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Probability, 2ADV S1 2007 HSC 9b

A pack of 52 cards consists of four suits with 13 cards in each suit.

  1. One card is drawn from the pack and kept on the table. A second card is drawn and placed beside it on the table. What is the probability that the second card is from a different suit to the first?  (1 mark)

  2. The two cards are replaced and the pack shuffled. Four cards are chosen from the pack and placed side by side on the table. What is the probability that these four cards are all from different suits?  (2 marks)

Show Answers Only
  1. `13/17`
  2. `0.105\ \ \ text{(to 3 d.p.)}`
Show Worked Solution

i.  `text(After 1st card is drawn)`

`text(# Cards left from another suit) = 39`

`text(# Cards left in pack) = 51`

`:. P\ text{(2nd card from a different suit)}`

`= 39/51`

`= 13/17`

 

ii.  `P\ text{(all 4 cards from different suits)}`

`= 52/52 xx 39/51 xx 26/50 xx 13/49`

`= 2197/(20\ 825)`

`= 0.1054…`

`= 0.105\ \ \ text{(to 3 d.p.)}`

Calculus, EXT1* C3 2007 HSC 9a

2007 9a
  

The shaded region in the diagram is bounded by the curve  `y = x^2 + 1`, the `x`-axis, and the lines  `x = 0`  and  `x = 1.`

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

Show Answers Only

`(28 pi)/15\ \ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 y^2\ dx`
  `= pi int_0^1 (x^2 + 1)^2\ dx`
  `= pi int_0^1 x^4 + 2x^2 + 1\ dx`
  `= pi [1/5 x^5 + 2/3 x^3 + x]_0^1`
  `= pi[(1/5 + 2/3 + 1) – 0]`
  `= pi [3/15 + 10/15 + 1]`
  `= (28 pi)/15\ \ text(u³)`

Plane Geometry, 2UA 2007 HSC 8b

In the diagram, `AE` is parallel to `BD`, `AE = 27`, `CD = 8`, `BD = p`, `BE = q` and `/_ABE`, `/_BCD` and `/_BDE` are equal.

Copy or trace this diagram into your writing booklet.

  1. Prove  that `Delta ABE\ text(|||)\ Delta BCD`.  (2 marks)
  2. Prove  that `Delta EDB\ text(|||)\ Delta BCD`.  (2 marks)
  3. Show that `8`, `p`, `q`, `27` are the first four terms of a geometric series.  (1 mark)
  4. Hence find the values of `p` and `q`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solution)}`
  2. `text(Proof)\ \ text{(See Worked Solution)}`
  3. `text(Proof)\ \ text{(See Worked Solution)}`
  4. `p = 12, q = 18`
Show Worked Solution

(i)

`text(Prove)\ Delta ABE\ text(|||)\  Delta BCD`

`/_ ABE = /_ BCD\ \ \ text{(given)}`

`/_ EAB = /_ DBC\ \ \ text{(corresponding angles,}\ AE\ text(||)\ BD text{)}`

`:. Delta ABE\ text(|||)\ Delta BCD\ \ \ text{(equiangular)}`

 

(ii)  `text(Prove)\ Delta EDB\ text(|||)\ Delta BCD`

`/_ EDB = /_ BCD\ \ \ text{(given)}`

`/_ CDB = 180° – (/_ BCD + /_ DBC)\ \ \ text{(Angle sum of}\ Delta BCD text{)}`

`/_ EBD` `= 180° – (/_ ABE + /_ DBC)\ \ \ text{(}/_ ABC\ text{is a straight angle)}`
  `= 180° – (/_ BCD + /_ DBC)\ \ \ text{(}/_ ABE = /_ BCD,\ text{given)}`
  `= /_ CDB`

`:. Delta EDB\ text(|||)\ Delta BCD\ \ \ text{(equiangular)}`

 

(iii)  `text(In a GP,)\ \ r = T_n/T_(n-1)`

`text(If)\ \ \ 8, p, q, 27\ \ \ text(are 1st 4 terms of a GP)`

`=> p/8 = q/p = 27/q .`

`text(S)text(ince corresponding sides of similar)`

`text(triangles are in the same ratio)`

`(BD)/(DC) = (EB)/(BD) = (AE)/(EB)`

`p/8 = q/p = 27/q .`

 

`:. 8, p, q, 27\ \ text(are 1st 4 terms of a GP.)`

 

(iv)  `8, p, q, 27`

`a = 8`

`text(Using)\ \ T_n = ar^(n-1)`

`T_4=ar^3`

`27` `= 8 xx r^3`
`r^3` `= 27/8`
`r` `= 3/2`

 

`T_2` `=ar`
`:.p` `= 8 * 3/2`
  `= 12`

 

`T_3` `=ar^2`
`:.q` `= 8 * (3/2)^2`
  `= 18`

Calculus, EXT1* C1 2007 HSC 8a

One model for the number of mobile phones in use worldwide is the exponential growth model,

`N = Ae^(kt)`,

where `N` is the estimate for the number of mobile phones in use (in millions), and `t` is the time in years after 1 January 2008.

  1. It is estimated that at the start of 2009, when  `t = 1`, there will be 1600 million mobile phones in use, while at the start of 2010, when  `t = 2`, there will be 2600 million. Find `A` and `k`.  (3 marks)

  2. According to the model, during which month and year will the number of mobile phones in use first exceed 4000 million?  (2 marks)

Show Answers Only
  1. `A = (12\ 800)/13, k = log_e\ 13/8`
  2. `text(November 2010)`
Show Worked Solution

i.  `N = Ae^(kt)`

`text(When)\ t = 1,\ N = 1600`

`:. 1600` `= Ae^k`
`A` `= 1600/e^k`

 
`text(When)\ t = 2, N = 2600`

`:. 2600` `= A e^(2k)`
  `= 1600/e^k xx e^(2k)`
  `= 1600 e^k`
`e^k` `= 2600/1600 = 13/8`
`log_e e^k` `= log_e\ 13/8`
`k` `= log_e\ 13/8`
  `=0.4855…`
  `=0.49\ \ \ text{(to 2 d.p.)}`

 

`:. A` `= 1600/(e^(log_e\ 13/8)`
  `= 1600/(13/8)`
  `= (12\ 800)/13`

 

ii.  `text(Find)\ \ t\ \ text(such that)\ N > 4000`

`(12\ 800)/13 xx e^(kt)` `> 4000`
`e^(kt)` `> (13 xx 4000)/(12\800)`
`log_e e^(kt)` `> log_e\ 65/16`
`kt` `> log_e\ 65/16\ \ \ \ (k = log_e\ 13/8)`
`t` `> (log_e\ 65/16)/(log_e\ 13/8`
`t` `> 2.887…\ \ text(years)`
  `>\ text{2 years and 10.6 months (approx)}`

 

`:.\ text(The number of mobile phones in use will exceed)`

`text(4000 million in November 2010.)`

Calculus, 2ADV C4 2007 HSC 7b

2ua 2007 7b
 

The diagram shows the graphs of  `y = sqrt 3 cos x`  and  `y = sin x`. The first two points of intersection to the right of the `y`-axis are labelled  `A`  and  `B`.

  1. Solve the equation  `sqrt 3 cos x = sin x`  to find the `x`-coordinates of  `A`  and  `B`.  (2 marks)

  2. Find the area of the shaded region in the diagram.  (3 marks)

Show Answers Only
  1. `pi/3, (4 pi)/3`
  2. `4\ text(u²)`
Show Worked Solution
i.  `sqrt 3 cos x` `= sin x`
`tan x` `= sqrt 3`

 

`text(S)text(ince)\ tan\ pi/3 = sqrt 3 and tan\ text(is)`

`text(positive in 1st/3rd quadrants:)`

`x` `= pi/3 , pi + pi/3`
  `= pi/3, (4 pi)/3`

 

`:. text(The)\ x text(-coordinates of)\ A and B`

`text(are)\ \ x = pi/3, (4 pi)/3.`

 

ii.  `text(Shaded Area)`

`= int_(pi/3)^((4 pi)/3) sin\ x\ dx – int_(pi/3)^((4 pi)/3) sqrt 3\ cos\ x\ dx`

`= int_(pi/3)^((4 pi)/3) sin\ x – sqrt 3\ cos\ x\ dx`

`= [-cos\ x – sqrt 3\ sin\ x]_(pi/3)^((4 pi)/3)`

`= [(-cos\ (4 pi)/3 – sqrt 3\ sin\ (4 pi)/3) – (-cos\ pi/3 – sqrt 3\ sin\ pi/3)]`

`= [(1/2 + sqrt 3 xx (sqrt 3)/2) – (- 1/2 – sqrt 3 xx (sqrt 3)/2)]`

`= [2 – (-2)]`

`= 4\ text(u²)`

Quadratic, 2UA 2007 HSC 7a

  1. Find the coordinates of the focus, `S`, of the parabola  `y = x^2 + 4`.  (2 marks)
  2. The graphs of  `y = x^2 + 4`  and the line  `y = x + k`  have only one point of intersection, `P`. Show that the `x`-coordinate of `P` satisfies.
    1. `x^2 - x + 4 - k = 0`.  (1 mark)
  3. Using the discriminant, or otherwise, find the value of `k`.  (1 mark)
  4. Find the coordinates of `P`.  (2 marks)
  5. Show that `SP` is parallel to the directrix of the parabola.  (1 mark)
Show Answers Only
  1. `(0, 4 1/4)`

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

  3. `15/4`

  4. `(1/2, 4 1/4)`

  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   `y` `= x^2 + 4`
`x^2` `= y – 4`

`text(Using)\ (x – x_0) = 4a (y – y_0)`

`x_0` `= 0`
`a` `= 1/4`
`y_0` `= 4`

`(x – 0) = 4 xx 1/4 xx (y – 4)`

`text(Vertex)` `= (0, 4)`
`:.\ text(Focus)` `= (0, 4 1/4)`

 

(ii)  `y` `= x^2 + 4` `\ \ …\ \ (1)`
`y` `= x + k` `\ \ …\ \ (2)`

 

`text(Solving simultaneously)`

`x^2 + 4` `= x + k`
`x^2 – x + 4 – k` `= 0`

 

(iii)  `text(If only 1 point of intersection,)`

`b^2 – 4ac` `= 0`
`(–1)^2 – 4 xx 1 xx (4 – k)` `= 0`
`1 – 16 + 4k` `= 0`
`4k` `= 15`
`k` `= 15/4`

 

(iv)  `text(Finding)\ \ P`

`x^2 – x + 4 – 15/4 = 0`

`x^2 – x + 1/4 = 0`

`(x – 1/2)^2 = 0`

`x = 1/2`

`text(When)\ x = 1/2`

`y` `= (1/2)^2 + 4`
  `= 4 1/4`

`:. P\ text(has coordinates)\ \  (1/2, 4 1/4)`

 

(v)  `text(Focus)\ (S)` `= (0, 4 1/4)`
`P` `= (1/2, 4 1/4)`

 

`:.\ text(S)text(ince)\ y text(-values are the same,)\ SP\ text(is parallel)`

`text(with the)\ x text(-axis.)`

`=>text(Directrix has the equation)\ \ y = 3 3/4`

`:. SP\ text(is parallel with the directrix.)`

Calculus, 2ADV C3 2007 HSC 6b

Let  `f (x) =x^4 - 4x^3`.

  1. Find the coordinates of the points where the curve crosses the axes.  (2 marks)

  2. Find the coordinates of the stationary points and determine their nature.  (4 marks)

  3. Find the coordinates of the points of inflection.  (1 mark)

  4. Sketch the graph of  `y = f (x)`, indicating clearly the intercepts, stationary points and points of inflection.  (3 marks)

Show Answers Only
  1. `(0, 0)\ , \ (4, 0)`
  2. `text(Minimum S.P. at)\ (3,\ text(-27))`
  3. `(0, 0) and (2, 16)`
  4.  
Show Worked Solution

i.  `f (x) = x^4 – 4x^3`

`text(Cuts)\ x text(-axis when)\ f(x) = 0`

`x^4 – 4x` `= 0`
`x^3 (x – 4)` `= 0`

`x = 0 or 4`

`:. text(Cuts the)\ x text(-axis at)\ (0, 0)\ ,\ (4, 0)`

 

`text(Cuts the)\ y text(-axis when)\ x = 0`

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

 

ii.   `f(x) = x^4 – 4x^3`

`f prime (x) = 4x^3 – 12x^2`

`f″ (x) = 12x^2 – 24x`

 

`text(S.P.’s when)\ f prime (x) = 0`

`4x^3 – 12x^2` `= 0`
`4x^2 (x – 3)` `= 0`

`x = 0 or 3`

`text(When)\ x = 0`

`f(0) = 0`

`f″(0) = 0`

`text(S)text(ince concavity changes, a P.I.)`

`text(occurs at)\ (0, 0)`

 

`text(When)\ x = 3`

`f (3)` `= 3^4 – 4 xx 3^3`
  `= -27`
`f″ (3)` `= 12 xx 3^2 – 24 xx 3`
  `= 36 > 0`

 

`:. text(Minimum S.P. at)\ (3,\ text(–27))`

 

iii.  `text(P.I. when)\ f″(x) = 0`

`12x^2 – 24x` `= 0`
`12x(x – 2)` `= 0`

`x = 0 or 2`

`text(P.I. at)\ (0, 0)\ \ \ text{(from(ii))}`

 

`text(When)\ x = 2`

`text(S)text(ince concavity changes, a P.I.)`

`text(occurs when)\ x = 2`

`f (2)` `= 2^4 – 4 xx 2^3`
  `= 16`

 

`:. text(P.I.’s at)\ (0, 0) and (2, 16)`

 

iv.

 2UA HSC 2007 6b

L&E, 2ADV E1 2007 HSC 6a

Solve the following equation for `x`:

`2e^(2x)-e^x = 0`.   (2 marks)

Show Answers Only

`x = ln\ 1/2`

Show Worked Solution

`text(Solution 1)`

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

`text(Let)\ \ X = e^x:`

`2X^2-X` `= 0`
`X (2X-1)` `= 0`

 
`X = 0 or 1/2`

 
`text(When)\ e^x = 0\  =>\ text(no solution)`

`text(When)\ e^x = 1/2`

`ln e^x` `= ln\ 1/2`
`:. x` `= ln\ 1/2`

 

`text(Solution 2)`

`2e^(2x)-e^x` `=0`
`2e^(2x)` `=e^x`
`ln 2e^(2x)` `=ln e^x`
`ln 2 +ln e^(2x)` `=x`
`ln 2 + 2x` `=x`
`x` `=-ln2`
  `=ln\ 1/2`

Calculus, EXT1* C1 2007 HSC 5b

A particle is moving on the `x`-axis and is initially at the origin. Its velocity, `v` metres per second, at time `t` seconds is given by

`v = (2t)/(16 + t^2).`

  1. What is the initial velocity of the particle?  (1 mark)

  2. Find an expression for the acceleration of the particle.  (2 marks)

  3. Find the time when the acceleration of the particle is zero.  (1 mark)

  4. Find the position of the particle when `t = 4`.  (3 marks)

Show Answers Only
  1. `0`
  2. `{2(16 – t^2)}/(16 + t^2)^2`
  3. `4\ text(seconds)`
  4. `log_e 2\ \ text(metres)`
Show Worked Solution

i.  `v = (2t)/(16 + t^2)`

`text(When)\ t` `= 0`
`v` `= 0`

`:.\ text(Initial velocity is 0.)`

 

ii.  `a = d/(dt) ((2t)/(16 + t^2))`
 

`text(Using quotient rule)`

`u` `= 2t` `v` `= 16 + t^2`
`u prime` `= 2` `v prime` `= 2t`
`(dv)/(dt)` `= (u prime v – uv prime)/v^2`
  `= {2(16 + t^2) – 2t * 2t}/(16 + t^2)^2`
  `= (32 + 2t^2 – 4t^2)/(16 + t^2)^2`
  `= {2(16 – t^2)}/(16 + t^2)^2`

 

iii.  `text(Find)\ t\ text(when)\ (dv)/(dt) = 0`

`{2 (16 – t^2)}/(16 + t^2)^2` `= 0`
`2 (16 – t^2)` `= 0`
`t^2` `= 16`
`t` `= 4\ ,\ t >= 0`

 

`:.\ text(The acceleration is zero when)`

`t = 4\ text(seconds.)`

 

iv.  `v = (2t)/(16 + t^2)`

`x` `= int v\ dt`
  `= int (2t)/(16 + t^2)`
  `= log_e (16 + t^2) + c`

 

`text(When)\ \ t = 0\ ,\ x = 0`

`0 = log_e (16 + 0) + c`

`c = -log_e 16`

`:. x = log_e(16 + t^2) – log_e 16`
 

`text(When)\ t = 4,`

`x` `= log_e (16 + 4^2) – log_e 16`
  `= log_e 32 – log_e 16`
  `= log_e (32/16)`
  `= log_e 2\ \ text(metres)`

 

`:.\ text(When)\ t = 4\ , \ text(the position of the)`

`text(particle is)\ log_e 2 \ text(metres.)`

Plane Geometry, 2UA 2007 HSC 5a

In the diagram, `ABCDE` is a regular pentagon. The diagonals `AC` and `BD` intersect at `F`.

Copy or trace this diagram into your writing booklet.

  1. Show that the size of `/_ABC` is `108°`.  (1 mark)
  2. Find the size of `/_BAC`. Give reasons for your answer.  (2 marks)
  3. By considering the sizes of angles, show that `Delta ABF` is isosceles.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `36°`
  3. `text(See Worked Solutions)`
Show Worked Solution
(i)

`text(Sum of all internal angles`

`= (n – 2) xx 180°`

`= (5 – 2) xx 180°`

`= 540°`

`:. /_ABC` `= 540/5= 108°`

 

(ii)  `BA = BC`

`text{(equal sides of a regular pentagon)}`

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

`/_BAC` `= 1/2 (180 – 108)\ \ \ text{(base angle of}\ Delta BAC text{)}`
  `= 36°`

 

(iii)  `text(Consider)\ Delta BCD and Delta ABC`

`BC = CD = BA`

`text{(equal sides of a regular pentagon)}`

`/_BCD = /_ABC = 108°`

`text{(internal angles of a regular pentagon)}`

`:. Delta BCD -= Delta ABC\ \ \ text{(SAS)}`

`:. CBF` `= 36°` `text{(corresponding angles in}`
     `text{congruent triangles)}`
`/_FBA` `= 108 – 36`
  `= 72°`
`/_BFA` `= 180 – (72 + 36)\ \ \ \ text{(angle sum of}\ Delta ABF text{)}`
  `= 72°`

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

Trigonometry, 2ADV T1 2007 HSC 4c

 
An advertising logo is formed from two circles, which intersect as shown in the diagram.

The circles intersect at `A` and `B` and have centres at `O` and `C`.

The radius of the circle centred at `O` is 1 metre and the radius of the circle centred at `C` is `sqrt 3` metres. The length of `OC` is 2 metres.

  1. Use Pythagoras’ theorem to show that  `/_OAC = pi/2`.  (1 mark)

  2. Find  `/_ ACO`  and  `/_ AOC`.  (2 marks)

  3. Find the area of the quadrilateral  `AOBC`.  (1 mark)

  4. Find the area of the major sector  `ACB`.  (1 mark)

  5. Find the total area of the logo (the sum of all the shaded areas).  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `/_ ACO = pi/6\ ,\ /_ AOC = pi/3`
  3. `sqrt 3\ \ text(m²)`
  4. `(5 pi)/2\ text(m²)`
  5. `((19 pi + 6 sqrt 3)/6)\ text(m²)`
Show Worked Solution

i.

`text(In)\ Delta AOC`

`AO^2 + AC^2` `= 1^2 + sqrt 3^2`
  `=1 + 3`
  `= 4`
  `= OC^2`

 
`:. Delta AOC\ \ text(is right-angled and)\ \ /_OAC = pi/2`

 

ii.  `sin\  /_ACO` `= 1/2`
`:. /_ACO` `= pi/6`
`sin\  /_AOC` `= sqrt 3/2`
`:. /_AOC` `= pi/3`

 

iii.  `text(Area)\ AOBC`

`= 2 xx text(Area)\ Delta AOC`

`= 2 xx 1/2 xx b xx h`

`= 2 xx 1/2 xx 1 xx sqrt 3`

`= sqrt 3\ \ text(m²)`

 

iv.  `/_ACB = pi/6 + pi/6 = pi/3`

`:. /_ACB\ text{(reflex)}` `= 2 pi – pi/3`
  `= (5 pi)/3`

 
`text(Area of major sector)\ ACB`

`= theta/(2 pi) xx pi r^2`

`= {(5 pi)/3}/(2 pi) xx pi(sqrt 3)^2`

`= (5 pi)/6 xx 3`

`= (5 pi)/2\ text(m²)`

 

v.  `/_AOB = pi/3 + pi/3 = (2 pi)/3`

`:. /_AOB\ text{(reflex)}` `= 2 pi – (2 pi)/3`
  `= (4 pi)/3`

`text(Area of major sector)\ AOB`

`= {(4 pi)/3}/(2 pi) xx pi xx 1^2`

`= (2 pi)/3\ text(m²)`

 

`:.\ text(Total area of the logo)`

`= (5 pi)/2 + (2 pi)/3 + text(Area)\ AOBC`

`= (15 pi + 4 pi)/6 + sqrt 3`

`= ((19 pi + 6 sqrt 3)/6)\ text(m²)`

Probability, 2ADV S1 2007 HSC 4b

Two ordinary dice are rolled. The score is the sum of the numbers on the top faces.

  1. What is the probability that the score is 10?  (2 marks)

  2. What is the probability that the score is not 10?  (1 mark)

Show Answers Only
  1. `1/12`
  2. `11/12`
Show Worked Solution
i. 2UA HSC 2007 4b

`text{P (score = 10)}`

`= 3/36`

`= 1/12`

 

ii.  `text{P (score is not ten)}`

`= 1 – text{P (score is ten)}`

`= 1 – 1/12`

`= 11/12`

Geometry and Calculus, EXT1 2005 HSC 7b

Let  `f(x) = Ax^3 - Ax + 1`,  where  `A > 0`.

  1. Show that  `f(x)`  has stationary points at
    1. `x = +- (sqrt 3)/3.`  (1 mark)

  2. Show that  `f(x)`  has exactly one zero when
    1. `A < (3 sqrt 3)/2.`  (2 marks)

  3. By observing that  `f(-1) = 1`, deduce that  `f(x)`  does not have a zero in the interval  `-1 <= x <= 1`  when
    1. `0 < A < (3 sqrt 3)/2.`  (1 mark)

  4. Let  `g(theta) = 2 cos theta + tan theta`,  where  `-pi/2 < theta < pi/2.`
  6. By calculating  `g prime (theta)`  and applying the result in part (iii), or otherwise, show that  `g(theta)`  does not have any stationary points.  (3 marks)
  7. Hence, or otherwise, deduce that  `g(theta)`  has an inverse function.  (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)   `f(x) = Ax^3 – Ax + 1\ \ ,\ \ \ A > 0`

`f prime (x) = 3Ax^2 – A`

`text(S.P.’s when)\ \ f prime (x) = 0`

`3Ax^2 – A` `= 0`
`A (3x^2 – 1)` `= 0`
`3x^2` `= 1`
`x^2` `= 1/3`
`x` `= +- 1/sqrt3 xx (sqrt 3)/(sqrt 3)`
  `= +- (sqrt 3)/3\ \ text(…  as required)`

 

(ii)   `text(Consider)\ \ f(x)\ \ text(with only 1 zero.)`

`f(0) = 1`

`:.\ f(x)\ \ text(passes through)\ \ (0, 1)\ \ text(and has)`

`text(turning points either side at)\ \ x = +- (sqrt 3)/3.`

 

`text(Given 1 zero)`

`=> f((sqrt 3)/3) > 0`

`A ((sqrt 3)/3)^3 – A ((sqrt 3)/3) + 1` `> 0`
`A((3 sqrt 3)/27) – A ((9 sqrt 3)/27)` `> -1`
`A ((-6 sqrt 3)/27)` `> -1`
`A` `< 27/(6 sqrt 3)`
`A` `< 9/(2 sqrt 3) xx (sqrt 3)/(sqrt 3)`
`A` `< (9 sqrt 3)/6`
 `A` `< (3 sqrt 3)/2\ \ text(…  as required.)`

 

(iii)  `text(S)text(ince 1 zero occurs when)\ \ A < (3 sqrt 3)/2\ \ text{(part (ii))}`

`=> text(Minimum TP when)\ \ x = (sqrt3)/3`

`=> text(Maximum TP when)\ \ x = -(sqrt 3)/3\ \ text{(see graph)}`

 

`text(S)text(ince)\ \ f(-1) = 1`

`:. f(x)\ \ text(cannot have a zero for)\ \ \ -1 <= x <= 1`

 

(iv)   `g(theta) = 2 cos theta + tan theta,\ \ \ \ \ \ -pi/2 < theta < pi/2`

`g prime (theta) = -2 sin theta + sec^2 theta`

 

`text(S.P.’s when)\ \ g prime (theta) = 0`

`-2 sin theta + sec^2 theta` `= 0`
`-2 sin theta + 1/(cos^2 theta)` `= 0`
`-2 sin theta + 1/((1 – sin^2 theta))` `= 0`
`(-2 sin theta (1 – sin^2 theta) + 1)/((1 – sin^2 theta))` `= 0`
`-2 sin theta (1 – sin^2 theta) + 1` `= 0`
 `-2 sin theta + 2 sin^3 theta + 1` `= 0`
 `2 sin^3 theta – 2 sin theta + 1` `= 0\ \ text(…  (1))`

 

`text(Let)\ \ x = sin theta and A = 2`

`text{Equation (1) becomes}`

`Ax^3 – 2x + 1 = 0`

`text(S)text(ince)\ \ A = 2`

`0 < A < (3 sqrt 3)/2`

`text(S)text(ince)\ -pi/2` `< \ \ \ theta` `< pi/2`
`-1` `< sin theta` `< 1`
`-1` `< \ \ \ x` `< 1`

 

`:.\ text{Using Part (iii)} => g prime (theta)\ \ text(has no zeros and)`

`text(therefore)\ \ g (theta)\ \ text(has no S.P.’s for)\ \ -pi/2 < theta < pi/2.`

 

(v)   `text(Given)\ \ g(theta)\ \ text(has no stationary points, it will be)`

`text(either an increasing or decreasing function in the)`

`text(range)\ \ (-pi)/2 < theta < pi/2\ \ text(and therefore it has an inverse)`

`text(function.)`

Quadratic, 2UA 2005 HSC 10a

2005 10a

The parabola  `y = x^2`  and the line  `y = mx + b`  intersect at the points  `A(α,α^2)`  and  `B(β, β^2)`  as shown in the diagram.

  1. Explain why  `α + β = m`  and  `αβ = –b`.  (1 mark)
  2. Given that
  3. `(α − β)^2 + (α^2 − β^2)^2 = (α − β)^2[1 + (α + β)^2]`, show that the distance  `AB = sqrt((m^2 + 4b)(1 + m^2)).`  (2 marks)
  4. The point  `P(x, x^2)`  lies on the parabola between  `A`  and  `B`. Show that the area of the triangle  `ABP`  is given by  `1/2(mx − x^2 + b)sqrt(m^2 + 4b).`  (2 marks)
  5.  
  6. The point  `P`  in part (iii) is chosen so that the area of the triangle  `ABP`  is a maximum.
  7. Find the coordinates of  `P`  in terms of  `m`.  (2 marks)
  8.  
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `P(m/2, m^2/4)`
Show Worked Solution

(i)   `text(Instersection)`

`y = x^2` `\ \ …\ (1)`
`y = mx + b` `\ \ …\ (2)`

`text(Substitute)\ y = x^2\ text(into)\ \ (2)`

`x^2 = mx + b`

`x^2 − mx − b = 0`

`text(We know the intersection occurs when)`

`x = α\ \ text(and)\ \ x = β`

`:.α\ \ text(and)\ \ β\ \ text(are roots of)\ \ x^2 − mx − b = 0`

`:.α + β` `= (−b)/a` `= m`  
`αβ` `= c/a` `= −b` `\ \ …text(as required)`

 

 

(ii)   `A(α, α^2),\ \ B(β, β^2)`

`AB` `= sqrt((α − β)^2 + (α^2 − β^2)^2)`
  `= sqrt((α − β)^2[1 + (α + β)^2])`
  `= sqrt([(α + β)^2 − 4αβ][1 + (α + β)^2])`
  `= sqrt([m^2 − 4(−b)][1 + m^2])`
  `= sqrt((m^2 + 4b)(1 + m^2))\ \ …text(as required)`

 

(iii) `⊥\ text(distance of)\ \ (x, x^2)\ \ text(from)\ \ \ y=mx+b`

`text(i.e.)\ \ mx – y+b=0`

`= |(ax_1 + by_1 + c)/(sqrt(a^2 + b^2))|`

`= |(mx − 1(x^2) + b)/(sqrt(m^2 + (−1)^2))|`

`= |(mx − x^2 + b)/(sqrt(m^2 + 1))|`

 

`text(Area of)\ ΔABP`

`= 1/2 xx AB xx h`

`= 1/2 xx sqrt((m^2 + 4b)(1 + m^2)) xx (mx − x^2 + b)/(sqrt(m^2 + 1))`

`= 1/2 (mx − x^2 + b)sqrt(m^2 + 4b)`

 

(iv) `A` `= 1/2 sqrt(m^2 + 4b)(mx − x^2 + b)`
  `(dA)/(dx)` `= 1/2 sqrt(m^2 + 4b)(m − 2x)` 
  `(d^2A)/(dx^2)` `= 1/2 sqrt(m^2 + 4b)(−2)`
    `= −sqrt(m^2 + 4b)` 

 

`text(Max or min when)\ (dA)/(dx) = 0`

`1/2sqrt(m^2 + 4b)(m − 2x)` `= 0`
`m-2x` `=0`
`2x` `= m`
`x` ` = m/2`

`text(When)\ x = 2`

`(d^2A)/(dx^2) = −sqrt(m^2 + 4b) < 0\ \ \ text{(constant)}`

`:.\ text(Maximum when)\ x = m/2`

`:.P\ text(is)\ (m/2, m^2/4)` 

Mechanics, EXT2* M1 2015 HSC 14b

A particle is moving horizontally. Initially the particle is at the origin `O` moving with velocity `1 text(ms)^(−1)`.

The acceleration of the particle is given by  `ddot x = x − 1`, where `x` is its displacement at time  `t`.

  1. Show that the velocity of the particle is given by  `dot x = 1 − x`.  (3 marks)

  2. Find an expression for `x` as a function of `t`.  (2 marks)

  3. Find the limiting position of the particle.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `x = 1 − e^(-t)`
  3. `x = 1`
Show Worked Solution

i.   `ddot x = d/(dx)(1/2 v^2) = x − 1`

`1/2 v^2` `= int ddot x\ dx`
  `= int x − 1\ dx`
  `= 1/2x^2 − x + c`

 
`text(When)\ \ x = 0, \ v = 1:`

`1/2·1^2` `= 0 − 0 + c`
`c` `= 1/2`

 

`:.1/2  v^2` `= 1/2x^2 − x + 1/2`
`v^2` `= x^2 − 2x + 1`
  `= (x − 1)^2`
`:.dot x` `= ±(x − 1)`

 
`text(S)text(ince)\ \ dotx = 1\ \ text(when)\ \ x = 0,`

`dot x = 1 − x\ \ …\ text(as required)`

 

ii.   `(dx)/(dt)` `= 1 − x\ \ \ text{(from (i))}`
  `(dt)/(dx)` `= 1/(1 − x)`
  `t` `= int 1/(1 − x)\ dx`
    `= -ln(1 − x) + c`

 

`text(When)\ \ t = 0, x = 0:`

♦ Mean mark 49%.
`0` `= -ln1 + c`
`:.c` `= 0`
`t` `= -ln(1 − x)`
`-t` `= ln(1 − x)`
`1 − x` `= e^(-t)`
`:.x` `= 1 − e^(-t)`

 

♦ Mean mark 42%.
iii.   `text(As)\ t` `→ ∞`
  `e^(-t)` `→ 0`
  `x` `→ 1`

 

`:.\ text(Limiting position is)\ \ x = 1`

Calculus, EXT1 C2 2015 HSC 13d

Let  `f(x) = cos^(-1)\ (x) + cos^(-1)\ (-x)`, where  `-1 ≤ x ≤ 1`.

  1. By considering the derivative of  `f(x)`, prove that  `f(x)`  is constant.  (2 marks)

  2. Hence deduce that  `cos^(-1)\ (-x) = pi - cos^(-1)\ (x)`.  (1 mark)

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

i.  `text(Prove)\ f(x)\ text(is a constant)`

`f(x)` `= cos^(-1)(x) + cos^(-1)(-x), \ \ -1 ≤ x ≤ 1`
`f′(x)` `= (-1)/sqrt(1 – x^2) + (-1)/sqrt(1 -(-x)^2) xx d/(dx) (-x)`
  `= (-1)/sqrt(1 – x^2) + 1/sqrt(1 – x^2)`
  `= 0`

 
`:.\ text(S)text(ince)\ \ f′(x) = 0,  f(x)\ text(must be a constant.)`
 

♦ Mean mark 35%.
ii.   `f(0)` `= cos^(−1)(0) + cos^(−1)(0)`
    `= pi/2 + pi/2`
    `= pi`

 
`:.f(x) = pi`

`pi = cos^(−1)(x) + cos^(−1)(−x)`

`:.cos^(−1)(−x) = pi – cos^(−1)(x)\ \ …text(as required)`

Financial Maths, 2ADV M1 2007 HSC 3b

Heather decides to swim every day to improve her fitness level.

On the first day she swims 750 metres, and on each day after that she swims `100` metres more than the previous day. That is, she swims 850 metres on the second day, 950 metres on the third day and so on.

  1. Write down a formula for the distance she swims on the `n`th day.  (1 mark)

  2. How far does she swim on the 10th day?  (1 mark)

  3. What is the total distance she swims in the first 10 days?  (1 mark)

  4. After how many days does the total distance she has swum equal the width of the English Channel, a distance of 34 kilometres?  (2 marks)

Show Answers Only
  1. `T_n = 650 + 100n`
  2. `1650\ text(metres)`
  3. `12\ text(km)`
  4. `34\ text(km)`
Show Worked Solution
i.  `T_1` `= 750`
`T_2` `= 850`

`=>  text(AP)\ text(where)\ a = 750\ ,\ d = 100`

`:. T_n` `= a + (n-1) d`
  `= 750 + (n – 1) 100`
  `= 750 + 100n – 100`
  `= 650 + 100n`

 

ii.  `T_10` `= 650 + 100 xx 10`
  `= 1650\ text(metres)`

 

`:.\ text(She swims 1650 metres on the 10th day.)`

 

iii.  `S_n = n/2 [2a + (n – 1) d]`

`:. S_10` `= 10/2 [2 xx 750 + (10 -1) 100]`
  `= 5 [1500 + 900]`
  `= 12\ 000`

 

`:.\ text(She swims 12 km in the first 10 days.)`

 

iv.  `text(Find)\ n\ text(such that)\ S_n = 34\ text(km)`

`n/2 [2 xx 750 + (n – 1) 100]` `= 34\ 000`
`n/2 [1500 + 100n – 100]` `= 34\ 000`
`n/2 [1400 + 100n]` `= 34\ 000`
`700n + 50n^2` `= 34\ 000`
`50 n^2 + 700n – 34\ 000` `= 0`
`50 (n^2 + 14 n – 680)` `= 0`

 

`text(Using the quadratic formula)`

`n` `= {-b +- sqrt(b^2 – 4ac)}/(2a)`
  `= {-14 +- sqrt(14^2 – 4 xx 1 xx (-680))}/(2 xx 1)`
  `= (-14 +- sqrt 2916)/2`
  `= (-14 +- 54)/2`
  `= 20 or -34`
  `= 20\ ,\ n > 0`

 

`:.\ text(Her total distance equals 34 km after 20 days.)`

Calculus, 2ADV C3 2007 HSC 2c

The point  `P (pi, 0)`  lies on the curve  `y = x sinx`. Find the equation of the tangent to the curve at  `P`.  (3 marks)

Show Answers Only

`y = – pi x + pi^2`

Show Worked Solution

`y = x sin x`

`(dy)/(dx)` `= x xx d/(dx) (sin x) + d/(dx) x xx sin x`
  `= x  cos x + sin x`

 
`text(When)\ \ x = pi`

`(dy)/(dx)` `= pi xx cos pi + sin pi`
  `= pi (-1) + 0`
  `= – pi`

 
`text(Equation of line,)\ \ m = – pi,\ text(through)\ P(pi, 0):`

`y – y_1` `= m(x – x_1)`
`y – 0` `= – pi(x – pi)`
`:. y` `= – pi x + pi^2`

Combinatorics, EXT1 A1 2015 HSC 13b

Consider the binomial expansion
 

`(2x + 1/(3x))^18 = a_0x^(18) + a_1x^(16) + a_2x^(14) + …`
 

where `a_0, a_1, a_2`, . . . are constants.

  1. Find an expression for `a_2`.  (2 marks)

  2. Find an expression for the term independent of `x`.  (2 marks)

Show Answers Only
  1. `(\ ^18C_2 · 2^(16))/(3^2)`
  2. `(\ ^(18)C_9 · 2^9)/(3^9)`
Show Worked Solution

i.   `text(Need co-efficient of)\ x^(14)`

`text(General term of)\ (2x + 1/(3x))^(18)`

`T_k` `= \ ^(18)C_k(2x)^(18 − k) · (1/(3x))^k`
  `= \ ^(18)C_k · 2^(18 − k) · x^(18 − k) · 3^(−k) · x^(−k)`
  `= \ ^(18)C_k · 2^(18 − k) · 3^(−k) · x^(18 − 2k)`

 

`a_2\ text(occurs when:)`

`18 − 2k` `= 14`
`2k` `= 4`
`k` `= 2`

 

`:.a_2` `= \ ^(18)C_2 · 2^(18 − 2) · 3^(−2)`
  `= (\ ^(18)C_2 · 2^(16))/(3^2)`

 

ii.  `text(Independent term occurs when:)`

`18 − 2k` `= 0`
`2k` `= 18`
`k` `= 9`

 
`:.\ text(Independent term)`

`= \ ^(18)C_9 · 2^(18− 9) · 3^(−9)`

`= (\ ^(18)C_9 · 2^9)/(3^9)`

Mechanics, EXT2* M1 2015 HSC 13a

A particle is moving along the `x`-axis in simple harmonic motion. The displacement of the particle is `x` metres and its velocity is `v` ms`\ ^(–1)`. The parabola below shows `v^2` as a function of `x`.

2015 13a

  1. For what value(s) of `x` is the particle at rest?  (1 mark)

  2. What is the maximum speed of the particle?  (1 mark)

  3. The velocity `v` of the particle is given by the equation

         `v^2 = n^2(a^2 − (x −c)^2)`  where `a`, `c` and `n` are positive constants.

     

    What are the values of `a`, `c` and `n`?  (3 marks)

Show Answers Only
  1. `3\ text(or)\ 7`
  2. `V = sqrt11\ text(m/s)`
  3. `a = 2, c = 5,`
  4. `n = (sqrt11)/2`
Show Worked Solution

i.   `text(Particle is at rest when)\ v^2 = 0`

`:. x = 3\ \ text(or)\ \ 7`

 

ii.   `text(Maximum speed occurs when)`

`v^2` `= 11`
`v` `= sqrt11\ text(m/s)`

 

iii.  `v^2 = n^2(a^2 − (x − c)^2)`

♦ Mean mark 41%.

`text(Amplitude) = 2`

`:.a = 2`

`text(Centre of motion when)\ x = 5`

`:.c = 5`

`text(S)text(ince)\ \ v^2 = 11\ \ text(when)\ \ x = 5`

`11` `= n^2(2^2 − (5 − 5)^2)`
  `= 4n^2`
`n^2` `= 11/4`
`:.n`  `= sqrt11/2`

Trig Ratios, EXT1 2015 HSC 12d

A kitchen bench is in the shape of a segment of a circle. The segment is bounded by an arc of length 200 cm and a chord of length 160 cm. The radius of the circle is `r` cm and the chord subtends an angle `theta` at the centre `O` of the circle.
 

2015 12d
 

  1. Show that  `160^2 = 2r^2 (1 - cos\ theta)`.  (1 mark)
  2. Hence, or otherwise, show that  `8 theta^2 + 25 cos\ theta - 25 = 0`.  (2 marks)
  3. Taking  `theta_1 = pi`  as a first approximation to the value of  `theta`, use one application of Newton’s method to find a second approximation to the value of  `theta`. Give your answer correct to two decimal places.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `2.57\ \ text{(to 2 d.p.)}`
Show Worked Solution

(i)   `text(Using the cosine rule)`

`a^2` `= b^2 + c^2 − 2bc\ cos\ A`
`160^2` `= r^2 + r^2 −2 xx r xx r xx cos\ theta`
  `= 2r^2 − 2r^2\ cos\ theta`
  `= 2r^2(1 − cos\ theta)\ \ …\ text(as required)`

 

(ii)   `text(Show)\ \ 8theta^2 + 25\ cos\ theta − 25 = 0`

`text(Arc length)` `= 200`
`theta/(2pi) xx 2pir` `= 200`
`rtheta` `= 200`
`r` `= 200/theta`

 

`text(Substitute)\ \ r = 200/theta\ text{into part (i) equation}`

`2 xx (200/theta)^2(1 – cos\ theta)` `= 160^2`
`80\ 000(1 – cos\ theta)` `= 25\ 600\ theta^2`
`25(1 – cos\ theta)` `= 8theta^2`
`8theta^2 + 25\ cos\ theta – 25` `= 0\ \ …\ text(as required.)`

 

(iii)   `f(theta)` `= 8theta^2 + 25\ cos\ theta – 25`
  `f′(theta)` `= 16theta – 25\ sin\ theta`
  `f(pi)` `= 8pi^2 + 25\ cos\ pi – 25`
    `= 28.9568…`
  `f′(pi)` `= 16pi – 25\ sin\ pi`
    `= 50.2654…`
`theta_2` `= pi – (f(pi))/(f′(pi))`
  `= pi – (28.9568…)/(50.2654…)`
  `= 2.5655…`
  `= 2.57\ \ text{(to 2 d.p.)}`

Trig Ratios, EXT1 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)}`

Quadratic, EXT1 2015 HSC 12b

The points  `P(2ap, ap^2)`  and  `Q(2aq, aq^2)`  lie on the parabola  `x^2 = 4ay`.

The equation of the chord  `PQ`  is given by  `(p + q)x- 2y- 2apq = 0`.  (Do NOT prove this.)

  1. Show that if  `PQ`  is a focal chord then  `pq = –1`.  (1 mark)
  2. If  `PQ`  is a focal chord and  `P`  has coordinates  `(8a, 16a)`, what are the coordinates of  `Q`  in terms of  `a`?  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(−a/2, a/16)`
Show Worked Solution

(i)   `text(Show if)\ PQ\ text(is a focal chord)\ pq = −1`
 

Quadtratic, EXT1 2015 HSC 12b Answer1

 
`(p + q)x – 2y – 2apq = 0`

`text(If)\ PQ\ text(is a focal chord, it passes)\ (0, a)`

`(p + q)0 − 2a − 2apq` `= 0`
`2apq` `= -2a`
`pq` `= (−2a)/(2a)`
  `= −1\ \ …\ text(as required)`

 

(ii)  `text(If)\ P(2ap, ap^2) = (8a, 16a)`

`⇒ 2ap` `= 8a`
`p` `= 4`
`text(Using)\ \ q` `= −1/p\ \ \ text{(from part (i))}`
`q` `= −1/4`
`:.Q(2aq, aq^2)` `= (2a(−1/4), a(−1/4)^2)`
  `= (−a/2, a/16)`

Trig Calculus, EXT1 2015 HSC 10 MC

The graph of the function  `y = cos\ (2t − pi/3)`  is shown

2015 10 mc

What are the coordinates of the point `P`?

  1. `((5pi)/(12), 0)`
  2. `((2pi)/3, 0)`
  3. `((11pi)/(12), 0)`
  4. `((7pi)/6, 0)`
Show Answers Only

`C`

Show Worked Solution

`y = cos\ (2t − pi/3)`

`cos\ x = 0\ text(when)\ x = pi/2\ text(or)\ (3pi)/2`

`:.2t − pi/3` `= pi/2`    `text(or)` `2t − pi/3` `= (3pi)/2`
`2t` `= (5pi)/6`   `2t` `= (11pi)/6`
`t` `= (5pi)/12`   `t` `= (11pi)/12`

`text(From the graph,)\ P\ text(is where)\ y = cos\ (2t − pi/3)`

`text(hits the)\ xtext(-axis the second time.)`

`:.P\ text(has the coordinates)\ ((11pi)/12, 0)`

`⇒ C`

Mechanics, EXT2* M1 2015 HSC 9 MC

Two particles oscillate horizontally. The displacement of the first is given by  `x = 3\ sin\ 4t`  and the displacement of the second is given by  `x = a\ sin\ nt`. In one oscillation, the second particle covers twice the distance of the first particle, but in half the time.

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

  1. `a = 1.5,\ \ n = 2`
  2. `a = 1.5, \ \ n = 8`
  3. `a = 6,\ \ n = 2`
  4. `a = 6, \ \ n = 8`
Show Answers Only

`D`

Show Worked Solution
`x_1` `= 3\ sin\ 4t`
`x_2` `= a\ sin\ nt`

 

`x_2\ text(has twice the amplitude of)\ x_1`

`:. a = 2 xx 3 = 6`

`x_2\ text(has a period)\ (T)\ text(that is half of)\ x_1`

`T(x_1)` `= (2 pi)/n = (2pi)/4`
`:.T(x_2)` `= 1/2 xx (2 pi)/4 = (2pi)/8`
`:.n` `= 8`

 
`=> D`

Geometry and Calculus, EXT1 2015 HSC 5 MC

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

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

`A`

Show Worked Solution

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

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

`text(As)\ x → ∞, y → 0^+`

`text(As)\ x → −∞, y → 0^−`

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

`⇒ A`

Statistics, STD2 S4 2015 HSC 28e

The shoe size and height of ten students were recorded.

\begin{array} {|l|c|c|}
\hline \rule{0pt}{2.5ex} \text{Shoe size} \rule[-1ex]{0pt}{0pt} & \text{6} & \text{7} & \text{7} & \text{8} & \text{8.5} & \text{9.5} & \text{10} & \text{11} & \text{12} & \text{12} \\
\hline \rule{0pt}{2.5ex} \text{Height} \rule[-1ex]{0pt}{0pt} & \text{155} & \text{150} & \text{165} & \text{175} & \text{170} & \text{170} & \text{190} & \text{185} & \text{200} & \text{195} \\
\hline
\end{array}

  1. Complete the scatter plot AND draw a line of fit by eye.  (2 marks)
     
     
  2. Use the line of fit to estimate the height difference between a student who wears a size 7.5 shoe and one who wears a size 9 shoe.  (1 mark)

  3. A student calculated the correlation coefficient to be 1 for this set of data. Explain why this cannot be correct.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `13\ text{cm  (or close given LOBF drawn)}
  3. `text(A correlation co-efficient of 1 would)`
    `text(mean that all data points occur on the)`
    `text(line of best fit which clearly isn’t the case.)`
Show Worked Solution

i.    
      2UG 2015 28e Answer

ii.   `text{Shoe size 7½ gives a height estimate of 162 cm (see graph)}`

`text{Shoe size 9 gives a height estimate of 175 cm (see graph)}`

`:.\ text(Height difference)` `= 175-162`
  `= 13\ text{cm  (or close given LOBF drawn)}`

 

iii.   `text(A correlation co-efficient of 1 would mean)`

♦ Mean mark (c) 39%.

`text(that all data points occur on the line of best)`

`text(fit which clearly isn’t the case.)`

Algebra, STD2 A1 2015 HSC 28d

The formula  `C = 5/9 (F-32)`  is used to convert temperatures between degrees Fahrenheit `(F)` and degrees Celsius `(C)`.

Convert 3°C to the equivalent temperature in Fahrenheit.   (2 marks)

Show Answers Only

`37.4\ text(degrees)\ F`

Show Worked Solution
`C` `= 5/9(F-32)`
`F-32` `= 9/5C`
`F`  `= 9/5C + 32`

 
`text(When)\ \ C = 3,`

`F` `= (9/5 xx 3) + 32`
  `= 37.4\ text(degrees)\ F`

Measurement, 2UG 2015 HSC 28c

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

2UG 2015 29c

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

Show Answers Only

`4\ text(litres)`

Show Worked Solution

2UG 2015 29c Answer 

`V` `≈ h/3[y_0 + 4y_1 + y_2]`
  `≈ 15/3[45 + 4(180) + 35]`
  `≈ 5[800]`
  `≈ 4000\ text(cm³)`
  `~~4\ text(litres)\ \ \ text{(1 cm³ = 1 mL)}`

Measurement, STD2 M1 2015 HSC 28a

The diagram shows an annulus.
 

2015 28a

 
Calculate the area of the annulus.  (1 mark)

Show Answers Only

`≈ 50.26…\ text(cm²)`

Show Worked Solution

`text(Area of annulus)`

`= pi(R^2 − r^2)`

`= pi(5^2 − 3^2)`

`= pi(25 − 9)`

`= 16pi\ text(cm²)`

`≈ 50.26…\ text(cm²)`

Statistics, STD2 S1 2015 HSC 27d

In a small business, the seven employees earn the following wages per week:

\(\$300, \ \$490, \ \$520, \ \$590, \ \$660, \ \$680, \ \$970\)

  1.  Is the wage of $970 an outlier for this set of data? Justify your answer with calculations.  (3 marks)

  2.  Each employee receives a $20 pay increase.

     

     What effect will this have on the standard deviation?  (1 mark)

Show Answers Only

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

b.    \(\text{The standard deviation will remain the same.}\)

Show Worked Solution

a.    \(300, 490, 520, 590, 660, 680, 970\)

\(\text{Median}\) \(= 590\)
\(Q_1\) \(= 490\)
\(Q_3\) \(= 680\)
\(IQR\) \(= 680-490 = 190\)

 

\(\text{Outlier if \$970 is greater than:} \)

\(Q_3 + 1.5 x\times IQR = 680 + 1.5 \times 190 = \$965 \) 

\(\therefore\ \text{The wage \$970 per week is an outlier.}\)

♦ Mean mark (a) 39%.


b.    \(\text{All values increase by \$20, but so too does the mean.} \)

\(\text{Therefore the spread about the new mean will not change} \)

\(\text{and therefore the standard deviation will remain the same.} \)

Measurement, STD2 M7 2015 HSC 27a

At a particular time during the day, a tower of height 19.2 metres casts a shadow. At the same time, a person who is 1.65 metres tall casts a shadow 5 metres long.

  

What is the length of the shadow cast by the tower at that time?  (2 marks)

Show Answers Only

`58\ text{m}`

Show Worked Solution

`text(Both triangles have right angles and a common)`

`text(angle to the ground.)`

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

 

`text(Let)\ x =\ text(length of tower shadow)`

`x/5` `= 19.2/1.65\ \ text{(corresponding sides of similar triangles)}`

 
`x` `= (5 xx 19.2)/1.65`  
  `= 58.1818…`  
  `= 58\ text{m  (nearest m)}`  

FS Comm, 2UG 2015 HSC 26g

Pat’s mobile phone plan is shown.

2015 26g

Last month Pat:

• made calls to the value of `$561`
• sent `152` SMS messages
• sent `37` MMS messages
• used `1.7` GB of data.

What was the total of Pat’s phone bill for last month?  (3 marks)

Show Answers Only

`$96.36`

Show Worked Solution

`text(C)text(ost of excess calls)`

`= 561 − 550`

`= $11`

`text(Excess data)`

`1.7\ text(GB)` `= 1.7 xx 2^(10)\ text(MB)`
`text(Excess)` `= 1.7 xx 2^(10) − 500`
  `= 1240.8\ text(MB)`

 

`:.\ text(C)text(ost of excess data)`

`= 1240.8 xx $0.0293`

`= $36.355…`

`= $36.36`

 

`text(SMS and MMS messages within plan)`

`:.\ text(Total phone bill)`

`= $49 + $11 + $36.36`

`= $96.36`

Measurement, STD2 M1 2015 HSC 26f

Approximately 71% of Earth’s surface is covered by water. Assume Earth is a sphere with a radius of 6400 km.

Calculate the number of square kilometres covered by water.  (2 marks)

Show Answers Only

`3.7 xx 10^8\ text(km²)\ \ text{(nearest km²)}`

Show Worked Solution

`text(Surface area of Earth)`

`= 4pir^2`

`= 4pi xx 6400^2`

 

`:.\ text(Surface covered by water)`

`= text(71%) xx 4pi xx 6400^2`

`= 365\ 450\ 163.7…`

`= 3.7 xx 10^8\ text(km²)\ \ text{(nearest km²)}`

Probability, STD2 S2 2015 HSC 26e

The table shows the relative frequency of selecting each of the different coloured jelly beans from packets containing green, yellow, black, red and white jelly beans.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Colour} \rule[-1ex]{0pt}{0pt} & \textit{Relative frequency} \\
\hline
\rule{0pt}{2.5ex} \text{Green} \rule[-1ex]{0pt}{0pt} & 0.32 \\
\hline
\rule{0pt}{2.5ex} \text{Yellow} \rule[-1ex]{0pt}{0pt} & 0.13 \\
\hline
\rule{0pt}{2.5ex} \text{Black} \rule[-1ex]{0pt}{0pt} & 0.14 \\
\hline
\rule{0pt}{2.5ex} \text{Red} \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} \text{White} \rule[-1ex]{0pt}{0pt} & 0.24 \\
\hline
\end{array}

  1. What is the relative frequency of selecting a red jelly bean?  (1 mark)

  2. Based on this table of relative frequencies, what is the probability of NOT selecting a black jelly bean?  (1 mark)

Show Answers Only
  1. \(0.17\)
  2. \(0.86\)
Show Worked Solution

i.  \(\text{Relative frequency of red}\)

\(= 1-(0.32 + 0.13 + 0.14 + 0.24)\)

\(= 1-0.83\)

\(= 0.17\)

 

ii.  \(P\text{(not selecting black)}\)

\(= 1-P\text{(selecting black)}\)

\(= 1-0.14\)

\(= 0.86\)

Financial Maths, STD2 F4 2015 HSC 26d

A family currently pays $320 for some groceries.

Assuming a constant annual inflation rate of 2.9%, calculate how much would be paid for the same groceries in 5 years’ time.  (2 marks)

Show Answers Only

`$369.17\ \ text{(nearest cent)}`

Show Worked Solution
`FV` `= PV(1 + r)^n`
  `= 320(1.029)^5`
  `= $369.1703…`
  `= $369.17\ \ text{(nearest cent)}`

Data, 2UG 2015 HSC 26a

A farmer used the ‘capture‑recapture’ technique to estimate the number of chickens he had on his farm. He captured, tagged and released 18 of the chickens. Later, he caught 26 chickens at random and found that 4 had been tagged.

What is the estimate for the total number of chickens on this farm?  (2 marks)

Show Answers Only

`117`

Show Worked Solution

`text(Let)\ C = text(total chickens on farm)`

`text(Capture)` `= 18/C`
`text(Recapture)` `= 4/26`
`⇒ 18/C` `= 4/26`
`4C` `= 18 xx 26`
`C` `= (18 xx 26)/4`
  `= 117`

 

`:.\ text(There are an estimated 117 chickens)`

`text(on the farm.)`

Financial Maths, STD2 F1 2015 HSC 25 MC

An insurance company offers customers the following discounts on the basic annual premium for car insurance.

2015 25 mc

If a customer is eligible for more than one discount, subsequent discounts are applied to the already discounted premium. The combined compulsory third party (CTP) and comprehensive insurance discount is always applied last.

Jamie has three insurance policies, including combined CTP and comprehensive insurance, with this company. He has used this company for 8 years and he has never made a claim.

The basic annual premium for his car insurance is $870.

How much will Jamie need to pay after the discounts are applied?

  1.    $482.44
  2.    $515.50
  3.    $541.60
  4.    $557.60
Show Answers Only

`C`

Show Worked Solution

`text(Multi-policy discount)`

`text(New premium)` `= 870 − (text(15%) xx 870)`
  `= $739.50`

 
`text(No claim bonus)`

`text(New premium)` `= 739.50 − (text(20%) xx 739.50)`
  `= $591.60`

 
`text(Combined CTP bonus)`

`text(New premium)` `= 591.60 − 50`
  `= $541.60`

`⇒ C`

Algebra, STD2 A1 2015 HSC 24 MC

Consider the equation  `(2x)/3-4 = (5x)/2 + 1`.

Which of the following would be a correct step in solving this equation?

  1. `(2x)/3-3 = (5x)/2`
  2. `(2x)/3 = (5x)/2 + 5`
  3. `2x-4 = (15x)/2 + 3`
  4. `(4x)/6-8 = 5x + 2`
Show Answers Only

`B`

Show Worked Solution
`(2x)/3-4` `= (5x)/2 + 1`
`(2x)/3` `= (5x)/2 + 5`

 
`=>B`

Algebra, STD2 A1 2015 HSC 23 MC

The number of ‘standard drinks’ in various glasses of wine is shown.
 

A woman weighing 62 kg drinks three small glasses of white wine and two large glasses of red wine between 8 pm and 1 am.

Using the formula for calculating blood alcohol below, what would be her blood alcohol content (BAC) estimate at 1 am, correct to three decimal places?
 

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

where    `N` is the number of standard drinks consumed

`H` is the number of hours drinking

`M` is the person's mass in kilograms
 

  1. `0.030`
  2. `0.037`
  3. `0.046`
  4. `0.057`
Show Answers Only

`D`

Show Worked Solution
`N` `= 3 xx 0.9 + 2 xx 1.5`
  `= 5.7\ text(standard drinks)`
`H` `=\ text(5 hours)`
`M` `=\ text(62 kg)`
   
`:.BAC_f` `= (10 xx 5.7-7.5 xx 5)/ (5.5 xx 62)`
  `= 0.05718…`

`⇒ D`

Calculus, EXT1* C1 2015 HSC 14a

In a theme park ride, a chair is released from a height of  `110`  metres and falls vertically. Magnetic brakes are applied when the velocity of the chair reaches  `text(−37)`  metres per second.
 

2015 2ua 14a
 

The height of the chair at time `t` seconds is `x` metres. The acceleration of the chair is given by   `ddot x = −10`. At the release point,  `t = 0, x = 110 and dot x = 0`.

  1. Using calculus, show that  `x = -5t^2 + 110`.  (2 marks)

  2. How far has the chair fallen when the magnetic brakes are applied?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `68.45\ text(m)`
Show Worked Solution

i.  `text(Show)\ \ x = -5t^2 + 110`

`ddot x` `= -10`
`dot x` `= int ddot x\ dt`
  `= int -10\ dt`
  `= -10t + c`

 
`text(When)\ \ t = 0,\ dot x = 0`

`:.\ 0` `= -10 (0) + c`
`c` `= 0`
`dot x` `= -10t`
`x` `= int dot x\ dt`
  `= int -10t\ dt`
  `= -5t^2 + c`

 
`text(When)\ \ t = 0,\ x = 110`

`:.\ 110` `= -5 (0^2) + c`
`c` `= 110`

 
`:.\ x = -5t^2 + 110\ \ text(…  as required.)`
 

ii.  `text(Find)\ \t\ \text(when)\ \ dot x = -37`

`-37` `= -10t`
`t` `= 3.7\ \ text(seconds)`

 
`text(When)\ \ t = 3.7`

`x` `= -5 (3.7^2) + 110`
  `= -68.45 + 110`
  `= 41.55`

 
`:.\ text(Distance the chair has fallen)`

`= 110 – 41.55`

`= 68.45\ text(m)`

Financial Maths, STD2 F4 2015 HSC 17 MC

What amount must be invested now at 4% per annum, compounded quarterly, so that in five years it will have grown to  $60 000?

  1. $8919
  2. $11 156
  3. $49 173
  4. $49 316
Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ FV = PV(1 + r)^n`

`r` `= text(4%)/4` `= text(1%) = 0.01\ text(per quarter)`
`n` `= 5 xx 4` `= 20\ text(quarters)`

 

`60\ 000` `= PV(1 + 0.01)^(20)`
`:.PV` `= (60\ 000)/1.01^(20)`
  `= $49\ 172.66…`

`⇒ C`

Probability, STD2 S2 2015 HSC 16 MC

The probability of winning a game is `7/10`.

Which expression represents the probability of winning two consecutive games?

  1. `7/10 xx 6/9`
  2. `7/10 xx 6/10`
  3. `7/10 xx 7/9`
  4. `7/10 xx 7/10`
Show Answers Only

`D`

Show Worked Solution

`text{Since the two events are independent:}`

`P text{(W)}` `= 7/10`
`P text{(WW)}` `= 7/10 xx 7/10`

 
`=>D`

Measurement, 2UG 2015 HSC 14 MC

Stockholm is located at `text(59°N 18°E)` and Darwin is located at `text(13°S 131°E)`. 

What is the time difference between Stockholm and Darwin? (Ignore time zones and daylight saving.)

(A)   `184` minutes

(B)   `288` minutes

(C)   `452` minutes

(D)   `596` minutes

Show Answers Only

`C`

Show Worked Solution

`text(Stockholm is 59°N 18°E,  Darwin is 13°S 131°E)`

`text{Angular difference (longitude)}`

`= 131^@− 18^@`

`= 113^@`
 

`:.\ text(Time difference)` `= 113 xx 4`
  `= 452\ text(minutes)`

`⇒ C`

Financial Maths, STD2 F4 2015 HSC 10 MC

A piece of machinery, initially worth $56 000, depreciates at 8% per annum.

Which graph best shows the salvage value of this piece of machinery over time?
 

2015 10 mc1

2015 10 mc2

Show Answers Only

`A`

Show Worked Solution

`text(By Elimination)`

`text(A depreciation of 8% per annum depreciates the largest)`

`text(amount in year 1 and then gradually depreciates less each)`

`text(subsequent year.)`

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

`text(Consider when)\ t` `= 5`
`text(Salvage Value)` `= V_0(1 − r)^n`
  `= 56\ 000(1 − 0.08)^5`
  `= 36\ 908.5…`

`text(Graph B depreciates too quickly)`

`:.text(Cannot be)\ B`

`⇒ A`

Measurement, STD2 M1 2015 HSC 8 MC

The Louvre Pyramid in Paris has a square base with side length 35 m and a perpendicular height of 22 m.
 

What is the volume of this pyramid, to the nearest m³?

  1. `257\ text(m)^3`
  2. `1027\ text(m)^3`
  3. `8983\ text(m)^3`
  4. `26\ 950\ text(m)^3`
Show Answers Only

`C`

Show Worked Solution
`V` `= 1/3Ah`
`A` `= 35 xx 35`
  `= 1225\ text(m)^2`

 

`:.V` `= 1/3 xx 1225 xx 22`
  `= 8983.33…\ text(m)^3`

 
`=>C`

Statistics, STD2 S1 2015 HSC 6 MC

The times, in minutes, that a large group of students spend on exercise per day are presented in the box‑and‑whisker plot.
 

What percentage of these students spend between 40 minutes and 60 minutes per day on exercise?

  1. 17%
  2. 20%
  3. 25%
  4. 50%
Show Answers Only

`C`

Show Worked Solution

`text{Q}_1 = 40, \ text(Median) = 60`

`:.\ text(% Students between 40 and 60)`

`= 50text{%}-25text{%}`

`=25 text{%}`
 

`=>C`

Statistics, STD2 S1 2015 HSC 4 MC

On a school report, a student’s record of completing homework is graded using the following codes.

C = consistently
U = usually
S = sometimes
R = rarely
N = never

What type of data is this?

  1. Categorical, ordinal
  2. Categorical, nominal
  3. Numerical, continuous
  4. Numerical, discrete
Show Answers Only

`A`

Show Worked Solution

`text(The data has been grouped into categories and)`

`text(because each category can be ranked, it is ordinal.)`

`⇒ A`

Calculus, 2ADV C4 2015 HSC 15c

Water is flowing in and out of a rock pool. The volume of water in the pool at time `t` hours is `V` litres. The rate of change of the volume is given by

`(dV)/(dt) = 80 sin(0.5t)`

At time  `t = 0`, the volume of water in the pool is 1200 litres and is increasing.

  1. After what time does the volume of water first start to decrease?  (2 marks)

  2. Find the volume of water in the pool when  `t = 3`.  (2 marks)

  3. What is the greatest volume of water in the pool?  (1 mark)

Show Answers Only
  1. `2 pi\ text(hours)`
  2. `1349\ text(litres)`
  3. `1520\ text(litres)`
Show Worked Solution

i.   `(dV)/(dt) = 80 sin (0.5t)`

♦ Mean mark (i) 37%.

`text(Volume decreases when)\ \ (dV)/(dt) < 0`

`(dV)/(dt) < 0\ \ text(when)\ \ t > 2 pi\ text(hours)`

`:.\ text(After 2)\pi\ \text(hours, volume starts to decrease.)`

 

ii.   `V` `= int (dV)/(dt)\ dt`
  `= int 80 sin (0.5t)\ dt`
  `= -160 cos (0.5t) + c`

 

`text(When)\ \ t = 0,\ V = 1200`

`1200` `= -160 cos 0 + c`
`c` `= 1360`
`:.\ V` `= -160 cos (0.5t) + 1360`

 

`text(When)\ \ t = 3`

`V` `= -160 cos (0.5 xx 3) + 1360`
  `= 1348.68…\ \ text(litres)`
  `= 1349\ text{litres  (nearest litre)}`

 

iii.  `V = -160 cos (0.5t) + 1360`

♦♦ Mean mark (iii) 27%.

`=>\ text(Greatest volume occurs when)`

`cos (0.5t) = -1`

`:.\ text(Maximum volume)`

`= -160 (-1) + 1360`

`= 1520\ text(litres)`

Plane Geometry, 2UA 2015 HSC 15b

The diagram shows `Delta ABC` which has a right angle at `C`. The point `D` is the midpoint of the side `AC`. The point `E` is chosen on `AB` such that `AE = ED`. The line segment `ED` is produced to meet the line `BC` at `F`.

Copy or trace the diagram into your writing booklet.

  1. Prove that `Delta ACB` is similar to `Delta DCF.`  (2 marks)
  2. Explain why `Delta EFB` is isosceles.  (1 mark)
  3. Show that `EB = 3AE.`  (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(Prove)\ \ Delta ACB\ \ text(|||)\ \ Delta DCF`

`/_ EAD = /_ ADE = theta\ \ text{(base angles of isosceles}\ \ Delta AED text{)}`

`/_ CDF = /_ ADE = theta\ \ text{(vertically opposite angles)}`

`/_ DCF = /_ ACB = 90°\ \ text{(}/_ FCB\ text{is a straight angle)}`

`:.\ Delta ACB\ \ text(|||)\ \ Delta DCF\ \ text{(equiangular)}`

 

♦ Mean mark 45%.

(ii)  `/_ DFC = 90 – theta\ \ text{(angle sum of}\ \ Delta DCF text{)}`

`/_ ABC = 90 – theta\ \ text{(angle sum of}\ \ Delta ACB text{)}`

`:.\ Delta EFB\ \ text{is isosceles  (base angles are equal)}`

 

(iii)  `text(Show)\ \ EB = 3AE`

♦♦ Mean mark 23%.
`(DC)/(AC)` `= (DF)/(AB)`

`\ \ \ \ \ text{(corresponding sides of}`

`\ \ \ \ \ text{similar triangles)}`
`1/2` `= (DF)/(AB)`  
`2DF` `= AB`  
`2(EF – ED)` `= AE + EB`  
`2(EB – AE)` `= AE + EB` `\ \ \ \ \ text{(given}\ EF = EB, ED = AE text{)}`
`2EB – 2AE` `= AE + EB`  

 

`:. EB = 3AE\ \ text(…  as required)`

Calculus, EXT1* C1 2015 HSC 15a

The amount of caffeine, `C`, in the human body decreases according to the equation

`(dC)/(dt) = -0.14C,` 

where `C` is measured in mg and `t` is the time in hours.

  1. Show that  `C = Ae^(-0.14t)`  is a solution to  `(dC)/(dt) = -0.14C,` where ` A` is a constant.

     

    When `t = 0`, there are 130 mg of caffeine in Lee’s body.  (1 mark)

  2. Find the value of `A.`  (1 mark)

  3. What is the amount of caffeine in Lee’s body after 7 hours?   (1 mark)

  4. What is the time taken for the amount of caffeine in Lee’s body to halve?  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `130`
  3. `48.8\ text(mg)`
  4. `4.95\ text(hours)`
Show Worked Solution
i. `C` `= Ae^(-0.14t)`
  `(dC)/(dt)` `= d/(dt) (Ae^(-0.14t))`
    `= -0.14 xx Ae^(-0.14t)`
    `= -0.14\ C`

 
`:.\ C = Ae^(-0.14t)\ \ text(is a solution)`

 

ii.  `text(When)\ \ t = 0,\ C = 130`

`130` `= Ae^(-0.14 xx 0)`
`:.\ A` `= 130`

 

iii.  `text(Find)\ \ C\ \ text(when)\ \ t = 7`

`C` `= 130\ e^(-0.14 xx 7)`
  `= 130\ e^(-0.98)`
  `= 48.79…`
  `= 48.8\ text{mg  (to 1 d.p.)}`

 
`:.\ text(After 7 hours, Lee will have 48.8 mg)`

`text(of caffeine left in her body.)`

 

iv.  `text(Find)\ \ t\ \ text(when caffeine has halved.)`

`text(When)\ \ t = 0,\ \ C = 130`

`:.\ text(Find)\ \ t\ \ text(when)\ \ C = 65`

`65` `= 130 e^(-0.14 xx t)`
`e^(-0.14t)` `= 65/130`
`ln e^(-0.14t)` `= ln\ 65/130`
`-0.14t xx ln e` `= ln\ 65/130`
`t` `= (ln\ 65/130)/-0.14`
  `= 4.951…`
  `= 4.95\ text{hours  (to 2 d.p.)}`

 

`:.\ text(It will take 4.95 hours for Lee’s)`

`text(caffeine to halve.)`

Financial Maths, 2ADV M1 2015 HSC 14c

Sam borrows $100 000 to be repaid at a reducible interest rate of 0.6% per month. Let  `$A_n`  be the amount owing at the end of  `n`  months and  `$M`  be the monthly repayment.

  1. Show that  `A_2 = 100\ 000 (1.006)^2 - M (1 + 1.006).`  (1 mark)

  2. Show that  `A_n = 100\ 000 (1.006)^n - M (((1.006)^n - 1)/0.006).`  (2 marks)

  3. Sam makes monthly repayments of $780. Show that after making 120 monthly repayments the amount owing is $68 500 to the nearest $100.  (1 mark)

Immediately after making the 120th repayment, Sam makes a one-off payment, reducing the amount owing to $48 500. The interest rate and monthly repayment remain unchanged.

  1. After how many more months will the amount owing be completely repaid?  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(Show Worked Solutions)}`
  2. `text(Proof)\ \ text{(Show Worked Solutions)}`
  3. `text(Proof)\ \ text{(Show Worked Solutions)}`
  4. `79\ text(months)`
Show Worked Solution
i.  `A_1` `= 100\ 000 (1.006) – M`
`A_2` `= A_1 (1.006) – M`
  `= [100\ 000 (1.006) – M] (1.006) – M`
  `= 100\ 000 (1.006)^2 – M (1.006) – M`
  `= 100\ 000 (1.006)^2 – M (1 + 1.006)\ \ text(…  as required)`

 

ii.  `A_3 = 100\ 000 (1.006)^3 – M (1 + 1.006 + 1.006^2)`

`vdots`

`A_n = 100\ 000 (1.006)^n – M (1 + 1.006 + … + 1.006^(n-1))`

`=> text(S)text(ince)\ \ (1 + 1.006 + … + 1.006^(n-1))\ text(is a)`

`text(GP with)\ \ a = 1,\ r = 1.006`

`:.\ A_n` `= 100\ 000 (1.006)^n – M ((a (r^n – 1))/(r – 1))`
  `= 100\ 000 (1.006)^n – M ((1 (1.006^n – 1))/(1.006 – 1))`
  `= 100\ 000 (1.006)^n – M (((1.006)^n – 1)/0.006)`

`text(…  as required.)`

 

iii.  `text(If)\ \ M = 780 and n = 120`

`A_120` `= 100\ 000 (1.006)^120 – 780 ((1.006^120 – 1)/0.006)`
  `= 205\ 001.80… – 780 (175.0030…)`
  `= 205\ 001.80… – 136\ 502.34…`
  `= 68\ 499.45…`
  `= $68\ 500\ \ text{(to nearest $100)  …  as required}`

 

iv.  `text(After the one-off payment, amount owing)=$48\ 500`

`:. A_n = 48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006)`

`text(where)\ \ n\ \ text(is the number of months after)`

 `text(the one-off payment.)`
 

`text(Find)\ \ n\ \ text(when)\ \ A_n = 0`

`48\ 500 (1.006)^n – 780 ((1.006^n – 1)/0.006) = 0`

`48\ 500 (1.006)^n` `= 780 ((1.006^n – 1)/0.006)`
`48\ 500 (1.006)^n` `= 130\ 000 (1.006^n – 1)`
  `= 130\ 000 (1.006)^n – 130\ 000`
`81\ 500 (1.006)^n` `= 130\ 000`
`1.006^n` `= (130\ 000)/(81\ 500)`
`n xx ln 1.006` `= ln\ 1300/815`
`n` `= (ln\ 1300/815)/(ln\ 1.006)`
  `= 78.055…`

 
`:.\ text(The amount owing will be completely repaid after)`

`text(another 79 months.)`

Probability, 2ADV S1 2015 HSC 14b

Weather records for a town suggest that:

  • if a particular day is wet `(W)`, the probability of the next day being dry is  `5/6`
  • if a particular day is dry `(D)`, the probability of the next day being dry is  `1/2`.

In a specific week Thursday is dry. The tree diagram shows the possible outcomes for the next three days: Friday, Saturday and Sunday.
 

2015 2ua 14b
 

  1. Show that the probability of Saturday being dry is `2/3`.  (1 mark)

  2. What is the probability of both Saturday and Sunday being wet?  (2 marks)

  3. What is the probability of at least one of Saturday and Sunday being dry?  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1/(18)`
  3. `(17)/(18)`
Show Worked Solution

i.  `text{Show}\ \ P text{(Sat dry)} = 2/3`

`P text{(Sat dry)}`

`= P (W,D) + P (D, D)`

`=(1/2 xx 5/6) + (1/2 xx 1/2)`

`= 5/(12) + 1/4`

`= 2/3\ \ text(…  as required)`
 

ii.  `Ptext{(Sat and Sun wet)}`

`= P (WWW) + P (DWW)`

`= (1/2 xx 1/6 xx 1/6) + (1/2 xx 1/2 xx 1/6)`

`= 1/(72) + 1/(24)`

`= 1/(18)`
 

iii.  `Ptext{(At least Sat or Sun dry)}`

`= 1 – Ptext{(Sat and Sun both wet)}`

`= 1 – 1/(18)`

`= (17)/(18)`

Calculus, 2ADV C3 2015 HSC 13c

Consider the curve  `y = x^3 − x^2 − x + 3`.

  1. Find the stationary points and determine their nature.   (4 marks)

  2. Given that the point  `P (1/3, 70/27)`  lies on the curve, prove that there is a point of inflection at  `P`.  (2 marks)

  3. Sketch the curve, labelling the stationary points, point of inflection and `y`-intercept.  (2 marks)

Show Answers Only
  1. `text(MAX at)\ (-1/3, 86/27); \ text(MIN at)\ (1, 2)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3.  
Show Worked Solution
i. `y` `= x^3 – x^2 – x + 3`
  `(dy)/(dx)` `= 3x^2 – 2x – 1`
  `(d^2y)/(dx^2)` `= 6x – 2`

`text(S.P.’s when)\ (dy)/(dx) = 0`

`3x^2 – 2x – 1` `= 0`
`(3x + 1) (x – 1)` `= 0`

`x = -1/3 or 1`

 

`text(When)\ \ x = -1/3`

`f(-1/3)` `= (-1/3)^3 – (-1/3)^2 – (-1/3) + 3`
  `= -1/27 – 1/9 + 1/3 + 3`
  `= 86/27`
`f″(-1/3)` `= (6 xx -1/3) – 2 = -4 < 0`

`:.\ text(MAX at)\ \ (-1/3, 86/27)`

 

`text(When)\ \ x = 1`

`f(1)` `= 1^3 – 1^2 – 1 + 3 =2`
`f″(1)` `= (6 xx 1) – 2 = 4 > 0`

`:.\ text(MIN at)\ \ (1, 2)`

 

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

`6x-2` `=0`
`x` `=1/3`

 

`text(Checking change of concavity)`

`text(Concavity changes either side of)\ x = 1/3`

`:.\ (1/3, 70/27)\ \ text(is a P.I.)`

 

iii.  `text(When)\ \ x` `= 0`
`y` `= 3`

Functions, EXT1* F1 2015 HSC 13b

  1. Find the domain and range for the function  `f(x) = sqrt (9 - x^2)`.  (2 marks)

  2. On a number plane, shade the region where the points `(x, y)` satisfy both of the inequalities
     
      `qquad y <= sqrt (9 - x^2)`  and  `y >= x` .  (2 marks)

Show Answers Only
  1. `text(Domain)\ -3 <= x <= 3\ \ \ \ \ \ \ \ text(Range)\ 0<= y <= 3`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.  `f(x) = sqrt(9 – x^2)`

`text(Domain)`

`9 – x^2` `>= 0`
`x^2` `<= 9`

`-3 <= x <= 3`

`text(Range)`

`0 <= y <= 3`

 

♦ Mean mark 34%.
ii.   

Quadratic, 2UA 2015 HSC 12e

The diagram shows the parabola `y = x^2/2` with focus `S (0, 1/2).` A tangent to the parabola is drawn at `P (1, 1/2).`

  1. Find the equation of the tangent at the point `P`.   (2 marks)
  2. What is the equation of the directrix of the parabola?   (1 mark)
  3. The tangent and directrix intersect at `Q`.
    Show that `Q` lies on the `y`-axis.   (1 mark)

  4. Show that `Delta PQS` is isosceles.   (1 mark)
Show Answers Only
  1. `y = x – 1/2`
  2. `y = -1/2`
  3.  
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)   

`y = 1/2 x^2`

`(dy)/(dx) = x`
 

`text(When)\ \ x = 1,\ \ (dy)/(dx) = 1`

`text(Equation of tangent,)\ m = 1,\ text(through)\ (1, 1/2):`

`y – y_1` `= m (x – x_1)`
`y – 1/2` `= 1 (x – 1)`
`y – 1/2` `= x – 1`
`y` `= x – 1/2`

 

(ii)  `text(Directrix is)\ \ y = -1/2`
 

(iii)  `Q\ text(is at the intersection of)`

`y = x – 1/2\ \ …\ \ text{(1)}`

`y = -1/2\ \ \ \ …\ \ text{(2)}`

`text{(1) = (2)}`

`x – 1/2` `= -1/2`
`x` `= 0`

 
`:.\ Q\ text(lies on the)\ y text(-axis)\ \ …\ \ text(as required)`

 

(iv)  `text(Show)\ Delta PQS\ text(is isosceles.)`

`text(Distance)\ PS = 1 – 0 = 1`

`Q\ text(has coordinates)\ (0, -1/2)`

`text(Distance)\ SQ = 1/2 + 1/2 = 1`

`:. PS = SQ = 1`

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