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Calculus, SPEC2 2012 VCAA 19 MC

A body is moving in a straight line and, after  `t`  seconds, it is `x` metres from the origin and travelling at  `v` ms`\ ^(−1)`.

Given that  `v = x`, and that  `t = 3`  where  `x = −1`, the equation for `x` in terms of  `t`  is

  1. `x = e^(t - 3)`
  2. `x = −e^(3 - t)`
  3. `x = sqrt(2t - 5)`
  4. `x = −sqrt(2t - 5)`
  5. `x = −e^(t - 3)`
Show Answers Only

`E`

Show Worked Solution

`text(S)text(ince)\ \ v = x`

`(dx)/(dt)` `= x`
`(dt)/(dx)` `= 1/x`
`t` `= int 1/x\ dx`
`t` `= log_e|\ x\ | +c`

 
`text(When)\ \ x=-1,\ t=3`

`3` `=log_e|\ -1\ |+c`  
`c` `=3`  

 

`t` `=log_e|\ x\ |+3`
`log_e|\ x\ |` `= t-3`
`|\ x\ |` `= e^(t-3)`
`:.x` `= -e^(t-3)\ \ \ (e^(t-3)\ text{doesn’t satisfy}\ \ x=-1\ \ text{when}\ \ t=3)`

 
`=> E`

Calculus, SPEC2 2012 VCAA 12 MC

The volume of the solid of revolution formed by rotating the graph of  `y = sqrt (9-(x-1)^2)`  above the `x`-axis is given by

  1. `4 pi(3)^2`
  2. `pi int_(−3)^3(9-(x-1)^2)dx`
  3. `pi int_(−2)^(4)(sqrt(9-(x-1)^2))dx`
  4. `pi int_(−2)^4(9-(x-1)^2)^2dx`
  5. `pi int_(−4)^2(9-(x-1)^2)dx`
Show Answers Only

`A`

Show Worked Solution

`y = sqrt (9-(x-1)^2)\ \ => text(Semi-circle of circle)`

♦ Mean mark 48%.

`text{with centre (1,0), radius = 3}`
 

`:. V` `= 4/3 pi r^3`
  `= 4/3 pi (3)^3`
  `=4 pi (3)^2`

 
`=> A`

Complex Numbers, SPEC2 2011 VCAA 8 MC

On the argand diagram below, the twelve points  `P_1, P_2, P_3 …, P_12`  are evenly spaced around the circle of radius 3.
 

SPEC2 2011 VCAA 8 MC
 

The points which represent complex numbers such that  `z^3 = -27i`  are

A.   `P_10` only

B.   `P_4` only

C.   `P_2, P_6, P_10`

D.   `P_3, P_7, P_11`

E.   `P_4, P_8, P_12`

Show Answers Only

`E`

Show Worked Solution

`z^3 = 27text(cis)(−pi/2)`

♦ Mean mark 49%.

`:. z_1 = 3text(cis)(−pi/6)\ \ text(is one root.)`

`text(The other two are evenly spaced around the circle)`

`text(which separates them by)\ \ (2pi)/3.`

`=> E`

Graphs, SPEC2 2011 VCAA 1 MC

The number of straight line asymptotes of the graph of  `y = (2x^3 + x^2 - 1)/(x^2 - x - 2)`  is

A.   0

B.   1

C.   2

D.   3

E.   4

Show Answers Only

`D`

Show Worked Solution
`y` `= (2x^3 + x^2 – 1)/(x^2 – x – 2)`
  `= 2x + 3 + (7x + 5)/((x – 2)(x + 1))`

 
`:.\ text(Straight line asymptotes occur at:)`

♦♦ Mean mark 34%.

`y = 2x + 3, \ x = 2\ \ text(and)\ \ x = −1`

`=> D`

Trigonometry, SPEC1 2011 VCAA 8

Find the coordinates of the points of intersection of the graph of the relation

`y = text(cosec)^2 ((pi x)/6)`  with the line  `y = 4/3`, for  `0 < x < 12.`  (3 marks)

Show Answers Only

`(2, 4/3),(4, 4/3),(8, 4/3), (10, 4/3)`

Show Worked Solution

`text(Intersection occurs when:)`

♦ Mean mark 46%.

`text(cosec)^2((pix)/6)` `=4/3`
`text(cosec)((pix)/6)` `= ±2/sqrt3`
`sin((pix)/6)` `= ±sqrt3/2`

 
`text(Given:)\ \ 0 < x < 12 \ \ =>\ \ 0 < (pix)/6 < 2pi`
 

`(pix)/6` `= pi/3, pi – pi/3, pi + pi/3, 2pi – pi/3`
  `= pi/3, (2pi)/3, (4pi)/3,(5pi)/3`
`x` `= 2, 4, 8, 10`

  
`=> y = 4/3\ text(for each)`

`:.\ text(Intersection at:)\ \ (2, 4/3),(4, 4/3),(8, 4/3), (10, 4/3)`

Complex Numbers, SPEC2 2012 VCAA 6 MC

For any complex number `z`, the location on an Argand diagram of the complex number  `u = i^3 bar z`  can be found by

A.   rotating `z` through `(3 pi)/2` in an anticlockwise direction about the origin

B.   reflecting `z` about the `x`-axis and then reflecting about the `y`-axis

C.   reflecting `z` about the `y`-axis and then rotating anticlockwise through `pi/2` about the origin

D.   reflecting `z` about the `x`-axis and then rotating anticlockwise through `pi/2` about the origin

E.   rotating `z` through `(3 pi)/2` in a clockwise direction about the origin

Show Answers Only

`C`

Show Worked Solution

`z -> bar z:\ text(reflect in)\ x text(-axis)`

`x+iy\ \ =>\ \ x-iy`

`bar z -> i^3 bar z:\ text(rotate)\ 90^@ xx 3\ text(anticlockwise, or)\ 90^@\ text(clockwise)`

`i^3 bar z` `= i^3 (x – iy)`
  `= -i(x – iy)`
  `= -y – ix `

 
`text(Consider option)\ C:`

♦♦ Mean mark 37%.

`z\ text(reflected about)\ y text(-axis)`

`-x + iy`

`text(then rotated)\ pi/2\ text(anticlockwise)`

`i(-x + iy)` `= -ix + i^2y`  
  `= -y – ix`  ✔  

 
`=> C`

Graphs, SPEC1 2012 VCAA 10

Consider the functions with rules  `f(x) = arcsin (x/2) + 3/sqrt (25 x^2-1)`  and  `g(x) = arcsin (3x)-3/sqrt (25x^2-1).`

    1. Find the maximal domain of  `f_1(x) = arcsin (x/2).`   (1 mark)

    2. Find the maximal domain of  `f_2(x) = 3/sqrt (25x^2-1).`   (1 mark)

    3. Find the largest set of values of  `x in R`  for which  `f(x)`  is defined.   (1 mark)

  2. Given that  `h(x) = f(x) + g(x)`  and that  `theta = h(1/4)`, evaluate  `sin (theta).`

     

    Give your answer in the form  `(a sqrt b)/c, \ a, b, c in Z.`   (3 marks)

Show Answers Only
    1. `-2 <= x <= 2`
    2. `x < -1/5 uu x > 1/5`
    3. `-2 <= x < -1/5 uu 1/5 < x <= 2`
  2. `(5 sqrt 7)/16`
Show Worked Solution

a.i.   `text(Maximal Domain occurs when:)`

  `-1 <= x/2 <= 1`
  `{x: -2 <= x <= 2}`

 
a.ii.   `text(Maximal Domain occurs when:)`

♦♦ Mean mark part (a)(ii) 33%.

  `25x^2-1 > 0`
  `x^2 > 1/25`
  `{x: x < -1/5 uu x > 1/5}`

 

a.iii.  `text(Max domain for which)\ \ f(x)\ \ text(is defined:)`

♦♦ Mean mark part (a)(iii) 26%.

  `(-2 <= x <= 2) nn (x < -1/5 uu x > 1/5)`
  `{x: -2 <= x < -1/5 \ uu \ 1/5 < x <= 2}`

 

b.    `h(x)` `= sin^(-1) (x/2) + 3/sqrt(25x^2-1) + sin^(-1)(3x)-3/sqrt(25x^2-1)`
    `= sin^(-1)(x/2) + sin^(-1)(3x)`

 
`text(When)\ \ x=1/4,\ \ h(x)=theta`

♦♦♦ Mean mark part (b) 20%.

`theta=sin^(-1) (1/8) + sin^(-1) (3/4)`

`text(Let)\ \ theta_1 = sin^(-1) (1/8),\ \ theta_2 = sin^(-1) (3/4)`

`text(Using)\ \ sin(theta_1 + theta_2)=sin theta_1 cos theta_2 + cos theta_1 sin theta_2:`

`sin(theta)` `= 1/8 * sqrt7/4 + sqrt63/8 * 3/4`
  `=sqrt7/32 + (9sqrt7)/32`
  `=(5 sqrt7)/16`

Vectors, SPEC1 2012 VCAA 9

The position of a particle at time  `t`  is given by

`underset ~r (t) = (2 sqrt (t^2 + 2) - t^2) underset ~i + (2 sqrt (t^2 + 2) + 2t) underset ~j,\ \ t >= 0.`

  1. Find the velocity of the particle at time  `t.`   (1 mark)

  2. Find the speed of the particle at time  `t = 1`  in the form  `(a sqrt b)/c`, where `a, b` and `c` are positive integers.   (2 marks)

  3. Show that at time  `t = 1,\ \ (dy)/(dx) = (1 + sqrt 3)/(1 - sqrt 3).`   (2 marks)

  4. Find the angle in terms of `pi`, between the vector  `-sqrt 3 underset ~i + underset ~j`  and the vector  `underset ~r (t)`  at time  `t = 0.`   (2 marks)

Show Answers Only
  1. `((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`
  2. `(4 sqrt 6)/3`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `(7 pi)/12`
Show Worked Solution

a.  `underset ~r (t) = (2 sqrt (t^2 + 2) – t^2) underset ~i + (2 sqrt (t^2 + 2) +2t) underset ~j`

  `underset ~v(t)` `=dot underset ~r(t)`
    `= (2(2t)(1/2)(t^2 + 2)^(-1/2) – 2t) underset ~i + (2(2t)(1/2)(t^2 + 2)^(-1/2) + 2)underset ~j`
    `= ((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`

 

b.    `underset ~v(1)` `= (2/sqrt(1 + 2) – 2)underset ~i + (2/sqrt(1 + 2) + 2) underset ~j`
    `= (2/sqrt 3 – 2) underset ~i + (2/sqrt 3 + 2)underset ~j`
  `|\ underset ~v(1)\ |` `= sqrt((2/sqrt 3 – 2)^2 + (2/sqrt 3 + 2)^2)`
    `= sqrt(4/3 – 8/sqrt 3 + 4 + 4/3 + 8/sqrt 3 + 4)`
    `= sqrt(8/3 + 8)`
    `= sqrt(32/3)`
    `= (4sqrt2)/sqrt3`
    `= (4 sqrt 6)/3`

♦ Mean mark part (c) 41%.

c.    `(dy)/(dt)` `= (dy)/(dx) xx (dx)/(dt)`
  `(dy)/(dx)` `=((dy)/(dt))/((dx)/(dt))`
  `:. (dy)/(dx)|_(t=1)` `= (2/sqrt 3 + 2)/(2/sqrt 3 – 2)`
    `= ((2 + 2 sqrt 3)/sqrt 3) xx (sqrt 3/(2 – 2 sqrt 3))`
    `= (2 + 2 sqrt 3)/(2 – 2 sqrt 3)`
    `= (1 + sqrt 3)/(1 – sqrt 3)`

 

d.   `text(At)\ \ t=0:` 

♦♦ Mean mark part (d) 32%.

`underset ~r(0) = 2 sqrt 2 underset ~i + 2 sqrt 2 underset ~j`

`theta_1` `= tan^(-1)((2 sqrt 2)/(2 sqrt 2)) = pi/4`
`theta_2` `= tan^(-1)(1/sqrt 3)=pi/6`

 
`text(Let)\ \ theta=\ text(angle between the vectors:)`

`theta` `= pi – theta_1 – theta_2`
  `= pi – pi/4 – pi/6`
  `= (12 pi – 3 pi – 2 pi)/12`
  `= (7 pi)/12`

Complex Numbers, SPEC1 2012 VCAA 3

Consider the equation  `z^3-z^2-2z-12 = 0, \ z in C.`

  1. Given that  `z = 2 text(cis) ((2pi)/3)`  is a root of the equation, find the other two roots in the form  `a + ib`, where  `a, b in R.`   (3 marks)

  2. Plot all of the roots clearly on the Argand diagram below.   (1 mark)

     

     


Show Answers Only
  1. `z_2 = -1-i sqrt 3`
    `z_3 = 3 + 0i`
     

  2.  

Show Worked Solution

a.   `text(Given)\ \  z = 2 text(cis)((2 pi)/3)\ \ text(is a root, and)`

`z^3-z^2-2z-12 = 0\ \ text(has real coefficients,)`

`=> z= 2 text(cis)(-(2 pi)/3)\ \ text{is a root (conjugate).}`

`=> text(Roots:)\ \ -1+sqrt3i,\ \ -1-sqrt3i`
 

`(z-alpha) (z-beta) (z-gamma) = z^3-z^2-2z-12`
 

`text(Using)\ \ alpha beta gamma = 12:`

`(-1 + i sqrt 3) (-1-i sqrt 3) gamma` `=12`  
`((-1)^2-(sqrt3i)^2)gamma` `=12`  
`4gamma` `=12`  
`gamma` `=3`  

 
`:.\ text(Other two roots:)`

♦♦ Mean mark part (b) 33%.

`z_2` `= -1-i sqrt 3`
`z_3` `= 3 + 0i`

 

b.   

Trigonometry, SPEC1 2012 VCAA 2

Find all real solutions of the equation  `2 cos(x) = sqrt 3 cot (x).`  (3 marks)

Show Answers Only

`x=((2n+1)pi)/2, pi/3 + 2npi, (2pi)/3 + 2npi\ \ \ (n in ZZ)`

Show Worked Solution

`2 cos x = sqrt 3 cot x`

`2 cos x – sqrt 3 cot x` `= 0`  
`2 cos x – sqrt 3 (cos x)/(sin x)` `=0`  
`(2 – sqrt 3/sin x) cos x` `=0`  

 
`text(Solution 1:)\ \ cosx=0`

♦♦ Mean mark 28%.

`x=pi/2, (3pi)/2, (5pi)/2, …`

`x=((2n+1)pi)/2\ \ \ (n in ZZ)`
 

`text(Solution 2:)\ \ 2 – sqrt 3/sin x = 0\ \ =>\ \ sin x = sqrt 3/2`

`x = pi/3, (2pi)/3, (7pi)/3, (8pi)/3, …`

`x=pi/3 + 2npi, (2pi)/3 + 2npi\ \ (n in ZZ)`

Mechanics, SPEC2 2013 VCAA 22 MC

A parachutist of mass `m` kg falls towards Earth and is slowed by air resistance of  `kv^2` newtons, where `v` is the velocity of the parachutist `t` seconds after commencing the fall.

The equation of motion for the parachutist is

A.   `(dv)/(dt) = g - kv^2`

B.   `m(d^2x)/(dt^2) = g - kv^2`

C.   `m(d(v^2))/(dx) = (mg - kv^2)/2`

D.   `v(dv)/(dx) = 1 - (kv^2)/(mg)`

E.   `(dv)/(dx) = g/v - (kv)/m`

Show Answers Only

`E`

Show Worked Solution

`ma = mg – kv^2`

`mv(dv)/(dx) = mg – kv^2`

`:. (dv)/(dx) = g/v – (kv)/m`

`=> E`

 

—–NEW—–

`ma` `= mg – kv^2`
`a` `= g – (kv^2)/m`

`(dv)/(dt) = g – (kv^2)/m = v(dv)/(dx) = (d^2x)/(dt^2)`

`(dv)/(dx) = g/v – (kv)/m`

`=> E`

Graphs, SPEC2 2013 VCAA 4 MC

The graphs of  `y = ax`  and  `y = arctan(bx)`  intersect exactly three times if

A.   `0 < b < a`

B.   `a < b < 0`

C.   `a = b`

D.   `b < a < 0`

E.   `0 < b^2 < a^2`

Show Answers Only

`D`

Show Worked Solution

`text{Graphs intersect at (0,0).}`

♦ Mean mark 43%.

`y = tan^(−1)(bx)\ \ =>\ \ dy/dx = b/(1 + (bx)^2)`

`text(At)\ \ x=0,\ \ m=b`

`y=ax\ \ =>\ \ dy/dx = a`
 

`text(To intersect exactly 3 times:)`

`text(Gradient of)\ \ y=ax\ \ text(must be less than)\ b\ text(for positive gradients.)`

`0<a<b`

`text(Gradient of)\ \ y=ax\ \ text(must be greater than)\ b\ text(for negative gradients.)`

`b<a<0`

`=> D`

Algebra, SPEC2 2013 VCAA 3 MC

The graph of  `y = 1/(ax^2 + bx + c)`  has asymptotes at  `x = − 5`, `x = 3`  and  `y = 0`.

Given that the graph has one stationary point with a `y`-coordinate of  `−1/8`, it follows that

A.   `a = 1`, `b = 2`, `c = −15`

B.   `a = 1/2`, `b = −1`, `c = −15/2`

C.   `a = −1/2`, `b = −1`, `c = 15`

D.   `a = −1`, `b = −2`, `c = −15`

E.   `a = 1/2`, `b = 1`, `c = −15/2`

Show Answers Only

`E`

Show Worked Solution
`y` `= 1/(ax^2 + bx + c)`
  `= 1/(a(x + 5)(x – 3))`
  `= 1/(ax^2 + 2ax – 15a)`

 
`ax^2 + bx + c = ax^2 + 2ax – 15a`

♦ Mean mark 47%.

 
`text(Vertex:)\ \ ((− 5 + 3)/2, – 1/8) = (– 1,– 1/8)`

`- 1/8` `= 1/(a – 2a – 15a)`
`- 1/8` `= -1/(16a)`
`a` `= 1/2`

 

`:. a = 1/2, \ b = 2a = 1, \ c = − 15a = − 15/2`

`=> E`

Complex Numbers, SPEC1 2013 VCAA 8

Find all solutions of  `z^4 - 2z^2 + 4 = 0,\ \ z in C`  in cartesian form.  (4 marks)

Show Answers Only

`sqrt6/2 + i/sqrt2; −sqrt6/2 – i/sqrt2; sqrt6/2 – i/sqrt2; −sqrt6/2 + i/sqrt2`

Show Worked Solution
`(z^2)^2 – 2(z^2) + 4` `= 0`
`(z^2)^2 – 2z^2 + 1^2 – 1 + 4` `= 0`
`(z^2 – 1)^2 + 3` `= 0`
`(z^2 – 1)^2` `= −3`
`z^2 – 1` `= ±isqrt3`
`z^2` `= 1 ± isqrt3`

 

`text(If)\ \ z^2` `= 1 + isqrt3`
  `= sqrt(1 + 3)text(cis)(tan^(−1)(sqrt3))`
  `= 2text(cis)(pi/3)`

 
`z_1 = sqrt2 text(cis)(pi/6)`

♦ Mean mark 50%.

`z_2` `= sqrt2 text(cis)(pi/6 – pi)`
  `= sqrt2 text(cis)((−5pi)/6)`

 

`text(If)\ \ z^2` `= 1 – isqrt3`
  `= sqrt(1 + 3) text(cis)(−pi/3)`
  `= 2 text(cis)(−pi/3)`

 
`z_3 = sqrt2 text(cis)(−pi/6)`

`z_4` `= sqrt2 text(cis)(−pi/6 + pi)`
  `= sqrt2 text(cis)((5pi)/6)`

 
`:. z = sqrt2(sqrt3/2 + i/2), sqrt2(−sqrt3/2 – i/2), sqrt2(sqrt3/2 – i/2), sqrt2(−sqrt3/2 + i/2)`

`= sqrt6/2 +sqrt2/2 i; −sqrt6/2 – sqrt2/2 i; sqrt6/2 – sqrt2/2 i; −sqrt6/2 + sqrt2/2 i`

Vectors, SPEC1 2013 VCAA 7

The position vector  `underset ~r (t)`  of a particle moving relative to an origin `O` at time `t` seconds is given by

`underset ~r(t) = 4 sec (t) underset ~i + 2 tan (t) underset ~j,\ t in [0, pi/2)`

where the components are measured in metres.

  1. Show that the cartesian equation of the path of the particle is  `x^2/16-y^2/4 = 1.`   (2 marks)

  2. Sketch the path of the particle on the axes below, labelling any asymptotes with their equations.   (2 marks)

     

        
     VCAA 2013 spec 7b
     

  3. Find the speed of the particle, in `text(ms)^-1`, when `t = pi/4.`   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  
  3. `4 sqrt 3\ \ text(ms)^-1`
Show Worked Solution
a.   `x` `= 4sec(t)`
  `x/4` `= sec(t)`
  `y` `= 2tan(t)`
  `y/2` `= tan(t)`

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

`(x^2)/16 +1 ` `= y^2/4`
`:. (x^2)/16-(y^2)/4` `=1`

 

b.   `y^2/4 = x^2/16 -1\ \ =>\ \ y=+- sqrt(x^2/4 -4)`

♦ Mean mark part (b) 45%.

`lim_(x->oo) y = +- x/2`

 

♦ Mean mark part (c) 39%.

c.    `overset·underset~r(t)` `= d/(dt)(4(cos(t))^(−1))underset~i + d/(dt)(2tan(t)) underset~j`
    `= 4(−1)(−sin(t))(cos(t))^(−2)underset~i + 2sec^2(t)underset~j`
    `= 4sin(t)sec^2(t)underset~i + 2sec^2(t)underset~i`

 

`|overset·underset~r(pi/4)|` `= sqrt(16sin^2(pi/4)sec^4(pi/4) + 4sec^4(pi/4))`
  `= sqrt(16(1/sqrt2)^2(sqrt2)^4 + 4(sqrt2)^4)`
  `= sqrt(16(1/2)(4) + 4(4))`
  `= sqrt(48)`
  `= 4sqrt3\ \ text(ms)^(-1)`

Calculus, SPEC2 2014 VCAA 22 MC

The velocity–time graph below shows the motion of a body travelling in a straight line, where `v\ text(ms)^(−1)` is its velocity after `t` seconds.

The velocity of the body over the time interval  `t in [4,9]`  is given by  `v(t) = −9/16(t - 4)^2 + 9`.

The distance, in metres, travelled by the body over nine seconds is closest to

A.   45.6

B.   47.5

C.   48.6

D.   51.0

E.   53.4

Show Answers Only

`E`

Show Worked Solution

`text(Distance travelled = area between curve and)\ xtext(-axis.)`

`text(Between)\ \ t=0 and t=4:`

♦ Mean mark 49%.

`d` `= 1/2 xx 9 (4 + 2)`
  `= 27`

 
`text(Between)\ \ t=4 and t=8:`

`d` `= int_4^8 9 – 9/16(t – 4)^2\ dt`
  `= 24`

 
`text(Between)\ \ t=8 and t=9:`

`d` `= |\ int_8^9 9 – 9/16 (t – 4)^2\ dt\ |`
  `= 39/16`

 

`:.\ text(Total distance)` `= 27 + 24 + 39/16`
  `~~ 53.4`

 
`=> E`

Vectors, SPEC2 2014 VCAA 19 MC

The velocity vector of a 5 kg mass moving in the cartesian plane is given by  `underset ~v(t) = 3 sin(2t) underset ~i + 4 cos(2t) underset ~j`, where velocity components are measured in m/s.

During its motion, the maximum magnitude of the net force, in newtons, acting on the mass is

  1. 8
  2. 30
  3. 40
  4. 50
  5. 70
Show Answers Only

`C`

Show Worked Solution

`underset ~a(t) = underset ~dot v(t)= 6 cos(2t) underset ~i – 8 sin(2t) underset ~j`

`|\ underset ~a(t)\ |` `= sqrt(36 cos^2(2t) + 64 sin^2(2t))`
  `= sqrt(36 (cos^2(2t) + sin^2(2t)) + 28 sin^2(2t))`
  `= sqrt(36 + 28 sin^2 (2t))`

 

`|\ underset ~a(t)\ |_max` `= sqrt(36 + 28) qquad text(as max) (sin^2(2t)) = 1`
  `= 8`

 

♦ Mean mark 49%.

`:. |\ underset ~F\ |_max` `= 8 xx 5`
  `= 40`

`=> C`

Mechanics, SPEC2 2014 VCAA 18 MC

A body on a horizontal smooth plane is acted upon by four forces, `underset ~F_1`, `underset ~F_2`, `underset ~F_3` and `underset ~F_4` as shown.

The force `underset ~F_1` acts in a northerly direction and the force `underset ~F_4` acts in a westerly direction.
 

 

Given that  `|\ underset ~F_1\ | = 1`, `|\ underset ~F_2\ | = 2`, `|\ underset ~F_3\ | = 4` and `|\ underset ~F_4\ | = 5`, the motion of the body is such that it

A.   is in equilibrium.

B.   moves to the west.

C.   moves to the north.

D.   moves in the direction 30° south of west.

E.   moves to the east.

Show Answers Only

`E`

Show Worked Solution

`text(Resolving forces horizontally:)`

♦ Mean mark 45%.

`sum F` `= underset ~F_2 cos30 + underset ~F_3 sin60 –  underset ~F_4`
  `= 2 xx sqrt3/2 + 4 xx sqrt3/2 -5`
  `~~0.2\ \ text(East)`

 
`text(Resolving forces vertically:)`

`sum F` `= underset ~F_1 + underset ~F_2 sin30 – underset ~F_3 cos 60`
  `= 1 + 2 xx 1/2 -4 xx 1/2`
  `= 0`

 
`:.\ text(Body moves to the east.)`

`=> E`

Calculus, SPEC2 2015 VCAA 22 MC

A ball is thrown vertically up with an initial velocity of  `7sqrt6` ms`\ ^(–1)`, and is subject to gravity and air resistance.

The acceleration of the ball is given by  `ddotx = −(9.8 + 0.1v^2)`, where `x` metres is its vertical displacement, and `v` ms`\ ^(–1)` is its velocity at time `t` seconds.

The time taken for the ball to reach its maximum height is

A.   `pi/3`

B.   `(5pi)/(21sqrt2)`

C.   `log_e(4)`

D.   `(10pi)/(21sqrt2)`

E.   `10log_e(4)`

Show Answers Only

`D`

Show Worked Solution

`(dv)/(dt) = −(9.8 + 0.1v^2)`

♦ Mean mark 42%.

`(dt)/(dv) = −1/(9.8 + 0.1v^2)`
 

`text(Max height occurs when)\ \ v=0.`

`text(Find)\ \ t\ \ text(when)\ \ v=0:`

`t` `= int_(7sqrt6)^0 −1/(9.8 + v^2/10)\ dv`
  `= (10pi)/(21sqrt2)`

 
`=> D`

Vectors, SPEC2 2015 VCAA 17 MC

Points  `A`, `B` and `C` have position vectors  `underset~a = 2underset~i + underset~j`,  `underset~b = 3underset~i - underset~j + underset~k`  and  `underset~c = -3underset~j + underset~k`  respectively.

The cosine of angle `ABC` is equal to

  1. `5/(sqrt6sqrt10)`
  2. `7/(sqrt6sqrt13)`
  3. `-1/(sqrt6sqrt13)`
  4. `-7/(sqrt21sqrt6)`
  5. `-2/(sqrt6sqrt13)`
Show Answers Only

`C`

Show Worked Solution

`vec(BA) = vec(OA) – vec(OB) = -tildei + 2tildej – tildek`

♦ Mean mark 48%.

`=>\ |\ vec(BA)\ | = sqrt6`

`vec(BC) = vec(OC) – vec(OB) = -3tildei – 2tildej`

`=>\ |\ vec(BC)\ | = sqrt13`
 

`vec(BA).vec(BC)` `= |\ vec(BA)\ ||\ vec(BC)\ |cos angleABC`
`cos angleABC` `= (vec(BA).vec(BC))/(|\ vec(BA)\ ||\ vec(BC)\ |)`
  `= (-3xx-1 + 2 xx – 2)/(sqrt6sqrt13)`
  `= (-1)/(sqrt6sqrt13)`

`=> C`

Vectors, SPEC2 2015 VCAA 15 MC

The component of the force  `underset~F = aunderset~i + bunderset~j`, where `a` and `b` are non-zero real constants, in the direction of the vector  `underset~w = underset~i + underset~j`, is

A.   `((a + b)/2)underset~w`

B.   `underset~F/(a + b)`

C.   `((a + b)/(a^2 + b^2))underset~F`

D.   `(a + b)underset~w`

E.   `((a + b)/sqrt2)underset~w` 

Show Answers Only

`A`

Show Worked Solution
`hatw` `= tildew/sqrt(1+1)`
  `= (tildei + tildej)/sqrt2`

`tildeF*hatw = (a + b)/sqrt2`

♦ Mean mark 49%.

`(tildeF*hatw)hatw` `= ((a + b)/sqrt2) tildew/sqrt2`
  `= ((a + b)/2) tildew`

 
`=> A`

Calculus, SPEC2 2015 VCAA 13 MC

SPEC2 2015 VCAA 13 MC
 

The direction field for a certain differential equation is shown above.

The solution curve to the differential equation that passes through the point  `(–2.5, 1.5)`  could also pass through

A.   `(0, 2)`

B.   `(1, 2)`

C.   `(3, 1)`

D.   `(3, –0.5)`

E.   `(–0.5, 2)`

Show Answers Only

`C`

Show Worked Solution

`text{Draw a graph that goes through (–2.5, 1.5) such that all}`

♦ Mean mark 47%.

`text{gradient curve lines are tangential:}`

`(3, –0.5)`

`=> D`

Calculus, SPEC2 2016 VCAA 16 MC

A cricket ball is hit from the ground at an angle of 30° to the horizontal with a velocity of 20 ms¯¹. The ball is subject only to gravity and air resistance is negligible.

Given that the field is level, the horizontal distance travelled by the ball, in metres, to the point of impact is

A.   `(10 sqrt 3)/g`

B.   `20/g`

C.   `(100 sqrt 3)/g`

D.   `(200 sqrt 3)/g`

E.   `400/g`

Show Answers Only

`D`

Show Worked Solution

`u_x` `= 20 cos 30^@`
  `= 10 sqrt 3`
`u_y` `= 20 sin 30^@`
  `= 10`
`h` `= u_y t + 1/2 a_yt^2`
  `= 10t – (g t^2)/2`

 
`text(Find)\ \ t\ \ text(when)\ \ h=0:`

`0 = t(10 – (g t)/2)`

♦ Mean mark 46%.

`t = 0, or  t=20/g`

 
`text(Horizontal distance travelled:)`

`x` `= u_xt + 1/2 a_xt^2`
  `= 10 sqrt 3 xx 20/g +0`
  `= (200 sqrt 3)/g`

 
`=>  D`

Mechanics, SPEC2 2016 VCAA 14 MC

Two light strings of length 4 m and 3 m connect a mass to a horizontal bar, as shown below

The strings are attached to the horizontal bar 5 m apart.
 


 

Given the tension in the longer string is  `T_1`  and the tension in the shorter string is  `T_2`,  the ratio of the tensions  `T_1/T_2`  is

A.   `3/5`

B.   `3/4`

C.   `4/5`

D.   `5/4`

E.   `4/3`

Show Answers Only

`B`

Show Worked Solution

`text{Strings make up 3-4-5 (right-angled) triangle:}`

♦ Mean mark 41%.

`text(Let)\ \ theta =\ text(angle between 3m string and horizontal bar)`

`T_1/T_2` `= cot theta`
  `= 3/4`

  
`=>  B`

Calculus, SPEC2 2016 VCAA 7 MC

Given that  `x = sin(t) - cos(t)`  and  `y = 1/2 sin(2t)`, then  `(dy)/(dx)`  in terms of  `t`  is

A.  `cos(t) - sin(t)`

B.  `cos(t) + sin(t)`

C.  `sec(t) + text(cosec)(t)`

D.  `sec(t) - text(cosec)(t)`

E.  `(cos (2t))/(cos(t) - sin(t))` 

Show Answers Only

`A`

Show Worked Solution

`(dx)/(dt)= cos(t) + sin(t),\ \ dy/(dt)=cos(2t)`

♦ Mean mark 37%.

`(dy)/(dx)` `= (dy)/(dt) ⋅ (dt)/(dx)`
  ` = cos(2t) /(cos(t) + sin(t))`
  `= (2 cos^2(t) – 1)/(cos(t) + sin(t))`
  ` = (cos^2(t) – sin^2(t))/(cos(t) + sin(t))`
  `= {(cos(t) – sin(t))(cos(t) + sin(t))}/(cos(t) + sin(t))`
  `= cos(t) – sin(t)`

 
`=>  A`

Calculus, SPEC1 2016 VCAA 10

Solve the differential equation  `sqrt(2-x^2) (dy)/(dx) = 1/(2-y)`, given that  `y(1) = 0`. Express `y` as a function of  `x`.  (5 marks)

Show Answers Only

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

Show Worked Solution
`sqrt(2-x^2) *(dy)/(dx)` `= 1/(2-y)`
`(2-y)* (dy)/(dx)` `= 1/sqrt(2-x^2)`
`int 2-y\ dy` `= int 1/(sqrt(2-x^2))\ dx`
`2y-y^2/2` `= sin^(-1) (x/sqrt 2) + c`

 
`text(Given)\ \ y(1) = 0:`

♦ Mean mark 46%.

`0=sin^(-1) (1/sqrt 2) + c`

`c=-pi/4`

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

 
`text(Given)\ \ y=0\ \ text(when)\ \ x=1:`

`:. y=2-sqrt(4 + pi/2-2 sin^(-1) (x/sqrt 2))`

Statistics, SPEC1 2016 VCAA 2

A farmer grows peaches, which are sold at a local market. The mass, in grams, of peaches produced on this farm is known to be normally distributed with a variance of 16. A bag of 25 peaches is found to have a total mass of 2625 g.

Based on this sample of 25 peaches, calculate an approximate 95% confidence interval for the mean mass of all peaches produced on this farm. Use an integer multiple of the standard deviation in your calculations.   (3 marks)

Show Answers Only

`(103.4, 106.6)`

Show Worked Solution
`bar x` `= 2625/25`  
  `= (2500 + 125)/25`  
  `= 100 + 5`  
  `= 105`  

 
`σ_(bar x) = 4/sqrt25 = 4/5`

♦ Mean mark 44%.
  

`:. 95 text(%  CI)` `= (105 – 2 xx 4/5, 105 + 2 xx 4/5)`
  `= (105 – 1.6, 105 + 1.6)`
  `= (103.4, 106.6)`

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.)`

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

  5. 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)

  6. 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`

Calculus, SPEC2 2017 VCAA 2

A helicopter is hovering at a constant height above a fixed location. A skydiver falls from rest for two seconds from the helicopter. The skydiver is subject only to gravitational acceleration and air resistance is negligible for the first two seconds. Let downward displacement be positive.

  1. Find the distance, in metres, fallen in the first two seconds.   (2 marks)

  2. Show that the speed of the skydiver after two seconds is 19.6 ms–1.   (1 mark)

After two seconds, air resistance is significant and the acceleration of the skydiver is given by  `a = g -0.01v^2`.

  1. Find the limiting (terminal) velocity, in ms–1, that the skydiver would reach.   (1 mark)

  2. i.  Write down an expression involving a definite integral that gives the time taken for the skydiver to reach a speed of 30 ms–1.   (2 marks)

  3. ii. Hence, find the time, in seconds, taken to reach a speed of 30 ms–1, correct to the nearest tenth of a second.   (1 mark)

  4. Write down an expression involving a definite integral that gives the distance through which the skydiver falls to reach a speed of 30 ms–1. Find this distance, giving your answer in metres, correct to the nearest metre.   (3 marks)

Show Answers Only
  1. `19.6\ text(m)`
  2. `text(See Worked Solutions)`
  3. `v_t = 10sqrtg\ text(ms)^(−1)\ \ text(downwards)`
  4.  i.  `t(30) = int_19.6^30 100/(100g-v^2)dv + 2`
  5. ii. `5.8\ text(s)`
  6. `120\ text(m)`
Show Worked Solution

a.   `u = 0, \ t = 2, \ a = g = 9.8`

`x` `= ut + 1/2 at^2`
  `=0 xx 2 + 1/2 xx 9.8 xx 2^2`
  `= 19.6\ text(m)`

 

b.    `v` `= u+ at`
  `v(2)` `= 9.8 xx 2`
    `= 19.6\ text(ms)^(−1)\ \ \ text{(downward → positive)}`

 

c.   `text(Terminal velocity), v_t, text(occurs when)\ \ a = 0,`

♦ Mean mark 48%.

`g -0.01v_t^2` `=0`
`v_t^2` `= 100 xx 9.8`
`:. v_t` `= 14 sqrt5\ text(ms)^(−1)\ \ \ (v_t > 0\ \ text{as}\ \ v_t\ \ text{is downwards})`

 

d.i.    `(dv)/(dt)` `= g-0.01v^2`
  `(dt)/(dv)` `= 1/(g-0.01v^2)`
    `= 100/(100g-v^2)`
  `t` `= int 100/(100g-v^2)`

 
`text(Time taken until)\ \ v=19.6\ \ text(is 2 seconds.)`

♦ Mean mark part (d)(i) 39%.

 
`:.\ text(Time taken until)\ \ v=30`

`=int_19.6^30 100/(100g-v^2)\ dv + 2`

 

d.ii.   `5.8\ text(seconds)\ \ \ text{(by CAS)}`

♦♦ Mean mark part (d)(ii) 25%.

 

e.    `v *(dv)/(dx)` `= g-0.01 v^2`
  `(dv)/(dx)` `= g/v-(v^2)/(100v)`
    `= (100g-v^2)/(100v)`
  `(dx)/(dv)` `= (100v)/(100g-v^2)`
  `x` `= int (100v)/(100g-v^2)\ dv`

 
`text(Distance fallen in 1st 2 seconds)\ = 19.6\ text(m)`

♦♦ Mean mark part (e) 29%.

 
`:.\ text(Total distance fallen until)\ \ v=30`

`= int_19.6^30 (100v)/(100g-v^2)\ dv + 19.6`

`~~ 120\ text(m)`

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)

    [wa_lines lines="3" style="lined"
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)

     


 

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

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

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

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

Show Answers Only

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

Show Worked Solution

a.  

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

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

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

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

 

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

 

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

 

b.ii.   `V = (45 pi^2)/4 text(cm)^3`

 

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

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

`=> (dV)/(dh) = pi`
 

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

 

d.   `f(x) = 10cos^(-1)(2-2x)`

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

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

Calculus, SPEC2 2017 VCAA 1

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

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

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

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

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

     

     

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

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

Show Answers Only

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

Show Worked Solution

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

♦♦ Mean mark 36%.

`text(Vertical asymptote:)\ x = −1`

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

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

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

 
`text(S.P. when)\ \ f′(x)=0:\ `

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

 

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

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

`x != -1`

`text(Check concavity changes:)`

`f″(−1/2) = −1632/343`

`f″(1) = −3/4`

`f″(3) = 675/(10\ 976)`
 

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

`=> f(sqrt2) ~~ 0.42`

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

 

b.   

 

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

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

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

 

c.ii.  `a=0.98\ \ \ text{(by CAS)}`

Statistics, SPEC2 2018 VCAA 6

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

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

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

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

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

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

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

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

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

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

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

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

 

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

 

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

`p ~~ 0.0092`

 

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

`p~~ 0.0092 < 0.05`

 

e.  `text(Pr)(bar X < x) = 0.05`

♦ Mean mark part (e) 48%.

`x ~~ 146.511`

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

 

f.   `bar X_2\ ~\ N (145, 9/2)`

♦♦♦ Mean mark part (f) 11%.

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

 

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

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

`=>Z_99 ~~ 2.57583`

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

Mechanics, SPEC2 2018 VCAA 5

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

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

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

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

♦ Mean mark part (a) 44%.

 

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

 

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

 

♦ Mean mark part (c) 48%.

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

 

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

 
`text(When)\ \ x=0, v=0:`

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

 

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

 

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

`15 = -v + 4.9 ln (4.9/(4.9 – v))`

♦♦ Mean mark part (d) 26%.

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

 

♦ Mean mark 41%.

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

 

♦ Mean mark 41%.

e.ii.  `t=[-ln (9.8 -2v)]_0^4.5`

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

Vectors, SPEC2 2018 VCAA 4

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

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

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

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

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

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

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

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

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

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

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

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

 
`x_B = t^2,`

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

 

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

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

 
`text(When)\ \ t = (1+sqrt5)/2,`

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

`y_B= (sqrt5+9)/2  !=y_A`
 

`:.\ text(No collision)`

 

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

 

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

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

♦ Mean mark part (d) 41%.

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

 
`:. 0 < t < 5/2`

 

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

 

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

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

♦ Mean mark part (e) 37%.

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

`=> 1.52883 < t < 1.59734`

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

`~~ 1.59734-1.52883`

`~~0.068506\ text(hours)`

`~~ 4.1\ text(minutes)`

Statistics, SPEC2-NHT 2017 VCAA 20 MC

The mass of suspended matter in the air in a particular locality is normally distributed with a mean of `mu` micrograms per cubic metre and a standard deviation of  `sigma = 8`  micrograms per cubic metre. The mean of 100 randomly selected air samples was found to be 40 micrograms per cubic metre.

Based on this, a 90% confidence interval for `mu`, correct to two decimal places, is

A.   `(38.68, 41.32)`

B.   `(26.84, 53.16)`

C.   `(38.43, 41.57)`

D.   `(24.32, 55.68)`

E.   `(37.93, 42.06)`

Show Answers Only

`A`

Show Worked Solution

`M\ ~\ N (mu, 8^2)`

`bar M\ ~\ N (mu, 8^2/100)`
 

`text(Find 90%  CI),\ bar m =40:`

`=> (40 – Z_0.9 xx 8/sqrt 100,\ 40 + Z_0.9 xx 8/sqrt 100)`
 

`Z\ ~\ N(0, 1)`

`Pr(Z < Z_0.9) = 0.95\ \ => \ \ Z_0.9 ~~ 1.64485\ \ \ text{(by CAS)}`
 

`:.­text(CI)\ => (40 – 1.64485 xx 8/sqrt 100,\ 40 + 1.64485 xx 8/sqrt 100)`

`~~ (38.68, 41.32)`

`=>   A`

Calculus, SPEC2-NHT 2017 VCAA 17 MC

The acceleration,  `a\ text(ms)^(-2)`, of a particle moving in a straight line is given by  `a = v^2 + 1`,  where  `v`  is the velocity of the particle at any time `t`. The initial velocity of the particle when at origin `O` is  `2\ text(ms)^(-1)`.

The displacement of the particle from `O` when its velocity is  `3\ text(ms)^(-1)`  is

  1. `log_e(2)`
  2. `1/2 log_e(10/3)`
  3. `1/2 log_e(2)`
  4. `1/2 log_e(5/2)`
  5. `log_e(4/5)`
Show Answers Only

`C`

Show Worked Solution
`v* (dv)/(dx)` `= v^2 + 1`
`(dv)/(dx)` `= (v^2 + 1)/v`
`(dx)/(dv)` `= v/(v^2 + 1)`
`x` `= int (v/(v^2 + 1))\ dv`
  `=1/2 ln(v^2 + 1)+c`

 
`text(When)\ \ x=0, \ v=2:`

`c=-1/2 ln5`

 
`text(Find)\ \ x\ \ text(when)\ \ v=3:`

`x` `= 1/2ln(9 + 1) – 1/2 ln5`
  `= 1/2 ln (10/5)`
  `= 1/2 ln 2`

 
`=>   C`

Vectors, SPEC2-NHT 2017 VCAA 13 MC

Let  `OABCD`  be a right square pyramid where  `underset ~a = vec(OA),\ underset ~b = vec(OB),\ underset ~c = vec(OC)`  and  `underset ~d = vec(OD)`.

An equation correctly relating these vectors is

A.   `underset ~a + underset ~c = underset ~b + underset ~d`

B.   `(underset ~a - underset ~c) ⋅ (underset ~d - underset ~c) = 0`

C.   `underset ~a + underset ~d = underset ~b + underset ~c`

D.   `(underset ~a - underset ~d) ⋅ (underset ~c - underset ~b) = 0`

E.   `underset ~a + underset ~b = underset ~c + underset ~d`

Show Answers Only

`A`

Show Worked Solution

`|underset ~a| = |underset ~b| = |underset ~c| = |underset ~d|`

`text(Let)\ \ underset ~a` `= alpha underset ~i + beta underset ~j + gamma underset ~k`
`underset ~b` `= -alpha underset ~i + beta underset ~j + gamma underset ~k`
`underset ~c` `= -alpha underset ~i + beta underset ~j – gamma underset ~k`
`underset ~d` `= alpha underset ~i + beta underset ~j – gamma underset ~k`

 

`A:\ \ underset ~a + underset ~ c` `= 2 beta underset ~ j`
`underset ~b + underset ~d` `= 2 beta underset ~j`

 

`=>   A`

Vectors, SPEC2-NHT 2017 VCAA 12 MC

If  `underset ~u = 3 underset ~i + 6 underset ~j - 2 underset ~k`  and  `underset ~v = 2 underset ~i + 2 underset ~j - underset ~k`, then the vector resolute of  `underset ~u`  in the direction of  `underset ~v`  is

A.   `20/7 (3 underset ~i + 6 underset ~j - 2 underset ~k)`

B.   `20/9 (2 underset ~i + 2 underset ~j - underset ~k)`

C.   `20/49 (3 underset ~i + 6 underset ~j - 2 underset ~k)`

D.   `20/3 (2 underset ~i + 2 underset ~j - underset ~k)`

E.   `3/7 (3 underset ~i + 6 underset ~j - 2 underset ~k)`

Show Answers Only

`B`

Show Worked Solution

`text(Using CAS:)`

`(underset ~u ⋅ underset ~hat v) underset ~ hat v`

`= 20/9 (2 underset ~i + 2 underset ~j – underset ~k)`

 
`=>   B`

Complex Numbers, SPEC2 2015 VCAA 9 MC

Let  `z_1 = r_1 text(cis)(theta_1)`  and  `z_2 = r_2 text(cis)(theta_2)`, where  `z_1`  and  `z_1z_2`  are shown in the Argand diagram below;  `theta_1`  and  `theta_2`  are acute angles.
 

SPEC2 2015 VCAA 9 MC
 

A statement that is necessarily true is

A.   `r_2 > 1`

B.   `theta_1 < theta_2`

C.   `|\ (z_1)/(z_2)\ | > r_1`

D.   `theta_1 = theta_2`

E.   `r_1 > 1`

Show Answers Only

`C`

Show Worked Solution

`text(Consider each option:)`

♦ Mean mark 47%.

`A:\ \ r_1r_2 < r_1\ \ =>\ \ r_2 < 1`

`B:`

`alpha + pi/2 – theta_1 = theta_2 ~~ theta_1\ \ text(not necessarily true)`

`C:\ \ (r_1)/(r_2) = (|\ z_1\ |)/(|\ z_2\ |) = |(z_1)/(z_2)|`

`text(S)text(ince)\ \ r_2<1\ \ =>\ \ |(z_1)/(z_2)| > r_1`

`D:\ \ theta_1 = theta_2\ \ text(possible but no scale given)`

`E:\ \ r_1>1\ \ text(possible but no scale given.)`

 

`=> C`

Calculus, SPEC1-NHT 2017 VCAA 11

Find the length of the curve specified parametrically by  `x = a theta - a sin (theta), \ y = a - a cos (theta)`  from  `theta = (2 pi)/3`  to  `theta = 2 pi`, where  `a in R^+`. Give your answer in terms of `a`.  (4 marks)

Show Answers Only

`6a`

Show Worked Solution
`(dx)/(d theta)` `= a – a cos (theta)`
`(dy)/(d theta)` `= a sin (theta)`
   
`((dx)/(d theta))^2` `= (a(1 – cos (theta)))^2`
  `= a^2(1 – 2 cos (theta) + cos^2 theta)`
`((dy)/(d theta))^2` `= a^2 sin^2 (theta)`

 

`l` `= int_((2 pi)/3)^(2pi) sqrt(a^2 sin^2(theta) + a^2 – 2a^2 cos (theta) + a^2 cos^2(theta))\ d theta`
  `= int_((2 pi)/3)^(2pi) sqrt(a^2(sin^2(theta) + cos^2(theta) + 1 – 2 cos(theta)))\ d theta`
  `= int_((2 pi)/3)^(2pi) a sqrt(1 + 1 – 2 cos (theta))\ d theta, \ \ a > 0`
  `= a int_((2 pi)/3)^(2pi) underbrace{sqrt(2 – 2cos (theta))}_{cos (theta) = 2 cos^2(theta/2) – 1}\ d theta`
  `= a int_((2 pi)/3)^(2pi) sqrt(2 – 4 cos^2 (theta/2) + 2)\ d theta`
  `= a int_((2 pi)/3)^(2pi) sqrt(4 – 4 cos^2(theta/2))\ d theta`
  `= a int_((2 pi)/3)^(2pi) sqrt(4 sin^2 (theta/2))\ d theta`
  `= a int_((2 pi)/3)^(2pi) 2 sin (theta/2)\ d theta`
  `(theta in [(2 pi)/3, 2 pi]\ \ =>\ \  theta/2 in [pi/3, pi]\ \ =>\ \ sin(theta/2) > 0)`
  `= a [-4 cos (theta/2)]_((2 pi)/3)^(2pi)`
  `= a(-4 cos (pi) – (-4)cos (pi/3))`
  `= a(-4 xx (-1) + 4 xx 1/2)`
  `= 6a`

Calculus, SPEC2 2018 VCAA 3

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


 

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

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

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

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

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

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

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

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

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

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

a.   `V= pi int_0^h x^2\ dy`

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

 

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

 

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

 

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

  

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

 

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

 

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

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

 

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

♦♦ Mean mark part (e) 30%.

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

 

f.    `(dV)/(dt) = 0`

♦♦ Mean mark part (f) 32%.

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

  

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

Complex Numbers, SPEC2 2018 VCAA 2

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

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

     

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

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

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

a.   `(x-1)^2 + (y-2)^2 = 2^2`

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

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

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

 

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

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

 

d.   `|z-1| = |z-3|`

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

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

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

 

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

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

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

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

`:.\ text(Shaded Area)`

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

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

`=(4pi)/3-sqrt3`

Calculus, SPEC2 2018 VCAA 1

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

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

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


      

    `qquad` 
     

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

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

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

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

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


     

Show Answers Only

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

Show Worked Solution

a.   `-1 <= x^2-1 <= 1`

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

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

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

 

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

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

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

 

b.   

 

c.   `f(x) = 2 sin^(-1)(x^2-1)`

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

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

 

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

 

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

♦♦ Mean mark part (e)(i) 21%.

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

`:. x in (-sqrt2, 0) ∪ (0, sqrt 2)`

 

e.ii.  `g(x) = +- 4`

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

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

 

e.iii.   

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`

Trigonometry, SPEC1-NHT 2017 VCAA 6

Find all real solutions of  `tan(2x) = -tan(x)`.  (3 marks)

Show Answers Only

`x = k pi, \ pi/3 + k pi, \ (2 pi)/3 + k pi, k in ZZ`

Show Worked Solution

`(2 tan (x))/(1 – tan^2(x)) = -tan(x)`

 
`text(Let)\ \ tan(x) = k,`

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

 
`k(3 – k^2) = 0 \ \ =>\ \  k = 0, k^2 = 3`

`=> tan(x) = 0, tan(x) = +- sqrt 3`

`:. x = k pi, \ pi/3 + k pi, \ (2 pi)/3 + k pi\ \ (k in ZZ)`