Secondary 2 Mathematics Algebra Functions Quiz

Secondary 2 Mathematics Quiz — Algebra Functions

Answer Key

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Section A: Proportionality and Variation

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1. [2]

$y = kx$
Substitute $x = 8$, $y = 24$:
$24 = k \times 8$
$k = 3$

Answer: $\boxed{y = 3x}$

Marking: [1] for correct proportionality form, [1] for correct $k$ and final equation.

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2. (a) [2]

$P = \dfrac{k}{\sqrt{q}}$
Substitute $q = 16$, $P = 5$:
$5 = \dfrac{k}{\sqrt{16}} = \dfrac{k}{4}$
$k = 20$

Answer: $\boxed{P = \dfrac{20}{\sqrt{q}}}$

Marking: [1] for correct form, [1] for correct $k$.

(b) [1]

$P = \dfrac{20}{\sqrt{4}} = \dfrac{20}{2} = 10$

Answer: $\boxed{P = 10}$

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3. [3]

$A = kb^3$
Substitute $b = 3$, $A = 81$:
$81 = k \times 27$
$k = 3$

When $b = 5$:
$A = 3 \times 5^3 = 3 \times 125 = 375$

Answer: $\boxed{A = 375}$

Marking: [1] for correct form, [1] for finding $k = 3$, [1] for correct final answer.

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4. (a) [2]

$M = \dfrac{k}{n^2}$
Substitute $n = 2$, $M = 12$:
$12 = \dfrac{k}{4}$
$k = 48$

Answer: $\boxed{M = \dfrac{48}{n^2}}$

Marking: [1] for correct form, [1] for correct $k$.

(b) [1]

$3 = \dfrac{48}{n^2}$
$n^2 = 16$
$n = 4$ (positive value)

Answer: $\boxed{n = 4}$

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5. (a) [2]

$C = kp$
Substitute $p = 40$, $C = 18$:
$18 = k \times 40$
$k = \dfrac{18}{40} = 0.45$

Answer: $\boxed{C = 0.45p}$

Marking: [1] for correct form, [1] for correct $k$.

(b) [1]

$C = 0.45 \times 75 = 33.75$

Answer: $33.75

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Section B: Algebraic Manipulation and Factorisation

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6. [2]

$(3x - 4)(2x + 5)$
$= 3x(2x) + 3x(5) - 4(2x) - 4(5)$
$= 6x^2 + 15x - 8x - 20$
$= 6x^2 + 7x - 20$

Answer: $\boxed{6x^2 + 7x - 20}$

Marking: [1] for correct expansion, [1] for correct simplification.

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7. [2]

$6x^2 + 11x - 10$
Find two numbers with product $6 \times (-10) = -60$ and sum $11$: $15$ and $-4$.

$= 6x^2 + 15x - 4x - 10$
$= 3x(2x + 5) - 2(2x + 5)$
$= (3x - 2)(2x + 5)$

Answer: $\boxed{(3x - 2)(2x + 5)}$

Marking: [1] for correct splitting or method, [1] for correct factorisation.

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8. [2]

$4x^2 - 25$
This is a difference of squares: $a^2 - b^2 = (a - b)(a + b)$
$= (2x)^2 - 5^2$
$= (2x - 5)(2x + 5)$

Answer: $\boxed{(2x - 5)(2x + 5)}$

Marking: [1] for recognising difference of squares, [1] for correct answer.

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9. [3]

$\dfrac{3x^2 - 12x}{x^2 - 16}$

Numerator: $3x^2 - 12x = 3x(x - 4)$
Denominator: $x^2 - 16 = (x - 4)(x + 4)$

$= \dfrac{3x(x - 4)}{(x - 4)(x + 4)}$
$= \dfrac{3x}{x + 4}$, where $x \neq 4$

Answer: $\boxed{\dfrac{3x}{x + 4}}$

Marking: [1] for factorising numerator, [1] for factorising denominator, [1] for correct simplification.

Common mistake: Forgetting to state the condition $x \neq 4$ (not penalised here but good practice).

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10. (a) [1]

Width $= \dfrac{\text{Area}}{\text{Length}} = \dfrac{x^2 + 9x + 20}{x + 5}$

Factorise numerator: $x^2 + 9x + 20 = (x + 4)(x + 5)$

Width $= \dfrac{(x + 4)(x + 5)}{x + 5} = (x + 4)$ ✓

Answer: Width $= \boxed{(x + 4) \text{ m}}$

(b) [3]

$x^2 + 9x + 20 = 42$
$x^2 + 9x - 22 = 0$
$(x + 11)(x - 2) = 0$
$x = -11$ or $x = 2$

Since $x$ represents a measurement, $x = 2$ (reject $x = -11$).

Answer: $\boxed{x = 2}$

Marking: [1] for correct equation, [1] for correct factorisation, [1] for correct solution with rejection of negative value.

(c) [1]

Length $= x + 5 = 2 + 5 = 7$ m
Width $= x + 4 = 2 + 4 = 6$ m

Answer: Length $= \boxed{7 \text{ m}}$, Width $= \boxed{6 \text{ m}}$

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Section C: Quadratic Equations

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11. [2]

$x^2 - 7x + 10 = 0$
$(x - 2)(x - 5) = 0$
$x = 2$ or $x = 5$

Answer: $\boxed{x = 2 \text{ or } x = 5}$

Marking: [1] for correct factorisation, [1] for both correct values.

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12. [3]

$2x^2 + 5x - 3 = 0$

Using the quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 2$, $b = 5$, $c = -3$:

$x = \dfrac{-5 \pm \sqrt{25 + 24}}{4} = \dfrac{-5 \pm \sqrt{49}}{4} = \dfrac{-5 \pm 7}{4}$

$x = \dfrac{2}{4} = 0.5$ or $x = \dfrac{-12}{4} = -3$

Answer: $\boxed{x = 0.5 \text{ or } x = -3}$

Marking: [1] for correct substitution into formula, [1] for correct discriminant, [1] for both correct answers.

Note: This also factorises as $(2x - 1)(x + 3) = 0$, which is acceptable.

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13. [2]

$(x - 3)^2 = 16$
$x - 3 = \pm 4$
$x = 3 + 4 = 7$ or $x = 3 - 4 = -1$

Answer: $\boxed{x = 7 \text{ or } x = -1}$

Marking: [1] for taking square root correctly (both signs), [1] for both correct values.

Common mistake: Only taking the positive root and getting $x = 7$ only.

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14. (a) [3]

$h = 20t - 5t^2 = 15$
$20t - 5t^2 - 15 = 0$
Divide by 5: $4t - t^2 - 3 = 0$
$t^2 - 4t + 3 = 0$
$(t - 1)(t - 3) = 0$
$t = 1$ or $t = 3$

The ball is at 15 m at $t = 1$ s (going up) and $t = 3$ s (coming down).

Answer: $\boxed{t = 1 \text{ s and } t = 3 \text{ s}}$

Marking: [1] for correct equation, [1] for correct factorisation, [1] for both times.

(b) [1]

Ball hits ground when $h = 0$:
$20t - 5t^2 = 0$
$5t(4 - t) = 0$
$t = 0$ (start) or $t = 4$

Answer: $\boxed{t = 4 \text{ s}}$

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15. (a) [1]

Smaller integer $= x$, larger integer $= x + 1$
$x(x + 1) = 132$

Answer: $\boxed{x(x + 1) = 132}$ or $\boxed{x^2 + x - 132 = 0}$

(b) [3]

$x^2 + x - 132 = 0$
$(x + 12)(x - 11) = 0$
$x = -12$ or $x = 11$

Since the integers are positive, $x = 11$.
The integers are 11 and 12.

Answer: $\boxed{11 \text{ and } 12}$

Marking: [1] for correct equation, [1] for correct factorisation, [1] for correct positive integers.

Common mistake: Giving both $x = -12$ and $x = 11$ without selecting the positive solution.

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Section D: Functions and Graphs

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16. (a) [1]

$f(0) = 2(0)^2 - 3(0) + 1 = 1$

Answer: $\boxed{f(0) = 1}$

(b) [1]

$f(-2) = 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15$

Answer: $\boxed{f(-2) = 15}$

(c) [2]

$f(x) = 0$:
$2x^2 - 3x + 1 = 0$
$(2x - 1)(x - 1) = 0$
$x = \dfrac{1}{2}$ or $x = 1$

Answer: $\boxed{x = \dfrac{1}{2} \text{ or } x = 1}$

Marking: [1] for correct factorisation, [1] for both correct values.

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17. (a) [1]

$g(2) = \dfrac{6}{2} = 3$

Answer: $\boxed{g(2) = 3}$

(b) [1]

$g(-3) = \dfrac{6}{-3} = -2$

Answer: $\boxed{g(-3) = -2}$

(c) [1]

$g(a) = 1$:
$\dfrac{6}{a} = 1$
$a = 6$

Answer: $\boxed{a = 6}$

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18. (a) [2]

$y = 3x - 5$
Swap $x$ and $y$: $x = 3y - 5$
$3y = x + 5$
$y = \dfrac{x + 5}{3}$

Answer: $\boxed{h^{-1}(x) = \dfrac{x + 5}{3}}$

Marking: [1] for correct method (swapping variables), [1] for correct final answer.

(b) [2]

$h(x) = h^{-1}(x)$:
$3x - 5 = \dfrac{x + 5}{3}$
Multiply by 3: $9x - 15 = x + 5$
$8x = 20$
$x = \dfrac{5}{2} = 2.5$

Answer: $\boxed{x = \dfrac{5}{2}}$

Marking: [1] for correct equation, [1] for correct solution.

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19. (a) [2]

Using $y = x^2 - 2x - 3$:

• $x = -2$: $y = 4 + 4 - 3 = 5$
• $x = -1$: $y = 1 + 2 - 3 = 0$
• $x = 0$: $y = 0 - 0 - 3 = -3$
• $x = 1$: $y = 1 - 2 - 3 = -4$
• $x = 2$: $y = 4 - 4 - 3 = -3$
• $x = 3$: $y = 9 - 6 - 3 = 0$
• $x = 4$: $y = 16 - 8 - 3 = 5$

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$5$	$0$	$-3$	$-4$	$-3$	$0$	$5$

Marking: [2] for all correct, [1] for 5–6 correct.

(b) [2]

Graph should be a smooth U-shaped parabola passing through the points above, with vertex at $(1, -4)$.

Marking: [1] for correct shape (parabola), [1] for correct points plotted.

(c) [1]

Answer: $\boxed{(1, -4)}$

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20. (a) [1]

$x^2 - 4x + 3 = 0$
$(x - 1)(x - 3) = 0$

Answer: $\boxed{x = 1 \text{ or } x = 3}$

(b) [1]

$x^2 - 4x + 3 = 3$
$x^2 - 4x = 0$
$x(x - 4) = 0$

Answer: $\boxed{x = 0 \text{ or } x = 4}$

(c) [1]

$x^2 - 4x + 3 = -1$
$x^2 - 4x + 4 = 0$
$(x - 2)^2 = 0$

Answer: $\boxed{x = 2}$ (repeated root)

(d) [1]

$x^2 - 4x + 3 < 0$ when the graph is below the x-axis, i.e., between the roots $x = 1$ and $x = 3$.

Answer: $\boxed{1 < x < 3}$

Marking note: Accept $x \in (1, 3)$ or equivalent notation.

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End of Answer Key

Total: 40 marks