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

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Questions

TuitionGoWhere Practice Paper - Mathematics Secondary 2

TuitionGoWhere Secondary School (AI)

Subject: Mathematics
Level: Secondary 2 (G3)
Paper: SA2 Version 1
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

INSTRUCTIONS TO CANDIDATES

Write your name, class, and date in the spaces provided above.
Answer all questions.
Write your answers and working in the spaces provided.
Omission of essential working will result in loss of marks.
Calculators may be used where appropriate.
If the degree of accuracy is not specified, give answers to 3 significant figures.
The number of marks is given in brackets [ ] at the end of each question or part question.
The total number of marks for this paper is 60.

Section A [20 marks]

Answer all questions in this section.

1

$y$ is directly proportional to the square of $x$ . When $x = 3$ , $y = 36$ .

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

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

2

$P$ is inversely proportional to the cube root of $Q$ . When $Q = 8$ , $P = 12$ .

(a) Find an equation connecting $P$ and $Q$ . [2]

(b) Find the value of $Q$ when $P = 6$ . [1]

3

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

4

The function $g$ is defined as $g(x) = \frac{4x - 1}{3}$ for all real $x$ .

(a) Find $g(5)$ . [1]

(b) Find the value of $x$ such that $g(x) = 7$ . [2]

5

Solve the equation $3x^2 - 11x - 4 = 0$ . [3]

6

Solve the simultaneous equations:

\begin{cases} 3x + 2y = 13 \\ 5x - 4y = 3 \end{cases} $$ [3] --- ### 7 Factorise completely: $12x^2 - 27y^2$. [2] --- ### 8 Simplify the algebraic fraction:

\frac{x^2 - 9}{x^2 - 4x - 21}

--- ### 9 The diagram shows the graph of $y = x^2 - 4x - 5$ for $-2 \le x \le 6$. <image_placeholder> id: Q9-fig1 type: graph linked_question: Q9 description: Graph of quadratic function y = x^2 - 4x - 5 with x-axis from -2 to 6, y-axis from -10 to 10. Parabola opening upwards with vertex at (2, -9), x-intercepts at (-1, 0) and (5, 0), y-intercept at (0, -5). Grid lines at integer intervals. labels: x-axis, y-axis, vertex (2, -9), x-intercepts (-1, 0) and (5, 0), y-intercept (0, -5) values: x from -2 to 6, y from -10 to 10 must_show: Parabola shape, vertex, intercepts, grid lines, axis labels </image_placeholder> (a) Write down the coordinates of the minimum point of the graph. [1] (b) Write down the equation of the line of symmetry of the graph. [1] (c) Use the graph to find the solutions of $x^2 - 4x - 5 = 0$. [1] --- ### 10 The table below shows values of $x$ and the corresponding values of $y$ for the function $y = \frac{12}{x}$, $x \neq 0$. | $x$ | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | |-----|----|----|----|----|----|---|---|---|---|---| | $y$ | -2 | -3 | -4 | -6 | -12 | 12 | 6 | 4 | 3 | 2 | (a) On the grid, plot the points from the table and draw the graph of $y = \frac{12}{x}$ for $-6 \le x \le -1$ and $1 \le x \le 6$. [2] <image_placeholder> id: Q10-fig1 type: graph linked_question: Q10 description: Blank coordinate grid for reciprocal function y = 12/x. x-axis from -7 to 7, y-axis from -13 to 13. Grid lines at integer intervals. Two separate branches for negative and positive x. labels: x-axis, y-axis, grid lines values: x from -7 to 7, y from -13 to 13 must_show: Empty grid with labelled axes and scale </image_placeholder> (b) Write down the equations of the two asymptotes of the graph. [1] --- ## Section B [25 marks] Answer all questions in this section. ### 11 A rectangular piece of paper has length $(2x + 5)$ cm and width $(x - 2)$ cm. (a) Write down an expression, in terms of $x$, for the area of the paper. [1] (b) Given that the area of the paper is $72 \text{ cm}^2$, form an equation in $x$ and show that it reduces to $2x^2 + x - 82 = 0$. [2] (c) Solve the equation $2x^2 + x - 82 = 0$, giving your answers correct to 2 decimal places. [2] (d) Hence, find the perimeter of the paper. [2] --- ### 12 The diagram shows a right-angled triangle with sides $(x - 3)$ cm, $(x + 1)$ cm, and $(x + 5)$ cm, where $(x + 5)$ cm is the hypotenuse. <image_placeholder> id: Q12-fig1 type: diagram linked_question: Q12 description: Right-angled triangle with sides labelled (x-3) cm, (x+1) cm, and (x+5) cm as hypotenuse. Right angle between the two shorter sides. labels: Side lengths (x-3) cm, (x+1) cm, (x+5) cm, right angle symbol values: Algebraic expressions for side lengths must_show: Right-angled triangle with algebraic side labels and right angle marker </image_placeholder> (a) Form an equation in $x$ using Pythagoras' theorem. [1] (b) Show that the equation reduces to $x^2 - 14x - 15 = 0$. [2] (c) Solve this equation to find the value of $x$. [2] (d) Hence, find the area of the triangle. [2] --- ### 13 The variables $x$ and $y$ are connected by the equation $y = \frac{k}{x^2}$, where $k$ is a constant. The table below shows some values of $x$ and $y$. | $x$ | 1 | 2 | 3 | 4 | 5 | |-----|---|---|---|---|---| | $y$ | 100 | 25 | $p$ | 6.25 | 4 | (a) Find the value of $k$. [1] (b) Find the value of $p$. [1] (c) On the grid, plot the points and draw the graph of $y = \frac{100}{x^2}$ for $1 \le x \le 5$. [2] <image_placeholder> id: Q13-fig1 type: graph linked_question: Q13 description: Blank coordinate grid for y = 100/x^2. x-axis from 0 to 6, y-axis from 0 to 110. Grid lines at integer intervals. Curve decreasing in first quadrant only. labels: x-axis, y-axis, grid lines values: x from 0 to 6, y from 0 to 110 must_show: Empty grid with labelled axes and scale </image_placeholder> (d) Use your graph to estimate the value of $x$ when $y = 15$. [1] (e) The equation $y = \frac{100}{x^2}$ can be written as $x^2 = \frac{100}{y}$. On the same axes, draw the line $y = 15$ and use it to find the value of $x$ when $y = 15$. [1] --- ### 14 A function $h$ is defined by $h(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. The graph of $y = h(x)$ passes through the points $(1, 6)$, $(2, 11)$, and $(3, 18)$. (a) Form three equations in $a$, $b$, and $c$. [2] (b) Solve these equations to find the values of $a$, $b$, and $c$. [3] (c) Hence, find the minimum value of $h(x)$. [2] --- ### 15 The diagram shows the graph of $y = (x - 2)(x + 4)$ for $-5 \le x \le 5$. <image_placeholder> id: Q15-fig1 type: graph linked_question: Q15 description: Graph of quadratic function y = (x-2)(x+4) = x^2 + 2x - 8. x-axis from -5 to 5, y-axis from -10 to 10. Parabola opening upwards with vertex at (-1, -9), x-intercepts at (-4, 0) and (2, 0), y-intercept at (0, -8). labels: x-axis, y-axis, vertex (-1, -9), x-intercepts (-4, 0) and (2, 0), y-intercept (0, -8) values: x from -5 to 5, y from -10 to 10 must_show: Parabola shape, vertex, intercepts, grid lines, axis labels </image_placeholder> (a) Write down the coordinates of the points where the graph cuts the $x$-axis. [1] (b) Write down the equation of the line of symmetry. [1] (c) The line $y = k$ cuts the graph at two points. Find the range of values of $k$ for which this happens. [1] (d) The graph of $y = (x - 2)(x + 4)$ is translated 3 units to the right. Write down the equation of the new graph. [2] --- ## Section C [15 marks] Answer all questions in this section. ### 16 The cost $C$ dollars of producing $n$ items is given by the formula $C = an^2 + bn$, where $a$ and $b$ are constants. When 10 items are produced, the cost is $120. When 20 items are produced, the cost is $400. (a) Form two equations in $a$ and $b$. [2] (b) Solve these equations to find the values of $a$ and $b$. [2] (c) Find the cost of producing 15 items. [1] (d) The selling price of each item is $10. Find the minimum number of items that must be produced and sold to make a profit. [3] --- ### 17 The diagram shows a rectangular garden with a path of uniform width $x$ metres around it. The garden measures $12$ m by $8$ m. The total area of the garden and path is $192 \text{ m}^2$. <image_placeholder> id: Q17-fig1 type: diagram linked_question: Q17 description: Rectangle 12m by 8m with a uniform border of width x m around it. Outer rectangle dimensions (12+2x) by (8+2x). Shaded path area between inner and outer rectangles. labels: Inner rectangle 12m × 8m, path width x m, outer dimensions (12+2x) m × (8+2x) m values: 12 m, 8 m, x m, total area 192 m² must_show: Nested rectangles with dimensions labelled, path shaded </image_placeholder> (a) Show that $x$ satisfies the equation $x^2 + 10x - 24 = 0$. [3] (b) Solve this equation to find the width of the path. [2] (c) Find the perimeter of the outer edge of the path. [2] --- ### 18 A ball is thrown vertically upwards from a height of $2$ m above the ground. Its height $h$ metres above the ground after $t$ seconds is given by $h = -5t^2 + 20t + 2$. (a) Find the height of the ball after $1$ second. [1] (b) Find the maximum height reached by the ball. [3] (c) Find the time when the ball hits the ground, giving your answer correct to 2 decimal places. [3] --- ### 19 The function $f$ is defined by $f(x) = x^2 - 6x + 11$ for $x \in \mathbb{R}$. (a) Express $f(x)$ in the form $(x - a)^2 + b$, where $a$ and $b$ are constants. [2] (b) Write down the minimum value of $f(x)$ and the value of $x$ at which it occurs. [1] (c) The function $g$ is defined by $g(x) = f(x) + 4$ for $x \in \mathbb{R}$. Describe fully the transformation that maps the graph of $y = f(x)$ onto the graph of $y = g(x)$. [2] (d) The function $h$ is defined by $h(x) = f(x + 2)$ for $x \in \mathbb{R}$. Find the minimum value of $h(x)$. [1] --- ### 20 The diagram shows the graph of $y = x^2 - 6x + 5$ and the line $y = 2x - 3$. <image_placeholder> id: Q20-fig1 type: graph linked_question: Q20 description: Graph showing parabola y = x^2 - 6x + 5 (vertex at (3, -4), x-intercepts at (1,0) and (5,0), y-intercept at (0,5)) and straight line y = 2x - 3 (y-intercept at (0,-3), x-intercept at (1.5,0)). Two intersection points visible. labels: Parabola y = x^2 - 6x + 5, line y = 2x - 3, intersection points, axes values: x from -1 to 7, y from -5 to 10 must_show: Both curves clearly drawn with intersection points, axes labelled </image_placeholder> (a) Find the $x$-coordinates of the points of intersection of the curve and the line. [3] (b) Hence, solve the inequality $x^2 - 6x + 5 > 2x - 3$. [2] --- **END OF PAPER**

Answers

TuitionGoWhere Practice Paper - Mathematics Secondary 2

SA2 Version 1 - Answer Key and Marking Scheme

Total Marks: 60

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) $y = 100$

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) $Q = 64$

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: $21$

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 $5.5$

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 $x = 4$

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

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

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

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 $x = 5$

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 $x = 0$

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

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) $96 \text{ cm}^2$

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) $x \approx 2.6$

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 $= 2$

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 $y = x^2 - 4x - 5$

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

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

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

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 "

<stage3_exam_answers_md>

TuitionGoWhere Practice Paper - Mathematics Secondary 2

SA2 Version 1 - Answer Key and Marking Scheme

Total Marks: 60

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) $y = 100$

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) $Q = 64$

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: $21$

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 $5.5$

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 $x = 4$

6

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