SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

CORE, FUR2 SM-Bank VCE 4

Damon runs a swim school.

The value of his pool pump is depreciated over time using flat rate depreciation.

Damon purchased the pool pump for $28 000 and its value in dollars after `n` years, `P_n`, is modelled by the recursion equation below:

`P_0 = 28\ 000,qquad P_(n + 1) = P_n - 3500`

  1. Write down calculations, using the recurrence relation, to find the pool pump's value after 3 years.   (1 mark)

  2. After how many years will the pump's depreciated value reduce to $7000?   (1 mark)

The reducing balance depreciation method can also be used by Damon.

Using this method, the value of the pump is depreciated by 15% each year.

A recursion relation that models its value in dollars after `n` years, `P_n`, is:

`P_0 = 28\ 000, qquad P_(n + 1) = 0.85P_n`

  1. After how many years does the reducing balance method first give the pump a higher valuation than the flat rate method in part (a)?   (2 marks)

Show Answers Only
  1. `$17\ 500`
  2. `6\ text(years)`
  3. `4\ text(years)`
Show Worked Solution
a.    `P_1` `= 28\ 000-3500 = 24\ 500`
  `P_2` `= 24\ 500-3500 = 21\ 000`
  `P_3` `= 21\ 000-3500 = 17\ 500`

  
`:.\ text(After 3 years, the pump’s value is $17 500.)`
  

b.   `text(Find)\ n\ text(such that:)`

`7000` `= 28\ 000-3500n`
`3500n` `= 21\ 000`
`n` `= (21\ 000)/3500`
  `= 6\ text(years)`

  
c.   `text(Using the reducing balance method)`

`P_1` `= 0.85 xx 28\ 000 = 23\ 800`
`P_2` `= 0.85 xx 23\ 800 = 20\ 230`
`P_3` `= 0.85 xx 20\ 230 = 17\ 195`
`P_4` `= 0.85 xx 17\ 195 = 14\ 615.75`

  
`text{Using the flat rate method (see part (a))}`

`P_4 = 17\ 500-3500 = 14\ 000`

`14\ 615.75 > 14\ 000`
  

`:.\ text(After 4 years, the reducing balance method)`

`text(first values the pump higher.)`

Calculus, MET2 2016 VCAA 4

  1. Express  `(2x + 1)/(x + 2)`  in the form  `a + b/(x + 2)`,  where `a` and `b` are non-zero integers.   (2 marks)

  2. Let  `f: R text(\){−2} -> R,\ f(x) = (2x + 1)/(x + 2)`.
    1. Find the rule and domain of `f^(-1)`, the inverse function of `f`.   (2 marks)

    2. Part of the graphs of `f` and  `y = x`  are shown in the diagram below.
       

    3. Find the area of the shaded region.   (1 mark)

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

    5. Find the area of the shaded region.   (1 mark) 
  1. Part of the graph of `f` is shown in the diagram below.
     
       
     

     

    The point `P(c, d)` is on the graph of `f`.

     

    Find the exact values of `c` and `d` such that the distance of this point to the origin is a minimum, and find this minimum distance.   (3 marks)

Let  `g: (−k, oo) -> R, g(x) = (kx + 1)/(x + k)`, where `k > 1`.

  1. Show that  `x_1 < x_2`  implies that  `g(x_1) < g(x_2),` where  `x_1 in (−k, oo) and x_2 in (−k, oo)`.   (2 marks)

  2. Let `X` be the point of intersection of the graphs of  `y = g (x) and y = −x`.
    1. Find the coordinates of `X` in terms of `k`.   (2 marks)

    2. Find the value of `k` for which the coordinates of `X` are  `(-1/2, 1/2)`.   (2 marks)

    3. Let  `Ztext{(− 1, − 1)}, Y(1, 1)`  and `X` be the vertices of the triangle `XYZ`. Let  `s(k)`  be the square of the area of triangle `XYZ`.
       

       

         
       

       

      Find the values of `k` such that  `s(k) >= 1`.   (2 marks)

  3. The graph of `g` and the line  `y = x`  enclose a region of the plane. The region is shown shaded in the diagram below.
     

     

     

    Let  `A(k)`  be the rule of the function `A` that gives the area of this enclosed region. The domain of `A` is  `(1, oo)`.

    1. Give the rule for `A(k)`.   (2 marks)

    2. Show that  `0 < A(k) < 2`  for all  `k > 1`.   (2 marks)

Show Answers Only
  1. `2 + {(−3)}/(x + 2)`
  2.   i. `f^(-1) (x) = (-3)/(x-2)-2,\ \ x in R text(\){2}`
  3.  ii. `4-3 ln(3)\ text(units²)`
  4. iii. `8-6 log_e (3)\ text(units²)`
  5. `c = sqrt 3-2, \ d = 2-sqrt 3`
  6. `text(min distance) = (2 sqrt 2-sqrt 6)`
  7. `text(Proof)\ \ text{(See Worked Solutions)}`
  8.   i. `X (−k + sqrt (k^2-1), k-sqrt (k^2-1))`
  9.  ii. `k = 5/4`
  10. iii. `k in (1, 5/4]`
  11.  i. `A(k) = (k^2-1) log_e ((k-1)/(k + 1)) + 2k`
  12. ii. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

a.   `text(Solution 1)`

`(2x + 1)/(x + 2)` `= 2-3/(x + 2)`
  `= 2 + {(-3)}/(x + 2) qquad [text(CAS: prop Frac) ((2x + 1)/(x + 2))]`

 

`text(Solution 2)`

`(2x + 1)/(x + 2)` `=(2(x+2)-3)/(x+2)`
  `=2+ (-3)/(x+2)`

 

b.i.  `text(Let)\ \ y = f(x)`

`text(For Inverse: swap)\ \ x ↔ y`

`x` `=2-3/(y + 2)`
`(x-2)(y+2)` `=-3`
`y` `=(-3)/(x-2)-2`

 

`text(Range of)\ f(x):\ \ y in R text(\){2}`

`:. f^(-1) (x) = (-3)/(x-2)-2, \ x in R text(\){2}`

 

b.ii.   `text(Find intersection points:)`

`f(x)` `= x`
`(2x+1)/(x+2)` `=x`
`2x+1` `=x^2+2x`
`:. x` `= +- 1`
`:.\ text(Area)` `= int_(-1)^1 (f(x)-x)\ dx`
  `=int_(-1)^1 (2-3/(x + 2)-x)\ dx`
  `= 4-3 ln(3)\ text(u²)`

 

b.iii.    `text(Area)` `= int_(-1)^1 (f(x)-f^(-1) (x)) dx`
    `=2 xx (4-3 ln(3))\ \ \ text{(twice the area in (b)(ii))}`
    `= 8-6 log_e (3)\ text(u²)`
♦♦♦ Mean mark part (c) 25%.

 

c.   
  `text(Let)\ \ z` `= OP, qquad P(c, -3/(c + 2) + 2)`
  `z` `= sqrt (c^2 + (2-3/(c + 2))^2), \ c > -2`

`text(Stationary point when:)`

`(dz)/(dc) = 0, c > -2`

`:.\ c = sqrt 3-2 overset and (->) d = 2-sqrt 3`

`:. text(Minimum distance) = (2 sqrt 2-sqrt 6)`

 

d.   `text(Given:)\ \ -k < x_1 < x_2`

♦♦♦ Mean mark part (d) 8%.

`text(Must prove:)\ \ g(x_2)-g(x_1) > 0`

`text(LHS:)`

`g(x_2)-g(x_1)`

`= (kx_2 +1)/(x_2+k)-(kx_1 +1)/(x_1+k)`

`=((kx_2 +1)(x_1+k)-(kx_1 +1)(x_2+k))/((x_2+k)(x_1+k))`

`=(k^2(x_2-x_1)-(x_2-x_1))/((x_2+k)(x_1+k))`

`=((k^2-1)(x_2-x_1))/((x_2+k)(x_1+k))`

 

`x_2-x_1 >0,\ and \ k^2-1>0`

`text(S)text(ince)\ \ x_2>x_1> -k,`

`=> -x_2<-x_1<k`

`=>k+x_1 >0, \ and \ k+x_2>0`

 

`:.g(x_2)-g(x_1) >0`

`:. g(x_2) > g(x_1)`

 

♦♦ Mean mark part (e)(i) 32%.
e.i.    `g(x)` `= -x`
  `(kx +1)/(x+k)` `=-x`
  `kx+1` `=-x^2-xk`
  `x^2+2k+1` `=0`
  `:. x` `=(-2k +- sqrt(4k^2-4))/2`
    `= sqrt (k^2-1)-k\ \ text(for)\ \ x > -k`

 

`:. X (-k + sqrt(k^2-1),\ \ k-sqrt(k^2-1))`

 

e.ii.   `text(Equate)\ \ x text(-coordinates:)`

♦ Mean mark part (e)(ii) 44%.
`-k + sqrt(k^2-1)` `= -1/2`
`sqrt(k^2-1)` `=k-1/2`
`k^2-1` `=k^2-k+1/4`
`:. k` `= 5/4`

 

e.iii.   `s(k)` `= (1/2 xx YZ xx XO)^2`
    `= 1/4 xx (YZ)^2 xx (XO)^2`

 

`ZO = sqrt(1^2+1^2) = sqrt2`

♦♦♦ Mean mark (e)(iii) 6%.

`YZ=2 xx ZO = 2sqrt2`

`(YZ)^2 = 8`

`(XO)^2` `=(-k + sqrt(k^2-1))^2-(k-sqrt (k^2-1))^2`
  `=2(-k + sqrt(k^2-1))^2`
   

 `text(Solve)\ \ s(k) >= 1\ \ text(for)\ \ k >= 1,`

`1/4 xx 8 xx 2(-k + sqrt(k^2-1))^2` `>=1`
`(-k + sqrt(k^2-1))^2` `>=1/4`
`k-sqrt(k^2-1)` `>=1/2`
`k-1/2` `>= sqrt(k^2-1)`
`k^2-k+1/4` `>=k^2-1`
`:.k` `<= 5/4`

`:.  1<k<= 5/4`

 

♦♦ Mean mark (f)(i) 28%.
f.i.    `A(k)` `= int_(-1)^1 (g(x)-x)\ dx,\ \ k > 1`
    `= int_(-1)^1 ((kx+1)/(x+k) -x)\ dx`
    `= int_(-1)^1 (k+ (1-k^2)/(x+k) -x)`
    `=[kx + (1-k^2) log_e(x+k)-x^2/2]_(-1)^1`
    `=(k+(1-k^2)log_e(1+k)-1/2)-(-k+(1-k^2)log_e(k-1)-1/2)`
    `=2k+(1-k^2)log_e ((1+k)/(k-1))`
    `= (k^2-1) log_e ((k-1)/(k + 1)) + 2k`

 

♦♦♦ Mean mark part (f)(ii) 4%.
f.ii.  
  `0` `< A(k) < text(Area of)\ Delta ABC`
  `0` `< A(k) < 1/2 xx AC xx BO`
  `0` `< A(k) < 1/2 sqrt(2^2 + 2^2) xx (sqrt(1^2 + 1^2))`
  `0` `< A(k) < 1/2 xx 2 sqrt 2 xx sqrt 2`
  `:. 0` `< A(k) < 2`

Probability, MET2 2016 VCAA 3*

A school has a class set of 22 new laptops kept in a recharging trolley. Provided each laptop is correctly plugged into the trolley after use, its battery recharges.

On a particular day, a class of 22 students uses the laptops. All laptop batteries are fully charged at the start of the lesson. Each student uses and returns exactly one laptop. The probability that a student does not correctly plug their laptop into the trolley at the end of the lesson is 10%. The correctness of any student’s plugging-in is independent of any other student’s correctness.

  1. Determine the probability that at least one of the laptops is not correctly plugged into the trolley at the end of the lesson. Give your answer correct to four decimal places.   (2 marks)

  2. A teacher observes that at least one of the returned laptops is not correctly plugged into the trolley.
  3. Given this, find the probability that fewer than five laptops are not correctly plugged in. Give your answer correct to four decimal places.   (2 marks)

The time for which a laptop will work without recharging (the battery life) is normally distributed, with a mean of three hours and 10 minutes and standard deviation of six minutes. Suppose that the laptops remain out of the recharging trolley for three hours.

  1. For any one laptop, find the probability that it will stop working by the end of these three hours. Give your answer correct to four decimal places.   (2 marks)

A supplier of laptops decides to take a sample of 100 new laptops from a number of different schools. For samples of size 100 from the population of laptops with a mean battery life of three hours and 10 minutes and standard deviation of six minutes, `hat P` is the random variable of the distribution of sample proportions of laptops with a battery life of less than three hours.

  1. Find the probability that `text(Pr) (hat P >= 0.06 | hat P >= 0.05)`. Give your answer correct to three decimal places. Do not use a normal approximation.   (3 marks)

It is known that when laptops have been used regularly in a school for six months, their battery life is still normally distributed but the mean battery life drops to three hours. It is also known that only 12% of such laptops work for more than three hours and 10 minutes.

  1. Find the standard deviation for the normal distribution that applies to the battery life of laptops that have been used regularly in a school for six months, correct to four decimal places.   (2 marks)

The laptop supplier collects a sample of 100 laptops that have been used for six months from a number of different schools and tests their battery life. The laptop supplier wishes to estimate the proportion of such laptops with a battery life of less than three hours.

  1. Suppose the supplier tests the battery life of the laptops one at a time.
  2. Find the probability that the first laptop found to have a battery life of less than three hours is the third one.   (1 mark)

The laptop supplier finds that, in a particular sample of 100 laptops, six of them have a battery life of less than three hours.

  1. Determine the 95% confidence interval for the supplier’s estimate of the proportion of interest. Give values correct to two decimal places.   (1 mark)

  2. The supplier also provides laptops to businesses. The probability density function for battery life, `x` (in minutes), of a laptop after six months of use in a business is
     

     

    `qquad qquad f(x) = {(((210-x)e^((x-210)/20))/400, 0 <= x <= 210), (0, text{elsewhere}):}`
     

  3. Find the mean battery life, in minutes, of a laptop with six months of business use, correct to two decimal places.   (1 mark)

Show Answers Only

  1. `0.9015`
  2. `0.9311`
  3. `0.0478`
  4. `0.658`
  5. `8.5107`
  6. `1/8`
  7. `p in (0.01, 0.11)`
  8. `170.01\ text(min)`

Show Worked Solution

a.   `text(Solution 1)`

`text(Let)\ \ X = text(number not correctly plugged),`

`X ~ text(Bi) (22, .1)`

`text(Pr) (X >= 1) = 0.9015\ \ [text(CAS: binomCdf)\ (22, .1, 1, 22)]`

 

`text(Solution 2)`

`text(Pr) (X>=1)` `=1-text(Pr) (X=0)`
  `=1-0.9^22`
  `=0.9015\ \ text{(4 d.p.)}`

 

 b.   `text(Pr) (X < 5 | X >= 1)`

MARKER’S COMMENT: Early rounding was a common mistake, producing 0.9312.

`= (text{Pr} (1 <= X <= 4))/(text{Pr} (X >= 1))`

`= (0.83938…)/(0.9015…)\ \ [text(CAS: binomCdf)\ (22, .1, 1,4)]`

`= 0.9311\ \ text{(4 d.p.)}`

 

c.   `text(Let)\ \ Y = text(battery life in minutes)`

MARKER’S COMMENT: Some working must be shown for full marks in questions worth more than 1 mark.

`Y ~ N (190, 6^2)`

`text(Pr) (Y <= 180)= 0.0478\ \ text{(4 d.p.)}`

`[text(CAS: normCdf)\ (−oo, 180, 190,6)]`

 

d.   `text(Let)\ \ W = text(number with battery life less than 3 hours)`

♦ Mean mark part (d) 33%.

`W ~ Bi (100, .04779…)`

`text(Pr) (hat P >= .06 | hat P >= .05)` `= text(Pr) (X_2 >= 6 | X_2 >= 5)`
  `= (text{Pr} (X_2 >= 6))/(text{Pr} (X_2 >= 5))`
  `= (0.3443…)/(0.5234…)`
  `= 0.658\ \ text{(3 d.p.)}`

 

e.   `text(Let)\ \ B = text(battery life), B ~ N (180, sigma^2)`

`text(Pr) (B > 190)` `= .12`
`text(Pr) (Z < a)` `= 0.88`
`a` `dot = 1.17499…\ \ [text(CAS: invNorm)\ (0.88, 0, 1)]`
`-> 1.17499` `= (190-180)/sigma\ \ [text(Using)\ Z = (X-u)/sigma]`
`:. sigma` `dot = 8.5107`

 

f.    `text(Pr) (MML)` `= 1/2 xx 1/2 xx 1/2`
    `= 1/8`

 

g.   `text(95% confidence int:) qquad quad [(text(CAS:) qquad qquad 1-text(Prop)\ \ z\ \ text(Interval)), (x = 6), (n = 100)]`

`p in (0.01, 0.11)`

 

h.    `mu` `= int_0^210 (x* f(x)) dx`
  `:. mu` `dot = 170.01\ text(min)`

Calculus, MET2 2016 VCAA 2

Consider the function  `f(x) = -1/3 (x + 2) (x-1)^2.`

  1.  i. Given that  `g^{′}(x) = f (x) and g (0) = 1`,
  2.      show that  `g(x) = -x^4/12 + x^2/2-(2x)/3 + 1`.   (1 mark)

  3. ii. Find the values of `x` for which the graph of  `y = g(x)`  has a stationary point.   (1 mark)

The diagram below shows part of the graph of  `y = g(x)`, the tangent to the graph at  `x = 2`  and a straight line drawn perpendicular to the tangent to the graph at  `x = 2`. The equation of the tangent at the point `A` with coordinates  `(2, g(2))`  is  `y = 3-(4x)/3`.

The tangent cuts the `y`-axis at `B`. The line perpendicular to the tangent cuts the `y`-axis at `C`.
 


 

  1.   i. Find the coordinates of `B`.   (1 mark)

  2.  ii. Find the equation of the line that passes through `A` and `C` and, hence, find the coordinates of `C`.   (2 marks)

  3. iii. Find the area of triangle `ABC`.   (2 marks)

  4. The tangent at `D` is parallel to the tangent at `A`. It intersects the line passing through `A` and `C` at `E`.

     


     
     i. Find the coordinates of `D`.   (2 marks)

  5. ii. Find the length of `AE`.   (3 marks)

Show Answers Only
  1.  i. `text(Proof)\ \ text{(See worked solutions)}`
  2. ii. `x = -2, 1`
  3.   i. `(0, 3)`
  4. ii. `y = 3/4 x-7/6`
  5.      `(0, -7/6)`
  6. iii. `25/6\ text(u²)`
  7. `(-1, 25/12)`
  8. `27/20\ text(u)`
Show Worked Solution
a.i.    `g(x)` `= int f(x)\ dx`
    `=-1/3 int (x + 2) (x-1)^2\ dx`
    `=-1/3int(x^3-3x+2)\ dx`
  `:.g(x)` `= -x^4/12 + x^2/2-(2x)/3 + c`

 
`text(S)text(ince)\ \ g(0) = 1,`

`1` `= 0 + 0-0 + c`
`:. c` `= 1`

  
`:. g(x) = -x^4/12 + x^2/2-(2x)/3 + 1\ \ …\ text(as required)`
 

a.ii.   `text(Stationary point when:)`

`g^{′}(x) = f(x) = 0`

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

`:. x = -2, 1`
 

b.i.   

`B\ text(is the)\ y text(-intercept of)\ \ y = 3-4/3 x`

`:. B (0, 3)`
 

b.ii.   `m_text(norm) = 3/4, \ text(passes through)\ \ A(2, 1/3)`

   `text(Equation of normal:)`

`y-1/3` `=3/4(x-2)`
`y` `=3/4 x-7/6`
   

`:. C (0, -7/6)`
 

b.iii.   `text(Area)` `= 1/2 xx text(base) xx text(height)`
    `= 1/2 xx (3 + 7/6) xx 2`
    `= 25/6\ text(u²)`

 

♦ Mean mark part (c)(i) 43%.
MARKER’S COMMENT: Many students gave an incorrect `y`-value here. Be careful!
c.i.    `text(Solve)\ \ \ g^{′}(x)` `= -4/3\ \ text(for)\ \ x < 0`
  `=> x` `= -1`
  `g(-1)` `=-1/12+1/2+2/3+1`
    `=25/12`

  
`:. D (−1, 25/12)`
 

c.ii.   `text(T) text(angent line at)\ \ D:`

`y-25/12` `=-4/3(x+1)`
`y` `=-4/3x + 3/4`

 
`DE\ \ text(intersects)\ \ AE\ text(at)\ E:`

♦♦ Mean mark 32%.
`-4/3 x + 3/4` `= 3/4 x-7/6`
`25/12 x` `= 23/12`
`x` `=23/25`

 
`:. E (23/25, -143/300)`
 

`:. AE` `= sqrt((2-23/25)^2 + (1/3-(-143/300))^2)`
  `= 27/20\ text(units)`

Calculus, MET2 2016 VCAA 1

Let  `f: [0, 8 pi] -> R, \ f(x) = 2 cos (x/2) + pi`.

  1. Find the period and range of `f`.   (2 marks)

  2. State the rule for the derivative function `f^{′}`.   (1 mark)

  3. Find the equation of the tangent to the graph of `f` at  `x = pi`.   (1 mark)

  4. Find the equations of the tangents to the graph of  `f: [0, 8 pi] -> R,\ \ f(x) = 2 cos (x/2) + pi`  that have a gradient of 1.   (2 marks)

  5. The rule of  `f^{′}` can be obtained from the rule of `f` under a transformation `T`, such that
      
    `qquad T: R^2 -> R^2,\ T([(x), (y)]) = [(1, 0), (0, a)] [(x), (y)] + [(−pi), (b)]`
     

     

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

  6. Find the values of  `x, \ 0 <= x <= 8 pi`, such that  `f(x) = 2 f^{′} (x) + pi`.   (2 marks)

Show Answers Only
  1. `text(Period:)\ 4 pi; qquad text(Range:)\ [pi-2, pi + 2]`
  2. `f^{′} (x) =-sin (x/2)`
  3. `y =-x + 2 pi`
  4. `y = x-2 pi and y = x-6 pi`
  5. `a = 1/2 and b =-pi/2`
  6. `x = (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2`
Show Worked Solution

a.   `text(Period)= (2pi)/n = (2 pi)/(1/2) = 4pi`

MARKER’S COMMENT: Including round brackets rather than square ones was a common mistake.

`text(Range:)\ [pi-2, pi + 2]`
  

b.   `f^{′} (x) = text(−sin) (x/2)`
 

c.   `[text(CAS: tangentLine)\ (f(x), x, pi)]`

`y = -x + 2 pi`
 

d.   `text(Solve)\ \ f^{′} (x) = 1\ \ text(for)\ x in [0, 8 pi]`

♦ Mean mark part (d) 50%.

`-> x = 3 pi or 7 pi`

`:. y = x-2 pi and y = x-6 pi\ \ [text(CAS)]`

 

e.   `text(Using the transition matrix,)`

♦♦ Mean mark part (e) 27%.
`x_T` `=x-pi`
`x` `=x_T+pi`
`y_T` `=ay+b`
`y` `=(y_T-b)/a`
   

`f(x)= cos (x/2) + pi/2\ \ ->\ \ f{′}(x) = -sin(x/2)`

`(y_T-b)/a` `=2cos((x_T+pi)/2)+pi`
`y_T` `=2a cos((x_T+pi)/2)+a pi +b`
  `=-2a sin(x_T/2)+a pi + bqquad [text(Complementary Angles)]`
   
`-2a` `=-1`
`:. a` `=1/2`
`1/2 pi +b` `=0`
`:.b` `=-pi/2`

 

f.   `text(Solve)\ \ f(x) = 2 f^{′} (x) + pi\ \ text(for)\ \ x in [0, 8 pi]`

♦ Mean mark part (f) 50%.
`2 cos (x/2) + pi` `= -2 sin(x/2)+pi`
`tan(x/2)` `=-1`
`x/2` `=(3pi)/4, (7pi)/4, (11pi)/4, (15pi)/4`
`:.x` `= (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2`

Calculus, MET2 2010 VCAA 17 MC

The function `f` is differentiable for all  `x in R`  and satisfies the following conditions.

  • `f prime (x) < 0\ \ text(where)\ \ x < 2`
  • `f prime (x) = 0\ \ text(where)\ \ x = 2`
  • `f prime (x) = 0\ \ text(where)\ \ x = 4`
  • `f prime (x) > 0\ \ text(where)\ \ 2 < x < 4`
  • `f prime (x) > 0\ \ text(where)\ \ x > 4`

Which one of the following is true?

  1.  The graph of `f` has a local maximum point where `x = 4.`
  2. The graph of `f` has a stationary point of inflection where `x = 4.`
  3. The graph of `f` has a local maximum point where `x = 2.`
  4. The graph of `f` has a local minimum point where `x = 4.`
  5. The graph of `f` has a stationary point of inflection where `x = 2.`
Show Answers Only

`B`

Show Worked Solution

`text(A table summary of the gradients is:)`

 

`:.\ text(Point of Inflection at)\ \ x = 4`

`=>   B`

Probability, MET2 2010 VCAA 13 MC

The continuous random variable `X` has a normal distribution with mean 20 and standard deviation 6. The continuous random variable `Z` has the standard normal distribution.

The probability that `Z` is between – 2 and 1 is equal to

  1. `text(Pr) (18 < X < 21)`
  2. `text(Pr) (14 < X < 32)`
  3. `text(Pr) (14 < X < 26)`
  4. `text(Pr) (8 < X < 32)`
  5. `text(Pr) (X > 14) + text(Pr) (X < 26)`
Show Answers Only

`B`

Show Worked Solution

`text(When)\ \ Z=-2,`

`X=mu – 2σ = 20-2xx6 = 8`

`text(When)\ \ Z=1,`

`X=mu + σ = 20+6 = 26`

`text(Due to symmetry,)`

`text(Pr) (-2<Z<1)` `= text(Pr) (8<X<26)`
  `= text(Pr) (14<X<32)`

`=>   B`

Probability, MET2 2010 VCAA 11 MC

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

`f(x) = {(cos(2x), if (3 pi)/4 < x < (5 pi)/4), (qquad qquad quad 0,\ \ \ text(elsewhere)):}`
 

The value of `a` such that `text(Pr) (X < a) = 0.25`  is closest to

  1. `2.25`
  2. `2.75`
  3. `2.88`
  4. `3.06`
  5. `3.41`
Show Answers Only

`C`

Show Worked Solution

`text(Solve)\ \ int_((3 pi)/4)^a cos (2x)\ dx = 1/4\ \ text(for)\ \ a in ((3 pi)/4, (5 pi)/4)`

`:. a` `= (11 pi)/12\ \ \ text([by CAS.])`
  `~~ 2.88`

 
`=>   C`

Algebra, MET2 2010 VCAA 9 MC

The function  `f:\ (–oo, a] -> R`  with rule  `f(x) = x^3 - 3x^2 + 3`  will have an inverse function provided

  1. `a <= 0`
  2. `a >= 2`
  3. `a >= 0`
  4. `a <= 2`
  5. `a <= 1`
Show Answers Only

`A`

Show Worked Solution
`f(x)` `= x^3-3x^2 + 3`
`f′(x)` `=3x^2-6x`
  `=3x(x-2)`

 

`text(Stationary points at)\ \ x=0 and 2.`

`text{Local max at (0,3) and local min at (2,-1).}`

`text(Sketch the graph:)`

`text(Inverse exists if)\ \ f(x)\ \ text(is)\ \ 1-1.`

`:. x <= 0`

`=>   A`

Calculus, MET2 2010 VCAA 2 MC

For  `f(x) = x^3 + 2x`, the average  rate of change with respect to `x` for the interval  `[1, 5]`  is

A.   `18`

B.   `20.5`

C.   `24`

D.   `32.5`

E.   `33`

Show Answers Only

`E`

Show Worked Solution
`text(Average ROC)` `= (f(5) – f(1))/(5 – 1)`
  `=(135-3)/4`
  `= 33`

`=>   E`

Probability, MET2 2016 VCAA 18 MC

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

`qquad f(x) = {(1/4 cos (x/2), 3 pi <= x <= 5 pi), (0, text{elsewhere}):}`
 

The value of `a` such that  `text(Pr) (X < a) = (sqrt 3 + 2)/4`  is

  1. `(19 pi)/6`
  2. `(14 pi)/3`
  3. `(10 pi)/3`
  4. `(29 pi)/6`
  5. `(17 pi)/3`
Show Answers Only

`B`

Show Worked Solution
`int_(3 pi)^a f(x)\ dx` `= (sqrt 3 + 2)/4`
`[1/2 sin (x/2)]_(3pi)^a` `= (sqrt 3 + 2)/4`
`1/2 sin (a/2) +1/2` `= (sqrt 3 + 2)/4`
`sin (a/2)` `=sqrt3/2`
`a/2` `=pi/3, (2pi)/3, (4pi)/3, (5pi)/3, (7pi)/3, …`
`:. a` `= (14 pi)/3\ \ text(for)\ a in (3 pi, 5 pi)`

 
`=>   B`

Probability, MET2 2016 VCAA 17 MC

Inside a container there are one million coloured building blocks. It is known that 20% of the blocks are red.

A sample of 16 blocks is taken from the container. For samples of 16 blocks, `hat P` is the random variable of the distribution of sample proportions of red blocks. (Do not use a normal approximation.)

`text(Pr) (hat P >= 3/16)` is closest to

A.  `0.6482`

B.  `0.8593`

C.  `0.7543`

D.  `0.6542`

E.  `0.3211`

Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ X = text(Number of red blocks),`

`X ~\ text(Bi) (16, 0.2)`

`text(Pr) (hat P >= 3/16)`

`= text(Pr) (X >= 3) qquad [text(binomCdf) (16, .2, 3, 16)]`

`= 0.6482`

`=>   A`

Calculus, MET2 2016 VCAA 13 MC

Consider the graphs of the functions `f` and `g` shown below.

The area of the shaded region could be represented by

A.   `int_a^d (f(x) - g(x))\ dx`

B.   `int_0^d (f(x) - g(x))\ dx`

C.   `int_0^b (f(x) - g(x))\ dx + int_b^c (f(x) - g(x))\ dx`

D.   `int_0^a f(x)\ dx + int_a^c (f(x) - g(x))\ dx + int_b^d f(x)\ dx`

E.   `int_0^d f(x)\ dx - int_a^c g(x)\ dx`

Show Answers Only

`E`

Show Worked Solution

`text(Shaded Area:)`

`text(Area under)\ f(x)\ text(between)\ [0, d]\ text(less)`

`text(Area under)\ g(x)\ text(between)\ [a, c]`

`=>   E`

Graphs, MET2 2016 VCAA 12 MC

The graph of a function  `f` is obtained from the graph of the function `g` with rule  `g(x) = sqrt (2x - 5)`  by a reflection in the `x`-axis followed by a dilation from the `y`-axis by a factor of  `1/2`.

Which one of the following is the rule for the function  `f`?

  1. `f(x) = sqrt (5 - 4x)`
  2. `f(x) = - sqrt (x - 5)`
  3. `f(x) = sqrt (x + 5)`
  4. `f(x) = −sqrt (4x - 5)`
  5. `f(x) = −sqrt (4x - 10)`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ y=sqrt(2x-5)`

`text(1st transformation:)`

`y = – sqrt(2x-5)`
 

`text(2nd transformation:)`

`y` `=-sqrt(2(2x)-5)`
  `=- sqrt(4x-5)`
`:. f(x)` `= −sqrt(4x – 5)`

 
`=>   D`

Calculus, MET2 2016 VCAA 10 MC

For the curve  `y = x^2 - 5`, the tangent to the curve will be parallel to the line connecting the positive x-intercept and the y-intercept when `x` is equal to

A.   `sqrt 5`

B.   `5`

C.   `−5`

D.   `sqrt 5/2`

E.   `1/sqrt 5`

Show Answers Only

`D`

Show Worked Solution

`text{Intercepts: (0, −5) and}\ (sqrt 5, 0)`

`text(Gradient between intercepts:)`

`m = (0 – (−5))/(sqrt 5 – 0) = 5/sqrt 5`

`text(Solve:)`

`f prime (x)` `= 5/sqrt 5`
`2x` `= 5/sqrt 5`
`:. x` `=5/(2sqrt5) xx sqrt5/sqrt5`
  `= sqrt 5/2`

 
`=>   D`

Graphs, MET2 2016 VCAA 6 MC

Consider the graph of the function defined by  `f: [0, 2 pi] -> R,\ f(x) = sin (2x).`

The square of the length of the line segment joining the points on the graph for which  `x = pi/4 and x = (3 pi)/4` is

  1. `(pi^2 + 16)/4`
  2. `pi + 4`
  3. `4`
  4. `(3 pi^2 + 16 pi)/4`
  5. `(10 pi^2)/16`
Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=pi/4,\ \ f(x) = sin(pi/2)=1`

`text(When)\ \ x=(3pi)/4,\ \ f(x) = sin((3pi)/2)=-1`

`text(Let)\ \ z` `= text(distance between)\ (pi/4, 1) and ((3pi)/4, −1)`
`z^2` `= ((3pi)/4 – pi/4)^2 + (−1 −1)^2`
   `=pi^2/4 + 4`
  `= (pi^2 + 16)/4`

`=>   A`

Calculus, MET2 2011 VCAA 20 MC

A part of the graph of  `g: R -> R, g(x) = x^2 - 4`  is shown below.

met2-2011-vcaa-20-mc

The area of the region marked `A` is the same area of the region marked `B`.

The exact value of `a` is

A.   `0`

B.   `6`

C.   `sqrt6`

D.   `12`

E.   `2sqrt3` 

Show Answers Only

`=> E`

Show Worked Solution

`text(S)text(ince Area)\ A = text(Area)\ B,`

`int_0^a (x^2-4)\ dx` `=0`
`[x^3/3 – 4x]_0^a` `=0`
`a^3/3 – 4a` `=0`
`a^2` `=12`
`:.a` `=2sqrt3,\ \ text(for)\ a > 0`

`=> E`

Algebra, MET2 2011 VCAA 18 MC

The equation  `x^3 - 9x^2 + 15x + w = 0`  has only one solution for `x` when

A.   `−7 < w < 25`

B.   `w <= −7`

C.   `w >=25`

D.   `w < −7` or `w > 25`

E.   `w > 1`

Show Answers Only

`=> D`

Show Worked Solution
`y` `=x^3 – 9x^2 + 15x + w`
`y′` `=3x^2 – 18x + 15`
  `=3(x-5)(x-1)`

 

`text(Stationary points at)\ \ x=5 and 1.`

`text{Sketch the graph}\ \ (w=0),`

met2-2011-vcaa-18-mc-answer
 

`text(By inspection, one solution occurs when)`

`w > 25\ \ text{(shift curve up > 25), or}`

`w < −7\ \ text{(shift curve down > 7)`

`=> D`

Graphs, MET2 2011 VCAA 15 MC

met2-2011-vcaa-15-mc 
 

The graph shown could have equation

  1. `y = 2cos(x + pi/6) + 1`
  2. `y = 2cos4(x - pi/6) + 1`
  3. `y = 4sin2(x - pi/12) - 1`
  4. `y = 3cos(2x + pi/6) - 1`
  5. `y = 2sin(4x + (2pi)/3) - 1`
Show Answers Only

`=> B`

Show Worked Solution

`text{Amplitude = 2  (range from – 1 to 3)}`

`text(Median) = 1`

`:.\ text(Solution is)\ A\ text(or)\ B.`

`text(From graph:)`

`text(Period) = (2pi)/3 – pi/6 = pi/2`

`text(Consider option)\ B,`

`text(Period)= (2pi)/n= (2pi)/4 = pi/2`

`=> B`

Probability, MET2 2011 VCAA 12 MC

The continuous random variable `X` has a normal distribution with mean 30 and standard deviation 5. For a given number `a, text(Pr)(X > a) = 0.20`.

Correct to two decimal places, `a` is equal to

  1. `23.59`
  2. `24.00`
  3. `25.79`
  4. `34.21`
  5. `36.41`
Show Answers Only

`=> D`

Show Worked Solution

met2-2011-vcaa-12-mc-answer1

`X ∼\ N(30,5^2)`

`text(Pr)(X < a)` `= 0.8qquad[text(CAS: inv Norm) (.8,30,5)]`
`:. a` `= 34.21`

`=> D`

Graphs, MET2 2011 VCAA 8 MC

Consider the function  `f: R -> R, \ f(x) = x(x - 4)`  and the function

`g: [3/2,5) -> R, \ g(x) = x + 3`.

If the function  `h = f + g`, then the domain of the inverse function of `h` is

  1. `[0,13)`
  2. `[−3/4,10]`
  3. `(−3/4,15/4]`
  4. `[3/4,13)`
  5. `[3/2,13)` 
Show Answers Only

`=> D`

Show Worked Solution
`h(x)` `= x^2 – 3x + 3, \ x ∈[3/2,5)`
`h′(x)` `=2x-3`

 
`:.\ text(Minimum at)\ \ (3/2, 3/4)`

`h(5)=5^2 -3 xx 5 +3=13`

 

`text(Sketch the graph:)` 

met1-2011-vcaa-8-mc-answer

`text(Domain) (h^(−1)(x))` `= text(Range)\ (h)`
  `= [3/4,13)`

`=> D`

Calculus, MET2 2011 VCAA 4 MC

The derivative of  `log_e(2f(x))`  with respect to `x` is

  1. `(f′(x))/(f(x))`
  2. `2(f′(x))/(f(x))`
  3. `(f′(x))/(2f(x))`
  4. `log_e(2f′(x))`
  5. `2log_e(2f′(x))`
Show Answers Only

`A`

Show Worked Solution

`text(Chain Rule:)`

`text(If)\ \ h(x)` `= f(g(x))`
`h′(x)` `= f′(g(x)) xx g′(x)`
`d/(dx)(log_e(2f(x)))` `= 1/(2f(x)) xx 2f′(x)`
  `= (f′(x))/(f(x))`

`=> A`

Algebra, MET2 2011 VCAA 3 MC

If  `x + a`  is a factor of  `4x^3 - 13x^2 - ax`  where  `a ∈ R text(\{0})`, then the value of `a` is

A.   `−4`

B.   `−3`

C.   `−1`

D.      `1`

E.      `2`

Show Answers Only

`=> B`

Show Worked Solution

`text(Let)\ \ p(x) = 4x^3 – 13x^2 – ax`

`text(If)\ \ (x + a)\ \ text(is a factor,)`

`p(−a)` `= 0`
`0` `=4(-a)^3-13(-a)^2-a(-a)`
  `=-4a^3-13a^2+a^2`
  `=-4a^2(a+3)`

 

`:. a = −3,\ \ \ (a != 0)`

`=> B`

Algebra, MET2 2012 VCAA 2 MC

For the function with rule  `f(x) = x^3 - 4x`, the average rate of change of  `f(x)`  with respect to `x` on the interval `[1,3]` is

A.   `1`

B.   `3`

C.   `5`

D.   `6`

E.   `9`

Show Answers Only

`=> E`

Show Worked Solution
`text(Average ROC)` `= (f(3) – f(1))/(3 – 1)`
  `=(15-(-3))/2`
  `= 9`

`=> E`

CORE, FUR1 SM-Bank VCE 3 MC

The decreasing value of a depreciating asset is shown in the graph below.
 

 
 

Let `A_n` be the value of the asset after `n` years, in dollars.

What recurrence relation below models the value of `A_n`?

  1. `A_0 = 120\ 000,qquadA_n = 120\ 000 xx 1.125 xx n` 
  2. `A_0 = 120\ 000,qquadA_n = 120\ 000 xx (0.125)^n` 
  3. `A_0 = 120\ 000,qquadA_n = 120\ 000 xx (1 - 0.125) xx n` 
  4. `A_0 = 120\ 000,qquadA_n = 120\ 000 xx (1 - 0.125)^n` 
  5. `A_0 = 120\ 000,qquadA_n = 120\ 000 xx (1 + 1.125)^n` 
Show Answers Only

`D`

Show Worked Solution

`text(The asset is decreasing at 12.5% per year)`

`text(on a decreasing balance basis.)`

`A_1` `= 120\ 000(1 – 0.125)^1`
`vdots`  
`A_n` `= 120\ 000(1 – 0.125)^n`

`=> D`

CORE, FUR1 SM-Bank VCE 2 MC

Derek invests $48 000 in an investment that guarantees an annual interest rate of 4.8%, compounded monthly.

Let  `V_n`  be the value of the investment after `n` months.

Which recurrence relation below models the investment?

  1. `V_0 = 48\ 000,qquadV_(n + 1) = 1.048V_n`
  2. `V_0 = 48\ 000,qquadV_(n + 1) = 1.0048V_n`
  3. `V_0 = 48\ 000,qquadV_(n + 1) = 1.48V_n`
  4. `V_0 = 48\ 000,qquadV_(n + 1) = 1.004V_n`
  5. `V_0 = 48\ 000,qquadV_(n + 1) = 1.04V_n`
Show Answers Only

`D`

Show Worked Solution

`text(Monthly interest rate)`

`= (4.8%)/12`

`= 0.4%`

`= 0.004`
 

`:. V_(n + 1) = 1.004V_n`

`=> D`

CORE, FUR1 SM-Bank VCAA 1-2 MC

Part 1

Candy deposits $80 000 into a savings account that earns 5% interest per annum, compounded annually.

At the end of the first year, Candy withdraws $5600.

What is the balance of Candy's savings account at the start of the second year?

  1. `$74\ 400`
  2. `$77\ 600`
  3. `$78\ 400`
  4. `$85\ 600`
  5. `$89\ 600`

 

Part 2

If `B_n` is the balance of Candy's account at the start of year `n`, a recurrence equation that can be used to model the balance of the account over time is

  1. `B_(n + 1) = 1.5 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`
  2. `B_(n + 1) = 0.05 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`
  3. `B_(n + 1) = 1.05 (B_n - 5600)\ \ \ \ text(where)\ B_1 = 80\ 000`
  4. `B_(n + 1) = 0.05 (B_n - 5600)\ \ \ \ text(where)\ B_1 = 80\ 000`
  5. `B_(n + 1) = 1.05 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`
Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ E`

Show Worked Solution

`text(Part 1)`

`text(Balance at start of 2nd year)`

`= 80\ 000 xx 1.05 – 5600`

`= $78\ 400`

`=> C`

 

`text(Part 2)`

`B_2` `= 80\ 000 xx 1.05 – 5600`
  `= 1.05B_1 – 5600`
`B_3` `= 1.05B_2 – 5600`
`vdots`  
`B_(n + 1)` `= 1.05B_n – 5600`

 

`=> E`

CORE, FUR2 SM-Bank VCE 3

Luke purchased a new pizza oven for his restaurant for $23 500.

He can depreciate the pizza oven using the reducing balance method at a rate of 12.5% per year.

  1. If `V_n` represents the value of the pizza oven after `n` years, write a recurrence relation that models its value.   (1 mark)

  2. During what year will the pizza oven's value drop below $15 000?   (1 mark)

Luke has been advised that he can use flat rate depreciation at 10% of the purchase price.

  1. After 4 years, show which depreciation method gives the pizza oven the highest value?   (1 mark)

Show Answers Only
  1. `V_0 = 23\ 500,qquadV_(n + 1) = 0.875V_n`
  2. `text(year 4)`
  3. `text(See Worked Solutions)`
Show Worked Solution
a.    `V_0` `= 23\ 500`
  `V_1` `= 23\ 500-(12.5text(%) xx 23\ 500)`
    `= 0.875 V_0`
  `V_2` `= 0.875(0.875V_0)`
    `= 0.875 V_1`

  
`:.\ text(Recurrence relationship:)`

`V_0 = 23\ 500,qquadV_(n + 1) = 0.875V_n`
  

b.    `V_1` `= 0.875 xx 23\ 500 = 20\ 562.50`
  `V_2` `= 0.875 xx 20\ 562.50 = 17\ 992.1875`
  `V_3` `= 0.875 xx 17\ 992.1875 = 15\ 743.16…`
  `V_4` `= 0.875 xx 15\ 743.16… = 13\ 775.26…`

  
`:.\ text(The value drops below $15 000 in year 4.)`

  
c.   `text(Value after 4 years using reducing balance)`

`= 13\ 775.26`

`text(Depreciation each year for flat rate)`

`= 10text(%) xx 23\ 500`

`= $2350`

`text(Value of pizza oven after 4 years,)`

`= 23\ 500-(4 xx 2350)`

`= $14\ 100`
 

`:.\ text(The flat rate depreciation method)`

`text(values the pizza oven highest.)`

CORE, FUR2 SM-Bank VCE 2

Spiro is saving for a car. He has an account with $3500 in it at the start of the year.

At the end of each month, Spiro adds another $180 to the account.

The account pays 3.6% interest per annum, compounded monthly.

    1. What is the interest rate per month?   (1 mark)
    2. Write a recurrence relation that models Spiro's investment, with `V_n` representing the balance of his account after `n` months.   (1 mark)
  2. What will be the balance of Spiro's account after 3 months?
  3. Write your answer correct to the nearest cent.   (1 mark)

Show Answers Only

    1. `0.3text(% per month)`
    2. `V_0 = 3500,qquadV_(n + 1) = V_n xx 1.003 + 180`
  2. `$4073\ \ (text(nearest $))`

Show Worked Solution

a.i.    `text(Interest rate)` `= 3.6/12`
    `= 0.3text(% per month)`

 

a.ii.    `V_0` `= 3500`
  `V_1`  `= 3500 xx 1.003 + 180`
  `V_2`  `= V_1 xx 1.003 + 180`
  `vdots`   
  `V_(n + 1)`  `= V_n xx 1.003 + 180` 

  
`:.\ text(Recurrence relationship:)`

`V_0 = 3500,qquadV_(n + 1) = V_n xx 1.003 + 180`
  

b.    `V_1` `= 3500 xx 1.003 + 180 = $3690.50`
  `V_2` `= 3690.50 xx 1.003 + 180 = $3881.5715`
`V_3` `= 3881.5715 xx 1.003 + 180` `= $4073.216…`
    `= $4073.22\ \ text{(nearest cent)}`

CORE, FUR2 SM-Bank VCE 1

Joe buys a tractor under a buy-back scheme. This scheme gives Joe the right to sell the tractor back to the dealer through either a flat rate depreciation or unit cost depreciation.

  1. The recurrence relation below can be used to calculate the price Joe sells the tractor back to the dealer `(P_n)`, based on the flat rate depreciation, after `n` years
     
    `qquadP_0 = 56\ 000,qquadP_n = P_(n-1)-7000`
     

    1. Write the general rule to find the value of `P_n` in terms of `n`.?   (1 mark)
    2. Hence or otherwise, find the time it will take Joe's tractor to lose half of its value.   (1 mark)
  2. Joe uses the unit cost method to depreciate his tractor, he depreciates $2.75 per kilometre travelled.
    1. How many kilometres does Joe's tractor need to travel for half its value to be depreciated? Round your answer to the nearest kilometre?   (1 mark)
    2. Joe's tractor travels, on average, 2500 kilometres per year. Which method, flat rate depreciation or unit cost depreciation, will result in the greater annual depreciation? Write down the greater depreciation amount correct to the nearest dollar.   (1 mark)

Show Answers Only

  1. i.  `P_n = 56\ 000-7000n`
    ii. `4\ text(years)`
  2. i. `10\ 182\ text{km  (nearest km)}`
    ii. `text(The flat rate depreciation results in an extra)`
         `text($125 depreciation each year.)`

Show Worked Solution

a.i.    `P_1` `= P_0-7000`
  `P_2` `= P_0-7000-7000`
    `= 56\ 000-7000 xx 2`
  `vdots`  
  `P_n`  `= 56\ 000-7000n` 

 

a.ii.    `text(Depreciated value)` `= 56\ 000 ÷ 2=$28\ 000`

`text(Find)\ n,`

`28\ 000` `= 56\ 000-7000n`
`7000n` `= 28\ 000`
`:. n` `= 4\ text(years)`

 

b.i.    `text(Distance travelled)` `= ((56\ 000-28\ 000))/2.75`
    `= 10\ 181.81…`
    `= 10\ 182\ text{km  (nearest km)}`

  
b.ii.   `text(Annual depreciation of unit cost)`

`= 2500 xx $2.75`

`= $6875`

`text(Annual flat rate depreciation = $7000)`

`text(Difference)\ = 7000-6875 = $125`
 

`:.\ text(The flat rate depreciation results in an extra)`

 `text($125 depreciation each year.)`

NETWORKS, FUR2 2016 VCAA 3

A new skateboard park is to be built in Beachton.

This project involves 13 activities, `A` to `M`.

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


 

  1. Determine the earliest start time for activity `M`.   (1 mark)

  2. The minimum completion time for the skateboard park is 15 days.

     

    Write down the critical path for this project.   (1 mark)

  3. Which activity has a float time of two days?   (1 mark)

  4. The completion times for activities `E, F, G, I` and `J` can each be reduced by one day.

     

    The cost of reducing the completion time by one day for these activities is shown in the table below.
     

     

       

     

    What is the minimum cost to complete the project in the shortest time possible?   (1 mark)

  5. The original skateboard park project from part (a), before the reduction of time in any activity, will be repeated at another town named Campville, but with the addition of one extra activity.

     

    The new activity, `N`, will take six days to complete and has a float time of one day.

     

    Activity `N` will finish at the same time as the project.

     

     i.  Add activity `N` to the network below.   (1 mark) 


      
    ii.  What is the latest start time for activity `N`?   (1 mark)

Show Answers Only
  1. `11\ text(days)`
  2. `AEIK`
  3. `text(Activity)\ H`
  4. `text(Minimum cost of $2000 when activity)\ I\ text(is reduced by 1 day.)`

    2. `9\ text(days from the start)`
Show Worked Solution
a.    `text(EST)` `= 1 + 4 + 6`
    `= 11\ text(days)`

  
b.   `text(Critical Path:)\ AEIK`

♦♦ Mean mark part (c) 37%, part (d) 21%.
MARKER’S COMMENT: In part (d), `ADK` cannot be crashed, therefore shortest duration is 14 days. Activity `I` is cheapest to reduce.
  

c.   `text(Activity)\ H`
  

d.   `text(Minimum days to complete is 14 days by reducing)`

`text(either)\ E\ text(or)\ I\ text(by 1 day.)`

`:. text(Minimum cost of $2000 when activity)\ I\ text(is reduced)`

`text(by 1 day.)`
  

e.i.   

♦♦ Mean mark part (e)(i) 21%, (e)(ii) 27%.
MARKER’S COMMENT: In (e)(ii), activity `N` must have an arrow on it.
  

e.ii.    `text(LST)` `=\ text(critical path time − 6 days)`
    `= 15-6`
    `= 9\ text(days from the start.)`

NETWORKS, FUR2 2016 VCAA 2

The suburb of Alooma has a skateboard park with seven ramps.

The ramps are shown as vertices `T`, `U`, `V`, `W`, `X`, `Y` and `Z` on the graph below.
 

 

The tracks between ramps `U` and `V` and between ramps `W` and `X` are rough, as shown on the graph above.

  1. Nathan begins skating at ramp `W` and follows an Eulerian trail.

     

    At which ramp does Nathan finish?   (1 mark)

  2. Zoe begins skating at ramp `X` and follows a Hamiltonian path.

     

    The path she chooses does not include the two rough tracks.

     

    Write down a path that Zoe could take from start to finish.   (1 mark)

  3. Birra can skate over any of the tracks, including the rough tracks.

     

    He begins skating at ramp `X` and will complete a Hamiltonian cycle.

     

    In how many ways could he do this?   (1 mark) 

Show Answers Only
  1. `U`
  2. `XYTUZVW`
    `XYTUZWV`
  3. `4\ text(ways)`
Show Worked Solution

a.   `text{The Eulerian trail (visits each edge exactly once):}`

“XYZWVZUYTU`

`:. text(Finishes at ramp)\ U.`
  

b.   `text{Hamiltonian Paths (touch each vertex exactly once):}`

`XYTUZVW`

`XYTUZWV`
  

c.   `text(Hamiltonian cycles:)`

♦♦ Mean mark 31%.

`XYTUZVWX`

`XYTUVZWX`

`text(These two cycles can be reversed)`

`text(to add two more possibilities.)`

`XWVZUTYX`

`XWZVUTYX`

`:. 4\ text(ways.)`

NETWORKS, FUR2 2016 VCAA 1

A map of the roads connecting five suburbs of a city, Alooma (`A`), Beachton (`B`), Campville (`C`), Dovenest (`D`) and Easyside (`E`), is shown below.
 


  

  1. Starting at Beachton, which two suburbs can be driven to using only one road?   (1 mark)

A graph that represents the map of the roads is shown below.
 


 

One of the edges that connects to vertex `E` is missing from the graph.

  1.  i. Add the missing edge to the graph above.   (1 mark)

    (Answer on the graph above)
  2. ii. Explain what the loop at `D` represents in terms of a driver who is departing from Dovenest.   (1 mark)

Show Answers Only

a.    `text(Alooma and Easyside.)`

b.i. 

b.ii. `text(The loop represents that a driver can take a route out)`

`text(of Dovenest and return home without going through another)`

`text(suburb or turning back.)`

Show Worked Solution

a.   `text(Alooma and Easyside.)`

 

b.i.   

`text(Draw a third edge between Easyside and Dovenest.)`

 

b.ii. `text(The loop represents that a driver can take a)`

♦♦ Mean mark 30%.

`text(route out of Dovenest and return home without)`

`text(going through another suburb or turning back.)`

GRAPHS, FUR2 2016 VCAA 3

A company produces two types of hockey stick, the ‘Flick’ and the ‘Jink’.

Let `x` be the number of Flick hockey sticks that are produced each month.

Let `y` be the number of Jink hockey sticks that are produced each month.

Each month, up to 500 hockey sticks in total can be produced.

The inequalities below represent constraints on the number of each hockey stick that can be produced each month.

Constraint 1 `x >= 0` Constraint 2 `y >= 0`
Constraint 3 `x + y <= 500` Constraint 4 `y <= 2x`
  1. Interpret Constraint 4 in terms of the number of Flick hockey sticks and the number of Jink hockey sticks produced each month.  (1 mark)

There is another constraint, Constraint 5, on the number of each hockey stick that can be produced each month.

Constraint 5 is bounded by Line `A`, shown on the graph below.

The shaded region of the graph contains the points that satisfy constraints 1 to 5.

  1. Write down the inequality that represents Constraint 5.  (1 mark)

The profit, `P`, that the company makes from the sale of the hockey sticks is given by

`P = 62x + 86y`

  1. Find the maximum profit that the company can make from the sale of the hockey sticks.  (1 mark)
  2. The company wants to change the selling price of the Flick and Jink hockey sticks in order to increase its maximum profit to $42 000.

    All of the constraints on the numbers of Flick and Jink hockey sticks that can be produced each month remain the same.

    The profit, `Q`, that is made from the sale of hockey sticks is now given by

    `qquadQ = mx + ny`

     

    The profit made on the Flick hockey sticks is `m` dollars per hockey stick.

    The profit made on the Jink hockey sticks is `n` dollars per hockey stick.

    The maximum profit of $42 000 is made by selling 400 Flick hockey sticks and 100 Jink hockey sticks.

    What are the values of `m` and `n`?  (2 marks)

Show Answers Only
  1. `text(Constraint 4 means the number of Jink sticks produced each)`
    `text(month is less than or equal to twice the number of flick)`
    `text(sticks produced each month.)`
  2. `y <= 300`
  3. `$38\ 200`
  4. `84`
Show Worked Solution

a.   `text(Constraint 4 means the number of Jink sticks)`

♦♦♦ Mean mark 25%.

`text(produced each month is less than or equal to)`

`text(twice the number of flick sticks produced each)`

`text(month.)`

 

b.   `y <= 300`

 

c.   `text(From the equation, 1 Jink stick produces a)`

`text(higher profit than 1 Flick stick.)`

`text(Maximum profit at)\ (200,300)`

`P` `= (62 xx 200) + (86 xx 300)`
  `= $38\ 200`

 

d.   `Q = mx + ny`

♦♦♦ Mean mark 7%.
MARKER’S COMMENT: Very few students were able to identify the need for the sliding line concept and execute it correctly in this question.

`text(Max profit at)\ (400,100)`

`(400,100)\ text(lies on)\ x + y = 500`

`=>\ text(Max profit equation has the same)`

`text(gradient as the profit line.)`

`=> m = −1, m = n`

`text(Using the maximum profit = $42 000,)`

`400m + 100n` `= 42\ 000`
`500m` `= 42\ 000`
`m` `= (42\ 000)/500`
  `= 84`

`:. m = n = 84`

GRAPHS, FUR2 2016 VCAA 2

The bonus money is provided by a company that manufactures and sells hockey balls.

The cost, in dollars, of manufacturing a certain number of balls can be found using the equation

cost = 1200 + 1.5 × number of balls

  1. How many balls would be manufactured if the cost is $1650?  (1 mark)
  2. On the grid below, sketch the graph of the relationship between the manufacturing cost and the number of balls manufactured.  (1 mark)

     

  3. The company will break even on the sale of hockey balls when it manufactures and sells 200 hockey balls.

     

    Find the selling price of one hockey ball.  (1 mark) 

Show Answers Only
  1. `300`

  3. `$7.50\ text(per ball)`
Show Worked Solution
a.    `1200 + 1.5 xx n` `= 1650`
  `:. n` `= ((1650 – 1200))/1.5`
    `= 300`

 

b.   

 

c.   `text(When)\ n = 200,`

`C` `= 1200 + 1.5 xx 200`
  `= 1500`

 

`:.\ text(Selling price if break even)`

`= 1500/200`

`= $7.50\ text(per ball)`

GRAPHS, FUR2 2016 VCAA 1

Maria is a hockey player. She is paid a bonus that depends on the number of goals that she scores in a season.

The graph below shows the value of Maria’s bonus against the number of goals that she scores in a season.

  1. What is the value of Maria’s bonus if she scores seven goals in a season?  (1 mark) 
  2. What is the least number of goals that Maria must score in a season to receive a bonus of $2500?  (1 mark)

Another player, Bianca, is paid a bonus of $125 for every goal that she scores in a season.

  1. What is the value of Bianca’s bonus if she scores eight goals in a season?  (1 mark)
  2. At the end of the season, both players have scored the same number of goals and receive the same bonus amount.

     

    How many goals did Maria and Bianca each score in the season?  (1 mark)

Show Answers Only
  1. `$1500`
  2. `15`
  3. `$1000`
  4. `text(28 goals)`
Show Worked Solution

a.   `$1500`

 

b.   `15`

 

c.    `text(Bonus)` `= 8 xx 125`
    `= $1000`

 

d.   `text(Draw the graph of Bianca’s payments on)`

`text(the same graph.)`

`text(The intersection is where both players)`

`text(receive the same bonus amount.)`

`:.\ text(Maria and Bianca each score 28 goals.)`

MATRICES, FUR2 2016 VCAA 1

A travel company arranges flight (`F`), hotel (`H`), performance (`P`) and tour (`T`) bookings.

Matrix `C` contains the number of each type of booking for a month.

`C = [(85),(38),(24),(43)]{:(F),(H),(P),(T):}`

  1. Write down the order of matrix `C`.   (1 mark)

A booking fee, per person, is collected by the travel company for each type of booking.

Matrix `G` contains the booking fees, in dollars, per booking.

`{:((qquadqquadquadF,\ H,\ P,\ T)),(G = [(40,25,15,30)]):}`

  1.  i. Calculate the matrix product  `J = G × C`.   (1 mark)

  2. ii. What does matrix `J` represent?   (1 mark) 

Show Answers Only
  1. `4 xx 1`
    1. `[6000]`
    2. `J\ text(represents the total booking fees for the travel)`
      `text(company in the given month.)`
Show Worked Solution

a.   `text(Order:)\ 4 xx 1`
 

b.i.    `J = [(40,25,15,30)][(85),(38),(24),(43)]= [6000]`

 
b.ii.  `J\ text(represents the total booking fees for the)`

♦ Mean mark 42%.

 `text(travel company in the given month.)`

GEOMETRY, FUR2 2016 VCAA 4

During a game of golf, Salena hits a ball twice, from `P` to `Q` and then from `Q` to `R`.

The path of the ball after each hit is shown in the diagram below.

After Salena’s first hit, the ball travelled 80 m on a bearing of 130° from point `P` to point `Q`.

After Salena’s second hit, the ball travelled 100 m on a bearing of 054° from point `Q` to point `R`.

  1. Another ball is hit and travels directly from `P` to `R`.

     

    Use the cosine rule to find the distance travelled by this ball.

     

    Round your answer to the nearest metre.  (2 marks)

  2. What is the bearing of `R` from `P`?

     

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

Show Answers Only
  1. `142\ text{m  (nearest m)}`
  2. `0.87^@\ \ (text(nearest degree))`
Show Worked Solution
a.   
`PR^2` `= PQ^2 + QR^2 – 2 xx PQ xx QR xx cos104^@`
  `= 80^2 + 100^2 – 2 xx 80 xx 100 xx cos104^@`
  `= 20\ 270.75…`
  `= 142.37…`
  `= 142\ text{m  (nearest m)}`

 

b.   `text(Find)\ angle RPQ.`

`text(Using the sin rule),`

`(sin angleRPQ)/100` `= (sin104^@)/142`
`sin angle RPQ` `= (100 xx sin104^@)/142`
  `= 0.683…`
`angle RPQ` `= 43.1^@\ \ (text(1 d.p.))`

 

`:. text(Bearing of)\ R\ text(from)\ P`

`= 130 – 43.1`

`= 086.9`

`= 087^@\ \ (text(nearest degree))`

GEOMETRY, FUR2 2016 VCAA 2

Salena practises golf at a driving range by hitting golf balls from point  `T`.

The first ball that Salena hits travels directly north, landing at point  `A`.

The second ball that Salena hits travels 50 m on a bearing of 030°, landing at point  `B`.

The diagram below shows the positions of the two balls after they have landed.
  

  1. How far apart, in metres, are the two golf balls?  (1 mark)
  2. A fence is positioned at the end of the driving range.

     

    The fence is 16.8 m high and is 200 m from the point  `T`.


     
     
    What is the angle of elevation from  `T`  to the top of the fence?

     

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

Show Answers Only
  1. `25\ text(m)`
  2. `5^@\ text{(nearest degree)}`
Show Worked Solution

a.   `text(Let)\ \ d\ text(= distance apart)`

`sin30^@` `= d/50`
`:. d` `= 50 xx sin 30^@`
  `= 25\ text(m)`

 

b.   `text(Let)\ \ x^@\ text(= angle of elevation from)\ T`

`tanx` `= 16.8/200`
  `= 0.084`
`:. x` `= 4.801…`
  `= 5^@\ text{(nearest degree)}`

CORE, FUR2 2016 VCAA 5

Ken has opened a savings account to save money to buy a new caravan.

The amount of money in the savings account after `n` years, `V_n`, can be modelled by the recurrence relation shown below.

`V_0 = 15000, qquad qquad qquad V_(n + 1) = 1.04 xx V_n`

  1. How much money did Ken initially deposit into the savings account?  (1 mark)

  2. Use recursion to write down calculations that show that the amount of money in Ken’s savings account after two years, `V_2`, will be $16 224.  (1 mark)

  3. What is the annual percentage compound interest rate for this savings account?  (1 mark)

  4. The amount of money in the account after `n` years, `Vn` , can also be determined using a rule.
    i.     Complete the rule below by writing the appropriate numbers in the boxes provided.   (1 mark)

    `V_n =` 
     
    		 `­^n xx`
     
  5. ii. How much money will be in Ken’s savings account after 10 years?   (1 mark)

Show Answers Only
  1. `$15000`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `4 text(%)`
  4. i. `text(See Worked Solutions)`
    ii. `$22\ 203.66` 
Show Worked Solution
a.    `text(Initial deposit)` `= V_0`
    `= $15\ 000`
b.    `V_0` `= $15\ 000`
  `V_1` `= 1.04 xx 15\ 000`
    `= $15\ 600`
  `V_2` `= 1.04 xx 15\ 600`
    `= $16\ 224\ text(… as required.)`

MARKER’S COMMENT: (b) Stating `V_2 =1.04^2 xx 15\ 600` `=16\ 224` is not using recursion as required here and did not gain a mark.
c.    `text(Annual compound interest)` `= 0.04 xx 100`
    `= 4 text(%)`

 

d.i.   `V_n` `= 1.04^n xx V_0`
d.ii.    `V_10` `= 1.04^10 xx 15\ 000`
    `= $22\ 203.664…`
    `= $22\ 203.66\ text{(nearest cent)}`

♦ Mean mark (d)(ii) 47%.
MARKER’S COMMENT: Rounding to $22 203.70 lost a mark!

CORE, FUR2 2016 VCAA 3

The data in the table below shows a sample of actual temperatures and apparent temperatures recorded at a weather station. A scatterplot of the data is also shown.

The data will be used to investigate the association between the variables apparent temperature and actual temperature.
 

  1. Use the scatterplot to describe the association between apparent temperature and actual temperature in terms of strength, direction and form.   (1 mark)

  2.  i. Determine the equation of the least squares line that can be used to predict the apparent temperature from the actual temperature.
  3. Write the values of the intercept and slope of this least squares line in the appropriate boxes provided below.
  4. Round your answers to two significant figures.   (3 marks)
     apparent temperature `=`    
 
 `+`  
 
 `xx`   actual temperature
  1. ii. Interpret the intercept of the least squares line in terms of the variables apparent temperature and actual temperature.   (1 mark)

  2. The coefficient of determination for the association between the variables apparent temperature and actual temperature is 0.97
  3. Interpret the coefficient of determination in terms of these variables.   (1 mark)

  4. The residual plot obtained when the least squares line was fitted to the data is shown below.
     
     
  5.  i. A residual plot can be used to test an assumption about the nature of the association between two numerical variables.
  6.     What is this assumption?   (1 mark)
  7. ii. Does the residual plot above support this assumption? Explain your answer.   (1 mark)
Show Answers Only

a.   `text(Strong, positive and linear)`

b.i.  `text(apparent temperature) = -1.7 xx 0.94 xx text(actual temperature)`

b.ii.  `text(When actual temperature is 0°C, on average,)`

`text(the apparent temperature is)\ − 1.7^@\text(C.)`

c.  `text(97% of the variation in the apparent temperature can be explained)`

`text(by the variation in the actual temperature.)`

d.i.  `text(There is a linear relationship between the two variables.)`

d.ii.  `text(The random pattern supports the assumption.)`

`text{(Students should refer to randomness or a lack of pattern}`

`text{explicitly here).}`

Show Worked Solution

a.   `text(Strong, positive and linear)`
 

b.i.   `text(By calculator:)`

`text(apparent temperature) = -1.7 xx 0.94 xx text(actual temperature)`
 

♦♦ Mean mark of part (b)(ii) – 28%.
MARKER’S COMMENT: “the predicted apparent temp is -1.7°C” also gained a mark.
b.ii.    `text(When actual temperature is 0°C, on average,)`
 

`text(the apparent temperature is)\ − 1.7^@\text(C.)`

 

♦ Mean mark 49%.
IMPORTANT: Any mention of causality loses a mark!

c.  `text(97% of the variation in the apparent temperature can be explained)`

`text(by the variation in the actual temperature.)`
  

d.i.  `text(There is a linear relationship between the two variables.)`

♦ Mean mark of both parts of (d) was 46%.

d.ii.  `text(The random pattern supports the assumption.)`

`text{(Students should refer to randomness or a lack of pattern}`

`text{explicitly here).}`

CORE, FUR2 2016 VCAA 2

A weather station records daily maximum temperatures.

  1. The five-number summary for the distribution of maximum temperatures for the month of February is displayed in the table below.

 

  1. There are no outliers in this distribution.
  2.  i. Use the five-number summary above to construct a boxplot on the grid below.   (1 mark)

 

  1. ii. What percentage of days had a maximum temperature of 21°C, or greater, in this particular February?   (1 mark)

  2. The boxplots below display the distribution of maximum daily temperature for the months of May and July.
     

  3.   i. Describe the shapes of the distributions of daily temperature (including outliers) for July and for May.   (1 mark)

  4.  ii. Determine the value of the upper fence for the July boxplot.   (1 mark)

  5. iii. Using the information from the boxplots, explain why the maximum daily temperature is associated with the month of the year. Quote the values of appropriate statistics in your response.   (1 mark)

Show Answers Only
a.i.   

a.ii.   `text(75%)`

b.i.    `text(July – Positively skewed with an outlier.)`
  `text(May – Symmetrical with no outliers.)`

b.ii.  `15.5^@\text(C)`

b.iii. `text{The median temperature in May (14.5°C)}`

`text(differs from the median temperature in July)`

`text{(just over 9°C). This difference is why the}`

`text(maximum daily temperature is associated)`

`text(with the month.)`

Show Worked Solution
a.i.   

a.ii.   `text(75%)`

MARKER’S COMMENT: Incorrect May descriptors included “evenly or normally distributed”, “bell shaped” and “symmetrically skewed.”
b.i.    `text(July – Positively skewed with an outlier.)`
  `text(May – Symmetrical with no outliers.)`

 

b.ii.    `text(Upper fence)` `= Q_3 + 1.5 xx IQR`
    `= 11 + 1.5 xx (11 – 8)`
    `= 11 + 4.5`
    `= 15.5^@\text(C)`
♦♦ Mean mark (b)(iii) – 30%.
COMMENT: Refer to the difference in medians. Just quoting the numbers was not enough to gain a mark here.

b.iii. `text{The median temperature in May (14.5°C)}`

`text(differs from the median temperature in July)`

`text{(just over 9°C). This difference is why the}`

`text(maximum daily temperature is associated)`

`text(with the month.)`

CORE, FUR2 2016 VCAA 1

The dot plot below shows the distribution of daily rainfall, in millimetres, at a weather station for 30 days in September.
 

 

  1. Write down the
  2.  i. range   (1 mark)

  3. ii. median   (1 mark)

  1. Circle the data point on the dot plot above that corresponds to the third quartile `(Q_3).`   (1 mark)

  2. Write down the
  3.  i. the number of days on which no rainfall was recorded.   (1 mark)

  4. ii. the percentage of days on which the daily rainfall exceeded 12 mm.   (1 mark)

  1. Use the grid below to construct a histogram that displays the distribution of daily rainfall for the month of September. Use interval widths of two with the first interval starting at 0.   (2 marks) 

  

Show Answers Only
    1. `17.8\ text(mm)`
    2. `0`
  2. `text(See Worked Solutions)`

     

    1. `16\ text(days)`
    2. `10\text(%)`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.i.    `text(Range)` `=\ text(High) – text(Low)`
    `= 17.8 – 0`
    `= 17.8\ text(mm)`
     

a.ii.   `text(30 data points)`

`text(Median)` `= text{15th + 16th}/2`
  `= 0`

 

b.   

 

c.i.   `16\ text(days)`

c.ii.    `text(Percentage)` `= 3/30 xx 100`
    `= 10\ text(%)`

 

d.   

NETWORKS, FUR1 2016 VCAA 6-7 MC

The directed graph below shows the sequence of activities required to complete a project.

All times are in hours.
 

 
Part 1

The number of activities that have exactly two immediate predecessors is

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4

 
Part 2

There is one critical path for this project.

Three critical paths would exist if the duration of activity

  1. I were reduced by two hours.
  2. E were reduced by one hour.
  3. G were increased by six hours.
  4. K were increased by two hours.
  5. F were increased by two hours. 
Show Answers Only

`text(Part 1:)\ C`

`text(Part 1:)\ B`

Show Worked Solution

`text(Part 1)`

`I\ text(and)\ J`

`=> C`

 

`text(Part 2)`

♦ Mean mark of Part 2: 45%.

`text(The possible paths are:)`

`ADIL – 19\ text(mins)`

`BEIL – 20\ text(mins)`

`BFJL – 17\ text(mins)`

`CGJL – 13\ text(mins)`

`CHKL – 19\ text(mins)`

`text(If)\ E\ text(were reduced by 1 hour,)\ BEIL\ text(will)`

`text{take 19 minutes (i.e. 3 critical paths of 19}`

`text{minutes would exist).}`

`=> B`

NETWORKS, FUR1 2016 VCAA 2 MC

The following directed graph shows the flow of water, in litres per minute, in a system of pipes connecting the source to the sink.
 

 
The maximum flow, in litres per minute, from the source to the sink is

  1. `10`
  2. `14`
  3. `18`
  4. `20`
  5. `22`
Show Answers Only

`C`

Show Worked Solution
`text(Maximum flow)` `= 2 + 10 + 6`
  `= 18\ \ text(litres/minute)`

`=> C`

GRAPHS, FUR1 2016 VCAA 4 MC

The point  (3, –2)  satisfies the inequality

  1. `y >= 0`
  2. `y >= 2/3 x`
  3. `−2 < x < 3`
  4. `y < 5 -x`
  5. `y > x - 5` 
Show Answers Only

`D`

Show Worked Solution

`text(Consider option)\ D,`

`y` `< 5 – x`
`−2` `< 5 – 3`
`−2` `< 2\ \ \ text{(correct)}`

`=> D`

GEOMETRY, FUR1 2016 VCAA 6 MC

Marcus is on the opposite side of a large lake from a horse and its stable. The stable is 150 m directly east of the horse. Marcus is on a bearing of 170° from the horse and on a bearing of 205° from the stable.

The straight-line distance, in metres, between Marcus and the horse is closest to

  1.   `45`
  2.   `61`
  3.   `95`
  4. `192`
  5. `237`
Show Answers Only

`E`

Show Worked Solution

`text(Using the sine rule,)`

`(HM)/(sin65^@)` `= (HS)/(sin35^@)`
`HM` `= (150 xx sin 65^@)/(sin35^@)`
  `= 237.01…\ text(m)`

`=> E`

GEOMETRY, FUR1 2016 VCAA 5 MC

A water tank in the shape of a cylinder with a hemispherical top is shown below.

The volume of water that this tank can hold, in cubic metres, is closest to

  1.   `80`
  2.   `88`
  3.   `96`
  4. `105`
  5. `121`
Show Answers Only

`A`

Show Worked Solution
`V` `=\ text(cylinder + half sphere)`
  `= pir^2h + 1/2 xx 4/3 xx pir^3`
  `= pi xx 2^2 xx 5 + 1/2 xx 4/3 xx pi xx 2^3`
  `= 79.587…`
  `~~ 80\ text(m³)`

`=> A`

GEOMETRY, FUR1 2016 VCAA 3 MC

The diagram above shows the position of three cities, A, B and C.

According to the diagram

  1. A is located on a longitude of 60° N
  2. B is located on a longitude of 20° E
  3. C is located on a longitude of 40° S
  4. A and B are located on the same line of longitude
  5. A and C are located on the same line of longitude 
Show Answers Only

`E`

Show Worked Solution

`A\ text(and)\ C\ text(are on the same line)`

`text(of longitude.)`

`text(All other statements can be shown)`

`text(to be false.)`

`=> E`

MATRICES, FUR1 2016 VCAA 3 MC

The matrix equation below represents a pair of simultaneous linear equations.
 

`[(12,9),(m,3)][(x),(y)] = [(6),(6)]`
 

These simultaneous linear equations have no unique solution when `m` is equal to

  1. `−4`
  2. `−3`
  3.     `0`
  4.     `3`
  5.     `4`
Show Answers Only

`E`

Show Worked Solution
`[(x),(y)]` `= [(12,9),(m,3)]^(−1)[(6),(6)]`
  `= 1/((12 xx 3) – (9 xx m)) [(3,−9),(−m,12)][(6),(6)]`

 

`text(No unique solution occurs when)\ \ Δ=0 :`

`(12 xx 3) – (9 xx m)` `= 0`
`9m` `= 36`
`m` `= 4`

`=> E`

CORE, FUR1 2016 VCAA 20 MC

Consider the recurrence relation below.

`V_0 = 10\ 000,\ \ \ \ \ V_(n + 1) = 1.04 V_n + 500`

This recurrence relation could be used to model

  1. a reducing balance depreciation of an asset initially valued at $10 000.
  2. a reducing balance loan with periodic repayments of $500.
  3. a perpetuity with periodic payments of $500 from the annuity.
  4. an annuity investment with periodic additions of $500 made to the investment.
  5. an interest-only loan of $10 000. 
Show Answers Only

`D`

Show Worked Solution

`text(Each period, the model increases)`

♦ Mean mark 50%.

`text(the previous value by 4% and then)`

`text(adds 500 to the total.)`

`=> D`

CORE, FUR1 2016 VCAA 19 MC

The purchase price of a car was $26 000.

Using the reducing balance method, the value of the car is depreciated by 8% each year.

A recurrence relation that can be used to determine the value of the car after `n` years, `C_n` , is

  1. `C_0 = 26\ 000,\ \ \ C_(n + 1) = 0.92 C_n`
  2. `C_0 = 26\ 000,\ \ \ C_(n + 1) = 1.08 C_n`
  3. `C_0 = 26\ 000,\ \ \ C_(n + 1) = C_n + 8`
  4. `C_0 = 26\ 000,\ \ \ C_(n + 1) = C_n - 8`
  5. `C_0 = 26\ 000,\ \ \ C_(n + 1) = 0.92 C_n - 8`
Show Answers Only

`A`

Show Worked Solution

`C_0 = 26\ 000\ (text(given))`

`text(After 1 year:)`

`C_1` `= C_0 – (8text(%) xx C_0)`
  `= 0.92 C_0`

 
`text(After 2 years:)`

`C_2 = 0.92C_1`

`vdots`

`:.\ text(After)\ n\ text(years,)`

`C_(n + 1) = 0.92 C_n`

`=> A`

CORE, FUR1 2016 VCAA 14-16 MC

The table below shows the long-term average of the number of meals served each day at a restaurant. Also shown is the daily seasonal index for Monday through to Friday.

 

Part 1

The seasonal index for Wednesday is 0.84

This tells us that, on average, the number of meals served on a Wednesday is

  1. 16% less than the daily average.
  2. 84% less than the daily average.
  3. the same as the daily average.
  4. 16% more than the daily average.
  5. 84% more than the daily average.

 

Part 2

Last Tuesday, 108 meals were served in the restaurant.

The deseasonalised number of meals served last Tuesday was closest to

  1.   `93`
  2. `100`
  3. `110`
  4. `131`
  5. `152`

 

Part 3

The seasonal index for Saturday is closest to

  1. `1.22`
  2. `1.31`
  3. `1.38`
  4. `1.45`
  5. `1.49`
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ E`

`text(Part 3:)\ D`

Show Worked Solution

`text(Part 1)`

`1 – 0.84 = 0.16`

`:.\ text(A seasonal index of 0.84 tell us)`

`text(16% less meals are served.)`

`=> A`

 

`text(Part 2)`

`text{Deseasonalised number (Tues)}`

`= text(actual number)/text(seasonal index)`

`= 108/0.71`

`~~ 152`

`=> E`

 

`text(Part 3)`

`text(S)text(ince the same number of deseasonalised)`

`text(meals are served each day.)`

`text{S.I. (Sat)}/190` `= 1.10/145`
`text{S.I. (Sat)}` `= (1.10 xx 190)/145`
  `= 1.44…`

`=> D`

CORE, FUR1 2016 VCAA 11-12 MC

The table below gives the Human Development Index (HDI) and the mean number of children per woman (children) for 14 countries in 2007.

A scatterplot of the data is also shown. 
 

 

Part 1

The scatterplot is non-linear.

A log transformation applied to the variable children can be used to linearise the scatterplot.

With HDI as the explanatory variable, the equation of the least squares line fitted to the linearised data is closest to

  1. log(children) = 1.1 – 0.0095 × HDI
  2. children = 1.1 – 0.0095 × log(HDI)
  3. log(children) = 8.0 – 0.77 × HDI
  4. children = 8.0 – 0.77 × log(HDI)
  5. log(children) = 21 – 10 × HDI

 

Part 2

There is a strong positive association between a country’s Human Development Index and its carbon dioxide emissions.

From this information, it can be concluded that

  1. increasing a country’s carbon dioxide emissions will increase the Human Development Index of the country.
  2. decreasing a country’s carbon dioxide emissions will increase the Human Development Index of the country.
  3. this association must be a chance occurrence and can be safely ignored.
  4. countries that have higher human development indices tend to have higher levels of carbon dioxide emissions.
  5. countries that have higher human development indices tend to have lower levels of carbon dioxide emissions.
Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`=>A`

 

`text(Part 2)`

`text(A strong positive association does not mean)`

`text(an increase in one variable causes an increase)`

`text(in the other.)`

`=> D`