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HMS, BM EQ-Bank 358 MC

A student's ventilation rate increases during a 5km run. This occurs to:

  1. Increase oxygen intake and carbon dioxide removal
  2. Reduce oxygen delivery to the muscles
  3. Increase carbon dioxide in the bloodstream
  4. Decrease cardiac output
Show Answers Only

\(A\)

Show Worked Solution
  • A is correct: Increases oxygen intake and removes carbon dioxide

Other Options:

  • B is incorrect: Exercise increases oxygen delivery to muscles
  • C is incorrect: Exercise decreases carbon dioxide in bloodstream
  • D is incorrect: Ventilation increases with increased cardiac output

HMS, BM EQ-Bank 341

Outline the immediate response of cardiac output during aerobic training.   (3 marks)

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Sample Answer

  • Cardiac output increases immediately during aerobic training
  • This occurs due to an increase in both heart rate and stroke volume
  • The increase supports the delivery of oxygen and nutrients to working muscles
  • Cardiac output rises from approximately 5 L/min at rest to 15 L/min during moderate exercise
Show Worked Solution

Sample Answer

  • Cardiac output increases immediately during aerobic training
  • This occurs due to an increase in both heart rate and stroke volume
  • The increase supports the delivery of oxygen and nutrients to working muscles
  • Cardiac output rises from approximately 5 L/min at rest to 15 L/min during moderate exercise

HMS, BM EQ-Bank 327

Outline how ventilation rate responds to moderate aerobic exercise.   (3 marks)

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Sample Answer

  • Ventilation rate increases from resting levels during moderate aerobic exercise. Breathing frequency approximately doubles from baseline values.
  • The immediate physiological response supplies additional oxygen to working muscles. Deeper breaths also contribute to increased oxygen intake.
  • Carbon dioxide removal is enhanced during exercise. Increased ventilation clears metabolic waste products from energy production.
  • The response stabilises at a sustainable level during steady-state exercise. Recovery sees gradual return to resting ventilation rates.
Show Worked Solution

Sample Answer

  • Ventilation rate increases from resting levels during moderate aerobic exercise. Breathing frequency approximately doubles from baseline values.
  • The immediate physiological response supplies additional oxygen to working muscles. Deeper breaths also contribute to increased oxygen intake.
  • Carbon dioxide removal is enhanced during exercise. Increased ventilation clears metabolic waste products from energy production.
  • The response stabilises at a sustainable level during steady-state exercise. Recovery sees gradual return to resting ventilation rates.

HMS, BM EQ-Bank 320

Outline the immediate heart rate response when an individual begins aerobic exercise.   (3 marks)

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Sample Answer

  • When an individual begins aerobic exercise, there is an immediate increase in heart rate.
  • This occurs due to increased stimulation from the sympathetic nervous system.
  • The heart rate increases in proportion to exercise intensity to deliver more oxygen and nutrients to the working muscles.
Show Worked Solution

Sample Answer

  • When an individual begins aerobic exercise, there is an immediate increase in heart rate.
  • This occurs due to increased stimulation from the sympathetic nervous system.
  • The heart rate increases in proportion to exercise intensity to deliver more oxygen and nutrients to the working muscles.

HMS, BM EQ-Bank 318 MC

During an athletics training session, a coach observes a sprinter's heart rate. Which of the following is the most likely immediate physiological response to high-intensity sprint training?

  1. Decreased heart rate
  2. Increased heart rate
  3. Decreased cardiac output
  4. No change in heart rate
Show Answers Only

\(B\)

Show Worked Solution
  • B is correct: Heart rate increases immediately to supply oxygen to muscles

Other Options:

  • A is incorrect: Heart rate increases not decreases during exercise
  • C is incorrect: Cardiac output increases not decreases during exercise
  • D is incorrect: Heart rate always changes with exercise

HMS, BM EQ-Bank 311

Outline TWO immediate cardiac responses to a training session.   (3 marks)

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Sample Answer – Any 2 of the following

  • Heart rate increases immediately when training begins, causing the heart to beat faster to deliver more oxygen to working muscles.
  • Stroke volume increases during training, meaning more blood is pumped out of the heart with each beat.
  • Cardiac output (amount of blood pumped by the heart per minute) increases during training to supply more blood to muscles that are being used.
Show Worked Solution

Sample Answer – Any 2 of the following

  • Heart rate increases immediately when training begins, causing the heart to beat faster to deliver more oxygen to working muscles.
  • Stroke volume increases during training, meaning more blood is pumped out of the heart with each beat.
  • Cardiac output (amount of blood pumped by the heart per minute) increases during training to supply more blood to muscles that are being used.

HMS, BM EQ-Bank 277

Explain how you would apply each component of the FITT principle when designing an aerobic training program for a recreational tennis player.   (5 marks)

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Sample Answer

  • The FITT principle provides a framework using Frequency, Intensity, Time and Type for designing effective training programs.
  • For a recreational tennis player allowing 3-4 sessions weekly for frequency provides sufficient aerobic stimulus while still having 48-72 hours recovery between sessions. This helps prevent overtraining in recreational athletes.
  • Adopting an intensity of 65-75% MHR which is in the moderate zone, effectively develops aerobic capacity without excessive fatigue for non-competitive athletes.
  • Including one session at 75-80% MHR weekly so players develop their lactate buffering capacity, enhances rally endurance during longer matches.
  • Recreational players will benefit from sessions of 30-40 minutes for time because this duration stimulates cardiovascular improvements while maintaining quality.
  • Progressing session duration by 5-10% every two weeks provides gradual increases that allow fitness development without injury risk.
  • A combination of continuous running with tennis-specific movements is suitable for type and enables aerobic fitness development, while maintaining sport-specific patterns.
  • On-court drills at moderate intensity should be included as this integrates skill maintenance with aerobic development.
  • Moderate intensity training enables higher frequency because recovery demands remain manageable, optimising aerobic development for recreational tennis players.

Show Worked Solution

Sample Answer

  • The FITT principle provides a framework using Frequency, Intensity, Time and Type for designing effective training programs.
  • For a recreational tennis player allowing 3-4 sessions weekly for frequency provides sufficient aerobic stimulus while still having 48-72 hours recovery between sessions. This helps prevent overtraining in recreational athletes.
  • Adopting an intensity of 65-75% MHR which is in the moderate zone, effectively develops aerobic capacity without excessive fatigue for non-competitive athletes.
  • Including one session at 75-80% MHR weekly so players develop their lactate buffering capacity, enhances rally endurance during longer matches.
  • Recreational players will benefit from sessions of 30-40 minutes for time because this duration stimulates cardiovascular improvements while maintaining quality.
  • Progressing session duration by 5-10% every two weeks provides gradual increases that allow fitness development without injury risk.
  • A combination of continuous running with tennis-specific movements is suitable for type and enables aerobic fitness development, while maintaining sport-specific patterns.
  • On-court drills at moderate intensity should be included as this integrates skill maintenance with aerobic development.
  • Moderate intensity training enables higher frequency because recovery demands remain manageable, optimising aerobic development for recreational tennis players.

HMS, BM EQ-Bank 271

Explain how the FITT principle can be applied when designing an aerobic training program for a middle-distance runner.   (5 marks)

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Sample Answer

Frequency:

  • 4-5 sessions per week allowing for sufficient recovery between sessions to prevent overtraining while providing adequate stimulus for adaptation.

Intensity:

  • 70-85% of maximum heart rate to develop aerobic capacity and improve lactate threshold which is critical for middle-distance events.

Time:

  • 30-60 minutes per session to develop endurance without excessive fatigue that could lead to injury.

Type:

  • Combination of continuous running, tempo runs and fartlek training to develop aerobic capacity while simulating race conditions.

Progession:

  • The FITT principle ensures a structured approach to training that addresses specific requirements of middle-distance running while allowing for progressive overload.
Show Worked Solution

Sample Answe

Frequency:

  • 4-5 sessions per week allowing for sufficient recovery between sessions to prevent overtraining while providing adequate stimulus for adaptation.

Intensity:

  • 70-85% of maximum heart rate to develop aerobic capacity and improve lactate threshold which is critical for middle-distance events.

Time:

  • 30-60 minutes per session to develop endurance without excessive fatigue that could lead to injury.

Type:

  • Combination of continuous running, tempo runs and fartlek training to develop aerobic capacity while simulating race conditions.

Progession:

  • The FITT principle ensures a structured approach to training that addresses specific requirements of middle-distance running while allowing for progressive overload.

HMS, BM EQ-Bank 262 MC

A swimmer aims to improve their anaerobic capacity for 100 metre sprint events. Which training program based on the FITT principle is most appropriate?

  1. Frequency: daily; Intensity: 60% MHR; Time: 45 minutes; Type: continuous swimming
  2. Frequency: 3-4 times per week; Intensity: 85-95% MHR; Time: 20-30 minutes; Type: interval training
  3. Frequency: twice weekly; Intensity: 70% MHR; Time: 60 minutes; Type: fartlek training
  4. Frequency: 5 times per week; Intensity: 65-75% MHR; Time: 90 minutes; Type: long slow distance
Show Answers Only

\(B\)

Show Worked Solution
  • B is correct: High-intensity intervals, 85-95% MHR, with adequate recovery for anaerobic capacity.

Other Options:

  • A is incorrect: Moderate intensity, continuous → aerobic training.
  • C is incorrect: Low frequency, low intensity for anaerobic development.
  • D is incorrect: Endurance training parameters, not sprint-focused.

HMS, BM EQ-Bank 258 MC

A volleyball player completes an aerobic continuous training session by riding a stationary bike at a constant pace for 50 minutes at 65% of maximum heart rate.

What is the primary energy system being targeted by this type of training?

  1. ATP-PC system
  2. Lactic acid system
  3. Aerobic energy system
  4. All energy systems equally
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: Extended duration at moderate intensity targets aerobic system.

Other Options:

  • A is incorrect: ATP-PC system only operates for 10-15 seconds.
  • B is incorrect: Lactic acid system operates for 30 seconds to 2 minutes.
  • D is incorrect: This training specifically targets aerobic energy system.

HMS, BM EQ-Bank 251

Outline the key differences between continuous aerobic training and High Intensity Interval Training (HIIT).   (3 marks)

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Sample Answer

Continuous aerobic training

  • Sustained moderate-intensity exercise (60-75% MHR)
  • Extended periods without rest (20+ minutes)
  • Typically 30-60 minutes total duration
  • Predominantly aerobic energy system

HIIT:

  • Alternates high-intensity periods (85-95% MHR) with recovery
  • Varied work-to-rest ratios
  • Shorter total duration (15-30 minutes)
  • More intense overall
  • Engages both aerobic and anaerobic systems
Show Worked Solution

Sample Answer

Continuous aerobic training

  • Sustained moderate-intensity exercise (60-75% MHR)
  • Extended periods without rest (20+ minutes)
  • Typically 30-60 minutes total duration
  • Predominantly aerobic energy system

HIIT:

  • Alternates high-intensity periods (85-95% MHR) with recovery
  • Varied work-to-rest ratios
  • Shorter total duration (15-30 minutes)
  • More intense overall
  • Engages both aerobic and anaerobic systems

HMS, BM EQ-Bank 247

Explain how heart rate monitoring can be used to ensure appropriate intensity in both aerobic and anaerobic training sessions.  (3 marks)

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Sample Answer

  • Heart rate monitoring ensures aerobic training stays within 70-85% MHR zones. This works because sustained moderate intensity improves cardiovascular fitness without excessive fatigue. Runners maintain 140-160 bpm throughout sessions. As a result, athletes develop endurance efficiently.
  • For anaerobic training, monitoring shows peak rates (85-95% MHR) and recovery. The reason is reaching target zones ensures sufficient intensity for improvement. HIIT sessions need 170+ bpm during work periods. Therefore, athletes train at optimal intensity while avoiding overtraining.
Show Worked Solution

Sample Answer

  • Heart rate monitoring ensures aerobic training stays within 70-85% MHR zones. This works because sustained moderate intensity improves cardiovascular fitness without excessive fatigue. Runners maintain 140-160 bpm throughout sessions. As a result, athletes develop endurance efficiently.
  • For anaerobic training, monitoring shows peak rates (85-95% MHR) and recovery. The reason is reaching target zones ensures sufficient intensity for improvement. HIIT sessions need 170+ bpm during work periods. Therefore, athletes train at optimal intensity while avoiding overtraining.

HMS, BM EQ-Bank 153 MC

Which energy system produces ATP at the fastest rate?

  1. Glycolytic
  2. Aerobic
  3. ATP-PCr
  4. All systems produce ATP at the same rate
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: ATP-PCr has fastest production rate but limited stores

Other Options:

  • A is incorrect: Slower than ATP-PCr but faster than aerobic
  • B is incorrect: Slowest but most efficient system
  • D is incorrect: Systems have different production rates

HMS, BM EQ-Bank 138

Outline how inefficient jumping technique can affect the skeletal system and require first aid intervention.   (3 marks)

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Sample Answer

  • Incorrect landing mechanics – Poor technique causes excessive joint compression in ankles, knees and hips, potentially leading to acute injuries.
  • Abnormal force distribution – Impact forces travel through misaligned bones creating stress fractures, particularly in weight-bearing bones like tibia and metatarsals.
  • First aid requirements – RICER protocol needed for acute injuries, joint stabilisation to prevent further damage, and medical referral for suspected fractures.
Show Worked Solution

Sample Answer

  • Incorrect landing mechanics – Poor technique causes excessive joint compression in ankles, knees and hips, potentially leading to acute injuries.
  • Abnormal force distribution – Impact forces travel through misaligned bones creating stress fractures, particularly in weight-bearing bones like tibia and metatarsals.
  • First aid requirements – RICER protocol needed for acute injuries, joint stabilisation to prevent further damage, and medical referral for suspected fractures.

HMS, BM EQ-Bank 133

Explain how the circulatory and respiratory systems respond to dehydration during movement and outline appropriate first aid interventions.   (5 marks)

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Sample Answer

  • Dehydration affects the circulatory system because fluid loss reduces blood volume. This causes the heart to increase its rate to maintain adequate circulation. As a result, blood becomes thicker and more viscous, making it harder to pump efficiently through vessels.
  • The respiratory system responds to dehydration through increased breathing rates. This occurs due to the body’s attempt to maintain oxygen delivery despite reduced blood efficiency. Consequently, airways become dry and gas exchange becomes less efficient, further compromising oxygen delivery.
  • These system responses create a cycle of increasing stress during movement. This relationship demonstrates why dehydration severely impacts athletic performance. Therefore, recognizing early signs is crucial for intervention.
  • First aid interventions must address both immediate and ongoing needs. This involves ceasing activity immediately and moving to a cool environment. Following this, provide small sips of electrolyte solution rather than plain water.
  • Monitoring remains essential because vital signs indicate recovery progress. This process ensures gradual rehydration prevents shock while heart rate returns to normal. Hence, systematic first aid prevents serious complications.
Show Worked Solution

Sample Answer

  • Dehydration affects the circulatory system because fluid loss reduces blood volume. This causes the heart to increase its rate to maintain adequate circulation. As a result, blood becomes thicker and more viscous, making it harder to pump efficiently through vessels.
  • The respiratory system responds to dehydration through increased breathing rates. This occurs due to the body’s attempt to maintain oxygen delivery despite reduced blood efficiency. Consequently, airways become dry and gas exchange becomes less efficient, further compromising oxygen delivery.
  • These system responses create a cycle of increasing stress during movement. This relationship demonstrates why dehydration severely impacts athletic performance. Therefore, recognizing early signs is crucial for intervention.
  • First aid interventions must address both immediate and ongoing needs. This involves ceasing activity immediately and moving to a cool environment. Following this, provide small sips of electrolyte solution rather than plain water.
  • Monitoring remains essential because vital signs indicate recovery progress. This process ensures gradual rehydration prevents shock while heart rate returns to normal. Hence, systematic first aid prevents serious complications.

HMS, BM EQ-Bank 129

Describe how the digestive system can create undue stress on the body during physical activity and outline appropriate first aid responses.   (5 marks)

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Sample Answer

Digestive System Stress:

  • Exercise redirects blood flow away from digestive organs to working muscles, reducing digestive efficiency and causing painful abdominal cramping.
  • Eating within 2-3 hours of exercise leaves undigested food in stomach, leading to nausea, vomiting and uncomfortable bloating during activity.
  • Dehydration impairs digestive secretions and intestinal function, resulting in severe cramping and potential diarrhea during prolonged exercise.
  • High-intensity movement creates mechanical stress on abdominal organs, causing side stitches and acid reflux that impair performance.

First Aid Responses:

  • Stop activity immediately when severe cramping occurs to prevent further digestive distress and allow blood flow redistribution.
  • Place person in comfortable left side-lying position, which relieves pressure on stomach and reduces reflux symptoms.
  • Provide small sips of room-temperature water every 5-10 minutes if tolerated, avoiding cold fluids that may worsen cramping.
  • Monitor vital signs and observe for deterioration including persistent vomiting, severe dehydration or signs of heat illness.
  • Seek immediate medical attention if symptoms persist beyond 30 minutes or worsen despite first aid measures.
Show Worked Solution

Sample Answer

Digestive System Stress:

  • Exercise redirects blood flow away from digestive organs to working muscles, reducing digestive efficiency and causing painful abdominal cramping.
  • Eating within 2-3 hours of exercise leaves undigested food in stomach, leading to nausea, vomiting and uncomfortable bloating during activity.
  • Dehydration impairs digestive secretions and intestinal function, resulting in severe cramping and potential diarrhea during prolonged exercise.
  • High-intensity movement creates mechanical stress on abdominal organs, causing side stitches and acid reflux that impair performance.

First Aid Responses:

  • Stop activity immediately when severe cramping occurs to prevent further digestive distress and allow blood flow redistribution.
  • Place person in comfortable left side-lying position, which relieves pressure on stomach and reduces reflux symptoms.
  • Provide small sips of room-temperature water every 5-10 minutes if tolerated, avoiding cold fluids that may worsen cramping.
  • Monitor vital signs and observe for deterioration including persistent vomiting, severe dehydration or signs of heat illness.
  • Seek immediate medical attention if symptoms persist beyond 30 minutes or worsen despite first aid measures.

HMS, BM EQ-Bank 128 MC

During a cross-country run, an athlete experiences severe abdominal cramping. Which first aid response would be most appropriate?

  1. Continue running at race pace
  2. Increase fluid intake rapidly
  3. Start walking immediately
  4. Stop activity and lie in a comfortable position
Show Answers Only

\(D\)

Show Worked Solution
  • D is correct: Stopping activity and finding a comfortable position allows assessment of digestive system stress and prevents further complications.

Other Options:

  • A is incorrect: Would increase digestive system stress
  • B is incorrect: Rapid fluid intake may worsen symptoms
  • C is incorrect: Immediate walking may exacerbate cramping

HMS, BM EQ-Bank 123

Explain how the respiratory and circulatory systems respond to movement and describe appropriate first aid responses when these systems show signs of stress.   (5 marks)

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Sample Answer

  • The respiratory system responds to movement through increased oxygen demand. This occurs because working muscles require more oxygen for energy production. As a result, breathing rate and depth increase to supply adequate oxygen and remove carbon dioxide.
  • The circulatory system responds by increasing heart rate and stroke volume. This leads to greater cardiac output, which enables faster oxygen delivery to muscles. Consequently, blood flow is redirected from non-essential organs to working muscles.
  • This demonstrates how both systems work together during movement. The interaction allows efficient oxygen delivery and waste removal. This relationship results in sustained energy production for continued movement.
  • Signs of respiratory stress include abnormal breathing patterns, wheezing or gasping. This happens when oxygen demand exceeds supply. Therefore, first aid requires immediately stopping activity, sitting the person upright, and encouraging controlled breathing.
  • Circulatory stress presents as irregular pulse, chest pain or dizziness. This triggers the need for immediate intervention. The appropriate response involves lying the person down with elevated legs, monitoring vital signs, and implementing the STOP protocol if symptoms persist.
Show Worked Solution

Sample Answer

  • The respiratory system responds to movement through increased oxygen demand. This occurs because working muscles require more oxygen for energy production. As a result, breathing rate and depth increase to supply adequate oxygen and remove carbon dioxide.
  • The circulatory system responds by increasing heart rate and stroke volume. This leads to greater cardiac output, which enables faster oxygen delivery to muscles. Consequently, blood flow is redirected from non-essential organs to working muscles.
  • This demonstrates how both systems work together during movement. The interaction allows efficient oxygen delivery and waste removal. This relationship results in sustained energy production for continued movement.
  • Signs of respiratory stress include abnormal breathing patterns, wheezing or gasping. This happens when oxygen demand exceeds supply. Therefore, first aid requires immediately stopping activity, sitting the person upright, and encouraging controlled breathing.
  • Circulatory stress presents as irregular pulse, chest pain or dizziness. This triggers the need for immediate intervention. The appropriate response involves lying the person down with elevated legs, monitoring vital signs, and implementing the STOP protocol if symptoms persist.

HMS, BM EQ-Bank 122

Outline how the muscular and skeletal systems work together during movement and identify when first aid intervention is required.   (3 marks)

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Sample Answer

  • The muscular and skeletal systems work together as lever systems. Muscles attach to bones at origin and insertion points, creating movement through contraction and relaxation.
  • First aid intervention is required when movement causes sudden sharp pain or muscle inability to contract. Visible deformity or loss of joint stability indicates possible strain, sprain or fracture.
  • Warning signs include hearing a “pop” sound at time of injury, immediate swelling, or inability to bear weight on the affected area.
Show Worked Solution

Sample Answer

  • The muscular and skeletal systems work together as lever systems. Muscles attach to bones at origin and insertion points, creating movement through contraction and relaxation.
  • First aid intervention is required when movement causes sudden sharp pain or muscle inability to contract. Visible deformity or loss of joint stability indicates possible strain, sprain or fracture.
  • Warning signs include hearing a “pop” sound at time of injury, immediate swelling, or inability to bear weight on the affected area.

HMS, BM EQ-Bank 109

Outline how the skeletal and muscular systems work together during a squat movement.   (3 marks)

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Sample Answer

Descent:

  • Hip, knee and ankle joints flex while quadriceps and gluteal muscles lengthen eccentrically to control downward movement.
  • Skeletal system provides stable framework as muscles work to control lowering speed.

Bottom Position:

  • Skeletal joints maintain alignment while muscles sustain isometric contraction to hold position.
  • Bones bear body weight as muscles stabilise.

Rising:

  • Quadriceps and gluteal muscles contract concentrically to extend joints.
  • Skeleton provides mechanical leverage while muscles generate upward force through bone attachments.
Show Worked Solution

Sample Answer

Descent:

  • Hip, knee and ankle joints flex while quadriceps and gluteal muscles lengthen eccentrically to control downward movement.
  • Skeletal system provides stable framework as muscles work to control lowering speed.

Bottom Position:

  • Skeletal joints maintain alignment while muscles sustain isometric contraction to hold position.
  • Bones bear body weight as muscles stabilise.

Rising:

  • Quadriceps and gluteal muscles contract concentrically to extend joints.
  • Skeleton provides mechanical leverage while muscles generate upward force through bone attachments.

HMS, BM EQ-Bank 103 MC

During a netball game, a player performs a layup shot. Which body systems are working together to execute this movement?

  1. Skeletal and respiratory only
  2. Muscular and circulatory only
  3. Skeletal, muscular and nervous
  4. Respiratory and circulatory only
Show Answers Only

\(C\)

Show Worked Solution
  • C is correct: The skeletal (provides framework), muscular (produces force) and nervous systems (timing and precision) work together to execute coordinated movement.

Other Options:

  • A is incorrect: Misses muscular control needed
  • B is incorrect: Omits skeletal framework required
  • D is incorrect: Misses skeletal and muscular components

HMS, BM EQ-Bank 26 MC

The muscle group indicated in the image below is primarily responsible for:

  1. Hip flexion
  2. Knee extension
  3. Knee flexion
  4. Hip extension
Show Answers Only

\(B\)

Show Worked Solution
  • B is correct. The quadriceps are the primary knee extensors

Other Options:

  • A is Incorrect: Hip flexion is primarily performed by the iliopsoas
  • C is Incorrect: Knee flexion is performed by the hamstrings
  • D is Incorrect: Hip extension is primarily performed by the gluteus maximus and hamstrings

HMS, HAG 2022 HSC 4 MC

A person had knee surgery. They were able to choose their own doctor, hospital and the date for their surgery.

Which of the following enabled the person to make these choices?

  1. Medicare Safety Net
  2. Private health insurance
  3. Health care concession card
  4. Pharmaceutical Benefits Scheme
Show Answers Only

\(B\)

Show Worked Solution

  • B is correct: Private health insurance provides choice of doctor, hospital and surgery timing.

Other Options:

  • A is incorrect: Safety Net provides financial assistance but limited choice.
  • C is incorrect: Concession cards reduce costs but don’t increase choice.
  • D is incorrect: PBS covers medications, not surgical procedures.

HMS, HIC 2022 HSC 2 MC

To reduce the number of young people smoking, the sale of tobacco products to people under 18 years of age was made illegal.

Which action area of the Ottawa Charter is this strategy an example of?

  1. Developing personal skills
  2. Reorienting health services
  3. Building healthy public policy
  4. Strengthening community action
Show Answers Only

\( C\)

Show Worked Solution
  • C is correct: Legislation prohibiting tobacco sales creates healthy public policy through government regulation

Other Options:

  • A is incorrect: Focuses on individual education and skills development.
  • B is incorrect: Involves changing how health services operate.
  • D is incorrect: Community-led initiatives rather than government legislation.

Functions, MET1 2024 VCAA 5

The function  \(h:[0, \infty) \rightarrow R, \ h(t)=\dfrac{3000}{t+1}\)  models the population of a town after \(t\) years.

  1. Use the model \(h(t)\) to predict the population of the town after four years.   (1 mark)

  2. A new function, \(h_1\), models a population where  \(h_1(0)=h(0)\) but \(h_1\) decreases at half the rate of \(h\) at any point in time.
  3. State a sequence of two transformations that maps \(h\) to this new model \(h_1\).   (2 marks)

  4. In the town, 100 people were randomly selected and surveyed, with 60 people indicating that they were unhappy with the roads.
    1. Determine an approximate 95% confidence interval for the proportion of people in the town who are unhappy with the roads.
    2. Use  \(z=2\)  for this confidence interval.   (2 marks)
    3. A new sample of \(n\) people results in the same sample proportion.
    4. Find the smallest value of \(n\) to achieve a standard deviation of  \(\dfrac{\sqrt{2}}{50}\)  for the sample proportion.   (1 mark)

Show Answers Only

a.    \(600\)

b.   \(\text{Transformations:}\)

\(\text{1- Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}\)

\(\text{2- Translation of 1500 units upwards}\)

ci.   \(\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\ \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\)

cii.  \(300\)

Show Worked Solution

a.  \(h(4)=\dfrac{3000}{4+1}=600\)

b.     \(h(t)\) \(=3000(t+1)^{-1}\)
  \(h^{\prime}(t)\) \(=-\dfrac{3000}{(t+1)^2}\)
  \(h_1^{\prime}(t)\) \(=\dfrac{1}{2}h^{\prime}(t)=-\dfrac{1500}{(t+1)^2}\)
  \(h_1(t)\) \(=\dfrac{1500}{t+1}+C\)

♦♦♦ Mean mark (a) 17%.

\(\text{Given}\ \ h(0)=h_1(0):\)

\(\dfrac{1500}{0+1} +C= 3000\ \ \Rightarrow\ \ C=1500\)

\(h_1(t)=\dfrac{1500}{t+1}+1500\)
 

\(\text{Transformations:}\)

\(\text{1- Vertical dilation by a factor of }\frac{1}{2}\ \text{from the }t\ \text{axis}\)

\(\text{2- Translation of 1500 units upwards}\)
 

ci.    \(\hat{p}=\dfrac{60}{100}=\dfrac{3}{5},\quad 1-\hat{p}=\dfrac{2}{5},\quad z=2\)

\(\text{Approx CI}\) \(=\left(\dfrac{3}{5}-2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}},\ \dfrac{3}{5}+2\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{100}}\right)\)
  \(=\left(\dfrac{3}{5}-\dfrac{2\sqrt{6}}{50},\quad \dfrac{3}{5}+\dfrac{2\sqrt{6}}{50}\right)\)
  \(=\left(\dfrac{3}{5}-\dfrac{\sqrt{6}}{25},\quad \dfrac{3}{5}+\dfrac{\sqrt{6}}{25}\right)\)

  

cii.   \(\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\sqrt{\dfrac{\dfrac{3}{5}\times\dfrac{2}{5}}{n}}\) \(=\dfrac{\sqrt{2}}{50}\)
  \(\dfrac{6}{25n}\) \(=\dfrac{2}{2500}\)
  \(\dfrac{25n}{6}\) \(=1250\)
  \(n\) \(=\dfrac{6}{25}\times 1250\)
    \(=300\)
♦♦ Mean mark (c.ii.) 33%.

BIOLOGY, M8 2024 HSC 31

A study monitored the changes in the body temperature of a kookaburra (an Australian bird) and a human over a 24-hour period. The results of the study are shown in the graph.
 

  1. At what time was the kookaburra's body temperature the lowest?   (1 mark)

  1. Some endothermic organisms can display torpor (a significant decrease in physiological activity).

    With reference to the graph, explain whether the human or the kookaburra was displaying torpor and if so, state the time this occurred.   (3 marks)

  1. Outline an adaptation that may lead to an increase in the kookaburra's body temperature during the inactive period.   (2 marks)

Show Answers Only

a.   4 am

b.   Signs of torpor:

  • The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity
  • In contrast, the kookaburra exhibited classic torpor behaviour.
  • Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period.

c.   Kookaburra adaptation:

  • Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them.
  • This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier
  • The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.

Show Worked Solution

a.   4 am

b.   Signs of torpor:

  • The human maintained a steady body temperature throughout the observed period, showing no signs of torpor or reduced physiological activity
  • In contrast, the kookaburra exhibited classic torpor behaviour.
  • Its body temperature dropped significantly between 5 pm and 4 am, demonstrating the characteristic reduction in physiological functions during this period.
Mean mark (b) 56%.

c.   Kookaburra adaptation:

  • Kookaburras have an effective insulation mechanism where they puff out their feathers, creating space between them.
  • This fluffing action traps a layer of warm air between the feathers and the bird’s body, forming an insulating barrier
  • The trapped air pocket acts like natural insulation, minimising heat loss and helping the kookaburra maintain its body temperature efficiently.
♦ Mean mark (c) 47%.

Data Analysis, GEN2 2024 VCAA 1

Table 1 lists the Olympic year, \(\textit{year}\), and the gold medal-winning height for the men's high jump, \(\textit{Mgold}\), in metres, for each Olympic Games held from 1928 to 2020. No Olympic Games were held in 1940 or 1944, and the 2020 Olympic Games were held in 2021.

Table 1

\begin{array}{|c|c|}
\hline \quad \textit{year} \quad & \textit{Mgold}\,\text{(m)} \\
\hline 1928 & 1.94 \\
\hline 1932 & 1.97 \\
\hline 1936 & 2.03 \\
\hline 1948 & 1.98 \\
\hline 1952 & 2.04 \\
\hline 1956 & 2.12 \\
\hline 1960 & 2.16 \\
\hline 1964 & 2.18 \\
\hline 1968 & 2.24 \\
\hline 1972 & 2.23 \\
\hline 1976 & 2.25 \\
\hline 1980 & 2.36 \\
\hline 1984 & 2.35 \\
\hline 1988 & 2.38 \\
\hline 1992 & 2.34 \\
\hline 1996 & 2.39 \\
\hline 2000 & 2.35 \\
\hline 2004 & 2.36 \\
\hline 2008 & 2.36 \\
\hline 2012 & 2.33 \\
\hline 2016 & 2.38 \\
\hline 2020 & 2.37 \\
\hline
\end{array}

  1. For the data in Table 1, determine:
  2.  i. the maximum \(\textit{Mgold}\) in metres   (1 mark)

  3. ii. the percentage of \(\textit{Mgold}\) values greater than 2.25 m.   (1 mark)

  4. The mean of these \(\textit{Mgold}\) values is 2.23 m, and the standard deviation is 0.15 m.
  5. Calculate the standardised \(z\)-score for the 2000 \(\textit{Mgold}\) of 2.35 m.   (1 mark)

  6. Construct a boxplot for the \(\textit{Mgold}\) data in Table 1 on the grid below.   (2 marks)

     

  1. A least squares line can also be used to model the association between \(\textit{Mgold}\) and \(\textit{year}\).
  2. Using the data from Table 1, determine the equation of the least squares line for this data set.
  3. Use the template below to write your answer.
  4. Round the values of the intercept and slope to three significant figures.   (2 marks)

     

  1. The coefficient of determination is 0.857
  2. Interpret the coefficient of determination in terms of \(\textit{Mgold}\) and \(\textit{year}\).   (1 mark)

Show Answers Only

a.i.   \(2.39\)

a.ii.  \(50\%\)

b.    \(0.8\)

c.     

d.   
    

e.    \(\text{A coefficient of determination of 85.7% shows the variation in}\)

\(\text{the}\ Mgold\ \text{that is explained by the variation in the }year.\)

Show Worked Solution

a.i.   \(2.39\)

a.ii.  \(\dfrac{11}{22}=50\%\)

b.     \(z\) \(=\dfrac{x-\overline x}{s_x}\)
    \(=\dfrac{2.35-2.23}{0.15}\)
    \(=0.8\)

  
c.    \(Q_2=\dfrac{2.33+2.25}{2}=2.29\)

\(Q_1=2.12, \ Q_3=2.36\)

\(\text{Min}\ =1.94, \ \text{Max}\ =2.39\)
  

d.   \(\text{Using CAS:}\)


  
 

Mean mark (d) 52%.
Mean mark (e) 52%.

e.    \(\text{A coefficient of determination of 85.7% shows the variation in}\)

\(\text{the}\ Mgold\ \text{that is explained by the variation in the }year.\)

Vectors, EXT2 V1 2024 HSC 12a

The vector \(\underset{\sim}{a}\) is \(\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\) and the vector \(\underset{\sim}{b}\) is \(\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\).

  1. Find \(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}\).   (1 mark)

  2. Show that  \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}\)  is perpendicular to \(\underset{\sim}{b}\).   (2 marks)

Show Answers Only

1.     \(\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\)

ii.    \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)

\( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)

\(\therefore\ \text {Vectors are perpendicular.}\)

Show Worked Solution

i.    \(\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\)
 

\(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\)

 
ii.    \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)
 

\( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)

 
\(\therefore\ \text{Vectors are perpendicular.}\)

Complex Numbers, EXT2 N1 2024 HSC 11e

  1. Write the number  \(\sqrt{3}+i\)  in modulus-argument form.   (2 marks)

  2. Hence, or otherwise, write  \((\sqrt{3}+i)^7\)  in exact Cartesian form.   (2 marks)

Show Answers Only

i.     \(2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)

ii.    \(-64 \sqrt{3}-64 i\)

Show Worked Solution

i.     \(z=\sqrt{3}+i\)

\(|z|=\sqrt{3+1}=2\)

\(\arg (z)=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\dfrac{\pi}{6}\)

\(z=2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)
 

ii.     \((\sqrt{3}+i)^7\) \(=2^7\left(\cos \left(\dfrac{7 \pi}{6}\right)+i \sin \left(\dfrac{7 \pi}{6}\right)\right)\)
    \(=128\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} i\right)\)
    \(=-64 \sqrt{3}-64 i\)

Complex Numbers, EXT2 N1 2024 HSC 11b

Let  \(z=2+3 i\)  and  \(w=1-5 i\).

  1. Find  \(z+\bar{w}\).   (1 mark)

  2. Find  \(z^2\).   (1 mark)

Show Answers Only

i.    \(z+\bar{w}=3+8 i\)

ii.   \(z^2=-5+12 i\)

Show Worked Solution

i.     \(z=2+3 i\)

\(w=1-5 i \ \Rightarrow \ \bar{w}=1+5 i\)

\(z+\bar{w}=2+3 i+1+5 i=3+8 i\)
 

ii.    \(z^2\) \(=(2+3 i)^{2}\)
    \(=4+12 i+9 i^2\)
    \(=-5+12 i\)

Calculus, EXT2 C1 2024 HSC 11a

Find \(\displaystyle \int x e^x\, d x\)   (2 marks)

Show Answers Only

\(x e^x-e^x+c\)

Show Worked Solution

\(u=x \quad \ \ u^{\prime}=1\)

\(v^{\prime}=e^x \quad v=e^x\)

  \(\displaystyle\int x e^x \,d x\) \(=u v^{\prime}-\displaystyle \int v u^{\prime}\, d x\)
    \(=x e^x- \displaystyle \int e^x \cdot 1\, d x\)
    \(=x e^x-e^x+c\)

Vectors, EXT1 V1 2024 HSC 11a

Consider the vectors  \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}\)  and  \(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\).

  1. Find  \(2 \underset{\sim}{a}-\underset{\sim}{b}\).   (1 mark)

  2. Find  \(\underset{\sim}{a} \cdot \underset{\sim}{b}\).   (1 mark)

Show Answers Only

i.    \(\displaystyle \binom{7}{0}\)

ii.   \(5\)

Show Worked Solution

i.     \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}, \ \underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\)

\(2 \underset{\sim}{a}-\underset{\sim}{b}=2 \displaystyle \binom{3}{2}-\binom{-1}{4}=\binom{6}{4}-\binom{-1}{4}=\binom{7}{0}\)
 

ii.    \(\underset{\sim}{a} \cdot \underset{\sim}{b}=\displaystyle\binom{3}{2}\binom{-1}{4}=3 \times(-1)+2 \times 4=5\).

Measurement, STD1 M4 2024 HSC 2 MC

Four people completed the same fitness activity.

The graph shows the heart rate for each person before and after completing the activity.
 

Which person had the LEAST difference in heart rate?

  1. Jo
  2. Kim
  3. Lee
  4. Mal
Show Answers Only

\(B\)

Show Worked Solution
\(\text{Option A: }\)  \(\text{Jo}\)  \(=120-80=40\) 
\(\text{Option B: }\) \(\text{Kim}\) \(=120-100=20\ \checkmark\) 
\(\text{Option C: }\) \(\text{Lee}\) \(=120-90=30\) 
\(\text{Option D: }\) \(\text{Mal}\) \(=150-100=50\) 

  
\(\Rightarrow B\)

Proof, EXT2 P1 2024 HSC 3 MC

Consider the statement:

'If a polygon is a square, then it is a rectangle.'

Which of the following is the converse of the statement above?

  1. If a polygon is a rectangle, then it is a square.
  2. If a polygon is a rectangle, then it is not a square.
  3. If a polygon is not a rectangle, then it is not a square.
  4. If a polygon is not a square, then it is not a rectangle.
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Statement:}\  P \Rightarrow \ Q\)

\(\text{Converse of statement:}\  Q \Rightarrow \ P\)

\(\Rightarrow A\)

Proof, EXT2 P1 2024 HSC 2 MC

Consider the following statement written in the formal language of proof

\(\forall \theta \in\biggl(\dfrac{\pi}{2}, \pi\biggr) \exists\ \phi \in\biggl(\pi, \dfrac{3 \pi}{2}\biggr) ; \ \sin \theta=-\cos \phi\).

Which of the following best represents this statement?

  1. There exists a \(\theta\) in the second quadrant such that for all \(\phi\) in the third quadrant  \(\sin \theta=-\cos \phi\).
  2. There exists a \(\phi\) in the third quadrant such that for all \(\theta\) in the second quadrant  \(\sin \theta=-\cos \phi\).
  3. For all \(\theta\) in the second quadrant there exists a \(\phi\) in the third quadrant such that  \(\sin \theta=-\cos \phi\).
  4. For all \(\phi\) in the third quadrant there exists a \(\theta\) in the second quadrant such that  \(\sin \theta=-\cos \phi\).
Show Answers Only

\(C\)

Show Worked Solution

\(\Rightarrow C\)

Trigonometry, 2ADV T1 2024 HSC 20

A vertical tower \(T C\) is 40 metres high. The point \(A\) is due east of the base of the tower \(C\). The angle of elevation to the top \(T\) of the tower from \(A\) is 35°. A second point \(B\) is on a different bearing from the tower as shown. The angle of elevation to the top of the tower from \(B\) is 30°. The points \(A\) and \(B\) are 100 metres apart.
 

  1. Show that distance \(A C\) is 57.13 metres, correct to 2 decimal places.   (1 mark)
  1. Find the bearing of \(B\) from \(C\) to the nearest degree.   (3 marks)
Show Answers Only

\(194^{\circ}\)

Show Worked Solution

a.   \(\text{In}\ \Delta TCA:\)

\(\tan 35°\) \( =\dfrac{40}{AC}\)  
\(AC\) \( =\dfrac{40}{\tan 35°}=57.125…=57.13\ \text{m (2 d.p.)}\)  

 
b.   \(\text{In}\ \Delta TCB:\)

\(\tan 30°\) \( =\dfrac{40}{BC}\)  
\(BC\) \( =\dfrac{40}{\tan 30°}=69.28\ \text{m}\)  

  
\( \text{Find} \ \angle BCA \ \text{using cosine rule:}\)

\(\cos \angle B CA\) \( = \dfrac{57.13^2+69.28^2-100^2}{2 \times 57.13 \times 69.28}= -0.2446…\)  
\(\angle BCA\) \( = 104.2° \)  

\(\therefore\ \text{Bearing of}\ B\ \text{from}\ C= 90+104=194^{\circ} \text{(nearest degree)} \)

Networks, STD2 N2 2024 HSC 16

A network of towns and the distances between them in kilometres is shown.
 

  1. What is the shortest path from \(T\) to \(H\) ?   (2 marks)

  2. A truck driver needs to travel from \(Y\) to \(G\) but knows that the road from \(C\) to \(G\) is closed.
  3. What is the length of the shortest path the truck driver can take from \(Y\) to \(G\) after the road closure?   (2 marks)

Show Answers Only

a.   \(TYWH\)

b.  \(\text{Length of shortest path}\ (YWHMG) = 89 \text{km}\)

Show Worked Solution

a.    \(TYH=30+38=68, \quad TYWH=30+15+20=65\)

\(\therefore \text{ Shortest Path is}\ TYWH.\)
 

b.    \(Y W C M G=15+25+25+25=90\)

\(YWHMG=15+20+29+25=89\)

\(\Rightarrow \ \text{All other paths are longer.}\)

\(\therefore\text{ Length of shortest path = 89 km}\)

BIOLOGY, M1 EQ-Bank 4 MC

Which of the following structures is present in both prokaryotic and eukaryotic cells?

  1. Golgi apparatus
  2. Mitochondria
  3. Ribosomes
  4. Endoplasmic reticulum
Show Answers Only

\(C\)

Show Worked Solution
  • Both prokaryotic and eukaryotic cells contain ribosomes, which are responsible for protein synthesis.
  • Prokaryotes lack membrane-bound organelles like mitochondria or the Golgi apparatus.

\(\Rightarrow C\)

BIOLOGY, M6 2019 VCE 27 MC

Farmers and supermarkets agree that green beans are bought more frequently than yellow beans. A supermarket has asked a farmer to produce only green beans.

One way this could be achieved is by

  1. condensation polymerisation.
  2. DNA hybridisation.
  3. selective breeding.
  4. adaptive radiation.
Show Answers Only

\(C\)

Show Worked Solution
  • Selective breeding involves choosing parents with desirable traits and breeding them to produce offspring with those traits.
  • In this case, the farmer would selectively breed green bean plants to produce more green beans, as that is the preferred variety by consumers and supermarkets.
  • Selective breeding does not involve genetic engineering techniques like DNA hybridization (option B) and is a common agricultural practice used to enhance desirable traits in crops, unlike options A and D which are not relevant to this scenario.

\(\Rightarrow C\)

BIOLOGY, M5 2019 VCE 5 MC

Which one of the following statements about proteins is correct?

  1. The activity of a protein may be affected by the temperature and pH of its environment.
  2. The primary structure of a protein refers to its three-dimensional protein shape.
  3. Proteins are not involved in the human immune response.
  4. A protein with a quaternary structure will be an enzyme.
Show Answers Only

\(A\)

Show Worked Solution

Consider option A:

  • Changes in temperature or pH can disrupt the interactions that stabilise a protein’s higher-order structure.
  • This causes it to lose its three-dimensional shape and become denatured which usually results in a non-functional protein (Option A is Correct)

\(\Rightarrow A\)

BIOLOGY, M7 2021 VCE 36 MC

A study assessed the effectiveness and safety of a drug called doxycycline. One hundred and fifty adults hospitalised with malaria were involved. These adults were randomly placed into two groups of equal size. One group received doxycycline in addition to standard care. The other group received standard care only.

The group receiving standard care only was the

  1. control group.
  2. variable group.
  3. unsupported group.
  4. experimental group.
Show Answers Only

\(A\)

Show Worked Solution
  • A control group does not receive the experimental treatment or any intervention being tested.

\(\Rightarrow A\)

v1 Algebra, STD2 A2 2012 HSC 13 MC

Conversion graphs can be used to convert from one currency to another.  
  


  

Abbie converted 70 New Zealand dollars into Euros. She then converted all of these Euros into Australian dollars.

How much money, in Australian dollars, should Abbie have? 

  1. $30  
  2. $45
  3. $55
  4. $95
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Using the graphs:}\)

\($70\ \text{New Zealand}\) \(=40\ \text{Euro}\)
\(40\ \text{Euro}\) \(=$55\  \text{Australian}\)

 
\(\Rightarrow C\)

v1 Algebra, STD2 A2 2022 HSC 16

Rhonda is 38 years old, and likes to keep fit by doing cross-fit classes.

  1. Use this formula to find her maximum heart rate (bpm).   (1 mark)
             Maximum heart rate = 220 – age in years

  2. Rhonda will get the most benefit from this exercise if her heart rate is between 65% and 85% of her maximum heart rate.
  3. Between what two heart rates should Rhonda be aiming for to get the most benefit from her exercise?  (2 marks)

Show Answers Only

a.   \(182\ \text{bpm}\)

b.   \(118-155\ \text{bpm}\)

Show Worked Solution
a.     \(\text{Max heart rate}\) \(=220-38\)
    \(=182\ \text{bpm}\)

 

b.    \(\text{65% max heart rate}\ = 0.65\times 182 = 118.3\ \text{bpm}\)

\(\text{85% max heart rate}\ = 0.85\times 182 = 154.7\ \text{bpm}\)

\(\therefore\ \text{Rhonda should aim for between 118 and 155 bpm during exercise.}\)

v1 Algebra, STD2 A4 2022 HSC 22

The formula  \(C=80n+b\)  is used to calculate the cost of producing desktop computers, where \(C\) is the cost in dollars, \(n\) is the number of desktop computers produced and \(b\) is the fixed cost in dollars.

  1. Find the cost \(C\) when 2458 desktop computers are produced and the fixed cost is \($18\ 230\).  (1 mark)

  2. Some desktop computers have extra features added. The formula to calculate the production cost for these desktop computers is
  3.     \(C=80n+an+18\ 230\)
  4. where \(a\) is the additional cost in dollars per desktop computer produced.
  5. Find the number of desktop computers produced if the additional cost is $35 per desktop computer and the total production cost is \($103\ 330\).  (2 marks)

Show Answers Only
  1. \($214\ 870\)
  2. \(740\ \text{desktop computers}\)
Show Worked Solution

a.   \(\text{Find}\ C,\ \text{given}\ n=2458\ \text{and}\ b=18\ 230\)

\(C\) \(=80\times 2458+18\ 230\)  
  \(=$214\ 870\)  

 

b.   \(\text{Find}\ n,\ \text{given}\ C=18 \ 230\ \text{and}\ a=35\)

\(C\) \(=80n+an+18\ 230\)
\(103\ 330\) \(=80n+35n+18\ 230\)
\(115n\) \(=85\ 100\)
\(n\) \(=\dfrac{85\ 100}{115}\)
  \(=740\ \text{desktop computers}\)

v1 Algebra, STD2 A1 2006 HSC 2 MC

If  \(V=\dfrac{4}{3}\pi r^3\), what is the value of  \(V\) when  \(r = 5\), correct to two decimal places?

  1. \(20.94\)
  2. \(53.05\)
  3. \(104.72\)
  4. \(523.60\)
Show Answers Only

\(D\)

Show Worked Solution

\(V =\dfrac{4}{3}\pi r^3\)

\(\text{When}\  r = 2,\)

\(V\) \(=\dfrac{4}{3}\pi\times 5^3\)
  \(=523.598\dots\)

 
\(\Rightarrow D\)

v1 Algebra, STD2 A1 SM-Bank 2

If   \(A=P(1 + r)^n\), find  \(A\)  given  \(P=$500\),  \(r=0.09\) and  \(n=5\) (give your answer to the nearest cent).  (2 marks)

Show Answers Only

\($769.31\ \text{(nearest cent)}\)

Show Worked Solution
\(A\) \(=P(1 + r)^n\)
  \(=500(1 + 0.09)^5\)
  \(=500(1.09)^5\)
  \(=769.311\dots\)
  \(=$769.31\ \text{(nearest cent)}\)

v1 Algebra, STD2 A1 2017 HSC 7 MC

It is given that  \(I=\dfrac{3}{2}MR^2\).

What is the value of  \(I\) when  \(M =19.12\) and  \(R = 1.02\), correct to two decimal places?

  1. \(13.26\)
  2. \(29.84\)
  3. \(119.35\)
  4. \(570.52\)
Show Answers Only

\(B\)

Show Worked Solution
\(I\) \(=\dfrac{3}{2}\times 19.12\times 1.02^2\)
  \(=29.84\)

 

\(\Rightarrow B\)

PHYSICS, M6 2019 VCE 1 MC

Magnetic and gravitational forces have a variety of properties.

Which of the following best describes the attraction/repulsion properties of magnetic and gravitational forces?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\quad \textbf{Magnetic forces}\rule[-1ex]{0pt}{0pt}& \ \textbf{Gravitational forces} \\
\hline
\rule{0pt}{2.5ex}\text{either attract or repel}\rule[-1ex]{0pt}{0pt}&\text{only attract}\\
\hline
\rule{0pt}{2.5ex}\text{only repel}\rule[-1ex]{0pt}{0pt}& \text{neither attract nor repel}\\
\hline
\rule{0pt}{2.5ex}\text{only attract}\rule[-1ex]{0pt}{0pt}& \text{only attract} \\
\hline
\rule{0pt}{2.5ex}\text{either attract or repel}\rule[-1ex]{0pt}{0pt}& \text{either attract or repel} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • In magnets, like poles repel and opposite poles attract.
  • Gravitational forces are only attractive.

\(\Rightarrow A\)

Probability, MET2 2022 VCAA 3

Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.

Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.

Mika flips the coin five times.

    1. Find \(\text{Pr}(X=5)\).   (1 mark)
    2. Find \(\text{Pr}(X \geq 2).\) (1 mark)
    3. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)
    4. Find the expected value and the standard deviation for \(X\).   (2 marks)

The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
  

\(f(h)=\begin{cases} ah^2+bh+c         &\ \ 1.5\leq h\leq 3 \\ \\ 0       &\ \ \text{elsewhere} \\ \end{cases}\)
  

where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.

    1. State the value of the definite integral \(\displaystyle\int_{1.5}^3 f(h)\,dh\).   (1 mark)
    2. Given that  \(\text{Pr}(H \leq 2)=0.35\)  and  \(\text{Pr}(H \geq 2.5)=0.25\), find the values of \(a, b\) and \(c\).   (3 marks)

    3. The ceiling of Mika's room is 3 m above the floor. The minimum distance between the coin and the ceiling is a continuous random variable, \(D\), with probability density function \(g\).
    4. The function \(g\) is a transformation of the function \(f\) given by \(g(d)=f(rd+s)\), where \(d\) is the minimum distance between the coin and the ceiling, and \(r\) and \(s\) are real constants.
    5. Find the values of \(r\) and \(s\).   (1 mark)

  1. Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of \(p\) of landing on heads and \((1-p)\) of landing on tails, where \(p\) is a constant value between 0 and 1 .
  2. Bella flips her coin 25 times in order to estimate \(p\).
  3. Let \(\hat{P}\) be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.
    1. Is the random variable \(\hat{P}\) discrete or continuous? Justify your answer.   (1 mark)
    2. If \(\hat{p}=0.4\), find an approximate 95% confidence interval for \(p\), correct to three decimal places.   (1 mark)
    3. Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips.
    4. If \(\hat{p}=0.4\), how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?   (1 mark)

Show Answers Only

a. i.    `frac{1}{32}`    ii.    `frac{13}{16}`    iii.    `0.806` (3 d.p.)

a. iv    `text{E}(X)=5/2,  text{sd}(X)=\frac{\sqrt{5}}{2}`

b. i.   `1`   

b. ii.    `a=-frac{4}{5},  b=frac{17}{5},  c=-frac{167}{60}`

b. iii. `r=-1,  s=3`

c. i.   `text{Discrete}`   ii.   `(0.208,  0.592)`   iii.   `n=100`

Show Worked Solution

a.i  `X ~ text{Bi}(5 , frac{1}{2})`

`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
 

a.ii  By CAS:    `text{binomCdf}(5,0.5,2,5)`     `0.8125`

`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`

  
a.iii  `\text{Pr}(X \geq 2 | X<5)`

`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`

  
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
 

`= 0.806452 ~~ 0.806` (3 decimal places)
  

a.iv `X ~ text{Bi}(5 , frac{1}{2})`

`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`

`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`

  

b.i   `\int_{1.5}^3 f(h) d h = 1`


♦♦ Mean mark (b.i) 40%.
MARKER’S COMMENT: Many students did not evaluate the integral or evaluated incorrectly.

b.ii  By CAS:

`f(h):= a\·\h^2 + b\·\h +c`
 

`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`

`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\  c = =-2.78 \dot{3} = frac{-167}{60}`


♦♦ Mean mark (b.ii) 47%.
MARKER’S COMMENT: Many students did not give exact answers.

b.iii  `h + d = 3`

`:.\  f(h) = f(3  –  d) = f(- d + 3)`

`:.\  r = – 1 ` and ` s = 3`


♦♦♦♦ Mean mark (b.iii) 10%.
MARKER’S COMMENT: Many students did not attempt this question.

c.i  `\hat{p}`  is discrete.

The number of coin flips must be zero or a positive integer so  `\hat{p}`  is countable and therefore discrete.
 

c.ii  `\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`

`\left(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`

`\approx(0.208\ ,0.592)`

 

c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.

`:.\  \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`

`:.\  n = 100`

She would need to flip the coin 100 times


♦♦♦ Mean mark (c.iii) 30%.
MARKER’S COMMENT: Common incorrect answers were 0, 10, 11, 50 and 101.

Calculus, MET2 2022 VCAA 2

On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.

The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all `t \geq 0`, where time `t` is measured in weeks.

One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.

The graph has been drawn to scale.
 

The population of rabbits can be modelled by the rule `r(t)=1700 \sin \left(\frac{\pi t}{80}\right)+2500`.

  1.   i. State the initial population of rabbits.   (1 mark)

  2.  ii. State the minimum and maximum population of rabbits.   (1 mark)

  3. iii. State the number of weeks between maximum populations of rabbits.   (1 mark)

The population of foxes can be modelled by the rule `f(t)=a \sin (b(t-60))+1600`.

  1. Show that `a=900` and `b=\frac{\pi}{80}`.   (2 marks)

  2. Find the maximum combined population of foxes and rabbits. Give your answer correct to the nearest whole number.   (1 mark)

  3. What is the number of weeks between the periods when the combined population of foxes and rabbits is a maximum?   (1 mark)

The population of foxes is better modelled by the transformation of `y=\sin (t)` under `Q` given by
 

  1. Find the average population during the first 300 weeks for the combined population of foxes and rabbits, where the population of foxes is modelled by the transformation of `y=\sin(t)` under the transformation `Q`. Give your answer correct to the nearest whole number.   (4 marks)

Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.

The population of rabbits over a longer period of time can be modelled by the rule

`s(t)=1700cdote^(-0.003t)cdot sin((pit)/80)+2500,\qquad text(for all)\ t>=0`

  1. Find the average rate of change between the first two times when the population of rabbits is at a maximum. Give your answer correct to one decimal place.   (2 marks)

  2. Find the time, where `t>40`, in weeks, when the rate of change of the rabbit population is at its greatest positive value. Give your answer correct to the nearest whole number.   (2 marks)

  3. Over time, the rabbit population approaches a particular value.
  4. State this value.   (1 mark)

Show Answers Only

ai.   `r(0)=2500`

aii.  Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

aiii. `160`  weeks

b.    See worked solution.

c.    `~~ 5339` (nearest whole number)

d.    Weeks between the periods is 160

e.    `~~ 4142` (nearest whole number)

f.    Average rate of change `=-3.6` rabbits/week (1 d.p.)

g.    `t = 156` weeks (nearest whole number)  

h.    ` s → 2500`

Show Worked Solution

ai.  Initial population of rabbits

From graph when `t=0, \ r(0) = 2500`

Using formula when `t=0`

`r(t)` `= 1700\ sin \left(\frac{\pi t}{80}\right)+2500`  
`r(0)` `= 1700\ sin \left(\frac{\pi xx 0}{80}\right)+2500 = 2500`  rabbits  


aii. From graph,

Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

OR

Using formula

Minimum is when `t = 120`

`r(120) = 1700\ sin \left(\frac{\pi xx 120}{80}\right)+2500 = 1700 xx (-1) + 2500 = 800`

Maximum is when `t = 40`

`r(40) = 1700\ sin \left(\frac{\pi xx 40}{80}\right)+2500 = 1700 xx (1) + 2500 = 4200`

 
aiii. Number of weeks between maximum populations of rabbits `= 200-40 = 160`  weeks

 
b.  Period of foxes = period of rabbits = 160:

`frac{\2pi}{b} = 160`

`:.\  b = frac{\2pi}{160} = frac{\pi}{80}` which is the same period as the rabbit population.

Using the point `(100 , 2500)`

Amplitude when `b = frac{\pi}{80}`: 

`f(t)` `=a \ sin (pi/80(t-60))+1600`  
`f(100)` `= 2500`  
`2500` `= a \ sin (pi/80(100-60))+1600`  
`2500` `= a \ sin (pi/2)+1600`  
`a` `= 2500-1600 = 900`  

 
`:.\  f(t)= 900 \ sin (pi/80)(t-60) + 1600`

  

c.    Using CAS find `h(t) = f(t) + r(t)`:

`h(t):=900 \cdot \sin \left(\frac{\pi}{80} \cdot(t-60)\right)+1600+1700\cdot \sin \left(\frac{\pit}{80}\right) +2500`

  
`text{fMax}(h(t),t)|0 <= t <= 160`      `t = 53.7306….`
  

`h(53.7306…)=5339.46`

  
Maximum combined population `~~ 5339` (nearest whole number)


♦♦ Mean mark (c) 40%.
MARKER’S COMMENT: Many rounding errors with a common error being 5340. Many students incorrectly added the max value of rabbits to the max value of foxes, however, these points occurred at different times.

d.    Using CAS, check by changing domain to 0 to 320.

`text{fMax}(h(t),t)|0 <= t <= 320`     `t = 213.7305…`
 
`h(213.7305…)=5339.4568….`
 
Therefore, the number of weeks between the periods is 160.
  

e.    Fox population:

`t^{\prime} = frac{90}{pi}t + 60`   →   `t = frac{pi}{90}(t^{\prime}-60)`

`y^{\prime} = 900y+1600`   →   `y = frac{1}{900}(y^{\prime}-1600)`

`frac{y^{\prime}-1600}{900} = sin(frac{pi(t^{\prime}-60)}{90})`

`:.\  f(t) = 900\ sin\frac{pi}{90}(t-60) + 1600`

  
Average combined population  [Using CAS]   
  
`=\frac{1}{300} \int_0^{300} left(\900 \sin \left(\frac{\pi(t-60)}{90}\right)+1600+1700\ sin\ left(\frac{\pi t}{80}\right)+2500\right) d t`
  

`= 4142.2646…..  ~~ 4142` (nearest whole number)


♦ Mean mark (e) 40%.
MARKER’S COMMENT: Common incorrect answer 1600. Some incorrectly subtracted `r(t)`. Others used average rate of change instead of average value.

f.   Using CAS

`s(t):= 1700e^(-0.003t) dot\sin\frac{pit}{80} + 2500`
 

`text{fMax}(s(t),t)|0<=t<=320`        `x = 38.0584….`
 

`s(38.0584….)=4012.1666….`
 

`text{fMax}(s(t),t)|160<=t<=320`     `x = 198.0584….`
 

`s(198.0584….)=3435.7035….`
 

Av rate of change between the points

`(38.058 , 4012.167)`  and  `(198.058 , 3435.704)`

`= frac{4012.1666….-3435.7035….}{38.0584….-198.0584….} =-3.60289….`
 

`:.` Average rate of change `=-3.6` rabbits/week (1 d.p.)


♦ Mean mark (f ) 45%.
MARKER’S COMMENT: Some students rounded too early.
`frac{s(200)-s(40)}{200-40}` was commonly seen.
Some found average rate of change between max and min populations.

g.   Using CAS

`s^(primeprime)(t) = 0` , `t = 80(n-0.049) \ \forall n \in Z`

After testing `n = 1, 2, 3, 4` greatest positive value occurs for `n = 2` 

`t` `= 80(n-0.049)`  
  `= 80(2-0.049)`  
  `= 156.08`  

 
`:. \ t = 156` weeks (nearest whole number)


♦♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: Many students solved `frac{ds}{dt}=0`.
A common answer was 41.8, as `s(156.11…)=41.79`.
Another common incorrect answer was 76 weeks.

h.   As `t → ∞`, `e^(-0.003t) → 0`

`:.\ s → 2500`

Calculus, MET2 2022 VCAA 1

The diagram below shows part of the graph of `y=f(x)`, where `f(x)=\frac{x^2}{12}`.
 

  1. State the equation of the axis of symmetry of the graph of `f`.   (1 mark)

  2. State the derivative of `f` with respect to `x`.   (1 mark)

The tangent to `f` at point `M` has gradient `-2` .

  1. Find the equation of the tangent to `f` at point `M`.   (2 marks)

The diagram below shows part of the graph of `y=f(x)`, the tangent to `f` at point `M` and the line perpendicular to the tangent at point `M`.
 

 

  1.  i. Find the equation of the line perpendicular to the tangent passing through point `M`.   (1 mark)

  2. ii. The line perpendicular to the tangent at point `M` also cuts `f` at point `N`, as shown in the diagram above.
  3.     Find the area enclosed by this line and the curve `y=f(x)`.   (2 marks)

  4. Another parabola is defined by the rule `g(x)=\frac{x^2}{4 a^2}`, where `a>0`.
  5. A tangent to `g` and the line perpendicular to the tangent at `x=-b`, where `b>0`, are shown below.

  1. Find the value of `b`, in terms of `a`, such that the shaded area is a minimum.   (4 marks)

Show Answers Only

a.    `x=0`

b.    ` f^{\prime}(x)=1/6x`

c.    `x=-12`

di.    `y=1/2x + 18`

dii.   Area`= 375` units²

e.    `b = 2a^2`

Show Worked Solution

a.   Axis of symmetry:  `x=0`
  

b.    `f(x)`  `=\frac{x^2}{12}`  
  ` f^{\prime}(x)` `= 1/6x`  

  
c.   At `M` gradient `= -2`

`1/6x` `= -2`  
`x` `= -12`  

 
When  `x = -12`
 

`f(x) = (-12)^2/12 = 12`
 
Equation of tangent at `(-12 , 12)`:

`y-y_1` `=m(x-x_1)`  
`y-12` `= -2(x + 12)`  
`y` `= -2x-12`  


d.i   Gradient of tangent `= -2`

`:.`  gradient of normal `= 1/2` 

Equation at `M(- 12 , 12)`

`y -y_1` `=m(x-x_1)`  
`y-12` `= 1/2(x + 12)`  
`y` `=1/2x + 18`  


d.ii  Points of intersection of `f(x)` and normal are at  `M` and `N`.

So equate ` y = x^2/12` and  `y = 1/2x + 18` to find `N`

`x^2/12` `=1/2x + 18`  
`x^2-6x-216` `=0`  
`(x + 12)(x-18)` `=0`  

   
`:.\  x = -12` or `x = 18`

Area `= \int_{-12}^{18}\left(\frac{1}{2} x+18-\frac{x^2}{12}\right) d x`  
  `= [x^2/4 + 18x-x^3/36]_(-12)^18`  
  `= [18^2/4 +18^2-18^3/36] – [12^2/4 + 18 xx (-12)-(-12)^3/36]`  
  `= 375` units²  

 

e.   `g(x) = x^2/(4a^2)`   `a > 0`

At `x = -b`   `y = (-b)^2/(4a^2) = b^2/4a^2`

`g^{\prime}(x) = (2x)/(4a^2) = x/(2a^2)`

Gradient of tangent `= (-b)/(2a^2)`

Gradient of normal `= (2a^2)/b`

Equation of normal at `(- b , b^2/(4a^2))`

`y-y_1` `= m(x-x_1)`  
`y-b^2/(4a^2)` `= (2a^2)/b(x-(-b))`  
`y` `= (2a^2x)/b + 2a^2 + b^2/(4a^2)`  
`y` `= (2a^2x)/b +(8a^4 + b^2)/(4a^2)`  

 
Points of intersection of normal and parabola (Using CAS)

solve `((2a^2x)/b +(8a^4 + b^2)/(4a^2) = x^2/(4a^2),x)`

`x =-b`  or  `x = (8a^4+b^2)/b`

 
Calculate area using CAS

`A = \int_{-b}^{(8a^4+b^2)/b}\left(\frac{2a^2x}{b} +\frac{8a^4 + b^2}{4a^2}-frac{x^2}{4a^2} \right) dx`

`A = frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3}`
 
Using CAS Solve derivative of `A = 0`  with respect to `b` to find `b`

solve`(d/(db)(frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3})=0,b)`

 
`b =-2a^2`   and  `b = 2a^2`

Given `b > 0`

`b = 2a^2`


♦♦ Mean mark (e) 33%.
MARKER’S COMMENT: Common errors: Students found `\int_{-b}^{\frac{8 a^4+b^2}{b}}\left(y_n\right) d x` and failed to subtract `g(x)` or had incorrect terminals.

Calculus, MET2 2023 VCAA 3

Consider the function \(g:R \to R, g(x)=2^x+5\).

  1. State the value of \(\lim\limits_{x\to -\infty} g(x)\).   (1 mark)
  2. The derivative, \(g^{'}(x)\), can be expressed in the form \(g^{'}(x)=k\times 2^x\).
  3. Find the real number \(k\).   (1 mark)
  4.  i. Let \(a\) be a real number. Find, in terms of \(a\), the equation of the tangent to \(g\) at the point \(\big(a, g(a)\big)\).   (1 mark)

    ii. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places.   (2 marks)

  1.  

Let \(h:R\to R, h(x)=2^x-x^2\).

  1. Find the coordinates of the point of inflection for \(h\), correct to two decimal places.   (1 mark)
  2. Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
  3. Give your answer correct to two decimal places.   (1 mark)
  4. Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
  5. Write the estimates \(x_1, x_2,\) and \(x_3\) in the table below, correct to three decimal places.   (2 marks)
      

    \begin{array} {|c|c|}
    \hline
    \rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
    \hline
    \rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \end{array}
  6. For the function \(h\), explain why a solution to the equation \(\log_e(2)\times (2^x)-2x=0\) should not be used as an initial estimate \(x_0\) in Newton's method.   (1 mark)
  7. There is a positive real number \(n\) for which the function \(f(x)=n^x-x^n\) has a local minimum on the \(x\)-axis.
  8. Find this value of \(n\).   (2 marks)
Show Answers Only

a.    \(5\)

b.    \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci.  \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)

\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

cii. \(y=4.255x\)

d. \((2.06 , -0.07)\)

e. \([0.49, 3.21]\)

f. 

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.773 \\
\hline
\end{array}

g. \(\text{See worked solution.}\)

h. \(n=e\)

Show Worked Solution

a.    \(\text{As }x\to -\infty,\ \ 2^x\to 0\)

\(\therefore\ 2^x+5\to 5\)
  

b.     \(g(x)\) \(=2^x+5\)
    \(=\Big(e^{\log_{e}{2}}\Big)^x\)
    \(=e\ ^{x\log_{e}{2}}+5\)
  \(g^{\prime}(x)\) \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\)
     \(=\log_{e}{2}\times 2^x\)
  \(\therefore\ k\) \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)
   
ci.  \(\text{Tangent at}\ (a, g(a)):\)

\(y-(2^a+5)\) \(=\log_{e}{2}\times2^a(x-a)\)
\(\therefore\ y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

♦♦ Mean mark (c)(i) 50%.

cii.  \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)

\( y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

  
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)

\(\text{Equation of tangent when }\ a\approx 2.6178\)

\( y\) \(=2^{2.6178..}\ \log_{e}{(2)x}+0\)
\(\therefore\  y\) \(=4.255x\)

♦♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students did not substitute (0, 0) into the correct equation or did not find the value of \(a\) and used \(a=0\).
d.     \(h(x)\) \(=2^x-x^2\)
  \(h^{\prime}(x)\) \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\)
  \(h^{”}(x)\) \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\)

\(\text{Solving }h^{”}(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)

\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)

\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)

e.    \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)

\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)


♦♦ Mean mark (e) 40%.
MARKER’S COMMENT: Round brackets were often used which were incorrect as endpoints were included. Some responses showed the interval where the function was strictly increasing.

f.    \(\text{Newton’s Method }\Rightarrow\  x_a-\dfrac{h(x_a)}{h'(x_a)}\) 

\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)

\(h(x)=2x-x^2\ \text{and }h^{\prime}(x)=\ln{2}\times 2^x-2x\)

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}

g.    \(\text{The denominator in Newton’s Method is}\ h^{\prime}(x)=\log_{e}{(2)}\cdot 2^x-2x\)

\(\text{and the calculation will be undefined if }h^{\prime}(x)=0\ \text{as the tangent lines are horizontal}.\)

\(\therefore\ \text{The solution to }h^{\prime}(x)=0\ \text{cannot be used for }x_0.\)

♦♦♦ Mean mark (g) 20%.

h.    \(\text{For a local minimum }f(x)=0\)

\(\rightarrow\ n^x-x^n=0\)

\(\rightarrow\ n^x=x^n\ \ \ (1)\)

\(\text{Also for a local minimum }f^{\prime}(x)=0\)

\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)

\(\text{Substitute (1) into (2)}\)

\(\ln(n)\cdot x^n-nx^{n-1}=0\) 

\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)

\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)

\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)

\(\therefore\ n=e\)


♦♦♦ Mean mark (h) 10%.
MARKER’S COMMENT: Many students showed that \(f'(x)=0\) but failed to couple it with \(f(x)=0\). Ensure exact values are given where indicated not approximations.

Calculus, MET2 2023 VCAA 1

Let \(f:R \rightarrow R, f(x)=x(x-2)(x+1)\). Part of the graph of \(f\) is shown below.

  1. State the coordinates of all axial intercepts of \(f\).   (1 mark)
  2. Find the coordinates of the stationary points of \(f\).   (2 marks)
    1. Let \(g:R\rightarrow R, g(x)=x-2\).
    2. Find the values of \(x\) for which \(f(x)=g(x)\).   (1 mark)
    1. Write down an expression using definite integrals that gives the area of the regions bound by \(f\) and \(g\).  (2 marks)
    2. Hence, find the total area of the regions bound by \(f\) and \(g\), correct to two decimal places.   (1 mark)
  1. Let \(h:R\rightarrow R, h(x)=(x-a)(x-b)^2\), where \(h(x)=f(x)+k\) and \(a, b, k \in R\).
  2. Find the possible values of \(a\) and \(b\).   (4 marks)
Show Answers Only

a.    \((-1, 0), (0, 0), (2, 0)\)

b.    \(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)

c.i.  \(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

c.ii. \(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

 \(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)

c.iii. \(5.95\)

d.   \(\text{1st case }\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case }\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

Show Worked Solution

a.    \((-1, 0), (0, 0), (2, 0)\)
  

b.    \(\text{Using CAS solve for}\ x:\)

\(\dfrac{d}{dx}(x(x-2)(x+1))=0\)

\(\therefore\ x=\dfrac{1-\sqrt{7}}{3}\ \text{and }x=\dfrac{1+\sqrt{7}}{3}\)

\(\text{Substitute }x\ \text{values into }f(x)\ \text{using CAS to get}\ y\ \text{values}\)

\(\text{The stationary points of }f\ \text{are}:\)

\(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)
  

ci    \(\text{Given }f(x)=g(x)\)

\(x(x-2)(x+1)\) \(=x-2\)
\(x(x-2)(x+1)(x-2)\) \(=0\)
\((x-2)(x(x+1)-1)\) \(=0\)
\((x-2)(x^2+x-1)\) \(=0\)

  
\(\therefore\ \text{Using CAS: } \)

\(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

cii  \(\text{Area of bounded region:}\)

\(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

\(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)
  

ciii  \(\text{Solve the integral in c.ii above using CAS:}\)
  \(\text{Total area}=5.946045..\approx 5.95\)

  

d.   \(\text{Method 1 – Equating coefficients}\)

\((x-a)(x-b)^2=x(x-2)(x+1)+k\)

\(x^3-2bx^2-ax^2+b^2x+2abx-ab^2=x^3-x^2-2x+k\)

\((x^3-(a+2b)x^2+(2ab+b^2)x-ab^2=x^3-x^2-2x+k\)

\(\therefore\ -(a+2b)=-1\ \to\ a=1-2b …(1)\)

\(2ab+b^2=-2\ \ …(2)\)

\(\text{Substitute (1) into (2) and solve for }b.\)

\(2b(1-2b)+b^2\) \(=-2\)
\(3b^2-2b-2\) \(=0\)
\(b\) \(=\dfrac{1\pm \sqrt{7}}{3}\)
\(\text{When }b\) \(=\dfrac{1+\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1+\sqrt{7}}{3}\Bigg)=\dfrac{-2\sqrt{7}+1}{3}\)
\(\text{When }b\) \(=\dfrac{1-\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1-\sqrt{7}}{3}\Bigg)=\dfrac{2\sqrt{7}+1}{3}\)

  

\(\text{Method 2 – Using transformations}\)

\(\text{The squared factor in }(x-a)(x-b)^2=x(x-2)(x+1)+k,\)

\(\text{shows that the turning point is on the }x\ \text{axis}.\)

\(\therefore\ \text{Lowering }f(x)\ \text{by }\dfrac{2(7\sqrt{7}-10)}{27}\ \text{and raising }f(x)\ \text{by }\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\text{will give the 2 possible sets of values for }a\ \text{and}\ b.\)

\(\text{1st case – lowering using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)-\dfrac{2(7\sqrt{7}-10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case – raising using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)+\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)