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PHYSICS, M5 2019 HSC 30

A ball, initially at rest in position `P`, travels along a frictionless track to point `Q` and then falls to strike the floor below.
 

At the instant the ball leaves the track at `Q` it has a velocity of 1.5 `text{m s}^(-1)` at an angle of 50° to the horizontal.

  1. Calculate the difference in height between `P` and `Q`.   (3 marks)
  1. The ball takes 0.5 s to reach the floor after leaving the track at `Q`.
  2. Calculate the height of `Q` above the floor.   (3 marks)
Show Answers Only

a.   `h=0.11\ \text{m}`

b.   `h=1.8\ \text{m}`

Show Worked Solution

a. `Delta U` `=Delta E_(k)`
  `mgDeltah` `=(1)/(2)mv^2-(1)/(2)m u^2`
  `mgDeltah` `=(1)/(2)mv^2\ \ (u=0)`
  `Delta h` `=(v^2)/(2g)=(1.5^(2))/(2xx9.8)=0.1145\ \text{m}`


♦ Mean mark part (a) 41%.

b.   `u_(y)=u sin theta=1.50  sin 50^(@)=1.15\ text{m s}^(-1)`

  `s_(y)` `=ut+(1)/(2)at^(2)`
    `=1.15 xx0.5+(1)/(2)(9.80)xx0.5^(2)`
    `= 1.8\ text{m}`

 
`:.\ \text{Height of Q = 1.8 m}`

PHYSICS, M6 2019 HSC 29

A particle having mass `m` and charge `q` is accelerated from rest through a potential difference `V`. Assume that the only force acting on the particle is due to the electric field associated with this potential difference.

Show that the final velocity of the particle is given by  `v = sqrt((2qV)/m)`.   (3 marks)

Show Answers Only
`W` `=Delta E_(k)`  
  `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)mv^2` `(u=0)`
`qV` `=(1)/(2)mv^2` `(W=qV)`
`v^2` `=(2qV)/m`  
`∴v` `=sqrt((2qV)/(m))`  
Show Worked Solution

The work done on the particle is equal to its change in its kinetic energy:

`W` `=Delta E_(k)`  
  `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)mv^2` `(u=0)`
`qV` `=(1)/(2)mv^2` `(W=qV)`
`v^2` `=(2qV)/m`  
`∴v` `=sqrt((2qV)/(m))`  

PHYSICS, M7 2019 HSC 27

  1. Outline a thought experiment that relates to the prediction of time dilation.   (3 marks)
  1. Outline experimental evidence that validated the prediction of time dilation.   (3 marks)
Show Answers Only

a.    Consider a train moving at a high speed:

  • There are two observers, one on the train and one stationary outside the train. 
  • A pulse of light starts from one side of the carriage and then reflects off a mirror on the other side of the carriage returning to its source. 
  •  The stationary observer, outside the train, will observe the light travel in a triangular path. This is longer than the path observed by the observer on the train.
  • The speed of light is constant for both observers.
  • the observer outside the train measures a longer, dilated time for the light pulse to travel. This demonstrates time dilation. 

b.    Experiment predicting time dilation:

  • Muons are particles in the upper atmosphere produced by cosmic rays that travel at a high speed greater than 0.99c and have a short half-life. 
  • The amount of muons striking the ground in a particular area at the top of a mountain was measured.
  • Using this data the number of muons expected to reach the ground at sea level was predicted (assuming no relativistic effects).
  • The actual number observed at sea level was greater than predicted.
  • This is consistent with an increase in the muons half life due to time dilation.

Answers could also reference:

  • The Hafele-Keating atomic clock experiment.
  • Evidence from particle accelerators.
Show Worked Solution

a.    Consider a train moving at a high speed:

  • There are two observers, one on the train and one stationary outside the train. 
  • A pulse of light starts from one side of the carriage and then reflects off a mirror on the other side of the carriage returning to its source. 
  • The stationary observer, outside the train, will observe the light travel in a triangular path. This is longer than the path observed by the observer on the train.
  • The speed of light is constant for both observers.
  • the observer outside the train measures a longer, dilated time for the light pulse to travel. This demonstrates time dilation. 

b.    Experiment predicting time dilation:

  • Muons are particles in the upper atmosphere produced by cosmic rays that travel at a high speed greater than 0.99c and have a short half-life. 
  • The amount of muons striking the ground in a particular area at the top of a mountain was measured.
  • Using this data the number of muons expected to reach the ground at sea level was predicted (assuming no relativistic effects).
  • The actual number observed at sea level was greater than predicted.
  • This is consistent with an increase in the muons half life due to time dilation.

Answers could also reference:

  • The Hafele-Keating atomic clock experiment.
  • Evidence from particle accelerators.

PHYSICS, M5 2019 HSC 26

A student carried out an experiment to investigate the relationship between the torque produced by a force and the angle at which the force is applied. A 400 N force was applied to the same position on the handle of a spanner at different angles, as shown.
 

A high-precision device measured the torque applied to the bolt.

The data from the experiment is graphed below.
 

The student concluded that the torque `(tau)` was proportional to the angle `(theta)` and proposed the model

`tau=ktheta`

where  `k` = 1.7 Nm/degree.

  1. Justify the validity of the student's model using information from the graph.   (3 marks)
  1. What happens to the accuracy of this model's predictions as the angle increases beyond 25°? Justify your answer with reference to a different model.   (3 marks)
Show Answers Only

a.   The graph shows a linear relationship.

  • Constructing a line of best fit on the student’s graph shows the the gradient is approximately`=(34-17)/(20-10) =1.7`.
  • These findings are consistent with the student’s model of `tau=1.7 theta`
  • Therefore, the model is valid.

b.   Model accuracy:

  • The accuracy of this model decreases as the angle increases beyond 25°.
  • Torque produced by a force is more accurately described by `tau=rF sin theta.`
  • At small angles, `sin theta ~~ theta`, so the model is accurate. However at larger angles, `sin theta lt theta`, so the student’s model will predict values of torque greater than those predicted by `tau=rF sin theta`.
Show Worked Solution

a.   The graph shows a linear relationship.

  • Constructing a line of best fit on the student’s graph shows the the gradient is approximately`=(34-17)/(20-10) =1.7`.
  • These findings are consistent with the student’s model of `tau=1.7 theta`
  • Therefore, the model is valid.

b.   Model accuracy:

  • The accuracy of this model decreases as the angle increases beyond 25°.
  • Torque produced by a force is more accurately described by `tau=rF sin theta.`
  • At small angles, `sin theta ~~ theta`, so the model is accurate. However at larger angles, `sin theta lt theta`, so the student’s model will predict values of torque greater than those predicted by `tau=rF sin theta`.

BIOLOGY, M5 2021 HSC 22

In a population of rabbits, black fur colour is dominant over white fur. A black rabbit, whose mother has white fur, mates with a white rabbit.

Predict the phenotypic ratio for the offspring of this cross. Show your working.   (3 marks)

 

Show Answers Only

Black : White = 1 : 1

Show Worked Solution

`text{P: Bb × bb}`

\begin{array} {|c|c|c|}
\hline  & \text{b} & \text{b} \\
\hline \text{B} & \text{Bb} & \text{Bb} \\
\hline \text{b} & \text{bb} & \text{bb} \\
\hline \end{array}

 
Black : White = 1 : 1

BIOLOGY, M7 2021 HSC 21

  1. Label TWO features on the diagram below that would help to classify this pathogen as a bacterium.   (2 marks)
     
     

     
  2. A scientist followed Koch's postulates to confirm that this bacterium was causing diarrhoea in pigs on a local farm. 
  3. Complete the boxes in the flowchart provided to show the steps taken by the scientist.    (2 marks)
     
     

  4.  
  5. Two pig farmers on neighbouring farms noticed that their pigs were suffering from diarrhoea and gradually losing weight. The farmers each adopted a different strategy to deal with this disease, as shown in the table.  

\begin{array} {|c|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Farm} \quad \rule[-1ex]{0pt}{0pt} & \quad\quad\quad \quad \textit{Strategy} & \quad\quad\quad \quad \textit{Result}\\
\hline
\rule{0pt}{2.5ex} 1  & \text{Treatment with antibiotics} & \text{All pigs recovered after two}\\
\rule[-1ex]{0pt}{0pt} & \text{} & \text{weeks}\\
\hline
\rule{0pt}{2.5ex} 2  & \text{Elimination of rats and mice} & \text{Decrease in number of sick}\\
& \text{from pig sheds to improve} & \text{animals over three months}\\
\rule[-1ex]{0pt}{0pt} & \text{hygiene} & \\
\hline
\end{array}

Outline ONE benefit and ONE limitation of the strategies used on each farm.   (3 marks)

Show Answers Only

a.    Include two of the following labels:
 
       

b.    Box 2: Bacteria grown in pure culture and identified.

Box 4: Healthy pig became ill with diarrhoea. 

c.   Benefits and Limitations of the strategies used on each farm.

  • The use of antibiotics on farm 1 has resulted in a rapid elimination of diarrhoea cases, however may induce antibiotic resistance in the
    future, rendering the strategy less effective.
  • The removal of rats and mice from pig sheds to increase hygiene on farm 2 is slow to eliminate diarrhoea cases, however provides reassurance to prevent future outbreaks. 

Other correct answers:

  • Proper hygiene practices can reduce incidence of other diseases, not just diarrhoea.
Show Worked Solution

a.    Include two of the following labels:
 
       
 

b.    Box 2: Bacteria grown in pure culture and identified.

Box 4: Healthy pig became ill with diarrhoea. 
 

c.   Benefits and Limitations of the strategies used on each farm.

  • The use of antibiotics on farm 1 has resulted in a rapid elimination of diarrhoea cases, however may induce antibiotic resistance in the
    future, rendering the strategy less effective.
  • The removal of rats and mice from pig sheds to increase hygiene on farm 2 is slow to eliminate diarrhoea cases, however provides reassurance to prevent future outbreaks. 

Other correct answers:

  • Proper hygiene practices can reduce incidence of other diseases, not just diarrhoea.

Statistics, EXT1 S1 2022 HSC 13e

A chocolate factory sells 150-gram chocolate bars. There has been a complaint that the bars actually weigh less than 150 grams, so a team of inspectors was sent to the factory to check. They randomly selected 16 bars, weighed them and noted that 8 bars weighed less than 150 grams.

The factory manager claims 80% of the chocolate bars produced by the factory weigh 150 grams or more.

  1. The inspectors used the normal approximation to the binomial distribution to calculate the probability, `cc "P"`, of having at least 8 bars weighing less than 150 grams in a random sample of 16, assuming the factory manager's claim is correct.
  2. Using the attached probability table, calculate the value of `cc "P"`.   (2 marks)

  3. The factory manager disagrees with the method used by the inspectors as described in part (i).
  4. Explain why the method used by the inspectors might not be valid.   (1 mark)

Show Answers Only
  1. `P=0.0013`
  2. `text{See Worked Solutions}`
Show Worked Solution

i.   `text{Let}\ \ X=\ text{number of chocolate bars < 150 grams}`

`X\ ~\ text{Bin}(16, 0.2)`

`mu=np=16xx0.2=3.2`

`sigma^2=np(1-p)=16xx0.2xx0.8=2.56`

`ztext{-score}\ (X=8)=(8-3.2)/sqrt2.56=3`

 

`cc P(X>=8)` `=P(z>=3.0)`  
  `=1-P(z<=3.0)`  
  `=1-0.9987`  
  `=0.0013`  

 

ii.   `np=3.2,\ \ n(1-p)=16xx0.8=12.8`

`text{S}text{ince}\ \ np<5,\ text{the sample size is too small.}`


♦ Mean mark part (i) 37%.
 
Mean mark part (ii) 54%.

Calculus, EXT1 C3 2022 HSC 13b

A solid of revolution is to be found by rotating the region bounded by the `x`-axis and the curve  `y=(k+1) \sin (k x)`, where  `k>0`, between  `x=0`  and  `x=\frac{\pi}{2 k}`  about the `x`-axis.
 

     

Find the value of `k` for which the volume is `pi^2`.  (3 marks)

Show Answers Only

`k=1`

Show Worked Solution

`y=(k+1) \sin (k x)`

`V` `=pi int_0^((pi)/(2k)) (k+1)^2sin^(2)(kx)\ dx`  
  `=pi(k+1)^2 int_0^((pi)/(2k)) 1/2[1-cos(2kx)]\ dx`  
  `=(pi/2)(k+1)^2[x-(sin(2kx))/(2k)]_0^((pi)/(2k)) `  
  `=(pi/2)(k+1)^2[(pi/(2k)- sin(pi)/(2k))-(0-sin0/(2k))]`  
  `=(pi/2)(k+1)^2(pi/(2k))`  
  `=pi^2/(4k)(k+1)^2`  

 
`text{Given}\ \ V=pi^2:`

`pi^2/(4k)(k+1)^2` `=pi^2`  
`(k+1)^2` `=4k`  
`k^2+2k+1` `=4k`  
`k^2-2k+1` `=0`  
`(k-1)^2` `=0`  

 
`:.k=1`

Vectors, EXT1 V1 2022 HSC 13a

Three different points `A, B` and `C` are chosen on a circle centred at `O`.

Let  `underset~a= vec(OA),underset~b= vec(OB)`  and  `underset~c= vec(OC)`. Let  `underset~h=underset~a + underset~b+ underset~c`  and let `H` be the point such that  `vec(OH)= underset~h`, as shown in the diagram.
 

     
 

Show that  `vec(BH)`  and  `vec(CA)`  are perpendicular.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Prove}\ \ vec(BH) ⊥ vec(CA):`

`vec(BH)*vec(CA)=(-underset~b+underset~h)*(-underset~c+underset~a)`
 

`text{Given}\ \ underset~h=underset~a + underset~b+ underset~c\ \ =>\ \ underset~h-underset~b=underset~a + underset~c`

`vec(BH)*vec(CA)` `=(underset~a+underset~c)*(underset~a-underset~c)`  
  `=|underset~a|^2-|underset~c|^2`  

 
`underset~a and underset~c\ \ text{are radii}\ \ => \ \ |underset~a|=|underset~c|`

 
`:.\ text{S}text{ince}\ \ vec(BH)*vec(CA)=0\ \ =>\ \ vec(BH)⊥vec(CA)`

Proof, EXT1 P1 2022 HSC 12f

Use mathematical induction to prove that  `15 ^(n)+6^(2n+1)`  is divisible by 7 for all integers  `n >= 0`.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Prove}\ \ 15 ^(n)+6^(2n+1)\ \ text{is divisible by 7 for}\ \ n>=0`

`text{If}\ \ n=0:`

`15^0+6^1=7`

`:.\ text{True for}\ \ n=0`
 

`text{Assume true for}\ \ n=k`

`text{i.e.}\ \ 15^(k)+6^(2k+1)=7P\ \ text{(where}\ P\ text{is an integer)}`

`=>6^(2k+1)=7P-15^(k)\ \ …\ (1)`
 

`text{Prove true for}\ \ n=k+1`

`15^(k+1)+6^(2k+3)` `=15*15^(k)+6^2*6^(2k+1)`  
  `=15*15^(k)+36(7P-15^(k))\ \ text{(see (1))}`  
  `=15*15^(k)+36*7P-36*15^(k)`  
  `=36*7P-21*15^(k)`  
  `=7(36P-3*15^(k))\ \ text{(which is divisible by 7)}`  

 
`=>\ text{True for}\ \ n=k+1`

`:.\ text{S}text{ince true for}\ \ n=0, text{by PMI, true for integers}\ \ n>=0`

Calculus, EXT1 C1 2022 HSC 12d

In a room with temperature 12°C, coffee is poured into a cup. The temperature of the coffee when it is poured into the cup is 92°C, and it is far too hot to drink.

The temperature, `T`, in degrees Celsius, of the coffee, `t` minutes after it is made, can be modelled using the differential equation  `(dT)/(dt)=k(T-T_(1))`, where `k` is the constant of proportionality and `T_1` is a constant.

  1. It takes 5 minutes for the coffee to cool to a temperature of 76°C.
  2. Using separation of variables, solve the given differential equation to show that  `T=12+80e^((t)/(5)ln((4)/(5)))`.  (3 marks)

  3. The optimal drinking temperature for a hot beverage is 57°C.
  4. Find the value of `t` when the coffee reaches this temperature, giving your answer to the nearest minute.  (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{13 minutes}`
Show Worked Solution

i.   `T_1=12\ \ text{(Room temperature = 12°C)}`

`(dT)/(dt)` `=k(T-12)`  
`(dt)/(dT)` `=1/(k(T-12))`  
`int dt` `=1/k int(1/(T-12))\ dT`  
`t` `=1/k ln|T-12|+c`  

 
`text{When}\ \ t=0,\ \ T=92:`

`0` `=1/k ln|92-12|+c`  
`c` `=-1/k ln(80)`  

 

`t` `=1/k ln|T-12|-1/k ln(80)`  
`kt` `=ln((|T-12|)/80)`  
`T-12` `=80e^(kt)`  
`T` `=12+80e^(kt)`  

 
`text{When}\ \ t=5,\ \ T=76:`

`76` `=12+80e^(5k)`  
`e^(5k)` `=64/80`  
`5k` `=ln(4/5)`  
`k` `=1/5 ln(4/5)`  

  
`:.T=12+80e^((t)/(5)ln((4)/(5)))\ \ \ text{… as required}`
 

ii.   `text{Find}\ \ t\ \ text{when}\ \ T=57:`

`57` `=12+80e^((t)/(5)ln((4)/(5)))`  
`e^((t)/(5)ln((4)/(5)))` `=45/80`  
`t/5ln(4/5)` `=ln(9/16)`  
`t` `=(5ln(9/16))/ln(4/5)`  
  `=12.89…`  
  `~~13\ text{minutes}`  

Calculus, EXT1 C2 2022 HSC 12c

Find the equation of the tangent to the curve  `y=x  text{arctan}(x)`  at the point with coordinates `(1,(pi)/(4))`. Give your answer in the form  `y=mx+c`  (3 marks)

Show Answers Only

`y=((2+pi)/4)x-1/2`

Show Worked Solution
`y` `=xtan^(-1)(x)`  
`dy/dx` `=x xx 1/(1+x^2)+tan^(-1)(x)`  

 
`text{When}\ \ x=1:`

`dy/dx=1/2+tan^(-1)(1)=1/2+pi/4=(2+pi)/4`
 

`text{Equation of tangent}\ \ m=(2+pi)/4,\ text{through}\ \ (1,(pi)/(4)):`

`y-pi/4` `=(2+pi)/4 (x-1)`  
`y` `=((2+pi)/4)x-(2+pi)/4+pi/4`  
`y` `=((2+pi)/4)x-1/2`  

Trigonometry, EXT1 T3 2022 HSC 11e

Express  `sqrt3sin(x)-3cos(x)`  in the form  `R sin(x+alpha)`.  (3 marks)

Show Answers Only

`2sqrt3 sin(x-pi/3)`

Show Worked Solution

`R sin(x+alpha)=sqrt3sin(x)-3cos(x)`

`R sin(x)cos alpha +Rcos(x)sin alpha=sqrt3sin(x)-3cos(x)` 

`text{Equating coefficients:}`

`R cos alpha=sqrt3, \ \ R sin alpha=-3`

`R^2` `=(sqrt3)^2+(-3)^2=12`  
`:.R` `=sqrt12=2sqrt3`  

 
`text{Given}\ \ sin alpha<0\ \ text{and}\ \ cos alpha>0`

`=> alpha\ \ text{is in the 4th quadrant}`

`sqrt12 cos alpha` `=sqrt3`  
`cos alpha` `=sqrt3/sqrt12=1/2`  
`:.alpha` `=-pi/3\ \ (-pi<=alpha<=pi)`  

 
`:.sqrt3sin(x)-3cos(x)=2sqrt3 sin(x-pi/3)`

Combinatorics, EXT1 A1 2022 HSC 11c

Find the coefficients of `x^(2)` and `x^(3)` in the expansion of `(1-(x)/(2))^(8)`.  (2 marks)

Show Answers Only

`x^2: 7, \ x^3:-7`

Show Worked Solution

`text{General Term:}\ (1-(x)/(2))^(8) `

`T_k` `=((8),(k))(1)^(8-k)(- x/2)^k`  
  `=((8),(k))(- 1/2)^k x^k`  

 
`text{Coefficient of}\ \ x^2 = ((8),(2))(- 1/2)^2=7`

`text{Coefficient of}\ \ x^3 = ((8),(3))(- 1/2)^3=-7`

Functions, EXT1 F1 2022 HSC 5 MC

A curve is defined in parametric form by  `x=2+t`  and  `y=3-2t^(2)`  for `-1 <= t <= 0`.

Which diagram best represents this curve?
 


 

Show Answers Only

`B`

Show Worked Solution

`x=2+t\ \ =>\ \ t=x-2`

`text{Substitute into}\ \ y=3-2t^(2)`

`y=3-2(x-2)^2`

`text{Concave down parabola with maximum at (2, 3)}`

`=>B`

Trigonometry, EXT1 T1 2022 HSC 1 MC

It is given that  `cos((23 pi)/(12))=(sqrt6+sqrt2)/(4)`.

Which of the following is the value of  `cos^(-1)((sqrt6+sqrt2)/(4))`?

  1. `{23pi}/{12}`
  2. `{11pi}/{12}`
  3. `pi/12`
  4. `- {11pi}/{12}`
Show Answers Only

`C`

Show Worked Solution

`(23pi)/12\ \ =>\ \ text{4th quadrant}`

`cos((23pi)/12)=cos(2pi-(23pi)/12)=cos(pi/12)`

`:. cos^(-1)((sqrt6+sqrt2)/(4))=pi/12`

`=>C`


Mean mark 54%.

Financial Maths, 2ADV M1 2022 HSC 32

In a reducing-balance loan, an amount `$P` is borrowed for a period of `n` months at an interest rate of 0.25% per month, compounded monthly. At the end of each month, a repayment of `$M` is made. After the `n`th repayment has been made, the amount owing, `$A_n`, is given by

`A_(n)=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`

(Do NOT prove this.)

  1. Jane borrows $200 000 in a reducing-balance loan as described.

  2. The loan is to be repaid in 180 monthly repayments.

  3. Show that  `M` = 1381.16, when rounded to the nearest cent.  (2 marks)

  4. After 100 repayments of $1381.16 have been made, the interest rate changes to 0.35% per month.

  5. At this stage, the amount owing to the nearest dollar is $100 032. (Do NOT prove this.)

  6. Jane continues to make the same monthly repayments.

  7. For how many more months will Jane need to make full monthly payments of $1381.16?  (3 marks)

  8. The final payment will be less than $1381.16.
  9. How much will Jane need to pay in the final payment in order to pay off the loan?   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `83`
  3. `$931.54`
Show Worked Solution

a.   `text{Show}\ \ M=$1381.16`

`A_(n)` `=P(1.0025)^(n)-M(1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^(n-1))`  
`0` `=200\ 000(1.0025)^180-M underbrace((1+(1.0025)^(1)+(1.0025)^(2)+cdots+(1.0025)^179))_(text(GP where)\ a = 1,\ r = 1.0025,\ n = 180)`  
`0` `=200\ 000(1.0025)^180-M((1(1.0025^180- 1))/(1.0025-1))`  

`M((1.0025^180-1)/(1.0025-1))=200\ 000(1.0025)^180`

`:.M` `=200\ 000(1.0025)^180 -: (1.0025^180-1 )/(1.0025-1)`   
  `=1381.163…`  
  `=$1381.16\ \ text{… as required}`  

 

b.   `P=$100\ 032,\ \ r=1.0035  and M=$1381.16`

`text{Find}\ \ n\ \ text{when}\ \ A_n=0:`

`A_(n)` `=P(1.0035)^(n)-1381.16(1+(1.0035)^(1)+(1.0035)^(2)+cdots+(1.0035)^(n-1))`  
`0` `=100\ 032(1.0035)^n-1381.16 ((1.0035^n- 1)/(1.0035-1))`  
`0` `=100\ 032(1.0035)^n-1381.16/0.0035 (1.0035^n- 1)`  
  `=100\ 032(1.0035)^n-394\ 617(1.0035)^n- 394\ 617`  
`294\ 585(1.0035)^n` `=394\ 617`  
`1.0035^n` `=(394\ 617)/(294\ 585)`  
`n` `=ln((394\ 617)/(294\ 585))/ln1.0035`  
  `=83.674…`  
  `=83\ text{more months with full payment}`  

 


♦♦ Mean mark part (b) 38%.

c.   `text{Find}\ \ A_83:`

`A_83` `=100\ 032(1.0035)^83-1381.16 ((1.0035^83- 1)/(1.0035-1))`  
  `=928.291…`  

 
`text{Interest will be added for the last month:}`

`:.\ text{Final payment}` `=928.291… xx 1.0035`  
  `=$931.54`  

♦♦♦ Mean mark part (c) 14%.

Calculus, 2ADV C3 2022 HSC 27

Let  `f(x)=xe^(-2x)`.

It is given that  `f^(′)(x)=e^(-2x)-2xe^(-2x)`.

  1. Show that  `f^(″)(x)=4(x-1)e^(-2x)`.  (2 marks)

  2. Find any stationary points of  `f(x)`  and determine their nature.  (2 marks)

  3. Sketch the curve  `y=x e^{-2 x}`, showing any stationary points, points of inflection and intercepts with the axes.  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Max S.P. at}\ \ (1/2, 1/(2e))`
  3.  
Show Worked Solution

a.   `f^(′)(x)=e^(-2x)-2xe^(-2x)`

`f^(″)(x)` `=-2e^(-2x)-[2x(-2)e^(-2x)+2e^(-2x)]`  
  `=-2e^(-2x)+4xe^(-2x)-2e^(-2x)`  
  `=4xe^(-2x)-4e^(-2x)`  
  `=4(x-1)e^(-2x)\ \ text{… as required}`  

 

b.   `text{S.P.’s occur when}\ \ f^(′)(x)=0`

`e^(-2x)-2xe^(-2x)` `=0`  
`e^(-2x)(1-2x)` `=0`  

 
`e^(-2x)=0\ \ =>\ \ text{No solution}`

`1-2x=0\ \ =>\ \ x=1/2`

`f^(″)(1/2)` `=4(1/2-1)e^(-2xx1/2)`  
  `=-2e^(-1)<0\ \ =>\ text{MAX}`  

 
`f(1/2)=1/2e^(-1)=1/(2e)`
 

c.   `text{POI occurs when}\ \ f^(″)(x)=0`

`f^(″)(x)=4(x-1)e^(-2x)=0\ \ =>\ \ x=1`

`text{Test for change in concavity:}`

`f^(″)(0)=4(0-1)e^0=-4`

`f^(″)(2)=4(2-1)e^(-4)>0\ \ =>\ text{Concavity changes}`

`:.\ text{POI exists at}\ \ (1,1/e^2)`

`text{As}\ \ x->oo\ \ =>\ \ y->0^+`
 


Mean mark part (c) 54%.

Calculus, 2ADV C4 2022 HSC 29

  1. The diagram shows the graph of  `y=2^{-x}`. Also shown on the diagram are the first 5 of an infinite number of rectangular strips of width 1 unit and height  `y=2^{-x}`  for non-negative integer values of `x`. For example, the second rectangle shown has width 1 and height `(1)/(2)`. 
     


 

  1. The sum of the areas of the rectangles forms a geometric series.
  2. Show that the limiting sum of this series is 2. (1 mark)

  3. Show that `int_(0)^(4)2^(-x)\ dx=(15)/(16 ln 2)`.  (2 marks)

  4. Use parts (a) and (b) to show that  `e^(15) < 2^(32)`.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
Show Worked Solution

a.   `text{Consider the rectangle heights:}`

`2^0=1, \ 2^(-1)=1/2, \ 2^(-2)= 1/4, \ 2^(-3)= 1/8, …`

`=>\ text{Rectangle Areas}\ = 1, \ 1/2, \  1/4, \ 1/8, …`

`a=1,\ \ r=1/2`

`S_oo=a/(1-r)=1/(1-1/2)=2\ \ text{… as required}`
 

b.   `text{Show}\ \ int_0^4 2^(-x)\ dx = 15/(16ln2)`

`int_0^4 2^(-x)\ dx ` `=(-1)/ln2[2^(-x)]_0^4`  
  `=(-1)/ln2(1/16-1)`  
  `=1/ln2-1/(16ln2)`  
  `=(16-1)/(16ln2)`  
  `=15/(16ln2)\ \ text{… as required}`  

 


Mean mark (b) 56%.

c.   `text{Show}\ \ e^15<2^32`

`text{Area under curve < Sum of rectangle areas}`

`15/(16ln2)` `<2`  
`15` `<32ln2`  
`15/32` `<ln2`  
`e^(15/32)` `<e^(ln2)`  
`root(32)(e^15)` `<2`  
`e^15` `<2^32\ \ text{… as required}`  

♦♦♦ Mean mark (c) 9%.

Calculus, 2ADV C4 2022 HSC 28

The graph of the circle  `x^2+y^2=2`  is shown.

The interval connecting the origin, `O`, and the point `(1,1)` makes an angle `theta` with the positive `x`-axis.
 

  1. By considering the value of `theta`, find the exact area of the shaded region, as shown on the diagram.  (2 marks)

Part of the hyperbola  `y=(a)/(b-x)-1`  which passes through the points `(0,0)` and `(1,1)` is drawn with the circle  `x^2+y^2=2`  as shown.
 

  1. Show that  `a=b=2`.  (2 marks)

  2. Using parts (a) and (b), find the exact area of the region bounded by the hyperbola, the positive `x`-axis and the circle as shown on the diagram.   (3 marks)

Show Answers Only
  1. `(pi-2)/4\ text{u}^2`
  2. `text{Proof (See Worked Solutions)}`
  3. `(8ln2+pi-6)/4\ text{u}^2`
Show Worked Solution

a.   `tan theta=1\ \ =>\ \ theta = pi/4`

`text{Using Pythagoras,}`

`r=sqrt(1^2+1^2)=sqrt2`

`text{Shaded Area}` `=A_text{sector}-A_Delta`  
  `=(pi/4)/(2pi) xx pi r^2-1/2 xx b xx h`  
  `=1/8xxpixx(sqrt2)^2-1/2xx1xx1`  
  `=pi/4-1/2`  
  `=(pi-2)/4\ text{u}^2`  

 


Mean mark (a) 54%.

b.   `text{Show}\ \ a=b=2`

`y=(a)/(b-x)-1\ \ text{passes through}\ \ (0,0):`

`0` `=a/(b-0)-1`  
`a/b` `=1`  
`a` `=b`  

 
`y=(a)/(b-x)-1\ \ text{passes through}\ \ (1,1):`

`1` `=a/(b-1)-1`  
`a/(b-1)` `=2`  
`a` `=2(b-1)`  
`a` `=2b-2`  
`b` `=2b-2\ \ text{(using}\ a=b)`  
`b` `=2`  

 
`:.a=b=2`
 


♦ Mean mark (b) 47%.
c.    `int_0^1 2/(2-x)-1\ dx` `=int_0^1 -2 xx (-1)/(2-x)-1\ dx`
    `=[-2ln|2-x|-x]_0^1`
    `=[(-2ln1-1)-(-2ln2-0)]`
    `=-1+2ln2`

 

`:.\ text{Total Area}` `=2ln2-1 + (pi-2)/4`  
  `=(8ln2-4+pi-2)/4`  
  `=(8ln2+pi-6)/4\ text{u}^2`  

♦♦ Mean mark (c) 36%.

Calculus, 2ADV C2 2022 HSC 25

Let  `f(x)=sin(2x)`.

Find the value of `x`, for  `0 < x < pi`, for which  `f^(′)(x)=-sqrt3`  AND  `f^(″)(x)=2`.  (3 marks)

Show Answers Only

`(7pi)/12`

Show Worked Solution

`f^(′)(x)=2cos(2x)`

`2cos(2x)` `=-sqrt3`  
`cos(2x)` `=- sqrt3/2`  
`2x` `=pi-pi/6,\ \ pi+pi/6`  
  `=(5pi)/6,\ \ (7pi)/6`  
`x` `=(5pi)/12,\ \ (7pi)/12`  

 
`f^(″)(x)=-4sin(2x)`

`-4sin(2x)` `=2`  
`sin(2x)` `=- 1/2`  
`2x` `=pi+pi/6,\ \ 2pi-pi/6`  
  `=(7pi)/6,\ \ (11pi)/6`  
`x` `=(7pi)/12,\ \ (22pi)/12`  

 
`:.x=(7pi)/12\ \ text{(satisfies both equations)}`


Mean mark 53%.

Trigonometry, 2ADV T3 2022 HSC 23

The depth of water in a bay rises and falls with the tide. On a particular day the depth of the water, `d` metres, can be modelled by the equation

`d=1.3-0.6 cos((4pi)/(25)t)`,

where `t` is the time in hours since low tide.

  1. Find the depth of water at low tide and at high tide.  (2 marks)

  2. What is the time interval, in hours, between two successive low tides?  (1 mark)

  3. For how long between successive low tides will the depth of water be at least 1 metre?  (3 marks)

Show Answers Only
  1. `text{Low: 0.7 m,  High: 1.9 m}`
  2. `25/2\ text{hours}`
  3. `25/3\ text{hours}`
Show Worked Solution

a.   `text{S}text{ince}\ \ –1<=cos((4pi)/(25)t)<=1:`

`text{Low Tide}\ =1.3-0.6(1)=0.7\ text{m}`

`text{High Tide}\ =1.3-0.6(-1)=1.9\ text{m}`
 

b.   `text{Time between two low tides = Period of equation}\ (n)`

`(2pi)/n` `=(4pi)/25`  
`n/(2pi)` `=25/(4pi)`  
`n` `=25/2\ text{hours}`  

 


♦ Mean mark part (b) 47%.

c.   `text{Find}\ \ t\ \ text{when}\ \ d=1:`

`1.3-0.6 cos((4pi)/(25)t)` `=1`  
`-0.6 cos((4pi)/(25)t)` `=-0.3`  
`cos((4pi)/(25)t)` `=1/2`  

 

`(4pi)/(25)t` `=pi/3,\ \ (5pi)/3`  
`t` `=25/12,\ \ 125/12`  

 
`:.\ text{Time between low tides where water depth}\ >= 1\ text{m}`

`=125/12-25/12`

`=100/12`

`=25/3\ text{hours}`


♦ Mean mark part (c) 46%.

Calculus, 2ADV C3 2022 HSC 22

Find the global maximum and minimum values of  `y=x^(3)-6x^(2)+8`, where  `-1 <= x <= 7`.   (4 marks)

Show Answers Only

`text{Global max = 57}`

`text{Global min = – 24}`

Show Worked Solution
`y` `=x^3-6x^2+8`  
`dy/dx` `=3x^2-12x`  
`(d^2y)/(dx^2)` `=6x-12`  

 
`text{SP’s when}\ \ dy/dx=0:`

`3x^2-12x` `=0`  
`3x(x-4)` `=0`  

 
`x=0\ \ text{or}\ \ 4`

`text{When}\ \ x=0,\ \ y=8,\ \ (d^2y)/(dx^2)<0`

`=>\ text{Local Max at}\ \ (0,8)`

`text{When}\ \ x=4,\ \ y=4^3-6(4^2)+8=-24,\ \ (d^2y)/(dx^2)>0`

`=>\ text{Local Min at}\ \ (4,-24)`
 

`text{Check ends of domain:}`

`text{When}\ \ x=-1,\ \ y=-1-6+8=1`

`text{When}\ \ x=7,\ \ y=7^3-6(7^2)+8=57`

`:.\ text{Global max = 57}`

`:.\ text{Global min = – 24}`

Calculus, 2ADV C4 2022 HSC 18

  1. Differentiate  `y=(x^(2)+1)^(4)`.  (2 marks)

  2. Hence, or otherwise, find `int x(x^(2)+1)^(3)dx`.  (1 mark)

Show Answers Only
  1.  `dy/dx=8x(x^2+1)^3`
  2.  `1/8(x^2+1)^4+C`
Show Worked Solution

a.   `y=(x^2+1)^4`

`text{Using chain rule:}`

`dy/dx` `=4 xx 2x(x^2+1)^3`  
  `=8x(x^2+1)^3`  

 

b.    `intx(x^2+1)^3\ dx` `=1/8 int8x(x^2+1)^3\ dx`
    `=1/8(x^2+1)^4+C`

Calculus, 2ADV C4 2022 HSC 16

The parabola `y=x^2` meets the line `y=2 x+3` at the points `(-1,1)` and `(3,9)` as shown in the diagram.
 

Find the area enclosed by the parabola and the line.  (3 marks)

Show Answers Only

`32/3\ text{u}^2`

Show Worked Solution
`A` `=int_(-1)^(3) 2x+3-x^2\ dx`  
  `=[x^2+3x-1/3 x^3]_(-1)^3`  
  `=(9+9-27/3)-(1-3+1/3)`  
  `=9-(-5/3)`  
  `=32/3\ text{u}^2`  

Probability, 2ADV S1 2022 HSC 15

In a bag there are 3 six-sided dice. Two of the dice have faces marked  1, 2, 3, 4, 5, 6. The other is a special die with faces marked  1, 2, 3, 5, 5, 5.

One die is randomly selected and tossed.

  1. What is the probability that the die shows a 5?  (1 mark)

  2. Given that the die shows a 5, what is the probability that it is the special die?  (1 mark)

Show Answers Only
  1. `5/18`
  2. `3/5`
Show Worked Solution
a.    `P(5)` `=1/3 xx 1/6 + 1/3 xx 1/6 + 1/3 xx 1/2`
    `=1/18+1/18+1/6`
    `=5/18`

 

b.   `Ptext{(special die given a 5 is rolled)}`

`=(Ptext{(special die)} ∩  P(5))/(P(5))`

`=(1/3 xx 1/2)/(5/18)`

`=1/6 xx 18/5`

`=3/5`


Mean mark (a) 53%.
 
♦ Mean mark (b) 40%.

Trigonometry, 2ADV T3 2022 HSC 14

The graph of  `y=k sin (ax)`

What are the values of `k` and `a`?  (2 marks)

Show Answers Only

`k=4, \ \ a=1/3`

Show Worked Solution

`text{Amplitude = 4}`

`=> k=4`

`text{Period} = 6pi`

`(2pi)/a` `=6pi`  
 `6pia` `=2pi`  
 `=>a` `=1/3`  

Calculus, 2ADV C4 2022 HSC 13

Use two applications of the trapezoidal rule to find an approximate value of `int_(0)^(2)sqrt(1+x^(2))\ dx`. Give your answer correct to 2 decimal places.  (2 marks)

Show Answers Only

`3.03`

Show Worked Solution
`A` `~~h/2(y_0+2y_1+y_2)`  
  `~~1/2(1+2 xx sqrt2+sqrt5)`  
  `~~3.03\ \ text{(to 2 d.p.)}`  

Mean mark 59%.
COMMENT: A surprisingly low mean mark warrants attention.

Calculus, 2ADV C4 2022 HSC 8 MC

The graph of the even function  `y=f(x)`  is shown.

The area of the shaded region `A` is `1/2` and the area of the shaded region `B` is `3/2`.

What is the value of `int_(-2)^(2)f(x)\ dx`?

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

`C`

Show Worked Solution

`text{Areas under the}\ x text{-axis are negative}`

`int_0 ^2 f(x)\ dx = 1/2-3/2=-1`
 

`text{S}text{ince}\ \ f(x)\ \ text{is even:}`

`int_(-2) ^2 f(x)\ dx = 2int_0 ^2 f(x)\ dx = -2`

`=>C`


Mean mark 52%.

Functions, 2ADV F1 2022 HSC 12

A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.

  1. Find the equation relating `M` and `T`.    (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time from `T=5^@C` to `T=30^@ text{C}`.   (2 marks)
     

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M &  &  &  \\
\hline \end{array}

 
                   

Show Answers Only

a.    `M=180/T`

 b.    

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}       

 

Show Worked Solution
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`
 


♦ Mean mark (a) 49%.

b.   

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}

Calculus, 2ADV C1 2022 HSC 5 MC

Let  `h(x)=(f(x))/(g(x))`, where

`{:[f(1)=2, qquad f^{′}(1)=4],[g(1)=8, qquad g^{′}(1)=12]:}`

What is the gradient of the tangent to the graph of  `y=h(x)`  at  `x=1` ?

  1. `-8`
  2. `\ \ \ 8`
  3. `- 1/8`
  4. `\ \ \ 1/8`
Show Answers Only

`D`

Show Worked Solution

`text{Using the quotient rule:}`

`h^{′}(1)` `=(g(1)\ f^{′}(1)-f(1)\ g^{′}(1))/[g(1)]^2`  
  `=(8xx4-2xx12)/(8^2)`  
  `=(32-24)/64`  
  `=1/8`  

 
`=>D`

Trigonometry, 2ADV T1 2022 HSC 3 MC

A tower `B T` has height `h` metres.

From point `A`, the angle of elevation to the top of the tower is `26^@` as shown.
 


 

Which of the following is the correct expression for the length of `A B` ?

  1. `h  tan 26^(@)`
  2. `h  cot 26^(@)`
  3. `h  sin 26^(@)`
  4. `h\ text{cosec}\ 26^(@)`
Show Answers Only

`B`

Show Worked Solution
`tan 26^@` `=h/(AB)`  
`AB` `=h/(tan26^@)`  
  `=h\ cot26^@`  

 
`=>B`

Statistics, 2ADV S3 2022 HSC 26

The life span of batteries from a particular factory is normally distributed with a mean of 840 hours and a standard deviation of 80 hours.

It is known from statistical tables that for this distribution approximately 60% of the batteries have a life span of less than 860 hours.

What is the approximate percentage of batteries with a life span between 820 and 920 hours?  (3 marks)

Show Answers Only

`44text{%}`

Show Worked Solution

`mu=840, \ sigma=80`

`ztext{-score (860)}\ = (x-mu)/sigma=(860-840)/80=0.25` 

`ztext{-score (820)}\ =(820-840)/80=-0.25` 

`ztext{-score (920)}\ =(920-840)/80=1`
 

`text{50% of batteries have a life span below 840 hours (by definintion)}`

`=>\ text{10% lie between 840 and 860 hours}`

`=>\ text{By symmetry, 10% lie between 820 and 840 hours}`

`=> P(-0.25<=z<=0)=10text{%}`
 

`:.\ text{Percentage between 820 and 920}`

`=P(-0.25<=z<=1)`

`=P(-0.25<=z<=0) + P(0<=z<=1)`

`=10+34`

`=44text{%}`

Statistics, 2ADV S2 2022 HSC 24

Jo is researching the relationship between the ages of teenage characters in television series and the ages of actors playing these characters.

After collecting the data, Jo finds that the correlation coefficient is 0.4564.

A scatterplot showing the data is drawn. The line of best fit with equation  `y=-7.51+1.85 x`, is also drawn.
 


 

Describe and interpret the data and other information provided, with reference to the context given.  (4 marks)

Show Answers Only

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`
Show Worked Solution

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`

Financial Maths, 2ADV M1 2022 HSC 21

Eli is choosing between two investment options.

A table of future value interest factors for an annuity of $1 is shown.

  1. What is the value of Eli's investment after 10 years using Option 1 ?  (2 marks)

  2. What is the difference between the future values after 10 years using Option 1 and Option 2?  (2 marks)

Show Answers Only
  1. `$45\ 097.17`
  2. `$40.87`
Show Worked Solution

a.   `text{Monthly r/i}\ = 1.2/12=0.1text{%}\ \ =>\ \ r= 0.001`

`text{Compounding periods}\ (n)=12xx10=120`

`FV` `=PV(1+r)^n`  
  `=40\ 000(1+0.001)^120`  
  `=$45\ 097.17`  

 

b.   `text{Quarterly r/i}\ = 2.4/4=0.6text{%}\ \ =>\ \ r= 0.006`

`text{Compounding periods}\ (N) =4xx10=40`

`text{Annuity factor (from table) = 45.05630}`

`FV` `=1000xx45.05630`  
  `=45\ 056.30`  

 

`text{Difference}` `=45\ 097.17-45\ 056.30`  
  `=$40.87`  

Statistics, 2ADV S2 2022 HSC 11

The table shows the types of customer complaints received by an online business in a month.

  1. What are the values of `A` and `B`?  (2 marks)

  2. The data from the table are shown in the following Pareto chart.
     
     
  3. The manager will address 80% of the complaints.
  4. Which types of complaints will the manager address?  (1 mark)

Show Answers Only
  1. `A=160, \ B=96`
  2. `text{Stock shortages and delivery fees.}`
Show Worked Solution

a.   `A=98+62=160`

`text{% Damaged items}\ = 8/200 xx 100 = 4text{%}`

`text{Cumulative % after damaged items = 96%}`

`B = 92+4=96`
 

b.   `text{The right hand side cumulative frequency percentage}`

`text{shows that 80% of all complaints received concern}`

`text{stock shortages and delivery fees.}`

`:.\ text{The manager will address stock shortages and delivery fees.}`

Financial Maths, STD2 F4 2022 HSC 36

Frankie borrows $200 000 from a bank. The loan is to be repaid over 23 years at a rate of 7.2% per annum, compounded monthly. The repayments have been set at $1485 per month.

The interest charged and the balance owing for the first three months of the loan are shown in the spreadsheet below.
 

  1. What are the values of `A` and `B`?  (2 marks)
  2. After 50 months of repaying the loan, Frankie decides to make a lump sum payment of $ 40 000 and to continue making the monthly repayments of $1485. The loan will then be fully repaid after a further 146 monthly repayments.
  3. How much less will Frankie pay overall by making the lump sum payment?  (3 marks)
Show Answers Only
  1. `A=$1198.29,\ \ B=$199\ 139.86`
  2. `$78\ 800`
Show Worked Solution

a.   `text{Monthly interest rate}\ =7.2/12=0.6text{%}`

`A` `=199\ 715 xx 0.6/100`  
  `=$1198.29`  

 

`B` `=P+I-R`  
  `=199\ 428.29 + 1196.57-1485`  
  `=$199\ 139.86`  

 

b.   `text{Total payments if lump sum not paid}`

`= (23xx12) xx 1485`

`=$409\ 860`
 

`text{Total payments if lump sum paid}`

`=40\ 000 + (50 + 146) xx 1485`

`=$331\ 060`
 

`text{Savings by paying the lump sum}`

`=409\ 860-331\ 060`

`=$78\ 800`


♦♦♦ Mean mark (b) 17%.
 

Measurement, STD2 M1 2022 HSC 32

The diagram shows a park consisting of a rectangle and a semicircle. The semicircle has a radius of 100 m. The dimensions of the rectangle are 200 m and 250 m.

A lake occupies a section of the park as shown. The rest of the park is a grassed section. Some measurements from the end of the grassed section to the edge of the lake are also shown.
 

  1. Using two applications of the trapezoidal rule, calculate the approximate area of the grassed section.  (2 marks)

  2. Hence calculate the approximate area of the lake, to the nearest square metre.   (2 marks)

Show Answers Only
  1. `35\ 500\ text{m}^2`
  2. `30\ 208\ text{m}^2`
Show Worked Solution

a.   `h=100\ text{m}`

`A` `~~h/2(x_1+x_2)+h/2(x_2+x_3)`  
  `~~100/2(160+150)+100/2(150+250)`  
  `~~50xx310+50xx400`  
  `~~35\ 500\ text{m}^2`  

 

b.    `text{Total Area}` `=\ text{Area of Rectangle + Area of Semi-Circle}`
    `= (250 xx 200) + 1/2 xx pi xx 100^2`
    `=65\ 707.96\ text{m}^2`

 

`text{Area of Lake}` `=67\ 708-35\ 500`  
  `=30\ 208\ text{m}^2`  

♦ Mean mark (b) 44%.

Networks, STD2 N3 2022 HSC 31

A wildlife park has 5 main attractions `(A, B, C, D, E)` connected by directional paths. A simple network is drawn to represent the flow through the park's paths. The number of visitors who can access each path at any one time is also shown.
 

   

  1. What is the flow capacity of the cut shown?  (1 mark)

  2. By showing a suitable cut on the diagram below, explain why the network's current maximum flow capacity is less than 40 visitors.  (2 marks)
     

  3. One path is to be increased in capacity so that the overall maximum flow will be 40 visitors at any one time.
  4. Which path could be increased and by how much?  (2 marks)

Show Answers Only
  1. `40`
  2.  
     
  3. `text{Max flow = min cut = 35}`
  4. `AE\ text{or}\ DE\ text{could be increased by 5}`
Show Worked Solution

a.   `text{Flow capacity = 10 + 20 + 10 = 40}`

`text{(DE is not counted as it runs from sink → source)}`
 

b.  
       

`text{Min Cut = Max Flow}`

`text{Max Flow}` `=15+10+10`  
  `=35<40`  


♦ Mean mark part (a) 45%.
♦♦ Mean mark part (b) 33%.

 

c.   `text{Two strategies:}`

  • `AE\ text{could be increased by 5}`
  • `DE\ text{could be increased by 5}`

`text{(both strategies would increase the minimum cut to}`

  `text{40 by increasing the flow to vertex}\ E\ text{to 30)}`


Mean mark 52%.

Measurement, STD2 M2 2022 HSC 29

Sydney is 10 hours ahead of Coordinated Universal Time (UTC +10) and New York is 5 hours behind Coordinated Universal Time (UTC –5).

Tony travels from Sydney to New York. His plane leaves Sydney at 8:20 pm on Wednesday local time and flies non-stop to New York.

The flight takes 20 hours and 24 minutes.

What time and day is it in New York when the plane lands?  (3 marks)

Show Answers Only

`text{1:44 am (Thu)}`

Show Worked Solution

`text{Sydney is 15 hours ahead of New York.}`

`text{Plane’s arrival (Sydney time)}`

`=\ text{8:20 pm + 20h 24 min}`

`=\ text{4:44 pm (Thu)}`
 

`text{Plane’s arrival (New York time)}`

`=\ text{4:44 pm (Thu) – 15 hours}`

`=\ text{1:44 am (Thu)}`

Financial Maths, STD2 F5 2022 HSC 25

The table shows the future value of an annuity of $1.
 
     

Zal is saving for a trip and estimates he will need $15 000. He opens an account earning 3% per annum, compounded annually.

  1. How much does Zal need to deposit every year if he wishes to have enough money for the trip in 4 years time?  (2 marks)

  2. How much interest will Zal earn on his investment over the 4 years? Give your answer to the nearest dollar.  (2 marks)

Show Answers Only
  1. `$3589.09`
  2. `$660`
Show Worked Solution

a.   `text{Using the table:}\ r=3text{%},\ \ n=4`

`text{Annuity factor}\ = 4.184`

`text{Let}\ \ A=\ text{amount invested each year}`

`FV` `=A xx 4.184`  
`15\ 000` `=A xx 4.184`  
`:.A` `=(15\ 000)/4.184`  
  `=$3585.09`  

 

b.   `text{Total payments}\ = 4 xx 3585.09=$14\ 340.36`

`text{Interest earned}` `=FV-\ text{total payments}`  
  `=15\ 000-14\ 340.36`  
  `=659.64`  
  `=$660\ \ text{(nearest $)}`  

♦♦ Mean mark 33%.

Probability, STD2 S2 2022 HSC 17

The numbers 1, 2, 3, 4, 5 and 6 are each written on separate cards.

Amy has cards 1, 3 and 5 and Bob has cards 2, 4 and 6.

They play a game in which each person randomly chooses one of their own cards and compares it with the other person's card. The person with the higher card wins.

  1. A partially completed tree diagram is shown.
     
           
     
    Complete the tree diagram and find the probability that Bob wins.  (2 marks)
     
  2. Suppose Amy and Bob play this game 30 times.
  3. How many times would Bob be expected to win?  (1 mark)

Show Answers Only
  1. `6/9`
  2. `20`
Show Worked Solution

a.   
       

`P(text{Bob wins}) = 6/9=2/3`
 

b.    `text{Expected wins}` `=2/3 xx 30`
    `=20`

Algebra, STD2 A4 2022 HSC 22

The formula  `C=100 n+b`  is used to calculate the cost of producing laptops, where `C` is the cost in dollars, `n` is the number of laptops produced and `b` is the fixed cost in dollars.

  1. Find the cost when 1943 laptops are produced and the fixed cost is $20 180.  (1 mark)

  2. Some laptops have some extra features added. The formula to calculate the production cost for these is
  3.      `C=100 n+a n+20\ 180`
  4. where `a` is the additional cost in dollars per laptop produced.
  5. Find the number of laptops produced if the additional cost is $26 per laptop and the total production cost is $97 040.  (2 marks)

Show Answers Only
  1. `$214\ 480`
  2. `610\ text{laptops}`
Show Worked Solution

a.   `text{Find}\ \ C\ \ text{given}\ \ n=1943 and b=20\ 180`

`C` `=100 xx 1943 + 20\ 180`  
  `=$214\ 480`  

 

b.   `text{Find}\ \ n\ \ text{given}\ \ C=97\ 040 and a=26`

`C` `=100 n+a n+20\ 180`  
`97\ 040` `=100n + 26n +20\ 180`  
`126n` `=76\ 860`  
`n` `=(76\ 860)/126`  
  `=610 \ text{laptops}`  

Statistics, STD2 S1 2022 HSC 19

The table shows the types of customer complaints received by an online business in a month.
 

  1. What are the values of `A` and `B`?  (2 marks)

  2. The data from the table are shown in the following Pareto chart.
     
     
  3. The manager will address 80% of the complaints.
  4. Which types of complaints will the manager address?  (1 mark)

Show Answers Only
  1. `A=160, \ B=96`
  2. `text{Stock shortages and delivery fees.}`
Show Worked Solution

a.   `A=98+62=160`

`text{% Damaged items}\ = 8/200 xx 100 = 4text{%}`

`text{Cumulative % after damaged items = 96%}`

`B = 92+4=96`
 

b.   `text{The right hand side cumulative frequency percentage}`

`text{shows that 80% of all complaints received concern}`

`text{stock shortages and delivery fees.}`

`:.\ text{The manager will address stock shortages and delivery fees.}`


♦ Mean mark part (b) 41%.

Statistics, STD2 S5 2022 HSC 18

The marks in a test were normally distributed. The mean mark was 60 and the standard deviation was 15 .

What was the percentage of marks higher than 90?  (2 marks)

Show Answers Only

`2.5text{%}`

Show Worked Solution
`ztext{-score}` `=(x-mu)/sigma`  
  `=(90-60)/15`  
  `=2`  

 

`text{% marks > 90}\ = 2.5text{%}`


Mean mark 52%.
COMMENT: A poor State result that warrants attention.

Algebra, STD2 A2 2022 HSC 14 MC

Which of the following correctly expresses `x` as the subject of  `y=(ax-b)/(2)` ?

  1. `x=(2y+b)/(a)`
  2. `x=(y+b)/(2a)`
  3. `x=(2y)/(a)+b`
  4. `x=(y)/(2a)+b`
Show Answers Only

`A`

Show Worked Solution
`y` `=(ax-b)/(2)`  
`2y` `=ax-b`  
`ax` `=2y+b`  
`:.x` `=(2y+b)/a`  

 
`=>A`

Statistics, STD2 S4 2022 HSC 12 MC

For a particular course, the recorded data show a relationship between the number of hours of study per week and the marks achieved out of 100 .

A least-squares regression line is fitted to this dataset. The equation of this line is given by

`M=20+3 H,`

where `M` is the predicted mark and `H` is the number of hours of study per week.

Based on this regression equation, which of the following is correct regarding the predicted mark of a student?

  1. It will be 3 for zero hours of study per week.
  2. It will be 20 for zero hours of study per week.
  3. It will increase by 20 for every additional hour of study per week.
  4. It will increase by 1 for every 3 additional hours of study per week.
Show Answers Only

`B`

Show Worked Solution

`text{Consider Option}\ B:`

`text{If zero hours of study are done per week}\ \ → \ \ H=0`

`:. M=20+(3 xx 0) = 20`

`=>B`

Financial Maths, STD2 F4 2022 HSC 11 MC

In ten years, the future value of an investment will be $150 000. The interest rate is 4% per annum, compounded half-yearly.

Which equation will give the present value `(PV)` of the investment?

  1. `PV=(150\ 000)/((1+0.04)^(10))`
  2. `PV=(150\ 000)/((1+0.04)^(20))`
  3. `PV=(150\ 000)/((1+0.02)^(10))`
  4. `PV=(150\ 000)/((1+0.02)^(20))`
Show Answers Only

`D`

Show Worked Solution

`text{Compounding periods}\ = 10 xx 2 = 20`

`text{Compounding rate}\ = (4text{%})/2 = 2text{%} = 0.02`

`PV=(150\ 000)/((1+0.02)^(20))`

`=>D`

Statistics, STD2 S1 2022 HSC 5 MC

Consider the following dataset.

`{:[13,16,17,17,21,24]:}`

Which row of the table shows how the median and mean are affected when a score of 5 is added to the dataset?

Show Answers Only

`D`

Show Worked Solution

`text{Mean decreases.}`

`text{Median remains 17.}`

`=>D`

PHYSICS, M7 2019 HSC 23

A student investigated the photoelectric effect. The frequency of light incident on a metal surface was varied and the corresponding maximum kinetic energy of the photoelectrons was measured.

The following results were obtained.
 

\begin{array}{|l|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\textit{Frequency}\ \left(\times 10^{14} Hz\right) \rule[-1ex]{0pt}{0pt}& 11.2 & 13.5 & 15.2 & 18.6 & 20.0 \\
\hline \rule{0pt}{2.5ex}\textit{Maximum kinetic energy}\ (eV) \rule[-1ex]{0pt}{0pt}& 0.6 & 1.3 & 2.3 & 3.3 & 4.2 \\
\hline
\end{array}

Plot the results on the axes below and hence determine the work function of the metal in electron volts.   (3 marks)
 

Show Answers Only

Work function = 4 eV

Show Worked Solution

The metal has a work function of 4 eV (this is the negative of the \(y\)-intercept).

PHYSICS, M7 2019 HSC 22

Spectra can be used to determine the chemical composition and surface temperature of stars.

Describe how spectra provide information about OTHER features of stars.   (3 marks)

Show Answers Only
  • Spectra provide information about the linear velocity of a star relative to Earth. A shift towards higher wavelengths indicates that a star is moving away from Earth. A shift towards lower wavelengths indicates a star is moving towards earth.
  • Darkened, broader spectral lines indicate that a star has a high density due to high pressures in the star’s atmosphere. Narrower spectral lines indicate a lower density star.
  • Blurring and broadening of the edges of spectral lines indicate a star has a high rotational velocity due to simultaneous red and blue doppler shifting. More defined, narrower spectral lines indicate a star with lower rotational velocity.
Show Worked Solution
  • Spectra provide information about the linear velocity of a star relative to Earth. A shift towards higher wavelengths indicates that a star is moving away from Earth. A shift towards lower wavelengths indicates a star is moving towards earth.
  • Darkened, broader spectral lines indicate that a star has a high density due to high pressures in the star’s atmosphere. Narrower spectral lines indicate a lower density star.
  • Blurring and broadening of the edges of spectral lines indicate a star has a high rotational velocity due to simultaneous red and blue doppler shifting. More defined, narrower spectral lines indicate a star with lower rotational velocity.

PHYSICS, M8 2019 HSC 21

Outline de Broglie's contribution to quantum mechanics. Support your answer with a relevant equation.   (2 marks)

Show Answers Only
  • de Broglie proposed the dual wave-particle model of matter which postulated that particles, such as electrons possessed wave properties.
  • The equation to predict the wavelength of such particles is:
  •    `lambda=(h)/(mv)`
Show Worked Solution
  • de Broglie proposed the dual wave-particle model of matter which postulated that particles, such as electrons possessed wave properties.
  • The equation to predict the wavelength of such particles is:
  •    `lambda=(h)/(mv)`

PHYSICS, M7 2019 HSC 15 MC

Monochromatic light passes through two slits 1 \(\mu \text{m}\) apart. The resulting diffraction pattern is measured at a distance of 0.3 m.
 


 

This diffraction pattern can be analysed using the equation  \(d \sin \theta=\lambda\).

What values of \(d\) and \(theta\) should be used in the equation?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-2.5ex]{0pt}{0pt}& \\
\rule{0pt}{3ex}\textbf{A.}\rule[-2.5ex]{0pt}{0pt}\\
\rule{0pt}{3ex}\textbf{B.}\rule[-3ex]{0pt}{0pt}\\
\rule{0pt}{3ex}\textbf{C.}\rule[-3ex]{0pt}{0pt}\\
\rule{0pt}{3ex}\textbf{D.}\rule[-3ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad d \quad \rule[-1ex]{0pt}{0pt}& \theta \\
\hline
\rule{0pt}{2.5ex}0.3 \ \text{m}\rule[-1ex]{0pt}{0pt}&\tan ^{-1}\left(\dfrac{0.08}{0.3}\right)\\
\hline
\rule{0pt}{2.5ex}0.3 \ \text{m}\rule[-1ex]{0pt}{0pt}& \sin ^{-1}\left(\dfrac{0.08}{0.3}\right) \\
\hline
\rule{0pt}{2.5ex}1 \ \mu \text{m}\rule[-1ex]{0pt}{0pt}& \tan ^{-1}\left(\dfrac{0.08}{0.3}\right)\\
\hline
\rule{0pt}{2.5ex}1 \ \mu \text{m} \rule[-1ex]{0pt}{0pt}&\sin ^{-1}\left(\dfrac{0.08}{0.3}\right) \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution
  • \(d\) is the distance between the slits, in this case,  \(d=1 \mu \text{m}\).
  • \(\theta\) is the angular deviation of a maximum, measured from the centre of the slits .
  •    \(\theta=\tan ^{-1}\left(\dfrac{0.08}{0.3}\right)\)

\(\Rightarrow C\)