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Mechanics, SPEC2 2019 VCAA 5

A mass of  `m_1`  kilograms is initially held at rest near the bottom of a smooth plane inclined at `theta` degrees to the horizontal. It is connected to a mass of  `m_2`  kilograms by a light inextensible string parallel to the plane, which passes over a smooth pulley at the end of the plane. The mass  `m_2`  is 2 m above the horizontal floor.

The situation is shown in the diagram below.
 


 

  1. After the mass  `m_1`  is released, the following forces, measured in newtons, act on the system:

     

    • weight forces  `W_1`  and  `W_2`
    •  the normal reaction force  `N`
    •  the tension in the string  `T`

     

    On the diagram above, show and clearly label the forces acting on each of the masses.  (1 mark) 

  2. If the system remains in equilibrium after the mass  `m_1`  is released, show that  `sin(theta) = (m_2)/(m_1)`.  (1 mark)
  3. After the mass  `m_1`  is released, the mass  `m_2`  falls to the floor.

     

    1. For what values of  `theta`  will this occur? Express your answer as an inequality in terms of  `m_1`  and  `m_2`.  (1 mark)
    2. Find the magnitude of acceleration, in ms−2, of the system after the mass  `m_1`  is released and before the mass  `m_2`  hits the floor. Express your answer in terms of  `m_1, \ m_2`  and  `theta`.  (2 marks)
  4. After the mass  `m_1`  is released, it moves up the plane.
    Find the maximum distance, in metres, that the mass  `m_1`  will move up the plane if  `m_1 = 2m_2`  and  `sin(theta) = 1/4`.  (5 marks)
Show Answers Only
  1.   
  2. `text(See Worked Solutions)`
    1. `theta < sin^(−1)\ ((m_2)/(m_1)), theta ∈ (0, pi/2)`
    2. `a = (g(m_2 – m_1 · sin(theta)))/(m_1 + m_2)`
  3. `10/3 \ text(m)`
Show Worked Solution
a.   

 

b.   `T = m_2g\ \ … \ (1)`

`T = m_1sin(theta)\ \ … \ (2)`

`text{Solve:  (1) = (2)}`

`m_1g sin(theta)` `= m_2g`
`sin(theta)` `= (m_2)/(m_1)`

 

c.i.   `m_2g > m_1gsin(theta)`

`sin(theta) < (m_2)/(m_1)`

`theta < sin^(−1)\ ((m_2)/(m_1)), \ \ theta ∈ (0, pi/2)`

 

c.ii.   `text(Net force)\ (F) = m_2g – m_1gsin(theta)`

`(m_1 + m_2) · a` `= g(m_2 – m_1 sin(theta))`
`:. a` `= (g(m_2 – m_1 sin(theta)))/(m_1 + m_2)`

 

d.   `text(Motion:)\ m_1\ text(will accelerate up the plane for 2 m.)`

 `m_1\ text(will then decelerate up plane until)\ v = 0.`

`text(Find)\ v_(m_1)\ \ text(given)\ \ s_(m_1) = 2, \ u = 0, \ m_1=2m_2`

`a = (g(m_2 – m_1  sintheta))/(m_1 + m_2) = (g(m_2 – 2m_2 · 1/4))/(2m_2 + m_2) = g/6`
 

`text(Using)\ \ v^2 = u^2 + 2as,`

`v_(m_1)^2 = 0 + 2 · g/6 · 2 = (2g)/3`

`text(Find distance)\ (s_2)\ text(for)\ m_1\ text(to decelerate until)\ v = 0:`

`a = −gsintheta = −g/4, \ u = sqrt((2g)/3)`

`0` `= (2g)/3 – 2 · g/4 · s_2`
`s_2` `= (2g)/3 xx 2/g`
  `= 4/3`

 

`:.\ text(Maximum distance)` `= 2 + 4/3`
  `= 10/3\ text(m)`

Vectors, SPEC2 2019 VCAA 4

The base of a pyramid is the parallelogram  `ABCD`  with vertices at points  `A(2,−1,3),  B(4,−2,1),  C(a,b,c)`  and  `D(4,3,−1)`. The apex (top) of the pyramid is located at  `P(4,−4,9)`.

  1. Find the values of  `a, b`  and  `c`.  (2 marks)

  2. Find the cosine of the angle between the vectors  `overset(->)(AB)`  and  `overset(->)(AD)`.  (2 marks)

  3. Find the area of the base of the pyramid.  (2 marks)

  4. Show that  `6underset~i + 2underset~j + 5underset~k`  is perpendicular to both  `overset(->)(AB)`  and  `overset(->)(AD)`, and hence find a unit vector that is perpendicular to the base of the pyramid.  (3 marks)

  5. Find the volume of the pyramid.  (2 marks)

Show Answers Only

  1. `a = 6, b = 2, c = −3`
  2. `4/9`
  3. `2sqrt65 u^2`
  4. `text(See Worked Solutions)`
  5. `24\ text(u³)`

Show Worked Solution

a.    `overset(->)(AB)` `= (4-2)underset~i + (−2 + 1)underset~j + (1-3)underset~k`
    `= 2underset~i-underset~j-2underset~k`

 
`text(S)text(ince)\ ABCD\ text(is a parallelogram)\ => \ overset(->)(AB)= overset(->)(DC)`

`overset(->)(DC) = (a-4)underset~i + (b-3)underset~j + (c + 1)underset~k`

`a-4 = 2 \ => \ a = 6`

`b-3 = −1 \ => \ b = 2`

`c + 1 = −2 \ => \ c = −3`

 

b.   `overset(->)(AB) = 2underset~i-underset~j-2underset~k`

`overset(->)(AD) = 2underset~i + 4underset~j-4underset~k`

`cos angleBAD` `= (overset(->)(AB) · overset(->)(AD))/(|overset(->)(AB)| · |overset(->)(AD)|)`
  `= (4-4 + 8)/(sqrt(4 + 1 + 4) · sqrt(4 + 16 + 16))`
  `= 4/9`

 

c.    `1/2 xx text(Area)_(ABCD)` `= 1/2 ab sin c`
  `text(Area)_(ABCD)` `= |overset(->)(AB)| · |overset(->)(AD)| *sin(cos^(−1)\ 4/9)`

 


 

`:. text(Area)_(ABCD)` `= 3 · 6 · sqrt65/9`
  `= 2sqrt65\ text(u²)`

 

d.   `text(Let)\ \ underset~r = 6underset~i + 2underset~j + 5underset~k`

`underset~r · overset(->)(AB)` `= 6 xx 2 + 2 xx −1 + 5 xx −2 = 0`
`underset~r · overset(->)(AD)` `= 6 xx 2 + 2 xx 4 + 5 xx −2 = 0`

 
`:. underset~r\ \ text(is ⊥ to)\ \ overset(->)(AB)\ \ text(and)\ \ overset(->)(AD)`

`text(Let)\ \ hatr\ = text(unit vector ⊥ to pyramid base)`

`underset~overset^r = 1/sqrt(6^2 + 2^2 + 5^2) *underset~r = 1/sqrt65(6underset~i + 2underset~j + 5underset~k)`

 

e.   `text(Find height)\ (h)\ text(of pyramid:)`

`overset(->)(AP)` `= (4-2)underset~i + (−4 + 1)underset~j + (9-3)underset~k`
  `= 2underset~i-3underset~j + 6underset~k`

 

`h` `= overset(->)(AP) · underset~overset^r`
  `= (2 xx 6/sqrt65)-(3 xx 2/sqrt65) + (6 xx 5/sqrt65)`
  `= 36/sqrt65`

 

`:. V` `= 1/3 b xx h`
  `= 1/3 xx 2sqrt65 xx 36/sqrt65`
  `= 24\ text(u³)`

Calculus, SPEC2 2019 VCAA 3

  1. The growth and decay of a quantity `P` with respect to time `t` is modelled by the differential equation
     
    `qquad qquad(dP)/(dt) = kP`
     
    where `t >= 0`.

    1. Given that  `P(a) = r`  and  `P(b) = s`, where `P` is a function of `t`,

       

      show that  `k = 1/(a-b)log_e(r/s)`.   (2 marks)

    2. Specify the condition(s) for which  `k >0`.   (2 marks)

  2. The growth of another quantity `Q` with respect to time `t` is modelled by the differential equation
     
       `qquad qquad (dQ)/(dt) = e^(t-Q)`
      
    where  `t >= 0`  and  `Q = 1`  when  `t = 0`.

    1. Express this differential equation in the form  `int f(Q)\ dQ = int h(t)\ dt`.   (1 mark)

    2. Hence, show that  `Q = log_e(e^t + e-1)`.   (2 marks)

    3. Show that the graph of `Q` as a function of `t` does not have a point of inflection.   (2 marks)

Show Answers Only
    1. `text(See Worked Solutions)`
    2. `a-b\ text(and)\ s > r > 0`
    1. `int e^Q\ dQ =  int e^t\ dt`
    2. `text(See Worked Solutions)`
    3. `text(See Worked Solutions)`
Show Worked Solution
a.i.    `(dP)/(dt)` `= kP`
  `(dt)/(dP)` `= 1/(kP)`

`t = 1/k int 1/P\ dP = 1/k log_e P + c`

`P(a) = r`

`a = 1/k · log_e r + c`

`P(b) = r`

`b = 1/k · log_e s + c`

`a-b` `= 1/k(log_e r + c-log_e s + c)`
`:.k` `= 1/(a-b)log_e (r/s)`

 

a.ii.   `text(Conditions for)\ \ k > 0,`

`text(Scenario 1:)`

`a-b > 0\ \ text(and)\ \ log_e(r/s) > 0`

`=> a > b\ \ text(and)\ \ r > s > 0`
 

`text(Scenario 2:)`

`a-b < 0\ \ text(and)\ \ log_e(r/s) < 0`

`=> a < b\ \ text(and)\ \ s > r > 0`
 

b.i.    `(dQ)/(dt)` `= (e^t)/(e^Q)`
  `int e^Q dQ` `= int e^t\ dt`

 

b.ii.    `int e^Q dQ` `= int e^t dt`
  `e^Q` `= e^t + c`

 
`text(When)\ \ Q = 1,  t = 0`

`e` `= e^0 + c`
`c` `= e-1`
`e^Q` `= e^t + e-1`

 
`:. Q = log_e(e^t + e-1)`

 

b.iii.   `Q = log_e(e^t + e-1)`

`(dQ)/(dt)` `= (e^t)/(e^t + e-1)`
`(d^2Q)/(dt^2)` `= (e^t(e-1))/((e^t + e-1)^2)`

 
`(d^2Q)/(dt^2) > 0\ text(for)\ t >= 0`

`:.\ text(No POI)`

Complex Numbers, SPEC2 2019 VCAA 2

  1. Show that the solutions of  `2z^2 + 4z + 5 = 0`, where  `z ∈ C`, are  `z = −1 ± sqrt6/2 i`.   (1 mark)

  2. Plot the solutions of  `2z^2 + 4z + 5 = 0`  on the Argand diagram below.   (1 mark)

     

     

Let  `|z + m| = n`, where  `m, n ∈ R`, represent the circle of minimum radius that passes through the solutions of  `2z^2 + 4z + 5 = 0`.

    1. Find  `m`  and  `n`.   (2 marks)

    2. Find the cartesian equation of the circle  `|z + m| = n`.   (1 mark)

    3. Sketch the circle on the Argand diagram in part a.ii. Intercepts with the coordinate axes do not need to be calculated or labelled.   (1 mark)

  2. Find all values of  `d`, where  `d ∈ R`, for which the solutions of  `2z^2 + 4z + d = 0`  satisfy the relation  `|z + m| <= n`.   (2 marks)

  3. All complex solutions of  `az^2 + bz + c = 0`  have non-zero real and imaginary parts.

     

    Let  `|z + p| = q`  represent the circle of minimum radius in the complex plane that passes through these solutions, where  `a, b, c, p, q ∈ R`.

     

    Find  `p`  and  `q`  in terms of  `a, b`  and  `c`.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2.   
  3. `m = 1, n = sqrt6/2`
  4. `(x + 1)^2 + y^2 = 3/2`
  5. `−1 <= d <= 5\ \ (text(by CAS))`
  6. `p = b/(2a), \ q = |sqrt(b^2-4ac)/(2a)|`
Show Worked Solution
a.i.    `z` `= (−b ± sqrt(b^2-4ac))/(2a)`
    `= (−4 ± sqrt(16-4 · 2 · 5))/(4)`
    `= (−4 ± 2sqrt6 i)/(4)`
    `= −1 ± sqrt6/2 i\ \ …\ text(as required)`

 

a.ii.   

 

b.i.   `text(Radius of circle = )sqrt6/2`

 `text(Centre) = (0, −1)`

`:. m = 1, \ n = sqrt6/2`

 

b.ii.    `|z + 1|` `= sqrt6/2`
  `|x + iy + 1|` `= sqrt6/2`
  `(x + 1)^2 + y^2` `= 3/2`

 

b.iii.   

 

c.   `text(Solve:)\ 2z^2 + 4z + d = 0`

`z = −1 ± sqrt(4-2d)/2 = −1 ± sqrt((2-d)/2)`

`z + 1 = ± sqrt((2-d)/2)`
 

`text(Solve for)\ d\ text(such that:)`

`|sqrt((2-d)/2)| <= sqrt6/2`

`−1 <= d <= 5\ \ (text(by CAS))`

 

d.   `z = (−b ± sqrt(b^2-4ac))/(2a) = (−b)/(2a) ± sqrt(b^2-4ac)/(2a)`

`z + b/(2a)` `= ± sqrt(b^2-4ac)/(2a)`
`|z + b/(2a)|` `= |sqrt(b^2-4ac)/(2a)|`

 
`:. p = b/(2a), \ q = |sqrt(b^2-4ac)/(2a)|`

Calculus, SPEC2 2019 VCAA 1

A curve is defined parametrically by  `x = sec(t) + 1, \ y = tan(t)`, where  `t ∈ [0, pi/2)`.

  1. Show that the curve can be represent in cartesian form by the rule  `y = sqrt(x^2-2x)`.   (2 marks)

  2. State the domain and range of the relation given by  `y = sqrt(x^2-2x)`.  (2 marks)

  3.  i. Express  `(dy)/(dx)`  in terms of  `sin(t)`.   (2 marks)

  4. ii. State the limiting value of  `(dy)/(dx)`  as  `t`  approaches  `pi/2`.   (1 mark)

  1. Sketch the curve  `y = sqrt(x^2-2x)`  on the axes below for  `x ∈ [2, 4]`, labelling the endpoints with their coordinates.   (2 marks)

     

     

  2. The portion of the curve given by  `y = sqrt(x^2-2x)`  for  `x ∈ [2, 4]`  is rotated about the `y`-axis to form a solid of revolution.
  3. Write down, but do not evaluate, a definite integral in terms of  `t`  that gives the volume of the solid formed.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Domain:)\ x ∈ [2, ∞)`

     

    `text(Range:)\ y ∈ [0, ∞)`

    1. `(dy)/(dx) = ((dy)/(dt))/((dt)/(dx)) = (sec^2(t))/(sin(t)sec^2(t)) = 1/(sin(t))`
    2. `(dy)/(dx) -> 1`
  4.  
  5. ` V = pi int_0^(tan^(−1)(2sqrt2)) (sec(t) + 1)^2sec^2(t)dt`
Show Worked Solution

a.   `x = sec(t) + 1 \ => \ sec(t) = x-1`

`y = tan(t)`

`text(Using)\ \ tan^2(t) + 1 = sec^2(t):`

`y^2 + 1` `= (x-1)^2`
`y^2 + 1` `= x^2-2x + 1`
`y^2` `= x^2-2x`
`y` `= sqrt(x^2-2x), \ y >= 0\ \ text(as)\ \ t ∈ [0, pi/2)`

 

b.   `text(Sketch:)\ \ x = sec(t) + 1, \ y = tan(t)\ \ text(for)\ \ t ∈ [0, pi/2)`

`text(Domain:)\ \ x ∈ [2, ∞)`

`text(Range:)\ \ y ∈ [0, ∞)`

 

c.i.   `(dy)/(dt) = sec^2(t), \ (dx)/(dt) = sin(t)sec^2(t)\ \ \ (text(by CAS))`

`(dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (sec^2(t))/(sin(t)sec^2(t)) = 1/(sin(t))`

 

c.ii.   `text(As)\ \ t -> pi/2:`

`(dy)/(dx) -> 1`

 

d.   

 

e.   `V = pi int_0^(2sqrt2) x^2\ dy`

`x^2 = (sec(t) + 1)^2`

`(dy)/(dt) = sec^2(t) \ => \ dy = sec^2(t)\ dt`

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

`text(When)\ y = 2sqrt2, t = tan^(−1)(2sqrt2)`
 

`:. V = pi int_0^(tan^(−1)(2sqrt2)) (sec(t) + 1)^2sec^2(t)\ dt`

Statistics, SPEC2 2019 VCAA 19 MC

`X` and `Y` are independent random variables where each has a mean of 4 and a variance of 9.

If the random variable  `Z = aX + bY`  has a mean of 8 and a variance of 90, possible values of `a` and `b` are

  1. `a = 1, \ b = 1`
  2. `a = 4, \ b = −2`
  3. `a = 3, \ b = −1`
  4. `a = 1, \ b = 3`
  5. `a = −2, \ b = 4`
Show Answers Only

`C`

Show Worked Solution
`E(aX + bY)` `= aE(X) + bE(Y)`
`8` `= 4a + 4b\ \ …\ (1)`
`text(Var)(aX + bY)` `= a^2text(Var)(X) + b^2text(Var)(Y)`
`9a^2 + 9b^2` `= 90\ \ …\ (2)`

 
`text{Solve simultaneous equations (by CAS):}`

`a = 3, b = −1\ \ text(or)\ \ a = −1, b = 3`

`=>C`

Statistics, SPEC2 2019 VCAA 18 MC

The masses of a random sample of 36 track athletes have a mean of 65 kg. The standard deviation of the masses of all track athletes is known to be 4 kg.

A 98% confidence interval for the mean of the masses of all track athletes, correct to one decimal place, would be closest to

  1. (51.0, 79.0)
  2. (63.6, 66.4)
  3. (63.3, 66.7)
  4. (63.4, 66.6)
  5. (64.3, 65.7)
Show Answers Only

`=>D`

Show Worked Solution

`text(Solution 1)`

`mu = 65, \ sigma = 4, \ n = 36`

`text(Confidence level 98%:)`

`(63.4, 66.6)\ \ \ text{(by CAS)}`

 

`text(Solution 2)`

`text(Find)\ z\ text(such that)\ \ text(Pr)(z < 2) = 0.99`

`z ~~ 2.3263,\ \ \ text{(by CAS)}`
 

`:.\ text(Confidence Interval)`

`= (65 – 2.3263 xx 4/sqrt36, 65 + 2.3263 xx 4/sqrt36)`

`= (63.4, 66.6)`

`=>D`

Mechanics, SPEC2 2019 VCAA 17 MC

A particle is held in equilibrium by three coplanar forces of magnitudes  `F_1, F_2`  and  `F_3`.

The angles between these forces are  `alpha, beta`  and  `gamma`  as shown in the diagram below.
 

If  `beta = 2alpha`, then  `(F_1)/(F_2)`  is equal to

  1. `1/2 sin(alpha)`
  2. `2sin(alpha)`
  3. `1/2text(cosec)(alpha)`
  4. `1/2cos(alpha)`
  5. `1/2sec(alpha)`
Show Answers Only

`E`

Show Worked Solution

`text(Using Lami’s theorem:)`

`(F_1)/(sin alpha)` `= (F_2)/(sinbeta)`
`(F_1)/(F_2)` `= (sin alpha)/(sin beta)`
  `= (sin alpha)/(sin 2alpha)`
  `= (sin alpha)/(2sin alphacos alpha)`
  `= 1/2 sec alpha`

 
`=>E`

Calculus, SPEC2 2019 VCAA 16 MC

A variable force acts on a particle, causing it to move in a straight line. At time `t` seconds, where  `t >= 0`, its velocity `v` metres per second and position `x` metres from the origin are such that  `v = e^x sin(x)`.

The acceleration of the particle, in ms−2, can be expressed as

  1. `e^(2x)(sin^2(x) + 1/2sin(2x))`
  2. `e^x sin(x)(sin(x) + cos(x))`
  3. `e^x(sin(x) + cos(x))`
  4. `1/2 e^(2x) sin^2(x)`
  5. `e^x cos(x)`
Show Answers Only

`A`

Show Worked Solution

`v = e^x sin(x)`

`(dv)/(dx) = e^x cos(x) + e^x sin(x)`

`v · (dv)/(dx)` `= e^x  sin(x)(e^x  cos(x) + e^x  sin(x))`
  `= e^(2x)(sin(x)cos(x) + sin^2(x))`
  `= e^(2x)(1/2 xx 2sin(x)cos(x) + sin^2(x))`
  `= e^(2x)(sin^2(x) + 1/2sin(2x))`

 
`=>A`

Mechanics, SPEC2 2019 VCAA 14 MC

A 4 kg mass is held at rest on a smooth surface. It is connected by a light inextensible string that passes over a smooth pulley to a 2 kg mass, which in turn is connected by the same type of string to a 1 kg mass. This is shown in the diagram below.
 


 

When the 4 kg mass is released, the tension in the string connecting the 1 kg and 2 kg masses is `T` newtons. The value of `T` is

  1. `(4g)/7`
  2. `(3g)/7`
  3. `g/7`
  4. `(6g)/7`
  5. `g`
Show Answers Only

`A`

Show Worked Solution

`text(Considering the whole system:)`

`text(Total mass = 7 kg)`

`text(Net Force ↓ =)\ 3g`

`F` `= ma`
`3g` `= 7a`
`:.a` `= (3g)/7`

 
`text(Consider the forces on the 1 kg mass:)`

`g – T` `= a`
`g – T` `= (3g)/7`
`T` `= g – (3g)/7`
  `= (4g)/7`

 
`=>A`

Calculus, MET2 2019 VCAA 5

Let  `f: R -> R, \ f(x) = 1-x^3`. The tangent to the graph of `f` at  `x = a`, where  `0 < a < 1`, intersects the graph of `f` again at `P` and intersects the horizontal axis at `Q`. The shaded regions shown in the diagram below are bounded by the graph of `f`, its tangent at  `x = a`  and the horizontal axis.
 

  1. Find the equation of the tangent to the graph of `f` at  `x = a`, in terms of `a`.   (1 mark)

  2. Find the `x`-coordinate of `Q`, in terms of `a`.   (1 mark)

  3. Find the `x`-coordinate of `P`, in terms of `a`.   (2 marks)

Let  `A`  be the function that determines the total area of the shaded regions.

  1. Find the rule of `A`, in terms of `a`.   (3 marks)

  2. Find the value of `a` for which `A` is a minimum.   (2 marks)

Consider the regions bounded by the graph of `f^(-1)`, the tangent to the graph of `f^(-1)` at  `x = b`, where  `0 < b < 1`, and the vertical axis.

  1. Find the value of `b` for which the total area of these regions is a minimum.   (2 marks)

  2. Find the value of the acute angle between the tangent to the graph of `f` and the tangent to the graph of `f^(-1)` at  `x = 1`.   (1 mark)

Show Answers Only
  1. `y = 2a^3-3a^2x + 1`
  2. `(2a^3 + 1)/(3a^2)`
  3. `-2a`
  4. `(80a^6 + 8a^3-9a^2 + 2)/(12a^2)`
  5. `10^(-1/3)`
  6. `9/10`
  7. `tan^(-1)(1/3)`
Show Worked Solution

a.   `f(x) = 1-x^3,\ \ f^{\prime}(x) = -3x^2`

`m_text(tang) = -3a^2\ \ text(through)\ \ (a, 1-a^3)`

`y = 2a^3-3a^2x + 1`
 

b.   `text(Solve for)\ \x:`

`2a^3-3a^2x + 1 = 0`

`x = (2a^3 + 1)/(3a^2)`
 

c.   `text(Solve for)\ \ x:`

`2a^3-3a^2x + 1 = 1-x^3`

`x=-2a`

`:. x text(-coordinate of)\ P = -2a`
 

d.   `P(-2a, 8a^3 + 1),\ \ Q((2a^3 + 1)/(3a^2), 0)`

`A` `= text(Area of triangle)-int_(-2a)^1 f(x)\ dx`
  `= 1/2((2a^3 + 1)/(3a^2) + 2a)(8a^3 + 1)-int_(-2a)^1 1-x^3\ dx`
  `= (80a^6 + 8a^3-9a^2 + 2)/(12a^2)\ \ text{(by CAS)}`

 

e.   `text(Solve for)\ a: \ (dA)/(da) = 0\ \ text{(by CAS)}`

`a = 10^(-1/3)`
 

f.   `text(Consider)\ f(x):\ \ f(10^(-1/3)) = 9/10`

`A_text(min)\ text(for)\ \ f(x)\ \ text(occurs at)\ \ (10^(-1/3), 9/10)`

`=> A_text(min)\ text(for)\ \ f^(-1)(x)\ \ text(occurs when)\ \ x=9/10`

`:. b = 9/10`
 

g.   `f^{\prime} (1)  = -3`

`text(Gradient of)\ \  f^(-1)(x)\ \ text(at)\ \ x = 1\ \ text(is vertical line.)`

`tan theta` `= 1/3`
`theta` `= tan^(-1)(1/3)`
  `=18.4°\ \ text{(to 1 d.p.)}`

Statistics, MET2 2019 VCAA 4

The Lorenz birdwing is the largest butterfly in Town A.

The probability density function that describes its life span, `X`, in weeks, is given by
 

`f(x) = {(4/625 (5x^3-x^4), quad 0 <= x <= 5),(0, quad text(elsewhere)):}`
 

  1. Find the mean life span of the Lorenz birdwing butterfly.  (2 marks)

  2. In a sample of 80 Lorenz birdwing butterflies, how many butterflies are expected to live longer than two weeks, correct to the nearest integer?  (2 marks)

  3. What is the probability that a Lorenz birdwing butterfly lives for at least four weeks, given that it lives for at least two weeks, correct to four decimal places?  (2 marks)

The wingspans of Lorenz birdwing butterflies in Town A are normally distributed with a mean of 14.1 cm and a standard deviation of 2.1 cm.

  1. Find the probability that a randomly selected Lorenz birdwing butterfly in Town A has a wingspan between 16 cm and 18 cm, correct to four decimal places.  (1 mark)

  2. A Lorenz birdwing butterfly is considered to be very small if its wingspan is in the smallest 5% of all the Lorenz birdwing butterflies in Town A.

     

    Find the greatest possible wingspan, in centimetres, for a very small Lorenz birdwing butterfly in Town A, correct to one decimal place.  (1 mark)

Each year, a detailed study is conducted on a random sample of 36 Lorenz birdwing butterflies in Town A.

A Lorenz birdwing butterfly is considered to be very large if its wingspan is greater than 17.5 cm. The probability that the wingspan of any Lorenz birdwing butterfly in Town A is greater than 17.5 cm is 0.0527, correct to four decimal places.

    1. Find the probability that three or more of the butterflies, in a random sample of 36 Lorenz birdwing butterflies from Town A, are very large, correct to four decimal places.  (1 mark)

    2. The probability that `n` or more butterflies, in a random sample of 36 Lorenz birdwing butterflies from Town A, are very large is less than 1%.

       

      Find the smallest value of `n`, where `n` is an integer.  (2 marks)

    3. For random samples of 36 Lorenz birdwing butterflies in Town A, `hat p` is the random variable that represents the proportion of butterflies that are very large.
    4. Find the expected value and the standard deviation of `hat p`, correct to four decimal places.  (2 marks)

    5. What is the probability that a sample proportion of butterflies that are very large lies within one standard deviation of 0.0527, correct to four decimal places? Do not use a normal approximation.  (2 marks)

  2. The Lorenz birdwing butterfly also lives in Town B.

     

    In a particular sample of Lorenz birdwing butterflies from Town B, an approximate 95% confidence interval for the proportion of butterflies that are very large was calculated to be (0.0234, 0.0866), correct to four decimal places.

     

    Determine the sample size used in the calculation of this confidence interval.  (2 marks)

Show Answers Only

  1. `10/3`
  2. `73`
  3. `0.2878`
  4. `0.1512`
  5. `10.6\ text(cm)`
    1. `0.2947`
    2. `7`
    3. `0.0372`
    4. `0.7380`
  7. `200`

Show Worked Solution

a.    `mu` `= 4/625 int_0^5 x(5x^3-x^4)\ dx`
    `= 4/625[x^5-1/6 x^6]_0^5`
    `= 10/3\ \ \ text{(by CAS)}`

 

b.    `text(Pr)(X > 2)` `= 4/625 int_2^5 5x^3-x^4\ dx`
    `= 4/625[5/4x^4-x^5/5]_2^5`
    `= 0.9129…`

 

`:.\ text(Expected number)` `= 80 xx 0.9129…`
  `~~ 73.03`
  `~~ 73`

 

c.    `text(Pr)(X = 4|X> 2)` `= (text(Pr)(X >= 4))/(text(Pr)(X >= 2))`
    `= 0.26272/0.91296`
    `= 0.2878`

 

d.   `W\ ~\ N (14.1, 2.1^2)`

`text(Pr)(16 < W < 18) = 0.1512\ \ \ text{(by CAS)}`

 

e.   `text(Solution 1:)`

`text(Pr)(W < w) = 0.05`

`text(Pr)(Z < z) = 0.05\ \ =>\ \ z = -1.6449\ \ text{(by CAS)}`

`(w-14.1)/2.1` `= -1.6449`
`w` `= 10.6\ text(cm)`

 
`text(Solution 2:)`

`text(invNorm)`

`text(Tail setting: left)`

`text(prob: 0.05)`

`sigma: 2.1`

`mu: 14.1`

`=> 10.6\ \ \ text{cm (by CAS)}`

 

f.i.   `L\ ~\ text(Bi)(n, p)\ ~\ text(Bi) (36, 0.0527)`

`text(Pr)(L >= 3) ~~ 0.2947`

 

f.ii.    `text(Pr)(L >= n) < 0.01`
 

`text(CAS: binomialCdf) (x, 36, 36, 0.0527)`

`text(Pr)(L >= 0) = 0.011 > 0.01`

`text(Pr)(L >= 7) = 0.002 < 0.01`

`:.\ text(Smallest)\ n = 7`

 

f.iii.    `E(hat p)` `= p = 0.0527`
  `sigma(hat p)` `= sqrt((p(1-p))/n) = sqrt((0.0527(1-0.0527))/36) ~~ 0.0372`

 

f.iv.        `hat p +- 1 sigma: (0.0527-0.0372, 0.0527 + 0.0372) = (0.0155, 0.0899)`

`text(Pr)(0.0155 < hat p < 0.0899)`

`= text(Pr)(36 xx 0.0155 < L < 36 xx 0.0899)`

`= text(Pr)(0.56 < L < 3.24)`

`= text(Pr)(1 <= L <= 3)`

`~~ 0.7380`

 

g.   `0.0234 = hat p-1.96 sqrt((hat p(1-hat p))/n) qquad text{… (1)}`

`0.0866 = hat p + 1.96 sqrt((hat p(1-hat p))/n) qquad text{… (2)}`

`text(Solve simultaneous equations:)`

`hat p ~~ 0.055, quad n ~~ 199.96`

`:.\ text(Sample size) = 200`

Calculus, EXT1 C3 2019 SPEC2 9 MC

The differential equation that has the diagram above as its direction field is

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

`B`

Show Worked Solution

`text(By elimination:)`

`text(Along line)\ y = x,\ text(gradient = 1)`

`:.\ text(Eliminate A and D.)`
 

`text{At (1, 0),  0 < gradient < 1}`

`1/(cos(-1)) > 1`

`:.\ text(Eliminate C)`

`=>B`

Calculus, SPEC2 2019 VCAA 9 MC

The differential equation that has the diagram above as its direction field is

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

`B`

Show Worked Solution

`text(Along line)\ y = x,\ text(gradient = 1)`

`:.\ text(Eliminate A, C, and E.)`
 

`text{At (1, 0),  0 < gradient < 1}`

`1/(cos(-1)) > 1`

`:.\ text(Eliminate D)`

`=>B`

Calculus, MET2 2019 VCAA 3

During a telephone call, a phone uses a dual-tone frequency electrical signal to communicate with the telephone exchange.

The strength, `f`, of a simple dual-tone frequency signal is given by the function  `f(t) = sin((pi t)/3) + sin ((pi t)/6)`, where  `t`  is a measure of time and  `t >= 0`.

Part of the graph of `y = f(t)`  is shown below

  1. State the period of the function.   (1 mark)

  2. Find the values of  `t`  where  `f(t) = 0`  for the interval  `t in [0, 6]`.   (1 mark)

  3. Find the maximum strength of the dual-tone frequency signal, correct to two decimal places.   (1 mark)

  4. Find the area between the graph of  `f`  and the horizontal axis for  `t in [0, 6]`.   (2 marks)

Let  `g`  be the function obtained by applying the transformation  `T`  to the function  `f`, where
 

`T([(x), (y)]) = [(a, 0), (0, b)] [(x), (y)] + [(c), (d)]`
 

and `a, b, c` and `d` are real numbers.

  1. Find the values of `a, b, c` and `d` given that  `int_2^0 g(t)\ dt + int_2^6 g(t)\ dt`  has the same area calculated in part d.   (2 marks)

  2. The rectangle bounded by the line  `y = k, \ k in R^+`, the horizontal axis, and the lines  `x = 0`  and  `x = 12`  has the same area as the area between the graph of  `f`  and the horizontal axis for one period of the dual-tone frequency signal.

     

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

Show Answers Only
  1. `12`
  2. `0, 4, 6`
  3. `1.760`
  4. `15/pi\ text(u²)`
  5. `a = 1,\ b =-1,\ c =-6,\ d = 0`
  6. `5/(2pi)`
Show Worked Solution

a.   `text(Period) = 12`
  

b.   `t = 0, 4, 6`
  

c.   `f(t) = sin ((pi t)/3) + sin ((pi t)/6)`

`f(t)_max ~~ 1.76\ \ text{(by CAS)}`

 

d.    `text(Area)` `= int_0^4 sin((pi t)/3) + sin ((pi t)/6) dt-int_4^6 sin ((pi t)/3) + sin ((pi t)/6) dt`
    `= 15/pi\ text(u²)`

 

e.   `text(Same area) => f(t)\ text(is reflected in the)\ x text(-axis and)`

`text(translated 6 units to the left.)`

`x′=ax+c`

`y′=by+d`

`text(Reflection in)\ xtext(-axis) \ => \ b=-1, \ d=0`

`text(Translate 6 units to the left) \ => \ a=1, \ c=-6`

`:. a = 1,\ b = -1,\ c = -6,\ d = 0`
  

f.    `text(Area of rectangle)` `= 2 xx text(Area between)\ f(t) and x text(-axis)\ \ t in [0, 6]`
  `12k` `= 2 xx 15/pi`
  `:. k` `=5/(2pi)`

Calculus, MET2 2019 VCAA 2

An amusement park is planning to build a zip-line above a hill on its property.

The hill is modelled by  `y = (3x(x-30)^2)/2000,  x in [0, 30]`, where  `x`  is the horizontal distance, in metres, from an origin and  `y`  is the height, in metres, above this origin, as shown in the graph below.
  

  1. Find  `(dy)/(dx)`.   (1 mark)

  2. State the set of values for which the gradient of the hill is strictly decreasing.   (1 mark)

The cable for the zip-line is connected to a pole at the origin at a height of 10 m and is straight for  `0 <= x <= a`, where  `10 <= a <= 20`. The straight section joins the curved section at  `A(a, b)`. The cable is then exactly 3 m vertically above the hill from  `a <= x <= 30`, as shown in the graph below.

  1. State the rule, in terms of `x`, for the height of the cable above the horizontal axis for  `x in [a, 30]`.   (1 mark)

  2. Find the values of `x` for which the gradient of the cable is equal to the average gradient of the hill for  `x in [10, 30]`.   (3 marks)

The gradients of the straight and curved sections of the cable approach the same value at  `x = a`, so there is a continuous and smooth join at `A`.

    1. State the gradient of the cable at `A`, in terms of `a`.   (1 mark)

    2. Find the coordinates of `A`, with each value correct to two decimal places.   (3 marks)

    3. Find the value of the gradient at `A`, correct to one decimal place.   (1 mark)

Show Answers Only
  1. `(9(x-10)(x-30))/2000`
  2. `x in [0, 20]`
  3. `h(x) = 3 + (3x(x-30)^2)/2000`
  4. `x = (60 +- 10 sqrt 3)/3`
    1. `(9(a-10)(a-30))/2000`
    2. `A(11.12, 8.95)`
    3. `-0.1`
Show Worked Solution

a.   `d/(dx)((3x(x-30)^2)/2000) = (9(x-10)(x-30))/2000\ \ text{(by CAS)}`
 

b.   `text(Sketch)\ \ dy/dx\ \ text{(by CAS)}:`

`(dy)/(dx)\ \ text(strictly decreasing for)\ \ x in [0, 20]`
 

c.   `text(Cable height is a vertical shift of +3 of)\ \ y:`

`h(x) = 3 + (3x(x-30)^2)/2000\ \ text(for)\ \ x in [a, 30]`
 

d.   `text(Let)\ \ f(x) = (9(x-10)(x-30))/2000`

`text(Average gradient) = (f(30)-f(10))/(30-10) = -3/10\ \ text{(by CAS)}`

`text(Solve for)\ \ x\ \ text{(by CAS):}`

`(9(x-10)(x-30))/2000 = -3/10`

`x = (60 +- 10 sqrt 3)/3`
 

e.i.   `text(S)text(ince)\ \ h(x)\ \ text(is a vertical shift of)\ \ f(x),`

`=> h^{′}(a) = f^{′}(a)`

`h^{′}(a) = (9(a-10)(a-30))/2000`
 

e.ii.   `h(a) = (3a(a-30)^2)/2000 + 3`

`:. A(a, (3a(a-30)^2)/2000 + 3),\ B(0, 10)`
 

`m_(AB) = ((3a(a-30)^2)/2000 + 3-10)/a = (3a(a-30)^2-14000)/(2000 a)`

 
`text(Equating gradients:)`

`text(Solve): \ f^{′}(a) = m_(AB)\ \ text{(by CAS)}`

`(9(a-10)(a-30))/2000 = (3a(a-30)^2-14000)/(2000 a)`

`a ~~ 11.12, quad b ~~ 8.95`

`:. A(11.12, 8.95)`

 

e.iii.    `M_A` `= h^{′}(11.12)`
    `~~ -0.1\ text{(by CAS)}`

Calculus, EXT1 C2 SM-Bank 1 MC

With a suitable substitution, `int_1^5(2x - 1)sqrt(2x + 1)\ dx` can be expressed as

  1. `1/2 int_1^5 (u^(3/2) + u^(1/2))\ du`
  2. `2 int_3^11 (u^(3/2) + u^(1/2))\ du`
  3. `2 int_1^5 (u^(3/2) - 2u^(1/2))\ du`
  4. `1/2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ u = 2x + 1 \ => \ u – 2 = 2x – 1`

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

`text(When)\ \ x = 5, \ u = 11`

`text(When)\ \ x = 1, \ u = 3`

`:. int_1^5 (2x – 1)sqrt(2x + 1)\ dx`

`= 1/2 int_3^11 (u – 2)u^(1/2)\ du`

`= 1/2 int_3^11 u^(3/2) – 2u^(1/2)\ du`

 
`=>D`

Calculus, SPEC2 2019 VCAA 8 MC

With a suitable substitution, `int_1^5(2x - 1)sqrt(2x + 1)\ dx` can be expressed as

  1. `1/2 int_1^5 (u^(3/2) + u^(1/2))\ du`
  2. `2 int_3^11 (u^(3/2) + u^(1/2))\ du`
  3. `2 int_1^5 (u^(3/2) - 2u^(1/2))\ du`
  4. `2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
  5. `1/2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
Show Answers Only

`E`

Show Worked Solution

`text(Let)\ \ u = 2x + 1 \ => \ u – 2 = 2x – 1`

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

`text(When)\ \ x = 5, \ u = 11`

`text(When)\ \ x = 1, \ u = 3`

`:. int_1^5 (2x – 1)sqrt(2x + 1)\ dx`

`= 1/2 int_3^11 (u – 2)u^(1/2)\ du`

`= 1/2 int_3^11 u^(3/2) – 2u^(1/2)\ du`

 
`=>E`

Calculus, SPEC2 2019 VCAA 7 MC

The length of the curve defined by the parametric equations  `x = 3sin(t)`  and  `y = 4cos(t)`  for  `0 <= t <= pi`  is given by

  1. `int_0^pi sqrt(9cos^2(t) - 16sin^2(t))\ dt`
  2. `int_0^pi sqrt(9 + 7sin^2(t))\ dt`
  3. `int_0^pi sqrt(1 + 16sin^2(t))\ dt`
  4. `int_0^pi (3cos(t) - 4sin(t))\ dt`
  5. `int_0^pi sqrt(3cos^2(t) + 4sin^2(t))\ dt`
Show Answers Only

`B`

Show Worked Solution

`x = 3sin(t) \ => \ (dx)/(dt) = 3cos(t)`

`y = 4cos(t) \ => \ (dy)/(dt) = −4sin(t)`

`text(Length)` `= int_0^pi sqrt((3cos(t))^2 + (−4sin(t))^2)\ dt`
  `= int_0^pi sqrt(9cos^2(t) + 16sin^2(t))\ dt`
  `= int_0^pi sqrt(9cos^2(t) + 9sin^2(t) + 7sin^2(t))\ dt`
  `= int_0^pi sqrt(9 + 7sin^2(t))\ dt`

 
`=>B`

Complex Numbers, SPEC2 2019 VCAA 6 MC

Let  `z, w ∈ C`, where  `text(Arg)(z) = pi/2`  and  `text(Arg)(w) = pi/4`.

The value of  `text(Arg)((z^5)/(w^4))`  is

  1. `−pi/2`
  2. `pi/2`
  3. `pi`
  4. `(5pi)/2`
  5. `(7pi)/2`
Show Answers Only

`A`

Show Worked Solution
`text(Arg)((z^5)/(w^4))` `= text(Arg)(z^5) – text(Arg)(w^4)`
  `= 5text(Arg)(z) – 4text(Arg)(w)`
  `= (5pi)/2 – (4pi)/4`
  `= (3pi)/2`

 
`:. text(Arg)((z^5)/(w^4)) = −pi/2`

`=>A`

Complex Numbers, SPEC2 2019 VCAA 5 MC

Let  `z = x + yi`, where  `x, y ∈ R`. The rays  `text(Arg)(z - 2)=pi/4`  and  `text(Arg)(z - (5 + i)) = (5pi)/6`, where  `z ∈ C`, intersect on the complex plane at a  point  `(a,b)`.

The value of `b` is

  1. `−sqrt3`
  2. `2 - sqrt3`
  3. `0`
  4. `sqrt3`
  5. `2 + sqrt3`
Show Answers Only

`D`

Show Worked Solution

`text(Convert to cartesian equations)`

`text(Arg)(z – 2) = pi/4`

`y = x – 2\ …\ (1)`

`text(Arg)(z – (5 + i)) = (5pi)/6`

`y – 1 = −1/sqrt3 (x – 5)\ …\ (2)`

`text{Solve (1) and (2) simultaneously:}`

`text(Intersection)\ (a,b) :\ \ x = 2+sqrt3, \ y = sqrt3`

`:. b = sqrt3`

`=>D`

Complex Numbers, EXT2 N1 SM-Bank 7

Calculate  `i^(1!) + i^(2!) + i^(3!) + …+ i^(100!)`

giving your answer in the form  `x+iy`  (3 marks)

Show Answers Only

`95+i`

Show Worked Solution

`i^(1!) =i^1 =  i`

`i^(2!) =i^2= −1`

`i^(3!) = i^6=-1`

`i^(4!) = i^24 = 1`

`i^(5!) = i^(24 xx 5)= 1^5 = 1`

`=> i^(n!) = 1\ \ text(for)\ \ n >= 4`

`:.\ text(Sum)` `= i – 1 – 1 + 97`
  `= 95 + i`

Graphs, SPEC2 2019 VCAA 3 MC

The implied domain of the function with rule  `f(x) = 1 - sec(x + pi/4)`  is

  1. `R`
  2. `[0,2]`
  3. `R\ text(\)\ {((4n - 1)pi)/4}, n ∈ ZZ`
  4. `R\ text(\)\ {((4n + 1)pi)/4}, n ∈ ZZ`
  5. `R\ text(\)\ {((2n - 1)pi)/4}, n ∈ ZZ`
Show Answers Only

`D`

Show Worked Solution

`y = sec (x)=1/cos(x)\ \ text(has asymptotes when)`

`x = −pi/2, pi/2, (3pi)/2, …`

`=> y = sec (x + pi/4)\ \ text(has asymptotes at when)`

`x = (−3pi)/4, pi/4, (5pi)/4, …`

`:.\ text(Domain:)\ R\ text(\)\ {((4n + 1)pi)/4}, n ∈ ZZ`

`=>D`

Mechanics, SPEC1 2019 VCAA 9

  1. A light inextensible string is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs in equilibrium from a smooth ring on the string, as shown in the diagram below. The string makes an angle `alpha` with the ceiling.
      
    `qquad qquad`
     
    Express the tension, `T` newtons, in the string in terms of `m`, `g` and `alpha`.  (1 mark)
  2. A different light inextensible sting is connected at each end to a horizontal ceiling. A mass of `m` kilograms hangs from a smooth ring on the string. A horizontal force of `F` newtons is applied to the ring. The tension in the sting has a constant magnitude and the system is in equilibrium. At one end the string makes an angle `beta` with the ceiling and at the other end the string makes an angle `2beta` with the ceiling, as shown in the diagram below.
     

     
    Show that  `F = mg((1 - cos(beta))/(sin(beta)))`.  (3 marks)
Show Answers Only
  1. `T = (mg)/(2sinalpha)`
  2. `text(See Worked Solutions)`
Show Worked Solution
a.   
`2 xx Tsinalpha` `= mg`
`:.T` `= (mg)/(2sinalpha)`

 

b.   

 
`text(Resolving forces vertically:)`

`Tsin(beta) + Tsin(2beta)` `= mg`
`T` `= (mg)/(sin(beta) + sin(2beta))`

 
`text(Resolving forces horizontally:)`

`F + Tcos(2beta)` `= Tcos(beta)`
`F` `= Tcos(beta) – Tcos(2beta)`
  `= T(cos(beta) – cos(2beta))`
  `= T[cos(beta) – (2cos^2beta – 1)]`
  `= T(−2cos^2(beta) + cos(beta) + 1)`
  `= T(−2cos(beta) – 1)(cos(beta) – 1)`
  `= (mg(1 -cos(beta))(2cosbeta + 1))/(sin(beta) + 2sin(beta)cos(beta))`
  `= (mg(1 – cos(beta))(2cos(beta) + 1))/(sin(beta)(1+2cos(beta)))`
  `= mg((1 – cos(beta))/(sin(beta)))`

Calculus, EXT1 C3 SM-Bank 3

Find the volume of the solid of revolution formed when the graph of  `y = sqrt((1 + 2x)/(1 + x^2))`  is rotated about the `x`-axis over the interval  `[0,1]`.  (3 marks)

Show Answers Only

`pi(pi/4 + ln2)\ \ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 (1 + 2x)/(1 + x^2)\ dx`
  `= pi int_0^1 1/(1 + x^2)\ dx + pi int_0^1 (2x)/(1 + x^2)\ dx`
  `= pi [tan^(−1)(x)]_0^1 + pi [ln(1 + x^2)]_0^1`
  `= pi(tan^(−1)1 – tan^(−1)0) + pi(ln2 – ln1)`
  `= pi(pi/4) + pi(ln2)`
  `= pi(pi/4 + ln2)\ \ text(u³)`

Calculus, SPEC1 2019 VCAA 8

Find the volume of the solid of revolution formed when the graph of  `y = sqrt((1 + 2x)/(1 + x^2))`  is rotated about the `x`-axis over the interval  `[0,1]`.  (4 marks)

Show Answers Only

`pi(pi/4 + ln2)\ \ text(u³)`

Show Worked Solution
`V` `= pi int_0^1 (1 + 2x)/(1 + x^2)\ dx`
  `= pi int_0^1 1/(1 + x^2)\ dx + pi int_0^1 (2x)/(1 + x^2)\ dx`
  `= pi [tan^(−1)(x)]_0^1 + pi [ln(1 + x^2)]_0^1`
  `= pi(tan^(−1)1 – tan^(−1)0) + pi(ln2 – ln1)`
  `= pi(pi/4) + pi(ln2)`
  `= pi(pi/4 + ln2)\ \ text(u³)`

Graphs, SPEC2 2019 VCAA 2 MC

The asymptote(s) of the graph of  `f(x) = (x^2 + 1)/(2x - 8)`  has equation(s)

  1. `x = 4`
  2. `x = 4 and y = x/2`
  3. `x = 4 and y = x/2 + 2`
  4. `x = 8 and y = x/2`
  5. `x = 8 and y = 2x + 2`
Show Answers Only

`C`

Show Worked Solution
`(x^2 + 1)/(2x – 8)` `= ((x – 4)(x + 4) + 17)/(2(x – 4))`
  `= (x + 4)/2 + 17/(2x – 8)`
  `= x/2 + 2 + 17/(2x – 8)`

 
`:.\ text(Asymptotes at)\ \ x = 4 and y = x/2 + 2`

`=>C`

Complex Numbers, SPEC1 2019 VCAA 7

  1. Show that  `3-sqrt3 i = 2sqrt3 text(cis)(-pi/6)`.   (1 mark)

  2. Find  `(3-sqrt3 i)^3`, expressing your answer in the form  `x + iy`, where  `x`,  `y ∈ R`.   (2 marks)

  3. Find the integer values of  `n`  for which  `(3-sqrt3 i)^n`  is real.   (1 mark)

  4. Find the integer values of  `n`  for which  `(3-sqrt3 i)^n = ai`, where  `a`  is a real number.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0-i 24sqrt3`
  3. `n = 6k\ \ (k ∈ ZZ)`
  4. `n = 3 + 6k\ \ (k ∈ ZZ)`
Show Worked Solution

a.  `|3-sqrt3 i|= sqrt(3^2 + (-sqrt3)^2)= sqrt12= 2sqrt3`

`text(Arg)(3-sqrt3 i)` `= tan^(-1)(-(sqrt3)/3)= -pi/6`
   

`:. 3-sqrt3 i = 2sqrt3\ text(cis)(-pi/6)`

b.    `(3-sqrt3 i)^3` `= (2sqrt3)^3\ text(cis)(3 xx-pi/6)`
    `= 24sqrt3\ text(cis)(-pi/2)`
    `= 24sqrt3(cos(-pi/2) + isin(-pi/2))`
    `= 0-i 24sqrt3`

 

c.   `(3-sqrt3 i)^n = (2sqrt3)^n\ text(cis)((-npi)/6)`

`text(Real when)\ \ sin(-(npi)/6) = -sin((npi)/6) = 0`

`(npi)/6 = 0, pi, 2pi, …, kpi\ \ (k ∈ ZZ)`

`:. n = 6k\ \ (k ∈ ZZ)`

 

d.  `(3-sqrt3 i)^n = ai\ \ text(when)\ \ cos(-(npi)/6) = cos((npi)/6) = 0`

`(npi)/6 = pi/2, (3pi)/2, …, pi/2 + kpi\ \ (k ∈ ZZ)`

`:. n = 3 + 6k\ \ (k ∈ ZZ)`

Vectors, SPEC1 2019 VCAA 6

Find value of  `d`  for which the vectors  `underset~a = 2underset~i - 3underset~j + 4underset~k`,  `underset~b = −2underset~i + 4underset~j - 8underset~k`  and  `underset~c = −6underset~i + 2underset~j + dunderset~k`  are linearly dependent.  (3 marks)

Show Answers Only

`16`

Show Worked Solution

`text(Linear dependent:)\ \ m underset~a + n underset~b = underset~c`

`2m – 2n` `= −6\ \ …\ (1)`
`−3m + 4n` `= −2\ \ …\ (2)`
`4m – 8n` `= d\ \ …\ (3)`

 
`(1) xx 2`

`4m – 4n = −12\ \ …\ (1′)`

`(2) + (1′)`

`m = −10`

`text(Substitute)\ \ m = −10\ \ text(into (1))`

`−20 – 2n` `= −6`
`n` `= −7`

 

`:.d` `= 4 xx −10 – 8 xx −7`
  `= 16`

Calculus, SPEC1 2019 VCAA 5

The graph of  `f(x) = cos^2(x) + cos(x) + 1`  over the domain  `0 <= x <= 2pi`  is shown below.

  1.  i.  Find `f^{′}(x)`.  (1 mark)

  2. ii. Hence, find the coordinates of the turning points of the graph in the interval  `(0, 2pi)`.  (2 marks)

  3. Sketch the graph of  `y = 1/(f(x))`  on the set of axes above. Clearly label the turning points and endpoints of this graph with their coordinates.  (3 marks)
Show Answers Only
  1. i.  `−sin(x)(2cos(x) + 1)`
  2. ii. `(pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`
  3.  

Show Worked Solution
a.i.    `f(x)` `= cos^2(x) + cos(x) + 1`
  `f^{′}(x)` `= -2sin(x)cos(x)-sin(x)`
    `= -sin(x)(2cos(x) + 1)`

 

a.ii.   `text(SP when)\ \ sin(x) = 0\ \ text(or)\ \ 2cos(x) + 1 = 0`

`sin(x) = 0 \ => \ x = pi\ \ (x = 0\ \ text{not in domain})`

`2cos(x) + 1` `= 0`
`cos(x)` `= -1/2`
`x` `= (2pi)/3, (4pi)/3`

 
`text(When)\ \ cos(x) = −1/2 \ => \ f(x) = 1/4-1/2 + 1 = 3/4`

`:.\ text(Turning Points:)\ (pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`

 

b.   

Calculus, MET2 2019 VCAA 1

Let  `f: R -> R,\ \ f(x) = x^2e^(-x^2)`.

  1. Find `f^{\prime}(x)`.  (1 mark)

  2. i.  State the nature of the stationary point on the graph of  `f`  at the origin.  (1 mark)

  3. ii.  Find the maximum value of the function  `f`  and the values of  `x`  for which the maximum occurs.  (2 marks)

  4. iii. Find the values of  `d in R`  for which  `f(x) + d`  is always negative.  (1 mark)

  5. i.  Find the equation of the tangent to the graph of  `f`  at  `x = –1`.  (1 mark)

  6. ii. Find the area enclosed by the graph of  `f`  and the tangent to the graph of  `f`  at  `x = –1`, correct to four decimal places.  (2 marks)

  7. Let  `M(m, n)`  be a point on the graph of  `f`, where  `m in [0, 1]`.
  8. Find the minimum distance between  `M`  and the point  `(0, e)`, and the value of  `m`  for which this occurs, correct to three decimal places.  (3 marks)

Show Answers Only

  1. `2x e^(-x^2)(1-x^2)`
  2. i.  `text(Local minimum)`
  3. ii. `f(x)_max= 1/e\ \ text(when) \ x = -1 and 1`
  4. iii. `d<1/e`
  5. i.  `y = 1/e` 
  6. ii. `0.3568\ text{(to 4 d.p.)}`
  7. `D_(min) = 2.511\ \ text(when)\ \ m ~~ 0.783`

Show Worked Solution

a.    `f^{\prime}(x)` `= x^2 ⋅ -2x ⋅ e^(-x^2) + e^(-x^2) ⋅ 2x`
    `= 2x e^(-x^2) (1-x^2)`

​

b.i.    `f ^{″}(0) = 2 > 0\ \ \ text{(by CAS)}`
  `:.\ text(Local minimum)`

 

b.ii.    `text(SP’s occur when)\ \ x = –1, 0, 1`
  `f(x)_max = 1/e\ \ text(when)\ \ x = –1  and  1`

 

b.iii.    `f(x)_max + d` `< 0`
  `d` `< 1/e`

 

c.i.    `text(At)\ \ x = –1, \ f(x)\ text(has a max turning point).`
  `:.\ text(T)text(angent:)\ \ y = 1/e`

 

c.ii.    `text(Area)` `= int_(_1)^1 1/e-x^2e^(-x^2) dx`
    `~~ 0.3568\ text{(to 4 d.p.)}`

 

d.   `text(When)\ x = m,\ \ n = m^2 e^(-m^2)`

`text(Find distance between)\ \ M(m, m^2 e^(-m^2)) and P(0, e):`

`D = sqrt((m-0)^2 + (m^2 e^(-m^2)-e)^2)`

`=>\ text(MIN distance when)\ \ (dD)/(dm) = 0`

`D_(min) = 2.511\ \ text(when)\ \ m ~~ 0.783\ \ \ text{(by CAS)}`

Vectors, SPEC1 2019 VCAA 4

The position vectors of two particles  `A`  and  `B`  at time  `t` seconds after they have started moving are given by  `underset~r_A(t) = (t^2-1)underset~i + (a + t/3)underset~j`  and  `underset~r_B(t) = (t^3-t)underset~i + (text(arccos)(t/2))underset~j`  respectively, where  `a`  is a real constant and  `0 <= t <= 2`.

Find the value of  `a`  if the particles collide after they have started moving.   (3 marks)

Show Answers Only

`(pi-1)/3`

Show Worked Solution

`text(Collision occurs when)\ underset~i\ text(and)\ underset~j\ text(components are equal).`

`text(Equating)\ underset~i\ text(components:)`

`t^2-1` `= t^3-t`
`t^3-t^2-t + 1` `= 0`
`t^2(t-1)-(t-1)` `= 0`
`(t^2-1)(t-1)` `= 0`

 
`:. t = 1\ \ (t > 0)`
 

`text(Equating)\ underset~j\ text{components (when}\ t = 1text{):}`

`a + 1/3` `= cos^(−1)(1/2)`
`:. a` `= (pi-1)/3`

Statistics, SPEC1 2019 VCAA 3

A machine produces chocolate in the form of a continuous cylinder of radius 0.5 cm. Smaller cylindrical pieces are cut parallel to its end, as shown in the diagram below.

The lengths of the pieces vary with a mean of 3 cm and a standard deviation of 0.1 cm.
 


 

  1. Find the expected volume of a piece of chocolate in cm³.   (1 mark)

  2. Find the variance of the volume of a piece of chocolate in cm6.   (1 mark)

  3. Find the expected surface area of a piece of chocolate in cm².   (1 mark)

Show Answers Only
  1. `0.75pi\ text(cm³)`
  2. `0.000625pi^2\ text(cm)^6`
  3. `3.5pi\ text(cm²)`
Show Worked Solution
a.    `E(V)` `= pir^2 xx E(h)`
    `= pi xx 0.5^2 xx 3`
    `= 0.75pi\ text(cm³)`

 

b.    `text(Var)(V)` `= text(Var)(pir^2h)`
    `= pi^2 xx 0.5^4 xx text(Var)(h)`
    `= 0.0625pi^2 xx 0.1^2`
    `= 0.000625pi^2\ text(cm)^6`

 

c.    `E text{(Surface Area)}` `= 2pir^2 + 2pir xx E(h)`
    `= 2pir(r +E(h))`
    `= pi(0.5 + 3)`
    `= 3.5pi\ text(cm²)`

Functions, 2ADV F2 SM-Bank 35

  1. Sketch the function  `y = f(x)`  where  `f(x) = (x - 1)^3`  on a number plane, labelling all intercepts.  (1 mark)

  2. On the same graph, sketch  `y = −f(−x)`. Label all intercepts.  (2 marks)

Show Answers Only
  1.   
  2.   
Show Worked Solution

i.   `y = (x – 1)^3 => y = x^3\ text(shifted 1 unit to the right.)`
 

 

ii.   `y = −f(x) \ => \ text(reflect)\ \ y = (x – 1)^3\ \ text(in)\ xtext(-axis).`

`y = −f(−x) \ => \ text(reflect)\ \ y = −f(x)\ \ text(in)\ ytext(-axis).`

 

Calculus, SPEC1 2019 VCAA 1

Solve the differential equation  `(dy)/(dx) = (2ye^(2x))/(1 + e^(2x))`  given that  `y(0) = pi`.  (4 marks)

Show Answers Only

`y = (pi(1 + e^(2x)))/2`

Show Worked Solution
`int 1/y\ dy` `= (2e^(2x))/(1 + e^(2x))\ dx`
`log_e |y|` `= log_e |1 + e^(2x)| + c`

 
`text(When)\ \ x=0, \ y= pi:`

`log_e pi` `= log_e |1 + e^0| + c`
`c` `= log_e pi – log_e 2`
  `= log_e\ pi/2`
`log_e |y|` `= log_e(1 + e^(2x)) + log_e\ pi/2`
`log_e |y|` `= log_e\ (pi(1 + e^(2x)))/2`
`:. y` `= (pi(1 + e^(2x)))/2`

Functions, 2ADV F1 SM-Bank 36

Consider the function  `f(x) = 1/(x + 2)`.
 

 
 

  1. Sketch the graph  `y = f(−x)`.   (2 marks)

  2. On the same graph, sketch  `y = −f(x)`.   (2 marks)

Show Answers Only

i.

ii.   

Show Worked Solution

i.   `text(Sketch)\ \ y = 1/(x + 2)`

`y = f(−x) \ =>\ text(reflect)\ \ y = 1/(x + 2)\ \ text(in the)\ ytext(-axis).`
 

 

ii.   `y = −f(x) \ =>\ text(reflect)\ \ y = 1/(x + 2)\ \ text(in the)\ xtext(-axis).`
 

Calculus, MET2 2019 VCAA 16 MC

Part of the graph of  `y = f(x)`  is shown below.
 

The corresponding part of the graph of  `y = f^{\prime}(x)`  is best represented by

A.    B.   
C.    D.   
E.       
Show Answers Only

`A`

Show Worked Solution

`text(By Elimination):`

`text(S.P. at)\ \ x = 5`

`=> f^{\prime}(x) = 0\ \ text(when)\ \ x = 0`

`text(Eliminate  C, E)`

`text(Gradient is negative when)\ \ x < 5,\ \ text(and positive when)\ \ x >5`

`text(Eliminate  B, D)`
 

`=>   A`

Statistics, MET2 2019 VCAA 14 MC

The weights of packets of lollies are normally distributed with a mean of 200 g.

If 97% of these packets of lollies have a weight of more than 190 g, then the standard deviation of the distribution, correct to one decimal place, is

  1.  3.3 g
  2.  5.3 g
  3.  6.1 g
  4.  9.4 g
  5. 12.1 g
Show Answers Only

`B`

Show Worked Solution

`text(Pr)(X > 190)` `= text(Pr)(Z > (190 – 200)/sigma) = 0.97`
`text(Pr)(Z < (-10)/sigma)` `= 0.03`
`text(z-score)` `= -1.8808\ \ text{(by CAS)}`
`-1.8808` `= (-10)/sigma`
`sigma` `= 5.31…`

 
`=>   B`

Probability, 2ADV S1 2019 MET2 11 MC

`A` and `B` are events from a sample space such that  `P(A) = p`, where  `p > 0, \ P(B\ text{|}\ A) = m`  and  `P(B\ text{|}\ A^{′}) = n`.

`A` and `B` are independent events when

  1. `m = n`
  2. `m = 1-p`
  3. `m + n = 1`
  4. `m = p`
Show Answers Only

`A`

Show Worked Solution

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

`P(B\ text{|}\ A) = P(B\ text{|}\ A^{′})`

`:. m = n`
 

`=>   A`

Probability, MET2 2019 VCAA 11 MC

`A` and `B` are events from a sample space such that  `text(Pr)(A) = p`, where  `p > 0, \ text(Pr)(B\ text{|}\ A) = m`  and  `text(Pr)(B\ text{|}\ A prime) = n`.

`A` and `B` are independent events when

  1. `m = n`
  2. `m = 1 - p`
  3. `m + n = 1`
  4. `m = p`
  5. `m + n = 1 - p`
Show Answers Only

`A`

Show Worked Solution

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

`text(Pr)(B\ text{|}\ A) = text(Pr)(B\ text{|}\ A prime)`

`:. m = n`
 

`=>   A`

Graphs, MET2 2019 VCAA 10 MC

Which one of the following statements is true for  `f: R -> R, \ f(x) = x + sin(x)`?

  1. The graph of  `f` has a horizontal asymptote
  2. There are infinitely many solutions to  `f(x) = 4`
  3. `f` has a period of  `2 pi`
  4. `f^{\prime}(x) >= 0`  for  `x in R`
  5. `f^{\prime}(x) = cos(x)`
Show Answers Only

`D`

Show Worked Solution

`text(By CAS, sketch)\ \ f(x) = x + sin(x):`

`text(By inspection, eliminate A, B, C)`

`text(By CAS, sketch)\ \ d/(dx)\ f(x):`

` text(Graph range) >= 0\ \ text(for)\ \ x in R`
 

`=>   D`

Graphs, MET2 2019 VCAA 9 MC

The point  `(a, b)`  is transformed by

`T([(x), (y)]) = [(1/2, 0), (0, -2)][(x), (y)]+[(-1/2), (-2)]`

If the image of  `(a, b)`  is  `(0, 0)`, then  `(a, b)`  is

  1. `(1, 1)`
  2. `(-1, 1)`
  3. `(-1, 0)`
  4. `(0, 1)`
  5. `(1, -1)`
Show Answers Only

`E`

Show Worked Solution

`x^{\prime} = 1/2x – 1/2`

`=> 0` `= 1/2a – 1/2`
`a` `= 1`

 
`y^{\prime}= -2y – 2`

`=> 0` `= -2b – 2`
`b` `= -1`

 
`:. (a, b) -= (1, -1)`

`=>   E`

Probability, MET2 2019 VCAA 8 MC

An archer can successfully hit a target with a probability of 0.9. The archer attempts to hit the target 80 times. The outcome of each attempt is independent of any other attempt.

Given that the archer successfully hits the target at least 70 times, the probability that the archer successfully hits the target exactly 74 times, correct to four decimal places, is

A.   0.3635

B.   0.8266

C.   0.1494

D.   0.3005

E.   0.1701

Show Answers Only

`C`

Show Worked Solution

`text(Pr)(H) = 0.9, \ text(Pr)(H prime) = 0.1`

`text(Let)\ \ X = text(number of times archer hits)`

`X\ ~\ text(Bi)(80, 0.9)`

`text(Pr)(X = 74|X >= 70)` `= (text(Pr)(X = 74))/(text(Pr)(X >= 70))`
  `~~ 0.1494`

 
`=>   C`

Calculus, MET2 2019 VCAA 6 MC

A rectangular sheet of cardboard has a length of 80 cm and a width of 50 cm. Squares, of side length `x` centimetres, are cut from each of the corners, as shown in the diagram below.
 

 

A rectangular box with an open top is then constructed, as shown in the diagram below.
 

 

The volume of the box is a maximum when `x` is equal to

  1. `10`
  2. `20`
  3. `25`
  4. `100/3`
  5. `200/3`
Show Answers Only

`A`

Show Worked Solution
`text(Base of box)` `= (80-2x) xx (50-2x)`
  `= 4000-260x + 4x^2`
`text(Volume)` `= 4x^3-260x^2 + 4000x`

 
`(dV)/(dx) = 12x^2-520x + 4000`

`(dV)/(dx) = 0\ \ text(when)\ \ x = 10`

`=>   A`

Calculus, MET2 2019 VCAA 4 MC

`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`  is equal to

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

`C`

Show Worked Solution

`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`

`= [-a cos(x) + b sin(x)]_0^(pi/6)`

`= [-a ⋅ sqrt 3/2 + b/2 – (-a + 0)]`

`= (2a – sqrt 3 a + b)/2`

`= ((2 – sqrt 3) a + b)/2`

`=>   C`

Calculus, MET2 2019 VCAA 3 MC

Let  `f: R\ text(\){4} -> R, \ f(x) = a/(x - 4),\ \ text(where)\ \ a > 0`.

The average rate of change of  `f`  from  `x = 6`  to  `x = 8`  is

A.   `a log_e(2)`

B.   `a/2 log_e(2)`

C.   `2a`

D.   `-a/4`

E.   `-a/8`

Show Answers Only

`E`

Show Worked Solution

`f(6) = a/2`

`f(8) = a/4`

`:.\ text(Average ROC)` `= (a/4 – a/2)/(8 – 6)`
  `= -a/8`

`=>   E`

Statistics, 2ADV S3 EQ-Bank 16

A probability density function is defined by
 

`f(x) = {(a, \ text(for)\ \ 0 <= x <= 4),(3a, \ text(for)\  4 < x <= 8):}`
 

  1. Find the value of  `a`.  (2 marks)

  2. Sketch the probability density function.  (1 mark)

  3. Find an expression for the cumulative distribution function.  (2 marks)

Show Answers Only
  1. `1/16`
  2.  
  3. `F(x) = {(x/16, text(for)\ 0 <= x <= 4),((3x)/16 – 1/2, text(for)\ 4 < x <= 8):}`
Show Worked Solution

i.   `int_0^4 a\ dx = [ax]_0^4 = 4a`

`int_4^8 3a\ dx = [3ax]_4^8 = 24a – 12a = 12a`

`4a + 12a` `= 1`
`a` `= 1/16`

 

ii.   

 

iii.   `text(When)\ \ 0 <= x <= 4:`

`F(x) = int 1/16\ dx = x/16 + C`

`F(0) = 0 \ => \ C = 0`

`F(x) = x/16\ \ …\ (1)`
 

`text(When)\ \ 4 < x <= 8:`

`F(x) = int 3/16\ dx = (3x)/16 + C`

`F(4) = 1/4\ \ (text{see (1) above})`

COMMENT: Understand why you can check your equation here by confirming  `F(8)=1`

`(3 xx 4)/16 + C` `= 1/4`
`C` `= −1/2`

 
`F(x) = (3x)/16 – 1/2`
 

`:. F(x) = {(x/16, \ text(for)\ \ 0 <= x <= 4),((3x)/16 – 1/2, \ text(for)\ \ 4 < x <= 8):}`

Statistics, 2ADV S2 SM-Bank 14

A probability density function  `f(x)`  is given by
 

`f(x) = {(px(3 - x), \ text(if)\ \ 0 <= x <= 3),(0, \ text(if)\ \ x < 0\ \ text(or if)\ \ x > 3):}`
 

where  `p`  is a positive constant.

  1. Find the value of  `p`.  (2 marks)

  2. Find the mode of  `f(x)`.  (2 marks)

Show Answers Only
  1. `2/9`
  2. `3/2`
Show Worked Solution

i.   `text(Total Area under curve = 1)`

`p int_0^3 3x-x^2\ dx` `= 1`
`p[3/2 x^2-(x^3)/3]_0^3` `= 1`
`p[27/2-27/3-0]` `= 1`
`(9p)/2` `= 1`
`p` `= 2/9`

 

ii.   `f(x) = 2/9(3x-x^2)`

`f^{′}(x) = 2/9(3-2x)`

`text(S.P. when)\ \ f^{′}(x) = 0:`

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

 
`f^{″}(x) = -4/9 < 0 \ =>\ text(MAX)`

`:.\ text(Mode) = 3/2`

Statistics, 2ADV S2 SM-Bank 15

A function  \(f(x)\)  is given by

 \(f(x)= \begin{cases}
\dfrac{3}{4}(x-2)(4-x) & \text {if } 2 \leq x \leq 4 \\
\ \\
0 & \text{if } x<2 \text { or if } x>4
\end{cases}\)

  1. Show this curve is a probability density function.  (2 marks)

  2. Find the mode.  (2 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(3\)
Show Worked Solution
i.    \(f(x)\) \( =\dfrac{3}{4}(x-2)(4-x)\)
  \(\displaystyle \int_2^4 f(x) \) \( =\dfrac{3}{4} \displaystyle\int_2^4-x^2+6 x-8 d x\)
    \( =\dfrac{3}{4}\left[\dfrac{-x^3}{3}+3 x^2-8 x\right]_2^4\)
    \( =\dfrac{3}{4}\left[\left(-\dfrac{64}{3}+48-32\right)-\left(-\dfrac{8}{3}+12-16\right)\right] \)
    \( =\dfrac{3}{4}\left(-\dfrac{16}{3}+\dfrac{20}{3}\right) \)
    \(= 1\)

 

\(f(2)=0, f(4)=0\)

\(f(x)>0 \text { for } 2<x<4\)

\(f(2) \geq 0 \text { for } 2 \leq x \leq 4\)

 
\(\therefore f(x) \ \text{is a probability density function.}\)
 

ii.    \(f(x\) \(=\dfrac{3}{4}\left(-x^2+6 x-8\right)\)
  \(f^{\prime}(x)\) \(=\dfrac{3}{4}(-2 x+6)\)
  \(f^{\prime\prime}(x)\)  \(=-\dfrac{3}{2}\)

 
\(\text{SP when}\ f'(x) = 0\)

\(\begin{array}{r}-2 x+6=0 \\
x=3\end{array}\)

\(f^{\prime \prime}(x)<0 \Rightarrow \text {MAX}\)

\(\therefore \text {Mode}=3\)

Statistics, 2ADV S3 EQ-Bank 17

The diastolic measurement for blood pressure in 35-year-old people is normally distributed, with a mean of 75 and a standard deviation of 12.

  1. A person is considered to have low blood pressure if their diastolic measurement is 63 or less.What percentage of 35-year-olds have low blood pressure?  (1 mark)

  2. Calculate the `z`-score for a diastolic measurement of 57.  (1 mark)

  3. The probability that a 35-year-old person has a diastolic measurement for blood pressure between 57 and 63 can be found by evaluating
     
    `qquad qquad int_a^b f(x)\ dx`
     
    where `a` and `b` are constants and where
     
    `qquad qquad f(x) = 1/(sqrt(2pi)) e^((−x^2)/2)`
     
    is the normal probability density function with mean 0 and standard deviation 1.

     

    By first finding the values `a` and `b`, calculate an approximate value for this probability by using the trapezoidal rule with 3 function values.  (3 marks)

  4. Hence, find the approximate probability that a 35-year-old person chosen at random has a diastolic measurement of 57 or less.  (1 mark)

Show Answers Only
  1. `16text(% have low blood pressure)`
  2. `−1.5`
  3. `9.2text(%)`
  4. `6.8text(%)`
Show Worked Solution
i.    `ztext(-score)\ (63)` `= (x – mu)/σ`
    `= (63 – 75)/12`
    `= −1`

 

`:. 16text(% have low blood pressure)`

 

ii.    `ztext(-score)` `= (57 – 75)/12`
    `= −1.5`

 

iii.  `y = 1/sqrt(2pi) e^((−x^2)/2)`

 
`text(Area)` `= h/2(y_0 + 2y_1 + y_2)`
  `~~ 0.25/2 (0.1295 + 2 xx 0.1826 + 0.2420)`
  `~~ 0.0920`
  `~~ 9.2text(%)`

 

iv.   

 

`P(text{blood pressure}\ <= 57)` `= 16 – 9.2`
  `~~ 6.8text(%)`

Functions, 2ADV F1 SM-Bank 40

A factory makes boots and sandals. In any week

• the total number of pairs of boots and sandals that are made is 200
• the maximum number of pairs of boots made is 120
• the maximum number of pairs of sandals made is 150.

The factory manager has drawn a graph to show the numbers of pairs of boots (`x`) and sandals (`y`) that can be made.
 

2UG-2009-24d
 

  1. Find the equation of the line `AD`.   (1 mark)
  2. Explain why this line is only relevant between `B` and `C` for this factory.     (1 mark)
  3. The profit per week, `$P`, can be found by using the equation
     
  4.       `P = 24x + 15y`.
     
    Compare the profits at `B` and `C`.     (2 marks)

 

 

Show Answers Only
  1. `x + y = 200`
  2. `text(S)text(ince the max amount of boots = 120)`
  3. `=> x\ text(cannot)\ >120`
  4.  
  5. `text(S)text(ince the max amount of sandals = 150`
  6.  
  7. `=> y\ text(cannot)\ >150`
  8.  
  9. `:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`
  10.  
  11. `text(The profits at)\ C\ text(are $630 more than at)\ B.`

 

Show Worked Solution

(i)   `text{We are told the number of boots}\ (x),`

Question taken from Gen2 exam – now common content with Advanced course.

`text{and shoes}\  (y),\ text(made in any week = 200)`

`=>text(Equation of)\ AD\ text(is)\ \ x + y = 200`

 

(ii)  `text(S)text(ince the max amount of boots = 120)`

`=> x\ text(cannot)\ >120`

`text(S)text(ince the max amount of sandals = 150`

`=> y\ text(cannot)\ >150`

`:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`

 

(iii)  `text(At)\ B,\ \ x = 50,\ y = 150`

`=>$P  (text(at)\ B)` `= 24 xx 50 + 15 xx 150`
  `= 1200 + 2250`
  ` = $3450`

`text(At)\ C,\ \  x = 120 text(,)\ y = 80`

`=> $P  (text(at)\ C)` `= 24 xx 120 + 15 xx 80`
  `= 2880 + 1200`
  `= $4080`

 

`:.\ text(The profits at)\ C\ text(are $630 more than at)\ B.`

Calculus, MET1 2019 VCAA 9

Consider the functions  `f: R -> R,\ \ f(x) = 3 + 2x-x^2`  and  `g: R -> R,\ \ g(x) = e^x`

  1. State the rule of `g(f(x))` .  (1 mark) 

  2. Find the values of `x` for which the derivative of `g(f(x))` is a negative.  (2 marks)

  3. State the rule of `f(g(x))`.  (1 mark)

  4. Solve  `f(g(x)) = 0`.  (2 marks)

  5. Find the coordinates of the stationary point of the graph of `f(g(x))`.  (2 marks)

  6. State the number of solutions to  `g(f(x)) + f(g(x)) = 0`.  (1 mark)

Show Answers Only
  1. `g(f(x)) = e^(3 + 2x-x^2)`
  2. `x > 1`
  3. `f(g(x)) = 3 + 2e^x-e^(2x)`
  4. `x = ln 3`
  5. `text(S.P. at)\ (0, 4)`
  6. `text(1 solution only)`
Show Worked Solution

a.   `g(f(x)) = e^(3 + 2x-x^2)`
 

b.   `d/(dx) g(f(x)) = (2-2x)e^(3 + 2x-x^2)`

`e^(3 + 2x-x^2) > 0\ \ text(for all)\ \ x`

`=> d/(dx) g(f(x)) < 0\  \text(when):`

`2-2x` `< 0`
`x` `> 1`

 
c.   `f(g(x)) = 3 + 2e^x-e^(2x)`
 

d.   `e^(2x)-2e^x-3 = 0`

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

`X^2-2X-3` `= 0`
`(X-3)(X + 1)` `= 0`

 
`X = 3 or -1`

`text(When)\ \ X = 3,\ \ e^x = 3 => x = ln 3`

`text(When)\ \ X = -1,\ \ e^x = -1 \ => \ text(no solution)`

`:. x = ln 3`
 

e.    `d/(dx)\ f(g(x))` `= 2e^x-2e^(2x)`
    `= 2e^x (1-e^x)`

 
`text(S.P. occurs when:)`

`2e^x(1-e^x)` `= 0`
`e^x` `= 1`
`x` `= 0`

 
`text(When)\ \ x = 0,`

`f(g(x))` `= 3 + 2-1=4`

 
`:.\ text(S.P. at)\ \ (0, 4)`
 

f.    `text(Solutions occur when)\ \ g(f(x)) = -f(g(x))`

`text{Sketch both graphs (using parts a-e):}`
 


 

`:. 1\ text(solution only)`

Algebra, MET1 2019 VCAA 8

The function  `f: R -> R, \ f(x)`  is a polynomial function of degree 4. Part of the graph of  `f`  is shown below.

The graph of  `f`  touches the `x`-axis at the origin.
 


 

  1. Find the rule of  `f`.   (1 mark) 

Let  `g`  be a function with the same rule as  `f`.

Let  `h: D -> R, \ h(x) = log_e (g(x))-log_e (x^3 + x^2)`, where  `D`  is the maximal domain of  `h`.

  1. State  `D`.   (1 mark)

  2. State the range of  `h`.   (2 marks)

Show Answers Only
  1. `f(x) = -4x^2(x^2-1)`
  2. `x in (-1, 1) text(\{0})`
  3. `h(x) in (-oo, 3 log_e 2) text(\ {)2 log_e 2 text(})`
Show Worked Solution
a.    `y` `= ax^2 (x-1)(x + 1)`
    `= ax^2 (x^2-1)\ …\ (1)`

 
`text(Substitute)\ (1/ sqrt 2, 1)\ text{into  (1):}`

`1 = a ⋅ (1/sqrt 2)^2 ((1/sqrt 2)^2-1)`

`1 = a(1/2)(-1/2)`

`a = -4`

`:. f(x) = -4x^2(x^2-1)`

 

b.    `g(x) > 0` `=> -4x^2 (x^2-1) > 0`
    `=> x in (-1, 1)\  text(\{0})`

`text(and)`

`x^3 + x^2 > 0 => text(true for)\ \ x in (-1 , 1)\ text(\{0})`

`:. D:\ x in (-1, 1)\ text(\{0})`

 

c.    `h(x)` `= log_e ((-4x^2(x^2-1))/(x^3 + x^2))`
    `= log_e ((-4x^2(x + 1)(x-1))/(x^2(x + 1)))`
    `= log_e (4(1-x))\ \ text(where)\ \ x in (-1, 1)\ text(\{0})`

  
`text(As)\ \ x -> -1,\ \ h(x) -> log_e 8 = 3 log_e 2`

`text(As)\ \ x -> 1,\ \ h(x) -> -oo`

`text(As)\ \ x -> 0,\ \ h(x) -> log_e 4 = 2 log_e 2`

`text{(}h(x)\ text(undefined when)\ \ x = 0 text{)}`
 

`:.\ text(Range)\ \ h(x) in (-oo, 3 log_e 2)\ text(\{) 2 log_e 2 text(})`

Calculus, MET1 2019 VCAA 5

Let  `f: R\ text(\{1}) -> R, \ f(x) = 2/(x-1)^2 + 1`.

    1. Evaluate  `f(-1)`.   (1 mark)

    2. Sketch the graph of `f` on the axes below, labelling all asymptotes with their equations.   (2 marks)

        

  2. Find the area bounded by the graph of `f`, the `x`-axis, the line  `x = -1`  and the line  `x = 0`.   (2 marks)

Show Answers Only
    1. `3/2`
    2. `text(See Worked Solutions)`
  2. `2`
Show Worked Solution
a.i.    `f(-1)` `= 2/(-1-1)^2 + 1`
    `= 3/2`

​

a.ii.     

 

b.    `text(Area)` `= int_(-1)^0 2/(x-1)^2 + 1\ dx`
    `= int_(-1)^0 2(x-1)^(-2) + 1\ dx`
    `= [-2(x-1)^(-1) + x]_(-1)^0`
    `= [((-2)/-1 + 0)-((-2)/-2-1)]`
    `= 2`

Graphs, MET1 2019 VCAA 4

  1. Solve  `1-cos (x/2) = cos (x/2)`  for  `x in [-2 pi, pi]`.   (2 marks) 

  2. The function  `f: [-2pi, pi] -> R,\ \ f(x) = cos (x/2)`  is shown on the axes below.
     

     

     

    Let  `g: [-2pi, pi] -> R,\ \ g(x) = 1-f(x)`.

     

     

    Sketch the graph of  `g`  on the axes above. Label all points of intersection of the graphs of  `f`  and  `g`, and the endpoints of  `g`, with their coordinates.   (2 marks)

Show Answers Only
  1. `(2 pi)/3, (-2 pi)/3`
  2. `text(See Worked Solutions)`
Show Worked Solution
a.    `2 cos (x/2)` `= 1`
  `cos (x/2)` `= 1/2`

 
`=>\ text(Base angle)\ \ pi/3`

`x/2 = pi/3, -pi/3, (-5 pi)/3`

`x = (2 pi)/3, (-2 pi)/3 qquad (x in [-2pi, pi])`

 

b.   `text(Plot points for)\ \ g(x):`

`text(- reflect)\ f(x)\ text(in)\ x text(-axis, then)`

`text(- translate 1 upwards.)`

Probability, MET1 2019 VCAA 3

The only possible outcomes when a coin is tossed are a head or a tail. When an unbiased coin is tossed, the probability of tossing a head is the same as the probability of tossing a tail.

Jo has three coins in her pocket; two are unbiased and one is biased. When the biased coin is tossed, the probability of tossing a head is `1/3`.

Jo randomly selects a coin from her pocket and tosses it.

  1. Find the probability that she tosses a head.  (2 marks) 

  2.  Find the probability that she selected an unbiased coin, given that she tossed a head.  (1 mark)

Show Answers Only

  1. `4/9`
  2. `3/4`

Show Worked Solution

a.    `text(Pr)(B) = 1/3,\ \text(Pr)(B^{\prime}) = 2/3`

`text(Pr)(H|B) = 1/3,\ \ text(Pr)(H|B^{\prime}) = 1/2`

 

`:. text(Pr)(H)` `= text(Pr)(B) xx text(Pr)(H nn B) + text(Pr)(B^{\prime}) xx text(Pr)(H nn B^{\prime})`
  `= 1/3 xx 1/3 + 2/3 xx 1/2`
  `= 1/9 + 1/3`
  `= 4/9`

 

b.    `text(Pr)(B^{\prime}|H)` `= (text(Pr)(B^{\prime} nn H))/(text(Pr)(H))`
    `= (2/3 xx 1/2)/(4/9)`
    `= 3/4`

Algebra, MET1 2019 VCAA 2

  1. Let  `f: R\{1/3} -> R,\ f(x) = 1/(3x-1)`.

     

     

    Find the rule of  `f^(-1)`.   (2 marks) 

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

  3. Let  `g`  be the function obtained by applying the transformation  `T`  to the function  `f`, where
  4. `qquad qquad qquad T ([(x), (y)]) = [(x), (y)] + [(c), (d)]`
  5. and  `c, d in R`.
  6. Find the values of  `c`  and  `d`  given that  `g = f^(-1)`.   (1 mark)

Show Answers Only
  1. `y = 1/(3x) + 1/3`
  2. `x in R\ text(\){0}`
  3. `c = -1/3,\ d = 1/3`
Show Worked Solution

a.   `text(Let)\ \ y = 1/(3x-1)`

`text(Inverse:  swap)\ \ x↔ y`

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

 

b.   `x in R\ text(\){0}`

 

c.   `f(x) = 1/(3x-1) \ -> \ f(x) = 1/(3x) + 1/3`

`f(x + 1/3) = 1/(3(x + 1/3)-1) = 1/(3x)`

`:. c = -1/3,\ \ d = 1/3`

Calculus, MET1 2019 VCAA 1b

Let  `g: R text(\ {−1}) -> R,\ \ g(x) = (sin(pi x))/(x + 1)`.

Evaluate  `g^{prime}(1)`.   (2 marks)

Show Answers Only

`-pi/2`

Show Worked Solution
  `u = sin(pi x)` `v = x + 1`
  `u^{prime}=pi cos(pi x)` `v^{prime}=1`

 

`g^{prime}(x)` `=(vu^{prime}-uv^{prime})/v^2`
  `= ((x + 1) ⋅ pi cos(pi x)-sin (pi x))/(x + 1)^2`
`g^{prime}(1)` `= (2 pi cos(pi)-sin(pi))/2^2`
  `= (2 pi(-1)-0)/4`
  `= -pi/2`

L&E, 2ADV E1 SM-Bank 14

The spread of a highly contagious virus can be modelled by the function

`f(x) = 8000/(1 + 1000e^(−0.12x))`

Where `x` is the number of days after the first case of sickness due to the virus is diagnosed and `f(x)` is the total number of people who are infected by the virus in the first `x` days.

  1. Calculate `f(0)`.   (1 mark)

  2. Find the value of `f(365)` and interpret it result.   (2 marks)

Show Answers Only
  1. `7.99…`
  2. `text(After 1 year, the model predicts the total number)`
    `text(of people infected by the virus is 8000.)`
Show Worked Solution
i.   `f(0)` `= 8000/(1 + 1000e^0)`
    `= 8000/1001`
    `= 7.99…`

 

ii.    `f(365)` `= 8000/(1 + 1000e^(−0.12 xx 365))`
    `= 8000/(1 + 1000e^(−43.8))`
    `~~ 8000`

 
`text(After 1 year, the model predicts the total number)`

`text(of people infected by the virus is 8000.)`

Vectors, EXT1 V1 SM-Bank 23

A fireworks rocket is fired from an origin `O`, with a velocity of 140 metres per second at an angle of  `theta`  to the horizontal plane.
 


 

The position vector `underset~s(t)`, from `O`, of the rocket after  `t`  seconds is given by

`underset~s = 140tcosthetaunderset~i + (140tsintheta - 4.9t^2)underset~j`

The rocket explodes when it reaches its maximum height.

  1. Show the rocket explodes at a height of  `1000sin^2theta`  metres.  (2 marks)

  2. Show the rocket explodes at a horizontal distance of  `1000sin 2theta`  metres from `O`.  (1 mark)

  3. For best viewing, the rocket must explode at a horizontal distance of 500 m and 800 m from `O`, and at least 600 m above the ground.

     

    For what values of  `theta`  will this occur.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `63.4° <= theta <= 75°`
Show Worked Solution

i.    `underset~s = 140tcosthetaunderset~i + (140tsintheta – 4.9t^2)underset~j`

`underset~v = 140costhetaunderset~i + (140sintheta – 9.8t)underset~j`

`text(Max height occurs when)\ underset~j\ text(component of)\ underset~v = 0`

`0` `= 140sintheta – 9.8t`
`t` `= (140sintheta)/9.8`

 
`text(Max height:)\ \ underset~j\ text(component of)\ underset~s\ text(when)\ t = (140sintheta)/9.8`

`text(Max height)` `= 140sintheta · (140sintheta)/9.8 – (4.9 · 140^2sin^2theta)/(9.8^2)`
  `= 2000sin^2theta – 1000sin^2theta`
  `= 1000sin^2theta`

 

ii.   `text(Horizontal distance)\ (d):`

`=>\ underset~i\ text(component of)\ underset~s\ text(when)\ \ t = (140sintheta)/9.8`

`:.d` `= 140costheta · (140sintheta)/9.8`
  `= (140 xx 70 xx sin2theta)/9.8`
  `= 1000sin2theta`

 

iii.   `text(Using part ii),`

`500<=1000sin2theta<=800`
`0.5<=sin2theta<=0.8`

 

`text(In the 1st quadrant:)`

`30° <=` `2theta` `<= 53.13°`
`15° <=` `theta` `<= 26.6°`

 
`text(In the 2nd quadrant:)`

`126.87°<=` `2theta` `<= 150°`
`63.4°<=` `theta` `<= 75°`

 
`text(When)\ theta = 26.6°:`

`text(Max height)` `= 1000 · sin^2 26.6°`
  `= 200.5\ text(metres)\ (< 600\ text(m))`

 
`=>\ text(Highest max height for)\ \ 15° <= theta < 26.6°\ \ text(does not satisfy.)`
 

`text(When)\ theta = 63.4°:`

`text(Max height)` `= 1000 · sin^2 63.4°`
  `= 799.5\ text(metres)\ (> 600\ text(m))`

 
`=>\ text(Lowest max height for)\ \ 63.4° <= theta <= 75°\ \ text(satisfies).`

`:. 63.4° <= theta <= 75°`