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Statistics, MET2 2019 VCAA 14 MC

The weights of packets of lollies are normally distributed with a mean of 200 g.

If 97% of these packets of lollies have a weight of more than 190 g, then the standard deviation of the distribution, correct to one decimal place, is

  1.  3.3 g
  2.  5.3 g
  3.  6.1 g
  4.  9.4 g
  5. 12.1 g
Show Answers Only

`B`

Show Worked Solution

`text(Pr)(X > 190)` `= text(Pr)(Z > (190 – 200)/sigma) = 0.97`
`text(Pr)(Z < (-10)/sigma)` `= 0.03`
`text(z-score)` `= -1.8808\ \ text{(by CAS)}`
`-1.8808` `= (-10)/sigma`
`sigma` `= 5.31…`

 
`=>   B`

Probability, 2ADV S1 2019 MET2 11 MC

`A` and `B` are events from a sample space such that  `P(A) = p`, where  `p > 0, \ P(B\ text{|}\ A) = m`  and  `P(B\ text{|}\ A^{′}) = n`.

`A` and `B` are independent events when

  1. `m = n`
  2. `m = 1-p`
  3. `m + n = 1`
  4. `m = p`
Show Answers Only

`A`

Show Worked Solution

`text(S) text(ince)\ A and B\ text(are independent:)`

`P(B\ text{|}\ A) = P(B\ text{|}\ A^{′})`

`:. m = n`
 

`=>   A`

Probability, MET2 2019 VCAA 11 MC

`A` and `B` are events from a sample space such that  `text(Pr)(A) = p`, where  `p > 0, \ text(Pr)(B\ text{|}\ A) = m`  and  `text(Pr)(B\ text{|}\ A prime) = n`.

`A` and `B` are independent events when

  1. `m = n`
  2. `m = 1 - p`
  3. `m + n = 1`
  4. `m = p`
  5. `m + n = 1 - p`
Show Answers Only

`A`

Show Worked Solution

`text(S) text(ince)\ A and B\ text(are independent),`

`text(Pr)(B\ text{|}\ A) = text(Pr)(B\ text{|}\ A prime)`

`:. m = n`
 

`=>   A`

Graphs, MET2 2019 VCAA 10 MC

Which one of the following statements is true for  `f: R -> R, \ f(x) = x + sin(x)`?

  1. The graph of  `f` has a horizontal asymptote
  2. There are infinitely many solutions to  `f(x) = 4`
  3. `f` has a period of  `2 pi`
  4. `f^{\prime}(x) >= 0`  for  `x in R`
  5. `f^{\prime}(x) = cos(x)`
Show Answers Only

`D`

Show Worked Solution

`text(By CAS, sketch)\ \ f(x) = x + sin(x):`

`text(By inspection, eliminate A, B, C)`

`text(By CAS, sketch)\ \ d/(dx)\ f(x):`

` text(Graph range) >= 0\ \ text(for)\ \ x in R`
 

`=>   D`

Graphs, MET2 2019 VCAA 9 MC

The point  `(a, b)`  is transformed by

`T([(x), (y)]) = [(1/2, 0), (0, -2)][(x), (y)]+[(-1/2), (-2)]`

If the image of  `(a, b)`  is  `(0, 0)`, then  `(a, b)`  is

  1. `(1, 1)`
  2. `(-1, 1)`
  3. `(-1, 0)`
  4. `(0, 1)`
  5. `(1, -1)`
Show Answers Only

`E`

Show Worked Solution

`x^{\prime} = 1/2x – 1/2`

`=> 0` `= 1/2a – 1/2`
`a` `= 1`

 
`y^{\prime}= -2y – 2`

`=> 0` `= -2b – 2`
`b` `= -1`

 
`:. (a, b) -= (1, -1)`

`=>   E`

Probability, MET2 2019 VCAA 8 MC

An archer can successfully hit a target with a probability of 0.9. The archer attempts to hit the target 80 times. The outcome of each attempt is independent of any other attempt.

Given that the archer successfully hits the target at least 70 times, the probability that the archer successfully hits the target exactly 74 times, correct to four decimal places, is

A.   0.3635

B.   0.8266

C.   0.1494

D.   0.3005

E.   0.1701

Show Answers Only

`C`

Show Worked Solution

`text(Pr)(H) = 0.9, \ text(Pr)(H prime) = 0.1`

`text(Let)\ \ X = text(number of times archer hits)`

`X\ ~\ text(Bi)(80, 0.9)`

`text(Pr)(X = 74|X >= 70)` `= (text(Pr)(X = 74))/(text(Pr)(X >= 70))`
  `~~ 0.1494`

 
`=>   C`

Calculus, MET2 2019 VCAA 6 MC

A rectangular sheet of cardboard has a length of 80 cm and a width of 50 cm. Squares, of side length `x` centimetres, are cut from each of the corners, as shown in the diagram below.
 

 

A rectangular box with an open top is then constructed, as shown in the diagram below.
 

 

The volume of the box is a maximum when `x` is equal to

  1. `10`
  2. `20`
  3. `25`
  4. `100/3`
  5. `200/3`
Show Answers Only

`A`

Show Worked Solution
`text(Base of box)` `= (80-2x) xx (50-2x)`
  `= 4000-260x + 4x^2`
`text(Volume)` `= 4x^3-260x^2 + 4000x`

 
`(dV)/(dx) = 12x^2-520x + 4000`

`(dV)/(dx) = 0\ \ text(when)\ \ x = 10`

`=>   A`

Calculus, MET2 2019 VCAA 4 MC

`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`  is equal to

  1. `((2 - sqrt 3)a - b)/2`
  2. `(b - (2 - sqrt 3) a)/2`
  3. `((2 - sqrt 3)a + b)/2`
  4. `((2 - sqrt 3) b - a)/2`
  5. `((2 - sqrt 3) b + a)/2`
Show Answers Only

`C`

Show Worked Solution

`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`

`= [-a cos(x) + b sin(x)]_0^(pi/6)`

`= [-a ⋅ sqrt 3/2 + b/2 – (-a + 0)]`

`= (2a – sqrt 3 a + b)/2`

`= ((2 – sqrt 3) a + b)/2`

`=>   C`

Calculus, MET2 2019 VCAA 3 MC

Let  `f: R\ text(\){4} -> R, \ f(x) = a/(x - 4),\ \ text(where)\ \ a > 0`.

The average rate of change of  `f`  from  `x = 6`  to  `x = 8`  is

A.   `a log_e(2)`

B.   `a/2 log_e(2)`

C.   `2a`

D.   `-a/4`

E.   `-a/8`

Show Answers Only

`E`

Show Worked Solution

`f(6) = a/2`

`f(8) = a/4`

`:.\ text(Average ROC)` `= (a/4 – a/2)/(8 – 6)`
  `= -a/8`

`=>   E`

Statistics, 2ADV S3 EQ-Bank 16

A probability density function is defined by
 

`f(x) = {(a, \ text(for)\ \ 0 <= x <= 4),(3a, \ text(for)\  4 < x <= 8):}`
 

  1. Find the value of  `a`.  (2 marks)

  2. Sketch the probability density function.  (1 mark)

  3. Find an expression for the cumulative distribution function.  (2 marks)

Show Answers Only
  1. `1/16`
  2.  
  3. `F(x) = {(x/16, text(for)\ 0 <= x <= 4),((3x)/16 – 1/2, text(for)\ 4 < x <= 8):}`
Show Worked Solution

i.   `int_0^4 a\ dx = [ax]_0^4 = 4a`

`int_4^8 3a\ dx = [3ax]_4^8 = 24a – 12a = 12a`

`4a + 12a` `= 1`
`a` `= 1/16`

 

ii.   

 

iii.   `text(When)\ \ 0 <= x <= 4:`

`F(x) = int 1/16\ dx = x/16 + C`

`F(0) = 0 \ => \ C = 0`

`F(x) = x/16\ \ …\ (1)`
 

`text(When)\ \ 4 < x <= 8:`

`F(x) = int 3/16\ dx = (3x)/16 + C`

`F(4) = 1/4\ \ (text{see (1) above})`

COMMENT: Understand why you can check your equation here by confirming  `F(8)=1`

`(3 xx 4)/16 + C` `= 1/4`
`C` `= −1/2`

 
`F(x) = (3x)/16 – 1/2`
 

`:. F(x) = {(x/16, \ text(for)\ \ 0 <= x <= 4),((3x)/16 – 1/2, \ text(for)\ \ 4 < x <= 8):}`

Statistics, 2ADV S2 SM-Bank 14

A probability density function  `f(x)`  is given by
 

`f(x) = {(px(3 - x), \ text(if)\ \ 0 <= x <= 3),(0, \ text(if)\ \ x < 0\ \ text(or if)\ \ x > 3):}`
 

where  `p`  is a positive constant.

  1. Find the value of  `p`.  (2 marks)

  2. Find the mode of  `f(x)`.  (2 marks)

Show Answers Only
  1. `2/9`
  2. `3/2`
Show Worked Solution

i.   `text(Total Area under curve = 1)`

`p int_0^3 3x-x^2\ dx` `= 1`
`p[3/2 x^2-(x^3)/3]_0^3` `= 1`
`p[27/2-27/3-0]` `= 1`
`(9p)/2` `= 1`
`p` `= 2/9`

 

ii.   `f(x) = 2/9(3x-x^2)`

`f^{′}(x) = 2/9(3-2x)`

`text(S.P. when)\ \ f^{′}(x) = 0:`

`3-2x` `= 0`
`x` `= 3/2`

 
`f^{″}(x) = -4/9 < 0 \ =>\ text(MAX)`

`:.\ text(Mode) = 3/2`

Statistics, 2ADV S2 SM-Bank 15

A function  \(f(x)\)  is given by

 \(f(x)= \begin{cases}
\dfrac{3}{4}(x-2)(4-x) & \text {if } 2 \leq x \leq 4 \\
\ \\
0 & \text{if } x<2 \text { or if } x>4
\end{cases}\)

  1. Show this curve is a probability density function.  (2 marks)

  2. Find the mode.  (2 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(3\)
Show Worked Solution
i.    \(f(x)\) \( =\dfrac{3}{4}(x-2)(4-x)\)
  \(\displaystyle \int_2^4 f(x) \) \( =\dfrac{3}{4} \int_2^4-x^2+6 x-8 d x\)
    \( =\dfrac{3}{4}\left[\dfrac{-x^3}{3}+3 x^2-8 x\right]_2^4\)
    \( =\dfrac{3}{4}\left[\left(-\dfrac{64}{3}+48-32\right)-\left(-\dfrac{8}{3}+12-16\right)\right] \)
    \( =\dfrac{3}{4}\left(-\dfrac{16}{3}+\dfrac{20}{3}\right) \)
    \(= 1\)

 

\(f(2)=0, f(4)=0\)

\(f(x)>0 \text { for } 2<x<4\)

\(f(2) \geq 0 \text { for } 2 \leq x \leq 4\)

 
\(\therefore f(x) \ \text{is a probability density function.}\)
 

ii.    \(f(x\) \(=\dfrac{3}{4}\left(-x^2+6 x-8\right)\)
  \(f^{\prime}(x)\) \(=\dfrac{3}{4}(-2 x+6)\)
  \(f^{\prime\prime}(x)\)  \(=-\dfrac{3}{2}\)

 
\(\text{SP when}\ f'(x) = 0\)

\(\begin{array}{r}-2 x+6=0 \\
x=3\end{array}\)

\(f^{\prime \prime}(x)<0 \Rightarrow \text {MAX}\)

\(\therefore \text {Mode}=3\)

Statistics, 2ADV S3 EQ-Bank 17

The diastolic measurement for blood pressure in 35-year-old people is normally distributed, with a mean of 75 and a standard deviation of 12.

  1. A person is considered to have low blood pressure if their diastolic measurement is 63 or less.What percentage of 35-year-olds have low blood pressure?  (1 mark)

  2. Calculate the `z`-score for a diastolic measurement of 57.  (1 mark)

  3. The probability that a 35-year-old person has a diastolic measurement for blood pressure between 57 and 63 can be found by evaluating
     
    `qquad qquad int_a^b f(x)\ dx`
     
    where `a` and `b` are constants and where
     
    `qquad qquad f(x) = 1/(sqrt(2pi)) e^((−x^2)/2)`
     
    is the normal probability density function with mean 0 and standard deviation 1.

     

    By first finding the values `a` and `b`, calculate an approximate value for this probability by using the trapezoidal rule with 3 function values.  (3 marks)

  4. Hence, find the approximate probability that a 35-year-old person chosen at random has a diastolic measurement of 57 or less.  (1 mark)

Show Answers Only
  1. `16text(% have low blood pressure)`
  2. `−1.5`
  3. `9.2text(%)`
  4. `6.8text(%)`
Show Worked Solution
i.    `ztext(-score)\ (63)` `= (x – mu)/σ`
    `= (63 – 75)/12`
    `= −1`

 

`:. 16text(% have low blood pressure)`

 

ii.    `ztext(-score)` `= (57 – 75)/12`
    `= −1.5`

 

iii.  `y = 1/sqrt(2pi) e^((−x^2)/2)`

 
`text(Area)` `= h/2(y_0 + 2y_1 + y_2)`
  `~~ 0.25/2 (0.1295 + 2 xx 0.1826 + 0.2420)`
  `~~ 0.0920`
  `~~ 9.2text(%)`

 

iv.   

 

`P(text{blood pressure}\ <= 57)` `= 16 – 9.2`
  `~~ 6.8text(%)`

Functions, 2ADV F1 SM-Bank 40

A factory makes boots and sandals. In any week

• the total number of pairs of boots and sandals that are made is 200
• the maximum number of pairs of boots made is 120
• the maximum number of pairs of sandals made is 150.

The factory manager has drawn a graph to show the numbers of pairs of boots (`x`) and sandals (`y`) that can be made.
 

2UG-2009-24d
 

  1. Find the equation of the line `AD`.   (1 mark)
  2. Explain why this line is only relevant between `B` and `C` for this factory.     (1 mark)
  3. The profit per week, `$P`, can be found by using the equation
     
  4.       `P = 24x + 15y`.
     
    Compare the profits at `B` and `C`.     (2 marks)

 

 

Show Answers Only
  1. `x + y = 200`
  2. `text(S)text(ince the max amount of boots = 120)`
  3. `=> x\ text(cannot)\ >120`
  4.  
  5. `text(S)text(ince the max amount of sandals = 150`
  6.  
  7. `=> y\ text(cannot)\ >150`
  8.  
  9. `:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`
  10.  
  11. `text(The profits at)\ C\ text(are $630 more than at)\ B.`

 

Show Worked Solution

(i)   `text{We are told the number of boots}\ (x),`

Question taken from Gen2 exam – now common content with Advanced course.

`text{and shoes}\  (y),\ text(made in any week = 200)`

`=>text(Equation of)\ AD\ text(is)\ \ x + y = 200`

 

(ii)  `text(S)text(ince the max amount of boots = 120)`

`=> x\ text(cannot)\ >120`

`text(S)text(ince the max amount of sandals = 150`

`=> y\ text(cannot)\ >150`

`:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`

 

(iii)  `text(At)\ B,\ \ x = 50,\ y = 150`

`=>$P  (text(at)\ B)` `= 24 xx 50 + 15 xx 150`
  `= 1200 + 2250`
  ` = $3450`

`text(At)\ C,\ \  x = 120 text(,)\ y = 80`

`=> $P  (text(at)\ C)` `= 24 xx 120 + 15 xx 80`
  `= 2880 + 1200`
  `= $4080`

 

`:.\ text(The profits at)\ C\ text(are $630 more than at)\ B.`

Calculus, MET1 2019 VCAA 9

Consider the functions  `f: R -> R,\ \ f(x) = 3 + 2x-x^2`  and  `g: R -> R,\ \ g(x) = e^x`

  1. State the rule of `g(f(x))` .  (1 mark) 

  2. Find the values of `x` for which the derivative of `g(f(x))` is a negative.  (2 marks)

  3. State the rule of `f(g(x))`.  (1 mark)

  4. Solve  `f(g(x)) = 0`.  (2 marks)

  5. Find the coordinates of the stationary point of the graph of `f(g(x))`.  (2 marks)

  6. State the number of solutions to  `g(f(x)) + f(g(x)) = 0`.  (1 mark)

Show Answers Only
  1. `g(f(x)) = e^(3 + 2x-x^2)`
  2. `x > 1`
  3. `f(g(x)) = 3 + 2e^x-e^(2x)`
  4. `x = ln 3`
  5. `text(S.P. at)\ (0, 4)`
  6. `text(1 solution only)`
Show Worked Solution

a.   `g(f(x)) = e^(3 + 2x-x^2)`
 

b.   `d/(dx) g(f(x)) = (2-2x)e^(3 + 2x-x^2)`

`e^(3 + 2x-x^2) > 0\ \ text(for all)\ \ x`

`=> d/(dx) g(f(x)) < 0\  \text(when):`

`2-2x` `< 0`
`x` `> 1`

 
c.   `f(g(x)) = 3 + 2e^x-e^(2x)`
 

d.   `e^(2x)-2e^x-3 = 0`

`text(Let)\ \ X = e^x`

`X^2-2X-3` `= 0`
`(X-3)(X + 1)` `= 0`

 
`X = 3 or -1`

`text(When)\ \ X = 3,\ \ e^x = 3 => x = ln 3`

`text(When)\ \ X = -1,\ \ e^x = -1 \ => \ text(no solution)`

`:. x = ln 3`
 

e.    `d/(dx)\ f(g(x))` `= 2e^x-2e^(2x)`
    `= 2e^x (1-e^x)`

 
`text(S.P. occurs when:)`

`2e^x(1-e^x)` `= 0`
`e^x` `= 1`
`x` `= 0`

 
`text(When)\ \ x = 0,`

`f(g(x))` `= 3 + 2-1=4`

 
`:.\ text(S.P. at)\ \ (0, 4)`
 

f.    `text(Solutions occur when)\ \ g(f(x)) = -f(g(x))`

`text{Sketch both graphs (using parts a-e):}`
 


 

`:. 1\ text(solution only)`

Algebra, MET1 2019 VCAA 8

The function  `f: R -> R, \ f(x)`  is a polynomial function of degree 4. Part of the graph of  `f`  is shown below.

The graph of  `f`  touches the `x`-axis at the origin.
 


 

  1. Find the rule of  `f`.   (1 mark) 

Let  `g`  be a function with the same rule as  `f`.

Let  `h: D -> R, \ h(x) = log_e (g(x))-log_e (x^3 + x^2)`, where  `D`  is the maximal domain of  `h`.

  1. State  `D`.   (1 mark)

  2. State the range of  `h`.   (2 marks)

Show Answers Only
  1. `f(x) = -4x^2(x^2-1)`
  2. `x in (-1, 1) text(\{0})`
  3. `h(x) in (-oo, 3 log_e 2) text(\ {)2 log_e 2 text(})`
Show Worked Solution
a.    `y` `= ax^2 (x-1)(x + 1)`
    `= ax^2 (x^2-1)\ …\ (1)`

 
`text(Substitute)\ (1/ sqrt 2, 1)\ text{into  (1):}`

`1 = a ⋅ (1/sqrt 2)^2 ((1/sqrt 2)^2-1)`

`1 = a(1/2)(-1/2)`

`a = -4`

`:. f(x) = -4x^2(x^2-1)`

 

b.    `g(x) > 0` `=> -4x^2 (x^2-1) > 0`
    `=> x in (-1, 1)\  text(\{0})`

`text(and)`

`x^3 + x^2 > 0 => text(true for)\ \ x in (-1 , 1)\ text(\{0})`

`:. D:\ x in (-1, 1)\ text(\{0})`

 

c.    `h(x)` `= log_e ((-4x^2(x^2-1))/(x^3 + x^2))`
    `= log_e ((-4x^2(x + 1)(x-1))/(x^2(x + 1)))`
    `= log_e (4(1-x))\ \ text(where)\ \ x in (-1, 1)\ text(\{0})`

  
`text(As)\ \ x -> -1,\ \ h(x) -> log_e 8 = 3 log_e 2`

`text(As)\ \ x -> 1,\ \ h(x) -> -oo`

`text(As)\ \ x -> 0,\ \ h(x) -> log_e 4 = 2 log_e 2`

`text{(}h(x)\ text(undefined when)\ \ x = 0 text{)}`
 

`:.\ text(Range)\ \ h(x) in (-oo, 3 log_e 2)\ text(\{) 2 log_e 2 text(})`

Calculus, MET1 2019 VCAA 5

Let  `f: R\ text(\{1}) -> R, \ f(x) = 2/(x-1)^2 + 1`.

    1. Evaluate  `f(-1)`.   (1 mark)

    2. Sketch the graph of `f` on the axes below, labelling all asymptotes with their equations.   (2 marks)

        

  2. Find the area bounded by the graph of `f`, the `x`-axis, the line  `x = -1`  and the line  `x = 0`.   (2 marks)

Show Answers Only
    1. `3/2`
    2. `text(See Worked Solutions)`
  2. `2`
Show Worked Solution
a.i.    `f(-1)` `= 2/(-1-1)^2 + 1`
    `= 3/2`

​

a.ii.     

 

b.    `text(Area)` `= int_(-1)^0 2/(x-1)^2 + 1\ dx`
    `= int_(-1)^0 2(x-1)^(-2) + 1\ dx`
    `= [-2(x-1)^(-1) + x]_(-1)^0`
    `= [((-2)/-1 + 0)-((-2)/-2-1)]`
    `= 2`

Graphs, MET1 2019 VCAA 4

  1. Solve  `1-cos (x/2) = cos (x/2)`  for  `x in [-2 pi, pi]`.   (2 marks) 

  2. The function  `f: [-2pi, pi] -> R,\ \ f(x) = cos (x/2)`  is shown on the axes below.
     

     

     

    Let  `g: [-2pi, pi] -> R,\ \ g(x) = 1-f(x)`.

     

     

    Sketch the graph of  `g`  on the axes above. Label all points of intersection of the graphs of  `f`  and  `g`, and the endpoints of  `g`, with their coordinates.   (2 marks)

Show Answers Only
  1. `(2 pi)/3, (-2 pi)/3`
  2. `text(See Worked Solutions)`
Show Worked Solution
a.    `2 cos (x/2)` `= 1`
  `cos (x/2)` `= 1/2`

 
`=>\ text(Base angle)\ \ pi/3`

`x/2 = pi/3, -pi/3, (-5 pi)/3`

`x = (2 pi)/3, (-2 pi)/3 qquad (x in [-2pi, pi])`

 

b.   `text(Plot points for)\ \ g(x):`

`text(- reflect)\ f(x)\ text(in)\ x text(-axis, then)`

`text(- translate 1 upwards.)`

Probability, MET1 2019 VCAA 3

The only possible outcomes when a coin is tossed are a head or a tail. When an unbiased coin is tossed, the probability of tossing a head is the same as the probability of tossing a tail.

Jo has three coins in her pocket; two are unbiased and one is biased. When the biased coin is tossed, the probability of tossing a head is `1/3`.

Jo randomly selects a coin from her pocket and tosses it.

  1. Find the probability that she tosses a head.  (2 marks) 

  2.  Find the probability that she selected an unbiased coin, given that she tossed a head.  (1 mark)

Show Answers Only

  1. `4/9`
  2. `3/4`

Show Worked Solution

a.    `text(Pr)(B) = 1/3,\ \text(Pr)(B^{\prime}) = 2/3`

`text(Pr)(H|B) = 1/3,\ \ text(Pr)(H|B^{\prime}) = 1/2`

 

`:. text(Pr)(H)` `= text(Pr)(B) xx text(Pr)(H nn B) + text(Pr)(B^{\prime}) xx text(Pr)(H nn B^{\prime})`
  `= 1/3 xx 1/3 + 2/3 xx 1/2`
  `= 1/9 + 1/3`
  `= 4/9`

 

b.    `text(Pr)(B^{\prime}|H)` `= (text(Pr)(B^{\prime} nn H))/(text(Pr)(H))`
    `= (2/3 xx 1/2)/(4/9)`
    `= 3/4`

Algebra, MET1 2019 VCAA 2

  1. Let  `f: R\{1/3} -> R,\ f(x) = 1/(3x-1)`.

     

     

    Find the rule of  `f^(-1)`.   (2 marks) 

  2. State the domain of  `f^(-1)`.   (1 mark)

  3. Let  `g`  be the function obtained by applying the transformation  `T`  to the function  `f`, where
  4. `qquad qquad qquad T ([(x), (y)]) = [(x), (y)] + [(c), (d)]`
  5. and  `c, d in R`.
  6. Find the values of  `c`  and  `d`  given that  `g = f^(-1)`.   (1 mark)

Show Answers Only
  1. `y = 1/(3x) + 1/3`
  2. `x in R\ text(\){0}`
  3. `c = -1/3,\ d = 1/3`
Show Worked Solution

a.   `text(Let)\ \ y = 1/(3x-1)`

`text(Inverse:  swap)\ \ x↔ y`

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

 

b.   `x in R\ text(\){0}`

 

c.   `f(x) = 1/(3x-1) \ -> \ f(x) = 1/(3x) + 1/3`

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

`:. c = -1/3,\ \ d = 1/3`

Calculus, MET1 2019 VCAA 1b

Let  `g: R text(\ {−1}) -> R,\ \ g(x) = (sin(pi x))/(x + 1)`.

Evaluate  `g^{prime}(1)`.   (2 marks)

Show Answers Only

`-pi/2`

Show Worked Solution
  `u = sin(pi x)` `v = x + 1`
  `u^{prime}=pi cos(pi x)` `v^{prime}=1`

 

`g^{prime}(x)` `=(vu^{prime}-uv^{prime})/v^2`
  `= ((x + 1) ⋅ pi cos(pi x)-sin (pi x))/(x + 1)^2`
`g^{prime}(1)` `= (2 pi cos(pi)-sin(pi))/2^2`
  `= (2 pi(-1)-0)/4`
  `= -pi/2`

L&E, 2ADV E1 SM-Bank 14

The spread of a highly contagious virus can be modelled by the function

`f(x) = 8000/(1 + 1000e^(−0.12x))`

Where `x` is the number of days after the first case of sickness due to the virus is diagnosed and  `f(x)`  is the total number of people who are infected by the virus in the first  `x`  days.

  1. Calculate  `f(0)`.  (1 mark)

  2. Find the value of  `f(365)`  and interpret it result.  (2 marks)

Show Answers Only
  1. `7.99…`
  2. `text(After 1 year, the model predicts the total number)`
    `text(of people infected by the virus is 8000.)`
Show Worked Solution
i.   `f(0)` `= 8000/(1 + 1000e^0)`
    `= 8000/1001`
    `= 7.99…`

 

ii.    `f(365)` `= 8000/(1 + 1000e^(−0.12 xx 365))`
    `= 8000/(1 + 1000e^(−43.8))`
    `~~ 8000`

 
`text(After 1 year, the model predicts the total number)`

`text(of people infected by the virus is 8000.)`

Vectors, EXT1 V1 SM-Bank 23

A fireworks rocket is fired from an origin `O`, with a velocity of 140 metres per second at an angle of  `theta`  to the horizontal plane.
 


 

The position vector `underset~s(t)`, from `O`, of the rocket after  `t`  seconds is given by

`underset~s = 140tcosthetaunderset~i + (140tsintheta - 4.9t^2)underset~j`

The rocket explodes when it reaches its maximum height.

  1. Show the rocket explodes at a height of  `1000sin^2theta`  metres.  (2 marks)

  2. Show the rocket explodes at a horizontal distance of  `1000sin 2theta`  metres from `O`.  (1 mark)

  3. For best viewing, the rocket must explode at a horizontal distance of 500 m and 800 m from `O`, and at least 600 m above the ground.

     

    For what values of  `theta`  will this occur.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `63.4° <= theta <= 75°`
Show Worked Solution

i.    `underset~s = 140tcosthetaunderset~i + (140tsintheta – 4.9t^2)underset~j`

`underset~v = 140costhetaunderset~i + (140sintheta – 9.8t)underset~j`

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

`0` `= 140sintheta – 9.8t`
`t` `= (140sintheta)/9.8`

 
`text(Max height:)\ \ underset~j\ text(component of)\ underset~s\ text(when)\ t = (140sintheta)/9.8`

`text(Max height)` `= 140sintheta · (140sintheta)/9.8 – (4.9 · 140^2sin^2theta)/(9.8^2)`
  `= 2000sin^2theta – 1000sin^2theta`
  `= 1000sin^2theta`

 

ii.   `text(Horizontal distance)\ (d):`

`=>\ underset~i\ text(component of)\ underset~s\ text(when)\ \ t = (140sintheta)/9.8`

`:.d` `= 140costheta · (140sintheta)/9.8`
  `= (140 xx 70 xx sin2theta)/9.8`
  `= 1000sin2theta`

 

iii.   `text(Using part ii),`

`500<=1000sin2theta<=800`
`0.5<=sin2theta<=0.8`

 

`text(In the 1st quadrant:)`

`30° <=` `2theta` `<= 53.13°`
`15° <=` `theta` `<= 26.6°`

 
`text(In the 2nd quadrant:)`

`126.87°<=` `2theta` `<= 150°`
`63.4°<=` `theta` `<= 75°`

 
`text(When)\ theta = 26.6°:`

`text(Max height)` `= 1000 · sin^2 26.6°`
  `= 200.5\ text(metres)\ (< 600\ text(m))`

 
`=>\ text(Highest max height for)\ \ 15° <= theta < 26.6°\ \ text(does not satisfy.)`
 

`text(When)\ theta = 63.4°:`

`text(Max height)` `= 1000 · sin^2 63.4°`
  `= 799.5\ text(metres)\ (> 600\ text(m))`

 
`=>\ text(Lowest max height for)\ \ 63.4° <= theta <= 75°\ \ text(satisfies).`

`:. 63.4° <= theta <= 75°`

Vectors, EXT1 V1 SM-Bank 22

Consider the rhombus  `OPQR`  shown below, where  `overset(->)(OP) = punderset~i`  and  `overset(->)(OR) = underset~i + 2underset~j`  and  `p`  is a real constant.
 

  1. Find  `p`.  (1 mark)

  2. Show that the diagonals of rhombus  `OPQR`  are perpendicular.  (2 marks)

Show Answers Only
  1. `sqrt5`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `|overset(->)(OP)| = |overset(->)(OR)|`

`p` `= sqrt(1^2 + 2^2)`
  `= sqrt5`

 

ii.   `overset(->)(RQ) = overset(->)(OP),\ \ overset(->)(OR) = overset(->)(PQ)`

`overset(->)(OQ)` `= overset(->)(OR) + overset(->)(RQ)`
  `= [(1),(2)] + [(sqrt5),(0)]`
  `= (1 + sqrt5)underset~i + 2underset~j`

 

`overset(->)(PR)` `= overset(->)(OR)-overset(->)(RQ)`
  `= [(1),(2)]-[(sqrt5),(0)]`
  `= (1-sqrt5)underset~i + 2underset~j`

 

`overset(->)(OQ) · overset(->)(PR)` `= [(1 + sqrt5),(2)] · [(1-sqrt5),(2)]`
  `= (1 + sqrt5)(1-sqrt5) + 2 xx 2`
  `= 1-5 + 4`
  `= 0`

 
`:. overset(->)(OQ) ⊥ overset(->)(PQ)`

Vectors, EXT1 V1 SM-Bank 18

If  `theta`  is the angle between  `underset~a = underset~i + 3j`  and  `underset~b = 3underset~i + underset~j`, then find the exact value of  `cos 2theta`.  (2 marks)

Show Answers Only

`−7/25`

Show Worked Solution

`underset~a = [(1),(3)],\ \ underset~b = [(3),(1)]`

`|underset~a| = sqrt(1^2 + 3^2) = sqrt10`

`|underset~b| = sqrt(3^2 + 1^2) = sqrt10`

`cos theta` `= (1 xx 3 + 3 xx 1)/(sqrt10 sqrt10)`
  `= 3/5`

 

`cos2theta` `= 2cos^2theta – 1`
  `= 2(3/5)^2 – 1`
  `= −7/25`

Vectors, EXT1 V1 SM-Bank 17

Two vectors are given by `underset~a = 2underset~i + m underset~j`  and  `underset~b = −5underset~i + n underset~j`  where  `m, n > 0`.

If  `|underset~a| = 3`  and  `underset~a`  is perpendicular to  `underset~b`, find the values of  `m`  and  `n`.  (2 marks)

Show Answers Only

`2sqrt5`

Show Worked Solution

`underset~a = [(2),(m)],\ \ underset~b = [(−5),(n)]`

 
`text(Using)\ |underset~a| = 3:`

`3` `= sqrt(2^2 + m^2)`
`m^2` `= 5`
`:.m` `= sqrt5,\ \ \ (m > 0)`

 
`text(S)text(ince)\ underset~a ⊥ underset~b:`

`a · b` `= 0`
`2xx −5 + mn` `= 0`
`sqrt5 n` `= 10`
`n` `= 10/sqrt5`
  `= 2sqrt5`

Vectors, EXT1 V1 SM-Bank 21

Relative to a fixed origin, the points `A`, `B` and `C` are defined respectively by the position vectors  `underset~a = −underset~i - underset~j, \ underset~b = 3underset~i + 2underset~j`  and  `underset~c = −aunderset~i + 2underset~j`, where  `a`  is a real constant.

If the magnitude of angle `ABC`  is  `pi/3`, find `a`.  (3 marks)

Show Answers Only

`−3`

Show Worked Solution

`text(Angle between)\ overset(->)(BA)\ text(and)\ overset(->)(BC) = pi/3`

`overset(->)(BA)` `= overset(->)(OA) – overset(->)(OB)`
  `= [(−1),(−1)] – [(3),(2)] = [(−4),(−3)]`

 

`overset(->)(BC)` `= overset(->)(OC) – overset(->)(OB)`
  `= [(−a),(2)] – [(3),(2)] = [(−a−3),(0 )]`

 

`overset(->)(BA) · overset(->)(BC)` `= [(−4),(−3)] · [(−a −3),(0 )]`
  `= 4a + 12`

 
`overset(->)(BA) · overset(->)(BC) = |overset(->)(BA)| · |overset(->)(BC)|costheta`

`4a + 12` `= sqrt((−4)^2 + (−3)^2) · sqrt((−a – 3)^2) · cos\ pi/3`
`4a + 12` `= 5(−a – 3) · 1/2`
`4a + 12` `= −(5a)/2 – 15/2`
`(13a)/2` `= −39/2`
`:.a` `= −3`

Vectors, EXT1 V1 SM-Bank 20

Consider the vector  `underset~a = underset~i + sqrt3underset~j`, where  `underset~i`  and  `underset~j`  are unit vectors in the positive direction of the `x` and `y` axes respectively.

  1. Find the unit vector in the direction of  `underset~a`.    (1 mark)

  2. Find the acute angle that  `underset~a`  makes with the positive direction of the `x`-axis.   (1 mark)

  3. The vector  `underset~b = m underset~i - 2underset~j`.

     

    Given that  `underset~b`  is perpendicular to  `underset~a`, find the value of  `underset~m`.   (1 mark)

Show Answers Only
  1. `1/2(underset~i + sqrt3underset~j)`
  2. `60°`
  3. `2sqrt3`
Show Worked Solution

i.   `underset~a = underset~i + sqrt3underset~j`

`|underset~a| = sqrt(1 + (sqrt(3))^2) = 2`

`overset^a = (underset~a)/(|underset~a|) = 1/2(underset~i + sqrt3underset~j)`

 

ii.   `text(Solution 1)`

`underset~a\ =>\ text(Position vector from)\ \ O\ \ text{to}\ \ (1, sqrt3)`

`tan theta` `=sqrt3`  
`:. theta` `=60°`  
     

`text(Solution 2)`

`text(Angle with)\ xtext(-axis = angle with)\ \ underset~b = underset~i`

`underset~a · underset~i = 1 xx 1 = 1`

`underset~a · underset~i` `= |underset~a||underset~i|costheta`
`1` `= 2 xx 1 xx costheta`
`costheta` `= 1/2`
`:. theta` `= 60°`

 

iii.   `underset~b = m underset~i – 2underset~j`

`underset~a · underset~b = [(1),(sqrt3)] · [(m),(−2)] = m – 2sqrt3`

`text(S)text(ince)\ underset~a ⊥ underset~b:`

`m – 2sqrt3` `= 0`
`m` `= 2sqrt3`

Vectors, EXT1 V1 SM-Bank 19

Consider the following vectors

`overset(->)(OA) = 2underset~i + 2underset~j,\ \  overset(->)(OB) = 3underset~i - underset~j,\ \ overset(->)(OC) = 5underset~i + 3underset~j`

  1. Find  `overset(->)(AB)`.  (1 mark)

  2. The points `A`, `B` and `C` are vertices of a triangle. Prove that the triangle has a right angle at `A`.  (2 marks)

  3. Find the length of the hypotenuse of the triangle.  (1 mark)

Show Answers Only
  1. `underset~i – 3underset~j`
  2. `text(See Worked Solutions)`
  3. `2sqrt5`
Show Worked Solution

i.  `text(Find)\ overset(->)(AB):`

COMMENT: Many teachers recommend column vector notation to simplify calculations and minimise errors – we agree!

`overset(->)(OA) = [(2),(2)],\ \ overset(->)(OB)[(3),(−1)]`

`overset(->)(AB)` `= overset(->)(OB) – overset(->)(OA)`
  `= [(3),(−1)] – [(2),(2)]`
  `= [(1),(−3)]`
  `= underset~i – 3underset~j`

 

ii.    `overset(->)(AC)` `= overset(->)(OC) – overset(->)(OA)`
    `= [(5),(3)] – [(2),(2)]`
    `= [(3),(1)]`
    `= 3underset~i + underset~j`

 

`overset(->)(AB) · overset(->)(AC)` `= 1 xx 3 + −3 xx 1=0`

`=> AB ⊥ AC`

`:. DeltaABC\ text(has a right angle at)\ A.`

 

iii.   `overset(->)(BC)\ text(is the hypotenuse)`

`overset(->)(BC)` `= overset(->)(OC) – overset(->)(OB)`
  `= [(5),(3)] – [(3),(−1)]`
  `= [(2),(4)]`
`|overset(->)(BC)|` `=\ text(length of hypotenuse)`
  `= sqrt(2^2 + 4^2)`
  `= sqrt(20)`
  `= 2sqrt5`

Vectors, EXT1 V1 SM-Bank 15

Consider the vectors

`underset~a = 6underset~i + 2underset~j,\ \ underset~b = 2underset~i - m underset~j`

  1. Calculate  `2underset~a - 3underset~b`.  (1 mark)

  2. Find the values of  `m`  for which  `|underset~b| = 3sqrt2`.  (2 marks)

  3. Find the value of  `m`  such that  `underset~a`  is perpendicular to  `underset~b`.  (1 mark)

Show Answers Only
  1. `[(6),(4 + 3m)]`
  2. `±sqrt14`
  3. `6`
Show Worked Solution
i.    `2underset~a – 3underset~b` `= 2[(6),(2)] – 3[(2),(−m)]`
    `= [(12),(4)] – [(6),(−3m)]`
    `= [(6),(4 + 3m)]`

 

ii.   `underset~a = [(6),(2)], \ \ underset~b = [(2),(−m)]`

`|underset~b|` `= sqrt(4 + m^2)`
`3sqrt2` `= sqrt(4 + m^2)`
`18` `= 4 + m^2`
`m^2` `= 14`
`m` `= ±sqrt14`

 

iii.   `text(If)\ \ underset~a ⊥ underset~b \ => \ underset~a · underset~b = 0`

`6 xx 2 + 2 xx – m` `= 0`
`2m` `= 12`
`:. m` `= 6`

Vectors, EXT1 V1 SM-Bank 14

Consider the vectors,  `underset~a = overset(->)(OA)`  where  `|OA| = 5`  and  `underset~b = overset(->)(OB)`  where  `|OB| = 7`.

If  `angleAOB = 30°`, find  `text(proj)_(underset~b)underset~a`  as a multiple of  `underset~b`.  (2 marks)

Show Answers Only

`(5sqrt3)/14 · underset~b`

Show Worked Solution

`underset~overset^b = (underset~b)/(|OB|) = (underset~b)/7`

`text(proj)_underset~bunderset~a` `= (|underset~a|\ cos30°) · underset~overset^b`
  `= 5 xx sqrt3/2 xx (underset~b)/7`
  `= (5sqrt3)/14 · underset~b`

Vectors, EXT1 V1 SM-Bank 13

Given  `underset~a = 4underset~i - 3underset~j`  and  `underset~b = 7underset~i - underset~j`, what is the magnitude of the projection of  `underset~a`  onto  `underset~b`. Give your answer in simplest form.  (3 marks)

Show Answers Only

`(31sqrt2)/10`

Show Worked Solution

`underset~a = [(4),(−3)],\ \ underset~b = [(7),(−1)]`

`text(proj)_(underset~b) underset~a` `= (28 + 3)/(49 + 1)(7underset~i – underset~j)`
  `= 31/50(7underset~i – underset~j)`
  `= 217/50 underset~i – 31/50 underset~j`

 

`|\ text(proj)_(underset~b) underset~a\ |` `= sqrt((217/50)^2 + (31/50)^2)`
  `= sqrt((217^2 + 31^2))/50`
  `= sqrt(48\ 050)/50`
  `= (155sqrt2)/50`
  `= (31sqrt2)/10`

Vectors, EXT1 V1 SM-Bank 11

Points  `A`  and  `B`  have position vectors  `underset~a = 2underset~i - 2underset~j`  and  `underset~b =  4i + sqrt2 underset~j` respectively.

Find the angle between  `underset~a`  and  `underset~b`  to the nearest minute.  (2 marks)

Show Answers Only

`64°28′`

Show Worked Solution

`underset~a = 2underset~i – 2underset~j \ => \ |underset~a| = sqrt(2^2 + 2^2) = sqrt8`

`underset~b = 4 i + sqrt2 j \ => \ |underset~b| = sqrt(4^2 + (sqrt2)^2) = sqrt18`

`underset~a · underset~b = |underset~a||underset~b|costheta`

`costheta` `= (underset~a · underset~b)/(|underset~a||underset~b|)`
  `= (8 – 2sqrt2)/(sqrt8sqrt18)`
  `= (8 – 2sqrt2)/12`
`:. theta` `=64.471…`
  `=64°28′\ \ \ text{(nearest minute)}`

Calculus, EXT1 C3 SM-Bank 2

The parabola with equation  `y = 9 - x^2`  cuts the `y`-axis at  `P(0,9)`  and the `x`-axis at  `Q(3,0)`.

Find the exact volume of the solid of revolution formed when the area between the line  `PQ`  and the parabola is rotated about the `y`-axis.  (4 marks)

Show Answers Only

`(27pi)/2\ text(units³)`

Show Worked Solution

`text(Equation of)\ PQ:`

`m = −3,\ \ ytext(-intercept) = 9`

`y = 9 – 3x`
 

`text(Rotating about the)\ ytext(-axis:)`

`x_1^(\ 2)` `= 9 – y\ \ \ text{(parabola)}`
`y` `= 9 – 3x_2\ \ \ (text{line}\ PQ)`
`3x_2` `= 9 – y`
`x_2` `= 3 – y/3`
`x_2^(\ 2)` `= (3 – y/3)^2`

 

`V` `= pi int_0^9 x_1^(\ 2) – x_2^(\ 2)\ dy`
  `= pi int_0^9 9 – y – (3 – y/3)^2\ dy`
  `= pi int_0^9 9 – y – (9 – 2y + (y^2)/9)\ dy`
  `= pi int_0^9 y – (y^2)/9\ dy`
  `= pi [(y^2)/2 – (y^3)/27]_0^9`
  `= pi(81/2 – 27)`
  `= (27pi)/2\ text(units³)`

Number, NAP-L3-CA03

Con packs oranges into boxes for the market.

Each box holds 20 oranges.

Con fills 3 boxes and has 6 oranges left over.
 


 

How many oranges does Con have altogether?

`46` `60` `66` `78`
 
 
 
 
Show Answers Only

`66`

Show Worked Solution

`text(One full box holds 20 oranges.)`

`:.\ text(Total oranges)` `= 3 xx 20`
  `= 60 + 6`
  `= 66`

Statistics, NAP-L3-CA02

Kim is a bricklayer and he recorded the number of bricks he layed each day for five days in a picture graph.
 


 

What is the difference between the number of bricks Kim laid on Monday and the number of bricks he laid on Thursday?

`2500` `3000` `3500` `4000`
 
 
 
 
Show Answers Only

`3500`

Show Worked Solution

`text(Bricks laid on Monday = 1500)`

`text(Bricks laid on Thursday = 5000)`

`text(Difference)` `= 5000 – 1500`
  `= 3500`

Statistics, STD2 S1 SM-Bank 2

A survey question is shown.

Give TWO reasons why this survey may be considered to be poorly designed.  (2 marks)

Show Answers Only

`text(Only 3 choices of many colours are given.)`

Show Worked Solution

`text(Poor design reason:)`

`text(- only 3 choices of many colours are given.)`

`text(- two colours might be someone’s equal favourite colours.)`

Calculus, EXT2 C1 2019 HSC 15c

  1. Show that  `int_0^1 x/(x + 1)^2\ dx = ln 2 - 1/2`.  (2 marks)


  2. Let  `I_n = int_0^1 x^n/(x + 1)^2\ dx`.

     


    Show that  `I_n = 1/(2(n - 1)) - n/(n - 1) I_(n - 1)\ \ text(for)\ \ n >= 2`.  (3 marks)


  3. Evaluate  `I_3`.  (2 marks)

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

i.    `text(Show)\ \ int_0^1 x/(x + 1)^2\ dx = ln 2 – 1/2`

`text(Let)\ \ u = x + 1 \ => \ x = u – 1`

`(du)/(dx) = 1 \ => \ du = dx`
 

`text(When)\ \ x = 1,\ \ u = 2`

`text(When)\ \ x = 0,\ \ u = 1`

`int_0^1 x/(x + 1)^2\ dx` `= int_1^2 (u – 1)/(u^2)\ du`
  `= int_1^2 1/u\ du – int_1^2 1/u^2\ du`
  `= [ln u]_1^2 + [1/u]_1^2`
  `= ln 2 – ln 1 + 1/2 – 1`
  `= ln 2 – 1/2`

 
ii.    `I_u = int_0^1 x^n/(x + 1)^2\ dx`

`u = x^n` `v prime = 1/(x + 1)^2`
`u prime = nx^(n – 1)` `v = -1/{(x + 1)}`

 

`I_n` `= [uv]_0^1 – int_0^1 u prime v\ dx`
  `= [(-x^n)/(x + 1)]_0^1 + int_0^1 (n x^(n – 1))/(x + 1)\ dx`
  `= (-1/2 – 0) + n int_0^1 (x^(n – 1) (x + 1))/(x + 1)^2\ dx`
  `= -1/2 + n int_0^1 x^n/(x + 1)^2 + (x^(n – 1))/(x + 1)^2\ dx`
  `= -1/2 + n I_n + n I_(n – 1)`
`nI_n – I_n` `= 1/2 – n I_(n – 1)`
`I_n(n – 1)` `= 1/2 – n I_(n – 1)`
`:. I_n` `= 1/(2(n – 1)) – n/(n – 1) I_(n – 1)`

 

iii.    `I_1` `= ln 2 – 1/2`
  `I_2` `= 1/(2(2 – 1)) – 2/{(2 – 1)} I_1`
    `= 1/2 – 2 I_1`
    `= 1/2 – 2 ln 2 + 1`
    `= 3/2 – 2 ln 2`

 

`:. I_3` `= 1/4 – 3/2(3/2 – 2 ln 2)`
  `= 1/4 – 9/4 + 3 ln 2`
  `= 3 ln 2 – 2`

Calculus, EXT2 C1 2019 HSC 15a

  1. Show that
     
    `qquad int_(-a)^a (f(x))/(f(x) + f(-x))\ dx = int_(-a)^a (f(-x))/(f(x) + f(-x))\ dx`.  (2 marks)

  2. Hence, or otherwise, evaluate
     
    `qquad int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx`.  (2 marks)

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

i.    `text(Let)\ \ u = -x \ => \ du = -dx`

`text(When)\ \ x=a, \ u=-a`

`text(When)\ \ x=-a, \ u=a`

`int_(-a)^a (f(x))/(f(x) + f(-x))\ dx` `= -int_a^(-a) (f(-u))/(f(-u) + f(u))\ du`
  `= int_(-a)^a (f(-u))/(f(u) + f(-u))\ du`
  `= int_(-a)^a (f(-x))/(f(x) + f(-x))\ dx`

 

ii.    `text(Let)\ \ I = int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx`

`2I` `= int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx + int_(-1)^1 (e^(-x))/(e^x + e^(-x))\ dx`
`2I` `= int_(-1)^1 (e^x + e^(-x))/(e^x + e^(-x))\ dx`
`2I` `= [x]_(-1)^1`
`2I` `= 2`

 
`:. int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx = 1`

Mechanics, EXT2 M1 2019 HSC 14b

A parachutist jumps from a plane, falls freely for a short time and then opens the parachute. Let t be the time in seconds after the parachute opens, `x(t)`  be the distance in metres travelled after the parachute opens, and  `v(t)`  be the velocity of the parachutist in `text(ms)^(-1)`.

The acceleration of the parachutist after the parachute opens is given by

`ddot x = g - kv,`

where `g\ text(ms)^(-2)` is the acceleration due to gravity and `k` is a positive constant.

  1. With an open parachute the parachutist has a terminal velocity of  `w\ text(ms)^(-1)`.

     

    Show that  `w = g/k`.  (1 mark)

    At the time the parachute opens, the speed of descent is `1.6 w\ text(ms)^(-1)`.

  2. Show that it takes  `1/k log_e 6`  seconds to slow down to a speed of `1.1w\ text(ms)^(-1)`.  (4 marks)

  3. Let  `D`  be the distance the parachutist travels between opening the parachute and reaching the speed `1.1w\ text(ms)^(-1)`.

     

     

    Show that  `D = g/k^2 (1/2 + log_e 6)`.  (3 marks)

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

i.   `v_T=w\ \  text(when)\ \ ddot x = 0`

`0` `= g – kw`
`w` `= g/k`

 

ii.   `text(Show)\ \ t = 1/k log_e 6\ \ text(when)\ \ v = 1.1w`

`(dv)/(dt)` `= g – kv`
`(dt)/(dv)` `= 1/(g – kv)`
`t` `= int 1/(g – kv)\ dv`
  `= -1/k ln(g – kv) + C`

 
`text(When)\ \ t = 0,\ \ v = 1.6w`

`0` `= -1/k ln(g – 1.6 kw) + C`
`C` `= 1/k ln(g – 1.6 kw)`
`t` `= 1/k ln (g – 1.6kw) – 1/k ln(g – kv)`
  `= 1/k ln((g – 1.6 kw)/(g – kv))`

 
`text(Find)\ \ t\ \ text(when)\ \ v = 1.1w`

`t` `= 1/k ln((g – 1.6 k xx g/k)/(g – 1.1k xx g/k))`
  `=1/k ln((g – 1.6 g)/(g – 1.1g))`
  `=1/k((-0.6g)/(-0.1g))`
  `= 1/k ln 6`

 

iii.    `v ⋅ (dv)/(dx)` `= g – kv`
  `(dv)/(dx)` `= (g – kv)/v`
  `(dx)/(dv)` `= v/(g – kv)`
  `x` `= int v/(g – kv)\ dv`
    `= 1/k int (kv)/(g – kv)\ dv`
    `= -1/k int 1 – g/(g – kv)\ dv`

 

`:. D` `= -1/k int_(1.6w)^(1.1w) 1 – g/(g – kv)\ dv`
  `= 1/k int_(1.1w)^(1.6w) 1 – g/(g – kv)\ dv`
  `= 1/k[v + g/k ln (g – kv)]_(1.1w)^(1.6w)`
  `= g/k^2[(kv)/g + ln (g – kv)]_(1.1w)^(1.6w)`
  `= g/k^2[((1.6kw)/g + ln (g – 1.6kw)) – ((1.1 kw)/g + ln (g – 1.1kw))]`
  `= g/k^2[1.6 + ln ((g – 1.6kw)/(g – 1.1kw)) – 1.1]`
  `= g/k^2(0.5 + ln 6)`

Combinatorics, EXT1 A1 2019 13b

In the expansion of  `(5x + 2)^20`, the coefficients of  `x^r`  and  `x^(r + 1)`  are equal.

What is the value of  `r` ?  (3 marks)

Show Answers Only

`14`

Show Worked Solution

`(5x + 2)^20`

COMMENT: Arithmetic becomes easier by expanding  `(2 + 5x)^20`.

`text(General term:)\ (2 + 5x)^20`

`T_r` `= \ ^20C_r · 2^(20 – r) · (5x)^r`
  `= \ ^20C_r · 2^(20 – r) · 5^r · x^r`
`T_(r + 1)` `=\ ^20C_(r + 1) · 2^(19 – r) · 5^(r + 1) · x^(r + 1)`

 

`text(Equating co-efficients:)`

`(20!)/(r!(20 – r)!) · 2^(20 – r) · 5^r` `= (20!)/((r + 1)!(19 – r)!) · 2^(19 – r) · 5 ^(r + 1)`
`2/(20 – r)` `= 5/(r + 1)`
`2r + 2` `= 100 – 5r`
`7r` `= 98`
`r` `= 14`

Functions, EXT1′ F1 2019 HSC 12d

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

  1.  Sketch the graph  `y = |\ f(x)\ |`.  (1 mark)

  2.  Sketch the graph  `y = 1/(f(x))`.  (2 marks)

  3.  Without using calculus, sketch the graph  `y = x/(f(x))`.  (2 marks)

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

i.   `y = |\ x^3 – 1\ |`

 

ii.   `y = 1/(x^3 – 1)`

 

iii.   `y = x/(x^3 – 1)`

Calculus, EXT1′ C3 2019 HSC 12b

The diagram shows two straight railway tracks that meet at an angle of  `(2 pi)/3`  at the point `P`.

Trains  `A`  and  `B`  are joined by a cable which is 70 m long.

At time  `t`  seconds, train  `A`  is  `x`  metres from  `P`  and train  `B`  is  `y`  metres from `P`.

Train  `B`  is towing train  `A`  and is moving at a constant speed of  `4\ text(ms)^(-1)` away from `P`.

  1. Show that  `x^2 + xy + y^2 = 70^2`.  (1 mark)
  2. What is the value of  `(dx)/(dt)`  when train  `A`  is 30 metres from  `P`  and train  `B`  is 50 metres from `P`?  (3 marks)

 

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `-52/11`
Show Worked Solution

(i)   `text(Using cosine rule):`

`x^2 + y^2 – 2xy cos ((2 pi)/3)` `= 70^2`
`x^2 + y^2 – 2xy xx -1/2` `= 70^2`
`x^2 + xy + y^2` `= 70^2`

 

(ii)    `(dy)/(dt)` `= 4`
  `(dx)/(dt)` `= (dy)/(dt) ⋅ (dx)/(dy)`

 
`text{Differentiate part (i)}:`

`2x + x ⋅ (dy)/(dx) + y + 2y ⋅ (dy)/(dx)` `= 0`
`(dy)/(dx) (x + 2y)` `= -2x – y`
`(dy)/(dx)` `= (-2x – y)/(x + 2y)`

 
`text(When)\ \ x = 30,\ \ y = 50`

`(dx)/(dt)` `= 4 xx (30 + 2(50))/(-2(30) – 50)`
  `= 4 xx -130/110`
  `= -52/11`

Complex Numbers, EXT2 N1 2019 HSC 8 MC

Let  `z`  be a complex number such that  `z^2 = -i bar z`.

Which of the following is a possible value for `z`?

  1. `1/2 - sqrt 3/2 i`
  2. `1/2 + sqrt 3/2 i`
  3. `sqrt 3/2 - 1/2 i`
  4. `sqrt 3/2 + 1/2 i`
Show Answers Only

`C`

Show Worked Solution

`text(Solution 1)`

`text(Consider)\ C:`

`z` `=sqrt3/2 – 1/2 i`  
`barz` `=sqrt3/2 + 1/2 i`  
`-i barz` `=1/2-sqrt3/2 i`  

 

`z^2` `=(sqrt3/2 – 1/2 i)^2`  
  `=3/4 – 2 sqrt3/2 * 1/2 i -1/4`  
  `=1/2 – sqrt3/2 i`  

 
`=>C`

 

`text(Solution 2)`

`text(Let)\ \ text(arg)(z)` `= theta`
`text(arg)(z^2)` `= 2 theta`

 

`text(arg)(-i bar z)` `=text(arg)(i bar z) – pi`
  `= text(arg) (bar z) – pi + pi/2`
  `= -theta – pi/2`

 

`:. 2 theta` `= -theta – pi/2`
`3 theta` `= -pi/2`
`theta` `= -pi/6`

 
`=>   C`

Calculus, EXT2 C1 2019 HSC 7 MC

Which of these integrals has the largest value?

  1. `int_0^(pi/4) tan x\ dx`
  2. `int_0^(pi/4) tan^2 x\ dx`
  3. `int_0^(pi/4) 1 - tan x\ dx`
  4. `int_0^(pi/4) 1 - tan^2 x\ dx`
Show Answers Only

`D`

Show Worked Solution

`text(Consider options A and B:)`

`text(Consider options C and D:)`


 

`:. int_0^(pi/4) 1 – tan^2 x\ dx\ \ text(is the largest)`

`text{(largest area under the curve}`

  `text(between)\ \ x=0 \ and \ x=pi/4. text{)}`

 
`=>   D`

Measurement, STD2 M6 SM-Bank 1

The diagram shows a triangle with side lengths 25 cm and 47 cm and angle 30° and `theta`.
 

Find  `theta`  given  it is an obtuse angle. Give your answer to the nearest minute.  (3 marks)

Show Answers Only

`109°57′`

Show Worked Solution

`text(Using sine rule:)`

`(sintheta)/47` `= (sin30°)/25`
`sintheta` `=47 xx (1/2)/25`
`sintheta` `= 47/50`
`theta` `= sin^(−1)\ 47/50`
  `= 70.05\ text(or)\ 109.94`
  `= 109.948…\ \ (theta\ text(is obtuse))`
  `= 109°57′`

Functions, EXT1′ F2 2019 HSC 4 MC

The polynomial  `2x^3 + bx^2 + cx + d`  has roots 1 and – 3, with one of them being a double root.

What is a possible value of  `b`?

  1. – 10
  2. – 5
  3. 5
  4. 10
Show Answers Only

`D`

Show Worked Solution

`f(x) = 2x^3 + bx^2 + cx + d`

`f prime (x) = 6x^2 + 2bx + c`

`text(Roots at 1 and −3:)`

`f(1) = 0`

`2 + b + c + d` `= 0`
`b + c + d` `= -2\ \ text{… (1)}`

 
`f(-3) = 0`

`-54 + 9b – 3c + d` `= 0`
`9b – 3c + d` `= 54\ \ text{… (2)}`

 
`(2) – (1)`

`8b – 4c` `= 56`
`2b – c` `= 14\ \ text{… (3)}`

 
`text(If double root at 1:)`

`f prime(1) = 0`

`6 + 2b + c` `= 0`
`2b + c` `= -6\ \ text{… (4)}`

 
`(3) + (4)`

`4b` `= 8`
`b` `= 2`

 
`text(If double root at – 3:)`

`f prime(-3) = 0`

`54 – 6b + c` `= 0`
`-6b + c` `= -54\ \ text{… (5)}`

 
`(3) + (5)`

`-4b` `= -40`
`b` `= 10`

 
`=>   D`

Calculus, EXT2 C1 2019 HSC 2 MC

Which of the following is a primitive of  `(sin x)/(cos^3 x)`?

  1. `1/2 sec^2 x`
  2. `-1/2 sec^2 x`
  3. `1/4 sec^4 x`
  4. `-1/4 sec^4 x`
Show Answers Only

`A`

Show Worked Solution

`text(Solution 1)`

`int(sin x)/(cos^3 x)` `= int sinx cos^(-3) x\ dx`
  `=1/2 cos^(-2)x + C`
  `= 1/2 sec^2 x + C`

 
`text(Solution 2)`

`int(sin x)/(cos^3 x)` `= int tanx sec^2x\ dx`
  `= 1/2 tan^2 + C_1`
  `= 1/2 (sec^2 x -1) + C_1`
  `=1/2 sec^2 x + C_2`

 
`=>A`

Algebra, STD1 A3 2019 HSC 30

A small business makes and sells bird houses.

Technology was used to draw straight-line graphs to represent the cost of making the bird houses `(C)` and the revenue from selling bird houses `(R)`. The `x`-axis displays the number of bird houses and the `y`-axis displays the cost/revenue in dollars.
 


 

  1. How many bird houses need to sold to break even?  (1 mark)

  2. By first forming equations for cost `(C)` and revenue `(R)`, determine how many bird houses need to be sold to earn a profit of $1900.  (3 marks)

Show Answers Only
  1. `20`
  2. `96`
Show Worked Solution

a.   `20\ \ (xtext(-value at intersection))`

 

b.   `text(Find equations of both lines):`

♦ Mean mark part (b) 47%.

`(0, 500)\ text(and)\ (20, 800)\ text(lie on)\ \ C`

`m_C = (800 – 500)/(20 – 1) = 15`

`=> C = 500 + 15x`
 

`(0,0)\ text(and)\ (20, 800)\ text(lie on)\ \ R`

`m_R = (800 – 0)/(20 – 0) = 40`

`=> R = 40x`
 

`text(Profit) = R – C`

`text(Find)\ \ x\ \ text(when Profit = $1900:)`

`1900` `= 40x – (500 + 15x)`
`25x` `= 2400`
`x` `= 96`

Statistics, STD1 S3 2019 HSC 27

A set of bivariate data is collected by measuring the height and arm span of eight children. The graph shows a scatterplot of these measurements.
 

  1. On the graph, draw a line of best fit by eye.  (1 mark)

  2. Robert is a child from the class who was absent when the measurements were taken. He has an arm span of 147 cm. Using your line of best fit from part (a), estimate Robert’s height.  (1 mark)

Show Answers Only
  1.   
  2. `text(Robert’s height ≈ 151.1 cm)`
Show Worked Solution

a.     
       

♦ Mean mark (a) 38%.

b.   `text(Robert’s height ≈ 151.1 cm)`

`text{(Answers can vary slightly depending on line of best fit drawn).}`

Financial Maths, STD1 F1 2019 HSC 26

Scott decided to have a shed built at his home. He agreed to the following costs:

  • Materials
– Timber $5400
– Roof $1800
– Nails $160
– Paint $375
  • $70 per hour for the builder when working Monday to Friday
  • $30 per hour for the labourer when working Monday to Friday
  • Builder and labourer paid time-and-a-half when working on Saturday.

It took six days to build the shed. The builder and labourer both worked from 8 am until 4 pm each day from Monday to Friday. On each of these days, they both took a 1 hour unpaid lunch break at 12 noon.

The builder and labourer also both worked 4 hours on Saturday, without a break, to finish the job.

What was the total cost to build the shed?  (4 marks)

Show Answers Only

`$11\ 835`

Show Worked Solution
`text(Material cost)` `= 5400 + 1800 + 160 + 375`
  `= $7735`

 
`text(Labour cost:)`

`text{Hours per weekday = 8 am to 4 pm (less 1 hour) = 7}`

`text(Builder cost)` `= 5 xx 7 xx 70 + 4 xx 1.5 xx 70`
  `= $2870`
`text(Labourer cost)` `= 5 xx 7 xx 30 + 4 xx 1.5 xx 30`
  `= $1230`

 

`:.\ text(Total cost)` `= 7735 + 2870 + 1230`
  `= $11\ 835`

Probability, STD1 S2 2019 HSC 24

The faces on a biased six-sided die are labelled 1, 2, 3, 4, 5 and 6. The die was rolled twenty times. The relative frequency of rolling a 6 was 30% and the relative frequency of rolling a 2 was 15%. The number 3 was the only other number rolled in the twenty rolls.

How many times was the number 3 rolled in the twenty rolls of the die?  (3 marks)

 
Show Answers Only

`11`

Show Worked Solution

`text(Number of 6’s) = 30/100 xx 20 = 6`

`text(Number of 2’s) = 15/100 xx 20 = 3`

`:.\ text(Number of 3’s)` `= 20 – (6 + 3)`
  `= 11`

Measurement, STD1 M5 2019 HSC 20

The plan of the lower level of a small house is shown.


 

  1. How many windows are shown on the plan?  (1 mark)

  2. What is the actual perimeter, in metres, of the shaded part of the kitchen floor?  (2 marks)

Show Answers Only
  1. `text(6 windows)`
  2. `6.5\ text(metres)`
Show Worked Solution

a.   `text(6 windows)`

 

b.    `text(Plan Perimeter)` `= 2 xx 3\ text(units) + 2 xx 3.5\ text(units)`
    `= 13\ text(units)`

 

`:.\ text(Actual perimeter)` `= 13 xx 0.5`
  `= 6.5\ text(metres)`