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Statistics, STD2 S5 2019 HSC 38

In a particular country, the birth weight of babies is normally distributed with a mean of 3000 grams. It is known that 95% of these babies have a birth weight between 1600 grams and 4400 grams.

One of these babies has a birth weight of 3497 grams. What is the `z`-score of this baby's birth weight?  (2 marks)

Show Answers Only

`0.71`

Show Worked Solution

`text(95% babies within)\ 1600 – 4400`

♦ Mean mark 44%.

`text(i.e.)\ \ 3000 +- \ 2\ text(σ = 1600 – 4400)`

`2sigma` `= 4400-3000`
  `= 1400`
`sigma` `= 700`
`:. ztext(-score)\ (3497)` `= (x-mu)/σ`
  `= (3497 – 3000)/700`
  `= 0.71`

Algebra, STD2 A4 2019 HSC 36

A small business makes and sells bird houses.

Technology was used to draw straight-line graphs to represent the cost of making the bird houses `(C)` and the revenue from selling bird houses `(R)`. The `x`-axis displays the number of bird houses and the `y`-axis displays the cost/revenue in dollars.
 


 

  1. How many bird houses need to sold to break even?  (1 mark)

  2. By first forming equations for cost `(C)` and revenue `(R)`, determine how many bird houses need to be sold to earn a profit of $1900.  (3 marks)

Show Answers Only
  1. `20`
  2. `96`
Show Worked Solution

a.   `20\ \ (xtext(-value at intersection))`

 

b.   `text(Find equations of both lines):`

♦♦ Mean mark 28%.

`(0, 500)\ text(and)\ (20, 800)\ text(lie on)\ \ C`

`m_C = (800-500)/(20-0) = 15`

`=> C = 500 + 15x`
 

`(0,0)\ text(and)\ (20, 800)\ text(lie on)\ \ R`

`m_R = (800-0)/(20-0) = 40`

`=> R = 40x`
 

`text(Profit) = R-C`

`text(Find)\ \ x\ \ text(when Profit = $1900:)`

`1900` `= 40x-(500 + 15x)`
`25x` `= 2400`
`x` `= 96`

Measurement, STD2 M6 2019 HSC 35

A compass radial survey shows the position of four towns  `A`, `B`, `C` and `D`  relative to point `O`.
 


 

The area of triangle  `BOC`  is 198 km².

Calculate the bearing of town  `C`  from point  `O`, correct to the nearest degree.  (3 marks)

Show Answers Only

`207°`

Show Worked Solution

`text(In)\ DeltaCOB,`

♦♦ Mean mark 26%.

`text(Area) = 1/2 ab sinc`

`198` `= 1/2 xx 16 xx 25 xx sinangleCOB`
`sinangleCOB` `= (2 xx 198)/(16 xx 25)`
  `= 0.99`
`angleCOB` `= sin^(−1)0.99`
  `= 81°53′`

 

`:. text(Bearing of)\ C\ text(from)\ O` `= 125° + 81°53′`
  `= 206°53′`
  `= 207°\ \ (text(nearest degree))`

Algebra, STD2 A2 2019 HSC 34

The relationship between British pounds `(p)` and Australian dollars `(d)` on a particular day is shown in the graph.
 

  1. Write the direct variation equation relating British pounds to Australian dollars in the form  `p = md`. Leave `m` as a fraction.  (1 mark)

  2. The relationship between Japanese yen `(y)` and Australian dollars `(d)` on the same day is given by the equation  `y = 76d`.

     

    Convert 93 100 Japanese yen to British pounds.  (2 marks)

Show Answers Only
  1. `p = 4/7 d`
  2. `93\ 100\ text(Yen = 700 pounds)`
Show Worked Solution

a.   `m = text(rise)/text(run) = 4/7`

♦ Mean mark 42%.

`p = 4/7 d`

 

b.   `text(Yen to Australian dollars:)`

`y` `=76d`
`93\ 100` `= 76d`
`d` `= (93\ 100)/76`
  `= 1225`

 
`text(Aust dollars to pounds:)`

`p` `= 4/7 xx 1225`
  `= 700\ text(pounds)`

 
`:. 93\ 100\ text(Yen = 700 pounds)`

Algebra, STD2 A4 2019 HSC 31

A rectangle has width `w` centimetres. The area of the rectangle, `A`, in square centimetres, is  `A = 2w^2 + 5w`.

The graph  `A = 2w^2 + 5w`  is shown.
 


 

  1. Explain why, in this context, the model  `A = 2w^2 + 5w`  only makes sense for the bold section of the graph.  (1 mark)

  2. The area of the rectangle is 18 cm². Calculate the perimeter of the rectangle.  (2 marks)

Show Answers Only
  1. `text(The width of a rectangle cannot be negative.)`
  2. `22\ text(cm)`
Show Worked Solution

a.   `text(The width of a rectangle cannot be negative.)`

♦ Mean mark 44%.

 

b.   `text(When)\ A = 18, w = 2`

♦ Mean mark 42%.

`text(Let)\ h =\ text(height of rectangle)`

`18` `= 2 xx h`
`h` `= 9\ text(cm)`

 

`:.\ text(Perimeter)` `= 2 xx (2 + 9)`
  `= 22\ text(cm)`

Calculus, 2ADV C1 2019 HSC 14d

The equation of the tangent to the curve  `y = x^3 + ax^2 + bx + 4`  at the point where  `x = 2`  is  `y = x - 4`.

Find the values of  `a`  and  `b`.  (3 marks)

Show Answers Only

`b = -3,\ \ a = -2`

Show Worked Solution
`y ` `= x^3 + ax^2 + bx + 4`
`(dy)/(dx)` `= 3x^2 + 2ax + b`

 
`text(When)\ \ x = 2,\ \ (dy)/(dx) = 1`

♦ Mean mark 46%.

`12 + 4a + b` `= 1`
`4a + b` `= -11\ …\ (1)`

 
`text(The point)\ (2, -2)\ text(lies on)\ y:`

`8 + 4a + 2b + 4` `=-2`
`4a + 2b` `= -14\ …\ (2)`

  
`text(Subtract)\ \ (2) – (1)`

`b = -3`

`text(Substitute into)\ (1)`

`4a – 3` `= -11`
`4a` `= -8`
`a` `= -2`

Calculus, 2ADV C3 2019 HSC 14b

The derivative of a function  `y = f(x)`  is given by  `f prime(x) = 3x^2 + 2x - 1`.

  1. Find the `x`-values of the two stationary points of  `y = f(x)`, and determine the nature of the stationary points.  (2 marks)

  2. The curve passes through the point  `(0, 4)`.

     

    Find an expression for  `f(x)`.  (2 marks)

  3. Hence sketch the curve, clearly indicating the stationary points.  (2 marks)

  4. For what values of `x` is the curve concave down?  (1 mark)

Show Answers Only
  1. `x = 1/3\ \ text{(min)}`
    `x = -1\ \ text{(max)}`
  2. `f(x) = x^3 + x^2 – x + 4`
  3. `text(See Worked Solution)`
  4. `x < -1/3`
Show Worked Solution

a.    `f prime(x) = 3x^2 + 2x – 1`

`f″(x) = 6x + 2`

`text(S.P.’s when)\ \ f prime(x) = 0`

`3x^2 + 2x – 1` `= 0`
`(3x – 1)(x + 1)` `= 0`

 
`x = 1/3 or -1`

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

`f″(x) = 4 > 0 =>\ text(MIN)`
 

`text(When)\ x = -1,`

`f″(x)= -4 < 0 =>\ text(MAX)`

 

b.    `f(x)` `= int f prime(x)\ dx`
    `= int 3x^2 + 2x – 1\ dx`
    `= x^3 + x^2 – x + c`

 
`(0, 4)\ \ text(lies on)\ \ f(x)\ \ =>\ \ c = 4`

`:. f(x) = x^3 + x^2 – x + 4`

 

c.    `text(When)\ \ x = -1,\ \ y = 5`
  `text(When)\ \ x = 1/3,\ \ y = 103/27`

 

 

d.   `text(Concave down when)\ f″(x) < 0`

♦ Mean mark 36%.

`6x + 2` `< 0`
`6x` `< -2`
`x` `< -1/3`

Calculus, 2ADV C4 2019 HSC 13c

  1.  Differentiate  `(ln x)^2`.  (2 marks)

  2.  Hence, or otherwise, find  `int(ln x)/x\ dx`.  (1 mark)

Show Answers Only
  1. `(2 ln x)/x`
  2. `1/2 (ln x)^2 + C`
Show Worked Solution
i.    `y` `= (ln x)^2`
  `(dy)/(dx)` `= 2 ⋅ 1/x ⋅ ln x`
    `= (2 ln x)/x`

♦ Mean mark part (ii) 49%.

ii.    `int (ln x)/x\ dx` `=1/2 int (2 ln x)/x dx`
    `= 1/2 (ln x)^2 +C`

Trigonometry, 2ADV T2 2019 HSC 13a

Solve  `2 sin x cos x = sin x`  for  `0 <= x <= 2pi`.  (3 marks)

Show Answers Only

`x = 0, quad pi/3, quad pi, quad (5 pi)/3`

Show Worked Solution

♦ Mean mark 49%.

`2 sin x cos x – sin x` `= 0`
`sin x (2 cos x – 1)` `= 0`
`sin x` `= 0`
`=> x` `= 0,\  pi,\  2pi`
`cos x` `= 1/2`
`=> x` `= pi/3, (5 pi)/3`

 
`:. x = 0, quad pi/3, quad pi, quad (5 pi)/3,quad 2pi`

Financial Maths, STD2 F1 2019 HSC 29

Part of a supermarket receipt is shown.

Determine the missing values, `A` and `B`, to complete the receipt.  (2 marks)

Show Answers Only

`$9.00`

Show Worked Solution

`text(Chocolate is the only item where GST applies.)`

♦♦ Mean mark 25%.

`text(GST on chocolate = 0.70`

`=> text(C)text(ost of chocolate) = $7.00`

`:. A = 7.00 + 0.70 = $7.70`

`:. B` `= 36.25 – (7.70 + 5.00 + 8.50 + 3.20 + 2.85)`
  `= $9.00`

Financial Maths, STD2 F4 2019 HSC 27

Ashley has a credit card with the following conditions:

  • There is no interest-free period.
  • Interest is charged at the end of each month at 18.25% per annum, compounding daily, from the purchase date (included) to the last day of the month (included).

Ashley's credit card statement for April is shown, with some figures missing.

The minimum payment is calculated as 2% of the closing balance on 30 April.

Calculate the minimum payment.  (3 marks)

Show Answers Only

`$74.40`

Show Worked Solution
`text(Daily interest)` `= 18.25/(100 xx 365)`
  `= 0.0005`

 

♦ Mean mark 39%.

`text(Closing balance)` `= 3700(1.0005)^11`
  `= 3720.40`

 

`:.\ text(Minimum payment)` `= 3720.40 xx 0.02`
  `= $74.408…`
  `= $74.41\ \ text{(nearest cent)}`

Networks, STD2 N3 2019 HSC 26

A project requires activities `A` to `F` to be completed. The activity chart shows the immediate prerequisite(s) and duration for each activity.

  1. By drawing a network diagram, determine the minimum time for the project to be completed.  (3 marks)

  2. Determine the float time of the non-critical activity.  (1 mark)

Show Answers Only
  1. `text(15 hours)`
  2. `text(3 hours)`
Show Worked Solution
a.   

 

`text(Scanning forwards:)`

`text(Minimum time = 2 + 6 + 2 + 4 + 1 = 15 hours)`

`text{(Scanning forwards and backwards is highly recommended but not}`

 `text{required in the network diagram.)}`

 

b.   `text(Critical Path is)\ ABDEF.`

♦♦ Mean mark 30%.

`text(Non-critical activity is)\ C.`

`text(Float time)` `= 5 – 2`
  `= 3\ text(hours)`

Probability, STD2 S2 2019 HSC 25

A bowl of fruit contains 17 apples of which 9 are red and 8 are green.

Dennis takes one apple at random and eats it. Margaret also takes an apple at random and eats it.

By drawing a probability tree diagram, or otherwise, find the probability that Dennis and Margaret eat apples of the same colour.  (3 marks)

Show Answers Only

`8/17`

Show Worked Solution

`P(text(same colour))` `= P(R R) + P(GG)`
  `= 9/17 xx 8/16 + 8/17 xx 7/16`
  `= 72/272 + 56/272`
  `= 8/17`
♦♦ Mean mark 35%.

Calculus, EXT1* C1 2019 HSC 10 MC

A particle is moving along a straight line with displacement `x` at time `t`.

The particle is stationary when  `t = 11`  and when  `t = 13`.

Which of the following MUST be true in this case?

A.  The particle changes direction at some time between  `t = 11`  and  `t = 13`.
B.   The displacement function of the particle has a stationary point at some time between  `t = 11`  and  `t = 13`.
C.   The acceleration of the particle is 0 at some time between  `t = 11`  and  `t = 13`.
D.   The acceleration function of the particle has a stationary point at some time between  `t = 11`  and  `t = 13`.
Show Answers Only

`C`

Show Worked Solution

`text(Given)\ \ v = 0\ \ text(at)\ \ t = 11 and t = 13,`

♦ Mean mark 37%.

`text(Consider option)\ C:`

`text(If)\ v\ text(remains)\ 0, \ a = 0.`

`text(If)\ v\ text(increases)\ (a > 0 )\ text(or decreases)\ (a < 0)\ text(after)\ \ t = 11,`

`text(it must then decrease)\ (a < 0)\ text(or increase)\ (a > 0)`

`text(respectively so that)\ \ v = 0\ \ text(when)\ \ t = 13.`

`text(In each scenario,)\ a = 0\ text(between)\ \ t = 11 and t = 13.`

`=>  C`

Calculus, 2ADV C4 2019 HSC 9 MC

Which expression is equal to  `int tan^2 x\ dx`?

  1. `tan x - x + C`
  2. `tan x - 1 + C`
  3. `(tan^3 x^2)/6 + C`
  4. `(tan^3 x)/3 + C`
Show Answers Only

`A`

Show Worked Solution

`text(Consider option)\ A:`

♦♦ Mean mark 35%.

`d/(dx) (tan x – x + C)`

`= sec^2 x – 1`

`= tan^2 x`

`:. int tan^2 x\ dx = tan x – x + C`

`=>  A`

Measurement, STD2 M6 2019 HSC 12 MC

An owl is 7 metres above ground level, in a tree. The owl sees a mouse on the ground at an angle of depression of 32°.

How far must the owl fly in a straight line to catch the mouse, assuming the mouse does not move?

  1.  3.7 m
  2.  5.9 m
  3.  8.3 m
  4.  13.2 m
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ OM = text(Flight distance)`

♦ Mean mark 36%.

`sin32°` `= 7/(OM)`
`:. OM` `= 7/(sin32°)`
  `= 13.2\ text(m)`

 
`=> D`

Algebra, STD2 A1 2019 HSC 11 MC

Which of the following correctly expresses `y` as the subject of the formula  `3x-4y-1 = 0`?

  1.  `y = 3/4 x-1`
  2.  `y = 3/4 x + 1`
  3.  `y = (3x-1)/4`
  4.  `y = (3x + 1)/4`
Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 50%.

`3x-4y-1` `= 0`
`4y` `= 3x-1`
`:. y` `= (3x-1)/4`

 
`=> C`

Measurement, STD2 M1 2019 HSC 8 MC

A person's weight is measured as 79.3 kg.

What is the absolute error of this measurement?

  1. 10 grams
  2. 50 grams
  3. 100 grams
  4. 500 grams
Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 46%.

`text(A)text(bsolute error)` `= 1/2 xx\ text(precision)`
  `= 1/2 xx 0.1\ text(kg)`
  `= 1/2 xx 100\ text(grams)`
  `= 50\ text(grams)`

 
`=> B`

Statistics, STD2 S4 2019 HSC 23

A set of bivariate data is collected by measuring the height and arm span of seven children. The graph shows a scatterplot of these measurements.
 


 

  1. Calculate Pearson's correlation coefficient for the data, correct to two decimal places.  (1 mark)

  2. Identify the direction and the strength of the linear association between height and arm span.  (1 mark)

  3. The equation of the least-squares regression line is shown.
     
               Height = 0.866 × (arm span) + 23.7
     
    A child has an arm span of 143 cm.

     

    Calculate the predicted height for this child using the equation of the least-squares regression line.  (1 mark)

Show Answers Only
  1. `0.98\ \ (text(2 d.p.))`
  2. `text(Direction: positive)`
    `text(Strength: strong)`
  3. `147.538\ text(cm)`
Show Worked Solution

a.   `text{Use  “A + Bx”  function (fx-82 calc):}`

♦ Mean mark 40%.
COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 correlation” .

`r` `= 0.9811…`
  `= 0.98\ \ (text(2 d.p.))`

 

b.   `text(Direction: positive)`

`text(Strength: strong)`

 

c.    `text(Height)` `= 0.866 xx 143 + 23.7`
    `= 147.538\ text(cm)`

Vectors, EXT1 V1 SM-Bank 9

The diagram shows a projectile fired at an angle  `theta`  to the horizontal from the origin `O` with initial velocity  `V\ text(ms)^(−1)`.
 

The position vector for the projectile is given by
 

`qquad underset~s(t) = Vtcosthetaunderset~i + (Vtsintheta - 1/2 g t^2)underset~j`     (DO NOT prove this)
 

where `g` is the acceleration due to gravity.

  1.  Show the horizontal range of the projectile is

    `qquad (V^2sin2theta)/g`  (2 marks)

The projectile is fired so that  `theta = pi/3`.

  1.  State whether the projectile is travelling upwards or downwards when

    `qquad t = (2V)/(sqrt3g)`  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Downwards – See Worked Solutions)`
Show Worked Solution

i.   `text(Time of flight when)`

`underset~j\ text(component of)\ underset~v = 0`

`Vtsintheta – 1/2 g t^2` `= 0`
`t(Vsintheta – 1/2 g t)` `= 0`
`1/2g t` `= Vsintheta`
`t` `= (2Vsintheta)/g`

 
`text(Range) \ => \ underset~i\ text(component of)\ underset~s`

`text(when) \ \ t = (2Vsintheta)/g`

`text(Range)` `= V · ((2Vsintheta)/g) · costheta`
  `= (V^2)/g · 2sinthetacostheta`
  `= (V^2sin2theta)/g`

 

ii.   `text(Time of flight) = (2Vsin\ pi/3)/g = (sqrt3 V)/g`

`text(S)text(ince parabolic path is symmetrical,)`

`=>\ text(Upwards if)\ \ t < (sqrt3 V)/(2g)`

`=>\ text(Downwards if)\ \ t > (sqrt3 V)/(2g)`

`:. \ text(At)\ \ t = (2V)/(sqrt3 g), text(travelling downwards)`

`text(as) \ \ 2/sqrt3 · V/g > sqrt3/2 · V/g`

Vectors, EXT1 V1 SM-Bank 8

A projectile is fired horizontally off a cliff at an initial speed of  `V`  metres per second.
 

 

The projectile strikes the water, `l`  metres from the base of the cliff.

Let `g` be the acceleration due to gravity and assume air resistance is negligible.

  1.  Show the projectile hits the water when
     
    `qquadt = sqrt((2d)/g)`  (2 marks)

  2.  If  `l`  equals twice the height of the cliff, at what angle does the projectile hit the water?  (2 marks)

  3.  Show that the speed at which the projectile hits the water is
     
    `qquad2sqrt(dg)`  metres per second.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `45°`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `underset~v` `= Vcosthetaunderset~i + (Vsintheta – g t)underset~j`
    `= Vcos0 underset~i + (Vsin0 – g t)underset~j`
    `= Vunderset~i – g tunderset~j`
`underset~s` `= intunderset~v\ dt`
  `= Vtunderset~i – 1/2g t^2 underset~j + c`

 
`text(When)\ t = 0, underset~s = 0, c = 0`
 

`text(Time of flight:)`

`underset~j\ text(component of)\ underset~s = −d`

`−1/2 g t^2` `= −d`
`t^2` `= (2d)/g`
`t` `= sqrt((2d)/g)`

 

ii.   `l = 2d\ \ (text(given))`

`text(Projectile hits water at)\ theta:`
 

`overset.y` `= underset~j\ text(component of)\ underset~v`
  `= −g t`
  `= −g · sqrt((2d)/g)`
  `= −sqrt(2dg)`
   
`overset.x` `= underset~i\ text(component of)\ underset~v`
  `= V`

 
`text(When)\ \ t = sqrt((2d)/g),`

`underset~i\ text(component of)\ underset~s = 2d`

`2d` `= V · sqrt((2d)/g)`
`V` `= (2dsqrtg)/sqrt(2d) = sqrt(2dg)`

 

`tantheta` `= (|overset.y|)/(|overset.x|)= sqrt(2dg)/sqrt(2dg)=1`
`:. theta` `= 45°`

 

iii.   `text(Speed = magnitude of velocity)`

`|underset~v|` `= sqrt((sqrt(2dg))^2 + (sqrt(2dg))^2)`
  `= sqrt(4dg)`
  `= 2sqrt(dg)`

Vectors, EXT1 V1 SM-Bank 7

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 ~s(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.

For the player’s first shot at goal, `V = 7\ text(ms)^(-1)` and  `alpha = 45^@`

  1. 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) 

  2. Find the maximum height, in metres, above floor level, reached by the centre of the ball.  (2 marks)

  3. 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)

Show Answers Only
  1. `(5 sqrt 2)/14`
  2. `3\ text(m)`
  3. `128\ text(m)`
Show Worked Solution
a.i.    `underset ~s(t)` `=7t\ cos 45^@ underset ~i+(7t\ sin 45^@ – 4.9t^2) underset ~j`
    `=(7sqrt2 t)/2 underset ~i + ((7sqrt2)/2 t – 4.9t^2) underset ~j`
     

 `text(Maximum height  ⇒  find)\ \t\ text(when)\ \ underset~j\ \ text(component of)\ \ underset ~V(t) = 0:`

`underset~V = (7sqrt2)/2 underset ~i + ((7sqrt2)/2 – 9.8t) underset ~j`

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

 

a.ii.  `text(Max height)\ =>\ text(Find)\ \ underset~j\ \ text(component of)\ \ underset~s(t)\ \ text(when)\ \ t=(5 sqrt 2)/14`

  `underset ~s_(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(m)`

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

 

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

`text(After 1 second,)`

`underset~s_text(ball) = 7/sqrt 2 underset ~i + ((7sqrt2)/2 – 4.9)underset ~j`

  `:.d` `= |underset ~s_text(ring) – underset ~s(1)|`
    `= sqrt((4.5 – 7/sqrt 2)^2 + (1.25 – 7/sqrt 2 + 4.9)^2)`
    `~~ 1.28\ text(m)`

Vectors, EXT1 V1 SM-Bank 6

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 ~s(t)`, from `O`, of the ball after `t` seconds is given by
 
  `qquad underset ~s(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 s (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 symmetrical parabolic path.)`

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

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

`= (15 sqrt 3)/4.9`

`~~ 5.302\ text(s)`
 

d.  `text(Range)\ =>underset~i\ \ text(component of)\ \ underset~s(t)\ \ text(when)\ \ t= (15 sqrt 3)/4.9`

`:.\ text(Range)` `= 15 xx (15 sqrt 3)/4.9`
  `= (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=(15 sqrt 3 +- sqrt ((15 sqrt 3)^2 – 4 xx 4.9 xx2))/(2 xx 4.9)`  
     

`t ~~ 0.078131\ \ text(or)\ \ t ~~ 5.22406`

 
`text(When)\ \ t=0.0781,`

`x= 15 xx 0.0781 = 1.17\ text(m)\ \ 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.`

Vectors, EXT1 V1 SM-Bank 5

Points `A` and `B` are on the number plane. The vector  `overset(->)(AB)`  is  `((4),(1))`.

Point `C` is chosen so that the area of  `DeltaABC`  is  `17/2`  square units and  `|overset(->)(AC)| = sqrt34`.

Find all possible vectors  `overset(->)(AC)`.  (4 marks)

Show Answers Only

`((3),(5)),((5),(−3)),((−3),(−5))\ text(and)\ ((−5),(3))`

Show Worked Solution

`|overset(->)(AB)| = sqrt(4^2 + 1^2) = sqrt17`

`text(Let)\ \ angleCAB = theta`

`text(Area)\ DeltaABC` `= 1/2 · |overset(->)(AB)| · |overset(->)(AC)| · sintheta`
`17/2` `= 1/2 · sqrt17 · sqrt34 sintheta`
`17` `= 17sqrt2 sintheta`
`sintheta` `= 1/sqrt2`
`theta` `= pi/4\ text(or)\ (3pi)/4\ \ (theta>0)`

 
`=> costheta = ±1/sqrt2`

`overset(->)(AB) · overset(->)(AC)` `= |overset(->)(AB)| · |overset(->)(AC)| · costheta`
  `= ±sqrt17 · sqrt34 · 1/sqrt2`
  `= ±17`

 
`text(Let)\ \ overset(->)(AC) = ((x),(y))`

`overset(->)(AB) · overset(->)(AC)` `= ((4),(1))((x),(y))`
  `= 4x + y`

 
`4x + y = ±17…\ \ (1)`

`|overset(->)(AC)|` `= sqrt(x^2 + y^2) = sqrt34`
`x^2 + y^2` `= 34\ …\ \ (2)`

 
`text(Solving simultaneously:)`

`4x + y = 17 \ => \ y = 17 – 4x`

`text{Substitute into (2)}`

`x^2 + (17 – 4x)^2` `= 34`
`x^2 + 289 – 136x + 16x^2` `= 34`
`17x^2 – 136x + 255` `= 0`
`x^2 – 8x + 15` `= 0`
`(x – 3)(x – 5)` `= 0`

 
`:. x = 3, y = 5\ \ text(or)\ \ x = 5, y = −3`

 
`text(Similarly for)\ \ 4x + y = −17 \ => \ y = −4x – 17`

COMMENT: A diagram is highly recommended. Here, the possibility of 4 solutions is clearly shown.

`x^2 + (−4x – 17)^2` `= 34`
`x^2 + 8x + 15` `= 0`
`(x + 3)(x + 5)` `= 0`

 
`:. x = −3, y = −5\ \ text(or)\ \ x = −5, y = 3`
 

 

 
`:. text(Possible vectors)\ overset(->)(AC)\ text(are)`

`((3),(5)),((5),(−3)),((−3),(−5))\ text(and)\ ((−5),(3))`

Statistics, EXT1 S1 SM-Bank 1

Shoddy Ltd produces statues that are classified as Superior or Regular and are entirely made by machines, on a construction line. The quality of any one of Shoddy’s statues is independent of the quality of any of the others on its construction line. The probability that any one of Shoddy’s statues is Regular is 0.8.

Shoddy Ltd wants to ensure that the probability that it produces at least one Superior statues in a day’s production run is at least 0.95.

Calculate the minimum number of statues that Shoddy would need to produce in a day to achieve this aim.  (3 marks)

Show Answers Only

`14`

Show Worked Solution

`text(Let)\ \ X = text(Number of superior statues)`

♦♦ Mean mark 27%.

`X∼\ text(Bin) (n, 0.2)`

`P(X >= 1)` `>= 0.95`
`1 – P(X = 0)` `>= 0.95`
`1 – ((n),(0)) (0.8^n) (0.2^0)` `>= 0.95`
`1-0.8^n` `>=0.95`
`0.8^n` `<=0.05`
`n ln 0.8` `<=ln 0.05`
`n` `>= ln 0.05/ln 0.8,\ \ \ text{(ln 0.8 < 0)}`
`n` `>= 13.4…`
`:. n_min` `= 14`

Statistics, EXT1 S1 2011 MET1 7

A biased coin tossed three times. The probability of a head from a toss of this coin is `p.`

  1. Find, in terms of `p`, the probability of obtaining

    1. three heads from the three tosses  (1 mark)

    2. two heads and a tail from the three tosses.  (1 mark)

  2. If the probability of obtaining three heads equals the probability of obtaining two heads and a tail, find `p`.  (2 marks)

Show Answers Only
    1. `p^3`
    2. `3p^2 (1 – p)`
  2. `0 or 3/4`
Show Worked Solution
a.i.    `text(P) (HHH)` `= p xx p xx p`
    `= p^3`

 

  ii.   `text(P) text{(2 Heads and 1 Tail from 3 tosses)}`

♦ Part (a)(ii) mean mark 41%.

`= ((3), (2)) xx p^2 xx (1 – p)^1`

`= 3 p^2 (1 – p)`

 

b.   `text(If probabilities are equal:)`

♦ Mean mark 48%.
MARKER’S COMMENT: Many students incorrectly assumed `p` could not be zero.
`p^3` `= 3p^2 – 3p^3`
`4p^3 – 3p^2` `= 0`
`p^2 (4p – 3)` `= 0`

 
`:. p = 0 or p = 3/4`

Statistics, EXT1 S1 2017 MET2 18

Let  `X`  be a discrete random variable with binomial distribution  `X ~\ text(Bin)(n, p)`. The mean and the standard deviation of this distribution are equal.

Given that  `0 < p < 1`, what is the smallest number of trials, `n`, such that  `p ≤ 0.01`.   (2 marks)

Show Answers Only

`99`

Show Worked Solution

`mu = np,\ \ text(Var)(X) = np(1-p) = σ_x ^2`

`text(Given)\ \ mu = σ_x,`

`np` `= sqrt(np(1 – p))`
`n^2p^2` `= np(1 – p)`
`np(np-1+p)` `=0`
`np-1+p` `=0,\ \ \ (np!=0)`
`p(n+1)` `=1`
`p` `=1/(n+1)`

 

`1/(n + 1)` `<= 1/100`
`n+1` `>=100`
`n` `>= 99`

 
`:. n_text(min) = 99`

Calculus, EXT1 C3 2017 SPEC1 8

A slope field representing the differential equation  `dy/dx = −x/(1 + y^2)`  is shown below.

  1. Sketch the solution curve of the differential equation corresponding to the condition  `y(−1) = 1`  on the slope field above and, hence, estimate the positive value of `x` when  `y = 0`. Give your answer correct to one decimal place.  (2 marks)
  2. Solve the differential equation  `(dy)/(dx) = (−x)/(1 + y^2)`  with the condition  `y(−1) = 1`. Express your answer in the form  `ay^3 + by + cx^2 + d = 0`, where `a`, `b`, `c` and `d` are integers.  (2 marks)

Show Answers Only
  1.  

  2. `2y^3 + 6y + 3x^2 – 11 = 0`
Show Worked Solution
a.   

♦♦ Mean mark part (a) 32%.
MARKER’S COMMENT: Solution curve should follow slope ticks and not cross them.

 

b.    `(1 + y^2)(dy)/(dx)` `= −x`
  `int 1 + y^2 dy` `= −int x\ dx`
  `y + (y^3)/3` `= −(x^2)/2 + C, C ∈ R`

 
`text(Substituting)\ (-1,1):`

`1 + (1^3)/3` `= −((−1)^2)/2 + C`
`1 + 1/3` `= −1/2 + C`
`:. C` `= 11/6`

 

`y + 1/3y^3` `= −1/2x^2 + 11/6`
`6y + 2y^3` `= −3x^2 + 11`

 
`:. 2y^3 + 6y + 3x^2 – 11 = 0`

Calculus, EXT1 C3 2015 SPEC2 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)`

Show Answers Only

`D`

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, EXT1 C3 2016 SPEC1 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`.  (4 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))`

Calculus, EXT1 C3 2018 VCE 8

A tank initially holds 16 L of water in which 0.5 kg of salt has been dissolved. Pure water then flows into the tank at a rate of 5 L per minute. The mixture is stirred continuously and flows out of the tank at a rate of 3 L per minute.

  1.  Show that the differential equation for `Q`, the number of kilograms of salt in the tank after `t` minutes, is given by
    `qquad (dQ)/(dt) = -(3Q)/(16 + 2t)`  (1 mark)
      
  2.  Solve the differential equation given in part a. to find `Q` as a function of `t`.
      
    Express your answer in the form  `Q = a/(16 + 2t)^(b/c)`, where `a, b` and `c` are positive integers.  (3 marks)
Show Answers Only
  1.  `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  `Q = 32/(16 + 2t)^(3/2)`
Show Worked Solution

a. `Q_0 = 0.5, \ V_0 = 16`

♦ Net mean mark of both parts 44%.

`V(t)` `= 16 + (5 – 3) t`
  `= 16 + 2t`

 
`text(Concentration)\ (C)= Q/V= Q/(16 + 2t)\ text(kg/L)`

`(dQ)/(dt)` `= 0 xx 5 – 3C`
  `= -(3Q)/(16 + 2t)`

MARKER’S COMMENT: Many students took the common factor of 2 from  `16+2t`. This wasn’t necessary and complicated the arithmetic in part b.

 

b.   `-1/(3Q) * (dQ)/(dt) = 1/(16 + 2t)`

`int -1/(3Q)\ dQ` `= int 1/(16 + 2t) dt`
`-1/3 int 1/Q\ dQ` `= 1/2 int 2/(16 + 2t)\ dt`
`-1/3 ln Q` ` = [1/2 ln(16 + 2t)] + c`

 
`text(When)\ \ t=0,\ \ Q=0.5,`

COMMENT: A very challenging test of using exponential and log laws!

`-1/3 ln (1/2)` `= 1/2 ln (16) +c`
`c` `= -1/2 ln (16) -1/3 ln (1/2)`

 

`-1/3 ln Q` `= 1/2 ln(16 + 2t) -1/2 ln(16) – 1/3 ln (1/2)`
`-1/3 ln Q` `= 1/2 ln ((16 + 2t)/16) – 1/3 ln (1/2)`
`-1/3 ln Q` `= ln (((16 + 2t)^(1/2))/4) – ln (2^(-1/3))`
`ln (Q^(-1/3))` `= ln (((16 + 2t)^(1/2))/(2^2 ⋅ 2^(-1/3)))`
`Q^(-1/3)` `= ((16 + 2t)^(1/2))/(2^(5/3))`
`Q` `= (((16 + 2t)^(1/2))/(2^(5/3)))^-3`
  `= (16 + 2t)^(- 3/2)/(2^(-5))`
`:. Q` `= 32/((16 + 2t)^(3/2))`

Vectors, EXT2 V1 2017 SPEC1 5

Relative to a fixed origin, the points `B`, `C` and `D` are defined respectively by the position vectors  `underset~b = underset~i - underset~j + 2underset~k, \ underset~c = 2underset~i - underset~j + underset~k`  and  `underset~d = aunderset~i - 2underset~j`  where `a` is a real constant.

Given that the magnitude of angle `BCD` is  `pi/3`, find `a`.  (4 marks)

Show Answers Only

`a = −2`

Show Worked Solution

`text(Angle between)\ overset(->)(CB)\ text(and)\ overset(->)(CD) = pi/3`

♦ Mean mark 45%.

`overset(->)(CB)` `= (1 – 2)underset~i + (−1 – −1)underset~j + (2 – 1)underset~k`
  `= −underset~i + underset~k`
`overset(->)(CD)` `= (a – 2)underset~i + (−2 – −1)underset~j + (−1 + 0)underset~k`
  `= (a – 2)underset~i – underset~j – underset~k`

 

`overset(->)(CD) · overset(->)(CB)` `= −(a – 2) + 0 – 1`
  `= 1 – a`
  `= |overset(->)(CD)||overset(->)(CB)|cos(pi/3)`

 

`1 – a` `= sqrt((a – 2)^2 + (−1)^2 + (−1)^2)sqrt((−1)^2 + (1)^2)cos(pi/3)`
`1 – a` `= sqrt(a^2 – 4a + 4 + 1 + 1) xx sqrt2 xx 1/2`
`2(1 – a)` `= sqrt(2a^2 – 8a + 12), \ \ a < 1`
`4(1 – a)^2` `= 2a^2 – 8a + 12, \ \ a < 1`
`4(1 – 2a + a^2)` `= 2a^2 – 8a + 12`
`4 – 8a + 4a^2` `= 2a^2 – 8a + 12`
`2a^2` `= 8`
`a^2` `= 4`
`:. a` `= −2\ \ \ (a<1)`

Vectors, EXT2 V1 SM-Bank 9

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.

Find the cosine of angle `ABC`.   (2 marks)

Show Answers Only

`(-1)/(sqrt6sqrt13)`

Show Worked Solution

`vec(BA) = vec(OA) – vec(OB) = – underset~ i + 2underset~j – underset~k`

♦ Mean mark 48%.

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

`vec(BC) = vec(OC) – vec(OB) = -3underset~i – 2underset~j`

`=>\ |\ 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)`

Vectors, EXT1 V1 2015 SPEC2 15 MC

The projection 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)/sqrt2)underset~w` 

Show Answers Only

`A`

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

`tildeF*hat w = (a + b)/sqrt2`

♦ Mean mark 49%.

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

 
`=> A`

Calculus, MET1 2018 VCAA 9

Consider a part of the graph of  `y = xsin(x)`, as shown below.

    1. Given that  `int(xsin(x))\ dx = sin(x) - xcos(x) + c`,
      evaluate  `int_(npi)^((n + 1)pi)(xsin(x))\ dx`  when `n` is a positive even integer or 0.

       

      Give your answer in simplest form.  (2 marks)

    2. Given that  `int(xsin(x))\ dx = sin(x) - xcos(x) + c`,
      evaluate  `int_(npi)^((n + 1)pi)(xsin(x))\ dx`  when `n` is a positive odd integer.

       

      Give your answer in simplest form.  (1 mark)

  2. Find the equation of the tangent to  `y = xsin(x)`  at the point  `(−(5pi)/2,(5pi)/2)`.  (2 marks)
  3. The translation `T` maps the graph of  `y = xsin(x)`  onto the graph of  `y = (3pi - x)sin(x)`, where
     
    `qquad T: R^2 -> R^2, T([(x),(y)]) = [(x),(y)] + [(a),(0)]`
     
    and `a` is a real constant.

     

    State the value of `a`.  (1 mark)

  4. Let  `f:[0,3pi] -> R, f(x) = (3pi - x)sin(x)`  and  `g:[0,3pi] -> R, g(x) = (x - 3pi)sin(x)`.
    The line  `l_1` is the tangent to the graph of  `f`  at the point  `(pi/2,(5pi)/2)`  and the line `l_2` is the tangent to the graph of  `g`  at  `(pi/2,−(5pi)/2)`, as shown in the diagram below.
     
         
     
    Find the total area of the shaded regions shown in the diagram above.  (2 marks)
Show Answers Only
  1.   
    1. `(2n + 1)pi`
    2. `−(2n + 1)pi`
  2. `:. y = −x`
  3. `−3pi`
  4. `9pi(pi – 2)`
Show Worked Solution

a.i.  `text(Given)\ \ n\ \ text(is a positive even integer:)`

♦♦ Mean mark 31%.

`int_(npi)^((n + 1)pi)(xsin(x))dx`

`= [sin(x) – xcos(x)]_(npi)^((n + 1)pi)`

`= [sin((n + 1)pi) – (n + 1)pi · cos((n + 1)pi)] – [sin(npi) – npi · cos(npi)]`

`= [0 – (n + 1)pi (−1)] – [0 – npi]`

`= (n + 1)pi + npi`

`= (2n + 1)pi`

 

a.ii. `text(Given)\ \ n\ \ text(is a positive odd integer:)`

♦♦♦ Mean mark 19%.

`int_(npi)^((n + 1)pi)(xsin(x))dx`

`= [sin((n + 1)pi) – (n + 1)pi · cos((n + 1)pi)] – [sin(npi) – npi · cos(npi)]`

`= [0 – (n + 1)pi(1)] – [0 – npi(−1)]`

`= −(n + 1)pi – npi`

`= −(2n + 1)pi`

 

b.    `y` `= xsin(x)`
  `(dy)/(dx)` `= x · cos(x) + sin(x)`

 

♦ Mean mark part (b) 49%.

`(dy)/(dx)` `= −(5pi)/2 · cos(−(5pi)/2) + sin(−(5pi)/2)`
  `= −(5pi)/2 · (0) + (−1)`
  `= −1`

 
`text(T)text(angent has equation)\ \ y = −x + c\ \ text(and passes through)\ \ (−(5pi)/2, (5pi)/2):`

`(5pi)/2` `= +(5pi)/2 + c`
`c` `= 0`

 
`:. y = −x`

 

♦♦ Mean mark part (c) 34%.

c.    `y` `= (3pi – x)sin(x)`
    `= – (x – 3pi)sin(x)`

 
`:. a = 3pi`
 

d.   `f(x) = (3pi – x)sin(x)`

♦♦♦ Mean mark 7%.

 
`-> l_1\ text(is the tangent)\ \ y = −x\ \ (text{using part (b)})`

`-> g(x)\ text(is)\ \ y=xsin(x)\ \ text(translated 3π to the right.)`

`-> f(x)\ text(is)\ \ g(x)\ \ text(reflected in the)\ \ x text(-axis.)`

 
`text(Area between)\ f(x)\ text(and)\ xtext(-axis)`

`= int_0^pi xsin(x)\ dx + | int_pi^(2pi) xsin(x)\ dx | + int_(2pi)^(3pi) xsin(x)\ dx`

`= (2 xx 0 + 1)pi + | −1(2 xx 1 + 1)pi | + (2 xx 2 + 1)pi\ \ (text{using part (a)})`

`= pi + 3pi + 5pi`

`= 9pi`
 

`:.\ text(Shaded Area)`

`= 2 xx (1/2 xx 3pi xx 3pi) – 2 xx 9pi`

`= 9pi^2 – 18pi`

`= 9pi(pi – 2)`

Functions, 2ADV F2 SM-Bank 8 MC

The transformation that maps the graph of  `y = sqrt(8x^3 + 1)`  onto the graph of  `y = sqrt(x^3 + 1)`  is a

  1. dilation by a factor of `2` from the `y`-axis.
  2. dilation by a factor of `2` from the `x`-axis.
  3. dilation by a factor of `1/2` from the `x`-axis.
  4. dilation by a factor of `1/2` from the `y`-axis.
Show Answers Only

`A`

Show Worked Solution
`text(Let)\ f(x)` `= sqrt(8x^3 + 1)`
`f(1/2 x)` `= sqrt(8(1/2 x)^3 + 1)`
  `= sqrt(x^3 + 1)`

 

`:.\ text(Transformation correct when)\ \ x\ \ text(is swapped for)\ \ x/2`

`text(i.e. graph is dilated by factor of 2 from)\ ytext(-axis)`

`=> A`

Functions, 2ADV F2 SM-Bank 7 MC

The point  `A (3, 2)`  lies on the graph of the function  `f(x)`. A transformation maps the graph of  `f(x)`  to the graph of  `g(x)`,

where  `g(x) = 1/2 f(x - 1)`. The same transformation maps the point `A` to the point `P`.

The coordinates of the point `P` are

A.  `(2, 1)`

B.  `(2, 4)`

C.  `(4, 1)`

D.  `(4, 2)`

Show Answers Only

`C`

Show Worked Solution

`g(x) = 1/2 f(x – 1),\ A(3, 2)`

`text(Dilate by a factor of)\ 1/2\ text(from)\ x text(-axis:)`

`A(3, 2) -> A′(3, 1)`
 

`text(Translate 1 unit to right:)`

`A′(3, 1) -> P(4, 1)`
 

`=>   C`

Statistics, 2ADV S3 SM-Bank 13

The time Jennifer spends on her homework each day varies, but she does some homework every day.

The continuous random variable \(T\), which models the time, \(t\), in minutes, that Jennifer spends each day on her homework, has a probability density function \(f\), where

\(f(t)= \begin{cases}
\dfrac{1}{625}(t-20) & 20 \leq t<45 \\
\ \\
\dfrac{1}{625}(70-t) & 45 \leq t \leq 70 \\
\ \\
0 & \text {elsewhere }
\end{cases}\)

  1. Sketch the graph of  \(f(t)\) on the axes provided below.  (3 marks)

     

        
     

  2. Find the mode.  (1 mark)

  3. Find  \(P(25 \leq T \leq 55)\).  (2 marks)

Show Answers Only

  1.  

  2. \(45\)
  3. \(\dfrac{4}{5}\)
Show Worked Solution
i.   

MARKER’S COMMENT: Many did not draw graph along \(t\)-axis between 0 and 20 and for  \(t>70\).

 
ii.     \(\text{Mode \(=45\) (value of \(t\) at highest value of \(f(t))\)}\)
 

iii.    \(P(25 \leq T \leq 55)\)

\(\begin{aligned}
& =\int_{25}^{45} \dfrac{1}{625}(t-20) d t+\int_{45}^{55} \dfrac{1}{625}(70-t) d t \\
& =\dfrac{1}{625}\left[\dfrac{t^2}{2}-20 t\right]_{25}^{45}+\dfrac{1}{625}\left[70 t-\dfrac{t^2}{2}\right]_{45}^{55} \\
& =\dfrac{1}{625}[112.5-(-187.5)]+\dfrac{1}{625}(2337.5-2137.5) \\ & =\dfrac{300}{625}+\dfrac{200}{625} \\
& =\dfrac{4}{5}
\end{aligned}\)

Statistics, 2ADV S3 SM-Bank 7 MC

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

\(f(x)= \begin{cases}
\dfrac{1}{12}\left(8 x-x^3\right) & 0 \leq x \leq 2 \\
\ \\
0 & \text{elsewhere }
\end{cases}\)

The median  \(m\)  of this function satisfies the equation

  1. \(-m^4+16 m^2-6=0\)
  2. \(m^4-16 m^2=0\)
  3. \(m^4-16 m^2+24=0.5\)
  4. \(m^4-16 m^2+24=0\)
Show Answers Only

\(D\)

Show Worked Solution

\(\begin{aligned} \int_0^m \dfrac{1}{12}\left(8 x-x^3\right) d x & =0.5 \\
{\left[\dfrac{1}{12}\left(4 x^2-\dfrac{x^4}{4}\right)\right]_0^m } & =0.5 \\
4 m^2-\dfrac{m^4}{4} & =6 \\
16 m^2-m^4 & =24 \\ m^4-16 m^2+24 & =0
\end{aligned}\)

\(\Rightarrow D\)

Statistics, 2ADV S3 SM-Bank 6 MC

The continuous random variable, \(X\), has a probability density function given by

\(f(x)= \begin{cases}
\dfrac{1}{4} \cos \left(\dfrac{x}{2}\right) & 3 \pi \leq x \leq 5 \pi \\
\ & \ \\
0 & \text {elsewhere}
\end{cases}\)

The value of \(a\) such that  \(P(X<a)=\dfrac{\sqrt{3}+2}{4}\)  is

  1. \(\dfrac{19 \pi}{6}\)
  2. \(\dfrac{14 \pi}{3}\)
  3. \(\dfrac{10 \pi}{3}\)
  4. \(\dfrac{29 \pi}{6}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\begin{aligned}
\int_{3 \pi}^a \dfrac{1}{4}\, \cos \left(\dfrac{x}{2}\right) d x & =\dfrac{\sqrt{3}+2}{4} \\
{\left[\dfrac{1}{2}\, \sin \left(\dfrac{x}{2}\right)\right]_{3 \pi}^a } & =\dfrac{\sqrt{3}+2}{4} \\
\dfrac{1}{2}\left[\sin \left(\dfrac{a}{2}\right)-\sin \left(\dfrac{3 \pi}{2}\right)\right] & =\dfrac{\sqrt{3}+2}{4} \\
\dfrac{1}{2}\, \sin \left(\dfrac{a}{2}\right)+\dfrac{1}{2} & =\dfrac{\sqrt{3}+2}{4} \\
\sin \left(\dfrac{a}{2}\right) & =\dfrac{\sqrt{3}}{2} \\
\dfrac{a}{2} & =\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{7 \pi}{3}, \ldots \\
\therefore a & =\dfrac{14 \pi}{3} \quad(3 \pi \leq a \leq 5 \pi)
\end{aligned}\)

\(\Rightarrow B\)

Statistics, 2ADV S3 SM-Bank 3

The continuous random variable `X` has a probability density function given by
 

`f(x) = {(pi sin (2 pi x), text(if)\ \ 0 <= x <= 1/2), (0, text(elsewhere)):}`
 

Find the value of  `a`  such that  `P(X > a) = 0.2`. Give your answer to 2 decimal places.   (3 marks)

Show Answers Only

`0.35`

Show Worked Solution

`int_a^(1/2) pi sin (2 pi x)\ dx = 0.2`

`-1/2 [cos(2 pi x)]_a^(1/2)` `=0.2`  
`-1/2[cos(pi) -cos(2 pia)]`  `=0.2`  
`-1/2(-1-cos(2pia))` `=0.2`  
`-1-cos(2pia)` `=-0.4`  
`cos(2pia)` `=-0.6`  
`2pia` `=cos^(-1)(-0.6)`  
`:.a` `=cos^(-1)(-0.6)/(2pi)`  
  `=0.3524…`  
  `=0.35\ \ \ text{(to 2 d.p.)}`  

Trigonometry, 2ADV T3 SM-Bank 16

Sammy visits a giant Ferris wheel. Sammy enters a capsule on the Ferris wheel from a platform above the ground. The Ferris wheel is rotating anticlockwise. The capsule is attached to the Ferris wheel at point `P`. The height of `P` above the ground, `h`, is modelled by  `h(t) = 65 - 55cos((pit)/15)`, where `t` is the time in minutes after Sammy enters the capsule and `h` is measured in metres.

Sammy exits the capsule after one complete rotation of the Ferris wheel.
 


 

  1. State the minimum and maximum heights of `P` above the ground.  (1 mark)

  2. For how much time is Sammy in the capsule?  (1 mark)

  3. Find the rate of change of `h` with respect to `t` and, hence, state the value of `t` at which the rate of change of `h` is at its maximum.  (2 marks)

Show Answers Only
  1. `h_text(min) = 10\ text(m), h_text(max) = 120\ text(m)`
  2. `30\ text(min)`
  3. `t = 7.5`
Show Worked Solution
i.    `h_text(min)` `= 65 – 55` `h_text(max)` `= 65 + 55`
    `= 10\ text(m)`   `= 120\ text(m)`

 

ii.   `text(Period) = (2pi)/(pi/15) = 30\ text(min)`

 

iii.    `h′(t)` `=65 – 55cos((pit)/15)`
  `h′(t)` `=pi/15 xx 55sin(pi/15 t)`
    `= (11pi)/3\ sin(pi/15 t)`
     

`text(S)text(ince)\ \ sin(pi/15 t)_text(max) = sin (pi/2),`

 
`:. h′(t)_text(max)\ \ text(occurs when)`

`(pi t)/15` `=pi/2`  
`:. t` `=pi/2 xx 15/pi`  
  `=15/2\ text(minutes)\ \ (0<=t<=30)`  

Trigonometry, 2ADV T2 SM-Bank 40

Let  `(tantheta - 1) (sin theta - sqrt 3 cos theta) (sin theta + sqrt 3 costheta) = 0`.

  1. State all possible values of  `tan theta`.  (1 mark)

  2. Hence, find all possible solutions for  `(tan theta - 1) (sin^2 theta - 3 cos^2 theta) = 0`, where  `0 <= theta <= pi`.  (2 marks)

Show Answers Only
  1. `tan theta = 1 or tan theta = +- sqrt 3`
  2. `theta = pi/4, pi/3 or (2 pi)/3`
Show Worked Solution

i.  `(tantheta – 1) (sin theta – sqrt 3 cos theta) (sin theta + sqrt 3 costheta) = 0`

`=> tan theta = 1`

♦ Mean mark 42%.
`=>sin theta – sqrt 3 cos theta` `=0`
`sin theta` `=sqrt3 cos theta`
`tan theta` `=sqrt3`

 

`=>sin theta + sqrt 3 cos theta` `=0`
`sin theta` `=-sqrt3 cos theta`
`tan theta` `=-sqrt3`

 
`:. tan theta = 1 or tan theta = +- sqrt 3`

 

ii.  `(tan theta – 1) (sin^2 theta – 3 cos^2 theta) = 0`

`text(Using part a:)`

♦ Mean mark 42%.

`(tan theta – 1) (sin theta – sqrt 3 cos theta) (sin theta + sqrt 3 cos theta) = 0`

`=> tan theta` `= 1` `qquad or qquad` `tan theta` `= +- sqrt 3`
`theta` `= pi/4`   `theta` `= pi/3, (2 pi)/3`

 
`:. theta = pi/4, pi/3 or (2 pi)/3\ \ \ \ (0<=theta<=pi)`

Trigonometry, 2ADV T3 SM-Bank 15

The graphs of  `y = cos (x) and y = a sin (x)`,  where `a` is a real constant, have a point of intersection at  `x = pi/3.`

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

  2. Find the  `x`-coordinate of the other point of intersection of the two graphs, given  `0<=x<= 2 pi`  (1 mark)

Show Answers Only
  1. `1/sqrt 3`
  2. `(4 pi)/3`
Show Worked Solution

i.   `text(Intersection occurs when)\ \ x=pi/3,`

`a sin(pi/3)` `= cos (pi/3)`
`tan(pi/3)` `= 1/a`
`sqrt 3` `=1/a`
`:. a` `=1/sqrt3`

 

ii.   `tan (x)` `= sqrt 3`
  `x` `= pi/3, (4 pi)/3, 2pi+ pi/3, …`
  `:. x` `= (4 pi)/3\ \ \ (0<= x<= 2 pi)`

Trigonometry, 2ADV T3 SM-Bank 14

For the function  `f(x) = 5 cos (2 (x + pi/3)),\ \ \ -pi<=x<=pi`

  1. Write down the amplitude and period of the function  (2 marks)

  2. Sketch the graph of the function  `f(x)`  on the set of axes below. Label axes intercepts with their coordinates.

     

    Label endpoints of the graph with their coordinates.  (3 marks)

VCAA 2006 meth 4b

Show Answers Only
  1. `text(Amplitude) = 5;\ \ \ text(Period) = pi`
  2.  
Show Worked Solution

a.   `text(Amplitude) = 5`

`text(Period) = (2 pi)/2 = pi`

 

b.   `text(Shift)\ \ y = 5 cos (2x)\ \ text(left)\ \ pi/3\ \ text(units).`

`text(Period) = pi`

`text(Endpoints are)\ \ (-pi, -5/2) and (pi,-5/2)`

Trigonometry, 2ADV T3 SM-Bank 10

The population of wombats in a particular location varies according to the rule  `n(t) = 1200 + 400 cos ((pi t)/3)`, where `n` is the number of wombats and `t` is the number of months after 1 March 2018.

  1. Find the period and amplitude of the function `n`.  (2 marks)

  2. Find the maximum and minimum populations of wombats in this location.  (2 marks)

  3. Find  `n(10)`.  (1 mark)

  4. Over the 12 months from 1 March 2018, find the fraction of time when the population of wombats in this location was less than  `n(10)`.  (2 marks)

Show Answers Only
  1. `text(Period) = text(6 months);\ text(Amplitude) = 400`
  2. `text(Max) = 1600;\ text(Min) = 800`
  3. `1000`
  4. `1/3`
Show Worked Solution

i.   `text(Period) = (2pi)/n = (2pi)/(pi/3) = 6\ text(months)`

`text(A)text(mplitude) = 400`
 

ii.   `text(Max:)\ 1200 + 400 = 1600\ text(wombats)`

`text(Min:)\ 1200 – 400 = 800\ text(wombats)`
 

iii.    `n(10)` `=1200 + 400 cos ((10 pi)/3)`
    `=1200 + 400 cos ((2 pi)/3)`
    `=1200-400 xx 1/2`
    `= 1000\ text(wombats)`

 

iv.  `text(Find)\ \ t\ \ text(when)\ \ n(t)=1000`

`1000` `=1200 + 400 cos((pit)/3)`  
`cos((pit)/3)` `=- 1/2`  
`(pit)/3` `=(2pi)/3, (4pi)/3, (8pi)/3, (10pi)/3, … `  
`t` `=2,4,8,10`  

 
`text(S)text(ince)\ \ n(0)=1600,`

`=> n(t)\ \ text(drops below 1000 between)\ \ t=2\ \ text(and)\ \ t=4,`

`text(and between)\ \ t=8\ \ text(and)\ \ t=10.`
 

`:.\ text(Fraction)` `= (2 + 2)/12`
  `= 1/3\ \ text(year)`

Trigonometry, 2ADV T3 SM-Bank 8

`f(x) = 2 sin (2x)`  is defined in the domain  `{x: \ pi/8 <= x < pi/3)`

What is the range of the function  `f(x)`?  (2 marks)

Show Answers Only

`text(Range) = [sqrt 2, 2]`

Show Worked Solution

♦ Mean mark 45%.

`sin(2x)_text(max)\ \ text(occurs when)\ \ x=pi/4\ \ text{(within domain)}`

`=> f(x)_text(max)  = 2 sin(pi/2) = 2`

`text(Checking endpoints:)`

`text(When)\ \ x=pi/8\ \ =>\ \ y=2 sin(pi/4) = sqrt2`

`text(When)\ \ x=pi/3\ \ =>\ \ y=2 sin((2pi)/3) = sqrt3`
 


 

`:.\ text(Range) = [sqrt 2, 2],`

Measurement, STD2 M7 SM-Bank 11

On a map, the distance between two towns is measured at 54 millimetres.

  1. The actual distance between the two towns is 16.2 kilometres. What is the scale of the map?  (1 mark)

  2. If two other towns on the same map are 9.2 centimetres apart, what is the actual distance between the two towns in kilometres?  (1 mark)

Show Answers Only
  1. `1: 300\ 000`
  2. `27.6\ text(km)`
Show Worked Solution

i.   `text{Convert both distance to the same unit (cm).}`

`text(54 mm)` `= 5.4\ text(cm)`
`text(16.2 km)` `= 16\ 200\ text(m)`
  `= 1\ 620\ 000\ text(cm)`

 

`:.\ text(Scale) \ \ 5.4\ ` `: 1\ 620\ 000`
`1\ ` `: 300\ 000`

 

ii.    `text(Actual distance)` `= 9.2 xx 300\ 000`
    `= 2\ 760\ 000\ text(cm)`
    `= 27\ 600\ text(m)`
    `= 27.6\ text(km)`

Measurement, STD2 M7 SM-Bank 9

A farmer sells 216 sheep at a livestock auction.

Three buyers purchased all the sheep in the ratio  7:3:2

How many sheep did the buyer of the highest number of sheep purchase?  (2 marks)

Show Answers Only

`126\ text(sheep)`

Show Worked Solution

`7:3:2 => 7/12: 3/12:2/12`

`:.\ text(Buyer of the most sheep)`

`= 7/12 xx 216`

`= 126\ text(sheep)`

Networks, STD2 N3 SM-Bank 45

An oil pipeline network is drawn below that shows the flow capacity of oil pipelines in kilolitres per hour.
 


 

A cut is shown.

  1. What is the capacity of the cut.  (1 mark)

  2. Calculate the minimum cut of this network?  (2 marks)

  3. Copy the network diagram, showing the maximum flow capacity of the network by labelling the flow of each edge.  (2 marks)

Show Answers Only
  1. `35`
  2. `text(See Worked Solutions)`
  3.  
Show Worked Solution
i.    `text(Capacity of cut)` `= 7 + 15 + 13`
    `= 35\ text(kL/h)`

 

ii. 

♦♦ COMMENT: Be very careful! RS is not included as it goes from sink to source.
 


  

`text(Minimum cut)` `= 7 + 14 + 9`
  `= 30\ text(kL/h)`

 

iii.   

Networks, STD2 N3 SM-Bank 39

Bianca is designing a project for producing an advertising brochure. It involves activities A-M.

The network below shows these activities and their completion time in hours.
 


 

  1.  What is the earliest starting time of activity J?  (1 mark)

  2.  What is the latest starting time of activity H?  (1 mark)

  3.  What is the float time of activity I?  (1 mark)

  4.  What is the minimum time required to produce the brochure?  (2 marks)

Show Answers Only
  1. `text(11 hours)`
  2. `text(18 hours)`
  3. `text(2 hours)`
  4. `text(34 hours)`
Show Worked Solution

i.   `text(Scanning forwards and backwards)`
 

 

`text{EST (activity}\ J) = 11\ text(hours)`

 

ii.    `text{LST (activity}\ H)` `=\ text{LST}\ (H)-text(weight)\ (H)`
    `= 20-2`
    `=18\ text(hours)`

 

iii.    `text{Float time}\ (I)` `=\ text(LST)\ (I)-text(EST)\ (I)`
    `=12-10`
    `= 2\ text(hours)`

 

iv.   `text(Critical Path is)\ CGJLM`

`:.\ text(Minimum time)` `= 2 + 9 + 6 + 7 + 10`
  `= 34\ text(hours)`

Algebra, STD2 A4 SM-Bank 10

The number of trees that can be planted along the fence line of a paddock varies inversely with the distance between each tree.

There will be 108 trees if the distance between them is 5 metres.

  1. How many trees can be planted if the distance between them is 6 metres?  (2 marks)

  2. What is the distance between the trees if 120 trees are planted.  (1 mark)

Show Answers Only
  1. `90`
  2. `4.5\ text(metres)`
Show Worked Solution

i.   `t ∝ 1/d`

`t` `= k/d`
`108` `= k/5`
`k` `= 540`

 

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

`t` `= 540/6`
  `= 90`

 

ii.   `text(Find)\ \ d\ \ text(when)\ \ t = 120:`

`120` `= 540/d`
`d` `= 540/120`
  `= 4.5\ text(metres)`

Calculus, SPEC2 2012 VCAA 5

At her favourite fun park, Katherine’s first activity is to slide down a 10 m long straight slide. She starts from rest at the top and accelerates at a constant rate, until she reaches the end of the slide with a velocity of `6\ text(ms)^(-1)`.

  1. How long, in seconds, does it take Katherine to travel down the slide?  (1 mark)

When at the top of the slide, which is 6 m above the ground, Katherine throws a chocolate vertically upwards. The chocolate travels up and then descends past the top of the slide to land on the ground below. Assume that the chocolate is subject only to gravitational acceleration and that air resistance is negligible.

  1. If the initial speed of the chocolate is 10 m/s, how long, correct to the nearest tenth of a second, does it take the chocolate to reach the ground?  (2 marks)
  2. Assume that it takes Katherine four seconds to run from the end of the slide to where the chocolate lands.

     

    At what velocity would the chocolate need to be propelled upwards, if Katherine were to immediately slide down the slide and run to reach the chocolate just as it hits the ground?

     

    Give your answer in `text(ms)^(-1)`, correct to one decimal place.  (2 marks)

Katherine’s next activity is to ride a mini speedboat. To stop at the correct boat dock, she needs to stop the engine and allow the boat to be slowed by air and water resistance.

At time `t` seconds after the engine has been stopped, the acceleration of the boat, `a\ text(ms)^(-2)`, is related to its velocity, `v\ text(ms)^(-1)`, by

`a = -1/10 sqrt(196 - v^2)`.

Katherine stops the engine when the speedboat is travelling at `7\ text(m/s)`.

  1. i.  Find an expression for `v` in terms of `t`.  (3 marks)
  2. ii. Find the time it takes the speedboat to come to rest.Give your answer in seconds in terms of `pi`.  (2 marks)
  3. iii. Find the distance it takes the speedboat to come to rest, from when the engine is stopped.Give your answer in metres, correct to one decimal place.  (3 marks)
Show Answers Only
  1. `10/3\ text(s)`
  2. `2.5\ text(s)`
  3. `35.1\ text(m s)^(−1)`
  4. i.  `v = 14 sin(pi/6 – t/10)`
  5. ii. `(5 pi)/3`
  6. iii. `18.8`
Show Worked Solution

a.   `u = 0, quad v = 6, quad s = 10`

COMMENT: Exact form required. 3.3 seconds was marked incorrect!

`text(Solve for)\ \ t:\ \ \ ((u + v)/2)t` `= s`
`((0 + 6)/2)t` `= 10`
`t` `= 10/3\ text(s)`

 

b.   `u = 10, quad a = -9.8, quad s = -6`

COMMENT: Many students showed a “lack of understanding” of displacement here.

`text(Solve for)\ \ t:`

`s` `= ut + 1/2 at^2`
`-6` `= 10t – 4.9 t^2`
`:. t` `~~2.5\ text(s)\ \ \ text{(by CAS)}`

♦♦ Mean mark 27%.

c.   `text(Time of chocolate in air)`

`= 10/3 + 4`

`=22/3`
 

`text(Solve for)\ \ u:`

`-6` `= 22/3 u -4.9(22/3)^2`
`:. u` `~~35.1\ text(m s)^(−1)\ \ \ text{(by CAS)}`

 

d.i.    `(dv)/(dt)` `= -1/10 sqrt(196 – v^2)`
  `(dt)/(dv)` `= (-10)/sqrt(196 – v^2)`
  `t` `= int(-10)/sqrt(196 – v^2)\ dv`
  `-t/10` `= int 1/sqrt(14^2 – v^2) dv`
  `-t/10` `= sin^(-1)(v/14) + c`

 
`text(When)\ \ t = 0, v = 7`

`=> c=-sin^(-1)(1/2) = -pi/6`
 

`-t/10` `=sin^(-1)(v/14) – pi/6`  
`sin^(-1) (v/14)` `= pi/6-t/10`  
`:. v` `=14sin(pi/6-t/10)`  

 

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

  `-t/10` `=sin^(-1)(0) – pi/6`
  `t` `= -10 sin^(-1)(0) + (10 pi)/6`
    `= (5 pi)/3\ text(s)`

 

d.iii.   `v = 14sin(pi/6-t/10)`

`(dx)/(dt)` `=14sin(pi/6-t/10)`
`x` `= int_0^((5pi)/3)14sin(pi/6-t/10)\ dt`
  `~~ 18.8\ text(m)\ \ \ text{(by CAS)}`

Vectors, SPEC2 2012 VCAA 4

The position vector of the International Space Station `text{(S)}`, when visible above the horizon from a radar tracking location `text{(O)}` on the surface of Earth, is modelled by

`underset ~r(t) = 6800 sin(pi(1.3t - 0.1))underset~i + (6800 cos(pi(1.3t - 0.1)) - 6400)underset ~j`,

for  `t in [0, 0.154]`,

where  `underset ~i`  is a unit vector relative to `text(O)` as shown and  `underset ~j`  is a unit vector vertically up from point `text(O)`. Time  `t`  is measured in hours and displacement components are measured in kilometres.
 

 

  1. Find the height, `h` km, of the space station above the surface of Earth when it is at point `text(P)`, directly above point `text(O)`.

     

    Give your answer correct to the nearest km  (1 mark)

  2. Find the acceleration of the space station, and show that its acceleration is perpendicular to its velocity.  (3 marks)
  3. Find the speed of the space station in km/h.

     

    Give your answer correct to the nearest integer.  (2 marks)

  4. Find the equation of the path followed by the space station in cartesian form.  (2 marks)
  5. Find the times when the space station is at a distance of 1000 km from the radar tracking location `text(O)`.

     

    Give your answers in hours, correct to two decimal places.  (3 marks)

Show Answers Only

  1. `text(400 km)`

  2. `text(See Worked Solutions.)`

  3. `27\ 772\ text(km/h)`

  4. `x^2 + (y + 6400)^2 = 46\ 240\ 000`

  5. `t ~~ 0.04, 0.11`

Show Worked Solution

a.   `text(At)\ \ P,\ \ x(t)=0`

♦ Mean mark 49%.

  `text{Solve (by CAS):}\ \ 6800 sin (pi(1.3t – 0.1))` `= 0`

 
`=> t=0.076923…`
 

`:. h` `= 6800 cos (pi (1.3(0.076923) – 0.1)) – 6400`
  `= 400`

 

b.    `underset ~(dot r) (t)` `= 8840pi cos (pi(1.3t – 0.1)) underset ~i – 8840pi sin (pi(1.3t – 0.1)) underset ~j`
  `underset ~(ddot r) (t)` `= -11\ 492 pi^2 sin (pi(1.3t – 0.1))underset ~i – 11\ 492 pi^2 cos (pi(1.3t – 0.1)) underset ~j`

 
`text(Let)\ \ u=pi(1.3t-0.1),`

`underset ~(ddot r) (t) ⋅ underset ~(dot r) (t)` `= -(8840)(11\492) pi^3 cos (u) sin (u) + (8840)(11\492) pi^3 cos(u)sin(u)`

 
  `= 0`

 
`:. underset ~(ddot r) (t) _|_ underset ~(dot r) (t)`

 

c.   `text(Speed)\ = |\ dotr(t)\ |`

♦ Mean mark part (c) 45%.

`|\ dotr(t)\ |` `= sqrt((8840pi)^2 cos^2(pi(1.3t – 0.1)) + (8840pi)^2 sin^2 (pi(1.3t – 0.1)))`
  `= 8840 pi`
  `~~ 27\ 772`

 

d.    `x/6800` `= sin(pi(1.3t – 0.1))`
  `(6400 + y)/6800` `= cos(pi (1.3t – 0.1))`

 
`text(Using)\ \ sin^2theta + cos^2theta = 1`

`x^2/6800^2 + (6400 + y)^2/6800^2` `=1`  
`x^2 + (y + 6400)^2` `= 6800^2`  

 

e.  `underset ~r(t) = 6800 sin(u)underset~i + (6800 cos(u) – 6400)underset ~j`

♦ Mean mark part (e) 37%.

`text(Find)\ \ t\ \ text(such that)\ \ |\ underset ~r(t)\ | = 1000:`

`6800^2sin^2(u) + (6800 cos(u) -6400)^2 = 1000^2,\ \ \ (u=pi(1.3t-0.1))`

`6800^2 – 2 xx 6400 xx 6800 cos(u) + 6400^2 = 1000^2`

`=>cos(pi(1.3t-0.1)) = 0.9903\ \ \ text{(by CAS)}`

`:. t=0.04 or 0.11`

Calculus, SPEC2 2012 VCAA 3

A car accelerates from rest. Its speed after `T` seconds is `V\ text(ms)^(−1)`, where

`V = 17 tan^(−1)((pi T)/6), T >= 0`

  1. Write down the limiting speed of the car as  `T -> oo`.  (1 mark)
  2. Calculate, correct to the nearest `0.1\ text(ms)^(−2)`, the acceleration of the car when  `T = 10`.  (1 mark)
  3. Calculate, correct to the nearest second, the time it takes for the car to accelerate from rest to `25\ text(ms)^(−1)`.  (2 marks)

After accelerating to `25\text(ms)^(−1)`, the car stays at this speed for 120 seconds and then begins to decelerate while braking. The speed of the car  `t`  seconds after the brakes are first applied is `v\ text(ms)^(−1)` where

`(dv)/(dt) = - 1/100 (145 - 2t),`

until the car comes to rest.

  1. i.  Find `v` in terms of `t`.  (2 marks)
  2. ii. Find the time, in seconds, taken for the car to come to rest while braking.  (2 marks)
  3. i.  Write down the expressions for the distance travelled by the car during each of the three stages of its motion. (2 marks)
  4. ii. Find the total distance travelled from when the car starts to accelerate to when it comes to rest.

     

        Give your answer in metres correct to the nearest metre.  (1 mark)

Show Answers Only
  1. `(17 pi)/2`
  2. `0.3`
  3. `19\ text(s)`
  4. i.  `V = t^2/100 – (145 t)/100 + 25`
  5. ii. `t_1 = 20\ text(s)`
  6. i.  `d_1 = int_0^19 17 tan^(-1) ((pi T)/6)\ dT`

     

        `d_2 = 25 xx 120`

     

        `d_3 = int_0^20 -1/100 [145 t – t^2] + 25\ dt`

  7. ii. `3637\ text(m)`
Show Worked Solution

a.   `text(As)\ \ T->oo,\ \ tan^(-1)((piT)/6)->pi/2`

  `:.underset (T->oo) (limV)` `= (17 pi)/2`

 

b.  `(dV)/(dt)= (102 pi)/(36 + pi^2 T^2),\ \ \ text{(by CAS)}`
 

`text(When)\ \ T=10:`

`(dV)/(dt)` `= (102 pi)/(36 + 100 pi^2)`
  `~~ 0.3`

 
c.   `text(Find)\ \ T\ \ text(when)\ \ V=25:`

  `17 tan^(-1) ((pi T)/6)` `=25 `
  `T` `= 18.995\ \ \ text{(by CAS)}`
    `~~ 19\ text(seconds)`

 

d.i.    `v` `= -1/100 int_0^t 145 – 2t\ dt`
  `v` `= -1/100 [145 t – t^2] + c`

 
`text(When)\ \ t=0, \ v=25:`

`=> c=25`

`:. v= -1/100 [145 t – t^2] + 25`
 

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

`-1/100[145t – t^2] + 25=0`

`:. t=20\ text(seconds)\ \ \ text{(by CAS)}`

 

e.i.   `text(Stage 1: car travels from rest to 25 m/s)`

`d_1 = int_0^19 17 tan^(-1) ((pi T)/6)\ dT`
  

`text(Stage 2: car travels at 25 m/s for 120 seconds)`

`d_2` `= 25 xx 120`
  `= 3000`

 
`text(Stage 3: car decelerates for 20 seconds`

`d_3 = int_0^20 -1/100 [145 t – t^2] + 25\ dt`

 

e.ii.    `d_1` `~~ 400.131`
  `d_2` `= 3000`
  `d_3` `= 236.6`

 

`text(Total distance)` `= d_1 + d_2 + d_3`
  `~~ 3637\ text(m)`

Complex Numbers, SPEC2 2012 VCAA 2

  1.  Given that  `cos(pi/12) = (sqrt (sqrt 3 + 2))/2`, show that  `sin(pi/12) = (sqrt (2 - sqrt 3))/2`.  (2 marks)
  2. Express  `z_1 = (sqrt(sqrt3 + 2))/2 + i(sqrt(2 - sqrt3))/2`  in polar form.  (1 mark)
  3. i.  Write down  `z_1^4`  in polar form.  (1 mark)
  4. ii. On the Argand diagram below, shade the region defined by

`{z: text(Arg)(z_1) <= text(Arg)(z) <= text(Arg)(z_1^4)} ∩ {z: 1 <= |\ z\ | <= 2}, z ∈ C`.  (2 marks)
 


 

  1.  Find the area of the shaded region in part c.  (2 marks) 
  2. i.  Find the value(s) of `n` such that  `text(Re)(z_1^n) = 0`, where  `z_1 = (sqrt(sqrt3 + 2))/2 + i(sqrt(2 - sqrt3))/2`.  (3 marks)
  3. ii. Find  `z_1^n`  for the value(s) of `n` found in part i.  (1 mark)
Show Answers Only
  1. `text(See Worked Solutions.)`
  2. i.  `z_1 = text(cis) pi/12`

     

    ii.  `z_1^4 = text(cis) pi/3`

  3. `text(See Worked Solutions.)`
  4. `(3pi)/8`
  5. i.  `n = (2k + 1)6`

     

    ii.  `z_1^n = ±i\ text(for)\ k ∈ Z`

Show Worked Solution
a.    `cos^2(pi/12) + sin^2(pi/12)` `= 1`
  `sin^2(pi/12)` `= 1-(sqrt 3 + 2)/4`
    `= (2 – sqrt 3)/4`

 
`:. sin (pi/12) = sqrt(2 – sqrt 3)/2,\ \ \ (sin (pi/12) > 0)`

 

b.i.    `z_1` `= cos(pi/12) + i sin (pi/12)`
    `= text(cis)(pi/12)`

 

b.ii.    `z_1^4` `= 1^4 text(cis) ((4 pi)/12)`
    `= text(cis)(pi/3)`

 

c.   

 

d.  `text(Area of large sector)`

Mean mark 51%.

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

`=(pi/3-pi/12)/(2pi) xx pi xx 2^2`

`=pi/2`
  

`text(Area of small sector)`

`=1/2 xx pi/4 xx 1`

`=pi/8`

 
`:.\ text(Area shaded)`

`= pi/2 – pi/8`

`= (3 pi)/8`
 

♦ Mean mark (e)(i) 34%.

e.i.    `z_1^n` `= 1^n text(cis) ((n pi)/12)`
  `text(Re)(z_1^n)` `= cos((n pi)/12) = 0`
  `(n pi)/12` `=cos^(-1)0 +2pik,\ \ \ k in ZZ`
  `(n pi)/12` `= pi/2 + k pi`
  `n` `= 6 + 12k,\ \ \ k in ZZ`

 

e.ii.  `text(When)\ \ n=(6 + 12k):`

♦ Mean mark (e)(ii) 36%.

  `z_1^n` `= 1^(6 + 12k) text(cis) (((6 + 12k)pi)/12), quad k in ZZ`
    `= i xx sin (((6 + 12k)pi)/12)`
    `= i xx sin (pi/2 + k pi)`

 
`text(If)\ \ k=0  or text(even,)\ \ z_1^n=i`

`text(If)\ \ k\ text(is odd,)\ \ z_1^n=-i`

`:. z_1^n = +-i,\ \ \ k in ZZ`

Mechanics, SPEC2 2012 VCAA 21 MC

A particle of mass 3 kg is acted on by a variable force, so that its velocity `v` m/s when the particle is `x` m from the origin is given by   `v = x^2`.

The force acting on the particle when  `x = 2`, in newtons, is

A.     4

B.   12

C.   16

D.   36

E.   48

Show Answers Only

`E`

Show Worked Solution
`v` `=x^2`
`a` `= v *(dv)/(dx)`
  `= x^2 xx 2x`
  `= 2x^3`

 

♦ Mean mark 50%.

`text(When)\ \ x=2:`

`a` `= 2 xx 2^3`
  `=16`
`:.F` `= 3 xx 16`
  `=48`

 
`=> E`

Mechanics, SPEC2 2012 VCAA 20 MC

Particles of mass 3 kg and `m` kg are attached to the ends of a light inextensible string that passes over a smooth pulley, as shown.
 


 

If the acceleration of the 3 kg mass is  `4.9\ text(m/s)^2`  upwards, then

A.   `m` = 4.5

B.   `m` = 6.0

C.   `m` = 9.0

D.   `m` = 13.5

E.   `m` = 18.0

Show Answers Only

`C`

Show Worked Solution

`sum F` `=\ text(total mass × acceleration)`
`mg – 3g` `= (m + 3) xx 4.9`
`9.8m-3xx9.8` `= 4.9m + 14.7`
`:. m` `= 9`

 
`=> C`

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`