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Functions, 2ADV F2 EQ-Bank 9

Consider the function  `f(x) = 1/(4x - 1)`.

  1. Find the domain of  `f(x)`.  (1 mark)

  2. Sketch  `f(x)`, showing all asymptotes and intercepts?  (2 marks)

Show Answers Only
  1. `{text(all real)\ x,  x!=1/4}`
  2.   
Show Worked Solution
i.    `4x – 1` `!= 0`
  `x` `!= 1/4`

 
`:.\ text(Domain:)\ {text(all real)\ x,  x!=1/4}`

 

ii.   `text(When)\ \ x = 0, \ y = −1`

`text(As)\ \ x -> ∞, \ y -> 0^+`

`text(As)\ \ x -> −∞, \ y -> 0^−`

Functions, 2ADV’ F2 2012 HSC 13b

  1. Find the horizontal asymptote of the graph  `y=(2x^2)/(x^2 + 9)`.   (1 mark)

  2. Without the use of calculus, sketch the graph  `y=(2x^2)/(x^2 + 9)`, showing the asymptote found in part (i).    (2 marks)

Show Answers Only
  1. `text(Horizontal asymptote at)\ y = 2`
  2.  
    Geometry and Calculus, EXT1 2012 HSC 13b Answer
Show Worked Solution
i.    `y` `= (2x^2)/(x^2 +9)`
    `= 2/(1 + 9/(x^2))`

 

`text(As)\ \ x -> oo,\ y ->2`

`text(As)\ \ x -> – oo,\ y -> 2`

`:.\ text(Horizontal asymptote at)\ y = 2`

 

ii.    `text(At)\ \ x = 0,\ y = 0`

`f(x) = (2x^2)/(x^2 + 9) >= 0\ text(for all)\ x`

`f(–x) = (2(–x)^2)/((–x)^2 + 9) = (2x^2)/(x^2 + 9) = f(x)`

`text(S)text(ince)\ \ f(x) = f(–x) \ \ =>\ text(EVEN function)`

Geometry and Calculus, EXT1 2012 HSC 13b Answer

Calculus, EXT1 C3 2015 SPEC2 12

Find  `y`  given  `dy/dx = 1 - y/3`  and  `y = 4`  when  `x = 2`.   (2 marks)

Show Answers Only

`y= 3 + e^((2 – x)/3)`

Show Worked Solution
`(dy)/(dx)` `= (3 – y)/3`
`(dx)/(dy)` `= 3/(3 – y)`
`x` `= int 3/(3 – y)\ dy`
`x/3` `= -ln |3 – y| + c`

 
`text(Given)\ \ y=4\ \ text(when)\ \ x=2:`

`2/3= -ln|-1| + c`

`c=2/3`
 

` x/3` `=-ln |3 – y| +2/3`
`ln|3-y|` `= (2-x)/3`
`3-y` `= ±e^((2 – x)/3)`
`:. y` `= 3 + e^((2 – x)/3)`

Vectors, EXT2 V1 2014 SPEC1 1

Consider the vector  `underset ~a = sqrt 3 underset ~i - underset ~j - sqrt 2 underset ~k`, where  `underset ~i, underset ~j`  and  `underset ~k`  are unit vectors in the positive directions of the `x, y` and `z` 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.  (2 marks)

  3.  The vector  `underset ~b = 2 sqrt 3 underset ~i + m underset ~j - 5 underset ~k`.

     

     Given that  `underset ~b`  is perpendicular to  `underset ~a,` find the value of `m`.  (2 marks)

Show Answers Only
  1. `1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`
  2. `theta = 45^@`
  3. `m = 6 + 5 sqrt 2`
Show Worked Solution
i.    `|underset ~a|` `= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)`
    `= sqrt 6`
`:. hat underset ~a` `= underset ~a/|underset ~a|`
  `= 1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`

 

Mean mark part (b) 51%.

ii.    `underset ~a ⋅ underset ~i` `= sqrt 3 xx 1 = sqrt 3`
  `underset ~a ⋅ underset ~i` `= |underset ~a||underset ~i| cos theta`
    `= sqrt 6 cos theta`
  `sqrt 3` `= sqrt 6 cos theta`
`cos theta` `= 1/sqrt 2`
`:. theta` `= pi/4 = 45^@`

 

iii.   `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6 – m + 5 sqrt 2` `=0`  
`:. m` `=6 + 5 sqrt 2`  

Vectors, EXT2 V1 2013 SPEC1 3

The coordinates of three points are  `A\ ((– 1), (2), (4)), \ B\ ((1), (0), (5)) and C\ ((3), (5), (2)).`

  1. Find  `vec (AB).`  (1 mark)

  2. The points `A, B` and `C` are the 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. `2underset~i – 2underset~j + underset~k`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `sqrt 38`
Show Worked Solution
i.   `vec(AB)` `=(1 – −1)underset~i + (0 – 2)underset~j + (5 – 4)underset~k`
    `= 2underset~i – 2underset~j + underset~k`

 

ii.    `overset(->)(AC)` `= (3 – −1)underset~i + (5 – 2)underset~j + (2 – 4)underset~k`
    `= 4underset~i + 3underset~j – 2underset~k`

 

`overset(->)(AB) · overset(->)(AC)` `= 2 xx 4 + (−2) xx 3 + 1 xx (−2)`
  `= 8 – 6 – 2`
  `= 0`

 
`=>  overset(->)(AB) ⊥ overset(->)(AC)`

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

iii.    `overset(->)(BC)` `= (3 – 1)underset~i + (5 – 0)underset~j + (2 – 5)underset~k`
    `= 2underset~i + 5underset~j – 3underset~k`

 

`|overset(->)(BC)|` `= sqrt(2^2 + 5^2 + (−3)^2)`
  `= sqrt(4 + 25 + 9)`
  `= sqrt38`

Vectors, EXT2 V1 SM-Bank 12

Two vectors are given by  `underset ~a = 4 underset ~i + m underset ~j - 3 underset ~k`  and  `underset ~b = −2 underset ~i + n underset ~j - underset ~k`, where  `m`, `n in R^+`.

If  `|\ underset ~a\ | = 10`  and  `underset ~a`  is perpendicular to  `underset ~b`, then find `m` and `n`.   (2 marks)

Show Answers Only

`m=5sqrt3, \ n=sqrt3/3`

Show Worked Solution

`text(Using)\ \ |\ underset ~a\ | = 10:`

`10` `= sqrt(4^2 + m^2 + (-3)^2)`
`100` `=m^2 +25`
`m` `= 5sqrt3`

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

`underset ~a ⋅ underset ~b` `= 0`
`0` `=4 xx (−2) + mn + (−3) xx (−1)`
`0` `= mn-5`
`n` `=5/(5sqrt3)`
  `=sqrt3/3`

 

Vectors, EXT2 V1 SM-Bank 5

Find the value(s) of  `m`  so that the vectors  `underset~a = 2underset~i + m underset~j - 3underset~k`  and  `underset~b = m^2underset~i - underset~j + underset~k`  are perpendicular.   (2 marks)

Show Answers Only

`m = 3/2, quad m = -1`

Show Worked Solution

`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`

`underset ~a ⋅ underset ~b` `= 2m^2 + m(-1) + (-3)(1)`
`0` `= 2m^2 – m – 3`
`0` `= (2m – 3)(m + 1)`

 
`:. m = 3/2, quad m = -1`

Vectors, EXT2 V1 SM-Bank 4

Consider the three vectors

`underset ~a = underset ~i - underset ~j + 2underset ~k,\ underset ~b = underset ~i + 2 underset ~j + m underset ~k`  and  `underset ~c = underset ~i + underset ~j - underset ~k`, where  `m in R.`

  1. Find the value(s) of `m` for which  `|\ underset ~b\ | = 2 sqrt 3.`  (2 marks)
  2. Find the value of `m` such that  `underset ~a`  is perpendicular to  `underset ~b.`  (1 mark)
Show Answers Only
  1. `+- sqrt 7`
  2. `1/2`
Show Worked Solution
i.    `|underset~b|` `= sqrt(1^2 + 2^2 + m^2)` `= 2sqrt3`
    `1 + 4 + m^2` `= 4 xx 3`
    `m^2` `= 7`
    `m` `= ±sqrt7`

 

ii.    `underset~a * underset~b` `= 1 xx 1 + (−1) xx 2 + 2 xx m`  
  `0` `= 1 – 2 + 2m\ \ ( underset~a ⊥ underset~b)`  
  `2m` `= 1`  
  `m` `= 1/2`  

Vectors, EXT2 V1 2011 SPEC2 12 MC

The angle between the vectors  `3underset~i + 6underset~j - 2underset~k`  and  `2underset~i - 2underset~j + underset~k`, correct to the nearest tenth of a degree, is

A.       2.0°

B.     91.0°

C.   112.4°

D.   121.3°

Show Answers Only

`C`

Show Worked Solution

`|3underset~i + 6underset~j – 2underset~k| = sqrt(9 + 36 + 4) = sqrt49 = 7`

`|2underset~i – 2underset~j + underset~k| = sqrt(4 + 4 + 1) = sqrt9 = 3`

`(3underset~i + 6underset~j – 2underset~k) * (2underset~i – 2underset~j + underset~k)`

`= 3 xx 2 + 6 xx (−2) + (−2) xx 1`

`= 6 – 12 – 2`

`= -8`
  

`costheta` `= ((3tildei + 6tildej – 2tildek).(2tildei – 2tildej + tildek))/(|\ 3tildei + 6tildej – 2tildek\ ||\ 2tildei – 2tildej + tildek\ |)`
  `= −8/21`
`:. theta` `= cos^(−1)(−8/12)`
  `~~ 112.4^@`

 
`=> C`

Functions, EXT1 F1 SM-Bank 12

Given  `f(x) = x^3 - x^2 - 2x`, without calculus sketch a separate half page graph of the following functions, showing all asymptotes and intercepts.

  1.   `y = |\ f(x)\ |`  (1 mark)

  2.   `y = f(|x|)`  (2 marks)

  3.    `y = 1/(f(x))`  (2 marks)

Show Answers Only
  1.  
  2.   
  3.   
Show Worked Solution
i.    `f(x)` `= x^3 – x^2 – 2x`
    `= x(x^2 – x – 2)`
    `= x(x – 2)(x + 1)`

 

 

ii.   

 
`y = f(|x|)\ text(is a reflection of)\ y = f(x)\ text(for)\ x > 0`

`text(is the)\ ytext(-axis.)`

 

iii.

Functions, EXT1 F1 EQ-Bank 11

  1. Find the function described by the following parametric equations

     

        `x = 3t^2` 

     

        `y = 9t, \ \ t > 0`  (1 mark)

  2. Sketch the function.  (1 mark)

Show Answers Only
  1. `y = 3sqrt(3x), (t > 0)`
  2.   
Show Worked Solution

i.   `y = 9t \ \ => \ \ t = y/9`

`x` `= 3t^2`
`x` `= 3 · (y/9)^2`
`x` `= y^2/27`
`y^2` `= 27x`
`y` `= 3sqrt(3x), \ \ (t > 0)`

 
ii. 

Functions, EXT1 F1 SM-Bank 9

  1. Sketch the graph of the function described by the parametric equations

     

          `x = 4t - 7`

     

          `y = 2t^2 + t`  (2 marks)

  2. State the domain and range of the function.  (1 mark)

Show Answers Only
  1.  
  2. `text(Domain:  all)\ x`
    `text(Range)\ {y: −1/8 <= y < ∞}`
Show Worked Solution

i.   `x = 4t – 7 \ \ => \ \ t = (x + 7)/4`

`y` `= 2t^2 + t`
`y` `= 2((x + 7)/4)^2 + ((x + 7)/4)`
`16y` `= 2(x + 7)^2 + 4(x + 7)`
`16y` `= 2x^2 + 28x + 98 + 4x + 28`
`16y` `= 2x^2 + 32x + 126`
`8y` `= x^2 + 16x + 63`
`y` `= 1/8(x + 7)(x + 9)`

 
`=>\ text(Equation is a concave up quadratic with)`

`text(zeros at)\ \ x = −9\ text(and)\ \ x = −7.`
  

 

ii.   `text(Axis at)\ \ x = −8`

`:.\ y_text(min)` `= 1/8(−1)(1)`
  `= −1/8`

 
`text(Domain: all)\ x`

`text(Range:)\ −1/8 <= y < ∞`

Functions, EXT1 F1 EQ-Bank 10

An equation can be expressed in the parametric form

`x = 2costheta - 1`

`y = 2 + 2sintheta`

  1.  Express the equation in Cartesian form.  (2 marks)

  2.  Sketch the graph.  (1 mark)

Show Answers Only
  1. `(x + 1)^2 + (y – 2)^2 = 4`
  2.    
Show Worked Solution
i.    `2costheta` `= x + 1`
  `costheta` `= (x + 1)/2`
`2sintheta` `= y – 2`
`sintheta` `= (y – 2)/2`

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

`((x + 1)/2)^2 + ((y – 2)/2)^2` `= 1`
`(x + 1)^2 + (y – 2)^2` `= 4`

 

ii.  `text{Sketch circle with centre (−1, 2),  radius = 2}`
 

Functions, EXT1 F1 SM-Bank 8

A circle has the equation  `x^2 - 10x + y^2 + 6y +25 = 0`

  1.  Express the circle in parametric form.  (2 marks)

  2.  Sketch the circle.  (1 mark)

Show Answers Only
  1. `x = 5 + 3costheta`
    `y = −3 + 3sintheta`
  2.   
Show Worked Solution
i.    `x^2 – 10x + y^2 + 6y+25` `= 0`
  `(x – 5)^2 + (y + 3)^2 – 9` `= 0`
  `(x – 5)^2 + (y + 3)^2` `= 9`

 
`=>\ text{Circle centre (5, −3),  radius 3}`
 

`:.\ text(Parametric form is:)`

`x = 5 + 3costheta`

`y = −3 + 3sintheta`

 

ii.  

Functions, 2ADV F2 SM-Bank 5 MC

The point  `P\ text{(4, −3)}`  lies on the graph of a function  `f(x)`. The graph of  `f(x)`  is translated four units vertically up and then reflected in the `y`-axis.

The coordinates of the final image of `P` are

  1. `(-4, 1)`
  2. `(-4, 3)`
  3. `(0, -3)`
  4. `(4, -6)`
Show Answers Only

`A`

Show Worked Solution

`text(1st transformation:)`

`P(4,−3)\ ->\ (4,1)`
 

`text(2nd transformation:)`

`(4,1)\ ->\ (-4,1)`
 

`=>   A`

Financial Maths, 2ADV M1 SM-Bank 7

Joe buys a tractor under a buy-back scheme. This scheme gives Joe the right to sell the tractor back to the dealer.

The recurrence relation below can be used to calculate the price Joe sells the tractor back to the dealer `(P_n)`, after `n` years
 

`qquad\ \ \ P_0 = 56\ 000,qquadP_n = P_(n - 1) - 7000`
  

  1. Write the general rule to find the value of  `P_n`  in terms of  `n`.  (1 mark)

  2. After how many years will the dealer offer to buy back Joe's tractor at half of its original value.  (1 mark)

Show Answers Only

i.  `P_n = 56\ 000 – 7000n`

ii.  `4\ text(years)`

Show Worked Solution
i.    `P_1` `= P_0 – 7000`
  `P_2` `= P_0 – 7000 – 7000`
    `= 56\ 000 – 7000 xx 2`
  `vdots`  
  `P_n`  `= 56\ 000 – 7000n` 

 

ii.    `text(Half original value)` `= 56\ 000 ÷ 2=$28\ 000`

 

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

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

Financial Maths, 2ADV M1 SM-Bank 6

Julie deposits some money into a savings account that will pay compound interest every month.

The balance of Julie’s account, in dollars, after `n` months, `V_n` , can be modelled by the recurrence relation shown below.
 

`V_0 = 12\ 000, qquad V_(n + 1) = 1.0062\ V_n`
 

  1.  Recursion can be used to calculate the balance of the account after one month.

     

    1. Write down a calculation to show that the balance in the account after one month, `V_1`, is  $12 074.40. (1 mark)

    2. After how many months will the balance of Julie’s account first exceed $12 300  (1 mark)

  2.  A rule of the form  `V_n = a xx b^n`  can be used to determine the balance of Julie's account after `n` months.

    1. Complete this rule for Julie’s investment after `n` months by writing the appropriate numbers in the boxes provided below. (1 mark)
       
balance = 
 
  × 
 
  n

 

    1. What would be the value of `n` if Julie wanted to determine the value of her investment after three years?  (1 mark)

Show Answers Only

a.i.   `text(Proof)\ \ text{(See Worked Solutions)}`

a.ii.  `4\ text(months)`

b.i  `text(balance) = 12\ 000 xx 1.0062^n`

b.ii.  `36`

Show Worked Solution
a.i.   `V_1` `= 1.0062 xx V_0`
    `= 1.0062 xx 12000`
    `= $12\ 074.40\ text(… as required.)`

 

a.ii.   `V_2` `= 1.0062 xx 12\ 074.40 = 12\ 149.26`
  `V_3` `= 1.0062 xx 12\ 149.26 = 12\ 224.59`
  `V_4` `= 1.0062 xx 12\ 224.59 = 12\ 300.38`

 
`:.\ text(After 4 months)`

 
b.i.  `text(balance) = 12\ 000 xx 1.0062^n`

 
b.ii.  `n = 12 xx 3 = 36`

Financial Maths, 2ADV M1 SM-Bank 1 MC

On day 1, Vikki spends 90 minutes on a training program.

On each following day, she spends 10 minutes less on the training program than she did the day before.

Let  `t_n`  be the number of minutes that Vikki spends on the training program on day  `n`.

A recursive equation that can be used to model this situation for  `1 ≤ n ≤ 10`  is

A.   `t_(n + 1) = 0.90t_n` `t_1 = 90`
B.   `t_(n + 1) = 1.10 t_n` `t_1 = 90`
C.   `t_(n + 1) = 1 - 10 t_n` `t_1 = 90`
D.   `t_(n + 1) = t_n - 10` `t_1 = 90`

 

Show Answers Only

`D`

Show Worked Solution

`text(Difference equation where each term is 10 minutes)`

`text(less than the preceding term.)`

`∴\ text(Equation)\ \ \t_(n+1) = t_n-10, \ \ t_1 = 90`

`=>  D`

Functions, 2ADV F1 SM-Bank 33

  1.  State the domain and range of  `y = -sqrt(12-x^2)`.   (2 marks)

  2.  Sketch the graph.  (1 mark)

Show Answers Only

i.  `text(Domain:)\ -sqrt12<=x<= sqrt12`

   `text(Range:)\ -sqrt12<=y<= 0`

ii.   

Show Worked Solution

i.   `y = -sqrt(12-x^2)`

`text(Domain:)\ -sqrt12<=x<= sqrt12`

`text(Range:)\ -sqrt12<=y<= 0`
 

ii.  

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 T3 SM-Bank 15

The graphs of  `y = cos (x) and y = a sin (x)`,  where `a` is a real constant, have a point of intersection at  `x = pi/3.`

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

  2. Find the  `x`-coordinate of the other point of intersection of the two graphs, given  `0<=x<= 2 pi`  (1 mark)

Show Answers Only
  1. `1/sqrt 3`
  2. `(4 pi)/3`
Show Worked Solution

i.   `text(Intersection occurs when)\ \ x=pi/3,`

`a sin(pi/3)` `= cos (pi/3)`
`tan(pi/3)` `= 1/a`
`sqrt 3` `=1/a`
`:. a` `=1/sqrt3`

 

ii.   `tan (x)` `= sqrt 3`
  `x` `= pi/3, (4 pi)/3, 2pi+ pi/3, …`
  `:. x` `= (4 pi)/3\ \ \ (0<= x<= 2 pi)`

Trigonometry, 2ADV T3 SM-Bank 14

For the function  `f(x) = 5 cos (2 (x + pi/3)),\ \ \ -pi<=x<=pi`

  1. Write down the amplitude and period of the function  (2 marks)

  2. Sketch the graph of the function  `f(x)`  on the set of axes below. Label axes intercepts with their coordinates.

     

    Label endpoints of the graph with their coordinates.  (3 marks)

VCAA 2006 meth 4b

Show Answers Only
  1. `text(Amplitude) = 5;\ \ \ text(Period) = pi`
  2.  
Show Worked Solution

a.   `text(Amplitude) = 5`

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

 

b.   `text(Shift)\ \ y = 5 cos (2x)\ \ text(left)\ \ pi/3\ \ text(units).`

`text(Period) = pi`

`text(Endpoints are)\ \ (-pi, -5/2) and (pi,-5/2)`

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 3 MC

Let  `f (x) = 5sin(2x) - 1`.

The period and range of this function are respectively

  1. `π\ text(and)\ [−1, 4]`
  2. `2π\ text(and)\ [−1, 5]`
  3. `π\ text(and)\ [−6, 4]`
  4. `2π\ text(and)\ [−6, 4]`
Show Answers Only

`C`

Show Worked Solution

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

`text(Range)` `= [−1 – 5, −1 + 5]`
  `= [−6 ,4]`

 
`=> C`

Trigonometry, 2ADV T3 SM-Bank 2 MC

Let   `f(x) = 1 - 2 cos ({pi x}/2).`

The period and range of this function are respectively

  1. `4 and [−2, 2]`
  2. `4 and [−1, 3]`
  3. `1 and [−1, 3]`
  4. `4 pi and [−2, 2]`
Show Answers Only

`B`

Show Worked Solution
`text(Period)` `= (2 pi)/n = (2pi)/(pi/2)=4`
   

`text(Amplitude = 2)`

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

`:.\ text(Range)` `= [1 – 2, quad 1 + 2]`
  `= [−1, 3]`

`=>   B`

Trigonometry, 2ADV T3 SM-Bank 1 MC

`f(x) = 2sin(3x) - 3`

The period and range of this function are respectively

  1. `text(period) = (2 pi)/3 and text(range) = text{[−5, −1]}`
  2. `text(period) = (2 pi)/3 and text(range) = text{[−2, 2]}`
  3. `text(period) = pi/3 and text(range) = text{[−1, 5]}`
  4. `text(period) = 3 pi and text(range) = text{[−1, 5]}`
Show Answers Only

`A`

Show Worked Solution

`text(Range:)\ [−3 -2, −3 + 2]`

`= [−5,−1]`

`text(Period) = (2pi)/n = (2pi)/3`

`=>   A`

Networks, STD2 N3 SM-Bank 45

An oil pipeline network is drawn below that shows the flow capacity of oil pipelines in kilolitres per hour.
 


 

A cut is shown.

  1. What is the capacity of the cut.  (1 mark)

  2. Calculate the minimum cut of this network?  (2 marks)

  3. Copy the network diagram, showing the maximum flow capacity of the network by labelling the flow of each edge.  (2 marks)

Show Answers Only
  1. `35`
  2. `text(See Worked Solutions)`
  3.  
Show Worked Solution
i.    `text(Capacity of cut)` `= 7 + 15 + 13`
    `= 35\ text(kL/h)`

 

ii. 

♦♦ COMMENT: Be very careful! RS is not included as it goes from sink to source.
 


  

`text(Minimum cut)` `= 7 + 14 + 9`
  `= 30\ text(kL/h)`

 

iii.   

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`

Vectors, SPEC2 2012 VCAA 15 MC

The vectors  `underset~a = 2underset~i + m underset~j - 3underset~k`  and  `underset~b = m^2underset~i - underset~j + underset~k`  are perpendicular for

  1. `m = −2/3`  and  `m = 1`
  2. `m = −3/2`  and  `m = 1`
  3. `m = 2/3`  and  `m = −1`
  4. `m = 3/2`  and  `m = −1`
  5. `m = 3`  and  `m = −1`
Show Answers Only

`D`

Show Worked Solution

`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`

`underset ~a ⋅ underset ~b` `= 2m^2 + m(-1) + (-3)(1)`
`0` `= 2m^2 – m – 3`
`0` `= (2m – 3)(m + 1)`

 
`:. m = 3/2, quad m = -1`

`=> D`

Mechanics, SPEC2 2011 VCAA 21 MC

A constant force of magnitude  `F`  newtons accelerates a particle of mass 2 kg in a straight line from rest to 12 ms`\ ^(−1)` over a distance of 16 m.

It follows that

  1. `F` = 4.5
  2. `F` = 9.0
  3. `F` = 12.0
  4. `F` = 18.0
  5. `F` = 19.6
Show Answers Only

`B`

Show Worked Solution

`u = 0, \ v = 12, \ s = 16`

`v^2` `= u^2 + 2as`
`144` `= 0 + 32a`
`a` `= 9/2\ \ \ text{(by CAS)}`

 

`:. F` `= 2(9/2)`
  `= 9`

 
`=> B`

Vectors, SPEC2 2011 VCAA 12 MC

The angle between the vectors  `3underset~i + 6underset~j - 2underset~k`  and  `2underset~i - 2underset~j + underset~k`, correct to the nearest tenth of a degree, is

A.       2.0°

B.     91.0°

C.   112.4°

D.   121.3°

E.   124.9°

Show Answers Only

`C`

Show Worked Solution

`|3underset~i + 6underset~j – 2underset~k| = sqrt(9 + 36 + 4) = sqrt49 = 7`

`|2underset~i – 2underset~j + underset~k| = sqrt(4 + 4 + 1) = sqrt9 = 3`

`(3underset~i + 6underset~j – 2underset~k) * (2underset~i – 2underset~j + underset~k)`

`= 3 xx 2 + 6 xx (−2) + (−2) xx 1`

`= 6 – 12 – 2`

`= -8\ \ text{(do calculations on CAS)}`
  

`costheta` `= ((3tildei + 6tildej – 2tildek).(2tildei – 2tildej + tildek))/(|\ 3tildei + 6tildej – 2tildek\ ||\ 2tildei – 2tildej + tildek\ |)`
  `= −8/21`
`:. theta` `= cos^(−1)(−8/12)`
  `~~ 112.4^@`

 
`=> C`

Trigonometry, SPEC2 2011 VCAA 9 MC

The number of distinct solutions of the equation

`xsin(x)sec(2x) = 0, \ x ∈ [0,2pi]`  is

  1. 3
  2. 4
  3. 5
  4. 6
  5. 7
Show Answers Only

`A`

Show Worked Solution
`xsin(x)sec(2x)` `= (xsin(x))/(cos(2x))=0`

 
`text(Find)\ \ x\ \ text(that satisfies:)`

`x = 0\ \ text(or)\ \ sin(x) = 0\ \ text(and)\ \ cos(2x) != 0`

`:. x = 0, \ pi\ \ text(or)\ \ 2pi`

`=> A`

Graphs, SPEC2 2011 VCAA 3 MC

The implied domain of the function with rule  `f(x) = b + cos^(−1)(ax)`  where  `a > 0`  is

A.   `(−pi/a,pi/a)`

B.   `[−1,1]`

C.   `[−pi/a,pi/a]`

D.   `(−1/a,1/a)`

E.   `[−1/a,1/a]`

Show Answers Only

`E`

Show Worked Solution

`−1 <= ax <= 1,\ \ a > 0`

`−1/a <= x <= 1/a`

`=> E`

Algebra, SPEC2 2011 VCAA 2 MC

A circle with centre  `(a,−2)`  and radius 5 units has equation

`x^2 - 6x + y^2 + 4y = b`  where `a` and `b` are real constants.

The values of `a` and `b` are respectively

A.   −3 and 38

B.   3 and 12

C.   −3 and −8

D.   −3 and 0

E.   3 and 18

Show Answers Only

`B`

Show Worked Solution
`x^2 – 6x + y^2 + 4y` `=b`
`x^2 – 6x + 3^2 – 9 + y^2 + 4y + 2^2 – 4` `= b`
`(x – 3)^2 + (y + 2)^2 – 13` `= b`
`(x – 3)^2 + (y + 2)^2` `= b + 13`

 
`:. a=3`

`:. b+13=25\ \ =>\ \ b=12`

`=> B`

Calculus, SPEC1 2011 VCAA 10

Consider the relation  `y log_e (x) = e^(2y) + 3x - 4.`

Evaluate  `(dy)/(dx)`  at the point  `(1, 0).`  (4 marks)

Show Answers Only

`-3/2`

Show Worked Solution

`y log_e x = e^(2y) + 3x – 4`

`text(Using implicit differentiation:)`

`d/(dx)(y ln(x))` `= d/(dx)(e^(2y)) + d/(dx)(3x) – d/(dx)(4)`
`dy/dx*ln(x) + y(1/x)` `= 2e^(2y)*dy/dx + 3`

 
`text(At)\ \ (1,0):`

`dy/dx xx ln1 + 0` `= 2 e^0 * dy/dx+ 3`
`2*dy/dx` `= -3`
`:. dy/dx` `=- 3/2`

Vectors, SPEC1 2011 VCAA 9

Consider the three vectors

`underset ~a = underset ~i - underset ~j + 2underset ~k,\ underset ~b = underset ~i + 2 underset ~j + m underset ~k`  and  `underset ~c = underset ~i + underset ~j - underset ~k`, where `m in R.`

  1. Find the value(s) of `m` for which  `|\ underset ~b\ | = 2 sqrt 3.`  (2 marks)
  2. Find the value of `m` such that  `underset ~a`  is perpendicular to  `underset ~b.`  (1 mark)
  3. i.  Calculate  `3 underset ~c - underset ~a.`  (1 mark)
  4. ii. Hence find a value of `m` such that  `underset ~a, underset ~b`  and  `underset ~c`  are linearly dependent.  (1 mark)
Show Answers Only
  1. `+- sqrt 7`
  2. `1/2`
  3. i.  `2 tilde i + 4 tilde j – 5 tilde k`
  4. ii. `-5/2`
Show Worked Solution
a.    `|underset~b|` `= sqrt(1^2 + 2^2 + m^2)` `= 2sqrt3`
    `1 + 4 + m^2` `= 4 xx 3`
    `m^2` `= 7`
    `m` `= ±sqrt7`

 

b.    `underset~a * underset~b` `= 1 xx 1 + (−1) xx 2 + 2 xx m`  
  `0` `= 1 – 2 + 2m\ \ ( underset~a ⊥ underset~b)`  
  `2m` `= 1`  
  `m` `= 1/2`  

 

c.i.   `3 underset ~c – underset ~a`

`=3underset~i – underset~i + 3underset~j + underset~j -3underset~k – 2underset~k`

`=2underset~i + 4underset~j-5underset~k`

 

c.ii.   `text(Linear dependence)\ \ =>\ \ 3underset~c – underset~a = t underset~b,\ \ t in RR`

`2 tilde i + 4 tilde j – 5 tilde k= t (tilde i + 2 tilde j + m tilde k)`

`=> t = 2 and tm = -5,`

`:. m=-5/2`

Graphs, SPEC2 2012 VCAA 4 MC

The domain and range of the function with rule  `f(x) = arccos(2x - 1) + pi/2`  are respectively

A.   `[−2,0]`  and  `[0,pi]`

B.   `[−2,0]`  and  `[pi/2,(3pi)/2]`

C.   `[0,1]`  and  `[0,pi]`

D.   `[0,1]`  and  `[pi/2,(3pi)/2]`

E.   `[0,pi]`  and  `[0,1]`

Show Answers Only

`D`

Show Worked Solution

`text(Domain:)`

`-1 <= 2x – 1 <= 1`

`0 <= 2x <= 2`

`0 <= x <= 1`

`text(Range:)`

`0 <= cos^(-1) (2x – 1) <= pi`

`pi/2 <= cos^(_1) (2x – 1) + pi/2 <= (3 pi)/2`

 
`=> D`

Graphs, SPEC2 2012 VCAA 1 MC

The graph with equation  `y = 1/(2x^2 - x - 6)` has asymptotes given by

A.  `x = -3/2,\ x = 2 and y = 1`

B.   `x = -3/2 and x = 2`  only

C.   `x = 3/2,\ x = -2 and y = 0`

D.   `x = -3/2,\ x = 2 and y = 0`

E.   `x = 3/2 and x = -2`  only

Show Answers Only

`D`

Show Worked Solution
`y` `=1/(2x^2 – x – 6)`  
  `=1/((2x + 3)(x – 2))`  

 

`:.\ text(Asymptotes:)`   `x = 2`
  `x = -3/2`
  `y = 0`

 
`=> D`

Vectors, SPEC1 2012 VCAA 9

The position of a particle at time  `t`  is given by

`underset ~r (t) = (2 sqrt (t^2 + 2) - t^2) underset ~i + (2 sqrt (t^2 + 2) + 2t) underset ~j,\ \ t >= 0.`

  1. Find the velocity of the particle at time  `t.`  (1 mark)
  2. Find the speed of the particle at time  `t = 1`  in the form  `(a sqrt b)/c`, where `a, b` and `c` are positive integers.  (2 marks)
  3. Show that at time  `t = 1,\ \ (dy)/(dx) = (1 + sqrt 3)/(1 - sqrt 3).`  (2 marks)
  4. Find the angle in terms of `pi`, between the vector  `-sqrt 3 underset ~i + underset ~j`  and the vector  `underset ~r (t)`  at time  `t = 0.`  (2 marks)
Show Answers Only
  1. `((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`
  2. `(4 sqrt 6)/3`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `(7 pi)/12`
Show Worked Solution

a.  `underset ~r (t) = (2 sqrt (t^2 + 2) – t^2) underset ~i + (2 sqrt (t^2 + 2) +2t) underset ~j`

  `underset ~v(t)` `=dot underset ~r(t)`
    `= (2(2t)(1/2)(t^2 + 2)^(-1/2) – 2t) underset ~i + (2(2t)(1/2)(t^2 + 2)^(-1/2) + 2)underset ~j`
    `= ((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`

 

b.    `underset ~v(1)` `= (2/sqrt(1 + 2) – 2)underset ~i + (2/sqrt(1 + 2) + 2) underset ~j`
    `= (2/sqrt 3 – 2) underset ~i + (2/sqrt 3 + 2)underset ~j`
  `|\ underset ~v(1)\ |` `= sqrt((2/sqrt 3 – 2)^2 + (2/sqrt 3 + 2)^2)`
    `= sqrt(4/3 – 8/sqrt 3 + 4 + 4/3 + 8/sqrt 3 + 4)`
    `= sqrt(8/3 + 8)`
    `= sqrt(32/3)`
    `= (4sqrt2)/sqrt3`
    `= (4 sqrt 6)/3`

♦ Mean mark part (c) 41%.

c.    `(dy)/(dt)` `= (dy)/(dx) xx (dx)/(dt)`
  `(dy)/(dx)` `=((dy)/(dt))/((dx)/(dt))`
  `:. (dy)/(dx)|_(t=1)` `= (2/sqrt 3 + 2)/(2/sqrt 3 – 2)`
    `= ((2 + 2 sqrt 3)/sqrt 3) xx (sqrt 3/(2 – 2 sqrt 3))`
    `= (2 + 2 sqrt 3)/(2 – 2 sqrt 3)`
    `= (1 + sqrt 3)/(1 – sqrt 3)`

 

d.   `text(At)\ \ t=0:` 

♦♦ Mean mark part (d) 32%.

`underset ~r(0) = 2 sqrt 2 underset ~i + 2 sqrt 2 underset ~j`

`theta_1` `= tan^(-1)((2 sqrt 2)/(2 sqrt 2)) = pi/4`
`theta_2` `= tan^(-1)(1/sqrt 3)=pi/6`

 
`text(Let)\ \ theta=\ text(angle between the vectors:)`

`theta` `= pi – theta_1 – theta_2`
  `= pi – pi/4 – pi/6`
  `= (12 pi – 3 pi – 2 pi)/12`
  `= (7 pi)/12`

Algebra, SPEC1 VCE SM-Bank 7

Given that  `(16x - 43)/((x - 3)^2 (x + 2))`  can be written as

 
`(16x - 43)/((x - 3)^2 (x + 2)) = a/(x - 3)^2 + b/(x - 3) + c/(x + 2)`,

  
where  `a, b` and `c in RR`, find  `a, b and c.`  (3 marks)

    1.  
Show Answers Only

`a = 1, b = 3, c = -3`

 

Show Worked Solution

`(16x – 43)/((x – 3)^2 (x + 2)) = a/(x – 3)^2 + b/(x – 3) + c/(x + 2)`

`16x – 43 = a (x + 2) + b (x – 3) (x + 2) + c (x – 3)^2`
 

`text(When)\ \ x = 3,\ \ 5a =5\ \ =>a=1`

`text(When)\ \ x=-2,\ \ 25c=-75\ \ =>c=-3`

`text(When)\ \ x=0`

`-43` `= 2(1) – 6b + (-3)(-3)^2`
`6b` `= 18`
`b` `=3`

 
`:.a=1, b=3, c=-3`

Algebra, SPEC1 VCE SM-Bank 4

Find  `A, B`  and  `C in RR`, such that

`1/(x(x^2 + 2)) = A/x + (Bx + C)/(x^2 + 2).`   (2 marks)

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`A = 1/2,\ \ \ B = -1/2,\ \ \ C = 0`

Show Worked Solution
`1/(x(x^2 + 2))` `= A/x + (Bx + C)/(x^2 + 2)`
`1` `= A (x^2 + 2) + (Bx + C) x`
`1` `= (A+B)x^2 + Cx + 2A`

 

`2A = 1,\ \ =>A=1/2`

`C=0`

`A + B = 0,\ \ =>B=-1/2`

`:.A = 1/2,\ \ \ \ B = -1/2,\ \ \ \ C = 0`

Algebra, SPEC1 VCE SM-Bank 1

Find numbers  `A, B` and `C in RR`, such that

`(x^2 + 8x + 11)/((x -3)(x^2 + 2)) = A/(x - 3) + (Bx + C)/(x^2 + 2).`  (2 marks)

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`A = 4,\ B = -3,\ C = -1`

Show Worked Solution

`(x^2 + 8x + 11)/((x -3)(x^2 + 2)) = A/(x – 3) + (Bx + C)/(x^2 + 2)`

`x^2 + 8x + 11` `=A(x^2 + 2)+(Bx+C)(x-3)`
  `=(A+B)x^2+(C-3B)x+(2A-3C)`

 

`A+B=1\ \ \ => B=1-A`

`C-3B=8\ \ \ =>C=11-3A`

`2A-3C=11 \ \ \ =>2A-33+9A=11\ \ \ =>A=4`

`B=1-4=-3`

`C=11-12=-1`

 `:.A=4, B=-3, C=-1`

Mechanics, SPEC2 2013 VCAA 21 MC

A particle of mass 2 kg moves in a straight line with an initial velocity of 20 m/s. A constant force opposing the direction of the motion acts on the particle so that after 4 seconds its velocity is 2 m/s.

The magnitude of the force, in newtons, is

A.     4.5

B.     6

C.     9

D.   18

E.   36

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`C`

Show Worked Solution

`u = 20, \ t = 4, \ v = 2`

`v = u + at`

`v` `=u + at`
`2` `= 20 + 4a`
`a` `= −9/2`

 

`|underset~F|` `= 2 xx |−9/2|=9`

 
`=> C`

Calculus, SPEC2 2013 VCAA 9 MC

The definite integral  `int_(e^3)^(e^4)1/(xlog_e(x))\ dx`  can be written in the form  `int_a^b1/u\ du`  where
 

A.   `u = log_e(x), a = log_e(3), b = log_e(4)`

B.   `u = log_e(x), a = 3, b = 4`

C.   `u = log_e(x), a = e^3, b = e^4`

D.   `u = 1/x, a = e^(−3), b = e^(−4)`

E.   `u = 1/x, a = e^3, b = e^4`

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`B`

Show Worked Solution

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

`(du)/(dx) = 1/x\ \ =>\ \ du=1/x dx`

`text(When)\ \ x=e^3\ \ =>\ \ u=lne^3=3`

`text(When)\ \ x=e^4\ \ =>\ \ u=lne^4=4`
 

`:. int_(e^3)^(e^4)1/(xlog_e(x))\ dx = int_3^4 1/udu`

`:.  a = 3, b = 4, u = lnx`

`=> B`

Complex Numbers, SPEC2 2013 VCAA 7 MC

If  `z = r text(cis)(theta)`, then  `(z^2)/barz`  is equivalent to

A.   `r^3text(cis)(3theta)`

B.   `r^3text(cis)(−theta)`

C.   `2text(cis)(3theta)`

D.   `r^3text(cis)(theta)`

E.   `rtext(cis)(3theta)`

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`E`

Show Worked Solution
`z^2` `= r^2text(cis)(2theta)`
`barz` `= rtext(cis)(−theta)`
`(z^2)/z` `= (r^2text(cis)(2theta))/(rtext(cis)(−theta))`
  `= rtext(cis)(2theta – −theta)`
  `= rtext(cis)(3theta)`

 
`=> E`

Trigonometry, SPEC2 2013 VCAA 2 MC

The rule of the relation determined by the parametric equations  `x = 2text(cosec)(t) + 1`  and  `y = 3cot(t) -1`  is

  1. `((x - 1)^2)/4 - ((y + 1)^2)/9 = 1`
  2. `((y + 1)^2)/9 - ((x - 1)^2)/4 = 1`
  3. `((x - 1)^2)/4 + ((y + 1)^2)/4 = 1`
  4. `((y + 1)^2)/3 - ((x - 1)^2)/2 = 1`
  5. `((x - 1)^2)/2 - ((y + 1)^2)/3 = 1`
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`A`

Show Worked Solution

`(x – 1)/2 = text(cosec)(t)qquad(y + 1)/3 = cot(t)`

`text(Using)\ \ 1 + cot^2(t)= text(cosec)^2(t)`

`1 + ((y + 1)/3)^2` `= ((x – 1)/2)^2`
`((x – 1)^2)/4 – ((y + 1)^2)/9` `=1`

 
`=> A`

Algebra, SPEC2 2012 VCAA 2 MC

A rectangle is drawn so that its sides lie on the lines with equations   `x = −2`, `x = 4`, `y = –1`  and  `y = 7`.

An ellipse is drawn inside the rectangle so that it just touches each side of the rectangle.

The equation of the ellipse could be

A.   `(x^2)/9 + (y^2)/16 = 1`

B.   `((x + 1)^2)/9 + ((y + 3)^2)/16 = 1`

C.   `((x - 1)^2)/9 + ((y - 3)^2)/16 = 1`

D.   `((x + 1)^2)/36 + ((y + 3)^2)/64 = 1`

E.   `((x - 1)^2)/36 + ((y - 3)^2)/64 = 1`

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`C`

Show Worked Solution

SPEC2 2012 VCAA 2 MC Answer 1_1

`text(Equation of the ellipse:)\ ((x – 1)^2)/9 + ((y – 3)^2)/16 = 1`

`=> C`

Vectors, SPEC1 2013 VCAA 7

The position vector  `underset ~r (t)`  of a particle moving relative to an origin `O` at time `t` seconds is given by

`underset ~r(t) = 4 sec (t) underset ~i + 2 tan (t) underset ~j,\ t in [0, pi/2)`

where the components are measured in metres.

  1. Show that the cartesian equation of the path of the particle is  `x^2/16 - y^2/4 = 1.`  (2 marks)
  2. Sketch the path of the particle on the axes below, labelling any asymptotes with their equations.  (2 marks)

     

        
     VCAA 2013 spec 7b
     

  3. Find the speed of the particle, in `text(ms)^-1`, when `t = pi/4.`  (2 marks)
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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. “
  3. `4 sqrt 3\ \ text(ms)^-1`
Show Worked Solution
a.   `x` `= 4sec(t)`
  `x/4` `= sec(t)`
  `y` `= 2tan(t)`
  `y/2` `= tan(t)`

  
`text(Using)\ \ tan^2 (t) + 1 = sec^2(t),`

`(x^2)/16 +1 ` `= y^2/4`
`:. (x^2)/16 – (y^2)/4` `=1`

 

b.   `y^2/4 = x^2/16 -1\ \ =>\ \ y=+- sqrt(x^2/4 -4)`

♦ Mean mark part (b) 45%.

`lim_(x->oo) y = +- x/2`

 

♦ Mean mark part (c) 39%.

c.    `overset·underset~r(t)` `= d/(dt)(4(cos(t))^(−1))underset~i + d/(dt)(2tan(t)) underset~j`
    `= 4(−1)(−sin(t))(cos(t))^(−2)underset~i + 2sec^2(t)underset~j`
    `= 4sin(t)sec^2(t)underset~i + 2sec^2(t)underset~i`

 

`|overset·underset~r(pi/4)|` `= sqrt(16sin^2(pi/4)sec^4(pi/4) + 4sec^4(pi/4))`
  `= sqrt(16(1/sqrt2)^2(sqrt2)^4 + 4(sqrt2)^4)`
  `= sqrt(16(1/2)(4) + 4(4))`
  `= sqrt(48)`
  `= 4sqrt3\ \ text(ms)^(-1)`

Vectors, SPEC1 2013 VCAA 3

The coordinates of three points are  `A (– 1, 2, 4), \ B(1, 0, 5) and C(3, 5, 2).`

  1. Find  `vec (AB)`  (1 mark)
  2. The points `A, B` and `C` are the 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)
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  1. `2underset~i – 2underset~j + underset~k`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `sqrt 38`
Show Worked Solution
a.   `vec(AB)` `=(1 – −1)underset~i + (0 – 2)underset~j + (5 – 4)underset~k`
    `= 2underset~i – 2underset~j + underset~k`

 

b.    `overset(->)(AC)` `= (3 – −1)underset~i + (5 – 2)underset~j + (2 – 4)underset~k`
    `= 4underset~i + 3underset~j – 2underset~k`

 

`overset(->)(AB) · overset(->)(AC)` `= 2 xx 4 + (−2) xx 3 + 1 xx (−2)`
  `= 8 – 6 – 2`
  `= 0`

 
`=>  overset(->)(AB) ⊥ overset(->)(AC)`

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

c.    `overset(->)(BC)` `= (3 – 1)underset~i + (5 – 0)underset~j + (2 – 5)underset~k`
    `= 2underset~i + 5underset~j – 3underset~k`

 

`|overset(->)(BC)|` `= sqrt(2^2 + 5^2 + (−3)^2)`
  `= sqrt(4 + 25 + 9)`
  `= sqrt38`