SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Mechanics, EXT2 M1 2014 HSC 14c

A high speed train of mass `m` starts from rest and moves along a straight track. At time `t` hours, the distance travelled by the train from its starting point is `x` km, and its velocity is `v` km/h.

The train is driven by a constant force  `F`  in the forward direction. The resistive force in the opposite direction is  `Kv^2`, where  `K`  is a positive constant. The terminal velocity of the train is 300 km/h.

  1. Show that the equation of motion for the train is
     
         `m ddot x = F[1 − (v/300)^2]`.  (2 marks)

  2. Find, in terms of  `F`  and  `m`, the time it takes the train to reach a velocity of 200 km/h.  (4 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `(150 m ln5)/F`
Show Worked Solution

i.   `m ddot x = F – Kv^2,\ \ v_T = 300`

`text(At)\ \ v_T,\ \ ddotx=0`

`m(0)` `=F-K(300^2)`
`=>K` `=F/300^2`
`:. m ddot x` `= F – F/300^2 v^2`
  `=F[1 − (v/300)^2]\ \ \ text(… as required)`

 

ii. `m*(dv)/(dt)` `= F[1 − (v/(300))^2]`
  `(dv)/(1-(v/300)^2)` `=F/m\ dt`
  `(dv)/(300^2-v^2)` `=F/(300^2 m)\ dt`
  `dt` `=(300^2 m)/F xx (dv)/(300^2-v^2)`

 

`int_0^t dt` `=(300 m)/F  int_0^200  300/((300+v)(300-v))\ dv` 
 `:. t` `=(300 m)/F  int_0^200  (1/2)/(300+v) + (1/2)/(300-v)\ dv`
  `=(150 m)/F [ln(300+v) – ln(300-v)]_0^200`
  `=(150 m)/F [ln ((300+v)/(300-v))]_0^200`
  `=(150 m)/F  (ln5 – ln1)`
  `=(150 m ln5)/F`

Volumes, EXT2 2014 HSC 13b

The base of a solid is the region bounded by  `y = x^2`,  `y = –x^2`  and  `x = 2`. Each cross-section perpendicular to the `x`-axis is a trapezium, as shown in the diagram. The trapezium has three equal sides and its base is twice the length of any one of the equal sides.

Volumes, EXT2 2014 HSC 13b1

Find the volume of the solid.  (4 marks) 

Show Answers Only

`(24sqrt3)/5\ \ text(u³)`

Show Worked Solution

`text(Trapezium cross section)`

Volumes, EXT2 2014 HSC 13b Answer

`h^2` `=a^2-(a/2)^2`
  `=a^2-a^2/4`
  `=(3a^2)/4`
`:.h` `=(sqrt3 a)/2`

 

`text(S)text(ince)\ \ a=x^2`

`=>text(Base length)\ = 2x^2`

`=>text(Height)\ =(sqrt3 x^2)/2`

`:.\ text(Area of trapezium)`

`= 1/2h(a+b)`

`=1/2 xx (sqrt3 x^2)/2 xx(2x^2 + x^2)`

`= (3sqrt3 x^4)/4`

 

`:.V` `= int_0^2 (3sqrt3 x^4)/4 dx`
  `= (3sqrt3)/4[(x^5)/5]_0^2`
  `= (3sqrt3)/4 xx 32/5`
  `= (24sqrt3)/5\ \ text(u³)`

Calculus, EXT2 C1 2014 HSC 13a

Using the substitution  `t = tan\ x/2`, or otherwise, evaluate

`int_(pi/3)^(pi/2) 1/(3sinx - 4cosx + 5)\ dx`.  (3 marks)

Show Answers Only

`(2sqrt3 − 3)/6`

Show Worked Solution

`t = tan\ x/2,\ \ \ sinx=(2t)/(1+t^2)`

`cos x = (1-t^2)/(1+t^2),\ \ \ dx=(2 dt)/(1+t^2`

`text(When)\ \ \ x = pi/3,\ \ t = tan\ pi/6 =1/sqrt3`

`text(When)\ \ \ x = pi/2, \ \ t = tan\ pi/4=1`.

`int_(pi/3)^(pi/2) 1/(3\ sin\ x − 4\ cos\ x + 5)`
`= int_(1/sqrt3)^1 1/((6t)/(1 + t^2) − (4(1 − t^2))/(1 + t^2) + 5) xx (2dt)/(1 + t^2)`
`= int_(1/sqrt3)^1 2/(6t − 4 + 4t^2 + 5 + 5t^2)dt`
`= int_(1/sqrt3)^1 2/(9t^2 + 6t + 1)dt`
`= 2 int_(1/sqrt3)^1 (dt)/((3t + 1)^2)`
`= -2/3[1/(3t + 1)]_(1/sqrt3)^1`
`= -2/3(1/4 − 1/(sqrt3 + 1))` 
`= -2/3(1/4 − (sqrt3 − 1)/2)`
`= -1/6+sqrt3/3-1/3`
`= (2sqrt3 − 3)/6`

Calculus, EXT2 C1 2014 HSC 12d

Let  `I_n = int_0^1 (x^(2n))/(x^2 + 1)\ dx`, where  `n`  is an integer and  `n ≥ 0`.

  1. Show that  `I_0 = pi/4`.  (1 mark)

  2. Show that
     
         `I_n + I_(n − 1) = 1/(2n − 1)`.  (2 marks)

  3. Hence, or otherwise, find
     
         `int_0^1 (x^4)/(x^2 + 1)\ dx`.  (2 marks)

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

i.   `I_n = int_0^1 (x^(2n))/(x^2 + 1)\ dx, \ \ n ≥ 0`

  `I_0` `= int_0^1 (x^0)/(x^2 + 1)\ dx`
  `= [tan^(−1) x]_0^1` 
  ` = tan^(−1) 1 − tan^(−1)0`
  `= pi/4`

 

♦ Mean mark 41%.
STRATEGY: The denominator of the proof, `(2n-1)`, should flag the potential existence of `x^(2n-2)` in the integral.
ii.  `I_(n − 1)` `= int_0^1 x^(2n-2)/(x^2 + 1)\ dx`
  `I_n` `= int_0^1 x^(2n)/(x^2 + 1)\ dx`

 

`I_n + I_(n − 1)` `= int_0^1(x^(2n))/(x^2 + 1)\ dx + int_0^1(x^(2n − 2))/(x^2 + 1)\ dx`
  `= int_0^1(x^(2n) + x^(2n − 2))/(x^2 + 1)\ dx`
  `= int_0^1(x^(2n − 2)(x^2 + 1))/(x^2 + 1)\ dx`
  `= int_0^1x^(2n − 2)\ dx`
  `= [(x^(2n − 1))/(2n − 1)]_0^1`
  `= 1/(2n − 1)`

 

iii.  `I_2` `= int_0^1(x^4)/(x^2 + 1)\ dx`
  `I_1` `= int_0^1(x^2)/(x^2 + 1)\ dx`
`I_n + I_(n − 1)` `= 1/(2n − 1)`
`I_2 + I_1` `= 1/(2(2) − 1) = 1/3\ \ …\ (1)`
`I_1 + I_0` `= 1/(2(1)-1)=1`
`I_1+pi/4` `= 1`
`:.I_1` `=1- pi/4`
`text(Substitute into (1))`
`I_2+1- pi/4` `= 1/3`
 `:.I_2` `= pi/4 − 2/3`

Harder Ext1 Topics, EXT2 2014 HSC 12b

It can be shown that  `4\ cos^3 theta - 3 cos theta = cos 3theta`.   (Do NOT prove this.)

Assume that  `x = 2cos theta` is a solution of `x^3 − 3x= sqrt3`. 

  1. Show that
    1. `cos 3theta = sqrt3/2`.  (1 mark)
  2. Hence, or otherwise, find the three real solutions of  `x^3 - 3x = sqrt3`.  (2 marks)
Show Answers Only
  1. `text{Proof (See Worked Solutions)`
  2. `2cos\ pi/18, 2cos\ (11pi)/18, 2cos\ (13pi)/18`
Show Worked Solution

(i)   `4\ cos^3 theta − 3\ cos theta = cos 3theta`

`text(Substitute)\ \ x = 2 cos theta\ \ text(into)`

`x^3 − 3x` `= sqrt3`
  `(2\ cos\ theta)^3 − 6\ cos\ theta` `= sqrt3`
`8\ cos^3\ theta − 6\ cos\ theta` `= sqrt3`
`4\ cos^3\ theta − 3\ cos\ theta` `= sqrt3/2`
`:.cos\ 3theta` `= sqrt3/2\ \ \ text{ … as required}`

 

MARKER’S COMMENT: Many students failed to complete the final step of part (ii) and lost an easy mark!
(ii)   `3theta` `= pi/6, 2pi − pi/6, 2pi + pi/6`
  `3theta` `= pi/6, (11pi)/6, (13pi)/6`
  `theta` `= pi/18, (11pi)/18, (13pi)/18`
  `:.x` `= 2cos\ pi/18, \ 2cos\ (11pi)/18, \ 2 cos\ (13pi)/18`

Functions, EXT1′ F1 2014 HSC 11d

Without the use of calculus, sketch the graph  `y = x^2 - 1/(x^2)`, showing all intercepts.  (2 marks)

Show Answers Only

`text(See Worked Solutions.)`

Show Worked Solution

`y = x^2 − 1/(x^2)\ \ =>text(even function)`

MARKER’S COMMENT: The best solutions showed a curve approaching the asymptote `y=x^2`, as shown in the solution.

`text(Domain:  All)\ x, \ x≠0`

`y=x^2\ \ text(is an asymptote)`
 

`x text(-intercepts at)\ \ x=+-1`

`lim_(x→0^+) (x^2 − 1/(x^2))=-oo`
 

Graphs, EXT2 2014 HSC 11d Answer 

Complex Numbers, EXT2 N2 2014 HSC 11c

Sketch the region in the Argand diagram where `|\ z\ | ≤ |\ z − 2\ |`  and  `−pi/4 ≤ text(arg)\ z ≤ pi/4`.  (3 marks)

Show Answers Only

`text(See Worked Solutions.)`

Show Worked Solution

`text(Let)\ \ z=x+iy`

`|\ x + iy\ |` `≤ |\ x − 2 + iy\ |`
`sqrt(x^2 + y^2)` `≤ sqrt((x − 2)^2 + y^2)`
`x^2 + y^2` `≤ x^2 − 4x + 4 + y^2`
`4x − 4` `≤ 0`
`x` `≤ 1`

 

`−pi/4 ≤ text(arg)\ z ≤ pi/4`
 

Complex Numbers, EXT2 2014 HSC 11c Answer4

Mechanics, EXT2* M1 2004 HSC 7a

The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours. A ship is to sail from a wharf to the harbour entrance and then out to sea. On the morning the ship is to sail, high tide at the wharf occurs at 2 am. The water depths at the wharf at high tide and low tide are 10 metres and 4 metres respectively.

  1. Show that the water depth, `y` metres, at the wharf is given by
     
         `y = 7 + 3 cos\ ((4pit)/(25))`, where  `t`  is the number of hours after high tide.  (2 marks)  

  2. An overhead power cable obstructs the ship’s exit from the wharf. The ship can only leave if the water depth at the wharf is 8.5 metres or less.

     

    Show that the earliest possible time that the ship can leave the wharf is 4:05 am.  (2 marks)

  3. At the harbour entrance, the difference between the water level at high tide and low tide is also 6 metres. However, tides at the harbour entrance occur 1 hour earlier than at the wharf. In order for the ship to be able to sail through the shallow harbour entrance, the water level must be at least 2 metres above the low tide level.

     

    The ship takes 20 minutes to sail from the wharf to the harbour entrance and it must be out to sea by 7 am. What is the latest time the ship can leave the wharf?  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `4:28\ text(am)`
Show Worked Solution

i.   `text(Period)`

`(2pi)/n =` ` 12.5`
`12.5n =` ` 2pi`
`25n =` ` 4pi`
`n =` ` (4pi)/25`

 
`text(Amplitude)`

`a` `= 1/2(10 − 4)`
  `= 3`

 
`=>\ text(Motion centres around)\ \ x = 7\ \ text(with)`

`y = 10\ \ text(when)\ \ t= 0`

 

`:.\ text(Water depth is given by)`

`y` `= 7+a cos nt`
  `= 7 + 3 cos\ ((4pit)/(25))\ \ …\ text(as required.)`

  

ii.  

 Calculus in the Physical, EXT1 2004 HSC 7a Answer

`text(Find)\ \ t\ \ text(when)\ \ y = 8.5:`

`8.5` `= 7 + 3\ cos\ ((4pit)/25)`
`3\ cos\ ((4pit)/25)` `= 1.5`
`cos\ ((4pit)/25)` `= 1/2`
`(4pit)/25` `=pi/3`
`t` `=(25pi)/(3 xx 4pi)`
  `=2 1/12`
  `= 2\ text(hrs 5 mins)`

 

`:.\ text(The earliest time the ship can leave)`

`text(is 4:05 am  … as required.)`

 

iii.  `text(2 metres above low tide = 6 m)`

`text(Find)\ t\ text(when)\ y = 6:`

`6 = 7 + 3\ cos\ ((4pit)/(25))`

`3\ cos\ ((4pit)/(25))` `= -1`
`cos\ ((4pit)/(25))` `= -1/3`
`(4pit)/25` `= 1.9106…`
`:.t` `= (25 xx 1.9106…)/(4pi)`
  `= 3.801…`
  `= 3\ text{hr 48 min  (nearest min)}`

 

`:.\ text(At the harbour entrance, water depth)`

`text(of 6 m occurs when)\ \ t = 2\ text(hr 48 min.)`

 

`:.\ text(Given 20 mins sailing time, the latest the ship can)`

`text(leave the wharf is at)\ \ t = 2\ text(hr 28 min, or 4:28 am.)`

Mechanics, EXT2* M1 2007 HSC 7b

A small paintball is fired from the origin with initial velocity `14` metres per second towards an eight-metre high barrier. The origin is at ground level, `10` metres from the base of the barrier.

The equations of motion are

`x = 14t\ cos\ theta`

`y = 14t\ sin\ theta – 4.9t^2`

where  `theta`  is the angle to the horizontal at which the paintball is fired and  `t`  is the time in seconds. (Do NOT prove these equations of motion)
 

  1. Show that the equation of trajectory of the paintball is
     
         `y = mx − ((1 + m^2)/40)x^2`, where  `m = tan\ theta`.  (2 mark)

  2. Show that the paintball hits the barrier at height  `h`  metres when
     
         `m = 2 ± sqrt(3 − 0.4h)`.

     

    Hence determine the maximum value of  `h`.  (2 marks)

  3. There is a large hole in the barrier. The bottom of the hole is `3.9` metres above the ground and the top of the hole is `5.9` metres above the ground. The paintball passes through the hole if  `m`  is in one of two intervals. One interval is  `2.8 ≤ m ≤ 3.2`.

     

    Find the other interval.  (2 marks)

  4. Show that, if the paintball passes through the hole, the range is 
     
         `(40m)/(1 + m^2)\ \ text(metres.)`
     
    Hence find the widths of the two intervals in which the paintball can land at ground level on the other side of the barrier.  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `0.8 ≤ m ≤ 1.2`
  4. `text{1.3 m  (to 1 d.p.)  and  0.5 m  (to 1 d.p.)}`
Show Worked Solution
i.    `x` `= 14t\ cos\ theta` `\ \ …\ (1)`
  `y` `= 14t\ sin\ theta-4.9t^2` `\ \ …\ (2)`

 

`text(Substitute)\ \ t = x/(14\ cos\ theta)\ \ text{from (1) into (2)}`

`y` `= 14(x/(14\ cos\ theta))\ sin\ theta − 4.9(x/(14\ cos\ theta))^2`
  `= x\ tan\ theta − (4.9/(14^2))((x^2)/(cos^2\ theta))`
  `= x\ tan\ theta − (x^2)/40\ sec^2\ theta`
  `= x\ tan\ theta − (x^2)/40(1 + tan^2\ theta)`
  `= mx − ((1 + m^2)/40)x^2\ \ \ \ \ (text(Given)\ \ m = tan\ theta)`

 

ii.  `text(Show paintball hits at)\ \ h\ \ text(when)`

`m = 2 ± sqrt(3 − 0.4h)`

`text(i.e.)\ \ y = h\ \ text(when)\ \ x = 10`

`10m − ((1 + m^2)/40) · 10^2` `= h`
`10m − 5/2(1 + m^2)` `= h`
`20m − 5 − 5m^2` `= 2h`
`5m^2 − 20m + 2h + 5` `= 0`

 

`text(Using the quadratic formula)`

`m` `=(20 ± sqrt((-20)^2 − 4 · 5 · (2h + 5)))/(2 · 5)`
  `= (20 ± sqrt(400 − 40h −100))/10`
  `= (20 ± sqrt(300 − 40h))/10`
  `= (20 ± 10sqrt(3 − 0.4h))/10`
  `= 2 ± sqrt(3 − 0.4h)\ \ \ text(… as required)`

 

`text(Find maximum)\ \ h`

`sqrt(3 − 0.4h)` `≥ 0`
`3 − 0.4h` `≥ 0`
`0.4h` `≤ 3`
`h` `≤ 7.5`

 

`:.\ text(Maximum)\ \ h = 7.5\ text(m)`

 

iii.   EXT1 2007 7bi

`text{Using part (ii)}`

`text(When)\ \ h = 3.9`

`m` `= 2 ± sqrt(3 − 0.4(3.9))`
  `= 2 ± sqrt(1.44)`
  `= 2 ± 1.2`
  `= 3.2\ \ text(or)\ \ 0.8`

 

`text(When)\ \ h = 5.9`

`m` `= 2 ± sqrt(3 − 0.4(5.9))`
  `= 2 ± sqrt(0.64)`
  `= 2 ± 0.8`
  `= 2.8\ \ text(or)\ \ 1.2`

 

`:.\ text(The other interval is)\ \ \ 0.8 ≤ m ≤ 1.2`

 

iv.  `text(Find)\ \ x\ \ text(when)\ \ y = 0`

`mx − ((1 + m^2)/40)x^2` `= 0`
`x[m − ((1 + m^2)/40)x]` `= 0`
`((1 + m^2)/40)x` `= m,\ \  \ \ x ≠ 0`
`:. x` `= (40m)/(1 + m^2)\ \ …\ text(as required)`

 

`text(Consider the interval)\ \ \ 2.8 ≤ m ≤ 3.2`

`text(When)\ \ m = 2.8`

`=> x = (40(2.8))/(1 + 2.8^2) = 12.669…\ text(m)`

`text(When)\ \ m = 3.2`

`=>x = (40(3.2))/(1 + 3.2^2) = 11.387…\ text(m)`

 

`:.\ text(Landing width interval)`

`= 12.669… − 11.387…`

`= 1.281…`

`= 1.3\ text(m)\ \ text{to 1 d.p.}`

 

`text(Consider the interval)\ \ \ 0.8 ≤ m ≤ 1.2`

`text(When)\ \ m = 0.8`

`=>x = (40(0.8))/(1 + 0.8^2) = 19.512…\ text(m)`

`text(When)\ \ m = 1.2`

`=>x = (40(1.2))/(1 + 1.2^2) = 19.672…`

 

`text(S)text(ince interval includes)\ \ m = 1\ \ text(where the)`

`text(paintball has maximum range.)`

`x_(text(max)) = (40(1))/(1 + 1^2) = 20\ text(m)`

 

`:.\ text(Landing width interval)`

`= 20 − 19.512…`

`= 0.487…`

`= 0.5\ text(m)\ \ \ text{(to 1 d.p.)}`

L&E, EXT1 2007 HSC 7a

2007 7a

The graphs of the functions  `y = kx^n`  and  `y = log_e x`  have a common tangent at  `x = a`, as shown in the diagram.

  1. By considering gradients, show that 
    1. `a^n = 1/(nk)`. (1 mark)
  2.  
  3. Express  `k`  as a function of  `n`  by eliminating `a`.   (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `1/(en)`
Show Worked Solution
(i)    `y_1` `= kx^n`
  `(dy_1)/(dx)` `= nkx^(n − 1)`

 

`text(At)\ \ x = a,\ \ (dy_1)/(dx) = nka^(n − 1)`

 

`y_2` `= log_e x`
`(dy_2)/(dx)` `= 1/x`

 

`text(At)\ \ x = a,\ \ (dy_2)/dx = 1/a`

 

`text(S)text(ince tangents have the same gradient at)\ \ x=a,` 

`:. nka^(n − 1)` `= 1/a`
`a^n nk` `= 1`
`a^n` `= 1/(nk)\ \ …\ text(as required)`

 

(ii)  `text(At)\ \ x = a`

`y_1` `= ka^n`
`y_2` `= log_e a`

 

`text(Given a common tangent)`

♦♦♦ “Vast majority” could not eliminate `a` in part (ii).
`ka^n` `= log_e a`
`k(1/(nk))` `= log_e a\ \ \ text{(from (i))}`
`1/n` `= log_e a`
`:.a` `= e^(1/n)`

 

`text(Substitute)\ \ a = e^(1/n)\ \ text(into)\ \ a^n = 1/(nk)`

`(e^(1/n))^n` `= 1/(nk)`
`e` `= 1/(nk)`
`:.k` `= 1/(en)`

Functions, EXT1 F1 2007 HSC 6b

Consider the function  `f(x) = e^x − e^(-x)`.

  1. Show that  `f(x)`  is increasing for all values of `x`.  (1 mark)

  2. Show that the inverse function is given by
     
    `qquad qquad f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)` (3 marks)

  3. Hence, or otherwise, solve  `e^x - e^(-x) = 5`. Give your answer correct to two decimal places.  (1 mark)

Show Answers Only
  1. `text{Proof (See Worked Solutions.)}`
  2. `text{Proof (See Worked Solutions.)}`
  3. `1.65\ \ text{(to 2 d.p.)}`
Show Worked Solution
i.    `f(x)` `= e^x − e^(-x)`
  `f′(x)` `= e^x + e^(-x)`

 

`text(S)text(ince)\ \ ` `e^x` `> 0\ \ text(for all)\ x`
  `e^(-x)` `> 0\ \ text(for all)\ x`
  `f′(x)` `> 0\ \ text(for all)\ x`

 

`:.f(x)\ \ text(is an increasing function for all)\ x.`

 

ii.  `y = e^x − e^(-x)`

`text(Inverse function)`

`x` `= e^y − 1/(e^y)`
`xe^y` `= e^(2y) − 1`
`e^(2y) − xe^y − 1` `= 0`

 

`text(Let)\ \ A = e^y`

`:.A^2 − xA − 1 = 0`

 

`text(Using the quadratic formula)`

`A` `=(x ± sqrt((-x)^2 − 4 · 1 · (-1)))/(2 · 1)`
  `=(x ± sqrt(x^2 + 4))/2`

 

`text(S)text(ince)\ \ (x – sqrt(x^2 + 4))/2<0\ \ text(and)\ \ e^y>0,`

`:.e^y` `= (x + sqrt(x^2 + 4))/2`
`log_e e^y` ` = log_e((x + sqrt(x^2 + 4))/2)`
`y` `= log_e((x + sqrt(x^2 + 4))/2)`
`:.f^(-1)(x)` `= log_e((x + sqrt(x^2 + 4))/2)\ \  …\ text(as required)`

  

iii.   `e^x − e^(-x)` `= 5`
  `f(x)` `= 5`
  `f^(-1)(5)` `= x`

 

`f^(-1)(5)` `= log_e((5 + sqrt(5^2 + 4))/2)`
  `= log_e((5 + sqrt29)/2)`
  `= 1.647…`
  `= 1.65\ \ text{(to 2 d.p.)}`

Mechanics, EXT2* M1 2007 HSC 6a

A particle moves in a straight line. Its displacement, `x` metres, after `t` seconds is given by

`x = sqrt3\ sin\ 2t − cos\ 2t + 3`.

  1. Prove that the particle is moving in simple harmonic motion about  `x = 3`  by showing that   `ddot x = -4(x − 3)`.  (2 marks)

  2. What is the period of the motion?  (1 mark)

  3. Express the velocity of the particle in the form  `dotx = A\ cos\ (2t − α)`, where  `α`  is in radians.  (2 marks)

  4. Hence, or otherwise, find all times within the first  `pi`  seconds when the particle is moving at `2` metres per second in either direction.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions.)}`
  2. `pi\ text(seconds)`
  3. `dot x = 4\ cos\ (2t − pi/6)`
  4. `t = pi/4, (5pi)/12, (3pi)/4, (11pi)/12\ text(seconds.)`
Show Worked Solution

i.   `text(Show)\ \ ddot x = -4(x − 3)`

`x` `= sqrt3\ sin\ 2t − cos\ 2t + 3`
`dot x` `= 2sqrt3\ cos\ 2t + 2\ sin\ 2t`
`ddot x` `= -4sqrt3\ sin\2t + 4\ cos\ 2t`
  `= -4(sqrt3\ sin\ 2t − cos\ 2t)`
  `= -4(sqrt3\ sin\ 2t − cos\ 2t + 3 − 3)`
  `= -4(x − 3)\ \ …text(as required)`

 

ii.  `text(Period)\ = (2pi)/n`

`n^2 = 4 ⇒ n = 2\ \ text{(part (i))}`

`:.\ text(Period)` `= (2pi)/2`
  `= pi\ \ text(seconds)`

 

iii.  `text(Write)\ \ dot x = 2sqrt3\ cos\ 2t + 2\ sin\ 2t`

`text(in form)\ \ \ A\ cos\ (2t − α)`

`A(cos\ 2t\ cos\ α + sin\ 2t\ sin\ α)` `= 2sqrt3\ cos\ 2t + 2\ sin\ 2t`
`cos\ 2t\ cos\ α + sin\ 2t\ sin\ α` `= (2sqrt3)/A\ cos\ 2t + 2/A\ sin\ 2t`
`⇒ cos\ α` `= (2sqrt3)/A`
`⇒ sin\ α` `= 2/A`
`((2sqrt3)/A)^2 + (2/A)^2` `= 1`
`(2sqrt3)^2 + 2^2` `= A^2`
`:.A` `= sqrt16`
  `= 4`

 

`:.cos\ α` `= (2sqrt3)/4 = sqrt3/2`
`α` `= pi/6`

`:. dot x = 4\ cos\ (2t − pi/6)`

 

(iv)  `text(Find)\ \ t\ \ text(when)\ \ dot x = ±2`

`text(If)\ \ dot x = 2`

`4\ cos\ (2t − pi/6)` `= 2`
`cos\ (2t − pi/6)` `= 1/2`
`2t − pi/6` `= pi/3, 2pi − pi/3`
`2t` `= pi/2, (11pi)/6`
`t` `= pi/4, (11pi)/12`

 

`text(If)\ \ dot x = -2`

`cos\ (2t − pi/6)` `= – 1/2`
`2t − pi/6` `= (2pi)/3, (4pi)/3`
`2t` `= (5pi)/6, (3pi)/2`
`t` `= (5pi)/12, (3pi)/4`

 

`:.t = pi/4, (5pi)/12, (3pi)/4, (11pi)/12\ \ text(seconds.)`

Quadratic, EXT1 2007 HSC 5d

2007 5d

The diagram shows a point  `P(2ap, ap^2)`  on the parabola  `x^2= 4ay`. The normal to the parabola at  `P`  intersects the parabola again at  `Q(2aq,aq^2)`.

The equation of  `PQ`  is  `x + py − 2ap − ap^3 = 0`. (Do NOT prove this.)

  1. Prove that  `p^2+ pq + 2 = 0`.  (1 mark)
  2. If the chords  `OP`  and  `OQ`  are perpendicular, show that  `p^2 = 2`.  (2 marks)
Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

(i)   `text(Solution 1)`

`text(Prove)\ \ p^2 + pq + 2=0`

`PQ\ \ \ x+py-2ap-ap^3=0`

`=>y=-1/p x +2a+ap^2`

`:. m_(PQ)=-1/p`

`text(Also,)\ \ m_(PQ)` `=(y_2-y_1)/(x_2-x_1)`
  `=(ap^2-aq^2)/(2ap-2aq)`
  `=(a(p+q)(p-q))/(2a(p-q))`
  `=(p+q)/2`
`:. (p+q)/2` `=-1/p`
`p^2+pq` `=-2`
`p^2+pq+2` `=0\ \ \ text(… as required)`

 

`text(Solution 2)`

`PQ\ \ \ \ x + py − 2ap − ap^3 = 0`

`Q(2aq, aq^2)\ text(lies on)\ PQ`

`:.2aq + p(aq^2) − 2ap − ap^3` `= 0`
`2q − 2p + pq^2 − p^3` `= 0`
`2(q − p) + p(q^2 − p^2)` `= 0`
`2(q − p) + p(q − p)(q + p)` `= 0`
`p(q + p) + 2` `= 0`
`p^2 + pq + 2` `= 0\ \ \ text(… as required)`

 

(ii)  `text(If)\ OP ⊥ OQ, \ text(show)\ p^2 = 2`

`m_(OP) xx m_(OQ) = -1`

`m_(OP)` `= (ap^2 − 0)/(2ap − 0)`  `= p/2`
`m_(OQ)` `= (aq^2 − 0)/(2aq − 0)` `= q/2`
`:.p/2 * q/2` `= -1`
`pq` `= -4`

 

`text(Substitute)\ \ pq = -4\ \ text{into part (i)}`

`p^2 − 4 + 2` `= 0`
`:.p^2` `= 2\ \ …\ text(as required)`

Trigonometry, EXT1 T1 2007 HSC 5c

Find the exact values of `x` and `y` which satisfy the simultaneous equations

`sin^(-1)\ x + 1/2\ cos^(-1)\ y = pi/3`   and

`3\ sin^(-1)\ x-1/2\ cos^(-1)\ y = (2pi)/3`. (3 marks)

Show Answers Only

`x = 1/sqrt2, \ \ \ y = sqrt3/2`

Show Worked Solution
`sin^(-1)\ x + 1/2\ cos^(-1)\ y` `= pi/3` `\ \ …\ (1)`
`3\ sin^(-1)\ x-1/2\ cos^(-1)\ y` `= (2pi)/3` `\ \ …\ (2)`

 

MARKER’S COMMENT: The use of ther elimination method here proved much more successful than substitution.

`text(Add)\ \ (1) + (2)`

`4\ sin^(-1)\ x` `= pi`
`sin^(-1)\ x` `= pi/4`
`:.x` `= 1/sqrt2`

 

`text(Substitute)\ \ x = 1/sqrt2\ \ text(into)\ (1)`

`sin^(-1)\ 1/sqrt2 + 1/2\ cos^(-1)\ y` `= pi/3`
`pi/4 + 1/2\ cos^(-1)\ y` `= pi/3`
`1/2\ cos^(-1)\ y` `= pi/12`
`cos^(-1)\ y` `= pi/6`
`:.y` `= sqrt3/2`

Combinatorics, EXT1 A1 2007 HSC 5b

Mr and Mrs Roberts and their four children go to the theatre. They are randomly allocated six adjacent seats in a single row.

What is the probability that the four children are allocated seats next to each other?  (2 marks)

Show Answers Only

`1/5`

Show Worked Solution

`text(Total combinations possible)`

`= 6! = 720`
 

`text(Consider the children as 4 individuals in a group)`

`=>\ text(Combinations)\ = 4! = 24`
 

`text(Consider the two parents and a group of 4)`

`text(children as 3 elements.)`

`=>\ text(Combinations)\ = 3! = 6`
 

`:.\ text{P(children all sit next to each other)}`

`= (6 xx 24)/720`

`= 1/5`

Trig Calculus, EXT1 2006 HSC 7

A gutter is to be formed by bending a long rectangular metal strip of width `w` so that the cross-section is an arc of a circle.

Let `r` be the radius of the arc and `2 theta` the angle at the centre, `O`, so that the cross-sectional area, `A`, of the gutter is the area of the shaded region in the diagram on the right.

  1. Show that, when  `0 < theta <= pi/2`, the cross-sectional area is
    2. `A = r^2 (theta - sin theta cos theta).`  (2 marks)

  2. The formula in part (i) for  `A`  is true for  `0 < theta < pi.`      (Do NOT prove this.)
  3. By first expressing  `r`  in terms of  `w`  and  `theta`, and then differentiating, show that
    2. `(dA)/(d theta) = (w^2 cos theta (sin theta - theta cos theta))/(2 theta^3).`
  4. for  `0 < theta < pi.`  (3 marks)

  5. Let  `g(theta) = sin theta - theta cos theta.`
  6. By considering  `g prime(theta)`, show that  `g(theta) > 0`  for   `0 < theta < pi.`  (3 marks)

  8. Show that there is exactly one value of  `theta`  in the interval  `0 < theta < pi` for which
    1. `(dA)/(d theta) = 0.`  (2 marks)

  9. Show that the value of  `theta`  for which  `(dA)/(d theta) = 0`  gives the maximum cross-sectional area. Find this area in terms of `w.`  (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)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
  5. `w^2/(2 pi)\ \ text(u²)`
Show Worked Solution

(i)   `text(Show)\ \ A = r^2(theta – sin theta cos theta)`

`text(Area of segment)\ \ OBC`

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

`= r^2 theta`

`text(Area of)\ \ Delta OBC` `= 1/2 ab sin C`
  `= 1/2 r * r * sin 2 theta`
  `= 1/2 r^2 * 2 sin theta cos theta`
  `= r^2 sin theta cos theta`

 

`:.\ text(Shaded Area (A))`

`= text(Area of segment)\ \ OBC – text(Area of)\ \ Delta OBC`

`= r^2 theta – r^2 sin theta cos theta`

`= r^2 (theta – sin theta cos theta)\ \ text(…  as required.)`

 

(ii)   `text(Consider Arc length)\ \ BC`

`w` `= (2 theta)/(2 pi) xx 2 pi r`
  `= 2 theta r`
`:.\ r` `= w/(2 theta)`

 

`:. A` `= (w/(2 theta))^2 (theta – sin theta cos theta)`
  `= w^2/(4 theta) – (w^2 sin theta cos theta)/(4 theta^2)`

 

`:. (dA)/(d theta)` `= (-w^2)/(4 theta^2) – (w^2/4) [((sin theta xx -sin theta + cos theta * cos theta) theta^2 – 2 theta sin theta cos theta)/theta^4]`
  `= (-w^2)/(4 theta^2) – (w^2/(4 theta^4))[(cos^2 theta – sin^2 theta) theta^2 – 2 theta sin theta cos theta]`
  `= (-w^2)/(4 theta^2) – (w^2/(4 theta^3))[(2 cos^2 theta – 1) theta – 2 sin theta cos theta]`
  `= (-w^2)/(4 theta^2) – (w^2/(4 theta^3))[(2 cos^2 theta – 1) theta – 2 sin theta cos theta]`
  `= w^2/(4 theta^3) [-theta – 2 theta cos^2 theta + theta + 2 sin theta cos theta]`
  `= w^2/(4 theta^3) [2 sin theta cos theta – 2 theta cos^2 theta]`
  `= w^2/(2 theta^3) (sin theta cos theta – theta cos^2 theta)`
  `= (w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3)\ \ text(… as required)`

 

(iii)   `g(theta)`  `= sin theta- theta cos theta`
  `g prime (theta)` `= cos theta – (theta xx -sin theta + cos theta * 1)`
    `= cos theta + theta sin theta – cos theta`
    `= theta sin theta`

 

`text(S)text(ince)\ \ 0 < theta < pi,`

`=> sin theta > 0`

`:.\ g prime (theta) > 0`

`g(0) = sin 0 – 0 * cos 0 = 0`

 

`:.\ text(S)text(ince)\ \ g(0) = 0 and g(theta)\ \ text(is an increasing function)`

`(g prime (theta) > 0)\ \ text(for)\ \ 0 < theta < pi , text(then)\ \ g(theta) > 0.`

 

(iv)   `text(If)\ \ (dA)/(d theta) = 0\ ,\ \ \ 0 < theta < pi`

`(w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3) = 0`

`text(Consider)`

`(w^2 cos theta)/(2 theta^3) = 0\ ,\ \ theta != 0`

`=>theta = pi/2`

`text(Consider)`

`sin theta – theta cos theta=` ` 0`
`text(i.e.)\ \ g(theta)=` ` 0`

 

`=>\ text(No solution for)\ \ 0 < theta < pi\ \ text{(using part (iii))}`

 `:.\ text(There is only one value of)\ \ theta.`

 

(v)   `(dA)/(d theta)` `= (w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3)`
  `= (w^2 cos theta * g(theta))/(2 theta^3)`

 

`text(If)\ \ theta = pi/4\ ,\ \ g(pi/4) > 0\ ,\ \ cos (pi/4) > 0`

`:.\ (dA)/(d theta) > 0`

`text(If)\ \ theta = (3 pi)/4\ ,\ \ g((3 pi)/4) > 0\ ,\ \ cos ((3 pi)/4) < 0`

`:.\ (dA)/(d theta) < 0`

`:.\ text(Maximum when)\ \ theta = pi/2`

 

`text(When)\ \ theta = pi/2`

`A` `= r^2 (theta – sin theta cos theta)`
  `= (w/(2 theta))^2 (theta – sin theta cos theta)`
  `= w^2/(2^2 xx (pi/2)^2) (pi/2 – sin\ pi/2 cos\ pi/2)`
  `= w^2/pi^2 (pi/2)`
  `= w^2/(2 pi)\ \ text(u²)`

Statistics, EXT1 S1 2006 HSC 6b

In an endurance event, the probability that a competitor will complete the course is  `p`  and the probability that a competitor will not complete the course is  `q = 1 - p.` Teams consist of either two or four competitors. A team scores points if at least half its members complete the course.

  1. Show that the probability that a four-member team will have at least three of its members not complete the course is  `4pq^3 + q^4.`  (1 mark)

  2. Hence, or otherwise, find an expression in terms of  `q`  only for the probability that a four-member team will score points.  (2 marks)

  3. Find an expression in terms of  `q`  only for the probability that a two-member team will score points.  (1 mark)

  4. Hence, or otherwise, find the range of values of  `q`  for which a two-member team is more likely than a four-member team to score points.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1  – 4q^3 + 3q^4`
  3. `1 – q^2 `
  4. `1/3 < q < 1`
Show Worked Solution

i.  `P text{(at least 3 don’t complete)}`

`= P\ text{(3 don’t complete)} + P\ text{(4 don’t complete)}`

`= \ ^4C_3 q^3 p + \ ^4C_4 q^4`

`= 4pq^3 + q^4`

 

ii.  `P text{(4-member team scores)}`

`= 1 – P\ text{(at least 3 don’t complete)}`

`= 1 – 4pq^3 + q^4`

`= 1 – [4 (1 – q) q^3 + q^4]`

`= 1 – (4q^3 – 4q^4 + q^4)`

`= 1  – 4q^3 + 3q^4`

 

iii.   `P text{(2-member team scores)}`

`= 1 – P\ text{(both don’t complete)}`

`= 1 – q*q`

`= 1 – q^2`

 

iv.   `text(A 2-member team is more likely to score when)`

`1 – q^2` `> 1 -4q^3 + 3q^4`
`3q^4 – 4q^3 + q^2` `< 0`
`q^2 (3q^2 – 4q + 1)` `< 0`
`q^2 (3q – 1) (q – 1)` `< 0`

 

`text(Consider)\ \ (3q – 1) (q – 1) < 0`

`:.\ text(S)text(ince)\ \ q\ \ text(is positive and)\ != 1`

`q^2 (3q – 1) (q – 1) < 0\ \ text(when)`

`1/3 < q < 1.`

Mechanics, EXT2* M1 2006 HSC 6a

Two particles are fired simultaneously from the ground at time  `t = 0.`

Particle 1 is projected from the origin at an angle  `theta, \ \ 0 < theta < pi/2`, with an initial velocity  `V.`

Particle 2 is projected vertically upward from the point  `A`, at a distance  `a`  to the right of the origin, also with an initial velocity of  `V.`
 


 

It can be shown that while both particles are in flight, Particle 1 has equations of motion:

`x = Vt cos theta`

`y = Vt sin theta -1/2 g t^2,`

and Particle `2` has equations of motion:

`x = a`

`y = Vt -1/2 g t^2.`   Do NOT prove these equations of motion.

Let  `L`  be the distance between the particles at time  `t.`

  1. Show that, while both particles are in flight,
     
         `L^2 = 2V^2t^2 (1 - sin theta) - 2aVt cos theta + a^2.`  (2 marks)

  2. An observer notices that the distance between the particles in flight first decreases, then increases.

     

    Show that the distance between the particles in flight is smallest when
     
         `t = (a cos theta)/(2V(1 - sin theta))`  and that this smallest distance is  `a sqrt ((1 - sin theta)/2).`  (3 marks)

  3. Show that the smallest distance between the two particles in flight occurs while Particle 1 is ascending if  
     
         `V > sqrt((a g cos theta)/(2 sin theta \ (1 - sin theta))).`  (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)}`
Show Worked Solution
i.   

`text(Show)\ \ L^2 = 2V^2t^2 (1 – sin theta) – 2aVt cos theta + a^2`

`text(Consider)\ \ P_1`

`x_1 = Vt cos theta`

`y_1 = Vt sin theta – 1/2 g t^2`

`text(Consider)\ \ P_2`

`x_2 = a`

`y_2 = Vt -1/2 g t^2`

`text(Let)\ \ d=\ text(Vertical distance between particles)`

`d= Vt -1/2 g t^2 – (Vt sin theta – 1/2 g t^2)`

`d= Vt (1 – sin theta)`

 

`text(Using Pythagoras:)`

`L^2` `= (a – x_1)^2 + d^2`
  `= (a – Vt cos theta)^2 + V^2t^2 (1 – sin theta)^2`
  `= a^2 – 2aVt cos theta + V^2t^2 cos ^2 theta + V^2t^2 (1 – 2 sin theta + sin^2 theta)`
  `= a^2 – 2aVt cos theta + V^2 t^2 (cos^2 theta + sin^2 theta + 1 – 2 sin theta)`
  `= a^2 – 2aVt cos theta + V^2 t^2 (2 – 2 sin theta)`
  `= 2 V^2 t^2 (1 – sin theta) – 2aVt cos theta + a^2\ \ text(…  as required.)`

 

ii.   `L^2 = 2V^2 t^2 (1 – sin theta) – 2a Vt cos theta + a^2`

`(d(L^2))/(dt) = 4 V^2 t\ (1 – sin theta) – 2aV cos theta`

`text(Max or min when)\ \ (d(L^2))/(dt) = 0`

`4V^2t\ (1 – sin theta)` `= 2aV cos theta`
 `t` `= (2a V cos theta)/(4V^2 (1 – sin theta))`
  `= (a cos theta)/(2V(1 – sin theta)`

 

`(d^2(L^2))/(dt^2) = 4V^2(1 – sin theta) > 0\ \ text(for)\ \ V > 0,\ \ 0 < theta < pi/2`

`:.\ L^2\ \ text(is a minimum)`

`:.\ L\ \ text(is a minimum when)\ \ t = (a cos theta)/(2V (1 – sin theta)`

 

`text(Show minimum distance is)\ \ a sqrt {(1 – sin theta)/2}`

`text(When)\ \ t = (a cos theta)/(2V(1 – sin theta))`

`L^2` `= 2V^2 ((a^2 cos ^2 theta)/(4V^2 (1 – sin theta)^2)) (1 – sin theta)`
  `\ \ – 2aV ((a cos theta)/(2V (1 – sin theta))) cos theta + a^2`
  `= (a^2 cos^2 theta)/(2 (1 – sin theta)) – (a^2 cos^2 theta)/((1 – sin theta)) + a^2`
  `= a^2[(cos^2 theta – 2 cos^2 theta + 2 (1 – sin theta))/(2(1 – sin theta))]`
  `= a^2[(-cos^2 theta + 2 – 2 sin theta)/(2(1 – sin theta))]`
  `= a^2[(-(1 – sin^2 theta) + 2 – 2 sin theta)/(2 (1 – sin theta))]`
  `= a^2 [(sin^2 theta – 2 sin theta + 1)/(2(1 – sin theta))]`
  `= a^2 [(1 – sin theta)^2/(2 (1 – sin theta))]`
  `= a^2 [((1 – sin theta))/2]`
`:.\ L` `= sqrt ((a^2(1 – sin theta))/2)`
  `= a sqrt ((1 – sin theta)/2)\ \ text(…  as required.)`

 

iii.   `text(Smallest distance occurs when)`

`t = (a cos theta)/(2V (1 – sin theta)`

`text(If)\ \ P_1\ \ text(is ascending,)\ \ dot y_1 > 0`

`y_1 = Vt sin theta – 1/2 g t^2`

`dot y_1 = V sin theta – g t`

`:.\ V sin theta – g ((a cos theta)/(2V (1 – sin theta)))` `> 0`
`2V^2 sin theta\ (1 – sin theta) – a g cos theta` `> 0`
`2V^2 sin theta\ (1 – sin theta)` `> ag cos theta`
`V^2` `> (a g cos theta)/(2 sin theta\ (1 – sin theta))`
`V` `> sqrt ((a g cos theta)/(2 sin theta\ (1 – sin theta)))\ \ text(…  as required.)`

Plane Geometry, EXT1 2007 HSC 4c

EXT1 2007 4c

The diagram shows points `A`, `B`, `C` and `D` on a circle. The lines  `AC`  and  `BD`  are perpendicular and intersect at  `X`. The perpendicular to  `AD`  through  `X`  meets  `AD`  at  `P`  and  `BC`  at  `Q`.

Copy or trace this diagram into your writing booklet.

  1. Prove that  `∠QXB =∠QBX`.  (3 marks)
  2. Prove that  `Q`  bisects  `BC`.  (2 marks)
Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution
(i)   

 `text(Prove)\ \ ∠QXB =∠QBX`

MARKER’S COMMENT: Many students did not copy the diagram into their answer! Efficient solutions labelled angles with a symbol.

`∠ADX = ∠ACB = theta`

`text{(angles in the same segment on arc}\ AB)`

 

`text(In)\ ΔDPX`

`∠DPX` `= 90^@\ \ (PQ ⊥ AD)`
`∠PXD` `= (90 – theta)^@\ \ text{(angle sum of}\ Δ DPX)`
`∠QXB` `= (90 – theta)^@\ \ text{(vertically opposite angle)}`

 

`text(In)\ Δ BXC`

`∠BXC` `= 90^@\ \ (∠AXC\ text{is a straight angle)}`
`∠QBX` `= (90 – theta)^@\ \ text{(angle sum of}\ ΔBXC)`
`:.∠QXB` `= ∠QBX`

 

(ii)  `text(Prove)\ \ Q\ \ text(bisects)\ \ BC`

`BQ = QX\ \ ` `text{(sides opposite equal angles}`
  `text{of isosceles}\ Delta BXQ text{)}`
`∠QXC` `= 180 − 90 − (90 − theta)\ \ (∠AXC\ text{is a straight angle)}`
  `= theta`
`∠XCB` `=theta\ \ \ text{(from part (i))}`
`:. ΔXQC\ text(is isosceles)`

 

`QX = QC\ \ ` `text{(sides opposite base angles}`
  `text{of isosceles}\ ΔQXC)`

`:. BQ = QC`

`:. Q\ text(bisects)\ BC`

Proof, EXT1 P1 2007 HSC 4b

Use mathematical induction to prove that  `7^(2n – 1) + 5`  is divisible by 12, for all integers  `n ≥ 1`.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Prove)\ \ 7^(2n – 1) + 5\ \ text(is divisible by)\ 12,\ text(for)\ \ n ≥ 1`

`text(If)\ \ n = 1`

`7^(2 – 1) + 5 = 7 + 5 = 12\ \ \ text{(divisible by 12)}`

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

 

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

`text(i.e.)\ 7^(2k − 1) + 5` `= 12text{N   (N is an integer)}`
 `7^(2k-1)` `= 12text(N) – 5\ \ …\ (1)`

 

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

`7^(2(k + 1) − 1) + 5` `= 7^(2k + 1) + 5`
  `= 7^2 · 7^(2k − 1) + 5`
  `= 49(12text(N) − 5) + 5\ \ \ text{(from (1) above)}`
  `= 49*12text(N) − 245 + 5`
  `= 49*12text(N) − 240`
  `= 12(49text(N) − 20)`

`…text(which is divisible by)\ 12`

 
`=>\ text(True for)\ n = k + 1`

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

Statistics, EXT1 S1 2007 HSC 4a

In a large city, 10% of the population has green eyes.

  1. What is the probability that two randomly chosen people both have green eyes?  (1 mark)

  2. What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal places.  (1 mark)

  3. What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places.  (2 marks)

Show Answers Only
  1. `0.01`
  2. `0.285\ \ \ text{(to 3 d.p.)}`
  3. `0.32\ \ \ text{(to 2 d.p.)}`
Show Worked Solution
i.       `P(text(G))` `= 0.1`
  `P(text(GG))` `= 0.1 xx 0.1`
    `= 0.01`

 

ii.  `P(text(not G)) = 1 − 0.1 = 0.9`

`:. P(text(2 out of 20 have green eyes))`

`= \ ^(20)C_2 · (0.1)^2 · (0.9)^(18)`

`= 0.2851…`

`= 0.285\ \ \ text{(to 3 d.p.)}`

 

iii. `P(text(more than 2 have green eyes))`

`= 1 − [P(0) + P(1) + P(2)]`

`= 1 − [0.9^20 + \ ^20C_1(0.1)^1(0.9)^19 + \ ^20C_2(0.1)^2(0.9)^(18)]`

`= 1 − [0.1215… + 0.2701… + 0.2851…]`

`= 1 − 0.6769…`

`= 0.3230`

`= 0.32\ \ \ text{(to 2 d.p.)}`

Mechanics, EXT2* M1 2007 HSC 3c

A particle is moving in a straight line with its acceleration as a function of  `x`  given by  `ddot x = -e^(-2x)`. It is initially at the origin and is travelling with a velocity of 1 metre per second.

  1. Show that  `dot x = e^(-x)`.  (2 marks)

  2. Hence show that  `x = log_e(t + 1)`.  (2 marks)

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

i.   `text(Show that)\ \ dot x = e^(−x)`

`ddot x` `= d/(dx)\ (1/2 (dot x)^2) = −e^(−2x)`
`1/2 (dot x)^2` `= int −e^(−2x)\ dx`
  `= 1/2 e^(−2x) + c`

 
`text(When)\ \ x = 0, \ dot x = 1,`

`1/2 · 1^2` `= 1/2 e^0 + c`
`1/2` `= 1/2 + c`
`:.c` `= 0`

 

MARKER’S COMMENT: Most candidates “neglected” to consider the two cases that  `dot x=+-e^(-x)`.
`1/2 (dot x)^2` `= 1/2 e^(−2x)`
`(dot x)^2` `= e^(−2x)`
`dot x` `= +-e^(−x)`

 
`text(Given initial conditions:)\ \ x=0, dot x = 1,`

`dot x = e^(-x)\ \ text(… as required)`

 

ii.   `text(Show)\ \ x = log_e(t + 1)`

`(dx)/(dt)` `= e^(−x)`
`(dt)/(dx)` `= e^x`
`t` `= int e^x`
  `= e^x + c`

 

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

`0` `= e^0 + c`
`c` `= −1`
`:.t` `= e^x − 1`
`e^x` `= t + 1`
`log_ee^x` `= log_e(t + 1)`
`x` `= log_e(t + 1)\ …\ text(as required.)`

Functions, EXT1 F1 2007 HSC 3b

  1. Find the vertical and horizontal asymptotes of the hyperbola  `y = (x − 2)/(x − 4)`  

     

    and hence sketch the graph of  `y = (x − 2)/(x − 4)`.  (3 marks)

  2. Hence, or otherwise, find the values of  `x`  for which
     
    `qquad qquad (x − 2)/(x − 4) ≤ 3`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `x < 4\ text(and)\ x ≥ 5`
Show Worked Solution

i.    `y = (x − 2)/(x − 4)`

`text(Vertical asymptote at)\ x = 4`

`lim_(x → ∞) (x − 2)/(x − 4)` `= lim_(x → ∞) (1 − 2/x)/(1 − 4/x)=1`

`ytext(–intercept)\ = 1/2`

`xtext(–intercept)\ = 2`

 

Geometry and Calculus, EXT1 2007 HSC 3b Answer

 

ii.  `text(Find)\ \ x\ \ text(so that)\ \ (x − 2)/(x − 4) ≤ 3`

`text(When)\ \ (x − 2)/(x − 4)` `= 3`
`x − 2` `= 3x − 12`
`2x` `= 10`
`x` `= 5`

 
`=>(5, 3)\ \ text(is the intersection of)\ \ y = 3\ and\ y = (x − 2)/(x − 4)`

`:. (x − 2)/(x − 4) ≤ 3\ \ text(when)\ \ x < 4\ \ text(and)\ \ x ≥ 5.`

Calculus, EXT1* C3 2007 HSC 3a

Find the volume of the solid of revolution formed when the region bounded by the curve  `y = 1/(sqrt(9 + x^2))`, the `x`-axis, the `y`-axis and the line  `x = 3`, is rotated about the `x`-axis.  (3 marks)

Show Answers Only

`(pi^2)/(12)\ \ text(u³)`

Show Worked Solution

`y = 1/(sqrt(9 + x^2))`

`:.\ text(Volume)` `= pi int_0^3 y^2\ dx`
  `= pi int_0^3 (1/(sqrt(9 + x^2)))^2\ dx`
  `= pi int_0^3 1/(9 + x^2)\ dx`
  `= pi [1/3 tan^(−1)\ x/3]_0^3`
  `= pi [1/3 tan^(−1)\ 1 − 1/3 tan^(−1)\ 0]`
  `= pi [(1/3 xx pi/4) − 0]`
  `= (pi^2)/(12)\ \ text(u³)`

Mechanics, EXT2* M1 2007 HSC 2d

A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, `v` metres per second, at which she is falling  `t`  seconds after jumping is given by  `v =50(1 - e^(-0.2t))`.

  1. Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place.  (2 marks)

  2. Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre.  (2 marks)

Show Answers Only
  1. `1.4\ text(ms)^(−2)\ \ text{(to 1 d.p.)}`
  2. `284\ text{m  (nearest m)}`
Show Worked Solution
i.     `v` `= 50(1 − e^(-0.2t))`
    `=50-50e^(-0.2t)`
  `ddot x` `= (dv)/(dt)`
    `= −0.2 xx 50 xx −e^(−0.2t)`
    `= 10 e^(−0.2t)`

 
`text(When)\ \ t = 10:`

`ddot x` `= 10 e^(−0.2 xx 10)`
  `= 10 e ^(−2)`
  `= 1.353…`
  `= 1.4\ text(ms)^(−2)\ \ \ text{(to 1 d.p.)}`

 

ii.  `text(Distance travelled)`

`= int_0^10 v\ dt`

`= 50 int_0^10 1 − e^(−0.2t) \ dt`

`= 50 [t + 1/0.2 · e^(−0.2t)]_0^10`

`= 50 [t + 5e^(−0.2t)]_0^10`

`= 50 [(10 + 5e^(−2)) − (0 + 5e^0)]`

`= 50 [10 + 5e^(−2) − 5]`

`= 50 [5 + 5e^(−2)]`

`= 283.833…`

`= 284\ text{m  (nearest m)}`

Trigonometry, EXT1 T1 2007 HSC 2b

Let  `f(x) = 2 cos^(-1)x`.

  1. Sketch the graph of  `y = f(x)`, indicating clearly the coordinates of the endpoints of the graph.  (2 marks)

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

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `0 ≤ y ≤ 2pi`
Show Worked Solution

i.   `y= 2\ cos^(-1)x`

`text(Domain:)\ -1 ≤ x ≤ 1`

`text{Range:}\ \0 ≤`  `y/2 ≤ pi`  
`0 ≤` `y ≤ 2pi`  

 

 

ii.  `text(Range:)\ \ 0 ≤ y ≤ 2pi`

Trigonometry, EXT1 T3 2007 HSC 2a

By using the substitution  `t = tan\ theta/2`, or otherwise, show that  `(1 − cos\ theta)/(sin\ theta) = tan\ theta/2`.  (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution
COMMENT: The values of `sin theta` and `cos theta` can be committed to memory or quickly derived using double angle formulae (as shown in the worked solution).

EXT1 2007 2a

`=>tan\ theta/2 = t`

`=>sin theta` `= 2 · t/(sqrt(1 + t^2)) · 1/(sqrt(1 + t^2)) = (2t)/(1 + t^2)`
`=>cos theta` `= 1/(1 + t^2) − t^2/(1 + t^2) = (1 − t^2)/(1 + t^2)`

 

`text(Show)\ \ (1 − cos theta)/(sin theta) = tan\ theta/2 :`

`(1 − cos\ theta)/(sin theta)` `= (1 − ((1 − t^2)/(1 +  t^2)))/((2t)/(1 + t^2)) xx (1+t^2)/(1+t^2)`
  `= (1 + t^2 − (1 − t^2))/(2t)`
  `= (2t^2)/(2t)`
  `= t`
  `= tan\ theta/2\ \ …text(as required)`

Calculus, EXT1 C2 2007 HSC 1e

Use the substitution  `u = 25 - x^2`  to evaluate  `int_3^4 (2x)/(sqrt(25 - x^2))\ dx`.  (3 marks)

Show Answers Only

`2`

Show Worked Solution
`u` `= 25 − x^2`
`(du)/(dx)` `= -2 x`
`du`  `= -2x\ dx`
`text(If)`   `x = 4,`   `u = 9`
    `x = 3,`   `u = 16`

 

`:. int_3^4 (2x)/(sqrt(25 − x^2))\ dx`

MARKER’S COMMENT: A “significant number” of students put the integral limits in the wrong order.

`= − int_16^9 u^(−1/2) du`

`= − [1/(1/2) u^(1/2)]_16^9`

`= − [2sqrtu]_16^9`

`= − [2sqrt9 − 2sqrt16]`

`= − [6 − 8]`

`= 2`

Mechanics, EXT2* M1 2004 HSC 6b

A fire hose is at ground level on a horizontal plane. Water is projected from the hose. The angle of projection, `theta`, is allowed to vary. The speed of the water as it leaves the hose, `v` metres per second, remains constant. You may assume that if the origin is taken to be the point of projection, the path of the water is given by the parametric equations

`x = vt\ cos\ theta`

`y = vt\ sin\ theta − 1/2 g t^2`

where  `g\ text(ms)^(−2)`  is the acceleration due to gravity.  (Do NOT prove this.)

  1. Show that the water returns to ground level at a distance`(v^2\ sin\ 2theta)/g`  metres from the point of projection.   (2 marks)

This fire hose is now aimed at a 20 metre high thin wall from a point of projection at ground level 40 metres from the base of the wall. It is known that when the angle  `theta`  is 15°, the water just reaches the base of the wall.  
 

Calculus in the Physical World, EXT1 2004 HSC 6b
 

  1. Show that  `v^2 = 80g`.  (1 mark)

  2. Show that the cartesian equation of the path of the water is given by
     
         `y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`.  (2 marks)

  3. Show that the water just clears the top of the wall if
     
         `tan^2\ theta − 4\ tan\ theta + 3 = 0`.  (2 marks)

  4. Find all values of  `theta`  for which the water hits the front of the wall.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
  5. `15^@ ≤ theta ≤ 45^@ \ \ text(and)\ \ 71.6^@ ≤ theta ≤ 75^@`
Show Worked Solution
i.    `x` `= vt\ cos\ theta`
  `y` `= vt\ sin\ theta − 1/2 g t^2`

 

`text(Find)\ \ t\ \ text(when)\ \ y = 0`

`vt\ sin\ theta − 1/2 g t^2` `= 0`
`t(v\ sin\ theta − 1/2 g t)` `= 0`
`v\ sin\ theta − 1/2 g t` `= 0, \ \ t ≠ 0`
`1/2 g t` `= v\ sin\ theta`
`t` `= (2v\ sin\ theta)/g`

 

`text(Find)\ \ x\ \ text(when)\ \ t = (2v\ sin\ theta)/g`

`x` `= v · (2v\ sin\ theta)/g\ cos\ theta`
  `= (v^2 · \ 2\ sin\ theta\ cos\ theta)/g`
  `= (v^2\ sin\ 2theta)/g`

 

`:.\ text(When)\ \ x = (v^2\ sin\ 2theta)/g,\ \ \text(the water returns)`

`text(to ground level  … as required.)`

 

ii.   `text(Show)\ \ v^2 = 80\ text(g)`

`text(When)\ \ theta = 15^@, \ x = 40`

`text{From part (i)}`

`40` `= (v^2\ sin\ 30^@)/g`
`v^2 xx 1/2` `= 40g`
`v^2` `= 80g\ \ \ …text(as required)`

 

iii.  `text(Show)\ \ y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`

`x` `= vt\ cos\ theta`
`:.t` `= x/(v\ cos\ theta)`

 

`text(Substitute into)`

`y` `= vt\ sin\ theta − 1/2 g t^2`
  `= v · x/(v\ cos\ theta) · sin\ theta −1/2 g  (x/(v\ cos\ theta))^2`
  `= x\ tan\ theta − 1/2  g (x^2/(v^2\ cos^2\ theta))`
  `= x\ tan\ theta − 1/2  g · x^2/(80g\ cos^2\ theta)`
  `= x\ tan\ theta − (x^2\ sec^2\ theta)/160\ \ \ …text(as required.)`

 

iv.   `text(Water clears the top if)\ \ y = 20\ \ text(when)\ \ x = 40`

`text{Substitute into equation from (iii)}`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160` `= 20`
`40\ tan\ theta − 10\ sec^2\ theta` `= 20`
`40\ tan\ theta − 10(1 + tan^2\ theta)` `= 20`
`40\ tan\ theta − 10 − 10\ tan^2\ theta` `= 20`
`10\ tan^2\ theta − 40\ tan\ theta\ + 30` `= 0`
`tan^2\ theta − 4\ tan\ theta + 3` `= 0\ \ \ text(… as required)`

 

v.   

Calculus in the Physical World, EXT1 2004 HSC 6b Answer

`text(Water hits the bottom of the wall when)`

`x = 40\ \ text(and)\ \ y = 0`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160` `= 0`
`40\ tan\ theta − 10\ sec^2\ theta` `= 0`
`4\ tan\ theta − (1 + tan^2\ theta)` `= 0`
`tan^2\ theta − 4\ tan\ theta + 1` `= 0`

 

`text(Using the quadratic formula)`

`tan\ theta` `= (+4 ± sqrt(16 − 4 · 1 · 1))/ (2 xx 1)`
  `= (4 ± sqrt12)/2`
  `= 2 ± sqrt3`
`theta` `= 15^@\ \ text(or)\ \ 75^@`

 

`text(Water hits the top of the wall when)`

`x = 40\ text(and)\ \ y = 20`

`tan^2\ theta − 4\ tan\ theta + 3` `= 0\ \ \ text{from (iv)}`
`(tan\ theta − 1)(tan\ theta − 3)` `= 0`
`tan\ theta` `= 1` `text(or)` `tan\ theta` `= 3`
`theta` `= 45^@`   `theta` `= tan^(−1)\ 3`
        `= 71.565…`
        `= 71.6^@\ \ \text{(to 1 d.p.)}`

 

`:.\ text(Following the motion of the water as)\ \ theta\ \ text(increases,)`

`text(water hits the wall when)`

`15^@ ≤ theta ≤ 45^@` `\ text(and)`
`71.6^@ ≤ theta ≤ 75^@`  

Plane Geometry, EXT1 2004 HSC 6a

Plane Geometry, EXT1 2004 HSC 6a

The points  `A, \ B, \ C`  and  `D`  are placed on a circle of radius  `r`  such that  `AC`  and  `BD`  meet at  `E`. The lines  `AB`  and  `DC`  are produced to meet at  `F`, and  `BECF`  is a cyclic quadrilateral.

Copy or trace this diagram into your writing booklet.

  1. Find the size of  `∠DBF`, giving reasons for your answer.  (2 marks)

  2. Find an expression for the length of  `AD`  in terms of  `r`.  (1 mark)
Show Answers Only
  1. `90^@`
  2. `2r`
Show Worked Solution
(i)   

Plane Geometry, EXT1 2004 HSC 6a Answer

 `∠ACD = ∠ABD = theta\ \ \ text{(angles in the same segment on arc}\ AD)`

`text(Consider cyclic quad)\ \ BECF`

`∠DBF = ∠ECD = theta` `\ \ \ text{(exterior angle of cycle quad}`
  `\ \ \ text{equals its interior opposite)}`
`2theta` `= 180^@\ \ \ (∠ABF\ text{is a straight angle)}`
`theta` `= 90^@`
`:.∠DBF` `= 90^@`

 

(ii)   `text(S)text(ince)\ AD\ text(subtends right angles)\ \ ∠ABD\ \ text(and)\ \ ∠ACD`

`text{(from part (i))}`

`⇒ AD\ \ text(is a diameter)`

`:.AD = 2r`

 

Inverse Functions, EXT1 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. Copy or trace this diagram into your writing booklet.
    On the same set of axes, sketch the graph of the inverse function,  `y = f^(−1)(x)`.  (1 mark)
  2.  
  3. State the domain of  `f^(−1)(x)`.  (1 mark)

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

  5. The graphs of  `y = f(x)`  and  `y = f^(−1)(x)`  meet at exactly one point  `P`.
  6. Let  `α`  be the `x`-coordinate of  `P`. Explain why  `α`  is a root of the equation
     

    1. `x^3 + x − 1 = 0`.  (1 mark)
    2.  
  7. Take 0.5 as a first approximation for  `α`. Use one application of Newton’s method to find a second approximation for  `α`.  (2 marks) 

 

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0 < x ≤ 1`
  3. `y = sqrt((1 − x)/x), y > 0`
  4.  
  5. `text(See Worked Solutions)`
  6. `0.714\ \ \ text{(to 3 d.p.)}`
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`

 

(v)    `f(x)` `= x^3 + x − 1`
  `f′(x)` `= 3x^2 + 1`
  `f(0.5)` `= 0.5^3 + 0.5 − 1`
    `= −0.375`
  `f′(0.5)` `= 3 xx 0.5^2 + 1`
    `= 1.75`

 

`α_2` `= α_1 − (f(0.5))/(f′(0.5))`
  `= 0.5 − ((−0.375))/1.75`
  `= 0.5 − (−0.2142…)`
  `= 0.7142…`
  `= 0.714\ \ \ text{(to 3 d.p.)}`

Mechanics, EXT2* M1 2004 HSC 5a

A particle is moving along the `x`-axis, starting from a position  `2`  metres to the right of the origin (that is,  `x = 2`  when  `t = 0`) with an initial velocity of  `5\ text(ms)^(−1)`  and an acceleration given by

`ddot x = 2x^3 + 2x`.

  1. Show that  `dot x = x^2 + 1`.  (2 marks)

  2. Hence find an expression for  `x`  in terms of  `t`.  (3 marks)

Show Answers Only
  1. `text(Show Worked Solutions)`
  2. `tan\ (t + tan^(−1)2)`
Show Worked Solution

i.   `text(Show)\ \ dot x = x^2 + 1`

`ddot x` `= d/(dx)\ (1/2 v^2) = 2x^3 + 2x`
`:.1/2 v^2` `= int2x^3 + 2x \ dx`
  `= 2/4x^4 + x^2 + c`
`v^2` `= x^4 + 2x^2 + c`

 
`text(When)\ \ x = 2, \ v = 5`

`5^2` `= 2^4 + (2 xx 2^2) + c`
`25` `= 16 + 8 + c`
`c` `= 1`
`:.v^2` `= x^4 + 2x^2 + 1`
  `= (x^2 + 1)^2`
`v` `= sqrt((x^2 + 1)^2)`
 `:.dot x` `= x^2 + 1\ \ \ …\ text(as required)`

  

ii.    `(dx)/(dt)` `= x^2 + 1`
  `(dt)/(dx)` `= 1/(x^2 + 1)`
  `:.t` `= int1/(x^2 + 1)\ dx`
    `= tan^(−1)\ x + c`

 
`text(When)\ \ t = 0, \ x = 2`

`0` `= tan^(−1)\ 2 + c`
`c` `= −tan^(−1)\ 2`
`:.t` `=tan^(−1)\ x − tan^(−1)\ 2`

 

`tan^(−1)\ x` `= t + tan^(−1)2`
`:.x` `= tan (t + tan^(−1)2)`

Combinatorics, EXT1 A1 2004 HSC 4c

Katie is one of ten members of a social club. Each week one member is selected at random to win a prize.

  1. What is the probability that in the first 7 weeks Katie will win at least 1 prize?  (1 mark)

  2. Show that in the first 20 weeks Katie has a greater chance of winning exactly 2 prizes than of winning exactly 1 prize.  (2 marks)

  3. For how many weeks must Katie participate in the prize drawing so that she has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes?  (2 marks)

Show Answers Only
  1. `0.52`
  2. `text(See Worked Solutions)`
  3. `text(30 weeks)`
Show Worked Solution

i.  `Ptext{(wins at least 1 prize)}`

`= 1 − Ptext{(wins no prize)}`

`= 1 − (9/10)^7`

`= 0.5217…`

`= 0.52\ \ \ text{(2 d.p.)}`
 

ii.  `text(In 1st 20 weeks,)`

`Ptext{(winning exactly 1 prize)}`

`=\ ^(20)C_1 · (1/10) · (9/10)^19`

`= 0.2701…`
 

`Ptext{(winning exactly 2 prizes)}`

`=\ ^(20)C_2 · (1/10)^2 · (9/10)^18`

`= 0.2851…`
 

`:.\ text(Katie has a greater chance of winning)`

`text(exactly 2 prizes.)`

 

iii. `Ptext{(winning exactly 3 prizes)}`

`=\ ^nC_3 · (1/10)^3 · (9/10)^(n − 3)`

`Ptext{(winning exactly 2 prizes)}`

`=\ ^nC_2 · (1/10)^2 · (9/10)^(n − 2)`
 

`text(If greater chance of winning exactly 3 than exactly 2:)`

`\ ^nC_3 · (1/10)^3 · (9/10)^(n − 3)` `>\ ^nC_2 · (1/10)^2 · (9/10)^(n − 2)`
`(n!)/(3!(n − 3)) · 1/10` `> (n!)/(2!(n − 2)!) · 9/10`
`(2!(n − 2)!)/(3!(n − 3)!)` `> 9`
`(n − 2)/3` `> 9`
`n − 2` `> 27`
`n` `> 29`

 
`:.\ text(Katie must participate for 30 weeks.)`

Induction, EXT1 2004 HSC 4a

Use mathematical induction to prove that for all integers  `n ≥ 3`,

`(1 − 2/3)(1 − 2/4)(1 − 2/5)…(1 − 2/n) = 2/(n(n − 1))`.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Prove)`

`(1 − 2/3)(1 − 2/4)(1 − 2/5)…(1 − 2/n) = 2/(n(n − 1))\ text(for)\ n ≥ 3`

`text(If)\ \ n = 3`

`text(LHS) = (1 − 2/3) = 1/3`

`text(RHS)\ = 2/(3(3 − 1)) = 2/6 = 1/3 =\ text(LHS)`

`:.\ text(True for)\ n = 3`

 
`text(Assume true for)\ \ n = k`

`(1 − 2/3)(1 − 2/4)…(1 − 2/k) = 2/(k(k − 1))`

 
`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ (1 − 2/3)(1 − 2/4)…(1 − 2/k)(1 − 2/(k + 1)) = 2/((k + 1)k)`

`text(LHS)` `= (1 − 2/3)(1 − 2/4)…(1 − 2/k) (1 − 2/(k + 1))`
  `= 2/(k(k − 1))(1 − 2/(k + 1))`
  `= 2/(k(k − 1))·((k + 1 − 2)/(k + 1))`
  `= 2/(k(k − 1))·((k − 1)/(k + 1))`
  `= (2(k − 1))/(k(k − 1)(k + 1))`
  `= 2/(k(k + 1))`
  `=\ text(RHS)`

 
`=>\ text(True for)\ n = k + 1`

`:.\ text(S)text(ince true for)\ \ n = 3, text(by PMI, true for integral)\ n ≥ 3.`

Trig Ratios, EXT1 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)}`

Calculus, EXT1 C1 2004 HSC 3c

A ferry wharf consists of a floating pontoon linked to a jetty by a 4 metre long walkway. Let  `h`  metres be the difference in height between the top of the pontoon and the top of the jetty and let  `x`  metres be the horizontal distance between the pontoon and the jetty.
 

2004 3c
 

  1. Find an expression for  `x`  in terms of  `h`.  (1 mark)

When the top of the pontoon is 1 metre lower than the top of the jetty, the tide is rising at a rate of 0.3 metres per hour.

  1. At what rate is the pontoon moving away from the jetty?  (3 marks)

Show Answers Only
  1. `x=sqrt(16 − h^2)`
  2. `0.077\ text(m/hr)`
Show Worked Solution

i.   `text(Using Pythagoras,)`

`x^2 + h^2` `= 4^2`
`x^2` `= 16 − h^2`
`x` `= sqrt(16 − h^2)`

 

ii.   `text(Find)\ \ (dx)/(dt)\ \ text(when)\ \ h = 1`

`(dx)/(dt)` `= (dx)/(dh)·(dh)/(dt)`
`x` `= (16 − h^2)^(1/2)`
`(dx)/(dh)` `= 1/2 xx (16 − h^2)^(−1/2) xx d/(dh)(16 − h^2)`
  `= 1/2 (16 − h^2)^(−1/2) xx −2h`
  `= (−h)/(sqrt(16 − h^2))`

 

`text(When)\ \ h = 1, (dh)/(dt)= −0.3\ text(m/hr)`

`text{(}h\ \ text{decreases when the tide is rising)}`

`(dx)/(dt)` `= (−h)/(sqrt(16 − h^2)) xx −0.3`
  `= (−1)/sqrt(16 − 1^2) xx −0.3`
  `= 0.3/sqrt15`
  `= 0.0774…`
  `= 0.077\ \ \ text{metres per hr (to 2 d.p.)}`

 

`:.\ text(When)\ \ h = 1,\ text(the pontoon is moving away)`

`text(at 0.077 metres per hr.)`

Functions, EXT1 F2 2004 HSC 3b

Let  `P(x) = (x + 1) (x − 3)Q(x) + a(x + 1) + b`, where  `Q(x)`  is a polynomial and  `a`  and  `b`  are real numbers.

When  `P(x)`  is divided by  `(x + 1)`  the remainder is  `−11`.

When  `P(x)`  is divided by  `(x − 3)`  the remainder is  `1`.

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

  2. What is the remainder when  `P(x)`  is divided by  `(x + 1)(x − 3)`?  (2 marks)

Show Answers Only
  1. `−11`
  2. `3x − 8`
Show Worked Solution

i.   `P(x)= (x + 1) (x − 3)Q(x) + a(x + 1) + b`

`P(-1)=-11`

`-11=(−1 + 1)(−1 − 3)Q(x) + a(−1 + 1) + b`

`-11=(0)(-4)Q(x)+a(0)+b`

`:.b = −11`

 

 (ii)   `P(3) = 1`

`:.(3 + 1)(3 − 3)Q(x) + a(3 + 1) −11 = 1`

`4a` `= 12`
`a` `= 3`

 

`text(When)\ \ P(x)\ \ text(is divided by)\ \ (x + 1)(x − 3)`

`R(x)` `= a(x + 1) + b`
  `= 3(x + 1) − 11`
  `= 3x + 3 − 11`
  `= 3x − 8`

Combinatorics, EXT1 A1 2004 HSC 2e

A four-person team is to be chosen at random from nine women and seven men.

  1. In how many ways can this team be chosen?  (1 mark)

  2. What is the probability that the team will consist of four women?  (1 mark)

Show Answers Only
  1. `1820`
  2. `9/130`
Show Worked Solution

i.   `text(# Team combinations)`

`=\ ^(16)C_4`

`= (16!)/((16 − 4)!\ 4!)`

`= 1820`
 

 ii.  `text{P(4 women)}`

`= (\ ^9C_4)/1820`

`= 126/1820`

`= 9/130`

Trigonometry, EXT1 T3 2004 HSC 2d

  1. Write  `8 cos x + 6 sin x`  in the form  `A cos(x −α)`, where  `A > 0`  and  `0 ≤α ≤ π/2`.  (2 marks)

  2. Hence, or otherwise, solve the equation  `8cos x + 6 sin x = 5`  for  `0 ≤ x ≤ 2π`. 

     

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

Show Answers Only
  1. `10\ cos\ (x − 0.6435…)`
  2. `1.691\ \ text(or)\ 5.879\ \ text{(to 3 d.p.)}`
Show Worked Solution

i.   `text(Write)\ \ 8 cos x + 6 sin x\ \ text(in the form)\ \ A cos (x − α)`

`A(cos\ x\ cos\ α + sin\ x\ sin\ α)` `= 8\ cos\ x + 6\ sin\ x`
`(cos\ x\ cos\ α + sin\ x\ sin\ α)` `= 8/A\ cos\ x + 6/A\ sin\ x`

`⇒ cos\ α = 8/A`

`⇒ sin\ α = 6/A`

`(8/A)^2 + (6/A)^2` `= 1`
`8^2 + 6^2` `= A^2`
`:.A` `= sqrt100`
  `= 10`
`=>cos\ α` `= 8/10`
`:.α` `= 0.6435…\ text(radians)`

 
`:.8 cos x + 6 sin x = 10 cos (x − 0.6435…)`

 

ii.    `8\ cos\ x + 6\ sin\ x` `= 5`
  `10\ cos\ (x − 0.6435…)` `= 5`
  `cos\ (x − 0.6435…)` `= 1/2`
  `x − 0.6435…` `= pi/3\ \ text(or)\ \ 2pi − pi/3`
  `x` `= pi/3 + 0.6435…\ \ text(or)\ \ (5pi)/3 + 0.6435…`
    `= 1.691\ \ text(or)\ \ 5.879\ \ text{radians  (to 3 d.p.)}`

Functions, EXT1 F1 2006 HSC 5b

Let  `f(x) = log_e (1 + e^x)`  for all  `x`.

Show that  `f(x)`  has an inverse.  (2 marks)

Show Answers Only

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

Show Worked Solution
`f(x)` `= log_e (1 + e^x)`
`f prime (x)` `= e^x/(1 + e^x)`

 

`=> e^x > 0\ \ text(for all)\ \ x`

`:.  e^x/(1 + e^x) > 0\ \ text(for all)\ \ x`

 

`:.\ f(x)\ \ text(is a monotonic increasing function)`

`text(for all)\ \ x.`

`:.\ f(x)\ \ text(has an inverse for all)\ \ x.`

Mechanics, EXT2* M1 2006 HSC 4c

A particle is moving so that  `ddot x = 18x^3 + 27x^2 + 9x.`

Initially  `x = – 2`  and the velocity, `v`, is  `– 6.`

  1. Show that  `v^2 = 9x^2 (1 + x)^2.`  (2 marks)

  2. Hence, or otherwise, show that 
     
         `int 1/(x(1 + x)) \ dx = -3t.`  (2 marks)

  3. It can be shown that for some constant  `c`,
     
         `log_e (1 + 1/x) = 3t + c.`       (Do NOT prove this.)
     
    Using this equation and the initial conditions, find  `x`  as a function of  `t.`  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `x = 2/(e^(3t) – 2)`
Show Worked Solution

i.   `text(Show)\ \ v^2 = 9x^2 (1 + x)^2`

`ddot x` `= d/(dx) (1/2 v^2)`
  `= 18x^3 + 27x^2 + 9x`
`1/2 v^2` `= int 18x^3 + 27x^2 + 9x\ dx`
  `= 18/4 x^4 + 27/3 x^3 + 9/2 x^2 + c`
`v^2` `= 9x^4 + 18x^3 + 9x^2 + c`

 

`text(When)\ \ t = 0,\ \ x = -2,\ \ v = -6`

`:.\ (-6)^2` `= 9 (-2)^4 + 18 (-2)^3 + 9 (-2)^2 + c`
`36` `= 144 – 144 + 36 + c`
`c` `= 0`

 

`:.\ v^2` `= 9x^4 + 18x^3 + 9x^2`
  `= 9x^2 (x^2 + 2x + 1)`
  `= 9x^2 (1 + x)^2\ \ text(…  as required)`

 

ii.   `text(Show)\ \ int 1/(x (1 + x)) \ dx = -3t`

`v^2 = 9x^2 (1 + x)^2`

`v = +- sqrt (9x^2 (1 + x)^2)`
 

`text(When)\ \ x = -2,\ \ v = -6:`

`:.\ v` `= -sqrt (9x^2 (1 + x)^2)`
  `= -3x (1 + x)`

 

`(dx)/(dt)` `= -3x (1 + x)`
`(dt)/(dx)` `= -1/(3x (1 + x))`
`t` `= -1/3 int 1/(x (1 + x)) \ dx`
`-3t` `= int 1/(x (1 + x)) \ dx\ \ text(…  as required)`

 

iii.  `text(Given)\ \ log_e (1 + 1/x) = 3t + c`

`text(When)\ \ t = 0,\ \ x = -2:`

`log_e (1 – 1/2)` `= 3(0) + c`
`log_e (1/2)` `= c`
`c` `= log_e 2^-1`
  `= -log_e 2`

 

`log_e (1 + 1/x)` `= 3t – log_e 2`
`1 + 1/x` `= e^(3t – log_e 2)`
`1/x` `= e^(3t – log_e 2) – 1`
  `= (e^(3t))/(e^(log_e 2)) – 1`
  `= e^(3t)/2 – 1`
  `= (e^(3t) – 2)/2`
`:.\ x` `= 2/(e^(3t) – 2)`

Functions, EXT1 F2 2006 HSC 4a

The cubic polynomial  `P(x) = x^3 + rx^2 + sx + t`. where  `r, \ s`  and  `t`  are real numbers, has three real zeros,  `1, alpha`  and  `-alpha.`

  1. Find the value of  `r.`  (1 mark)

  2. Find the value of  `s + t.`  (2 marks)

Show Answers Only
  1. `-1`
  2. `0`
Show Worked Solution

i.  `P(x) = x^3 + rx^2 + sx + t`

`text(Roots are)\ \ 1, alpha, -alpha`

`1 + alpha + -alpha` `= – b/a`
`1` `= – r/1`
`:.\ r` `= -1`

 

ii.  `P(x) = x^3 – x^2 + sx + t`

`P(1) = 0`

`0` `= 1 – 1 + (s xx 1) + t`
`0` `= s + t`

 
`:.\ text(Value of)\ \ s + t = 0`

Plane Geometry, EXT1 2006 HSC 3d

The points  `P, Q`  and  `T`  lie on a circle. The line  `MN`  is tangent to the circle at  `T`  with  `M`  chosen so that  `QM`  is perpendicular to  `MN`. The point  `K`  on  `PQ`  is chosen so that  `TK`  is perpendicular to  `PQ`  as shown in the diagram.

  1. Show that  `QKTM`  is a cyclic quadrilateral.  (1 mark)
  2. Show that  `/_KMT = /_KQT.`  (1 mark)
  3. Hence, or otherwise, show that  `MK`  is parallel to  `TP.`  (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)   

`/_ QMT = 90^@\ \ \ (QM _|_ MN)`

`/_ QKT = 90^@\ \ \ (PQ _|_ TK)`

`:.\ /_ QMT + /_ QKT = 180^@`

 

`:.\ QKTM\ \ text(is cyclic)\ \ text{(opposite angles are supplementary)}`

`text(…  as required.)`

 

(ii)  `text(Show)\ \ /_ KMT = /_ KQT`

`/_ KMQ = /_ KTQ = theta`

`text{(angles in the same segment on arc}\ \ KQ text{)}`

 

`/_ KQT` `= 90 – theta\ \ text{(angle sum of}\ \ Delta KQT text{)}`
`/_ KMT` `= /_ QMT – /_ KMQ`
  `= 90 – theta`

 

`:.\ /_ KMT = /_ KQT\ \ text(…  as required.)`

 

(iii)   `text(Show)\ \ MK\ text(||)\ TP`

`/_ NTP` `= /_ KQT\ \ text{(angle in alternate segment)`
  `= 90 – theta`

 

`:.\ /_ NTP = /_ KMT\ \ text{(from part (ii))}`

`:.\ MK\ text(||)\ TP\ \ text{(corresponding angles are equal)}`

Combinatorics, EXT1 A1 2006 HSC 3c

Sophie has five coloured blocks: one red, one blue, one green, one yellow and one white. She stacks two, three, four or five blocks on top of one another to form a vertical tower.

  1. How many different towers are there that she could form that are three blocks high?  (1 mark)

  2. How many different towers can she form in total?  (2 marks)

Show Answers Only
  1. `60`
  2. `320`
Show Worked Solution
i.   `text(Towers)` `= \ ^5P_3`
  `= 60`

 
ii.   `text(Number of different towers)`

`= \ ^5P_2 + \ ^5P_3 + \ ^5P_4 + \ ^5P_5`

`= 20 + 60 + 120 + 120`

`= 320`

Polynomials, EXT1 2006 HSC 3b

  1. By considering  `f(x) = 3log_e x - x`, show that the curve  `y = 3 log_e x`  and the line  `y = x`  meet at a point  `P`  whose `x`-coordinate is between  `1.5`  and  `2`.  (1 mark)

  2. Use one application of Newton’s method, starting at  `x = 1.5`, to find an approximation to the `x`-coordinate of  `P`. Give your answer correct to two decimal places.  (2 marks)

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.78\ \ text{(to 2 d.p.)}`
Show Worked Solution
(i)     `f(x)` `= 3 log_e x – x`
  `f(1.5)` `= 3 log_e 1.5 – 1.5`
    `= -0.283… <0`
  `f(2)` `= 3 log_e 2 – 2`
    `= 0.079… >0`

 

`:.\ text(S)text(ince the sign changes, a zero must)`

`text(exist between 1.5 and 2.)`

 

(ii) `f(x)` `= 3 log_e x – x`
  `f prime(x)` `= 3/x – 1`
  `f(1.5)` `= -0.283…`
  `f prime (1.5)` `= 3/1.5 – 1 = 1`

 

`x_1` `= x_0 – (f(x_0))/(f prime (x_0))`
  `= 1.5 – {(-0.283…)}/1`
  `= 1.783…`
  `= 1.78\ \ text{(to 2 d.p.)}`

Quadratic, EXT1 2006 HSC 2c

2006 2c

The points  `P(2ap, ap^2), Q(2aq, aq^2)`  and  `R(2ar, ar^2)`  lie on the parabola  `x^2 = 4ay`. The chord  `QR`  is perpendicular to the axis of the parabola. The chord  `PR`  meets the axis of the parabola at  `U`.

The equation of the chord  `PR`  is  `y = 1/2(p + r)x - apr.`     (Do NOT prove this.)

The equation of the tangent at  `P`  is  `y = px - ap^2.`            (Do NOT prove this.) 

  1. Find the coordinates of  `U.`  (1 mark)
  2. The tangents at  `P`  and  `Q`  meet at the point  `T`. Show that the coordinates of  `T`  are  `(a(p + q), apq).`  (2 marks)
  3. Show that  `TU`  is perpendicular to the axis of the parabola.  (1 mark)
Show Answers Only
  1. `(0, –apr)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

(i)   `U\ \ text(is the)\ \ y\ \ text(intercept of)\ \ PR`

`y = 1/2 (p + r)x – apr`

`text(When)\ \ x = 0`

`y = -apr`

`:.\ U\ \ text(has coordinates)\ \ (0, –apr)`

 

(ii)   `text(Show)\ \ T\ \ text(is)\ \ (a(p + q), apq)`

`text(T)text(angents at)\ \ P and Q\ \ text(are)`

`y` `= px – ap^2\ \ \ \ text{…  (1)}`
`y` `= qx – aq^2\ \ \ \ text{…  (2)}`

 

`T\ \ text(occurs when)\ \ \ (1) = (2)`

`px – ap^2` `= qx – aq^2`
`px – qx` `= ap^2 – aq^2`
`x(p – q)` `= a (p^2 – q^2)`
  `= a (p – q) (p + q)`
`:.\ x` `= a (p + q)`

 

`text(Substitute)\ \ x = a (p + q)\ \ text{into  (1)}`

`y` `= p * a (p + q) – ap^2`
  `= ap^2 + apq – ap^2`
  `= apq`

 

`:.\ T (a (p + q), apq)\ \ text(…  as required.)`

 

(iii)  `text(Axis of parabola)\ \ x^2 = 4ay\ \ text(is)\ \ y text(-axis)`

`=>TU\ \ text(will be perpendicular to the)\ \ y text(-axis if it has a)`

`text(gradient of)\ \ 0\ \ text{(i.e.}\ \ T and U\ text{have same} y text{-values)}`

`:.\ text(Need to show)\ \ \ -apr = apq.`

 

`text(Consider)\ \ QR`

`text(S)text(ince it is perpendicular to)\ \ y text(-axis)`

`text(and the parabola is symmetrical)`

`=>\ \ 2aq` `= -2ar`
`q`  `= -r`
`:.\ apq` `= ap (-r)`
`apq` `= -apr`

 

`:.\ TU\ \ text(is perpendicular.)`

Binomial, EXT1 2006 HSC 2b

  1. By applying the binomial theorem to `(1 + x)^n` and differentiating, show that
  2. `n(1 + x)^(n - 1) = ((n), (1)) + 2((n), (2)) x + … + r((n), (r)) x^(r - 1) + … + n((n), (n)) x^(n - 1).`   (1 mark)
  4. Hence deduce that
  5. `n3^(n - 1) = ((n), (1)) + … + r((n), (r)) 2^(r - 1) + … + n((n), (n)) 2^(n - 1).`  (1 mark)

 

 

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

(i)  `text(Show)`

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

`+ … + n((n), (n)) x^(n – 1)`

`text(Using binomial expansion)`

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

`+ … + ((n), (n)) x^n`

`text(Differentiate both sides)`

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

`+ … + n((n), (n)) x^(n – 1)\ \ \ …\ text(as required)`

 

(ii)  `text(Substitute)\ \ x = 2\ \ text{into above equation}`

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

`+ … + n((n), (n)) 2^(n – 1)`

`n3^(n – 1) = ((n), (1)) + … + r((n), (r)) 2^(r – 1) + … + n((n), (n)) 2^(n – 1)`

`text(…  as required.)`

Calculus, EXT1 C2 2006 HSC 2a

Let  `f(x) = sin^-1 (x + 5).`

  1. State the domain and range of the function  `f(x).`  (2 marks)

  2. Find the gradient of the graph of  `y = f(x)`  at the point where  `x = -5.`  (2 marks)

  3. Sketch the graph of  `y = f(x).`  (2 marks)

Show Answers Only
  1. `text(Domain):\ -6 <= x <= -4, \ \ \ text(Range):\ -pi/2 <= y <= pi/2`
  2. `1`
  3.  

Show Worked Solution

i.  `f(x) = sin^-1 (x + 5)`

`text(Domain)`

`-1 <= x + 5 <= 1`

`-6 <= x <= -4`

`text(Range)`

`-pi/2 <= y <= pi/2`

 

ii.  `y = sin^-1 (x + 5)`

`(dy)/(dx) = 1/sqrt(1-(x + 5)^2)`
 

`text(When)\ \ x = -5`

`(dy)/(dx)` `= 1/sqrt(1-(-5 + 5)^2)`
  `= 1/sqrt(1-0)`
  `= 1`

 
`:.\ text(Gradient of)\ \ y = f(x)\ \ text(at)\ \ x = -5\ \ text(is)\ \ 1.`

 

iii.

 EXT1 2006 2a

Differentiation, EXT1 2006 HSC 1e

For what values of  `b`  is the line  `y =12x + b`  tangent to  `y =x^3`?  (3 marks)

Show Answers Only

`b = +- 16`

Show Worked Solution
`y_1` `= x^3`
`(dy_1)/(dx)` `= 3x^2`
`y_2` `= 12x + b`
`(dy_2)/(dx)` `= 12`

 

`text(T)text(angent occurs when)`

`(dy_1)/(dx)` `= (dy_2)/(dx)`
`3x^2` `= 12`
`x^2` `= 4`
`x` `= +- 2`

 

`text(When)\ \ x = +- 2,\ \ y_1 = +- 8` 

`:.\ y_2\ \ text(must satisfy)\ \ (2, 8) or (–2, –8)`

 

`text(Consider)\ \ (2, 8)`

`8 = (12 xx 2) + b`

`b = -16`

`text(Consider)\ \ (–2, –8)`

`-8 = (12 xx -2) + b`

`b = 16`

`:.\ b = +- 16`

Mechanics, EXT2* M1 2005 6b

An experimental rocket is at a height of  5000 m, ascending with a velocity of  ` 200 sqrt 2\ text(m s)^-1`  at an angle of  45°  to the horizontal, when its engine stops.
 

 
After this time, the equations of motion of the rocket are:

`x = 200t`

`y = -4.9t^2 + 200t + 5000,`

where `t` is measured in seconds after the engine stops. (Do NOT show this.)

  1. What is the maximum height the rocket will reach, and when will it reach this height?  (2 marks)

  2. The pilot can only operate the ejection seat when the rocket is descending at an angle between 45° and 60° to the horizontal. What are the earliest and latest times that the pilot can operate the ejection seat?  (3 marks)

  3. For the parachute to open safely, the pilot must eject when the speed of the rocket is no more than  `350\ text(m s)^-1`. What is the latest time at which the pilot can eject safely?  (2 marks)

Show Answers Only
  1. `7041\ text(metres)`

     

    `20.4\ text(seconds.)`

  2. `40.8\ text(seconds)\ ;\ 55.8\ text(seconds)`
  3. `49.7\ text(seconds)`
Show Worked Solution

i.

    

`y = -4.9t^2 + 200t + 5000`

`dot y = -9.8t + 200`

`text(Max height occurs when)\ \ dot y = 0`

`9.8t` `= 200`
`t` `= 20.408…`
  `= 20.4\ text{seconds  (to 1 d.p.)}`

 

`text(When)\ t = 20.408…`

`y` `= -4.9 (20.408…)^2 + 200 (20.408…) + 5000`
  `= 7040.816…`
  `= 7041\ text{m  (to nearest metre)}`

 

`:.\ text(The rocket will reach a maximum height of)`

`text(7041 metres when)\ \ t = 20.4\ text(seconds.)`

 

ii.  `text(The rocket is descending at 45° at point)\ A\ text(on the graph.)`

`text(The symmetry of the parabolic motion means that)\ \ A`

`text(occurs when)\ \ t = 2 xx text(time to reach max height, or)`

`40.8\ text(seconds.)`

 

`text(Point)\ B\ text(occurs when the rocket is descending at 60°)`

`text(to the horizontal.)`

 

`text(At)\ \ B,`

 

`-tan 60° = (dot y)/(dot x)`

`text{(The gradient of the projectile becomes}`

`text{negative after reaching its max height.)}`

`dot y = -9.8t + 200`

`dot x = d/(dt) (200t) = 200`

`:.\ – sqrt 3` `= (-9.8t + 200)/200`
`-200 sqrt 3` `= -9.8t + 200`
`9.8t` `= 200 + 200 sqrt 3`
`t` `= (200 + 200 sqrt 3)/9.8`
  `= (546.410…)/9.8`
  `= 55.756…`
  `= 55.8\ text{seconds  (nearest second)}`

 

`:.\ text(The pilot can operate the ejection seat)`

`text(between)\ \ t = 40.8 and t = 55.8\ text(seconds.)`

 

iii.

`v^2 = (dot x)^2 + (dot y)^2`

`text(When)\ \ v = 350`

`350^2` `= 200^2 + (-9.8t + 200)^2`
`122\ 500` `= 40\ 000 + (-9.8t + 200)^2`
`(-9.8t + 200)^2` `= 82\ 500`
`-9.8t + 200` `= +- sqrt (82\ 500)`
`9.8t` `= 200 +- sqrt (82\ 500)`
`t` `= (200 +- sqrt (82\ 500))/9.8`
  `= 49.717…\ \ \ (t > 0)`
  `= 49.7\ text{seconds  (to 1 d.p.)}`

 

`:.\ text(The latest time the pilot can eject safely)`

`text(is when)\ \ t = 49.7\ text(seconds.)`

Statistics, EXT1 S1 2005 HSC 6a

There are five matches on each weekend of a football season. Megan takes part in a competition in which she earns one point if she picks more than half of the winning teams for a weekend, and zero points otherwise. The probability that Megan correctly picks the team that wins any given match is `2/3`.

  1. Show that the probability that Megan earns one point for a given weekend is  0.7901, correct to four decimal places.  (2 marks)

  2. Hence find the probability that Megan earns one point every week of the eighteen-week season. Give your answer correct to two decimal places.  (1 mark)

  3. Find the probability that Megan earns at most 16 points during the eighteen-week season. Give your answer correct to two decimal places.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `0.01\ \ text{(to 2 d.p.)}`
  3. `0.92\ \ text{(to 2 d.p.)}`
Show Worked Solution

i.  `P\ text{(earns a point)}`

`= P\ text{(picks 3, 4 or 5 winners)}`

`=\ ^5 C_3  (2/3)^3 * (1/3)^2 + \ ^5 C_4  (2/3)^4 * (1/3)^1`

`+\ ^5 C_5 (2/3)^5*(1/3)^0`

`= 10 * 8/27 * 1/9 + 5 * 16/81 * 1/3 + 1 * 32/243*1`

`= 80/243 + 80/243 + 32/243`

`= 192/243`

`= 0.790123…`

`= 0.7901\ \ text{(to 4 d.p.)  …  as required.}`

 

ii.  `P\ text{(earns a point 18 weeks in a row)}`

`= (0.7901…)^18`

`= 0.01440…`

`= 0.01\ \ text{(to 2 d.p.)}`

 

iii.  `P\ text{(earns at most 16 points)}`

`= 1 – P\ text{(earns 17 or 18 points)}`
 

`P\ text{(earns 17)}`

`=\ ^18 C_17 * (0.7901…)^17 xx (1 – 0.7901…)`

`= 0.0688…`

`P\ text{(earns 18)} = 0.01440…\ \ \ \ \ text{(from (ii))}`
 

`:.\ P\ text{(earns at most 16 points)}`

`= 1 – (0.0688… + 0.0144…)`

`= 0.916…`

`= 0.92\ \ \ text{(to 2 d.p.)}`

Mechanics, EXT2* M1 2005 HSC 5c

A particle moves in a straight line and its position at time  `t`  is given by

`x = 5 + sqrt 3 sin3t - cos 3t.`

  1. Express  `sqrt 3 sin3t − cos 3t`  in the form  `R sin(3t - alpha)`  where  `alpha`  is in radians.  (2 marks)

  2. The particle is undergoing simple harmonic motion. Find the amplitude and the centre of the motion.  (2 marks)

  3. When does the particle first reach its maximum speed after time  `t = 0`?  (1 mark)

Show Answers Only
  1. `2 sin ( 3t – pi/6)`
  2. `text(Amplitude) = 2\ text(units),\ \ \ text(Centre of the motion at)\ \ x = 5`
  3. `pi/18`
Show Worked Solution

i.    `x = 5 + sqrt 3 sin 3t – cos 3t`

`sqrt 3 sin 3t – cos 3t` `= R sin (3t – alpha)`
  `= R sin 3t cos alpha – R cos 3t sin alpha`

 
`=> R cos alpha = sqrt 3\ \ \ \ \ R sin alpha = 1`

`R^2 = sqrt 3^2 + 1^2 = 4`

`R = 2`

`=> 2 cos alpha` `= sqrt3`
`cos alpha` `= (sqrt3)/2`
`alpha` `= pi/6`

 
`:.\ 2 sin ( 3t – pi/6) = sqrt 3 sin 3t – cos 3t`

 

ii.  `x` `= 5 + sqrt 3 sin 3t – cos 3t`
  `= 5 + 2 sin (3t – pi/6)`

 
`text(Amplitude) = 2\ text(units)`

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

 

iii.  `text(Solution 1)`

`x = 5 + 2 sin (3t – pi/6)`

`dot x = 6 cos (3t – pi/6)`

 
`text(Maximum speed occurs when)`

`cos (3t – pi/6)` `= 1`
`3t – pi/6` `= 0`
`3t` `= pi/6`
`t` `= pi/18`

 
`text(Solution 2)`

`text(Maximum speed occurs at the)`

`text(centre of motion,)\ \ x = 5`

`5 + 2 sin (3t – pi/6)` `= 5`
`2 sin (3t – pi/6)` `= 0`
`sin (3t – pi/6)` `= 0`
`3t – pi/6` `= 0`
`t` `= pi/18`

Plane Geometry, EXT1 2005 HSC 5b

Two chords of a circle, `AB`  and  `CD`, intersect at  `E`. The perpendiculars to  `AB`  at  `A` and  `CD`  at  `D`  intersect at  `P`. The line  `PE`  meets  `BC`  at  `Q`, as shown in the diagram.

  1. Explain why  `DPAE`  is a cyclic quadrilateral.  (1 mark)
  2. Prove that  `/_ APE = /_ ABC.`  (2 marks)
  3. Deduce that  `PQ`  is perpendicular to  `BC.`  (1 mark)
Show Answers Only

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

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

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

Show Worked Solution

(i)

 

  

`/_ PAE = /_ PDE = 90°\ \ text{(given)}`

`:. DPAE\ \ text(is a cyclic quadrilateral)`

`text{(opposite angles are supplementary)}`

 

(ii)  `text(Prove)\ \ /_ APE = /_ ABC`

`/_ ABC = /_ ADE`

`text{(angles in the same segment on arc}\ AC text{)}`

`text(S) text(ince)\ \ DPAE\ \ text(is a cyclic quad)\ \ text{(from (i))}`

`/_ APE = /_ ADE`

`text{(angles in the same segment on arc}\ AE text{)}`

`:.\ /_ APE = /_ ABC\ \ text(…  as required.)`

 

(iii)   `/_ APE = /_ ABC\ \ \ text{(from part (ii))}`

`/_ AEP = /_ BEQ\ \ \ text{(vertically opposite angles)}`

`:.\ Delta APE\ \ text(|||)\ \ Delta QBE\ \ text{(equiangular)}`

`:.\ /_ EQB = /_ EAP = 90°`

`text{(corresponding angles of similar triangles)}`

`:.\ PQ _|_ BC\ \ text(…  as required.)`

Calculus, EXT1 C3 2005 HSC 5a

Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve  `y = sin 2x`, the `x`-axis and the line  `x = pi/8`  is rotated about the `x`-axis.  (3 marks)

Show Answers Only

`pi/16 (pi – 2)\ \ \ text(u³)`

Show Worked Solution
`y` `= sin 2x`
`y^2` `= sin^2 2x` 

 
`text(Using:)\ \ sin^2 x= 1/2 (1 – cos 2x)`

COMMENT: Michael Wells (1st in state Ext2) would derive this formula in his working from the `sin^2x+cos^2=1` identity in ~5 seconds every time he used it in an exam.
  

`:. V` `=pi int_0^(pi/8) y^2 \ dx`
  `= pi int_0^(pi/8) sin^2 2x \ dx`
  `= pi/2 int_0^(pi/8) 1 – cos\ 4x\ dx`
  `= pi/2 [x – 1/4 sin\ 4x]_0^(pi/8)`
  `= pi/2 [(pi/8 – 1/4 sin\ pi/2) – 0]`
  `= pi/2 (pi/8 – 1/4)`
  `= pi/2 ((pi – 2)/8)`
  `= pi/16 (pi – 2)\ \ \ text(u³)`

Quadratic, EXT1 2005 HSC 4c

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

The equation of the normal to the parabola at  `P`  is  `x + py = 2ap + ap^3`  and the equation of the normal at  `Q`  is similarly given by  `x + qy = 2aq + aq^3.`

  1. Show that the normals at  `P`  and  `Q`  intersect at the point  `R`  whose coordinates are
    1. `(–apq[p + q], a[p^2 + pq + q^2 + 2]).`  (2 marks)

  2. The equation of the chord  `PQ`  is
    1. `y = 1/2 (p + q) x - apq.` (Do NOT show this.)
  4. If the chord  `PQ`  passes through  `(0, a)`, show that  `pq = –1.`  (1 mark)

  6. Find the equation of the locus of  `R`  if the chord  `PQ`  passes through  `(0, a).`  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `y = x^2/a + 3a`
Show Worked Solution

(i)   `text(Show)\ \ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`

 

ext1 2005 4c

`text(Equations of normals through)\ \  P and Q`

`x + py` `= 2ap + ap^3` `\ \ text{… (1)}`
`x + qy` `= 2aq + aq^3` `\ \ text{… (2)}`

 

`text{Multiply (1)} xx q\ ,\ \ text{(2)} xx p`

`qx + pqy` `= 2apq + ap^3q` `\ \ text{… (3)}`
`px + pqy` `= 2apq + apq^3` `\ \ text{… (4)}`

 

`text{Subtract  (4) – (3)}`

`x (p – q)` `= apq^3 – ap^3q`
  `= apq (q^2 – p^2)`
  `= apq (q – p) (q + p)`
  `= -apq (p -q) (p + q)`
`x` `= -apq [p + q]`

 

`text(Substitute)\ \ x = -apq [p + q]\ \ text{into  (1)}`

`py – apq[p + q]` `= 2ap + ap^3`
`py` `= 2ap + ap^3 + apq (p + q)`
`y` `= 2a + ap^2 + aq (p + q)`
  `= 2a + ap^2 + apq + aq^2`
  `= a[p^2 + pq + q^2 + 2]`

 

`:.\ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`

`text(…  as required.)`

 

(ii)  `PQ\ \ text(is)\ \ y = 1/2 (p + q)x – apq`

`text(Passes)\ \ (0, a)`

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

 

(iii)   `R\ \ text(has coordinates)`

`x` `= -apq [p + q]`
`x` `= a(p + q)\ \ \ text{(using part (ii))}`
 `x/a` `=(p + q)`

 

`y` `= a [p^2 + pq + q^2 + 2]`
  `= a (p^2 – 1 + q^2 + 2)`
  `= a (p^2 + q^2 +1)`
  `= a [(p + q)^2 – 2pq + 1]`
  `= a [(p + q)^2 + 3]`
  `= a [(x/a)^2 + 3]`
  `= x^2/a + 3a`

 

`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = x^2/a + 3a`