O Level Additional Mathematics Practice Paper 1

TuitionGoWhere Practice Paper - Additional Mathematics O-Level (Marking Scheme)

Total Marks: 90

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Question 1 [6 marks]

(a) Find the coordinates of intersection points. [4 marks]

Solution: Set $2x - 5 = x^2 - 3x + 1$ $0 = x^2 - 5x + 6$ $0 = (x - 2)(x - 3)$ $x = 2$ or $x = 3$

When $x = 2$: $y = 2(2) - 5 = -1$ When $x = 3$: $y = 2(3) - 5 = 1$

Answer: $(2, -1)$ and $(3, 1)$

Marking:

• 1 mark: Setting equations equal
• 2 marks: Solving quadratic equation correctly
• 1 mark: Finding both y-coordinates

(b) Find distance between intersection points. [2 marks]

Solution: Distance $= \sqrt{(3-2)^2 + (1-(-1))^2} = \sqrt{1 + 4} = \sqrt{5}$

Marking:

• 1 mark: Correct distance formula
• 1 mark: Correct calculation

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Question 2 [7 marks]

(a) Find centre and radius. [4 marks]

Solution: $x^2 + y^2 - 8x + 6y - 11 = 0$ $(x^2 - 8x) + (y^2 + 6y) = 11$ $(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9$ $(x - 4)^2 + (y + 3)^2 = 36$

Centre: $(4, -3)$, Radius: $6$

Marking:

• 2 marks: Completing the square correctly
• 1 mark: Centre coordinates
• 1 mark: Radius

(b) Find equation of tangent at $P(7, 1)$. [3 marks]

Solution: Centre is $(4, -3)$, so gradient of radius $OP = \frac{1-(-3)}{7-4} = \frac{4}{3}$ Gradient of tangent $= -\frac{3}{4}$ (perpendicular to radius) Equation: $y - 1 = -\frac{3}{4}(x - 7)$ $y = -\frac{3}{4}x + \frac{25}{4}$ or $4y + 3x = 25$

Marking:

• 1 mark: Finding gradient of radius
• 1 mark: Using perpendicular gradient
• 1 mark: Correct equation

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Question 3 [8 marks]

(a) Express in partial fractions. [5 marks]

Solution: $\frac{3x^2 + 7x + 2}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2+1}$

$3x^2 + 7x + 2 = A(x^2 + 1) + (Bx + C)(x + 1)$

When $x = -1$: $3 - 7 + 2 = A(2) \Rightarrow A = -1$

Expanding: $3x^2 + 7x + 2 = Ax^2 + A + Bx^2 + Bx + Cx + C$ $3x^2 + 7x + 2 = (A + B)x^2 + (B + C)x + (A + C)$

Comparing coefficients: $x^2$: $A + B = 3 \Rightarrow B = 4$ $x^1$: $B + C = 7 \Rightarrow C = 3$ Constant: $A + C = 2$ ✓

Answer: $\frac{-1}{x+1} + \frac{4x + 3}{x^2+1}$

Marking:

• 1 mark: Correct partial fraction setup
• 2 marks: Finding A correctly
• 2 marks: Finding B and C correctly

(b) Hence find the integral. [3 marks]

Solution: $\int \frac{3x^2 + 7x + 2}{(x+1)(x^2+1)} dx = \int \left(\frac{-1}{x+1} + \frac{4x + 3}{x^2+1}\right) dx$

$= -\ln|x+1| + 2\ln(x^2+1) + 3\tan^{-1}x + c$

Marking:

• 1 mark: $-\ln|x+1|$
• 1 mark: $2\ln(x^2+1)$
• 1 mark: $3\tan^{-1}x + c$

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Question 4 [6 marks]

(a) Find equation of curve. [3 marks]

Solution: $y = \int (6x^2 - 4x + 3) dx = 2x^3 - 2x^2 + 3x + c$

Using $(1, 4)$: $4 = 2(1) - 2(1) + 3(1) + c = 3 + c$ So $c = 1$

Answer: $y = 2x^3 - 2x^2 + 3x + 1$

Marking:

• 2 marks: Correct integration
• 1 mark: Finding constant using given point

(b) Explain why no stationary points. [2 marks]

Solution: For stationary points, $\frac{dy}{dx} = 0$ $6x^2 - 4x + 3 = 0$ Discriminant $= (-4)^2 - 4(6)(3) = 16 - 72 = -56 < 0$ Since discriminant is negative, there are no real solutions.

Marking:

• 1 mark: Setting derivative equal to zero
• 1 mark: Correct explanation using discriminant

(c) State whether increasing or decreasing. [1 mark]

Solution: Since $6x^2 - 4x + 3 > 0$ for all real $x$ (as shown above), the function is always increasing.

Marking:

• 1 mark: Correct answer with reason

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Question 5 [9 marks]

(a) Find constants $a$, $b$, $c$, $d$. [6 marks]

Solution: $f(x) = ax^3 + bx^2 + cx + d$ $f'(x) = 3ax^2 + 2bx + c$

From conditions: $f(0) = 2 \Rightarrow d = 2$ $f'(0) = -3 \Rightarrow c = -3$ $f(1) = 0 \Rightarrow a + b - 3 + 2 = 0 \Rightarrow a + b = 1$ $f'(1) = 6 \Rightarrow 3a + 2b - 3 = 6 \Rightarrow 3a + 2b = 9$

Solving: From $a + b = 1$ and $3a + 2b = 9$ $3a + 2b = 9$ and $2a + 2b = 2$ Subtracting: $a = 7$, so $b = -6$

Answer: $a = 7$, $b = -6$, $c = -3$, $d = 2$

Marking:

• 1 mark each for finding $c$ and $d$
• 2 marks for setting up equations for $a$ and $b$
• 2 marks for solving correctly

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

Solution: $f'(x) = 21x^2 - 12x - 3 = 3(7x^2 - 4x - 1)$ $7x^2 - 4x - 1 = 0$ $x = \frac{4 \pm \sqrt{16 + 28}}{14} = \frac{4 \pm \sqrt{44}}{14} = \frac{2 \pm \sqrt{11}}{7}$

When $x = \frac{2 + \sqrt{11}}{7}$: Calculate $y$-coordinate When $x = \frac{2 - \sqrt{11}}{7}$: Calculate $y$-coordinate

Marking:

• 2 marks: Solving $f'(x) = 0$ correctly
• 1 mark: Finding y-coordinates (accept in terms of surds)

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Question 6 [8 marks]

(a) State amplitude and period. [2 marks]

Solution: Amplitude = 1 Period = $\frac{2\pi}{2} = \pi$

Marking:

• 1 mark each for amplitude and period

(b) Find x-intercepts. [3 marks]

Solution: $\sin(2x + \frac{\pi}{3}) = 0$ $2x + \frac{\pi}{3} = n\pi$ where $n$ is an integer $x = \frac{n\pi - \frac{\pi}{3}}{2} = \frac{(3n-1)\pi}{6}$

For $0 \leq x \leq \pi$: $n = 1$: $x = \frac{2\pi}{6} = \frac{\pi}{3}$ $n = 2$: $x = \frac{5\pi}{6}$

Answer: $(\frac{\pi}{3}, 0)$ and $(\frac{5\pi}{6}, 0)$

Marking:

• 1 mark: Setting function equal to zero
• 2 marks: Finding correct x-values in given domain

(c) Find maximum and minimum points. [3 marks]

Solution: Maximum when $2x + \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi$ $x = \frac{\pi}{12} + n\pi$ For $0 \leq x \leq \pi$: $x = \frac{\pi}{12}$

Minimum when $2x + \frac{\pi}{3} = \frac{3\pi}{2} + 2n\pi$ $x = \frac{7\pi}{12} + n\pi$ For $0 \leq x \leq \pi$: $x = \frac{7\pi}{12}$

Answer: Maximum: $(\frac{\pi}{12}, 1)$, Minimum: $(\frac{7\pi}{12}, -1)$

Marking:

• 1 mark: Method for finding extrema
• 1 mark: Maximum point
• 1 mark: Minimum point

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Question 7 [10 marks]

(a) Find velocity and acceleration. [2 marks]

Solution: $v = \frac{ds}{dt} = 3t^2 - 12t + 9$ $a = \frac{dv}{dt} = 6t - 12$

Marking:

• 1 mark each for velocity and acceleration

(b) Find times when particle is at rest. [3 marks]

Solution: $v = 0 \Rightarrow 3t^2 - 12t + 9 = 0$ $t^2 - 4t + 3 = 0$ $(t - 1)(t - 3) = 0$ $t = 1$ or $t = 3$

Marking:

• 1 mark: Setting velocity equal to zero
• 2 marks: Solving correctly

(c) Find displacement when $t = 4$. [1 mark]

Solution: $s = (4)^3 - 6(4)^2 + 9(4) + 2 = 64 - 96 + 36 + 2 = 6$ metres

Marking:

• 1 mark: Correct substitution and calculation

(d) Find total distance in first 4 seconds. [4 marks]

Solution: Need to find displacement at $t = 0, 1, 3, 4$: $s(0) = 2$ $s(1) = 1 - 6 + 9 + 2 = 6$ $s(3) = 27 - 54 + 27 + 2 = 2$ $s(4) = 6$

Distance = $|6-2| + |2-6| + |6-2| = 4 + 4 + 4 = 12$ metres

Marking:

• 1 mark: Identifying need to check turning points
• 2 marks: Finding displacements at key times
• 1 mark: Correct total distance

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Question 8 [8 marks]

(a) Show $OABC$ is a parallelogram. [3 marks]

Solution: $\vec{OA} = (4, 2)$ $\vec{CB} = (6, 6) - (2, 4) = (4, 2)$ $\vec{OC} = (2, 4)$ $\vec{AB} = (6, 6) - (4, 2) = (2, 4)$

Since $\vec{OA} = \vec{CB}$ and $\vec{OC} = \vec{AB}$, opposite sides are equal and parallel.

Marking:

• 1 mark: Finding two pairs of vectors
• 1 mark: Showing they are equal
• 1 mark: Conclusion

(b) Find area using cross product. [3 marks]

Solution: Area = $|\vec{OA} \times \vec{OC}| = |(4)(4) - (2)(2)| = |16 - 4| = 12$ square units

Marking:

• 1 mark: Setting up cross product
• 1 mark: Correct calculation
• 1 mark: Final answer

(c) Find equation of circle through all vertices. [2 marks]

Solution: For a parallelogram, the circle through all vertices has its centre at the intersection of diagonals. Centre = midpoint of diagonal $OB = (\frac{0+6}{2}, \frac{0+6}{2}) = (3, 3)$ Radius = distance from centre to any vertex = $\sqrt{(3-0)^2 + (3-0)^2} = 3\sqrt{2}$

Answer: $(x-3)^2 + (y-3)^2 = 18$

Marking:

• 1 mark: Finding centre
• 1 mark: Finding radius and equation

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Question 9 [9 marks]

(a) Find $P_0$ and $k$. [4 marks]

Solution: $P_0 = 500$ (initial population) When $t = 3$, $P = 2000$: $2000 = 500e^{3k}$ $4 = e^{3k}$ $\ln 4 = 3k$ $k = \frac{\ln 4}{3} = 0.462$ (3 s.f.)

Marking:

• 1 mark: $P_0 = 500$
• 2 marks: Setting up equation with $t = 3$
• 1 mark: Solving for $k$

(b) Find time for population to reach 10000. [2 marks]

Solution: $10000 = 500e^{0.462t}$ $20 = e^{0.462t}$ $\ln 20 = 0.462t$ $t = \frac{\ln 20}{0.462} = 6.49$ hours (3 s.f.)

Marking:

• 1 mark: Setting up equation
• 1 mark: Solving correctly

(c) Find rate of increase when $t = 6$. [3 marks]

Solution: $\frac{dP}{dt} = P_0 k e^{kt} = 500 \times 0.462 \times e^{0.462 \times 6}$ $= 231 \times e^{2.772} = 231 \times 16 = 3696$ bacteria per hour (3 s.f.)

Marking:

• 1 mark: Differentiating correctly
• 1 mark: Substituting $t = 6$
• 1 mark: Correct calculation

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Question 10 [8 marks]

(a) Show the derivative. [3 marks]

Solution: $y = \frac{x^2 + 4}{x - 1}$ Using quotient rule: $\frac{dy}{dx} = \frac{(x-1)(2x) - (x^2+4)(1)}{(x-1)^2}$ $= \frac{2x^2 - 2x - x^2 - 4}{(x-1)^2} = \frac{x^2 - 2x - 4}{(x-1)^2}$ ✓

Marking:

• 1 mark: Using quotient rule
• 2 marks: Correct algebraic manipulation

(b) Find stationary points. [4 marks]

Solution: For stationary points: $\frac{dy}{dx} = 0$ $x^2 - 2x - 4 = 0$ $x = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}$

When $x = 1 + \sqrt{5}$: $y = \frac{(1+\sqrt{5})^2 + 4}{(1+\sqrt{5}) - 1} = \frac{10 + 2\sqrt{5}}{\sqrt{5}} = 2\sqrt{5} + 2$

When $x = 1 - \sqrt{5}$: $y = \frac{(1-\sqrt{5})^2 + 4}{(1-\sqrt{5}) - 1} = \frac{10 - 2\sqrt{5}}{-\sqrt{5}} = 2\sqrt{5} - 2$

Answer: $(1 + \sqrt{5}, 2\sqrt{5} + 2)$ and $(1 - \sqrt{5}, 2\sqrt{5} - 2)$

Marking:

• 2 marks: Solving $x^2 - 2x - 4 = 0$
• 2 marks: Finding corresponding y-coordinates

(c) Determine nature of stationary points. [1 mark]

Solution: Using second derivative test or considering the sign of $\frac{dy}{dx}$ around each point: $(1 + \sqrt{5}, 2\sqrt{5} + 2)$ is a minimum $(1 - \sqrt{5}, 2\sqrt{5} - 2)$ is a maximum

Marking:

• 1 mark: Correct identification of both natures

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Question 11 [6 marks]

(a) Expand $(2 + 3x)^4$. [3 marks]

Solution: $(2 + 3x)^4 = \binom{4}{0}2^4 + \binom{4}{1}2^3(3x) + \binom{4}{2}2^2(3x)^2 + \binom{4}{3}2(3x)^3 + \binom{4}{4}(3x)^4$ $= 16 + 4 \times 8 \times 3x + 6 \times 4 \times 9x^2 + 4 \times 2 \times 27x^3 + 81x^4$ $= 16 + 96x + 216x^2 + 216x^3 + 81x^4$

Marking:

• 1 mark: Correct binomial coefficients
• 1 mark: Correct powers
• 1 mark: Correct final expansion

(b) Find coefficient of $x^3$ in $(1-x)(2+3x)^4$. [3 marks]

Solution: $(1-x)(16 + 96x + 216x^2 + 216x^3 + 81x^4)$ $= 16 + 96x + 216x^2 + 216x^3 + 81x^4 - 16x - 96x^2 - 216x^3 - 216x^4 - 81x^5$ $= 16 + 80x + 120x^2 + 0x^3 + ...$

Coefficient of $x^3 = 216 - 216 = 0$

Marking:

• 2 marks: Correct multiplication
• 1 mark: Identifying coefficient of $x^3$

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Question 12 [5 marks]

(a) Sketch region $R$. [1 mark]

Solution: [Sketch showing parabola $y = x^2 + 1$, horizontal line $y = 5$, and $y$-axis, with shaded region between them]

Marking:

• 1 mark: Correct sketch with region clearly indicated

(b) Find area of region $R$. [4 marks]

Solution: Intersection points: $x^2 + 1 = 5 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$ Since region is bounded by $y$-axis, we use $x = 0$ to $x = 2$.

Area $= \int_0^2 (5 - (x^2 + 1)) dx = \int_0^2 (4 - x^2) dx$ $= [4x - \frac{x^3}{3}]_0^2 = 8 - \frac{8}{3} = \frac{16}{3}$ square units

Marking:

• 1 mark: Finding intersection points
• 1 mark: Setting up correct integral
• 1 mark: Integrating correctly
• 1 mark: Evaluating and final answer

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Total: 90 marks