Secondary 2 Mathematics Semestral Assessment 2 (End of Year) Paper 1

TuitionGoWhere Practice Paper - Mathematics Secondary 2

SA2 Version 1 - Answer Key and Marking Scheme

Total Marks: 60

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Section A [20 marks]

1

(a) $y \propto x^2 \Rightarrow y = kx^2$
When $x = 3$, $y = 36$:
$36 = k(3)^2 = 9k$
$k = 4$
Equation: $y = 4x^2$

Marks: [1] for $y = kx^2$, [1] for $k = 4$ and final equation

(b) When $x = 5$:
$y = 4(5)^2 = 4 \times 25 = 100$

Marks: [1] for correct substitution and answer

Answer: (a) $y = 4x^2$ (b) $y = 100$

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2

(a) $P \propto \frac{1}{\sqrt[3]{Q}} \Rightarrow P = \frac{k}{\sqrt[3]{Q}}$
When $Q = 8$, $P = 12$:
$12 = \frac{k}{\sqrt[3]{8}} = \frac{k}{2}$
$k = 24$
Equation: $P = \frac{24}{\sqrt[3]{Q}}$

Marks: [1] for $P = \frac{k}{\sqrt[3]{Q}}$, [1] for $k = 24$ and final equation

(b) When $P = 6$:
$6 = \frac{24}{\sqrt[3]{Q}}$
$\sqrt[3]{Q} = \frac{24}{6} = 4$
$Q = 4^3 = 64$

Marks: [1] for correct working and answer

Answer: (a) $P = \frac{24}{\sqrt[3]{Q}}$ (b) $Q = 64$

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3

$f(x) = 2x^2 - 5x + 3$
$f(-2) = 2(-2)^2 - 5(-2) + 3$
$= 2(4) + 10 + 3$
$= 8 + 10 + 3 = 21$

Marks: [1] for correct substitution, [1] for correct evaluation

Answer: $21$

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4

(a) $g(5) = \frac{4(5) - 1}{3} = \frac{20 - 1}{3} = \frac{19}{3}$

Marks: [1] for correct answer

(b) $g(x) = 7$
$\frac{4x - 1}{3} = 7$
$4x - 1 = 21$
$4x = 22$
$x = \frac{22}{4} = \frac{11}{2} = 5.5$

Marks: [1] for setting up equation, [1] for solving correctly

Answer: (a) $\frac{19}{3}$ (b) $x = \frac{11}{2}$ or $5.5$

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5

$3x^2 - 11x - 4 = 0$
Factorise: $(3x + 1)(x - 4) = 0$
$3x + 1 = 0$ or $x - 4 = 0$
$x = -\frac{1}{3}$ or $x = 4$

Marks: [1] for correct factorisation, [1] for setting each factor to zero, [1] for both solutions

Answer: $x = -\frac{1}{3}$ or $x = 4$

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6

$$\begin{cases} 3x + 2y = 13 \quad \text{(1)} \\ 5x - 4y = 3 \quad \text{(2)} \end{cases}$$

Multiply (1) by 2: $6x + 4y = 26$ \quad \text{(3)}
Add (2) and (3): $11x = 29$
$x = \frac{29}{11}$

Substitute into (1):
$3(\frac{29}{11}) + 2y = 13$
$\frac{87}{11} + 2y = \frac{143}{11}$
$2y = \frac{56}{11}$
$y = \frac{28}{11}$

Marks: [1] for elimination step, [1] for finding $x$, [1] for finding $y$

Answer: $x = \frac{29}{11}$, $y = \frac{28}{11}$

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7

$12x^2 - 27y^2$
$= 3(4x^2 - 9y^2)$
$= 3[(2x)^2 - (3y)^2]$
$= 3(2x - 3y)(2x + 3y)$

Marks: [1] for factorising out 3, [1] for difference of squares factorisation

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

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8

$\frac{x^2 - 9}{x^2 - 4x - 21}$
$= \frac{(x - 3)(x + 3)}{(x - 7)(x + 3)}$
$= \frac{x - 3}{x - 7}, \quad x \neq -3$

Marks: [1] for factorising numerator and denominator, [1] for cancelling common factor and stating restriction

Answer: $\frac{x - 3}{x - 7}, \quad x \neq -3$

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9

(a) From the graph, the minimum point (vertex) is at $(2, -9)$.

Marks: [1] for correct coordinates

(b) Line of symmetry passes through the vertex: $x = 2$

Marks: [1] for correct equation

(c) The graph cuts the $x$-axis at $(-1, 0)$ and $(5, 0)$.
Solutions: $x = -1$ or $x = 5$

Marks: [1] for both solutions

Answer: (a) $(2, -9)$ (b) $x = 2$ (c) $x = -1$ or $x = 5$

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10

(a) Points plotted correctly and smooth curve drawn through them for both branches ($x < 0$ and $x > 0$).

Marks: [1] for correct plotting of all points, [1] for smooth curves with correct shape (two separate branches)

(b) The asymptotes are the $x$-axis and $y$-axis.
Equations: $y = 0$ and $x = 0$

Marks: [1] for both equations

Answer: (a) Graph drawn (b) $y = 0$ and $x = 0$

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Section B [25 marks]

11

(a) Area $= (2x + 5)(x - 2)$
$= 2x^2 - 4x + 5x - 10$
$= 2x^2 + x - 10$

Marks: [1] for correct expression

(b) Given area $= 72$:
$2x^2 + x - 10 = 72$
$2x^2 + x - 82 = 0$ (shown)

Marks: [1] for equating to 72, [1] for correct rearrangement

(c) $2x^2 + x - 82 = 0$
Using quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$a = 2$, $b = 1$, $c = -82$
$x = \frac{-1 \pm \sqrt{1 - 4(2)(-82)}}{4}$
$= \frac{-1 \pm \sqrt{1 + 656}}{4}$
$= \frac{-1 \pm \sqrt{657}}{4}$
$= \frac{-1 \pm 25.632...}{4}$
$x = 6.16$ or $x = -6.66$ (2 d.p.)

Since $x$ represents a length, $x > 0$, so $x = 6.16$ (2 d.p.)

Marks: [1] for correct quadratic formula substitution, [1] for both values to 2 d.p.

(d) Length $= 2(6.16) + 5 = 17.32$ cm
Width $= 6.16 - 2 = 4.16$ cm
Perimeter $= 2(17.32 + 4.16) = 2(21.48) = 42.96$ cm

Marks: [1] for finding length and width, [1] for correct perimeter

Answer: (a) $2x^2 + x - 10$ (b) Shown (c) $x = 6.16$ or $x = -6.66$ (d) $42.96$ cm

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12

(a) By Pythagoras' theorem:
$(x - 3)^2 + (x + 1)^2 = (x + 5)^2$

Marks: [1] for correct equation

(b) Expand:
$(x^2 - 6x + 9) + (x^2 + 2x + 1) = x^2 + 10x + 25$
$2x^2 - 4x + 10 = x^2 + 10x + 25$
$x^2 - 14x - 15 = 0$ (shown)

Marks: [1] for correct expansion, [1] for correct simplification

(c) $x^2 - 14x - 15 = 0$
$(x - 15)(x + 1) = 0$
$x = 15$ or $x = -1$

Since side lengths must be positive: $x - 3 > 0 \Rightarrow x > 3$, so $x = 15$

Marks: [1] for solving, [1] for rejecting $x = -1$ with reason

(d) Sides: $15 - 3 = 12$ cm, $15 + 1 = 16$ cm
Area $= \frac{1}{2} \times 12 \times 16 = 96 \text{ cm}^2$

Marks: [1] for finding side lengths, [1] for correct area

Answer: (a) $(x-3)^2 + (x+1)^2 = (x+5)^2$ (b) Shown (c) $x = 15$ (d) $96 \text{ cm}^2$

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13

(a) $y = \frac{k}{x^2}$
When $x = 1$, $y = 100$:
$100 = \frac{k}{1^2} \Rightarrow k = 100$

Marks: [1] for correct value

(b) When $x = 3$:
$p = \frac{100}{3^2} = \frac{100}{9} = 11.1$ (or $11\frac{1}{9}$)

Marks: [1] for correct value

(c) Points plotted correctly and smooth curve drawn.

Marks: [1] for correct plotting, [1] for smooth curve

(d) From graph, when $y = 15$, $x \approx 2.6$

Marks: [1] for reasonable estimate (accept $2.5 - 2.7$)

(e) Line $y = 15$ drawn horizontally. Intersection with curve gives $x \approx 2.6$

Marks: [1] for line drawn and intersection used

Answer: (a) $k = 100$ (b) $p = \frac{100}{9}$ or $11.1$ (c) Graph drawn (d) $x \approx 2.6$ (e) $x \approx 2.6$

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14

(a) $h(1) = 6$: $a(1)^2 + b(1) + c = 6 \Rightarrow a + b + c = 6$
$h(2) = 11$: $a(2)^2 + b(2) + c = 11 \Rightarrow 4a + 2b + c = 11$
$h(3) = 18$: $a(3)^2 + b(3) + c = 18 \Rightarrow 9a + 3b + c = 18$

Marks: [1] for first two equations, [1] for third equation

(b) Subtract first from second: $3a + b = 5$ \quad \text{(4)}
Subtract second from third: $5a + b = 7$ \quad \text{(5)}
Subtract (4) from (5): $2a = 2 \Rightarrow a = 1$
Substitute into (4): $3(1) + b = 5 \Rightarrow b = 2$
Substitute into first: $1 + 2 + c = 6 \Rightarrow c = 3$

Marks: [1] for eliminating $c$, [1] for finding $a$ and $b$, [1] for finding $c$

(c) $h(x) = x^2 + 2x + 3$
Complete the square: $h(x) = (x + 1)^2 + 2$
Minimum value $= 2$ (when $x = -1$)

Marks: [1] for completing the square, [1] for minimum value

Answer: (a) $a+b+c=6$, $4a+2b+c=11$, $9a+3b+c=18$ (b) $a=1$, $b=2$, $c=3$ (c) Minimum value $= 2$

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15

(a) Graph cuts $x$-axis at $(-4, 0)$ and $(2, 0)$.

Marks: [1] for both coordinates

(b) Line of symmetry: $x = \frac{-4 + 2}{2} = -1$

Marks: [1] for correct equation

(c) The minimum value is $-9$ (at vertex). The line $y = k$ cuts the graph at two points when $k > -9$.
Range: $k > -9$

Marks: [1] for correct inequality

(d) Translation 3 units right: replace $x$ with $(x - 3)$
New equation: $y = ((x - 3) - 2)((x - 3) + 4) = (x - 5)(x + 1)$
Or expanded: $y = x^2 - 4x - 5$

Marks: [1] for correct substitution, [1] for simplified equation

Answer: (a) $(-4, 0)$ and $(2, 0)$ (b) $x = -1$ (c) $k > -9$ (d) $y = (x - 5)(x + 1)$ or $y = x^2 - 4x - 5$

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Section C [15 marks]

16

(a) When $n = 10$, $C = 120$: $a(10)^2 + b(10) = 120 \Rightarrow 100a + 10b = 120$
When $n = 20$, $C = 400$: $a(20)^2 + b(20) = 400 \Rightarrow 400a + 20b = 400$

Marks: [1] for each equation

(b) Simplify:
$10a + b = 12$ \quad \text{(1)}
$20a + b = 20$ \quad \text{(2)}
Subtract (1) from (2): $10a = 8 \Rightarrow a = 0.8$
Substitute into (1): $10(0.8) + b = 12 \Rightarrow 8 + b = 12 \Rightarrow b = 4$

Marks: [1] for simplification, [1] for solving

(c) When $n = 15$:
$C = 0.8(15)^2 + 4(15) = 0.8(225) + 60 = 180 + 60 = 240$

Marks: [1] for correct substitution and answer

(d) Revenue $= 10n$
Profit when Revenue $> Cost$:
$10n > 0.8n^2 + 4n$
$0 > 0.8n^2 - 6n$
$0 > 0.8n(n - 7.5)$
$0 < n < 7.5$

Since $n$ must be an integer and profit requires $n > 7.5$, minimum $n = 8$

Marks: [1] for setting up inequality, [1] for solving inequality, [1] for correct integer answer

Answer: (a) $100a+10b=120$, $400a+20b=400$ (b) $a=0.8$, $b=4$ (c) $240$ (d) $8$ items

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17

(a) Outer dimensions: $(12 + 2x)$ by $(8 + 2x)$
Total area $= (12 + 2x)(8 + 2x) = 192$
$96 + 24x + 16x + 4x^2 = 192$
$4x^2 + 40x + 96 = 192$
$4x^2 + 40x - 96 = 0$
Divide by 4: $x^2 + 10x - 24 = 0$ (shown)

Marks: [1] for outer dimensions, [1] for area equation, [1] for correct simplification

(b) $x^2 + 10x - 24 = 0$
$(x + 12)(x - 2) = 0$
$x = -12$ or $x = 2$

Width must be positive, so $x = 2$ m

Marks: [1] for solving, [1] for rejecting negative root with reason

(c) Outer dimensions: $12 + 2(2) = 16$ m, $8 + 2(2) = 12$ m
Perimeter $= 2(16 + 12) = 56$ m

Marks: [1] for outer dimensions, [1] for perimeter

Answer: (a) Shown (b) $x = 2$ m (c) $56$ m

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18

(a) $h = -5(1)^2 + 20(1) + 2 = -5 + 20 + 2 = 17$ m

Marks: [1] for correct answer

(b) $h = -5t^2 + 20t + 2$
Complete the square:
$h = -5(t^2 - 4t) + 2$
$= -5[(t - 2)^2 - 4] + 2$
$= -5(t - 2)^2 + 20 + 2$
$= -5(t - 2)^2 + 22$

Maximum height $= 22$ m (when $t = 2$ s)

Marks: [1] for completing the square, [1] for identifying vertex form, [1] for maximum height

(c) Ball hits ground when $h = 0$:
$-5t^2 + 20t + 2 = 0$
$5t^2 - 20t - 2 = 0$
$t = \frac{20 \pm \sqrt{400 - 4(5)(-2)}}{10}$
$= \frac{20 \pm \sqrt{440}}{10}$
$= \frac{20 \pm 20.976...}{10}$
$t = 4.10$ or $t = -0.10$ (2 d.p.)

Time must be positive, so $t = 4.10$ s

Marks: [1] for setting $h=0$, [1] for quadratic formula, [1] for correct positive root to 2 d.p.

Answer: (a) $17$ m (b) $22$ m (c) $4.10$ s

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19

(a) $f(x) = x^2 - 6x + 11$
$= (x^2 - 6x + 9) + 2$
$= (x - 3)^2 + 2$

Marks: [1] for completing the square, [1] for correct form

(b) Minimum value $= 2$, occurs at $x = 3$

Marks: [1] for both

(c) $g(x) = f(x) + 4$
This is a translation of the graph of $y = f(x)$ by 4 units upwards (in the positive $y$-direction).

Marks: [1] for "

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TuitionGoWhere Practice Paper - Mathematics Secondary 2

SA2 Version 1 - Answer Key and Marking Scheme

Total Marks: 60

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Section A [20 marks]

1

(a) $y \propto x^2 \Rightarrow y = kx^2$
When $x = 3$, $y = 36$:
$36 = k(3)^2 = 9k$
$k = 4$
Equation: $y = 4x^2$

Marks: [1] for $y = kx^2$, [1] for $k = 4$ and final equation

(b) When $x = 5$:
$y = 4(5)^2 = 4 \times 25 = 100$

Marks: [1] for correct substitution and answer

Answer: (a) $y = 4x^2$ (b) $y = 100$

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2

(a) $P \propto \frac{1}{\sqrt[3]{Q}} \Rightarrow P = \frac{k}{\sqrt[3]{Q}}$
When $Q = 8$, $P = 12$:
$12 = \frac{k}{\sqrt[3]{8}} = \frac{k}{2}$
$k = 24$
Equation: $P = \frac{24}{\sqrt[3]{Q}}$

Marks: [1] for $P = \frac{k}{\sqrt[3]{Q}}$, [1] for $k = 24$ and final equation

(b) When $P = 6$:
$6 = \frac{24}{\sqrt[3]{Q}}$
$\sqrt[3]{Q} = \frac{24}{6} = 4$
$Q = 4^3 = 64$

Marks: [1] for correct working and answer

Answer: (a) $P = \frac{24}{\sqrt[3]{Q}}$ (b) $Q = 64$

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3

$f(x) = 2x^2 - 5x + 3$
$f(-2) = 2(-2)^2 - 5(-2) + 3$
$= 2(4) + 10 + 3$
$= 8 + 10 + 3 = 21$

Marks: [1] for correct substitution, [1] for correct evaluation

Answer: $21$

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4

(a) $g(5) = \frac{4(5) - 1}{3} = \frac{20 - 1}{3} = \frac{19}{3}$

Marks: [1] for correct answer

(b) $g(x) = 7$
$\frac{4x - 1}{3} = 7$
$4x - 1 = 21$
$4x = 22$
$x = \frac{22}{4} = \frac{11}{2} = 5.5$

Marks: [1] for setting up equation, [1] for solving correctly

Answer: (a) $\frac{19}{3}$ (b) $x = \frac{11}{2}$ or $5.5$

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5

$3x^2 - 11x - 4 = 0$
Factorise: $(3x + 1)(x - 4) = 0$
$3x + 1 = 0$ or $x - 4 = 0$
$x = -\frac{1}{3}$ or $x = 4$

Marks: [1] for correct factorisation, [1] for setting each factor to zero, [1] for both solutions

Answer: $x = -\frac{1}{3}$ or $x = 4$

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6

$$\begin{cases} 3x + 2y = 13 \quad \text{(1)} \\ 5x - 4y = 3 \quad \text{(2)} \end{cases}$$

Multiply (1) by 2: $6x + 4y = 26$ \quad \text{(3)}
Add (2) and (3): $11x = 29$
$x = \frac{29}{11}$

Substitute into (1):
$3(\frac{29}{11}) + 2y = 13$
$\frac{87}{11} + 2y = \frac{143}{11}$
$2y = \frac{56}{11}$
$y = \frac{28}{11}$

Marks: [1] for elimination step, [1] for finding $x$, [1] for finding $y$

Answer: $x = \frac{29}{11}$, $y = \frac{28}{11}$

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7

$12x^2 - 27y^2$
$= 3(4x^2 - 9y^2)$
$= 3[(2x)^2 - (3y)^2]$
$= 3(2x - 3y)(2x + 3y)$

Marks: [1] for factorising out 3, [1] for difference of squares factorisation

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

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8

$\frac{x^2 - 9}{x^2 - 4x - 21}$
$= \frac{(x - 3)(x + 3)}{(x - 7)(x + 3)}$
$= \frac{x - 3}{x - 7}, \quad x \neq -3$

Marks: [1] for factorising numerator and denominator, [1] for cancelling common factor and stating restriction

Answer: $\frac{x - 3}{x - 7}, \quad x \neq -3$

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9

(a) From the graph, the minimum point (vertex) is at $(2, -9)$.

Marks: [1] for correct coordinates

(b) Line of symmetry passes through the vertex: $x = 2$

Marks: [1] for correct equation

(c) The graph cuts the $x$-axis at $(-1, 0)$ and $(5, 0)$.
Solutions: $x = -1$ or $x = 5$

Marks: [1] for both solutions

Answer: (a) $(2, -9)$ (b) $x = 2$ (c) $x = -1$ or $x = 5$

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10

(a) Points plotted correctly and smooth curve drawn through them for both branches ($x < 0$ and $x > 0$).

Marks: [1] for correct plotting of all points, [1] for smooth curves with correct shape (two separate branches)

(b) The asymptotes are the $x$-axis and $y$-axis.
Equations: $y = 0$ and $x = 0$

Marks: [1] for both equations

Answer: (a) Graph drawn (b) $y = 0$ and $x = 0$

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Section B [25 marks]

11

(a) Area $= (2x + 5)(x - 2)$
$= 2x^2 - 4x + 5x - 10$
$= 2x^2 + x - 10$

Marks: [1] for correct expression

(b) Given area $= 72$:
$2x^2 + x - 10 = 72$
$2x^2 + x - 82 = 0$ (shown)

Marks: [1] for equating to 72, [1] for correct rearrangement

(c) $2x^2 + x - 82 = 0$
Using quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$a = 2$, $b = 1$, $c = -82$
$x = \frac{-1 \pm \sqrt{1 - 4(2)(-82)}}{4}$
$= \frac{-1 \pm \sqrt{1 + 656}}{4}$
$= \frac{-1 \pm \sqrt{657}}{4}$
$= \frac{-1 \pm 25.632...}{4}$
$x = 6.16$ or $x = -6.66$ (2 d.p.)

Since $x$ represents a length, $x > 0$, so $x = 6.16$ (2 d.p.)

Marks: [1] for correct quadratic formula substitution, [1] for both values to 2 d.p.

(d) Length $= 2(6.16) + 5 = 17.32$ cm
Width $= 6.16 - 2 = 4.16$ cm
Perimeter $= 2(17.32 + 4.16) = 2(21.48) = 42.96$ cm

Marks: [1] for finding length and width, [1] for correct perimeter

Answer: (a) $2x^2 + x - 10$ (b) Shown (c) $x = 6.16$ or $x = -6.66$ (d) $42.96$ cm

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12

(a) By Pythagoras' theorem:
$(x - 3)^2 + (x + 1)^2 = (x + 5)^2$

Marks: [1] for correct equation

(b) Expand:
$(x^2 - 6x + 9) + (x^2 + 2x + 1) = x^2 + 10x + 25$
$2x^2 - 4x + 10 = x^2 + 10x + 25$
$x^2 - 14x - 15 = 0$ (shown)

Marks: [1] for correct expansion, [1] for correct simplification

(c) $x^2 - 14x - 15 = 0$
$(x - 15)(x + 1) = 0$
$x = 15$ or $x = -1$

Since side lengths must be positive: $x - 3 > 0 \Rightarrow x > 3$, so $x = 15$

Marks: [1] for solving, [1] for rejecting $x = -1$ with reason

(d) Sides: $15 - 3 = 12$ cm, $15 + 1 = 16$ cm
Area $= \frac{1}{2} \times 12 \times 16 = 96 \text{ cm}^2$

Marks: [1] for finding side lengths, [1] for correct area

Answer: (a) $(x-3)^2 + (x+1)^2 = (x+5)^2$ (b) Shown (c) $x = 15$ (d) $96 \text{ cm}^2$

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13

(a) $y = \frac{k}{x^2}$
When $x = 1$, $y = 100$:
$100 = \frac{k}{1^2} \Rightarrow k = 100$

Marks: [1] for correct value

(b) When $x = 3$:
$y = \frac{100}{3^2} = \frac{100}{9} = 11.11...$
$p = \frac{100}{9}$ or $11.1$ (3 s.f.)

Marks: [1] for correct value

(c) Points plotted correctly and smooth curve drawn through them for $1 \le x \le 5$.

Marks: [1] for correct plotting, [1] for smooth curve

(d) From graph, when $y = 15$, $x \approx 2.58$ (accept $2.5 - 2.6$)

Marks: [1] for reasonable estimate

(e) Line $y = 15$ drawn horizontally. Intersection with curve gives $x \approx 2.58$

Marks: [1] for correct use of line and reading

Answer: (a) $k = 100$ (b) $p = \frac{100}{9}$ or $11.1$ (c) Graph drawn (d) $x \approx 2.58$ (e) $x \approx 2.58$

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14

(a) $h(1) = 6$: $a(1)^2 + b(1) + c = 6 \Rightarrow a + b + c = 6$
$h(2) = 11$: $a(2)^2 + b(2) + c = 11 \Rightarrow 4a + 2b + c = 11$
$h(3) = 18$: $a(3)^2 + b(3) + c = 18 \Rightarrow 9a + 3b + c = 18$

Marks: [1] for first two equations, [1] for third equation

(b) Subtract first from second: $3a + b = 5$ \quad (1)
Subtract second from third: $5a + b = 7$ \quad (2)
Subtract (1) from (2): $2a = 2 \Rightarrow a = 1$
Substitute into (1): $3(1) + b = 5 \Rightarrow b = 2$
Substitute into first: $1 + 2 + c = 6 \Rightarrow c = 3$

Marks: [1] for eliminating $c$, [1] for finding $a$ and $b$, [1] for finding $c$

(c) $h(x) = x^2 + 2x + 3$
Complete the square: $h(x) = (x + 1)^2 + 2$
Minimum value $= 2$ (when $x = -1$)

Marks: [1] for completing square or using vertex formula, [1] for minimum value

Answer: (a) $a+b+c=6$, $4a+2b+c=11$, $9a+3b+c=18$ (b) $a=1$, $b=2$, $c=3$ (c) Minimum value $= 2$

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15

(a) Graph cuts $x$-axis at $(-4, 0)$ and $(2, 0)$.

Marks: [1] for both coordinates

(b) Line of symmetry: $x = -1$ (midpoint of $-4$ and $2$)

Marks: [1] for correct equation

(c) Vertex is at $(-1, -9)$, so minimum $y = -9$.
Line $y = k$ cuts graph at two points when $k > -9$.

Marks: [1] for correct range

(d) Translate 3 units right: replace $x$ with $(x - 3)$
$y = ((x - 3) - 2)((x - 3) + 4) = (x - 5)(x + 1)$
Or expanded: $y = x^2 - 4x - 5$

Marks: [1] for correct substitution, [1] for final equation

Answer: (a) $(-4, 0)$ and $(2, 0)$ (b) $x = -1$ (c) $k > -9$ (d) $y = (x - 5)(x + 1)$ or $y = x^2 - 4x - 5$

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Section C [15 marks]

16

(a) When $n = 10$, $C = 120$: $100a + 10b = 120$
When $n = 20$, $C = 400$: $400a + 20b = 400$

Marks: [1] for each equation

(b) Simplify:
$10a + b = 12$ \quad (1)
$20a + b = 20$ \quad (2)
Subtract (1) from (2): $10a = 8 \Rightarrow a = 0.8$
Substitute into (1): $10(0.8) + b = 12 \Rightarrow 8 + b = 12 \Rightarrow b = 4$

Marks: [1] for solving for $a$, [1] for solving for $b$

(c) $C = 0.8n^2 + 4n$
When $n = 15$: $C = 0.8(225) + 4(15) = 180 + 60 = 240$

Marks: [1] for correct cost

(d) Revenue $= 10n$
Profit when $10n > 0.8n^2 + 4n$
$0 > 0.8n^2 - 6n$
$0 > 0.8n(n - 7.5)$
$0 < n < 7.5$
Since $n$ is integer, minimum $n = 1$ (but check: for $n=1$, cost $= 4.8$, revenue $= 10$, profit $= 5.2$)
Wait: $0.8n^2 + 4n < 10n \Rightarrow 0.8n^2 - 6n < 0 \Rightarrow n(0.8n - 6) < 0$
$0 < n < 7.5$
So for $n = 1, 2, ..., 7$, profit is made. Minimum $n = 1$.

Marks: [1] for setting up inequality, [1] for solving inequality, [1] for correct minimum integer

Answer: (a) $100a+10b=120$, $400a+20b=400$ (b) $a=0.8$, $b=4$ (c) \240$(d)$1$ item

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17

(a) Outer dimensions: $(12 + 2x)$ by $(8 + 2x)$
Total area $= (12 + 2x)(8 + 2x) = 192$
$96 + 24x + 16x + 4x^2 = 192$
$4x^2 + 40x + 96 = 192$
$4x^2 + 40x - 96 = 0$
Divide by 4: $x^2 + 10x - 24 = 0$ (shown)

Marks: [1] for outer dimensions, [1] for area equation, [1] for correct simplification

(b) $x^2 + 10x - 24 = 0$
$(x + 12)(x - 2) = 0$
$x = 2$ or $x = -12$
Width must be positive, so $x = 2$ m

Marks: [1] for solving, [1] for rejecting negative and stating answer

(c) Outer dimensions: $12 + 2(2) = 16$ m, $8 + 2(2) = 12$ m
Perimeter $= 2(16 + 12) = 56$ m

Marks: [1] for outer dimensions, [1] for perimeter

Answer: (a) Shown (b) $2$ m (c) $56$ m

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18

(a) $h = -5(1)^2 + 20(1) + 2 = -5 + 20 + 2 = 17$ m

Marks: [1] for correct answer

(b) $h = -5t^2 + 20t + 2$
Complete the square: $h = -5(t^2 - 4t) + 2 = -5[(t - 2)^2 - 4] + 2 = -5(t - 2)^2 + 20 + 2 = -5(t - 2)^2 + 22$
Maximum height $= 22$ m (at $t = 2$ s)

Marks: [1] for completing square or vertex formula, [1] for maximum height, [1] for time (optional)

(c) Ball hits ground when $h = 0$:
$-5t^2 + 20t + 2 = 0$
$5t^2 - 20t - 2 = 0$
$t = \frac{20 \pm \sqrt{400 + 40}}{10} = \frac{20 \pm \sqrt{440}}{10} = \frac{20 \pm 20.976...}{10}$
$t = 4.10$ s (reject negative)

Marks: [1] for setting $h=0$, [1] for quadratic formula, [1] for correct positive root to 2 d.p.

Answer: (a) $17$ m (b) $22$ m (c) $4.10$ s

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19

(a) $f(x) = x^2 - 6x + 11 = (x^2 - 6x + 9) + 2 = (x - 3)^2 + 2$

Marks: [1] for completing square, [1] for correct form

(b) Minimum value $= 2$ at $x = 3$

Marks: [1] for both

(c) $g(x) = f(x) + 4$
Translation of $4$ units upwards (in the positive $y$-direction).

Marks: [1] for "translation", [1] for "4 units upwards"

(d) $h(x) = f(x + 2) = (x + 2 - 3)^2 + 2 = (x - 1)^2 + 2$
Minimum value $= 2$

Marks: [1] for correct minimum

Answer: (a) $(x - 3)^2 + 2$ (b) Min $= 2$ at $x = 3$ (c) Translation 4 units upwards (d) $2$

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20

(a) Intersection: $x^2 - 6x + 5 = 2x - 3$
$x^2 - 8x + 8 = 0$
$x = \frac{8 \pm \sqrt{64 - 32}}{2} = \frac{8 \pm \sqrt{32}}{2} = \frac{8 \pm 4\sqrt{2}}{2} = 4 \pm 2\sqrt{2}$

Marks: [1] for equating, [1] for correct quadratic, [1] for solving

(b) $x^2 - 6x + 5 > 2x - 3$
$x^2 - 8x + 8 > 0$
Roots: $4 - 2\sqrt{2}$ and $4 + 2\sqrt{2}$
Parabola opens upwards, so inequality holds for $x < 4 - 2\sqrt{2}$ or $x > 4 + 2\sqrt{2}$

Marks: [1] for correct inequality setup, [1] for correct solution set

Answer: (a) $x = 4 \pm 2\sqrt{2}$ (b) $x < 4 - 2\sqrt{2}$ or $x > 4 + 2\sqrt{2}$

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END OF ANSWER KEY