O Level Additional Mathematics Practice Paper 2

TuitionGoWhere Practice Paper - Additional Mathematics O-Level

Answer Key and Marking Scheme

Paper: Practice Paper (Graphs & Coordinate Geometry)
Version: 2 of 5
Total Marks: 60

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Section A: Straight Lines and Basic Coordinate Geometry (20 marks)

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Question 1

(a) Gradient of $AB$: $m = \frac{7 - (-1)}{8 - 2} = \frac{8}{6} = \frac{4}{3}$ [M1] Correct substitution into gradient formula.
[A1] $\frac{4}{3}$ (1 mark)

(b) Using point $A(2, -1)$ and gradient $\frac{4}{3}$: $y - (-1) = \frac{4}{3}(x - 2)$ $y + 1 = \frac{4}{3}x - \frac{8}{3}$ Multiply by 3: $3y + 3 = 4x - 8$ $4x - 3y - 11 = 0$ [M1] Correct use of point-gradient form.
[A1] Correct equation $4x - 3y - 11 = 0$ (2 marks)

(c) Midpoint of $AB$: $\left(\frac{2 + 8}{2}, \frac{-1 + 7}{2}\right) = (5, 3)$ [A1] $(5, 3)$ (1 mark)

(d) Perpendicular bisector:

• Passes through midpoint $(5, 3)$
• Gradient perpendicular to $AB$: $m_\perp = -\frac{3}{4}$ $y - 3 = -\frac{3}{4}(x - 5)$ $4y - 12 = -3x + 15$ $3x + 4y - 27 = 0$ [M1] Correct perpendicular gradient.
  [M1] Correct use of midpoint.
  [A1] $3x + 4y - 27 = 0$ (3 marks)

Total: 7 marks

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Question 2

(a) $L_1: 3x - 4y + 12 = 0$ Rearrange: $4y = 3x + 12 \implies y = \frac{3}{4}x + 3$ Gradient $= \frac{3}{4}$ [A1] $\frac{3}{4}$ (1 mark)

(b) $L_2$ is parallel to $L_1$, so gradient $= \frac{3}{4}$. Passes through $P(5, -2)$: $y - (-2) = \frac{3}{4}(x - 5)$ $y + 2 = \frac{3}{4}x - \frac{15}{4}$ $y = \frac{3}{4}x - \frac{23}{4}$ [M1] Correct use of point-gradient form.
[A1] $y = \frac{3}{4}x - \frac{23}{4}$ (2 marks)

(c) At $x$-axis, $y = 0$: $0 = \frac{3}{4}x - \frac{23}{4}$ $\frac{3}{4}x = \frac{23}{4}$ $x = \frac{23}{3}$ Coordinates: $\left(\frac{23}{3}, 0\right)$ [A1] $\left(\frac{23}{3}, 0\right)$ (1 mark)

Total: 4 marks

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Question 3

(a) Gradient of $L$: $m = \frac{-2 - 4}{1 - (-3)} = \frac{-6}{4} = -\frac{3}{2}$ [A1] Correct working and $-\frac{3}{2}$ shown (1 mark)

(b) Midpoint of $CD$: $\left(\frac{-3 + 1}{2}, \frac{4 + (-2)}{2}\right) = (-1, 1)$ Gradient of $M$ (perpendicular to $L$): $m_M = \frac{2}{3}$ Equation of $M$: $y - 1 = \frac{2}{3}(x - (-1))$ $y - 1 = \frac{2}{3}(x + 1)$ $3y - 3 = 2x + 2$ $2x - 3y + 5 = 0$ [M1] Correct midpoint.
[M1] Correct perpendicular gradient.
[A1] $2x - 3y + 5 = 0$ (3 marks)

(c) Line $L$: gradient $-\frac{3}{2}$, passes through $C(-3, 4)$: $y - 4 = -\frac{3}{2}(x + 3)$ $2y - 8 = -3x - 9$ $3x + 2y + 1 = 0$ $x$-intercept ($y = 0$): $3x + 1 = 0 \implies x = -\frac{1}{3}$ $y$-intercept ($x = 0$): $2y + 1 = 0 \implies y = -\frac{1}{2}$ Area of triangle $= \frac{1}{2} \times \left|-\frac{1}{3}\right| \times \left|-\frac{1}{2}\right| = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{12}$ square units. [M1] Finding intercepts.
[A1] $\frac{1}{12}$ (2 marks)

Total: 6 marks

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Section B: Circles (20 marks)

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Question 4

(a) $x^2 + y^2 - 6x + 10y + 9 = 0$ Complete the square: $(x^2 - 6x) + (y^2 + 10y) = -9$ $(x - 3)^2 - 9 + (y + 5)^2 - 25 = -9$ $(x - 3)^2 + (y + 5)^2 = 25$ Centre: $(3, -5)$, Radius: $5$ [M1] Completing the square for $x$ terms.
[M1] Completing the square for $y$ terms.
[A1] $(x - 3)^2 + (y + 5)^2 = 25$, centre $(3, -5)$, radius $5$ (3 marks)

(b) Distance from $P(7, -2)$ to centre $(3, -5)$: $d = \sqrt{(7 - 3)^2 + (-2 - (-5))^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ Since $d = 5 = r$, point $P$ lies on the circle. [M1] Correct distance calculation.
[A1] Correct conclusion with justification (2 marks)

Total: 5 marks

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Question 5

(a) Radius $= QR = \sqrt{(1 - 4)^2 + (3 - (-1))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ [A1] $5$ (1 mark)

(b) Equation: $(x - 4)^2 + (y + 1)^2 = 25$ [A1] $(x - 4)^2 + (y + 1)^2 = 25$ (1 mark)

(c) Centre $Q(4, -1)$, point $R(1, 3)$. Gradient of radius $QR = \frac{3 - (-1)}{1 - 4} = \frac{4}{-3} = -\frac{4}{3}$ Gradient of tangent at $R$: $m_\perp = \frac{3}{4}$ Equation of tangent: $y - 3 = \frac{3}{4}(x - 1)$ $4y - 12 = 3x - 3$ $3x - 4y + 9 = 0$ [M1] Correct gradient of radius.
[M1] Correct perpendicular gradient.
[M1] Correct use of point-gradient form.
[A1] $3x - 4y + 9 = 0$ (4 marks)

Total: 6 marks

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Question 6

(a) Centre is midpoint of $AB$: $\left(\frac{2 + 8}{2}, \frac{1 + 9}{2}\right) = (5, 5)$ [A1] $(5, 5)$ (1 mark)

(b) Radius $= \frac{1}{2} \times AB = \frac{1}{2}\sqrt{(8 - 2)^2 + (9 - 1)^2} = \frac{1}{2}\sqrt{36 + 64} = \frac{1}{2}\sqrt{100} = 5$ [M1] Correct distance formula for diameter.
[A1] $5$ (2 marks)

(c) Centre $(5, 5)$, radius $5$: $(x - 5)^2 + (y - 5)^2 = 25$ Expand: $x^2 - 10x + 25 + y^2 - 10y + 25 = 25$ $x^2 + y^2 - 10x - 10y + 25 = 0$ [M1] Correct expansion.
[A1] $x^2 + y^2 - 10x - 10y + 25 = 0$ (2 marks)

Total: 5 marks

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Question 7

(a) $x^2 + y^2 - 4x + 2y - 20 = 0$ $(x^2 - 4x) + (y^2 + 2y) = 20$ $(x - 2)^2 - 4 + (y + 1)^2 - 1 = 20$ $(x - 2)^2 + (y + 1)^2 = 25$ Centre: $(2, -1)$, Radius: $5$ [M1] Completing the square.
[A1] Centre $(2, -1)$, radius $5$ (2 marks)

(b) Substitute $y = 2x + k$ into circle equation: $x^2 + (2x + k)^2 - 4x + 2(2x + k) - 20 = 0$ $x^2 + 4x^2 + 4kx + k^2 - 4x + 4x + 2k - 20 = 0$ $5x^2 + 4kx + k^2 + 2k - 20 = 0$ For tangency, discriminant $= 0$: $(4k)^2 - 4(5)(k^2 + 2k - 20) = 0$ $16k^2 - 20k^2 - 40k + 400 = 0$ $-4k^2 - 40k + 400 = 0$ $k^2 + 10k - 100 = 0$ $k = \frac{-10 \pm \sqrt{100 + 400}}{2} = \frac{-10 \pm \sqrt{500}}{2} = \frac{-10 \pm 10\sqrt{5}}{2} = -5 \pm 5\sqrt{5}$ [M1] Substituting line into circle equation.
[M1] Forming quadratic in $x$.
[M1] Setting discriminant to zero.
[A1] $k = -5 \pm 5\sqrt{5}$ (4 marks)

Total: 6 marks

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Section C: Coordinate Geometry Applications and Linear Law (20 marks)

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Question 8

(a) Intersection: $x^2 - 5x + 6 = 2x - 4$ $x^2 - 7x + 10 = 0$ $(x - 2)(x - 5) = 0$ $x = 2$ or $x = 5$ When $x = 2$: $y = 2(2) - 4 = 0$, point $(2, 0)$ When $x = 5$: $y = 2(5) - 4 = 6$, point $(5, 6)$ [M1] Equating and forming quadratic.
[M1] Solving quadratic.
[A1] $(2, 0)$ and $(5, 6)$ (3 marks)

(b) Length $= \sqrt{(5 - 2)^2 + (6 - 0)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$ [M1] Correct distance formula.
[A1] $3\sqrt{5}$ (2 marks)

Total: 5 marks

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Question 9

(a) Area of $\triangle PQR$ using shoelace formula: $\frac{1}{2}\left|(-2)(-1) + (4)(5) + (1)(3) - (3)(4) - (-1)(1) - (5)(-2)\right|$ $= \frac{1}{2}\left|2 + 20 + 3 - 12 + 1 + 10\right|$ $= \frac{1}{2}\left|24\right| = 12$ [M1] Correct application of shoelace formula.
[A1] $12$ square units (2 marks)

(b) Gradient of $QR = \frac{5 - (-1)}{1 - 4} = \frac{6}{-3} = -2$ Line through $P(-2, 3)$ parallel to $QR$: $y - 3 = -2(x + 2)$ $y - 3 = -2x - 4$ $y = -2x - 1$ [M1] Correct gradient of $QR$.
[A1] $y = -2x - 1$ (2 marks)

(c) Equation of $QR$: gradient $-2$, passes through $Q(4, -1)$: $y + 1 = -2(x - 4)$ $y + 1 = -2x + 8$ $2x + y - 7 = 0$ Perpendicular distance from $P(-2, 3)$ to $QR$: $d = \frac{|2(-2) + 1(3) - 7|}{\sqrt{2^2 + 1^2}} = \frac{|-4 + 3 - 7|}{\sqrt{5}} = \frac{|-8|}{\sqrt{5}} = \frac{8}{\sqrt{5}} = \frac{8\sqrt{5}}{5}$ [M1] Finding equation of $QR$.
[M1] Correct use of perpendicular distance formula.
[A1] $\frac{8\sqrt{5}}{5}$ (3 marks)

Total: 7 marks

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Question 10

(a) Taking logarithms of both sides: $\log y = \log(ax^n) = \log a + n\log x$ Plot $\log y$ (on vertical axis) against $\log x$ (on horizontal axis). The graph will be a straight line with gradient $n$ and vertical intercept $\log a$. [M1] Correct logarithmic transformation.
[A1] Clear statement of axes (2 marks)

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

$x$	$y$	$\log x$	$\log y$
2	4.8	0.301	0.681
3	16.2	0.477	1.210
5	75.0	0.699	1.875
8	307.2	0.903	2.487
12	1036.8	1.079	3.016

From the graph (best-fit line): Gradient $n = \frac{3.016 - 0.681}{1.079 - 0.301} = \frac{2.335}{0.778} \approx 3.0$ Vertical intercept $\log a \approx 0.08$, so $a \approx 10^{0.08} \approx 1.2$ [M1] Correct calculation of log values.
[M1] Plotting points and drawing best-fit line.
[M1] Correct method for finding gradient and intercept.
[A1] $n \approx 3$, $a \approx 1.2$ (4 marks)

(c) $y = 1.2 \times 10^3 = 1.2 \times 1000 = 1200$ [A1] $1200$ (1 mark)

Total: 7 marks

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Question 11

(a) $A(2, 5)$ lies on $y = \frac{k}{x^2} + 3$: $5 = \frac{k}{4} + 3 \implies \frac{k}{4} = 2 \implies k = 8$ [A1] $k = 8$ (1 mark)

(b) $y = 8x^{-2} + 3$ $\frac{dy}{dx} = -16x^{-3} = -\frac{16}{x^3}$ At $x = 2$: $\frac{dy}{dx} = -\frac{16}{8} = -2$ Tangent at $(2, 5)$: $y - 5 = -2(x - 2)$ $y - 5 = -2x + 4$ $y = -2x + 9$ [M1] Correct differentiation.
[M1] Substituting $x = 2$ for gradient.
[M1] Correct point-gradient form.
[A1] $y = -2x + 9$ (4 marks)

(c) At $x$-axis, $y = 0$: $0 = -2x + 9 \implies 2x = 9 \implies x = 4.5$ Point: $(4.5, 0)$ [A1] $(4.5, 0)$ (1 mark)

Total: 6 marks

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Question 12

(a) Gradient of $OP = \frac{5 - 2}{7 - 3} = \frac{3}{4}$ [A1] $\frac{3}{4}$ (1 mark)

(b) Tangent is perpendicular to radius, so gradient $= -\frac{4}{3}$ Equation of tangent at $P(7, 5)$: $y - 5 = -\frac{4}{3}(x - 7)$ $3y - 15 = -4x + 28$ $4x + 3y - 43 = 0$ [M1] Correct perpendicular gradient.
[M1] Correct use of point-gradient form.
[A1] $4x + 3y - 43 = 0$ (3 marks)

(c) $Q$ ($x$-intercept, $y = 0$): $4x - 43 = 0 \implies x = \frac{43}{4} = 10.75$ $R$ ($y$-intercept, $x = 0$): $3y - 43 = 0 \implies y = \frac{43}{3}$ Area of $\triangle OQR = \frac{1}{2} \times \frac{43}{4} \times \frac{43}{3} = \frac{1849}{24} \approx 77.0$ square units. [M1] Finding $Q$ and $R$.
[M1] Correct area formula.
[A1] $\frac{1849}{24}$ or $77.0$ (3 marks)

Total: 7 marks

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END OF ANSWER KEY

Total marks: 60