SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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.
  4. 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^{prime} = 28 + 18 sin ({pi(x^{prime}-k)}/{7})`

`x^{prime} = ax + c` `\ \ \ \ \ \ y^{prime} = by + d`

 

`text(Using) \ \ y^{prime} = by + d`

`28 + 18 sin ({pi(x^{prime}-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^{prime}-k)/(7)` `= (x)/(14)`
`x^{prime}` `= (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
    1. `f^{'}(x) > 0`   (1 mark)

    2. `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 4

The current flowing through an electrical circuit can be modelled by the function
 
`qquad f(t) = 6sin 0.05t + 8cos 0.05t, \ \ t >= 0`

  1.  Express the function in the form  `f(t) = Asin(at + b),\ \ text(for)\ \ 0<=b<=pi/2`.  (2 marks)

  2.  Find the time at which the current first obtains it maximum value.   (1 mark)

  3.  Sketch the graph of  `f(t)`. Clearly show its range and label the coordinates of its first maximum value. Do not label `x`-intercepts.   (1 mark)

Show Answers Only
  1. `f(t) = 10sin(0.05t + 0.927)`
  2.   `12.9\ \ (text(to 1 d.p.))`
  3.  
Show Worked Solution

i.   `f(t) = 6sin 0.05t + 8cos 0.05t`

`Asin(at + b)` `=Asin(0.05t + b)`  
  `= Asin(0.05t)\ cos(b) + Acos(0.05t)\ sin(b)`  

 

`=> Acos(b) = 6, \ Asin(b) = 8`

`A^2` `= 6^2 + 8^2`
`A` `= 10`

 

`=> 10cos(b)` `= 6`
`cos(b)` `= 6/10`
`b` `= cos^(−1) 0.06 ~~ 0.927\ text(radians)`

 
`:. f(t) = 10sin(0.05t + 0.927)`

 

ii.   `text(Max occurs at)\ \ sin(0.05t + 0.927)= sin\ pi/2`

`0.05t + 0.927` `= pi/2`
`0.05t` `= 0.643…`
`:.t` `= 12.87`
  `= 12.9\ \ (text(to 1 d.p.))`

 

iii.   

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`

Trigonometry, EXT1 T3 EQ-Bank 3

Find all values of  `theta`  that satisfy the equation  `sqrt3 cos theta = sin(2theta)`.  (3 marks)

Show Answers Only

`theta = 90° + m xx 180°, \ 60° + m xx 360°\ or\ 120° + m xx 360°`

Show Worked Solution
`sqrt3 cos theta` `= sin(2 theta)`
`2 cos theta sin theta – sqrt3 cos theta` `= 0`
`cos theta(sin theta – sqrt3/2)` `= 0`

COMMENT: Expressing “all values” is specifically mentioned in Topic Guidance as an application of arithmetic sequences.

`cos theta = 0 \or \  sin theta = sqrt3/2`

`theta = 90°, 270°\ or\ theta = 60°, 120° \ \ \ (0°<=theta<=360°)`

`:. theta = 90° + m xx 180°, \ 60° + m xx 360°\ or\ 120° + m xx 360°`

`text(where)\ \ m\ \ text(is an arbitrary integer).`

Probability, 2ADV S1 2019 MET2-N 9 MC

At the start of a particular week, Kim has three red apples and two green apples. She eats one apple everyday. On Monday, Tuesday and Wednesday of that week, she randomly selects an apple to eat. In this three-day period, the probability that Kim does not eat an apple of the same colour on any two consecutive days is

  1.  `(1)/(5)`
  2.  `(3)/(10)`
  3.  `(2)/(5)`
  4.  `(6)/(25)`
Show Answers Only

`B`

Show Worked Solution

`P text{(alternate colours)}`

`= P(RGR) + P(GRG)`

`= (3)/(5) ·(2)/(4) ·(2)/(3) + (2)/(5) ·(3)/(4) ·(1)/(3)`

`= (12)/(60) + (6)/(60)`

`= (3)/(10)`
 

`=> \ B`

Probability, 2ADV S1 2019 MET2-N 5 MC

Consider the probability distribution for the discrete random variable `X` shown in the table below.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ -1\ \ \ \  & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ \ \ 1\ \ \ \ \  & \ \ \ \ \ 2\ \ \ \ \  &\ \ \ \ \ 3\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & b & b & b & \dfrac{3}{5}-b & \dfrac{3b}{5} \\
\hline
\end{array}

The value of `E(X)` is

  1.  `(76)/(65)`
  2.  `1`
  3.  `0`
  4.  `(2)/(13)`
  5.  `(86)/(65)`
Show Answers Only

`A`

Show Worked Solution
`1` `= b + b + b + (3)/(5) – b + (3b)/(5)`
`(2)/(5)` `= (13b)/(5)`
`b` `= (2)/(13)`

 

`E(X)` `= -(2)/(13) + 0 + (2)/(13) + 2((3)/(5) – (2)/(13)) +3 ((6)/(65))`
  `= (76)/(65)`

 

\(\Rightarrow A\)

CORE, FUR1-NHT 2019 VCAA 19 MC

Consider the recurrence relation shown below.

`V_0 = 125\ 000,quadqquadV_(n + 1) = 1.013V_n - 2000`

This recurrence relation could be used to determine the value of

  1. a perpetuity with a payment of $2000 per quarter.
  2. an annuity with withdrawals of $2000 per quarter.
  3. an annuity investment with additional payments of $2000 per quarter.
  4. an item depreciating at a flat rate of 1.3% of the purchase price per quarter.
  5. a compound interest investment earning interest at the rate of 1.3% per annum.
Show Answers Only

`B`

Show Worked Solution

`text(The investment earns interest of 13% each period)`

`text(and a withdrawal of $2000 is made at the end of)`

`text(each period.)`

`=>\ B`

CORE, FUR1-NHT 2019 VCAA 18 MC

A truck was purchased for $134 000.

Using the reducing balance method, the value of the truck is depreciated by 8.5% each year.

Which one of the following recurrence relations could be used to determine the value of the truck after `n` years, `V_n`?

  1. `V_0 = 134\ 000,quadqquad V_(n + 1) = 0.915 xx V_n`
  2. `V_0 = 134\ 000,quadqquad V_(n + 1) = 1.085 xx V_n`
  3. `V_0 = 134\ 000,quadqquad V_(n + 1) = V_n - 11\ 390`
  4. `V_0 = 134\ 000,quadqquad V_(n + 1) = 0.915 xx V_n - 8576`
  5. `V_0 = 134\ 000,quadqquad V_(n + 1) = 1.085 xx V_n - 8576`
Show Answers Only

`A`

Show Worked Solution

`text(Each year, value decreases by 8.5%)`

`:. V_1` `= V_0 – 0.085 xx V_0`
  `= 0.915 xx V_0`

 
`=>\ A`

CORE, FUR1-NHT 2019 VCAA 16 MC

A small cafe is open every day of the week except Tuesday.

The table below shows the daily seasonal indices for labour costs at this cafe.

Excluding the day that the cafe is closed, the long-term average weekly labour cost is $9786.

The expected daily labour cost on a Wednesday is closest to

  1.  $1127
  2.  $1315
  3.  $1631
  4.  $1733
  5.  $2022
Show Answers Only

`E`

Show Worked Solution

`text(Average daily labour cost)`

`= 9786/6`

`= 1631`

 

`:.\ text(Wednesday’s expected cost)`

`= 1631 xx 1.24`

`= $2022.44`
 

`=>\ E`

CORE, FUR1-NHT 2019 VCAA 15 MC

The table below shows the long-term monthly average heating cost, in dollars, for a small office.

The seasonal index for April is closest to

  1.  0.80
  2.  0.87
  3.  0.96
  4.  1.25
  5.  1.56
Show Answers Only

`B`

Show Worked Solution

`text(Average cost per month = 125)`

`text{S.I. (April)}` `= 109/125`
  `= 0.872`

 
`=>\ B`

CORE, FUR1-NHT 2019 VCAA 14 MC

The time series plot below shows the daily number of visitors to a historical site over a two-week period.
 

This time series plot is to be smoothed using seven-median smoothing.

The smoothed number of visitors on day 4 is closest to

  1. 120
  2. 140
  3. 145
  4. 150
  5. 160
Show Answers Only

`D`

Show Worked Solution

`text{The seven data points about day 4 (ascending):}`

`120, 140, 147, 150, 160, 170, 183`

`:.\ text{7 median smoothing (Day 4)} = 150`

`=> D`

CORE, FUR1-NHT 2019 VCAA 13 MC

The association between amount of protein consumed (in grams/day) and family income (in dollars) is best displayed using

  1. a scatterplot.
  2. a time series plot.
  3. parallel boxplots.
  4. back-to-back stem plots.
  5. a two-way frequency table.
Show Answers Only

`A`

Show Worked Solution

`text(Both variables are numerical and continuous.)`

`:.\ text(A scatterplot is best.)`

`=>\ A`

CORE, FUR1-NHT 2019 VCAA 10 MC

A student uses the data in the table below to construct the scatterplot shown below.
 


 

A squared transformation is applied to the variable `x` to linearise the data.

A least squares line is fitted to this linearised data with `x^2` as the explanatory variable.

The equation of this least squares line is closest to

  1. `y = 94.1 - 12.3x^2`
  2. `y = 95.8 - 1.57x^2`
  3. `y = 8.76 - 0.0768x^2`
  4. `y = 107 - 111x^2`
  5. `y = 107 - 0.0768x^2`
Show Answers Only

`B`

Show Worked Solution

`text(Input all data points into CAS:)`

`y = 95.8 – 1.57x^2`

`=> B`

CORE, FUR1-NHT 2019 VCAA 9 MC

A least squares line of the form  `y = a + bx`  is fitted to a set of bivariate data for the variables `x` and `y`.

For this set of bivariate data,  `barx = 5.50`,  `bary = 5.60`,  `s_x = 3.03`,  `s_y =1.78`  and  `a = 3.1`

The slope of the least squares line, `b`, is closest to

  1.  0.44
  2.  0.45
  3.  0.58
  4.  0.59
  5.  0.76
Show Answers Only

`B`

Show Worked Solution
`bar y` `=a+b barx`
`5.60` `= 3.1 – b(5.50)`
`b` `= (5.60 – 3.1)/5.50`
  `= 04545…`

 
`=> B`

CORE, FUR1-NHT 2019 VCAA 8 MC

The variables recovery time after exercise (in minutes) and fitness level (below average, average, above average) are

  1. both numerical.
  2. both categorical.
  3. an ordinal variable and a nominal variable respectively.
  4. a numerical variable and a nominal variable respectively.
  5. a numerical variable and an ordinal variable respectively.
Show Answers Only

`E`

Show Worked Solution

`text{Recovery time in minutes → numerical variable}`

`text{Fitness level → ordinal (categories that can be ordered)}`

`=> E`

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`

CORE, FUR1-NHT 2019 VCAA 4 MC

The boxplot and dot plot shown below both display the distribution of the gestation period, in weeks, for 117 baby girls.
 


 

The median gestation period, in weeks, of these baby girls is

  1.  38
  2.  38.5
  3.  39
  4.  39.5
  5.  40
Show Answers Only

`E`

Show Worked Solution

`text(Median =)  (117 + 1)/2 = 54text(th datapoint)`

`:.\ text(Median = 40)`

`=>\ E`

Algebra, MET2-NHT 2019 VCAA 14 MC

If  `2log_e(x) - log_e(x + 2) = log_e(y)`, then  `x`  is equal to

  1.  `(y + sqrt{y^2 + 8y})/(2)`
  2.  `(y ± sqrt{y^2 + 8y})/(2)`
  3.  `(y ± sqrt{y^2 - 8y})/(2)`
  4.  `(-1 ± sqrt{4y - 7})/(2)`
  5.  `(-1 + sqrt{4y - 7})/(2)`
Show Answers Only

`A`

Show Worked Solution
`2log_e x – log_e(x + 2)` `= log_e y`
`log_e ({x^2}/{x + 2})`  `= log_e y`

 
`y = (x^2)/(x + 2)`
 
`text(S) text(olve for) \ x :`

`x = (y + sqrt(y^2 + 8y))/(2) \ \ text(only,) \ \ ((y – sqrt{y^2 + 8y})/(2) < 0) `
 
`=> \ A`

Calculus, MET2-NHT 2019 VCAA 13 MC

The graph of  `f(x) = x^3 - 6(b - 2)x^2 + 18x + 6`  has exactly two stationary points for

  1.  `1 < b < 2`
  2.  `b = 1`
  3.  `b = (4 ± sqrt6)/(2)`
  4.  `(4 - sqrt6)/(2) ≤  b ≤ (4 + sqrt6)/(2)`
  5.  `b < (4 - sqrt6)/(2) \ or \ b > (4 + sqrt6)/(2)`
Show Answers Only

`E`

Show Worked Solution
`f(x)` `= x^3 – 6(b – 2) x^2 + 18x + 6`
`f′(x)` `= 3x^3 – 12(b – 2) x + 18`
`Δ` `= [-12(b – 2)]^2 – 4 . 3 . 18`
  `= 144(b – 2)^2 – 216`

 
`text(If 2 solutions) \ => \ Δ > 0`

`text(S) text(olve for) \ b \ text{(by CAS):}`
 
`b < (4 – sqrt6)/(2) \ or \ b > (4 + sqrt6)/(2)`
 
`=> E`

Probability, MET2-NHT 2019 VCAA 10 MC

Let  `f`  be the probability density function  `f : [ 0, (2)/(3)] → R, \ f(x) = kx(2x + 1)(3x -2)(3x + 2)`.
The value of  `k`  is

  1.     `(308)/(405)`
  2.  `–(308)/(405)`
  3.  `–(405)/(308)`
  4.     `(405)/(308)`
  5.     `(960)/(133)`
Show Answers Only

`C`

Show Worked Solution

`text(Solve for)\ k :`

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

`k = -(405)/(308)`
 

`=> \ C`

Probability, MET2-NHT 2019 VCAA 9 MC

At the start of a particular week, Kim has three red apples and two green apples. She eats one apple everyday. On Monday, Tuesday and Wednesday of that week, she randomly selects an apple to eat. In this three-day period, the probability that Kim does not eat an apple of the same colour on any two consecutive days is

  1.  `(1)/(10)`
  2.  `(1)/(5)`
  3.  `(3)/(10)`
  4.  `(2)/(5)`
  5.  `(6)/(25)`
Show Answers Only

`C`

Show Worked Solution

`text{Pr (alternate colours)}`

`= text(Pr)(RGR) + text(Pr)(GRG)`

`= (3)/(5) ·(2)/(4) ·(2)/(3) + (2)/(5) ·(3)/(4) ·(1)/(3)`

`= (12)/(60) + (6)/(60)`

`= (3)/(10)`
 

`=> \ C`

Algebra, MET2-NHT 2019 VCAA 8 MC

The simultaneous linear equations  `2y + (m - 1) x = 2`  and  `my + 3x = k`  have infinitely many solutions for

  1.  `m = 3`  and  `k = –2`
  2.  `m = 3`  and  `k = 2`
  3.  `m = 3`  and  `k = 4`
  4.  `m = –2`  and  `k = –2`
  5.  `m = –2`  and  `k = 3`
Show Answers Only

`D`

Show Worked Solution
`2y + (m – 1)x` `= 2\ \ =>\ \ y= -((m-1)/(2)) x + 1`
`my + 3x` `= k\ \ =>\ \ y=-(3)/(m) x + (k)/(m)`

 

`text(Infinite solutions) \ => \ text(gradients and y-intercepts equal)`

`(m – 1)/(2)` `= (3)/(m)`
`m^2 – m – 6` `= 0`
`(m – 3)(m + 2)` `= 0`

 
`m = 3 \ \ text(or)\ \ –2`
 

`text(If) \ \ m = 3 ,`

`(k)/(3) = 1 \ => \ k = 3`
 
`text(If) \ \ m = –2,`

`(k)/(–2) = 1 \ => \ k = –2`
 

`=> \ D`

Algebra, MET2-NHT 2019 VCAA 6 MC

Let  `f : [0, ∞) → R, \ f(x) = x^2 + 1`.

The equation  `f(f(x)) = (185)/(16)`  has real solution(s)

  1.  `x = ± (sqrt13)/(4)`
  2.  `x = (sqrt13)/(4)`
  3.  `x = ± (sqrt13)/(2)`
  4.  `x = (3)/(2)`
  5.  `x = ± (3)/(2)`
Show Answers Only

`D`

Show Worked Solution
`f(f(x))` `= (x^2 + 1)^2 + 1`
`(185)/(16)` `= x^4 + 2x^2 + 2`
`0` `= x^4 + 2x^2 – (153)/(16)`

 
`text{Solve (by CAS):}`

`x = (3)/(2) , \ x ∈ [0, ∞)`

Probability, MET2-NHT 2019 VCAA 5 MC

Consider the probability distribution for the discrete random variable `X` shown in the table below.

The value of  `text(E)(X)`  is

  1.  `(76)/(65)`
  2.  `1`
  3.  `0`
  4.  `(2)/(13)`
  5.  `(86)/(65)`
Show Answers Only

`A`

Show Worked Solution
`1` `= b + b + b + (3)/(5) – b + (3b)/(5)`
`(2)/(5)` `= (13b)/(5)`
`b` `= (2)/(13)`

 

`text(E)(X)` `= -(2)/(13) + 0 + (2)/(13) + 2((3)/(5) – (2)/(13)) +3 ((6)/(65))`
  `= (76)/(65)`

Graphs, MET2-NHT 2019 VCAA 4 MC

The graph of the function  `ƒ : D → R, \ f(x) = (2x -3)/(4 + x)`, where `D` is the maximal domain, has asymptotes

  1.  `x = –4, \ y = 2`
  2.  `x = (3)/(2), \ y = –4`
  3.  `x = –4, \ y = (3)/(2)`
  4.  `x = (3)/(2), \ y = 2`
  5.  `x = 2, \ y = 1`
Show Answers Only

`A`

Show Worked Solution
`f(x)` `= (2x + 8 – 11)/(x +4)`
  `= 2 – (11)/(x + 4)`

 
`:. \ text(Asymptotes at) \ \ x = –4, y = 2 `

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))`

Calculus, 2ADV C4 2019 MET1 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.

Show that  `A(a) = 1/pi - 1/pi cos(a pi) - a sin (a pi)`.  (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

    `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)`

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)`

Functions, EXT1 F1 2019 MET1 5b

Given the function  `h(x) = sqrt(2x + 3) - 2`  for  `h>=-3/2`, find the inverse function  `h^(-1)`  and its domain.  (3 marks)

Show Answers Only

`x>=–2  or  [–2, oo)`

Show Worked Solution

 `y = sqrt (2x + 3) – 2`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= sqrt(2y + 3) – 2`
`sqrt(2y + 3)` `= x + 2`
`2y + 3` `= (x + 2)^2`
`y` `= 1/2(x + 2)^2 – 3/2`
`:. h^(-1)` `= 1/2(x + 2)^2 – 3/2`

 

`text(Domain)\ \ h^(-1)(x)` `= text(Range)\ \ h(x)`
  `= x>=–2  or [–2, oo)`

Calculus, 2ADV C3 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.  

Calculus, 2ADV C4 2019 MET1 2

Find  `f(x)`  given that  `f(1) = -7/4`  and  `f ^{\prime}(x) = 2x^2 - 1/4x^(-2/3)`.  (2 marks)

Show Answers Only

`f(x) = 2/3x^3 – 3/4x^(1/3) – 5/3`

Show Worked Solution
`f(x)` `= int 2x^2 – 1/4x^(-2/3) dx`
  `= 2/3x^3 – 1/4 ⋅ 1/(1/3) x^(1/3) + c`
  `= 2/3x^3 – 3/4x^(1/3) + c`

 
`text(Given)\ \ f(1) = -7/4:`

`-7/4` `= 2/3 – 3/4 + c`
`c` `= -5/3`
`:. f(x)` `= 2/3x^3 – 3/4x^(1/3) – 5/3`

Calculus, MET1-NHT 2019 VCAA 1b

Let  `f(x) = x^2 cos(3x)`.
 
Find  `f ^{\prime} (pi/3)`.   (2 marks)

Show Answers Only

`-(2pi)/3`

Show Worked Solution
  `f(x)` `= x^2 cos 3x`
  `f^{\prime}(x)` `= x^2 ⋅ 3(-sin 3x) + 2x cos 3x`
  `f^{\prime}(pi/3)` `= (pi/3)^2 ⋅ 3 (-sin pi) + 2 (pi/3) cos pi`
    `= -(2pi)/3`

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)`

Probability, MET1-NHT 2019 VCAA 6a

Jacinta tosses a coin five times.

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)

Show Answers Only

`1/2`

Show Worked Solution

`text(After 2 tosses, 2 heads.)`

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

Algebra, MET1-NHT 2019 VCAA 5b

Let  `h:[-3/2, oo) -> R,\ h(x) = sqrt(2x + 3)-2.`

Find the domain and the rule of the inverse function  `h^(-1)`.   (3 marks)

Show Answers Only

`[–2, oo)`

Show Worked Solution

 `y = sqrt (2x + 3)-2`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= sqrt(2y + 3)-2`
`sqrt(2y + 3)` `= x + 2`
`2y + 3` `= (x + 2)^2`
`y` `= 1/2(x + 2)^2 -3/2`
`:. h^(-1)` `= 1/2(x + 2)^2-3/2`

 

`text(Domain)\ \ h^(-1)(x)` `= text(Range)\ h(x)`
  `= [–2, oo)`

Calculus, MET1-NHT 2019 VCAA 4

A function `g` has rule  `g(x) = log_e (x-3) + 2`.

  1. State the maximal domain of `g` and the range of `g` over its maximal domain.   (2 marks)

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

     

    ii. On the axes below, sketch the graph of the function `g`, labelling any asymptote with its equation.

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

Show Answers Only
  1. `x in (3, oo)`

     

    `y in R`

  2. i.  `y = x-2`
    ii. 
     
Show Worked Solution
a.   `text(Domain)` `: \ x in (3, oo)`
  `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.  

Calculus, MET1-NHT 2019 VCAA 3

  1. Evaluate  `int_2^7 1/(x + sqrt 3)\ dx`  and  `int_2^7 1/(x-sqrt 3)\ dx`.   (2 marks)

  2. Show that  `1/2 (1/(x-sqrt 3) + 1/(x + sqrt 3)) = x/(x^2-3)`.   (1 mark)

  3. Use your answers to part a. and part b. to evaluate  `int_2^7 x/(x^2-3)\ dx`  in the form  `1/a log_e(b)`, where  `a` and `b` are positive integers.   (1 mark)

Show Answers Only
  1. `log_e((7-sqrt 3)/(2-sqrt 3))`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `1/2 log_e 46`
Show Worked Solution
a.   `int_2^7 1/(x + sqrt 3)\ dx` `= [log_e (x + sqrt 3)]_2^7`
    `= log_e(7 + sqrt 3)-log_e (2 + sqrt 3)`
    `= log_e ((7 + sqrt 3)/(2 + sqrt 3))`

 

`int_2^7 1/(x-sqrt 3)\ dx` `= [log_e (x – sqrt 3)]_2^7`
  `= log_e (7-sqrt 3)-log_e (2-sqrt 3)`
  `= log_e ((7-sqrt 3)/(2-sqrt 3))`

 

b.   `1/2(1/(x-sqrt 3) + 1/(x + sqrt 3))` `= 1/2 ((x + sqrt 3 + x-sqrt 3)/((x-sqrt 3)(x + sqrt 3)))`
    `= 1/2 ((2x)/(x^2-3))`
    `= x/(x^2-3)\ \ text(… as required)`

 

c.   `int_2^7 x/(x^2-3)` `= 1/2 int_2^7 1/(x-sqrt 3) + 1/(x + sqrt 3)\ dx`
    `= 1/2[log_e ((7-sqrt 3)/(2-sqrt 3)) + log_e ((7 + sqrt 3)/(2 + sqrt 3))]`
    `= 1/2 log_e (((7-sqrt 3)(7 + sqrt 3))/((2-sqrt 3)(2 + sqrt 3)))`
    `= 1/2 log_e ((49-3)/(4-3))`
    `= 1/2 log_e 46`

Calculus, MET1-NHT 2019 VCAA 2

Find  `f(x)`  given that  `f(1) = -7/4`  and  `f ^{\prime}(x) = 2x^2 - 1/4x^(-2/3)`.  (2 marks)

Show Answers Only

`f(x) = 2/3x^3 – 3/4x^(1/3) – 5/3`

Show Worked Solution
`f(x)` `= int 2x^2 – 1/4x^(-2/3) dx`
  `= 2/3x^3 – 1/4 ⋅ 1/(1/3) x^(1/3) + c`
  `= 2/3x^3 – 3/4x^(1/3) + c`

 
`text(Given)\ \ f(1) = -7/4:`

`-7/4` `= 2/3 – 3/4 + c`
`c` `= -5/3`
`:. f(x)` `= 2/3x^3 – 3/4x^(1/3) – 5/3`

Statistics, EXT1 S1 EQ-Bank 24

The proportion of a population that have brown eyes is 0.35.

Using  `overset^p`  to represent the sample proportion, calculate the smallest sample size, `n`, such that the standard deviation of  `overset^p`  is below 0.02.  (2 marks)

Show Answers Only

`569`

Show Worked Solution

`E(overset^p) = p = 0.35`

`text(Var)(overset^p) = (0.35(1 – 0.35))/n = 0.2275/n`

`sigma(overset^p) = sqrt(0.2275/n)`
 

`text(If)\ \ sigma(overset^p) < 0.02,`

`sqrt(0.2275/n)` `< 0.02`
`0.0004n` `> 0.0275`
`n` `> 0.2275/0.0004`
`n` `> 568.75`

 

`:. text(Smallest)\ \ n = 569`

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 22

It is known that 43% of voters in a city voted for Labor in the election.

If 250 voters from the city are selected at random, use a suitable approximation and the attached probability table to find the probability that at least 120 of those chosen voted for Labor.  (3 marks)

Show Answers Only

`0.0571`

Show Worked Solution

`E(overset^p) = p = 0.43`

`text(Var)(overset^p) = (0.43(1 – 0.43))/250 = 0.0009804`

`sigma(overset^p) = 0.0313`
 

`text(Let)\ \ X = text(number who voted Labor)`

`text(If)\ \ X = 120 \ => \ overset^p = 120/250 = 0.48`

`P(X >= 120)` `= P(z >= (0.48 – 0.43)/0.0313)`
  `= P(z >= 1.58)`
  `= 1 – 0.9429`
  `= 0.0571`

Statistics, EXT1 S1 EQ-Bank 20

Netball Australia records show that 10% of all registered players are over the age of 25.

  1. A random survey of 100 netball players was carried out to find out how many were over 25 years of age.

     

    Assuming the sample proportion is normally distributed, calculate the expected mean and standard deviation of this group.  (2 marks)

  2. Using the probability table attached, estimate the probability that at least 15 players surveyed will be over 25 years of age.  (2 marks)

Show Answers Only
  1. `0.03`
  2. `0.0475`
Show Worked Solution
i.   `E(hat p)` `= p = 0.1`
  `text(Var)(hat p)` `= (p(1 – p))/n`
    `= (0.1(0.9))/100`
    `= 0.0009`
  `sigma(hat p)` `= sqrt(0.0009)`
    `= 0.03`

 

ii.  `Ptext{(at least 15 players are over 25)} = P(hat p >= 0.15)`

`hat p\ ~\ N(mu,sigma)\ ~\ N(0.1, 0.03)`

`P(hat p >= 0.15)` `= P(z >= (0.15 – 0.10)/0.03)`
  `= P(z >= 1.67)`
  `= 1 – 0.9525`
  `= 0.0475`

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.)`

Combinatorics, EXT1 A1 EQ-Bank 14

A delivery company has 1095 packages to deliver on a given day.

It has 17 delivery vans that will deliver all packages. If one van delivers more packages than all other vans, the company pays the driver a $100 bonus.

What is the minimum number of packages a van could deliver and still win the $100 bonus.  (2 marks)

Show Answers Only

`66`

Show Worked Solution

`text(Pigeonholes)\ (k) = 17`

COMMENT: Note that “By PHP” refers to by pigeonhole principle.

`text(Pigeons)\ (n) = 1095`
 
`(n)/(k) = (1095)/(17) = 64\ text(remainder 7)`

`text(S)text(ince 7 vans could deliver 65 packages and the rest 64 packages)`

`=> \ text(By PHP, the minimum packages to win the $100 bonus = 66)`

Combinatorics, EXT1 A1 EQ-Bank 13

A sock drawer contains blue, white and green socks.

If individual socks are randomly chosen from the drawer, what is the minimum number that must be selected to ensure there are at least three pairs?   (2 marks)

Show Answers Only

`8`

Show Worked Solution

`text(Consider the worst-case scenario:)`

COMMENT: Note that “By PHP” refers to by pigeonhole principle.

`text{(i.e. the most socks chosen without 3 pairs)}`

`text(It is possible to choose 7 socks and only have 2 pairs)`

`=> \ text(5B, 1W, 1G  (2 pairs))`

`=> \ text(3B, 3W, 1G  (2 pairs))`
 
`text(Choosing the 8th sock produces 3 pairs in any scenario.)`

`:. \ text(By PHP, a minimum of 8 selections ensures 3 pairs.)`

Combinatorics, EXT1 A1 EQ-Bank 12

Eleven numbers are randomly chosen from the set of integers, `S`, where

`S = {1, 2, 3, 4, ..., 20}`

Prove that the sum of two of the eleven numbers randomly selected must equal 21.   (2 marks)

Show Answers Only

`text(Proof (Show Worked Solution))`

Show Worked Solution

`text(Rearranging)\ S\ text(into 10 pairs that sum to 21:)`

COMMENT: Note that “By PHP” refers to by pigeonhole principle.

`{1, 20}, {2, 19}, {3, 18}, …. , {10, 11}`
 
`text(Select one number from each pair)`

`=>\ text(10 numbers where two do not sum to 21`
 
`text(The 11th number chosen must complete a pair that sums to 21)`

`:. \ text(By PHP, two of the eleven numbers must sum to 21.)`

Combinatorics, EXT1 A1 SM-Bank 11

A multiple choice quiz asks students 4 questions. Each question has three possible answers, a, b or c, and students must attempt each question.

How many students must do the quiz to ensure that at least two sets of answers are identical?    (2 marks)

Show Answers Only

`82`

Show Worked Solution

`text(Total possible answer combinations)`

COMMENT: Note that “By PHP” refers to by pigeonhole principle.

`= 3 xx 3 xx 3 xx 3`

`= 81`
 

`:. \ text(By PHP, 82 students must do the quiz. `

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`

Statistics, EXT1 S1 EQ-Bank 13

On average, batsmen playing cricket in a T20 competition play a scoring shot two out of every three balls.

In a regular season, a total of 1200 overs that each contain six balls, are bowled.

Estimate how many overs would have at least five scoring shots.  (3 marks)

Show Answers Only

`421`

Show Worked Solution

`text(Let)\ \ X = text(number of scoring shots in a six ball over)`

`X\ ~\ text(Bin) (6, 2/3)`

`P(X >= 5)` `= P(X = 5) + P(X = 6)`
  `=\ ^6 C_5 ⋅ (2/3)^5 (1/3) + \ ^6 C_6 ⋅ (2/3)^6`
  `= 64/243 + 64/729`
  `= 256/729`

 

`text(Let)\ \ Y=\ text(number of overs with at least 5 scoring shots)`

`Y\ ~\ text(Bin)(1200, 256/729)`

`E(Y)` `=np`  
  `=1200 xx 256/729`  
  `=421.39…`  
  `=421\ \ text{(nearest over)}`