SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

MATRICES, FUR2 2017 VCAA 3

Senior students at a school choose one elective activity in each of the four terms in 2018.

Their choices are communication (`C`), investigation (`I`), problem-solving (`P`) and service (`S`).

The transition matrix `T` shows the way in which senior students are expected to change their choice of elective activity from term to term.
 

`{:(qquadqquadqquadqquadquadtext(this term)),(qquadqquadqquad\ CqquadquadIqquadquadPqquad\ S),(T = [(0.4,0.2,0.3,0.1),(0.2,0.4,0.1,0.3),(0.2,0.3,0.3,0.4),(0.2,0.1,0.3,0.2)]{:(C),(I),(P),(S):}qquadtext(next term)):}`
 

Let `S_n` be the state matrix for the number of senior students expected to choose each elective activity in Term `n`.

For the given matrix `S_1`, a matrix rule that can be used to predict the number of senior students in each elective activity in Terms 2, 3 and 4 is
 

`S_1 = [(300),(200),(200),(300)],qquadS_(n + 1) = TS_n`
 

  1. How many senior students will not change their elective activity from Term 1 to Term 2?   (1 mark)

  2. Complete `S_2`, the state matrix for Term 2, below.   (1 mark)


             

  3. Of the senior students expected to choose investigation (`I`) in Term 3, what percentage chose service (`S`) in Term 2?   (2 marks)

  4. What is the maximum number of senior students expected in investigation (`I`) at any time during 2018?   (1 mark)

Show Answers Only
  1. `320`
  2.  
    `S_2 = [(250),(250),(300),(200)]`
  3. `text(25%)`
  4. `250`
Show Worked Solution

a.   `text(Students who do not change)`

♦ Mean mark 47%.

`= 0.4 xx 300 + 0.4 xx 200 + 0.3 xx 200 + 0.2 xx 300`

`= 120 + 80 + 60 + 60`

`= 320`

 

b.    `S_2 = TS_1` `= [(0.4,0.2,0.3,0.1),(0.2,0.4,0.1,0.3),(0.2,0.3,0.3,0.4),(0.2,0.1,0.3,0.2)][(300),(200),(200),(300)]`
    `= [(250),(250),(300),(200)]`

 

c.    `S_3` `= TS_2`
    `= [(260),(240),(295),(205)]`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: A poorly understood and answered question worthy of careful attention.

`text(Number)\ (I)\ text(in Term 3 = 240)`

`text(Number)\ (S)\ text(in Term 2 = 200)`

`text(S)text(ince 30% move from)\ S\ text(to)\ I\ text(each term:)`

`text(Percentage)` `= (0.3 xx 200)/240`
  `= 60/240`
  `= 25text(%)`

 

d.    `S_4` `= TS_3`
    `= [(261),(239),(294.5),(205.5)]`

♦ Mean mark 43%.

`:. text(Max number of)\ (I)\ text(students is 250.)`

`(text(During term 2))`

CORE, FUR2 2017 VCAA 6

Alex sends a bill to his customers after repairs are completed.

If a customer does not pay the bill by the due date, interest is charged.

Alex charges interest after the due date at the rate of 1.5% per month on the amount of an unpaid bill.

The interest on this amount will compound monthly.

  1. Alex sent Marcus a bill of $200 for repairs to his car.

     

    Marcus paid the full amount one month after the due date.

     

    How much did Marcus pay?   (1 mark)

Alex sent Lily a bill of $428 for repairs to her car.

Lily did not pay the bill by the due date.

Let `A_n` be the amount of this bill `n` months after the due date.

  1. Write down a recurrence relation, in terms of `A_0`, `A_(n + 1)` and `A_n`, that models the amount of the bill.   (2 marks)

  2. Lily paid the full amount of her bill four months after the due date.

     

    How much interest was Lily charged?

     

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

Show Answers Only
  1. `$203`
  2. `A_o = 428,qquadA_(n + 1) = 1.015A_n`
  3. `$26.26\ \ (text(nearest cent))`
Show Worked Solution
a.    `text(Amount paid)` `= 200 + 200 xx 1.5text(%)`
    `= 1.015 xx 200`
    `= $203`

♦ Mean mark part (b) 47%.
MARKER’S COMMENT: A recurrence relation has the initial value written first. Know why  `A_n=428 xx 1.015^n`  is incorrect.

 

b.   `A_o = 428,qquadA_(n + 1) = 1.015A_n`

 

c.    `text(Total paid)\ (A_4)` `= 1.015^4 xx 428`
    `= $454.26`

♦♦ Mean mark part (c) 29%.

`:.\ text(Total Interest)` `= 454.26-428`
  `= $26.26\ \ (text(nearest cent))`

CORE, FUR2 2017 VCAA 4

The eggs laid by the female moths hatch and become caterpillars.

The following time series plot shows the total area, in hectares, of forest eaten by the caterpillars in a rural area during the period 1900 to 1980.

The data used to generate this plot is also given.
 

The association between area of forest eaten by the caterpillars and year is non-linear.

A log10 transformation can be applied to the variable area to linearise the data.

  1. When the equation of the least squares line that can be used to predict log10 (area) from year is determined, the slope of this line is approximately 0.0085385
  2. Round this value to three significant figures.   (1 mark)
  3. Perform the log10 transformation to the variable area and determine the equation of the least squares line that can be used to predict log10 (area) from year.
  4. Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.
  5. Round your answers to three significant figures.  (2 marks)

The least squares line predicts that the log10 (area) of forest eaten by the caterpillars by the year 2020 will be approximately 2.85

  1. Using this value of 2.85, calculate the expected area of forest that will be eaten by the caterpillars by the year 2020.
  2.  i. Round your answer to the nearest hectare.   (1 mark)

  3. ii. Give a reason why this prediction may have limited reliability.   (1 mark)

Show Answers Only

a.  `0.00854\ (text(3 sig fig))`

b.  `log_10(text(area)) = −14.4 + 0.000854 xx text(year)`

c.i.  `708\ text(hectares)`

c.ii. `text(This prediction extrapolates significantly from the given)`
        `text(data range and as a result, its reliability decreases.)`

Show Worked Solution

a.   `0.0085385 = 0.00854\ (text(3 sig fig))`

♦ Mean marks of part (a) and (b) 44%.

 

b.    `log_10(text(area))` `= −14.4 + 0.000854 xx text(year)`

 

♦♦ Mean mark part (c)(i) 29%.
COMMENT: When the question specifies using the value 2.85, use it!

c.i.    `log_10(text(Area))` `= 2.85`
  `:.\ text(Area)` `= 10^2.85`
    `= 707.94…`
    `= 708\ text(hectares)`

 

c.ii.   `text(This prediction extrapolates significantly from the given)`

  `text(data range and as a result, its reliability decreases.)`

Calculus, 2ADV C3 SM-Bank 7

The graph of  `f(x) = sqrt x (1 - x)`  for  `0<=x<=1`  is shown below.
 


 

  1. Calculate the area between the graph of  `f(x)` and the `x`-axis.  (2 marks)

  2. For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of  `f(x)`  is  `(1 - 3x)/(2 sqrt x)`.  (1 mark)

The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of  `f(x)`, and the line segment `AB`, which is part of the horizontal axis, as shown below.

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis.
 


 

  1. Find the equation of the line through `B` and `C` in the form  `y = mx + c`, for  `theta = 45^@`.  (3 marks)

Show Answers Only
  1. `4/15\ text(units)^2`
  2. `text(Proof)\ \ text{(See Workes Solutions)}`
  3. `y = -x + 1`
Show Worked Solution
i.   `text(Area)` `= int_0^1 (sqrt x – x sqrt x)\ dx`
    `= int_0^1 (x^(1/2) – x^(3/2))\ dx`
    `= [2/3 x^(3/2) – 2/5 x^(5/2)]_0^1`
    `= (2/3 – 2/5) – (0 – 0)`
    `= 10/15 – 6/15`
    `= 4/15\ text(units)^2`

 

ii.   `f (x)` `= x^(1/2) – x^(3/2)`
  `f prime (x)` `= 1/2 x^(-1/2) – 3/2 x^(1/2)`
    `= 1/(2 sqrt x) – (3 sqrt x)/2`
    `= (1 – 3x)/(2 sqrt x)\ \ text(.. as required.)`

 

iii.  `m_(AC) = tan 45^@=1`

♦♦♦ Mean mark (Vic) part (iii) 20%.
MARKER’S COMMENT: Most successful answers introduced a pronumeral such as  `a=sqrtx`  to solve.

`=> m_(BC) = -1\ \ (m_text(BC) _|_ m_(AC))`

 
`text(At point of tangency of)\ BC,\  f prime(x) = -1`

`(1 – 3x)/(2 sqrt x)` `=-1`
`1-3x` `=-2sqrtx`
`3x-2sqrt x-1` `=0`

 
`text(Let)\ \ a=sqrtx,`

`3a^2-2a-1` `=0`
`(3a+1)(a-1)` `=0`
`a=1 or -1/3`   
`:. sqrt x` `=1` `or`   `sqrt x=- 1/3\ \ text{(no solution)}`
`x` `=1`    

 
`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`
 

`text(Equation of)\ \ BC, \ m=-1, text{through (1,0):}`

`y-0` `=-1(x-1)`
`y` `=-x+1`

Algebra, MET2 2017 VCAA 4

Let  `f : R → R :\  f (x) = 2^(x + 1)-2`. Part of the graph of  `f` is shown below.
 

  1. The transformation  `T: R^2 -> R^2, \ T([(x),(y)]) = [(x),(y)] + [(c),(d)]`  maps the graph of  `y = 2^x`  onto the graph of  `f`.

     

    State the values of `c` and `d`.   (2 marks)

  2. Find the rule and domain for  `f^(-1)`, the inverse function of  `f`.   (2 marks)

  3. Find the area bounded by the graphs of  `f` and  `f^(-1)`.   (3 marks)

  4. Part of the graphs of  `f` and  `f^(-1)` are shown below.
     

         
     
    Find the gradient of  `f` and the gradient of  `f^(-1)`  at  `x = 0`.   (2 marks)

The functions of  `g_k`, where  `k ∈ R^+`, are defined with domain `R` such that  `g_k(x) = 2e^(kx)-2`.

  1. Find the value of `k` such that  `g_k(x) = f(x)`.  (1 mark)

  2. Find the rule for the inverse functions  `g_k^(-1)` of  `g_k`, where  `k ∈ R^+`.   (1 mark)

  3. i. Describe the transformation that maps the graph of  `g_1` onto the graph of  `g_k`.   (1 mark)

    ii. Describe the transformation that maps the graph of  `g_1^(-1)` onto the graph of  `g_k^(-1)`.   (1 mark)

  4. The lines `L_1` and `L_2` are the tangents at the origin to the graphs of  `g_k`  and  `g_k^(-1)`  respectively.
  5. Find the value(s) of `k` for which the angle between `L_1` and `L_2` is 30°.   (2 marks)

  6. Let `p` be the value of `k` for which  `g_k(x) = g_k^(−1)(x)`  has only one solution.
  7.  i. Find `p`.   (2 marks)

  8. ii. Let  `A(k)`  be the area bounded by the graphs of  `g_k`  and  `g_k^(-1)`  for all  `k > p`.
  9.     State the smallest value of `b` such that  `A(k) < b`.   (1 mark)

Show Answers Only

a.  `c =-1, \ d =-2`

b.  `f^(-1)(x) = log_2 (x + 2)-1, \ x ∈ (-2,∞)`

c.  `3-2/(log_e(2))\ text(units)²`

d. `f^{prime}(0)= 2log_e(2) and f^(-1)^{prime}(0)= 1/(2log_e(2))`

e.  `k = log_e(2)`

f.  `g_k^(-1)(x)= 1/k log_e((x + 2)/2)`

g.i.  `text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

g.ii.  `text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

h.  `k=sqrt3/6\ or\ sqrt3/2`

i.i.  `p = 1/2`

i.ii.  `b=4`

Show Worked Solution

a.   `text(Using the matrix transformation:)`

`x^{prime}` `=x+c\ \ => x=x^{prime}-c`
`y^{prime}` `=y+d\ \ =>y= y^{prime}-d`
   
`y^{prime}-d` `=2^((x)^{prime}-c)`
`y^{prime}` `= 2^((x)^{prime}-c)+d`

 

`:. c = -1, \ d = -2`

 

b.   `text(Let)\ \ y = f(x)`

`text(Inverse : swap)\ x\ ↔ \ y`

`x` `= 2^(y + 1)-2`
`x + 2` `= 2^(y + 1)`
`y + 1` `= log_2(x + 2)`
`y` `= log_2(x + 2)-1`

 

`text(dom)(f^(-1)) = text(ran)(f)`

`:. f^(-1)(x) = log_2 (x + 2)-1, \ x ∈ (-2,∞)`

 

c.   `text(Intersection points occur when)\ \ f(x)=f^(-1)(x)`

MARKER’S COMMENT: Specifically recommends using  `f^(-1)(x)-f(x)`  in this type of integral to avoid errors in stating the integral.

`x` `= -1, 0`

 

`text(Area)` `= int_(-1)^0 (f^(-1)(x)-f(x))\ dx`
  `= 3-2/(log_e(2))\ text(units)²`

 

d.    `f^{prime}(0)` `= 2log_e(2)`
  `f^{(-1)^prime}(0)` `= 1/(2log_e(2))`

 

e.   `g_k(x) = 2e^(kx)-2`

`text(Solve:)\ \ g_k(x) = f(x)quad text(for)\ k ∈ R^+`

`:. k = log_e(2)`

 

f.   `text(Let)\ \ y = g_k(x)`

`text(Inverse : swap)\ x\ text(and)\ y`

`x` `= 2e^(ky)-2`
`e^(ky)` `=(x+2)/2`
`ky` `=log_e((x+2)/2)`
`:. g_k^(-1)(x)` `= 1/k log_e((x + 2)/2)`

 

♦♦ Mean mark part (g)(i) 31%.

g.i.    `g_1(x)` `= 2e^x-2`
  `g_k(x)` `= 2e^(kx)-2`

 
`:. text(Dilation by factor of)\ 1/k\ text(from the)\ ytext(-axis)`

 

♦♦ Mean mark part (g)(ii) 30%.

g.ii.    `g_1^(-1)(x)` `= log_e((x + 2)/2)`
  `g_k^(-1)(x)` `= 1/klog_e((x + 2)/2)`

 
`:. text(Dilation by factor of)\ 1/k\ text(from the)\ xtext(-axis)`

 

h.   `text(When)\ \ x=0,`

♦♦♦ Mean mark part (h) 13%.

`g_k^{prime}(0)=2k\ \ => m_(L_1)=2k`

`g_k^{(-1)^prime}(0)=1/(2k)\ \ => m_(L_2)=1/(2k)`

 

`text(Using)\ \ tan 30^@=|(m_1-m_2)/(1+m_1m_2)|,`

`text(Solve:)\ \ 1/sqrt3` `=+- ((2k-1/(2k)))/2\ \ text(for)\ k`

 
`:. k=sqrt3/6\ or\ sqrt3/2`

 

i.i   `text(By inspection, graphs will touch once if at)\ \ x=0,`

♦♦♦ Mean mark part (i)(i) 4%.

`m_(L_1)` `=m_(L_2)`
`2k` `=1/(2k)`
`k` `=1/2,\ \ (k>0)`

 
`:. p = 1/2`

 

i.ii  `text(As)\ k→oo, text(the graph of)\ g_k\ text(approaches)\ \ x=0`

♦♦♦ Mean mark part (i)(ii) 2%.

`text{(vertically) and}\ \ y=-2\ \ text{(horizontally).}`

`text(Similarly,)\ \ g_k^(-1)\ \ text(approaches)\ \ x=-2`

`text{(vertically) and}\ \ y=0\ \ text{(horizontally).}`

 

`:. lim_(k→oo) A(k) = 4`

`:.b=4`

Probability, MET2 2017 VCAA 3

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

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

 

`f(t) = {{:(1/625 (t - 20)),(1/625 (70 - t)),(0):}qquad{:(20 <= t < 45),(45 <= t <= 70),(text(elsewhere)):}:}`

 

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


     

       
     

  2. Find  `text(Pr)(25 ≤ T ≤ 55)`.  (2 marks)

  3. Find  `text(Pr)(T ≤ 25 | T ≤ 55)`.  (2 marks)

  4. Find `a` such that  `text(Pr)(T ≥ a) = 0.7`, correct to four decimal places.  (2 marks)

  5. The probability that Jennifer spends more than 50 minutes on her homework on any given day is `8/25`. Assume that the amount of time spent on her homework on any day is independent of the time spent on her homework on any other day.

     

    1. Find the probability that Jennifer spends more than 50 minutes on her homework on more than three of seven randomly chosen days, correct to four decimal places.  (2 marks)

    2. Find the probability that Jennifer spends more than 50 minutes on her homework on at least two of seven randomly chosen days, given that she spends more than 50 minutes on her homework on at least one of those days, correct to four decimal places.  (2 marks)

Let `p` be the probability that on any given day Jennifer spends more than `d` minutes on her homework.

Let `q` be the probability that on two or three days out of seven randomly chosen days she spends more than `d` minutes on her homework.

  1. Express `q` as a polynomial in terms of `p`.  (2 marks)

    1. Find the maximum value of `q`, correct to four decimal places, and the value of `p` for which this maximum occurs, correct to four decimal places.  (2 marks)

    2. Find the value of `d` for which the maximum found in part g.i. occurs, correct to the nearest minute.  (2 marks)


Show Answers Only

  1.  

  2. `4/5`
  3. `1/41`
  4. `39.3649`
    1. `0.1534`
    2. `0.7626`
  6. `q =7p^2(1-p)^4(2p+3)`
    1. `p = 0.3539quadtext(and)quadq = 0.5665`
    2. `49\ text(min)`

Show Worked Solution

a.   

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

 

b.   `text(Pr)(25 <= T <= 55)`

`= int_25^45 1/625 (t – 20)\ dt + int_45^55 1/625 (70 – t)\ dt`

`= 4/5`

 

c.   `text(Pr)(T ≤ 25 | T ≤ 55)`

`=(text(Pr)(T <= 25))/(text(Pr)( T <= 55))`

`= (int_20^25 1/625(t – 20)\ dt)/(1 – int_55^70 1/625(70 – t)\ dt)`

`= (1/50)/(1 – 9/50)`

`= 1/41`

 

d.   `text(Pr)(T ≥ a) = 0.7`

♦ Mean mark part (d) 36%.

`=>\ text(Pr)(T <= a) = 0.3`

`text(Solve:)`

`int_20^a 1/625(t – 20)\ dt` `= 0.3quadtext(for)quada ∈ (20, 45)`

 

`:. a == 39.3649`

 

e.i.   `text(Let)\ X =\ text(Number of days Jenn studies more than 50 min)`

`X ~\ text(Bi) (7, 8/25)`

`text(Pr)(X >= 4) = 0.1534`

 

e.ii.    `text(Pr)(X >= 2 | X >= 1)` `= (text(Pr)(X >= 2))/(text(Pr)(X >= 1))`
    `= (0.7113…)/(0.9327…)`
    `= 0.7626\ \ text{(to 4 d.p.)}`

 

f.   `text(Let)\ Y =\ text(Number of days Jenn spends more than)\ d\ text(min)`

`Y ~\ text(Bi)(7,p)`

♦ Mean mark part (f) 36%.

`q` `= text(Pr)(Y = 2) + text(Pr)(Y = 3)`
  `= ((7),(2))p^2(1 – p)^5 + ((7),(3))p^3(1 – p)^4`
  `= 21p^2(1 – p)^5 + 35p^3(1 – p)^4`
   `=7p^2(1-p)^4[3(1-p)+5p]`
  `=7p^2(1-p)^4(2p+3)`

 

g.i.   `text(Solve)\ \ q′(p) = 0,`

♦♦ Mean mark part (g)(i) 30%.

`p` `=0.35388…`
  `=0.3539\ \ text{(to 4 d.p.)}`

`:. q_text(max)= 0.5665\ \ text{(to 4 d.p.)}`

 

g.ii.   `text(Pr)(T > d) = p= 0.35388…`

♦♦♦ Mean mark part (g)(ii) 8%.

  `text(Solve:)`

`int_d^70 (1/625(70 – t))dt` `= 0.35388… quadtext(for)quadd ∈ (45,70)`

 

`:. d` `=48.967…`
  `=49\ text(mins)`

Algebra, MET2 2017 VCAA 2

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

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


 

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

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

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

As the Ferris wheel rotates, a stationary boat at `B`, on a nearby river, first becomes visible at point `P_1`. `B` is 500 m horizontally from the vertical axis through the centre `C` of the Ferris wheel and angle `CBO = theta`, as shown below.
 

   
 

  1. Find `theta` in degrees, correct to two decimal places.   (1 mark)

Part of the path of `P` is given by  `y = sqrt(3025-x^2) + 65, x ∈ [-55,55]`, where `x` and `y` are in metres.

  1. Find `(dy)/(dx)`.   (1 mark)

As the Ferris wheel continues to rotate, the boat at `B` is no longer visible from the point  `P_2(u, v)` onwards. The line through `B` and `P_2` is tangent to the path of `P`, where angle `OBP_2 = alpha`.
 

   
 

  1. Find the gradient of the line segment `P_2B` in terms of `u` and, hence, find the coordinates of `P_2`, correct to two decimal places.   (3 marks)

  2. Find `alpha` in degrees, correct to two decimal places.   (1 mark)

  3. Hence or otherwise, find the length of time, to the nearest minute, during which the boat at `B` is visible.   (2 marks)

Show Answers Only
  1. `h_text(min) = 10\ text(m), h_text(max) = 120\ text(m)`
  2. `30\ text(min)`
  3. `t = 7.5`
  4. `7.41^@`
  5. `(-x)/(sqrt(3025-x^2))`
  6. `P_2(13.00, 118.44)`
  7. `13.67^@`
  8. `7\ text(min)`
Show Worked Solution
a.    `h_text(min)` `= 65-55` `h_text(max)` `= 65 + 55`
    `= 10\ text(m)`   `= 120\ text(m)`

 

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

 

c.   `h^{prime}(t) = (11pi)/3\ sin(pi/15 t)`

♦ Mean mark 50%.
MARKER’S COMMENT: A number of commons errors here – 2 answers given, calc not in radian mode, etc …

 

`text(Solve)\ h^{primeprime}(t) = 0, t ∈ (0,30)`

`t = 15/2\ \ text{(max)}`   `text(or)`   `t = 45/2\ \ text{(min – descending)}`

`:. t = 7.5`

 

d.   

♦ Mean mark 36%.
MARKER’S COMMENT: Choosing degrees vs radians in the correct context was critical here.

`tan(theta)` `= 65/500`
`:. theta` `=7.406…`
  `= 7.41^@`

 

e.    `(dy)/(dx)` `= (-x)/(sqrt(3025-x^2))`

 

f.   

`P_2(u,sqrt(3025-u^2) + 65),\ \ B(500,0)`

`:. m_(P_2B)` `= (sqrt(3025-u^2) + 65)/(u-500)`

 

`text{Using part (e), when}\ \ x=u,`

♦♦♦ Mean mark part (f) 18%.
MARKER’S COMMENT: Many students were unable to use the rise over run information to calculate the second gradient.

`dy/dx=(-u)/(sqrt(3025-u^2))`

 

`text{Solve (by CAS):}`

`(sqrt(3025-u^2) + 65)/(u-500)` `= (-u)/(sqrt(3025-u^2))\ \ text(for)\ u`

 

`u=12.9975…=13.00\ \ text{(2 d.p.)}`

 

`:. v` `= sqrt(3025-(12.9975…)^2) + 65`
  `= 118.4421…`
  `= 118.44\ \ text{(2 d.p.)}`

 

`:.P_2(13.00, 118.44)`

 

♦♦♦ Mean mark part (g) 7%.

g.    `tan alpha` `=v/(500-u)`
    `= (118.442…)/(500-12.9975…)`
  `:. alpha` `= 13.67^@\ \ text{(2 d.p.)}`

 

h.   

♦♦♦ Mean mark 5%.

`text(Find the rotation between)\ P_1 and P_2:`

`text(Rotation to)\ P_1 = 90-7.41=82.59^@`

`text(Rotation to)\ P_2 = 180-13.67=166.33^@`

`text(Rotation)\ \ P_1 → P_2 = 166.33-82.59 = 83.74^@`

 

`:.\ text(Time visible)` `= 83.74/360 xx 30\ text(min)`
  `=6.978…`
  `= 7\ text{min  (nearest degree)}`

Calculus, MET2 2017 VCAA 17 MC

The graph of a function  `f`, where  `f(−x) = f (x)`, is shown below.

The graph has `x`-intercepts at `(a, 0)`, `(b, 0)`, `(c, 0)` and `(d, 0)` only.

The area bound by the curve and the `x`-axis on the interval `[a, d]` is

  1. `int_a^d f(x)\ dx`
  2. `int_a^b f(x)\ dx - int_c^b f(x)\ dx + int_c^d f(x)\ dx`
  3. `2int_a^b f(x)\ dx + int_b^c f(x)\ dx`
  4. `2int_a^b f(x)\ dx - 2int_b^(b + c) f(x)\ dx`
  5. `int_a^b f(x)\ dx + int_c^b f(x)\ dx + int_d^c f(x)\ dx`
Show Answers Only

`D`

Show Worked Solution

`text(S)text(ince)\ \ f(x)\ \ text(is an odd function,)`

♦♦♦ Mean mark 21%.

`=> b=-c`

`:. 2int_a^b f(x)\ dx – 2int_b^(b + c) f(x)\ dx`

`= 2int_a^b f(x)\ dx – 2int_b^(0) f(x)\ dx`

`=> D`

Calculus, MET1 2017 VCAA 9

The graph of  `f: [0, 1] -> R,\ f(x) = sqrt x (1-x)`  is shown below.
 

  1. Calculate the area between the graph of `f` and the `x`-axis.   (2 marks)

  2. For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of `f` is  `(1-3x)/(2 sqrt x)`.   (1 mark)

The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of `f`, and the line segment `AB`, which is part of the horizontal axis, as shown below.

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis, where  `45^@ <= theta < 90^@`.

  1. Find the equation of the line through `B` and `C` in the form  `y = mx + c`, for  `theta = 45^@`.   (2 marks)

  2. Find the coordinates of `C` when  `theta = 45^@`.   (4 marks)

Show Answers Only
  1. `4/15\ text(units)^2`
  2. `text(Proof)\ \ text{(See Workes Solutions)}`
  3. `y = -x + 1`
  4. `C (11/27, 16/27)`
Show Worked Solution
a.   `text(Area)` `= int_0^1 (sqrt x-x sqrt x)\ dx`
    `= int_0^1 (x^(1/2)-x^(3/2))\ dx`
    `= [2/3 x^(3/2)-2/5 x^(5/2)]_0^1`
    `= (2/3-2/5)-(0-0)`
    `= 10/15-6/15`
    `= 4/15\ text(units)^2`

 

♦♦ Mean mark part (b) 35%.
MARKER’S COMMENT: Establishing the common denominator in the working was required!
b.   `f (x)` `= x^(1/2)-x^(3/2)`
  `f^{′}(x)` `= 1/2 x^(-1/2)-3/2 x^(1/2)`
    `= 1/(2 sqrt x)-(3 sqrt x)/2`
    `= (1-3x)/(2 sqrt x)\ \ text(.. as required.)`

 

c.  `m_(AC) = tan 45^@=1`

♦♦♦ Mean mark part (c) 20%.
MARKER’S COMMENT: Most successful answers introduced a pronumeral such as `a=sqrtx` to solve.

`=> m_(BC) = -1\ \ (m_text(BC) _|_ m_(AC))`

 

`text(At point of tangency of)\ BC,\  f^{prime}(x) = -1`

`(1-3x)/(2 sqrt x)` `=-1`
`1-3x` `=-2sqrtx`
`3x-2sqrt x-1` `=0`

 

`text(Let)\ \ a=sqrtx,`

`3a^2-2a-1` `=0`
`(3a+1)(a-1)` `=0`
`a=1 or -1/3`   
`:. sqrt x` `=1` `or`   `sqrt x=- 1/3\ \ text{(no solution)}`
`x` `=1`    

 
`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`
 

`:.\ text(Equation of)\ \ BC, \ m=-1, text{through (1,0) is:}`

`y-0` `=-1(x-1)`
`y` `=-x+1`

 

d.  `text(Find Equation)\ AC:`

♦♦♦ Mean mark part (d) 17%.

`m_(AC) =1`

`text(At point of tangency of)\ AC,\  f^{prime}(x) = 1`

`(1-3x)/(2 sqrt x)` `=1`
`1-3x` `=2sqrtx`
`3x+2sqrt x-1` `=0`
`(3 sqrtx-1)(sqrtx+1)` `=0`
   
`:. sqrt x` `=1/3` `or`   `sqrt x=-1\ \ text{(no solution)}`
`x` `=1/9`    

 
`f(1/9)=sqrt(1/9)(1-1/9)=1/3 xx 8/9 = 8/27\ \ =>P(1/9,8/27)`
 

`:.\ text(Equation of)\ AC, m=1, text(through)\ \ P\ \ text(is):`

`y-8/27` `= 1 (x-1/9)`
`y` `= x + 5/27`

 
`C\ text(is at intersection of)\ AC and CB:`

`-x + 1` `= x + 5/27`
`2x` `= 22/27`
`:. x` `= 11/27`
`y` `= -11/27 + 1 = 16/27`

 
`:. C (11/27, 16/27)`

Functions, MET1 2017 VCAA 7

Let  `f: [0, oo) -> R,\ f(x) = sqrt(x + 1)`.

  1.  State the range of `f`.   (1 mark)

  2.  Let  `g: (-oo, c] -> R,\ \ g(x) = x^2 + 4x + 3`.
    1. Find the largest possible value of `c` such that the range of `g` is a subset of the domain of `f`.   (2 marks)

    2. For the value of `c` found in part b.i., state the range of `f(g(x))`.   (1 mark) 

  3. Let  `h: R -> R,\ \ h(x) = x^2 + 3`.
  4. State the range of `f(h(x))`.   (1 mark)

Show Answers Only
  1. `[1, oo)`
    1. `-3`
    2. `[1, oo)`
  2. `[2, oo)`
Show Worked Solution

a.  `text(Sketch of)\ \ f(x):`
 

`:.\ text(Range)\ \ (f) = [1, oo)`

 

b.i.  `text(Sketch)\ \ g(x) = (x + 1) (x + 3)`

♦ Mean mark 38%.
 


 

`text(Domain of)\ \ f(x)=[0,oo)`

`text(Find domain of)\ \ g(x)\ \ text(such that Range)\ (g) = [0, oo)`

`text(Graphically, this occurs when)\ \ g(x)\ \ text(has domain:)`

`=> x ∈ (–oo, –3] ∪ [–1,oo)`

`:. c = -3`
 

b.ii.  `text(Range)\ g(x) = [0, oo) = text(Domain)\ \ f(x)`

♦♦♦ Mean mark 20%.

`:.\ text(Range)\ \ f(g(x)) = [1, oo)`
 

c.  `text(Range)\ h(x) = [3, oo)`

♦♦ Mean mark 30%.
`f(3)` `= sqrt (3 + 1)`
  `= sqrt 4`
  `= 2`

 

`:.\ text(Range)\ \ f(h(x)) = [2, oo)`

Financial Maths, STD2 F1 SM-Bank 5

Alex is buying a used car which has a sale price of  $13 380. In addition to the sale price there are the following costs:

2014 27a1

  1. Stamp Duty for this car is calculated at $3 for every $100, or part thereof, of the sale price.  
    Calculate the Stamp Duty payable.   (1 mark)

  2. Alex wishes to take out comprehensive insurance for the car for 12 months.

     

    The cost of comprehensive insurance is calculated using the following:
     
          2014 27a2
    Find the total amount that Alex will need to pay for comprehensive insurance.   (3 marks)

  3. Alex has decided he will take out the comprehensive car insurance rather than the less expensive non-compulsory third-party car insurance.

  4.  

    What extra cover is provided by the comprehensive car insurance?   (1 mark)

Show Answers Only
  1. `$402`
  2. `$985.74`
  3. `text(Comprehensive insurance covers Alex)`
  4. `text(for damage done to his own car as well.)`
Show Worked Solution
♦♦♦ Mean mark 12%
IMPORTANT: “or part thereof ..” in the question requires students to round up to 134 to get the right multiple of $3 for their calculation.
a.    `($13\ 380)/100 = 133.8`
`:.\ text(Stamp duty)` `= 134 xx $3`
  `= $402`

 

b.   `text(Base rate) = $845`

 
`text(FSL) =\ text(1%) xx 845 = $8.45`
 

`text(Stamp)` `=\ text(5.5%) xx(845 + 8.45)`
  `= 46.9397…`
  `= $46.94\ text{(nearest cent)}`

 

`text(GST)` `= 10 text(%) xx(845 + 8.45)`
  `= 85.345`
  `= $85.35`

 

`:.\ text(Total cost)` `= 845 + 8.45 + 46.94 + 85.35`
  `= $985.74`

 

♦ Mean mark 34%.
c.   `text(Comprehensive insurance covers Alex)`
  `text(for damage done to his own car as well.)`

Measurement, STD2 M1 2017 HSC 30e

A solid is made up of a sphere sitting partially inside a cone.

The sphere, centre `O`, has a radius of 4 cm and sits 2 cm inside the cone. The solid has a total height of 15 cm. The solid and its cross-section are shown.
 


 

Using the formula  `V=1/3 pi r^2h`  where `r`  is the radius of the cone's circular base and `h` is the perpendicular height of the cone, find the volume of the cone, correct to the nearest cm³?  (3 marks)

Show Answers Only

`113\ text{cm}^3`

Show Worked Solution

`V = 1/3 xx text(base of cone × height)`

`text(Consider the circular base area of the cone,)`

`text(Find)\ x\ \ text{(using Pythagoras):}`

`x^2` `= 4^2-2^2 = 16-4 = 12`
`x` `= sqrt12\ text(cm)`

 

`:. V` `= 1/3 xx pi xx (sqrt12)^2 xx (15-6)`
  `= 1/3 xx pi xx 12 xx 9`
  `= 113.097…`
  `= 113\ text{cm}^3\ text{(nearest cm}^3 text{)}`

Proof, EXT2 P2 2017 HSC 16c

A 2 by `n` grid is made up of two rows of `n` square tiles, as shown.
 

The tiles of the 2 by `n` grid are to be painted so that tiles sharing an edge are painted using different colours. There are `x` different colours available, where  `x ≥ 2`.

It is NOT necessary to use all the colours.

Consider the case of the 2 by 2 grid with tiles labelled A, B, C and D, as shown.
 

There are `x(x - 1)` ways to choose colours for the first column containing tiles A and B. Do NOT prove this.

  1. Assume the colours for tiles A and B have been chosen. There are two cases to consider when choosing colours for the second column. Either tile C is the same colour as tile B, or tile C is a different colour from tile B.

     

    By considering these two cases, show that the number of ways of choosing colours for the second column is  `x^2 - 3x +3`.  (2 marks)

  2. Prove by mathematical induction that the number of ways in which the 2 by `n` grid can be painted is  `x(x - 1)(x^2 - 3x + 3)^(n - 1)`, for  `n ≥ 1`.  (2 marks)

  3. In how many ways can a 2 by 5 grid be painted if 3 colours are available and each colour must now be used at least once?  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `480`
Show Worked Solution
i.   

`text(If Colour)\ C =\ text(Colour)\ B:`

♦♦♦ Mean mark 20%.

`text(Column 2 combinations)`

`= 1 xx (x – 1)`

`= x – 1`

 

`text(If Colour)\ C !=\ text(Colour)\ B:`

`text(Column 2 combinations)`

`= (x – 2)(x – 2)`

 

`:.\ text(Total number of ways for column 2)`

`= x – 1 + (x – 2)^2`

`= x^2 – 3x + 3\ \ …\ text(as required.)`

 

ii.   `P(n) = x(x – 1)(x^2 – 3x + 3)^(n – 1)`

♦♦♦ Mean mark 22%.

`text(If)\ \ n = 1\ \ (text(i.e. 1 column)),`

`text(Combinations) = x(x – 1)\ \ (text(given))`

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

`:. text(True for)\ n = 1`

 

`text(Assume true for)\ n = k:`

`text(i.e. Possible combinations for)\ k\ text(columns)`

`P(k) = x(x – 1)(x^2 – 3x + 3)^(k – 1)`

`text(Prove true for)\ \ n = k + 1`

`text(i.e.)\ \ P(k + 1) = x(x – 1)(x^2 – 3x + 3)^k`

 

`text(Consider a grid)\ 2 xx k\ text(and add an extra two)`

`text(tiles to make a)\ 2 xx (k + 1)\ text(grid.)`

`text(Total possible combinations)`

`= text(combinations of)\ (2 xx k)\ text(grid) xx (x^2 – 3x + 3)\ \ (text{from (i)})`

`= x(x – 1)(x^2 – 3x + 3)^(k – 1) xx (x^2 – 3x + 3)`

`= x(x – 1)(x^2 – 3x + 3)^k`

`=> text(True for)\ \ n = k + 1`

`:. text(S)text(ince true for)\ n = 1,\ text(by PMI, true for integral)\ n >= 1.`

 

iii.   `text(If)\ \ n = 5, \ x = 3:`

♦♦♦ Mean mark 22%.

`text{Total combinations (includes using only 2 colours)}`

`P(5)` `= 3(3 – 1)(3^2 – 3.3 + 3)^4`
  `= 3(2)(3)^4`
  `= 486`

 

`text(Consider pattern when only 2 colours used:)`

`text(Combinations that only use 2 colours)`

`=\ text(Colours in box 1 × possible colours in box 2)`

`= 3 xx 2`

`= 6`

`:.\ text(Combinations if each colour used at least once)`

`= 486 – 6`
`= 480`

Harder Ext1 Topics, EXT2 2017 HSC 16a

  Let  `alpha = costheta + i sintheta`, where  `0 < theta < 2pi`.

  1. Show that  `alpha^k + alpha^(−k) = 2 cos ktheta`, for any integer `k`.  (1 mark)
  3. Let  `C = alpha^(−n) + … + alpha^(−1) + 1 + alpha + … + alpha^n`, where `n` is a positive integer.
  4. By summing the series, prove that  

    `C = (alpha^n + alpha^(−n) - (a^(n + 1) + alpha^(−(n + 1))))/((1 - alpha)(1 - baralpha))`.  (3 marks)

  6. Deduce, from parts (i) and (ii), that`1 + 2(costheta + cos2theta + … + cosntheta) = (cosntheta - cos(n + 1)theta)/(1 - costheta)`.  (2 marks)

  7. Show that  
    `cos\ pi/n + cos\ (2pi)/n + …  + cos\ (npi)/n`  is independent of `n`.  (1 mark)

 

 

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

(i)   `alpha = costheta – isintheta`

`text(Using de Moivre:)`

`alpha^k` `= cos(ktheta) + isin(ktheta)`
`alpha^(−k)` `= cos(−ktheta) + isin(−ktheta)`
  `= cos(ktheta) – isin(ktheta)`
`:. alpha^k + alpha^(−k)` `= cos(ktheta) + isin(ktheta) + cos(ktheta) – isin(ktheta)`
  `= 2cos(ktheta)\ …\ \ text(as required.)`

 

(ii)   `C = alpha^(−n) + … + alpha^(−1) + 1 + alpha + … + alpha^n`

♦♦ Mean mark 34%.

`=> text(GP where)\ \ a = alpha^(−n), r = alpha`

`=> text(Number of terms) = 2n + 1`

`:. C` `= (a(1 – r^n))/(1 – r)`
  `= (alpha^(−n)(1 – alpha^(2n + 1)))/(1 – alpha)`
  `= (alpha^(−n)(1 – alpha^(2n + 1)))/(1 – alpha) xx (1 – baralpha)/(1 – baralpha)\ \ \ \ text{(where}\ baralpha = alpha^(−1) text{)}`
  `= ((alpha^(−n) – alpha^(n + 1))(1 – alpha^(−1)))/((1 – alpha)(1 – baralpha))`
  `= (alpha^(−n) – alpha^(−n – 1) – alpha^(n + 1) + alpha^n)/((1 – alpha)(1 – baralpha))`
  `= (alpha^n + alpha^(−n) – (alpha^(n + 1) + alpha^(−(n + 1))))/((1 – alpha)(1 – baralpha))\ …\ \ text(as required.)`
♦ Mean mark part (iii) 48%.

 

(iii)    `C` `= alpha^(−n) + a^(−n + 1) … + alpha^(−1) + 1 + alpha + … + alpha^n `
    `= 1 + (alpha + alpha^(−1)) + (alpha^2 + alpha^(−2)) + … + (alpha^n + alpha^(−n))`
    `= 1 + 2costheta + 2cos2theta + … + 2cosntheta\ \ \ \ (text{using part (i)})`
    `= 1 + 2(costheta + 2cos2theta + … + cosntheta)`

 

`text{Also, using part (ii)}`

`C` `= (alpha^n + alpha^(−n) – (alpha^(n + 1) + alpha^(−(n + 1))))/((1 – alpha)(1 – baralpha))`
  `= (2cosntheta – 2cos(n + 1)theta)/(1 – baralpha – alpha + 1)\ \ (text{part (i)})`
  `= (2cosntheta – 2cos(n + 1)theta)/(2 – (alpha + alpha^(−1)))`
  `= (2cosntheta – 2cos(n + 1)theta)/(2 – 2costheta)`
  `= (cosntheta – cos(n + 1)theta)/(1 – costheta)`

`:. 1 + 2(costheta, cos2theta + … + cosntheta) = (cosntheta – cos(n + 1)theta)/(1 – costheta)`

♦♦♦ Mean mark part (iv) 28%.

 

(iv)   `text{Rearrange the result from (iii):}`

`costheta + cos2theta + … + cosntheta = 1/2((cosntheta – cos(n + 1)theta)/(1 – costheta) – 1)`

`text(Let)\ \ theta = pi/n`

`:. cos\ pi/n + cos\ (2pi)/n + … + cos\ (npi)/n`

`= 1/2((cos((npi)/n) – cos(((n + 1)pi)/n))/(1 – cos(pi/n)) – 1)`
`= 1/2((cospi – cos(pi + pi/n))/(1 – cos(pi/n)) – 1)`
`= 1/2((−(1 – cos(pi/n)))/(1 – cos(pi/n)) – 1)`
`= 1/2 (−1 – 1)`
`= −1`

 

`=> text(which is independent of)\ n.`

Conics, EXT2 2017 HSC 15c

The ellipse with equation  `(x^2)/(a^2) + (y^2)/(b^2) = 1`, where  `a > b`, has eccentricity `e`.

The hyperbola with equation  `(x^2)/(c^2) - (y^2)/(d^2) = 1`, has eccentricity `E`.

The value of `c` is chosen so that the hyperbola and the ellipse meet at  `P(x_1, y_1)`, as shown in the diagram.

  1. Show that  
  2. `(x_1^(\ 2))/(y_1^(\ 2)) = (a^2c^2)/((a^2 - c^2)) xx ((b^2 + d^2))/(b^2d^2)`.  (2 marks)
  4. If the two conics have the same foci, show that their tangents at `P` are perpendicular.  (3 marks)

 

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

(i)   `text(At)\ \ P(x_1,y_1):`

`(x_1^(\ 2))/(a^2) + (y_1^(\ 2))/(b^2)` `= (x_1^(\ 2))/(c^2) – (y_1^(\ 2))/(d^2) = 1`
`(x_1^(\ 2))/(a^2) – (x_1^(\ 2))/(c^2)` `= −(y_1^(\ 2))/(b^2) – (y_1^(\ 2))/(d^2)`
`x_1^(\ 2)(1/(a^2) – 1/(c^2))` `= −y_1^(\ 2)(1/(b^2) + 1/(d^2))`
`x_1^(\ 2)((c^2 – a^2)/(a^2c^2))` `= −y_1^(\ 2)((d^2 + b^2)/(b^2d^2))`
`(x_1^(\ 2))/(y_1^(\ 2))` `= −((a^2c^2)/(c^2 – a^2))((b^2 + d^2)/(b^2d^2))`
  `= (a^2c^2)/((a^2 – c^2)) xx ((b^2 + d^2))/(b^2d^2)\ …\ text(as required).`

 

(ii)   `text(Find gradient of ellipse at)\ \ P(x_1,y_1):`

♦ Mean mark 51%.
`(x^2)/(a^2) + (y^2)/(b^2)` `= 1`
`(2x)/(a^2) + (2y)/(b^2)·(dy)/(dx)` `= 0`
`(dy)/(dx)` `= −x/(a^2) xx (b^2)/y`

`text(At)\ P,\ \ m = (−x_1b^2)/(y_1a^2)`

 

`text(Similarly for gradient of hyperbola at)\ \ P(x_1,y_1):`

`(x^2)/(c^2) – (y^2)/(d^2)` `= 1`
`(2x)/(c^2) – (2y)/(d^2)·(dy)/(dx)` `= 0`
`(dy)/(dx)` `= x/(c^2) xx (d^2)/y`

`text(At)\ P,\ \ m = (x_1d^2)/(y_1c^2)`

`:. m_1m_2` `= −(x_1b^2)/(y_1a^2) xx (x_1d^2)/(y_1c^2)`
  `= −(x_1^2)/(y_1^2) xx (b^2d^2)/(a^2c^2)`

 

`text{Rearranging part (i)}`

`(x_1^(\ 2))/(y_1^(\ 2))` `= ((b^2 + d^2))/((a^2 – c^2)) xx (a^2c^2)/(b^2d^2)`
`(a^2c^2)/(b^2d^2)` `= (x_1^(\ 2))/(y_1^(\ 2))·((a^2 – c^2))/((b^2 + d^2))`
`(b^2d^2)/(a^2c^2)` `= (y_1^(\ 2)(b^2 + d^2))/(x_1^(\ 2)(a^2 – c^2))`
`:. m_1m_2` `= −(x_1^(\ 2))/(y_1^(\ 2)) xx (y_1^(\ 2)(b^2 + d^2))/(x_1^(\ 2)(a^2 – c^2))`
  `= −((b^2 + d^2))/(a^2 – c^2)\ \ …(1)`

 

`text(S)text{ince conics have same foci (given),}`

`ae` `= cE`    
`a^2e^2` `= c^2E^2`    
`a^2 – b^2` `= c^2 + d^2` `\ \ \ text{(using}` `\ \ b^2=a^2(1-e^2) and` 
`a^2 – c^2` `= b^2 + d^2`   `\ \ d^2=c^2(E^2-1) text{)}`

 

`text{Substituting into (1)}`

`m_1m_2` `= −((b^2 + d^2))/(b^2 + d^2)`
  `= −1`

 

`:.\ text(The two tangents are perpendicular.)`

MATRICES, FUR1 2017 VCAA 8 MC

Consider the matrix recurrence relation below.
 

`S_0 = [(40),(15),(20)], \ S_(n + 1) = TS_n`     where `T = [(0.3,0.2,V),(0.2,0.2,W),(X,Y,Z)]`
 

Matrix `T` is a regular transition matrix.

Given the above and that  `S_1 = [(29),(13),(33)]`, which of the following expressions is not true?

  1. `W > Z`
  2. `Y > X`
  3. `V > Y`
  4. `V + W + Z = 1`
  5. `X + Y + Z > 1`
Show Answers Only

`A`

Show Worked Solution

`TS_0 = S_1`

`[(0.3,0.2,V),(0.2,0.2,W),(X,Y,Z)][(40),(15),(20)] = [(29),(13),(33)]`

`(0.3 xx 40) + (0.2 xx 15) + 20V` `= 29`
`20V` `= 14`
`V` `= 0.7`

`text(Similarly)`

`8 + 3 + 20W` `= 13`
`20W` `= 2`
`W` `= 0.1`

 

`text(S)text(ince each column sums to 1:)`

`X` `= 1 – (0.3 + 0.2) = 0.5`
`Y` `= 1 – (0.2 + 0.2) = 0.6`
`Z` `= 1 – (0.7 + 0.1) = 0.2`

 

`:. W > Z\ \ text(is not true.)`

`=> A`

MATRICES, FUR1 2017 VCAA 7 MC

At a fish farm:

    • young fish (`Y`) may eventually grow into juveniles (`J`) or they may die (`D`)
    •  juveniles (`J`) may eventually grow into adults (`A`) or they may die (`D`)
    • adults (`A`) eventually die (`D`).

The initial state of this population, `F_0`, is shown below.
 

`F_0 = [(50\ 000),(10\ 000),(7000),(0)]{:(Y),(J),(A),(D):}`

 

Every month, fish are either sold or bought so that the number of young, juvenile and adult fish in the farm remains constant.

The population of fish in the fish farm after `n` months, `F_n`, can be determined by the recurrence rule
 

`F_(n + 1) = [(0.65,0,0,0),(0.25,0.75,0,0),(0,0.20,0.95,0),(0.10,0.05,0.05,1)]\ F_n + B`
 

where `B` is a column matrix that shows the number of young, juvenile and adult fish bought or sold each month and the number of dead fish that are removed.

Each month, the fish farm will

  1. sell 1650 adult fish.
  2. buy 1750 adult fish.
  3. sell 17 500 young fish.
  4. buy 50 000 young fish.
  5. buy 10 000 juvenile fish.
Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince the number of each type of fish remains constant,)`

`F_0 = F_1 = F_n = F_(n + 1)`

 

`[(50\ 000),(10\ 000),(7000),(0)]` `= [(0.65,0,0,0),(0.25,0.75,0,0),(0,0.20,0.95,0),(0.10,0.05,0.05,1)][(50\ 000),(10\ 000),(7000),(0)] + B`
`:. B` `= [(50\ 000),(10\ 000),(7000),(0)] – [(32\ 500),(20\ 000),(8650),(5850)]`
  `= [(17\ 500),(−10\ 000),(−1650),(−5850)]{:(Y),(J),(A),(D):}`

 
`=> A`

GRAPHS, FUR1 2017 VCAA 8 MC

The shaded area in the graph below shows the feasible region for a linear programming problem.

The objective function is given by

`Z = mx + ny`

Which one of the following statements is not true?

  1. When `m = 4` and `n = 1`, the minimum value of `Z` is at point `A`.
  2. When `m = 1` and `n = 6`, the maximum value of `Z` is at point `B`.
  3. When `m = 2` and `n = 5`, the minimum value of `Z` is at point `C`.
  4. When `m = 2` and `n = 6`, the maximum value of `Z` is at point `D`.
  5. When `m = 12` and `n = 1`, the maximum value of `Z` is at point `E`.
Show Answers Only

`D`

Show Worked Solution

`text(By trial and error: Consider option)\ D`

`Z` `= 2x + 6y`
`6y` `= −2x + Z`
`y` `= −1/3x + Z/6`

 

`text(By applying the sliding rule technique for function)`

`text(with gradient of)\ −1/3, text(the maximum value)`

`text(occurs at point)\ B,\ text(not point)\ D.`

`=> D`

NETWORKS, FUR1 2017 VCAA 8 MC

The flow of oil through a series of pipelines, in litres per minute, is shown in the network below.
 

 
The weightings of three of the edges are labelled `x`.

Five cuts labelled A–E are shown on the network.

The maximum flow of oil from the source to the sink, in litres per minute, is given by the capacity of

  1. `text(Cut A if)\ x = 1`
  2. `text(Cut B if)\ x = 2`
  3. `text(Cut C if)\ x = 2`
  4. `text(Cut D if)\ x = 3`
  5. `text(Cut E if)\ x = 3`
Show Answers Only

`B`

Show Worked Solution

`=> B`

GEOMETRY, FUR1 2017 VCAA 8 MC

Three circles of radius 50 mm are placed so that they just touch each other.
The region enclosed by the circles is shaded in the diagram below.

The area of the shaded region, in square millimetres, is closest to

  1.   `403`
  2.   `436`
  3. `1309`
  4. `2844`
  5. `4330`
Show Answers Only

`A`

Show Worked Solution

`text(Connect the centres of each circle to form)`

`text(an equilateral triangle.)`

`text(Shaded area)` `=\ text(Area of triangle − Area of sectors)`
  `=1/2 ab sinC – 3 xx (60/360 xx pi r^2)`
  `= 1/2 xx 100^2 xx sin60° – 3 xx (60/360 xx pi xx 50^2)`
  `= 403.13…`

`=> A`

GEOMETRY, FUR1 2017 VCAA 7 MC

A triangle `ABC` has:

• one side, `bar(AB)`, of length 4 cm
• one side, `bar(BC)`, of length 7 cm
• one angle, `∠ACB`, of 26°.

Which one of the following angles, correct to the nearest degree, could not be another angle in triangle `ABC`?

  1.   `24°`
  2.   `50°`
  3. `104°`
  4. `130°`
  5. `144°`
Show Answers Only

`E`

Show Worked Solution

`text(Using the sine rule,)`

`(sin x)/7` `= (sin 26)/4`
`sin x` `= (7 xx sin 26)/4`
  `= 0.767…`
   
`:. x` `= 50° text{or 130°  (nearest degree)}`

`text(If)\ \ x = 50°,\ text(other angle = 104°)`

`text(If)\ \ x = 130°,\ text(other angle = 24°)`

`=> E`

CORE, FUR1 2017 VCAA 24 MC

Xavier borrowed $245 000 to pay for a house.

For the first 10 years of the loan, the interest rate was 4.35% per annum, compounding monthly.

Xavier made monthly repayments of $1800.

After 10 years, the interest rate changed.

If Xavier now makes monthly repayments of $2000, he could repay the loan in a further five years.

The new annual interest rate for Xavier’s loan is closest to

  1.  0.35%
  2.  4.1%
  3.  4.5%
  4.  4.8%
  5.  18.7%
Show Answers Only

`B`

Show Worked Solution

`text(Find principal left after 10 years:)`

`text(By TVM Solver,)`

`N` `= 10 xx 12 = 120`
`I(%)` `= 4.35`
`PV` `= 245\ 000`
`PMT` `= −1800`
`FV` `= ?`
`text(P/Y)` `= text(C/Y) = 12`

 
`=> FV = −108\ 219.1611`
 

`text(Loan can be repaid in 5 years at $2000/month.)`

`text(Find interest rate by TVM Solver;)`

`N` `= 5 xx 12 = 60`
`I(%)` `= ?`
`PV` `= 108\ 219.16`
`PMT` `= −2000`
`FV` `= 0`
`text(P/Y)` `= text(C/Y) = 12`

 

`=> I(%)` `= 4.1427…`
  `= 4.14text{%  (2 d.p.)}`

`=> B`

Statistics, NAP-I3-CA09

The graph shows the origin and type of all vehicles in a city.
 

 
Which statement is most accurate based on the graph?

 
 There are more utility vehicles than trucks and vans.
 
 Utility vehicles are the most common type of vehicles
 
 There are more Australian vehicles than European vehicles.
 
 There are more Asian vehicles than European vehicles.
Show Answers Only

`text(There are more Asian vehicles)`

`text(than European vehicles.)`

Show Worked Solution

`text(Consider the 4th option:)`

`text(Total Asian vehicles)`

`=5+7+7=19`

`text(Total European vehicles)`

`=4+3+3=10`

`:.\ text(There are more Asian vehicles than European vehicles.)`

Statistics, NAP-A3-NC09

This graph shows the UV index for the town of Coonamble during one day.
 

 
When was the UV index always in the high range?

 
 between 8 am and 6 pm
 
 between 11 am and 12 pm
 
 after 2 pm
 
 between 2 pm and 4 pm

​

Show Answers Only

`text(between 11 am and 12 pm)`

Show Worked Solution

`text(UV index is on high)`

`text(– between 11 am – 12 pm and)`

`text(– between 2 pm – 3 pm)`

`:.\ text(between 11 am and 12 pm)`

Probability, NAP-A3-NC08

Ryan has white and black marbles in his bag.

If he chooses one without looking he is likely, but not certain, to get a white marble.

Which is Ryan's bag?

 
 
 
 

​

Show Answers Only

Show Worked Solution

`text(The 1st option gives a)\ \ 3/5\ \ text(chance for a)`

`text{white marble (likely but not certain).}`

Algebra, NAP-A3-NC05

Flynn is making a pattern with soccer balls.
 

 
This table shows the number of balls he needs for each shape in the pattern.
 

  Pattern # 1 2 3 4 5
  Number of balls 1 4 9 16 ?

 
How many balls will Flynn need for Pattern #5?

`17` `23` `25` `30`
 
 
 
 
Show Answers Only

`25`

Show Worked Solution

`text(Solution 1)`

`text(By drawing the next shape, pattern #5 is)`

`:. 25\ text(balls needed for Pattern #5.)`

 

`text(Solution 2)`

`text(The table shows a pattern where the balls in each)`

`text(shape are the square of the pattern number:)`

`:.\ text(Balls in pattern #5) = 5^2 = 25`

Statistics, NAP-A3-CA04

This table summarises the time Tutty spent training her parrot over five days.
 

 
What was the average (mean) time for training the parrot each day?

`text(30 minutes)` `text(56 minutes)` `text(66 minutes)` `text(280 minutes)`
 
 
 
 
Show Answers Only

`text(56 minutes)`

Show Worked Solution
`text(Average)` `= (25 + 55 + 60 + 94 + 46)/5`
  `= 280/5`
  `= 56\ text(minutes)`

Algebra, NAP-A4-NC01

Flynn is making a pattern with soccer balls.
 

 
This table shows the number of balls he needs for each shape in the pattern.
 

  Pattern # 1 2 3 4 5
  Number of balls 1 4 9 16 ?

 
How many balls will Flynn need for Pattern #5?

`17` `23` `25` `30`
 
 
 
 
Show Answers Only

`25`

Show Worked Solution

`text(Pattern #5 is)`

`:. 25\ text(balls needed for Pattern #5.)`

Statistics, NAP-A4-CA04

This table summarises the time Tutty spent training her parrot over five days.
 

 
What was the average (mean) time for training the parrot each day?

`text(30 minutes)` `text(56 minutes)` `text(66 minutes)` `text(280 minutes)`
 
 
 
 
Show Answers Only

`text(56 minutes)`

Show Worked Solution
`text(Average)` `= (25 + 55 + 60 + 94 + 46)/5`
  `= 280/5`
  `= 56\ text(minutes)`

Statistics, NAP-A4-CA03

A dealership sells new and used cars.

The graph shows the price of 2 similar cars and their age in years
 

 
Which one of these statements is true?

 
 Car Q is older and less expensive than Car P.
 
 Car P is newer and less expensive than Car Q.
 
 Car P is older and more expensive than Car Q.
 
 Car Q is newer and more expensive than Car P.
Show Answers Only

`text(Car Q is older and less expensive than Car P.)`

Show Worked Solution

`text(Car)\ Q\ text(is further right on)\ xtext(-axis)`

`=> Q\ text(is older)`

`text(Car)\ Q\ text(is lower on)\ ytext(-axis)`

`=> Q\ text(is less expensive)`

 

`:. text(Car)\ Q\ text(is older and less expensive than Car)\ P.`

Number, NAP-A4-CA01

At 7 am the temperature in Merewether was 15.9°C.

At midday it was 12.8°C warmer.

At 7 pm it was 13.9°C cooler than at midday.

What was the temperature at 7 pm?

`14.8^@text(C)` `15^@text(C)` `17^@text(C)` `42.6^@text(C)`
 
 
 
 
Show Answers Only

`14.8^@text(C)`

Show Worked Solution

`text(Temperature at 6 pm)`

`= 15.9 + 12.8 – 13.9`

`= 14.8^@text(C)`

Number, NAP-B4-CA04

Shelly and Carly collect dolls.

The ratio of the number of dolls Shelly owns compared to Carly is  3 : 2.

Shelley owns 12 dolls.

How many dolls does Carly own?

`2` `6` `8` `18`
 
 
 
 
Show Answers Only

`8`

Show Worked Solution

`text(Ratio 3 : 2)`

`text(Let)\ \ x=\ text(Number of Carly’s dolls)`

`x/12` `= 2/3`
`x` `= (2 xx 12)/3`
  `= 8\ text(dolls)`

Quadratic, EXT1 2017 HSC 14b

Let  `P(2p, p^2)`  be a point on the parabola  `x^2 = 4y`.

The tangent to the parabola at `P` meets the parabola  `x^2 = −4ay`, `a > 0`, at `Q` and `R`. Let `M` be the midpoint of `QR`.

  1. Show that the `x` coordinates of `R` and `Q` satisfy
  2. `qquadx^2 + 4apx - 4ap^2 = 0`.  (2 marks)

  3. Show that the coordination of `M` are  `(−2ap, −p^2(2a + 1))`.  (2 marks)
  4. Find the value of `a` so that the point `M` always lies on the parabola  `x^2 = −4y`.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `1 + sqrt2, a > 0`
Show Worked Solution

(i)   `x^2 = 4y,\ \ =>y = (x^2)/4`

`(dy)/(dx) = x/2`

`text(At)\ P(2p, p^2),`

`(dy)/(dx) = p`

`text(Equation of tangent:)`

`y = px – p^2`

 

`R and Q\ text(at intersection)`

`y` `= px – p^2\ …\ (1)`
`y` `= −(x^2)/(4a)\ …\ (2)`

 

`text(Subtract)\ (1) – (2)`

`px – p^2 + (x^2)/(4a)` `= 0`
`x^2 + 4apx – 4ap^2` `= 0\ …\ text(as required)`

 

(ii)   `x^2 + 4apx – 4ap^2 = 0`

♦♦ Mean mark 30%.
`x=` `\ (−4ap ± sqrt((4ap)^2 – 4 · 1 · (−4ap^2)))/2`
`=`  `\ (−4ap ± sqrt(16ap^2(a + 1)))/2`
`=`  `\ −2ap ± 2psqrt(a(a + 1))`

 

`=> xtext(-coordinate of)\ M\ text(is)\ −2ap.`

 

`M\ text(lies on tangent)\ \ y = px – p^2`

`:.y` `= p(−2ap) – p^2`
  `= −2ap^2 – p^2`
  `= −p^2(2a + 1)`

`:. M\ text(has coordinates)\ \ (−2ap, −p^2(2a + 1))`

 

(iii)   `text(If)\ M\ text(always lies on)\ \ x^2 = −4y`

♦♦ Mean mark 23%.
`(−2ap)^2` `= −4(−p^2(2a + 1))`
`4a^2p^2` `= 4p^2(2a + 1)`
`a^2` `= 2a + 1`
`a^2 – 2a – 1` `= 0`
`:. a` `= (+2 ± sqrt(4 + 4 · 1· 1))/2`
  `= 1 ± sqrt2`
  `= 1 + sqrt2, \ a > 0`

Mechanics, EXT2* M1 2017 HSC 13c

A golfer hits a golf ball with initial speed `V\ text(ms)^(−1)` at an angle `theta` to the horizontal. The golf ball is hit from one side of a lake and must have a horizontal range of 100 m or more to avoid landing in the lake.
 

     

Neglecting the effects of air resistance, the equations describing the motion of the ball are

`x = Vt costheta`

`y = Vt sintheta - 1/2 g t^2`,

where `t` is the time in seconds after the ball is hit and `g` is the acceleration due to gravity in `text(ms)^(−2)`. Do NOT prove these equations.

  1. Show that the horizontal range of the golf ball is
     
         `(V^2sin 2theta)/g` metres.  (2 marks)

  2. Show that if  `V^2 < 100 g`  then the horizontal range of the ball is less than 100 m.  (1 mark)

It is now given that  `V^2 = 200 g`  and that the horizontal range of the ball is 100 m or more.

  1. Show that  `pi/12 <= theta <= (5pi)/12`.  (2 marks)

  2. Find the greatest height the ball can achieve.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `text(See Worked Solution)`
  4. `25(2 – sqrt 3)\ text(metres)`
Show Worked Solution

i.   `text(Find)\ \ t\ \ text(when)\ \ y = 0:`

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

 

`text(Horizontal range)\ (x)\ text(when)\ \ t = (2Vsintheta)/g :`

`x` `= V · (2Vsintheta)/g costheta`
  `= (V^2 2sintheta costheta)/g`
  `= (V^2sin2theta)/g\ … text(as required)`

 

ii.   `text(If)\ \ V^2 < 100 g`

♦ Mean mark 44%.
`x` `< (100 g sin2theta)/g`
`x` `< 100 sin2theta`

 

`text(S)text(ince)\ −1 <= 2theta <= 1,`

`x < 100\ text(metres)`

 

iii.   `V^2 = 200g,\ \ x >= 100`

`(200 g · sin2theta)/g` `>= 100`
`sin2theta` `>= 1/2`

`:. pi/6 <= 2theta <= (5pi)/6`

`:. pi/12 <= theta <= (5pi)/12\ …\ text(as required)`

 

iv.   `text(Max height occurs when)`

♦♦ Mean mark 35%.
`t` `= 1/2 xx text(time of flight)`
  `= (Vsintheta)/g`

 
`text(Find)\ \ y\ \ text(when)\ \ t = (Vsintheta)/g`

`y` `= V · (Vsintheta)/g · sintheta – 1/2 g ((Vsintheta)/g)^2`
  `= (V^2 sin^2theta)/g – 1/2 · (V^2 sin^2 theta)/g`
  `= (V^2 sin^2theta)/(2g)`

 

`text(Max height when)\ theta = (5pi)/12\ (text(steepest angle)), V^2 = 200 g\ (text(given))`

`y_text(max)` `= (200 g · sin^2 ((5pi)/12))/(2g)`
  `= 100 sin^2 ((5pi)/12)`
  `= 50(1 – cos((5pi)/6))`
  `= 50(1 + sqrt3/2)`
  `= 25(2 + sqrt 3)\ text(metres)`

Plane Geometry, 2UA 2017 HSC 16c

In the triangle `ABC`, the point `M` is the mid-point of `BC`. The point `D` lies on `AB` and

`BD = DA + AC`.

The line that passes through the point `C` and is parallel to `MD` meets `BA` produced at `E`.

Copy or trace this diagram into your writing booklet.

  1. Prove that `Delta ACE` is isosceles.  (3 marks)
  2. The point `F` is chosen on `BC` so that `AF` is parallel to `DM`.
    `qquad` 
    Show that `AF` bisects `/_ BAC`.  (2 marks)

 

 

 

 

 

 

 

 

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
(i)  

`text(Prove)\ Delta ACE\ text(is isosceles)`

♦♦♦ Mean mark 21%.

`text(S) text(ince)\ DM text(||) EC,`

`∠BDM = ∠BEC\ \ text{(corresponding angles)}`

`∠DBM\ \ text(is common)`

`:. Delta BDM\ text(|||)\ Delta BEC\ \ text{(AAA)}`

`text(Using ratios of similar)\ Delta text(s):`

`(BM)/(BD)` `= (BC)/(BE)`
`(BM)/(BC)` `= (BD)/(BE) = 1/2\ \ text{(}M\ text(is midpoint of)\ BC text{)}`

 

`=> D\ text(is the midpoint of)\ BE.`

`=> BD = DE`

`AE` `= DE – DA`
`AC` `= BD – DA\ text{(given)}`
  `= DE – DA`
`:.  AE` `= AC`

`:. Delta ACE\ text(is isosceles)`

 

(ii)  

`text(Show)\ AF\ text(bisects)\ /_ BAC`

♦♦ Mean mark 31%.

`/_ BAF = /_ AEC\ \ \ text{(corresponding angles)}`

`text{From part (i)}`

`/_ AEC` `= /_ ECA\ \ \ text{(}Delta AEC\ text{is isosceles)}`
`/_ FAC` `= /_ ECA\ \ \ text{(alternate angles)}`
`:. /_ BAF` `= /_ FAC`

`:. AF\ text(bisects)\  /_ BAC\ \ \ text(… as required)`

Calculus, 2ADV C3 2017 HSC 16a

John’s home is at point `A` and his school is at point `B`. A straight river runs nearby.

The point on the river closest to `A` is point `C`, which is 5 km from `A`.

The point on the river closest to `B` is point `D`, which is 7 km from `B`.

The distance from `C` to `D` is 9 km.

To get some exercise, John cycles from home directly to point `E` on the river, `x` km from `C`, before cycling directly to school at `B`, as shown in the diagram.
 


  

The total distance John cycles from home to school is `L` km.

  1. Show that  `L = sqrt (x^2 + 25) + sqrt (49 + (9 - x)^2)`.   (1 mark)

  2. Show that if  `(dL)/(dx) = 0`, then  `sin alpha = sin beta`.   (3 marks)

  3. Find the value of `x` that makes  `sin alpha = sin beta`.   (2 marks)

  4. Explain why this value of `x` gives a minimum for `L`.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `45/12`
  4. `text(See Worked Solutions)`
Show Worked Solution
i.  

`text(Using Pythagoras:)`

`L` `= AE + EB`
  `= sqrt (5^2 + x^2) + sqrt (7^2 + (9 – x)^2)`
  `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)\ text(… as required)`

 

ii.  `text(From diagram):`

♦ Mean mark 41%.

`sin alpha = x/sqrt(25 + x^2) and sin beta = (9 – x)/sqrt(49 + (9 – x)^2)`

`L` `= sqrt(25 + x^2) + sqrt (49 + (9 – x)^2)`
`(dL)/(dx)` `= (2x)/sqrt(25 + x^2) – (2(9 – x))/sqrt(49 + (9 – x)^2)`

 

`text(If)\ \ (dL)/(dx) = 0,`

`=> (2x)/sqrt(25 + x^2)` `= (2(9 – x))/sqrt(49 + (9 – x)^2)`
`x/sqrt(25 + x^2)` `= (9 – x)/sqrt(49 + (9 – x)^2)`
`sin alpha` `= sin beta\ text(… as required)`

 

iii.  `text(If)\ sin alpha = sin beta,\ text(then)\ alpha = beta and`

♦♦ Mean mark 29%.

`Delta ACE\ text(|||)\ Delta BDE`

`text(Using corresponding sides of similar triangles:)`

`x/5` `= (9 – x)/7`
`7x` `= 45 – 5x`
`12x` `= 45`
`:. x` `= 45/12\ text(km)`
♦♦♦ Mean mark 9%.

 

iv.  

`text(If point)\ B\ text(is reflected across the)`

`text(river),\ AEB\ text(will be a straight line.)`

`text(If any other point is chosen,)\ AEB`

`text(would not be straight and the distance)`

`text(would be longer.)`

Measurement, 2UG 2017 HSC 30e

A solid is made up of a sphere sitting partially inside a cone.

The sphere, centre `O`, has a radius of 4 cm and sits 2 cm inside the cone. The solid has a total height of 15 cm. The solid and its cross-section are shown.
 


 

What is the volume of the cone, correct to the nearest cm³?  (3 marks)

Show Answers Only

`113\ text{cm³  (nearest cm³)}`

Show Worked Solution

`V = 1/3 xx text(base of cone × height)`

`text(Consider the base area of the cone,)`

♦♦ Mean mark 34%.

`text(Need to find)\ x:`

`x^2` `= 4^2 – 2^2 = 16 – 4 = 12`
`x` `= sqrt12\ text(cm)`

 

`:. V` `= 1/3 xx pi xx (sqrt12)^2 xx (15 – 6)`
  `= 1/3 xx pi xx 12 xx 9`
  `= 113.097…`
  `= 113\ text{cm³  (nearest cm³)}`

Statistics, STD2 S5 2017 HSC 29d

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.
 


 

  1. Find the median test mark.  (1 mark)

  2. The mean test mark is 5.4. The standard deviation of the test marks is 4.22.

     

    Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean.  (2 marks)

  3. A student states that for any data set, 68% of the scores should lie within one standard deviation of the mean. With reference to the dot plot, explain why the student’s statement is NOT relevant in this context.  (1 mark)

Show Answers Only
  1. `6`
  2. `text(43%)`
  3. `text(The statement assumes the data is normally)`
    `text(distributed which is not the case here.)`
Show Worked Solution
♦ Mean mark 50%.
i.    `text(Median)` `= text(15th + 16th score)/2`
    `= (4 + 8)/2`
    `= 6`

 

ii.   `text(Lower limit) = 5.4 – 4.22 = 1.18`

♦♦ Mean mark 34%.

`text(Upper limit) = 5.4 + 4.22 = 9.62`

`:.\ text(Percentage in between)`

`= 13/30 xx 100`

`= 43.33…`

`= 43text{%  (nearest %)}`

 

iii.   `text(The statement assumes the data is normally)`

♦♦♦ Mean mark 13%.

`text(distributed which is not the case here.)`

Measurement, 2UG 2017 HSC 29a

A new 200-metre long dam is to be built.
The plan for the new dam shows evenly spaced cross-sectional areas.
 


 

  1. Using TWO applications of Simpson’s rule, show that the volume of the dam is approximately  44 333 m³.  (2 marks)
  2. It is known that the catchment area for this dam is 2 km².
    Calculate how much rainfall is needed, to the nearest mm, to fill the dam.  (2 marks)

 

Show Answers Only
  1. `text(See Worked Solution)`
  2. `22\ text{mm  (nearest mm)}`
Show Worked Solution
(i)    `text(Volume)` `~~ h/3 (y_0 + 4y_1 + y_2)\ \ …\ text(applied twice)`
    `~~ 50/3(0 + 4 xx 140 + 270) + 50/3(270 + 4 xx 300 + 360)`
    `~~ 50/3(830) + 50/3(1830)`
    `~~ 44\ 333\ text(m³)\ …\ \ text(as required)`

 

(ii)   `text(Convert 2km² to m²:)`

♦♦♦ Mean mark 11%.
`text(2 km²)` `= 2 xx 1000 xx 1000`
  `= 2\ 000\ 000\ text(m²)`
`:.\ text(Rainfall needed)` `= (44\ 333)/(2\ 000\ 000)`
  `= 0.0221…\ text(m)`
  `= 22.1…\ text(mm)`
  `= 22\ text{mm  (nearest mm)}`

Algebra, 2UG 2017 HSC 28a

Temperature can be measured in degrees Celsius (`C`) or degrees Fahrenheit (`F`).

The two temperature scales are related by the equation  `F = (9C)/5 + 32`.

  1. Calculate the temperature in degrees Fahrenheit when it is  −20 degrees Celsius.  (1 mark)
  2. Solve the following equations simultaneously, using either the substitution method or the elimination method.  (2 marks)
    `qquadF = (9C)/5 + 32`

    `qquadF = C`

     

  3. The graphs of  `F = (9C)/5 + 32`  and  `F = C`  are shown below.
  4.  


  5. What does the result from part (ii) mean in the context of the graph?  (1 mark)     

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Show Answers Only
  1. `−4^@F`
  2. `C = −40, F = −40`
  3. `text(It means the two graphs intersect)`
    `text(at)\ (−40,−40).`

 

 

Show Worked Solution
(i)   `F` `= (9(−20))/5 + 32`
    `= −4^@F`

 

(ii)   `F` `= (9C)/5 + 32` `…\ (1)`
  `F` `= C` `…\ (2)`

 

♦♦ Mean mark 31%.
MARKER’S COMMENT: An area that requires attention.

`text(Substitute)\ \ F = C\ \ text{from (2) into (1)}`

`C` `= (9C)/5 + 32`
`(9C)/5 – C` `= −32`
`(4C)/5 – C` `= −32`
`C` `= −32 xx 5/4 = −40`

 

`text{From (2),}`

`F = −40`

`text{(i.e. when}\ C = −40, F = −40)`

♦♦♦ Mean mark 20%.

 

(iii)   `text(It means the two graphs intersect)`

`text{at (−40,−40).}`

Probability, STD2 S2 2017 HSC 24 MC

A deck of 52 playing cards contains 12 picture cards. Two cards from the deck are drawn at random and placed on a table.

What is the probability, correct to four decimal places, that exactly one picture card is on the table?

A.     `0.0498`

B.     `0.1810`

C.     `0.3550`

D.     `0.3620`

Show Answers Only

`text(D)`

Show Worked Solution

`P(text(exactly 1 picture card))`

`= P(text(picture)) xx P(text(no picture)) + P(text(no picture)) xx P(text(picture))`

`= 12/52 xx 40/51quad+quad40/52 xx 12/51`

♦♦♦ Mean mark 21%.

`= 2((12 xx 40)/(52 xx 51))`

`= 0.36199…`

`=>\ text(D)`

FS Comm, 2UG 2017 HSC 23 MC

How many bits are there in 2 terabytes?

A.     `2^40`

B.     `2^41`

C.     `2^43`

D.     `2^44`

Show Answers Only

`text(D)`

Show Worked Solution

`text(Bytes in 2 terabytes)`

♦♦ Mean mark 34%.

`= 2 xx 2^40\ text(bytes)`

`:.\ text(Bits)` `= 8 xx 2 xx 2^40`
  `= 16 xx 2^40`
  `= 2^4 xx 2^40`
  `= 2^44\ text(bits)`

`=>\ text(D)`

Number and Algebra, NAP-J2-19

A company has 3706 computers in a warehouse.

Another 423 computers are delivered to the warehouse.

Which of these could be used to calculate the correct number of computers in the warehouse?

 
 3000 + 4000 + 700 + 20 + 6 + 3
 
 3000 + 700 + 400 + 20 + 60 + 3
 
 3000 + 700 + 400 + 20 + 6 + 3
 
 3000 + 400 + 70 + 20 + 6 + 3
Show Answers Only

`3000 + 700 + 400 + 20 + 6 + 3`

Show Worked Solution

`3706 + 423`

`= 3000 + 700 + 6 + 400 + 20 + 3`

`=3000 + 700 + 400 + 20 + 6 + 3`