Mathematics Extension 1

A C E EXAM PAPER Student name: ______________________

PAPER 1

Mathematics Extension 1

– Working time - 120 minutes – Write using black pen

– NESA approved calculators may be used

– A reference sheet is provided at the back of this paper

– In questions 11-14, show relevant mathematical reasoning and/or calculations

Total marks:

70 Section I – 10 marks – Attempt Questions 1-10

– Allow about 15 minutes for this section Section II – 60 marks

– Attempt questions 11-14

– Allow about 1 hour and 45 minutes for this section

2 Section I 10 marks Attempt questions 1 - 10 Allow about 15 minutes for this section Use the multiple-choice answer sheet for questions 1-10 1. A coin is biased such that the probability of a head is 0.7. The probability that exactly three heads will be observed when the coin is tossed five times is: (A) 0.73_0.342 (B) 10 × 0.7#_0.3$ (C) & 𝐶# 0.7# (D) & 𝐶_{# 0.7}& 2. Which of the following is an expression for ; sin$ _{2𝑥𝑑𝑥 ?} (A) 𝑥 2− 1 8sin 4𝑥 + 𝐶 (B) 𝑥 2+ 1 8sin 4𝑥 + 𝐶 (C) 𝑥 −1 4sin 4𝑥 + 𝐶 (D) 𝑥 +1 4sin 4𝑥 + 𝐶 3. A stone is thrown at an angle of 𝜃 to the horizontal. The position of the stone at time

t seconds is given by 𝑥 = 𝑉𝑡cos𝜃 and 𝑦 = 𝑉𝑡sin𝜃 −'

$𝑔𝑡$ where g m/s2 is the acceleration due to gravity and v m/s is the initial velocity of projection. What is the maximum height reached by the stone? (A) 𝑉sin𝜃 𝑔 (B) 𝑔sin𝜃_𝑔 (C) 𝑉$sin$𝜃 2𝑔 (D) 𝑔sin$𝜃 2𝑉$

4. A parabola 𝑦 = 6𝑥 − 𝑥$ _{meets the line 𝑦 = 2𝑥 at (0, 0) and (4, 8).}

Which expression gives the area of the shaded region bounded by the parabola and the line? (A) ; 𝑥( $_{− 4𝑥} ) 𝑑 (B) ; 4𝑥 − 𝑥( $ ) 𝑑𝑥 (C) ; 𝑥* $_{− 4𝑥} ) 𝑑𝑥 (D) ; 4𝑥 − 𝑥* $ ) 𝑑𝑥 5. The number, N, of koalas in a population at time t years is given by 𝑁 = 135 + 𝐴𝑒+, _for constants A > 0 and k > 0. Which of the following is the correct differential equation? (A) 𝑑𝑁 𝑑𝑡 = −𝑘(𝑁 + 135) (B) 𝑑𝑁 𝑑𝑡 = −𝑘(𝑁 − 135) (C) 𝑑𝑁 𝑑𝑡 = 𝑘(𝑁 + 135) (D) 𝑑𝑁 𝑑𝑡 = 𝑘(𝑁 − 135)

4 6. What is the value of ;- _𝑥ln𝑥1 ! -𝑑𝑥 ? Use the substitution 𝑢 = ln𝑥. (A) ln 0.5 (B) ln 2 (C) ln 4 (D) 1 7. Vectors 𝑢̰, 𝑣̰ and 𝑤̰ are shown below. Which of the following statements is true? (A) |𝑤̰|$_{= |𝑢̰|}$_{+ |𝑣̰|}$ (B) |𝑤̰|$_{= |𝑢̰|}$_{+ |𝑣̰|}$_{− |𝑢̰𝑣̰|} (C) |𝑤̰|$_{= |𝑢̰|}$_{+ |𝑣̰|}$_{+ |𝑢̰𝑣̰|} (D) |𝑤̰|$_{= |𝑢̰|}$_{+ |𝑣̰|}$_{− |𝑢̰||𝑣̰|} 8. Mathematical induction is used to prove for 𝑛 ≥2, _{2 × 1 + 3 × 2+ . . . + 𝑛(𝑛 − 1) =}1 3𝑛(𝑛$− 1) Which of the following has an incorrect expression for part of the induction proof? (A) Step 1: To prove the statement true for n = 1 LHS = 2 × 1 = 2 RHS =1 3× 1 × (1$− 1) = 2 Result true for n = 1 (B) Step 2: Assume the result true for n = k 2 × 1 + 3 × 2 + . . . + 𝑘(𝑘 − 1) =1 3𝑘(𝑘$− 1) (C) To prove the result true for n = k + 1 2 × 1 + . . . + 𝑘(𝑘 − 1) + (𝑘 + 1)𝑘 =1 3(𝑘 + 1)((𝑘 + 1)$− 1) (D) LHS = 2 × 1 + 3 × 2 + ⋯ + 𝑘(𝑘 − 1) + (𝑘 + 1)𝑘 =1 3𝑘(𝑘$− 1) + (𝑘 + 1)𝑘 =1 3[𝑘(𝑘 + 1)(𝑘 − 1) + 3(𝑘 + 1)𝑘] =1 3𝑘(𝑘 + 1)[(𝑘 − 1) + 3] =1 3(𝑘 + 1)[𝑘$+ 2𝑘] =1 3(𝑘 + 1)((𝑘 + 1)$− 1) = RHS

9. What is sin𝑥 − cos𝑥 in the form 𝑅sin(𝑥 + α), where R > 0? (A) √2sin f𝑥 +π 4h (B) _{√2sin i𝑥 +}3π 4j (C) _{√2sin i𝑥 +}5π 4j (D) √2sin i𝑥 +7π 4j 10. The diagram below shows the trajectory of a ball thrown horizontally, at a speed of 50 ms–1, from the top of a tower 90 metres above ground level. After what time does the ball strike the ground?

(A) 3k_.& seconds (B) 3k&)_. seconds (C) 6k_.& seconds (D) 6k&)_. seconds

6 Section II 60 marks Attempt questions 11 - 14 Allow about 1 hour and 45 minutes for this section Answer each question in the spaces provided. Your responses should include relevant mathematical reasoning and/or calculations. Question 11 (15 marks) Marks

(a) If 𝑃 = f3₄h and 𝑄 = f8₀h, what is vector PQppppp⃗ in terms of 𝚤̰ and 𝚥̰? 2

(b) Show that sin f𝑥 +π 4h = sin𝑥 + cos𝑥 √2 2

Hence or otherwise, solve sin𝑥 + cos𝑥

√2 = √3 2 for 0 ≤ 𝑥 ≤ 2 π. 2 (c) A box contains seven identical balls except for their colour. Four are blue, two are white and one is red. Two balls are selected at random. After replacing the two balls the selection is repeated several times. What is the probability of getting two blue balls on at least one occasion from five selections of two balls? Answer correct to three decimal places. 2 What is the probability of getting two blue balls on exactly three occasions from five selections of two balls? Answer correct to three decimal places. 2

(d) The function 𝑓(𝑥) is given by 𝑓(𝑥) = 4tan/'_{𝑥 . Find the slope of the tangent}

to the curve where the function 𝑦 = 𝑓(𝑥) cuts the y-axis.

2

(e) Show that 1 + cos2𝑥

sin2𝑥 = cot𝑥 2

Hence find the exact value of cot15°. 1

Question 12 (14 marks) Marks

(a) In the diagram, the shaded region bounded by the curve 𝑦 = ln𝑥, x-axis, y-axis and the line 𝑦 = ln4, is rotated about the y-axis. Find the exact volume of the solid of revolution. 2

(b) In ∆𝑂𝐴𝐵, 𝑂𝐴ppppp⃗ = 𝑢̰ and 𝑂𝐵ppppp⃗ = 𝑣̰ . Point P is the midpoint of 𝐴𝐵ppppp⃗. Find the following vectors in terms of 𝑢̰ and 𝑣̰. _{𝐴𝐵ppppp⃗} 1 _{𝑂𝑃ppppp⃗} 2 _{𝐴𝑃ppppp⃗} 1 _{𝐵𝑃ppppp⃗} 1 (c) Find ; 𝑑𝑥 √36 − 𝑥$ 2 (d) Evaluate ; cos$1 2𝑥𝑑𝑥 0 ( ) 2

(e) Prove by mathematical induction that 31_{− 2𝑛 − 1 is divisible by 4 for all}

positive integers greater than 1.

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8 Question 13 (15 marks) Marks (a) A bottle of water has a temperature of 20˚C and is placed in a refrigerator whose temperature is 2˚C. The cooling rate of the bottle of water is proportional to the difference between the temperature of the refrigerator and the temperature T of the bottle of water. This is expressed by the equation: _𝑑𝑇 𝑑𝑡 = −𝑘(𝑇 − 2) where k is a constant of proportionality and t is the number of minutes after the bottle of water is placed in the refrigerator. Show that 𝑇 = 2 + 𝐴𝑒/+, _{satisfies the above equation. 1} After 20 minutes in the refrigerator the temperature of the bottle of water is 10˚C. What is the value of A and k in the above equation? 3 How long will it take for the bottle of water to cool down to 5˚C? 2 (b) Use the substitution 𝑢 = 𝑥 + 1 to evaluate ; 𝑥 √𝑥 + 1𝑑𝑥. '& ) 3

(c) A binomial distribution is 𝑋~Bin(40, 𝑝). If 𝐸(𝑋) = 5 find:

p 2 Var(𝑋) 2 (d) What is the unit vector in the direction 𝑢̰ = 2𝚤̰ + 3𝚥̰ ? 2

Question 14 (16 marks) Marks (a) A ball is thrown from the origin O with a velocity V and angle of elevation of 𝜃, where 𝜃 ≠2 $. You may assume that: _{𝑥 = 𝑉𝑡cos𝜃, 𝑦 = 𝑉𝑡sin𝜃 −}1 2𝑔𝑡$ where x and y are the horizontal and vertical displacements of the ball in metres from O at time t seconds after being thrown. Let ℎ =𝑉$ 2𝑔 and show that the equation of flight of the ball is: 3 𝑦 = 𝑥tan𝜃 − 1 4ℎ𝑥$(1 + tan$𝜃) The point of intersection when two balls are thrown with an angle of

elevation of 𝜃' and 𝜃$ is (a, b). Show that:

3 𝑎$_{> 4ℎ(ℎ − 𝑏)} (b) Use mathematical induction to prove that: 3 2 1 × 2+ 2 2 × 3+ 2 3 × 4+. . . + 2 𝑛 × (𝑛 + 1)= 2𝑛 𝑛 + 1 for all positive integers 𝑛. (c) Find ; 𝑒#3 1 + 𝑒3𝑑𝑥 using the substitution 𝑢 = 1 + 𝑒3. 2

(d) Where do the curves 𝑦 = 4𝑥 − 𝑥$_{+ 8 and 𝑦 = 𝑥}$_{− 2𝑥 intersect? 1}

Calculate the area between the two curves. 2 (e) Solve the differential equation below using the method of separation of variables. 2 𝑑𝑦 𝑑𝑥= 𝑒43(1 + 𝑦$) End of paper

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1 ACE Examination Paper 1 Year 12 Mathematics Extension 1 Yearly Examination Worked solutions and marking guidelines Section I Solution Criteria 1. Three heads with n = 5 𝑃(𝑋 = 5) = 𝐶# ! 0.7!(1 − 0.7)#$! = 10 × 0.7!_0.3% 1 Mark: B 2. _0sin% _{2𝑥𝑑𝑥 =}1 20(1 − cos 4𝑥)𝑑𝑥 =1 2:𝑥 − 1 4sin 4𝑥; + 𝐶 =𝑥 2− 1 8sin 4𝑥 + 𝐶 1 Mark: A 3. The maximum height, for angle  θ, occurs when 𝑦̇ = 0, that is 𝑦 = 𝑉𝑡sın𝜃 −1 2𝑔𝑡% ̇ 𝑦̇ = 𝑉sin𝜃 − 𝑔𝑡 0 = 𝑉sin𝜃 − 𝑔𝑡 𝑡 =𝑉sin𝜃 𝑔 Maximum height 𝑦 = 𝑉𝑡sin𝜃 −1 2𝑔𝑡% 𝑦 = E𝑉%sin%𝜃 𝑔 − 𝑉%_𝑠𝑖𝑛%_𝜃 2𝑔 I =𝑉%sin%𝜃 2𝑔 1 Mark: C 4. _{𝐴 = 0 6𝑥 − 𝑥}& _{%− 2𝑥} ' 𝑑𝑥 = 0 4𝑥 − 𝑥& % ' 𝑑𝑥 1 Mark: B 5. 𝑁 = 135 + 𝐴𝑒() 𝑑𝑁 𝑑𝑡 = 𝑘 × 𝐴𝑒() = 𝑘(𝑁 − 135) 1 Mark: D 6. _{𝑢 = ln𝑥 and 𝑑𝑢 = 1} 𝑥𝑑𝑥 𝑢 = ln𝑒 = 1, 𝑢 = ln𝑒%= 2 0 1 𝑥ln𝑥 *! * 𝑑𝑥 = 0 1 𝑢 % + du = [lnu] = ln 2 − ln 1 = ln 2 1 Mark: B

_{Solution Criteria} 7. 𝑤̰ + 𝑣̰ = 𝑢̰ 𝑤̰ = 𝑢̰ − 𝑣̰ 𝑤̰ · 𝑤̰ = (𝑢̰ − 𝑣̰) · (𝑢̰ − 𝑣̰) = 𝑢̰%_{− 2𝑢̰𝑣̰ + 𝑣̰}% |𝑤̰|%_{= |𝑢̰|}%_{+ |𝑣̰|}%_{− 2|𝑢̰||𝑣̰|cos60} ∴ |𝑤̰|%_{= |𝑢̰|}%_{+ |𝑣̰|}%_{− |𝑢̰||𝑣̰|} 1 Mark: D 8. The correct step 1 is shown below Step 1: To prove the statement true for n = 2 LHS = 2 × 1 = 2 RHS =1 3× 2 × (2%− 1) = 2 Result true for n = 2 1 Mark: A

9. sin𝑥 − cos𝑥 = 𝑅sin(𝑥 + α)

= 𝑅sin𝑥cosα + 𝑅cos𝑥sinα 𝑅cosα = 1 ① 𝑅sinα = −1 ② Equation ② divided by equation ① tanα = −1 α =7π 4 Squaring and adding the equations 𝑅%_(sin%_{α + cos}%_{α) = 1 + 1} 𝑅%_{= 2} 𝑅 = √2 (𝑅 > 0) ∴ sin𝑥 − cos𝑥 = √2sin :𝑥 +7π₄; 1 Mark: D 10. Particle reaches the ground when y = 0 𝑦 = −1 2𝑔𝑡%+ 𝑉𝑡sin𝜃 + 90 0 = −1 2𝑔𝑡%+ 50𝑡sin0 + 90 1 2𝑔𝑡% = 90 𝑡 = j180 𝑔 = j 36 × 5 𝑔 = 6j5 𝑔 seconds 1 Mark: C

3 Section II 11(a) 𝑃𝑄_{mmmmm⃗ = 5𝚤̰ − 4𝚥̰} 2 Marks: Correct answer. 1 Mark: Shows some understanding. 11(b) (i) LHS = sin q𝑥 + π 4r = sin𝑥cosπ 4+ cos𝑥sin π 4 = sin𝑥 × 1 √2+ cos𝑥 × 1 √2 =sin𝑥 + cos𝑥 √2 = RHS 2 Marks: Correct answer. 1 Mark: Uses the sum of angles formula or exact values. 11(b)

(ii) sin𝑥 + cos𝑥

√2 = √3 2 sin q𝑥 +π 4r = √3 2 𝑥 +π 4 = π 3 or 2π 3 𝑥 = π 12 or 5π 12 2 Marks: Correct answer. 1 Mark: Finds one solution or shows some understanding. 11(c)

(i) 𝑃(Two blue balls at least once) = 1 − 𝑃(No blue ball 5 times) = 1 − :1 −4₇×3 6; # = 0.8140 … ≈ 0.814 2 Marks: Correct answer. 1 Mark: Use of complementary event. 11(c) (ii) Let p be the probability of getting two blue balls. 𝑝 =4 7× 3 6= 2 7 𝑛 = 5 𝑃(𝑋 = 𝑥) = 𝐶# _,:2 7; , :5₇; #'$, 𝑃(𝑋 = 3) = 𝐶# _!:2 7; ! :5₇; #$! = 2000 16 807≈ 0.119 2 Marks: Correct answer. 1 Mark: Shows some understanding. 11(d) 𝑓(𝑥) = 4tan$+ _𝑥 𝑓′(𝑥) = 4 1 + 𝑥% Th curve cuts the y-axis when x = 0 𝑓′(0) = 4 1 + 0% = 4 \Slope of the tangent is 4. 2 Marks: Correct answer. 1 Mark: Differentiates the inverse function.

11(e) (i) LHS = 1 + cos 2𝑥 sin 2𝑥 = 2cos%𝑥 2sin𝑥cos𝑥 =cos𝑥 sin𝑥 = cot𝑥 = RHS 2 Marks: Correct answer. 1 Mark: Uses a correct and appropriate trig identity. 11(e) (ii) cot 15° = 1 + cos(2 × 15°) sin(2 × 15°) = 1 + E√_{2 I}3 1 2 = 2 + √3 1 Mark: Correct answer. 12(a) 𝑉 = π 0 𝑥-.& % ' 𝑑𝑦 = π 0 (e-.& /₎% ' 𝑑𝑦 = π •1 2e%/€+' -.& =π 2•e%-.&− e'‚ =15π 2 cubic units 2 Marks: Correct answer. 1 Mark: Uses volume formula with at least one correct value. 12(b) (i) 𝑢̰ + 𝐴𝐵_{𝐴𝐵mmmmm⃗ = 𝑣̰ − 𝑢̰} mmmmm⃗ = 𝑣̰ 1 Mark: Correct answer. 12(b) (ii) 𝑢̰ + 1 2𝐴𝐵mmmmm⃗ = 𝑂𝑃mmmmm⃗ 𝑂𝑃 mmmmm⃗ = 𝑢̰ +1₂(𝑣̰ − 𝑢̰) =1 2(𝑢̰ + 𝑣̰) 2 Marks: Correct answer. 1 Mark: Shows some understanding. 12(b) (iii) 𝐴𝑃mmmmm⃗ = 1 2𝐴𝐵mmmmm⃗ =1 2(𝑣̰ − 𝑢̰) 1 Mark: Correct answer. 12(b) (iv) 𝐵𝑃mmmmm⃗ = −𝐴𝑃mmmmm⃗ = −1 2(𝑣̰ − 𝑢̰) =1 2(𝑢̰ − 𝑣̰) 1 Mark: Correct answer. 12(c) 0 𝑑𝑥 √36 − 𝑥%= sin $+𝑥 6+ 𝐶 2 Marks: Correct answer.

5 12(d) 0 cos% 1 2𝑥𝑑𝑥 = 0 1 2(1 + cos 𝑥)𝑑𝑥 0 & ' 0 & ' =1 2[𝑥 + sin 𝑥]' 0 & =1 2•: 𝜋 4+ 1 √2; − (0 + 0)€ =1 2: 𝜋 4+ 1 √2; 2 Marks: Correct answer. 1 Mark: Uses double angle formula to simplify the integral. 12(e) Step 1: To prove true for n = 2 31_{− 2𝑛 − 1 = 3}%_{− 2 × 2 − 1 = 4} Result is true for n = 2 Step 2: Assume true for n = k 3(_{− 2𝑘 − 1 = 4𝑚} where m is an integer Step 3: To prove true for n = k + 1 3(2+_{− 2(𝑘 + 1) − 1 = 4𝑝} where p is an integer LHS = 3(2+_{− 2(𝑘 + 1) − 1} = 3•3(_{‚ − 2𝑘 − 2 − 1} = 3•3(_{− 2𝑘 − 1‚ + 4𝑘} = 3(4𝑚) + 4𝑘 = 4(3𝑚 + 𝑘) = 4𝑝 = RHS Step 4: True by induction 3 marks: Correct answer. 2 marks: Proves the result true for n = 1 and attempts to use the result of n = k to prove the result for n = k + 1 1 mark: Proves the result true for n = 1. 13(a) (i) 𝑇 = 2 + 𝐴𝑒 $() (𝐴𝑒$() _{= 𝑇 − 2)} 𝑑𝑇 𝑑𝑡 = −𝑘 × 𝐴𝑒$() = −𝑘(𝑇 − 2) 1 Mark: Correct answer. 13(a) (ii) Initially t = 0 and T = 20 _{𝑇 = 2 + 𝐴𝑒}$() 20 = 2 + 𝐴𝑒$(×' 𝐴 = 18 Now t = 20 and T = 10 𝑇 = 2 + 18𝑒$() 10 = 2 + 18𝑒$(×%' 𝑒$%'( ₌ 8 18= 4 9 −20𝑘 = ln4 9 𝑘 = − 1 20 ln 4 9 = 1 20 ln 9 4 = 0.0405 … 3 Marks: Correct answer. 2 Marks: Finds the value of A and an expression for k. 1 Mark: Finds the value of A.

13(a) (iii) We need to find t when T = 5 𝑇 = 2 + 18𝑒$() 5 = 2 + 18𝑒$() 𝑒$() ₌ 3 18= 1 6 −𝑘𝑡 = ln1 6 𝑡 = −1 𝑘ln 1 6 =1 𝑘ln6 = 20 ln6 ln9₄ = 44.1902. . . ≈ 44 minutes \It will take about 44 minutes for the bottle to cool to 5˚C. 2 Marks: Correct answer. 1 Mark: Makes some progress towards the solution. 13(b) 𝑢 = 𝑥 + 1 𝑑𝑢 = 𝑑𝑥 when x = 15, u = 16 and x = 0, u = 1 0 𝑥 √𝑥 + 1𝑑𝑥 = 0 𝑢 − 1 √𝑢 𝑑𝑢 +4 + +# ' = 0 𝑢+%− 𝑢$+%𝑑𝑢 +4 + = ˆ2𝑢 ! % 3 − 2𝑢 + % _‰ + +4 =2 3:16 ! %− 1; − 2 :16+%− 1; = 36 3 Marks: Correct answer. 2 Marks: Makes significant progress towards the solution. 1 Mark: Sets up the integration using the substitution. 13(c) (i) _{40 × 𝑝 = 5} 𝐸(𝑋) = 𝑛𝑝 𝑝 = 0.125 2 Marks: Correct answer. 1 Mark: Makes some progress. 13(c) (ii) 𝑉𝑎𝑟(𝑋) = 𝑛𝑝(1 − 𝑝) = 40 × 0.125(1 − 0.125) = 4.375 2 Marks: Correct answer. 1 Mark: Makes some progress. 13(d) |𝑢̰| = •2%_{+ 3}% = √13 𝑢̰Ž =_|𝑢̰|𝑢̰ = 1 √13(2𝚤̰ + 3𝚥̰) 2 Marks: Correct answer. 1 Mark: Shows some understanding.

7 14(a) (i) 𝑥 = 𝑉𝑡cos𝜃 ① 𝑦 = 𝑉𝑡sin𝜃 −1 2𝑔𝑡%② From equation ① 𝑡 = 𝑥 𝑉cos𝜃 substitute into equation ② 𝑦 = 𝑉 × 𝑥 𝑉cos𝜃× sin𝜃 − 1 2𝑔 × 𝑥% 𝑉%_cos%_𝜃 = 𝑥tan𝜃 −𝑔𝑥% 2𝑉%sec%𝜃 Given ℎ =𝑉%

2𝑔 and sec%𝜃 = (1 + tan%𝜃) then 𝑦 = 𝑥tan𝜃 − 1 4ℎ𝑥%(1 + tan%𝜃) 3 Marks: Correct answer. 2 Marks: Makes significant progress towards the solution. 1 Mark: Makes t the subject of equation ①. 14(a)

(ii) Now (𝑎, 𝑏) satisfies the equation 𝑦 = 𝑥tan𝜃 −_4ℎ1 𝑥%(1 + tan%𝜃) 𝑏 = 𝑎tan𝜃 − 1 4ℎ𝑎%(1 + tan%𝜃) 4ℎ𝑏 = 4ℎatan𝜃 − 𝑎%_{(1 + tan}%_𝜃) (1 + tan%_𝜃)𝑎%_{− 4ℎatan𝜃 + 4ℎ𝑏 = 0} 𝑎% _tan%_{𝜃 − 4ℎatan𝜃 + 4ℎ𝑏 + 𝑎}%_{= 0} Quadratic equation has 2 solutions if ∆> 0 𝑏%_{− 4𝑎𝑐 > 0} (−4ℎ𝑎)%_{− 4𝑎}%_{(4ℎ𝑏 + 𝑎}%_{) > 0} 16ℎ% _𝑎%_{− 16𝑎}%_{ℎ𝑏 − 4𝑎}& _{> 0} 4𝑎%_(4ℎ%_{− 4ℎ𝑏 − 𝑎}%_{) > 0} 4ℎ%_{− 4ℎ𝑏 − 𝑎}%_{> 0} 𝑎%_{< 4ℎ}%_{− 4ℎ𝑏} 𝑎%_{< 4ℎ(ℎ − 𝑏)} 3 Marks: Correct answer. 2 Marks: Makes significant progress towards the solution. 1 Mark: Substitutes (a, b). into equation of flight and simplifies. 14(b) Step 1: To prove true for n = 1 LHS = 2 1 × 2= 1 RHS =2 × 1 1 + 1= 1 Result is true for n = 1 Step 2: Assume true for n = k 𝑆₍ = 2𝑘 𝑘 + 1 Step 3: To prove true for n = k + 1 𝑆₍₂₊=2(𝑘 + 1) 𝑘 + 2 𝑆₍+ 𝑇₍₂₊= 𝑆₍₂₊ LHS = 2𝑘 𝑘 + 1+ 2 (𝑘 + 1)(𝑘 + 2) = 2𝑘(𝑘 + 2) + 2 (𝑘 + 1)(𝑘 + 2) =2(𝑘%+ 2𝑘 + 1) (𝑘 + 1)(𝑘 + 2) = 2(𝑘 + 1)(𝑘 + 1) (𝑘 + 1)(𝑘 + 2) =2(𝑘 + 1) 𝑘 + 2 = RHS Step 4: True by induction 3 marks: Correct answer. 2 marks: Proves the result true for n = 1 and attempts to use the result of n = k to prove the result for n = k + 1 1 mark: Proves the result true for n = 1.

14(c) 𝑢 = 1 + 𝑒, _{or 𝑒}, _{= 𝑢 − 1} 𝑑𝑢 𝑑𝑥 = 𝑒, or 𝑑𝑢 = 𝑒,𝑑𝑥 0 𝑒!, 1 + 𝑒,𝑑𝑥 = 0 𝑒%,_{× 𝑒}, 1 + 𝑒, 𝑑𝑥 = 0(𝑢 − 1)% 𝑢 𝑑𝑢 = 0𝑢%− 2𝑢 + 1 𝑢 𝑑𝑢 = 0(𝑢 − 2 +1 𝑢)𝑑𝑢 =𝑢% 2 − 2𝑢 + ln𝑢 + 𝐶 =(1 + 𝑒,)% 2 − 2(1 + 𝑒,) + ln (1 + 𝑒,) + 𝐶 2 Marks: Correct answer. 1 Mark: Sets up the integral in terms of u. 14(d) (i) 𝑦 = 4𝑥 − 𝑥 %_{+ 8 ①} 𝑦 = 𝑥%_{− 2𝑥 ②} Substitute 𝑥%_{− 2𝑥 for y into equation ①} 𝑥%_{− 2𝑥 = 4𝑥 − 𝑥}%_{+ 8} 2𝑥%_{− 6𝑥 − 8 = 0} 𝑥%_{− 3𝑥 − 4 = 0} (𝑥 − 4)(𝑥 + 1) = 0 𝑥 = 4 and 𝑦 = 8 𝑥 = −1 and 𝑦 = 3 \Points of intersection are (4, 8) and (–1, 3) 1 Mark: Correct answer. 14(d) (ii) 0 (4𝑥 − 𝑥%+ 8) − (𝑥%− 2𝑥) & $+ 𝑑𝑥 = 0 (−2𝑥& %_{+ 6𝑥 + 8} $+ )𝑑𝑥 = −2 0 (𝑥& %_{− 3𝑥 − 4)} $+ 𝑑𝑥 = −2 ›𝑥! 3 − 3𝑥% 2 − 4𝑥œ_$+ & = −2 •:64₃ −48 2 − 16; − :− 1 3− 3 2+ 4;€ = 412 3 square units 2 Marks: Correct answer. 1 Mark: Shows some understanding. 14(e) 𝑑𝑦 𝑑𝑥 = 𝑒4,(1 + 𝑦%) 0 𝑒4,_{𝑑𝑥 = 0} 1 1 + 𝑦%𝑑𝑦 1 6𝑒4,+ 𝐶 = tan$+ 𝑦 𝑦 = tan :1 6𝑒4,+ 𝐶; 2 Marks: Correct answer. 1 Mark: Separates the variables and attempts to integrate.