SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, 2ADV C4 2019 MET-N 15 MC

The area bounded by the graph of  `y = f(x)`, the line  `x = 2`, the line  `x = 8`  and the `x`-axis, as shaded in the diagram below, is  `3log_e(13)`

The value of  `int_4^10 3 f(x - 2)\ dx`  is

  1.  `3log_e(13)`
  2.  `9log_e(13)`
  3.  `6log_e(39)`
  4.  `9log_e(11)`
Show Answers Only

`B`

Show Worked Solution

`A_1 = int_2^8 f(x)\ dx = 3 log_e 13`

`f(x – 2) = f(x) \ text(shifted 2 units to right.)`

`:. \ int_2^8 f(x)\ dx = int_4^10 f(x – 2)\ dx`

`:. \ 3 int_4^10 f(x – 2)\ dx` `= 3 xx 3log_e 13`
  `= 9 log_e 13`

 
`=> \ B`

Functions, EXT1 F1 2019 MET2-N 11 MC

The function  `f(x) = 5x^3 + 10x^2 + 1`  will have an inverse function for the domain

  1.  `D = (–2, ∞)`
  2.  `D = (–∞ , (1)/(2)]`
  3.  `D = (–∞ , –1]`
  4.  `D = [0 , ∞)`
Show Answers Only

`D`

Show Worked Solution

`y = 5x^3 + 10x^2 + 1`

`y^{′} = 15x^2 + 20x`

`y^{″}=30x+20`

`text(SP’s when)\ \ y^{′}=0\ \ =>\ \ x= – 4/3\ text{(max)}, \ 0\ text{(min)}`

`text(Sketch graph:)`
 

`=> \ D`

Calculus, MET2-NHT 2019 VCAA 5

Let  `f: R → R, \ f(x) = e^((x/2))`  and  `g: R → R, \ g(x) = 2log_e(x)`.

  1. Find  `g^-1 (x)`.   (1 mark)
  2. Find the coordinates of point  `A`, where the tangent to the graph of  `f` at  `A` is parallel to the graph of  `y = x`.   (2 marks)
  3. Show that the equation of the line that is perpendicular to the graph of  `y = x`  and goes through point  `A` is  `y = -x + 2log_e(2) + 2`.   (1 mark)

Let `B` be the point of intersection of the graphs of `g` and  `y = -x + 2log_e(2) + 2`, as shown in the diagram below.
 

         
 

  1. Determine the coordinates of point `B`.   (1 mark)
  2. The shaded region below is enclosed by the axes, the graphs of  `f` and `g`, and the line  `y = -x + 2log_e(2) + 2`.
     
     
         
     
    Find the area of the shaded region.   (2 marks)

Let  `p : R→ R, \ p(x) = e^(kx)`  and  `q : R→ R, \ q(x) = (1)/(k) log_e(x)`.

  1. The graphs of `p`, `q` and  `y = x`  are shown in the diagram below. The graphs of `p` and `q` touch but do not cross.
     
     
         
     

     Find the value of  `k`.   (2 marks)
  2. Find the value of  `k, k > 0`, for which the tangent to the graph of `p` at its `y`-intercept and the tangent to the graph of `q` at its `x`-intercept are parallel.   (1 mark)
Show Answers Only
  1. `e^{(x)/(2)}`
  2. `(2log_e 2, 2)`
  3. `text(Proof(See Worked Solution))`
  4. `(2, 2log_e 2)`
  5. `6-2(log_e 2)^2-4 log_e 2`
  6. `(1)/(e)`
  7. `k =1`
Show Worked Solution

a.    `g(x) = 2log_e x`

`text(Inverse: swap) \ x ↔ y`

`x` `= 2log_e y`
`log_e y` `= (x)/(2)`
`y` `= e^{(x)/(2)}`

 

b.    `f(x) = e^{(x)/(2)}`

`f′(x) = (1)/(2) e^{(x)/(2)}`
 
`text(S) text(olve) \ \ f′(x) = 1 \ text(for) \ x:`

`x = 2log_e 2`

`y = e^(log_e 2) = 2`
 
`:. \ A\ text(has coordinates)\  (2log_e 2, 2)`

 

c.    `m_(⊥) = -1`

`text(Equation of line) \ \ m = -1 \ \ text(through)\ \ (2log_e 2, 2) :`

`y-2` `= -(x-2log_e 2)`
`y` `= -x +  2log_e 2 + 2`

 

d.    `text(Method 1)`

`text(S) text(olve for) \ x :`

`-x + 2log_e 2 + 2 = 2log_e x`

`=> x = 2 , \ y = 2log_e 2`

`:. B ≡ (2, 2log_e 2)`
 

`text(Method 2)`

`text(S) text(ince) \ \ f(x) = g^-1 (x)`

`B \ text(is the reflection of) \ \ A(2log_e 2, 2) \ \ text(in the) \ \ y=x \ \ text(axis)`

`:. \ B ≡ (2, 2log_e 2)`

 

e.
             
 

`y = g(x) \ \ text(intersects) \ x text(-axis at) \ \ x = 1`

`text(Dividing shaded area into 3 sections:)`

`A` `= int_0^1 f(x)\ dx \ + \ int_1^(2log_e 2) f(x)-g(x)\ dx`  
  ` \ + \ int_(2log_e 2)^2 (-x + 2log_e 2 + 2)-g(x)\ dx`  
  `= 6-2(log_e 2)^2-4 log_e 2`  

 

f.   `p(x) = e^(kx) \ , \ q(x) = (1)/(k) log_e x`

`p′(x) = k e^(kx) \ , \ q′(x) = (1)/(kx)`
 
`text(S) text(ince graphs touch on) \ y = x`

`k e^(kx) = 1\ …\ (1)`

`(1)/(kx) = 1\ …\ (2)`

`text(Substitute) \ x = (1)/(k) \ text{from (2) into (1)}`

`k e^(k xx 1/k)` `= 1`
`:. k` `= (1)/(e)`

 

g.   `text(Consider)\ \ p(x):`

`text(When) \ \ x = 0 , \ p(0) = 1 , \ p′(0) = k`
 

 `text(Consider) \ \ q(x):`

`text(When) \ \ y= 0, \ (1)/(k) log_e x = 0 \ => \ x = 1`

`q′(1) = (1)/(k)`

`text(If lines are parallel), \ k = (1)/(k)`
 
`:. \ k = 1`

Calculus, MET2-NHT 2019 VCAA 4

A mining company has found deposits of gold between two points, `A` and `B`, that are located on a straight fence line that separates Ms Pot's property and Mr Neg's property. The distance between `A` and `B` is 4 units.

The mining company believes that the gold could be found on both Ms Pot's property and Mr Neg's property.

The mining company initially models he boundary of its proposed mining area using the fence line and the graph of 
 

`f : [0, 4] → R, \ f(x) = x(x - 2)(x - 4)`
 

where `x` is the number of units from point `A` in the direction of point `B` and `y` is the number of units perpendicular to the fence line, with the positive direction towards Ms Pot's property. The mining company will only mine from the boundary curve to the fence line, as indicated by the shaded area below.
 

 

  1. Determine the total number of square units that will be mined according to this model.   (2 marks)

The mining company offers to pay Mr Neg $100 000 per square unit of his land mined and Ms Pot $120 000 per square unit of her land mined.

  1. Determine the total amount of money that the mining company offers to pay.   (1 mark)

The mining company reviews its model to use the fence line and the graph of
 
    `p : [0, 4] → R, \ p(x) = x(x - 4 + (4)/(1 + a)) (x - 4)`
 
where `a > 0`.

  1. Find the value of  `a`  for which  `p(x) = f(x)`  for all  `x`.   (1 mark)
  2. Solve  `p′(x) = 0`  for  `x`  in terms of  `a`.   (2 marks)

Mr Neg does not want his property to be mined further than 4 units measured perpendicular from the fence line.

  1. Find the smallest value of  `a`, correct to three decimal places, for this condition to be met.   (2 marks)
  2. Find the value of  `a`  for which the total area of land mined is a minimum.   (3 marks)
  3. The mining company offers to pay Ms Pot $120 000 per square unit of her land mined and Mr Neg $100 000 per square unit of his land mined.

     

    Determine the value of `a` that will minimize the total cost of the land purchase for the mining company. Give your answer correct to three decimal places.   (2 marks)

Show Answers Only
  1. `8`
  2. `$880\ 000`
  3. `1`
  4. `x = ((8a + 4) ± 4 sqrt(a^2 + a + 1))/(3a + 3)`
  5. `0.716`
  6. `1`
  7. `0.886`
Show Worked Solution
a.    `A` `= int_0^2 x(x – 2)(x – 4)\ dx  –  int_2^4 x(x – 2)(x – 4)\ dx`
  `= 4 – (-4)`
  `= 8`

 

b.    `text(Total payment)` `= 4 xx 100 000 + 4 xx 120 000`
  `= $880\ 000`

 

c.    `text(If) \ \ p(x) = f(x)`

`x – 2` `= x – 4 + (4)/(1 + a)`  
`(4)/(1 + a)` `=2`  
`:. a` `=1`  

 

d.    `p′(x) = (3(a + 1) x^2 – 8(2a + 1) x + 16a)/(a + 1) \ \ \ text{(by CAS)}`
 

`text(S) text(olve) \ \ p′(x) = 0 \ \ text(for) \ \ x :`

`x = ((8a + 4) ± 4 sqrt(a^2 + a + 1))/(3a +3) \ , \ a > 0`

 

e.    `text(If no mining further than 4 units,) \ \ p(x) ≥ – 4`

`text(Max distance from Mr Neg’s fence occurs when) \ \ 2< x <4 .`

`text((i.e. the higher) \ x text(-value when) \ \ p′(x) = 0)`
 

`text(At) \ \ x = ((8a + 4) + 4 sqrt(a^2 + a + 1))/(3a + 3) \ ,`
 
`text(S) text(olve for) \ \ a \ \ text(such that) \ \ p(x) = -4`

`=> a = 0.716 \ \ text((to 3 d.p.))`

 

f.    `p(x) \ text(intersects) \ x text(-axis at) \ \ x = 4 – (4)/(1 + a) = (4a)/(1 + a)`

`A` `= int_0^{(4a)/(1 +a)} p(x)\ dx – int_{(4a)/(1 +a)}^4 p(x)\ dx`
  `= (64(1 + 2a + 2a^3 + a^4))/(3(1 + a)^4)`

 
`text(S) text(olve) \ \ A′(a) = 0 \ \ text(for) \ \ a :`

`a = 1`

 

g.    `C(a) = 120\ 000 int_0^{(4a)/(1 +a)} p(x)\ dx – 100\ 000 int_{(4a)/(1 +a)}^4 p(x)\ dx`

`text(S) text(olve) \ \ C′(a) = 0 \ \ text(for) \ \ a :`

`a = 0.886 \ \ text((to 3 d.p.))`

Statistics, MET2-NHT 2019 VCAA 3

Concerts at the Mathsland Concert Hall begin `L` minutes after the scheduled starting time. `L` is a random variable that is normally distributed with a mean of 10 minutes and a standard deviation of four minutes.

  1. What proportion of concerts begin before the scheduled starting time, correct to four decimal places?   (1 mark)
  2. Find the probability that a concert begins more than 15 minutes after the scheduled starting time, correct to four decimal places.   (1 mark)

If a concert begins more than 15 minutes after the scheduled starting time, the cleaner is given an extra payment of $200. If a concert begins up to 15 minutes after the scheduled starting time, the cleaner is given an extra payment of $100. If a concert begins at or before the scheduled starting time, there is no extra payment for the cleaner.

Let `C` be the random variable that represents the extra payment for the cleaner, in dollars.

    1. Using your responses from part a. and part b., complete the following table, correct to three decimal places.   (1 mark)
       

    2. Calculate the expected value of the extra payment for the cleaner, to the nearest dollar.   (1 mark)
    3. Calculate the standard deviation of `C`, correct to the nearest dollar.   (1 mark)

The owners of the Mathsland Concert Hall decide to review their operation. They study information from 1000 concerts at other similar venues, collected as a simple random sample. The sample value for the number of concerts that start more than 15 minutes after the scheduled starting time is 43.

    1. Find the 95% confidence interval for the proportion of the concerts that begin more than 15 minutes after the scheduled starting time. Give values correct to three decimal places.   (1 mark)
    2. Explain why this confidence interval suggests that the proportion of concerts that begin more than 15 minutes after the scheduled starting time at the Mathsland Concert Hall is different from the proportion at the venue in the sample.   (1 mark)

The owners of the Mathsland Concert Hall decide that concerts must not begin before the scheduled starting time. They also make changes to reduce the number of concerts that begin after the scheduled starting time. Following these changes, `M` is the random variable that represents the number of minutes after the scheduled starting time that concerts begin. The probability density function for `M` is
 

`qquad qquad f(x) = {(8/(x + 2)^3), (0):} qquad {:(x ≥ 0), (x < 0):}`
 

where `x` is the the time, in minutes, after the scheduled starting time.

  1. Calculate the expected value of `M`.   (2 marks) 
    1. Find the probability that a concert now begins more than 15 minutes after the scheduled starting time.   (1 mark)
    2. Find the probability that each of the next nine concerts begins more than 15 minutes after the scheduled starting time and the 10th concert begins more than 15 minutes after the scheduled starting time. Give your answer correct to four decimal places.   (2 marks)
    3. Find the probability that a concert begins up to 20 minutes after the scheduled starting time, given that it begins more than 15 minutes after the scheduled starting time. Give your answer correct to three decimal places.   (2 marks)
Show Answers Only
  1. `0.0062`
  2. `0.1056`
  3. i.   

     ii.   `$110 \ \ text((nearest dollar))`
     iii.  `$32 \ \ text(nearest dollar))`
  4. i.   `(0.030, 0.056)`
    ii.   `text(The proportion of concerts that begin more than 15 minutes)` 

     

    `qquad text(late is not within the sample 95% confidence interval.)`

  5.  `2`
  6. i.   `(4)/(289)`
    ii.   `0.0122 \ \ text((to 4 decimal places))`
    iii.  `0.403 \ \ text((to 3 decimal places))`
Show Worked Solution

a.    `L\ ~\ N (10, 4^2)`

`text(Pr) (L < 0)` `= P(z < –2.5)`
  `= 0.0062`

 

b.    `text(Pr) (L > 15)` `= text(Pr) ( z > 1.25)`
  `= 0.1056`

 

c.i.

ii.  `E(C)` `= 0.8882 xx 100 + 0.1056 xx 200`
  `= 109.94`
  `= $110 \ \ text((nearest dollar))`

 

iii.  `E(C^2)` `= 100^2 xx 0.8882 + 200^2 xx 0.1056`
  `= 13\ 106`

 

`text(s.d.) (C)` `= sqrt(E(C^2) – [E(C)]^2)`
  `= sqrt(13\ 106 – (109.94)^2)`
  `= sqrt(1019.1 …)`
  `= 31.92 …`
  `= $32 \ \ text((nearest dollar))`

 

 
d.i.  `overset^p = (43)/(1000) \ \ , \ n = 1000`

`95% \ text(C. I.)` `= overset^p ± 1.96 sqrt((overset^p(1 – overset^p))/(n))`
  `= (0.030, 0.056)`

 
ii.   `text(The proportion of concerts that begin more than 15 minutes)`

`text(late is not within the sample 95% confidence interval.)`

 

e.    `E(M)` `= int_0^∞ x ((8)/((x +2)^3))\ dx`
  `= 2`

 

f.i.    `text(Pr)(M > 15)` `=int_0^∞ x ((8)/((x +2)^3))\ dx`
  `= (4)/(289)`

 
ii.  `text(Pr)(M > 15) = (4)/(289) \ \ , \ \ text(Pr)(M ≤ 15) = (285)/(289)`
 

`:. \ text(Pr) text{(9 concerts}\ \ M ≤ 15 ,\ text{10th concert}\ \ M > 15)`

`= ((205)/(289))^9 xx (4)/(289)`

`= 0.0122 \ \ text((to 4 decimal places))`

 
iii.  `text(Pr)(15 < M < 20)= int_15^20 (8)/((x + 2)^3)\ dx = (195)/(34\ 969)`

 `text(Pr)(M > 15)= int_15^oo (8)/((x + 2)^3)\ dx = (4)/(289)`
 

`text(Pr)(M < 20 | M>15)` `= (text(Pr)(15 < M < 20))/(text(Pr)(M > 15))`
  `= ((195)/(34\ 969))/((4)/(289))`
  `= (195)/(484)`
  `= 0.403 \ \ text((to 3 decimal places))`

Trigonometry, EXT1 T3 EQ-Bank 2

θθ

A particular energy wave can be modelled by the function

`f(t) = sqrt5 sin 0.2t + 2 cos 0.2t, \ \ t ∈ [0, 50]`

  1. Express this function in the form  `f(t) = Rsin(nt - alpha), \ \ alpha ∈ [0, 2pi]`.  (2 marks)

  2. Find the time the wave first attains its maximum value. Give your answer to one decimal place.  (2 marks)

Show Answers Only
  1. `f(t) = 3sin(0.2t – 5.553)`
  2. `t = 4.2`
Show Worked Solution

i.   `f(t) = sqrt5 sin 0.2t + 2cos 0.2t`

`Rsin(nt – alpha)` `= Rsin(0.2t – alpha)`
  `= Rsin 0.2t cosalpha – Rcos 0.2t sinalpha`

 

`=> Rcosalpha = sqrt5,\ \ R sinalpha = −2`

`R^2` `= (sqrt5)^2 + (−2)^2 = 9`
`R` `= 3`

 
`cosalpha = sqrt5/3, sinalpha = −2/3`

`=> alpha\ text(is in 4th quadrant)`
 

`text(Base angle) = cos^(−1)(sqrt5/3) = 0.7297`

`:.alpha` `= 2pi – 0.7297`
  `= 5.553…`

 
`:. f(t) = 3sin(0.2t – 5.553)`

 

ii.   `text(Max value occurs when)\ sin(0.2t – 5.553) = 1`

`0.2t – 5.553` `= pi/2`
`0.2t` `= 7.124…`
`t` `= 35.62…`

 
`text(Test if)\ \ t > 0\ \ text(for:)`

`0.2t – 5.553` `= -(3pi)/2`
`0.2t` `= 0.8406`
`t` `= 4.20…`

 
`:. text(Time of 1st maximum:)\ \ t = 4.2`

Trigonometry, MET2-NHT 2019 VCAA 2

The wind speed at a weather monitoring station varies according to the function

`v(t) = 20 + 16sin ((pi t)/(14))`

where `v` is the speed of the wind, in kilometres per hour (km/h), and  `t`  is the time, in minutes, after 9 am.

  1. What is the amplitude and the period of  `v(t)`?   (2 marks)
  2. What are the maximum and minimum wind speeds at the weather monitoring station?   (1 mark)
  3. Find  `v(60)`, correct to four decimal places.   (1 mark)
  4. Find the average value of  `v(t)`  for the first 60 minutes, correct to two decimal places.   (2 marks)

A sudden wind change occurs at 10 am. From that point in time, the wind speed varies according to the new function

                    `v_1(t) = 28 + 18 sin((pi(t - k))/(7))`

where  `v_1`  is the speed of the wind, in kilometres per hour, `t` is the time, in minutes, after 9 am and  `k ∈ R^+`. The wind speed after 9 am is shown in the diagram below.
 

  1. Find the smallest value of `k`, correct to four decimal places, such that  `v(t)`  and  `v_1(t)`  are equal and are both increasing at 10 am.   (2 marks)
  2. Another possible value of `k` was found to be 31.4358

     

    Using this value of `k`, the weather monitoring station sends a signal when the wind speed is greater than 38 km/h.

     

    i.  Find the value of `t` at which a signal is first sent, correct to two decimal places.   (1 mark)

     

    ii. Find the proportion of one cycle, to the nearest whole percent, for which  `v_1 > 38`.   (2 marks)

  3. Let  `f(x) = 20 + 16 sin ((pi x)/(14))`  and  `g(x) = 28 + 18 sin ((pi(x - k))/(7))`.
     
    The transformation  `T([(x),(y)]) = [(a \ \ \ \ 0),(0 \ \ \ \ b)][(x),(y)] + [(c),(d)]`  maps the graph of  `f`  onto the graph of  `g`.

     

    State the values of  `a`, `b`, `c` and `d`, in terms of `k` where appropriate.   (3 marks)

Show Answers Only
  1. `28`
  2. `v_text(max) \ = 36 \ text(km/h)`

     

    `v_text(min) \ = 4 \ text(km/h)`

  3. `32.5093 \ text(km/h)`
  4. `20.45 \ text(km/h)`
  5. `3.4358`
  6. i. `60.75 \ text(m)`

     

    ii. `31text(%)`

  7. `a = (1)/(2) \ , \ b = (9)/(8) \ , \ c = k \ , \ d = (11)/(2)`
Show Worked Solution

a.    `text(Amplitude) = 16`

`text{Find Period (n):}`

`(2 pi)/(n)` `= (pi)/(14)`
`n` `= 28`

 

b.    `v_text(max) = 20 + 16 = 36 \ text(km/h)`

`v_text(min) = 20 – 16 = 4 \ text(km/h)`

 

c.    `v(60)` `= 20 + 16 sin ((60 pi)/(14))`
  `= 32.5093 \ \ text(km/h)`

 

d.    `v(t)\ \ text(is always positive.)`

`s(t) = int_0^60 v(t) \ dt`

`v(t)_(avg)` `= (1)/(60) int_0^60 20 + 16 sin ((pi t)/(14))\ dt`
  `= 20.447`
  `= 20.45 \ text(km/h) \ \ text((to 2 d.p.))`

 

e.    `text(S) text{olve (for}\ k text{):} \ \ v(60) = v_1(60)`

`k = 3.4358 \ \ text((to 4 d.p.))`

 

f.i.  `text(S) text(olve for) \ t , \ text(given) \ \ v_1(t) = 38 \ \ text(and) \ \ k = 31.4358`

`=> t = 60.75 \ text(minutes)`
 

f.ii.  `text(S) text(olving for) \ \ v_1(t) = 38 \ , \ k = 31.4358`

`t_1 = 60.75 \ text{(part i)}, \ t_2 = 65.123`

`text(Period of) \ \ v_1 = (2 pi)/(n) = (pi)/(7)\ \ => \ n = 14`

`:. \ text(Proportion of cycle)` `= (65.123 – 60.75)/(14)`
  `= 0.312`
  `= 31 text{%  (nearest %)}`

 

g.    `f(x) → g(x)`

`y’ = 28 + 18 sin ({pi(x′ – k)}/{7})`

`x’ = ax + c` `\ \ \ \ \ \ y′ = by + d`

 

`text(Using) \ \ y’ = by + d`

`28 + 18 sin ({pi(x’ – k)}/{7}) = b (20 + 16 sin ({pi x}/{14})) + d`
 

`text(Equating coefficients of) \ \ sin theta :`

`16b = 18 \ \ \ => \ b = (9)/(8)`
 

`text(Equating constants:)`

`20 xx (9)/(8) + d = 28 \ \ \ => \ \ d = (11)/(2)`

`(x’ – k)/(7)` `= (x)/(14)`
`x’` `= (x)/(2) + k \ \ => \ \ a = (1)/(2) \ , \ c = k`

 

`a = (1)/(2) \ , \ b = (9)/(8) \ , \ c = k \ , \ d = (11)/(2)`

Graphs, MET2-NHT 2019 VCAA 1

Parts of the graphs of  `f(x) = (x - 1)^3(x + 2)^3`  and  `g(x) = (x - 1)^2(x + 2)^3`  are shown on the axes below.
 


 

The two graphs intersect at three points,  (–2, 0),  (1, 0)  and  (`c`, `d`). The point  (`c`, `d`)  is not shown in the diagram above.

  1. Find the values of `c` and `d`.   (2 marks)
  2. Find the values of `x` such that  `f(x) > g(x)`.   (1 mark)
  3. State the values of `x` for which

     

    i.   `f' (x) > 0`   (1 mark)

     

    ii.  `g' (x) > 0`   (1 mark)

  4. Show that  `f(1 + m) = f(–2 - m)`  for all  `m`.   (1 mark)
  5. Find the values of `h` such that  `g(x + h) = 0`  has exactly one negative solution.   (2 marks)
  6. Find the values of `k` such that  `f(x) + k = 0`  has no solutions.   (1 mark)
Show Answers Only
  1. `c = 2 \ , \ d = 64`
  2. `(–∞, –2) \ ∪ \ (2, ∞)`
  3. i.  `(–(1)/(2), 1) \ ∪ \ (1, ∞)`
    ii. `(–∞, –2) \ ∪ \ (–2, –(1)/(5)) \ ∪ \ (1, ∞)`
  4. `text{Proof (Show Worked Solution)}`
  5.  `-2< h <=1`
  6. `(729)/(64)`
Show Worked Solution

a.    `text(Solve:) \ \ f(x) = g(x)`

`x = 1 , \ –2 \ text(and) \ 2`
 
`f(2) = 1^3 xx 4^3 = 64`
 
`text(Intersection at) \ (2, 64)`

`:. \ c = 2 \ , \ d = 64`

 

b.    `text(Using the graph and intersection at) \ (2, 64):`

`f(x) > g(x) \ \ text(for) \ \ (–∞, –2) \ ∪ \ (2, ∞)`

 

c.i.  `f'(x) > 0 \ \ text(for) \ \ (–(1)/(2), 1) \ ∪ \ (1, ∞)`

c.ii.  `g'(x) > 0 \ \ text(for) \ \ (–∞, –2) \ ∪ \ (–2, –(1)/(5)) \ ∪ \ (1, ∞)`

 

d.     `f(1 + m)` `= (1 + m – 1)^3 (1 + m + 2)^3`
  `= m^3 (m + 3)^3`

 

`f(–2 – m)` `= (–2 – m -1)^3 (-2 – m + 2)^3`
  `= (-m – 3)^3 (-m)^3`
  `= (–1)^3 (m + 3)^3 (–1)^3 m^3`
  `= m^3 (m + 3)^3`

 

e.     `g(x + h)` `= (x + h – 1)^2(x + h + 2)^3`
  `= underbrace{(x – (1 -h))^2}_{text(+ve solution)} * underbrace{(x – (h – 2))^3}_{text(–ve solution)}`

 
`1 – h ≥ 0 \ \ => \ \ h ≤ 1`

`-h – 2 < 0 \ \ =>\ \ h > -2`

`:. -2< h <=1`

 

f.    `f(x) \ \ text(minimum S.P. when) \ \ f ′(x) = 0 \ =>  \ x =-(1)/(2)`

`text(S.P. at) \ \ (-(1)/(2) \ , \ -(729)/(64))`

`:. \ text(No solution if) \ \ k > (729)/(64)`

Trigonometry, EXT1 T3 EQ-Bank 1

  1. Show that `sinx + sin3x = 2sin2xcosx`.  (2 marks)

  2. Hence or otherwise, find all values of `x` that satisfy
     
    `qquad sinx + sin2x + sin3x = 0,\ \ \ x in [0,2pi]`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `x = 0, pi/2, pi, (3pi)/2, 2pi, (2pi)/3, (4pi)/3`
Show Worked Solution
i.   `sinx + sin3x` `= sinx + sin2xcosx + cos2xsinx`
  `= sinx + 2sinxcos^2x + (cos^2x – sin^2x)sinx`
  `= sinx + 2sinxcos^2x + cos^2xsinx – sin^3x`
  `= sinx + 3sinx(1 – sin^2x) – sin^3x`
  `= sinx + 3sinx – 3sin^3x- sin^3x`
  `= 4sinx(1 – sin^2x)`
  `= 4sinxcos^2x`
  `= 2sin2xcosx`
  `=\ text(RHS)`

 

ii.    `sinx + sin2x + sin3x` `= 0`
  `sin2x + 2sin2xcosx` `= 0`
  `sin2x(1 + 2cosx)` `= 0`

 

`text(If)\ sin2x` `= 0:`
`2x` `= 0, pi, 2pi, 3pi, 4pi`
`:.x` `= 0, pi/2, pi, (3pi)/2, 2pi`
`text(If)\ \ cosx` `= -1/2:`
`:.x` `= (2pi)/3, (4pi)/3`

CORE, FUR1-NHT 2019 VCAA 12 MC

Which one of the following statements could be true when written as part of the results of a statistical investigation?

  1. The correlation coefficient between height (in centimetres) and foot length (in centimetres) was found to be  `r = 1.24`
  2. The correlation coefficient between height (below average, average, above average) and arm span (in centimetres) was found to be  `r = 0.64`
  3. The correlation coefficient between blood pressure (low, normal, high) and age (under 25, 25–49, over 50) was found to be  `r = 0.74`
  4. The correlation coefficient between the height of students of the same age (in centimetres) and the money they spent on snack food (in dollars) was found to be  `r = 0.22`
  5. The correlation coefficient between height of wheat (in centimetres) and grain yield (in tonnes) was found to be  `r = –0.40`  and the coefficient of determination was found to be  `r^2 = –0.16`
Show Answers Only

`D`

Show Worked Solution

`text(Consider each option,)`

`A: \ r <= 1\ (text(incorrect))`

`B: \ text(height categories are ordinal) :. r\ text(not applicable)`

`C: \ text(both variables are ordinal) :. r\ text(not applicable)`

`D: \ text(possibly true)`

`E: \ text(Coefficient of determination cannot be negative (incorrect))`

`=>\ D`

CORE, FUR1-NHT 2019 VCAA 11 MC

The Human Development Index (HDI) and the mean number of children per woman for 13 countries are related.

This relationship is non-linear.

To linearise the data, a  `log_10`  transformation is applied to the response variable children.

A least squares line is then fitted to the linearised data.

The equation of this least squares line is

`log_10 (text(children)) = 1.06 - 0.00674 xx HDI`

Using this equation, the mean number of children per woman for a country with a HDI of 95 is predicted to be closest to

  1.  0.4
  2.  1.5
  3.  2.6
  4.  2.9
  5.  3.1
Show Answers Only

`C`

Show Worked Solution
`log_10(text(children))` `= 1.06 – 0.00674 xx 95`
  `= 0.4197`
`:.\ text(children)` `= 10^(0.4197)`
  `= 2.62…`

`=>\ C`

CORE, FUR1-NHT 2019 VCAA 5-7 MC

The birth weights of a large population of babies are approximately normally distributed with a mean of 3300 g and a standard deviation of 550 g.

Part 1

A baby selected at random from this population has a standardised weight of  `z = – 0.75`

Which one of the following calculations will result in the actual birth weight of this baby?
 

  1. `text(actual birth weight)\ = 550 - 0.75 × 3300`
  2. `text(actual birth weight)\ = 550 + 0.75 × 3300`
  3. `text(actual birth weight)\ = 3300 - 0.75 × 550`
  4. `text(actual birth weight)\ = 3300 + 0.75/550`
  5. `text(actual birth weight)\ = 3300 - 0.75/550`

 

Part 2

Using the 68–95–99.7% rule, the percentage of babies with a birth weight of less than 1650 g is closest to

  1. 0.14%
  2. 0.15%
  3. 0.17%
  4. 0.3%
  5. 2.5%

 

Part 3

A sample of 600 babies was drawn at random from this population.

Using the 68–95–99.7% rule, the number of these babies with a birth weight between 2200 g and 3850 g is closest to

  1. 111
  2. 113
  3. 185
  4. 408
  5. 489
Show Answers Only

`text(Part 1:)\ \ C`

`text(Part 2:)\ \ B`

`text(Part 3:)\ \ E`

Show Worked Solution

`text(Part 1)`

`text(Actual weight)` `= text(mean) + z xx text(std dev)`
  `= 3300 – 0.75 xx 550`

`=> C`

 

`text(Part 2)`

`z-text(score)` `= (x – barx)/5`
  `= (1650 – 3300)/550`
  `= −3`

 

`:. P(x < 1650)` `= P(z < −3)`
  `= 0.3/2`
  `= 0.15\ text(%)`

`=>B`
 

`text(Part 3)`

`ztext(-score)\ (2200) = (2200 – 3300)/550 = −2`

`ztext(-score)\ (3850) = (3850 – 3300)/550 = 1`
 


 

`text(Percentage)` `= (47.5 + 34)text(%) xx 600`
  `= 81.5text(%) xx 600`
  `= 48text(%)`

`=>\ E`

Probability, MET2-NHT 2019 VCAA 19 MC

A random sample of computer users was surveyed about whether the users had played a particular computer game. An approximate 95% confidence interval for the proportion of computer users who had played this game was calculated from this random sample to be (0.6668, 0.8147).

The number of computer users in the sample is closest to

  1.      5
  2.    33
  3.  135
  4.  150
  5.  180
Show Answers Only

`C`

Show Worked Solution

`95% text(C.I.) \ => \ z = 1.96`

`overset^p = (0.6668 + 0.8147)/(2) = 0.74075`

`text(S) text(olve for) \ \n\ \ text{(by CAS):}`

`2 xx 1.96 xx sqrt((0.74075(1 – 0.74075))/(n)) = \ 0.8147 – 0.6668`

`n ≈ 134.9`
 
`=> \ C`

Calculus, MET2-NHT 2019 VCAA 18 MC

Part of the graph of the function `f`, where  `f(x) = 8 - 2^(x-1)`, is shown below. It intersects the axes at the points `A` and `B`. The line segment joining `A` and `B` is also shown on the graph.
 


 

The area of the shaded region is

  1.  `16 - (15)/(log_e (2))`
  2.  `17 - (15)/(2log_e (2))`
  3.  `(7)/(log_e (2)) - (159)/(16)`
  4.  `16 - (15)/(2log_e (2))`
  5.  `(17)/(2log_e (2)) - 15`
Show Answers Only

`B`

Show Worked Solution

`text(When) \ \ x = 0 \ => \ y = 8 – (1)/(2) = 15/2`

`text(When) \ \ y = 0,\ => \ 8 – 2^(x-1)=0 \ => x=4`

`text(Area)_(AOB)` `= (1)/(2) xx 4 xx (15)/(2)`
  `= 15 \ text(u²)`

 

`text(Shaded Area)` `= int_0^4 8 – 2^(x -1)\ dx – (15)/(2)`
  `= 17 – (15)/(2 log_e 2)`

 
`=> \ B`

Functions, MET2-NHT 2019 VCAA 17 MC

The graph of the function `g` is obtained from the graph of the function `f` with rule  `f(x) = cos(x) - (3)/(8)`  by a dilation of factor  `(4)/(pi)`  from the `y`-axis, a dilation of factor  `(4)/(3)`  from the `x`-axis, a reflection in the  `y`-axis and a translation of  `(3)/(2)`  units in the positive `y` direction, in that order.

The range and period of `g` are respectively

  1. `[–(1)/(3) , (7)/(3)] \ text(and) \ 2`
  2. `[–(1)/(3) , (7)/(3)] \ text(and) \ 8`
  3. `[–(7)/(3) , (1)/(3)] \ text(and) \ 2`
  4. `[–(7)/(3) , (1)/(3)] \ text(and) \ 8`
  5. `[–(4)/(3) , (4)/(3)] \ text(and) \ (pi^2)/(2)`
Show Answers Only

`B`

Show Worked Solution

`text(Dilation of) \ \ (4)/(pi) \ \ text(from) \ y text(-axis:)`

`cos x -(3)/(8) \ => \ cos ((pi x)/(4)) – (3)/(8)`
 
`text(Dilation of) \ \ (4)/(3) \ \ text(from) \ x text(-axis:)`

`cos((pi x)/(4)) – (3)/(8) \ => \ (4)/(3) (cos({pi x}/{4}) -(3)/(8)) = (4)/(3) \ cos((pi x)/(4)) – (1)/(2)`
 
`text(Reflection in) \ y text(-axis and translation up) \ (3)/(2) :`

`(4)/(3) cos((pi x)/(4)) – (1)/(2) \ => \ (4)/(3) \ cos((–pi x)/(4)) + 1`

`:. y = (4)/(3) \ cos ((–pi x)/(4)) + 1`
 
`text(Range:) \ [1 – (4)/(3) , 1 + (4)/(3)] = [–(1)/(3) , (7)/(3)]`
 
`text(Period:) \ (2 pi)/(n) = (pi)/(4) \ => \ n = 8`

`=> \ B`

Graphs, MET2-NHT 2019 VCAA 16 MC

Parts of the graphs of  `y = f(x)`  and  `y = g(x)`  are shown below.
 


 

The corresponding part of the graph of `y = g(f(x))` is best represented by

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

`E`

Show Worked Solution

`text(Possible graphs are:)`
 
`g(x) = -(1)/(x)`

`f(x) = (x – 1)^2 + 1`
 
`text{Graph (by CAS):}`

`g(f(x)) = -(1)/((x – 1)^2 + 1)`
 
`=>  E`

Calculus, MET2-NHT 2019 VCAA 15 MC

The area bounded by the graph of  `y = f(x)`, the line  `x = 2`, the line  `x = 8`  and the `x`-axis, as shaded in the diagram below, is  `3log_e(13)`

The value of  `int_4^10 3 f(x - 2)\ dx`  is

  1.  `3log_e(13)`
  2.  `9log_e(13)`
  3.  `6log_e(39)`
  4.  `log_e(13)`
  5.  `9log_e(11)`
Show Answers Only

`B`

Show Worked Solution

`A_1 = int_2^8 f(x)\ dx = 3 log_e 13`

`f(x – 2) = f(x) \ text(shifted 2 units to right.)`

`:. \ int_2^8 f(x)\ dx = int_4^10 f(x – 2)\ dx`

`:. \ 3 int_4^10 f(x – 2)\ dx` `= 3 xx 3log_e 13`
  `= 9 log_e 13`

 
`=> \ B`

Graphs, MET2-NHT 2019 VCAA 12 MC

The transformation  `T : R^2 → R^2`,  which maps the graph of  `y = -sqrt(2x + 1)-3`  onto the graph of  `y = sqrtx`, has rule

  1. `T ([(x),(y)]) = [((1)/(2) \ \ \ \ \ 0),(\ \ 0\ \ \ \–1)] [(x),(y)] + [(–1),(–3)]`
  2. `T ([(x),(y)]) = [((1)/(2) \ \ \ \ \ 0),(\ \ 0\ \ \ \–1)] [(x),(y)] + [(–1),(3)]`
  3. `T ([(x),(y)]) = [((1)/(2) \ \ \ \ \ 0),(\ \ 0 \ \ \ \ –1)] [(x),(y)] + [(1),(–3)]`
  4. `T ([(x),(y)]) = [(2 \ \ \ \ \ 0),(\ \ 0 \ \ \ –1)] [(x),(y)] + [(1),(–3)]`
  5. `T ([(x),(y)]) = [(2 \ \ \ \ \ 0),(\ \ 0 \ \ \ –1)] [(x),(y)] + [(–1),(3)]`
Show Answers Only

`D`

Show Worked Solution
`y` `= -sqrt(2x + 1)-3`
`y + 3` `= -sqrt(2x + 1)`
`-y-3` `= -sqrt(2x + 1)`

 
`text(S) text(ince) \ \ y^{′} = sqrt(x^{′})`

`=> x^{′} = 2x + 1 \ , \ \ y^{′} = -y-3` 
 

`[(x^{′}),(y^{′})] = [(2 \ \ \ 0),(0 \ \–1)] [(x),(y)] + [(1),(–3)]` 
 

`=> D`

Algebra, MET2-NHT 2019 VCAA 11 MC

The function  `f : D → R, \ f(x) = 5x^3 + 10x^2 + 1`  will have an inverse function for

  1.  `D = R`
  2.  `D = (–2, ∞)`
  3.  `D = (–∞ , (1)/(2)]`
  4.  `D = (–∞ , –1]`
  5.  `D = [0 , ∞)`
Show Answers Only

`E`

Show Worked Solution

`text(Graph:) \ \ y = 5x^3 + 10x^2 + 1 \ \ text{(by CAS)}`
 


 

`text(Max when) \ \ x = -(4)/(3)`

`text(Min when) \ \ x = 0`

`=> \ E`

Algebra, MET2-NHT 2019 VCAA 7 MC

If  `m = int_1^3 (2)/(x)\ dx`, then the value of  `e^m`  is

  1.  `log_e (9)`
  2.  `–9`
  3.  `(1)/(9)`
  4.  `9`
  5.  `–(1)/(9)`
Show Answers Only

`D`

Show Worked Solution
`int_1^3 (2)/(x)\ dx` `= [2 log_e x]_1^3`
`m` `= 2 log_e 3 – 2 log_e 1`
`m` `= 2 log_e 3`
`e^(2 log_e 3)` `= e^(log_e 9)`
  `= 9`

 
`=>D`

Combinatorics, EXT1 A1 2019 MET1 8

A fair standard die is rolled 50 times. Let `W` be a random variable with binomial distribution that represents the number of times the face with a six on it appears uppermost.

  1. Write down the expression for  `P(W = k)`, where  `k in {0, 1, 2, …, 50}`.  (1 mark)

  2. Show that  `(P(W = k + 1))/(P(W = k)) = (50 - k)/(5(k + 1))`.  (2 marks)

Show Answers Only
  1. `\ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.  `P(W = k) = \ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`

 

b.   `(P(W = k + 1))/(P(W = k))` `= (\ ^50C_(k+1) ⋅ (1/6)^(k+1) ⋅ (5/6)^(49 – k))/(\ ^ 50C_k ⋅ (1/6)^k ⋅ (5/6)^(50-k))`
    `= ((50!)/((49 – k)!(k + 1)!) ⋅ (1/6))/((50!)/((50 – k)! k!) ⋅ (5/6))`
    `= ((50 – k)!k!)/(5(49 – k)!(k + 1)!)`
    `= (50 – k)/(5(k + 1))`

Statistics, EXT1 S1 2019 MET1 6

Jacinta tosses a coin five times.

  1. Assuming that the coin is fair and given that Jacinta observes a head on the first two tosses, find the probability that she observes a total of either four or five heads.  (2 marks)

  2. Albin suspects that a coin is not actually a fair coin and he tosses it 18 times.

     

    Albin observes a total of 12 heads from the 18 tosses.

  3. Let  `X` = probability of obtaining a head.
  4. Find the range of `X` in which 95% of observations are expected to lie within.  (2 marks)

Show Answers Only
  1.  `1/2`
  2.  `(4/9, 8/9)`
Show Worked Solution

a.   `text(After 2 tosses, 2 heads.)`

`Ptext{(4 or 5}\ H)` `= HHT + HTH + THH + HHH`
  `= (1/2)^3 xx 4`
  `= 1/2`

 

b.    `E(hat p) = p=12/18 = 2/3`

`text(Var)(hatp) = (2/3(1-2/3))/18=1/81`

`sigma(hatp)=1/9`

`text(95% interval)\  (z= +-2)`

`=(hatp-z xx sigma(hatp), \ hatp+z xx sigma(hatp))`

`= (2/3 – 2 xx 1/9, 2/3 + 2 xx 1/9)`

`= (4/9, 8/9)`

Calculus, 2ADV E1 2019 MET1 4

Given the function  `f(x) = log_e (x-3) + 2`,

  1. State the domain and range of `f(x)`.  (1 mark)

  2. i.  Find the equation of the tangent to the graph of  `f(x)` at  `(4, 2)`.  (2 marks)

     

    ii. On the axes below, sketch the graph of the function  `f(x)`, labelling any asymptote with its equation.

     

        Also draw the tangent to the graph of  `f(x)`  at  `(4, 2)`.  (4 marks)
     

Show Answers Only
  1. `x >3`

     

    `y in R`

  2. i.  `y = x-2`

     

    ii. `text(See Worked Solutions)`

Show Worked Solution
a.   `text(Domain)` `: \ x > 3`
  `text(Range)` `: \ y in R`

 

b.i.   `g(x)` `= log_e (x-3) + 2`
  `g^{\prime}(x)` `= 1/(x-3)`
  `g^{\prime}(4)` `= 1`

 
`text(Equation of tangent),\ m = 1\ \ text(through)\ (4, 2):`

`y-2` `= 1(x-4)`
`y` `= x-2`

 

b.ii.  

Probability, MET1-NHT 2019 VCAA 8

A fair standard die is rolled 50 times. Let `W` be a random variable with binomial distribution that represents the number of times the face with a six on it appears uppermost.

  1. Write down the expression for  `text(Pr)(W = k)`, where  `k in {0, 1, 2, …, 50}`.  (1 mark)
  2. Show that  `(text(Pr)(W = k + 1))/(text(Pr)(W = k)) = (50 - k)/(5(k + 1))`.  (2 marks)
  3. Hence, or otherwise, find the value of `k` for which  `text(Pr)(W = k)`  is the greatest.  (2 marks)
Show Answers Only
  1. `\ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `8`
Show Worked Solution

a.  `text(Pr)(W = k) = \ ^50 C_k ⋅ (1/6)^k ⋅ (5/6)^(50 – k)`

 

b.   `(text(Pr)(W = k + 1))/(text(Pr)(W = k))` `= (\ ^50C_(k+1) ⋅ (1/6)^(k+1) ⋅ (5/6)^(49 – k))/(\ ^ 50C_k ⋅ (1/6)^k ⋅ (5/6)^(50-k))`
    `= ((50!)/((49 – k)!(k + 1)!) ⋅ (1/6))/((50!)/((50 – k)! k!) ⋅ (5/6))`
    `= ((50 – k)!k!)/(5(49 – k)!(k + 1)!)`
    `= (50 – k)/(5(k + 1))`

 

c.  `text(Find)\ k\ text(such that)`

`text(Pr)(W = k + 1)` `< text(Pr)(W = k)`
`50 – k` `< 5(k + 1)`
`6k` `> 45`
`k` `> 7 1/2`

 
`=> text(Pr)(W = 8) > text(Pr)(W = 9)`

`=> text(Pr)(W = 9) > text(Pr)(W = 10)\ …`

`:. text(Pr)(W = 8)\ text(is the greatest.)`

Calculus, MET1-NHT 2019 VCAA 7

The shaded region in the diagram below is bounded by the vertical axis, the graph of the function with rule  `f(x) = sin(pix)`  and the horizontal line segment that meets the graph at  `x = a`, where  `1 <= a <= 3/2`.
 


 

Let  `A(a)`  be the area of the shaded region.

  1. Show that  `A(a) = 1/pi - 1/pi cos(a pi) - a sin (a pi)`.  (3 marks)
  2. Determine the range of values of  `A(a)`.  (2 marks)
    1. Express in terms of  `A(a)`, for a specific value of `a`, the area bounded by the vertical axis, the graph of  `y = 2(sin(pi x) + sqrt 3/2)`  and the horizontal axis.  (2 marks)
    2. Hence, or otherwise, find the area described in part c.i.  (1 mark)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `[2/pi, (2 + 3 pi)/(2pi)]`
    1. `2 A(a)`
    2. `(9 + 4 sqrt 3 pi)/(3 pi)`
Show Worked Solution

a.   `text(Lower border of shaded area:)\ y = f(a) = sin(a pi)`

`text(Area)` `= int_0^a sin(pi x) – sin (a pi)\ dx`
  `= [-1/pi cos (pi x) – x sin(a pi)]_0^a`
  `= [-1/pi cos(pi a) – a sin (a pi) – (-1/pi – 0)]`
  `= 1/pi – 1/pi cos (a pi) – a sin(a pi)`

 

b.   `text(S) text(ince)\ 1 <= a <= 3/2,`

`A(1) = 1/pi – 1/pi cos(pi) – 1 sin (pi) = 2/pi`

`A(3/2) = 1/pi – 1/pi cos ((3 pi)/2) – 3/2 sin ((3pi)/2) = 1/pi + 3/2 = (2 + 3pi)/(2 pi)`
 

`:.\ text(Range:)\ \ [2/pi, (2 + 3pi)/(2 pi)]`

 

c.i.  `A(a) = int_0^a sin(pi x) – sin (a pi)\ dx`

`A_1` `=2int _0^a sin(pi x) + sqrt 3/2\ dx`   
  `=2int _0^a sin(pi x) – sin((4pi)/3)\ dx,\ \ \ (a=4/3)`  
  `=2int _0^(4/3) sin(pi x) – sin((4pi)/3)\ dx`  
  `=2 xx A(a)`  

 
`:.\ text(When)\ \ a = 4/3,\ \ text(Area) = 2 xx A(a)`

 

c.ii.  `text(When)\ \ a = 4/3`

`text(Area)` `= 2 xx (1/pi – 1/pi cos ((4 pi)/3) – 4/3 sin ((4 pi)/3))`
  `= 2 xx (1/pi + 1/(2 pi) + 4/3 xx sqrt 3/2)`
  `= 2(3/(2pi) + (2 sqrt 3)/3)`
  `= 3/pi + (4 sqrt 3)/3`
  `= (9 + 4 sqrt 3 pi)/(3 pi)`

Statistics, MET1-NHT 2019 VCAA 6b

Albin suspects that a coin is not actually a fair coin and he tosses it 18 times.

Albin observes a total of 12 heads from the 18 tosses.

Based on this sample, find the approximate 90% confidence interval for the probability of observing a head when this coin is tossed. Use the `z` value `33/20`.  (2 marks)

Show Answers Only

`(29/60, 51/60)`

Show Worked Solution

`hat p = 12/18 = 2/3`

`text(90% confidence interval)`

`= (2/3 – 33/20 sqrt((2/3 xx 1/3)/18), 2/3 + 33/20 sqrt((2/3 xx 1/3)/18))`

`= (2/3 – 33/20 sqrt(1/81), 2/3 + 33/20 sqrt(1/81))`

`= (2/3 – 11/60, 2/3 + 11/60)`

`= (29/60, 51/60)`

Statistics, EXT1 S1 EQ-Bank 23

A light manufacturer knows that 6% of the light bulbs it produces are defective.

Light bulbs are supplied in boxes of 20 bulbs. Boxes are supplied in pallets of 120 boxes.

Calculate the probability that

  1. A box of light bulbs contains exactly 3 defective bulbs.  (1 mark)

  2. A box of light bulbs contains at least 1 defective bulb.  (1 mark)

  3. A pallet contains between 90 and 95 (inclusive) boxes with at least 1 defective bulb (use the probability table attached).  (3 marks)

Show Answers Only
  1. `0.086\ \ (text(to 3 d.p.))`
  2. `0.710\ \ (text(3 d.p.))`
  3. `0.1416`
Show Worked Solution

i.   `P(D) = 0.06, \ P(barD) = 0.94`

`P(D = 3)` `= \ ^20C_3(0.06)^3(0.94)^17`
  `= 0.086\ \ (text(to 3 d.p.))`

 

ii.    `P(D >= 1)` `= 1 – P(D = 0)`
    `= 1 – \ ^20C_0(0.06)^0(0.94)^20`
    `= 0.710\ \ (text(to 3 d.p.))`

 

iii.   `text(Let)\ \ X = text(number of boxes where)\ \ D >= 1`

`text(Let)\ \ overset^p = text(proportion of boxes where)\ \ D >= 1`

`E(overset^p) = p = 0.710`

`text(Var)(overset^p) = (0.710(1 – 0.710))/120 = 0.0017158`

`sigma(overset^p) = 0.04142`
 

`overset^p\ ~\ N(0.710, 0.04142)`

`text(If)\ \ X = 90 \ => \ overset^p = 90/120 = 0.75`

`text(If)\ \ X = 95 \ => \ overset^p = 95/120 = 0.79167`
 

`ztext(-score)\ (X = 90) = (0.75 – 0.710)/(0.04142) = 0.9657`

`ztext(-score)\ (X = 95) = (0.79167 – 0.710)/(0.04142) = 1.972`
 

`P(90 <= X <= 95)` `= P(0.97 <= z <= 1.97)`
  `= 0.9756 – 0.8340`
  `= 0.1416`

Statistics, EXT1 S1 EQ-Bank 21

A biased coin has a 0.6 chance of landing on heads. The coin is tossed 15 times.

  1. Calculate the probability of obtaining 7, 8 or 9 heads using binomial probability distribution.
  2. Give your answer correct to 3 decimal places.  (2 marks)

  3. Calculate the probability of obtaining 7, 8 or 9 heads using normal approximation to the binomial distribution and the probability table attached.
  4. Give your answer correct to 3 decimal places.  (2 marks)

  5. Could this binomial distribution be reasonably approximated with a normal distribution? Support your answer with a brief calculation.  (1 mark)

Show Answers Only
  1. `0.502\ (text(to 3 d.p.))`
  2. `0.509`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `P(7, 8\ text(or)\ 9)` `= \ ^15C_7(0.6)^7(0.4)^8  \ ^15C_8(0.6)^8(0.4)^7 + \ ^15C_9(0.6)^9(0.4)^6`
    `= 0.118056 + 0.177084 + 0.206600`
    `= 0.502\ \ (text(to 3 d.p.))`

 

ii.   `E(overset^p) = p = 0.6`

`text(Var)(overset^p) = (p(1 – p))/n = (0.6 xx 0.4)/15 = 0.016`

`sigma (overset^p) = 0.12649`

 
`text(Let)\ \ X = text(number of heads)`

`P(7, 8, 9) = P(6.5 < X < 9.5)`

`text(If)\ \ X = 6.5 \ => \ overset^p = 6.5/15 = 0.4333`

`text(If)\ \ X = 9.5 \ => \ overset^p = 9.5/15 = 0.6333`

`ztext(-score)\ (X = 6.5) = (0.4333 – 0.6)/0.12649 = −1.32`

`ztext(-score)\ (X = 9.5) = (0.6333 – 0.6)/0.12649 = 0.26`

COMMENT: A quick sketch of the normal distribution curve is often helpful when using probability tables in this context.
 

`P(7, 8, 9)` `= P(−1.32 < z < 0.26)`
  `= P(z < 1.32) – P(z < −0.26)`
  `= 0.9066 – 0.3974`
  `= 0.509`

 

iii.   `np = 15 xx 0.6 = 9`

COMMENT: `np>10, nq>10` is also common in assessing if normal approximation is reasonable.

`nq=n(1 – p) = 15 xx 0.4 = 6`

`text(S)text(ince)\ \ np > 5 and nq > 5,`

`=>\ text(Normal approximation is reasonable.)`

Statistics, SPEC2-NHT 2019 VCAA 6

A paint company claims that the mean time taken for its paint to dry when motor vehicles are repaired is 3.55 hours, with a standard deviation of 0.66 hours.

Assume that the drying time for the paint follows a normal distribution and that the claimed standard deviation value is accurate.

  1. Let the random variable  `barX`  represent the mean time taken for the paint to dry for a random sample of 36 motor vehicles.

     

     Write down the mean and standard deviation of  `barX`.   (2 marks)

At a car crash repair centre, it was found that the mean time taken for the paint company's paint to dry on randomly selected vehicles was 3.85 hours. The management of this crash repair centre was not happy and believed that the claim regarding the mean time taken for the paint to dry was too low. To test the paint company's claim, a statistical test was carried out.

  1. Write down suitable null and alternative hypotheses `H_0` and `H_1` respectively to test whether the mean time taken for the paint to dry is longer than claimed.   (1 mark)
  2. Write down an expression for the  `p`  value of the statistical test and evaluate it correct to three decimal places.   (2 marks)
  3. Using a 1% level of significance, state with a reason whether the crash repair centre is justified in believing that the paint company's claim of mean time taken for its paint to dry of 3.55 hours is too low.  (1 mark)
  4. At the 1% level of significance, find the set of sample mean values that would support the conclusion that the mean time taken for the paint to dry exceeded 3.55 hours. Give your answer in hours, correct to three decimal places.  (2 marks)
  5. If the true time taken for the paint to dry is 3.83 hours, find the probability that the paint company's claim is not rejected at the 1% level of significance, assuming the standard deviation for the paint to dry is still 0.66 hours. Give your answer correct to two decimal places.  (1 mark)
Show Answers Only
  1. `0.11`
  2. `H_0 : \ mu = 3.55`
    `H_1 : \ mu > 3.55`
  3. `0.003 \ text{(to 3 decimal places)}`
  4. `text(S) text(ince) \ \ p < 0.01 , H_0 \ text(should be rejected at the 1% level.)`
  5. `barX > 3.806`
  6. `0.41`
Show Worked Solution
a.   `E(barX)` `= 3.55`
`sigma(barX)` `= (sigma)/(sqrtn)`
  `= (0.66)/(sqrt36)`
  `= 0.11`

 

b.   `H_0 : \ mu = 3.55`

`H_1 : \ mu > 3.55`

 

c.   `p` `= text(Pr) (barX > 3.85)`
  `= text(Pr) (z > (3.85 – 3.55)/(0.11))`
  `= text(Pr) (z >2.326)`
  `= 0.003 \ text{(to 3 decimal places)}`

 

d.   `text(S) text(ince)\ \ p < 0.01 , H_0 \ text(should be rejected at the 1% level.)`

`text(i.e. repair centre is justified that the mean time 3.55 hours is too low.)`

 

e.   `text(If) \ \ mu = 3.55`

`(barX – mu)/(sigma)` `> 2.3263`
`barX` `> 2.3263 xx 0.11 + 3.55`
`barX` `> 3.806`

 

f.   `text(Pr) (barX< 3.806 | mu = 3.83)` `= text(Pr) (z < (3.806 – 3.83)/(0.11))`
  `= text(Pr) (z < -0.21818)`
  `= 0.41`

Mechanics, SPEC2-NHT 2019 VCAA 5

A pallet of bricks weighing 500 kg sits on a rough plane inclined at an angle of  `α°` to the horizontal, where  `tan(α°) = (7)/(24)`. The pallet is connected by a light inextensible cable that passes over a smooth pulley to a hanging container of mass `m` kilograms in which there is 10 L of water. The pallet of bricks is held in equilibrium by the tension `T` newtons in the cable and a frictional resistance force of 50 `g` newtons acting up and parallel to the plane. Take the weight force exerted by 1 L of water to be `g` newtons.
  


 

  1. Label all forces acting on both the pallet of bricks and the hanging container on the diagram above, when the pallet of bricks is in equilibrium as described.   (1 mark)
  2. Show that the value of `m` is 80.  (3 marks)

Suddenly the water is completely emptied from the container and the pallet of bricks begins to slide down the plane. The frictional resistance force of 50 `g` newtons acting up the plane continues to act on the pallet.

  1. Find the distance, in metres, travelled by the pallet after 10 seconds.  (3 marks)
  2. When the pallet reaches a velocity of  `3\ text(ms)^-1`, water is poured back into the container at a constant rate of 2 L per second, which in turn retards the motion of the pallet moving down the plane. Let  `t`  be the time, in seconds, after the container begins to fill. 
  3.   i. Write down, in terms of  `t`, an expression for the total mass of the hanging container and the water it contains after `t` seconds. Give your answer in kilograms.  (1 mark)
  4.  ii. Show that the acceleration of the pallet down the plane is given by  `(text(g)(5 - t))/(t + 290)\ text(ms)^-2`  for  `t ∈[0, 5)`.  (2 marks)
  5. iii. Find  the velocity of the pallet when  `t = 4`. Give your answer in metres per second, correct to one decimal place.  (2 marks)
Show Answers Only
  1.  
    `qquad`
  2. `text(Proof(Show Worked Solution))`
  3. `(25 text(g))/(29)`
  4.   i. `80 + 2t`
     ii. `text(Proof (Show Worked Solution))`
    iii. `3.4\ text(ms)^-1`
Show Worked Solution

a.

 
b.   `text(Resolving vertical forces on container:)`

`T – (m + 10)g = 0 \ …\ (1)`

`text(Resolving forces on plane:)`
 


 

`tan α = (7)/(24) \ => \ sin α = (7)/(25)`
 

`text(Solve for m:)`

`(m + 10)g` `= 500 text(g) · (7)/(25) – 50 text(g)`
`m + 10` `= 140 – 50`
`:. \ m` `= 80`

 

c.   `text(Resolving vertical forces on container:)`

`T – 80 g = 80 a \ …\ (1)`

`text(Resolving forces on plane:)`

`500 g sin α – (T + 50 g) = 500 a`

`90 g – T = 500 a \ …\ (2)`

`text(Add) \ (1) + (2)`

`10 g` `= 580 a`
`a` `= (g)/(58)`
`s` `= ut + (1)/(2) at^2`
  `= 0 + (1)/(2) · (g)/(58) + 10^2`
  `= (25g)/(29)`

 

d.i.   `m = 80 + 2t`
 

d.ii.   `text(Resolving vertical forces on container:)`

`T – (80+2t)g = (80+2t)a \ …\ (1)`

`text(Resolving forces on plane:)`

`90g – T = 500a \ …\ (2)`

`text(Add)\ (1) + (2)`

`(90 – 80 – 2t)g` `= (500 + 80 + 2t)a`
`(10 – 2t)g` `=(580 + 2t)a`
`a` `= (g(5 – t))/(t + 290) ms^-2`

 

d.iii.   `(dv)/(dt) = (g(5 – t))/(t + 290)`

`v = int (dv)/(dt)\ dt = 295 log_e ((t + 290)/(290)) – g t + c`

 
`text(When)\ t = 0, v = 3\ \ text{(given)} \ => \ c = 3`

`:. \ v = 295 log_e ((t+290)/(290)) – g t + 3`

`:. \ v(4) = 3.4\ text(ms)^-1`

Calculus, SPEC2-NHT 2019 VCAA 3


 

The vertical cross-section of a barrel is shown above. The radius of the circular base (along the `x`-axis) is 30 cm and the radius of the circular top is 70 cm. The curved sides of the cross-section shown are parts of the parabola with rule  `y = (x^2)/(80) - (45)/(4)`. The height of the barrel is 50 cm.
 
a.  i.   Show that the volume of the barrel is given by  `pi int_0^50 (900 + 80 y)\ dy`.  (1 marks)
     ii.  Find the volume of the barrel in cubic centimetres.  (1 marks)    

The barrel is initially full of water. Water begins to leak from the bottom of the barrel such that  `(dV)/(dt) = (-8000pi sqrth)/(A)`  cubic centimetres per second, where after  `t`  seconds the depth of the water is  `h`  centimetres, the volume of water remaining in the barrel is  `V`  cubic centimetres and the uppermost surface area of the water is  `A`  square centimetres.
 
b.     Show that  `(dV)/(dt) = (-400 sqrth)/(4h + 45)`?  (2 marks)
c.      Find  `(dh)/(dt)`  in terms of  `h`. Express your answer in the form  `(-a sqrth)/(pi(b + ch)^2)`, where  `a, b`  and  `c`  are positive integers.  (3 marks)
d.     Using a definite integral in terms of  `h`, find the time, in hours, correct to one decimal place, taken for the barrel to empty.  (2 marks)

Show Answers Only

a.  i. `text(Proof(Show Worked Solution))`

      ii.  `145000 pi \ text(cm)^3`

b.    `text(Proof (Show Worked Solution))`

c.    `(-20 sqrth)/(pi(45 + 4h)^2)`

d.    `9.9\ text(hours)`

Show Worked Solution
a.i. `V` `= pi int x^2 dy`
  `y` `= (x^2)/(80) – (45)/(4)`
  `x^2` `= 80y + 900`
     
  `:. \ V` `= pi int_0^50 (900 + 80y)dy`

 

a.ii. `V` `= pi int_0_50 (900 + 80y) dy`
    `= pi [900y + 40y^2]_0^50`
    `= 145000pi \ text(cm)^3`

 

b.    `A = pi x^2 = pi (900 + 80h)`

`(dV)/(dt)` `= (-8000pi sqrth)/(pi(900 + 80h))`
  `= (-8000 sqrth)/(20(4h + 45))`
  `= (-400 sqrth)/(4h + 45)`

 

c.    `(dV)/(dh) = pi(900 + 80h)`

`(dh)/(dt)` `= (dh)/(dV) ⋅ (dV)/(dt)`
  `= (1)/(pi(900 +80h)) xx (-400 sqrth)/(4h +45)`
  `= (-400 sqrth)/(20pi(4h +45)^2)`
  `= (-20 sqrth)/(pi(45 +4h)^2)`

  
 
d.    `(dt)/(dh) = (-pi(45 + 4h)^2)/(20 sqrth)`

`t` `= -pi int_50^0 ((45 + 4h)^2)/(20 sqrth)\ dh`
  `≈ 35\ 598.6 \ text(seconds)`
  `≈ 9.9 \ text{hours  (to 1 d.p.)}`

Calculus, SPEC2-NHT 2019 VCAA 2

Consider the function `f` with rule  `f(x) = (x^2 + x + 1)/(x^2-1)`.

  1. State the equations of the asymptotes of the graph of `f`.  (2 marks)
  2. State the coordinates of the stationary points and the point of inflection. Give your answers correct to two decimal places.  (2 marks)
  3. Sketch the graph of `f` from  `x = -6`  to  `x = 6`  (endpoint coordinates are not required) on the set of axes below, labeling the turning points and the point of inflection with their coordinates correct to two decimal places. Label the asymptotes with their equations.  (3 marks)

Consider the function `f_k` with rule  `f_k(x) = (x^2 + x + k)/(x^2-1)`  where  `k ∈ R`.

  1. For what values of `k` will `f_k` have no stationary points?  (2 marks)
  2. For what value of `k` will the graph of `f_k` have a point of  inflection located on the `y`-axis?  (1 marks)
Show Answers Only
i.     `x = 1, x = -1, y = 1`

 ii.   `(-3.73, 0.87) \ text(and) \ (-0.27, -0.87)`

`(-5.52, 0.88)`
iii.

iv.  `-2`

v.  `-1`

Show Worked Solution

i.    `f(x) = 1 + (x + 2)/((x + 1)(x-1))`

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

ii.   `f′(x) = (-(x^2 + 4x + 1))/(x^2-1)^2`

`text(Solve:) \ \ f^{′}(x) = 0 \ => \ x = -2 ± sqrt3`

`text(SP’s at) \ (-3.73, 0.87) \ \ text(and)\ \ (-0.27, -0.87)`
 
`f^{″}(x) = (-2x^3-12x^2-6x-4)/(x^2-1)^3`

`text(Solve:) \ \ f^{″}(x) = 0 \ => \ x = -5.52`

`text(POI at) \ \ (-5.52, 0.88)`
 

iii.

 

iv.    `f_k^{′}(x) = (-x^2-2(k+1) x-1)/(x^2-1)^2`

`text(If no SP’s,) \ \ Δ < 0`

`[-2( k + 1)]^2-4(-1)(-1)` `< 0`
`4k^2 + 8k + 4-4` `< 0`
`4k(k + 2)` `< 0`

 
`:. \ -2 < k < 0`
 

v.    `f_k^{″}(x) = (2(x^3 + 3(k + 1) x^2 + 3x + k + 1))/((x^2-1)^3)`

`text(Solve:)\ \ f_k^{″}(x) = 0`

`k = -1`

Vectors, EXT1 V1 SM-Bank 30

A canon ball is fired from a castle wall across a horizontal plane at `V` ms−1.

Its position vector  `t` seconds after it is fired from its origin is given by  `underset~s(t) = V tunderset~i - 1/2g t^2 underset~j`.

  1. If the projectile hits the ground at a distance 8 times the height at which it was fired, show that it initial velocity is given by
     
          `V = 4sqrt(2hg)`  (2 marks)

  2. Show that the total distance the canon ball travels can be expressed as
     
          `int_0^sqrt((2h)/g) sqrt(g(32h + g t^2))\ dt`      (2 marks)

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

i.   `text(Time of flight  ⇒  find)\ t\ text(when)\ \ y = −h`

`−1/2g t^2` `= −h`
`t^2` `= (2h)/g`
`t` `= sqrt((2h)/g),\ \ t > 0`

 

`text(S)text(ince the canon ball impacts when)\ \ x = 8h:`

`Vt` `= 8h`
`Vsqrt((2h)/g)` `= 8h`
`V` `= (8sqrt(hg))/sqrt2`
  `= 4sqrt(2hg)`

 

ii.   `underset~v = 4sqrt(2hg) underset~i – g t underset~j`

`|underset~v|` `= sqrt((4sqrt(2hg))^2 + (−g t)^2)`
  `= sqrt(16 xx 2hg  + g^2 t^2)`
  `=sqrt(g(32h + g t^2)`

 

`text(Distance)` `= int_0^sqrt((2h)/g) |underset~v|\ dt`
  `= int_0^sqrt((2h)/g) sqrt(g(32h + g t^2))\ dt`

Vectors, EXT1 V1 2019 SPEC2-N 4

A snowboarder at the Winter Olympics leaves a ski jump at an angle of `theta` degrees to the horizontal, rises up in the air, performs various tricks and then lands at a distance down a straight slope that makes an angle of 45° to the horizontal, as shown below.

Let the origin `O` of a cartesian coordinate system be at the point where the snowboarder leaves the jump, with a unit vector in the positive `x` direction being represented by  `underset~i`  and a unit vector in the positive `y` direction being represented by  `underset~j`. Distances are measured in metres and time is measured in seconds.

The position vector of the snowboarder  `t`  seconds after leaving the jump is given by

`underset~r (t) = (6t - 0.01t^3) underset~i + (6 sqrt3 t - 4.9t^2 + 0.01t^3) underset~j , \ t ≥ 0`
 


 

  1. Find the angle  `theta °`.    (2 marks)

  2. Find the speed, in metres per second, of the snowboarder when she leaves the jump at `O`.    (1 mark)

  3. Find the maximum height above `O` reached by the snow boarder. Give your answer in metres, correct to one decimal place.    (2 marks)

  4. Show that the time spent in the air by the snowboarder is  `(60(sqrt3 + 1))/(49)`  seconds.    (3 marks)

Show Answers Only
  1. `60°`
  2. `12 \ text(ms)^-1`
  3. `5.5 \ text(m)`
  4. `text(See Worked Solutions)`
Show Worked Solution

a.   `v(t) = (6 – 0.03t^2)underset~i + (6 sqrt3 – 9.8t + 0.03t^2) underset~j`

`text(When) \ t =0,`

`v(t) = 6underset~i + 6 sqrt3 underset~j`

`tan theta = (6 sqrt3)/(6) = sqrt3`

`:. \ theta` `= tan^-1 sqrt3`
  `= 60°`

 

b.    `text(Speed)` `= |v(0)|`
    `= sqrt(6^2 + (6 sqrt3)^2)`
    `= 12 \ text(ms)^-1`

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

`text(Solve): \ \ 6 sqrt3 – 9.8t + 0.03t^2 = 0`

`=> t =  1.064 \ text(seconds)`

`text(Max height)` `= 6 sqrt3 (1.064) – 4.9(1.064)^2 + 0.01(1.064)^3`  
  `~~5.5\ text(m)`  

 
d.   `text(Time of Flight  ⇒  Solve for)\ \ t\ \ text(when)\ \ y=-x:`

`6 sqrt 3 t – 4.9t^2 + 0.01 t^3` `= -(6t-0.01t^3)`
`(6 + 6 sqrt3)t – 4.9 t^2` `= 0`
`t(6 + 6 sqrt3 – 4.9 t)` `= 0`
`4.9 t` `= 6 + 6 sqrt3`
`t` `= (6 + 6 sqrt3)/(4.9)`
  `= (60(sqrt3 + 1))/(49)\ text(seconds)`

Vectors, SPEC2-NHT 2019 VCAA 4

A snowboarder at the Winter Olympics leaves a ski jump at an angle of `theta` degrees to the horizontal, rises up in the air, performs various tricks and then lands at a distance down a straight slope that makes an angle of 45° to the horizontal, as shown below.

Let the origin `O` of a cartesian coordinate system be at the point where the snowboarder leaves the jump, with a unit vector in the positive `x` direction being represented by  `underset~i`  and a unit vector in the positive `y` direction being represented by  `underset~j`. Distances are measured in metres and time is measured in seconds.

The position vector of the snowboarder  `t`  seconds after leaving the jump is given by

`underset~r (t) = (6t - 0.01t^3) underset~i + (6 sqrt3 t - 4.9t^2 + 0.01t^3) underset~j , \ t ≥ 0`
 


 

  1. Find the angle  `theta °`.    (2 marks)
  2. Find the speed, in metres per second, of the snowboarder when she leaves the jump at `O`.    (1 mark)
  3. Find the maximum height above `O` reached by the snow boarder. Give your answer in metres, correct to one decimal place.    (2 marks)
  4. Show that the time spent in the air by the snowboarder is  `(60(sqrt3 + 1))/(49)`  seconds.    (3 marks)
  5. Find the total distance the snowboarder travels while airborne. Give your answer in metres, correct to two decimal places.     (2 marks)
Show Answers Only
  1. `60°`
  2. `12 \ text(ms)^-1`
  3. `5.5 \ text(m)`
  4. `text(See Worked Solutions)`
  5. `38.51 \ text{m  (to 2 decimal places)}`
Show Worked Solution

a.   `v(t) = (6 – 0.03t^2)underset~i + (6 sqrt3 – 9.8t + 0.03t^2) underset~j`

`text(When) \ t =0,`

`v(t) = 6underset~i + 6 sqrt3 underset~j`

`tan theta = (6 sqrt3)/(6) = sqrt3`

`:. \ theta` `= tan^-1 sqrt3`
  `= 60°`

 

b.    `text(Speed)` `= |v(0)|`
    `= sqrt(6^2 + (6 sqrt3)^2)`
    `= 12 \ text(ms)^-1`

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

`text(Solve): \ \ 6 sqrt3 – 9.8t + 0.03t^2 = 0`

`=> t =  1.064 \ text(seconds)`

`text(Max height)` `= 6 sqrt3 (1.064) – 4.9(1.064)^2 + 0.01(1.064)^3`  
  `~~5.5\ text(m)`  

 
d.   `text(Time of Flight  ⇒  Solve for)\ \ t\ \ text(when)\ \ y=-x:`

`6 sqrt 3 t – 4.9t^2 + 0.01 t^3` `= -(6t-0.001t^3)`
`(6 + 6 sqrt3)t – 4.9 t^2` `= 0`
`t(6 + 6 sqrt3 – 4.9 t)` `= 0`
`4.9 t` `= 6 + 6 sqrt3`
`t` `= (6 + 6 sqrt3)/(4.9)`
  `= (60(sqrt3 + 1))/(49)\ text(seconds)`


 
e.   `text(Total distance) \ = text(Area under) \ v(t) \ text(graph from)\ \ t = 0 \ \ text(to)\ \ t_1 = (60(sqrt3 + 1))/(49)`

`|v(t)| = sqrt{(6 – 0.03 t^2)^2 + (6 sqrt3 –  9.8t + 0.03 t^2)^2}`

`text(Distance)` `= int_0^(t_1) |v(t)| \ dt`
  `= 38.51 \ text{m (to 2 d.p.)}`

Complex Numbers, SPEC2-NHT 2019 VCAA 1

In the complex plane, `L` is the with equation `|z + 2| = |z - 1 - sqrt3 i|`.

  1.  Verify that the point (0, 0) lies on `L`.  (1 marks)
  2.  Show that the cartesian form of the equation of  `L`  is  `y = - sqrt3 x`.  (2 marks)
  3.  The line  `L`  can also be expressed in the form  `|z - 1| = |z - z_1|`, where  `z_1 ∈ C`.

     

     Find  `z_1` in cartesian form.  (2 marks)

  4.  Find, in cartesian form, the points(s) of intersection of  `L`  and the graph of  `|z| = 4`.  (2 marks)
  5.  Sketch  `L`  and the graph of  `|z| = 4`  on the Argand diagram below.  (2 marks)
     
     
         
     
  6.  Find the area of the sector defined by the part of  `L`  where  `text(Re)(z) ≥ 0`, the graph of  `|z| = 4`  where  `text(Re)(z) ≥ 0`, and imaginary axis where  `text(Im)(z) > 0`.  (1 marks)
Show Answers Only
  1. `text(Proof(See Worked Solution))`
  2. `text(Proof(See Worked Solution))`
  3. `-(1)/(2) – (sqrt3)/(2) i`
  4. `(2,-2 sqrt3) text(and) (-2, 2 sqrt3)`
  5.  

     
  6. `(20pi)/(3)`
Show Worked Solution

a.   `text(Substitute)\ \ z = 0 + 0i\ \  text(into both sides:)`

`text(LHS) = |2| = 2`

`text(RHS) = |-1 – sqrt3i| = sqrt{(-1)^2 + (-sqrt3)^2} = 2`

`:. (0,0)\ \ text(lies on both sides.)`

 

b.  `|x + yi + 2|` `= |x + yi – 1 – sqrt3 i|`
  `|(x + 2) + yi|` `= |(x – 1) + (y – sqrt3) i|`
  `sqrt(x^2 + 4x + 4 + y^2` `= sqrt(x^2 – 2x + 1 + y^2 -2 sqrt3 y + 3`
  `x^2 + 4x + 4 + y^2` `= x^2 – 2x + 4 – 2 sqrt3 y + y^2`
  `6x` `= -2 sqrt3 y`
  `y` `= -(3)/(sqrt3) x`
  `y` `= -sqrt3 x`


c.

`m_text(perp) = (sqrt2)/(3) , text(through)\ (1, 0)`

`y = (sqrt3)/(3) (x – 1)\ …\ L_1`

`text(Intersection) \ L\ text(and) \ L_1,`

`text(Solve:) \ (sqrt3)/3 (x-1) = -sqrt3 x\ \ \ text{(by CAS)}`

`=> x = (1)/(4) , y = -(sqrt3)/(4) \ \ \ text{(point}\ Ptext{)}`
 

`P(x_1,y_1) \ text(is midpoint of) \ \ z_1 \ text(and) \ \ (1, 0):`

`(x_1 + 1)/(2) = (1)/(4) \ => \ x_1 = -(1)/(2)`

`(y_1 + 0)/(2) = -sqrt3/(4) \ => \ y_1 = -sqrt3/(2)`

`:. \ z_1 = -(1)/(2) – (sqrt3)/(2) i`

 

d.   `|z| = 4 => \ text(circle, centre) \ (0,0), \ text(radius) = 4`

`x^2 + y^2 = 16\ …\ (1)`

`y = – sqrt3 x\ …\ (2)`

`text(Substitute)\ (2) \ text(into) \ (1)`

`x^2 + 3x^2 = 16`

`x = ±2`

`:. \ text(Intersection at)\ (2, – 2 sqrt3) \ text(and) \ (-2, 2 sqrt3)`

 

e.


 

f.    

`text(Area)` `= (5)/(12) xx  pi xx 4^2`
  `= (20pi)/(3)`

Calculus, 2ADV C4 SM-Bank 1 MC

A lift accelerates from rest at a constant rate until it reaches a speed of 3 ms−1. It continues at this speed for 10 seconds and then decelerates at a constant rate before coming to rest. The total travel time for the lift is 30 seconds.

The total distance, in metres, travelled by the lift is

  1.  45
  2.  60
  3.  75
  4.  90
Show Answers Only

`B`

Show Worked Solution

`text(Consider the velocity graph:)`
 

`t_1 -> t_2 = text(10 seconds)`
 

`:.\ text(Total distance travelled)`

`=\ text(Area of trapezium)`

`= 1/2 xx 3(10 + 30)`

`= 60\ text(m)`
 

`=>\ B`

Mechanics, SPEC2-NHT 2019 VCAA 15 MC

A lift accelerates from rest at a constant rate until it reaches a speed of 3 ms−1. It continues at this speed for 10 seconds and then decelerates at a constant rate before coming to rest. The total travel time for the lift is 30 seconds.

The total distance, in metres, travelled by the lift is

  1.  30
  2.  45
  3.  60
  4.  75
  5.  90
Show Answers Only

`C`

Show Worked Solution

`text(Consider the velocity graph:)`
 

`t_1 -> t_2 = text(10 seconds)`

`text(Distance = area of trapezium)`

`= 1/2 xx 3(10 + 30)`

`= 60\ text(m)`
 

`=>\ C`

Statistics, SPEC2-NHT 2019 VCAA 19 MC

Bags of peanuts are packed by a machine. The masses of the bags are normally distributed with a standard deviation of three grams.

The minimum size of a sample required to ensure that the manufacturer can be 98% confident that the sample mean is within one gram of the population mean is

  1.  37
  2.  38
  3.  48
  4.  49
  5.  60
Show Answers Only

`D`

Show Worked Solution

`M\ ~\ N(mu, 3^2)`

`text(Confidence interval 98%) => \ z = 2.326\ \ \ text{(by CAS)}`
 


 

`mu ± z sigma/sqrtn\ \ text(is within 1 gram of)\ mu:`

`text(Solve:)\ \ 2.326 xx 3/sqrtn <=1`

`n >= 48.7`

`n = 49`

`=>\ D`

Statistics, SPEC2-NHT 2019 VCAA 20 MC

Nitrogen oxide emissions for a certain type of car are known to be normally distributed with a mean of 0.875 g/km and a standard deviation of 0.188 g/km.

For two randomly selected cars, the probability that their nitrogen oxide emissions differ by more than 0.5 g/km is closest to

  1. 0.030
  2. 0.060
  3. 0.960
  4. 0.970
  5. 0.977
Show Answers Only

`B`

Show Worked Solution

`C\ ~\ N(0.875, 0.188^2)`

`text(Let)\ \ C_1 = text(emissions of car 1,)\ C_2 = text(emissions of car 2)`

`E(C_1-C_2) = E(C_1)-E(C_2) = 0`

`text(Var)(C_1-C_2) = text(Var)(C_1) + text(Var)(C_2) = 2 xx 0.188^2`

`sigma(C_1-C_2) = sqrt2 xx 0.188`

`text(Pr)(|C_1-C_2| > 0.5) = 0.060\ \ \ text{(by CAS)}`

`=>\ B`

Mechanics, SPEC2-NHT 2019 VCAA 17 MC

A ball is thrown vertically upwards with an initial velocity of  `7sqrt6` ms−1, and is subject to gravity and air resistance. The acceleration of the ball is given by  `overset(¨)x = −(9.8 + 0.1v^2)`, where `v` ms−1 is its velocity when it is at a height of `x` metres above ground level.

The maximum height, in metres, reached by the ball is

  1.  `5log_e(4)`
  2.  `log_e(sqrt31)`
  3.  `(5pisqrt2)/21`
  4.  `5log_e(2)`
  5.  `(7pisqrt2)/3`
Show Answers Only

`A`

Show Worked Solution

`overset(¨)x = v · (dv)/(dx) = −(9.8 + 0.2v^2)`

`(dv)/(dx)` `= (−(9.8 + 0.2v^2))/v`
`(dx)/(dv)` `= −(10v)/(98 + v^2)`
`x` `= −int_(7sqrt6)^0 (10v)/(98 + v^2)\ dv`
  `= 5log_e(4)`

 
`=>\ A`

Mechanics, SPEC2-NHT 2019 VCAA 16 MC

An object of mass 2 kg is travelling horizontally in a straight line at a constant velocity of magnitude 2 ms−1.  The object is hit in such a way that it deflects 30° from its original path, continuing at the same speed in a straight line.

The magnitude, correct to two decimal places, of the change of momentum, in kg ms−1, of the object is

  1. 0.00
  2. 0.24
  3. 1.04
  4. 1.46
  5. 2.07
Show Answers Only

`E`

Show Worked Solution

`text(Assume initial direction is along the)\ underset~i\ text(axis:)`

`underset~(p_1) = 2 xx 2underset~i = 4underset~i`

`underset~(p_2)` `= 2 xx (2cos30^@underset~i + 2sin30^@underset~j)`
  `= 2sqrt3underset~i + 2underset~j`

`underset~(p_2) – underset~(p_1) = (2sqrt3 – 4)underset~i + 2underset~j`

`|underset~(p_2) – underset~(p_1)|` `= sqrt((2sqrt3 – 4)^2 + 4)`
  `~~ 2.07`

`=>\ E`

Complex Numbers, SPEC2-NHT 2019 VCAA 5 MC

The circle defined by  `|z + a| = 3 |z + i|`, where  `a ∈ R`, has a centre and radius respectively given by

  1. `(a/8, −9/8), \ 3/8sqrt(a^2 + 1)`
  2. `(a/8, −9/8), \ (9a^2 + 9)/64`
  3. `(a/8, −9/8), \ 1/8sqrt(153 - 7a^2)`
  4. `(−a/8, 9/8), \ (9a^2 + 9)/64`
  5. `(−a/8, 9/8), \ 3/8sqrt(a^2 + 1)`
Show Answers Only

`A`

Show Worked Solution
`|z + a|` `= 3|z + i|`
`|(x + a) + yi|` `= 3|x + (y + 1)i|`
`sqrt((x + a)^2 + y^2)` `= 3sqrt(x^2 + (y + 1)^2)`
`x^2 + 2ax + a^2 + y^2` `= 9(x^2 + y^2 + 2y + 1)`
`8x^2 – 2ax + 8y^2 + 18y` `= a^2 – 9`
`8(x^2 – a/4x + (a^2)/64) + 8(y^2 + 9/4y + 81/64)` `= a^2 – 9 + (a^2)/8 + 81/8`
`(x – a/8)^2 + (y + 9/8)^2` `= (9a^2)/64 + 9/64`
  `= 9/64(a^2 + 1)`

 
`:. text(Centre)\ (a/8, −9/8),\ text(radius) = 3/8sqrt(a^2 + 1)`

`=>\ A`

Calculus, EXT1 C3 2019 SPEC1-N 9

i.  Show that  `tan((5pi)/(12)) = sqrt3 + 2`.   (2 marks)

ii. 
       
 
Hence, find the area bounded by the graph of  `f(x) = (2)/(x^2 - 4x + 8)`  shown above, the `x`-axis and the lines  `x = 0`  and  `x = 2 sqrt3 +6`.   (4 marks)

Show Answers Only
  1. `text(Proof (See Worked Solution))`
  2. `(2pi)/(3)`
Show Worked Solution

i.    `text(Method 1:)`

`tan \ (5pi)/(12)` `= tan ((pi)/(4) + (pi)/(6))`
  `= (tan \ (pi)/(4) + tan\ (pi)/(6))/(1 – tan \ (pi)/(4) · tan \ (pi)/(6))`
  `= (1 + (1)/(sqrt3))/(1 – (1)/(sqrt3))`
  `= (sqrt3+1)/(sqrt3-1) xx (sqrt3+1)/(sqrt3+1)`
  `= (3+ 2 sqrt3 + 1)/(3 – 1)`
  `= sqrt3 +2`

  
`text(Method 2:)`

`tan \ (5pi)/(6)` `= (2tan \ (5pi)/(12))/(1 – tan^2 \ (5pi)/(12))`
`- 1/sqrt3` `=(2tan \ (5pi)/(12))/(1 – tan^2 \ (5pi)/(12))`
`-2 sqrt3 tan \ (5pi)/(12)` `= 1 – tan^2 \ (5pi)/(12)`

 

`tan^2 \ (5pi)/(12) – 2 sqrt(3) tan \ (5pi)/(12) – 1 = 0`

`tan \ (5pi)/(12)` `= (2 sqrt3 ± sqrt(12 + 4))/(2)`
  `= sqrt3 + 2 \ \ \ (tan theta > 0)`

 

ii.   `text(Area)` `= int_0 ^(2 sqrt3 + 6) \ (2)/(x^2 – 4x + 8)\ dx`
  `= int_0 ^(2 sqrt3 + 6) \ (2)/((x -2)^2 + 2^2)`
  `= [tan^-1 ((x – 2)/(2))]_0 ^(2 sqrt3 + 6)`
  `= tan^-1 (sqrt3 + 2) – tan^-1 (-1)`
  `= (5pi)/(12) – (-(pi)/(4))`
  `= (2pi)/(3)`

Calculus, SPEC1-NHT 2019 VCAA 9

  1.   Show that  `tan((5pi)/(12)) = sqrt3 + 2`.   (2 marks)
  2.  
     
         
     
    Hence, find the area bounded by the graph of  `f(x) = (2)/(x^2 - 4x + 8)`  shown above, the `x`-axis and the lines  `x = 0`  and  `x = 2 sqrt3 +6`.   (4 marks)
Show Answers Only
  1. `text(Proof (See Worked Solution))`
  2. `(2pi)/(3)`
Show Worked Solution

a.    `text(Method 1:)`

`tan \ (5pi)/(12)` `= tan ((pi)/(4) + (pi)/(6))`
  `= (tan \ (pi)/(4) + tan\ (pi)/(6))/(1 – tan \ (pi)/(4) · tan \ (pi)/(6))`
  `= (1 + (1)/(sqrt3))/(1 – (1)/(sqrt3))`
  `= (sqrt3+1)/(sqrt3-1) xx (sqrt3+1)/(sqrt3+1)`
  `= (3+ 2 sqrt3 + 1)/(3 – 1)`
  `= sqrt3 +2`

  
`text(Method 2:)`

`tan \ (5pi)/(6)` `= (2tan \ (5pi)/(12))/(1 – tan^2 \ (5pi)/(12))`
`- 1/sqrt3` `=(2tan \ (5pi)/(12))/(1 – tan^2 \ (5pi)/(12))`
`-2 sqrt3 tan \ (5pi)/(12)` `= 1 – tan^2 \ (5pi)/(12)`

 

`tan^2 \ (5pi)/(12) – 2 sqrt(3) tan \ (5pi)/(12) – 1 = 0`

`tan \ (5pi)/(12)` `= (2 sqrt3 ± sqrt(12 + 4))/(2)`
  `= sqrt3 + 2 \ \ \ (tan theta > 0)`

 

b.   `text(Area)` `= int_0 ^(2 sqrt3 + 6) \ (2)/(x^2 – 4x + 8)\ dx`
  `= int_0 ^(2 sqrt3 + 6) \ (2)/((x -2)^2 + 2^2)`
  `= [tan^-1 ((x – 2)/(2))]_0 ^(2 sqrt3 + 6)`
  `= tan^-1 (sqrt3 + 2) – tan^-1 (-1)`
  `= (5pi)/(12) – (-(pi)/(4))`
  `= (2pi)/(3)`

Vectors, EXT2 V1 2019 SPEC2 4

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

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

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

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

Show Answers Only
  1. `a = 6, b = 2, c = −3`
  2. `4/9`
  3. `2sqrt65\ text(u²)`
Show Worked Solution
i.    `overset(->)(AB)` `= (4 – 2)underset~i + (−2 + 1)underset~j + (1 – 3)underset~k`
    `= 2underset~i – underset~j – 2underset~k`

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

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

`a – 4 = 2 \ => \ a = 6`

`b – 3 = −1 \ => \ b = 2`

`c + 1 = −2 \ => \ c = −3`

 

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

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

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

 

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

  

  

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

Statistics, SPEC1-NHT 2019 VCAA 3

The number of cars per day making a U-turn at a particular location is known to be normally distributed with a standard deviation of 17.5. In a sample of 25 randomly selected days, a total of 1450 cars were observed making the U-turn.

  1. Based on this sample, calculate an approximate 95% confidence interval for the number of cars making the U-turn each day. Use an integer multiple of the standard deviation in your calculations.  (3 marks)
  2. The average number of U-turns made at the location is actually 60 per day.

     

    Find an approximation, correct to two decimal places, for the probability that on 25 randomly selected days the average number of U-turns is less than 53.   (1 mark)

Show Answers Only
  1. `(51, 65)`
  2. `0 .02`
Show Worked Solution

a.      `barx = (1450)/(25) = 58`

`(σ)/(sqrtn) = (17.5)/(sqrt25) = (35)/(2xx5) = (7)/(2)`

`text(Limit:)\ ` `(58 – 2 xx (7)/(2) \ , \ 58 + 2 xx (7)/(2)) `
  `= (51, 65)`

 

b.     `μ = 60 \ , \ (σ)/(sqrtn) = (7)/(2)`
 
          `\ barX ∼ N (60, ((7)/(2))^2)`
 
 
         
 

`text(Pr) (barX < 53)` `= text(Pr) (z <–2)`
  `= (0.05)/(2)`
  `= 0.025`
  `≈ 0.02 \ \ text{(round down)}`

 
`text{(0.03 was also accepted as a correct answer)}`

Trigonometry, 2ADV* T1 2011 HSC 24c

A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`.  The
size of angle `ABC` is 114°.
 

2UG 2011 24c

Copy the diagram into your writing booklet and show all the information on it.

  1. What is the bearing of `C` from `B`?   (1 mark)

  2. Find the distance `AC`. Give your answer correct to the nearest kilometre.   (2 marks)

  3. What is the bearing of `A` from `C`? Give your answer correct to the nearest degree.   (3 marks)

Show Answers Only
  1.  `055^@`
  2.  `13\ text(km)`
  3.  `261^@`
Show Worked Solution
STRATEGY: This deserves repeating again: Draw North-South parallel lines through major points to make the angle calculations easier!
i.     2011 HSC 24c

 `text(Let point)\ D\ text(be due North of point)\ B`

`/_ABD` `=180-121\ text{(cointerior with}\ \ /_A text{)}`
  `=59^@`
`/_DBC` `=114-59`
  `=55^@`

   
`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`

 

ii.    `text(Using cosine rule:)`

`AC^2` `=AB^2+BC^2-2xxABxxBCxxcos/_ABC`
  `=6^2+9^2-2xx6xx9xxcos114^@`
  `=160.9275…`
`:.AC` `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}`
  `=13\ text(km)\ text{(nearest km)}`

 

iii.    `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`

MARKER’S COMMENT: The best responses showed clear working on the diagram.
`cos/_ACB` `=(AC^2+BC^2-AB^2)/(2xxACxxBC)`
  `=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)`
  `=0.9018…`
`/_ACB` `=25.6^@\ text{(to 1 d.p.)}`

 

`text(From diagram,)`

`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`

`:.\ text(Bearing of)\ A\ text(from)\ C`

  `=180+55+25.6`
  `=260.6`
  `=261^@\ text{(nearest degree)}`

Trigonometry, 2ADV* T1 2017 HSC 30c

The diagram shows the location of three schools. School `A` is 5 km due north of school `B`, school `C` is 13 km from school `B` and `angleABC` is 135°.
 


 

  1. Calculate the shortest distance from school `A` to school `C`, to the nearest kilometre.  (2 marks)

  2. Determine the bearing of school `C` from school `A`, to the nearest degree.  (3 marks)

Show Answers Only
  1. `17\ text{km  (nearest km)}`
  2. `213^@`
Show Worked Solution

(i)   `text(Using cosine rule:)`

`AC^2` `= AB^2 + BC^2 – 2 xx AB xx BC xx cos135^@`
  `= 5^2 + 13^2 – 2 xx 5 xx 13 xx cos135^@`
  `= 285.923…`
`:. AC` `= 16.909…`
  `= 17\ text{km  (nearest km)}`

 

(ii)   

`text(Using sine rule, find)\ angleBAC:`

♦♦ Mean mark part (ii) 31%.
`(sin angleBAC)/13` `= (sin 135^@)/17`
`sin angleBAC` `= (13 xx sin 135^@)/17`
  `= 0.5407…`
`angleBAC` `= 32.7^@`

 

`:. text(Bearing of)\ C\ text(from)\ A`

`= 180 + 32.7`

`= 212.7^@`

`= 213^@`

Trigonometry, 2ADV* T1 2007 26a

The diagram shows information about the locations of towns  `A`,  `B`  and  `Q`.
 

 
 

  1. It takes Elina 2 hours and 48 minutes to walk directly from Town  `A`  to Town  `Q`.

     

    Calculate her walking speed correct to the nearest km/h.    (1 mark)

  2. Elina decides, instead, to walk to Town  `B`  from Town  `A`  and then to Town  `Q`.

     

    Find the distance from Town  `A`  to Town  `B`. Give your answer to the nearest km.   (2 marks)

  3. Calculate the bearing of Town  `Q`  from Town  `B`.   (1 mark)

Show Answers Only
  1. `5\ text(km/hr)\ text{(nearest km/hr)}`
  2. `18\ text(km)\ text{(nearest km)}`
  3. `236^@`
Show Worked Solution

i.  `text(2 hrs 48 mins) = 168\ text(mins)`

`text(Speed)\ text{(} A\ text(to)\ Q text{)}` `= 15/168`
  `= 0.0892…\ text(km/min)`

 

`text(Speed)\ text{(in km/hr)}` `= 0.0892… xx 60`
  `= 5.357…\ text(km/hr)`
  `= 5\ text(km/hr)\ text{(nearest km/hr)}`

 

ii.  

`text(Using cosine rule)`

`AB^2` `= 15^2 + 10^2 – 2 xx 15 xx 10 xx cos 87^@`
  `= 309.299…`
`AB` `= 17.586…`
  `= 18\ text(km)\ text{(nearest km)}`

 

`:.\ text(The distance from Town)\ A\ text(to Town)\ B\ text(is 18 km.)`

 

iii.  
`/_CAQ` `= 31^@\ \ \ text{(} text(straight angle at)\ A text{)}`
`/_AQD` `= 31^@\ \ \ text{(} text(alternate angle)\ AC\ text(||)\ DQ text{)}`
`/_DQB` `= 87 – 31 = 56^@`
`/_QBE` `= 56^@\ \ \ text{(} text(alternate angle)\ DQ \ text(||)\ BE text{)}`

 

`:.\ text(Bearing of)\ Q\ text(from)\ B`

`= 180 + 56`

`= 236^@`

Vectors, SPEC2 2009 VCAA 17 MC

Vectors `underset ~a, underset ~b` and `underset ~c` are shown below.
 

VCAA 2009 spec2 17mc
 

From the diagram it follows that

A.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2`

B.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 - |\ underset ~a\ | |\ underset ~b\ |`

C.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a * underset ~b\ |`

D.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a\ | |\ underset ~b\ |`

E.   `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 - |\ underset ~a * underset ~b\ |`

Show Answers Only

`D`

Show Worked Solution

`underset~ c + underset~ b = underset~ a\ \ => \ underset~ c = underset~ a – underset~ b`

♦♦ Mean mark 30%.

`:. underset~ c * underset~ c` `= (underset~ a – underset~ b) * (underset~ a – underset~ b)`
  `= underset~ a * underset~ a – underset~ a * underset~ b -underset~b*underset~a + underset~ b * underset~ b`
  `= underset~ a * underset~ a  + underset~ b * underset~ b-2underset~b*underset~a `

`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\ underset~ b\ |^2 – 2 |\ underset~ a\ | |\ underset~ b\ | cos 120^@`

`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\underset~ b\ |^2 + |\ underset~ a\ | |\ underset~ b\ |`

`=>   D`

Vectors, EXT1 V1 SPEC2 2009 17 MC

Vectors  `underset ~a, underset ~b`  and  `underset ~c`  are shown below.

VCAA 2009 spec2 17mc
From the diagram it follows that

  1. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2`
  2. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2-|\ underset ~a\ | |\ underset ~b\ |`
  3. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a * underset ~b\ |`
  4. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a\ | |\ underset ~b\ |`
Show Answers Only

`D`

Show Worked Solution

`underset~ c + underset~ b = underset~ a\ \ => \ underset~ c = underset~ a-underset~ b`

`:. underset~ c * underset~ c` `= (underset~ a-underset~ b) * (underset~ a-underset~ b)`
  `= underset~ a * underset~ a-underset~ a * underset~ b -underset~b*underset~a + underset~ b * underset~ b`
  `= underset~ a * underset~ a  + underset~ b * underset~ b-2underset~b*underset~a `

`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\ underset~ b\ |^2-2 |\ underset~ a\ | |\ underset~ b\ | cos 120^@`

`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\underset~ b\ |^2 + |\ underset~ a\ | |\ underset~ b\ |`

`=>   D`

Statistics, STD2 S1 EQ-Bank 4

A high school conducted a survey asking students what their favourite Summer sport was.

The Pareto chart shows the data collected.
 


 

  1. What percentage of students chose Hockey as their favourite Summer sport?  (1 mark)

  2. What percentage of students chose Touch Football as their favourite Summer sport?  (1 mark)

Show Answers Only
  1. `text(2%)`
  2. `text(22%)`
Show Worked Solution

i.   `text(Method 1)`

`text(Percentage who chose hockey)`

`=\ text{cum freq (at hockey column) – cum freq (at tennis column)}` 

`=100-98`

`=2 text(%)`
 

`text(Method 2)`

`text(Column 1 → 180 people = 30%)`

`text(Total surveyed) = 180 -: 0.3 = 600`

`:.\ text(Hockey %)` `=12/600 xx 100`  
  `=2\text(%)`  

 

ii.   `text(Percentage who chose touch football)`

`=\ text{cum freq (at touch football column) – cum freq (at cricket column)}` 

`=76-54`

`=22 text(%)`

Statistics, STD2 S1 EQ-Bank 6

The Pareto chart below shows the data collected from a survey where people were asked to choose their favourite overseas holiday destination.

Using the chart, how many people were surveyed?  (2 marks)
 

Show Answers Only

`500`

Show Worked Solution

`text(Percentage of people whose chose NZ)`

`= 80 – 60`

`= 20text(%)`
 

`text(Number of people who chose NZ)`

`= 100`
 

`text(20%)\ xx\ text(Total surveyed = 100)`

`:.\ text(Total surveyed = 500)`

GRAPHS, FUR2 2019 VCAA 3

Members of the association will travel to a conference in cars and minibuses:

  • Let `x` be the number of cars used for travel.
  • Let `y` be the number of minibuses used for travel.
  • A maximum of eight cars and minibuses in total can be used.
  • At least three cars must be used.
  • At least two minibuses must be used.

The constraints above can be represented by the following three inequalities.
 

`text(Inequality 1) qquad qquad x + y <= 8`

`text(Inequality 2) qquad qquad x >= 3`

`text(Inequality 3) qquad qquad y >= 2`
 

  1. Each car can carry a total of five people and each minibus can carry a total of 10 people.

      

    A maximum of 60 people can attend the conference.

      

    Use this information to write Inequality 4.   (1 mark)

The graph below shows the four lines representing Inequalities 1 to 4.

Also shown on this graph are four of the integer points that satisfy Inequalities 1 to 4. Each of these integer points is marked with a cross (✖).
 


 

  1. On the graph above, mark clearly, with a circle (o), the remaining integer points that satisfy Inequalities 1 to 4.  (1 mark)

Each car will cost $70 to hire and each minibus will cost $100 to hire.

  1. What is the cost for 60 members to travel to the conference?  (1 mark)
  2. What is the minimum cost for 55 members to travel to the conference?  (1 mark)
  3. Just before the cars were booked, the cost of hiring each car increased.

      

    The cost of hiring each minibus remained $100.

      

    All original constraints apply.

      

    If the increase in the cost of hiring each car is more than `k` dollars, then the maximum cost of transporting members to this conference can only occur when using six cars and two minibuses.

      

    Determine the value of  `k`.  (1 mark)

Show Answers Only
  1. `5x + 10y <= 60`
  2. `text(See Worked Solutions)`
  3. `$680`
  4. `$610`
  5. `30`
Show Worked Solution

a.  `5x + 10y <= 60`

 

b.    

 

c.    `text(Consider the line)\ \ 5x + 10y = 60\ \ text(on the graph)`

`text(Touches the feasible region at)\ (4, 4)\ text(only)`

`:.\ text (C) text(ost of 60 members)`

`= 4 xx 70 + 4 xx 100`

`= $680`

 

d.    `text(Coordinates that allow 55 to travel) \ => \ (5, 3) and (3, 4)`

`text(C) text(ost)\ (5, 3) = 5 xx 70 + 3 xx 100 = $650`

`text(C) text(ost)\ (3, 4) = 3 xx 70 + 4 xx 100 = $610`

`:.\ text(Minimum cost is $610)`

 

e.    `text(Objective function): \ C = ax + 100y`

`text(Max cost occurs at)\ \ (6, 2)\ \ text(when)\ \ a > 100`

`text{(i.e. graphically when the slope of}\ \ C = ax + 100y`

   `text(is steeper than)\ \ x + y= 8 text{)}`

`:. k + 70` `= 100`
`k` `= 30`

GRAPHS, FUR2 2019 VCAA 2

Each branch within the association pays an annual fee based on the number of members it has.

To encourage each branch to find new members, two new annual fee systems have been proposed.

Proposal 1 is shown in the graph below, where the proposed annual fee per member, in dollars, is displayed for branches with up to 25 members.
 


 

  1. What is the smallest number of members that a branch may have?  (1 mark)
  2. The incomplete inequality below shows the number of members required for an annual fee per member of $10.

      

    Complete the inequality by writing the appropriate symbol and number in the box provided.   (1 mark)
     

3 ≤ number of members  
 

 

Proposal 2 is modelled by the following equation.

annual fee per member = – 0.25 × number of members + 12.25

  1. Sketch this equation on the graph for Proposal 1, shown below.  (1 mark)

 

 

  1. Proposal 1 and Proposal 2 have the same annual fee per member for some values of the number of members.

      

    Write down all values of the number of members for which this is the case.  (1 mark)

Show Answers Only
  1. `3`
  2. `3 <= text(number of members) <= 10`
  3. `text(See Worked Solutions)`
  4. `9, 17, 25`
Show Worked Solution

a.  `3`
 

b.  `3 <= text(number of members) <= 10`
 

c.  

 

d.    `text(Same annual fee occurs when graphs intersect.)`

`text(Member numbers): 9, 17, 25`

GEOMETRY, FUR2 2019 VCAA 2

A cargo ship travels from Magadan (60° N, 151° E) to Sydney (34° S, 151° E).

  1. Explain, with reference to the information provided, how we know that Sydney is closer to the equator than Magadan.   (1 mark)
  2. Assume that the radius of Earth is 6400 km.

      

    Find the shortest great circle distance between Magadan and Sydney.

      

    Round your answer to the nearest kilometre.  (1 mark)

  3. The cargo ship left Sydney (34° S, 151° E) at 6 am on 1 June and arrived in Perth (32° S, 116° E) at 10 am on 11 June.

      

    There is a two-hour time difference between Sydney and Perth at that time of year.

      

    How many hours did it take the cargo ship to travel from Sydney to Perth?  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `10\ 500\ text(km)`
  3. `246\ text(hours)`
Show Worked Solution
a.  

`text{Sydney’s angle with the equator (34°S) is smaller than Magadan’s}`

`text{(60°N) resulting in a shorter distance to the equator.)`

 

b.   `text(Distance)` `= (34 + 60)/360 xx 2pi xx 6400`
    `~~ 10\ 500\ text(km)`

 

c.    `text(Perth is 2 hours behind Sydney)`

`text(6 am Sydney = 4 am Perth)`

`:.\ text(Travel time)` `= 4\ text{am (1 Jun)} – 10\ text{am (11 Jun)}`
  `= 6\ text(hours) + 10\ text(days)`
  `= 6 + 10 xx 24`
  `= 246\ text(hours)`

Statistics, STD2 S1 EQ-Bank 5

An island resort surveyed 400 guests by asking them on which continent they lived.

The table below shows the data collected.
 


 

Complete the Pareto chart below to show the data collected.  (3 marks)
 

Show Answers Only

Show Worked Solution

`text(% Asia)\ = 168/400=42 text(%)`

`text(% North America)\ = 120/400=30 text(%)`

`text(% Australia)\ = 88/400=22 text(%)`

`text(% Europe)\ = 24/400=6 text(%)`
 

Networks, STD2 N3 2019 FUR2 3

Fencedale High School is planning to renovate its gymnasium.

This project involves 12 activities, `A` to `L`.

The directed network below shows these activities and their completion times, in weeks.
 


 

The minimum completion time for the project is 35 weeks.

  1. Identify the critical path and state how many activities are on it?  (2 marks)

  2. Determine the latest start time of activity `E`.  (1 mark)

  3. Which activity has the longest float time?  (1 mark)

It is possible to reduce the completion time for activities `C, D, G, H` and `K` by employing more workers.

  1.  The completion time for each of these five activities can be reduced by a maximum of two weeks.

      

    What is the minimum time, in weeks, that the renovation project could take?  (1 mark)

Show Answers Only
  1. `8\ text(activities)`
  2. `12\ text(weeks)`
  3. `text(Activity)\ J`
  4. `29\ text(weeks)`
Show Worked Solution

a.  `text(Scanning forwards and backwards:)`
 

 

 

`text(Critical path:)\ ABDFGIKL`

`:. 8\ text(activities)`
 

b.  `text(LST for activity)\ E = 12\ text{weeks  (i.e. start of 13th week)}`
 

c.   `text(Consider float times of all activities not on critical path.)`

`J-5, H-1, E-1, C-1`

`:.\ text(Activity)\ J\ text(has the largest float time.)`
 

d.   `text(Critical path after reducing)\ CDGHK\ text(by 2 weeks is)`

`ABDFGIKL.`
 

`:.\ text(Minimum time)` `= 2 + 4 + 7 + 1 + 2 + 2 + 5 + 6`
  `= 29\ text(weeks)`

NETWORKS, FUR2 2019 VCAA 3

Fencedale High School is planning to renovate its gymnasium.

This project involves 12 activities, `A` to `L`.

The directed network below shows these activities and their completion times, in weeks.
 


 

The minimum completion time for the project is 35 weeks.

  1. How many activities are on the critical path?  (1 mark)
  2. Determine the latest start time of activity `E`.  (1 mark)
  3. Which activity has the longest float time?  (1 mark)

It is possible to reduce the completion time for activities `C, D, G, H` and `K` by employing more workers.

  1.  The completion time for each of these five activities can be reduced by a maximum of two weeks.

      

    What is the minimum time, in weeks, that the renovation project could take?  (1 mark)

  2. The reduction in completion time for each of these five activities will incur an additional cost to the school.

     

    The table below shows the five activities that can have their completion times reduced and the associated weekly cost, in dollars.
      

           Activity                    Weekly cost ($)             
    `C` 3000
    `D` 2000
    `G` 2500
    `H` 1000
    `K` 4000

      
    The completion time for each of these five activities can be reduced by a maximum of two weeks.

      

    Fencedale High School requires the overall completion time for the renovation project to be reduced by four weeks at minimum cost.

     

    Complete the table below, showing the reductions in individual activity completion times that would achieve this.  (2 marks)
      

           Activity       

    		              Reduction in completion time             
            (0, 1 or 2 weeks)
    `C`  
    `D`  
    `G`  
    `H`  
    `K`  
Show Answers Only
  1. `8\ text(activities)`
  2. `12\ text(weeks)`
  3. `text(Activity)\ J`
  4. `29\ text(weeks)`
  5.  
             Activity       

    		          Reduction in completion time         
            (0, 1 or 2 weeks)
      `C` 0
      `D` 1
      `G` 2
      `H` 1
      `K` 1
Show Worked Solution

a.  `text(Scanning forwards and backwards:)`
 

 

 

`text(Crirical path:)\ ABDFGIKL`

`:. 8\ text(activities)`
 

b.  `text(LST for activity)\ E = 12\ text{weeks  (i.e. start of 13th week)}`
 

c.   `text(Consider float times of all activities not on critical path.)`

`J-5, H-1, E-1, C-1`

`:.\ text(Activity)\ J\ text(has the largest float time.)`
 

d.   `text(Critical path after reducing)\ CDGHK\ text(by 2 weeks is)`

`ABDFGIKL.`
 

`:.\ text(Minimum time)` `= 2 + 4 + 7 + 1 + 2 + 2 + 5 + 6`
  `= 29\ text(weeks)`

 

e.   `text(Reduce cheapest activities on the critical path by 1 week)`

`↓ 1 =>\ text(Activity)\ D\ text(and)\ G`
 

`text{Possibilities for reducing by a further week (choose 2)}`

`C and D:\ text(cost $5000)\ \ text{(too expensive)}`

`G and H:\ text(cost $3500)\ \ text{(yes)}`

`K:\ text(cost $4000)\ \ text{(yes)}`
 

         Activity       

          Reduction in completion time         
        (0, 1 or 2 weeks)
  `C` 0
  `D` 1
  `G` 2
  `H` 1
  `K` 1