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Financial Maths, 2ADV M1 SM-Bank 4 MC

Each trading day, a share trader buys and sells shares according to the rule
 

 `T_(n+1)=0.6 T_n + 50\ 000`
 

where `T_n` is the number of shares the trader owns at the start of the `n`th trading day.

From this rule, it can be concluded that each day

A.    the trader sells 60% of the shares that she owned at the start of the day and then buys another 50 000 shares.

B.    the trader sells 40% of the shares that she owned at the start of the day and then buys another 50 000 shares.

C.    the trader sells 50 000 of the shares that she owned at the start of the day.

D.    the trader sells 60% of the 50 000 shares that she owned at the start of the day.

Show Answers Only

`B`

Show Worked Solution

`T_(n+1)=0.6\ \T_n + 50 000`

`text(The difference equation describes a rule)`

`text(where a trader sells 40% of shares owned on)`

`text{the day before (left with 60% or 0.6}\ T_n text{)} `

`text(and then buys another 50 000 each day.)`

`=> B`

Financial Maths, 2ADV M1 SM-Bank 3 MC

The first four terms of a sequence are

`12, 18, 30, 54`

A recursive equation that generates this sequence is

A. `t_(n+1)` `= t_n + 6` `t_1` `=12`
B. `t_(n+1)` `= 1.5t_n` `t_1` `= 12`
C. `t_(n+1)` `= 0.5t_n + 12` `t_1` `= 12`
D. `t_(n+1)` `= 2t_n - 6` `t_1` `= 12`
Show Answers Only

`D`

Show Worked Solution

`text(Calculating)` `t_3` `text(in each given option)`

`text(eliminates)\ A, B\ text(and)\ C.`

`rArr D`

Financial Maths, 2ADV M1 SM-Bank 2 MC

The values of the first five terms of a sequence are plotted on the graph shown below.
 

 
The recursion equation that could describe the sequence is

A.   `t_(n+1) = t_n + 5,` `\ \ \ \ \ t_1 = 4`
B.   `t_(n+1) = 2t_n + 1,` `\ \ \ \ \ t_1 = 4`
C.   `t_(n+1) = t_n - 3,` `\ \ \ \ \ t_1 = 4`
D.   `t_(n+1) = 3t_n,` `\ \ \ \ \ t_1 = 4`
Show Answers Only

`B`

Show Worked Solution

`text(By elimination,)`

`text(There is no common difference between terms,)`

`:.\ text(Cannot be A, or C)`

`text(The equation in B has)\ \ t_2=9,\ \ text(while the equation)`

`text(in C has)\ \ t_2=12.`

`rArr B`

Probability, 2ADV S1 SM-Bank 4 MC

In a classroom, students are asked what sports club they are members of and the results are shown in the Venn diagram.
 


 

A student who is a member of a soccer club is chosen at random. What is the probability that he/she is also a member of a surf club?

  1. `2/5`
  2. `4/11`
  3. `2/9`
  4. `7/18`
Show Answers Only

`D`

Show Worked Solution
`P(text(Surf | Soccer))` `= (n(text(Surf) ∩ text(Soccer)))/(n(text(Soccer)))`
  `= (3 + 4)/(3 + 4 + 5 + 6)`
  `= 7/18`

 
`=>\ D`

Statistics, 2ADV S3 SM-Bank 20

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

`f(x) = {{:(Cx + D),(0):}\ \ \ \ {:(2 <= x <= 5),(text(elsewhere)):}:}`
 

where `C` and `D` are constants.

Find the exact values of `C` and `D`, given the median of  `X`  is 4.  (4 marks)

Show Answers Only

`C = 1/6`

`D = −1/4`

Show Worked Solution

`int_2^5 Cx + D\ dx = 1`

`[C/2 x^2 + Dx]_2^5 = 1`

`[(25/2 C + 5D) – (2C + 2D)]` `= 1`
`21/2C + 3D` `= 1`
`21C + 6D` `= 2\ \ …\ (1)`

 
`text(Using median)\ \ X = 4:`

`[C/2 x^2 + Dx]_2^4 = 0.5`

`[(8C + 4D) – (2C + 2D)]` `= 0.5`
`6C + 2D` `= 0.5\ \ …\ (2)`

  
`text(Multiply:)\ (2) xx 3`

`18C + 6D = 1.5\ \ …\ (3)`
 

`text(Subtract:)\ \ (1) – (3)`

`3C` `= 1/2`
`:. C` `= 1/6`

 
`text{Substitute into (1):}`

`21/6 + 6D` `= 2`
`6D` `= −9/6`
`:. D` `= −1/4`

Statistics, 2ADV S3 SM-Bank 12

The function
 

`f(x) = {{:(k),(0):}{:(sin(pix)qquad\ \ 0<=x<=1),(qquadqquadqquadqquadquadtext(otherwise)):}`
 

is a probability density function for the continuous random variable `X`.

Show that  `k = pi/2`.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Total Area under curve) = 1\ text(u²)`

`int_0^1 k sin(pix)\ dx` `= 1`
`- k/pi [cos(pix)]_0^1` `= 1`
`- k/pi[cos(pi) – cos(0)]` `= 1`
`- k/pi[-1 – 1]` `= 1`
`2k` `= pi`
`:.k` `= pi/2`

Statistics, 2ADV S3 SM-Bank 11

The probability density function of a continuous random variable \(X\) is given by

\(f(x)=\begin{cases}
\dfrac{x}{12} & 1 \leq x \leq 5 \\
\ \\
0 & \text {otherwise }
\end{cases}\)

Find  \(P(X < 3)\)  (2 marks)

Show Answers Only

\(\dfrac{1}{3}\)

Show Worked Solution

\(\begin{aligned}
P(X & <3)\\
& =\int_1^3 \dfrac{1}{12} x d x \\
& =\dfrac{1}{12}\left[\frac{1}{2} x^2\right]_1^3 \\
& =\dfrac{1}{24}\left[3^2-1^2\right] \\
& =\dfrac{1}{3}
\end{aligned}\)

Statistics, 2ADV S3 SM-Bank 10

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

\(f(x)= \begin{cases}\dfrac{1}{5}\,e^{-\frac{x}{5}} & x \geq 0 \\
\ \\
0 & x<0
\end{cases}\) 

The median of \(X\) is  \(m\).

Determine the value of  \(m\).  (3 marks)

Show Answers Only

\(-5 \log _e\left(\dfrac{1}{2}\right)\) or \(5 \log _e(2)\) or \(\log _e 32\)

Show Worked Solution

\(\begin{aligned} \dfrac{1}{5} \int_0^m e^{-\frac{x}{5}} d x & =\dfrac{1}{2} \\
\dfrac{1}{5} \times(-5)\left[e^{-\frac{x}{5}}\right]_0^m & =\dfrac{1}{2} \\
{\left[-e^{-\frac{x}{5}}\right]_0^m } & =\dfrac{1}{2} \\
-e^{-\frac{m}{5}}+1 & =\frac{1}{2} \\ e^{-\frac{m}{5}} & =\dfrac{1}{2} \\
-\frac{m}{5} & =\log _e\left(\dfrac{1}{2}\right)
\end{aligned}\)

\(\therefore m=-5 \log _e\left(\dfrac{1}{2}\right)\) (or \(5 \log _e(2)\), or \(\log _e 32\) )

Statistics, 2ADV S3 SM-Bank 9

The probability density function  \(f(x)\)  of a random variable \(X\) is given by

\(f(x)=\begin{cases}
\dfrac{x+1}{12} & 0 \leq x \leq 4 \\
\ \\
0 & \text{otherwise }
\end{cases}\)

Find the value of \(b\) such that \(P(X \leq b)=\dfrac{5}{8}\).  (3 marks)

Show Answers Only

\(3\)

Show Worked Solution

\(\begin{aligned}
\dfrac{1}{12} \int_0^b(x+1) d x & =\dfrac{5}{8} \\
{\left[\dfrac{1}{2} x^2+x\right]_0^b } & =\dfrac{15}{2} \\
\dfrac{1}{2} b^2+b & =\dfrac{15}{2} \\
b^2+2 b-15 & =0 \\ (b+5)(b-3) & =0
\end{aligned}\)

\(\therefore b=3 \quad(0 \leq b \leq 4)\)

Statistics, 2ADV S3 SM-Bank 8

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

`f(x) = {(ax(5 - x), \ text(if)\ \ 0 <= x <= 5), (0,\ text (if)\ \ x < 0\ \ text(or if)\ \ x > 5):}`
 

where `a` is a positive constant.

  1. Find the value of  `a`.  (3 marks)

  2. Express  `P(X < 3)`  as a  definite integral. (Do not evaluate the definite integral.)  (1 mark)

Show Answers Only
  1. `6/125`
  2. `int_0^3 ax(5 – x)\ dx`
Show Worked Solution
a.   `text(Total Area under curve)` `= 1`
  `a int_0^5 (5x – x^2)\ dx` `= 1`
  `a [5/2 x^2 – 1/3 x^3]_0^5` `= 1`
  `a [(125/2 – 125/3) – (0)]` `= 1`
  `125/6 a` `= 1`
  `:. a` `= 6/125`

 

b.   `P(X < 3) = int_0^3 ax(5 – x)\ dx`

Statistics, 2ADV S3 SM-Bank 5 MC

The function  `f(x)`  is a probability density function of a continuous random variable with the rule
 

`f(x) = {(ae^x, 0 <= x <= 1), (ae, 1 < x <= 2), (\ 0, text(otherwise)):}`
 

The value of `a` is

A.   `1`

B.   `1/e`

C.   `1/(2e)`

D.   `1/(2e - 1)`

Show Answers Only

`D`

Show Worked Solution

`text(Total area) = 1`

`int_0^1 ae^x\ dx + int_1^2 ae\ dx` `= 1`
`[ae^x]_0^1 + [ae*x]_1^2` `=1`
`[ae-a] + [2ae-ae]` `=1`
`2ae-a` `=1`
`a(2e-1)` `=1`
`:. a` `= 1/(2e – 1)`

 
`=>   D`

Statistics, 2ADV S3 SM-Bank 2

If a continuous random variable  \(X\)  has probability density function

\(f(x)=
\begin{cases}
\dfrac{x}{2} & \text{if } \quad 0 \leq x \leq 2 \\
\ \\
0 & \text {otherwise }
\end{cases}\)

Find the exact value of  \(p\)  such that  \(P(X>p) = 0.4\).   (3 marks)

Show Answers Only

\(\dfrac{2 \sqrt{15}}{5}\)

Show Worked Solution

\(\displaystyle \int_p^2 \dfrac{x}{2} d x=0.4\)

\(\begin{aligned} {\left[\dfrac{x^2}{4}\right]_p^2 } & =\dfrac{2}{5} \\
1-\dfrac{p^2}{4} & =\dfrac{2}{5} \\
\dfrac{p^2}{4} & =\dfrac{3}{5} \\ p^2 & =\dfrac{12}{5} \\
\therefore p & =\dfrac{2 \sqrt{3}}{\sqrt{5}} \\
& =\dfrac{2 \sqrt{15}}{5} \quad(0 \leq p \leq 2)
\end{aligned}\)

Probability, 2ADV S1 EQ-Bank 40

One bag contains red and green balls.

Kalyn randomly chooses one ball from the bag. Without replacement, he then chooses a second ball from the bag.

Complete the tree diagram below and then draw a probability distribution table for the number of red balls that could be drawn out of the bag.  (3 marks)
 

Show Answers Only

Show Worked Solution

 

`x = 2: \ P(R R) = 4/9 xx 3/8 = 1/6`

`x = 1: \ P(R and G) = 4/9 xx 5/8 + 5/9 xx 4/8 = 5/9`

`x = 0: \ P(G G) = 5/9 xx 4/8 = 5/18`
 

Probability, 2ADV S1 SM-Bank 41

Evaluate `p` and `q` in the discrete probability distribution table below, given that  `E(X) = 3`.   (3 marks)

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  & \ \ \ 4\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p & q & 0.2 & 0.4 \\
\hline
\end{array}

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution
`p + q + 0.2 + 0.4` `= 1`
`p + q` `= 0.4\ \ …\ (1)`

 
`text(Given)\ \ E(X) = 3,`

`p + 2q + 0.6 + 1.6` `= 3`
`p + 2q` `= 0.8\ \ …\ (2)`

 
`text{Subtract: (2) – (1)}`

`q` `= 0.4`
`:.p` `= 0`

Trigonometry, 2ADV T2 SM-Bank 42

Prove that

`(1 - sin^2 x cos^2 x)/(sin^2 x) = cot^2 x + sin^2 x`.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution
`text(RHS)` `= (cos^2 x)/(sin^2 x) + sin^2 x`
  `= (cos^2 x + sin^4 x)/(sin^2 x)`
  `= (cos^2 x + sin^2 x(1 – cos^2 x))/(sin^2 x)`
  `= (cos^2 x + sin^2 x – sin^2 x cos^2 x)/(sin^2 x)`
  `= (1 – sin^2 x cos^2 x)/(sin^2 x)`
  `= \ text(LHS)`

Trigonometry, 2ADV T3 SM-Bank 16

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

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


 

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

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

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

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

 

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

 

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

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

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

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

Functions, 2ADV F1 SM-Bank 32

Find the centre and radius of the circle with the equation

`x^2-12x + y^2 + 2y-12 = 0`  (2 marks)

Show Answers Only

`text(Centre)\ (6, −1)`

`text(Radius = 7)`

Show Worked Solution
`x^2-12x + y^2 + 2y-12` `= 0`
`(x-6)^2 + (y + 1)^2-36-1-12` `= 0`
`(x-6)^2 + (y + 1)^2` `= 49`

 
`:.\ text(Centre)\ (6, −1)`

`:.\ text(Radius = 7)`

Functions, 2ADV F1 SM-Bank 31

Find the domain and range of  `f(g(x))`  given

`f(x) = 2x^2 - 8x`  and  `g(x) = x + 2`.   (2 marks)

Show Answers Only

`text(Domain: all)\ x`

`text(Range:)\ −8<=y< ∞`

Show Worked Solution
`f(g(x))` `= 2(x + 2)^2 – 8(x + 2)`
  `= 2(x^2 + 4x + 4) – 8x – 16`
  `= 2x^2 + 8x + 8 – 8x – 16`
  `= 2(x^2 – 4)`

 
`:. text(Domain: all)\ x`

`:. text(Range:)\ −8<=y< ∞`

Functions, 2ADV F1 SM-Bank 30

Given  `f(x) = sqrtx`  and  `g(x) = 25 - x^2`

  1. Find  `g(f(x))`.  (1 mark)

  2. Find the domain and range of  `f(g(x))`.  (2 marks)

Show Answers Only
  1. `25 – x`
  2. `text(Domain:)\ −5<= x <= 5`

     

    `text(Range:)\ 0<=y<= 5`

Show Worked Solution
i.    `g(f(x))` `= 25 – (f(x))^2`
    `= 25 – (sqrtx)^2`
    `= 25 – x`

 

ii.    `f(g(x))` `= sqrt(g(x))`
    `= sqrt(25 – x^2)`

 
`:.\ text(Domain:)\ −5<= x <= 5`

`:.\ text(Range:)\ 0<=y<= 5`

Trigonometry, 2ADV T2 SM-Bank 39

Let  `f(x) = sin((2pix)/3)`.

Solve the equation  `sin((2pix)/3) = -sqrt3/2`  for  `0<=x<=3`  (2 marks)

Show Answers Only

`x = 2, 5/2`

Show Worked Solution

`sin((2pix)/3) = -sqrt3/2`

`=>\ text(Base angle)\ = pi/3`

`(2 pi x)/3` `=(4pi)/3, (5pi)/3, (10pi)/3, …` 
`:.x` `=2 or 5/2, \ \ \ (0<=x<=3)`

Trigonometry, 2ADV T3 SM-Bank 13

On any given day, the depth of water in a river is modelled by the function

`h(t) = 14 + 8sin((pit)/12),\ \ 0 <= t <= 24`

where `h` is the depth of water, in metres, and  `t`  is the time, in hours, after 6 am. 

  1. Find the minimum depth of the water in the river.  (1 mark)

  2. Find the values of  `t`  for which  `h(t) = 10`.  (2 marks)

Show Answers Only
  1. `6\ text(m)`
  2. `14quadtext(or)quad22`
Show Worked Solution

i.   `h_(text(min))\ text(occurs when)\ \ sin((pit)/12)=-1`

MARKER’S COMMENT: Students who used calculus to find the minimum were less successful.
`:. h_(text(min))` `= 14 – 8`
  `= 6\ text(m)`

 

ii.    `14 + 8sin(pi/12t)` `= 10`
  `sin(pi/12t)` `= – 1/2`

 

`text(Solve in general:)`

`pi/12t` `=(7pi)/6 + 2pi n\ \ \ \ text(or)\ \ \ `  `pi/12t` `= (11t)/6 + 2pi n,`
`t` `= 14 + 24n` `t` `=22 + 24n`

 

`text(Substitute integer values for)\ n,`

`:. t = 14quadtext(or)quad22,\ \ \ (0<=t<=24)`

Trigonometry, 2ADV T2 SM-Bank 38

Solve  `2cos(2x) = −sqrt3`  for  `x`, where  `0 <= x <= pi`.  (2 marks) 

Show Answers Only

`x = (5pi)/12, (7pi)/12`

Show Worked Solution
`cos(2x)` `= – sqrt3/2`
`2x` `= (5pi)/6, 2pi – (5pi)/6, 2pi+(5pi)/6`
  `=(5pi)/6, (7pi)/6, (17pi)/6,\ …`
`:. x` `=(5pi)/12, (7pi)/12\ \ \ (0 <= x <= pi)`

Trigonometry, 2ADV T2 SM-Bank 37

Solve the equation  `cos((3x)/2) = 1/2`  for  `−pi/2<=x<=pi/2`.  (2 marks)

Show Answers Only

`x = ± (2pi)/9`

Show Worked Solution

`cos((3x)/2) = 1/2`

`=>\ text(Base angle)\ =pi/3`

`(3x)/2` `= (-pi)/3, pi/3, (5pi)/3, …`
`:. x` `=(-2pi)/9, (2pi)/9, (10pi)/9`
  `= (-2pi)/9, (2pi)/9\ \ \ (−pi/2<=x<=pi/2)`

Trigonometry, 2ADV T2 SM-Bank 36

Solve the equation  `sin (x/2) = -1/2`  for  `2 pi<=x<= 4 pi`  (2 marks)

Show Answers Only

`x = (7 pi)/3, (11 pi)/3`

Show Worked Solution
`x/2` `=pi/6 + pi, 2pi – pi/6, 2pi + (pi/6 +pi), …`
  `=(7pi)/6, (11pi)/6, (19pi)/6, …`
`:. x` `=(7pi)/3, (11pi)/3, (19pi)/3, …`
   

`text(Given)\ \ \2 pi<=x<= 4 pi`

`:. x = (7 pi)/3, (11 pi)/3`

Trigonometry, 2ADV T2 SM-Bank 35

Solve the equation

`sin (2x + pi/3) = 1/2\ \ text(for)\ \ 0<= x <=pi`  (2 marks)

Show Answers Only

`x = pi/4, (11 pi)/12`

Show Worked Solution

`sin (2x + pi/3) = 1/2`

`=>\ text(Base angle is)\ \ pi/6`

`(2x + pi/3)` `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, …`
`2x` `= – pi/6, pi/2, (11pi)/6, (15pi)/6, …`
`x` `= – pi/12, pi/4, (11pi)/12, (15pi)/12, …`

 
`:. x = pi/4, (11 pi)/12\ \ (0<=x<=pi)`

Trigonometry, 2ADV T3 SM-Bank 12

State the range and period of the function

`h(x) = 4 + 3 cos ((pi x)/2).`  (2 marks)

Show Answers Only

`text(Range:)\ \ 1<=y<= 7`

`text(Period) = 4`

Show Worked Solution
  `-1` `<= cos ((pi x)/2)<=1`
  `-3` `<=3cos ((pi x)/2)<=3`
  `1` `<= 4+ 3cos ((pi x)/2)<=7`

 
`:.\ text(Range:)\ \ 1<=y<= 7`

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

Trigonometry, 2ADV T2 SM-Bank 34

Solve the equation  `sqrt 3 sin x = cos x`  for  `– pi<=x<= pi`.  (2 marks)

Show Answers Only

`x = pi/6,\ \ \ – (5 pi)/6`

Show Worked Solution

`text(Divide both sides by)\ \ cos x :`

MARKER’S COMMENT: Many students who found the base angle correctly could not solve within the restrictions.
`sqrt 3 sin x` `=cos x`
`sqrt 3 tan x` `= 1`
`tan x` `= 1/sqrt 3`
`=>\ text(Base angle)\ = pi/6`

 
`:. x = pi/6\ \ text(or)\ – (5 pi)/6,\ \ \ (– pi <=x<= pi)`

Trigonometry, 2ADV T3 SM-Bank 10

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

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

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

  3. Find  `n(10)`.  (1 mark)

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

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

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

`text(A)text(mplitude) = 400`
 

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

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

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

 

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

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

 
`text(S)text(ince)\ \ n(0)=1600,`

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

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

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

Trigonometry, 2ADV T3 SM-Bank 9

Let   `f(x) = 2cos(x) + 1`  for  `0<=x<=2pi`.

  1. Solve the equation  `2cos(x) + 1 = 0`  for  `0 <= x <= 2pi`.  (2 marks)

  2. Sketch the graph of the function  `f(x)`  on the axes below. Label the endpoints and local minimum point with their coordinates.  (3 marks)

 

Show Answers Only
  1. `(2pi)/3, (4pi)/3`
  2.  
Show Worked Solution
i. `2cos(x) + 1` `= 0`
  `cos(x)` `= −1/2`

`=> cos\ pi/3 = 1/2\ text(and cos is negative)`

`text(in 2nd/3rd quadrant)`

`:.x` `= pi – pi/3, pi + pi/3`
  `= (2pi)/3, (4pi)/3`

 

ii.   

Trigonometry, 2ADV T3 SM-Bank 7 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`
Show Answers Only

`=> B`

Show Worked Solution

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

`text{Graph centre line (median):}\ \ y= 1`

`:.\ text(Eliminate)\ \ C\ \ text(and)\ \ D.`
 

`text(Period) = (2pi)/3 – pi/6 = pi/2\ \ text{(from graph)}`

`text(Consider option)\ B,`

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

`=> B`

Trigonometry, 2ADV T3 SM-Bank 5 MC

The UV index, `y`, for a summer day in Newcastle East is illustrated in the graph below, where  `t`  is the number of hours after 6 am.
 

 
 

The graph is most likely to be the graph of

  1. `y = 5 + 5 cos ((pi t)/7)`
  2. `y = 5 - 5 cos ((pi t)/7)`
  3. `y = 5 + 5 cos ((pi t)/14)`
  4. `y = 5 - 5 cos ((pi t)/14)`
Show Answers Only

`B`

Show Worked Solution

`text{Centre line (median):}  \ y = 5`

`text(Amplitude) = 5`

`text(Period:)\ \ 14` `= (2 pi)/n`
`n` `= pi/7`

 

`:.\ text(Graph:)\ \ y = 5 – 5 cos ((pi t)/7)`

`=>   B`

Trigonometry, 2ADV T3 SM-Bank 4 MC

A section of the graph of  `f(x)`  is shown below.
 

VCAA 2012 6mc
 

The equation of  `f(x)`  could be

  1. `f (x) = tan (x)`
  2. `f (x) = tan (x - pi/4)`
  3. `f (x) = tan(2 (x - pi/4))`
  4. `f (x) = tan(2 (x - pi/2))`
Show Answers Only

`C`

Show Worked Solution

`text(Period) = pi/2`

`=>\ text(must be)\ C\ text(or)\ D`

`text(Shift)\ \ y = tan(x)\ \ text(right)\ \ pi/4.`

`=>   C`

Measurement, STD2 M7 SM-Bank 12

Francis is buying a fridge. The energy usage of his two choices are below:

Fridge 1 (5 star rating): 227 kWh per year

Fridge 2 (2 star rating): 492 kWh per year

Each fridge has an expected life of 8 years.

If electricity costs 32.4 cents per kWh, how much will Francis save in electricity over the expected life of the more energy efficient fridge, to the nearest dollar?  (2 marks)

Show Answers Only

`$687`

Show Worked Solution

`text(Energy cost of fridge 1)`

`= 227 xx 0.324 xx 8`

`= $588.38`

`text(Energy cost of fridge 2)`

`= 492 xx 0.324 xx 8`

`= $1275.26`

`:.\ text(Energy saving)` `= 1275.26 – 588.38`
  `= 686.88`
  `= $687`

Measurement, STD2 M7 SM-Bank 10

A potting mix is made up of two ingredients, potting soil and manure, in the ratio 5:2.

21 kilograms of potting mix are used to fertilise the  rectangular vegetable garden pictured below.
 


 

  1. What rate must the whole 21 kilograms of potting mix be applied to be spread evenly over the garden. Give your answer in grams per square metre.  (2 marks)

  2. How much manure is needed to make the required potting mix?  (1 mark)

Show Answers Only
  1. `150\ text(g/m²)`
  2. `text(6 kg)`
Show Worked Solution

i.   `text(Potting mix: 21 kg  ⇒  21, 000 grams)`

`text(Area of garden)` `= 20 xx 7`  
  `= 140\ text(m²)`  

 

`:.\ text(Rate)` `= (21\ 000)/140`
  `= 150\ text(g/m²)`


ii.   `text(Ratio  5 : 2)\ \ =>\ \ 5/7 : 2/7`

`:.\ text(Amount of manure)` `= 2/7 xx 21`
  `= 6\ text(kg)`

Networks, STD2 N3 SM-Bank 42 MC

The network diagram below flows from the source (S) to sink (T).

Which of the edges is not at maximum capacity?
 


 

  1. `AC`
  2. `BC`
  3. `CT`
  4. `DT`
Show Answers Only

`=>\ text(C)`

Show Worked Solution

`text(Consider the minimum cut line: all edges are saturated.)`
 


 

`:.\ text(Edge)\ CT\ text(is not at maximum capacity.)`

`=>\ C`

Networks, STD2 N3 SM-Bank 46

A network diagram is drawn below.
 

 
 

  1. Calculate the maximum flow through this network.  (2 marks)

  2. Copy the network above and illustrate the maximum flow capacity.  (2 marks)

Show Answers Only
  1. `28`
  2.  
Show Worked Solution
i.   

 

`text(Maximum flow)` `=\ text(minimum cut)`
  `= 8 + 8 + 7 + 5`
  `= 28`

 

ii.   

Networks, STD2 N3 SM-Bank 39

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

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


 

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

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

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

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

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

i.   `text(Scanning forwards and backwards)`
 

 

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

 

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

 

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

 

iv.   `text(Critical Path is)\ CGJLM`

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

Calculus, SPEC2 2012 VCAA 5

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

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

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

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

     

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

     

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

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

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

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

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

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

a.   `u = 0, quad v = 6, quad s = 10`

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

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

 

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

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

`text(Solve for)\ \ t:`

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

♦♦ Mean mark 27%.

c.   `text(Time of chocolate in air)`

`= 10/3 + 4`

`=22/3`
 

`text(Solve for)\ \ u:`

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

 

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

 
`text(When)\ \ t = 0, v = 7`

`=> c=-sin^(-1)(1/2) = -pi/6`
 

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

 

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

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

 

d.iii.   `v = 14sin(pi/6-t/10)`

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

Vectors, SPEC2 2012 VCAA 4

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

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

for  `t in [0, 0.154]`,

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

 

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

     

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

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

     

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

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

     

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

Show Answers Only

  1. `text(400 km)`

  2. `text(See Worked Solutions.)`

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

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

  5. `t ~~ 0.04, 0.11`

Show Worked Solution

a.   `text(At)\ \ P,\ \ x(t)=0`

♦ Mean mark 49%.

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

 
`=> t=0.076923…`
 

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

 

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

 
`text(Let)\ \ u=pi(1.3t-0.1),`

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

 
  `= 0`

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

 

c.   `text(Speed)\ = |\ dotr(t)\ |`

♦ Mean mark part (c) 45%.

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

 

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

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

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

 

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

♦ Mean mark part (e) 37%.

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

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

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

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

`:. t=0.04 or 0.11`

Calculus, SPEC2 2012 VCAA 3

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

`V = 17 tan^(−1)((pi T)/6), T >= 0`

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

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

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

until the car comes to rest.

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

     

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

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

     

        `d_2 = 25 xx 120`

     

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

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

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

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

 

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

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

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

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

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

 

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

 
`text(When)\ \ t=0, \ v=25:`

`=> c=25`

`:. v= -1/100 [145 t – t^2] + 25`
 

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

`-1/100[145t – t^2] + 25=0`

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

 

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

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

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

`d_2` `= 25 xx 120`
  `= 3000`

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

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

 

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

 

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

Complex Numbers, SPEC2 2012 VCAA 2

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

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


 

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

     

    ii.  `z_1^4 = text(cis) pi/3`

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

     

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

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

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

 

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

 

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

 

c.   

 

d.  `text(Area of large sector)`

Mean mark 51%.

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

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

`=pi/2`
  

`text(Area of small sector)`

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

`=pi/8`

 
`:.\ text(Area shaded)`

`= pi/2 – pi/8`

`= (3 pi)/8`
 

♦ Mean mark (e)(i) 34%.

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

 

e.ii.  `text(When)\ \ n=(6 + 12k):`

♦ Mean mark (e)(ii) 36%.

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

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

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

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

Mechanics, SPEC2 2012 VCAA 22 MC

A 12 kg mass is suspended in equilibrium from a horizontal ceiling by two identical light strings. Each string makes an angle of 60° with the ceiling, as shown.
 


 

The magnitude, in newtons, of the tension in each string is equal to

A.     `6 g`

B.   `12 g`

C.   `24 g`

D.   `4 sqrt 3 g`

E.   `8 sqrt 3 g`

Show Answers Only

`D`

Show Worked Solution

`T/(sin 30^@)` `= (12 g)/(sin 120^@)`
`2T` `= (24 g)/sqrt3`
`T` `= (12 g)/sqrt3`
  `= 4g sqrt 3`

 
`=> D`

Vectors, SPEC2 2012 VCAA 17 MC

If  `underset ~u = 2 underset ~i - 2 underset ~j + underset ~k`  and  `underset ~v = 3 underset ~i - 6 underset ~j + 2 underset ~k`, the vector resolute of  `underset ~v`  in the direction of  `underset ~u`  is

A.   `20/49(3 underset ~i - 6 underset ~j + 2 underset ~k)`

B.   `20/3(2 underset ~i - 2underset ~j + underset ~k)`

C.   `20/7(3 underset ~i - 6 underset ~j + 2 underset ~k)`

D.   `20/9(2 underset ~i - 2 underset ~j + underset ~k)`

E.   `1/9(−2 underset ~i + 2 underset ~j - underset ~k)`

Show Answers Only

`D`

Show Worked Solution

`tildeu = 2tildei – 2tildej + tildek,qquadtildev = 3tildei – 6tilde j + 2tildek`

`|\ tildeu\ |` `= sqrt(2^2 + (−2)^2 + 1^2)=3`
`hatu` `= = 1/3(2tildei – 2tildej + tildek)`

 

`tildev*hatu\ \ (text(Scalar resolute of)\ tildev\ text(in the direction of)\ tildeu)`

`= (3tildei – 6tildej + 2tildek)* 1/3(2tildei – 2tildej + tildek)`

`= 20/3`
 

`(tildev*hatu)hatu\ \ (text(Vector resolute of)\ tildev\ text(in the direction of)\ tildeu)`

`= 20/3 xx 1/3(2tildei – 2tildej + tildek)`

`= 20/9(2tildei – 2tildej + tildek)`
 

`=> D`

Mechanics, SPEC2 2012 VCAA 14 MC

A particle is acted on by two forces, one of 6 newtons acting due south, the other of 4 newtons acting in the direction N60° W.

The magnitude of the resultant force, in newtons, acting on the particle is

  1. `10`
  2. `2sqrt7`
  3. `2sqrt19`
  4. `sqrt(52 - 24sqrt3)`
  5. `sqrt(52 + 24sqrt3)`
Show Answers Only

`B`

Show Worked Solution

Mean mark 51%.

`x^2` `= 6^2 + 4^2 – 2(6)(4) cos 60^@`
  `= 28`
   
`:. x` `= sqrt 28`
  `= 2 sqrt 7`

 
`=> B`

Calculus, SPEC2 2012 VCAA 13 MC

Using a suitable substitution, `int_0^(pi/3) sin^3 (x) cos^4 (x)\ dx`  can be expressed in terms of `u` as

A.   `int_0^(pi/3) (u^6 - u^4)\ du`

B.   `int_1^(1/2) (u^6 - u^4)\ du`

C.   `int_(1/2)^1 (u^6 - u^4)\ du`

D.   `int_0^(sqrt3/2) (u^6 - u^4)\ du`

E.   `int_0^(sqrt3/2) (u^4 - u^6)\ du`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ u = cos(x)`

`(du)/(dx) = -sin(x)\ \ =>\ \ du=-sin(x)\ dx`

`sin^2(x)` `= 1 – cos^2(x)`
  `= 1 – u^2`

 
`text(When)\ \ x = 0, \ u = 1`

`text(When)\ \ x = pi/3, \ u = 1/2`
 

`int_0^(pi/3) sin^3 (x) cos^4 (x)\ dx`

`=-int_1^(1/2) (1 – u^2)u^4\ du`

`= int_1^(1/2) (u^6 – u^4)\ du`

 
`=> B`

Calculus, SPEC2 2012 VCAA 11 MC

If  `(d^2y)/(dx^2) = x^2 - x`  and  `(dy)/(dx) = 0`  at  `x = 0`, then the graph of `y` will have

  1. a local minimum at  `x = 1/2`
  2. a local maximum at  `x = 0`  and a local minimum at  `x = 1`
  3. stationary points of inflection at  `x = 0`  and  `x = 1`, and a local minimum at  `x = 3/2`
  4. a stationary points of inflection at  `x = 0`, no other points of inflection and a local minimum at  `x = 3/2`
  5. a stationary point of inflection at  `x = 0`, a non-stationary point of inflection at  `x = 1`  and a local minimum at  `x = 3/2` 
Show Answers Only

`E`

Show Worked Solution

`(d^2y)/(dx^2)=x^2-x`

♦ Mean mark 51%.

`(d^2y)/(dx^2)=0\ \ text(when)\ \ x=0 or 1`

`(dy)/(dx)` ` = (x^3)/3 – (x^2)/2 + c`
  `= x^2(x/3 – 1/2) + c`

 
`text(Given)\ \ (dy)/(dx) = 0\ \ text(when)\ \ x = 0\ \ =>\ \ c=0`

`(dy)/(dx) = 0\ \ text(when)\ \ x = 0 or 3/2`

`(d^2y)/(dx^2)|_(x = 3/2) = 3/4>0`

`:.\ text(Local minimum at)\ \ x = 3/2`
 

`text(At)\ \ x=0, dy/dx<0\ \ text(either side)`

`:.\ text(Stationary point of inflection at)\ x=0`

`text(At)\ \ x=1, \ dy/dx!=0, \ dy/dx<0\ \ text(either side)`

`:.\ text(Non-stationary point of inflection at)\ \ x = 1`

`=> E`

Calculus, SPEC2 2012 VCAA 10 MC

The diagram that best represents the direction field of the differential equation  `(dy)/(dx) = xy`  is

A. B.
C. D.
Show Answers Only

`A`

Show Worked Solution

`(dy)/(dx) = xy`

`text(When)\ \ x=0 \ or\  y=0\ \ =>\ text(gradient = 0)`

`text(In 1st and 3rd quartile)\ \ =>\ \ text(gradients positive)`

`text(In 2nd and 4th quartile)\ \ =>\ \ text(gradients negative)`

`=> A`

Calculus, SPEC2 2012 VCAA 9 MC

Euler's formula is used to find  `y_2`, where  `(dy)/(dx) = cos(x),\ x_0 = 0, \ y_0 = 1`  and  `h = 0.1`.

The value of `y_2` correct to four decimal places is

A.   1.1000 and this is an underestimate of  `y(0.2)`

B.   1.1995 and this is an overestimate of  `y(0.2)`

C.   1.1995 and this is an underestimate of  `y(0.2)`

D.   1.2975 and this is an underestimate of  `y(0.2)`

E.   1.2975 and this is an overestimate of  `y(0.2)`

Show Answers Only

`B`

Show Worked Solution

`x_0 = 0,\ \ y_0 = 1,\ \ h=0.1`

`y_1` `= y_0 + h* (dy)/(dx)|_{(0,1)}`
  `= 1 + 0.1 xx cos(0)`
  `= 1.1`
   
`y_2` `= 1.1 + 0.1 xx cos(0.1)`
  `~~ 1.1995`

 
`text(S)text(ince)\ \ sin(x)\ \ text(is concave down in the region)\ \ x=0.2,`

`1.1995\ text(is an overestimate.)`

`=> B`

Calculus, SPEC2 2011 VCAA 22 MC

A particle moves in a straight line. At time  `t`  seconds, where  `t >= 0`, its displacement `x` metres from the origin and its velocity  `v`  metres per second are such that  `v = 25 + x^2`.

If  `x = 5`  initially, then  `t`  is equal to

  1. `25x + (x^3)/3`
  2. `25x + (x^3)/3 - 500/3`
  3. `1/5tan^(−1)(x/5) + 5`
  4. `tan^(−1)(x/5) - pi/4`
  5. `1/5tan^(−1)(x/5) - pi/20`
Show Answers Only

`E`

Show Worked Solution

`(dx)/(dt) = 25 + x^2`

`(dt)/(dx) = 1/(25 + x^2)`

`t` `= int 1/(25 + x^2)\ dx`
  `= 1/5 int 5/(5^2 + x^2)\ dx`
  `= 1/5 tan^(−1)(x/5) + c`

 
`text(When)\ \ t=0,\ \ x=5:`

`0` `= 1/5 tan^(−1)(5/5) +c`
`c` `= – pi/20`

 
`:. t= 1/5 tan^(−1)(x/5) – pi/20`

`=> E`

Calculus, SPEC2 2011 VCAA 20 MC

A body moves in a straight line such that its velocity  `v\ text(ms)^(-1)`  is given by  `v = 2sqrt(1 - x^2)`, where `x` metres is its displacement from the origin.

The acceleration of the body in `text(ms)^(-2)`  is given by

  1. `(−2x)/(sqrt(1 - x^2))`
  2. `−2x`
  3. `2/(sqrt(1 - x^2))`
  4. `2(1 - 2x)`
  5. `−4x`
Show Answers Only

`E`

Show Worked Solution

`v = 2(sqrt(1 – x^2))`

`a` `=v *(dv)/(dx)`  
  `= 2sqrt(1 – x^2) xx 2 xx (−2x) xx 1/(2sqrt(1 – x^2))`  
  `= −(4x sqrt(1 – x^2))/sqrt( 1 – x^2)`  
  `= −4x`  

  
`=> E`

Calculus, SPEC2 2011 VCAA 19 MC

The motion of a lift (elevator) in a shopping centre is given by the velocity-time graph below, where time  `t`  is in seconds, and the velocity of the lift is `v` metres per second. For  `v > 0`  the lift is moving upwards.
 

SPEC2 2011 VCAA 19 MC
 

The graph shows that at the end of 30 seconds, the position of the lift is

A.   17.5 metres above its starting level.

B.   5 metres above its starting level.

C.   at the same position as its starting level.

D.   5 metres below its starting level.

E.   17.5 metres below its starting level.

Show Answers Only

`A`

Show Worked Solution
`text(Distance)` `= text(Area of trap) – text(Area of triangle)`
  `= 1/2(5 + 15)3 – 1/2 xx 10 xx 2.5`
  `= 17.5\ text(m)`

 
`text{Net area is above the}\ xtext{-axis (lift is higher).}`

`=> A`

Calculus, SPEC2 2011 VCAA 18 MC

The amount of chemical `x` in a tank at time  `t`  is given by the differential equation  `dx/dt = -10/(10 - t)`  and when  `t = 0`, `\ x_0 = 5`. Euler's method is used with a step size of 0.5 in the values of  `t`.

The value of `x` correct to two decimal places when  `t = 1`  is found to be

A.   3.95

B.   3.97

C.   4.50

D.   5.50

E.   6.03

Show Answers Only

`B`

Show Worked Solution
`x_1 ~~ x(0.5)` `= 5 + 0.5 xx (−10/(10 – 0))`
  `= 5 – 0.5`
  `= 4.5`

 

Mean mark 51%.

`x_2 ~~ x(1)` `= 4.5 + 0.5 xx (−10/(10 – 0.5))`
  `= 4.5 – 10/19`
  `~~ 3.97`

`=> B`

Calculus, SPEC2 2011 VCAA 17 MC

SPEC2 2011 VCAA 17 MC

The differential equation which best represents the above direction field is

A.   `(dy)/(dx) = (y - 2x)/(2y + x)`

B.   `(dy)/(dx) = (2x - y)/(y - 2x)`

C.   `(dy)/(dx) = (2y - x)/(y + 2x)`

D.   `(dy)/(dx) = (y - 2x)/(2y - x)`

E.   `(dy)/(dx) = (x - 2y)/(2y + x)`

Show Answers Only

`A`

Show Worked Solution

`text(When)\ \ x=0\ \ => \ \ text(gradients are all positive)`

Almost half of all students answered incorrectly – mean mark 52%.

`text(Eliminate B and E.)`

`text(When)\ \ y=0\ \ => \ \ text(gradients are all negative)`

`text(Eliminate D.)`

`text(Option A will have zero gradient along)\ \ y=2x\ \ text{(correct)}`

`text(Option C will have zero gradient along)\ \ y=1/2 x\ \ text{(incorrect)}`

`=> A`

Calculus, SPEC2 2011 VCAA 15 MC

Using a suitable substitution, the definite integral  `int_0^(pi/24)tan(2x)sec^2(2x)\ dx`  is equivalent to

A.   `1/2int_0^(pi/24)(u)\ du`

B.   `2int_0^(pi/24)(u)\ du`

C.   `int_0^(2 - sqrt3)(u)\ du`

D.   `1/2int_0^(2 - sqrt3)(u)\ du`

E.   `2int_0^(2 - sqrt3)(u)\ du`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ u = tan(2x)`

`(du)/dx = 2 sec^2(2x)\ \ =>\ \ 1/2 du = sec^2(2x)\ dx`
 

`text(When)\ \ x=pi/24\ \ =>\ \ u=tan(pi/12)=2-sqrt3\ \ text{(by CAS)}`

`text(When)\ \ x=0\ \ =>\ \ u=0`
 

`:. int_0^(pi/24)tan(2x)sec^2(2x)\ dx`

`= 1/2int_0^(2 – sqrt3)(u)\ du`

`=> D`

Calculus, SPEC2 2011 VCAA 14 MC

If  `f″(x) = 2e^xsin(x)`,  `f′(0) = 0`  and  `f(0) = 0`, the  `f(x)`  equals

A.   `−e^x(cos(x) + sin(x))`

B.   `−e^x(cos(x) - sin(x)) + 1`

C.   `−e^xcos(x)`

D.   `x - e^xcos(x) + 1`

E.   `x - e^xcos(x)`

Show Answers Only

`D`

Show Worked Solution

`text(Consider:)\ \ f(0)=0,`

`=>\ text(Eliminate A, C and E.)`

`text(By CAS, B and D satisfy)\ \ f′(0) = 0.`

`text(By CAS, only D satisfies)\ \ f″(x) = 2e^x sin(x)`

`=> D`

Vectors, SPEC2 2011 VCAA 13 MC

The position of a particle at time  `t`  is given by  `underset~r(t) = (sqrt(t - 2))underset~i + (2t)underset~j`  for  `t >= 2`.

The cartesian equation of the path of the particle is

A.   `y = 2x^2 + 4`, `x >= 2`
B.   `y = 2x^2 + 2`, `x >= 2`
C.   `y = 2x^2 + 4`, `x >= 0`
D.   `y = sqrt((x - 4)/2)`, `x >= 4`
E.   `y = 2x^2 + 2`, `x >= 0`

 

Show Answers Only

`C`

Show Worked Solution

`tilder(t) = (sqrt(t – 2)tildei + (2t)tildej)`

`x = sqrt(t – 2)`

`text(Given)\ \ t >= 2\ \ =>\ \ x >=0`

`y = 2t\ \ =>\ \ t = y/2`

`:. x` `= sqrt(y/2 – 2)`
`x^2` `= y/2 – 2`
`y/2` `= x^2 + 2`
`y` `= 2x^2 + 4`

 
`=> C`

Vectors, SPEC2 2011 VCAA 11 MC

Consider the three forces

`underset~F_1 = −sqrt3underset~j`,  `underset~F_2 = underset~i + sqrt3underset~j`  and  `underset~F_3 = −1/2underset~i + sqrt3/2underset~j`.

The magnitude of the sum of these three forces is equal to

  1. the magnitude of `(underset~F_3 - underset~F_1)`
  2. the magnitude of `(underset~F_2 - underset~F_1)`
  3. the magnitude of `underset~F_1`
  4. the magnitude of `underset~F_2`
  5. the magnitude of `underset~F_3`
Show Answers Only

`E`

Show Worked Solution
`underset~(F_1) + underset~(F_2) + underset~(F_3)` `= (1 – 1/2)underset~i + (−sqrt3 + sqrt3 + sqrt3/2)underset~j`
  `= 1/2 underset~i + sqrt3/2underset~j`
`|underset~(F_1) + underset~(F_2) + underset~(F_3)|` `= sqrt((1/2)^2 + (sqrt3/2)^2)`
  `= |underset~(F_3)|`

`=> E`

Vectors, SPEC2 2011 VCAA 10 MC

The diagram below shows a rhombus, spanned by the two vectors  `underset~a`  and  `underset~b`.
 

SPEC2 2011 VCAA 10 MC
 

It follows that

A.   `underset~a.underset~b = 0`

B.   `underset~a = underset~b`

C.   `(underset~a + underset~b).(underset~a - underset~b) = 0`

D.   `|\ underset~a + underset~b\ | = |\ underset~a - underset~b|`

E.   `2underset~a + 2underset~b = underset~0`

Show Answers Only

`C`

Show Worked Solution

`text(Consider A:)`

`text(If)\ \ underset~a · underset~b = 0\ \ =>\ \ underset~a ⊥ underset~b\ \ text{(incorrect)}`

`text(Consider B:)`

`underset~a != underset~b\ \ text{(incorrect)}`

`text(Consider C:)`

`underset~a + underset~b` `= overset(->)(OC)`
`underset~a – underset~b` `= overset(->)(BA)`

 
`overset(->)(OC) · overset(->)(BA)=0`

`text{The diagonals of a rhombus are perpendicular  (correct)}`

`=> C`