$

Working: $T = \frac{k}{n}$. When $n = 6$, $T = 8$: $8 = \frac{k}{6}$, so $k = 48$

Marking: M1 for correct form, A1 for correct constant

(b) How long will it take 4 workers to complete the same job? [1 mark]

Answer: $12$ hours

Working: $T = \frac{48}{4} = 12$

(c) How many workers are needed to complete the job in 3 hours? [1 mark]

Answer: $16$ workers

Working: $3 = \frac{48}{n}$, so $n = 16$

20. A quadratic function has the form $f(x) = x^2 + bx + c$. The function passes through the points $(1, 6)$ and $(2, 11)$. [3 marks]

(a) Set up two equations using the given information. [1 mark]

Equations: $1 + b + c = 6$ and $4 + 2b + c = 11$

Working: Substitute the points into $f(x) = x^2 + bx + c$

(b) Solve to find the values of $b$ and $c$. [2 marks]

Answer: $b = 2$, $c = 3$

Working: From the equations: $b + c = 5$ and $2b + c = 7$ Subtracting: $b = 2$, so $c = 3$

Marking: M1 for correct method to solve simultaneous equations, A1 for both correct values

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Total: 40 marks

Common Student Errors to Watch For:

• Sign errors when expanding brackets or solving equations
• Forgetting to reject negative solutions in real-world contexts
• Confusing direct and inverse proportionality
• Arithmetic errors in fraction calculations
• Not showing sufficient working for method marks