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Algebra, NAP-I3-NC07

Point `Q` is translated down 3 units.

 
What are the new coordinates of  `Q`?

`(4,5)` `(1,2)` `(7,2)` `(4,−1)`
 
 
 
 
Show Answers Only

`(4,−1)`

Show Worked Solution

`Q\ text(has coordinates)\ (4,2)`

`text(When translated 3 down, the)`

`ytext(-value decreases by 3.)`

`:.\ text{New coordinates are  (4, −1)}.`

Probability, NAP-I3-NC06 SA

Peter has a marble bag that contains 20 marbles that are either red or green in colour.

The probability of randomly picking a green marble is 70%.

What is the probability of randomly picking a red marble?

  %
Show Answers Only

`text(30%)`

Show Worked Solution
`P(text(green)) + P(text(red))` `= 100text(%)`
`text(70%)\ + P(text(red))` `= 100text(%)`
`:. P(text(red))` `= 30text(%)`

Geometry, NAP-C3-NC06

An orientation course is set up with 4 drinking stations.

Runners start and then progress through the course in the order of the drinking stations labelled 1-4 in the diagram below.
 

 
At which drinks station do the runners make the greatest change of direction?

Station 1 Station 2 Station 3 Station 4
 
 
 
 
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`text(Station 2)`

Show Worked Solution

`text(Station 2)`

Geometry, NAP-C3-NC04

Rahul draws a net, which is pictured below.
 

 
What 3D object will it make?

 
 Pentagonal prism
 
 Pentagonal pyramid
 
 Hexagonal pyramid
 
 Hexagonal prism
Show Answers Only

`text(Pentagonal pyramid)`

Show Worked Solution

`text(The net will have a pentagon shaped base, and)`

`text(converge to a point.)`

`:.\ text(Net will make a pentagonal pyramid.)`

Algebra, NAP-C3-CA03

At an apple orchard, apples are picked and put in a basket.

The table below shows the total number of apples in the basket after each minute.

\begin{array} {|c|c|c|}
\hline \textbf{Minutes} & \textbf{Total number of apples} \\
\hline 1 & 4 \\
\hline 2 & 8 \\
\hline 3 & 12 \\
\hline 4 & 16 \\
\hline \end{array}

How many apples are in the basket after 10 minutes? 

`20` `30` `35` `40`
 
 
 
 
Show Answers Only

 `40\ text(apples)`

Show Worked Solution

`text(4 apples are put into the basket each minute.)`

`:.\ text(Apples in basket after 10 minutes)`

`=4 xx 10`

`= 40\ text(apples)`

Statistics, NAP-C3-CA02

For three days, Terry counted the colour of cars he saw drive past his house.
 

   

 
Which column on the graph below shows the total number of Red cars? 

 
 
 
 
Show Answers Only

 `text(The second bar from the left.)`

Show Worked Solution

`text(Option 2: 4 red cars passed.)`

 `text{(The second bar from the left)}`

Number, NAP-E4-CA05

Zilda has 4 children.

Each child needs 6 pens when they start school for the new year.

Zilda has no pens when she goes to the shop.

The shop sells its pens in packets of 5.

How many packets does Zilda need to buy?

`2`    `4`   `5`  `6`
 
 
 
 
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`5\ text(packets)`

Show Worked Solution
`text(Pens required)` `= 4 xx 6`
  `= 24`

 

`:.\ text(Packets required)` `= 24/5`
  `= 4.8`
  `= 5`

Statistics, NAP-E4-CA03

Anne taught a kindergarten class and asked her students what their favorite fruit is.

Their answers were used to draw the pie chart below.
 

 

 
Around a quarter of her students preferred which fruit?

`text(Banana)` `text(Apple)` `text(Mango)` `text(Grapes)`
 
 
 
 
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`text(Apple)`

Show Worked Solution

`text(A quarter of students will represent a sector that)`

`text(is the size of a quarter of the circle.)`

`:.\ text(Apple)`

Number, NAP-E4-CA01

The airforce buys 8 new fighter jets for $882.0 million.

Each jet costs the same amount.

What is the cost of 1 fighter jet?

`$110.25\ text(million)` `$441.0\ text(million)` `$874.0\ text(million)` `$7056\ text(million)`
 
 
 
 
Show Answers Only

`$110.25\ text(million)`

Show Worked Solution

`{$882.0\ text(million)}/ 8 = $110.25\ text(million)`

Number, NAP-F4-CA07

David buys a surfboard cover online that costs him $128.

The postage costs involved are listed in the table below.
 

 
The surfboard cover weighs 1.1 kg

What does David have to pay, in total, to purchase the surfboard and have it delivered?

`$140` `$145.50` `$149.75` `$153`
 
 
 
 
Show Answers Only

`$149.75`

Show Worked Solution

`text(Surfboard cover cost)\ +\ text(postage)`

`= $128 + $21.75`

`= $149.75`

Number, NAP-F4-NC04

Nigella placed a leg of lamb into her oven in the morning.

The 1st measurement of the lamb's temperature was  −6°C.

Fifteen minutes later, a second temperature reading was taken, which measured  16°C.

What was the change in temperature between the two measurements?

`text(decrease of)\ 22^@text(C)` `text(decrease of)\ 10^@text(C)` `text(increase of)\ 10^@text(C)` `text(increase of)\ 22^@text(C)`
 
 
 
 
Show Answers Only

`text(increase of)\  22^@text(C)`

Show Worked Solution
`text(Change in temperature)` `=16-(-6)`
  `=22^@ text(C)`

 

`:.\ text(An increase of  22°C.)`

Number, NAP-F4-NC03

A circle is divided into 5 equal areas and labelled, as shown in the diagram below.
 

 
What percentage of the circle's area has been labelled with an odd number?

`text(3%)` `text(15%)` `text(30%)` `text(40%)` `text(60%)`
 
 
 
 
 
Show Answers Only

`text(60%)`

Show Worked Solution

`text(S)text(ince all areas are equal,)`

`text(Percentage labelled with an odd number)`

`=3/5`

`=60text(%)`

Number, NAP-F4-CA04

The first three days of the Brisbane cricket test had the following attendances:

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex}\text{Day 1}\rule[-1ex]{0pt}{0pt} & 20\ 156\\
\hline
\rule{0pt}{2.5ex}\text{Day 2}\rule[-1ex]{0pt}{0pt} & 18\ 397\\
\hline
\rule{0pt}{2.5ex}\text{Day 3}\rule[-1ex]{0pt}{0pt} & 29\ 981\\
\hline
\end{array}

What was the total crowd over the first 3 days, to the nearest `1000?`

`68\ 000` `69\ 000` `70\ 000` `71\ 000`
 
 
 
 
Show Answers Only

`69\ 000\ text{(nearest 1000)}`

Show Worked Solution

`20\ 156 + 18\ 397 + 29\ 981`

`= 68\ 534`

`= 69\ 000\ text{(nearest 1000)}`

Number, NAP-G4-NC02

Ben spent twice as much money as Mark.

If they spent a total of $90, how much did Ben spend?

`$15` `$30` `$60` `$180`
 
 
 
 
Show Answers Only

`$60` 

Show Worked Solution

`text(Strategy 1)`

`text(Ratio)`

`text{Ben : Mark = 2 : 1  (3 parts)}`

`text(1 part)\ = $90-:3=$30`

`:.\ text(Ben spent  2 × $30 = $60)`

 

`text(Strategy 2)`

`text(Let)\ \ $x = text(Amount Mark spent)`

`2x + x` `= 90`
`x` `= $30`

 

`:.\ text(Ben spent $60.)`

Measurement and Geometry, NAP-G4-CA04

Petunia walks along the path from `A` to `B.`

How far does she walk in kilometres?

`0.14` `1.4` `14` `1400`
 
 
 
 
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`1.4\ text(km)`

Show Worked Solution
`text(Distance)` `= 14 xx 100\ text(m)`
  `= 1.4\ text (km)`

Measurement, NAP-I4-CA06

Peter left home at 9:15 in the morning and did not return until 5:25 in the afternoon.

How long was Peter away from his house?

 
 3 hours 50 minutes
 
 4 hours 10 minutes
 
 7 hours 50 minutes
 
 8 hours 10 minutes
 
 13 hours 20 minutes
Show Answers Only

`8\ text(hours)\ 10\ text(minutes)`

Show Worked Solution
`text(Time away)` `= 2 text(h)\ 45\ text(min) + 5 text(h)\ 25\ text(min)`
  `= 8\ text(hours)\ 10\ text(minutes)`

Quadratic, EXT1 2016 HSC 14c

The point  `T(2at,at^2)` lies on the parabola `P_1` with the equation  `x^2=4ay`.

The tangent to the parabola `P_1` at `T` meets the directrix at `D`.

The normal to the parabola `P_1` at `T` meets the vertical line through `D` at the point `R`, as shown in the diagram.

 

ext1-2016-hsc-q14

  1. Show that the point `D` has coordinates `(at - a/t, −a)`.  (1 mark)
  2. Show that the locus of `R` lies on another parabola `P_2`.  (3 marks)
  3. State the focal length of the parabola `P_2`.  (1 mark)

It can be shown that the minimum distance between `R` and `T` occurs when the normal to `P_1` at `T` is also the normal to `P_2` at `R`.  (Do NOT prove this.)

  1. Find the values of `t` so that the distance between `R` and `T` is a minimum.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `a/4`
  4. `+- sqrt 2`
Show Worked Solution

i.     `text(Show)\ \ D (text{at}\ -a/t, -a)`

`text(T)text(angent equation at)\ \ T:`

`y = tx – at^2`

`D\ \ text(occurs when)\ \ y = -a,`

`tx – at^2` `= -a`
`tx` `= at^2 – a`
`x` `= at – a/t`

`:. D\ text(has coordinates)\ \ (text{at}\ -a/t, -a)`

 

ii.    `text(Normal equation at)\ \ T:`

♦♦ Mean mark 32%.
MARKER’S COMMENT: Normal equation is found in the Reference Sheet. Save time by using it!

`x + ty = 2at + at^3`

`R\ text(occurs when)\ \ x = at – a/t`

`at – a/t + ty` `= 2at + at^3`
`ty` `= at + a/t + at^3`
`y` `= a + a/t^2 + at^2`
  `= a(1 + 1/t^2 + t^2)\ \ …\ text{(*)}`

 

`:. R (a (t – 1/t), a (1 + 1/t^2 + t^2))`

`x^2` `= a^2 (t – 1/t)^2`
  `= a^2 (t^2 – 2 + 1/t^2)`
  `= a + (1+1/t^2 + t^2 – 3)`
  `= a^2 (y/a – 3)\ \ text{(see (*) above)}`
  `= ay – 3a^2`

 

`:.\ text(Locus of)\ R\ text(is)\ \ x^2 = ay – 3a^2`

 

iii.   `text(In the form)\ \ x^2 = 4ay,`

♦♦♦ Mean mark 15%.
`x^2` `= a(y – 3a)`
  `= 4 · a/4 (y – 3a)`

 

`:.\ text(Focal length) = a/4`

 

iv.   `text(Equation of)\ \ P_2`

♦♦♦ Mean mark 11%.
`x^2` `= ay – 3a^2`
`y` `= x^2/a + 3a`
`y prime` `= (2x)/a`

`text(At)\ \ R,\ \ x = a (t – 1/t)`

`:.\ text(Gradient of normal at)\ \ R`

`=(-a)/(2x)`

`= (-a)/(2a(t – 1/t)) xx t/t`

`= (-t)/(2(t^2 – 1))`

 

`text(Gradient of normal at)\ \ T:`

`x + ty` `= 2at + at^3`
`y` `= -1/t x + 2a + at^2`
`:. m` `= -1/t`

 

`text(Distance)\ \ RT\ \ text(is a minimum when)`

`(-t)/(2(t^2 – 1))` `= -1/t`
`t^2` `= 2t^2 – 2`
`t^2` `= 2`
`:. t` `= +- sqrt 2`

Binomial, EXT1 2016 HSC 14b

Consider the expansion of  `(1 + x)^n`, where `n` is a positive integer.

  1. Show that  `2^n = ((n),(0)) + ((n),(1)) + ((n),(2)) + ((n),(3)) + … + ((n),(n))`.  (1 mark)
  2. Show that  `n2^(n - 1) = ((n),(1)) + 2((n),(2)) + 3((n),(3)) + … + n((n),(n))`.  (1 mark)
  3. Hence, or otherwise, show that  `sum_(r = 1)^n ((n),(r))(2r - n) = n`.  (2 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)}`
Show Worked Solution

i.     `text(Show)`

`2^n = ((n),(0)) + ((n),(1)) + ((n),(2)) + ((n),(3)) + … + ((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,`

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

`text(… as required.)`

 

ii.   `text(Differentiate both sides of expansion,)`

`n (1 + x)^(n – 1) = ((n), (1)) + 2 ((n), (2))x + 3 ((n), (3)) x^2 + … + n ((n), (n)) x^(n – 1)`

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

`n2^(n – 1) = ((n), (1)) + 2 ((n), (2)) + 3 ((n), (3)) + … + n ((n), (n))`

`text(… as required.)`

 

iii.  `text{Multiply part (i)} xx n`

♦♦♦ Mean mark 14%.
`n2^n` `= n[((n), (0)) + ((n), (1)) + … + ((n), (n))]`
  `= sum_(r = 0)^n ((n), (r)) n\ \ text{…  (1)}`

 

`text{Multiply part (ii)} xx 2`

`2 xx n2^(n – 1)` `= 2[((n), (1)) + 2 ((n), (2)) + … + n ((n), (n))]`
`n2^n` `= sum_(r = 1)^n ((n), (r)) 2r\ \ text{…  (2)}`

 

 `text(Subtract) qquad (2) – (1)`

`sum_(r = 1)^n ((n), (r)) 2r – sum_(r = 0)^n ((n), (r))n` `= n2^n – n2^n`
`sum_(r = 1)^n ((n), (r)) 2r – sum_(r = 1)^n ((n), (r)) n – n` `= 0`
`sum_(r = 1)^n ((n), (r)) (2r – n)` `= n\ \ text(…  as required)`

Plane Geometry, EXT1 2016 HSC 13c

The circle centred at `O` has a diameter `AB`. From the point `M` outside the circle the line segments `MA` and `MB` are drawn meeting the circle at `C` and `D` respectively, as shown in the diagram. The chords `AD` and `BC` meet at `E`. The line segment `ME` produced meets the diameter `AB` at `F`.

ext1-2016-hsc-q13_1

Copy or trace the diagram into your writing booklet.

  1. Show that `CMDE` is a cyclic quadrilateral.  (2 marks)
  2. Hence, or otherwise, prove that `MF` is perpendicular to `AB`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.  
`/_ ACB` `= /_ ADB = 90^@ qquad text{(angle in semi-circle)}`
`/_ MCE` `= 90^@ qquad (/_ MCA\ text{is a straight angle)}`
`/_ MDA` `= 90^@ qquad (/_ MDB\ text{is a straight angle)}`

 
`:. CMDE\ text(is a cyclic quad)`

`qquad text{(opposite angles are supplementary)}`
 

ii.   `text(Consider)\ \ Delta MAB,`

♦♦♦ Mean mark 14%.

`CB\ \ text(is an altitude.)`

`AD\ \ text(is an altitude.)`

`text(S) text(ince the altitudes of a triangle are)`

`text(concurrent)\ \ text{(in}\ Delta MAB, text{at}\ Etext{),}`

`=> MF\ \ text(must be an altitude.)`

`:. MF _|_ AB`

Calculus, 2ADV C3 2016 HSC 16b

Some yabbies are introduced into a small dam. The size of the population,  `y`, of yabbies can be modelled by the function

`y = 200/(1 + 19e^(-0.5t)),`

where  `t`  is the time in months after the yabbies are introduced into the dam.

  1. Show that the rate of growth of the size of the population is
  2. `qquad qquad (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2`.   (2 marks)

  3. Find the range of the function `y`, justifying your answer.   (2 marks)

  4. Show that the rate of growth of the size of the population can be written as
  5. `qquad qquad y/400 (200-y)`.   (1 mark)

  6. Hence, find the size of the population when it is growing at its fastest rate.   (2 marks)

Show Answers Only

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

b.    `10 <= y < 200`

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

d.    `100`

Show Worked Solution
a.     `y` `= 200/(1 + 19 e^(-0.5t))`
  `(dy)/(dt)` `= 200/(1 + 19 e^(-0.5t))^2 xx d/(dt) (1 + 19 e^(-0.5t))`
    `= (-200)/(1 + 19 e^(-0.5t))^2 xx -0.5 xx 19 e^(-0.5t)`
    `= (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2\ \ text(… as required)`

 

b.    `text(When)\ \ t = 0,`

♦♦♦ Mean mark (b) 21%.

`y = 200/(1 + 19) = 10`

`text(As)\ \ t -> oo,\ \ (1 + 19^(-0.5t)) -> 1`

`:. y -> 200`

`:.\ text(Range)\ \ \ 10 <= y < 200`

 

c.    `(dy)/(dt) = (1900 e^(-0.5t))/(1 + 19 e^(-0.5t))^2`

♦♦♦ Mean mark (c) 18%.

`text(S) text(ince)\ \ y = 200/(1 + 19 e^(-0.5t))`

`-> (1 + 19 e^(-0.5t)) = 200/y`

`-> 19 e^(-0.5t) = 200/y-1 = (200-y)/y`

`text(Substituting into)\ \ (dy)/(dt):`

`(dy)/(dt)` `= (100 ((200-y)/y))/(200/y)^2`
  `= 100 ((200-y)/y) xx y^2/200^2`
  `= y/400 (200-y)\ \ text(… as required)`

 

d.    `(dy)/(dt) = -y^2/400 + y/2`

♦♦♦ Mean mark (d) 14%.

`text(Sketching the parabola:)`

`(-y^2)/400 + y/2` `= 0`
`-y^2 + 200y` `= 0`
`y (200-y)` `= 0`

 

hsc-2016-16bi

`:.\ text(Maximum)\ \ (dy)/(dt)\ \ text(occurs when)\ \ y = 100.`

Proof, EXT2 P2 2016 HSC 16c

In a group of `n` people, each has one hat, giving a total of `n` different hats. They place their hats on a table. Later, each person picks up a hat, not necessarily their own.

A situation in which none of the `n` people picks up their own hat is called a derangement.

Let  `D(n)`  be the number of possible derangements.

  1. Tom is one of the `n` people. In some derangements Tom finds that he and one other person have each other's hat.

     

    Show that, for  `n > 2`, the number of such derangements is  `(n - 1) D (n - 2).`  (1 mark)

  2. By also considering the remaining possible derangements, show that, for  `n > 2,`

    `qquad qquad D(n) = (n - 1) [D(n - 1) + D(n - 2)].`  (2 marks)

  3. Hence, show that  `D(n) - nD(n - 1) = -[D(n - 1) - (n - 1) D(n - 2)]`,  for  `n > 2.`  (1 mark)

  4. Given  `D(1) = 0`  and  `D(2) = 1`,  deduce that  `D(n) - n D(n - 1) = (-1)^n`,  for  `n > 1.`  (1 mark)

  5. Prove by mathematical induction, or otherwise, that for all integers  `n >= 1,\ D(n) = n! sum_(r = 0)^n (-1)^r/(r!).`  (2 marks)

Show Answers Only

i.    `text(See Worked Solutions)`

ii.    `text(See Worked Solutions)`

iii.    `text(See Worked Solutions)`

iv.    `text(See Worked Solutions)`

v.    `text(See Worked Solutions)`

Show Worked Solution

i.    `text(Tom and one other person can have each)`

♦♦♦ Mean mark 22%.

`text(other’s hat in)\ (n – 1)\ text(combinations.)`

`text(There are)\ \ D(n – 2)\ \ text(possibilities for the)`

`text(rest of the people selecting the wrong hats.)`

`:. text(Number of derangements) = (n – 1)D(n – 2)`

 

ii.    `text(Consider all possible derangements:)`

♦♦♦ Mean mark 6%.

`text(Any of)\ n\ text(people choose the wrong hat.)`

`text(Remainder)\ (n – 1)\ text(people can select the)`

`text(wrong hat in)\ D(n – 1)\ text(ways.)`

`:. nD(n – 1)\ text(derangements.)`

`text{This includes part (i) combinations.}`

 

`:.\ text(Remaining possible derangements)`

`= nD(n – 1) – (n – 1)D(n – 2)`

`= nD(n – 1) – D(n – 1)`

`= (n – 1)D(n – 1)`

`:. D(n)` `= (n – 1)D(n – 1) + (n – 1)D(n – 2)`
  `= (n – 1)[D(n – 1) + D(n – 2)]`

 

♦♦ Mean mark part (iii) 37%.
iii.     `D(n)` `= (n – 1)[D(n – 1) + D(n – 2)]`
    `= nD(n – 1) – D(n – 1) + (n – 1)D(n – 2)`

 

`:. Dn – nD(n – 1) = −[D(n – 1) – (n – 1)D(n – 2)]`

 

iv.    `D(1) = 0, D(2) = 1`

♦♦♦ Mean mark 1%!

`D(2) – 2D(1) = 1`

`D(3) – 3D(2) = −[D(2) – 2D(1)] = −1`

`D(4) – 4D(3) = −[D(3) – 3D(2)] = −(−1) = 1`

`D(5) – 5D(4) = −[D(4) – 4D(3)] = −1`

 

`:. D(n) – nD(n – 1) = (−1)^n\ text(for)\ n > 1`

 

v.    `text(Prove)\ D(n) = n! sum_(r = 0)^n ((−1)^r)/(r!)\ text(for)\ n >= 1`

`text(When)\ n = 1,`

`D(1) = 1! sum_(r = 0)^1 ((−1)^r)/(r!) = 1 – 1 = 0`

`text(S)text(ince)\ D(1) = 0\ (text(given)),`

`:.\ text(True for)\ n = 1`

♦♦♦ Mean mark 13%.

 

`text(Assume true for)\ \ n = k,`

`text(i.e.)\ \ D(k) = k! sum_(r = 0)^k ((−1)^r)/(r!)`

`text(Prove true for)\ \ n=k+1,`

`text(i.e.)\ \ D(k+1) = (k+1)!sum_(r = 0)^(k+1) ((−1)^r)/(r!)`

`D(k+1)`

`= (k + 1)D(k) + (−1)^(k + 1)qquad(text{from part (iv)})`
`= (k + 1) · k! sum_(r = 0)^k ((−1)^r)/(r!) + (−1)^(k + 1)`
`= (k + 1)!(1 – 1/(1!) + 1/(2!) – 1/(3!) + … + ((−1)^k)/(k!)) + (−1)^(k + 1) · ((k + 1)!)/((k + 1)!)`
`= (k + 1)!(1 – 1/(1!) + 1/(2!) – 1/(3!) + … + ((−1)^k)/(k!) + ((−1)^(k+1))/((k+1)!))`
`= (k + 1)! sum_(r = 0)^(k + 1) ((−1)^r)/(r!)`

 

`=> text(True for)\ n = k + 1.`

`:. text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

Complex Numbers, EXT2 N2 2016 HSC 16b

  1. The complex numbers `0, \ u` and `v` form the vertices of an equilateral triangle in the Argand diagram.
  2. Show that  `u^2 + v^2 = uv.`   (2 marks)
  3. Give an example of non-zero complex numbers `u` and `v`, so that `0, \ u` and `v` form the vertices of an equilateral triangle in the Argand diagram.   (1 mark)

Show Answers Only

a.    `text(Show Worked Solutions)`

b.    `u = 1/2 + sqrt3/2 i`

Show Worked Solution

a.    `0,u,v\ text(are vertices of an equilateral triangle.)`

♦♦ Mean mark (a) 23%.

 

ext2-2016-hsc-16b-answer

`=> u` `= v text(cis)(±pi/3)`
`u^3` `= v^3text(cis)(±pi)`
`u^3` `= − v^3`
`u^3 + v^3` `= 0`

 
`(u + v)(u^2-uv + v^2) = 0`

`text(S)text(ince)\ u != v:`

`u^2-uv + v^2` `= 0`
`:. u^2 + v^2` `= uv`

 

b.    `text(Let)\ \ v = 1,`

♦♦ Mean mark (b) 33%.
`:. u` `= text(cis)(pi/3)`
  `= cos\ pi/3 + isin\ pi/3`
  `= 1/2 + sqrt3/2 i`

Complex Numbers, EXT2 N2 2016 HSC 16a

  1. The complex numbers  `z = cos theta + i sin theta`  and  `w = cos alpha + i sin alpha`, where  `-pi < theta < pi`  and  `-pi < alpha <= pi`, satisfy
  2.    `1 + z + w = 0.`
  3. By considering the real and imaginary parts of  `1 + z + w`, or otherwise, show that `1,  z` and `w` form the vertices of an equilateral triangle in the Argand diagram.   (3 marks)

  4. Hence, or otherwise, show that if the three non-zero complex numbers `2i, z_1` and `z_2` satisfy
  5.    `|\ 2i\ | = |\ z_1\ | = |\ z_2\ |`  AND  `2i + z_1 + z_2 = 0.`
  6. then they form the vertices of an equilateral triangle in the Argand diagram.   (2 marks)

Show Answers Only

i.    `text(See Worked Solutions)`

ii.   `text(See Worked Solutions)`

Show Worked Solution

i.    `z= costheta + isintheta, \ |\ z\ | = 1`

`w= cosalpha + isinalpha, \ |\ w\ | = 1`

`text(S)text(ince)\ 1 + z + w = 0:`

♦♦♦ Mean mark (i) 25%.
STRATEGY: A clear graphical image simplifies both parts of this question significantly.
`text(Im)(1 + z + w)` `= 0`
`sintheta + sinalpha` `= 0`
`sintheta` `= −sinalpha`
`:. theta` `= −alpha\ …\ (1)`

 

`text(Re)(1 + z + w)` `= 0`
`1 + costheta + cosalpha` `= 0`
`2costheta` `= −1qquad(text(S)text(ince)\ cosalpha  = cos(−theta) = costheta)`
`costheta` `= −1/2`
`:. theta` `= (2pi)/3`
`alpha` `= −(2pi)/3`

 
ext2-2016-hsc-16a-answer2 

`=>\ text(All points are on the unit circle separated by)\ (2pi)/3\ text(radians.)`

`:.\ text(They are vertices of an equilateral triangle.)`

 

ii.   `|\ 2i\ | = 2`

`text(Let)\ \ z_1` `= 2(costheta + isintheta), \  |\ z_1\ | = 2`
`z_2`  `= 2(cosalpha + isinalpha), \  |\ z_2\ | = 2`

 

♦♦♦ Mean mark (ii) 5%.
`text(Re)(2i + z_1 + z_2)` `= 0`
`2(costheta + cosalpha)` `= 0`
`:. costheta` `= −cosalpha`
`:. theta` `= pi-alpha`

 

`text(Im)(2i + z_1 + z_2)` `= 0`
`2(1 + sintheta + sinalpha)` `= 0`
`sintheta + sinalpha` `= −1`
`2sintheta` `= −1qquad(text(S)text(ince)\ sinalpha = sin(pi-theta) = sintheta)`
`sintheta` `= −1/2`
`:. theta` `= (7pi)/6`
`:. alpha` `= −pi/6`

 
ext2-2016-hsc-16a-answer3 
 

`=>\ text(All points are on the 2 unit circle separated by)\ (2pi)/3\ text(radians.)`

`:. 2i, z_1\ text(and)\ z_2\ text(are vertices of an equilateral triangle.)`

Polynomials, EXT2 2016 HSC 15c

  1. Use partial fractions to show that

    `qquad (3!)/(x(x + 1) (x + 2) (x + 3)) = 1/x - 3/(x + 1) + 3/(x + 2) - 1/(x + 3).`  (2 marks)

  2. Suppose that for `n` a positive integer

     

    `qquad qquad (n!)/(x(x + 1) … (x + n)) = a_0/x + a_1/(x + 1) + … + a_k/(x + k) + … + a_n/(x + n).`

    Show that  `a_k = (-1)^k ((n), (k)).`  (3 marks)

  3. Hence, or otherwise, find the limiting sum of

     

    `qquad qquad 1 - 1/2 ((n), (1)) + 1/3 ((n), (2)) - 1/4 ((n), (3)) + … + (-1)^n/(n + 1).`  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `1/(n + 1)`
Show Worked Solution

i.   `text(Show)`

`(3!)/(x(x + 1)(x + 2)(x + 3)) = 1/x – 3/(x + 1) + 3/(x + 2) – 1/(x + 3)`

`text(LHS)` `= (a_0)/x + (a_1)/(x + 1) + (a_2)/(x + 2) + (a_3)/(x + 3)`
  `= [(a_0(x + 1)(x + 2)(x + 3) + a_1x(x + 2)(x + 3) + a_2x(x + 1)(x + 3) +`
  ` a_3x(x + 1)(x + 2)]//(x(x + 1)(x + 2)(x + 3))`

 

`a_0 = lim_(x -> 0) (3!)/((x + 1)(x + 2)(x + 3)) = 6/6 = 1`

`a_1 = lim_(x -> −1) (3!)/(x(x + 2)(x + 3)) = 6/((−1)(1)(2)) = −3`

`a_2 = lim_(x -> −2) (3!)/(x(x + 1)(x + 3)) = 6/((−2)(−1)(1)) = 3`

`a_3 = lim_(x -> − 3)(3!)/(x(x + 1)(x + 2)) = 6/((−3)(−2)(−1)) = −1`

 

ii.   `text(Given that)`

`(n!)/(x(x + 1) … (x + n)) = (a_0)/x + (a_1)/(x + 1) + … + (a_k)/(x + k) + … +  (a_n)/(x + n)`

 

`text(Show)\ a_k = (−1)^k((n),(k))`

`a_k` `= lim_(x -> −k) (n!(x + k))/(x(x + 1) … (x + k – 1)(x + k)(x + k + 1) …(x + n))`
  `= (n!)/((−k)(−k + 1) … (−k + k – 1)(−k + k + 1) … (−k + n))`
   
  `=>\ text(S)text(ince negative for)\ k\ text(odd,)`
 `a_k` `= ((−1)^kn!)/(k(k – 1) … (2)(1)(1)(2) … (n – k))`
  `= ((−1)^kn!)/(k!(n – k)!)`
  `= (−1)^k ((n),(k))`

 

iii.   `(n!)/(x(x + 1) … (x + n))`

`= (a_0)/x + (a_1)/(x + 1) + … + (a_n)/(x + n)`

`= ((n),(0)) – 1/2((n),(1)) + 1/3((n),(2)) – … + ((−1)^n)/(x + n)((n),(n))`

 

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

`(n!)/((1)(2)(3) … (n + 1))`

`= ((n),(0)) – 1/2((n),(1)) + 1/3((n),(2)) – … + ((−1)^n)/(n + 1)`

`text(LHS)` `= (n!)/((n + 1)!)`
  `= 1/(n + 1)`

 

`:.\ text(Limiting sum) = 1/(n + 1)`

Functions, EXT1′ F2 2016 HSC 13d

Suppose  `p(x) = ax^3 + bx^2 + cx + d`  with `a, b, c` and `d` real, `a != 0.`

  1. Deduce that if  `b^2-3ac < 0`  then `p(x)` cuts the `x`-axis only once.   (2 marks)

  2. If  `b^2-3ac = 0`  and  `p(-b/(3a)) = 0`, what is the multiplicity of the root  `x = -b/(3a)?`   (2 marks)

Show Answers Only

a.    `text(See Worked Solutions)`

b.    `3`

Show Worked Solution

a.    `p(x) = ax^3 + bx^2 + cx + d`

`p^{′}(x) = 3ax^2 + 2bx + c`

`=> p(x)\ text(will cut the)\ xtext(-axis once only if)\ \ Delta(p^{′}(x)) < 0`

`(2b)^2-4(3a)c` `< 0`
`4b^2-12ac` `< 0`
`b^2-3ac` `< 0`

 

b.    `p(-b/(3a)) = 0`

`p^{′}(−b/(3a))` `=3a(- b/(3a))^2 + 2b (- b/(3a))+c`
  `=- b^2/(3a)+c`
  `=0\ \ \ (text{given}\ \ b^2-3ac = 0)`

 
`:.\ text(Multiplicity at least 2.)`

♦♦ Mean mark 32%.

 
`p^{″}(x) = 6ax + 2b`

`p^{″}(−b/(3a))=6a(−b/(3a)) + 2b=0`

`:. text(Multiplicity of)\ \ x=− b/(3a)\ \ text(is 3.)`

Algebra, STD2 A4 2016 HSC 29b

The mass `M` kg of a baby pig at age `x` days is given by  `M = A(1.1)^x`  where `A` is a constant. The graph of this equation is shown.
 

2ug-2016-hsc-q29_1

  1. What is the value of `A`?   (1 mark)

  2. What is the daily growth rate of the pig’s mass? Write your answer as a percentage.   (1 mark)

Show Answers Only

a.    `1.5\ text(kg)`

b.    `10text(%)`

Show Worked Solution

a.    `text(When)\ x = 0,`

♦ Mean mark (a) 48%.
♦♦♦ Mean mark part (b) 6%!
`1.5` `= A(1.1)^0`
`:. A` `= 1.5\ text(kg)`

 
b.    `text(Daily growth rate) = 0.1 = 10text(%)`

Probability, STD2 S2 2016 HSC 23 MC

A group of 485 people was surveyed. The people were asked whether or not they smoke. The results are recorded in the table.
 

A person is selected at random from the group.

What is the approximate probability that the person selected is a smoker OR is male?

  1. 33%
  2. 18%
  3. 68%
  4. 87%
Show Answers Only

`=> C`

Show Worked Solution

`P(text(Smoker or a male))`

`= (text(Total males + female smokers))/(text(Total surveyed))`

`= (264 + 68)/485`

`= 0.684…`
 

`=> C`

♦♦ Mean mark 34%.

Algebra, 2UG 2016 HSC 18 MC

The value of `E` varies directly with the square of `S`.

It is known that  `E = 20`  when  `S = 10`.

What is the value of `E` when  `S = 40`?

  1. `40`
  2. `80`
  3. `320`
  4. `400`
Show Answers Only

`=> C`

Show Worked Solution

`E ∝ S^2`

♦♦♦ Mean mark 22%.
STRATEGY: A well structured answer is important in solving this question type. See solution.

`E = kS^2`

`text(Find)\ k:`

`20` `= k(10)^2`
`:. k` `= 0.2`

 

`text(When)\ S = 40,`

`E` `= 0.2 xx 40^2`
  `= 320`

 
`=> C`

Statistics, STD2 S1 2016 HSC 7 MC

Which set of data is classified as categorical and nominal?

  1. blue, green, yellow
  2. small, medium, large
  3. 5.2 cm, 6 cm, 7.21 cm
  4. 4 people, 5 people, 9 people 
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Categorical data is qualitative (eliminate C and D)}\)

♦♦ Mean mark 26%.

\(\text{Nominal data not ordered (eliminate B)}\)

\(\Rightarrow A\)

Trigonometry, 2ADV T2 2016 HSC 8 MC

How many solutions does the equation  `|\ cos (2x)\ | = 1`  have for  `0 <= x <= 2 pi?`

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

`D`

Show Worked Solution

`|\ cos (2x)\ | = 1`

♦♦♦ Mean mark 23%.

`cos (2x) = +- 1`

`text(When)\ \ cos (2x)` `= 1`
`2x` `= 0, 2pi, 4 pi, …`
`:. x` `= 0, pi, 2 pi, …`

 

`text(When)\ \ cos (2x)` `= – 1`
`2x` `= pi, 3 pi, 5 pi, …`
`:. x` `= pi/2, (3 pi)/2, (5 pi)/2, …`

 

`:. x = 0, pi/2, pi, (3 pi)/2, 2 pi\ \ \ text(for)\ \ \ 0 <= x <= 2pi`

`=>  D`

Probability, MET1 2007 VCAA 6

Two events, `A` and `B`, from a given event space, are such that  `text(Pr)(A) = 1/5`  and  `text(Pr)(B) = 1/3`.

  1. Calculate  `text(Pr)(A^{′} ∩ B)`  when  `text(Pr)(A ∩ B) = 1/8`.  (1 mark)

  2. Calculate  `text(Pr)(A^{′} ∩ B)`  when `A` and `B` are mutually exclusive events.  (1 mark)

Show Answers Only

  1. `5/24`
  2. `1/3`

Show Worked Solution

a.   `text(Sketch Venn diagram:)`

♦♦ Mean mark (a) 31%.
MARKER’S COMMENTS: Students who drew a Venn diagram or Karnaugh map were the most successful.

met1-2007-vcaa-q6-answer3

`:. text(Pr)(A^{′} ∩ B)` `=text(Pr)(B)-text(Pr)(A ∩B)`
  `=1/3-1/8`
  `=5/24`

 

♦♦ Mean mark (b) 23%.

b.    met1-2007-vcaa-q6-answer4

`text(Mutually exclusive means)\ \ text(Pr)(A ∩ B)=0,`

`:. text(Pr)(A^{′} ∩ B) = 1/3`

Functions, MET1 2008 VCAA 10

Let  `f: R -> R,\ \ f(x) = e^(2x)-1`.

  1. Find the rule and domain of the inverse function  `f^(-1)`.   (2 marks)

  2. On the axes provided, sketch the graph of  `y = f(f^(-1)(x))`  for its maximal domain.   (1 mark)


     

    met1-2008-vcaa-q2
     

  3. Find  `f(-f^(-1)(2x))`  in the form  `(ax)/(bx + c)`  where `a`, `b` and `c` are real constants.   (2 marks)

Show Answers Only
  1. `f^(-1)(x) = 1/2log_e(x + 1), x ∈ (-1,∞)`
  2.  
    met1-2008-vcaa-q10-answer
  3. `f(-f^(-1)(2x)) = (-2x)/(2x + 1)`
Show Worked Solution

a.   `text(Let)\ \ y = f(x)`

`text(For Inverse, swap)\ x ↔ y`

`x` `= e^(2y)-1`
`x + 1` `= e^(2y)`
`2y` `= log_e(x + 1)`
`y` `= 1/2 log_e(x + 1)`
`text(Domain)(f^(-1))` `=\ text(Range)\ (f)`
  `= (-1,∞)`

 

`:. f^(-1)(x) = 1/2log_e(x + 1),quadx ∈ (-1,∞)`
  

b.   `f(f^(-1)(x)) = x`

♦ Mean mark part (b) 19%.
MARKER’S COMMENT: Few students were aware that a function of its own inverse function is the line  `y=x`  over the appropriate domain.

`text(Domain is)\ \ (-1, oo)`

met1-2008-vcaa-q10-answer

 

♦ Mean mark 34%.
c.    `-f^(-1)(2x)` `= -1/2 ln(2x + 1)`
  `:. f(-f^(-1)(2x))` `= e^(-log_e(2x + 1))-1`
    `=(2x+1)^-1-1`
    `= 1/(2x + 1)-1`
    `= (-2x)/(2x + 1)`

Calculus, MET1 2011 VCAA 10

The figure shown represents a wire frame where `ABCE` is a convex quadrilateral. The point `D` is on line segment `EC` with  `AB = ED = 2\ text(cm)` and  `BC = a\ text(cm)`, where `a` is a positive constant.

`/_ BAE = /_ CEA = pi/2`

Let  `/_ CBD = theta`  where  `0 < theta < pi/2.`

 vcaa-2011-meth-10a

  1. Find `BD` and `CD` in terms of `a` and `theta`.   (2 marks)

  2. Find the length, `L` cm, of the wire in the frame, including length `BD`, in terms of `a` and `theta`.   (1 mark)

  3. Find  `(dL)/(d theta)`, and hence show that  `(dL)/(d theta) = 0`  when  `BD = 2CD`.   (2 marks)

  4. Find the maximum value of `L` if  `a = 3 sqrt 5`.   (1 mark)

Show Answers Only
  1. `BD = a cos theta,\ \ \ CD = a sin theta`
  2. `L = 4 + a + 2 a cos theta + a sin theta`
  3. `text(Proof)\ text{(See Worked Solutions)}`
  4. `L_max = 19 + 3 sqrt 5`
Show Worked Solution

a.   `text(In)\ \ Delta BCD,`

`cos theta` `= (BD)/a`
`:. BD` `= a cos theta`
`sin theta` `= (CD)/a`
`:. CD` `= a sin theta`

 

b.   `L` `= 4 + 2 BD + CD + a`
    `= 4 + 2a cos theta + a sin theta + a`
    `= 4 + a + 2a cos theta + a sin theta`

 

c.   `text(Noting that)\ a\ text(is a constant:)`

♦ Mean mark 35%.

`(dL)/(d theta)= – 2 a sin theta + a cos theta`

`text(When)\ \ (dL)/(d theta) = 0`,

`- 2 a sin theta+ a cos theta` `= 0`
`a cos theta` `= 2 a sin theta`
`:.  BD` `= 2CD\ \ text{(using part (a))}`

 

d.   `text(SP’s when)\ \ (dL)/(d theta)=0,`

♦♦♦ Mean mark 5%.
`- 2 a sin theta+ a cos theta` `= 0`
`sin theta` `=1/2 cos theta`
`tan theta` `=1/2`

vcaa-2011-meth-10ai

`text(If)\ \ tan theta=1/2,\ \ cos theta = 2/sqrt5,\ \ sin theta = 1/sqrt5`

`L_(max)` `= 4 + a + 2a cos theta + a sin theta`
  `= 4 + (3 sqrt 5) + 2 (3 sqrt 5) (2/sqrt 5) + (3 sqrt 5) (1/sqrt 5)`
  `= 4 + 3 sqrt 5 + 12 + 3`
  `= 19 + 3 sqrt 5\ text(cm)`

Functions, MET1 2011 VCAA 4

If the function `f` has the rule  `f(x) = sqrt (x^2-9)`  and the function `g` has the rule  `g(x) = x + 5`

  1.  find integers `c` and `d` such that  `f(g(x)) = sqrt {(x + c) (x + d)}`  (2 marks)

  2.  state the maximal domain for which `f(g(x))` is defined.   (2 marks)

Show Answers Only
  1. `c = 2, d = 8 or c = 8, d = 2`
  2. `x in (– oo, – 8] uu [– 2, oo)`
Show Worked Solution
a.   `f(g(x))` `= sqrt {(x + 5)^2-9}`
    `= sqrt (x^2 + 10x + 16)`
    `= sqrt {(x + 2) (x + 8)}`

 
 `:. c = 2, d = 8 or c = 8, d = 2`

 

b.   `text(Find)\ x\ text(such that:)`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: “Very poorly answered” with a common response of `[-3,3)` that ignored the information from part (a).

`(x+2)(x+8) >= 0`
 

 vcaa-2011-meth-4ii

`(x + 2) (x + 8) >= 0\ \ text(when)`

`x <= -8 or x >= -2`

`:.\ text(Maximal domain is:)`

`x in (– oo, – 8] uu [– 2, oo)`

Calculus, MET1 2012 VCAA 10

Let  `f: R -> R,\ f(x) = e^(-mx) + 3x`,  where `m` is a positive rational number.

  1.  i. Find, in terms of `m`, the `x`-coordinate of the stationary point of the graph of  `y = f(x).`   (2 marks)

  2. ii. State the values of `m` such that the `x`-coordinate of this stationary point is a positive number.   (1 mark)

  3. For a particular value of `m`, the tangent to the graph of  `y = f(x)`  at  `x =-6`  passes through the origin.
  4. Find this value of `m`.   (3 marks)

Show Answers Only
  1.  i. `1/m log_e(m/3)`
  2. ii. `m > 3`
  3. `1/6`
Show Worked Solution

a.i.   `text(SP’s occur when)\ \ f^{prime}(x)=0`

`-me^(-mx) + 3` `= 0`
`me^(-mx)` `3`
`-mx` `= log_e (3/m)`
`:. x` `=-1/m log_e (3/m)`
  `= 1/m log_e (m/3), \ m>0`

 

♦♦♦ Part (a.ii.) mean mark 18%.
  ii.    `1/m log_e (m/3)` `> 0`
  `log_e (m/3)` `> 0`
  `m/3` `> 1`
  `:. m` `> 3`

 

b.   `text(Point of tangency:)\ \ (-6, e^(-6m)-18)`

♦♦ Mean mark (b) 33%.
MARKER’S COMMENT: Many confused `m` with the more common use of `m` for gradient in  `y=mx+c`.

`text(At)\ \ x=-6,`

`m_tan= f^{prime} (-6)= -me^(-6m) + 3`

`:.\ text(Equation of tangent:)`

`y-y_1` `= m (x-x_1)`
`y-(e^(-6m)-18)` `= (-me^(-6m) + 3) (x-(-6))`

 

`text(S)text{ince tangent passes through (0, 0):}`

`-e^(-6m) + 18` `= (-me^(-6m) + 3)(6)`
`-e^(-6m) + 18` `=-6 me^(-6m) + 18`
`e^(-6m)-6me^(-6m)` `= 0`
`e^(-6m) (1-6m)` `= 0`
`1-6m` `=0`
`:.m` `=1/6`

Calculus, MET1 2014 VCAA 10

A line intersects the coordinate axes at the points `U` and `V` with coordinates `(u, 0)` and `(0, v)`, respectively, where `u` and `v` are positive real numbers and  `5/2 <= u <= 6`.

  1. When `u = 6`, the line is a tangent to the graph of  `y = ax^2 + bx`  at the point `Q` with coordinates `(2, 4)`, as shown.

     


    met1-2014-vcaa-q10
     

     

    If `a` and `b` are non-zero real numbers, find the values of `a` and `b`.   (3 marks)

  2. The rectangle `OPQR` has a vertex at `Q` on the line. The coordinates of `Q` are `(2, 4)`, as shown.

     
    met1-2014-vcaa-q10_1
     

    1. Find an expression for `v` in terms of `u`.   (1 mark)

    2. Find the minimum total shaded area and the value of `u` for which the area is a minimum.   (2 marks)

    3. Find the maximum total shaded area and the value of `u` for which the area is a maximum.   (1 mark)

Show Answers Only
  1. `a = −3/2, b = 5`
    1. `v=(4u)/(u-2)`
    2. `8\ text{u²   (when}\ u=4)`
    3. `17\ text{u²}\ \  (text{when}\ u=5/2)`
Show Worked Solution

a.   `text(S)text{ince (2, 4) lies on parabola:}`

♦ Mean mark 48%.
MARKER’S COMMENT: The most common error incorrectly stated that (6,0) lay on the parabola.
`4` `= a(2)^2 + b(2)`
`4` `= 4a + 2b`
`2` `= 2a + b\ …\ (1)`

 

`y` `=ax^2+bx`
`dy/dx` `=2ax+b`

 

`text(At)\ \ x=2,`

`m_(QU)` `=m_(text(tang))`
`(4-0)/(2-6)` `= 2a(2) + b`
`−1` `= 4a + b\ …\ (2)`

 
`text(Subtract)\ \ (2)-(1)`

`−3 = 2a`

`:. a = −3/2, b = 5`

 

b.i.   `text(Solution 1)`

`text(Using similar triangles:)`

met1-2014-vcaa-q10-answer

`Delta VOU\ text(|||)\ Delta QPU`

♦♦ Mean mark (b.i.) 32%.
`v/4` `= u/(u -2)`
`:. v` `= (4u)/(u -2)`

 

`text(Solution 2)`

`m_(VQ)` `=m_(UQ)`
`(v-4)/(0-2)` `=(0-4)/(u-2)`
`v` `=8/(u-2) + 4`
  `=(8+4(u-2))/(u-2)`
  `=(4u)/(u-2)`

 

♦♦♦ Mean mark (b.ii.) 18%.
b.ii.    `text(Area)` `= 1/2uv-2 xx 4`
    `= 1/2u((4u)/(u -2))-8`
    `= (2u^2)/(u -2)-8`

 
`text(SP occurs when)\ \ (dA)/(du)=0,`

`(4u(u-2)-2u^2)/((u-2)^2)` `= 0`
`2u^2-8u` `= 0`
`2u(u-4)` `= 0`

 

`u = 0quadtext(or)quadu = 4`

`:. u = 4, \ \ u ∈ [5/2,6]`

`text(When)\ u=4,\ \ A` `=(2 xx 4^2)/(4-2)-8`
  `= 8\ text(u²)`

 
`text(Test areas at end points,)`

`text(When)\ u=5/2,\ A=17`

`text(When)\ u=6,\ A=10`

`:. A_(text(min))=8\ text{u²   (when}\ u=4)`
 

b.iii.   `text{As only one (local minimum) stationary point}`

♦♦♦ Mean mark (b.iii.) 8%.

`text(exists over)\ 5/2 <= u <= 6, text(the maximum area)`

`text(must occur at an endpoint.)`

`:. A_(text(max)) = 17\ text{u²}\ \  (text{when}\ u=5/2)`

Calculus, MET1 2015 VCAA 10

The diagram below shows a point, `T`, on a circle. The circle has radius 2 and centre at the point `C` with coordinates `(2, 0)`. The angle `ECT` is `theta`, where  `0 < theta <= pi/2`.
  

met1-2015-vcaa-q10
  

The diagram also shows the tangent to the circle at `T`. This tangent is perpendicular to `CT` and intersects the `x`-axis at point `X` and the `y`-axis at point `Y`.

  1. Find the coordinates of `T` in terms of `theta`.   (1 mark)

  2. Find the gradient of the tangent to the circle at `T` in terms of `theta`.   (1 mark)

  3. The equation of the tangent to the circle at `T` can be expressed as
  4. `qquad cos(theta)x + sin(theta)y = 2 + 2cos(theta)`
  5.  i. Point `B`, with coordinates `(2, b)`, is on the line segment `XY`.
  6.     Find `b` in terms of `theta`.   (1 mark)

  7. ii. Point `D`, with coordinates `(4, d)`, is on the line segment `XY`.
  8.     Find `d` in terms of `theta`.   (1 mark)

  9. Consider the trapezium `CEDB` with parallel sides of length `b` and `d`.
  10. Find the value of `theta` for which the area of the trapezium `CEDB` is a minimum. Also find the minimum value of the area.   (3 marks)

Show Answers Only
  1. `T(2 + 2costheta, 2 sintheta)`
  2. `(-1)/(tan(theta))`
  3.  i. `2/(sintheta)`
  4. ii. `(2-2 costheta)/(sintheta)`
  5. `theta = pi/3`
  6. `A_text(min) = 2sqrt3\ text(u²)`
Show Worked Solution
a.    `cos theta` `= (CM)/(CT)`
    `=(CM)/2`
  `CM` `= 2costheta`
♦♦♦ Part (a) mean mark 20%.
MARKER’S COMMENT: Many students did not include the “+2” in the `x`-coordinate.
`sintheta` `= (TM)/2`
`TM` `= 2sintheta`

 

`:. T\ text(has coordinates)\ \ (2 + 2costheta, 2 sintheta)`

 

♦♦♦ Part (b) mean mark 16%.
b.    `m_(CT)` `=(TM)/(CM)`
    `=(2 sin theta)/(2 cos theta)`
    `=tan theta`
     
  `:.m_(XY)` `=-1/tan theta,\ \ \ (CT ⊥ XY)`

 

c.i.   `text(Substitute)\ \ (2,b)\ \ text(into equation:)`

`2costheta + bsintheta` `= 2 + 2costheta`
`:. b` `= 2/(sintheta)`

 

c.ii.   `text(Substitute)\ \ (4,d)\ \ text(into equation:)`

♦ Part (c)(ii) mean mark 47%.
`4costheta + dsintheta` `= 2 + 2costheta`
`d sin theta` `=2-cos theta`
`:.d`  `= (2-2 costheta)/(sintheta)`

 

♦♦♦ Part (d) mean mark 19%.
d.    `text(A)_text(trap)` `= 1/2 xx 2 xx (b + d)`
    `= 2/(sintheta) + (2-2costheta)/(sintheta)`
    `= (4-2costheta)/(sintheta)`

 

`text(Stationary point when)\ \ (dA)/(d theta)=0`

`(2sin^2theta-costheta(4-2costheta))/(sin^2theta)` `= 0`
`2sin^2theta-4costheta + 2cos^2theta` `= 0`
`2[sin^2theta + cos^2theta]-4costheta` `= 0`
`2-4costheta` `= 0`
`costheta` `= 1/2`
`theta` `= pi/3,\ \ \ \ theta ∈ (0, pi/2)`

 

`A(pi/3)` `= (4-2(1/2))/(sqrt3/2)`
  `=3 xx 2/sqrt3`
  `= 2sqrt3`

 

`:. A_text(min) = 2sqrt3\ text(u²)`

Calculus, MET2 2010 VCAA 4

Consider the function  `f: R -> R,\ f(x) = 1/27 (2x-1)^3 (6-3x) + 1.`

  1. Find the `x`-coordinate of each of the stationary points of  `f` and state the nature of each of these stationary points.   (4 marks)

In the following, `f` is the function  `f: R -> R,\ f(x) = 1/27 (ax-1)^3 (b-3x) + 1` where `a` and `b` are real constants.

  1. Write down, in terms of `a` and `b`, the possible values of `x` for which `(x, f (x))` is a stationary point of `f`.   (3 marks)

  2. For what value of `a` does `f` have no stationary points?   (1 mark)

  3. Find `a` in terms of `b` if `f` has one stationary point.   (2 marks)

  4. What is the maximum number of stationary points that `f` can have?  (1 mark)

  5. Assume that there is a stationary point at `(1, 1)` and another stationary point `(p, p)` where  `p != 1`.
  6. Find the value of `p`.   (3 marks)

Show Answers Only
  1. `text(Point of inflection at)\ \ x = 1/2`
    `text(Local max at)\ \ x = 13/8`
  2. `x = (ab + 1)/(4a) or x = 1/a`
  3. `0`
  4. `3/b`
  5. `2`
  6. `4`
Show Worked Solution

a.   `text(S.P. occurs when)\ \ f^{′}(x) = 0`

`f(x)` `=1/27 (2x-1)^3 (6-3x) + 1`
`f^{′}(x)` `=- 1/9 (2x-1)^2 (8x-13) `

 
`text(Solve:)\ \ f^{′}(x)=0\ \ text(for)\ x,`

`:. x = 1/2 or x = 13/8`
 

 `text(Sketch the graph:)`

vcaa-graphs-fur2-2010-4ai

`=>\ text(Point of inflection at)\ \ x = 1/2`

`=>\ text(Local max at)\ \ x = 13/8`
 

b.   `text(S.P. occurs when)\ \ f prime (x) = 0`

♦ Mean mark (b) 46%.
MARKER’S COMMENT: Issues with use of CAS caused significant difficulties in this question.
`f(x)` `=1/27 (ax-1)^3 (b-3x) + 1`
`f^{′}(x)` `=1/9 (ax-1)^2 (ab+1-4ax)`

 
`text(Solve:)\ \ f^{′}(x) = 0\ \ text(for)\ \ x,`

`:. x = (ab + 1)/(4a)\ \ \text{or}\  \ x = 1/a`
 

c.   `text(For)\ \ x = (ab + 1)/(4a) or x = 1/a\ \ text(to exist,)`

♦ Mean mark part (c) 47%.

`a != 0`

`:.\ text(No stationary points when)\ \ a = 0`
 

d.   `text(If there is 1 S.P.,)`

♦♦♦ Mean mark (d) 16%.
`(ab + 1)/(4a)` `= 1/a`
`:. a` `= 3/b`

 

♦ Mean mark (e) 37%.

e.   `text(The maximum number of SP’s for a quartic)`

`text(polynomial is 3. In the function given, one of)`

`text(the SP’s is a point of inflection.)`

`:.f(x)\ \ text(has a maximum of 2 SP’s.)`

 

f.   `text{Solution 1 [by CAS]}` 

`text(Define)\ \ f(x) = 1/27 (x-1)^3 (b-3x) + 1`

`text(Solve:)\ \ f(p) = p, f^{′}(1) = 1  and f^{′}(p) = p\ \ text(for)\ \ p,`

`:. p = 4,\  \ \ (p != 1)`
 

`text(Solution 2)`

`text(SP’s occur at)\ \ (1,1) and (p,p),\ \ text(where,)`

♦♦♦ Mean mark (f) 9%.

`x = (ab + 1)/(4a) or x = 1/a`

`text(Consider)\ \ p=1/a,`

`f(p)` `=f(1/a)`
 

`=1/27 (a*1/a-1)^3(b-3*1/a)+1=1`
   

`f(p)=1,\ \ text(SP at)\ (1,1) and p!=1`

`=> p!=1/a`
 

`text(Consider)\ \ 1=1/a,`

`=> a=1 and  b=4p-1`

`f(1)=1`

`f(p)=p`

`1/27 (p-1)^3(4p-1-3p)+1` `=p`
`1/27(p-1)^4-(p-1)` `=0`
`(p-1)(1/27(p-1)^3-1)` `=0`
`(p-1)^3` `=27`
`p` `=4`