Secondary 4 Additional Mathematics Practice Paper 3

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

TuitionGoWhere Practice Paper (AI)

Field	Details
Subject:	Additional Mathematics
Level:	Secondary 4
Paper:	Practice Paper — Graphs & Coordinate Geometry
Version:	3 of 5
Duration:	1 hour 30 minutes
Total Marks:	80

Name: _________________________ Class: _________________________ Date: _________________________

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Instructions to Candidates

1. This paper consists of 20 questions covering the topic of Graphs and Coordinate Geometry.
2. Answer all questions in the spaces provided.
3. Marks for each question are indicated in brackets [ ].
4. You are reminded of the need for clear presentation in your answers.
5. Unless otherwise stated, give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees.
6. Solutions by accurate drawing will not be accepted.

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Section A: Straight Lines and Linear Relations (Questions 1–6)

[Total: 24 marks]

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1. The points $A(2, 5)$ and $B(8, -3)$ lie on a straight line.

(a) Find the gradient of the line $AB$. [2]

(b) Find the equation of the line $AB$, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [2]

(c) The line $AB$ meets the $x$-axis at point $C$. Find the coordinates of $C$. [2]

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2. A line $L_1$ passes through the point $P(3, -1)$ and is perpendicular to the line $2x - 5y + 10 = 0$.

(a) Find the gradient of $L_1$. [2]

(b) Hence find the equation of $L_1$ in the form $y = mx + c$. [2]

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3. The line $L_2$ has equation $3x + 4y = 24$. The line $L_3$ is parallel to $L_2$ and passes through the point $Q(-2, 5)$.

(a) Write down the gradient of $L_3$. [1]

(b) Find the equation of $L_3$. [2]

(c) Find the perpendicular distance from the origin to the line $L_2$. [3]

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4. The points $D(1, 2)$, $E(5, 6)$ and $F(9, 2)$ are given.

(a) Show that $DE = EF$. [2]

(b) Find the coordinates of the midpoint of $DF$. [2]

(c) Hence, or otherwise, determine the area of triangle $DEF$. [2]

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5. The straight line $y = 2x + k$ intersects the curve $y = x^2 - 3x + 1$ at two distinct points.

Find the set of possible values of $k$. [4]

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6. The variables $x$ and $y$ are related by the equation $y = ab^x$, where $a$ and $b$ are constants. The table below shows experimental values of $x$ and $y$.

$x$	1	2	3	4	5
$y$	6.0	10.8	19.4	35.0	63.0

(a) Explain how a straight line graph may be drawn to represent this data, stating clearly what should be plotted on each axis. [2]

(b) Using the data, estimate the values of $a$ and $b$. [4]

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Section B: Circles (Questions 7–12)

[Total: 26 marks]

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7. A circle $C_1$ has equation $x^2 + y^2 - 6x + 4y - 12 = 0$.

(a) Express the equation of $C_1$ in the form $(x - a)^2 + (y - b)^2 = r^2$, stating the coordinates of the centre and the radius. [3]

(b) Determine whether the point $P(5, 1)$ lies inside, on, or outside the circle $C_1$. [2]

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8. A circle passes through the points $A(2, 1)$ and $B(8, 1)$. The centre of the circle lies on the line $y = x - 2$.

Find the equation of the circle in the form $(x - a)^2 + (y - b)^2 = r^2$. [5]

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9. A circle $C_2$ has its centre at the point $(4, -3)$ and touches the $x$-axis.

(a) Write down the radius of $C_2$. [1]

(b) Find the equation of $C_2$. [2]

(c) Find the length of the tangent from the point $T(10, 5)$ to the circle $C_2$. [3]

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10. The circle $C_3$ has equation $x^2 + y^2 - 4x + 6y - 3 = 0$.

(a) Find the centre and radius of $C_3$. [3]

(b) The line $y = 2x + c$ is a tangent to $C_3$. Find the possible values of $c$. [4]

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11. A circle has diameter $PQ$ where $P$ is $(1, -2)$ and $Q$ is $(7, 4)$.

(a) Find the coordinates of the centre of the circle. [1]

(b) Find the radius of the circle, leaving your answer in surd form. [2]

(c) Write down the equation of the circle. [2]

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12. Two circles $C_4$ and $C_5$ have equations: $C_4: (x - 2)^2 + (y + 1)^2 = 9$ $C_5: (x - 8)^2 + (y + 1)^2 = r^2$

The circles touch externally.

(a) State the centre and radius of $C_4$. [1]

(b) Find the value of $r$. [2]

(c) Write down the coordinates of the point where the circles touch. [2]

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Section C: Curves, Intersections, and Applications (Questions 13–20)

[Total: 30 marks]

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13. The curve $y = x^3 - 6x^2 + 9x + 2$ has two stationary points.

(a) Find $\frac{dy}{dx}$. [1]

(b) Find the coordinates of the two stationary points. [3]

(c) Determine the nature of each stationary point. [3]

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14. The curve $y = \frac{4}{x} + x$ is defined for $x > 0$.

(a) Find $\frac{dy}{dx}$. [2]

(b) Find the coordinates of the stationary point on the curve. [2]

(c) Determine whether this stationary point is a maximum or a minimum. [2]

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15. The variables $x$ and $y$ are related by the equation $y = px^q$, where $p$ and $q$ are constants. The table below shows values of $x$ and $y$.

$x$	2	4	6	8	10
$y$	5.6	22.6	50.9	90.5	141.4

(a) Using a suitable transformation, explain how a straight line graph can be drawn to represent this data. State clearly what should be plotted on each axis. [2]

(b) Plot the transformed data and use your graph to estimate the values of $p$ and $q$. [4]

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16. The line $y = mx + 2$ is a tangent to the curve $y = x^2 + 3x + 1$.

(a) Form a quadratic equation in $x$ by eliminating $y$. [2]

(b) Hence, or otherwise, find the possible values of $m$. [3]

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17. A curve has equation $y = \frac{2x + 1}{x - 1}$, where $x \neq 1$.

(a) Find the coordinates of the points where the curve crosses the coordinate axes. [2]

(b) Show that the curve has no stationary points. [3]

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18. The curve $y = x^2 - 4x + 7$ and the line $y = 2x + k$ intersect at two distinct points.

Find the range of values of $k$. [4]

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19. The point $A(3, 1)$ lies on the curve $y = \frac{1}{2}x^2 - 2x + \frac{7}{2}$.

(a) Find the equation of the tangent to the curve at $A$. [3]

(b) Find the equation of the normal to the curve at $A$. [2]

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20. A particle moves along a straight line such that its displacement, $s$ metres, from a fixed point $O$ at time $t$ seconds is given by $s = t^3 - 6t^2 + 9t + 4$, for $t \geq 0$.

(a) Find expressions for the velocity, $v$, and acceleration, $a$, of the particle at time $t$. [2]

(b) Find the times when the particle is instantaneously at rest. [2]

(c) Find the acceleration of the particle when it is instantaneously at rest. [2]

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— END OF PAPER —

Check your work carefully. Total marks: 80

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TuitionGoWhere Practice Paper - Additional Mathematics Secondary 4

Answer Key and Marking Scheme — Version 3

Paper: Graphs & Coordinate Geometry Total Marks: 80

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Section A: Straight Lines and Linear Relations (24 marks)

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1. (a) Gradient of $AB$: $m = \frac{-3 - 5}{8 - 2} = \frac{-8}{6} = -\frac{4}{3}$ [M1] Correct substitution into gradient formula [A1] $m = -\frac{4}{3}$

(b) Using point $A(2, 5)$: $y - 5 = -\frac{4}{3}(x - 2)$ $3y - 15 = -4x + 8$ $4x + 3y - 23 = 0$ [M1] Correct use of point-gradient form [A1] $4x + 3y - 23 = 0$

(c) At $x$-axis, $y = 0$: $4x + 3(0) - 23 = 0 \implies 4x = 23 \implies x = \frac{23}{4}$ Coordinates of $C$ are $\left(\frac{23}{4}, 0\right)$ or $(5.75, 0)$. [M1] Setting $y = 0$ and solving [A1] $\left(\frac{23}{4}, 0\right)$

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2. (a) Line $2x - 5y + 10 = 0$: $5y = 2x + 10 \implies y = \frac{2}{5}x + 2$ Gradient of given line is $\frac{2}{5}$. For perpendicular lines, $m_1 \cdot m_2 = -1$: $m_{L_1} = -\frac{5}{2}$ [M1] Finding gradient of given line and using perpendicular condition [A1] $m_{L_1} = -\frac{5}{2}$

(b) Using $P(3, -1)$: $y - (-1) = -\frac{5}{2}(x - 3)$ $y + 1 = -\frac{5}{2}x + \frac{15}{2}$ $y = -\frac{5}{2}x + \frac{13}{2}$ [M1] Correct substitution [A1] $y = -\frac{5}{2}x + \frac{13}{2}$

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3. (a) $3x + 4y = 24 \implies 4y = -3x + 24 \implies y = -\frac{3}{4}x + 6$ Gradient of $L_2$ is $-\frac{3}{4}$. Since $L_3 \parallel L_2$, gradient of $L_3 = -\frac{3}{4}$. [A1] $-\frac{3}{4}$

(b) Using $Q(-2, 5)$: $y - 5 = -\frac{3}{4}(x - (-2))$ $y - 5 = -\frac{3}{4}(x + 2)$ $4y - 20 = -3x - 6$ $3x + 4y - 14 = 0$ [M1] Correct substitution [A1] $3x + 4y - 14 = 0$ or $y = -\frac{3}{4}x + \frac{7}{2}$

(c) Perpendicular distance from $(0, 0)$ to $3x + 4y - 24 = 0$: $d = \frac{|3(0) + 4(0) - 24|}{\sqrt{3^2 + 4^2}} = \frac{24}{5} = 4.8$ [M1] Correct formula [M1] Correct substitution [A1] $4.8$ units

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4. (a) $DE = \sqrt{(5 - 1)^2 + (6 - 2)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ $EF = \sqrt{(9 - 5)^2 + (2 - 6)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ Therefore $DE = EF$. [M1] Correct distance calculations [A1] Both equal $4\sqrt{2}$, shown

(b) Midpoint of $DF$: $\left(\frac{1 + 9}{2}, \frac{2 + 2}{2}\right) = (5, 2)$ [M1] Correct midpoint formula [A1] $(5, 2)$

(c) Triangle $DEF$ is isosceles with $DE = EF$. The midpoint of $DF$ is $(5, 2)$, which is point $M$. Height from $E(5, 6)$ to base $DF$: $6 - 2 = 4$ units. Length of base $DF = 9 - 1 = 8$ units. Area $= \frac{1}{2} \times 8 \times 4 = 16$ square units. [M1] Identifying height and base [A1] $16$ square units

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5. Substitute $y = 2x + k$ into $y = x^2 - 3x + 1$: $2x + k = x^2 - 3x + 1$ $x^2 - 5x + (1 - k) = 0$ For two distinct intersection points, discriminant $\Delta > 0$: $\Delta = (-5)^2 - 4(1)(1 - k) = 25 - 4 + 4k = 21 + 4k$ $21 + 4k > 0 \implies 4k > -21 \implies k > -\frac{21}{4}$ [M1] Substituting and forming quadratic [M1] Computing discriminant [M1] Setting $\Delta > 0$ [A1] $k > -\frac{21}{4}$ or $k > -5.25$

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6. (a) Taking logarithms (base 10 or natural): $y = ab^x \implies \log y = \log a + x \log b$ Plot $\log y$ on the vertical axis against $x$ on the horizontal axis. The graph should be a straight line with gradient $\log b$ and vertical intercept $\log a$. [A1] Correct transformation stated [A1] Correct axes identified

(b) Using $\log y = \log a + x \log b$: Let $Y = \log y$. Compute $\log y$ values:

$x$	1	2	3	4	5
$\log y$	$\log 6.0 \approx 0.778$	$\log 10.8 \approx 1.033$	$\log 19.4 \approx 1.288$	$\log 35.0 \approx 1.544$	$\log 63.0 \approx 1.799$

Gradient $\log b \approx \frac{1.799 - 0.778}{5 - 1} = \frac{1.021}{4} \approx 0.2553$ $b \approx 10^{0.2553} \approx 1.80$

Intercept $\log a \approx 0.778 - 0.2553(1) \approx 0.523$ $a \approx 10^{0.523} \approx 3.33$

[M1] Computing $\log y$ values [M1] Finding gradient [M1] Finding intercept [A1] $a \approx 3.33$, $b \approx 1.80$ (accept values within reasonable range)

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Section B: Circles (26 marks)

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7. (a) $x^2 + y^2 - 6x + 4y - 12 = 0$ $(x^2 - 6x) + (y^2 + 4y) = 12$ $(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$ $(x - 3)^2 + (y + 2)^2 = 25$ Centre: $(3, -2)$, Radius: $5$ [M1] Completing the square for $x$ terms [M1] Completing the square for $y$ terms [A1] $(x - 3)^2 + (y + 2)^2 = 25$, centre $(3, -2)$, radius $5$

(b) Distance from $P(5, 1)$ to centre $(3, -2)$: $d = \sqrt{(5 - 3)^2 + (1 - (-2))^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$ Since $\sqrt{13} < 5$, point $P$ lies inside the circle. [M1] Computing distance [A1] Inside, with correct reasoning

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8. Let centre be $(a, a - 2)$ (since centre lies on $y = x - 2$). Distance from centre to $A(2, 1)$ equals distance to $B(8, 1)$: $(a - 2)^2 + (a - 2 - 1)^2 = (a - 8)^2 + (a - 2 - 1)^2$ $(a - 2)^2 = (a - 8)^2$ $a^2 - 4a + 4 = a^2 - 16a + 64$ $12a = 60 \implies a = 5$ Centre: $(5, 3)$ Radius: $\sqrt{(5 - 2)^2 + (3 - 1)^2} = \sqrt{9 + 4} = \sqrt{13}$ Equation: $(x - 5)^2 + (y - 3)^2 = 13$ [M1] Letting centre be $(a, a - 2)$ [M1] Equating distances [M1] Solving for $a$ [M1] Finding radius [A1] $(x - 5)^2 + (y - 3)^2 = 13$

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9. (a) Centre $(4, -3)$. Touches $x$-axis, so distance from centre to $x$-axis equals radius. $r = |-3| = 3$ [A1] $r = 3$

(b) $(x - 4)^2 + (y + 3)^2 = 9$ [M1] Correct form [A1] $(x - 4)^2 + (y + 3)^2 = 9$

(c) Length of tangent from $T(10, 5)$ to circle: Distance $CT = \sqrt{(10 - 4)^2 + (5 - (-3))^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ Length of tangent $= \sqrt{CT^2 - r^2} = \sqrt{100 - 9} = \sqrt{91}$ [M1] Finding distance from $T$ to centre [M1] Using tangent length formula [A1] $\sqrt{91}$ units

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10. (a) $x^2 + y^2 - 4x + 6y - 3 = 0$ $(x^2 - 4x) + (y^2 + 6y) = 3$ $(x - 2)^2 - 4 + (y + 3)^2 - 9 = 3$ $(x - 2)^2 + (y + 3)^2 = 16$ Centre: $(2, -3)$, Radius: $4$ [M1] Completing square for $x$ [M1] Completing square for $y$ [A1] Centre $(2, -3)$, radius $4$

(b) Substitute $y = 2x + c$ into circle equation: $(x - 2)^2 + (2x + c + 3)^2 = 16$ $x^2 - 4x + 4 + 4x^2 + 4x(c + 3) + (c + 3)^2 = 16$ $5x^2 + (-4 + 4c + 12)x + 4 + (c + 3)^2 - 16 = 0$ $5x^2 + (4c + 8)x + (c^2 + 6c + 9 - 12) = 0$ $5x^2 + (4c + 8)x + (c^2 + 6c - 3) = 0$

For tangency, discriminant $= 0$: $(4c + 8)^2 - 4(5)(c^2 + 6c - 3) = 0$ $16c^2 + 64c + 64 - 20c^2 - 120c + 60 = 0$ $-4c^2 - 56c + 124 = 0$ $c^2 + 14c - 31 = 0$ $c = \frac{-14 \pm \sqrt{196 + 124}}{2} = \frac{-14 \pm \sqrt{320}}{2} = \frac{-14 \pm 8\sqrt{5}}{2} = -7 \pm 4\sqrt{5}$ [M1] Substituting line into circle [M1] Forming quadratic and computing discriminant [M1] Setting $\Delta = 0$ and solving [A1] $c = -7 \pm 4\sqrt{5}$

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11. (a) Centre is midpoint of $PQ$: $\left(\frac{1 + 7}{2}, \frac{-2 + 4}{2}\right) = (4, 1)$ [A1] $(4, 1)$

(b) Radius $= \frac{1}{2}PQ = \frac{1}{2}\sqrt{(7 - 1)^2 + (4 - (-2))^2} = \frac{1}{2}\sqrt{36 + 36} = \frac{1}{2}\sqrt{72} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$ [M1] Finding length of $PQ$ [A1] $3\sqrt{2}$

(c) $(x - 4)^2 + (y - 1)^2 = (3\sqrt{2})^2 = 18$ [M1] Correct form [A1] $(x - 4)^2 + (y - 1)^2 = 18$

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12. (a) $C_4$: Centre $(2, -1)$, radius $3$ [A1] Centre $(2, -1)$, $r = 3$

(b) $C_5$: Centre $(8, -1)$, radius $r$ Distance between centres $= 8 - 2 = 6$ For external tangency: $6 = 3 + r \implies r = 3$ [M1] Using external tangency condition [A1] $r = 3$

(c) The circles touch on the line joining centres. Since centres are $(2, -1)$ and $(8, -1)$, the point of contact divides the segment in ratio $3:3 = 1:1$. Point of contact: $(5, -1)$ [M1] Correct reasoning [A1] $(5, -1)$

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Section C: Curves, Intersections, and Applications (30 marks)

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13. (a) $\frac{dy}{dx} = 3x^2 - 12x + 9$ [A1] $3x^2 - 12x + 9$

(b) $\frac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0 \implies x^2 - 4x + 3 = 0 \implies (x - 1)(x - 3) = 0$ $x = 1$ or $x = 3$ At $x = 1$: $y = 1 - 6 + 9 + 2 = 6$ → $(1, 6)$ At $x = 3$: $y = 27 - 54 + 27 + 2 = 2$ → $(3, 2)$ [M1] Setting $\frac{dy}{dx} = 0$ [M1] Solving quadratic [A1] $(1, 6)$ and $(3, 2)$

(c) $\frac{d^2y}{dx^2} = 6x - 12$ At $x = 1$: $\frac{d^2y}{dx^2} = 6 - 12 = -6 < 0$ → maximum point $(1, 6)$ At $x = 3$: $\frac{d^2y}{dx^2} = 18 - 12 = 6 > 0$ → minimum point $(3, 2)$ [M1] Finding second derivative [M1] Evaluating at each stationary point [A1] $(1, 6)$ maximum, $(3, 2)$ minimum

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14. (a) $y = 4x^{-1} + x$ $\frac{dy}{dx} = -4x^{-2} + 1 = 1 - \frac{4}{x^2}$ [M1] Correct differentiation of $4x^{-1}$ [A1] $1 - \frac{4}{x^2}$

(b) $\frac{dy}{dx} = 0$: $1 - \frac{4}{x^2} = 0 \implies \frac{4}{x^2} = 1 \implies x^2 = 4 \implies x = 2$ (since $x > 0$) At $x = 2$: $y = \frac{4}{2} + 2 = 4$ Stationary point: $(2, 4)$ [M1] Solving $\frac{dy}{dx} = 0$ [A1] $(2, 4)$

(c) $\frac{d^2y}{dx^2} = 8x^{-3} = \frac{8}{x^3}$ At $x = 2$: $\frac{d^2y}{dx^2} = \frac{8}{8} = 1 > 0$ Therefore $(2, 4)$ is a minimum point. [M1] Finding second derivative [A1] Minimum, with correct reasoning

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15. (a) $y = px^q \implies \log y = \log p + q \log x$ Plot $\log y$ on the vertical axis against $\log x$ on the horizontal axis. The graph will be a straight line with gradient $q$ and vertical intercept $\log p$. [A1] Correct transformation [A1] Correct axes identified

(b) Compute $\log x$ and $\log y$:

$x$	$\log x$	$y$	$\log y$
2	0.301	5.6	0.748
4	0.602	22.6	1.354
6	0.778	50.9	1.707
8	0.903	90.5	1.957
10	1.000	141.4	2.150

Gradient $q \approx \frac{2.150 - 0.748}{1.000 - 0.301} = \frac{1.402}{0.699} \approx 2.01 \approx 2$

Intercept $\log p \approx 0.748 - 2(0.301) = 0.748 - 0.602 = 0.146$ $p \approx 10^{0.146} \approx 1.40$

[M1] Computing log values [M1] Finding gradient [M1] Finding intercept [A1] $p \approx 1.40$, $q \approx 2$

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16. (a) Substitute $y = mx + 2$ into $y = x^2 + 3x + 1$: $mx + 2 = x^2 + 3x + 1$ $x^2 + (3 - m)x - 1 = 0$ [M1] Correct substitution [A1] $x^2 + (3 - m)x - 1 = 0$

(b) For tangency, discriminant $= 0$: $(3 - m)^2 - 4(1)(-1) = 0$ $(3 - m)^2 + 4 = 0$ $(3 - m)^2 = -4$ No real solutions. Therefore there is no real value of $m$ for which the line is a tangent. [M1] Computing discriminant [M1] Setting $\Delta = 0$ [A1] No real values of $m$ (or equivalent conclusion)

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17. (a) $y = \frac{2x + 1}{x - 1}$ Crosses $y$-axis ($x = 0$): $y = \frac{1}{-1} = -1$ → $(0, -1)$ Crosses $x$-axis ($y = 0$): $2x + 1 = 0 \implies x = -\frac{1}{2}$ → $\left(-\frac{1}{2}, 0\right)$ [A1] $(0, -1)$ [A1] $\left(-\frac{1}{2}, 0\right)$

(b) $\frac{dy}{dx} = \frac{(x - 1)(2) - (2x + 1)(1)}{(x - 1)^2} = \frac{2x - 2 - 2x - 1}{(x - 1)^2} = \frac{-3}{(x - 1)^2}$ Since $(x - 1)^2 > 0$ for all $x \neq 1$, $\frac{dy}{dx} = -\frac{3}{(x - 1)^2} < 0$ for all $x \neq 1$. Therefore $\frac{dy}{dx} \neq 0$ for any $x$, so the curve has no stationary points. [M1] Correct differentiation using quotient rule [M1] Simplifying correctly [A1] Concluding no stationary points with valid reasoning

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18. Substitute $y = 2x + k$ into $y = x^2 - 4x + 7$: $2x + k = x^2 - 4x + 7$ $x^2 - 6x + (7 - k) = 0$ For two distinct intersection points, $\Delta > 0$: $(-6)^2 - 4(1)(7 - k) > 0$ $36 - 28 + 4k > 0$ $4k > -8$ $k > -2$ [M1] Substituting and forming quadratic [M1] Computing discriminant [M1] Setting $\Delta > 0$ [A1] $k > -2$

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19. (a) $y = \frac{1}{2}x^2 - 2x + \frac{7}{2}$ $\frac{dy}{dx} = x - 2$ At $A(3, 1)$: $\frac{dy}{dx} = 3 - 2 = 1$ Equation of tangent: $y - 1 = 1(x - 3) \implies y = x - 2$ [M1] Finding derivative [M1] Evaluating gradient at $A$ [A1] $y = x - 2$

(b) Gradient of normal $= -1$ (negative reciprocal of tangent gradient) Equation of normal: $y - 1 = -1(x - 3) \implies y = -x + 4$ [M1] Finding normal gradient [A1] $y = -x + 4$

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20. (a) $s = t^3 - 6t^2 + 9t + 4$ $v = \frac{ds}{dt} = 3t^2 - 12t + 9$ $a = \frac{dv}{dt} = 6t - 12$ [A1] $v = 3t^2 - 12t + 9$ [A1] $a = 6t - 12$

(b) Instantaneously at rest when $v = 0$: $3t^2 - 12t + 9 = 0 \implies t^2 - 4t + 3 = 0 \implies (t - 1)(t - 3) = 0$ $t = 1$ or $t = 3$ [M1] Setting $v = 0$ and solving [A1] $t = 1$ and $t = 3$

(c) At $t = 1$: $a = 6(1) - 12 = -6$ m/s² At $t = 3$: $a = 6(3) - 12 = 6$ m/s² [M1] Substituting $t$ values into acceleration [A1] $-6$ m/s² at $t = 1$; $6$ m/s² at $t = 3$

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— END OF ANSWER KEY —