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Functions, EXT1 F1 2010 HSC 3b*

Let  `f(x) = e^(-x^2)`.  The diagram shows the graph  `y = f(x)`.
 

 Inverse Functions, EXT1 2010 HSC 3b

  1. The graph has two points of inflection.  

     

    Find the  `x`  coordinates of these points.   (3 marks)

  2. Explain why the domain of  `f(x)`  must be restricted if  `f(x)`  is to have an inverse function.    (1 mark)

  3. Find a formula for  `f^(-1) (x)`  if the domain of  `f(x)`  is restricted to  `x ≥ 0`.   (2 marks)

  4. State the domain of  `f^(-1) (x)`.    (1 mark)

  5. Sketch the curve  `y = f^(-1) (x)`.    (1 mark)

Show Answers Only
  1. `x = +- 1/sqrt2` 
  2. `text(There can only be 1 value of)\ y\ text(for each value of)\ x.`
  3. `f^(-1)x = sqrt(ln(1/x))`
  4. `0 <= x <= 1`
  5.  
  6. Inverse Functions, EXT1 2010 HSC 3b Answer
Show Worked Solution
i.    `y` `= e^(-x^2)`
  `dy/dx` `= -2x * e^(-x^2)`
  `(d^2y)/(dx^2)` `= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
    `= 4x^2 e^(-x^2)\ – 2e^(-x^2)`
    `= 2e^(-x^2) (2x^2\ – 1)`

 

`text(P.I. when)\ \ (d^2y)/(dx^2) = 0`

`2e^(-x^2) (2x^2\ – 1)` `= 0` 
 `2x^2\ – 1` `= 0` 
 `x^2` `= 1/2`
 `x` `= +- 1/sqrt2` 
COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.
`text(When)\ \ ` `x < 1/sqrt2,` `\ (d^2y)/(dx^2) < 0`
  `x > 1/sqrt2,` `\ (d^2y)/(dx^2) > 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = 1/sqrt2`

`text(When)\ \ ` `x < – 1/sqrt2,` `\ (d^2y)/(dx^2) > 0`
  `x > – 1/sqrt2,` `\ (d^2y)/(dx^2) < 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = – 1/sqrt2`

 

ii.   `text(In)\ f(x), text(there are 2 values of)\ y\ text(for)`
  `text(each value of)\ x.`
  `:.\ text(The domain of)\ f(x)\ text(must be restricted)`
  `text(for)\ \ f^(-1) (x)\ text(to exist).`

 

iii.   `y = e^(-x^2)`

`text(Inverse function can be written)` 

`x` `= e^(-y^2),\ \ \ x >= 0`
`lnx` `= ln e^(-y^2)`
`-y^2` `= lnx`
`y^2` `= -lnx`
  `=ln(1/x)`
`y` `= +- sqrt(ln (1/x))`

 

`text(Restricting)\ \ x>=0,\ \ =>y>=0`

`:.  f^(-1) (x)=sqrt(ln (1/x))`
 

♦ Parts (iv) and (v) were poorly answered with mean marks of 39% and 49% respectively.
iv.   `f(0) = e^0 = 1`

`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`

`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`

 

v. 

Inverse Functions, EXT1 2010 HSC 3b Answer

Functions, EXT1 F1 2004 HSC 5b*

The diagram below shows a sketch of the graph of  `y = f(x)`, where  `f(x) = 1/(1 + x^2)`  for  `x ≥ 0`.
 

Inverse Functions, EXT1 2004 HSC 5b

  1. On the same set of axes, sketch the graph of the inverse function,  `y = f^(−1)(x)`.  (1 mark)
  2. State the domain of  `f^(−1)(x)`.  (1 mark)

  3. Find an expression for  `y = f^(−1)(x)`  in terms of  `x`.  (2 marks)

  4. The graphs of  `y = f(x)`  and  `y = f^(−1)(x)`  meet at exactly one point  `P`.

     

    Let  `α`  be the `x`-coordinate of  `P`. Explain why  `α`  is a root of the equation  `x^3 + x − 1 = 0`.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0 < x ≤ 1`
  3. `y = sqrt((1 − x)/x), \ y > 0`
  4. `text(See Worked Solutions)`
Show Worked Solution
i.  

Inverse Functions, EXT1 2004 HSC 5b Answer

ii.   `text(Domain of)\ \ f^(−1)(x)\ text(is)`

`0 < x ≤ 1`
 

iii.  `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\ \ x↔y`

`x` `= 1/(1 + y^2)`
`x(1 + y^2)` `= 1`
`1 + y^2` `= 1/x`
`y^2` `= 1/x − 1`
  `= (1 − x)/x`
`y` `= ± sqrt((1 − x)/x)`

 

`:.y = sqrt((1 − x)/x), \ \ y >= 0`

 

iv.   `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`

`text(i.e. where)`

`1/(1 + x^2)` `= x`
`1` `= x(1 + x^2)`
`1` `= x + x^3`
`x^3 + x − 1` `= 0`

 

`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`

 `text(it is a root of)\ \ \ x^3 + x − 1 = 0`

Functions, 2ADV F1 SM-Bank 15 MC

If the equation  `f(2x) - 2f(x) = 0`  is true for all real values of `x`, then `f(x)` could equal

A.   `x^2/2`

B.   `sqrt (2x)`

C.   `2x`

D.   `x - 2`

Show Answers Only

`C`

Show Worked Solution

`text(We need)\ \ f(2x)=2\ f(x),`

`text(Consider)\ C,`

`f(x)` `=2x,`
`f(2x)` `= 2(2x)`
  `= 2\ f(x)`

 
`=>   C`

Functions, 2ADV F1 SM-Bank 13 MC

Which one of the following functions satisfies the functional equation  `f (f(x)) = x`?

A.   `f(x) = 2 - x`

B.   `f(x) = x^2`

C.   `f(x) = 2 sqrt x`

D.   `f(x) = x - 2`

Show Answers Only

`A`

Show Worked Solution

`text(By trial and error,)`

`text(Consider:)\ \ f(x)=2-x`

`f(f(x))` `=2-(2-x)`
  `=x`

 
`=>  A`

Functions, 2ADV F1 SM-Bank 12 MC

If  `f(x - 1) = x^2 - 2x + 3`, then  `f(x)` is equal to

A.   `x^2 - 2`

B.   `x^2 + 2`

C.   `x^2 - 2x + 4`

D.   `x^2 - 4x + 6`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ g(x) = f(x – 1)`

`g(x+1) = f(x)`

`g(x + 1)` `= (x + 1)^2 – 2(x + 1) + 3`
  `=x^2+2x+1-2x-2+3`
  `= x^2 + 2`

`=>   B`

Functions, 2ADV F1 SM-Bank 8 MC

Let  `f (x) = x^2`

Which one of the following is not true?

A.   `f(xy) = f (x) f (y)`

B.   `f(x) - f(-x) = 0`

C.   `f (2x) = 4 f (x)`

D.   `f (x - y) = f(x) - f(y)`

Show Answers Only

`D`

Show Worked Solution

`text(By trial and error,)`

`text(Consider option)\ D:`

`f(x-y)` `=(x-y)^2`
  `=x^2 -2xy+y^2`
`f(x)-f(y)` `= x^2-y^2`
   
`:.f(x-y)` `!=f(x)-f(y)`

 
`=>D`

Functions, 2ADV F1 SM-Bank 7

Let  `f(x) = log_e(x)`  for  `x>0,`  and  `g (x) = x^2 + 1`  for  all `x`.

  1. Find  `h(x)`, where  `h(x) = f (g(x))`.  (1 mark)

  2. State the domain and range of  `h(x)`.  (2 marks)

  3. Show that  `h(x) + h(−x) = f ((g(x))^2 )`.  (2 marks)
Show Answers Only
  1. `log_e(x^2 + 1)`
  2. `text(Domain)\ (h):\ text(all)\ x`

     

    `text(Range)\ h(x):\ \  h>=0`

  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `h(x)` `= f(x^2 + 1)`
    `= log_e(x^2 + 1)`

 

ii.   `text(Domain)\ (h) =\ text(Domain)\ (g):\ text(all)\ x`

♦♦ Mean mark part (a)(ii) 30%.
`=> x^2 + 1 >= 1`
`=> log_e(x^2 + 1) >= 0`

 
`:.\ text(Range)\ h(x):\ \  h>=0`

 

MARKER’S COMMENT: Many students were unsure of how to present their working in this question. Note the layout in the solution.
iii.   `text(LHS)` `= h(x) + h(−x)`
    `= log_e(x^2 _ 1) + log_e((−x)^2 + 1)`
    `= log_e(x^2 + 1) + log_e(x^2 + 1)`
    `= 2log_e(x^2 + 1)`

 

`text(RHS)` `= f((x^2 + 1)^2)`
  `= 2log_e(x^2 + 1)`

 
`:. h(x) + h(−x) = f((g(x))^2)\ \ text(… as required)`

Functions, 2ADV F1 SM-Bank 3

Let  `f(x) = sqrt(x + 1)`   for   `x>=0`

  1. State the range of  `f(x)`.   (1 mark)

  2. Let  `g(x)=x^2+4x+3`,  where  `x<=c`  and  `c<=0`.
  3. Find the largest possible value of `c` such that the range of `g(x)` is a subset of the domain of `f(x)`.   (2 marks)

Show Answers Only
  1.   `y>= 1`
  2.  `-3`
Show Worked Solution

i.  `text(Sketch of)\ \ f(x):`
 

`:.\ text(Range:)\ \ y>= 1`

 

ii.  `text(Sketch)\ \ g(x) = (x + 1) (x + 3)`

 

 

 

 

 

 

`text(Domain of)\ \ f(x):\ \ x>=0`

`text(Find domain of)\ g(x)\ text(such that range)\ g(x):\ y>=0`

`text(Graphically, this occurs when)\ \ g(x)\ \ text(has domain:)`

`x <= –3\ \ text(and)\ \ x>=-1`
 

`:. c = -3`

Calculus, 2ADV C4 2007* HSC 10a

An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are  `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for  `t >= 6`.
 


 

  1. The object is initially at the origin. During which time(s) is the displacement of the object decreasing?  (1 mark)

  2. If the object travels 7 units in the first 4 seconds, estimate the time at which the object returns to the origin. Justify your answer.  (2 marks)

  3. Sketch the displacement, `x`, as a function of time.  (2 marks)

Show Answers Only
  1. `t > 5\ \ text(seconds)`
  2. `7.2\ \ text(seconds)`
  3.    
Show Worked Solution

i.  `text(Displacement is reducing when the velocity is negative.)`

`:. t > 5\ \ text(seconds)`

 

ii.  `text(At)\ B,\ text(the displacement) = 7\ text(units)`

`text(Considering displacement from)\ B\ text(to)\ D:`

`text(S)text(ince the area below the graph from)`

`B\ text(to)\ C\ text(equals the area above the)`

`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`

`text(in displacement from)\ B\ text(to)\ D.`

 

`text(Considering)\ t >= 6`

`text(Time required to return to origin)`

`t` `= d/v`
  `= 7/5`
  `= 1.4\ \ text(seconds)`

 

`:.\ text(The particle returns to the origin after 7.4 seconds.)`
   

iii.   

Statistics, NAPX-K2-04 SA

Sharnie records the number of items she recycles in one week.

The tally table shows Sharnie's results.
 

 
How many items does Sharnie recycle in total?
Show Answers Only

`25`

Show Worked Solution
`text(Total items)` `= 9 + 12 + 3 + 1`
  `= 25`

Probability, 2ADV S1 2017 MET1 8

For events `A` and `B` from a sample space, `P\ (A text(|)B) = 1/5`  and  `P\ (B text(|)A) = 1/4`.  Let  `P\ (A nn B) = p`.

  1. Find  `P\ (A)` in terms of `p`.  (1 mark)

  2. Find  `P\ (A prime nn B prime)` in terms of `p`.  (2 marks)

  3. Given that  `P\ (A uu B) <= 1/5`, state the largest possible interval for `p`.  (2 marks)

Show Answers Only
  1. `P (A) = 4p,\ \ p > 0`
  2. `1 – 8p`
  3. `0 < p <= 1/40`
Show Worked Solution
i.    `P\ (A)` `=(P\ (A nn B))/(P\ (B text(|) A))`
    `=p/(1/4)`
    `=4p`

 

ii.  `text(Consider the Venn diagram:)`

♦ Mean mark 40%.
MARKER’S COMMENT: The most successful answers used a Venn diagram or table.
 

`P\ (A′ nn B′) = 1 – 8p`

 

iii.  `text(Given)\ P(A uu B) = 8p`

♦ Mean mark 37%.

`=> 0 < 8p <= 1/5`

`:. 0 < p <= 1/40`

Probability, 2ADV S1 2014 MET1 9

Sally aims to walk her dog, Mack, most mornings. If the weather is pleasant, the probability that she will walk Mack is `3/4`, and if the weather is unpleasant, the probability that she will walk Mack is `1/3`.

Assume that pleasant weather on any morning is independent of pleasant weather on any other morning.

  1. In a particular week, the weather was pleasant on Monday morning and unpleasant on Tuesday morning.

     

     

    Find the probability that Sally walked Mack on at least one of these two mornings.  (2 marks)

  2. In the month of April, the probability of pleasant weather in the morning was `5/8`.
    1. Find the probability that on a particular morning in April, Sally walked Mack.  (2 marks)

    2. Using your answer from part b.i., or otherwise, find the probability that on a particular morning in April, the weather was pleasant, given that Sally walked Mack that morning.  (2 marks)

Show Answers Only
  1. `5/6`
  2. i. `19/32`
     
    ii. `15/19`
Show Worked Solution
a.    `Ptext{(at least 1 walk)}` `= 1 – Ptext{(no walk)}`
    `= 1 – 1/4 xx 2/3`
    `= 5/6`

 

b.i.   `text(Construct tree diagram:)`
 

met1-2014-vcaa-q9-answer1 
 

`P(PW) + P(P′W)` `= 5/8 xx 3/4 + 3/8 xx 1/3`
  `= 19/32`

 

♦ Part (b)(ii) mean mark 38%.
b.ii.    `P(P | W)` `= (P(P ∩ W))/(P(W))`
    `= (5/8 xx 3/4)/(19/32)`
    `= 15/32 xx 32/19`
    `= 15/19`

Probability, 2ADV S1 2011 MET1 8

Two events, `A` and `B`, are such that  `P(A) = 3/5`  and  `P(B) = 1/4.`

If  `A^{′}` denotes the compliment of `A`, calculate  `P (A^{′} nn B)` when

  1. `P (A uu B) = 3/4`  (2 marks)

  2. `A` and `B` are mutually exclusive.  (1 mark)

Show Answers Only
  1. `3/20`
  2. `1/4`
Show Worked Solution

i.   `text(Sketch Venn Diagram)`

vcaa-2011-meth-8i

`P (A uu B)` `= P (A) + P (B)-P (A nn B)`
`3/4` `= 3/5 + 1/4-P (A nn B)`
`P (A nn B)` `= 1/10`

 

 `:.\ P(A^{′} nn B) = 1/4-1/10 = 3/20`

 

ii.   vcaa-2011-meth-8ii

`P (A∩ B)=0\ \ text{(mutually exclusive)},`

`:.\ P (A^{′} nn B) = P (B) = 1/4`

Probability, 2ADV S1 2007 MET1 11

There is a daily flight from Paradise Island to Melbourne. The probability of the flight departing on time, given that there is fine weather on the island, is 0.8, and the probability of the flight departing on time, given that the weather on the island is not fine, is 0.6.

In March the probability of a day being fine is 0.4.

Find the probability that on a particular day in March

  1. the flight from Paradise Island departs on time  (2 marks)

  2. the weather is fine on Paradise Island, given that the flight departs on time.  (2 marks)

Show Answers Only
  1. `0.68`
  2. `8/17`
Show Worked Solution
i.   

 
`Ptext{(FT)}\ +\ Ptext{(F′T)}`

`= 0.4 xx 0.8 + 0.6 xx 0.6`

`= 0.32 + 0.36`

`= 0.68`

 

ii.   `text(Conditional probability:)`

♦♦ Mean mark 29%.
MARKER’S COMMENT: Students continue to struggle with conditional probability. Attention required here.
`P(F\ text(|)\ T)` `= (P(F ∩ T))/(P(T))`
  `= 0.32/0.68`

 

`:. P(F\ text(|)\ T) = 8/17`

Probability, 2ADV S1 2015 MET1 8

For events `A` and `B` from a sample space, `P(A | B) = 3/4`  and  `P(B) = 1/3`.

  1.  Calculate  `P(A ∩ B)`.  (1 mark)

  2.  Calculate  `P(A′ ∩ B)`, where `A′` denotes the complement of `A`.  (1 mark)

  3.  If events `A` and `B` are independent, calculate  `P(A ∪ B)`.  (1 mark)

Show Answers Only
  1. `1/4`
  2. `1/12`
  3. `5/6`
Show Worked Solution

i.   `text(Using Conditional Probability:)`

`P(A | B)` `= (P(A ∩ B))/(P(B))`
`3/4` `= (P(A ∩ B))/(1/3)`
`:. P(A ∩ B)` `= 1/4`

 

ii.    met1-2015-vcaa-q8-answer
`P(A′ ∩ B)` `= P(B) – P(A ∩B)`
  `= 1/3 – 1/4`
  `= 1/12`

 

iii.   `text(If)\ A, B\ text(independent)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: A lack of understanding of independent events was clearly evident.
`P(A ∩ B)` `= P(A) xx   P(B)`
`1/4` `= P(A) xx 1/3`
`:. P(A)` `= 3/4`

 

`P(A ∪ B)` `= P(A) + P(B) – P(A ∩ B)`
  `= 3/4 + 1/3 – 1/4`
`:. P(A ∪ B)` `= 5/6`

Probability, 2ADV S1 2009 MET1 5

Four identical balls are numbered 1, 2, 3 and 4 and put into a box. A ball is randomly drawn from the box, and not returned to the box. A second ball is then randomly drawn from the box.

  1. What is the probability that the first ball drawn is numbered 4 and the second ball drawn is numbered 1?  (1 mark)

  2. What is the probability that the sum of the numbers on the two balls is 5?  (1 mark)

  3. Given that the sum of the numbers on the two balls is 5, what is the probability that the second ball drawn is numbered 1?  (2 marks)

Show Answers Only
  1. `1/12`
  2. `1/3`
  3. `1/4`
Show Worked Solution
i.    `P (4, 1)` `= 1/4 xx 1/3`
    `= 1/12`

 

ii.   `P (text(Sum) = 5)`

`= P (1, 4) + P (2, 3) + P (3, 2) + P (4, 1)`

`= 4 xx (1/4 xx 1/3)`

`= 1/3`

 

iii.   `text(Conditional Probability)`

♦ Mean mark 46%.

`P (2^{text(nd)} = 1\ |\ text(Sum) = 5)`

`= (P (4, 1))/(P (text(Sum) = 5))`

`= (1/12)/(1/3)`

`= 1/4`

Probability, 2ADV S1 2013 MET2 10 MC

For events `A` and `B,\ P(A ∩ B) = p,\ P(A^{′}∩ B) = p -1/8`  and  `P(A ∩ B^{′}) = (3p)/5.`

If `A` and `B` are independent, then the value of  `p`  is

  1. `0`
  2. `1/4`
  3. `3/8`
  4. `1/2`
Show Answers Only

`C`

Show Worked Solution
`P(A)` `= P(A ∩ B) + P(A ∩ B^{′})`
  `= p + (3p)/5`
  `= (8p)/5`

 

`P(B)` `= P(B ∩ A) + P(B ∩ A^{′})`
  `= p + p-1/8`
  `= 2p – 1/8`

 
`text(S)text(ince)\ A and B\ text(are independent events,)`

`P(A ∩ B)` ` = P(A) xx P(B)`
`p` `=(8p)/5 (2p-1/8)`
`5p` `=16p^2-p`
`16p^2-6p` `=0`
`2p(8p-3)` `=0`
`:.p` `=3/8,\ \ \ p!=0`

 
`=>   C`

Probability, 2ADV S1 2012 MET2 13 MC

`A` and `B` are events of a sample space `S.`

`P(A nn B) = 2/5`  and  `P(A nn B^(′)) = 3/7`

`P(B^(′) | A)`  is equal to

  1. `6/35`
  2. `15/29`
  3. `14/35`
  4. `29/35`
Show Answers Only

`B`

Show Worked Solution

met2-2012-vcaa-13-mc-answer

`P(B ^(′) | A)` `= (P(B^(′) nn A))/(P(A))`
  `= (P(B^(′) nn A))/(P(B^(′) nn A)+ P(A nn B))`
  `= (3/7)/(3/7 + 2/5)`
  `= 15/29`

`=>   B`

Probability, 2ADV S1 2013 MET1 7

The probability distribution of a discrete random variable, `X`, is given by the table below.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \ &\ \ \ 4\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.6p^{2} & 0.1 & 1-p & 0.1 \\
\hline
\end{array}

  1. Show that  `p = 2/3 or p = 1`.  (3 marks)

  2. Let  `p = 2/3`.

    1. Calculate  `E(X)`.  Answer in exact form.  (2 marks)

    2. Find  `P(X >= E(X))`.  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
    1. `28/15`
    2. `8/15`
Show Worked Solution

a.   `text(S)text(ince probabilities must sum to 1:)`

`0.2 + 0.6p^2 + 0.1 + 1-p + 0.1` `= 1`
`0.6p^2-p + 0.4` `= 0`
`6p^2-10p + 4` `= 0`
`3p^2-5p + 2` `= 0`
`(p-1) (3p-2)` `= 0`

`:. p = 1 or p = 2/3`

 

b.i.   `E(X)` `= sum x * P (X = x)`
    `= 1 xx (3/5 xx 2^2/3^2) + 2 (1/10) + 3 (1-2/3) + 4 (1/10)`
    `= 4/15 + 1/5 + 1 + 2/5`
    `= 28/15`

 

♦♦ Part (b)(ii) mean mark 32%.
  ii.   `P(X >= 28/15)` `=P(X = 2) + P(X = 3) + P(X = 4)`
    `= 1/10 + 1/3 + 1/10`
    `= 8/15`

Probability, 2ADV S1 2012 MET1 4

On any given day, the number `X` of telephone calls that Daniel receives is a random variable with probability distribution given by

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.2 & 0.5 & 0.1 \\
\hline
\end{array}

  1. Find the mean of  `X`.   (2 marks)

  2. What is the probability that Daniel receives only one telephone call on each of three consecutive days?   (1 mark)

  3. Daniel receives telephone calls on both Monday and Tuesday.

     

    What is the probability that Daniel receives a total of four calls over these two days?   (3 marks)

Show Answers Only
  1. `1.5`
  2. `0.008`
  3. `29/64`
Show Worked Solution
i.    `E(X)` `= 0 xx 0.2 + 1 xx 0.2 + 2 xx 0.5 + 3 xx 0.1`
    `= 0 + .2 + 1 + 0.3`
    `= 1.5`

 

ii.   `P(1, 1, 1)` `= 0.2 xx 0.2 xx 0.2`
    `= 0.008`

 

iii.   `text(Conditional Probability:)`

♦ Mean mark 36%.

`P(x = 4 | x >= 1\ text{both days})`

`= (P(1, 3) + P(2, 2) + P(3, 1))/(P(x>=1\ text{both days}))`

`= (0.2 xx 0.1 + 0.5 xx 0.5 + 0.1 xx 0.2)/(0.8 xx 0.8)`

`= (0.02 + 0.25 + 0.02)/0.64`

`= 0.29/0.64`

`= 29/64`

Probability, 2ADV S1 2010 MET1 8

The discrete random variable `X` has the probability distribution

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ -1\ \ \ \  & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ \ \ 1\ \ \ \ \  & \ \ \ \ \ 2\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p^{2} & p^{2} & \dfrac{p}{4} & \dfrac{4p+1}{8} \\
\hline
\end{array}

Find the value of `p.`  (3 marks)

Show Answers Only

`1/2`

Show Worked Solution
`text(Sum of probabilities)` `= 1`
`p^2 + p^2 + p/4 + (4p + 1)/8` `= 1`
`16p^2 + 2p + 4p + 1` `= 8`
`16p^2 + 6p – 7` `= 0`
`(2p – 1) (8p + 7)` `= 0`

 

`:. p = 1/2,\ \ \ (p>0)`

Probability, 2ADV S1 2009 MET1 7

The random variable `X` has this probability distribution.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  &\ \ \ 4\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \\
\hline
\end{array}

Find

  1.  `P (X > 1 | X <= 3)`  (2 marks)

  2.  `P (X),` the variance of  `X.`  (3 marks)
Show Answers Only
  1. `2/3`
  2. `1.2`
Show Worked Solution

i.   `P(X > 1 | X <= 3)`

`= (P(X = 2) + P(X = 3))/(1-P(X = 4))`

`= (0.4 + 0.2)/(1-0.1)`

`= 0.6/0.9`

`= 2/3`

 

ii.   `E(X)` `= 0.1 (0) + 1 (0.2) + 2 (0.4) + 3 (0.2) + 4 (0.1)`
  `= 0 + 0.2 + 0.8 + 0.6 + 0.4`
  `= 2`

 

`E(X^2)` `= 0^2 (0.1) + 1^2 (0.2) + 2^2 (0.4) + 3^2 (0.2) + 4^2 (0.1)`
  `= 0 + 0.2 + 1.6 + 1.8 + 1.6`
  `= 5.2`

 

`:.\ text(Var) (X)` `= E(X^2)-[E(X)]^2`
  `= 5.2-(2)^2`
  `= 1.2`

Probability, 2ADV S1 2008 MET1 7

Jane drives to work each morning and passes through three intersections with traffic lights. The number `X` of traffic lights that are red when Jane is driving to work is a random variable with probability distribution given by

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.1 & 0.2 & 0.3 & 0.4 \\
\hline
\end{array}

  1. What is the mode of  `X`?  (1 mark)

  2. Jane drives to work on two consecutive days. What is the probability that the number of traffic lights that are red is the same on both days?  (2 marks)

Show Answers Only
  1. `3`
  2. `0.3`
Show Worked Solution

i.   `3`

 

ii.   `P(0,0) + P(1,1) + P(2,2) + P(3,3)`

`= 0.1^2 + 0.2^2 + 0.3^2 + 0.4^2`

`= 0.3`

Probability, 2ADV S1 2009 MET2 10 MC

The discrete random variable `X` has a probability distribution as shown.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.4 & 0.2 & 0.3 & 0.1 \\
\hline
\end{array}

The median of `X` is

  1. `0`
  2. `1`
  3. `1.1`
  4. `2`
Show Answers Only

`B`

Show Worked Solution
`P(X <= 0)` `= 0.4`
`P(X <= 1)` `= 0.6`

 
`:.\ text(Median) = 1`

`=>   B`

Probability, 2ADV S1 2017 MET2 14 MC

The random variable `X` has the following probability distribution, where  `0 < p < 1/3`.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ -1\ \ \  & \ \ \ \ 0\ \ \ \  & \ \ \ \ 1\ \ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p & 2p & 1-3p \\
\hline
\end{array}

The variance of `X` is

  1. `2p(1-3p)`
  2. `p(5-9p)`
  3. `(1-3p)^2`
  4. `6p-16p^2`
Show Answers Only

`D`

Show Worked Solution
`text(Var)(X)` `= E(X^2)-[E(X)]^2`
  `= [(-1)^2p + 0^2 xx 2p + 1^2(1-3p)]-[-p + 0 + 1-3p]^2`
  `= 6p-16p^2`

 
`=> D`

Calculus, 2ADV C4 2018* HSC 15c

The shaded region is enclosed by the curve  `y = x^3 - 7x`  and the line  `y = 2x`, as shown in the diagram. The line  `y = 2x`  meets the curve  `y = x^3 - 7x`  at  `O(0, 0)`  and  `A(3, 6)`. Do NOT prove this.
 


 

  1.  Use integration to find the area of the shaded region.  (2 marks)

  2. Use the Trapezoidal rule and four function values to approximate the area of the shaded region.  (2 marks)

The point `P` is chosen on the curve  `y = x^3 − 7x`  so that the tangent at `P` is parallel to the line  `y = 2x`  and the `x`-coordinate of `P` is positive

  1.  Show that the coordinates of `P` are  `(sqrt 3, -4 sqrt 3)`.  (2 marks)

  2.  Using the perpendicular distance formula  `|ax_1 + by_1 + c|/sqrt(a^2 + b^2)`,  find the area of  `Delta OAP`.  (2 marks)

Show Answers Only
  1. `81/4\ text(units²)`
  2. `18\ text(u²)`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `9 sqrt 3\ text(units²)`
Show Worked Solution
i.   `text(Area)` `= int_0^3 2x – (x^3 – 7x)\ dx`
    `= int_0^3 9x – x^3\ dx`
    `= [9/2 x^2 – 1/4 x^4]_0^3`
    `= [(9/2 xx 3^2 – 1/4 xx 3^4) – 0]`
    `= 81/2 – 81/4`
    `= 81/4\ text(units²)`

 

ii.  `f(x) = 9x – x^3`

`text(Area)` `~~ 1/2[0 + 2(8 + 10) + 0]`
  `~~ 1/2(36)`
  `~~ 18\ text(u²)`

 

iii.   `y = x^3 – 7x`

`(dy)/(dx) = 3x^2 – 7`

`text(Find)\ \ x\ \ text(such that)\ \ (dy)/(dx) = 2`

`3x^2 – 7` `= 2`
`3x^2` `= 9`
`x^2` `= 3`
`x` `= sqrt 3 qquad (x > 0)`

 

`y` `= (sqrt 3)^3 – 7 sqrt 3`
  `= 3 sqrt 3 – 7 sqrt 3`
  `= -4 sqrt 3`

 
`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`

 

iv.  

 

`text(dist)\ OA` `= sqrt((3 – 0)^2 + (6 – 0)^2)`
  `= sqrt 45`
  `= 3 sqrt 5`

 
`text(Find)\ _|_\ text(distance of)\ \ P\ \ text(from)\ \ OA`

`P(sqrt 3, -4 sqrt 3),\ \ 2x – y=0`

`_|_\ text(dist)` `= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|`
  `= (6 sqrt 3)/sqrt 5`
   
`:.\ text(Area)` `= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5`
  `= 9 sqrt 3\ text(units²)`

Calculus, 2ADV C4 2017* HSC 14b

  1. Find the exact value of  `int_0^(pi/3) cos x\ dx`.  (1 mark)

  2. Using the Trapezoidal rule with three function values, find an approximation to the integral  `int_0^(pi/3) cos x\ dx,` leaving your answer in terms of `pi` and `sqrt 3`.  (2 marks)

  3. Using parts (i) and (ii), show that  `pi ~~ (12 sqrt 3)/(3 + 2 sqrt 3)`.  (1 mark)

Show Answers Only
  1. `sqrt 3/2`
  2. `((2sqrt3 + 3)pi)/24`
  3. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i.   `int_0^(pi/3) cos x\ dx` `= [sin x]_0^(pi/3)`
    `= sin\ pi/3-0`
    `= sqrt 3/2`

 

ii.  

\begin{array} {|l|c|c|c|}\hline
x & \ \ \ 0\ \ \  & \ \ \ \dfrac{\pi}{6}\ \ \  & \ \ \ \dfrac{\pi}{3}\ \ \ \\ \hline
\text{height} & 1 & \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ \hline
\text{weight} & 1 & 2 & 1  \\ \hline \end{array}

`int_0^(pi/3) cos x\ dx` `~~ 1/2 xx pi/6[1 + 2(sqrt3/2) + 1/2]`
  `~~ pi/12((3 + 2sqrt3)/2)`
  `~~ ((3+2sqrt3)pi)/24`

 

♦ Mean mark part (iii) 49%.

(iii)    `((3+2sqrt3)pi)/24` `~~ sqrt3/2`
  `:. pi` `~~ (24sqrt3)/(2(3+2sqrt3))`
    `~~ (12sqrt3)/(3 + 2sqrt3)`

Calculus, 2ADV C4 2011* HSC 5c

The table gives the speed `v` of a jogger at time `t` in minutes over a  20-minute period. The speed `v` is measured in metres per minute, in intervals of 5 minutes.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ t\ \ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ &\ \ \ 5\ \ \ &\ \ \ 10\ \ \ &\ \ \ 15\ \ \ &\ \ \ 20\ \ \  \\
\hline
\rule{0pt}{2.5ex} \ \ \ v\ \ \  \rule[-1ex]{0pt}{0pt} & 173 & 81 & 127 & 195 & 168 \\
\hline
\end{array}

The distance covered by the jogger over the 20-minute period is given by  `int_0^20 v\ dt`.

Use the Trapezoidal rule and the speed at each of the five time values to find the approximate distance the jogger covers in the 20-minute period.   (3 marks)

Show Answers Only

 `text(2867.5 metres)`

Show Worked Solution

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ t\ \ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ &\ \ \ 5\ \ \ &\ \ \ 10\ \ \ &\ \ \ 15\ \ \ &\ \ \ 20\ \ \  \\
\hline
\rule{0pt}{2.5ex} \ \ \ v\ \ \  \rule[-1ex]{0pt}{0pt} & 173 & 81 & 127 & 195 & 168 \\
\hline
\rule{0pt}{2.5ex} \text{weight} \rule[-1ex]{0pt}{0pt} & 1 & 2 & 2 & 2 & 1 \\
\hline
\end{array}

`int_0^20 v\ dt`

 `~~ 5/2[173 + 2(81 + 127 + 195) + 168]`
  `~~ 5/2(1147)`
  `~~ 2867.5\ text(metres)`

Calculus, 2ADV C4 SM-Bank 07

The diagram shows a block of land and its dimensions, in metres. The block of land is bounded on one side by a river. Measurements are taken perpendicular to the line  `AB`, from  `AB`  to the river, at equal intervals of 50 m.
 

Integration, 2UA 2009 HSC 3d

 
Use the Trapezoidal rule with six subintervals to find an approximation to the area of the block of land.   (3 marks)

Show Answers Only

 `text(64 250 m²)`

Show Worked Solution

 

`A` `~~ 50/2[210 + 2(220 + 200 + 190 + 210 + 240) + 240]`
  `~~ 25(2570)`
  `~~ 64\ 250\ text(m²)`

 
`:.\ text(The block of land)\ ~~64 250 m²`

Calculus, 2ADV C4 2006* HSC 10a

Use the Trapezoidal rule with three function values to find an approximation to the value of
 
          `int_0.5^1.5 (log_e x )^3\ dx`.
 

Give your answer correct to three decimal places.  (2 marks)

Show Answers Only

`text(– 0.067)\ \ \ text{(to 3 d.p.)}`

Show Worked Solution

`text(Let)\ \ f(x) = (log_e x )^3`
 

`int_0.5^1.5(log_e x)^3 dx` `~~ 0.5/2[−0.3330… + 2(0) + 0.0666…]`
  `~~ 0.25(−0.2663…)`
  `~~ −0.06659…`
  `~~ −0.067\ \ \ text{(to 3 d.p.)}`

Calculus, 2ADV C4 2005* HSC 6a

Five values of the function  `f(x)`  are shown in the table.
 

Integration, 2UA 2005 HSC 6a

Use the Trapezoidal rule with the five values given in the table to estimate

`int_0^20 f(x)\ dx`.  (3 marks)

Show Answers Only

`387.5`

Show Worked Solution

`:. int_0^20 f(x)\ dx` `~~ 5/2[15 + 2(25 + 22 + 18) + 10]`
  `~~ 5/2(155)`
  `~~ 387.5`

Trigonometry, 2ADV’ T1 2004 HSC 3d

Trig Ratios, EXT1 2004 HSC 3d
 

The length of each edge of the cube  `ABCDEFGH`  is 2 metres. A circle is drawn on the face  `ABCD`  so that it touches all four edges of the face. The centre of the circle is  `O`  and the diagonal  `AC`  meets the circle at  `X`  and  `Y`.

  1. Explain why  `∠FAC = 60^@`.  (1 mark)

  2. Show that  `FO = sqrt6` metres.  (1 mark)

  3. Calculate the size of  `∠XFY`  to the nearest degree.  (1 mark)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `44^@\ text{(nearest degree)}`
Show Worked Solution
i.   

Trig Ratios, EXT1 2004 HSC 3d Answer

`text(S)text(ince)\ \ FA, \ AC\ \ text(and)\ \ FC\ \ text(are all)`

`text(diagonals of sides of a cube,)`

`FA = AC = FC`

`:.ΔFAC\ \ text(is equilateral)`

`:.∠FAC = 60^@`

 

ii.   

Trig Ratios, EXT1 2004 HSC 3d Answer2

`text(In)\ \ ΔAEF`

`AF^2` `= EF^2 + EA^2`
  `= 2^2 + 2^2`
  `= 8`
`AF` `= sqrt8`
  `= 2sqrt2`

 

`text(In)\ \ ΔAFO`

`sin\ 60^@` `= (FO)/(AF)`
`sqrt3/2` `= (FO)/(2sqrt2)`
`FO` `= sqrt3/2 xx 2sqrt2`
  `= sqrt6\ text(metres … as required.)`

 

iii.

Trig Ratios, EXT1 2004 HSC 3d Answer3

`XY\ \ text(is the diameter of a circle AND the width)`

`text(of the cube.)`

`:.XY` `= 2`
`:.OX` `= OY = 1`
`tan\ ∠OFX` `=1 /sqrt6`
`∠OFX` `= 22.207…^@`

 

`:.∠XFY` `= 2 xx 22.407…`
  `= 44.415…`
  `= 44^@\ text{(nearest degree)}`

Trigonometry, 2ADV’ T1 2010 HSC 5a

A boat is sailing due north from a point  `A`  towards a point  `P`  on the shore line.

The shore line runs from west to east.

In the diagram,  `T`  represents a tree on a cliff vertically above  `P`, and  `L`  represents a landmark on the shore. The distance  `PL`  is 1 km.

From  `A`  the point  `L`  is on a bearing of 020°, and the angle of elevation to  `T`  is 3°.

After sailing for some time the boat reaches a point  `B`, from which the angle of elevation to  `T`  is 30°.
 

5a
 

  1. Show that  `BP = (sqrt3 tan 3°)/(tan20°)`.   (3 marks) 

  2. Find the distance `AB`. Give your answer to 1 decimal place.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2.5\ text(km)\ \ text{(to 1 d.p.)}`
Show Worked Solution

i.   `text(Show)\ BP = (sqrt3 tan 3°)/(tan 20°)` 

`text(In)\ Delta ATP`
`tan 3°` `= (TP)/(AP)`
`=> AP` `= (TP)/(tan 3)`

 
`text(In)\ Delta APL:`

`tan 20°` `= 1/(AP)`
`=> AP` `= 1/tan 20`

 

`:. (TP)/(tan3)` `= 1/(tan20)`
`TP` `= (tan3°)/(tan20°)\ \ \ text(…)\ text{(} text(1)text{)}`

 

`text(In)\ \ Delta BTP:`

`tan 30°` `= (TP)/(BP)`
`1/sqrt3` `= (TP)/(BP)`
`BP` `= sqrt3 xx TP\ \ \ \ \ text{(using (1) above)}`
  `= (sqrt3 tan3°)/(tan20°)\ \ \ text(… as required)`

 

ii.    `AB` `= AP\ – BP`
  `AP` `= 1/(tan20°)\ \ \ text{(} text(from part)\ text{(i)} text{)}`
`:.\ AB` `= 1/(tan 20°)\ – (sqrt3 tan3°)/(tan20°)`
  `= (1\ – sqrt3 tan 3)/(tan20°)`
  `= 2.4980…`
  `= 2.5\ text(km)\ text{(to 1 d.p.)`

Trigonometry, 2ADV’ T1 2008 HSC 6a

From a point  `A`  due south of a tower, the angle of elevation of the top of the tower  `T`, is 23°. From another point  `B`, on a bearing of 120° from the tower, the angle of elevation of  `T`  is 32°. The distance  `AB`  is 200 metres.
 

Trig Ratios, EXT1 2008 HSC 6a 
 

  1. Copy or trace the diagram into your writing booklet, adding the given information to your diagram.  (1 mark)

  2. Hence find the height of the tower. Give your answer to the nearest metre.  (3 marks)

Show Answers Only
  1.  
    Trig Ratios, EXT1 2008 HSC 6a Answer

  2. `96\ text(m)`
Show Worked Solution

i.

Trig Ratios, EXT1 2008 HSC 6a Answer 

 

ii.  `text(Find)\ \ OT = h`

`text(Using the cosine rule in)\ Delta AOB :`

`200^2 = OA^2 + OB^2 – 2 * OA * OB * cos 60\ …\ text{(*)}`

 `text(In)\ Delta OAT,\tan 23^@= h/(OA)`

`=> OA= h/(tan 23^@)\  …\ (1)`

 `text(In)\ Delta OBT,\ tan 32^@= h/(OB)`

`=> OB= h/(tan 32^@)\ \ \ …\ (2)`
 

`text(Substitute)\ (1)\ text(and)\ (2)\ text(into)\ text{(*):}`

`200^2` `= (h^2)/(tan^2 23^@) + (h^2)/(tan^2 32^@) – 2 * h/(tan 23^@) * h/(tan 32^@) * 1/2`
  `= h^2 (1/(tan^2 23^@) + 1/(tan^2 32^@) + 1/(tan23^@ * tan32^@) )`
  `= h^2 (4.340…)`
`h^2` `= (40\ 000)/(4.340…)`
  `= 9214.55…`
`:. h` `= 95.99…`
  `= 96\ text(m)\ \ \ text{(to nearest m)}`

Trigonometry, 2ADV’ T1 2015 HSC 12c

A person walks 2000 metres due north along a road from point `A` to point `B`. The point `A` is due east of a mountain `OM`, where `M` is the top of the mountain. The point `O` is directly below point `M` and is on the same horizontal plane as the road. The height of the mountain above point `O` is `h` metres.

From point `A`, the angle of elevation to the top of the mountain is 15°.

From point `B`, the angle of elevation to the top of the mountain is 13°.
 

Trig Ratios, EXT1 2015 HSC 12c
 

  1. Show that  `OA = h\ cot\ 15°`.  (1 mark)

  2. Hence, find the value of  `h`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `910\ text{m  (nearest metre)}`
Show Worked Solution

i.   `text(Show)\ \ OA = h\ cot\ 15^@` 

Trig Ratios, EXT1 2015 HSC 12c Answer1

`text(In)\ \ Delta MOA,`

`tan\ 15^@` `= h/(OA)`
`OA` `= h/(tan\ 15^@)`
  `= h\ cot\ 15^@\ \ …text(as required)`

 

ii.   `text(In)\ \ ΔMOB`

`tan\ 13^@` `= h/(OB)`
`OB` `= h/(tan\ 13^@)`
  `= h\ cot\ 13^@`

 

Trig Ratios, EXT1 2015 HSC 12c Answer2 
 

`text(In)\ \ ΔAOB:`

`OA^2 + AB^2` `= OB^2`
`OB^2 − OA^2` `= AB^2`
`(h\ cot\ 13^@)^2 − (h\ cot\ 15^@)^2` `= 2000^2`
`h^2[(cot^2\ 13^@ − cot^2\ 15^@)]` `= 2000^2`
`h^2` `= (2000^2)/(cot^2\ 13^@ − cot^2\ 15^@)`
`:. h` `= sqrt((2000^2)/(cot^2\ 13^@ − cot^2\ 15^@))`
  `= 909.704…`
  `= 910\ text{m  (nearest metre)}`

Measurement, STD2 M6 2011 HSC 24c

A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`.  The
size of angle `ABC` is 114°.
 

  1. What is the bearing of `C` from `B`?   (1 mark)

  2. Find the distance `AC`. Give your answer correct to the nearest kilometre.   (2 marks)

  3. What is the bearing of `A` from `C`? Give your answer correct to the nearest degree.   (3 marks)

Show Answers Only
  1.  `055^@`
  2.  `13\ text(km)`
  3.  `261^@`
Show Worked Solution
STRATEGY: Important: Draw North-South parallel lines through major points to make the angle calculations easier!
i.     2011 HSC 24c

 `text(Let point)\ D\ text(be due North of point)\ B`

`/_ABD=180-121\ text{(cointerior with}\ \ /_A text{)}\ =59^@`

`/_DBC=114-59=55^@`   

`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`

 

ii.    `text(Using cosine rule:)`

`AC^2` `=AB^2+BC^2-2xxABxxBCxxcos/_ABC`
  `=6^2+9^2-2xx6xx9xxcos114^@`
  `=160.9275…`
`:.AC` `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}`
  `=13\ text(km)\ text{(nearest km)}`

 

iii.    `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`

MARKER’S COMMENT: The best responses clearly showed what steps were taken with working on the diagram. Note that all North/South lines are parallel.
`cos/_ACB` `=(AC^2+BC^2-AB^2)/(2xxACxxBC)`
  `=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)`
  `=0.9018…`
`/_ACB` `=25.6^@\ text{(to 1 d.p.)}`

 

`text(From diagram,)`

`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`

`:.\ text(Bearing of)\ A\ text(from)\ C`

  `=180+55+25.6`
  `=260.6`
  `=261^@\ text{(nearest degree)}`

Trigonometry, 2ADV* T1 2009 HSC 27b

A yacht race follows the triangular course shown in the diagram. The course from  `P`  to  `Q`  is 1.8 km on a true bearing of 058°.

At  `Q`  the course changes direction. The course from  `Q`  to  `R`  is 2.7 km and  `/_PQR = 74^@`.
 

 2009-2UG-27b
 

  1. What is the bearing of  `R`  from  `Q`?   (1 mark)

  2. What is the distance from  `R`  to  `P`?     (2 marks)

  3. The area inside this triangular course is set as a ‘no-go’ zone for other boats while the race is on.

     

    What is the area of this ‘no-go’ zone?   (1 mark)

Show Answers Only
  1. `312^@`
  2. `2.8\ text(km)\ \ \ text{(1 d.p.)}`
  3. `2.3\ text(km²)\ \ \ text{(1 d.p.)}`
Show Worked Solution
i.    2UG-2009-27b-Answer

`/_ PQS = 58^@ \ \ \ (text(alternate to)\ /_TPQ)`

TIP: Draw North-South parallel lines through relevant points to help calculate angles as shown in the Worked Solutions.

`text(Bearing of)\ R\ text(from)\ Q`

`= 180^@ + 58^@ + 74^@`
`= 312^@`

 

(ii)   `text(Using Cosine rule:)`

`RP^2` `=RQ^2` + `PQ^2` `- 2` `xx RQ` `xx PQ` `xx cos` `/_RQP`
  `= 2.7^2` + `1.8^2` `- 2` `xx 2.7` `xx 1.8` `xx cos74^@`
  `=7.29 + 3.24\ – 2.679…`
  `=7.851…`
`:.RP` `= sqrt(7.851…)`
  `=2.8019…`
  `~~ 2.8\ text(km)  (text(1 d.p.) )`

 

(iii)   `text(Using)\ A = 1/2 ab sinC`

`A` `= 1/2` `xx 2.7` `xx 1.8` `xx sin74^@`
  `= 2.3358…`
  `= 2.3\ text(km²)`

 

`:.\ text(No-go zone is 2.3 km²)`

Trigonometry, 2ADV* T1 2016 HSC 25 MC

The diagram shows towns `A`, `B` and `C`. Town `B` is 40 km due north of town `A`. The distance from `B` to `C` is 18 km and the bearing of `C` from `A` is 025°. It is known that  `∠BCA`  is obtuse.
 

2ug-2016-hsc-25-mc

 
What is the bearing of `C` from `B`?

  1. `070°`
  2. `095°`
  3. `110°`
  4. `135°`
Show Answers Only

`=> D`

Show Worked Solution

`text(Using the sine rule,)`

`(sin∠BCA)/40` `= (sin25^@)/18`
`sin angle BCA` `= (40 xx sin25^@)/18`
  `= 0.939…`
`angle BCA` `= 180 – 69.9quad(angleBCA > 90^@)`
  `= 110.1°`

 

`:. text(Bearing of)\ C\ text(from)\ B`

`= 110.1 + 25qquad(text(external angle of triangle))`

`= 135.1`

`=> D`

Trigonometry, 2ADV* T1 2014 HSC 23 MC

The following information is given about the locations of three towns `X`, `Y` and `Z`: 
 

• `X` is due east of  `Z`

• `X` is on a bearing of  `145^@`  from  `Y` 

• `Y` is on a bearing of  `060^@`  from  `Z`. 

 
Which diagram best represents this information?
 

HSC 2014 23mci

Show Answers Only

`C`

Show Worked Solution
COMMENT: Drawing a parallel North/South line through `Y` makes this question much simpler to solve.

`text(S)text(ince)\ X\ text(is due east of)\ Z`

`=> text(Cannot be)\ B\ text(or)\ D`
 

 
`text(The diagram shows we can find)`

`/_ZYX = 60 + 35^@ = 95^@`
 

`text(Using alternate angles)\ (60^@)\ text(and)`

`text(the)\ 145^@\ text(bearing of)\ X\ text(from)\ Y`

`=>  C`

Trigonometry, 2ADV* T1 2008 HSC 17 MC

The diagram shows the position of  `Q`,  `R`  and  `T`  relative to  `P`.
 

VCAA 2008 17 mc

 
In the diagram,

`Q`  is south-west of  `P`

`R`  is north-west  of  `P`

`/_QPT`  is 165°
 

What is the bearing of  `T`  from  `P`?

  1. `060^@`
  2. `075^@`
  3. `105^@` 
  4. `120^@`
Show Answers Only

`A`

Show Worked Solution

VCAA 2008 17 mci

`/_QPS=45^@\ \ \ text{(} Q\ text(is south west of)\ Ptext{)}`

`/_TPS = 165 – 45 = 120^@`

`:.\ /_NPT = 60^@\ \ text{(} 180^@\ text(in straight line) text{)}`

`=>  A`

Trigonometry, 2ADV* T1 2012 HSC 20 MC

Town `B` is 80 km due north of Town `A` and 59 km from Town `C`.

Town `A` is 31 km from Town `C`.
 

2012 20 mc
 

 What is the bearing of Town `C` from Town `B`?  

  1. `019^@`
  2. `122^@` 
  3. `161^@` 
  4. `341^@` 
Show Answers Only

`C`

Show Worked Solution

`text(Using the cosine rule:)`

`cos\ /_B` `= (a^2 + c^2 -b^2)/(2ac)`
  `= (59^2 + 80^2 -31^2)/(2 xx 59 xx 80)`
  `= 0.9449…`
`/_B` `= 19^@\  text((nearest degree))`

 

`:.\ text(Bearing of Town C from B) = 180-19= 161^@`

`=>  C`

Measurement, STD2 M1 2007 HSC 28c*

A piece of plaster has a uniform cross-section, which has been shaded, and has dimensions as shown.
 

 
 

  1. Use the Trapezoidal rule to approximate the area of the cross-section.    (3 marks)

  2. The total surface area of the piece of plaster is 7480.8 cm²
  3. Calculate the area of the curved surface as shown on the diagram. Give your answer to the nearest square centimetre   (2 marks)

Show Answers Only
  1. `48.96\ text(cm²)`
  2. `3502.88\ text(cm²)`
Show Worked Solution
i.   
`A` `~~ 3.6/2 [5 + 2(4.6 + 3.7 + 2.8) + 0]`
  `~~ 1.8(27.2)`
  `~~ 48.96\ text(cm²)`

 

ii.  `text(Total Area) = 7480.8\ text{cm²   (given)}`

`text(Area of Base)` `= 14.4 xx 200`
  `= 2880\ text(cm²)`
`text(Area of End)` `= 5 xx 200`
  `= 1000\ text(cm²)`
`text(Area of sides)` `= 2 xx 48.96`
  `= 97.92\ text(cm²)`

 

`:.\ text(Area of curved surface)`

`= 7480.8 – (2880 + 1000 + 97.92)`

`= 3502.88`

`=3503\ text{cm²  (nearest cm²)}`

Measurement, STD2 M1 2017 HSC 29a*

A new 200-metre long dam is to be built.

The plan for the new dam shows evenly spaced cross-sectional areas.
 
 


 

  1. Using the Trapezoidal rule, show that the volume of the dam is approximately 44 500 m³.  (2 marks)

  2. It is known that the catchment area for this dam is 2 km².

     

    Assuming no wastage, calculate how much rainfall is needed, to the nearest mm, to fill the dam.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `22\ text{mm  (nearest mm)}`
Show Worked Solution
i.   
`V` `~~ 50/2[360 + 2(300 + 270 + 140) + 0]`
  `~~ 25(1780)`
  `~~ 44\ 500\ text(m³)`

 

ii.   `text(Convert  2 km² → m²:)`

♦♦♦ Mean mark 11%.
`text(2 km²)` `= 2000\ text(m × 1000 m)`
  `= 2\ 000\ 000\ text(m²)`

 

`text(Using)\ \ V=Ah\ \ text(where)\ \ h= text(rainfall):`

`44\ 500` `= 2\ 000\ 000 xx h`
`:.h` `= (44\ 500)/(2\ 000\ 000)`
  `= 0.02225…\ text(m)`
  `= 22.25…\ text(mm)`
  `= 22\ text{mm  (nearest mm)}`

Measurement, STD2 M1 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 the Trapezoidal rule.  (2 marks)

  2. The area of the cross-section must be 600 m2. The tunnel is 80 m wide. 

     

    If the value of `a` increases by 2 metres, by how much will `b` change?   (2 marks)

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

 

ii.   `A = 600\ text(m²)`

`text(If tunnel is 80 metres wide)`

`4h` `= 80`
`h` `= 20`

 
`text{Using part (i):}`

`600 = 20(2a + b)`

`2a + b` `= 30`
`b` `= 30 – 2a`

 
`:.\ text(If)\ a\ text(increases by 2,)\ b\ text(must decrease by 4.)`

Measurement, STD2 M1 2014 HSC 28d*

An aerial diagram of a swimming pool is shown. 

The swimming pool is a standard length of 50 metres but is not in the shape of a rectangle.

In the diagram of the swimming pool, the five widths are measured to be: 
 

`CD = 21.88\ text(m)`

`EF = 25.63\ text(m)`

`GH = 31.88\ text(m)`

`IJ = 36.25\ text(m)`

`KL = 21.88\ text(m)` 
 

  1. Use four applications of the Trapezoidal Rule to calculate the surface area of the pool.  (2 marks)

  2.  The average depth of the pool is 1.2 m

     

    Calculate the approximate volume of the swimming pool, in litres.  (1 mark)

Show Answers Only
  1. `1445.5\ text(m²)`
  2. `1\ 734\ 600\ text(L)`
Show Worked Solution
i.   

`text(Surface Area of pool)`

`~~ 12.5/2[21.88 + 2(25.63 + 31.88 + 36.25) + 21.88]`

`~~ 1445.5\ text(m²)`
 

♦ Mean mark 50%. Be careful not to give away easy marks! 
ii.    `V` `= Ah`
    `~~ 1445.5 xx 1.2`
    `~~ 1734.6\ text(m³)`
    `~~ 1\ 734\ 600\ text(L)`

Measurement, STD2 M1 2009 HSC 25c*

There is a lake inside the rectangular grass picnic area  `ABCD`, as shown in the diagram.
 

2UG-2009-25c
  

  1. Use Trapezoidal’s Rule to find the approximate area of the lake’s surface.   (3 marks)

The lake is 60 cm deep. Bozo the clown thinks he can empty the lake using a four-litre bucket.

  1. How many times would he have to fill his bucket from the lake in order to empty the lake? (Note that 1 m³ = 1000 L).    (2 marks)

Show Answers Only
  1. 426 m²
  2. 63 900 times
Show Worked Solution
i.     `text(Area of lake = Area of rectangle)\ – text(Area of grass)`
`text(Area of rectangle)` `= 24 xx 55`
  `= 1320\ text(m²)`

 
`text{Area of grass (two applications)}`

`~~ 12/2(20 + 5) + 12/2(5 + 10) + 12/2(35 + 22) + 12/2(22 + 30)`
`~~ 6(25 + 15 + 57 + 52)`
`~~ 894\ text(m²)`

 

`:.\ text(Area of lake)` `~~ 1320\ – 894`
  `~~ 426\ text(m²)`

 

♦ Mean mark 44%
STRATEGY: Most students who did calculations in cm² and cm³ made errors. Keeping calculations in metres is much easier here.
ii.    `V` `= Ah`
    `= 426 xx 0.6`
    `= 255.6\ text(m³)`
    `= 255\ 600\ text(L)\ \ \ text{(1 m³ = 1000  L)}`

 

`:.\ text(Times to fill bucket)` `= 255\ 600 -: 4`
  `= 63\ 900`

Measurement, STD2 M1 2015 HSC 28c*

Three equally spaced cross-sectional areas of a vase are shown.
 

2UG 2015 29c

 
Use the Trapezoidal rule to find the approximate capacity of the vase in litres.  (3 marks)

Show Answers Only

`3.3\ text(litres)`

Show Worked Solution

`text(Solution 1)`

`V` `≈ 15/2(45 + 180) + 15/2(180 + 35)`
  `≈ 15/2(225 + 215)`
  `≈ 3300\ text{mL   (1 cm³ = 1 mL)}`
  `~~3.3\ text(L)`

 

`text(Solution 2)`

`V` `≈ 15/2(45 + 2 xx 180 + 35)`
  `≈ 15/2(440)`
  `≈ 3300\ text{mL}`
  `~~3.3\ text(L)`

Measurement, STD2 M1 SM-Bank 14

Cassius is a boxer and skips to keep fit.

The table below shows the average energy used, in kilojuoles per kilogram of body mass, by a person skipping for 10 minutes at different speeds.

   

Cassius eats a hamburger that contains 550 kilocalories.

If Cassuis weighs 72 kilograms, how long must he skip at 150 skips per minute to burn off the energy contained in the hamburger? (1 kilocalorie = 4.184 kJ) Give your answer to 1 decimal place.  (3 marks)

Show Answers Only

`13.2\ text{minutes  (to 1 d.p.)}`

Show Worked Solution

`text(Kilojoules in hamburger)`

`= 550 xx 4.184`

`= 2301.2`
 

`:.\ text(Skipping time to burn energy)`

`= 2301.2/((72 xx 24.18)) xx 10`

`= 1.321… xx 10`

`= 13.2\ text{minutes  (to 1 d.p.)}`

Measurement, STD2 M1 SM-Bank 13 MC

The table shows the average energy used, in kilojoules per kilogram of body mass, by a person walking for 30 minutes at different speeds.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Walking speed} \rule[-1ex]{0pt}{0pt} & \text{Energy used in 30 minutes} \\
\hline
\rule{0pt}{2.5ex} \text{3 km/h} \rule[-1ex]{0pt}{0pt} & \text{5.53 kJ/kg} \\
\hline
\rule{0pt}{2.5ex} \text{5 km/h} \rule[-1ex]{0pt}{0pt} & \text{7.37 kJ/kg} \\
\hline
\end{array}

Rob, who weighs 90 kg, drinks a large cappuccino made with full cream milk. It contains 146 kilocalories.

For approximately how long must Rob walk at 3 km/hr to burn off the energy contained in the cappuccino? (1 kilocalorie = 4.184 kilojoules)

  1. 28 minutes
  2. 37 minutes
  3. 83 minutes
  4. 110 minutes
Show Answers Only

`text(B)`

Show Worked Solution

`text(Kilojoules in cappuccino)`

`= 146 xx 4.184`

`= 610.864\ text(kJ)`
 

`text(Energy used by Rob in 30 minutes)`

`=90 xx 5.53`

`=497.7\ text(kJ)`
 

`:.\ text(Time of the walk)`

`= 610.864/(497.7) xx 30`

`= 36.8…\ text(minutes)`

 
`=>\ text(B)`

Measurement, STD2 M1 SM-Bank 12

The scale diagram shows the aerial view of a block of land bounded on one side by a road. The length of the block, `AB`, is known to be 90 metres.
 


 

Calculate the approximate area of the block of land, using three applications of the Trapezoidal rule.  (3 marks)

Show Answers Only

5175 m²

Show Worked Solution

`text(Solution 1)`

`text(6 cm → 90 metres)`

` text(1 cm → 15 metres)`
 

`text(Height) = 2 xx 15 = 30\ text(metres)`

`text(Area)` `~~ 30/2(75 + 60) + 30/2(60 + 45) + 30/2(45 + 60)`
  `~~ 15(135 + 105 + 105)`
  `~~ 5175\ text(m²)`

 

`text(Solution 2)`

`text(After converting from scale:)`

`text(Area)` `~~ 30/2(75 + 2 xx 60 + 2 xx 45 + 60)`
  `~~ 5175\ text(m²)`

Proof, EXT2 P1 2018 HSC 15c

Let  `n`  be a positive integer and let  `x`  be a positive real number.

  1.  Show that  `x^n - 1 - n(x - 1) = (x - 1)(1 + x + x^2 + … + x^(n - 1) - n)`.  (1 mark)

  2.  Hence, show that  `x^n >= 1 + n(x - 1)`.  (2 marks)

  3.  Deduce that for positive real numbers `a` and `b`,
     
          `a^nb^(1-n)>=na + (1-n)b`  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `text(RHS)` `= (x – 1)underbrace{(1 + x + x^2 + … + x^(n – 1) – n)}_{text(GP where)\ \ a = 1,\ r = x}`
    `= (x – 1) ((1(x^n – 1))/(x – 1) – n)`
    `= x^n – 1 – n(x – 1)`
    `=\ text(LHS)`

 

ii.   `text(Let)\ P(x) = (x – 1)(1 + x + x^2 + … + x^(n – 1) – n)`

♦♦♦ Mean mark 11%.

`text(If)\ \ x = 1, P(x) = 0`
 

`text(If)\ \ 0 < x < 1, \ (x – 1) < 0, \ (1 + x + x^2 + … + x^(n – 1) – n) < 0`

`=> P(x) > 0`
 

`text(If)\ \ x > 1, \ (x – 1) > 0, \ (1 + x + … + x^(n – 1) – n) > 0`

`=> P(x) > 0`
 

`x^n – 1 – n(x – 1) >= 0`

`:. x^n >= 1 + n(x – 1)`

 

iii.   `x^n >= 1 + n(x – 1)\ \ text(for)\ \ x ∈ R^+`

♦♦ Mean mark 23%.

`text(S)text(ince)\ a,b ∈ R^+`

`(a/b)^n` `>= 1 + n(a/b – 1)`
`(a^n)/(b^n) xx b` `>= b + na – nb,\ \ \ \ (b > 0)`
`a^n b^(1 – n)` `>= na + (1 – n)b`

Complex Numbers, EXT2 N2 2018 HSC 15b

  1. Use De Moivre's theorem and the expansion of `(costheta + isintheta)^8` to show that
     
    `sin8theta = ((8),(1)) cos^7thetasintheta - ((8),(3)) cos^5thetasin^3theta`
     
                        `+ ((8),(5)) cos^3thetasin^5theta - ((8),(7)) costhetasin^7theta`  (2 marks)

  2. Hence, show that
     
    `(sin8theta)/(sin2theta) = 4(1 - 10sin^2theta + 24sin^4theta - 16sin^6theta)`.  (3 marks)

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

i.   `text(By De Moivre)`

`costheta + isintheta^8 = cos8theta + isin8theta\ \ …\ (text{*})`
 

`text(Using Binomial Expansion)`

`(costheta + isintheta)^8`

`= cos^8theta + ((8),(1))cos^7theta * isintheta + ((8),(2)) cos^6theta *i^2sin^2theta`

`+ ((8),(3)) cos^5theta *i^3sin^3theta + ((8),(4)) cos^4theta *i^4sin^4theta + ((8),(5)) cos^3theta *i^5sin^5theta`

`+ ((8),(6)) cos^2theta *i^6sin^6theta + ((8),(7)) costheta *i^7sin^7theta + i^8sin^8theta`

 
`text(Equating imaginary parts of the expansion equation (*)):`

`isin8theta = ((8),(1)) cos^7theta* isintheta + ((8),(3)) icos^5theta* i^3sin^3theta`

`+ ((8),(5)) cos^3theta* i ^5sintheta + ((8),(7)) costheta *i^7sin^7theta`

`:. sin8theta = ((8),(1)) cos^7theta sintheta – ((8),(3)) cos^5theta sin^3theta`

`+ ((8),(5)) cos^3theta sin^5theta – ((8),(7)) costheta sin^7theta`
 

ii.    `sin8theta` `= 8cos^7theta sintheta – 56cos^5 sin^3theta + 56cos^3theta sin^5theta – 8costheta sin^7theta`
    `= 2sinthetacostheta (4cos^6theta – 28cos^4theta sin^2theta + 28cos^2theta sin^4theta – 4sin^6theta)`

 
`:. (sin8theta)/(sin2theta)`

`= 4cos^6theta – 28cos^4theta sin^2theta + 28cos^2theta sin^4theta – 4sin^6theta`

`= 4(1 – sin^2theta)^3 – 28(1 – sin^2theta)^2 sin^2theta + 28(1 – sin^2theta) sin^4theta – 4sin^6theta`

`= 4(1 – 3sin^2theta + 3sin^4theta + sin^6theta) – 28sin^2theta (1 – 2sin^2theta + sin^4theta)`

`+ 28sin^4theta (1 – sin^2theta) – 4sin^6theta`

`= 4 – 40sin^2theta + 96sin^4theta – 56sin^6theta`

`= 4(1 – 10sin^2theta + 24sin^4theta – 16sin^6theta)`

Harder Ext1 Topics, EXT2 2018 HSC 15a

The point  `P(acostheta, bsintheta)`, where  `0 < theta < pi/2`, lies on the ellipse  `(x^2)/(a^2) + (y^2)/(b^2) = 1`, where ` a > b`. The point  `A(acostheta, asintheta)`  lies vertically above `P` on the auxiliary circle  `x^2 + y^2 = a^2`. The point `B` lies on the auxiliary circle such that  `angleAOB = pi/2`  and the point `Q` lies on the ellipse vertically below `B`, as shown.
 

  1. Show that `Q` has coordinates  `(−asintheta, bcostheta)`.  (2 marks)
     

The line `QO` meets the ellipse again at  `Qprime(asintheta, −bcostheta)`.   (Do NOT prove this.)

  1. Show that the minimum size of  `anglePOQprime`  is  `tan^(−1) ((2ab)/(a^2 - b^2))`.  (3 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `x_Q = x_B`

`A(acostheta, asintheta)`
 

`text(S)text(ince)\ angleAOB = 90°,`

`=> x_B` `= acos(theta + pi/2)`
  `= −asintheta`

 
`text(Find)\ \ y_Q\ \ text(when)\ \ x = −asintheta`

`((−asintheta)^2)/(a^2) + (y^2)/(b^2) = 1`

`y^2` `= b^2(1 – sin^2theta)`
  `= b^2cos^2theta`
`=> y_Q` `= bcostheta\ \ \ \ (y > 0)`

 
`:.Q (−asintheta, bcostheta)\ \ …\ text(as required)`

 

ii.   `m_(OP) = (b sintheta)/(acostheta) = b/a tantheta`

♦ Mean mark 38%.

 

`m_(OQprime) = (−bcostheta)/(asintheta) = −b/a cottheta`
 

`tan anglePOQprime` `= |(m_1 – m_2)/(1 + m_1m_2)|`
  `= |(b/atantheta + b/acottheta)/(1 – b/atantheta b/acottheta)|`
  `= |(ab(tantheta + cottheta))/(a^2 – b^2(tanthetacottheta))|`
  `= |(ab)/(a^2 – b^2)| · |(sintheta)/(costheta) + (costheta)/(sintheta)|`
  `= |(ab)/(a^2 – b^2)| · |(sin^2theta + cos^2theta)/(sinthetacostheta)|`
  `= |(ab)/(a^2 – b^2)| · |2/(sin2theta)|`
  `= (ab)/(a^2 – b^2) · 2/(|sin2theta|)qquadtext{(given}\ \ a > b text{)}`

 

`=>\ text(Minimum angle when)\ tan anglePOQprime\ text(is a minimum.)`

`=>\ text(Minimum)\ \ (ab)/(a^2 – b^2) · 2/(|sin2theta|) = (2ab)/(a^2 – b^2) qquad (sin 2theta = 1)`
 

`:.\text(Minimum size of)\ anglePOQprime`

`= tan^(−1)((2ab)/(a^2 – b^2))`

Calculus, EXT2 C1 2018 HSC 14c

Let  `I_n = int_(−3)^0 x^n sqrt(x + 3)\ dx`  for  `n = 0, 1, 2…`

  1.  Show that, for  `n >= 1`,
     
         `I_n = (−6n)/(3 + 2n) I_(n - 1)`.   (3 marks)

  2.  Find the value of  `I_2`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(144sqrt3)/35`
Show Worked Solution

i.   `I_n = int_(−3)^0 x^n sqrt(x + 3)\ dx\ \ text(for)\ \ n = 0, 1, 2…`

`u` `= x^n` `vprime` `= (x + 3)^(1/2)`
`uprime` `= nx^(n – 1)` `v` `= 2/3 (x + 3)^(3/2)`

 

`I_n` `= [2/3 x^n(x + 3)^(3/2)]_(−3)^0 – 2/3int_(−3)^0  nx^(n – 1)(x + 3)sqrt(x + 3)\ dx`
`I_n` `= 0 – (2n)/3 int_(−3)^0 x^nsqrt(x + 3) + 3x^(n – 1)sqrt(x + 3)\ dx`
`I_n` `= −(2n)/3 (I_n + 3I_(n-1))`
`I_n + (2n)/3 I_n` `= −2nI_(n – 1)`
`I_n(1 + (2n)/3)` `= −2nI_(n – 1)`
`I_n((3 + 2n)/3)` `= −2nI_(n – 1)`
`:. I_n` `= (−6n)/(3 + 2n) I_(n – 1)\ \ \ text(… as required)`

 

ii.    `I_2` `= −12/7 I_1`
    `= −12/7 xx −6/5 I_0`
    `= 72/35 int_(−3)^0 (x + 3)^(1/2)dx`
    `= 72/35 xx 2/3 [(x + 3)^(3/2)]_(−3)^0`
    `= 48/35 (3^(3/2))`
    `= (144sqrt3)/35`

Mechanics, EXT2 2018 HSC 13c

A particle of mass `m` is attached to a light inextensible string of length `l`. The string makes an angle `theta` to the vertical.

The particle is moving in a circle of radius `r` on a smooth horizontal surface. The particle is moving with uniform angular velocity `ω`. The forces on the particle are the tension `T` in the string; the normal reaction `N` to the horizontal surface; and the gravitational force  `mg`.
 

 
The particle remains in contact with the horizontal surface.

By resolving the forces on the particle in the horizontal and vertical directions, show that

`ω^2 <= g/(lcostheta)`.   (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(Resolve forces horizontally,)`

`Tsintheta = mrω^2\ \ …\ (1)`

`text(Resolve forces vertically,)`

`Tcostheta + N = mg\ \ …\ (2)`

 

`text(Solving simultaneously for)\ \ N:`

`text{Divide (1) by}\ sintheta`

`T = mrω^2 text(cosec)theta`

`text{Divide (2) by}\ costheta`

`T = (mg – N) sectheta`
 

`(mg – N)sectheta` `= mrω^2 text(cosec)theta`
`mg – N` `= mrω^2 cottheta`
`N` `= mg – mrω^2 cottheta`

 

`text(Particle doesn’t leave table if)\ \ N >= 0`

`mg – mrω^2 cottheta` `>= 0`
`mrω^2 cottheta` `<= mg`
`ω^2` `<= g/(r(costheta)/(sintheta))\ \ \ \ (l = r/(sintheta))`
`:. ω^2` `<= g/(lcostheta)`

Complex Numbers, EXT2 N1 2018 HSC 13b

Let   `z = 1 - cos2theta + isin2theta`, where   `0 < theta <= pi`.

  1.  Show that  `|\ z\ | = 2sintheta`.  (2 marks)

  2.  Show that  `text(arg)(z) = pi/2 - theta`.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `z = 1 – cos2theta + isin2theta`

`|\ z\ |` `= sqrt((1 – cos2theta)^2 + sin^2 2theta)`
  `= sqrt(1 – 2cos2theta + cos^2 2theta + sin^2 2theta)`
  `= sqrt(2 – 2cos2theta)`
  `= sqrt(2(1 – cos2theta))`
  `= sqrt(2(2sin^2theta))`
  `= 2sintheta\ \ \ text(… as required)`

 

ii.  `z= 1 – cos2theta + isin2theta`

`text(arg)(z)` `= tan^(−1)((sin2theta)/(1 – cos2theta))`
  `= tan^(−1)((2sinthetacostheta)/(2sin^2theta))`
  `= tan^(−1)(cottheta)`
  `= tan^(−1)(tan(pi/2 – theta))`
  `= pi/2 – theta`

 
`(text(S)text(ince)\ \ 0 < theta < pi => \ −pi/2 <= pi/2 – theta < pi/2)`

Volumes, EXT2 2018 HSC 13a

The graph  `y^2 = x(1 - x)^2` is shown.
 

 
Use the method of cylindrical shells to find the volume of the solid formed when the shaded region is rotated about the line  `x = 1`.  (3 marks)

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`(64pi)/105\ \ text(u³)`

Show Worked Solution

`text(Cylindrical shell:)`

`text(Radius) = (1 – x),\ text(height) = 2y`

`y^2` `= x(1-x)^2`
`y` `= sqrtx(1 – x)`

 

`δV` `= 2pi(1-x)(2y)\ δx`
  `= 2pi (1 – x) · 2sqrtx (1 – x)\ δx`
  `= 4pi (1 – x)^2 sqrtx\ δx`

 

`:. V` `= int_0^1 4pi (1 – x)^2 sqrtx\ dx`
  `= 4pi int_0^1 (1 – 2x + x^2) sqrtx\ dx`
  `= 4pi int_0^1 x^(1/2) – 2x^(3/2) + x^(5/2)\ dx`
  `= 4pi [2/3x^(3/2) – 4/5x^(5/2) + 2/7x^(7/2)]_0^1`
  `= 4pi[(2/3 – 4/5 + 2/7) – 0]`
  `= 4pi(16/105)`
  `= (64pi)/105\ \ text(u³)`

Volumes, EXT2 2018 HSC 12a

The base of a solid is the region enclosed by the parabola  `x = 1 - y^2`  and the `y`-axis. Each cross-section perpendicular to the `y`-axis is an equilateral triangle, as shown in the diagram.
 

 
Find the volume of the solid.  (3 marks)

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`(4sqrt3)/15\ text(u³)`

Show Worked Solution

`text(Cross section of triangle)`

`sin60°` `= h/(1 – y^2)`
`:.h` `= sqrt3/2(1 – y^2)`

 

`δV` `= 1/2(1 – y^2) · sqrt3/2(1 – y^2)\ δy`
  `= sqrt3/4(1 – y^2)^2\ δy`

 

`:. V` `= int_(−1)^1 sqrt3/4(1 – y^2)\ dy`
  `= sqrt3/2 int_0^1 1 – 2y^2 + y^2\ dy`
  `= sqrt3/2 [y – 2/3y^3 + 1/5y^5]`
  `= sqrt3/2 [(1 – 2/3 + 1/5) – 0]`
  `= sqrt3/2(8/15)`
  `= (4sqrt3)/15\ text(u³)`

Harder Ext1 Topics, EXT2 2018 HSC 11e

The circle centred at `O` with radius `r` has a diameter `AB`. Points `C` and `D` are chosen on the circle as shown in the diagram. The chord `AC` has length `d`.
 

 
Show that  `d = 2r sin D`.  (2 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(Join)\ CB`

`angleB = angleD\ \ (text{angles on circumference on arc}\ AC)`

`angleACB = 90°\ \ (text{angle in a semi-circle})`
 

`text(In)\ DeltaABC,`

`sin B` `= d/(2r)`
`:. d` `= 2r sin B`
  `= 2r sin D\ \ \ text(.. as required)`