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Functions, 2ADV F1 2019 HSC 13e

  1. Sketch the graph of  `y = |\ x-1\ |`  for  `-4 <= x <= 4`.  (1 mark)

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  2. Using the sketch from part i, or otherwise, solve  `|\ x-1\ | = 2x + 4`.  (2 marks)

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Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(-1, 2)`
Show Worked Solution
i.   

 

ii.    `text(By inspection, intersection when)\ x = -1`

`text(Test:)`

`|-1-1|` `= -2 + 4`
`2` `= 2`

 
`:.\ text(Intersection at)\ (-1, 2)`

Filed Under: Further Functions and Relations (Y11), Graph Transformations (Adv-2027), Other Functions and Relations (Adv-2027) Tagged With: Band 4, smc-6218-10-Absolute Value, smc-6408-15-Absolute Value, smc-6408-60-Combinations, smc-987-10-Absolute Value, smc-987-30-Reflections and Other Graphs

Networks, STD2 N2 2019 HSC 30

The network diagram shows the tracks connecting 8 picnic sites in a nature park. The vertices `A` to `H` represents the picnic sites. The weights on the edges represent the distance along the tracks between the picnic sites, in kilometres.
 


 

  1. Each picnic site needs to provide drinking water. The main water source is at site `A`.

     

    Draw a minimum spanning tree and calculate the minimum length of water pipes required to supply water to all the sites if the water pipes can only be laid along the tracks.  (2 marks)

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  2. One day, the track between `C` and `H` is closed. State the vertices that identify the shortest path from `C` to `E` that avoids the closed track.  (1 mark)

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Show Answers Only
  1. `text(25 km)`
  2. `CGHE`
Show Worked Solution

a.   `text(One strategy – using Prim’s algorithm:)`

`text(Starting at)\ A`

`text(1st edge -)\ AB,\ \ text(2nd edge -)\ BC`

`text(3rd edge -)\ CH,\ \ text(4th edge -)\ HG`

`text(5th edge -)\ GF,\ \ text(6th edge -)\ HD`

`text(7th edge -)\ DE\ text(or)\ HE`
 

`text(Maximum length = 4 + 5 + 3 + 2 + 1 + 5 + 5 = 25 km)`

 

b.   `text(Shortest Path is)\ CGHE`

Filed Under: Minimum Spanning Trees, Shortest Paths, Shortest Paths (Std2-2027), Spanning Trees (Std2-2027) Tagged With: Band 4, smc-6308-10-Network Diagrams, smc-6320-20-Distance, smc-6320-40-Draw Tree, smc-913-10-Network Diagram, smc-914-20-Distance, smc-914-40-Draw Tree

Calculus, EXT1* C3 2019 HSC 13d

The diagram shows the region bounded by the curve  `y = x - x^3`, and the `x`-axis between  `x = 0`  and  `x = 1`. The region is rotated about the `x`-axis to form a solid.
 


 

Find the exact value of the volume of the solid formed.  (3 marks)

Show Answers Only

`(8 pi)/105\ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 (x – x^3)^2 dx`
  `= pi int_0^1 x^2 – 2x^4 + x^6\ dx`
  `= pi [x^3/3 – 2/5 x^5 + 1/7 x^7]_0^1`
  `= pi(1/3 – 2/5 + 1/7)`
  `= (8 pi)/105\ text(u³)`

Filed Under: Further Area and Solids of Revolution (Ext1) Tagged With: Band 4, smc-1039-10-Polynomial, smc-1039-60-x-axis Rotation

Trigonometry, 2ADV T1 2019 HSC 13b

The diagram shows a circle with centre `O` and radius 20 cm.

The points `A` and `B` lie on the circle such that  `∠AOB = 70^@`.
 


 

Find the perimeter of the shaded segment, giving your answer correct to one decimal place.  (3 marks)

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`47.4\ text{cm}`

Show Worked Solution
`text(Arc)\ AB` `= 70/360 xx 2pi xx 20`
  `= 24.43…`

 
`text(Using cosine rule,)`

`AB^2` `= 20^2 xx 20^2 – 2 ⋅ 20 ⋅ 20 xx cos 70`
  `= 526.383…`
`AB` `= 22.94…`

 

`:.\ text(Perimeter)` `= 24.43 + 22.94…`
  `= 47.37`
  `= 47.4\ text{cm (1 d.p.)}`

Filed Under: Circular Measure (Adv-2027), Circular Measure (Y11) Tagged With: Band 4, smc-6394-10-Arc Length/Perimeter, smc-978-10-Arc Length/Perimeter

Calculus, 2ADV C4 2019 HSC 12d

The diagram shows the graph of  `y = (3x)/(x^2 + 1)`.
 

 
The region enclosed by the graph, the `x`-axis and the line  `x = 3`  is shaded.

Calculate the exact value of the area of the shaded region.  (3 marks)

Show Answers Only

`3/2 ln 10\ text(u²)`

Show Worked Solution
`text(Area)` `= int_0^3 (3x)/(x^2 + 1)\ dx`
  `= 3/2 int_0^3 (2x)/(x^2 + 1)\ dx`
  `= 3/2[ln(x^2 + 1)]_0^3`
  `= 3/2(ln 10 – ln 1)`
  `= 3/2 ln10\ \ text(u²)`

Filed Under: Areas Under Curves (Y12) Tagged With: Band 4, smc-975-30-Hyperbola/Quotient

Calculus, EXT1* C1 2019 HSC 12c

The number of leaves, `L(t)`, on a tree `t` days after the start of autumn can be modelled by

`L(t) = 200\ 000e^(-0.14t)`

  1. What is the number of leaves on the tree when  `t = 31`?  (1 mark)

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  2. What is the rate of change of the number of leaves on the tree when  `t = 31`?  (2 marks)

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  3. For what value of `t` are there 100 leaves on the tree?  (2 marks)

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Show Answers Only
  1. `2607\ text(leaves)`
  2. `-365.02…`
  3. `54.3\ text{(1 d.p.)}`
Show Worked Solution

i.  `text(When)\ \ t = 31`

`L(t)` `= 200\ 000 xx e^(-0.14(31))`
  `=2607.305…`
  `= 2607\ text(leaves)`

 

ii.    `L` `= 2000\ 000^(-0.14t)`
  `(dL)/(dt)` `= -0.14 xx 200\ 000 e^(0.14t`
    `= -28\ 000e^(-0.14t)`

 
`text(When)\ \ t = 31,`

`(dL)/(dt)` `= -28\ 000 xx e^(-0.14(31))`
  `= -365.02…`

 
`:.\ text(365 leaves fall per day.)`

 

iii.   `text(Find)\ t\ text(when)\ \ L = 100:`

`100` `= 200\ 000 e^(-0.14t)`
`e^(-0.14t)` `= 0.0005`
`e^(-0.14t)` `= ln 0.0005`
`t` `= (ln 0.0005)/(-0.14)`
  `= 54.292…`
  `= 54.3\ text{(1 d.p.)}`

Filed Under: Standard Growth and Decay Tagged With: Band 2, Band 3, Band 4, smc-1081-20-Decay, smc-1081-50-Other Themes

Functions, EXT1* F1 2019 HSC 11g

The parabola  `y = x^2`  meets the line  `y = x + 2`  at the points  `(-1, 1)`  and  `(2, 4)`. Do NOT prove this.

By first sketching the graphs of  `y = x^2`  and  `y = x + 2`, shade the region which simultaneously satisfies the two inequalities  `y >= x^2`  and  `y >= x + 2`.  (2 marks)

Show Answers Only

Show Worked Solution

Filed Under: Inequalities (Ext1) Tagged With: Band 4, smc-1033-40-Regions

Algebra, STD2 A1 2019 HSC 28

The formula below is used to calculate an estimate for blood alcohol content `(BAC)` for females.

`BAC_text(female) = (10N - 7.5H)/(5.5M)`

The number of hours required for a person to reach zero `BAC` after they stop consuming alcohol is given by the following formula.

`text(Time) = (BAC)/0.015`

The number of standard drinks in a glass of wine and a glass of spirits is shown.
 

Hannah weighs 60 kg. She consumed 3 glasses of wine and 4 glasses of spirits between 6:15 pm and 12:30 am the following day. She then stopped drinking alcohol.

Using the given formulae, calculate the time in the morning when Hannah's `BAC` should reach zero.  (4 marks)

Show Answers Only

`text(6:23 am)`

Show Worked Solution

`text(Standard drinks consumed)\ (N) = 3 xx 1.2 + 4 = 7.6`

`text(Hours drinking)\ (H) = text(6 h 15 min = 6.25 hours)`

`BAC(text(Hannah))` `= (10 xx 7.6 – 7.5 xx 6.25)/(5.5 xx 60)`
  `= 0.08825…`

COMMENT: Convert a decimal answer into hours and minutes using the calculator degree/minute function.

`text(Time(to zero))` `= (0.08825…)/0.015`
  `= 5.883…\ text(hours)`
  `= 5\ text(hours 53 minutes)`

 
`:.\ text(Hannah should reach zero)\ BAC\ text(at 6:23 am)`

Filed Under: Applications: BAC, Medication and D=SxT (Std 2), Applications: BAC, Medicine and D=SxT (Std1-2027), Applications: BAC, Medicine and D=SxT (Std2-2027) Tagged With: Band 4, smc-6235-10-BAC, smc-6509-10-BAC, smc-791-10-BAC

Networks, STD2 N3 2019 HSC 26

A project requires activities `A` to `F` to be completed. The activity chart shows the immediate prerequisite(s) and duration for each activity.

  1. By drawing a network diagram, determine the minimum time for the project to be completed.  (3 marks)

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  2. Determine the float time of the non-critical activity.  (1 mark)

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  1. `text(15 hours)`
  2. `text(3 hours)`
Show Worked Solution
a.   

 

`text(Scanning forwards:)`

`text(Minimum time = 2 + 6 + 2 + 4 + 1 = 15 hours)`

`text{(Scanning forwards and backwards is highly recommended but not}`

 `text{required in the network diagram.)}`

 

b.   `text(Critical Path is)\ ABDEF.`

♦♦ Mean mark 30%.

`text(Non-critical activity is)\ C.`

`text(Float time)` `= 5 – 2`
  `= 3\ text(hours)`

Filed Under: Critical Paths Tagged With: Band 4, Band 5, smc-916-10-Table to Network, smc-916-30-Scanning Both Ways

Calculus, 2ADV C4 2019 HSC 11e

Evaluate  `int_0^1 1/(3x + 2)^2\ dx`.  (2 marks)

Show Answers Only

`1/10`

Show Worked Solution
`int_0^1 1/(3x + 2)^2\ dx` `= int_0^1 (3x + 2)^(-2)\ dx`
  `= [-1/3 (3x + 2)^(-1)]_0^1`
  `= [-1/3 ⋅ 1/5 – (-1/3 ⋅ 1/2)]`
  `= -1/15 + 1/6`
  `= 1/10`

Filed Under: Standard Integration Tagged With: Band 4, smc-1202-20-Definite Integrals

Algebra, STD2 A2 2019 HSC 14 MC

Last Saturday, Luke had 165 followers on social media. Rhys had 537 followers. On average, Luke gains another 3 followers per day and Rhys loses 2 followers per day.

If  `x`  represents the number of days since last Saturday and  `y`  represents the number of followers, which pair of equations model this situation?

A.  `text(Luke:)\ \ y = 165x + 3`

 

`text(Rhys:)\ \ y = 537x - 2`

B. `text(Luke:)\ \ y = 165 + 3x`

 

`text(Rhys:)\ \ y = 537 - 2x`

C. `text(Luke:)\ \ y = 3x + 165`

 

`text(Rhys:)\ \ y = 2x - 537`

D. `text(Luke:)\ \ y = 3 + 165x`

 

`text(Rhys:)\ \ y = 2 - 537x`

Show Answers Only

`B`

Show Worked Solution

`text(Luke starts with 165 and adds 3 per day:)`

`y = 165 + 3x`

`text(Rhys starts with 537 and loses 2 per day:)`

`y = 537 – 2x`

`=> B`

Filed Under: Applications: Currency, Fuel and Other Problems (Std 2), Linear Applications, Linear Functions (Adv-2027), Linear Functions (Y11) Tagged With: Band 4, common-content, num-title-ct-coreb, num-title-qs-hsc, smc-6214-50-Simultaneous Equations, smc-793-30-Other Linear Applications, smc-985-40-Simultaneous Equations

Financial Maths, STD2 F4 2019 HSC 13 MC

The graph show the future values over time of  `$P`, invested at three different rates of compound interest.
 


 

Which of the following correctly identifies each graph?

A. B.
C. D.
Show Answers Only

`C`

Show Worked Solution

`text(Values increase quicker)`

`text(- higher compounding interest rate)`

`text(- same rate but more frequent compounding period)`

`:. W = 10text(% quarterly)`

`X = 10text(% annually)`

`Y = 5text(% annually)`

 
`=> C`

Filed Under: Compound Interest, Compound Interest and Shares (Std2), Modelling Investments and Loans (Y12) Tagged With: Band 4, common-content, num-title-ct-coreb, num-title-qs-hsc, smc-1002-30-Interest Rate Graph Comparison, smc-4334-40-Find r, smc-4334-60-Graphs, smc-817-30-i/r comparisons (incl. graphs)

Measurement, STD2 M2 2019 HSC 5 MC

The Coordinated Universal Time (UTC) of Auckland is +12 hours and the UTC of Chicago is −5 hours.

When the time in Chicago is 2 pm, Thursday, what is the time in Auckland?

  1. 9 pm, Wednesday
  2. 7 am, Thursday
  3. 9 pm, Thursday
  4. 7 am, Friday
Show Answers Only

`D`

Show Worked Solution

`text(Auckland is 17 hours ahead of Chicago.)`

`text(2 pm Thursday + 17 hours)`

`text(= 2 am Friday + 5 hours)`

`text(= 7 am Friday)`

`=> D`

Filed Under: M2 Working with Time (Y11), Time and Time Difference (Std1-2027), Time and Time Difference (Std2-2027) Tagged With: Band 4, smc-6306-10-Time Differences, smc-6525-10-Time Differences, smc-776-10-Time Differences

Financial Maths, STD2 F4 2019 HSC 3 MC

Chris opens a bank account and deposits $1000 into it. Interest is paid at 3.5% per annum, compounding annually.

Assuming no further deposits or withdrawals are made, what will be the balance in the account at the end of two years?

  1. $1070.00
  2. $1071.23
  3. $1822.50
  4. $2070.00
Show Answers Only

`=> B`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 1000(1 + 0.035)^2`
  `= $1071.23`

 
`=> B`

Filed Under: Compound Interest, Compound Interest and Shares (Std2), Modelling Investments and Loans (Y12) Tagged With: Band 4, common-content, num-title-ct-coreb, num-title-qs-hsc, smc-1002-20-FV Formula, smc-4334-10-Find FV, smc-817-20-FV Formula

Statistics, STD2 S4 2019 HSC 23

A set of bivariate data is collected by measuring the height and arm span of seven children. The graph shows a scatterplot of these measurements.
 


 

  1. Calculate Pearson's correlation coefficient for the data, correct to two decimal places.  (1 mark)

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  2. Identify the direction and the strength of the linear association between height and arm span.  (1 mark)

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  3. The equation of the least-squares regression line is shown.
     
               Height = 0.866 × (arm span) + 23.7
     
    A child has an arm span of 143 cm.

     

    Calculate the predicted height for this child using the equation of the least-squares regression line.  (1 mark)

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Show Answers Only
  1. `0.98\ \ (text(2 d.p.))`
  2. `text(Direction: positive)`
    `text(Strength: strong)`
  3. `147.538\ text(cm)`
Show Worked Solution

a.   `text{Use  “A + Bx”  function (fx-82 calc):}`

♦ Mean mark 40%.
COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 correlation” .

`r` `= 0.9811…`
  `= 0.98\ \ (text(2 d.p.))`

 

b.   `text(Direction: positive)`

`text(Strength: strong)`

 

c.    `text(Height)` `= 0.866 xx 143 + 23.7`
    `= 147.538\ text(cm)`

Filed Under: Bivariate Data Analysis (Y12), S4 Bivariate Data Analysis (Y12) Tagged With: Band 3, Band 4, Band 5, common-content, smc-1001-30-Correlation, smc-1001-40-Pearson's, smc-1001-70-Calculator (Stats Mode), smc-785-30-Correlation, smc-785-40-Pearson's, smc-785-70-Calculator (Stats Mode)

Measurement, STD2 M6 2019 HSC 22

Two right-angled triangles, `ABC` and `ADC`, are shown.
 

Calculate the size of angle `theta`, correct to the nearest minute.  (3 marks)

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`41°4^{′}\ \ text{(nearest minute)}`

Show Worked Solution

`text(Using Pythagoras in)\ DeltaACD:`

Mean mark 51%.

`AC^2` `= 2.5^2 + 6^2`
  `= 42.25`
`:.AC` `= 6.5\ text(cm)`

 
`text(In)\ DeltaABC:`

`costheta` `= 4.9/6.5`
`theta` `= cos^(−1)\ 4.9/6.5`
  `= 41.075…`
  `= 41°4^{′}31^{″}`
  `= 41°5^{′}\ \ text{(nearest minute)}`

Filed Under: Pythagoras and Right-Angled Trig (Std2), Right-Angled Trig Tagged With: Band 4, num-title-ct-corea, num-title-qs-hsc, smc-4552-45-2-triangles, smc-802-10-Pythagoras, smc-802-20-Right-Angled Trig, smc-802-50-Rounding to the Minute

Financial Maths, STD2 F4 2019 HSC 21

A person owns 1526 shares with a market value of $8.75 per share. The total dividend received for these shares is $1068.20.

Calculate the percentage dividend yield.  (2 marks)

Show Answers Only

`8text(%)`

Show Worked Solution
`text(Value of shares)` `= 1526 xx 8.75`
  `= $13\ 352.50`

 

`:.\ text(Dividend yield)` `= 1068.20/(13\ 352.50)`
  `= 0.08`
  `= 8text(%)`

Filed Under: Compound Interest and Shares (Std2) Tagged With: Band 4, smc-817-40-Shares/Dividends

Probability, STD2 S2 2019 HSC 20

A roulette wheel has the numbers 0, 1, 2, …, 36 where each of the 37 numbers is equally likely to be spun.
 

 
If the wheel is spun 18 500 times, calculate the expected frequency of spinning the number 8.  (2 marks)

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Show Answers Only

`500`

Show Worked Solution

`P(8) = 1/37`

`:.\ text(Expected Frequency (8))`

`= 1/37 xx 18\ 500`

`= 500`

Filed Under: Probability, Relative Frequency (Std 1), Relative Frequency (Std 2), Relative Frequency (Y11) Tagged With: Band 4, common-content, num-title-ct-core, num-title-qs-hsc, smc-1133-20-Games of Chance, smc-4225-35-Relative frequency, smc-827-20-Games of Chance, smc-827-40-Expected Frequency (np), smc-990-20-Games of Chance, smc-990-40-Expected Frequency (np)

Statistics, STD2 S1 2019 HSC 19

The heights, in centimetres, of 10 players on a basketball team are shown.

170, 180, 185, 188, 192, 193, 193, 194, 196, 202

Is the height of the shortest player on the team considered an outlier? Justify your answer with calculations. (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`Q_1 = 185, \ Q_3 = 194`

Mean mark 51%.
COMMENT: The last statement must be made to achieve full marks here!

`IQR = 194 – 185 = 9`

`text(Shortest player = 170)`

`text(Outlier height:)`

`Q_1 – 1.5 xx IQR ` `= 185 – 1.5 xx 9`
  `= 171.5`

 
`:.\ text(S)text(ince 170 < 171.5, 170 is an outlier.)`

Filed Under: Measures of Centre and Spread (Std2-2027), Summary Statistics - No Graph (Std 2), Summary Statistics - No graph (Y12) Tagged With: Band 4, common-content, smc-6312-30-IQR and Outliers, smc-824-30-IQR and Outliers, smc-999-30-IQR and Outliers

Measurement, STD2 M7 2019 HSC 18

Andrew, Brandon and Cosmo are the first three batters in the school cricket team. In a recent match, Andrew scored 30 runs, Brandon scored 25 runs and Cosmo scored 40 runs.

  1. What is the ratio of Andrew's to Brandon's to Cosmo's runs scored, in simplest form?  (2 marks)

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  2. In this match, the ratio of the total number of runs scored by Andrew, Brandon and Cosmo to the total number of runs scored by the whole team is `19:36`.
  3. How many runs were scored by the whole team?  (2 marks)

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Show Answers Only
  1. `6:5:8`
  2. `180\ text(runs)`
Show Worked Solution
a.    `A:B:C` `= 30:25:40`
    `= 6:5:8`

 

b.    `text(Total runs by)\ A,B,C` `= 30 + 25 + 40`
    `= 95`

 

`text(Let)\ R` `=\ text(team runs)`
`R/95` `= 36/19`
`:. R` `= (36 xx 95)/19`
  `= 180\ text(runs)`

Filed Under: Ratio and Scale (Std2) Tagged With: Band 3, Band 4, smc-1187-10-Ratio (2 part), smc-1187-20-Ratio (3 part)

Measurement, STD2 M6 2019 HSC 17

The diagram shows a triangle with sides of length `x` cm, 11 cm and 13 cm and an angle of 80°.
 


 

Use the cosine rule to calculate the value of `x`, correct to two significant figures.  (3 marks)

Show Answers Only

`16\ text(cm  (2 sig. fig.))`

Show Worked Solution
`x^2` `= 11^2 + 13^2 – 2 xx 11 xx 13 xx cos80°`
  `= 240.336…`

 

`:.x` `= 15.502…`
  `= 16\ text(cm  (2 sig. fig.))`

Filed Under: Non-Right Angled Trig (Std2) Tagged With: Band 4, smc-804-10-Cosine Rule

Measurement, STD2 M1 2019 HSC 16

A bowl is in the shape of a hemisphere with a diameter of 16 cm.
 

What is the volume of the bowl, correct to the nearest cubic centimetre?  (2 marks)

Show Answers Only

`1072\ text(cm)^3`

Show Worked Solution
`V` `= 1/2 xx 4/3pir^3`
  `= 1/2 xx 4/3 xx pi xx 8^3`
  `= 1072.3…`
  `= 1072\ text{cm}^3\ text{(nearest cm}^3 text{)}`

Filed Under: Perimeter, Area and Volume (Std 2), Volume, Mass and Capacity (Std1-2027), Volume, Mass and Capacity (Std2-2027) Tagged With: Band 4, num-title-ct-pathb, num-title-qs-hsc, smc-4235-60-Spheres, smc-6304-50-Volume (Circular Measure), smc-6521-50-Volume (Circular Measure), smc-798-50-Volume (Circular Measure)

Calculus, 2ADV C1 2019 HSC 8 MC

A particle is moving along a straight line. The graph shows the acceleration of the particle.
 


 

For what value of `t` is the velocity `v` a maximum?

  1. `1`
  2. `2`
  3. `3`
  4. `5`
Show Answers Only

`C`

Show Worked Solution

`text(Velocity increases when)\ \ a > 0.`

`:. v_text(max)\ \ text(occurs when)\ \ t = 3.`

`=>  C`

Filed Under: Rates of Change (Adv-2027), Rates of Change (Y11) Tagged With: Band 4, smc-1083-10-Motion Graphs, smc-6438-10-Motion Graphs

Trigonometry, 2ADV T3 2019 HSC 7 MC

The diagram shows part of the graph of  `y = a sin(bx) + 4`.
 

What are the values of `a` and `b`?

  1. `a = 3 qquad qquad b = 1/2`
  2. `a = 3 qquad qquad b = 2`
  3. `a = 1.5 qquad \ b = 1/2`
  4. `a = 1.5 qquad \ b = 2`
Show Answers Only

`D`

Show Worked Solution

`a = 1/2 (5.5 – 2.5) = 1.5`

`text(S)text(ince graph passes through)\ \ (pi/4, 5.5):`

`5.5 = 1.5 sin(b xx pi/4) + 4`

`sin (b xx pi/4)` `= 1`
`b xx pi/4` `= pi/2`
`:. b` `= 2`

 
`=>  D`

Filed Under: Trig Graphs (Y12) Tagged With: Band 4, smc-977-10-sin

Probability, 2ADV S1 2019 HSC 6 MC

A game is played by tossing an ordinary 6-sided die and an ordinary coin at the same time. The game is won if the uppermost face of the die shows an even number or the uppermost face of the coin shows a tail (or both).

What is the probability of winning this game?

  1. `1/4`
  2. `1/2`
  3. `3/4`
  4. `1`
Show Answers Only

`C`

Show Worked Solution

`text(Game lost only if an odd and a head show.)`

`:. P(W)` `= 1 – P text{(odd)} ⋅ P text{(head)}`
  `= 1 – 3/6 ⋅ 1/2`
  `= 3/4`

 
`=>  C`

Filed Under: Multi-Stage Events (Adv-2027), Multi-Stage Events (Y11) Tagged With: Band 4, smc-6469-20-Other Multi-Stage Events, smc-6469-30-Complementary Probability, smc-989-20-Other Multi-Stage Events, smc-989-30-Complementary Probability

L&E, 2ADV E1 2019 HSC 5 MC

Which of the following is equal to  `(log_2 9)/(log_2 3)`?

  1. `2`
  2. `3`
  3. `log_2 3`
  4. `log_2 6`
Show Answers Only

`A`

Show Worked Solution
`(log_2 9)/(log_2 3)` `= (log_2 3^2)/(log_2 3)`
  `= (2 log_2 3)/(log_2 3)`
  `= 2`

 
`=>  A`

Filed Under: Log/Index Laws and Equations (Adv-2027), Log/Index Laws and Equations (Y11), Logarithms Tagged With: Band 4, num-title-ct-patha, num-title-qs-hsc, smc-4243-30-Power rule, smc-6455-20-Logs - Power Rule, smc-963-20-Log - power rule

L&E, 2ADV E1 2019 HSC 3 MC

What is the value of  `p` so that  `(a^2a^(-3))/sqrt a = a^p`?

  1. `-3`
  2. `-3/2`
  3. `-1/2`
  4. `12`
Show Answers Only

`B`

Show Worked Solution
`(a^2 a^(-3))/a^(1/2)` `= a^(-1) xx a^(-1/2)`
  `= a^(-3/2)`

 
`=>  B`

Filed Under: Indices, Log/Index Laws and Equations (Adv-2027), Log/Index Laws and Equations (Y11) Tagged With: Band 4, num-title-ct-pathc, num-title-qs-hsc, smc-4228-60-Fractional indices, smc-6455-50-Exponential Equations, smc-963-50-Exponential Equation

Vectors, EXT1 V1 SM-Bank 8

A projectile is fired horizontally off a cliff at an initial speed of  `V`  metres per second.
 

 

The projectile strikes the water, `l`  metres from the base of the cliff.

Let `g` be the acceleration due to gravity and assume air resistance is negligible.

  1.  Show the projectile hits the water when
     
    `qquadt = sqrt((2d)/g)`  (2 marks)

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  2.  If  `l`  equals twice the height of the cliff, at what angle does the projectile hit the water?  (2 marks)

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  3.  Show that the speed at which the projectile hits the water is
     
    `qquad2sqrt(dg)`  metres per second.  (1 mark)

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Show Answers Only
  1. `text(See Worked Solutions)`
  2. `45°`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `underset~v` `= Vcosthetaunderset~i + (Vsintheta – g t)underset~j`
    `= Vcos0 underset~i + (Vsin0 – g t)underset~j`
    `= Vunderset~i – g tunderset~j`
`underset~s` `= intunderset~v\ dt`
  `= Vtunderset~i – 1/2g t^2 underset~j + c`

 
`text(When)\ t = 0, underset~s = 0, c = 0`
 

`text(Time of flight:)`

`underset~j\ text(component of)\ underset~s = −d`

`−1/2 g t^2` `= −d`
`t^2` `= (2d)/g`
`t` `= sqrt((2d)/g)`

 

ii.   `l = 2d\ \ (text(given))`

`text(Projectile hits water at)\ theta:`
 

`overset.y` `= underset~j\ text(component of)\ underset~v`
  `= −g t`
  `= −g · sqrt((2d)/g)`
  `= −sqrt(2dg)`
   
`overset.x` `= underset~i\ text(component of)\ underset~v`
  `= V`

 
`text(When)\ \ t = sqrt((2d)/g),`

`underset~i\ text(component of)\ underset~s = 2d`

`2d` `= V · sqrt((2d)/g)`
`V` `= (2dsqrtg)/sqrt(2d) = sqrt(2dg)`

 

`tantheta` `= (|overset.y|)/(|overset.x|)= sqrt(2dg)/sqrt(2dg)=1`
`:. theta` `= 45°`

 

iii.   `text(Speed = magnitude of velocity)`

`|underset~v|` `= sqrt((sqrt(2dg))^2 + (sqrt(2dg))^2)`
  `= sqrt(4dg)`
  `= 2sqrt(dg)`

Filed Under: Vectors and Projectile Motion (Ext1) Tagged With: Band 4, Band 5, smc-1087-50-Angle at Impact, smc-1087-60-Speed at Impact

Vectors, EXT1 V1 EQ-Bank 1

A basketball player aims to throw a basketball through a ring, the centre of which is at a horizontal distance of 4.5 m from the point of release of the ball and 3 m above floor level. The ball is released at a height of 1.75 m above floor level, at an angle of projection `alpha` to the horizontal and at a speed of  `V\ text(ms)^(-1)`. Air resistance is assumed to be negligible.
 


 

The position vector of the centre of the ball at any time, `t` seconds, for  `t >= 0`, relative to the point of release is given by 
 
`qquad underset ~s(t) = Vt cos (alpha) underset ~i + (Vt sin(alpha) - 4.9t^2) underset ~j`,
 
where  `underset ~i`  is a unit vector in the horizontal direction of motion of the ball and  `underset ~j`  is a unit vector vertically up. Displacement components are measured in metres.

For the player’s first shot at goal, `V = 7\ text(ms)^(-1)` and  `alpha = 45^@`

  1. Find the time, in seconds, taken for the ball to reach its maximum height. Give your answer in the form  `(a sqrt b)/c`, where  `a, b` and `c` are positive integers.  (2 marks) 

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  2. Find the maximum height, in metres, above floor level, reached by the centre of the ball.  (2 marks)

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  3. Find the distance of the centre of the ball from the centre of the ring one second after release. Give your answer in metres, correct to two decimal places.  (2 marks)

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Show Answers Only
  1. `(5 sqrt 2)/14`
  2. `3\ text(m)`
  3. `128\ text(m)`
Show Worked Solution
a.i.    `underset ~s(t)` `=7t\ cos 45^@ underset ~i+(7t\ sin 45^@ – 4.9t^2) underset ~j`
    `=(7sqrt2 t)/2 underset ~i + ((7sqrt2)/2 t – 4.9t^2) underset ~j`
     

 `text(Maximum height  ⇒  find)\ \t\ text(when)\ \ underset~j\ \ text(component of)\ \ underset ~V(t) = 0:`

`underset~V = (7sqrt2)/2 underset ~i + ((7sqrt2)/2 – 9.8t) underset ~j`

`(7sqrt2)/2 – 9.8t` `= 0`
`t` `= (5 sqrt 2)/14\ \ text(seconds)`

 

a.ii.  `text(Max height)\ =>\ text(Find)\ \ underset~j\ \ text(component of)\ \ underset~s(t)\ \ text(when)\ \ t=(5 sqrt 2)/14`

  `underset ~s_(underset ~j)((5 sqrt 2)/14)` `= 7 xx (5 sqrt 2)/14 xx 1/sqrt 2 – 4.9 xx ((5 sqrt 2)/14)^2`
    `= 1.25\ text(m)`

 
`:.\ text(Height above floor) = 1.25 + 1.75 = 3\ text(m)`

 

a.iii.  `underset ~s_text(ring)= 4.5 underset ~i + 1.25 underset ~j`

`text(After 1 second,)`

`underset~s_text(ball) = 7/sqrt 2 underset ~i + ((7sqrt2)/2 – 4.9)underset ~j`

  `:.d` `= |underset ~s_text(ring) – underset ~s(1)|`
    `= sqrt((4.5 – 7/sqrt 2)^2 + (1.25 – 7/sqrt 2 + 4.9)^2)`
    `~~ 1.28\ text(m)`

Filed Under: Vectors and Projectile Motion (Ext1) Tagged With: Band 4, Band 5, smc-1087-20-Maximum Height

Vectors, EXT1 V1 SM-Bank 6

A cricketer hits a ball at time  `t = 0`  seconds from an origin `O` at ground level across a level playing field.

The position vector  `underset ~s(t)`, from `O`, of the ball after `t` seconds is given by
 
  `qquad underset ~s(t) = 15t underset ~i + (15 sqrt 3 t - 4.9t^2)underset ~j`,
 
where,  `underset ~i`  is a unit vector in the forward direction, `underset ~j`  is a unit vector vertically up and displacement components are measured in metres.

  1. Find the initial velocity of the ball and the initial angle, in degrees, of its trajectory to the horizontal.  (2 marks)

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  2. Find the maximum height reached by the ball, giving your answer in metres, correct to two decimal places.  (2 marks)

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  3. Find the time of flight of the ball. Give your answer in seconds, correct to three decimal places.  (1 mark)

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  4. Find the range of the ball in metres, correct to one decimal place.  (1 mark)

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  5. A fielder, more than 40 m from `O`, catches the ball at a height of 2 m above the ground.

     

    How far horizontally from `O` is the fielder when the ball is caught? Give your answer in metres, correct to one decimal place.  (2 marks)

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Show Answers Only
  1. `underset ~v (0) = 15 underset ~i + 15 sqrt 3 underset ~j`

     

    `theta = pi/3 = 60^@`

  2. `34.44\ text(m)`
  3. `5.302\ text(s)`
  4. `79.5\ text(m)`
  5. `78.4\ text(m)`
Show Worked Solution

a.   `underset ~v (t) = underset ~ dot s (t) = 15 underset ~i + (15 sqrt 3 – 9.8t)underset ~j`

`text(Initial velocity occurs when)\ \ t=0:`

`:. underset ~v (0) = 15 underset ~i + 15 sqrt 3 underset ~j`
 

`text(Let)\ \ theta = text(Initial trajectory,)`

`tan theta` `=(15sqrt3)/15`  
  `=sqrt3`  
`:. theta` `=pi/3\ \ text(or)\ \ 60^@`  

 

b.  `text(Max height)\ =>underset~j\ \ text(component of)\ \ underset ~v=0.`

`15 sqrt 3 – 9.8t` `=0`
`t` `=(15 sqrt 3)/9.8`
  `=2.651…`

 
`text(Find max height when)\ \ t = 2.651…`

`:.\ text(Max height)` `= 15 sqrt 3 xx 2.651 – 4.9 xx (2.651)^2`
  `~~ 34.44\ text(m)`

 

c.   `text(Ball travels in symmetrical parabolic path.)`

`:.\ text(Total time of flight)`

`= 2 xx (15 sqrt 3)/9.8`

`= (15 sqrt 3)/4.9`

`~~ 5.302\ text(s)`
 

d.  `text(Range)\ =>underset~i\ \ text(component of)\ \ underset~s(t)\ \ text(when)\ \ t= (15 sqrt 3)/4.9`

`:.\ text(Range)` `= 15 xx (15 sqrt 3)/4.9`
  `= (225 sqrt 3)/4.9`
  `~~ 79.5\ text(m)`


e.
   `text(Find)\ t\ text(when height of ball = 2 m:)`

`15 sqrt 3 t – 4.9t^2` `=2`  
`4.9t^2 – 15 sqrt 3 t + 2` `=0`  

 

  `t=(15 sqrt 3 +- sqrt ((15 sqrt 3)^2 – 4 xx 4.9 xx2))/(2 xx 4.9)`  
     

`t ~~ 0.078131\ \ text(or)\ \ t ~~ 5.22406`

 
`text(When)\ \ t=0.0781,`

`x= 15 xx 0.0781 = 1.17\ text(m)\ \ text{(no solution →}\ x<40 text{)}`
 

 `text(When)\ \ t=5.2241,`

`x=15 xx 5.2241 = 78.4\ text(m)`
 

`:.\ text(Ball is caught 78.4 m horizontally from)\ O.`

Filed Under: Vectors and Projectile Motion (Ext1) Tagged With: Band 3, Band 4, Band 5, smc-1087-10-Range, smc-1087-20-Maximum Height, smc-1087-30-Time of Flight, smc-1087-40-Initial Angle

Statistics, EXT1 S1 SM-Bank 5

A manufacturer makes torches that have a probability of 0.03 of being defective.

Let  `overset^p`  be the random variable that represents the sample proportion of torches for samples of size `n` drawn from production.

Find the smallest integer value of  `n`  such that the standard deviation of  `overset^p`  is less than  `1/50`.  (2 marks)

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Show Answers Only

`73`

Show Worked Solution

`text(Let)\ \ X =\ text(number of defective torches)`

`X\ ~\ text(Bin)(n, p)\ ~\ text(Bin)(n, 0.03)`

`text(Var)(overset^p)` `= (p(1 – p))/n`
`σ(overset^p)` `= sqrt((0.03(0.97))/n)`
  `= sqrt(291/(10\ 000n))`

 

`sqrt(291/(10\ 000n))` `< 1/50`
`291/(10\ 000n)` `< 1/2500`
`10\ 000n` `> 2500 xx 291`
`n` `> (2500 xx 291)/(10\ 000)`
  `> 72.75`

 
`:.n_text(min) = 73`

Filed Under: Normal Approximations of Sample Proportions (Ext1) Tagged With: Band 4, smc-1085-30-Find smallest n

Statistics, EXT1 S1 SM-Bank 4 MC

When a standard 6-sided die is thrown, the probability that it shows a prime number is  `2/3`.

If 10 standard dice are thrown, the number, `N`, of times a prime number is showing has a binomial distribution.

What is the standard deviation of  `N`, correct to 3 decimal places?

  1. 0.222
  2. 0.471
  3. 1.491
  4. 2.222
Show Answers Only

`C`

Show Worked Solution

`N\ ~\ text(Bin)(n, p)\ ~\ text(Bin)(10, 2/3)`

`text(Var)(N)` `= np(1 – p)`
  `= 10 · 2/3(1 – 2/3)`
  `= 20/9`

 

`:. σ_N` `= sqrt(20/9)`
  `= 1.4907…`

 
`=>\ C`

Filed Under: Statistics and Binomial Distributions (Ext1) Tagged With: Band 4, smc-1199-20-Calculate Var(X)/Std Dev

Statistics, EXT1 S1 2017 MET1 4

In a large population of fish, the proportion of angel fish is  `1/4`.

Let  `hat p`  be the random variable that represents the sample proportion of angel fish for samples of size `n` drawn from the population.

Find the smallest integer value of `n` such that the standard deviation of  `hat p`  is less than or equal to  `1/100`.  (2 marks)

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Show Answers Only

`1875`

Show Worked Solution

`text(Let)\ \ X =\ text(number of angel fish)`

`X\ ~\ text(Bin) (n, 1/4)`

`σ\ (hat p)` `= sqrt((p(1 – p))/n)`
  `= sqrt((1/4 xx 3/4)/n)`
  `= sqrt(3/(16n))`

 

`text(Solve:)\ \ \ sqrt(3/(16n))` `<= 1/100`
`3/(16 n)` `<= 1/(10\ 000)`
`(30\ 000)/16` `<= n`
`:. n` `>= 1875`

 
`:. n_text(min) = 1875`

Filed Under: Normal Approximations of Sample Proportions (Ext1) Tagged With: Band 4, smc-1085-30-Find smallest n

Statistics, EXT1 S1 2012 MET2 3

Steve and Jess are two students who have agreed to take part in a psychology experiment. Each has to answer several sets of multiple-choice questions. Each set has the same number of questions, `n`, where `n` is a number greater than 20. For each question there are four possible options A, B, C or D, of which only one is correct.

  1. Steve decides to guess the answer to every question, so that for each question he chooses A, B, C or D at random.

     

    Let the random variable `X` be the number of questions that Steve answers correctly in a particular set.

    1. What is the probability that Steve will answer the first three questions of this set correctly?  (1 mark)

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    2. Use the fact that the variance of `X` is `75/16` to show that the value of `n` is 25.  (1 mark)

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  1. The probability that Jess will answer any question correctly, independently of her answer to any other question, is  `p\ (p > 0)`. Let the random variable `Y` be the number of questions that Jess answers correctly in any set of 25.

    If   `P(Y > 23) = 6 xx P(Y = 25)`, show that the value of  `p=5/6`.  (2 marks)

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Show Answers Only

a.i.  `1/64`

a.ii.  `text(See Worked Solutions)`

b.  `text(See Worked Solutions)`

Show Worked Solution
a.i.    `Ptext{(3 correct in a row)}` `= (1/4)^3`
    `= 1/64`

 

a.ii.    `text(Var)(X)` `= np(1 – p)`
  `75/16` `= n(1/4)(3/4)`
  `75` `= 3n`
  `:. n` `= 25`

 

b.   `Y ∼\ text(Bin)(25,p)`

♦♦♦ Mean mark part (c) 19%.
`P(Y > 23)` `= 6xx P(Y = 25)`
`P(Y = 24) + P(Y = 25)` `= 6xx P(Y = 25)`
`P(Y = 24)` `= 5xx P(Y = 25)`
`((25),(24))p^24(1 – p)^1` `= 5p^25`
`25p^24(1 – p)` `= 5p^25`
`25p^24-25p^25-5p^25` `=0`
`25p^24-30p^25` `=0`
`5p^24(5 – 6p)` `= 0`

 
`:. p = 5/6,\ \ (p>0)\ \ text(… as required)`

Filed Under: Statistics and Binomial Distributions (Ext1) Tagged With: Band 3, Band 4, Band 6, smc-1199-30-Find n/p given E(X)/Var(X)

Statistics, EXT1 S1 SM-Bank 3

In a chocolate factory the material for making each chocolate is sent to a machines.

The time, `X` seconds, taken to produce a chocolate by machine is a binomial distribution where it can be shown that  `P(X <= 3) = 9/32`.

A random sample of 10 chocolates is chosen. Find the probability, correct to two decimal places, that exactly 4 of these 10 chocolates took 3 or less seconds to produce.  (2 marks)

Show Answers Only

`0.18`

Show Worked Solution

`text(Let)\ \ C = \ text(number of chocolates that take less than 3 seconds)`

COMMENT: Take care as  `P(X <= 3) = 9/32`  provides the equivalent of  `p` here.

`C ∼\ text(Bin)(n, p) ∼\ text(Bin)(10, 9/32)`

`P(C = 4)`  `=((10),(4)) (9/32)^4 (23/32)^6` 
  `=0.181…`
  `=0.18\ \ \ text{(2 d.p.)}`

Filed Under: Binomial Probability (Ext1) Tagged With: Band 4, smc-1084-10-General Case, smc-1084-15-Defective products

Statistics, EXT1 S1 SM-Bank 2

A school has a class set of 22 new laptops kept in a recharging trolley. Provided each laptop is correctly plugged into the trolley after use, its battery recharges.

On a particular day, a class of 22 students uses the laptops. All laptop batteries are fully charged at the start of the lesson. Each student uses and returns exactly one laptop. The probability that a student does not correctly plug their laptop into the trolley at the end of the lesson is 10%. The correctness of any student’s plugging-in is independent of any other student’s correctness.

Determine the probability that at least one of the laptops is not correctly plugged into the trolley at the end of the lesson. Give your answer correct to three decimal places.  (2 marks)

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Show Answers Only

`0.902`

Show Worked Solution

`text(Let)\ \ X = text(number not correctly plugged)`

`X\ ~\ text(Bin) (n,p)\ ~\ text(Bin) (22, 0.1)`

`P(X>=1)` `=1 – P(X=0)`
  `=1-((n),(0)) (0.1^0)(0.9^22)`
  `=1 – 0.9^22`
  `=0.9015…`
  `=0.902\ \ text{(3 d.p.)}`

Filed Under: Binomial Probability (Ext1) Tagged With: Band 4, smc-1084-10-General Case

Statistics, EXT1 S1 2007 MET1 5

It is known that 50% of the customers who enter a restaurant order a cup of coffee. If four customers enter the restaurant, what is the probability that more than two of these customers order coffee? (Assume that what any customer orders is independent of what any other customer orders.)  (2 marks)

Show Answers Only

`5/16`

Show Worked Solution

`X~\ text(Bin)(n,p)\ ~\ text(Bin)(4, 1/2)`

`P(X > 2)` `= P(X = 3) + P(X = 4)`
  `= ((4),(3))(1/2)^3(1/2) + ((4),(4))(1/2)^4(1/2)^0`
  `= 4 xx 1/16 + 1 xx 1/16` 
  `=5/16`

Filed Under: Binomial Probability (Ext1) Tagged With: Band 4, smc-1084-10-General Case

Statistics, EXT1 S1 2011 MET1 7

A biased coin tossed three times. The probability of a head from a toss of this coin is `p.`

  1. Find, in terms of `p`, the probability of obtaining

    1. three heads from the three tosses  (1 mark)

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    2. two heads and a tail from the three tosses.  (1 mark)

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  2. If the probability of obtaining three heads equals the probability of obtaining two heads and a tail, find `p`.  (2 marks)

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Show Answers Only
    1. `p^3`
    2. `3p^2 (1 – p)`
  1. `0 or 3/4`
Show Worked Solution
a.i.    `text(P) (HHH)` `= p xx p xx p`
    `= p^3`

 

  ii.   `text(P) text{(2 Heads and 1 Tail from 3 tosses)}`

♦ Part (a)(ii) mean mark 41%.

`= ((3), (2)) xx p^2 xx (1 – p)^1`

`= 3 p^2 (1 – p)`

 

b.   `text(If probabilities are equal:)`

♦ Mean mark 48%.
MARKER’S COMMENT: Many students incorrectly assumed `p` could not be zero.
`p^3` `= 3p^2 – 3p^3`
`4p^3 – 3p^2` `= 0`
`p^2 (4p – 3)` `= 0`

 
`:. p = 0 or p = 3/4`

Filed Under: Binomial Probability (Ext1) Tagged With: Band 4, Band 5, smc-1084-30-Algebraic examples

Statistics, EXT1 S1 2016 MET1 4

A paddock contains 10 tagged sheep and 20 untagged sheep. Four times each day, one sheep is selected at random from the paddock, placed in an observation area and studied, and then returned to the paddock.

  1. What is the probability that the number of tagged sheep selected on a given day is zero?  (1 mark)

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  2. What is the probability that at least one tagged sheep is selected on a given day?  (1 mark)

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  3. What is the probability that no tagged sheep are selected on each of six consecutive days?

     

    Express your answer in the form `(a/c)^c`, where `a`, `b` and `c` are positive integers.  (1 mark)

    --- 2 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `16/81`
  2. `65/81`
  3. `(2/3)^24`
Show Worked Solution

a.   `text(Let)\ \ X =\ text(Number of tagged sheep,)`

`X ~\ text(Bin)(n,p)\ ~\ text(Bin)(4,1/3)`

`P(X = 0)` `= ((4),(0)) xx (1/3)^0 xx (2/3)^4`
  `= 16/81`

 

b.    `P(X >= 1)` `= 1 – P(X = 0)`
    `= 1 – 16/81`
    `= 65/81`

 

c.   `text(Let)\ \ Y =\ text(Days that no tagged sheep selected,)`

`Y ~\ text(Bin)(6,16/81)`

`P(Y = 6)` `= ((6),(6)) xx (16/81)^6 xx (65/81)^0`
  `= (16/81)^6`
  `=(2/3)^24`

Filed Under: Binomial Probability (Ext1) Tagged With: Band 4, smc-1084-25-Compound Events

Statistics, EXT1 S1 2015 MET2 10

The binomial random variable,  `X`, has  `E(X) = 2`  and  `text(Var)( X ) = 4/3.`

Calculate  `P(X = 1)`.   (3 marks)

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Show Answers Only

`64/243`

Show Worked Solution

`np = 2\ …\ (1)`

`np(1 – p) = 4/3\ …\ (2)`

`text(Solve simultaneous equations:)`

`text(Substitute)\ \ np=2\ \ text(into)\ (2)`

`2(1-p)` `=4/3`  
`2-2p` `=4/3`  
`p` `=1/3`  

 
`n = 6,quadp = 1/3`

`:. X ∼\ text(Bin)(6, 1/3)`

 

`:. P(X=1)` `= ((6),(1)) xx (1/3)^1 xx (2/3)^5`
  `= 6 xx 1/3 xx (2/3)^5`
  `=64/243`

Filed Under: Statistics and Binomial Distributions (Ext1) Tagged With: Band 4, smc-1199-30-Find n/p given E(X)/Var(X)

Statistics, EXT1 S1 2008 MET2 5 MC

Let  `X`  be a discrete random variable with a binomial distribution. The mean of  `X`  is 1.2 and the variance of  `X`  is 0.72

The values of `n` (the number of independent trials) and `p` (the probability of success in each trial) are

A.   `n = 3,\ \ \ p = 0.6`

B.   `n = 2,\ \ \ p = 0.6`

C.   `n = 2,\ \ \ p = 0.4`

D.   `n = 3,\ \ \ p = 0.4`

Show Answers Only

`D`

Show Worked Solution

`X∼\ text(Bin) (n, p)`

`mu` `= 1.2`
`np` `= 1.2\ …\ (1)`

 

`text(Var) (X)` `= 0.72`
`np (1 – p)` `= 0.72\ …\ (2)`

 
`text(Solve simultaneous equations:)`

`:. n = 3,\ \ p = 0.4`

`=>   D`

Filed Under: Statistics and Binomial Distributions (Ext1) Tagged With: Band 4, smc-1199-30-Find n/p given E(X)/Var(X)

Statistics, EXT1 S1 MET2 2008 14 MC

The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is

A.     `9`

B.   `10`

C.   `11`

D.   `12`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ X = text(Number of heads)`

`X ∼ text(Bin) (n, p) ∼ text(Bin) (n, 1/2)`

`P(X = n)` `< 0.0005`
`((n), (n)) (1/2)^n (1/2)^0` `< 0.0005`
`1/2^n` `<5/(10\ 000)`
`2^n` `>2000`
`ln 2^n` `>ln 2000`
`n` `>ln2000/ln2`
`n` `> 10.97`

 
`:. n_min = 11`

`=>   C`

Filed Under: Binomial Probability (Ext1) Tagged With: Band 4, smc-1084-20-Games of Chance

Functions, 2ADV F2 SM-Bank 13

 

  1. Show that the function  `y = (1-e^x)/(1 + e^x)`  is an odd function?   (1 mark) 

    --- 5 WORK AREA LINES (style=lined) ---

  2. Sketch  `y = (1-e^x)/(1 + e^x)`, labelling all intercepts and asymptotes.   (2 marks)

    --- 8 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `{text(all real)\ x,  x!=1/4}`

 

 

 

 

 

 

Show Worked Solution

i.   `f(x) = (1-e^x)/(1 + e^x)`

`f(−x)` `= (1-e^(−x))/(1 + e^(−x)) xx (e^x)/(e^x)`
  `= (e^x-1)/(e^x + 1)`
  `= −(1-e^x)/(1 + e^x)`
  `= −f(x)`

 
`:. f(x)\ text(is ODD.)`

 

ii.   `y = (1-e^x)/(1 + e^x) xx (e^(−x))/(e^(−x)) = (e^(−x)-1)/(e^(−x) + 1) = 1-2/(e^(−x) + 1)`

`text(As)\ x -> ∞, \ 2/(e^(−x) + 1) -> 2, \ y -> −1`

`text(As)\ x ->-∞, \ 2/(e^(−x) + 1) -> 0, \ y -> 1`

`text(When)\ x = 0, \ y = 0`
 

Filed Under: Graphs and Applications (Adv-2027), Graphs and Applications (Y11), Non-Calculus Graphing (Y12) Tagged With: Band 4, smc-1009-20-Exponential, smc-1009-50-Odd Functions, smc-6456-20-Exponential Graphs, smc-6456-50-Odd/Even Functions, smc-966-10-Exponential graphs

Functions, 2ADV F2 EQ-Bank 9

Consider the function  `f(x) = 1/(4x - 1)`.

  1. Find the domain of  `f(x)`.  (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

  2. Sketch  `f(x)`, showing all asymptotes and intercepts?  (2 marks)

    --- 8 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `{text(all real)\ x,  x!=1/4}`
  2.   
Show Worked Solution
i.    `4x – 1` `!= 0`
  `x` `!= 1/4`

 
`:.\ text(Domain:)\ {text(all real)\ x,  x!=1/4}`

 

ii.   `text(When)\ \ x = 0, \ y = −1`

`text(As)\ \ x -> ∞, \ y -> 0^+`

`text(As)\ \ x -> −∞, \ y -> 0^−`

Filed Under: Non-Calculus Graphing (Y12), Reciprocal Functions (Adv-2027) Tagged With: Band 3, Band 4, smc-1009-10-Quotient Function, smc-6382-30-Sketch Graph

Functions, 2ADV F2 SM-Bank 12

Sketch the graph of  `f(x) = (2x+1)/(x-1)`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation.  (4 marks)

  

--- 0 WORK AREA LINES (style=lined) ---

Show Answers Only

Show Worked Solution
`(2x+1)/(x-1)` `=(2x-2+3)/(x-1)`  
  `=(2(x-1)+3)/(x-1)`  
  `=2 + 3/(x-1)`  

COMMENT: Manipulation of the equation makes graphing much easier.

 
`text(Asymptotes:)\ \ x = 1,\ \ y = 2`

`text(As)\ \ x->oo,\ \ y->2(+)`

`text(As)\ \ x->-oo,\ \ y->2(-)`

`text(As)\ \ x->-1 (-),\ \ y->-oo`

`text(As)\ \ x->-1 (+),\ \ y->oo`

 

 

Filed Under: Non-Calculus Graphing (Y12), Reciprocal Functions (Adv-2027) Tagged With: Band 4, smc-1009-10-Quotient Function, smc-6382-30-Sketch Graph

Functions, 2ADV F2 EQ-Bank 11

On the axes below, sketch the graph of   `f(x) = (2x-2)/(x + 1)`.

Label all axis intercepts. Label each asymptote with its equation.   (4 marks)

 
met1-2008-vcaa-q2

--- 0 WORK AREA LINES (style=lined) ---

Show Answers Only

met1-2008-vcaa-q2-answer1

Show Worked Solution
`(2x-2)/(x + 1)` `=(2(x+1) – 4)/(x+1)`
  `=2- 4/((x+1)`

 
`text(Asymptotes:)\ \ x = −1,\ \ y = 2`

`text(As)\ \ x->oo,\ \ y->2(-)`

`text(As)\ \ x->-oo,\ \ y->2(+)`

`text(As)\ \ x->-1 (-),\ \ y->oo`

`text(As)\ \ x->-1 (+),\ \ y->-oo`

 

met1-2008-vcaa-q2-answer1

Filed Under: Non-Calculus Graphing (Y12), Reciprocal Functions (Adv-2027) Tagged With: Band 4, smc-1009-10-Quotient Function, smc-6382-30-Sketch Graph

Functions, 2ADV F2 SM-Bank 10 MC

The graph of the function  `f(x) = (x - 3)/(2 - x)`  has asymptotes

  1. `x = 3,\ \ \ \ \ \ \ \ \ y = 2`
  2. `x = -2,\ \ \ \ \ y = 1`
  3. `x = 2,\ \ \ \ \ \ \ \ \ y = -1`
  4. `x = 2,\ \ \ \ \ \ \ \ \ y = 1`
Show Answers Only

`D`

Show Worked Solution
`y` `=-((3-x)/(2-x))`
  `=-((2-x)/(2-x) + 1/(2-x))`
  `=-1 -1/(2-x)`

 
`:.\ text(Asymptotes:)\ \ x = 2, \ y = – 1`

`=>   C`

Filed Under: Non-Calculus Graphing (Y12) Tagged With: Band 4, smc-1009-40-Identify Asymptotes

Functions, 2ADV F2 SM-Bank 10

Consider the function  `f(x) = x/(4 - x^2)`.

  1. Identify the domain of  `f(x)`.   (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

  2. Sketch the graph  `y = f(x)`, showing all intercepts and asymptotes.   (3 marks)

    --- 8 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `text(Domain:)\ \ text(all)\ x,\ x != +- 2`
  2.  
     
Show Worked Solution

i.   `4-x^2 !=0`

`=> x!= 2 or -2`

`:.\ text(Domain:)\ \ text(all)\ x,\ x != +- 2`

 

ii. `text(Asymptotes at)\ \ x = +- 2`

`text(Passes through)\ (0,0)`

COMMENT: Note that  `x->2^(–)`  is a short way of writing as `x` approaches 2 from the negative (or left-hand) side. This notation can save time when required.
`text(As)` `\ \ x -> 2^(-),` `\ y -> oo`
  `\ \ x -> 2 ^(+),` `\ y -> – oo`
`text(As)` `\ \ x -> -2^ (-),` `\ y -> oo`
  `\ \ x -> -2^ (+),` `\ y -> -oo`
`text(As)` `\ \ x -> oo,` `\ y -> 0`
  `\ \ x -> – oo,` `\ y -> 0`

 

  EXT1 2013 11d

 

Filed Under: Non-Calculus Graphing (Y12) Tagged With: Band 4, smc-1009-10-Quotient Function, smc-1009-50-Odd Functions

Functions, 2ADV’ F2 2007 HSC 3b

  1. Find the vertical and horizontal asymptotes of the hyperbola

     

    `qquad y = (x − 2)/(x − 4)`  and hence sketch the graph of  `y = (x − 2)/(x − 4)`.  (3 marks)

    --- 10 WORK AREA LINES (style=lined) ---

  2. Hence, or otherwise, find the values of  `x`  for which  `(x − 2)/(x − 4) ≤ 3`.  (2 marks)

    --- 5 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `x < 4\ text(and)\ x ≥ 5`
Show Worked Solution

i.    `y = (x − 2)/(x − 4)`

`text(Vertical asymptote at)\ \ x = 4`

`lim_(x → ∞) (x − 2)/(x − 4)` `= lim_(x → ∞) (1 − 2/x)/(1 − 4/x)=1`

`ytext(–intercept)\ = 1/2`

`xtext(–intercept)\ = 2`

 

Geometry and Calculus, EXT1 2007 HSC 3b Answer

 

ii.  `text(Find)\ \ x\ \ text(so that)\ \ (x − 2)/(x − 4) ≤ 3`

`(x − 2)/(x − 4)` `= 3`
`x − 2` `= 3x − 12`
`2x` `= 10`
`x` `= 5`

 
`=>(5, 3)\ \ text(is the intersection of)\ \ y = 3\ and\ y = (x − 2)/(x − 4)`
 

`:. (x − 2)/(x − 4) ≤ 3\ \ text(when)\ \ x < 4\ \ text(and)\ \ x ≥ 5.`

Filed Under: Non-Calculus Graphing (Y12), Reciprocal Functions (Adv-2027) Tagged With: Band 4, smc-1009-10-Quotient Function, smc-1009-40-Identify Asymptotes, smc-6382-30-Sketch Graph

Functions, 2ADV’ F2 2012 HSC 13b

  1. Find the horizontal asymptote of the graph  `y=(2x^2)/(x^2 + 9)`.   (1 mark)

    --- 1 WORK AREA LINES (style=lined) ---

  2. Without the use of calculus, sketch the graph  `y=(2x^2)/(x^2 + 9)`, showing the asymptote found in part (i).    (2 marks)

    --- 8 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `text(Horizontal asymptote at)\ y = 2`
  2.  
    Geometry and Calculus, EXT1 2012 HSC 13b Answer
Show Worked Solution
i.    `y` `= (2x^2)/(x^2 +9)`
    `= 2/(1 + 9/(x^2))`

 

`text(As)\ \ x -> oo,\ y ->2`

`text(As)\ \ x -> – oo,\ y -> 2`

`:.\ text(Horizontal asymptote at)\ y = 2`

 

ii.    `text(At)\ \ x = 0,\ y = 0`

`f(x) = (2x^2)/(x^2 + 9) >= 0\ text(for all)\ x`

`f(–x) = (2(–x)^2)/((–x)^2 + 9) = (2x^2)/(x^2 + 9) = f(x)`

`text(S)text(ince)\ \ f(x) = f(–x) \ \ =>\ text(EVEN function)`

Geometry and Calculus, EXT1 2012 HSC 13b Answer

Filed Under: Non-Calculus Graphing (Y12) Tagged With: Band 3, Band 4, smc-1009-10-Quotient Function, smc-1009-60-Even Functions

Functions, 2ADV’ F2 2015 HSC 5 MC

What are the asymptotes of `y = (3x)/((x + 1)(x + 2))`

A.    `y = 0,` `x = −1,` `x = −2`
B.    `y = 0,` `x = 1,` `x = 2`
C.    `y = 3,` `x = −1,` `x = −2`
D.    `y = 3,` `x = 1,` `x = 2`
Show Answers Only

`A`

Show Worked Solution

`y = (3x)/((x + 1)(x + 2))`

`text(Asymptotes at)\ \ x = −1\ \ text(and)\ \ x = −2`

`text(As)\ \ x → ∞, y → 0^+`

`text(As)\ \ x → −∞, y → 0^−`

`:.\ text(Horizontal asymptote at)\ \ y = 0`

`⇒ A`

Filed Under: Non-Calculus Graphing (Y12) Tagged With: Band 4, smc-1009-40-Identify Asymptotes

Functions, 2ADV’ F2 2017 HSC 5 MC

Which graph best represents the function  `y = (2x^2)/(1 - x^2)`?
 

A. B.
       
C. D.
Show Answers Only

`D`

Show Worked Solution
`y` `= (2x^2)/((1 – x^2))`
  `= −((2 – 2x^2 – 2))/((1 – x^2))`
  `= −(2(1 – x^2))/((1 – x^2)) – 2/((1 – x^2))`
  `= −2 – 2/((1 – x^2))`

 

`text(As)\ \ x -> oo,\ \ y -> −2`

`:. text(Horizontal asymptote at)\ \ y = −2`

`⇒D`

Filed Under: Non-Calculus Graphing (Y12) Tagged With: Band 4, smc-1009-30-Identify Graphs, smc-1009-60-Even Functions

Calculus, EXT1 C3 2017 SPEC1 8

A slope field representing the differential equation  `dy/dx = −x/(1 + y^2)`  is shown below.

  1. Sketch the solution curve of the differential equation corresponding to the condition  `y(−1) = 1`  on the slope field above and, hence, estimate the positive value of `x` when  `y = 0`. Give your answer correct to one decimal place.  (2 marks)
  2. Solve the differential equation  `(dy)/(dx) = (−x)/(1 + y^2)`  with the condition  `y(−1) = 1`. Express your answer in the form  `ay^3 + by + cx^2 + d = 0`, where `a`, `b`, `c` and `d` are integers.  (2 marks)

    --- 6 WORK AREA LINES (style=lined) ---

Show Answers Only
  1.  

  2. `2y^3 + 6y + 3x^2 – 11 = 0`
Show Worked Solution
a.   

♦♦ Mean mark part (a) 32%.
MARKER’S COMMENT: Solution curve should follow slope ticks and not cross them.

 

b.    `(1 + y^2)(dy)/(dx)` `= −x`
  `int 1 + y^2 dy` `= −int x\ dx`
  `y + (y^3)/3` `= −(x^2)/2 + C, C ∈ R`

 
`text(Substituting)\ (-1,1):`

`1 + (1^3)/3` `= −((−1)^2)/2 + C`
`1 + 1/3` `= −1/2 + C`
`:. C` `= 11/6`

 

`y + 1/3y^3` `= −1/2x^2 + 11/6`
`6y + 2y^3` `= −3x^2 + 11`

 
`:. 2y^3 + 6y + 3x^2 – 11 = 0`

Filed Under: Equations and Slope Fields Tagged With: Band 4, Band 5, smc-1197-10-Slope Fields

Calculus, EXT1 C3 SM-Bank 5

Bacteria are spreading over a Petri dish at a rate modelled by the differential equation

`(dP)/(dt) = P/2 (1 - P),\ 0 < P < 1`

where  `P`  is the proportion of the dish covered after  `t`  hours.

Given  `2/(P(1 - P)) = 2/P + 2/(1 - P),`

  1. Show by integration that  `(t - c)/2= log_e(P/(1 - P))`, where  `c`  is a constant of integration.  (2 marks)

    --- 5 WORK AREA LINES (style=lined) ---

  2. If half of the Petri dish is covered by the bacteria at  `t = 0`, express  `P`  in terms of  `t`.  (2 marks)

    --- 6 WORK AREA LINES (style=lined) ---

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `P = e^(t/2)/(1 + e^(t/2))`
Show Worked Solution

a.  `(dt)/(dP) = 2/(P(1 – P)) = 2/P + 2/(1 – P)`

`t` `= int 2/P + 2/(1 – P)\ dP`
  `= 2 log_e |P| – 2log_e|1 – P| + c`
`(t-c)/2` `=log_e|P| – log_e|1-P|`
  `=log_e |(P)/(1-P)|`
  `= log_e (P/(1 – P))`

  
`text(S)text(ince)\ \ 0 < P < 1 :\ |P| = P\ and\ |1 – P| = 1 – P`


b.
  `text(When)\ \ t=0,\ P=0.5`

`(-c)/2` `= log_e (0.5/0.5)`
`c` `= log_e (1)`
  `=0`

 

`t/2` `= ln (P/(1 – P))`
`e^(t/2)` `= P/(1 – P)`
`e^(t/2) (1 – P)` `= P`
`e^(t/2) – Pe^(t/2)` `= P`
`e^(t/2)` `= P(1 + e^(t/2))`
`:. P` `= e^(t/2)/(1 + e^(t/2))`

Filed Under: Applications of Differential Equations Tagged With: Band 4, smc-1198-30-Quantity

Calculus, EXT1 C3 2017 SPEC2 8 MC

The differential equation that best represents the direction field above is

A.   `(dy)/(dx) = x - y^2`

B.   `(dy)/(dx) = y - x`

C.   `(dy)/(dx) = y^2 - x^2`

D.   `(dy)/(dx) = y^2 - x`

Show Answers Only

`D`

Show Worked Solution

`text(Use CAS to graph the direction field of each option.)`

`=>   D`

Filed Under: Equations and Slope Fields Tagged With: Band 4, smc-1197-10-Slope Fields

Calculus, EXT1 C3 2018 SPEC2 10 MC

The differential equation that best represents the direction field above is

A.  `(dy)/(dx) = (2x + y)/(y - 2x)`

B.  `(dy)/(dx) = (x + 2y)/(2x - y)`

C.  `(dy)/(dx) = (2x - y)/(x + 2y)`

D.  `(dy)/(dx) = (x - 2y)/(y - 2x)`

Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=0, \ m=1`

`text(When)\ \ y=0, \ m=-1`

`=>  A`

Filed Under: Equations and Slope Fields Tagged With: Band 4, smc-1197-10-Slope Fields

Calculus, EXT1 C3 2016 SPEC2 10 MC

The direction field for the differential equation  `(dy)/(dx) + x + y = 0`  is shown above.

A solution to this differential equation that includes  `(0, -1)`  could also include

A.  `(3, –1)`

B.  `(3.5, –2.5)`

C.  `(–1.5, –2)`

D.  `(2.5, –1)`

Show Answers Only

`B`

Show Worked Solution

 
`=>  B`

Filed Under: Equations and Slope Fields Tagged With: Band 4, smc-1197-10-Slope Fields

Calculus, EXT1 C3 2014 SPEC2 14 MC

The differential equation that is best represented by the above direction field is

A.   `(dy)/(dx) = 1/(x - y)`

B.   `(dy)/(dx) = y - x`

C.   `(dy)/(dx) = 1/(y - x)`

D.   `(dy)/(dx) = x - y`

Show Answers Only

`C`

Show Worked Solution

`text(Consider quadrant 2,)`

`x < 0, \ \ y > 0, \ \  m > 0\ \ => text(Eliminate A and D)`
 

`text(Consider vertical gradients where)\ \ m=oo\ \ => text(Eliminate B)`

 
`=> C`

Filed Under: Equations and Slope Fields Tagged With: Band 4, smc-1197-10-Slope Fields

Calculus, EXT1 C3 2013 SPEC2 12 MC

SPEC2 2013 VCAA 12 MC

The differential equation that best represents the above direction field is

A.   `(dy)/(dx) = x^2 - y^2`

B.   `(dy)/(dx) = y^2 - x^2`

C.   `(dy)/(dx) = −x/y`

D.   `(dy)/(dx) = x/y`

Show Answers Only

`D`

Show Worked Solution

`text(By inspection:)`

`text(When)\ \ x=0\ \ =>\ \ (dy)/(dx) = 0`

`text(When)\ \ y=0\ \ => (dy)/(dx) -> oo`

`:.\ text(Eliminate A, and B)`
 

`text(Along)\ \ y = x\ \ =>\ \ (dy)/(dx) > 0`

`:.\ text(Eliminate C)`

`=> D`

Filed Under: Equations and Slope Fields Tagged With: Band 4, smc-1197-10-Slope Fields

Calculus, EXT1 C3 2012 SPEC2 10 MC

The diagram that best represents the direction field of the differential equation  `(dy)/(dx) = xy`  is

A. B. 
C. D.
Show Answers Only

`A`

Show Worked Solution

`(dy)/(dx) = xy`

`text(When)\ \ x=0 \ or\  y=0\ \ =>\ text(gradient = 0)`

`text(In 1st and 3rd quartile)\ \ =>\ \ text(gradients positive)`

`text(In 2nd and 4th quartile)\ \ =>\ \ text(gradients negative)`

`=> A`

Filed Under: Equations and Slope Fields Tagged With: Band 4, smc-1197-10-Slope Fields

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