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Calculus, SPEC2 2014 VCAA 10 MC

A large tank initially holds 1500 L of water in which 100 kg of salt is dissolved. A solution containing 2 kg of salt per litre flows into the tank at a rate of 8 L per minute. The mixture is stirred continuously and flows out of the tank through a hole at a rate of 10 L per minute.

The differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by

A.   `(dQ)/(dt) = 16 - (5Q)/(750 - t)`

B.   `(dQ)/(dt) = 16 - (5Q)/(750 + t)`

C.   `(dQ)/(dt) = 16 + (5Q)/(750 - t)`

D.   `(dQ)/(dt) = (100Q)/(750 - t)`

E.   `(dQ)/(dt) = 8 - Q/(1500 - 2t)`

Show Answers Only

`A`

Show Worked Solution

`(dQ_text(in))/(dV)= 2\ text(kg/L),\ (dV_text(in))/(dt) = 8\ text(L/min)`

`V_0 = 1500,\ Q_0 = 100`

`(dV_text(out))/(dt) = 10\ text(L/min)`
  

`V(t)` `= 1500 + (8 – 10)t`
  `= 1500 – 2t`
  `= 2(750 – t)`

 

`(dQ_text(in))/(dt)` `= 2 xx 8 = 16 text(kg/min)`
`(dQ_text(out))/(dt)` `= Q/(v(t)) xx 10`
  `= (10Q)/(2(750 – t))`
  `= (5Q)/(750 – t)`

 
`:.(dQ)/(dt)= 16 – (5Q)/(150 – t)`

`=> A`

Calculus, SPEC2 2016 VCAA 9 MC

If  `f(x) = (dy)/(dx) = 2x^2 - x`, where  `y_0 = 0 = y(2)`, then  `y_3`  using Euler’s formula with step size 0.1 is

A.  `0.1\ f(2)`

B.  `0.6 + 0.1\ f(2.1)`

C.  `1.272 + 0.1\ f(2.2)`

D.  `2.02 + 0.1\ f(2.3)`

E.  `2.02 + 0.1\ f(2.2)` 

Show Answers Only

`C`

Show Worked Solution
`y_1` `= y_0 + 0.1(2(2)^2 – 2)`
  `= 0.6`
`y_2` `= 0.6 + 0.1 (2(2.1)^2 – 2.1)`
  `= 1.272`
`:.  y_3` `= 1.272 + 0.1\ f(2.2)`

 
`=>  C`

Calculus, SPEC2 2016 VCAA 8 MC

Using a suitable substitution,  `int_a^b (x^3e^(2x^4))\ dx`, where  `a`  and  `b`  are real constants, can be written as

A.  `int_a^b (e^(2u))\ du`

B.  `int_(a^4)^(b^4) (e^(2u))\ du`

C.  `1/8 int_a^b (e^u)\ du`

D.  `1/4 int_(a^4)^(b^4) (e^(2u))\ du`

E.  `1/8 int_(8a^3)^(8b^3) (e^u)\ du` 

Show Answers Only

`D`

Show Worked Solution
`u` `= x^4`
`(du)/(dx)` `= 4x^3\ \ =>\ \ 1/4\ du = x^3\ dx`

  
`text(When)\ \ x=a, \ u=a^4`

`text(When)\ \ x=b, \ u=b^4`
 

`:. int_a^b x^3 e^(2x^4)\ dx`

`= 1/4 int_(a^4)^(b^4) e^(2u)\ du`
 

`=>  D`

Complex Numbers, SPEC2 2016 VCAA 6 MC

The points corresponding to the four complex numbers given by
 

`z_1 = 2 text(cis) (pi/3),\ z_2 = text(cis) ((3 pi)/4),\ z_3 = 2 text(cis)(-(2 pi)/3),\ z_4 = text(cis) (-pi/4)`
 

are the vertices of a parallelogram in the complex plane.

Which one of the following statements is not true?

A.  The acute angle between the diagonals of the parallelogram is  `(5 pi)/12`

B.  The diagonals of the parallelogram have lengths 2 and 4

C.  `z_1 z_2 z_3 z_4 = 0 `

D.  `z_1 + z_2 + z_3 + z_4 = 0`

E.  `1 <= |z| <= 2`  for all four of  `z_1, z_2, z_3, z_4` 

Show Answers Only

`C`

Show Worked Solution

`text(Consider each option:)`

`A: qquad theta_1` `= pi/3 + pi/4=(7 pi)/12`
`theta_2` `= pi – (7 pi)/12= (5 pi)/12\ \ text{(correct)}`

 

`B: qquad l_1` `= 2 + 2 = 4`
`l_2` `= 1+ 1 = 2\ \ text{(correct)}`

 

`C: qquad 2 text(cis) (pi/3) ⋅ text(cis) ((3 pi)/4) ⋅ 2 text(cis) (-(2 pi)/3) ⋅ text(cis) (-pi/4)`

`!= 0\ \ text{(by CAS → incorrect)}`

 
`=>  C`

Complex Numbers, SPEC2 2016 VCAA 4 MC

One of the roots of   `z^3 + bz^2 + cz = 0`  is  `3 - 2i`, where `b` and `c` are real numbers.

The values of `b` and `c` respectively are

A.  `6, 13`

B.  `3, -2`

C.  `-3, 2`

D.  `2, 3`

E.  `-6, 13` 

Show Answers Only

`E`

Show Worked Solution

`z(z^2 + bz + c) = 0,\ b, c in RR`

`text(Given)\ \ z= 3 – 2i \ \ text(is a root,)`

`=> z = 3 + 2i\ \ text{is a root (conjugate).}`
 

`z(z^2 + bz + c)` `= 0`
`z((z – 3) + 2i)((z – 3) – 2i)` `= 0`
`z((z – 3)^2 – 4i^2)` `= 0`
`z(z^2 – 6z + 9 + 4)` `= 0`
`z^3 – 6z^2 + 13z` `= 0`

 
`b = -6,\ \ c = 13` 

`=>  E`

Graphs, SPEC2 2016 VCAA 3 MC

The straight-line asymptote(s) of the graph of the function with rule  `f(x) = (x^3 - ax)/x^2`, where  `a`  is a non-zero real constant, is given by

A.  `x = 0`  only.

B.  `x = 0`  and  `y = 0`  only.

C.  `x = 0`  and  `y = x`  only.

D.  `x = 0, \ x = sqrt a`  and  `x = -sqrt a`  only.

E.  `x = 0`  and  `y = a`  only. 

Show Answers Only

`C`

Show Worked Solution

`f(x) = x – a/x,\ x != o`

`lim_(x -> oo) f(x) = x`

`:.\ text(Asymptotes):\ \ x = 0,\ \ y = x`

 
`=>  C`

Trigonometry, SPEC1 2016 VCAA 9

Given that  `cos (x - y) = 3/5`  and  `tan(x) tan (y) = 2`, find  `cos(x + y)`.  (3 marks)

Show Answers Only

`-1/5`

Show Worked Solution

`cos(x+y)= cos(x) cos(y) – sin(x) sin(y)`
 

`cos (x) cos (y) + sin (x) sin (y) = 3/5\ \ …\ (1)`

`(sin(x) sin(y))/(cos(x) cos(y))` `=2`  
`sin(x) sin(y)` `= 2 cos (x) cos(y)\ \ …\ (2)`  

  
`text{Substitute (2) into (1):}`

`cos(x) cos(y) + 2 cos(x) cos(y)` `= 3/5`
`3 cos(x) cos(y)` `= 3/5`
`cos(x) cos(y)` `= 1/5`

 
`text(Substitute)\ \ cos(x)cos(y)=1/5\ \ text{into (1):}`

`1/5 + sin(x) sin(y)` `= 3/5`
`sin(x) sin(y)` `= 2/5`

 

`:. cos(x + y)` `= cos(x) cos(y) – sin(x) sin(y)`
  `= 1/5 – 2/5`
  `= -1/5`

Vectors, SPEC1 2016 VCAA 8

The position of a body with mass 3 kg from a fixed origin at time  `t`  seconds, `t >= 0`, is given by  `underset ~r = (3 sin (2t) - 2)underset ~i + (3 - 2 cos(2t)) underset ~j`, where components are in metres.

  1. Find an expression for the speed, in metres per second, of the body at time  `t`.  (2 marks)
  2. Find the speed of the body, in metres per second, when  `t = pi/12`.  (1 mark)
  3. Find the maximum magnitude of the net force acting on the body in newtons.  (3 marks)
Show Answers Only
  1. `|underset ~ dot r| = sqrt(36 cos^2(2t) + 16 sin^2(2t))`
  2. `sqrt 31`
  3. `36`
Show Worked Solution
a.    `underset ~r ` `=(3 sin (2t) – 2)underset ~i + (3 – 2 cos(2t)) underset ~j`
   `underset ~ dot r` `= 6 cos (2t) underset ~i + 4 sin (2t) underset ~j`
  `|underset ~ dot r|` `= sqrt(36 cos^2(2t) + 16 sin^2(2t))`

 

b.   `text(Find)\ \ |underset ~ dot r|\ \ text(when)\ \ t=pi/12:`

`underset ~ dot r` `=(36 cos^2 (pi/6) + 16 sin^2 (pi/6))^(1/2)`  
  `= (36 (sqrt 3/2)^2 + 16 (1/2)^2)^(1/2)`  
  `= (36/4 xx 3 + 16/4)^(1/2)`  
  `= (27 + 4)^(1/2)`  
  `= sqrt 31`  

 

c.    `underset ~ ddot r` `= d/(dt) (underset ~dotr)= -12 sin (2t) underset ~i + 8 cos (2t) underset ~j`
  `underset ~F` `= 3 ddot r`
  `|underset ~F|` `= 3 |underset ~ ddot r|`
    `= 3 sqrt(144 sin^2 (2t) + 64 cos^2 (2t))`
    `= 3 sqrt(64 sin^2 (2t) + 64 cos^2(2t) + 80 sin^2 (2t))`
    `= 3 sqrt(64 + 80 sin^2 (2t))`

 
`text(Max occurs when)\ \ (sin^2 (2t)) = 1`

Mean mark 51%.

`:. max (|underset ~F|)` `= 3 sqrt(64 + 80)`
  `= 3 sqrt 144`
  `= 3 xx 12`
  `= 36`

Calculus, SPEC1 2016 VCAA 7

Find the arc length of the curve  `y = 1/3 (x^2 + 2)^(3/2)`  from  `x = 0`  to  `x = 2`.  (4 marks)

Show Answers Only

`14/3`

Show Worked Solution
`y` `= 1/3 (x^2 + 2)^(3/2)`
`(dy)/(dx)` `= 1/3 xx 3/2 xx 2x (x^2 + 2)^(3/2 – 1)`
  `= x(x^2 + 2)^(1/2)`
`((dy)/(dx))^2` `= x^2 (x^2 + 2)`
  `= x^4 + 2x^2`

 

`l` `= int_0^2 sqrt(1 + ((dy)/(dx))^2)\ dx`
  `= int_0^2 sqrt(x^4 + 2x^2 + 1)\ dx`
  `= int_0^2 sqrt((x^2 + 1)^2)\ dx`
  `= int_0^2 x^2 + 1\ dx`
  `= [x^3/3 + x]_0^2`
  `= 8/3 + 2`
  `= 14/3`

Complex Numbers, SPEC1 2016 VCAA 6

Write  `(1 - sqrt 3 i)^4/(1 + sqrt 3 i)`  in the form  `a + bi`, where  `a`  and  `b`  are real constants.  (3 marks)

Show Answers Only

`4 + 4 sqrt 3 i`

Show Worked Solution

`(1 – sqrt 3 i) \ \ => \ r_1=sqrt (1^2 + (sqrt 3)^2)=2`

`theta_1 = tan^(-1) (-sqrt 3)=- pi/3`

`(1 + sqrt 3 i) \ \ => \ r_2=2, \ theta_2 =pi/3`
 

`:. (1 – sqrt 3 i)^4/(1 + sqrt 3 i)` `= (2 text(cis) (-pi/3))^4/(2 text(cis) (pi/3))`
  `= (16 text(cis) (-(4 pi)/3))/(2 text(cis) (pi/3))`
  `= 8 text(cis) (-(5 pi)/3)`
  `= 8 text(cis) (pi/3)\ \ \ (-pi<=theta<=pi)`
  `= 8(1/2 + sqrt 3/2 i)`
  `= 4 + 4 sqrt 3 i`

Vectors SPEC1 2016 VCAA 5

Consider the vectors  `underset ~a = 3 underset ~i + 5 underset ~j - 2 underset ~k,\ underset ~b = underset ~i - 2 underset ~j + 3 underset ~k`  and  `underset ~c = underset ~i + d\  underset ~k`, where  `d`  is a real constant.

  1. Find the vector resolute of  `underset ~a`  in the direction of  `underset ~b`.  (2 marks)
  2. Find the value of  `d`  if the vectors are linearly dependent.  (2 marks)
Show Answers Only
  1.  `-13/14 (underset ~i – 2 underset ~j + 3 underset ~k)`
  2.  `d = 1`
Show Worked Solution

a.   `underset ~hat b = 1/sqrt14(underset ~i – 2 underset ~j + 3 underset ~k)`

`text(Scalar resolute)\ \ (underset ~a ⋅ underset ~hat b)`

`=1/sqrt14 (3xx1 – 5xx2 -2xx3)`

`=-13/sqrt14`
 

`text(Vector resolute)`

`(underset ~a ⋅ underset ~hat b) underset ~hat b` `= -13/sqrt14 xx 1/sqrt14(underset ~i – 2 underset ~j + 3 underset ~k)`  
   `=-13/14 (underset ~i – 2 underset ~j + 3 underset ~k)`  

 

b.    `text(If linearly dependent)\ \ =>\ \ alpha underset ~a + beta underset ~b = underset ~c`

`(3 alpha + beta) underset ~i + (5 alpha – 2 beta) underset ~j + (-2 alpha + 3 beta) underset ~k = underset ~i + d\ underset ~k`

`3 alpha + beta` `= 1\ \ …\ (1)`
`5 alpha – 2 beta` `=0\ \ …\ (2)`
`3 beta – 2 alpha` `= d\ \ …\ (3)`

 
`2 xx (1) + (2):`

`11 alpha = 2 => alpha = 2/11`

`text(Substitute)\ \ alpha = 2/11\ \ text{into (1):}`

`6/11 + beta = 1 \ => \ beta = 5/11`

`:.d` `=3(5/11) – 2(2/11)`  
  `=15/11 – 4/11`  
  `=1`  

Calculus, SPEC1 2016 VCAA 4

Chemicals are added to a container so that a particular crystal will grow in the shape of a cube. The side length of the crystal, `x`  millimetres, `t`  days after the chemicals were added to the container, is given by  `x = arctan(t)`.

Find the rate at which the surface area, `A` square millimetres, of the crystal is growing one day after the chemicals were added. Give your answer in square millimetres per day.  (4 marks)

Show Answers Only

`(3 pi)/2`

Show Worked Solution
`x` `=tan^(-1) x`
`(dx)/(dt)` `= 1/(1 + t^2)`

 
`text(Surface Area)\ \ (A) = 6x^2`

`(dA)/(dx)` `= 12x`
`(dA)/(dt)` `= (dA)/(dx) ⋅ (dx)/(dt)`
  `= (12 x)/(1 + t^2)`

 
`text(When)\ \ t=1,`

`x= tan^(-1) (1) = pi/4`

`:. (dA)/(dt) |_(x=pi/4,t=1)` `=(12 xx pi/4)/(1+1^2)`  
  `=(3pi)/2\ text(mm²/day)`  

Calculus, SPEC1 2016 VCAA 3

Find the equation of the line perpendicular to the graph of  `cos (y) + y sin(x) = x^2`  at  `(0, -pi/2)`.  (4 marks)

Show Answers Only

`y_N = (-2)/pi x – pi/2`

Show Worked Solution

`d/(dx) (cos(y)) + d/(dx) (y sin(x)) = d/(dx) (x^2)`

`-sin(y) *(dy)/(dx) + sin(x) *(dy)/(dx) + y cos (x) = 2x`

`dy/dx(sin(x)-sin(y))` `=2x-ycos(x)`  
`dy/dx` `=(2x-ycos(x))/(sin(x)-sin(y))`  

 
`dy/dx |_(x=0, y=pi/2) =(0-(pi/2)(1))/(0-1)=pi/2`

`m_T = pi/2\ \ =>\ \ m_⊥ = (-2)/pi`
 

`:.\ text(Equation of ⊥ line:)`

`y – ((-pi)/2)` `=(-2)/pi (x-0)`  
`y` `=((-2)/pi) x – pi/2`  

Statistics, SPEC2-NHT 2017 VCAA 6

A bank claims that the amount it lends for housing is normally distributed with a mean of $400 000 and a standard deviation of $30 000.

A consumer organisation believes that the average loan amount is higher than the bank claims.

To check this, the consumer organisation examines a random sample of 25 loans and finds the sample mean to be $412 000.

  1. Write down the two hypotheses that would be used to undertake a one-sided test.  (1 mark)
  2. Write down an expression for the `p` value for this test and evaluate it to four decimal places.  (2 marks)
  3. State with a reason whether the bank’s claim should be rejected at the 5% level of significance.  (1 mark)
  4. What is the largest value of the sample mean that could be observed before the bank’s claim was rejected at the 5% level of significance? Give your answer correct to the nearest 10 dollars.  (1 mark)
  5. If the average loan made by the bank is actually $415 000 and not $400 000 as originally claimed, what is the probability that a random selection of 25 loans has a sample mean that is at most $410 000? Give your answer correct to three decimal places.  (2 marks)
Show Answers Only
  1. `H_0: mu = 400\ 000`

     

    `H_1: mu > 400\ 000`

  2. `p ~~ 0.0228`
  3. `text(reject)`
  4. `x_max ~~ 409\ 860`
  5. `~~ 0.202`
Show Worked Solution

a.   `H_0: \ mu = 400\ 000`

`H_1: \ mu > 400\ 000`

 

b.   `L\ ~\ N (400\ 000, 30\ 000^2)`

`bar L\ ~\ N (400\ 000, (30\ 000^2)/25)`

`p = text(Pr)(bar L > $412\ 000)`
 

`:. p ~~ 0.0228`

 

c.   `text(The bank’s claim)\ (h_0: mu = 400\ 000)`

`text(should be rejected at the 5% level of)`

`text(significance as)\ \ p ~~ 0.0228 < 0.05.`

 

d.   `p = text(Pr)(bar L > x) = 0.05`

`=> text(Pr)(bar L < x) = 0.95`

`=> x ~~ 409\ 869.12`

`:. x_max ~~ $409\ 870\ \ text{(nearest $10)}`

 

e.   `text(New distribution):\ \ L_2\ ~\ N (415\ 000, 30\ 000^2)`

`bar L_2\ ~\ N (415\ 000, (30\ 000^2)/25)`

`text(Pr)(bar L_2 <= 410\ 000) ~~ 0.202`

Mechanics, SPEC2-NHT 2017 VCAA 5

A 5 kg mass is initially held at rest on a smooth plane that is inclined at 30° to the horizontal. The mass is connected by a light inextensible string passing over a smooth pulley to a 3 kg mass, which in turn is connected to a 2 kg mass.

The 5 kg mass is released from rest and allowed to accelerate up the plane.

Take acceleration to be positive in the directions indicated.
 

  1. Write down an equation of motion, in the direction of motion, for each mass.   (3 marks)
  2. Show that the acceleration of the 5 kg mass is  `g/4\ text(ms)^(-2)`.  (1 mark)
  3. Find the tensions  `T_1`  and  `T_2`  in the string in terms of  `g`.  (2 marks)
  4. Find the momentum of the 5 kg mass, in kg ms`­^(-1)`, after it has moved 2 m up the plane, giving your answer in terms of `g`.  (2 marks)
  5. A resistance force  `R`  acting parallel to the inclined plane is added to hold the system in equilibrium, as shown in the diagram below.
     

     

    `qquad`
     

     

    Find the magnitude of  `R`  in terms of  `g`.  (2 marks)

Show Answers Only
  1. `2a = 2g – T_2`

     

    `3a = 3g + T_2 – T_1`

     

    `5a = T_1 – (5g)/2`

  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `T_1 = (15g)/4`

     

    `T_2 = (3g)/2`

  4. `p = 5 sqrt g`
  5. `R = (5g)/2`
Show Worked Solution
a.    `2\ text(kg): \ 2a` `= 2g – T_2`
  `3\ text(kg): \ 3a` `= 3g + T_2 – T_1`
  `5\ text(kg): \ 5a` `= T_1 – 5g sin (30^@)`
  `5a` `= T_1 – (5g)/2`

 

 

b.   `sum F = 3g + 2g – 5g sin 30^@ = (5 + 3 + 2)a`

`5g – (5g)/2` `= 10a`
`a` `= (5g)/(2 xx 10)`
`:. a` `= g/4\ text(ms)^(-2)`

 

c.   ` T_1 – (5g)/2` `= 5a`
`:. T_1` `= (5g)/2 + 5(g/4)`
  `= (15g)/4`

 

`2g – T_2` `= 2a`
`:. T_2` `= 2g – 2(g/4)`
  `= (3g)/2`

 

d.   `u = 0,\ \ a = g/4,\ \ s = 2`

`text(Find)\ \ v\ \ text(when)\ \ s=2:`

`v^2` `= u^2 + 2as`  
  `=0 + 2 (g/4) xx 2`  
  `=g`  
`v` `=sqrtg\ \ \ (v>0)`  

 
`:. p = 5 sqrt g`

 

e.   `sum F = 2g + 3g – 5g sin 30^@ – R = 0`

`:. R` `= 2g + 3g – 5g sin 30^@`
  `= 5g – (5g)/2`
  `= (5g)/2`

Vectors, SPEC2-NHT 2017 VCAA 4

A cricketer hits a ball at time  `t = 0`  seconds from an origin `O` at ground level across a level playing field.

The position vector  `underset ~r(t)`, from `O`, of the ball after `t` seconds is given by
 
  `qquad underset ~r(t) = 15t underset ~i + (15 sqrt 3 t - 4.9t^2)underset ~j`,
 
where,  `underset ~i`  is a unit vector in the forward direction, `underset ~j`  is a unit vector vertically up and displacement components are measured in metres.

  1. Find the initial velocity of the ball and the initial angle, in degrees, of its trajectory to the horizontal.  (2 marks)
  2. Find the maximum height reached by the ball, giving your answer in metres, correct to two decimal places.  (2 marks)
  3. Find the time of flight of the ball. Give your answer in seconds, correct to three decimal places.  (1 mark)
  4. Find the range of the ball in metres, correct to one decimal place.  (1 mark)
  5. A fielder, more than 40 m from `O`, catches the ball at a height of 2 m above the ground.

     

    How far horizontally from `O` is the fielder when the ball is caught? Give your answer in metres, correct to one decimal place.  (2 marks)

Show Answers Only
  1. `underset ~v (0) = 15 underset ~i + 15 sqrt 3 underset ~j`

     

    `theta = pi/3 = 60^@`

  2. `34.44\ text(m)`
  3. `5.302\ text(s)`
  4. `79.5\ text(m)`
  5. `78.4\ text(m)`
Show Worked Solution

a.   `underset ~v (t) = underset ~ dot r (t) = 15 underset ~i + (15 sqrt 3 – 9.8t)underset ~j`

`text(Initial velocity occurs when)\ \ t=0:`

`:. underset ~v (0) = 15 underset ~i + 15 sqrt 3 underset ~j`
 

`text(Let)\ \ theta = text(Initial trajectory,)`

`tan theta` `=(15sqrt3)/15`  
  `=sqrt3`  
`:. theta` `=pi/3\ \ text(or)\ \ 60^@`  

 

b.  `text(Max height)\ =>underset~j\ \ text(component of)\ \ underset ~v=0.`

`15 sqrt 3 – 9.8t` `=0`
`t` `=(15 sqrt 3)/9.8`
  `=2.651…`

 
`text(Find max height when)\ \ t = 2.651…`

`:.\ text(Max height)` `= 15 sqrt 3 xx 2.651 – 4.9 xx (2.651)^2`
  `~~ 34.44\ text(m)`

 

c.   `text(Ball travels in parabolic path.)`

`:.\ text(Total time of flight)`

`= 2 xx (15 sqrt 3)/9.8`

`~~ 5.302\ text(s)`

 

d.   `text(Range)` `= x ((15 sqrt 3)/4.9) – x(0)`
  `= (225 sqrt 3)/4.9`
  `~~ 79.5\ text(m)`


e.   `text(Find)\ \ t\ \ text(when height of ball = 2 m:)`

`15 sqrt 3 t – 4.9t^2` `=2`  
`4.9t^2 – 15 sqrt 3 t + 2` `=0`  

  
`t_1 ~~ 0.078131,\ \ t_2 ~~ 5.22406`

 
`text(When)\ \ t=0.0781,`

`x= 15 xx 0.0781 = 1.17\ \ text{(no solution →}\ x<40 text{)}`

 `text(When)\ \ t=5.2241,`

`x=15 xx 5.2241 = 78.4\ text(m)`
 

`:.\ text(Ball is caught 78.4 m horizontally from)\ O.`

Calculus, SPEC2-NHT 2017 VCAA 3

Bacteria are spreading over a Petri dish at a rate modelled by the differential equation

`(dP)/(dt) = P/2 (1 - P),\ 0 < P < 1`

where  `P`  is the proportion of the dish covered after  `t`  hours.

    1. Express  `2/(P(1 - P))`  in partial fraction form.  (1 mark)
    2. Hence show by integration that  `(t - c)/2= log_e(P/(1 - P))`, where  `c`  is a constant of integration.  (2 marks)
    3. If half of the Petri dish is covered by the bacteria at  `t = 0`, express  `P`  in terms of  `t`.  (2 marks)

After one hour, a toxin is added to the Petri dish, which harms the bacteria and reduces their rate of growth. The differential equation that models the rate of growth is now

`(dP)/(dt) = P/2 (1 - P) - sqrt P/20`  for  `t >= 1`

  1. Find the limiting value of  `P`, which is the maximum possible proportion of the Petri dish that can now be covered by the bacteria. Give your answer correct to three decimal places.  (2 marks)
  2. The total time, `T`  hours, measured from time  `t = 0`, needed for the bacteria to cover 80% of the Petri dish is given by
     

     

    `qquad qquad T = int_q^r (1/(P/2(1 - P) - sqrt P/20)) dP + s`
     

     

    where  `q, r and s in R`.

     

     

    Find the values of  `q, r` and `s`, giving the value of `q` correct to two decimal places.  (2 marks)

  1. Given that  `P = 0.75`  when  `t = 3`, use Euler’s method with a step size of 0.5 to estimate the value of `P` when  `t = 3.5`. Give your answer correct to three decimal places.  (3 marks)
Show Answers Only
    1. `2/(P(1 – P)) = 2/P + 2/(1 – P)`
    2. `text(Proof)\ text{(See Worked Solutions)}`
    3. `P = e^(t/2)/(1 + e^(t/2))`
  2. `P ~~ 0.894`
  3. `r = 0.8`
    `s = 1`
    `q ~~ 0.62`
  4. `~~ 0.775`
Show Worked Solution

a.i.   `2/(P(1 – P)) = A/P + B/(1 ⋅ P)`

`A(1 – P) + BP = 2`

`text(If)\ \ P = 0\ \ =>\ \ A = 2`

`text(If)\ \ P = 1\ \ =>\ \ B = 2`

`:. 2/(P(1 – P)) = 2/P + 2/(1 – P)\ \ \ text{(can also solve by CAS)}`

 

a.ii.  `(dt)/(dP) = 2/(P(1 – P)) = 2/P + 2/(1 – P)`

`t` `= int 2/P + 2/(1 – P)\ dP`
  `= 2 ln |P| – 2ln|1 – P| + c`
`(t-c)/2` `=ln|P| – ln|1-P|`
  `=ln |(P)/(1-P)|`
  `= ln (P/(1 – P))`

  
`text(S)text(ince)\ \ 0 < P < 1 :\ |P| = P\ and\ |1 – P| = 1 – P`


a.iii.  `text(When)\ \ t=0,\ P=0.5`

`(-c)/2` `= ln (0.5/0.5)`
`c` `= ln (1) = 0`

 

`t/2` `= ln (P/(1 – P))\ \ \ text{(solve manually or by CAS)}`
`e^(t/2)` `= P/(1 – P)`
`e^(t/2) (1 – P)` `= P`
`e^(t/2) – Pe^(t/2)` `= P`
`e^(t/2)` `= P(1 + e^(t/2))`
`:. P` `= e^(t/2)/(1 + e^(t/2))`

 

b.   `(dP)/(dt) = P/2 (1 – P) – sqrt P/20`

`text(Limiting value occurs when)\ \ (dP)/(dt) = 0,`

`P ~~ 0.894\ \ \ text{(by CAS)}`
 

`=>\ text(Lower solution values at levels already exceeded)`

 `text(are ignored.)`

 

c.   `(dt)/(dP) = 1/(P/2 (1 – P) – sqrt P/20)\ \ text(for)\ \ t>=1`

`text(When)\ \ t=1\ \ =>\ \ P=0.622`
 

`T = int_0.62^0.8 1/(P/2 (1 – P) – sqrt P/20) dP + 1`
 

`:. r = 0.8,\ s = 1 and q = 0.62`

 

d.   `P(3.5) ~~ P(3) + h* (dP)/(dt)|_(P= 0.75)`

   `= 0.75 + 0.5 (0.75/2 (1 – 0.75) – sqrt 0.75/20)`

   `~~ 0.775`
 

`:. P~~0.775\ \ text(when)\ \ t=3.5`

Statistics, SPEC2 2017 VCAA 6

A dairy factory produces milk in bottles with a nominal volume of 2 L per bottle. To ensure most bottles contain at least the nominal volume, the machine that fills the bottles dispenses volumes that are normally distributed with a mean of 2005 mL and a standard deviation of 6 mL.

  1. Find the percentage of bottles that contain at least the nominal volume of milk, correct to one decimal place.  (1 marks)

Bottles of milk are packed in crates of 10 bottles, where the nominal total volume per crate is 20 L.

  1. Show that the total volume of milk contained in each crate varies with a mean of 20 050 mL and a standard deviation of  `6sqrt10`  mL.  (2 marks)
  2. Find the percentage, correct to one decimal place, of crates that contain at least the nominal volume of 20 L.  (1 mark)
  3. Regulations require at least 99.9% of crates to contain at least the nominal volume of 20 L.
  4. Assuming the mean volume dispensed by the machine remains 2005 mL, find the maximum allowable standard deviation of the bottle-filling machine needed to achieve this outcome. Give your answer in millilitres, correct to one decimal place.  (3 marks)
  5. A nearby dairy factory claims the milk dispensed into its 2 L bottles varies normally with a mean of 2005 mL and a standard deviation of 2 mL.
  6. When authorities visit the nearby dairy factory and check a random sample of 10 bottles of milk, they find the mean volume to be 2004 mL.
  7. Assuming that the standard deviation of 2 mL is correct, carry out a one-sided statistical test and determine, stating a reason, whether the nearby dairy’s claim should be accepted at the 5% level of significance.  (2 marks)
Show Answers Only
  1. `79.8text(%)`
  2. `text(See Worked Solutions)`
  3. `99.6text(%)`
  4. `text(max) (σ_x) ~~ 5.1`
  5. `text(The claim should be accepted at a 5% significance level.)`
Show Worked Solution

a.   `V_B~N(2005, 6^2)`

`text(Pr)(V_B > 2000)` `~~ 0.797672\ \ \ text{(by CAS)}`
  `~~ 79.8text(%)`

 

b.   `text(Let)\ \ V_C=\ text(Volume of a crate)`

`mu_(V_C)` `= 10 xx mu_(V_B)=20\ 050`  
     
  `σ_(V_C)^2` `= σ_(V_B)^2 xx 10`
    `= 360`
  `:. σ_(V_C)` `= 6sqrt(10)`

 

c.  `V_C~N(20\ 050, (6sqrt10)^2)`

`text(Pr)(V_C > 20\ 000)` `~~ 0.995796`
  `~~ 99.6text(%)`

 

d.   `text(Let)\ \ V_A =\ text(New distribution)`

`V_A~N(20\ 050, (σ_x  sqrt10)^2)`

`text(Pr)(V_A > 20\ 000)` `>= 0.999`  
`text(Pr)(Z=a)` `=0.999`  
`:.a` `=-3.0902`  

 

`(20\ 000-20\ 050)/(σ_x sqrt10)` `=-3.0902`
`σ_x` `=5.116…`

 

`:.\ text(max) (σ_x) ~~ 5.1`

 

e.   `H_0: mu=2005`

`H_1: mu<2005`

`D~N(2005, 2^2)\ \ =>\ \ barD~N(2005, (2^2)/10)`

`p` `= text(Pr)(barD < 2004)`
  `~~ 0.056923`

  

`:.\ text(S)text(ince)\ \ p>0.05,\ text(the claim should be accepted)`

`text(at a 5% significance level.)`

Trigonometry, SPEC2 2016 VCAA 1 MC

The cartesian equation of the relation given by  `x = 3 text(cosec)^2 (t)`  and  `y = 4 cot (t) - 1`  is

A.  `(y + 1)^2/16 - x^2/9 = 1`

B.  `(y + 1)^2 = (16 (x + 3))/3`

C.  `x^2/9 + (y + 1)^2/16 = 1`

D.  `4x - 3y = 15`

E.  `(y + 1)^2 = (16 (x - 3))/3` 

Show Answers Only

`E`

Show Worked Solution
`x/3` `= text(cosec)^2(t)`  
`(y+1)/4` `=cot (t)`  

 

`sin^2(t) + cos^2(t)` `= 1`
`1 + cot^2(t)` `= text(cosec)^2(t)`
`text(cosec)^2(t) – cot^2(t)` `= 1`
   
`(x/3) – ((y + 1)/4)^2` `= 1`
`(y + 1)^2/16` `= x/3 – 1`
`(y + 1)^2` `= (16 (x – 3))/3`

 
`=>  E`

Vectors, SPEC2 2017 VCAA 5

On a particular morning, the position vectors of a boat and a jet ski on a lake  `t`  minutes after they have started moving are given by  `underset~r_B(t) = (1 - 2cos(t)) underset~i + (3 + sin(t))underset~j`  and  `underset~r_J(t) = (1 - sin(t)) underset~i + (2 - cos(t))underset~j`  respectively for  `t >= 0`, where distances are measured in kilometres. The boat and the jet ski start moving at the same time. The graphs of their paths are shown below.

  1. On the diagram above, mark the initial positions of the boat and the jet ski, clearly identifying each of them. Use arrows to show the directions in which they move.  (2 marks)
    1. Find the first time for  `t > 0`  when the speeds of the boat and the jet ski are the same.  (2 marks)
    2. State the coordinates of the boat at this time.  (1 mark)
    1. Write down an expression for the distance between the jet ski and the boat at any time `t`.  (1 mark)
    2. Find the minimum distance separating the boat and the jet ski. Give your answer in kilometres, correct to two decimal places.  (1 mark)
  4. On another morning, the boat’s position vector remained the same but the jet skier considered starting from a different location with a new position vector given by  `underset~r(t) = (1 - 2sin(t)) underset~i + (a - cos(t))underset~j, \ t >= 0`, where `a` is a real constant. Both vessels are to start at the same time.
    Assuming the vessels would collide shortly after starting, find the time of the collision and the value of `a`.  (3 marks)
Show Answers Only
  1.  
  2.   
    1. `t = 0, t_2 = pi`
    2. `(3,3)`
  3.   
    1. `sqrt((sin(t) – 2cos(t))^2 + (1 + sin(t) + cos (t))^2)`
    2. `d_text(min) ~~ 0.33`
  4. `a = 3 + 3/sqrt5, t = tan^(−1)(2)`
Show Worked Solution
a.   

`text(Initial positions occur at)\ \ t=0.`

`text(Consider the graph)\ \ underset~r_B(t)\ \ text(at)\ \ t=pi/4:`

`(1 – 2cos0) < (1 – 2cos(pi/4)) =>\ text(moves higher)`
  

`text(Consider the graph)\ \ underset~r_J(t)\ \ text(at)\ \ t=pi/4:`

`(1 – sin(0)) > (1 – sin(pi/4)) =>\ text(moves left)`

 

b.i.    `underset~dotr_B(t)` `= 2sin(t)underset~i + cos(t)underset~j`
  `|underset~dotr_B(t)|` `= sqrt(4sin^2(t) + cos^2(t))`
`underset~dotr_J(t)` `= −cos(t)underset~i + sin(t)underset~j`
`|underset~dotr_J(t)|` `= sqrt(cos^2(t) + sin(t))=1`
   

`text(Find)\ \ t\ \ text(when)\ \ sqrt(4sin^2(t) + cos^2(t))=1:`

`t = pi\ text(seconds)\ \ \ (t!=0)`

 

♦ Mean mark (b)(ii) 48%.

b.ii.    `(underset~r)_B(pi)` `= (1 – 2cos(pi))underset~i + (3 + sin(pi))underset~j`
    `= 3underset~i + 3underset~j`

`:.\ text(Boat coordinates):\  (3,3)`

 

c.i.    `underset~r_B – underset~r_J` `=(sin(t) – 2 cos(t))i +(1+sin(t) + cos(t))`
  `:. d` `= |underset~r_B – underset~r_J|`
    `= sqrt((sin(t) – 2cos(t))^2 + (1 + sin(t) + cos (t))^2)`

 

c.ii.  `d_text(min) ~~ 0.33\ \ \ text{(by CAS)}`

♦♦♦ Mean mark part (c)(ii) 15%.

 

d.   `text(Equating coefficients for collision:)`

  `x:\ \ \ 1 – sin(t)` `= 1 – 2cos(t)\ …\ (1)`
  `sin(t)` `= 2cos(t)`
  `tan(t)` `= 2`
  `t` `=tan^(−1)(2)`

 
`y:\ \ \ a – cos(t) = 3 + sin(t)\ …\ (2)`

♦ Mean mark part (d) 41%.
 

 
`text(Using)\ \ tan(t)=2,`

`=> sin(t) = (2sqrt5)/5,\ \ cos(t) = sqrt5/5`

`text{Substitute into (1):}`

`a – 1/sqrt5` `= 3 + 2/sqrt5`  
`:. a` `= 3 + 3/sqrt5`  

 
`:.\ text(Collision occurs when)\ \ t=tan^(-1)2\ \ text(and)\ \ a=3 + 3/sqrt5`

Complex Numbers, SPEC2 2017 VCAA 4

  1. Express  `−2 - 2sqrt3 i`  in polar form.  (1 mark)
  2. Show that the roots of  `z^2 + 4z + 16 = 0`  are  `z = −2 - sqrt3 i`  and  `z = −2 + 2sqrt3 i`.  (1 mark)
  3. Express the roots of  `z^2 + 4z + 16 = 0`  in terms of  `2 - 2sqrt3 i`.  (1 mark)
  4. Show that the cartesian form of the relation  `|z| = |z - (2 - 2sqrt3 i)|`  is  `x - sqrt3 y - 4 = 0`  (2 marks)
  5. Sketch the line represented by  `x - sqrt3y  - 4 = 0`  and plot the roots of  `z^2 + 4z + 16 = 0`  on the Argand diagram below.  (2 marks)
     
       

  6. The equation of the line passing through the two roots of  `z^2 + 4z + 16 = 0`  can be expressed as  `|z - a| = |z - b|`, where  `a, b ∈ C`.

     

    Find `b` in terms of `a`.  (1 mark)

  7. Find the area of the major segment bounded by the line passing through the roots of  `z^2 + 4z + 16 = 0`  and the major arc of the circle given by  `|z| = 4`.  (2 marks)
Show Answers Only
  1. `4text(cis)((−2pi)/3)`
  2. `text(See Worked Solutions)`
  3. `−(2 – 2sqrt3 i) and – bar((2-2sqrt3 i))`
  4. `text(See Worked Solutions)`
  5.  
  6. `−4 – bara`
  7. `4sqrt3 + (32pi)/3`
Show Worked Solution
a.    `r` `= sqrt((−2)^2 + (−2sqrt3)^2)=4`

 

`theta` `= −pi + tan^(−1)((2sqrt3)/2)`
  `= −pi + pi/3`
  `=(-2pi)/3`

 
`:. −2 – 2sqrt3 i = 4text(cis)((−2pi)/3)`
 

b.   `z^2 + 4z + z^2 – 4 + 16` `=0`
`(z + 2)^2 + 12` `= 0`
`(z + 2)^2` `= -12`
`(z + 2)^2` `= 12i^2`
`z + 2` `= ±sqrt12 i`
`z + 2` `= ±2sqrt3 i`
`:. z` `= -2 ± 2sqrt3 i`

 

♦♦ Mean mark part (c) 31%.

c.    `z_1` `= -2 + 2sqrt3 i = -(2-2sqrt3 i)`
  `z_2` `= – 2 – 2sqrt3 i = – bar((2-2sqrt3 i))`

 

d. `|z|` `= |z – (2 – 2sqrt3 i)|`
     `x^2 + y^2` `= (x – 2)^2 + (y + 2sqrt3)^2`
    `x^2 + y^2` `= x^2 – 4x + 4+ y^2 + 4sqrt3 y + 12`
  `0` `= −4x + 4sqrt3 y + 16`
  `0` `= −x + sqrt3 y + 4`

 
`:. x – sqrt3 y – 4=0`

 

e.   

 

f.   `x = − 2\ \ text(is equidistant from)\ \ z_1 = a\ \ text(and)\ \ z_2 = b`

♦♦♦ Mean mark 1%!

`=> text(Im)(a) = text(Im)(a)`
 

`text(Let)\ \ a = alpha + betai, \ b = gamma + betaj`

`(alpha + gamma)/2` `= −2`
`alpha + gamma` `= −4`
`gamma` `= -4 – alpha`

  

`:. b` `= -4 – alpha + betaj`
  `= -4 – (alpha + betaj)`
  `= -4 – bara`

 

♦♦ Mean mark 31%.

g.    `text(Area)\ DeltaOAB` `= 1/2 xx (4sqrt3 xx 2)`
    `= 4sqrt3`

 

`text(Area of sector)\ AOB` `= pi xx 4^2 xx (2 xx pi/3)/(2pi)`
  `= (16pi)/3`

 
`:.\ text(Area of major segment area)`

`=pi(4)^2 – ((16pi)/3 – 4sqrt3)`

`= 4sqrt3 + (32pi)/3`

Calculus, SPEC2 2017 VCAA 3

A brooch is designed using inverse circular functions to make the shape shown in the diagram below.
 


 

The edges of the brooch in the first quadrant are described by the piecewise function
 

`f(x){(3text(arcsin)(x/2)text(,), 0 <= x <= sqrt2),(3text(arccos)(x/2)text(,), sqrt2 < x <= 2):}`
 

  1. Write down the coordinates of the corner point of the brooch in the first quadrant.  (1 mark)
  2. Specify the piecewise function that describes the edges in the third quadrant.  (1 mark)
  3. Given that each unit in the diagram represents one centimetre, find the area of the brooch.

     

    Give your answer in square centimetres, correct to one decimal place.  (3 marks)

  4. Find the acute angle between the edges of the brooch at the origin. Give your answer in degrees, correct to one decimal place.  (3 marks)
  5. The perimeter of the brooch has a border of gold.
    Show that the length of the gold border needed is given by a definite integral of the form  `int_0^2 (sqrt(a + b/(4 - x^2)))dx`, where  `a, b ∈ R`. Find the values of `a` and `b`.  (2 marks)
Show Answers Only
  1. `(sqrt2,(3pi)/4)`
  2. `text(See Worked Solutions)`
  3. `9.9\ text(cm²)`
  4. `67.4°`
  5. `a = 16, b = 144`
Show Worked Solution

a.   `text(Corner point occurs when)\ \ x=sqrt2.`

`y=3 sin^(-1) (sqrt2/2) = (3pi)/4`

`:.\ text(Coordinates are:)\ \ (sqrt2, (3pi)/4)`
 

b.    `text(Reflect in the)\ xtext(-axis and then the)\ ytext(-axis:)`

♦ Mean mark 49%.

`g(x){(-3text(arccos)(- x/2)text(,), -2 <= x < -sqrt2),(-3text(arcsin)(- x/2)text(,), -sqrt2 <= x <= 0):}`

 

c.    `A` `= 4 xx (int_0^sqrt2 3sin^(−1)(x/2)dx + int_sqrt2^2 3cos^(−1)(x/2)dx)`
    `~~ 9.9\ text(cm²)`

 

d.   `text(Find the gradient of graph at)\ \ x=0:`

`f′(0) = 3/2`

`alpha = tan^(−1)(3/2) = 56.31…°`

`beta = pi/2 – tan^(−1)(1.5) = 33.69°`


 
`:.\ text(Acute angle between the edges)`

`=2 xx 33.69`

`=67.4°`
  

e.  `f′(x)\ {(3/sqrt(4-x^2)text(,), 0<= x <= sqrt2),((-3)/sqrt(4-x^2)text(,), sqrt2 < x <= 2):}`
 

`:.\ text(Length of border)`

♦♦ Mean mark 27%.

`= 4 int_0^sqrt2 sqrt(1 + (3/sqrt(4 – x^2))^2)\ dx + 4 int_sqrt2^2 sqrt(1 + ((-3)/sqrt(4 – x^2))^2)\ dx`

`= 4 int_0^2 sqrt(1 + (3/sqrt(4 – x^2))^2)\ dx`

`= int_0^2 sqrt(16 + 144/(4 – x^2)\ dx`

  
`:. a=16, \ b=144`

Complex Numbers, SPEC2-NHT 2017 VCAA 2

One root of a quadratic equation with real coefficients is  `sqrt 3 + i`.

    1. Write down the other root of the quadratic equation.  (1 mark)
    2. Hence determine the quadratic equation, writing it in the form  `z^2 + bz + c = 0`.  (2 marks)
  1. Plot and label the roots of   `z^3 - 2 sqrt 3 z^2 + 4z = 0`  on the Argand diagram below.  (3 marks)

  

 

  1. Find the equation of the line that is the perpendicular bisector of the line segment joining the origin and the point  `sqrt 3 + i`. Express your answer in the form  `y = mx + c`.  (2 marks)
  1. The three roots plotted in part b. lie on a circle.

     

    Find the equation of this circle, expressing it in the form  `|z - alpha| = beta`,  where  `alpha, beta in R`.  (3 marks)

Show Answers Only
    1. `z_2 = sqrt 3 – i`
    2. `z^2 – 2 sqrt 3 z + 4 = 0`
  2. `text(See Worked Solutions)`
  3. `y = -sqrt 3 x + 2`
  4. `|z – 2/sqrt 3 | = 2/sqrt 3`
Show Worked Solution

a.i.   `z_1 = sqrt 3 + i`

 `z_2 = bar z_1 = sqrt 3 – i\ \ \ text{(conjugate root)}`

 

a.ii.    `(z – (sqrt 3 + i))(z – (sqrt 3 – i))` `= 0`
  `((z – sqrt 3) – i)((z – sqrt 3) + i)` `= 0`
  `(z – sqrt 3)^2 – i^2` `= 0`
  `z^2 – 2 sqrt 3 z + 3 + 1` `= 0`
  `z^2 – 2 sqrt 3 z + 4` `= 0`

 

b.    `z(z^2 – 2 sqrt 3 z + 4)` `= 0`
  `z(z – (sqrt 3 + i))(z – (sqrt 3 – i))` `= 0`

 
`text(Convert to polar form:)`

  `sqrt 3 + i` `= sqrt((sqrt 3)^2 + 1^2) * text(cis) (tan^(-1)(1/sqrt 3))`
    `= 2 text(cis) (pi/6)`
  `=> sqrt 3 – i` `= 2 text(cis) (-pi/6)`

 

   

 

c.   `text(Equidistant from)\ (0, 0) and (sqrt 3, 1)`

`text(Midpoint)\ (x_1, y_1)` `= (sqrt 3/2, 1/2)`

 
`m = (1 – 0)/(sqrt 3 – 0) = 1/sqrt 3`

`m_(_|_) = (-1)/m = -sqrt 3`
 

`:.\ text(Equation of ⊥ bisector:)`

`y-1/2` `=-sqrt3(x-sqrt3/2)`  
`y` `=-sqrt3 x +3/2+1/2`  
`:.y` `=-sqrt3 x +2`  

 

d.   `text(Let)\ O = (0, 0),\ P = (sqrt 3, 1),\ Q = (sqrt 3, -1)`

`text(⊥ bisector of two points on arc of a circle passes)`

`text(through the centre of the circle.)`
 

`OP = OQ = PQ = 2`

`=> Delta OPQ\ text(is equilateral)`
 

`text(⊥ bisector of)\ PQ\ text(is)\ y=0.`

`text(Centre of circle occurs when:)`

`0 = -sqrt 3 x + 2\ \ text{(using part c)`

`x=2/sqrt3`

`=>\ text(Radius)\ = 2/sqrt3`
 

`:. |z – 2/sqrt 3| = 2/sqrt 3`

Calculus, SPEC2-NHT 2017 VCAA 1

  1. i.  Use an appropriate double angle formula with  `t = tan((5 pi)/12)`  to deduce a quadratic equation of the form  `t^2 + bt + c = 0`, where `b` and `c` are real values.  (2 marks)
  2. ii. Hence show that  `tan((5 pi)/12) = 2 + sqrt 3`.  (1 mark)

Consider  `f: [sqrt 3, 6 + 3 sqrt 3] -> R,\ \ f(x) = arctan (x/3) - pi/6`.

  1. Sketch the graph of  `f`  on the axes below, labelling the end points with their coordinates.  (3 marks)


 

  1. The region between the graph of  `f`  and the `y`-axis is rotated about the `y`-axis to form a solid of revolution.
  2. i.  Write down a definite integral in terms of  `y`  that gives the volume of the solid formed.  (2 marks)
  3. ii. Find the volume of the solid, correct to the nearest integer.  (1 mark)
  1. A fish pond that has a shape approximately like that of the solid of revolution in part c. is being filled with water. When the depth is `h` metres, the volume, `V\ text(m)^3`, of water in the pond is given by

     

    `qquad V = tan(h + pi/6) - h - sqrt 3/3`

     

    If water is flowing into the pond at a rate of 0.03 m³ per minute, find the rate at which the depth is increasing when the depth is 0.6 m. Give your answer in metres per minute, correct to three decimal places.  (3 marks)

Show Answers Only
  1. i.  `t^2 – 2t sqrt 3 – 1 = 0`
  2. ii. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
  4. i.  `V = pi int_0^(pi/4) 9 tan^2 (y + pi/6)\ dy`
  5. ii. `67`
  6. `0.007\ text(m/min)`
Show Worked Solution

a.i.   `tan (theta/2) = tan ((5pi)/12) = t`

`tan (theta)` `=(2 tan (theta/2))/(1 – tan^2(theta/2)`
`tan((5 pi)/6)` `= (2tan ((5pi)/12))/(1 – tan^2((5pi)/2)`
`-1/sqrt 3` `= (2t)/(1 – t^2), quad (t != +- 1)`
`-(1 – t^2)` `= 2t sqrt 3`
`-1 + t^2` `= 2t sqrt 3`

 
`:. t^2 – 2 sqrt 3 t – 1 = 0`

 

a.ii.    `t^2 – 2 sqrt 3 t + (-sqrt 3)^2 +3 -4 = 0`
  `(t – sqrt 3)^2` `= 4`
  `t – sqrt 3` `= +- 2`
  `t` `= sqrt 3 +-2`

 
`=> tan ((5 pi)/12)\ \ text(is in 1st quadrant,)`

`:. t` `= tan ((5 pi)/12) = 2 + sqrt 3`

 

b.    `f(sqrt 3)` `= 0`
`f(6 + 3 sqrt 3)` `=pi/4`

`text(Graphing)\ \ f(x) = arctan (x/3) – pi/6\ \ text(on CAS will)`

`text(show the shape of the graph.)`
 

 

c.i.    `y` `= tan^(-1)(x/3) – pi/6`
  `y + pi/6` `= tan^(-1)(x/3)`
  `x/3` `= tan(y + pi/6)`
  `x` `= 3 tan (y + pi/6)`
  `x^2` `= 9 tan^2 (y + pi/6)`

 

`:. V` `= pi int_0^(pi/4) x^2\ dy`  
  `=pi int_0^(pi/4) 9 tan^2 (y + pi/6)\ dy`  

 

c.ii.  `pi int_0^(pi/4) 9 tan^2 (y + pi/6)\ dy`

`=66.99…`

`=67\ text(u³)`

 

d.    `(dV)/(dt)` `= 0.03`
  `V` `= tan(h + pi/6) – h – sqrt 3/3`
  `(dV)/(dh)` `= tan^2 (h+pi/6)\ \ \ text{(by CAS)}`
  `(dh)/(dt)` `= (dh)/(dV) * (dV)/(dt)`
    `= 1/(tan^2 (h+pi/6)) xx 3/100`

  
`(dh)/(dt)|_(h = 0.6)~~0.007\ text(m/min)`

Statistics, SPEC2-NHT 2018 VCAA 7

According to medical records, the blood pressure of the general population of males aged 35 to 45 years is normally distributed with a mean of 128 and a standard deviation of 14. Researchers suggested that male teachers had higher blood pressures than the general population of males.

To investigate this, a random sample of 49 male teachers from this age group was obtained and found to have a mean blood pressure of 133.

  1. State two hypotheses and perform a statistical test at the 5% level to determine if male teachers belonging to the 35 to 45 years age group have higher blood pressures than the general population of males. Clearly state your conclusion with a reason. (3 marks)
  2. Find a 90% confidence interval for the mean blood pressure of all male teachers aged 35 to 45 years using a standard deviation of 14. Give your answers correct to the nearest integer.  (1 mark)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(130, 136)`
Show Worked Solution

a.   `B\ ~\ N (128, 14^2)`

`H_0: mu = 128`

`H_1: mu > 128`

`bar B\ ~\ N (128, 14^2/49)`

`p` `= text(Pr) (bar B > 133 | mu=128)`
  `= text(Pr)(Z>(133-128)/(14/sqrt49))`
  `= text(Pr)(Z>5/2)`
  `~~ 0.00621`

 
`p < 0.05 => text(Reject)\ H_0\ text(at the 5% significance level):`

`text{evidence supports the contention that male teachers 35 to 45}`

`text(have higher blood pressure than the general male population.)`

 

b.  `text(If)\ \ text(Pr)(- z < Z < z) = 0.9`

`text(Pr)(Z < z) = 0.95\ \ =>\ \ z ~~ 1.64485`

`:.\ text(90% CI): (133 – (z xx 14)/sqrt 49, 133 + (z xx 14)/sqrt 49)`

`~~(130, 136)`

Statistics, SPEC2-NHT 2018 VCAA 6

A coffee machine dispenses coffee concentrate and hot water into a 200 mL cup to produce a long black coffee. The volume of coffee concentrate dispensed varies normally with a mean of 40 mL and a standard deviation of 1.6 mL.

Independent of the volume of coffee concentrate, the volume of water dispensed varies normally with a mean of 150 mL and a standard deviation of 6.3 mL.

  1. State the mean and the standard deviation, in millilitres, of the total volume of liquid dispensed to make a long black coffee.  (2 marks)
  2. Find the probability that a long black coffee dispensed by the machine overflows a 200 mL cup. Give your answer correct to three decimal places.  (1 mark)
  3. Suppose that the standard deviation of the volume of water dispensed by the machine can be adjusted, but that the mean volume of water dispensed and the standard deviation of the volume of coffee concentrate dispensed cannot be adjusted.
  4. Find the standard deviation of the volume of water dispensed that is needed for there to be only a 1% chance of a long black coffee overflowing a 200 mL cup. Give your answer in millilitres, correct to two decimal places.  (2 marks)
Show Answers Only
  1. `mu_V = 40 + 150 = 190\ text(mL)`

     

    `sigma_V = sqrt (1.6^2 + 6.3^2) = 6.5\ text(mL)`

  2. `0.062`
  3. `3.99\ text(mL)`
Show Worked Solution

a.   `C~N (40, 1.6^2)`

`W~N (150, 6.3^2)`

`V = C + W`

`mu_V = 40 + 150 = 190\ text(mL)`

`{:sigma^2:}_V` `= 1.6^2 + 6.3^2`  
`:.sigma_V ` `= sqrt (1.6^2 + 6.3^2)`  
  `= 6.5\ text(mL)`  

 

b.   `V~N (190, 6.5^2)`

`text(Pr)(V > 200) ~~ 0.062\ \ text{(by CAS)}`

 

c.  `C~N (40, 1.6^2)`

`W_2~N (150, sigma_w^2)`

`V_2 = C + W_2`

`V_2~N (190, 1.6^2 + sigma_w^2)`

`Z~N (0, 1)`
 

`text(Pr)(V_2 > 200) = 0.01`

`text(Pr)(Z > a) = 0.01\ \ =>\ \ a=2.326…`
  

`text(Find)\ \ sigma_w:`

`2.326…` `= (200-190)/sqrt(1.6^2 + sigma_w^2)`
`:. sigma_w` `~~ 3.99\ text(mL)\ \ \ text{(by CAS)}`

Calculus, SPEC2-NHT 2018 VCAA 5

A horizontal beam is supported at its endpoints, which are 2 m apart. The deflection `y` metres of the beam measured downwards at a distance `x` metres from the support at the origin `O` is given by the differential equation  `80 (d^2y)/(dx^2) = 3x - 4`.
 


 

  1. Given that both the inclination, `(dy)/(dx)`, and the deflection, `y`, of the beam from the horizontal at  `x = 2`  are zero, use the differential equation above to show that  `80 y = 1/2 x^3 - 2x^2 + 2x`.  (2 marks)
  2. Find the angle of inclination of the beam to the horizontal at the origin `O`. Give your answer as a positive acute angle in degrees, correct to one decimal place.  (2 marks)
  3. Find the value of `x`, in metres, where the maximum deflection occurs, and find the maximum deflection, in metres.  (3 marks)
  4. Find the maximum angle of inclination of the beam to the horizontal in the part of the beam where  `x >= 1`. Give your answer as a positive acute angle in degrees, correct to one decimal place.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.4^@`
  3. `x = 2/3; quad 1/135\ text(m)`
  4. `0.5^@`
Show Worked Solution
a.   `80 * int (d^2y)/(dx^2)\ dx` `= int 3x – 4\ dx`
  `80* (dy)/(dx)` `= (3x^2)/2 – 4x + c_0`

 
`text(When)\ \ x=2,\ \ dy/dx=0:`

`80 xx 0` `= (3xx2^2)/2 – 4xx2 + c_0`
`c_0` `= 2`

 

`80* (dy)/(dx)` `= (3x^2)/2 – 4x + 2`
`80y` `= int (3x^2)/2 – 4x + 2\ dx`
  `= 1/2 x^3 – 2x^2+2x +c_1`

 

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

`0` `= 1/2 2^3 – 2 xx 2^2 + 2 xx 2 + c_1`
`c_1` `= 0`

`:. 80y = 1/2x^3 – 2x^2 + 2x`

 

b.   `text(Find)\ \ dy/dx\ \ text(when)\ \ x=0:`

  `80 *(dy)/(dx)` `= 0 – 0 + 2`
  `(dy)/(dx)` `= 1/40`
  `:. theta` `= tan^(-1) (1/40) ~~ 1.4^@`

 

c.   `text(Find)\ \ x\ \ text(when)\ \ dy/dx=0:`

`x=2\ \ text(or)\ \ 2/3`

`=>y_text(max)\ \ text(occurs at)\ \ x=2/3\ \ (y=0\ \ text(at)\ \ x=2)`

 

`80*y_max` `= 1/2(2/3)^3 – 2(2/3)^2 + 2(2/3)`
`:. y_max` `= 1/80 xx 16/27`
  `= 1/135\ text(metres)`

 

d.   `text(Max inclination when convexity changes)\ => (d^2y)/(dx^2) =0`

`80*(d^2y)/(dx^2) = 3x – 4 = 0`

`=> x = 4/3`

`text(Find)\ \ dy/dx\ \ text(when)\ \ x = 4/3 :`

`(dy)/(dx)` `= 1/80 (3/2(4/3)^2 – 4(4/3) + 2)`
  `= (-1)/120`

 

`theta` `= tan^(-1) (|-1/120|)`
  `~~ 0.5^@`

Vectors, SPEC2-NHT 2018 VCAA 4

A basketball player aims to throw a basketball through a ring, the centre of which is at a horizontal distance of 4.5 m from the point of release of the ball and 3 m above floor level. The ball is released at a height of 1.75 m above floor level, at an angle of projection `alpha` to the horizontal and at a speed of  `V\ text(ms)^(-1)`. Air resistance is assumed to be negligible.
 


 

The position vector of the centre of the ball at any time, `t` seconds, for  `t >= 0`, relative to the point of release is given by 
 
`qquad underset ~r(t) = Vt cos (alpha) underset ~i + (Vt sin(alpha) - 4.9t^2) underset ~j`,
 
where  `underset ~i`  is a unit vector in the horizontal direction of motion of the ball and  `underset ~j`  is a unit vector vertically up. Displacement components are measured in metres.

  1. For the player’s first shot at goal, `V = 7\ text(ms)^(-1)` and  `alpha = 45^@`
  2.   i. Find the time, in seconds, taken for the ball to reach its maximum height. Give your answer in the form  `(a sqrt b)/c`, where  `a, b` and `c` are positive integers.  (2 marks) 
  3.  ii. Find the maximum height, in metres, above floor level, reached by the centre of the ball.  (2 marks)
  4. iii. Find the distance of the centre of the ball from the centre of the ring one second after release. Give your answer in metres, correct to two decimal places.  (2 marks)
  5. For the player’s second shot at goal, `V = 10\ text(ms)^(-1)`.
    Find the possible angles of projection, `alpha` , for the centre of the ball to pass through the centre of the ring. Give your answers in degrees, correct to one decimal place.  (3 marks)
  6. For the player’s third shot at goal, the angle of projection is  `alpha = 60^@`
  7. Find the speed `V` required for the centre of the ball to pass through the centre of the ring. Give your answer in metres per second, correct to one decimal place.  (2 marks)
Show Answers Only
    1. `(5 sqrt 2)/14`
    2. `3\ text(m)`
    3. `128\ text(m)`
  2. `alpha ~~ 29.7^@ or 75.8^@`
  3. `7.8\ text(ms)^(-1)`
Show Worked Solution
a.i.    `underset ~r(t)` `=7t\ cos 45^@ underset ~i+(7t\ sin 45^@ – 4.9t^2) underset ~j`
    `=(7sqrt2t)/2 underset ~i + ((7sqrt2)/2 t – 4.9t^2) underset ~j`
     

 `text(Maximum height when)\ \ underset ~V_(underset ~j) (t) = 0:`

`(7sqrt2)/2 – 9.8t` `= 0`
`t` `= (5 sqrt 2)/14\ \ text(seconds)`

 

a.ii.   `underset ~r_(underset ~j)((5 sqrt 2)/14)` `= 7 xx (5 sqrt 2)/14 xx 1/sqrt 2 – 4.9 xx ((5 sqrt 2)/14)^2`
    `= 1.25`

 
`:.\ text(Height above floor) = 1.25 + 1.75 = 3\ text(m)`

 

a.iii.  `underset ~r_text(ring)= 4.5 underset ~i + 1.25 underset ~j`

  `underset ~r_text(ring) – underset ~r_text(ball)` `= (4.5 – 7/sqrt 2) underset ~i + (1.25 – ((7sqrt2)/2 – 4.9)underset ~j`
  `d` `= |underset ~r_text(ring) – underset ~r(1)|`
    `= sqrt((4.5 – 7/sqrt 2)^2 + (1.25 – 7/sqrt 2 + 4.9)^2)`
    `~~ 1.28\ text(m)`

 

b.   `underset ~r(t) = 10 t cos (alpha) underset ~i + (10t sin (alpha) – 4.9t^2) underset ~j`

`=> 10t cos (alpha) = 4.5`

`=>10t sin (alpha) – 4.9t^2 = 1.25`
 

`:. alpha ~~ 29.7^@ or 75.8^@\ \ (text{solve by CAS for}\ \ 0<=alpha<=90)`

 

c.   `underset ~r(t)=Vt cos (60^@)underset ~i + (Vt sin (60^@) – 4.9t^2) underset ~j`

`Vt cos (60^@)` `=4.5`  
`(Vt)/2` `= 9/2`  
`=> Vt` `=9`  

 
`=> (Vt sqrt 3)/2 – 49t^2 = 1.25`
 

`:. V~~ 7.8\ text(ms)^(-1)\ \ \ text{(solve simultaneously by CAS)}`

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)

 

 

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)
  2. Water is poured into the vase at a rate of 20 cm³ s¯¹.
  3. 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)
  4. 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.

     

    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)`
    1.  `V = pi int_0^(10 pi) (1 – 1/2 cos (y/10))^2 dy`
    2. `V = (45 pi^2)/4 text(cm)^3`
  2. `20/pi text(cm s)^(-1)`
  3. `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. By expressing the circle given by  `|z + 1| = sqrt 2 |z - i|`  in cartesian form, show that this circle has the same centre and radius as the circle given by  `|z - (1 + 2i)| = 2`.  (2 marks)
  3. 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 prime(x)`  for  `x > 0`, expressing your answer in the form  `f prime(x) = A/sqrt(2 - x^2), \ A in R`.  (1 mark)
  4. Write down  `f prime(x)`  for  `x < 0`, expressing your answer in the form  `f prime(x) = B/sqrt(2 - x^2), \ B in R`.  (1 mark)
  5. The derivative  `f prime(x)`  can be expressed in the form  `f prime(x) = {g(x)}/sqrt(2 - x^2)` over its maximal domain.
  1. Find the maximal domain of  `f prime`.  (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 prime(x)= 4/sqrt(2 – x^2)\ \ \ (x > 0,\ \ text(by CAS))`

 

d.    `f prime(x)` `= (4x)/sqrt(x^2(2 – x^2))`
    `= (-4)/sqrt(2 – x^2)\ \ \ (x < 0)`

 

e.i.  `f prime(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)`