SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Combinatorics, EXT1 A1 2008 HSC 1d

Find an expression for the coefficient of  `x^8 y^4`  in the expansion of  `(2x + 3y)^12`.   (2 marks) 

Show Answers Only

`10\ 264\ 320`

Show Worked Solution

`text(Find co-efficient of)\ x^8 y^4 :`

MARKER’S COMMENT: More errors were made by students who used `T_(k+1)` as the general term rather than `T_k` (both are possible). The Worked Solution uses the more successful approach.

`T_k =\ text(General term of)\ (2x + 3y)^12 `

`T_k` `= ((12),(k)) (2x)^(12 – k) * (3y)^k`
  `= ((12),(k)) * 2^(12 – k) * 3^k * x^(12 – k) * y^k`

 

`x^8 y^4\ text(occurs when)\ k = 4`

`T_4` `= ((12),(4)) * 2^(12 – 4) * 3^4 * x^8 y^4`
   
`:.\ text(Co-efficient of)\ x^8y^4`
  `= ((12),(4)) * 2^8 * 3^4`
  `= 10\ 264\ 320`

Measurement, 2UG 2008 HSC 28b

A tunnel is excavated with a cross-section as shown.
 

 
 

  1. Find an expression for the area of the cross-section using TWO applications of Simpson’s rule.  (2 marks)
  2. The area of the cross-section must be 600 m2. The tunnel is 80 m wide.   
  3. If the value of `a` increases by 2 metres, by how much will `b` change?   (2 marks)

 

Show Answers Only
  1. `(2h)/3 (4a + b)`
  2.  
  3. `b\ text(decreases by 8 m.)`
Show Worked Solution
(i)    `A` `~~ h/3 [y_0 + 4y_1 + y_2] + h/3 [y_0 + 4y_1 + y_2]`
    `~~ h/3 [0 + 4a + b] + h/3 [b + 4a + 0]`
    `~~ (2h)/3 (4a + b)`

 

(ii)    `text(Given)\ \ A = 600\ text(m²)`
  `text(If 80 m wide) \ => h = 20`

 

`A` `= (2h)/3 (4a + b)`
`600` `= ((2 xx 20))/3 (4a + b)`
`4a + b` `= (600 xx 3)/40`
`b` `= 45 – 4a`

 
`:.\ text(If)\ a\ text(increases by 2 m,)\ b\ text(will)`

`text(decrease by 8 m.)`

Statistics, STD2 S5 2008 HSC 28a

The following graph indicates  `z`-scores of ‘height-for-age’ for girls aged  5 – 19 years.
 

 
 

  1. What is the  `z`-score for a six year old girl of height 120 cm? (1 mark)

  2. Rachel is 10 ½  years of age. 

     

    (1)  If  2.5% of girls of the same age are taller than Rachel, how tall is she?   (1 mark)

     

    (2)  Rachel does not grow any taller. At age 15 ½, what percentage of girls of the same age will be taller than Rachel?   (2 marks)

  3. What is the average height of an 18 year old girl?   (1 mark)

For adults (18 years and older), the Body Mass Index is given by

`B = m/h^2`  where  `m = text(mass)`  in kilograms and  `h = text(height)`  in metres.

The medically accepted healthy range for  `B`  is  `21 <= B <= 25`.

  1. What is the minimum weight for an 18 year old girl of average height to be considered healthy? (2 marks)

  2. The average height, `C`, in centimetres, of a girl between the ages of 6 years and 11 years can be represented by a line with equation
     
            `C = 6A + 79`   where `A` is the age in years. 
     
    (1)  For this line, the gradient is 6. What does this indicate about the heights of girls aged 6 to 11?   (1 mark)

     

    (2)  Give ONE reason why this equation is not suitable for predicting heights of girls older than 12.   (1 mark)

Show Answers Only
  1. `1`
  2. (1) `155\ text(cm)`

     

    (2) `text(84%)`

  3. `163\ text(cm)`
  4. `55.8\ text(kg)`
  5. (1) `text(It indicates that 6-11 year old girls)`

     

          `text(grow, on average, 6cm per year)`

     

    (2) `text(Girls eventually stop growing, and the)`

     

          `text(equation doesn’t factor this in.)`

Show Worked Solution
i.    `z text(-score) = 1`

 

ii. (1)   `text(If 2 ½ % are taller than Rachel)`
    `=> z text(-score of +2)`
    `:.\ text(She is 155 cm)`
     
   (2)   `text(At age)\ 15\ ½,\ 155\ text(cm has a)\ z text(-score of –1)`
    `text(68% between)\ z = 1\ text(and)\ –1`
    `=> text(34% between)\ z = 0\ text(and)\ –1`
    `text(50% have)\ z >= 0`
     
    `:.\ text(% Above)\ z text(-score of –1)`
    `= 50 + 34`
    `= 8text(4%)`

 
`:.\ text(84% of girls would be taller than Rachel at age)\ 15 ½.`

 

iii.   `text(Average height of 18 year old has)\ z text(-score = 0)`
  `:.\ text(Average height) = 163\ text(cm)`

 

iv.   `B = m/h^2`
  `h = 163\ text(cm) = 1.63\ text(m)`

 

`text(Given)\ \ 21 <= B <= 25,\ text(minimum healthy)`

`text(weight occurs when)\ B = 21`

`=> 21` `= m/1.63^2`
`m` `= 21 xx 1.63^2`
  `= 55.794…`
  `= 55.8\ text(kg)\ text{(1 d.p.)}`

 

v. (1)   `text(It indicates that 6-11 year old girls, on average, grow)`
    `text(6 cm per year.)`
  (2) `text(Girls eventually stop growing, and the equation doesn’t)`
    `text(factor this in.)`

Financial Maths, STD2 F4 2008 HSC 27c

A plasma TV depreciated in value by 15% per annum. Two years after it was purchased it had depreciated to a value of $2023, using the declining balance method.

What was the purchase price of the plasma TV?   (2 marks)

Show Answers Only

`$2800`

Show Worked Solution

`S = V_0 (1-r)^n`

`2023` `= V_0 (1-0.15)^2`
`2023` `= V_0 (0.85)^2`
`V_0` `= 2023/0.85^2`
  `= 2800`

 

`:.\ text(The purchase price) = $2800`

Measurement, 2UG 2008 HSC 27a

An aircraft travels at an average speed of  913 km/h. It departs from a town in Kenya  (0°, 38°E)  on Tuesday at 10 pm and flies east to a town in Borneo  (0°, 113°E).

  1. What is the distance, to the nearest kilometre, between the two towns? (Assume the radius of Earth is 6400 km.) (2 marks)
  2. How long will the flight take? (Answer to the nearest hour.)   (1 mark)
  3. What will be the local time in Borneo when the aircraft arrives? (Ignore time zones.)   (2 marks)

 

Show Answers Only
  1. `8378\ text(km)`
  2. `9\ text(hours)`
  3. `text(12 midday on Wednesday)`
Show Worked Solution
(i)   `text(Angular difference in longitude)`

`= 113 – 38`

`= 75^@`
 

`text(Arc length)` `= 75/360 xx 2 xx pi xx 6400`
  `= 8377.58…`
  `= 8378\ text(km)\ text{(nearest km)}`

 
`:.\ text(The distance between the two towns is 8378 km.)`

 

(ii)    `text(Flight time)` `= text(Distance)/text(Speed)`
    `= 8378/913`
    `= 9.176…`
    `= 9\ text(hours)\ text{(nearest hr)}`

 

(iii)   `text(Time Difference)` `= 75 xx 4`
    `= 300\ text(minutes)`
    `= 5\ text(hours)`

 
`text(Kenya is further East)`

`=>\ text(Kenya is +5 hours)`
 

`:.\ text(Arrival time in Kenya)`

`= text{10 pm (Tues) + 5 hrs + 9 hrs}\ text{(flight)}`

`= 12\ text(midday on Wednesday)`

Statistics, STD2 S1 2008 HSC 26d

The graph shows the predicted population age distribution in Australia in 2008.
 

 

  1. How many females are in the 0–4 age group?  (1 mark)

  2. What is the modal age group?   (1 mark)

  3. How many people are in the 15–19 age group?   (2 marks)

  4. Give ONE reason why there are more people in the 80+ age group than in the 75–79 age group.   (1 mark)

Show Answers Only
  1. `600\ 000`
  2. `35-39`
  3. `1\ 450\ 000`
  4. `text(The 80+ group includes all people over 80)`

  5.  

    `text(and is not restricted by a 5-year limit.)`

Show Worked Solution
i.    `text{# Females (0-4)}` `= 0.6 xx 1\ 000\ 000`
    `= 600\ 000`

 

ii.    `text(Modal age group)\ =` `text(35 – 39)`

 

iii.   `text{# Males (15-19)}` `= 0.75 xx 1\ 000\ 000`
    `= 750\ 000`

 

`text{# Females (15-19)}` `= 0.7 xx 1\ 000\ 000`
  `= 700\ 000`

 

`:.\ text{Total People (15-19)}` `= 750\ 000 + 700\ 000`
  `= 1\ 450\ 000`

 

iv.   `text(The 80+ group includes all people over 80)`
  `text(and is not restricted by a 5-year limit.)`

Probability, STD2 S2 2008 HSC 26b

The retirement ages of two million people are displayed in a table.
 

 
 

  1. What is the relative frequency of the 51–55 year retirement age?  (1 mark)

  2. Describe the distribution.   (1 mark)

Show Answers Only
  1. `7/400`
  2. `text(Distribution is negatively skewed because)`

     

    `text(as age increases, so does the number of)`

     

    `text(people in each age bracket.)`

Show Worked Solution

i.  `text(Relative frequency)\ (51-55)`

`= text{# People (51-55)}/text(Total People)`

`= (35\ 000)/(2\ 000\ 000)`

`= 7/400`

 

ii.  `text(Distribution is negatively skewed because)`

`text(as age increases, so does the number of)`

`text(people in each age bracket.)`

Probability, STD2 S2 2008 HSC 26a

Cecil invited 175 movie critics to preview his new movie. After seeing the movie, he conducted a survey. Cecil has almost completed the two-way table.
 

  1. Determine the value of  `A`.   (1 mark)

  2. A movie critic is selected at random.

     

    What is the probability that the critic was less than 40 years old and did not like the movie?  (2 marks)

  3. Cecil believes that his movie will be a box office success if 65% of the critics who were surveyed liked the movie.

     

    Will this movie be considered a box office success? Justify your answer.  (1 mark)

Show Answers Only
  1. `58`
  2. `6/25`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text{Critics liked and}\ >= 40`

`= 102-65`

`= 37`

`:. A = 37+31=68`

 
ii.  `text{Critics did not like and < 40}`

`= 175-65-37-31`

`= 42`
 

`:.\ P text{(not like and  < 40)}`

`= 42/175`

`= 6/25`
 

iii.   `text(Critics liked) = 102`

`text(% Critics liked)` `= 102/175 xx 100`
  `= 58.28…%`

 
`:.\ text{Movie NOT a box office success (< 65% critics liked)}`

Probability, STD2 S2 2008 HSC 25b

In a drawer there are 30 ribbons. Twelve are blue and eighteen are red.

Two ribbons are selected at random.

  1. Copy and complete the probability tree diagram.  (1 mark)
     

     
  2. What is the probability of selecting a pair of ribbons which are the same colour?  (2 marks)

Show Answers Only
  1.  
  2. `73/145`
Show Worked Solution

i. 

ii.  `Ptext{(same colour)}`

`=\ text{P(BB) + P(RR)}`

`= 12/30 xx 11/29 + 18/30 xx 17/29`

`= 132/870 + 306/870`

`= 73/145` 

Probability, STD2 S2 2008 HSC 24b

Three-digit numbers are formed from five cards labelled  1,  2,  3,  4  and  5.

  1. How many different three-digit numbers can be formed?  (1 mark)

  2. If one of these numbers is selected at random, what is the probability that it is odd?   (1 mark)

  3. How many of these three-digit numbers are even?   (1 mark)

  4. What is the probability of randomly selecting a three-digit number less than 500 with its digits arranged in descending order?   (2 marks)

Show Answers Only
  1. `60`
  2. `3/5`
  3. `24`
  4. `1/15`
Show Worked Solution

i.  `text(# Different numbers)`

♦ Mean mark 45%.

`= 5 xx 4 xx 3`

`= 60`

 

ii.  `text(The last digit must be one of the)`

`text(5 numbers, of which 3 are odd)`

`:.\ text{P(odd)} = 3/5`

 

iii. `text{P(even)} = 1- text{P(odd)} = 2/5`

♦ Mean mark 48%.

`:.\ text(Number of even numbers)`

`= 2/5 xx 60`

`= 24`

 

iv.  `text(The numbers that satisfy the criteria:)`

♦♦♦ Mean mark 10%.

`432, 431, 421, 321`

`:.\ text{P(selection)} = 4/60 = 1/15`

Financial Maths, STD2 F1 2008 HSC 24a

Bob is employed as a salesman. He is offered two methods of calculating his income.

\begin{array} {|l|}
\hline
\rule{0pt}{2.5ex}\text{Method 1: Commission only of 13% on all sales}\rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex}\text{Method 2: \$350 per week plus a commission of 4.5% on all sales}\rule[-1ex]{0pt}{0pt} \\
\hline
\end{array}

Bob’s research determines that the average sales total per employee per month is $15 670. 

  1. Based on his research, how much could Bob expect to earn in a year if he were to choose Method 1?   (2 marks)

  2. If Bob were to choose a method of payment based on the average sales figures, state which method he should choose in order to earn the greater income. Justify your answer with appropriate calculations.   (3 marks)

Show Answers Only
  1. `$24\ 445.20`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `text(Method 1)`

`text(Yearly sales)` `= 12 xx 15\ 670`
  `= 188\ 040`
`:.\ text(Earnings)` `= text(13%) xx 188\ 040`
  `= $24\ 445.20`

 

ii.  `text(Method 2)`

`text(In 1 Year, Weekly Wage)` `= 350 xx 52`
  `= 18\ 200`
`text(Commission)` `= text(4.5%) xx 188\ 040`
  `= 8461.80`
`text(Total earnings)` `= 18\ 200 + 8461.80`
  `= $26\ 661.80`

 

`:.\ text(Bob should choose Method 2.)`

Statistics, STD2 S1 2008 HSC 23e

In a survey, 450 people were asked about their favourite takeaway food. The results are displayed in the bar graph.

2008 23e
 
How many people chose pizza as their favourite takeaway food?   (2 marks)

Show Answers Only

`175`

Show Worked Solution

`text(Number of people who chose pizza)`

COMMENT: This question required measurement of the actual image on the exam. The same methodology works here.

`= text{Length of pizza section}/text{Total length of bar} xx 450`

`~~ 7/18 xx 450`

`~~ 175`
 

`:.\ 175\ text(people chose pizza.)`

Measurement, STD2 M7 2008 HSC 23c

An alcoholic drink has 5.5% alcohol by volume. The label on a 375 mL bottle says it contains 1.6 standard drinks.

  1. How many millilitres of alcohol are in a 375 mL bottle? (1 mark)

  2. It is recommended that a fully-licensed male driver should have a maximum of one standard drink every hour.

     

    Express this as a rate in millilitres per minute, correct to one decimal place.    (2 marks)

Show Answers Only
  1. `20.625\ text(mL)`
  2. `3.9\ text(mL/min)`
Show Worked Solution
i.    `text(Alcohol)` `= 5.5/100 xx 375`
    `= 20.625\ text(mL)`

 

ii.    `text(S)text(ince 1.6 standard drinks = 375 mL)`

`=>\ text(1 standard drink)`

`= 375/1.6`

`= 234.375\ text(mL)`
 

`:.\ text(Rate)` `= 234.375/60`
  `= 3.90625`
  `= 3.9\ text(mL/min)`

Measurement, STD2 M1 2008 HSC 23b

The capacity of a bottle is measured as 1.25 litres correct to the nearest 10 millilitres.

What is the percentage error for this measurement?   (1 mark)

Show Answers Only

`text(0.4%)`

Show Worked Solution

`text(A) text(bsolute error) = 5\ text(mL)`

`:.\ text(% error)` `= 5/1250 xx 100`
  `=\ text(0.4%)`

Measurement, STD2 M7 2008 HSC 20 MC

A point `P` lies between a tree, 2 metres high, and a tower, 8 metres high. `P` is 3 metres away from the base of the tree.

From `P`, the angles of elevation to the top of the tree and to the top of the tower are equal.
 

What is the distance, `x`, from `P` to the top of the tower?

  1. 9 m
  2. 9.61 m
  3. 12.04 m
  4. 14.42 m
Show Answers Only

`D`

Show Worked Solution

`text(Triangles are similar)\ \ text{(equiangular)}`

`text(In smaller triangle:)`

`h^2` `= 2^2 + 3^2`
  `= 13`
`h` `= sqrt 13`
   
`x/sqrt13` `= 8/2\ \ \ text{(sides of similar Δs in same ratio)}`
`x` `= (8 sqrt 13)/2`
  `= 14.422…`

 
`=>  D`

Probability, STD2 S2 2008 HSC 16 MC

A bag contains some marbles. The probability of selecting a blue marble at random from this bag is  `3/8`.

Which of the following could describe the marbles that are in the bag?

  1.    `3`  blue,  `8`  red
  2.    `6`  blue,  `11`  red
  3.    `3`  blue,  `4`  red,  `4`  green
  4.    `6`  blue,  `5`  red,  `5`  green 
Show Answers Only

`D`

Show Worked Solution

`P(B) = 3/8`

`text(In)\ A,\ \ ` `P(B) = 3/11`
`text(In)\ B,\ \ ` `P(B) = 6/17 `
`text(In)\ C,\ \ ` `P(B) = 3/11`
`text(In)\ D,\ \ ` `P(B) = 6/16 = 3/8`

`=>  D`

Financial Maths, STD2 F4 2008 HSC 15 MC

Ali is buying a speedboat at Betty’s Boats.
 

VCAA 2008 15 mc
 

What is the amount of interest Ali will have to pay if he chooses to buy the boat on terms?

  1.    $3200 
  2.    $5600
  3.    $19 200
  4.    $21 600
Show Answers Only

`B`

Show Worked Solution
`text(Deposit)` `= text(15%) xx 16\ 000`
  `= 2400`
   
`text(Payments)` `= 320 xx 5 xx 12`
  `= 19\ 200`
   
`text(Total paid)` `= 2400 + 19\ 200`
  `= 21\ 600`

 

`:.\ text(Interest)` `= 21\ 600 – 16\ 000`
  `= 5600`

`=>  B`

Statistics, STD2 S1 2008 HSC 13 MC

The height of each student in a class was measured and it was found that the mean height was 160 cm.

Two students were absent. When their heights were included in the data for the class, the mean height did not change.

Which of the following heights are possible for the two absent students?

  1.    155 cm and 162 cm
  2.    152 cm and 167 cm
  3.    149 cm and 171 cm
  4.    143 cm and 178 cm
Show Answers Only

`C`

Show Worked Solution

`text(S) text(ince the mean doesn’t change)`

`=>\ text(2 absent students must have a)`

`text(mean height of 160 cm.)`

`text(Considering each option given,)`

`(149 + 171) -: 2 = 160`

`=>  C`

Statistics, STD2 S4 2008 HSC 12 MC

A scatterplot is shown.
 

Which of the following best describes the correlation between  \(R\)  and  \(T\)?

  1. Positive
  2. Negative 
  3. Positively skewed
  4. Negatively skewed
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Correlation is positive.}\)

\(\text{NB. The skew does not directly relate to correlation.}\)

\(\Rightarrow  A\)

Measurement, STD2 M1 2008 HSC 11 MC

The diagram shows the floor of a shower. The drain in the floor is a circle with a diameter of 10 cm.

What is the area of the shower floor, excluding the drain?
 

 
 

  1. 9686 cm²
  2. 9921 cm²
  3. 9969 cm²
  4. 10 000 cm²
Show Answers Only

`B`

Show Worked Solution
COMMENT: Students should see that answers are all in cm², and therefore use cm as the base unit for their calculations. 
`text(Area)` `=\ text(Square – Circle)`
  `= (100 xx 100)-(pi xx 5^2)`
  `= 10\ 000-78.5398…`
  `= 9921.46…\ text(cm²)`

 
`=>  B`

Statistics, STD2 S1 2008 HSC 10 MC

The marks for a Science test and a Mathematics test are presented in box-and-whisker plots.
 

 Which measure must be the same for both tests?

  1. Mean
  2. Range
  3. Median
  4. Interquartile range
Show Answers Only

`D`

Show Worked Solution

`text(IQR)=text(Upper Quartile)-text(Lower Quartile)`

`text{In both box plots, IQR = 3 intervals (against bottom scale)}`

`=>  D`

Algebra, STD2 A1 2008 HSC 9 MC

What is the value of  `sqrt ( (x + 2y)/(8y) )`  if  `x = 5.6`  and  `y = 3.1`, correct to 2 decimal places? 

  1. `0.69`
  2. `2.62`
  3. `2.83` 
  4. `4.77`  
Show Answers Only

`A`

Show Worked Solution
`sqrt ( (x + 2y)/(8y) )` `= sqrt ( (5.6 + (2 xx 3.1))/((8 xx 3.1)) )`
  `= sqrt (11.8/24.8)`
  `= 0.6897…`

`=>  A`

Statistics, STD2 S1 2008 HSC 8 MC

What is the median of the following set of scores?
 

 
 

  1.    12
  2.    13
  3.    14
  4.    15
Show Answers Only

`C`

Show Worked Solution
`text(Median` `=(n+1)/2`
  `=(33+1)/2`
  `=\ text (17th score)`

 

`:.\ text(Median is 14)`

`=>  C`

Measurement, STD2 M6 2008 HSC 5 MC

What is the size of the smallest angle in this triangle?
 

  1. `29^@` 
  2. `47^@`
  3. `58^@`
  4. `76^@`
Show Answers Only

`B`

Show Worked Solution

`text(Smallest angle is opposite smallest side.)`

` cos A` `= (b^2 + c^2-a^2)/(2bc)`
  `= (7^2 + 8^2-6^2)/(2 xx 7 xx 8)`
  `= 0.6875`
`A` `=cos ^(-1)(0.6875)`
`:.\ A` `= 46.567…^@`

 
`=>  B`

Statistics, STD2 S1 2008 HSC 3 MC

The stem-and-leaf plot represents the daily sales of soft drink from a vending machine.

If the range of sales is 43, what is the value of  2008 3 mc  ?

 
 

  1.    `4` 
  2.    `5`
  3.    `24`
  4.    `25`
Show Answers Only

`A`

Show Worked Solution

`text(Range = High) – text(Low) = 43`

`:.\ 67 – text(Low)` `= 43`
`text(Low)` `= 24`

`:.\ N = 4`

`=>  A`

Probability, 2ADV S1 2008 HSC 9a

It is estimated that 85% of students in Australia own a mobile phone.

  1. Two students are selected at random. What is the probability that neither of them owns a mobile phone?  (2 marks)

  2. Based on a recent survey, 20% of the students who own a mobile phone have used their mobile phone during class time. A student is selected at random. What is the probability that the student owns a mobile phone and has used it during class time?   (1 mark)

Show Answers Only
  1. `9/400`
  2. `17/100`
Show Worked Solution

i.     `P(M) = 0.85`

COMMENT: `M^c` is syllabus notation for the complement of event `M`.

`P(M^c) = 1-0.85 = 0.15`

`:.\ P(M^c, M^c)` `= 15/100 * 15/100`
  `= 225/(10\ 000)`
  `= 9/400`

 

ii.  `text{P(owns mobile and used it)}`

`= P(M) xx  P\text{(used it)}`

`= 17/20 xx 20/100`

`= 17/100`

Plane Geometry, 2UA 2008 HSC 8b

In the diagram,  `ABCD`  is a parallelogram and  `ABEF`  and  `BCGH`  are both squares.

Copy or trace the diagram into your writing booklet.

  1. Prove that  `CD = BE`.  (1 mark)
  2. Prove that  `BD = EH`.   (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)  `text(Prove)\ CD = BE`

`CD` `= AB\ \ text{(opposite sides of parallelogram)}`
`AB` `= BE\ \ text{(sides of square)}`
`:.\ CD` `= BE\ \ \ text(… as required)`

 

(ii)

`text(Prove)\ BD = EH`

`CD` `= BE\ \ text{(from part (i))}`
`BC` `= BH\ \ text{(sides of square)}`

`/_BDC = /_ABD = alpha\ \ text{(} text(alternate,)\ AB\ text(||)\ CD text{)}`

`text(Let)\ /_DBC = beta`

`/_BCD` `= 180 – alpha – beta\ \ \ \ text{(angle sum of}\ Delta BDCtext{)}`
`/_EBH` `= 360 – 90 – 90 – alpha – beta\ \ \ `
   `text{(angles about a point)}`
  `= 180 – alpha – beta`
  `= /_BCD`

`=> Delta DCB ~= Delta EBH\ \ text{(SAS)}`

`:.\ BD = EH\ \ \ ` `text{(corresponding sides of)`
  `\ \ text{congruent triangles)}`

Calculus, 2ADV C3 2008 HSC 8a

Let  `f(x) = x^4 − 8x^2`.

  1. Find the coordinates of the points where the graph of  `y = f(x)`  crosses the axes.  (2 marks)

  2. Show that  `f(x)`  is an even function.   (1 mark)

  3. Find the coordinates of the stationary points of  `f(x)`  and determine their nature.   (4 marks)

  4. Sketch the graph of  `y = f(x)`.   (1 mark)

Show Answers Only
  1. `(-2 sqrt 2,0), (2 sqrt 2, 0)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(S.P.s at)\ (0,0), (2,-16), (-2,-16)`
  4.  
  5. 2UA HSC 2008 8ai
Show Worked Solution
i.    `f(x) = x^4 – 8x^2`

`y text(-intercept when)\ x = 0`

`:.\ text(Cuts)\ y\ text(axis at)\ (0,0)`

`x\ text(-intercept when)\ f(x) = 0`

`x^4 – 8x^2` `= 0`
`x^2 (x^2 – 8)` `= 0`

`x^2 – 8 = 0\ \ \ \ \ text(or)\ \ \ \ \ x = 0`

`x^2` `= 8`
`x` `= +- sqrt 8 = +- 2 sqrt 2`

 

`:.\ text(Cuts)\ x text(-axis at)\ (-2 sqrt 2,0),\ \ text(and)\ \ (2 sqrt 2, 0)`

`text{(Note that it only touches}\ xtext{-axis at (0,0))}`

 

ii.    `f(x)` `= x^4 – 8x^2`
  `f(–x)` `= (–x)^4 – 8(–x)^2`
    `= x^4 – 8x^2`
    `= f(x)`

 

`:.\ f(x)\ text(is an even function.)`

 

iii.   `f(x)` `= x^4 – 8x^2`
  `f'(x)` `= 4x^3 – 16x`
  `f″(x)` `=12x^2-16`

 

`text(S.P.  when)\ \ f'(x) = 0`

`4x^3 – 16x` `= 0`
`4x (x^2 – 4)` `= 0`
`x^2 – 4` `=0\ \ \ \ \ x = 0`
`x^2` `= 4`
`x` `= +-2`
`text(At)\ x=2\ \ `  `f(x)`  `=(2)^4 – 8(2)^2 = -16`
  `f″(x)` `=12(2^2)-16>0`

`:.\ text{MIN at (2, –16)}`

`:.\ text{MIN at (–2, –16)},\ \ \ (f(x)\ text(is even))`

 

`text(At)\ x=0\ \ `  `f(x)`  `=(0)^4 – 8(0)^2 = 0`
  `f″(x)` `=12(0^2)-16<0`

`:.\ text{MAX at (0,0)}`

 

iv. 

Probability, 2ADV S1 2008 HSC 7c

Xena and Gabrielle compete in a series of games. The series finishes when one player has won two games. In any game, the probability that Xena wins is  `2/3`  and the probability that Gabrielle wins is  `1/3`.

Part of the tree diagram for this series of games is shown.
 

 
 

  1. Complete the tree diagram showing the possible outcomes.  (1 mark)
  2. What is the probability that Gabrielle wins the series?   (2 marks)

  3. What is the probability that three games are played in the series?   (2 marks)

Show Answers Only
  1.    
  2. `7/27`
  3. `4/9`
Show Worked Solution
i. 2UA HSC 2008 7ci

 

MARKER’S COMMENT: A tree diagram with 8 outcomes is incorrect (i.e. no third game is played if 1 player wins the first 2 games). If outcomes cannot occur, do not draw them on a tree diagram.

 

ii.  `P text{(} G\ text(wins) text{)}`

`= P(XGG) + P (GXG) + P (GG)`

`= 2/3 * 1/3 * 1/3 + 1/3 * 2/3 * 1/3 + 1/3 * 1/3`

`= 2/27 + 2/27 + 1/9`

`= 7/27`

 

iii.  `text(Method 1:)`

`P text{(3 games played)}`

`= P (XG) + P(GX)`

`= 2/3 * 1/3 + 1/3 * 2/3`

`= 4/9`

 

`text(Method 2:)`

`P text{(3 games)}`

`= 1 – [P(XX) + P(GG)]`

`= 1 – [2/3 * 2/3 + 1/3 * 1/3]`

`= 1 – 5/9`

`= 4/9`

Trigonometry, 2ADV T1 2008 HSC 7b

2008 7b

The diagram shows a sector with radius  `r`  and angle  `theta`  where  `0 < theta <= 2pi`.

The arc length is  `(10pi)/3`. 

  1.  Show that  `r >= 5/3`.   (2 marks)

  2.  Calculate the area of the sector when  `r = 4`.    (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(20pi)/3\ text(u²)`
Show Worked Solution
i.    `text(Show)\ r >= 5/3`
`text(Arc length)\ ` `= r theta\ \ text(where)\ \ 0 < theta <= 2pi`
`r theta` `= (10pi)/3`
`:.theta` `= (10pi)/(3r)`

 

`text(Using)\ \ \ 0 <= theta <= 2 pi`

`0 <= (10pi)/(3r)` `<= 2pi`
`(10pi)/3` `<= 2 pi r`
`5/3` `<= r`

 

`:.\ r >= 5/3\ \ \ text(… as required.)`

 

ii.   `text(Area)` `= 1/2 r^2 theta`
    `= 1/2 xx 4^2 xx (10pi)/(3 xx 4)`
    `= (20pi)/3\ text(u²)`

Calculus, EXT1* C3 2008 HSC 6c

The graph of  `y = 5/(x - 2)`  is shown below.
 

2008 6c
 

The shaded region in the diagram is bounded by the curve  `y = 5/(x - 2)`, the  `x`-axis and the lines  `x = 3`  and  `x = 6`.

Find the volume of the solid of revolution formed when the shaded region is rotated about the  `x`-axis.  (3 marks)

Show Answers Only

`(75pi)/4\ text(u³)`

Show Worked Solution
`y` `= 5/(x – 2)`
`V` `= pi int_3^6 y^2\ dx`
  `= pi int_3^6 (5/(x – 2))^2\ dx`
  `= 25 pi int_3^6 1/((x – 2)^2)\ dx`
  `= 25 pi [(-1)/(x – 2)]_3^6`
  `= 25 pi [-1/4 – (-1)]`
  `=25 pi [3/4]`
  `= (75pi)/4\ text(u³)`

Calculus, 2ADV C4 2008 HSC 5a

The gradient of a curve is given by  `dy/dx = 1-6 sin 3x`. The curve passes through the point  `(0, 7)`.

What is the equation of the curve?   (3 marks)

Show Answers Only

`y = x + 2 cos 3x + 5`

Show Worked Solution
`dy/dx` `= 1-6 sin 3x`
`y` `= int 1-6 sin 3x\ dx`
  `= x + 2 cos 3x + c`

 

`text(Passes through)\ (0,7):`

`=> 0 + 2 cos 0 + c` `= 7`
`2 + c` `= 7`
`c` `= 5`

 

`:.\ text(Equation is)\ \ \ y = x + 2 cos 3x + 5`

Quadratic, 2UA 2008 HSC 4c

Consider the parabola  `x^2 = 8(y\ – 3)`.

  1. Write down the coordinates of the vertex.  (1 mark)
  2. Find the coordinates of the focus.   (1 mark)
  3. Sketch the parabola.   (1 mark)
  4. Calculate the area bounded by the parabola and the line  `y = 5`.   (3 marks)
Show Answers Only

(i)   `(0,3)`

(ii)   `(0,5)`

(iii)  

(iv)  `10 2/3\ text(u²)`

Show Worked Solution
(i)    `text(Vertex)\ = (0,3)`

 

(ii)   `text(Using)\ \ \ x^2` `= 4ay`
  `4a` `= 8`
  `a` `= 2`

`:.\ text(Focus) = (0,5)`

 

(iii) 2UA HSC 2008 4c 

 

(iv)   `text(Intersection when)\ y = 5`
`=> x^2` `= 8 (5-3)`
`x^2` `= 16`
`x` `= +- 4`

`text(Find shaded area)`

`x^2` `= 8 (y-3)`
`y – 3` `= x^2/8`
`y` `= x^2/8 +3`

 

`text(Area)` `= int_-4^4 5\ dx – int_-4^4 x^2/8 + 3\ dx`
  `= int_-4^4 5 – (x^2/8 + 3)\ dx`
  `= int_-4^4 2 – x^2/8\ dx`
  `= [2x – x^3/24]_-4^4`
  `= [(8 – 64/24) – (-8 + 64/24)]`
  `= 5 1/3 – (- 5 1/3)`
  `= 10 2/3\ text(u²)`

Plane Geometry, 2UA 2008 HSC 4a

In the diagram, `XR` bisects `/_PRQ` and `XY\ text(||)\ QR`.

Prove that `Delta XYR` is an isosceles triangle.   (2 marks)

Show Answers Only

`text(Proof)\ text{(See Worked Solutions)}`

Show Worked Solution

`text{Since}\ XR\ text{bisects}\ /_PRQ`

`/_XRQ` `= /_YRX = theta`
`/_RXY` `= theta\ \ \ text{(} text(alternate angles,)\ XY\ text(||)\ QR text{)}`

 
`:.\ Delta XYR\ \ text(is isosceles)`

Calculus, 2ADV C4 2008 HSC 3b

  1. Differentiate  `log_e (cos x)`  with respect to  `x`.  (2 marks)

  2. Hence, or otherwise, evaluate  `int_0^(pi/4) tan x\ dx`.    (2 marks)

Show Answers Only
  1. `- tan x`
  2. `- log_e (1/sqrt2)\ \ text(or)\ \ 0.35\ \ text{(2 d.p.)}`
Show Worked Solution
i.    `y` `= log_e (cos x)`
  `dy/dx` `= (- sin x)/(cos x)`
    `= – tan x`

 

ii.    `int_0^(pi/4) tan x\ dx`
  `= – [log_e (cos x)]_0^(pi/4)`
  `= – [log_e(cos (pi/4)) – log_e (cos 0)]`
  `= – [log_e (1/sqrt2) – log_e 1]`
  `= – [log_e (1/sqrt2) – 0]`
  `= – log_e (1/sqrt2)`
  `= 0.346…`
  `= 0.35\ \ text{(2 d.p.)}`

Linear Functions, 2UA 2008 HSC 3a

2008 3a

In the diagram,  `ABCD`  is a quadrilateral. The equation of the line  `AD`  is  `2x- y- 1 = 0`. 

  1. Show that  `ABCD`  is a trapezium by showing that  `BC`  is parallel to  `AD`.  (2 marks)
  2. The line  `CD`  is parallel to the  `x`-axis. Find the coordinates of  `D`.   (1 mark)
  3. Find the length of  `BC`.   (1 mark)
  4. Show that the perpendicular distance from  `B`  to  `AD`  is  `4/sqrt5`.   (2 marks)
  5. Hence, or otherwise, find the area of the trapezium  `ABCD`.   (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(3,5)`
  3. `sqrt 5\ text(units)`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `8\ text(u²)`
Show Worked Solution
(i)    `text(Show)\ BC \ text(||)\ AD`

`B(0,3),\ \ C(1,5)`

`m_(BC)` `= (y_2 – y_1)/(x_2 – x_1)`
  `= (5 – 3)/(1 – 0)`
  `= 2`

 

`text(Equation)\ \ AD\ \ text(is)\ \ 2x – y – 1 = 0`

`y` `= 2x – 1`
`m_(AD)` `= 2`

`:. BC\  text(||) \ AD`

`:. ABCD\ text(is a trapezium)`

 

(ii)    `text(Given)\ CD\  text(||) \ x text(-axis)`
  `text(Equation)\ CD\ text(is)\ y = 5`
  `D\ text(is intersection of)`
`y` `= 5,\ \ and`
`2x – y – 1` `= 0`
`:. 2x – 5 – 1` `=0`
`2x` `=6`
`x` `=3`
`:.\ D` `= (3,5)`

 

(iii)   `B(0,3),\ \ C(1,5)`
`text(dist)\ BC` `= sqrt ( (x_2 – x_1)^2 + (y_2 – y_1)^2 )`
  `= sqrt ( (1-0)^2 + (5-3)^2 )`
  `= sqrt (1 + 4)`
  `= sqrt 5\ text(units)`

 

(iv)   `text(Show)\ _|_\ text(dist of)\ B\ text(to)\ AD\ text(is)\ 4/sqrt5`

`B (0,3)\ \ \ \ \ 2x – y – 1 = 0` 

`_|_\ text(dist)` `= | (ax_1 + by_1 + c)/sqrt (a^2 + b^2) |`
  `= |( 2(0) – 1(3) -1 )/sqrt (2^2 + (-1)^2) |`
  `= | -4/sqrt5 |`
  `= 4/sqrt 5\ \ \ text(… as required.)`

 

(v)    `text(Area)` `= 1/2 h (a + b)`
    `= 1/2 xx 4/sqrt5 (BC + AD)`

 
`BC = sqrt5\ \ text{(part (iii))}`
 

`A(0,–1),\ \ D(3,5)`

`text(dist)\ AD` `= sqrt ( (3-0)^2 + (5+1)^2 )`
  `= sqrt (9 + 36)`
  `= sqrt 45`
  `= 3 sqrt 5`
`:.\ text(Area)\ ABCD` `= 1/2 xx 4/sqrt5 (sqrt5 + 3 sqrt 5)`
  `= 2 / sqrt5 (4 sqrt 5)`
  `= 8\ text(u²)`

Financial Maths, STD2 F5 SM-Bank 2

The table below shows the present value of an annuity with a contribution of  $1.
 

  1. Fiona pays $3000 into an annuity at the end of each year for 4 years at 2% p.a., compounded annually.   What is the present value of her annuity?  (1 mark)

  2. If John pays $6000 into an annuity at the end of each year for 2 years at 4% p.a., compounded annually, is he better off than Fiona?  Use calculations to justify your answer.    (2 marks)

Show Answers Only
  1. `$11\ 423.10`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(Table factor when)\ \ n = 4,\ r = text(2%) \ => \ 3.8077`

`:.\ PVA\ text{(Fiona)}` `= 3000 xx 3.8077`
  `= $11\ 423.10`

 

ii.  `text(Table factor when)\ n = 2, r = text(4%)`

`=> 1.8861`

`:.\ PVA\ text{(John)}` `= 6000 xx 1.8861`
  `= $11\ 316.60`

 
`:.\ text(Fiona will be better off because her)\ PVA`

`text(is higher.)`

Mechanics, EXT2* M1 2014 HSC 14a

The take-off point `O` on a ski jump is located at the top of a downslope. The angle between the downslope and the horizontal is  `pi/4`.  A skier takes off from `O` with velocity `V` m s−1 at an angle `theta` to the horizontal, where  `0 <= theta < pi/2`.  The skier lands on the downslope at some point `P`, a distance `D` metres from `O`.
 

2014 14a
 

The flight path of the skier is given by

`x = Vtcos theta,\ y = -1/2 g t^2 + Vt sin theta`,      (Do NOT prove this.)

where  `t`  is the time in seconds after take-off.

  1. Show that the cartesian equation of the flight path of the skier is given by
     
         `y =  x tan theta - (gx^2)/(2V^2) sec^2 theta`.   (2 marks)

  2. Show that  
     
         `D = 2 sqrt 2 (V^2)/(g) cos theta (cos theta + sin theta)`.   (3 marks)

  3. Show that  
     
         `(dD)/(d theta) = 2 sqrt 2 (V^2)/(g) (cos 2 theta - sin 2 theta)`.   (2 marks)

  4. Show that `D` has a maximum value and find the value of `theta` for which this occurs.   (3 marks)

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

i.  `text(Show)\ \ y = x tan theta – (gx^2)/(2V^2) sec^2 theta`

`x` `= Vt cos theta`
`t` `= x/(V cos theta)\ \ \ …\ text{(1)}`

`text(Subst)\ text{(1)}\ text(into)\ y = -1/2 g t^2 + Vt sin theta`

`y` `= -1/2 g (x/(Vcostheta))^2 + V sin theta (x/(Vcostheta))`
  `= (-gx^2)/(2V^2 cos^2 theta) + x * (sin theta)/(cos theta)`
  `= x tan theta – (gx^2)/(2V^2) sec^2 theta\ \ \ text(… as required.)`

 

ii.  `text(Show)\ D = 2 sqrt 2 (V^2)/g\ cos theta (cos theta + sin theta)`

♦♦ Mean mark 28%
MARKER’S COMMENT: The key to finding `P` and solving for `D` is to realise you need the intersection of the Cartesian equation in (i) and the line `y=–x`.

`text(S)text(ince)\ P\ text(lies on line)\ y = -x`

`-x` `=x tan theta – (gx^2)/(2V^2) sec^2 theta`
`-1` `=tan theta – (gx)/(2V^2) sec^2 theta`
`(gx)/(2V^2) sec^2 theta` `= tan theta + 1`
`x (g/(2V^2))` `=(sin theta)/(cos theta) * cos^2 theta + 1 * cos^2 theta`
`x` `=(2V^2)/g\ (sin theta cos theta + cos^2 theta)`
  `=(2V^2)/g\ cos theta (cos theta + sin theta)`

 

`text(Given that)\ \ cos(pi/4)` `= x/D = 1/sqrt2`
`text(i.e.)\ \ D` `= sqrt 2 x`

 
`:.\ D = 2 sqrt 2 (V^2)/g\ cos theta (cos theta + sin theta)`

`text(… as required.)`

 

iii.  `text(Show)\ (dD)/(d theta) = 2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)`

`D` `= 2 sqrt 2 (V^2)/g\ (cos^2 theta + cos theta sin theta)`
`(dD)/(d theta)` `= 2 sqrt 2 (V^2)/g\ [2cos theta (–sin theta) + cos theta cos theta + (– sin theta) sin theta]`
  `= 2 sqrt 2 (V^2)/(g) [(cos^2 theta – sin^2 theta) – 2 sin theta cos theta]`
  `= 2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)\ \ \ text(… as required)`

 

iv.  `text(Max/min when)\ (dD)/(d theta) = 0`

♦ Mean mark 47%
IMPORTANT: Note that students can attempt every part of this question, even if they couldn’t successfully prove earlier parts.
`2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)` `= 0`
`cos 2 theta – sin 2 theta` `= 0`
`sin 2 theta` `= cos 2 theta`
`tan 2 theta` `= 1`
`2 theta` `= pi/4`
`theta` `= pi/8`
`(d^2D)/(d theta^2)` `= 2 sqrt 2 (V^2)/g\ [-2 sin 2theta – 2 cos 2 theta]`
  `= 4 sqrt 2 (V^2)/g\ (-sin 2 theta – cos 2 theta)`

 

`text(When)\ \ theta = pi/8:`

`(d^2 D)/(d theta^2)` `= 4 sqrt 2 (V^2)/g\ (- sin (pi/4) – cos (pi/4))`
  `= 4 sqrt 2 (V^2)/g\ (- 1/sqrt2 – 1/sqrt2) < 0`
  ` =>\ text(MAX)`

 

`:.\ D\ text(has a maximum value when)\ theta = pi/8`

Plane Geometry, EXT1 2014 HSC 13d

In the diagram,  `AB`  is a diameter of a circle with centre  `O`. The point  `C`  is chosen such that  `Delta ABC`  is acute-angled. The circle intersects  `AC`  and  `BC`  at  `P`  and  `Q`  respectively.

Copy or trace the diagram into your writing booklet.

  1. Why is   `/_BAC = /_CQP`?   (1 mark)
  2. Show that the line  `OP`  is a tangent to the circle through  `P`,  `Q`  and  `C`.    (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   

`APQB\ text(is a cyclic quad)`

`=> /_BAC + /_BQP = 180°`

`text{(opposite angles of cyclic quadrilateral}\ BAPQ text{)}`

`/_CQP + /_BQP = 180°\ text{(}/_ CQB\ text{is a straight angle)}`

`:.\ /_BAC = /_CQP`

 

(ii)  `text(Let)\ /_BAC = alpha`

♦ Mean mark 37%
COMMENT: Labelling angles `alpha` and `beta` etc.. can often make a proof clearer and less messy.

`/_OPA = alpha\ \ \ \ text{(angles opposite equal sides}`

`text(in)\ Delta OAP,\ OA=OP\ text{radii)}`

`/_TPC = alpha\ \ \ text{(vertically opposite)}`

`text(S)text(ince)\ /_TPC = /_CQP = alpha`

 

`:. OP\ text(is a tangent to circle)\ CPQ`

`text{(angles in alternate segments equal)}`

Calculus, EXT1 C1 2014 HSC 13b

One end of a rope is attached to a truck and the other end to a weight. The rope passes over a small wheel located at a vertical distance of  40 m above the point where the rope is attached to the truck.

The distance from the truck to the small wheel is `L\ text(m)`, and the horizontal distance between them is  `x\ text(m)`. The rope makes an angle `theta` with the horizontal at the point where it is attached to the truck.

The truck moves to the right at a constant speed of   `text(3 m s)^(-1)`, as shown in the diagram.
 


 

  1. Using Pythagoras’ Theorem, or otherwise, show that  `(dL)/(dx) = cos theta`.   (2 marks)

  2.  Show that  `(dL)/(dt) = 3 cos theta`.    (1 mark)

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

`text(Show)\ \ (dL)/(dx) = cos theta`

`text(Using Pythagoras,)`

`L^2` `=40^2 + x^2`
`L` `=(40^2 + x^2)^(1/2)`
`(dL)/(dx)` `=1/2 * 2x * (40^2 + x^2)^(-1/2)`
  `=x/ sqrt((40^2 + x^2))`
  `=x/L`
  `=cos theta\ \ \ text(… as required.)`

 

ii.   `text(Show)\ \ (dL)/(dt) = 3 cos theta`

`(dL)/(dt)` `= (dL)/(dx) * (dx)/(dt)`
  `= cos theta * 3`
  `= 3 cos theta\ \ \ text(… as required)`

Binomial, EXT1 2014 HSC 12d

Use the binomial theorem to show that
 

 `0 = ((n),(0)) - ((n),(1)) + ((n),(2)) - ... + (-1)^n ((n),(n))`.   (2 marks) 

 

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(Prove)`

`0 = ((n),(0)) – ((n),(1)) + ((n),(2)) – … + (-1)^n ((n),(n))`
 

`text(Using binomial expansion:)`

`(1 + x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + … + ((n),(n))x^n`
 

`text(Let)\ \ x = -1`

`:. 0 = ((n),(0)) – ((n),(1)) + ((n),(2)) – … + (-1)^n ((n),(n))`

Mechanics, EXT2* M1 2014 HSC 12c

A particle moves along a straight line with displacement  `x\ text(m)`  and velocity  `v\ \ text(ms)^(-1)`. The acceleration of the particle is given by

 `ddot x = 2 - e^(-x/2)`.

Given that  `v = 4`  when  `x = 0`, express  `v^2`  in terms of  `x`.   (3 marks)

Show Answers Only

`v^2 = 4x + 4e^(-x/2) + 12`

Show Worked Solution

`ddot x = d/(dx) (1/2 v^2) = 2 – e^(-x/2)`

`1/2 v^2` `= int (2 – e^(-x/2))\ dx`
  `= 2x + 2e^(-x/2) + c`
`v^2` `= 4x + 4e^(-x/2) + c`

 
`text(When)\ \ v = 4, x = 0:`

`:. 4^2` `= 0 + 4e^0 + c`
`16` `= 4 + c`
`c` `= 12`

 
`:.\ v^2 = 4x + 4e^(-x/2) + 12`

Calculus, EXT1 C3 2014 HSC 12b

The region bounded by  `y = cos 4x`  and the  `x`-axis, between  `x = 0`  and  `x = pi/8`, is rotated about the  `x`-axis to form a solid.   
 

2014 12b
 

Find the volume of the solid.   (3 marks)

Show Answers Only

`(pi^2)/16\ \ text(u³)`

Show Worked Solution
COMMENT: The identities `cos 2theta=` `cos^2 theta-sin^2 theta =` ` 2cos^2 theta-1=1-2sin^2 theta`  are tested every year – know them. 
`V` `= pi int_0^(pi/8) y^2\ dx`
  `= pi int_0^(pi/8) cos^2 4x\ dx`
  `= pi int_0^(pi/8) 1/2 (cos 8x + 1)\ dx`
  `= pi/2 [1/8 sin 8x + x]_0^(pi/8)`
  `= pi/2 [(1/8 sin pi + pi/8)] – 0]`
  `= (pi^2)/16\ \ text(u³)`

Mechanics, EXT2* M1 2014 HSC 12a

 A particle is moving in simple harmonic motion about the origin, with displacement `x` metres. The displacement is given by  `x = 2 sin 3t`, where `t` is time in seconds. The motion starts when  `t = 0`.

  1. What is the total distance travelled by the particle when it first returns to the origin?   (1 mark)

  2. What is the acceleration of the particle when it is first at rest?    (2 marks)

Show Answers Only
  1. `4\ text(m)`
  2. `-18\ text(m/s²)`
Show Worked Solution
i.    `x = 2 sin 3t`

 
`text(At)\ \ t = 0,\ x = 0`

`text(Amplitude) = 2`

`:.\ text(Distance travelled)`

`= 2 xx text(amplitude)`

`= 4\ text(m)`

 

ii.    `x` `= 2 sin 3t`
  `v` `= 6 cos 3t` 
  `ddot x` `= -18 sin 3t`

 
`text(Particle first comes to rest when)\ v = 0`

`6 cos 3t` `= 0`
`cos 3t` `= 0`
`3t` `= pi/2`
`t` `= pi/6`

 
`text(When)\ \  t = pi/6`

`ddot x` `= -18 * sin (3 xx pi/6)`
  `= -18 sin (pi/2)`
  `= -18\ text(m/s²)`

L&E, EXT1 2014 HSC 11f

Differentiate  `(e^x ln x)/x`.   (2 marks)

Show Answers Only

`(e^x (x * lnx + 1 – lnx))/(x^2)`

Show Worked Solution

`y = (e^x * lnx)/x`

`u` `= e^x lnx` `\ \ \ \ \ v` `= x`
`u prime` `= (e^x)/x + e^x lnx` `\ \ \ \ \ v prime` `= 1`
  `= e^x (1/x + lnx)`    

 

`dy/dx` `= (u prime v – u v prime)/(v^2)`
  `= (e^x (1/x + lnx) * x – e^x lnx * 1)/(x^2)`
  `= (e^x (x* lnx + 1 – lnx))/(x^2)`

Calculus, EXT1 C2 2014 HSC 11d

Evaluate  `int_2^5 x/(sqrt(x - 1))\ dx`  using the substitution  `x = u^2 + 1`.   (3 marks) 

Show Answers Only

`20/3`

Show Worked Solution
`x` `= u^2 + 1`
`u^2` `= x – 1`
`u` `= sqrt (x – 1)`
`(du)/(dx)` `= 1/2 (x – 1)^(-1/2)`
  `= 1/(2 sqrt(x – 1))`
`\ \ =>2du` `= dx/sqrt(x – 1)`

 

`text(When)\ \ \ x = 5,` `\ \ u = 2`
`x = 2,` `\ \ u = 1`

`:.\ int_2^5 x/(sqrt(x – 1))\ dx`

`= 2 int_1^2 u^2 + 1\ du`

`= 2 [ (u^3)/3 + u]_1^2`

`= 2 [(8/3 + 2) – (1/3 + 1)]`

`= 20/3`

Statistics, EXT1 S1 2014 HSC 11b

The probability that it rains on any particular day during the 30 days of November is 0.1.

Write an expression for the probability that it rains on fewer than 3 days in November.   (2 marks)

Show Answers Only

`\ ^30 C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1) (0.9)^29 + (0.9)^30`

Show Worked Solution
`P (R)` `= 0.1`
`P (bar R)` `= 1 – 0.1 = 0.9`

 
`text(Over 30 days:)`

`P (R<3)`

`= P (R=2) + P(R=1) + P (R=0)`

  
  `=\ ^30 C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1)^1 (0.9)^29`  
  `qquad  qquad +\ ^30C_0 (0.1)^0 (0.9)^30`  
  `=\ ^30C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1)(0.9)^29 + (0.9)^30`  

Functions, EXT1 F2 2014 HSC 9 MC

The remainder when the polynomial  `P(x) = x^4-8x^3-7x^2 + 3`  is divided by  `x^2 + x`  is  `ax + 3`.

What is the value of  `a`?

  1. `-14`
  2. `-11`
  3. `-2`
  4. `5`
Show Answers Only

`C`

Show Worked Solution

`P(x) = x^4-8x^3-7x^2 + 3`

`text(Given)\ \ P(x)` `= (x^2 + x) *Q(x) + ax + 3`
  `= x (x + 1) Q(x) + ax + 3`

 
`P(-1) = 1 + 8-7 + 3 = 5`

`:. -a + 3` `= 5`
`a` `= -2`

 
`=>  C`

Functions, EXT1 F2 2014 HSC 5 MC

Which group of three numbers could be the roots of the polynomial equation  `x^3 + ax^2 − 41x + 42 = 0`?

  1. `2, 3, 7`
  2. `1, −6, 7`
  3. `−1, −2, 21`
  4. `−1, −3, −14`
Show Answers Only

`B`

Show Worked Solution

`text(Let roots be)\ alpha, beta, gamma`

`alpha beta gamma = -d/a = -42`

`:.\ text(Cannot be)\ A\ text(or)\ C`

`alpha beta + beta gamma + gamma alpha = c/a = -41`

`:.\ text(Cannot be D)`

`=>  B`

Linear Functions, EXT1 2014 HSC 4 MC

The acute angle between the lines  `2x + 2y = 5`  and  `y = 3x + 1`  is  `theta`.

What is the value of  `tan theta`?

  1. `1/7` 
  2. `1/2`
  3. `1`
  4. `2`
Show Answers Only

`D`

Show Worked Solution

`y = 3x + 1`

`:. m_1 = 3`

`2x + 2y` `= 5`
`2y` `= -2x + 5`
`y` `= -x + 5/2` 

`:. m_2 = -1`

 

`tan theta` `= | (m_1 – m_2)/(1 + m_1 m_2)|`
  `= | (3 – (–1))/(1 + (3 xx –1))|`
  `= |4/(–2)|`
  `= 2`

`=>  D`

Calculus, EXT1 C1 2009 HSC 5b

The cross-section of a 10 metre long tank is an isosceles triangle, as shown in the diagram. The top of the tank is horizontal.
 

 
 

When the tank is full, the depth of water is 3 m. The depth of water at time `t` days is `h` metres.   

  1. Find the volume, `V`, of water in the tank when the depth of water is `h` metres.     (1 mark)

  2. Show that the area, `A`, of the top surface of the water is given by  `A = 20 sqrt3 h`.   (1 mark)

  3. The rate of evaporation of the water is given by  `(dV)/(dt) = - kA`, where `k` is a positive constant. 

     

    Find the rate at which the depth of water is changing at time `t`.   (2 marks)

  4. It takes 100 days for the depth to fall from 3 m to 2 m. Find the time taken for the depth to fall from 2 m to 1 m.   (1 mark)

Show Answers Only
  1. `10 sqrt 3 h^2\ \ text(m³)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `-k\ \ \ text(metres per day)`
  4. `100\ text(days)`
Show Worked Solution
MARKER’S COMMENT: Students who drew a diagram and included their working calculations on it were the most successful.
(i) 
  `text(Let)\ A = text(area of front)`
`tan 30^@` `= h/x`
`x` `= h/(tan 30^@)`
  `= sqrt3 h`
`:.\ A` `= 2 xx 1/2 xx sqrt 3 h xx h`
  `= sqrt 3 h^2\ \ text(m²)`

 

`V` `= Ah`
  `= sqrt 3 h^2 xx 10`
  `= 10 sqrt3 h^2\ \ text(m³)`

 

(ii)    `text(Area of surface)`
  `= 10 xx 2 sqrt 3 h`
  `= 20 sqrt 3 h\ \ text(m²)`

 

(iii)    `(dV)/(dt)` `= -kA`
    `= -k\ 20 sqrt3 h`
`V` `= 10 sqrt3 h^2`
`(dV)/(dh)` `= 20 sqrt 3 h`

 

`text(Find)\ (dh)/(dt)`

MARKER’S COMMENT: Half marks awarded for stating an appropriate chain rule, even if the following calculations were incorrect. Show your working!
`(dV)/(dt)` `= (dV)/(dh) * (dh)/(dt)`
`(dh)/(dt)` `= ((dV)/(dt))/((dV)/(dh))`
  `= (-k * 20 sqrt 3 h)/(20 sqrt 3 h)`
  `= -k`

 

`:.\ text(The water depth is changing at a rate)`

`text(of)\ -k\ text(metres per day.)`

 

♦♦♦ Exact data for part (iv) not available.
COMMENT: Interpreting a constant rate of change was very poorly understood!
(iv)    `text(S)text(ince)\ \ (dh)/(dt)\ \ text(is a constant, each metre)`
  `text(takes the same time.)`

 
`:.\ text(It takes 100 days to fall from 2 m to 1 m.)`

Mechanics, EXT2* M1 2009 HSC 5a

The equation of motion for a particle moving in simple harmonic motion is given by

`(d^2x)/(dt^2) = -n^2x`

where  `n`  is a positive constant,  `x`  is the displacement of the particle and  `t`  is time.  

  1. Show that the square of the velocity of the particle is given by
     
         `v^2 = n^2 (a^2\ - x^2)`

     

    where  `v = (dx)/(dt)`  and  `a`  is the amplitude of the motion.   (3 marks)

  2. Find the maximum speed of the particle.     (1 mark)

  3. Find the maximum acceleration of the particle.    (1 mark)

  4. The particle is initially at the origin. Write down a formula for  `x`  as a function of  `t`, and hence find the first time that the particle’s speed is half its maximum speed.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `na`
  3. `n^2 a`
  4. `pi/(3n)`
Show Worked Solution
i.    `text(Show)\ \ v^2 = n^2 (a^2\ – x^2)`

`(d^2x)/(dt^2) = -n^2x`

`d/(dx) (1/2 v^2)` `= -n^2 x`
`1/2 v^2` `= int -n^2x\ dx`
  `= (-n^2 x^2)/2 + c`
`v^2` `= -n^2 x^2 + c`

 
`text(When)\ \ v = 0,\ \ x = a`

`0` `= -n^2a + c`
`c` `= n^2 a^2`
`:.\ v^2` `= -n^2 x^2 + n^2 a^2`
  `= n^2 (a^2\ – x^2)\ \ text(… as required)`

 

ii.    `text(Max speed when)\ \ x = 0`
`v^2` `= n^2 (a^2\ – 0)`
  `= n^2 a^2`
`:.v_text(max)` `= na`

 

iii.   `(d^2x)/(dt^2)\ \ text(is maximum at limits)\ \ (x = +-a)`
`(d^2x)/(dt^2)` `= |n^2 (a)|`
  `= n^2 a`

 

iv.   `x` `= a sin nt`
  `dot x` `= an cos nt`

 

`text(Find)\ \ t\ \ text(when)\ \ dot x = (na)/2`

`(na)/2` `= an cos nt`
`cos nt` `= 1/2`
`nt` `= pi/3`
`:.t` `= pi/(3n)`

Geometry and Calculus, EXT1 2009 HSC 4b

Consider the function  `f(x) = (x^4 + 3x^2)/(x^4 + 3)`. 

  1. Show that  `f(x)`  is an even function.    (1 mark)
  2. What is the equation of the horizontal asymptote to the graph  `y = f(x)`?    (1 mark)
  3. Find the  `x`-coordinates of all stationary points for the graph  `y = f(x)`.   (3 marks)
  4. Sketch the graph  `y = f(x)`. You are not required to find any points of inflection.    (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `y = 1`
  3. `x = 0, – sqrt 3, sqrt 3`
  5.  
Show Worked Solution
(i)    `f(x)` `= (x^4 + 3x^2)/(x^4 + 3)`
  `f(–x)` `= ((–x)^4 + 3(-x)^2)/((–x)^4 + 3)`
    `= (x^4 + 3x^2)/(x^4 + 3)`
    `= f(x)`

 

`:.\ text(Even function.)`

 

(ii)    `y` `= (x^4 + 3x^2)/(x^4 + 3)`
    `= (1 + 3/(x^2))/(1 + 3/(x^4))`
`text(As)\ \ ` `x` `-> oo`
  `y` `-> 1`

 

`:.\ text(Horizontal asymptote at)\ \ y = 1`

 

(iii)   `f(x) = (x^4 + 3x^2)/(x^4 + 3)`
`u` `= x^4 + 3x^2\ \ \ \ \ ` `v` `= x^4 + 3`
`u prime` `= 4x^3 + 6x\ \ \ \ \ ` `v prime` `= 4x^3`
`f prime (x)` `= (u prime v\ – u v prime)/(v^2)`
  `= ((4x^3 + 6x)(x^4 + 3)\ – (x^4 + 3x^2)4x^3)/((x^4 + 3)^2)`
  `= (4x^7 + 12x^3 + 6x^5 + 18x\ – 4x^7\ – 12x^5)/((x^4 + 3)^2)`
  `= (-6x^5 + 12x^3 + 18x)/((x^4 + 3)^2)`
  `= (-6x(x^4\ – 2x^2\ – 3))/((x^4 + 3)^2)`

 

`text(S.P. when)\ x = 0\ \ \ text(or) \ \ x^4\ – 2x^2\ – 3 = 0`

`text(Let)\ X = x^2`

MARKER’S COMMENT: Many students did not realise the denominator of  `f′(x)` could be ignored when equating  `f′(x)=0`.
`X^2\ – 2X\ – 3` `= 0`
`(X\ – 3)(X + 1)` `= 0`

`X = 3\ \ text(or)\ \ -1`

`:. x^2` `= 3` `text(or)\ \ \ \ \ ` `x^2 = -1`
`x` `= +- sqrt3\ \ \ \ `   `text{(no solution)}`

 

`:.\ text(SPs occur when)\ \ x = 0, – sqrt3, sqrt3`

 

(iv)    `text(When)\ x = 0,\ ` `y = 0`
  `text(When)\ x = sqrt3,\ \ \ ` `y = ((sqrt3)^4 + 3(sqrt3)^2)/((sqrt3)^4 + 3) = 3/2`

 

EXT1 2009 4b