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Mechanics, SPEC2-NHT 2018 VCAA 3

A 200 kg crate rests on a smooth plane inclined at `theta` to the horizontal. An external force of `F` newtons acts up the plane, parallel to the plane, to keep the crate in equilibrium.

  1. On the diagram below, draw and label all forces acting on the crate.  (1 mark)

 

 

  1. Find `F` in terms of `theta`.  (1 mark)

The magnitude of the external force `F` is changed to 780 N and the plane is inclined at  `theta = 30^@`.

    1. Taking the direction down the plane to be positive, find the acceleration of the crate.  (2 marks)
    2. On the axes below, sketch the velocity–time graph for the crate in the positive direction for the first four seconds of its motion.  (1 mark)
      `qquad`
       

       
       
    3. Calculate the distance the crate travels, in metres, in its first four seconds of motion.  (1 mark)

Starting from rest, the crate slides down a smooth plane inclined at  `alpha`  degrees to the horizontal.

A force of  `295 cos(alpha)`  newtons, up the plane and parallel to the plane, acts on the crate.

  1. If the momentum of the crate is 800 kg ms¯¹ after having travelled 10 m, find the acceleration, in ms¯², of the crate.  (2 marks)
  2. Find the angle of inclination, `alpha`, of the plane if the acceleration of the crate down the plane is 0.75 ms¯².  Give your answer in degrees, correct to one decimal place.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `F = 200\ text(kg)\ sin(theta)`
    1. `a = 1\ text(ms)^(-2)`
    2. `text(See Worked Solutions)`
    3. `8 text(m)`
  4. `a = 0.8\ text(ms)^(-2) quad text(down the incline)`
  5. `a ~~ 12.9^@`
Show Worked Solution
a.   

 

b.   `F – 200g sin (theta) = 0`

`:. F = 200g sin (theta)`

 

c.i.   `sum F` `= -F + 200g sin (30^@)`
  `200a` `= -780 + 200g xx 1/2`
    `=- 780 + 980`
  `:.a` `=1\ text(ms)^(-1)`

 

c.ii.  

 

c.iii.  `text(Distance travelled in 4 seconds)`

`=\ text(Area under graph between)\ \ t=0 and t=4`

`=1/2 xx 4 xx 4`

`= 8\ text(m)`
 

d.   `p` `=mv`
  `800` `=200v`
  `:.v` `=4`

 
`text(Find)\ \a\ \ text(given)\ \ x = 10,\ \ v=4:`

`v^2` `= u^2 + 2ax`  
`16` `=20a`  
`:.a` `=0.8\ \ text(ms)^(-2) quad text(down the incline)`  

 

e.  `200g sin(alpha) – 295 cos (alpha) = 200 xx 0.75`

`alpha ~~ 12.9^@\ \ \ text{(by CAS)}`

Complex Numbers, SPEC2-NHT 2018 VCAA 2

In the complex plane, `L` is the line given by  `|z + 1| = |z + 1/2-sqrt 3/2 i|`.

  1. Show that the cartesian equation of `L` is given by  `y = -1/sqrt 3 x`.   (2 marks)

  2.  Find the point(s) of intersection of `L` and the graph of the relation  `z bar z = 4`  in cartesian form.   (2 marks)

  3. Sketch `L` and the graph of the relation  `z bar z = 4`  on the Argand diagram below.   (2 marks)

 

 

The part of the line `L` in the fourth quadrant can be expressed in the form  `text(Arg)(z) = a`.

  1. State the value of `a`.   (1 mark)

  2. Find the area enclosed by `L` and the graphs of the relations  `z bar z = 4, \ text(Arg)(z) = pi/3`  and  `text(Re)(z) = sqrt 3`.   (2 marks)

  3. The straight line `L` can be written in the form `z = k bar z`, where  `k in C`.

     

    Find `k` in the form  `r text(cis)(theta)`, where  `theta`  is the principal argument of `k`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `z = sqrt 3-i, quad -sqrt 3 + i`
  3. `text(See Worked Solutions)`
  4. `a = -pi/6`
  5. `pi/3 + sqrt 3`
  6. `k = cis (-pi/3)`
Show Worked Solution

a.   `(x + 1)^2 + y^2 = (x + 1/2)^2 + (y-sqrt 3/2)^2`

`x^2 + 2x + 1 + y^2 = x^2 + x + 1/4 + y^2-y sqrt 3 + 3/4`

`0` `=-x-y sqrt 3`
`y sqrt 3` `= -x`
`:. y` `= -1/sqrt 3 x`

 

b.   `(x + iy) (x-iy)` `=4`
  `x^2-i^2 y^2` `=4`
  `x^2 + y^2` `=4`

 

`text(Substitute)\ \ y = -1/sqrt 3 x\ \ text(into)\ \  x^2 + y^2 = 4:`

`x^2 + 1/3 x ^2` `=4`
`4/3 x^2` `=4`
`x^2` `=3`
`x` `=+- sqrt3`
`=>y` `= +- 1`

 

`:.\ text(Intersection at):\  (sqrt 3, -1), quad (-sqrt 3, 1)`

 

c.   

 

d.   `alpha = tan^(-1) (-1/sqrt 3), quad alpha in (-pi/2, 0)`

`alpha = -pi/6`
 

e.   

`text(Total Area)`

`=\ text(Area of sector + Area of triangle)`

`=pi xx 2^2 xx ((pi/3-pi/6)/(2 pi)) + 1/2 xx sqrt 3 xx 2`

`=4 pi (1/12) + sqrt3`

`=pi/3 + sqrt3\ \ text(u²)`

 

f.    `r\ text(cis)(theta)` `=k (r\ text(cis)(-theta))`
  `:. k` `=(r\ text(cis)(theta))/(r\ text(cis)(-theta))`
    `= text(cis)(2 xx ((-pi)/6))`
    `= text(cis)(- pi/3)`

Calculus, SPEC2-NHT 2018 VCAA 1

Consider the function  `f`  with rule  `f(x) = 10 arccos (2-2x)`.

  1.  Sketch the graph of  `f`  over its maximal domain on the set of axes below. Label the endpoints with their coordinates.   (3 marks)

     


 

  1. A vase is to be modelled by rotating the graph of  `f`  about the `y`-axis to form a solid of revolution, where units of measurement are in centimetres.
    1.  Write down a definite integral in terms of `y` that gives the volume of the vase.   (2 marks)

    2.  Find the volume of the vase in cubic centimetres.  (1 mark)

  3. Water is poured into the vase at a rate of 20 cm³ s¯¹.
  4. Find the rate, in centimetres per second, at which the depth of the water is changing when the depth is  `5 pi`  cm.  (3 marks)

  5. The vase is placed on a table. A bee climbs from the bottom of the outside of the vase to the top of the vase.
  6. What is the minimum distance the bee will need to travel? Give your answer in centimetres, correct to one decimal place.  (1 mark)

Show Answers Only

  1. `text(See Worked Solutions)`
  2.  i. `V = pi int_0^(10 pi) (1-1/2 cos (y/10))^2 dy`
  3. ii. `V = (45 pi^2)/4 text(cm)^3`
  4. `20/pi text(cm s)^(-1)`
  5. `31.4\ text(cm)`

Show Worked Solution

a.  

`2-2x in [-1, 1]`

`-2x in [-3, -1]`

`:. x in [1/2, 3/2]`

`f(1/2)` `= 10 cos^(-1) (1)=0`
`f(3/2)` `= 10 cos^(-1) (1)=10pi`

 

b.i.   `y` `= 10 cos^(-1) (2-2x)`
  `y/10` `= cos^(-1) (2-2x)`
  `cos (y/10)` `= 2-2x`
  `2x` `= 2-cos (y/10)`
  `x` `= 1-1/2 cos (y/10)`

 

  `:. V` `= pi int_0^(10 pi) x^2\ dy`
    `= pi int_0^(10 pi) (1-1/2 cos (y/10))^2\ dy`

 

b.ii.   `V = (45 pi^2)/4 text(cm)^3`

 

c.  `(dV)/(dt) = 20\ text(cm³/s)\ \ \ text{(given)}`

`(dV)/(dh) = pi (1-1/2 cos (y/10))^2\ \ text(when)\ \ y=5pi` 

`=> (dV)/(dh) = pi`
 

`:. (dh)/(dt)` `= (dh)/(dV)*(dV)/(dt)`
  `= 1/pi * 20`
  `= 20/pi\ text(cm s)^(-1)`

 

d.   `f(x) = 10cos^(-1)(2-2x)`

`l=int_(1/2)^(3/2) sqrt(1 + (f′(x))^2)\ dx`

  `~~31.4\ text(cm)\ \ \ text{(by CAS)}`

Calculus, SPEC2 2017 VCAA 1

Let  `f:D ->R, \ f(x) = x/(1 + x^3)`, where `D` is the maximal domain of  `f`.

  1.   i. Find the equations of any asymptotes of the graph of  `f`.   (1 mark)

  2.  ii. Find  `f′(x)`  and state the coordinates of any stationary points of the graph of  `f`, correct to two decimal places.  (2 marks)

  3. iii. Find the coordinates of any points of inflection of the graph of  `f`, correct to two decimal places.  (2 marks)

  4. Sketch the graph of  `f(x) = x/(1 + x^3)`  from  `x=–3`  and  `x = 3`  on the axes provided below, marking all stationary points, points of inflection and intercepts with axes, labelling them with their coordinates. Show any asymptotes and label them with their equations.  (3 marks)

     

     

  5. The region `S`, bounded by the graph of  `f`, the `x`-axis and the line  `x = 3`, is rotated about the `x`-axis to form a solid of revolution. The line  `x = a`, where  `0 < a < 3`, divides the region `S` into two regions such that, when the two regions are rotated about the `x`-axis, they generate solids of equal volume.
  6. i.  Write down an equation involving definite integrals that can be used to determine `a`.  (2 marks)

  7. ii. Hence, find the value of `a`, correct to two decimal places.  (1 mark)

Show Answers Only

  1. i.  `text(vertical asymptote:)\ \ x = −1“text(horizontal asymptote:)\ \ y = 0`
  2. ii. `text(S.P.)\ \ (0.79, 0.53)`
  3. iii. `text(P.O.I.)\ (1.26, 0.42)`
  4.  
  5. i.  `pi int_a^3 (x^2)/((1 + x^3)^2) dx`
  6. ii. `~~ 0.98`

Show Worked Solution

a.i.   `text(Graphing the function on CAS:`

♦♦ Mean mark 36%.

`text(Vertical asymptote:)\ x = −1`

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

a.ii.   `u = x, \ u′ = 1,\ \ v = 1 + x^3, \ v′ = 3x^2\ \ \ text{(manual or by CAS)}`

`f′(x)` `= (1(1 + x^3)-x(3x^2))/((1 + x^3)^2)`
  `= (1-2x^3)/((1 + x^3)^2)`

 
`text(S.P. when)\ \ f′(x)=0:\ `

`=>  (0.79, 0.53)\ \ \ text{(by CAS)}`

 

a.iii.  `text(When)\ \ f″(x)=0,\ \ \ text{(by CAS)}`

`=> x = 0, \ x = -1, \ x = 2`

`x != -1`

`text(Check concavity changes:)`

`f″(−1/2) = −1632/343`

`f″(1) = −3/4`

`f″(3) = 675/(10\ 976)`
 

`text(P.O.I. at)\ \ x = sqrt2 ~~ 1.26\ \ text{(concavity changes)}`

`=> f(sqrt2) ~~ 0.42`

`:. text(P.O.I.)\ (1.26, 0.42)`

 

b.   

 

c.i.    `V_1` `= pi int_0^a y^2\ dx`
  `V_2` `= pi int_a^3 y^2\ dx`

 
`:.\ text(Equation to solve for)\ a:`

`int_0^a x^2/(1 + x^3)^2\ dx = int_a^3 x^2/(1 + x^3)^2\ dx`

 

c.ii.  `a=0.98\ \ \ text{(by CAS)}`

Statistics, SPEC2 2018 VCAA 6

The heights of mature water buffaloes in northern Australia are known to be normally distributed with a standard deviation of 15 cm. It is claimed that the mean height of the water buffaloes is 150 cm.

To decide whether the claim about the mean height is true, rangers selected a random sample of 50 mature water buffaloes. The mean height of this sample was found to be 145 cm.

A one-tailed statistical test is to be carried out to see if the sample mean height of 145 cm differs significantly from the claimed population mean of 150 cm.

Let `bar X` denote the mean height of a random sample of 50 mature water buffaloes.

  1. State suitable hypotheses `H_0` and `H_1` for the statistical test.   (1 mark)

  2. Find the standard deviation of `bar X`.   (1 mark)

  3. Write down an expression for the `p` value of the statistical test and evaluate your answer to four decimal places.   (2 marks)

  4. State with a reason whether `H_0` should be rejected at the 5% level of significance.   (1 mark)

  5. What is the smallest value of the sample mean height that could be observed for `H_0` to be not rejected? Give your answer in centimetres, correct to two decimal places.   (1 mark)

  6. If the true mean height of all mature water buffaloes in northern Australia is in fact 145 cm, what is the probability that `H_0` will be accepted at the 5% level of significance? Give your answer correct to two decimal places.   (1 mark)

  7. Using the observed sample mean of 145 cm, find a 99% confidence interval for the mean height of all mature water buffaloes in northern Australia. Express the values in your confidence interval in centimetres, correct to one decimal place.   (1 mark)

Show Answers Only
  1.  `H_0: mu = 150; qquad H_1: mu < 150`
  2. `3/sqrt 2`
  3. `p = text(Pr)(bar X < 145); qquad p~~ 0.0092`
  4. `text(Yes, he should be rejected as)`
    `p~~ 0.0092 < 0.05`
  5. `bar X_min = 146.52`
  6. `0.24`
  7. `(139.5, 150.5)`
Show Worked Solution
a.    `H_0: mu =150`
  `H_1: mu < 150`

 

b.    `sigma_bar X` `= (sigma_X)/sqrt n`
    `= 15/sqrt 50`
    `= (3sqrt 2)/2`

 

c.   `p = text(Pr)(bar X < 145), qquad bar X\ ~\ N (150, 9/2)`

`p ~~ 0.0092`

 

d.   `text(Yes, he should be rejected as)`

`p~~ 0.0092 < 0.05`

 

e.  `text(Pr)(bar X < x) = 0.05`

♦ Mean mark part (e) 48%.

`x ~~ 146.511`

`text(NOT rejected:) quad bar X_min = 146.52`

 

f.   `bar X_2\ ~\ N (145, 9/2)`

♦♦♦ Mean mark part (f) 11%.

`text(Pr)(bar X_2 > x)` `= text(Pr)(bar X_2 > 146.51074)`  
  `~~ 0.24`  

 

g.   `(145-(Z_99 xx 3)/sqrt 2, 145 + (Z_99 xx 3)/sqrt 2)`

`text(Pr)(Z < Z_99) = 0.995, \ Z\ ~\ N(0, 1)`

`=>Z_99 ~~ 2.57583`

`:. 99%\ text(C.I:)\ (139.5, 150.5)`

Mechanics, SPEC2 2018 VCAA 5

Luggage at an airport is delivered to its owners via a 15 m ramp that is inclined at 30° to the horizontal. A 20 kg suitcase, initially at rest at the top of the ramp, slides down the ramp against a resistance of `v` newtons per kilogram, where `v\ text(ms)^(-1)` is the speed of the suitcase.

  1.  On the diagram below, show all forces acting on the suitcase during its motion down the ramp.  (1 mark)
     
         

  2.  i. By resolving forces parallel to the ramp, write down an equation of motion for the 20 kg suitcase.  (1 mark)
  3. ii. Hence, show that the magnitude of the acceleration, `a\ text(ms)^(-2)`, of the suitcase down the ramp is given by  `a = (g - 2v)/2`. (1 mark)
  4. By expressing `a` in an appropriate form, find the distance `x` metres that the suitcase has slid as a function of `v`. Give your answer in the form  `x = bv + c log_e(c/(c - v))`, where  `b, c in R`.  (2 marks)
  5. Find the velocity of the suitcase just before it reaches the end of the ramp. Give your answer in `text(ms)^(-1)`, correct to two decimal places.  (1 mark)
  6.  i. Write down a definite integral that gives the time taken for the suitcase to reach a speed of `4.5\ text(ms)^(-1)`.  (1 mark)
  7. ii. Find the time taken for the suitcase to reach a speed of `4.5\ text(ms)^(-1)`. Give your answer in seconds, correct to two decimal places.  (1 mark)
Show Answers Only
  1.    

    1. `10g – 20v = 20a`
    2. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = -v + 4.9 ln ((4.9)/(4.9 – v))`
  3. `v ~~4.81\ text(ms)^(-1)`
    1. `t(4.5) = int_0^4.5 2/(g – 2v) dv`
    2. `t(4.5) ~~ 2.51\ text(seconds)`
Show Worked Solution
a.   

♦ Mean mark part (a) 44%.

 

b.i.    `20g sin 30^@ – 20v` `= sumF`
  `10g – 20v` `= 20a`

 

b.ii.    `(10g – 20v)/20` `= a`
  `g/2 – v` `= a`
  `:.a` `= (g – 2v)/2`

 

♦ Mean mark part (c) 48%.

c.   `v *(dv)/(dx)` `= (g – 2v)/2`
  `(dv)/(dx)` `= (g – 2v)/(2v)`
  `(dx)/(dv)` `= (2v)/(g – 2v)`
  `(dx)/(dv)` `= -(2v)/(2v – g)`
    `= – ((2v – g + g))/(2v – g)`
    `= -1 – g/(2v – g)`

 

`x` `= int_0^v – 1 – g/(2v – g)\ dv`
  `= int_0^v – 1 – g/2 (2/(2v – g))\ dv`
  `= [-v – 4.9ln\ |2v – g|]_0^v`

 
`text(When)\ \ x=0, v=0:`

`2v – g < 0\ \ =>\ \ |2v – g| = g – 2v`

 

`x` `= [-v – 4.9 ln (g – 2v)]_0^v`
  `= -v – 4.9 ln (g – 2v) – (0 – 4.9 ln (g))`
  `= -v – 4.9 ln (g – 2v) + 4.9 ln (g)`
  `= -v + 4.9 ln (g/(g – 2v))`
  `=-v + 4.9 ln(9.8/(9.8-2v))`
`:. x` `= -v + 4.9 ln ((4.9)/(4.9 – v))`

 

d.   `text(Find)\ \ v\ \ text(when)\ \ x=15:`

`15 = -v + 4.9 ln (4.9/(4.9 – v))`

♦♦ Mean mark part (d) 26%.

`:. v ~~ 4.81\ text(ms)^(-1)\ \ \ (v>0)`

 

♦ Mean mark 41%.

e.i    `(dv)/(dt)` `= (g – 2v)/2`
  `(dt)/(dv)` `= 2/(g – 2v)`
  `:.t` `= int_0^4.5 2/(g – 2v)\ dv`

 

♦ Mean mark 41%.

e.ii.  `t=[-ln (9.8 -2v)]_0^4.5`

`=> t ~~ 2.51 text(s)`

Vectors, SPEC2 2018 VCAA 4

Two yachts, `A` and `B`, are competing in a race and their position vectors on a certain section of the race after time  `t`  hours are given by

`underset ~ r_A (t) = (t + 1) underset ~i + (t^2 + 2t) underset ~j \ and \ underset ~r_B (t) = t^2 underset ~i + (t^2 + 3) underset ~j, \ t >= 0`

where displacement components are measured in kilometres from a given reference buoy at origin `O`.

  1. Find the cartesian equation of the path for each yacht.   (2 marks)

  2. Show that the two yachts will not collide if they follow these paths.   (2 marks)

  3. Find the coordinates of the point where the paths of the two yachts cross. Give your coordinates correct to three decimal places.   (2 marks)

One of the rules for the race is that the yachts are not allowed to be within 0.2 km of each other. If this occurs there is a time penalty for the yacht that is travelling faster.

  1. For what values of `t` is yacht `A` travelling faster than yacht `B`?   (2 marks)

  2. If yacht `A` does not alter its course, for what period of time will yacht `A` be within 0.2 km of yacht `B`? Give your answer in minutes, correct to one decimal place.   (2 marks)

Show Answers Only
  1. `y_A = x_A^2-1, x_A > 1; qquad y_B = x_B + 3, x_B > 0`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `(x, y) ~~ (2.562, 5.562)`
  4. `0 < t < 5/2`
  5. `4.1\ text(minutes)`
Show Worked Solution

a.  `x_A = t + 1,\ \ y_A = = t^2 + 2t`

`y_A` `= t^2 + 2 t + 1-1`
  `= (t + 1)^2-1`
`:. y_A` `= x_A^2-1`

 
`x_B = t^2,`

`y_B` `= t^2+3`
`:.y_B` `= x_B+3`

 

b.  `text(Same)\ \ xtext(-coordinate occurs when:)`

`t+1` `= t^2`
`t` `= (1+sqrt5)/2,\ \ \ (t>0)`

 
`text(When)\ \ t = (1+sqrt5)/2,`

`y_A=(3sqrt5+5)/2`

`y_B= (sqrt5+9)/2  !=y_A`
 

`:.\ text(No collision)`

 

c.   `x + 3` `= x^2-1`
  `x^2-x-4` `= 0`
  `x` `= (1 + sqrt 17)/2,\ \ \ (t>0 \ =>\ x>1)`
  `y` `= (7 + sqrt 17)/2`
  `:. (x, y)` `= ((1 + sqrt 17)/2, (7 + sqrt 17)/2)`
    `=(2.562, 5.562)`

 

d.    `underset ~dot r_A (t)` `= underset ~i + (2t + 2)underset ~j`
  `|underset ~dot r_A (t)|` `= sqrt(1 + (2t + 2)^2)`
    `= sqrt(4t^2 + 8t + 5)`
     
  `underset ~dot r_B (t)` `= 2t underset ~i + 2t underset ~j`
  `|underset ~dot r_B (t)|` `= sqrt (4t^2 + 4t^2)`
    `= sqrt(8t^2)`

  
 `text(Yacht A is travelling faster when:)`

♦ Mean mark part (d) 41%.

`sqrt(4t^2 + 8t + 5)` `> sqrt (8t^2)`
`4t^2 + 8t + 5` `> 8t^2`
`4t^2-8t-5` `< 0`

 
`:. 0 < t < 5/2`

 

e.    `underset ~ r_B-underset ~r_A` `= (t^2-(t + 1)) underset ~ i + (t^2 + 3-(t^2 + 2t)) underset ~j`
    `= (t^2-t-1) underset ~i + (-2t + 3) underset ~j`

 

`d = |underset ~r_B-underset~r_A|= sqrt ((t^2-t-1)^2 + (3-2t)^2)`
 

`text(Find)\ \ t\ \ text(such that:)`

♦ Mean mark part (e) 37%.

`sqrt ((t^2-t-1)^2 + (3-2t)^2) < 0.2`

`=> 1.52883 < t < 1.59734`

 
`:.\ text(Period of time yachts are within 0.2 km)`

`~~ 1.59734-1.52883`

`~~0.068506\ text(hours)`

`~~ 4.1\ text(minutes)`

Statistics, SPEC2-NHT 2017 VCAA 19 MC

The petrol consumption of a particular model of car is normally distributed with a mean of 12 L/100 km and a standard deviation of 2 L/100 km.

The probability that the average petrol consumption of 16 such cars exceeds 13 L/100 km is closest to

A.   0.0104

B.   0.0193

C.   0.0228

D.   0.3085

E.   0.3648

Show Answers Only

`C`

Show Worked Solution

`P\ ~\ N (12, 2^2) \ => \ bar P\ ~\ N (12, 2^2/16)`

`Pr(bar P > 13) ~~ 0.0228\ \ \ text{(by CAS)}`

`=>   C`

Mechanics, SPEC2-NHT 2017 VCAA 16 MC

A person of mass `M` kg carrying a bag of mass `m` kg is standing in a lift that is accelerating downwards at  `a\ text(ms)^(-2)`.

The force of the lift floor acting on the person has magnitude

A.   `Mg + mg`

B.   `Mg + (M + m)a`

C.   `Mg - (M + m)a`

D.   `(M + m) (g + a)`

E.   `(M + m) (g - a)`

Show Answers Only

`E`

Show Worked Solution

`sum F` `= (M+m) g – R`
`(M+m) a` `=(M+m)g-R`
`R` `= (M+m) (g-a)`

 
`=>E`

Vectors, SPEC2-NHT 2017 VCAA 15 MC

A particle of mass 2 kg has an initial velocity of   `underset ~i - 6 underset ~j\ \ text(ms)^(-1)`.

After a change of momentum of  `6 underset ~i - 2 underset ~j\ \ text(kg ms)^(-1)`,  the particle's velocity, in `text(ms)^(-1)`,  is

  1. `3 underset ~i - underset ~j`
  2. `2 underset ~i - 12 underset ~j`
  3. `4 underset ~i - 7 underset ~j`
  4. `2 underset ~i + 5 underset ~j`
  5. `11 underset ~i + 2 underset ~j`
Show Answers Only

`C`

Show Worked Solution
`m underset ~(v_i)` `= 2(underset ~i – 6 underset ~j)`
`m Delta underset ~v` `= m underset ~(v_f) – m underset ~(v_i)`
  `= m(underset ~(v_f) – underset ~(v_i))`
`6 underset ~i – 2 underset ~j` `= 2(underset ~ (v_f )- (underset ~i – 6 underset ~j))`
`2 underset ~ (v_f )` `= 6 underset ~i – 2 underset ~j + 2 underset ~i + – 12 underset ~j`
  `= 8 underset ~i -14 underset ~j`
`:. underset ~ (v_f )` `= 4 underset ~i -7 underset ~j`

 

`=>   C`

Vectors, SPEC2-NHT 2017 VCAA 14 MC

Given that the vectors  `underset ~a = underset ~i + underset ~j - underset ~k,\ underset ~b = 2 underset ~i - underset ~j + 2 underset ~k`  and  `underset ~c = lambda underset ~i - underset ~j + 4 underset ~k`  are linearly dependent, the value of  `lambda`  is

A.  –10

B.  –8

C.    2

D.    4

E.    8

Show Answers Only

`E`

Show Worked Solution

`alpha underset ~a + beta underset ~j = underset ~c`

`(1):\ \ (alpha + 2 beta) = lambda,`

`(2):\ \ alpha – beta = -1,`

`(3):\ \ -alpha + 2 beta = 4`
 

`(2) + (3):`

`beta = 3`
 

`text(Substitute)\ \ beta = 3\ \ text(into)\ \ (2):`

`alpha – 3` `= -1`
`alpha` `= 2`

 
`text(Substitute)\ \ alpha = 2,\ beta = 3\ \ text(into)\ \ (1):`

`lambda` `= 2 + 6`
  `= 8`

 
`=>   E`

Vectors, SPEC2-NHT 2017 VCAA 11 MC

Two particles have positions given by  `underset ~r_1 = (3 - 4t^2) underset ~i + (t + b) underset ~j`  and  `underset ~r_2 = 5t^2 underset ~i + (t^2 - 1) underset ~j`, where  `t >= 0`  and `b` is a real constant.

The particles will collide if the value of `b` is

  1. `(2 - sqrt 3)/3`
  2. `sqrt 3 - 1`
  3. `(2 + sqrt 3)/3`
  4. `(-2 - sqrt 3)/3`
  5. `-sqrt 3 - 1`
Show Answers Only

`D`

Show Worked Solution
`3 – 4t^2` `= 5t^2`
`9t^2` `= 3`
`t^2` `= 1/3`
`t` `= 1/sqrt 3\ \ \ (t >= 0)`

 

`t + b` `= t^2 – 1`
`1/sqrt 3 + b` `= 1/3 – 1`
`b` `= – 2/3 – 1/sqrt 3`
  `= (-2 – sqrt 3)/3`

 
`=>   D`

Calculus, SPEC2-NHT 2017 VCAA 10 MC

A solution to the differential equation  `(dy)/(dx) = (cos(x + y)-cos(x-y))/(e^(x + y))`  can be obtained from

  1. `int e^y/(sin(y))\ dy = -int (2 sin(x))/e^x\ dx`
  2. `int e^y/(cos(y))\ dy = int 2/e^x\ dx`
  3. `int e^y/(cos(y))\ dy = -int (2 cos(x))/e^x\ dx`
  4. `int e^(-y)/(sin(y))\ dy = int 2e^(-x) sin(x)\ dx`
  5. `int e^y/(cos(y))\ dy = int (2 sin(x))/e^x\ dx`
Show Answers Only

`A`

Show Worked Solution
`dy/dx` `=(cos(x + y)-cos(x-y))/(e^(x + y))`
`(dy)/(dx)` `= (cos(x) cos(y)-sin(x) sin(y)-cos(x) cos(y)-sin(x) sin(y))/(e^x ⋅ e^y)`
`e^y *(dy)/(dx)` `= (-2 sin(x) sin(y))/(e^x)`
`e^y/(sin(y)) *(dy)/(dx)` `= (-2 sin(x))/(e^x)`
`:. int e^y/(sin(y))\ dy` `= -int (2 sin(x))/e^x\ dx`

 
`=>   A`

Calculus, SPEC2-NHT 2017 VCAA 9 MC

The gradient of the tangent to a curve at any point  `P(x, y)`  is half the gradient of the line segment joining `P` and the point  `Q(-1, 1)`.

The coordinates of points on the curve satisfy the differential equation

A.   `(dy)/(dx) = (y + 1)/(2(x - 1))`

B.   `(dy)/(dx) = (2(y - 1))/(x + 1)`

C.   `(dy)/(dx) = (x - 1)/(2(y + 1))`

D.   `(dy)/(dx) = (2(x - 1))/(y + 1)`

E.   `(dy)/(dx) = (y - 1)/(2(x + 1))`

Show Answers Only

`E`

Show Worked Solution
`m_text(tang)` `= 1/2 m_(PQ)`
`m_(PQ)` `= (y – 1)/(x – (-1))`
  `= (y – 1)/(x + 2)`

 
`:. m_text(tang) = (dy)/(dx) = (y – 1)/(2(x + 1))`

`=>   E`

Calculus, SPEC2-NHT 2017 VCAA 7 MC

`int_(pi/4)^(pi/3) (sin^3(x) cos^2(x))\ dx`  is equivalent to

A.   `int_(1/2)^(1/sqrt 2) (u^4 - u^2)\ du`  where  `u = cos(x)`

B.   `-int_(1/sqrt 2)^(1/2) (u^2 - u^4)\ du`  where  `u = cos(x)`

C.   `-int_(pi/4)^(pi/3) (u^2 - u^4)\ du`  where  `u = sin(x)`

D.   `int_(pi/4)^(pi/3) (u^2 - u^4)\ du`  where  `u = sin(x)`

E.   `- int_(1/sqrt 2)^(1/2) (u^2 - u^4)\ du`  where  `u = sin(x)`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ u = cos (x)`

`sin^3(x)` `= sin(x) ⋅ sin^2(x)`
  `= sin(x) (1 – cos^2(x))`

`(du)/dx = – sin(x)\ \ =>\ \ du = – sin(x)\ dx`

  
`u(pi/3) = 1/2,\ \ u(pi/4) = 1/sqrt 2`
 

`:. int_(pi/4)^(pi/3) sin(x)(1-cos^2(x)) cos^2(x)\ dx`

`=int_(1/sqrt 2)^(1/2) -(1 – u^2)u^2\ du`

`= -int_(1/sqrt 2)^(1/2) (u^2 – u^4)\ du`

 
`=>   B`

Complex Numbers, SPEC2-NHT 2017 VCAA 6 MC


 

The relation that defines the line `S` above is

A.   `|z + 2| = |z + 2i|`

B.   `text(Arg)(z) = (3 pi)/4`

C.   `|z - 2| = |z + 2i|`

D.   `text(Im)(z) = text(Arg) ((3 pi)/4) + text(Arg)(-pi/4)`

E.   `|z - 2| = |z - 2i|`

Show Answers Only

`C`

Show Worked Solution

`text(Consider the options:)`

`A:\ text(Need)\ z\ text(to be equidistant from)`

`x = -2 \ and\  y = -2\ \ =>\ text(not correct)`
 

`B: text(Arg)(z) = (3 pi)/4\ \ text(is a ray originating at the)`

`text(origin)\ =>\ text(not correct)`

 

`C: text(Need)\ z\ text(to be equidistant from)`

`x = 2 \ and\  y = -2`
 

`text(Midpoint): ((2 +0)/2, (0 + (-2))/2) ≡(1,-1)`

`m = (-2 – 0)/(0 – 2) = (-2)/(-2) = 1`

`m_⊥ = (-1)/m_1 = -1`
 

`:.\ text(Option)\ C\ text(describes the correct relation.)`

`=>   C`

Algebra, SPEC2-NHT 2017 VCAA 3 MC

For the function  `f: R -> R, \ f(x) = k arctan (ax - b) + c`, where  `k > 0, \ c > 0`  and  `a, b in R, \ f(x) > 0`  if

A.   `c < (k pi)/2`

B.   `c >= (k pi)/2`

C.   `x > b/a`

D.   `c + k > pi/2`

E.   `c >= pi/2`

Show Answers Only

`B`

Show Worked Solution
`tan^(-1)(ax -b)` `in ((-pi)/2, pi/2)`
`k tan^(-1) (ax – b)` `in ((- k pi)/2, (k pi)/2),\ text(as)\ k > 0`

 
`f(x) = k tan^(-1) (ax – b) + c in (c – (k pi)/2, (k pi)/2 + c)`

`text(When)\ \ f(x)>0:`

`c – (k pi)/2` `>= 0\ \ \ text{(domain doesn’t include end limits)}`
`:. c` `>= (k pi)/2`

 
`=>   B`

Graphs, SPEC2-NHT 2017 VCAA 1 MC

The number of asymptotes of the graph of the function with rule  `f(x) = (x^3 - 7x + 5)/(x^2 + 3x - 4)`  is

A.   0

B.   1

C.   2

D.   3

E.   4

Show Answers Only

`D`

Show Worked Solution

`text(By CAS:)`

`(x^3 – 7x + 5)/(x^2 + 3x – 4)`

`=x-3 +31/(5(x+4)) – 1/(5(x-1))`

 
`:. text(Asymptotes at:)\ \ x=1,\ \ x=-4,\ \ y=x-3`

`=>   D`

Calculus, SPEC2 2015 VCAA 10 MC

Using a suitable substitution, the definite integral  `int_0^1(x^2sqrt(3x + 1))\ dx`  is equivalent to

A.   `1/9int_0^1(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

B.   `1/27int_1^4(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

C.   `1/9int_1^4(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

D.   `1/27int_0^1(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

E.   `1/3int_1^4(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ \ u = 3x + 1`

`x=(u – 1)/3\ \ =>\ \ x^2=(u^2 – 2u + 1)/9`

`(du)/dx = 3\ \ =>\ \ 1/3 du = dx`
 

`text(When)\ \ x=1,\ u=4`

`text(When)\ \ x=0,\ u=1`
 

`:. int_0^1(x^2sqrt(3x + 1))\ dx`

`=int_1^4 1/9(u^2 – 2u + 1)*u^(1/2) xx 1/3\ du`

`= 1/27 int_1^4 (u^(5/2) – 2u^(3/2) + u^(1/2))\ du`

 
`=> B`

Complex Numbers, SPEC2 2015 VCAA 8 MC

A relation that does not represent a circle in the complex plane is

  1. `zbarz = 4`
  2. `|\ z + 3i\ | = 2|\ z − i\|`
  3. `|\ z − i\ | = |\ z + 2\ |`
  4. `|\ z − 1 + i\ | = 4`
  5. `|\ z\ | + 2|\ barz\ | = 4`
Show Answers Only

`C`

Show Worked Solution

`text(Consider each option:)`

`A:` `(x + iy)(x – iy)` `= 4`
  `x^2 + y^2` `= 4\ \ \ \ (text{circle equation)}`

 

`B:\ x^2+ (y + 3)^2=4(x^2 + (y – 1)^2)`

`x^2 + y^2 + 6y + 9` `= 4x^2 + 4y^2 – 8y + 4`
`3x^2 + 3y^2 – 14y -5` `=0\ \ \ \ (text{circle equation)}`

 

`C:` `x^2 + (y – 1)^2` `= (x + 2)^2 + y^2`
  `x^2 + y^2 – 2y + 1` `= x^2 + 4x + 4 + y^2`
  `4x+2y+3` `=0 \ \ \ \ (text{not a circle equation)}`

 
`text(Similarly, D and E can be shown to represent circle equations.)`

`=> C`

Vectors, SPEC1-NHT 2017 VCAA 10

Consider the vectors  `underset ~a = - underset ~i - 2 underset ~j + 3 underset ~k`  and  `underset ~b = 2 underset ~i + c underset ~j + underset ~k`.

Find the value of  `c, \ c in R`, if the angle between  `underset ~a`  and  `underset ~b`  is  `pi/3`.  (4 marks)

Show Answers Only

`c = -3`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= -1 xx 2 + (-2) xx c + 3 xx 1`
  `= -2 – 2c + 3`
  `= 1 – 2c`

 

`1-2c` `= sqrt((-1)^2 + (-2)^2 + 3^3) *sqrt(2^2 + c^2 + 1^2) xx cos (pi/3)`
`1 – 2c` `= 1/2(sqrt 14 ⋅ sqrt(5 + c^2))`
`2 – 4c` `= sqrt(14(5 + c^2))`
`(2 – 4c)^2` `= 14(5 + c^2)`
`4 – 16c + 16c^2` `= 70 + 14c^2`
`2c^2 – 16c – 66` `= 0`
`c^2 – 8c – 33` `= 0`
`(c – 11)(c + 3)` `= 0`

 
`c = 11 or c = -3`

`text(S)text(ince)\ \ 2 – 4c = sqrt(15(5 + c^2))`

`2 – 4c > 0\ \ =>\ \ c<2`

`:. c = -3`

Statistics, SPEC1-NHT 2017 VCAA 9

The random variables `X` and `Y` are independent with  `mu_X = 4,\ text(Var)(X) = 36`  and  `mu_Y = 3,\ text(Var)(Y) = 25`.

  1. The random variable `Z` is such that  `Z = 2X + 3Y`.
  2.  i. Find `E(Z)`.   (1 mark)

  3. ii. Find the standard deviation of `Z`.   (1 mark)

  1. Researchers have reason to believe that the mean of `X` has decreased. They collect a random sample of 64 observations of `X` and find that the sample mean is  `bar X = 3.8`
  2.  i. State the null hypothesis and the alternative hypothesis that should be used to test that the mean has decreased.   (1 mark)

  3. ii. Calculate the mean and standard deviation for a distribution of sample means, `bar X`, for samples of 64 observations.   (1 mark)

Show Answers Only
  1.  i. `17`
  2. ii. `3 sqrt 41`
  3.  i. `H_0: mu = 4,\ \ H_1: mu < 4`
  4. ii. `mu_bar X = 4,\ \ sigma_bar X = 3/4`
Show Worked Solution
a.i.    `E(Z)` `= 2E(X) + 3E(Y)`
    `= 2(4) + 3(3)`
    `= 8 + 9`
    `= 17`

 

a.ii.    `text(Var)(Z)` `= 2^2 text(var)(X) + 3^2 text(var)(Y)`
    `= 4(36) + 9(25)`
    `= (4xx9xx4) + 9(25)`
    `= 9(16 + 25)`

 

`:. sigma_Z` `= sqrt(9 (41))`
  `= 3 sqrt 41`

 

b.i.    `H_0: \ mu = 4`
  `H_1: \ mu < 4`

 

b.ii.    `bar X~N(mu, (sigma^2)/n)`
  `bar X~N(4, 36/64)`

 
`E(barX) = 4`

`sigma_bar X` `= sqrt(36/64)`
  `= 3/4`

Calculus, SPEC2 2018 VCAA 3

Part of the graph of  `y = 1/2 sqrt(4x^2-1)`  is shown below.
 


 

The curve shown is rotated about the `y`-axis to form a volume of revolution that is to model a fountain, where length units are in metres.

  1. Show that the volume, `V` cubic metres, of water in the fountain when it is filled to a depth of `h` metres is given by  `V = pi/4(4/3h^3 + h)`.   (2 marks)

  2. Find the depth `h` when the fountain is filled to half's its volume. Give your answer in metres, correct to two decimal places.   (2 marks)

The fountain is initially empty. A vertical jet of water in the centre fills the fountain at a rate of 0.04 cubic metres per second and, at the same time, water flows out from the bottom of the fountain at a rate of  `0.05 sqrt h`  cubic metres per second when the depth is `h` metres.

  1.  i. Show that  `(dh)/(dt) = (4-5sqrt h)/(25 pi (4h^2 + 1))`.   (2 marks)

  2. ii. Find the rate, in metres per second, correct to four decimal places, at which the depth is increasing when the depth is 0.25 m.   (1 mark)

  3. Express the time taken for the depth to reach 0.25 m as a definite integral and evaluate this integral correct to the nearest tenth of a second.   (2 marks)

  4. After 25 seconds the depth has risen to 0.4 m.
    Using Euler's method with a step size of five seconds, find an estimate of the depth 30 seconds after the fountain began to fill. Give your answer in metres, correct to two decimal places.   (2 marks)

  5. How far from the top of the fountain does the water level ultimately stabilise? Give your answer in metres, correct to two decimal places.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `h~~ 0.59\ text(m)`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `0.0153\ text(ms)^(-1)`
  1. `9.8\ text(seconds)`
  2. `0.43\ text(m)`
  3. `0.23\ text(m)`
Show Worked Solution

a.   `V= pi int_0^h x^2\ dy`

`y` `=1/2 sqrt(4x^2-1)`
`2y` `=sqrt(4x^2-1)`
`4y^2` `= 4x^2-1`
`4x^2` `= 4y^2 + 1`
`x^2` `= y^2 + 1/4`

 

`:. V` `= pi int_0^h y^2 + 1/4\ dy`
  `= pi[y^3/3 + y/4]_0^h`
  `= pi(h^3/3 + h/4-0)`
  `= pi(1/4((4h^3)/3 + h))`
  `= pi/4((4h^3)/3 + h)\ \ …\ text(as required)`

 

b.    `V_text(max)` `= pi/4 (4/3 xx (sqrt 3/2)^3 + sqrt 3/2)`
    `= pi/4 (sqrt 3/2 + sqrt 3/2)`
    `= (pi sqrt 3)/4`

 

`1/2 V_text(max)` `= (pi sqrt 3)/8`
`(pi sqrt 3)/8` `= pi/4 (4/3 h^3 + h)`
`sqrt 3/2` `= 4/3 h^3 + h`
`:. h` `~~0.59\ text(m)`

  

c.i.   `((dV)/(dt))_text(in)` `= 0.04`
  `((dV)/(dt))_text(out)` `= 0.05 sqrt h`
  `(dV)/(dt)` `= 0.04-0.05 sqrt h`
    `= (4-5 sqrt h)/100`
     
  `(dV)/(dh)` `= pi/4(4h^2 + 1)`
  `:. (dh)/(dt)` `= (dh)/(dV) ⋅ (dV)/(dt)`
    `= 4/(pi(4h^2 + 1)) xx (4-5 sqrt h)/100`
    `= (4-5 sqrt h)/(25 pi (4h^2 + 1))`

 

c.ii.   `(dh)/(dt)|_(h = 0.25)` `= (4-5 sqrt(0.25))/(25 pi (4(0.25)^2 + 1))`
    `~~ 0.0153\ text(ms)^(-1)`

 

d.   `(dt)/(dh) = (25 pi (4h^2 + 1))/(4-5 sqrt h)`

`:. t(0.25)` `= int_0^0.25 (25 pi (4h^2 + 1))/(4-5 sqrt h)\ dh`
  `~~9.8\ text(seconds)`

 

e.   `text(When)\ \ t=25,\ \ h=0.4\ \ text{(given)}`

♦♦ Mean mark part (e) 30%.

`:. h(30)` `~~ h(25) + 5 xx (dh)/(dt)|_(h = 0.4)`
  `~~ 0.4 + 5 xx ((4-5 sqrt 0.4)/(25 pi (4(0.4)^2 + 1)))`
  `~~ 0.43\ text(m)`

 

f.    `(dV)/(dt) = 0`

♦♦ Mean mark part (f) 32%.

`0.04-0.05 sqrt h` `= 0`
`0.04` `= 0.05 sqrt h“
`sqrt h` `= 4/5`
`h` `= 16/25`

  

`d` `= h_max-16/25`
  `= sqrt 3/2-16/25`
  `~~ 0.23\ text(m)`

Complex Numbers, SPEC2 2018 VCAA 2

  1. State the centre in the form  `(x, y)`, where  `x, y in R`, and the state the radius of the circle given by  `|z-(1 + 2i)| = 2`, where  `z in C`.   (1 mark)

  2. Graph the circle given by  `|z + 1| = sqrt 2 |z-i|`  on the Argand diagram below, labelling the intercepts with the vertical axis.   (2 marks)

     

The line given by  `|z-1| = |z-3|`  intersects the circle given by  `|z + 1| = sqrt 2 |z-i|`  in two places.

  1. Draw the line given by  `|z-1| = |z-3|`  on the Argand diagram in part c. Label the points of intersection with their coordinates.   (2 marks)

  2. Find the area of the minor segment enclosed by an arc of the circle given by  `|z + 1| = sqrt 2 |z-i|`  and part of the line given by  `|z-1| = |z-3|`.   (3 marks)
Show Answers Only
  1. `text(Centre): (1, 2), quad text(radius): 2`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
  5. `(4 pi)/3-sqrt 3`
Show Worked Solution

a.   `(x-1)^2 + (y-2)^2 = 2^2`

`:.\ text(Centre)\ (1, 2), quad text(radius) = 2`
 

b.    `y^2 + (x + 1)^2` `= (sqrt 2)^2 (x^2 + (y-1)^2)`
  `y^2 + x^2 + 2x + 1` `= 2(x^2 + y^2-2y + 1)`
  `y^2 + x^2 + 2x + 1` `= 2x^2 + 2y^2-4y + 2`
`0` `= 2x^2 + 2y^2-4y + 2-y^2-x^2-2x-1`
`0` `= x^2 + y^2-4y-2x + 1`
`0` `= (x^2-2x + 1) + (y^2-4y)`
`0` `= (x-1)^2 + (y^2-4y + 2^2)-4`
`0` `= (x-1)^2 + (y-2)^2-4`
`4` `= (x-1)^2 + (y-2)^2`

 
`:.\ text(Centre)\ (1, 2), quad text(radius) = 2`

 

c.   `ytext(-axis intercepts occur when)\ \ x=0:`

`(0-1)^2 + (y-2)^2` `= 4`
`1 + (y-2)^2` `= 4`
`(y-2)^2` `= 3`
`y-2` `= +- sqrt 3`
`y` `= 2 +- sqrt 3`

 

d.   `|z-1| = |z-3|`

`=>\ text(Graph is a line equidistant from:)\ \ 1 + 0i and 3 + 0i`

`text(Circle intercepts occur when)\ \ x=2:`

`(2- 1)^2 + (y-2)^2` `= 4`
`(y-2)^2` `= 3`
`y-2` `= +- sqrt 3`
`y` `= 2 +- sqrt 3`

 

e.   `sin theta = sqrt3/2\ \ =>\ \ theta = pi/3`

♦ Mean mark 39%.
MARKER’S COMMENT: Students who used standard formulae rather than definite integrals were more successful.

`text(Angle at centre of segment) = (2pi)/3`

`text(Height of triangle)\ (h) = 2^2-(sqrt3)^2 =1`
 

`:.\ text(Shaded Area)`

`=\ text(Area of segment – Area of triangle)`

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

`=(4pi)/3-sqrt3`

Calculus, SPEC2 2018 VCAA 1

Consider the function  `f: D -> R`, where  `f(x) = 2 text(arcsin)(x^2-1)`.

  1. Determine the maximal domain `D` and the range of  `f`.  (2 marks)

  2. Sketch the graph of  `y = f(x)`  on the axes below, labelling any endpoints and the `y`-intercept with their coordinates.  (3 marks)


      

    `qquad` 
     

  3. Find  `f^{′}(x)`  for  `x > 0`, expressing your answer in the form  `f^{′}(x) = A/sqrt(2-x^2), \ A in R`.  (1 mark)

  4. Write down  `f^{′}(x)`  for  `x < 0`, expressing your answer in the form  `f^{′}(x) = B/sqrt(2-x^2), \ B in R`.  (1 mark)

  5. The derivative  `f^{′}(x)`  can be expressed in the form  `f^{′}(x) = {g(x)}/sqrt(2-x^2)` over its maximal domain.
  1. Find the maximal domain of  `f^{′}`.  (1 mark)

  2. Find  `g(x)`, expressing your answer as a piecewise (hybrid) function.  (1 mark)

  3. Sketch the graph of  `g` on the axes below.  (2 marks)


     

Show Answers Only

  1. `D [- sqrt 2 , sqrt 2]; qquad f(x) in [-pi, pi]`
  2. `text(See Worked Solutions)`
  3. `4/sqrt(2-x^2)`
  4. `(-4)/sqrt(2-x^2)`
    1. `x in (-sqrt2, 0) ∪ (0, sqrt 2)`
    2. `g(x) = {(-4text(,), x < 0), (\ \ \ 4text(,), x > 0):}`
    3. `text(See Worked Solutions)`

Show Worked Solution

a.   `-1 <= x^2-1 <= 1`

`=>\ 0 <= x^2<= 1`

`=>\ -sqrt2 <= x <=sqrt2`

`:.\ text(Domain:)\ [- sqrt 2 , sqrt 2]`

 

`sin^(-1)(x) in [-pi/2, pi/2]`

`=> 2 sin^(-1)(x) in [-2xx pi/2, 2 xx pi/2]`

`:.\ text(Range:)\ [-pi, pi]`

 

b.   

 

c.   `f(x) = 2 sin^(-1)(x^2-1)`

`f(x)` `=2 sin^(-1)(x^2-1)`

 
`:. f^{′}(x)= 4/sqrt(2-x^2)\ \ \ (x > 0,\ \ text(by CAS))`

 

d.    `f^{′}(x)` `= (4x)/sqrt(x^2(2-x^2))`
    `= (-4)/sqrt(2-x^2)\ \ \ (x < 0)`

 

e.i.  `f^{′}(x)\ \ text(is defined when:)`

♦♦ Mean mark part (e)(i) 21%.

`2-x^2 > 0\ \ and\ \ x!=0`

`:. x in (-sqrt2, 0) ∪ (0, sqrt 2)`

 

e.ii.  `g(x) = +- 4`

♦ Mean mark part (e)(ii) 49%, part (e)(iii) 48%.

`:.  g(x) = {(-4text(,), x < 0), (\ \ \ 4text(,), x > 0):}`

 

e.iii.   

Mechanics, SPEC1-NHT 2017 VCAA 8

A 3 kg mass has velocity  `v\ text(ms)^(-1)`, where  `v = 2 arctan sqrt x`  when it has a displacement `x` metres from the origin,  `x > 0`.

Find the net force, `F` newtons, acting on the mass in terms of `x`.  (3 marks)

Show Answers Only

`F = (6 tan^(-1)(sqrt x))/(sqrt x(1 + x))`

Show Worked Solution
`v` `= 2 tan^(-1) (x^(1/2))`
`(dv)/(dx)` `= 1/2 xx x^(-1/2) xx 2/(1 + (sqrt x)^2)`
  `= 1/(sqrt x (1 + x))`

 

`a=v*(dv)/(dx)` `= 2 tan^(-1) (sqrt x) xx 1/(sqrt x (1 + x))`
  `= {2 tan^(-1) (sqrt x)}/{sqrt x (1 + x)}`

 

`:. F` `= ma`
  `= 3a`
  `= (6 tan^(-1)(sqrt x))/(sqrt x(1 + x))`

Calculus, SPEC1-NHT 2017 VCAA 7

Let  `(dy)/(dx) = (4-y)^2`.

  1.  Express `y` in terms of `x`, where  `y(0) = 3`.  (3 marks)

  2.  Express  `(d^2y)/(dx^2)`  in terms of `y`.  (2 marks)

Show Answers Only
  1. `y = 4-1/(x + 1)`
  2. `-2(4-y)^3`
Show Worked Solution
a.    `(dy)/(dx)` `=(4-y)^2`
  `(dx)/(dy)` `= 1/(4-y)^2`
  `x` `= int 1/(4-y)^2\ dy`
    `= int (4-y)^(-2) dy`
    `= (-1)(-1)(4-y)^(-1)+ c`
    `= 1/(4-y) + c`

 
`text(When)\ \ x=0,\ \ y=3:`

`0` `= 1/(4-3) + c`
`:.c` `= -1`

 

`x` `= 1/(4-y)-1`
`x + 1` `= 1/(4-y)`
`1/(x + 1)` `= 4-y`
`:. y` `= 4-1/(x + 1)`

 

b.   `d/(dx) ((4-y)^2)`

`= 2(-1)(4-y) ⋅ (dy)/(dx)`

`= -2(4-y)(4-y)^2`

`= -2(4-y)^3`

Calculus, SPEC1-NHT 2017 VCAA 5

Part of the graph of  `y = (sqrt(x + 1))/(root4(1 - x^2))`  is shown below.
 


 

Find the volume generated if the region bounded by the graph of  `y = (sqrt(x + 1))/(root4(1 - x^2))`, the lines  `x = -1/2`  and  `x = 1/2`, and the `x`-axis is rotated about the `x`-axis.  (4 marks)

Show Answers Only

`pi^2/3`

Show Worked Solution
`V` `= pi int_(-1/2)^(1/2) y^2\ dx`
  `= pi int_(-1/2)^(1/2) (x + 1)/sqrt(1 – x^2)\ dx`
  `= pi int_(-1/2)^(1/2) x/sqrt(1 – x^2)\ dx + int_(-1/2)^(1/2) 1/sqrt(1 – x^2)\ dx`

 
`text(Let)\ \ u = 1 – x^2`

`(du)/(dx) = -2x\ \ =>\ \ -1/2\ du = x\ dx`

`text(When)\ \ x=1/2\ \ =>\ \ u=3/4`

`text(When)\ \ x=- 1/2\ \ =>\ \ u=3/4`

`text(Same limit)\ =>\ text(1st integral = 0)`
 

`:. V` `= 0+pi int_(-1/2)^(1/2) 1/sqrt(1 – x^2)\ dx`
  `= pi [sin^(-1) (x)]_(-1/2)^(1/2)`
  `= pi (pi/6 – ((-pi)/6))`
  `= pi (pi/3)`
  `= pi^2/3`

Complex Numbers, SPEC1-NHT 2017 VCAA 4

Find the values of `a` and `b` given that  `z - 1 - i`  is a factor of  `z^3 + (a + b)z^2 + (b^2 - a)z - 4 = 0`, where `a` and `b` are real constants.  (4 marks)

Show Answers Only

`(a, b) = (-5, 1) or (-2, -2)`

Show Worked Solution

`text(All coefficients are real):`

`=> z – 1 + i \ \  text{is a factor (conjugate pair)}`

`(z – 1 – i) (z – 1 + i)` `= ((z – 1) – i)((z – 1) + i)`
  `= (z – 1)^2 – i^2`
  `= z^2 – 2z + 1 + 1`
  `= z^2 – 2z + 2`

 
`(z – alpha)(z^2 – 2z + 2)`

`=z^3 -2z^2 + 2z -alpha z^2 +2 alpha z – 2 alpha, \ \ \ alpha in RR`

`=z^3 + (-alpha – 2)z^2 + (2 alpha + 2)z – 2 alpha`

`= z^3 + (a + b)z^2 + (b^2 – a) z – 4`
 

`text(Equating coefficients:)`

`2 alpha = -4 \ \ =>\ \ alpha = 2`

`a + b = -4 \ \ …\ (1)`

`b^2 – a = 6 \ \ …\ (2)`

 
`text{Substitute}\ \ a=-b-4\ \ text{from (1) into (2):}`

`b^2 – (-4 – b)` `= 6`
`b^2 + b + 4` `= 6`
`b^2 + b – 2` `=0`
`(b + 2)(b – 1)` `=0`

 
`b = 1\ \ => \ \ a=-5, \ text(or)`

`b=-2\ \ => \ \ a = -4 + 2=-2`

`:. (a, b) = (-5, 1) or (-2, -2)`

Calculus, SPEC1-NHT 2017 VCAA 3

Find the gradient of the curve with equation  `x = sin (y/15)`  when  `x = 1/4`. Give your answer in the form  `a sqrt b`, where  `a, b \ in Z^+`.  (3 marks)

Show Answers Only

 `m = 4 sqrt 15`

Show Worked Solution
`x` `= sin (y/15)`
`sin^(-1)x` `= y/15`
`y` `= 15 sin^(-1)x`
`dy/dx` `= 15/(sqrt(1-x^2))`

 

`text(When)\ \ x=1/4:`

`dy/dx` `= 15/(sqrt(1-(1/4)^2))`
  `= 15/(sqrt(15/16))`
  `= 60/sqrt15`
  `=4sqrt15`

 
`:. m = 4 sqrt 15`

Calculus, SPEC1 2017 VCAA 10

  1.  Show that  `d/dx(x arccos(x/a)) = arccos(x/a)−x/(sqrt(a^2-x^2))`, where  `a > 0`.   (1 mark)

  2.  State the maximum domain and the range of  `f(x) = sqrt(arccos(x/2))`.   (2 marks)

  3.  Find the volume of the solid of revolution generated when the region bounded by the graph of  `y = f(x)`, and the lines  `x = −2`  and  `y = 0`, is rotated about the `x`-axis.   (4 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `x ∈ [−2, 2], \ y ∈ [0, sqrtpi]`
  3. `2pi^2`
Show Worked Solution
a.    `u` `= x,` `v` `= cos^(−1)(x/a)`
  `uprime` `= 1,`   `vprime` `= (−1)/sqrt(a^2-x^2)`
`d/(dx)(xcos^(−1)(x/a))` `= uprimev + vprimeu`
  `= cos^(−1)(x/a) + (x(−1))/sqrt(a^2-x^2)`
  `= arccos(x/a)-x/sqrt(a^2-x^2)`

 

b.   `arccos(x/2)>=0`

`text(Maximal domain:)\  x ∈ [−2, 2]`

`f(x) = (arccos(x/2))^(1/2)`

`text(Range:)\ \ y ∈ [0, sqrtpi]`

 

c.   `V` `= pi int_(−2)^2 y^2\ dx`
  `= pi int_(−2)^2 cos^(-1)(x/2)\ dx`
  `= pi int_(−2)^2 cos^(-1) (x/2)-x/sqrt(4-x^2) + x/sqrt(4-x^2)\ dx`
  `= pi [x cos^(-1)(x/2)]_(−2)^2 + pi int_(−2)^2 x/sqrt(4-x^2)\ dx`

 
`text(Let)\ \ u = 4-x^2`

♦ Mean mark part (c) 35%.

`(du)/(dx) = -2x\ \ =>\ \ -1/2 (du) = x\ dx`
 

`text(When)\ \ x=2\ \ => \ u=0`

`text(When)\ \ x=-2\ \ =>\ \ u=0`

`:. V` `= pi [2cos^(−1)(1)-(-2)cos^(−1)(−1)]-pi/2 int_0^0 1/sqrtu du`
  `= pi(2 xx 0 + 2 xx pi)`
  `= 2pi^2`

Mechanics, SPEC1-NHT 2017 VCAA 1

A 5 kg mass on a smooth plane inclined at 30° is held in equilibrium by a horizontal force of magnitude `P` newtons, as shown in the diagram below.
 


 

  1. On the diagram above, show all other forces acting on the mass and label them. (1 mark)
  2.  Find  `P`.  (2 marks)
Show Answers Only
  1.  `text(See Worked Solutions)`
  2. `P = (5g sqrt 3)/3`
Show Worked Solution
a.   

 

b.    `P cos 30^@` `= 5g sin 30^@`
  `(P sqrt 3)/2` `= (5g)/2`
  `P sqrt 3` `= 5g`
  `P` `= (5g)/sqrt 3`
  `P` `= (5g sqrt 3)/3`

Complex Numbers, SPEC2 2014 VCAA 9 MC

The circle  `| z - 3 - 2i | = 2`  is intersected exactly twice by the line given by

A.   `| z - i | = | z + 1 |`

B.   `| z - 3 - 2i | = | z - 5 |`

C.   `| z - 3 - 2i | = | z - 10i |`

D.   `text(Im)(z) = 0`

E.   `text(Re)(z) = 5`

Show Answers Only

`B`

Show Worked Solution

`text{Circle has centre (3,2) and radius 2.}`

`text{Consider each option (using CAS):}`

`text(Options A and C do not intersect.)`

`text(Options D and E are tangents.)`

`text(Only option B intersects twice).`

`=> B`

Complex Numbers, SPEC2 2015 VCAA 7 MC

If  `z = sqrt3 + 3i`, then  `z^63`  is

  1. real and negative
  2.  equal to a negative real multiple of `i`
  3. real and positive
  4. equal to a positive real multiple of `i`
  5. a positive real multiple of  `1 + isqrt3`
Show Answers Only

`A`

Show Worked Solution
`z` `= sqrt3 + 3i`
  `= 2sqrt3 text(cis)\ (pi/3)`

 

`:. z^63` `= (2sqrt3)^63 text(cis)\ ((63pi)/12)`
  `= (2sqrt3)^63 text(cis)\ (21pi)`
  `= (2sqrt3)^63 text(cis)\ (pi)`
  `= -(2sqrt3)^63`

 
`=> A`

Graphs, SPEC2 2015 VCAA 2 MC

The range of the function with rule  `f(x) = (2 - x)arcsin(x/2 - 1)`  is

A.   `[-pi,0]`

B.   `[-pi/2,pi/2]`

C.   `[-((2 - x)pi)/2,((2 - x)pi)/2]`

D.   `[0,4]`

E.   `[0,pi]`

Show Answers Only

`A`

Show Worked Solution

`text(By CAS, create a graph of)\ \ f(x) = (2 – x)sin^(-1)(x/2 – 1).`

`=> A`

Calculus, SPEC1 2015 VCAA 9

Consider the curve represented by  `x^2 - xy + 3/2 y^2 = 9.`

  1. Find the gradient of the curve at any point  `(x, y).`  (2 marks)
  2. Find the equation of the tangent to the curve at the point  `(3, 0)`  and find the equation of the tangent to the curve at the point `(0, sqrt 6).`

     

    Write each equation in the form  `y = ax + b.`  (2 marks)

  3. Find the acute angle between the tangent to the curve at the point  `(3, 0)`  and the tangent to the curve at the point  `(0, sqrt 6).`

     

    Give your answer in the form  `k pi`, where `k` is a real constant  (2 marks)

Show Answers Only
  1. `(dy)/(dx) = (2x – y)/(x – 3y)`
  2. `y = 2(x – 3);\ \ \ y = 1/3 x + sqrt 6`
  3. `theta = pi/4`
Show Worked Solution
a.    `d/(dx)(x^2) – d/(dx)(xy) + 3/2* d/(dx) (y^2)` `= 0`
  `2x – x*(dy)/(dx) – y + 3/2(2y)*(dy)/(dx)` `= 0`
  `(dy)/(dx)(−x + 3y)` `= y – 2x`
  `:. (dy)/(dx)` `= (y – 2x)/(3y – x)`

 

b.   `m_{(3,0)} = (0 – 6)/(0 – 3) = 2`

`:.\ text{Equation of tangent at (3, 0):}`

`y = 2(x – 3)`

  `= 2x – 6`

 

`m_{(0,sqrt6)} = (sqrt6 – 0)/(3sqrt6 – 0) = 1/3`

`:.\ text{Equation of tangent at}\ (0,sqrt6):`

`y -sqrt6` `= 1/3(x – 0)`  
`y` `=1/3 x + sqrt6`  

 

c.   `m_1 = 2 = tan(theta_1), \ \ m_2 = 1/3 = tan(theta_2)`

`alpha` `= theta_1 – theta_2`
  `= tan^(−1)(2) – tan^(−1)(1/3)`
`tan(alpha)` `= tan(tan^(−1)(2) – tan^(−1)(1/3))`
  `= (tan(tan^(−1)(2)) – tan(tan^(−1)(1/3)))/(1 + tan(tan^(−1)(2)tan(tan^(−1)(1/3))))`
  `= (2 – 1/3)/(1 + 2/3)`
  `= 1`

 
`:. alpha = pi/4\ \ \ (alpha ∈ (0, pi/2))`

Calculus, SPEC1 2015 VCAA 8

  1.  Show that  `int tan (2x)\ dx = 1/2 log_e |\ sec (2x)\ | + c.`  (2 marks)


  2.  The graph of  `f(x) = 1/2 arctan (x)`  is shown below
     
         
     
  3. i.  Write down the equations of the asymptotes.  (1 mark)

  4. ii. On the axes above, sketch the graph of  `f^-1`, labelling any asymptotes with their equations.  (1 mark)

  5.  Find  `f(sqrt 3).`  (1 mark)

  6.  Find the area enclosed by the graph of  `f`, the `x`-axis and the line  `x = sqrt 3.`   (2 marks)

Show Answers Only

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

b. (i)   `y = -pi/4,\ \ y = pi/4`
  (ii)  

c.   `pi/6`

d.   `(sqrt 3 pi)/6-log_e sqrt 2`

Show Worked Solution

a.   `text(Let)\ \ u = cos(2x)`

`(du)/(dx) = -2 sin (2x)\ \ =>\ \ -1/2 xx du = sin (2x)\ dx`

`int tan (2x)\ dx` `= int (sin(2x))/(cos(2x))\ dx`
  `= -1/2 int 1/u\ du`
  `= -1/2 log_e |\ u\ | + c`
  `= -1/2 log_e |\ cos (2x)\ | + c`
  `= 1/2 log_e |\ sec (2x)\ | + c`

 

b.i.  `text(Range:)\ \ tan^(-1)(x) in (- pi/2,pi/2)`

`=>1/2 tan^(-1)(x) in (- pi/4, pi/4)`

`:.\ text(Asymptotes:)\ \ y = -pi/4,\ \ y = pi/4`

 

b.ii.

 

c.   `f(sqrt 3)` `= 1/2 tan^(-1) (sqrt 3)`
    `= 1/2 xx pi/3`
    `= pi/6`

 

d.    `y` `= 1/2tan^(−1)(x)`
  `2y` `= tan^(−1)(x)`
  `x` `= tan(2y)`

 

 

♦ Mean mark part (d) 49%.

`text(Area)` `=\ text(Area of rectangle – Area between graph and y-axis)`
  `= sqrt3 xx pi/6-int_0^(pi/6) tan(2y)\ dy`
  `= (sqrt3 pi)/6 -1/2[ln\ | sec(2y) |]_0^(pi/6)`
  `= (sqrt3 pi)/6-1/2[ ln(sec(pi/3))-ln(sec 0)]`
  `= (sqrt3 pi)/6-1/2 (ln2 -ln1)`
  `= (sqrt3 pi)/6-1/2 ln2`

Trigonometry, SPEC1 2015 VCAA 7

  1. Solve \(\sin (2 x)=\sin (x), x \in[0,2 \pi]\).  (3 marks)

  2. Find \(\left\{x: \operatorname{cosec}(2 x)<\operatorname{cosec}(x), x \in\left(0, \dfrac{\pi}{2}\right) \cup\left(\dfrac{\pi}{2}, \pi\right)\right\}\)   (2 marks)

Show Answers Only
  1. \(x=0, \pi, 2 \pi\) or \(x=\dfrac{\pi}{3}, \dfrac{5 \pi}{3}\)
  2. \(x \in\left(0, \dfrac{\pi}{3}\right) \cup\left(\dfrac{\pi}{2}, \pi\right)\)
Show Worked Solution

a.  \(2 \sin (x) \cos (x)=\sin (x)\)

MARKER’S COMMENT: Many students expanded sin(2x) and then cancelled sin(x) on both sides which lost a set of solutions!

\(\sin (x)(2 \cos (x)-1)=0\)

\(\sin (x)=0, \cos (x)=\dfrac{1}{2}\)

\(x=0, \dfrac{\pi}{3}, \pi, 2 \pi-\dfrac{\pi}{3}, 2 \pi\)

\(x=0, \dfrac{\pi}{3}, \pi, \dfrac{5 \pi}{3}, 2 \pi\)
 

b.  \(\text {Solving} \ \operatorname{cosec}(2 x)=\operatorname{cosec}(x), x \in(0, \pi)\left\{\dfrac{\pi}{2}\right\}:\)

♦♦ Mean mark 26%.

\(\text{Using part (a): }\)

\(\sin (2 x)=\sin (x)\) when \(x=\dfrac{\pi}{3}\)

\(\therefore \operatorname{cosec}(x)=\operatorname{cosec}(2 x)\) at \(x=\dfrac{\pi}{3}\)

\(\text {Sketch graphs:}\)
 

 

\(\text {When} \ \operatorname{cosec}(2 x)<\operatorname{cosec}(x):\)

\(x \in\left(0, \dfrac{\pi}{3}\right) \cup\left(\dfrac{\pi}{2}, \pi\right)\)

Calculus, SPEC1 2015 VCAA 5

Find the volume generated when the region bounded by the graph of  `y = 2x^2 - 3`, the line  `y = 5`  and the `y`-axis is rotated about the `y`-axis.  (3 marks)

Show Answers Only

`16 pi`

Show Worked Solution

`V = pi int_(−3)^5 x^2\ dy`

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

 

`:. V` `= pi/2 int_(−3)^5 (y + 3)\ dy`
  `= pi/2[(y^2)/2 + 3y]_(−3)^5`
  `= pi/2(25/2 + 15 – (9/2 – 9))`
  `= pi/2(16/2 + 24)`
  `= 16pi`

Complex Numbers, SPEC1 2015 VCAA 4

  1. Find all solutions of  `z^3 = 8i,  \ z in C` in cartesian form.   (3 marks)

  2. Find all solutions of  `(z − 2i)^3 = 8i, \ z in C`  in cartesian form.   (1 mark)

Show Answers Only
  1. `-2i, +- sqrt 3 + i`
  2. `z = 0 or z = +- sqrt 3 + 3i`
Show Worked Solution

a.   `z^3 = 8text(cis)(pi/2)`

`z_1` `= 8^(1/3)text(cis)(pi/6)`
  `= 2text(cis)(pi/6)`
  `= sqrt3 + i`
   
`z_2` `= 2text(cis)(pi/6 + (2pi)/3)`
  `= 2text(cis)((5pi)/6)`
  `= −sqrt3 + i`
   
`z_3` `= 2text(cis)(pi/6-(2pi)/3)`
  `= 2text(cis)(−pi/2)`
  `= −2i`

 
`:. z = sqrt3 + i, \ −sqrt3 + i, \ −2i`
 

♦ Mean mark part (b) 44%.

b.    `z_1-2i` `= sqrt3 + i`
  `z_1` `= sqrt3 + 3i`
     
`z_2-2i` `= −sqrt3 + i`
`z_2` `= −sqrt3 + 3i`
   
`z_3-2i` `= −2i`
`z_3` `= 0`

 
`:. z = sqrt3 + 3i, \ −sqrt3 + 3i, \ 0`

Mechanics, SPEC1 2015 VCAA 2

A 20 kg parcel sits on the floor of a lift.

  1. The lift is accelerating upwards at 1.2 ms¯².

     

    Find the reaction force of the lift floor on the parcel in newtons.  (2 marks)

  2. Find the acceleration of the lift downwards in ms¯² so that the reaction of the lift floor on the parcel is 166 N.  (2 marks)
Show Answers Only
  1. `220`
  2. `1.5\ text(ms)^-2`
Show Worked Solution
a.    

`text(Acceleration:)\ \ ↑ 1.2\ text(ms)^(−2)`

`∑F` `= 20 xx 1.2`
  `= R – 20text(g)`

 

`R – 20 xx 9.8` `=20 xx 1.2`
`:. R` `= 20(1.2 + 9.8)`
  `= 20 xx 11`
  `= 220\ text(N  upwards)`

 

b.   `↓a\ text(ms)^(−2)`

`∑F` `= 20a`
  `= 20text(g) – R_2`

 

`20a` `= 20text(g) – 166`
  `= 20text(g) – 2 xx 83`
  `= 20text(g) – 20 xx 8.3`
`:. a` `= text(g) – 8.3`
  `= 9.8 – 8.3`
  `= 1.5\ text(ms)^(−2)`

Mechanics, SPEC1 2016 VCAA 1

A taut rope of length `1 2/3` m suspends a mass of 20 kg from a fixed point `O`. A horizontal force of `P` newtons displaces the mass by 1 m horizontally so that the taut rope is then at an angle of  `theta`  to the vertical.

  1. Show all the forces acting on the mass on the diagram below.  (1 mark)

 

 

  1. Show that  `sin (theta) = 3/5`.  (1 mark)
  2. Find the magnitude of the tension force in the rope in newtons.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `T = 245\ text(N)`
Show Worked Solution
a.   

 

b.    `sin(theta)` `= 1/(1 2/3)`
    `= 1/((5/3)`
    `= 3/5`

 

c.   

`cos theta = (20g)/T`

`sin theta = 3/5\ \ text{(using part b)}`

`underbrace{(400g^2)/T^2 + 9/25}_(cos^2 theta + sin^2 theta = 1)` `= 1`
`(400g^2)/T^2` `= 16/25`
`(20g)/T` `=4/5`
`T/(20g)` `= 5/4`
`T` `= 25g`
  `=245\ text(N)`

Complex Numbers, SPEC2 2014 VCAA 8 MC

The principal argument of  `(−3sqrt2 - isqrt6)/(2 + 2i)`  is

A.   `(−13pi)/12`

B.   `(7pi)/12`

C.   `(11pi)/12`

D.   `(13pi)/12`

E.   `(−11pi)/12`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ z_1=−3sqrt2 – isqrt6`

`=>\ text(Arg)\ z_1 = pi + tan^(-1) (sqrt6/(3 sqrt2)) = (7pi)/6`

`text(Let)\ \ z_2 = 2 + 2i`

`=>\ text(Arg)\ z_2 = pi/4`
 

`:.\ text(Arg)\ (z_1/z_2)` `= text(Arg)\ ((−3sqrt2 – isqrt6)/(2 + 2i))`
  `= (7pi)/6 – pi/4`
  `= (11pi)/12`

 
`=> C`

Complex Numbers, SPEC2 2014 VCAA 7 MC

The sum of the roots of  `z^3 - 5z^2 + 11z - 7 = 0`,  where  `z ∈ C`,  is

  1. `1 + 2sqrt3i`
  2. `5i`
  3. `4 - 2sqrt3i`
  4. `2sqrt3i`
  5. `5`
Show Answers Only

`E`

Show Worked Solution

`text(Coefficients are real)`

`=>\ text(Conjugate root theorem applies.)`

`z_1=alpha + beta i,\ \ z_2=alpha – beta i, \ \ z_3=gamma`

`z_1+z_2+z_3` `= alpha + beta i + alpha – beta i + gamma`
  `= 2 alpha + gamma\ \ \ (∈R)`

 

`=> E`

Graphs, SPEC2 2014 VCAA 3 MC

The features of the graph of the function with rule  `f(x) = (x^2 - 4x + 3)/(x^2 - x - 6)` include

A.   asymptotes at  `x = 1`  and  `x = −2`

B.   asymptotes at  `x = 3`  and  `x = −2`

C.   an asymptote at  `x = 1`  and a point of discontinuity at  `x = 3`

D.   an asymptote at  `x = −2`  and a point of discontinuity at  `x = 3`

E.   an asymptote at  `x = 3`  and a point of discontinuity at  `x = −2`

Show Answers Only

`D`

Show Worked Solution
`f(x)` `= (x^2 – 4x + 3)/(x^2 – x – 6)`
  `=(x^2 -x-6 -3x+9)/(x^2 – x – 6)`
  `= 1 – (3(x – 3))/((x + 2)(x – 3))`
  `=1 – 3/(x + 2)\ text(where)\ x != 3`

 
`=> D`

Mechanics, SPEC1 2014 VCAA 8

A body of mass 5 kg is held in equilibrium by two light inextensible strings. One string is attached to a ceiling at `A` and the other to a wall at `B`. The string attached to the ceiling is at an angle `theta` to the vertical and has tension `T_1` newtons, and the other string is horizontal and has tension `T_2` newtons. Both strings are made of the same material.

  1. i. Resolve the forces on the body vertically and horizontally, and express `T_1` in terms of  `theta`.  (2 marks)
  2. ii. Express `T_2` in terms of  `theta`.  (1 mark)
  3.  Show that  `tan (theta) < sec (theta)`  for  `0 < theta < pi/2`.  (1 mark)
  4.  The type of string used will break if it is subjected to a tension of more than 98 N.

     

     Find the maximum allowable value of  `theta`  so that neither string will break.  (3 marks)

Show Answers Only
  1.  i. `T_1 = 5g sec theta`
  2. ii. `T_2 = 5g tan theta`
  3.  `text(Proof)\ text{(See Worked Solutions)}`
  4.  `theta_max = pi/3`
Show Worked Solution
a.i.   

`text(Resolve vertically): \ 5g = T_1 cos theta`

`text(Resolve horizontally): \ T_1 sin theta = T_2`

`:. T_1 = 5g sec theta`

 

a.ii.    `T_2` `= 5g sec theta sin theta`
    `= 5g tan theta`

 

b.    `cos theta in (0, 1), \ theta in (0, pi/2)`
  `sin theta in (0, 1), \ theta in (0, pi/2)`

`:. sin theta < 1, \ theta in (0, pi/2)`

♦ Mean mark part (b) 40%.

 

`text(If)\ sin theta < 1 and cos theta > 0:`

`(sin theta)/(cos theta) < 1/(cos theta)`

`:.  tan theta < sec theta, quad 0 < theta < pi/2`

 

c.   `tan theta < sec theta \ \ =>\ \  T_2\ text(will be smaller) => \ T_1\ text(will break first)`

♦ Mean mark part (c) 46%.

`5text(g)\ sec theta_max` `= 98`
`sec theta_max` `= 98/(5 xx 9.8)`
`sec theta_max` `= 10/5`
`sec theta_max` `= 2`
`:.  theta_max` `= pi/3`

Calculus, SPEC1 2014 VCAA 7

Consider  `f(x) = 3x arctan (2x)`.

  1.  Write down the range of  `f`.  (1 mark)

  2.  Show that  `f prime(x) = 3 arctan (2x) + (6x)/(1 + 4x^2)`.  (1 mark)

  3.  Hence evaluate the area enclosed by the graph of  `g(x) = arctan (2x)`, the `x`-axis and the lines  `x = 1/2`  and  `x = sqrt 3/2`.  (3 marks)

Show Answers Only

  1. `f(x) in [0, oo)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `(pi sqrt 3)/6-pi/8-1/4 ln2`

Show Worked Solution

a.  `f(x) = 3x arctan (2x)`

`f(0) = 0`

♦♦♦ Mean mark 4%.

`text(When)\ \ x < 0, \ \ 3x<0,\ \ text(and)\ \ y = tan^(-1) (2x) < 0`

`text(When)\ \ x>0,\ \ 3x>0,\ \ text(and)\ \ y = tan^(-1) (2x) > 0`
 


 
 `:. f(x) in [0, oo)`

 

b.  `u = 3x, qquad v = tan^(-1)(2x)`

`u prime = 3, qquad v prime = 2/(1 + 4x^2)`

`:. f prime (x)` `= 3(arctan (2x)) + (2/(1 + 4x^2))(3x)`
  `= 3 arctan (2x) + (6x)/(1 + 4x^2)`

 

c.    `A` `= int_(1/2)^(sqrt 3/2) arctan (2x)\ dx`
    `= 1/3 int_(1/2)^(sqrt 3/2) 3arctan (2x)\ dx`
    `=1/3 int_(1/2)^(sqrt 3/2) (3 arctan (2x) + (6x)/(1 + 4x^2)-(6x)/(1 + 4x^2))\ dx`
    `= 1/3 int_(1/2)^(sqrt 3/2) 3 arctan (2x) + (6x)/(1 + 4x^2) dx-int_(1/2)^(sqrt 3/2) (2x)/(1 + 4x^2)\ dx`
    `= 1/3 [3x arctan (2x)]_(1/2)^(sqrt 3/2)-1/4 int_(1/2)^(sqrt 3/2) (8x)/(1 + 4x^2)\ dx`
    `= [sqrt 3/2 arctan (sqrt 3)-1/2 arctan (1)]-1/4[ln (1 + 4x^2)]_(1/2)^(sqrt 3/2)`
    `= sqrt 3/2 (pi/3)-1/2 (pi/4)-1/4 [ln (1 + 4(3/4))-ln(1 + 4 (1/4)]`
    `= (pi sqrt 3)/6-pi/8-1/4 (ln4-ln 2)`
    `= (pi sqrt 3)/6-pi/8-1/4 ln2`

Calculus, SPEC1 2014 VCAA 6

  1.  Verify that  `a/(a-4) = 1 + 4/(a-4)`.   (1 mark)

Part of the graph of  `y = x/sqrt(x^2-4)`  is shown below.
 


  

  1.  The region enclosed by the graph of  `y = x/sqrt(x^2-4)`  and the lines  `y = 0, \ x = 3`  and  `x = 4`  is rotated about the `x`-axis

     

     Find the volume of the resulting solid of revolution.   (4 marks)


Show Answers Only

  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `pi(1 + l n (5/3))`

Show Worked Solution

a.    `a/(a-4)` `= (a-4 + 4)/(a-4)`
    `= (a-4)/(a-4) + 4/(a-4)`
    `= 1 + 4/(a-4), quad a != 4`

 

b.  `r = y`

`V` `= pi int_3^4 y^2\ dx`
  `= pi int_3^4 underbrace{x^2/(x^2-4)}_text(using part a.)\ dx`
  `= pi int_3^4 1 + 4/(x^2-4) dx`
  `= pi int_3^4 1 + 4/((x-2)(x + 2)) dx`

 
`text(Using partial fractions:)`

`4/((x-2)(x + 2)) = A/(x-2) + B/(x + 2)`

`A(x + 2) + B(x-2) = 4`
 

`text(When)\ \ x = 2: \ 4A = 4\ \ =>\ \ A = 1`

`text(When)\ \ x = -2: -4B = 4\ \ =>\ \ B = -1`

`:. V` `= pi int_3^4 1 + 1/(x-2)-1/(x + 2)\ dx`
  `= pi [x + ln |x-2|-ln |x + 2|]_3^4`
  `= pi [4 + ln2-ln6-(3 + ln 1-ln5)]`
  `= pi (1 + ln2-ln6 + ln5)`
  `= pi (1 + ln ((2 xx 5)/6))`
  `= pi (1 + ln (5/3))`

Calculus, SPEC1 2014 VCAA 5

  1. For the function with rule  `f(x) = 96 cos (3x) sin (3x)`, Find the value of `a` such that  `f(x) = a sin (6x)`.   (1 mark)

  2. Use an appropriate substitution in the form  `u = g(x)`  to find an equivalent definite integral for

     

         `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`  in terms of `u` only.   (3 marks)

  3. Hence evaluate  `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`, giving your answer in the form  `sqrt k, \ k in Z`.   (1 mark)

Show Answers Only
  1. `48`
  2. `int_(sqrt 3/2)^0-8u^2 du`
  3. `sqrt 3`
Show Worked Solution

a.   `96 cos (3x) sin(3x) = 48 (2 cos(3x) sin(3x))= 48 sin (6x)`

`:. a = 48`
 

b. `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`

`= int_(pi/36)^(pi/12) 48 sin (6x) cos^2 (6x)\ dx`

`u` `= cos (6x)`
`(du)/(dx)` `= -6 sin (6x)\ \ =>\ \ du = -6 sin(6x)\ dx`
   
`u(pi/12)` `= cos (pi/2) = 0`
`u(pi/36)` `= cos (pi/6) = sqrt 3/2`

 
`:. int_(pi/36)^(pi/12) 48 sin (6x) cos^2 (6x)\ dx`

`= int_(sqrt 3/2)^0-8u^2\ du`

c. `:. int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`

`= -8 [u^3/3]_(sqrt 3/2)^0`
`= -8 (0-(sqrt 3/2)^3/3)`
`= 8/3 xx (3sqrt3)/8`
`= sqrt 3`

Calculus, SPEC1 2014 VCAA 4

Find the gradient of the line perpendicular to the tangent to the curve defined by  `y = -3e^(3x) e^y`  at the point  `(1, -3)`.  (3 marks)

Show Answers Only

`4/9`

Show Worked Solution
`(dy)/(dx)` `= -3 d/(dx) (e^(3x) e^y)`
`(dy)/(dx)` `= -9e^(3x) e^y + -3 e^(3x) e^y (dy)/(dx)`
` -9e^(3x) e^y` `=(dy)/(dx) (1 + 3e^(3x) e^y)`
`(dy)/(dx)` `= (-9e^(3x) e^y)/(1 + 3e^(3x) e^(-3))`

 

`text{At  (1, –3):}`  
`:. m_text(norm)` `= (1 + 3e^(3 xx 1) e^(-3))/(9e^(3 xx 1) e^(-3))`
  `= (1 + 3e^3 e^(-3))/(9e^3 e^(-3))`
  `= 4/9`

Complex Numbers, SPEC1 2014 VCAA 3

Let  `f` be a function of a complex variable, defined by the rule  `f(z) = z^4-4z^3 + 7z^2-4z + 6`.

  1.  Given that  `z = i`  is a solution of  `f(z) = 0`, write down a quadratic factor of  `f(z)`.   (2 marks)

  2.  Given that the other quadratic factor of  `f(z)`  has the form  `z^2 + bz + c`, find all solutions of  `z^4-4z^3 + 7z^2-4z + 6 = 0`  in a cartesian form.   (3 marks)

Show Answers Only
  1. `z^2 + 1`
  2. ` z = i, quad z = -i, quad z = 2 + i sqrt 2, quad z = 2-i sqrt 2`
Show Worked Solution

a.   `text(Real coefficients) \ => \ z = -i\ \ text(is also a solution)`

`:. (z-i)(z + i)= z^2 + 1\ \ text(is a quadratic factor)`

 

b.    `(z^2 + 1)(z^2 + bz + c)` `= z^4-4z^3 + 7z^2-4z + 6`
  `z^4 + bz^3 + (c + 1)z^2 + bz + c` `= z^4-4z^3 + 7z^2-4z + 6`

 
`text(Equating co-efficients:)`

`bz^3 = -4z^3\ \ =>\ \ b=-4`

`(c + 1)z^2 = 7z^2\ \ =>\ \ c = 6`

`(z^2 + 1)(z^2 + 4z + 6)` `= 0`
`(z^2 + 1)(z^2-4z + 2^2-4 + 6)` `= 0`
`(z^2 + 1)((z-2)^2 + 2)` `= 0`
`(z^2 + 1)((z-2)^2-2i^2)` `= 0`
`(z^2 + 1)(z-2-i sqrt 2)(z-2 + i sqrt 2)` `= 0`

 
`:.z = i, quad z = -i, quad z = 2 + i sqrt 2, quad z = 2-i sqrt 2`

Vectors, SPEC1 2014 VCAA 2

The position vector of a particle at time  `t >= 0`  is given by

`underset ~r (t) = (t-2) underset ~ i + (t^2-4t + 1) underset ~j`

  1.  Show that the cartesian equation of the path followed by the particle is  `y = x^2-3`.  (1 mark)

  2.  Sketch the path followed by the particle on the axes below, labelling all important features.  (2 marks)


 

 

  1.  Find the speed of the particle when  `t = 1`.  (2 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(See Worked Solutions)`
  3. `sqrt 5`

Show Worked Solution

a.    `x` `= t-2\ \ =>\ \ t=x+2`
`y` `= t^2-4t + 1`
  `= (x + 2)^2-4 (x + 2) + 1`
`y` `= x^2 + 4x + 4-4x-8 + 1`
  `= x^2-3`

 

b.   `t >= 0`

`x = t-2\ \ =>\ \ x >= -2`

`y(-2)` `= (-2)^2-3`
  `= 1`
`y(0)` `= -3`

 

`0` `= x^2-3`
`x^2` `= 3`
`x` `= +- sqrt 3`

 

c.    `underset ~ dot r (t)` `= underset ~ dot i + (2t-4) underset ~j`
  `underset ~ dot r (1)` `= underset ~ dot i + (2-4) underset ~j`
    `= underset ~ dot i-2 underset ~j`
  `|underset ~ dot r(1)|` `= sqrt (1^2 + (-2)^2)`
    `= sqrt 5`

Vectors, SPEC1 2014 VCAA 1

Consider the vector  `underset ~a = sqrt 3 underset ~i - underset ~j - sqrt 2 underset ~k`, where  `underset ~i, underset ~j`  and  `underset ~k`  are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

  1. Find the unit vector in the direction of  `underset ~a`.  (1 mark)

  2. Find the acute angle that  `underset ~a`  makes with the positive direction of the `x`-axis.  (2 marks)

  3. The vector  `underset ~b = 2 sqrt 3 underset ~i + m underset ~j - 5 underset ~k`.
  4. Given that  `underset ~b`  is perpendicular to  `underset ~a,` find the value of `m`.  (2 marks)

Show Answers Only

  1. `1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`
  2. `theta = 45^@`
  3. `m = 6 + 5 sqrt 2`

Show Worked Solution

a.    `|underset ~a|` `= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)`
    `= sqrt 6`
`:. hat underset ~a` `= underset ~a/|underset ~a|`
  `= 1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`

 

Mean mark part (b) 51%.

b.    `underset ~a ⋅ underset ~i` `= sqrt 3 xx 1 = sqrt 3`
  `underset ~a ⋅ underset ~i` `= |underset ~a||underset ~i| cos theta`
    `= sqrt 6 cos theta`
  `sqrt 3` `= sqrt 6 cos theta`
`cos theta` `= 1/sqrt 2`
`:. theta` `= pi/4 = 45^@`

 

c.   `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6 – m + 5 sqrt 2` `=0`  
`:. m` `=6 + 5 sqrt 2`  

Vectors, SPEC1 2015 VCAA 1

Consider the rhombus  `OABC`  shown below, where  `vec (OA) = a underset ~i`  and  `vec (OC) = underset ~i + underset ~j + underset ~k`, and `a` is a positive real constant.
 

VCAA 2015 spec 1a
 

  1. Find  `a.`  (1 mark)

  2. Show that the diagonals of the rhombus  `OABC`  are perpendicular.  (2 marks)

Show Answers Only

  1. `sqrt 3`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a.   `|\ vec(OA)\ | = |\ vec(OC)\ |,`

`:. a` `= sqrt (1^2 + 1^2 + 1^2)`
  `= sqrt 3`

 

b.   `overset(->)(CB) = overset(->)(OA), \ \ overset(->)(AB) = overset(->)(OC)`

MARKER’S COMMENT: Vector notation was poor in many answers.

`overset(->)(OB) = overset(->)(OC) + overset(->)(CB)`

`= underset~i + underset~j + underset~k + sqrt3 underset~i`

`= (sqrt3 + 1)underset~i + underset~j + underset~k`
 

`overset(->)(AC) = overset(->)(AO) + overset(->)(OC)`

`= −sqrt3underset~i + underset~i + underset~j + underset~k`

`= (1 – sqrt3)underset~i + underset~j + underset~k`
 

`overset(->)(AC) · overset(->)(OB)` `= (1 + sqrt3)(1 – sqrt3) + 1 + 1`
  `= 1 – 3 + 1 + 1`
  `= 0`

 
`:. overset(->)(AC) ⊥ overset(->)(OB)`

Mechanics, SPEC2 2017 VCAA 16 MC

An object of mass 20 kg, initially at rest, is pulled along a rough horizontal surface by a force of 80 N acting at an angle of 40° upwards from the horizontal. A friction force of 20 N opposes the motion.

After the pulling force has acted for 5 seconds, the magnitude of the momentum, in kg ms−1, of the object is closest to

  1.     10
  2.     40
  3.   160
  4.   210
  5. 4100
Show Answers Only

`D`

Show Worked Solution


 

`∑underset~F = 80cos(40°) – 20`

`ma` `= 80cos(40°) – 20`
`a` `=4 cos(40°) – 1`

 

`Deltav` `= 0+at`
  `= 5(4cos(40°) – 1)`

 

`m Deltav` `= 100(4cos(40°) – 1)`
  `~~ 206.4`

 
`=> D`

Vectors, SPEC2 2017 VCAA 15 MC

A body has displacement of  `3underset~i + underset~j`  metres at a particular time. The body moves with constant velocity and two seconds later its displacement is  `−underset~i + 5underset~j`  metres.

The velocity, in ms−1, of the body is

  1. `2underset~i + 6underset~j`
  2. `−2underset~i + 2underset~j`
  3. `−4underset~i + 4underset~j`
  4. `4underset~i - 4underset~j`
  5. `underset~i + 3underset~j`
Show Answers Only

`B`

Show Worked Solution
`Deltaunderset~s` `= (−1 – 3)underset~i + (5 – 1)underset~j`
  `= −4underset~i + 4underset~j`
`underset~v` `= (Deltaunderset~s)/t`
  `= (−4underset~i + 4underset~j)/2`
  `= −2underset~i + 2underset~j`

 
`=> B`

Mechanics, SPEC2 2017 VCAA 14 MC

Two particles with mass `m_1` kilograms and `m_2` kilograms are connected by a taut light string that passes over a smooth pulley. The particles sit on smooth inclined planes, as shown in the diagram below.
 


 

If the system is in equilibrium, then  `m_1/m_2`  is equal to

  1.  `(sec(theta))/2`
  2.  `2sec(theta)`
  3.  `2cos(theta)`
  4.  `1/2`
  5.  `1`
Show Answers Only

`A`

Show Worked Solution

Mean mark 51%.

`∑F` `=m_1gsin(2theta) – m_2gsin(theta) = 0`
   
`m_1gsin(2theta)` `= m_2gsin(theta)`
`m_1/m_2` `= (sin(theta))/(sin(2theta))`
  `= (sin(theta))/(2sin(theta)cos(theta))`
  `= 1/(2 cos(theta))`
  `= 1/2 sec(theta)`

 
`=> A`