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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

Show Answers Only

`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`

Show Answers Only

`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)`

Show Answers Only

`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`
Show Answers Only

`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`

Show Answers Only

`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)
Show Answers Only
  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)
Show Answers Only
  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`

Mechanics, SPEC1 2013 VCAA 1

A body of mass 10 kg is held in place on a smooth plane inclined at 30° to the horizontal by a tension force, `T` newtons, acting parallel to the plane.

  1. On the diagram below, show all other forces acting on the body and label them.  (1 mark)
     

     

                   VCAA 2013 spec 1a
     

  2. Find the value of `T.`  (2 marks)
Show Answers Only
  1.  

  2. `49\ \ text(N)`
Show Worked Solution
a.   

 

b.    `T – 10g\ sin30^@` `= 0`
  `T – (10g)/2` `= 0`
  `T` `= (10g)/2`
  `T` `= 5g`
    `=49\ \ text(N)`

Vectors, SPEC2 2014 VCAA 16 MC

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 `m` and `n` respectively are

  1. `5 sqrt 3, sqrt 3/3`
  2. `5 sqrt 3, sqrt 3`
  3. `−5 sqrt 3, sqrt 3`
  4. `sqrt 93, (5 sqrt 93)/93`
  5. `5, 1`
Show Answers Only

`A`

Show Worked Solution

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

`10` `= sqrt(4^2 + m^2 + (-3)^2)`
`m` `= 5sqrt3\ \ \ (text{by CAS,}\ \ m in R^+)`

 

`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 = sqrt 3/3\ \ \ (text{by CAS,}\ \ n in R^+)`

`=> A`

Calculus, SPEC2 2015 VCAA 14 MC

A differential equation that has  `y = xsin(x)`  as a solution is

  1. `(d^2y)/(dx^2) + y = 0`
  2. `x(d^2y)/(dx^2) + y = 0`
  3. `(d^2y)/(dx^2) + y = -sin(x)`
  4. `(d^2y)/(dx^2) + y = -2cos(x)`
  5. `(d^2y)/(dx^2) + y = 2cos(x)`
Show Answers Only

`E`

Show Worked Solution
`(dy)/(dx)` `= sin(x) + xcos(x)`
`(d^2y)/(dx^2)` `= cos(x) + cos(x)-xsin(x)`
  `= 2cos(x)-xsin(x)`
  `= 2cos(x) – y`

 
`:. (d^2y)/(dx^2) + y= 2cos(x)`

  
`=> E`

Calculus, SPEC2 2015 VCAA 12 MC

Given  `dy/dx = 1-y/3`  and  `y = 4`  when  `x = 2`, then

  1. `y = e^((-(x-2))/3)-3`
  2. `y = e^((-(x-2))/3) + 3`
  3. `y = 4e^((-(x-2))/3) `
  4. `y = e^((4(y-x-2))/3)`
  5. `y = e^(((x-2))/3) + 3`
Show Answers Only

`B`

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`
 

`text(Find)\ \ y\ \ text{by CAS or manually (see below):}`

` 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)`

 
`=> B`

Statistics, SPEC2 2016 VCAA 19 MC

A random sample of 100 bananas from a given area has a mean mass of 210 grams and a standard deviation of 16 grams.

Assuming the standard deviation obtained from the sample is a sufficiently accurate estimate of the population standard deviation, an approximate 95% confidence interval for the mean mass of bananas produced in this locality is given by

A.   `(178.7, 241.3)`

B.   `(206.9, 213.1)`

C.   `(209.2, 210.8)`

D.   `(205.2, 214.8)`

E.   `(194, 226)`

Show Answers Only

`B`

Show Worked Solution

`(210 – 1.96 xx 16/sqrt 100, 210 + 1.96 xx 16/sqrt 100)`

`~~ (206.9, 213.1)`

 
`=>  B`

Graphs, SPEC2 2016 VCAA 2 MC

The implied domain of  `y = arccos ((x - a)/b)`, where  `b > 0`  is

A.  `[-1, 1]`

B.  `[a - b, a + b]`

C.  `[a - 1, a + 1]`

D.  `[a, a + b pi]`

E.  `[-b, b]` 

Show Answers Only

`B`

Show Worked Solution
`(x- a)/b` `in [-1, 1]`
`x – a` `in [-b, b]`
`x` `in [a – b, a + b]`

 
`=>  B`

Vectors, SPEC1 2016 VCAA 8

The position of a body with mass 3 kg from a fixed origin at time  `t`  seconds, `t >= 0`, is given by  `underset ~r = (3 sin (2t) - 2)underset ~i + (3 - 2 cos(2t)) underset ~j`, where components are in metres.

  1. Find an expression for the speed, in metres per second, of the body at time  `t`.  (2 marks)
  2. Find the speed of the body, in metres per second, when  `t = pi/12`.  (1 mark)
  3. Find the maximum magnitude of the net force acting on the body in newtons.  (3 marks)
Show Answers Only
  1. `|underset ~ dot r| = sqrt(36 cos^2(2t) + 16 sin^2(2t))`
  2. `sqrt 31`
  3. `36`
Show Worked Solution
a.    `underset ~r ` `=(3 sin (2t) – 2)underset ~i + (3 – 2 cos(2t)) underset ~j`
   `underset ~ dot r` `= 6 cos (2t) underset ~i + 4 sin (2t) underset ~j`
  `|underset ~ dot r|` `= sqrt(36 cos^2(2t) + 16 sin^2(2t))`

 

b.   `text(Find)\ \ |underset ~ dot r|\ \ text(when)\ \ t=pi/12:`

`underset ~ dot r` `=(36 cos^2 (pi/6) + 16 sin^2 (pi/6))^(1/2)`  
  `= (36 (sqrt 3/2)^2 + 16 (1/2)^2)^(1/2)`  
  `= (36/4 xx 3 + 16/4)^(1/2)`  
  `= (27 + 4)^(1/2)`  
  `= sqrt 31`  

 

c.    `underset ~ ddot r` `= d/(dt) (underset ~dotr)= -12 sin (2t) underset ~i + 8 cos (2t) underset ~j`
  `underset ~F` `= 3 ddot r`
  `|underset ~F|` `= 3 |underset ~ ddot r|`
    `= 3 sqrt(144 sin^2 (2t) + 64 cos^2 (2t))`
    `= 3 sqrt(64 sin^2 (2t) + 64 cos^2(2t) + 80 sin^2 (2t))`
    `= 3 sqrt(64 + 80 sin^2 (2t))`

 
`text(Max occurs when)\ \ (sin^2 (2t)) = 1`

Mean mark 51%.

`:. max (|underset ~F|)` `= 3 sqrt(64 + 80)`
  `= 3 sqrt 144`
  `= 3 xx 12`
  `= 36`

Statistics, SPEC2-NHT 2017 VCAA 6

A bank claims that the amount it lends for housing is normally distributed with a mean of $400 000 and a standard deviation of $30 000.

A consumer organisation believes that the average loan amount is higher than the bank claims.

To check this, the consumer organisation examines a random sample of 25 loans and finds the sample mean to be $412 000.

  1. Write down the two hypotheses that would be used to undertake a one-sided test.  (1 mark)
  2. Write down an expression for the `p` value for this test and evaluate it to four decimal places.  (2 marks)
  3. State with a reason whether the bank’s claim should be rejected at the 5% level of significance.  (1 mark)
  4. What is the largest value of the sample mean that could be observed before the bank’s claim was rejected at the 5% level of significance? Give your answer correct to the nearest 10 dollars.  (1 mark)
  5. If the average loan made by the bank is actually $415 000 and not $400 000 as originally claimed, what is the probability that a random selection of 25 loans has a sample mean that is at most $410 000? Give your answer correct to three decimal places.  (2 marks)
Show Answers Only
  1. `H_0: mu = 400\ 000`

     

    `H_1: mu > 400\ 000`

  2. `p ~~ 0.0228`
  3. `text(reject)`
  4. `x_max ~~ 409\ 860`
  5. `~~ 0.202`
Show Worked Solution

a.   `H_0: \ mu = 400\ 000`

`H_1: \ mu > 400\ 000`

 

b.   `L\ ~\ N (400\ 000, 30\ 000^2)`

`bar L\ ~\ N (400\ 000, (30\ 000^2)/25)`

`p = text(Pr)(bar L > $412\ 000)`
 

`:. p ~~ 0.0228`

 

c.   `text(The bank’s claim)\ (h_0: mu = 400\ 000)`

`text(should be rejected at the 5% level of)`

`text(significance as)\ \ p ~~ 0.0228 < 0.05.`

 

d.   `p = text(Pr)(bar L > x) = 0.05`

`=> text(Pr)(bar L < x) = 0.95`

`=> x ~~ 409\ 869.12`

`:. x_max ~~ $409\ 870\ \ text{(nearest $10)}`

 

e.   `text(New distribution):\ \ L_2\ ~\ N (415\ 000, 30\ 000^2)`

`bar L_2\ ~\ N (415\ 000, (30\ 000^2)/25)`

`text(Pr)(bar L_2 <= 410\ 000) ~~ 0.202`

Mechanics, SPEC2-NHT 2017 VCAA 5

A 5 kg mass is initially held at rest on a smooth plane that is inclined at 30° to the horizontal. The mass is connected by a light inextensible string passing over a smooth pulley to a 3 kg mass, which in turn is connected to a 2 kg mass.

The 5 kg mass is released from rest and allowed to accelerate up the plane.

Take acceleration to be positive in the directions indicated.
 

  1. Write down an equation of motion, in the direction of motion, for each mass.   (3 marks)
  2. Show that the acceleration of the 5 kg mass is  `g/4\ text(ms)^(-2)`.  (1 mark)
  3. Find the tensions  `T_1`  and  `T_2`  in the string in terms of  `g`.  (2 marks)
  4. Find the momentum of the 5 kg mass, in kg ms`­^(-1)`, after it has moved 2 m up the plane, giving your answer in terms of `g`.  (2 marks)
  5. A resistance force  `R`  acting parallel to the inclined plane is added to hold the system in equilibrium, as shown in the diagram below.
     

     

    `qquad`
     

     

    Find the magnitude of  `R`  in terms of  `g`.  (2 marks)

Show Answers Only
  1. `2a = 2g – T_2`

     

    `3a = 3g + T_2 – T_1`

     

    `5a = T_1 – (5g)/2`

  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `T_1 = (15g)/4`

     

    `T_2 = (3g)/2`

  4. `p = 5 sqrt g`
  5. `R = (5g)/2`
Show Worked Solution
a.    `2\ text(kg): \ 2a` `= 2g – T_2`
  `3\ text(kg): \ 3a` `= 3g + T_2 – T_1`
  `5\ text(kg): \ 5a` `= T_1 – 5g sin (30^@)`
  `5a` `= T_1 – (5g)/2`

 

 

b.   `sum F = 3g + 2g – 5g sin 30^@ = (5 + 3 + 2)a`

`5g – (5g)/2` `= 10a`
`a` `= (5g)/(2 xx 10)`
`:. a` `= g/4\ text(ms)^(-2)`

 

c.   ` T_1 – (5g)/2` `= 5a`
`:. T_1` `= (5g)/2 + 5(g/4)`
  `= (15g)/4`

 

`2g – T_2` `= 2a`
`:. T_2` `= 2g – 2(g/4)`
  `= (3g)/2`

 

d.   `u = 0,\ \ a = g/4,\ \ s = 2`

`text(Find)\ \ v\ \ text(when)\ \ s=2:`

`v^2` `= u^2 + 2as`  
  `=0 + 2 (g/4) xx 2`  
  `=g`  
`v` `=sqrtg\ \ \ (v>0)`  

 
`:. p = 5 sqrt g`

 

e.   `sum F = 2g + 3g – 5g sin 30^@ – R = 0`

`:. R` `= 2g + 3g – 5g sin 30^@`
  `= 5g – (5g)/2`
  `= (5g)/2`

Calculus, SPEC2-NHT 2017 VCAA 3

Bacteria are spreading over a Petri dish at a rate modelled by the differential equation

`(dP)/(dt) = P/2 (1 - P),\ 0 < P < 1`

where  `P`  is the proportion of the dish covered after  `t`  hours.

    1. Express  `2/(P(1 - P))`  in partial fraction form.  (1 mark)
    2. Hence show by integration that  `(t - c)/2= log_e(P/(1 - P))`, where  `c`  is a constant of integration.  (2 marks)
    3. If half of the Petri dish is covered by the bacteria at  `t = 0`, express  `P`  in terms of  `t`.  (2 marks)

After one hour, a toxin is added to the Petri dish, which harms the bacteria and reduces their rate of growth. The differential equation that models the rate of growth is now

`(dP)/(dt) = P/2 (1 - P) - sqrt P/20`  for  `t >= 1`

  1. Find the limiting value of  `P`, which is the maximum possible proportion of the Petri dish that can now be covered by the bacteria. Give your answer correct to three decimal places.  (2 marks)
  2. The total time, `T`  hours, measured from time  `t = 0`, needed for the bacteria to cover 80% of the Petri dish is given by
     

     

    `qquad qquad T = int_q^r (1/(P/2(1 - P) - sqrt P/20)) dP + s`
     

     

    where  `q, r and s in R`.

     

     

    Find the values of  `q, r` and `s`, giving the value of `q` correct to two decimal places.  (2 marks)

  1. Given that  `P = 0.75`  when  `t = 3`, use Euler’s method with a step size of 0.5 to estimate the value of `P` when  `t = 3.5`. Give your answer correct to three decimal places.  (3 marks)
Show Answers Only
    1. `2/(P(1 – P)) = 2/P + 2/(1 – P)`
    2. `text(Proof)\ text{(See Worked Solutions)}`
    3. `P = e^(t/2)/(1 + e^(t/2))`
  2. `P ~~ 0.894`
  3. `r = 0.8`
    `s = 1`
    `q ~~ 0.62`
  4. `~~ 0.775`
Show Worked Solution

a.i.   `2/(P(1 – P)) = A/P + B/(1 ⋅ P)`

`A(1 – P) + BP = 2`

`text(If)\ \ P = 0\ \ =>\ \ A = 2`

`text(If)\ \ P = 1\ \ =>\ \ B = 2`

`:. 2/(P(1 – P)) = 2/P + 2/(1 – P)\ \ \ text{(can also solve by CAS)}`

 

a.ii.  `(dt)/(dP) = 2/(P(1 – P)) = 2/P + 2/(1 – P)`

`t` `= int 2/P + 2/(1 – P)\ dP`
  `= 2 ln |P| – 2ln|1 – P| + c`
`(t-c)/2` `=ln|P| – ln|1-P|`
  `=ln |(P)/(1-P)|`
  `= ln (P/(1 – P))`

  
`text(S)text(ince)\ \ 0 < P < 1 :\ |P| = P\ and\ |1 – P| = 1 – P`


a.iii.  `text(When)\ \ t=0,\ P=0.5`

`(-c)/2` `= ln (0.5/0.5)`
`c` `= ln (1) = 0`

 

`t/2` `= ln (P/(1 – P))\ \ \ text{(solve manually or by CAS)}`
`e^(t/2)` `= P/(1 – P)`
`e^(t/2) (1 – P)` `= P`
`e^(t/2) – Pe^(t/2)` `= P`
`e^(t/2)` `= P(1 + e^(t/2))`
`:. P` `= e^(t/2)/(1 + e^(t/2))`

 

b.   `(dP)/(dt) = P/2 (1 – P) – sqrt P/20`

`text(Limiting value occurs when)\ \ (dP)/(dt) = 0,`

`P ~~ 0.894\ \ \ text{(by CAS)}`
 

`=>\ text(Lower solution values at levels already exceeded)`

 `text(are ignored.)`

 

c.   `(dt)/(dP) = 1/(P/2 (1 – P) – sqrt P/20)\ \ text(for)\ \ t>=1`

`text(When)\ \ t=1\ \ =>\ \ P=0.622`
 

`T = int_0.62^0.8 1/(P/2 (1 – P) – sqrt P/20) dP + 1`
 

`:. r = 0.8,\ s = 1 and q = 0.62`

 

d.   `P(3.5) ~~ P(3) + h* (dP)/(dt)|_(P= 0.75)`

   `= 0.75 + 0.5 (0.75/2 (1 – 0.75) – sqrt 0.75/20)`

   `~~ 0.775`
 

`:. P~~0.775\ \ text(when)\ \ t=3.5`

Complex Numbers, SPEC2 2017 VCAA 4

  1. Express  `−2 - 2sqrt3 i`  in polar form.  (1 mark)
  2. Show that the roots of  `z^2 + 4z + 16 = 0`  are  `z = −2 - sqrt3 i`  and  `z = −2 + 2sqrt3 i`.  (1 mark)
  3. Express the roots of  `z^2 + 4z + 16 = 0`  in terms of  `2 - 2sqrt3 i`.  (1 mark)
  4. Show that the cartesian form of the relation  `|z| = |z - (2 - 2sqrt3 i)|`  is  `x - sqrt3 y - 4 = 0`  (2 marks)
  5. Sketch the line represented by  `x - sqrt3y  - 4 = 0`  and plot the roots of  `z^2 + 4z + 16 = 0`  on the Argand diagram below.  (2 marks)
     
       

  6. The equation of the line passing through the two roots of  `z^2 + 4z + 16 = 0`  can be expressed as  `|z - a| = |z - b|`, where  `a, b ∈ C`.

     

    Find `b` in terms of `a`.  (1 mark)

  7. Find the area of the major segment bounded by the line passing through the roots of  `z^2 + 4z + 16 = 0`  and the major arc of the circle given by  `|z| = 4`.  (2 marks)
Show Answers Only
  1. `4text(cis)((−2pi)/3)`
  2. `text(See Worked Solutions)`
  3. `−(2 – 2sqrt3 i) and – bar((2-2sqrt3 i))`
  4. `text(See Worked Solutions)`
  5.  
  6. `−4 – bara`
  7. `4sqrt3 + (32pi)/3`
Show Worked Solution
a.    `r` `= sqrt((−2)^2 + (−2sqrt3)^2)=4`

 

`theta` `= −pi + tan^(−1)((2sqrt3)/2)`
  `= −pi + pi/3`
  `=(-2pi)/3`

 
`:. −2 – 2sqrt3 i = 4text(cis)((−2pi)/3)`
 

b.   `z^2 + 4z + z^2 – 4 + 16` `=0`
`(z + 2)^2 + 12` `= 0`
`(z + 2)^2` `= -12`
`(z + 2)^2` `= 12i^2`
`z + 2` `= ±sqrt12 i`
`z + 2` `= ±2sqrt3 i`
`:. z` `= -2 ± 2sqrt3 i`

 

♦♦ Mean mark part (c) 31%.

c.    `z_1` `= -2 + 2sqrt3 i = -(2-2sqrt3 i)`
  `z_2` `= – 2 – 2sqrt3 i = – bar((2-2sqrt3 i))`

 

d. `|z|` `= |z – (2 – 2sqrt3 i)|`
     `x^2 + y^2` `= (x – 2)^2 + (y + 2sqrt3)^2`
    `x^2 + y^2` `= x^2 – 4x + 4+ y^2 + 4sqrt3 y + 12`
  `0` `= −4x + 4sqrt3 y + 16`
  `0` `= −x + sqrt3 y + 4`

 
`:. x – sqrt3 y – 4=0`

 

e.   

 

f.   `x = − 2\ \ text(is equidistant from)\ \ z_1 = a\ \ text(and)\ \ z_2 = b`

♦♦♦ Mean mark 1%!

`=> text(Im)(a) = text(Im)(a)`
 

`text(Let)\ \ a = alpha + betai, \ b = gamma + betaj`

`(alpha + gamma)/2` `= −2`
`alpha + gamma` `= −4`
`gamma` `= -4 – alpha`

  

`:. b` `= -4 – alpha + betaj`
  `= -4 – (alpha + betaj)`
  `= -4 – bara`

 

♦♦ Mean mark 31%.

g.    `text(Area)\ DeltaOAB` `= 1/2 xx (4sqrt3 xx 2)`
    `= 4sqrt3`

 

`text(Area of sector)\ AOB` `= pi xx 4^2 xx (2 xx pi/3)/(2pi)`
  `= (16pi)/3`

 
`:.\ text(Area of major segment area)`

`=pi(4)^2 – ((16pi)/3 – 4sqrt3)`

`= 4sqrt3 + (32pi)/3`

Complex Numbers, SPEC2-NHT 2017 VCAA 2

One root of a quadratic equation with real coefficients is  `sqrt 3 + i`.

    1. Write down the other root of the quadratic equation.  (1 mark)
    2. Hence determine the quadratic equation, writing it in the form  `z^2 + bz + c = 0`.  (2 marks)
  1. Plot and label the roots of   `z^3 - 2 sqrt 3 z^2 + 4z = 0`  on the Argand diagram below.  (3 marks)

  

 

  1. Find the equation of the line that is the perpendicular bisector of the line segment joining the origin and the point  `sqrt 3 + i`. Express your answer in the form  `y = mx + c`.  (2 marks)
  1. The three roots plotted in part b. lie on a circle.

     

    Find the equation of this circle, expressing it in the form  `|z - alpha| = beta`,  where  `alpha, beta in R`.  (3 marks)

Show Answers Only
    1. `z_2 = sqrt 3 – i`
    2. `z^2 – 2 sqrt 3 z + 4 = 0`
  2. `text(See Worked Solutions)`
  3. `y = -sqrt 3 x + 2`
  4. `|z – 2/sqrt 3 | = 2/sqrt 3`
Show Worked Solution

a.i.   `z_1 = sqrt 3 + i`

 `z_2 = bar z_1 = sqrt 3 – i\ \ \ text{(conjugate root)}`

 

a.ii.    `(z – (sqrt 3 + i))(z – (sqrt 3 – i))` `= 0`
  `((z – sqrt 3) – i)((z – sqrt 3) + i)` `= 0`
  `(z – sqrt 3)^2 – i^2` `= 0`
  `z^2 – 2 sqrt 3 z + 3 + 1` `= 0`
  `z^2 – 2 sqrt 3 z + 4` `= 0`

 

b.    `z(z^2 – 2 sqrt 3 z + 4)` `= 0`
  `z(z – (sqrt 3 + i))(z – (sqrt 3 – i))` `= 0`

 
`text(Convert to polar form:)`

  `sqrt 3 + i` `= sqrt((sqrt 3)^2 + 1^2) * text(cis) (tan^(-1)(1/sqrt 3))`
    `= 2 text(cis) (pi/6)`
  `=> sqrt 3 – i` `= 2 text(cis) (-pi/6)`

 

   

 

c.   `text(Equidistant from)\ (0, 0) and (sqrt 3, 1)`

`text(Midpoint)\ (x_1, y_1)` `= (sqrt 3/2, 1/2)`

 
`m = (1 – 0)/(sqrt 3 – 0) = 1/sqrt 3`

`m_(_|_) = (-1)/m = -sqrt 3`
 

`:.\ text(Equation of ⊥ bisector:)`

`y-1/2` `=-sqrt3(x-sqrt3/2)`  
`y` `=-sqrt3 x +3/2+1/2`  
`:.y` `=-sqrt3 x +2`  

 

d.   `text(Let)\ O = (0, 0),\ P = (sqrt 3, 1),\ Q = (sqrt 3, -1)`

`text(⊥ bisector of two points on arc of a circle passes)`

`text(through the centre of the circle.)`
 

`OP = OQ = PQ = 2`

`=> Delta OPQ\ text(is equilateral)`
 

`text(⊥ bisector of)\ PQ\ text(is)\ y=0.`

`text(Centre of circle occurs when:)`

`0 = -sqrt 3 x + 2\ \ text{(using part c)`

`x=2/sqrt3`

`=>\ text(Radius)\ = 2/sqrt3`
 

`:. |z – 2/sqrt 3| = 2/sqrt 3`

Statistics, SPEC2-NHT 2018 VCAA 6

A coffee machine dispenses coffee concentrate and hot water into a 200 mL cup to produce a long black coffee. The volume of coffee concentrate dispensed varies normally with a mean of 40 mL and a standard deviation of 1.6 mL.

Independent of the volume of coffee concentrate, the volume of water dispensed varies normally with a mean of 150 mL and a standard deviation of 6.3 mL.

  1. State the mean and the standard deviation, in millilitres, of the total volume of liquid dispensed to make a long black coffee.  (2 marks)
  2. Find the probability that a long black coffee dispensed by the machine overflows a 200 mL cup. Give your answer correct to three decimal places.  (1 mark)
  3. Suppose that the standard deviation of the volume of water dispensed by the machine can be adjusted, but that the mean volume of water dispensed and the standard deviation of the volume of coffee concentrate dispensed cannot be adjusted.
  4. Find the standard deviation of the volume of water dispensed that is needed for there to be only a 1% chance of a long black coffee overflowing a 200 mL cup. Give your answer in millilitres, correct to two decimal places.  (2 marks)
Show Answers Only
  1. `mu_V = 40 + 150 = 190\ text(mL)`

     

    `sigma_V = sqrt (1.6^2 + 6.3^2) = 6.5\ text(mL)`

  2. `0.062`
  3. `3.99\ text(mL)`
Show Worked Solution

a.   `C~N (40, 1.6^2)`

`W~N (150, 6.3^2)`

`V = C + W`

`mu_V = 40 + 150 = 190\ text(mL)`

`{:sigma^2:}_V` `= 1.6^2 + 6.3^2`  
`:.sigma_V ` `= sqrt (1.6^2 + 6.3^2)`  
  `= 6.5\ text(mL)`  

 

b.   `V~N (190, 6.5^2)`

`text(Pr)(V > 200) ~~ 0.062\ \ text{(by CAS)}`

 

c.  `C~N (40, 1.6^2)`

`W_2~N (150, sigma_w^2)`

`V_2 = C + W_2`

`V_2~N (190, 1.6^2 + sigma_w^2)`

`Z~N (0, 1)`
 

`text(Pr)(V_2 > 200) = 0.01`

`text(Pr)(Z > a) = 0.01\ \ =>\ \ a=2.326…`
  

`text(Find)\ \ sigma_w:`

`2.326…` `= (200-190)/sqrt(1.6^2 + sigma_w^2)`
`:. sigma_w` `~~ 3.99\ text(mL)\ \ \ text{(by CAS)}`

Mechanics, SPEC2-NHT 2018 VCAA 3

A 200 kg crate rests on a smooth plane inclined at `theta` to the horizontal. An external force of `F` newtons acts up the plane, parallel to the plane, to keep the crate in equilibrium.

  1. On the diagram below, draw and label all forces acting on the crate.  (1 mark)

 

 

  1. Find `F` in terms of `theta`.  (1 mark)

The magnitude of the external force `F` is changed to 780 N and the plane is inclined at  `theta = 30^@`.

    1. Taking the direction down the plane to be positive, find the acceleration of the crate.  (2 marks)
    2. On the axes below, sketch the velocity–time graph for the crate in the positive direction for the first four seconds of its motion.  (1 mark)
      `qquad`
       

       
       
    3. Calculate the distance the crate travels, in metres, in its first four seconds of motion.  (1 mark)

Starting from rest, the crate slides down a smooth plane inclined at  `alpha`  degrees to the horizontal.

A force of  `295 cos(alpha)`  newtons, up the plane and parallel to the plane, acts on the crate.

  1. If the momentum of the crate is 800 kg ms¯¹ after having travelled 10 m, find the acceleration, in ms¯², of the crate.  (2 marks)
  2. Find the angle of inclination, `alpha`, of the plane if the acceleration of the crate down the plane is 0.75 ms¯².  Give your answer in degrees, correct to one decimal place.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `F = 200\ text(kg)\ sin(theta)`
    1. `a = 1\ text(ms)^(-2)`
    2. `text(See Worked Solutions)`
    3. `8 text(m)`
  4. `a = 0.8\ text(ms)^(-2) quad text(down the incline)`
  5. `a ~~ 12.9^@`
Show Worked Solution
a.   

 

b.   `F – 200g sin (theta) = 0`

`:. F = 200g sin (theta)`

 

c.i.   `sum F` `= -F + 200g sin (30^@)`
  `200a` `= -780 + 200g xx 1/2`
    `=- 780 + 980`
  `:.a` `=1\ text(ms)^(-1)`

 

c.ii.  

 

c.iii.  `text(Distance travelled in 4 seconds)`

`=\ text(Area under graph between)\ \ t=0 and t=4`

`=1/2 xx 4 xx 4`

`= 8\ text(m)`
 

d.   `p` `=mv`
  `800` `=200v`
  `:.v` `=4`

 
`text(Find)\ \a\ \ text(given)\ \ x = 10,\ \ v=4:`

`v^2` `= u^2 + 2ax`  
`16` `=20a`  
`:.a` `=0.8\ \ text(ms)^(-2) quad text(down the incline)`  

 

e.  `200g sin(alpha) – 295 cos (alpha) = 200 xx 0.75`

`alpha ~~ 12.9^@\ \ \ text{(by CAS)}`

Complex Numbers, SPEC2-NHT 2018 VCAA 2

In the complex plane, `L` is the line given by  `|z + 1| = |z + 1/2 - sqrt 3/2 i|`.

  1. Show that the cartesian equation of `L` is given by  `y = -1/sqrt 3 x`.  (2 marks)
  2.  Find the point(s) of intersection of `L` and the graph of the relation  `z bar z = 4`  in cartesian form.  (2 marks)
  3. Sketch `L` and the graph of the relation  `z bar z = 4`  on the Argand diagram below.  (2 marks)

 

 

The part of the line `L` in the fourth quadrant can be expressed in the form  `text(Arg)(z) = a`.

  1. State the value of `a`.  (1 mark)
  2. Find the area enclosed by `L` and the graphs of the relations  `z bar z = 4, \ text(Arg)(z) = pi/3`  and  `text(Re)(z) = sqrt 3`.  (2 marks)
  3. The straight line `L` can be written in the form `z = k bar z`, where  `k in C`.

     

    Find `k` in the form  `r text(cis)(theta)`, where  `theta`  is the principal argument of `k`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `z = sqrt 3 – i, quad -sqrt 3 + i`
  3. `text(See Worked Solutions)`
  4. `a = -pi/6`
  5. `pi/3 + sqrt 3`
  6. `k = cis (-pi/3)`
Show Worked Solution

a.   `(x + 1)^2 + y^2 = (x + 1/2)^2 + (y – sqrt 3/2)^2`

`x^2 + 2x + 1 + y^2 = x^2 + x + 1/4 + y^2 – y sqrt 3 + 3/4`

`0` `=-x – y sqrt 3`
`y sqrt 3` `= -x`
`:. y` `= -1/sqrt 3 x`

 

b.   `(x + iy) (x – iy)` `=4`
  `x^2 – i^2 y^2` `=4`
  `x^2 + y^2` `=4`

 

`text(Substitute)\ \ y = -1/sqrt 3 x\ \ text(into)\ \  x^2 + y^2 = 4:`

`x^2 + 1/3 x ^2` `=4`
`4/3 x^2` `=4`
`x^2` `=3`
`x` `=+- sqrt3`
`=>y` `= +- 1`

 

`:.\ text(Intersection at):\  (sqrt 3, -1), quad (-sqrt 3, 1)`

 

c.   

 

d.   `alpha = tan^(-1) (-1/sqrt 3), quad alpha in (-pi/2, 0)`

`alpha = -pi/6`
 

e.   

`text(Total Area)`

`=\ text(Area of sector + Area of triangle)`

`=pi xx 2^2 xx ((pi/3 – pi/6)/(2 pi)) + 1/2 xx sqrt 3 xx 2`

`=4 pi (1/12) + sqrt3`

`=pi/3 + sqrt3\ \ text(u²)`

 

f.    `r\ text(cis)(theta)` `=k (r\ text(cis)(-theta))`
  `:. k` `=(r\ text(cis)(theta))/(r\ text(cis)(-theta))`
    `= text(cis)(2 xx ((-pi)/6))`
    `= text(cis)(- pi/3)`

Calculus, SPEC2-NHT 2018 VCAA 1

Consider the function  `f`  with rule  `f(x) = 10 arccos (2 - 2x)`.

  1.  Sketch the graph of  `f`  over its maximal domain on the set of axes below. Label the endpoints with their coordinates.  (3 marks)

 

 

A vase is to be modelled by rotating the graph of  `f`  about the `y`-axis to form a solid of revolution, where units of measurement are in centimetres.

    1.  Write down a definite integral in terms of `y` that gives the volume of the vase.  (2 marks)
    2.  Find the volume of the vase in cubic centimetres.  (1 mark)
  2. Water is poured into the vase at a rate of 20 cm³ s¯¹.
  3. Find the rate, in centimetres per second, at which the depth of the water is changing when the depth is  `5 pi`  cm.  (3 marks)
  4. The vase is placed on a table. A bee climbs from the bottom of the outside of the vase to the top of the vase.

     

    What is the minimum distance the bee will need to travel? Give your answer in centimetres, correct to one decimal place.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
    1.  `V = pi int_0^(10 pi) (1 – 1/2 cos (y/10))^2 dy`
    2. `V = (45 pi^2)/4 text(cm)^3`
  2. `20/pi text(cm s)^(-1)`
  3. `31.4\ text(cm)`
Show Worked Solution
a.  

`2 – 2x in [-1, 1]`

`-2x in [-3, -1]`

`:. x in [1/2, 3/2]`

`f(1/2)` `= 10 cos^(-1) (1)=0`
`f(3/2)` `= 10 cos^(-1) (1)=10pi`

 

b.i.   `y` `= 10 cos^(-1) (2 – 2x)`
  `y/10` `= cos^(-1) (2 – 2x)`
  `cos (y/10)` `= 2 – 2x`
  `2x` `= 2 – cos (y/10)`
  `x` `= 1 – 1/2 cos (y/10)`

 

  `:. V` `= pi int_0^(10 pi) x^2\ dy`
    `= pi int_0^(10 pi) (1 – 1/2 cos (y/10))^2\ dy`

 

b.ii.   `V = (45 pi^2)/4 text(cm)^3`

 

c.  `(dV)/(dt) = 20\ text(cm³/s)\ \ \ text{(given)}`

`(dV)/(dh) = pi (1 – 1/2 cos (y/10))^2\ \ text(when)\ \ y=5pi` 

`=> (dV)/(dh) = pi`
 

`:. (dh)/(dt)` `= (dh)/(dV)*(dV)/(dt)`
  `= 1/pi * 20`
  `= 20/pi\ text(cm s)^(-1)`

 

d.   `f(x) = 10cos^(-1)(2-2x)`

`l=int_(1/2)^(3/2) sqrt(1 + (f′(x))^2)\ dx`

  `~~31.4\ text(cm)\ \ \ text{(by CAS)}`

Calculus, SPEC2 2017 VCAA 1

Let  `f:D ->R, \ f(x) = x/(1 + x^3)`, where `D` is the maximal domain of  `f`.

  1. i.  Find the equations of any asymptotes of the graph of  `f`.  (1 mark)
  2. ii. Find  `f′(x)`  and state the coordinates of any stationary points of the graph of  `f`, correct to two decimal places.  (2 marks)
  3. iii. Find the coordinates of any points of inflection of the graph of  `f`, correct to two decimal places.  (2 marks)
  4. Sketch the graph of  `f(x) = x/(1 + x^3)`  from  `x=–3`  and  `x = 3`  on the axes provided below, marking all stationary points, points of inflection and intercepts with axes, labelling them with their coordinates. Show any asymptotes and label them with their equations.  (3 marks)

     

     

  5. The region `S`, bounded by the graph of  `f`, the `x`-axis and the line  `x = 3`, is rotated about the `x`-axis to form a solid of revolution. The line  `x = a`, where  `0 < a < 3`, divides the region `S` into two regions such that, when the two regions are rotated about the `x`-axis, they generate solids of equal volume.
  6. i.  Write down an equation involving definite integrals that can be used to determine `a`.  (2 marks)
  7. ii. Hence, find the value of `a`, correct to two decimal places.  (1 mark)
Show Answers Only
  1. i.  `text(vertical asymptote:)\ \ x = −1“text(horizontal asymptote:)\ \ y = 0`
  2. ii. `text(S.P.)\ \ (0.79, 0.53)`
  3. iii. `text(P.O.I.)\ (1.26, 0.42)`
  4.  
  5. i.  `pi int_a^3 (x^2)/((1 + x^3)^2) dx`
  6. ii. `~~ 0.98`
Show Worked Solution

a.i.   `text(Graphing the function on CAS:`

♦♦ Mean mark 36%.

`text(Vertical asymptote:)\ x = −1`

`text(Horizontal asymptote:)\ \ y = 0`
 

a.ii.   `u = x, \ u′ = 1,\ \ v = 1 + x^3, \ v′ = 3x^2\ \ \ text{(manual or by CAS)}`

`f′(x)` `= (1(1 + x^3) – x(3x^2))/((1 + x^3)^2)`
  `= (1 – 2x^3)/((1 + x^3)^2)`

 
`text(S.P. when)\ \ f′(x)=0:\ `

`=>  (0.79, 0.53)\ \ \ text{(by CAS)}`

 

a.iii.  `text(When)\ \ f″(x)=0,\ \ \ text{(by CAS)}`

`=> x = 0, \ x = -1, \ x = 2`

`x != -1`

`text(Check concavity changes:)`

`f″(−1/2) = −1632/343`

`f″(1) = −3/4`

`f″(3) = 675/(10\ 976)`
 

`text(P.O.I. at)\ \ x = sqrt2 ~~ 1.26\ \ text{(concavity changes)}`

`=> f(sqrt2) ~~ 0.42`

`:. text(P.O.I.)\ (1.26, 0.42)`

 

b.   

 

c.i.    `V_1` `= pi int_0^a y^2\ dx`
  `V_2` `= pi int_a^3 y^2\ dx`

 
`:.\ text(Equation to solve for)\ a:`

`int_0^a x^2/(1 + x^3)^2\ dx = int_a^3 x^2/(1 + x^3)^2\ dx`

 

c.ii.  `a=0.98\ \ \ text{(by CAS)}`

Statistics, SPEC2 2018 VCAA 6

The heights of mature water buffaloes in northern Australia are known to be normally distributed with a standard deviation of 15 cm. It is claimed that the mean height of the water buffaloes is 150 cm.

To decide whether the claim about the mean height is true, rangers selected a random sample of 50 mature water buffaloes. The mean height of this sample was found to be 145 cm.

A one-tailed statistical test is to be carried out to see if the sample mean height of 145 cm differs significantly from the claimed population mean of 150 cm.

Let `bar X` denote the mean height of a random sample of 50 mature water buffaloes.

  1. State suitable hypotheses `H_0` and `H_1` for the statistical test.  (1 mark)
  2. Find the standard deviation of `bar X`.  (1 mark)
  3. Write down an expression for the `p` value of the statistical test and evaluate your answer to four decimal places.  (2 marks)
  4. State with a reason whether `H_0` should be rejected at the 5% level of significance.  (1 mark)
  5. What is the smallest value of the sample mean height that could be observed for `H_0` to be not rejected? Give your answer in centimetres, correct to two decimal places.  (1 mark)
  6. If the true mean height of all mature water buffaloes in northern Australia is in fact 145 cm, what is the probability that `H_0` will be accepted at the 5% level of significance? Give your answer correct to two decimal places.  (1 mark)
  7. Using the observed sample mean of 145 cm, find a 99% confidence interval for the mean height of all mature water buffaloes in northern Australia. Express the values in your confidence interval in centimetres, correct to one decimal place.  (1 mark)
Show Answers Only
  1.  `H_0: mu = 150; qquad H_1: mu < 150`
  2. `3/sqrt 2`
  3. `p = text(Pr)(bar X < 145); qquad p~~ 0.0092`
  4. `text(Yes, he should be rejected as)`
    `p~~ 0.0092 < 0.05`
  5. `bar X_min = 146.52`
  6. `0.24`
  7. `(139.5, 150.5)`
Show Worked Solution
a.    `H_0: mu =150`
  `H_1: mu < 150`

 

b.    `sigma_bar X` `= (sigma_X)/sqrt n`
    `= 15/sqrt 50`
    `= (3sqrt 2)/2`

 

c.   `p = text(Pr)(bar X < 145), qquad bar X\ ~\ N (150, 9/2)`

`p ~~ 0.0092`

 

d.   `text(Yes, he should be rejected as)`

`p~~ 0.0092 < 0.05`

 

e.  `text(Pr)(bar X < x) = 0.05`

♦ Mean mark part (e) 48%.

`x ~~ 146.511`

`text(NOT rejected:) quad bar X_min = 146.52`

 

f.   `bar X_2\ ~\ N (145, 9/2)`

♦♦♦ Mean mark part (f) 11%.

`text(Pr)(bar X_2 > x)` `= text(Pr)(bar X_2 > 146.51074)`  
  `~~ 0.24`  

 

g.   `(145 – (Z_99 xx 3)/sqrt 2, 145 + (Z_99 xx 3)/sqrt 2)`

`text(Pr)(Z < Z_99) = 0.995, \ Z\ ~\ N(0, 1)`

`=>Z_99 ~~ 2.57583`

`:. 99%\ text(C.I:)\ (139.5, 150.5)`

Statistics, SPEC2-NHT 2017 VCAA 18 MC

`X` is a random variable with a mean of 5 and a standard deviation of 4, and `Y` is a random variable with a mean of 3 and a standard deviation of 2.

If `X` and `Y` are independent random variables and  `Z = X-2Y`, then `Z` will have mean `mu` and standard deviation `sigma` given by

  1. `mu = -1, sigma = 0`
  2. `mu = -1, sigma = 4 sqrt 2`
  3. `mu = 2, sigma = 8`
  4. `mu = 2, sigma = 4 sqrt 2`
  5. `mu = -1, sigma = 2 sqrt 6`
Show Answers Only

`B`

Show Worked Solution

`mu_z= mu_x-2 mu_y= 5-2(3)= -1`

`sigma_z^2= sigma_x^2 + (-2)^2 sigma_y^2= 4^2 + 4(2)^2= 32`

`sigma_z= sqrt 32= 4 sqrt 2` 

`=>   B`

Algebra, SPEC2-NHT 2017 VCAA 5 MC

Given that `A, B, C` and `D` are non-zero rational numbers, the expression  `(3x + 1)/(x(x - 2)^2)`  can be represented in partial fraction form as

A.   `A/x + B/((x - 2))`

B.   `A/x + B/(x - 2)^2`

C.   `A/x + B/((x - 2)) + C/(x - 2)^2`

D.   `A/x + B/x^2 + C/((x - 2))`

E.   `A/x + (Bx)/((x - 2)) + (Cx + D)/(x - 2)^2`

Show Answers Only

`C`

Show Worked Solution

`(3x + 1)/(x(x – 2)^2) = A/x + B/(x – 2) + C/(x – 2)^2`

 
`=>   C`

Trigonometry, SPEC2-NHT 2017 VCAA 4 MC

If  `sin(theta + phi) = a`  and  `sin(theta - phi) = b`, then  `sin(theta) cos(phi)`  is equal to

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

`E`

Show Worked Solution

`sin(theta + phi) = a`

`sin theta cos phi + sin phi cos theta` `= a\ \ …\ (1)`
`sin(theta – phi) = b`  
`sin theta cos phi – sin phi cos theta` `= b\ \ …\ (2)`

 
`(1) + (2):`

`2 sin theta cos phi` `= a + b`
`:. sin theta cos phi` `= (a + b)/2`

 
`=>   E`

Statistics, SPEC1-NHT 2017 VCAA 9

The random variables `X` and `Y` are independent with  `mu_X = 4,\ text(Var)(X) = 36`  and  `mu_Y = 3,\ text(Var)(Y) = 25`.

  1. The random variable `Z` is such that  `Z = 2X + 3Y`.
  2.  i. Find  `E(Z)`.  (1 mark)
  3. ii. Find the standard deviation of `Z`.  (1 mark)
  1. Researchers have reason to believe that the mean of `X` has decreased. They collect a random sample of 64 observations of `X` and find that the sample mean is  `bar X = 3.8`
  2.  i. State the null hypothesis and the alternative hypothesis that should be used to test that the mean has decreased.  (1 mark)
  3. ii. Calculate the mean and standard deviation for a distribution of sample means, `bar X`, for samples of 64 observations.  (1 mark)
Show Answers Only
  1.  i. `17`
  2. ii. `3 sqrt 41`
  3.  i. `H_0: mu = 4,\ \ H_1: mu < 4`
  4. ii. `mu_bar X = 4,\ \ sigma_bar X = 3/4`
Show Worked Solution
a.i.    `E(Z)` `= 2E(X) + 3E(Y)`
    `= 2(4) + 3(3)`
    `= 8 + 9`
    `= 17`

 

a.ii.    `text(Var)(Z)` `= 2^2 text(var)(X) + 3^2 text(var)(Y)`
    `= 4(36) + 9(25)`
    `= (4xx9xx4) + 9(25)`
    `= 9(16 + 25)`

 

`:. sigma_Z` `= sqrt(9 (41))`
  `= 3 sqrt 41`

 

b.i.    `H_0: \ mu = 4`
  `H_1: \ mu < 4`

 

b.ii.    `bar X~N(mu, (sigma^2)/n)`
  `bar X~N(4, 36/64)`

 
`E(barX) = 4`

`sigma_bar X` `= sqrt(36/64)`
  `= 3/4`

Calculus, SPEC2 2018 VCAA 3

Part of the graph of  `y = 1/2 sqrt(4x^2 - 1)`  is shown below.
 


 

The curve shown is rotated about the `y`-axis to form a volume of revolution that is to model a fountain, where length units are in metres.

  1. Show that the volume, `V` cubic metres, of water in the fountain when it is filled to a depth of `h` metres is given by  `V = pi/4(4/3h^3 + h)`.  (2 marks)
  2. Find the depth `h` when the fountain is filled to half's its volume. Give your answer in metres, correct to two decimal places.  (2 marks)

The fountain is initially empty. A vertical jet of water in the centre fills the fountain at a rate of 0.04 cubic metres per second and, at the same time, water flows out from the bottom of the fountain at a rate of  `0.05 sqrt h`  cubic metres per second when the depth is `h` metres.

  1.  i. Show that  `(dh)/(dt) = (4-5sqrt h)/(25 pi (4h^2 + 1))`.  (2 marks)
  2. ii. Find the rate, in metres per second, correct to four decimal places, at which the depth is increasing when the depth is 0.25 m. (1 mark)
  3. Express the time taken for the depth to reach 0.25 m as a definite integral and evaluate this integral correct to the nearest tenth of a second.  (2 marks)
  4. After 25 seconds the depth has risen to 0.4 m.
    Using Euler's method with a step size of five seconds, find an estimate of the depth 30 seconds after the fountain began to fill. Give your answer in metres, correct to two decimal places.  (2 marks)
  5. How far from the top of the fountain does the water level ultimately stabilise? Give your answer in metres, correct to two decimal places.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `h~~ 0.59\ text(m)`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `0.0153\ text(ms)^(-1)`
  1. `9.8\ text(seconds)`
  2. `0.43\ text(m)`
  3. `0.23\ text(m)`
Show Worked Solution

a.   `V= pi int_0^h x^2\ dy`

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

 

`:. V` `= pi int_0^h y^2 + 1/4\ dy`
  `= pi[y^3/3 + y/4]_0^h`
  `= pi(h^3/3 + h/4 – 0)`
  `= pi(1/4((4h^3)/3 + h))`
  `= pi/4((4h^3)/3 + h)\ \ …\ text(as required)`

 

b.    `V_text(max)` `= pi/4 (4/3 xx (sqrt 3/2)^3 + sqrt 3/2)`
    `= pi/4 (sqrt 3/2 + sqrt 3/2)`
    `= (pi sqrt 3)/4`

 

`1/2 V_text(max)` `= (pi sqrt 3)/8`
`(pi sqrt 3)/8` `= pi/4 (4/3 h^3 + h)`
`sqrt 3/2` `= 4/3 h^3 + h`
`:. h` `~~0.59\ text(m)`

  

c.i.   `((dV)/(dt))_text(in)` `= 0.04`
  `((dV)/(dt))_text(out)` `= 0.05 sqrt h`
  `(dV)/(dt)` `= 0.04 – 0.05 sqrt h`
    `= (4 – 5 sqrt h)/100`
     
  `(dV)/(dh)` `= pi/4(4h^2 + 1)`
  `:. (dh)/(dt)` `= (dh)/(dV) ⋅ (dV)/(dt)`
    `= 4/(pi(4h^2 + 1)) xx (4 – 5 sqrt h)/100`
    `= (4 – 5 sqrt h)/(25 pi (4h^2 + 1))`

 

c.ii.   `(dh)/(dt)|_(h = 0.25)` `= (4 – 5 sqrt(0.25))/(25 pi (4(0.25)^2 + 1))`
    `~~ 0.0153\ text(ms)^(-1)`

 

d.   `(dt)/(dh) = (25 pi (4h^2 + 1))/(4 – 5 sqrt h)`
 

`:. t(0.25)` `= int_0^0.25 (25 pi (4h^2 + 1))/(4 – 5 sqrt h)\ dh`
  `~~9.8\ text(seconds)`

 

e.   `text(When)\ \ t=25,\ \ h=0.4\ \ text{(given)}`

♦♦ Mean mark part (e) 30%.

`:. h(30)` `~~ h(25) + 5 xx (dh)/(dt)|_(h = 0.4)`
  `~~ 0.4 + 5 xx ((4 – 5 sqrt 0.4)/(25 pi (4(0.4)^2 + 1)))`
  `~~ 0.43\ text(m)`

 

f.    `(dV)/(dt) = 0`

♦♦ Mean mark part (f) 32%.

`0.04 – 0.05 sqrt h` `= 0`
`0.04` `= 0.05 sqrt h“
`sqrt h` `= 4/5`
`h` `= 16/25`

  

`d` `= h_max – 16/25`
  `= sqrt 3/2 – 16/25`
  `~~ 0.23\ text(m)`

Calculus, SPEC2 2018 VCAA 1

Consider the function  `f: D -> R`, where  `f(x) = 2 text(arcsin)(x^2 - 1)`.

  1. Determine the maximal domain `D` and the range of  `f`.  (2 marks)
  2. Sketch the graph of  `y = f(x)`  on the axes below, labelling any endpoints and the `y`-intercept with their coordinates.  (3 marks)
     
    `qquad` 
     
  3. Find  `f prime(x)`  for  `x > 0`, expressing your answer in the form  `f prime(x) = A/sqrt(2 - x^2), \ A in R`.  (1 mark)
  4. Write down  `f prime(x)`  for  `x < 0`, expressing your answer in the form  `f prime(x) = B/sqrt(2 - x^2), \ B in R`.  (1 mark)
  5. The derivative  `f prime(x)`  can be expressed in the form  `f prime(x) = {g(x)}/sqrt(2 - x^2)` over its maximal domain.
  1. Find the maximal domain of  `f prime`.  (1 mark)
  2. Find  `g(x)`, expressing your answer as a piecewise (hybrid) function.  (1 mark)
  3. Sketch the graph of  `g` on the axes below.  (2 marks)

 

Show Answers Only
  1. `D [- sqrt 2 , sqrt 2]; qquad f(x) in [-pi, pi]`
  2. `text(See Worked Solutions)`
  3. `4/sqrt(2 – x^2)`
  4. `(-4)/sqrt(2 – x^2)`
    1. `x in (-sqrt2, 0) ∪ (0, sqrt 2)`
    2. `g(x) = {(-4text(,), x < 0), (\ \ \ 4text(,), x > 0):}`
    3. `text(See Worked Solutions)`
Show Worked Solution

a.   `-1 <= x^2 – 1 <= 1`

`=>\ 0 <= x^2<= 1`

`=>\ -sqrt2 <= x <=sqrt2`

`:.\ text(Domain:)\ [- sqrt 2 , sqrt 2]`

 

`sin^(-1)(x) in [-pi/2, pi/2]`

`=> 2 sin^(-1)(x) in [-2xx pi/2, 2 xx pi/2]`

`:.\ text(Range:)\ [-pi, pi]`

 

b.   

 

c.   `f(x) = 2 sin^(-1)(x^2 – 1)`

`f(x)` `=2 sin^(-1)(x^2 – 1)`

 
`:. f prime(x)= 4/sqrt(2 – x^2)\ \ \ (x > 0,\ \ text(by CAS))`

 

d.    `f prime(x)` `= (4x)/sqrt(x^2(2 – x^2))`
    `= (-4)/sqrt(2 – x^2)\ \ \ (x < 0)`

 

e.i.  `f prime(x)\ \ text(is defined when:)`

♦♦ Mean mark part (e)(i) 21%.

`2 – x^2 > 0\ \ and\ \ x!=0`

`:. x in (-sqrt2, 0) ∪ (0, sqrt 2)`

 

e.ii.  `g(x) = +- 4`

♦ Mean mark part (e)(ii) 49%, part (e)(iii) 48%.

`:.  g(x) = {(-4text(,), x < 0), (\ \ \ 4text(,), x > 0):}`

 

e.iii.   

Calculus, SPEC1 2017 VCAA 10

  1.  Show that  `d/dx(x arccos(x/a)) = arccos(x/a)−x/(sqrt(a^2 - x^2))`, where  `a > 0`.  (1 mark)
  2.  State the maximum domain and the range of  `f(x) = sqrt(arccos(x/2))`.  (2 marks)
  3.  Find the volume of the solid of revolution generated when the region bounded by the graph of  `y = f(x)`, and the lines  `x = −2`  and  `y = 0`, is rotated about the `x`-axis.  (4 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `x ∈ [−2, 2], \ y ∈ [0, sqrtpi]`
  3. `2pi^2`
Show Worked Solution
a.    `u` `= x,` `v` `= cos^(−1)(x/a)`
  `uprime` `= 1,`   `vprime` `= (−1)/sqrt(a^2 – x^2)`
`d/(dx)(xcos^(−1)(x/a))` `= uprimev + vprimeu`
  `= cos^(−1)(x/a) + (x(−1))/sqrt(a^2 – x^2)`
  `= arccos(x/a) – x/sqrt(a^2 – x^2)`

 

b.   `arccos(x/2)>=0`

`text(Maximal domain:)\  x ∈ [−2, 2]`

`f(x) = (arccos(x/2))^(1/2)`

`text(Range:)\ \ y ∈ [0, sqrtpi]`

 

c.   `V` `= pi int_(−2)^2 y^2\ dx`
  `= pi int_(−2)^2 cos^(-1)(x/2)\ dx`
  `= pi int_(−2)^2 cos^(-1) (x/2) – x/sqrt(4 – x^2) + x/sqrt(4 – x^2)\ dx`
  `= pi [x cos^(-1)(x/2)]_(−2)^2 + pi int_(−2)^2 x/sqrt(4 – x^2)\ dx`

 
`text(Let)\ \ u = 4 – x^2`

♦ Mean mark part (c) 35%.

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

`text(When)\ \ x=2\ \ => \ u=0`

`text(When)\ \ x=-2\ \ =>\ \ u=0`

`:. V` `= pi [2cos^(−1)(1) – (-2)cos^(−1)(−1)] – pi/2 int_0^0 1/sqrtu du`
  `= pi(2 xx 0 + 2 xx pi)`
  `= 2pi^2`

Mechanics, SPEC1-NHT 2017 VCAA 1

A 5 kg mass on a smooth plane inclined at 30° is held in equilibrium by a horizontal force of magnitude `P` newtons, as shown in the diagram below.
 


 

  1. On the diagram above, show all other forces acting on the mass and label them. (1 mark)
  2.  Find  `P`.  (2 marks)
Show Answers Only
  1.  `text(See Worked Solutions)`
  2. `P = (5g sqrt 3)/3`
Show Worked Solution
a.   

 

b.    `P cos 30^@` `= 5g sin 30^@`
  `(P sqrt 3)/2` `= (5g)/2`
  `P sqrt 3` `= 5g`
  `P` `= (5g)/sqrt 3`
  `P` `= (5g sqrt 3)/3`

Complex Numbers, SPEC2 2015 VCAA 5 MC

Given  `z = (1 + isqrt3)/(1 + i)`, the modulus and argument of the complex number  `z^5`  are respectively

  1. `2sqrt2`  and  `(5pi)/6`
  2. `4sqrt2`  and  `(5pi)/12`
  3. `4sqrt2`  and  `(7pi)/12`
  4. `2sqrt2`  and  `(5pi)/12`
  5. `4sqrt2`  and  `-pi/12`
Show Answers Only

`B`

Show Worked Solution

`z_1 = 1 + isqrt3`

`r` `= sqrt(1 + 3)=2`
`theta` `= tan^(−1)(sqrt3)= pi/3`

 
`z_1 = 2\ text(cis)(pi/3)`
 

`z_2 = 1 + i`

`r` `= sqrt(1 + 1)=sqrt2`
`theta` `= tan^(−1)(1)=pi/4`

 

`z_1/z_2` `= 2/sqrt2\ text(cis)(pi/3 – pi/4)`
  `= sqrt2\ text(cis)(pi/12)`

 

`:. z^5` `= (sqrt2)^5\ text(cis)((5pi)/12)\ \ \ text{(De Moivre)}`
  `= 4sqrt2\ text(cis)((5pi)/12)`

 
`=> B`

Calculus, SPEC1 2015 VCAA 8

  1.  Show that  `int tan (2x)\ dx = 1/2 log_e |\ sec (2x)\ | + c.`  (2 marks)
  2.  The graph of  `f(x) = 1/2 arctan (x)`  is shown below
     
         
     
  3. i.  Write down the equations of the asymptotes.  (1 mark)
  4. ii. On the axes above, sketch the graph of  `f^-1`, labelling any asymptotes with their equations.  (1 mark)
  5.  Find  `f(sqrt 3).`  (1 mark)
  6.  Find the area enclosed by the graph of  `f`, the `x`-axis and the line  `x = sqrt 3.`   (2 marks)
Show Answers Only

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

b. (i)   `y = -pi/4,\ \ y = pi/4`
  (ii)  

c.   `pi/6`

d.   `(sqrt 3 pi)/6 – log_e sqrt 2`

Show Worked Solution

a.   `text(Let)\ \ u = cos(2x)`

`(du)/(dx) = -2 sin (2x)\ \ =>\ \ -1/2 xx du = sin (2x)\ dx`

`int tan (2x)\ dx` `= int (sin(2x))/(cos(2x))\ dx`
  `= -1/2 int 1/u\ du`
  `= -1/2 log_e |\ u\ | + c`
  `= -1/2 log_e |\ cos (2x)\ | + c`
  `= 1/2 log_e |\ sec (2x)\ | + c`

 

b.i.  `text(Range:)\ \ tan^(-1)(x) in (- pi/2,pi/2)`

`=>1/2 tan^(-1)(x) in (- pi/4, pi/4)`

`:.\ text(Asymptotes:)\ \ y = -pi/4,\ \ y = pi/4`

 

b.ii.

 

c.   `f(sqrt 3)` `= 1/2 tan^(-1) (sqrt 3)`
    `= 1/2 xx pi/3`
    `= pi/6`

 

d.    `y` `= 1/2tan^(−1)(x)`
  `2y` `= tan^(−1)(x)`
  `x` `= tan(2y)`

 

 

♦ Mean mark part (d) 49%.

`text(Area)` `=\ text(Area of rectangle – Area between graph and y-axis)`
  `= sqrt3 xx pi/6 – int_0^(pi/6) tan(2y)\ dy`
  `= (sqrt3 pi)/6 -1/2[ln\ | sec(2y) |]_0^(pi/6)`
  `= (sqrt3 pi)/6 – 1/2[ ln(sec(pi/3)) – ln(sec 0)]`
  `= (sqrt3 pi)/6 – 1/2 (ln2 -ln1)`
  `= (sqrt3 pi)/6 – 1/2 ln2`

Calculus, SPEC1 2014 VCAA 7

Consider  `f(x) = 3x arctan (2x)`.

  1.  Write down the range of  `f`.  (1 mark)
  2.  Show that  `f prime(x) = 3 arctan (2x) + (6x)/(1 + 4x^2)`.  (1 mark)
  3.  Hence evaluate the area enclosed by the graph of  `g(x) = arctan (2x)`, the `x`-axis and the lines  `x = 1/2`  and  `x = sqrt 3/2`.  (3 marks)
Show Answers Only
  1. `f(x) in [0, oo)`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `(pi sqrt 3)/6 – pi/8 – 1/4 ln2`
Show Worked Solution

a.  `f(x) = 3x arctan (2x)`

`f(0) = 0`

♦♦♦ Mean mark 4%.

`text(When)\ \ x < 0, \ \ 3x<0,\ \ text(and)\ \ y = tan^(-1) (2x) < 0`

`text(When)\ \ x>0,\ \ 3x>0,\ \ text(and)\ \ y = tan^(-1) (2x) > 0`
 


 
 `:. f(x) in [0, oo)`

 

b.  `u = 3x, qquad v = tan^(-1)(2x)`

`u prime = 3, qquad v prime = 2/(1 + 4x^2)`

`:. f prime (x)` `= 3(arctan (2x)) + (2/(1 + 4x^2))(3x)`
  `= 3 arctan (2x) + (6x)/(1 + 4x^2)`

 

c.    `A` `= int_(1/2)^(sqrt 3/2) arctan (2x)\ dx`
    `= 1/3 int_(1/2)^(sqrt 3/2) 3arctan (2x)\ dx`
    `=1/3 int_(1/2)^(sqrt 3/2) (3 arctan (2x) + (6x)/(1 + 4x^2) – (6x)/(1 + 4x^2))\ dx`
    `= 1/3 int_(1/2)^(sqrt 3/2) 3 arctan (2x) + (6x)/(1 + 4x^2) dx – int_(1/2)^(sqrt 3/2) (2x)/(1 + 4x^2)\ dx`
    `= 1/3 [3x arctan (2x)]_(1/2)^(sqrt 3/2) – 1/4 int_(1/2)^(sqrt 3/2) (8x)/(1 + 4x^2)\ dx`
    `= [sqrt 3/2 arctan (sqrt 3) – 1/2 arctan (1)] – 1/4[ln (1 + 4x^2)]_(1/2)^(sqrt 3/2)`
    `= sqrt 3/2 (pi/3) – 1/2 (pi/4) – 1/4 [ln (1 + 4(3/4)) – ln(1 + 4 (1/4)]`
    `= (pi sqrt 3)/6 – pi/8 – 1/4 (ln4 – ln 2)`
    `= (pi sqrt 3)/6 – pi/8 – 1/4 ln2`

Calculus, SPEC1 2014 VCAA 6

  1.  Verify that  `a/(a - 4) = 1 + 4/(a - 4)`.  (1 mark)

Part of the graph of  `y = x/sqrt(x^2 - 4)`  is shown below.
 


  

  1.  The region enclosed by the graph of  `y = x/sqrt(x^2 - 4)`  and the lines  `y = 0, \ x = 3`  and  `x = 4`  is rotated about the `x`-axis

     

     Find the volume of the resulting solid of revolution.  (4 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `pi(1 + l n (5/3))`
Show Worked Solution
a.    `a/(a – 4)` `= (a – 4 + 4)/(a – 4)`
    `= (a – 4)/(a – 4) + 4/(a – 4)`
    `= 1 + 4/(a – 4), quad a != 4`

 

b.  `r = y`

`V` `= pi int_3^4 y^2\ dx`
  `= pi int_3^4 underbrace{x^2/(x^2 – 4)}_text(using part a.)\ dx`
  `= pi int_3^4 1 + 4/(x^2 – 4) dx`
  `= pi int_3^4 1 + 4/((x – 2)(x + 2)) dx`

 
`text(Using partial fractions:)`

`4/((x – 2)(x + 2)) = A/(x – 2) + B/(x + 2)`

`A(x + 2) + B(x – 2) = 4`
 

`text(When)\ \ x = 2: \ 4A = 4\ \ =>\ \ A = 1`

`text(When)\ \ x = -2: -4B = 4\ \ =>\ \ B = -1`

`:. V` `= pi int_3^4 1 + 1/(x – 2) – 1/(x + 2)\ dx`
  `= pi [x + ln |x – 2| – ln |x + 2|]_3^4`
  `= pi [4 + ln2 – ln6 – (3 + ln 1 – ln5)]`
  `= pi (1 + ln2 – ln6 + ln5)`
  `= pi (1 + ln ((2 xx 5)/6))`
  `= pi (1 + ln (5/3))`

Calculus, SPEC1 2014 VCAA 5

  1. For the function with rule  `f(x) = 96 cos (3x) sin (3x)`, Find the value of `a` such that  `f(x) = a sin (6x)`.  (1 mark)
  2. Use an appropriate substitution in the form  `u = g(x)`  to find an equivalent definite integral for

     

         `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`  in terms of `u` only.  (3 marks)

  3. Hence evaluate  `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`, giving your answer in the form  `sqrt k, \ k in Z`.  (1 mark)
Show Answers Only
  1. `48`
  2. `int_(sqrt 3/2)^0 – 8u^2 du`
  3. `sqrt 3`
Show Worked Solution

a.   `96 cos (3x) sin(3x)`

`= 48 (2 cos(3x) sin(3x))`

`= 48 sin (6x)`

`:. a = 48`

 

b. `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`

`= int_(pi/36)^(pi/12) 48 sin (6x) cos^2 (6x)\ dx`

`u` `= cos (6x)`
`(du)/(dx)` `= -6 sin (6x)\ \ =>\ \ du = -6 sin(6x)\ dx`
   
`u(pi/12)` `= cos (pi/2) = 0`
`u(pi/36)` `= cos (pi/6) = sqrt 3/2`

 
`:. int_(pi/36)^(pi/12) 48 sin (6x) cos^2 (6x)\ dx`

`= int_(sqrt 3/2)^0 – 8u^2\ du`

c. `:. int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`

`= -8 [u^3/3]_(sqrt 3/2)^0`
`= -8 (0 – (sqrt 3/2)^3/3)`
`= 8/3 xx (3sqrt3)/8`
`= sqrt 3`

Vectors, SPEC1 2014 VCAA 2

The position vector of a particle at time  `t >= 0`  is given by

`underset ~r (t) = (t - 2) underset ~ i + (t^2 - 4t + 1) underset ~j`

  1.  Show that the cartesian equation of the path followed by the particle is  `y = x^2 - 3`.  (1 mark)
  2.  Sketch the path followed by the particle on the axes below, labelling all important features.  (2 marks)

 

 

  1.  Find the speed of the particle when  `t = 1`.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(See Worked Solutions)`
  3. `sqrt 5`
Show Worked Solution
a.    `x` `= t – 2\ \ =>\ \ t=x+2`
`y` `= t^2 – 4t + 1`
  `= (x + 2)^2 – 4 (x + 2) + 1`
`y` `= x^2 + 4x + 4 – 4x – 8 + 1`
  `= x^2 – 3`

 

b.   `t >= 0`

`x = t – 2\ \ =>\ \ x >= -2`

`y(-2)` `= (-2)^2 – 3`
  `= 1`
`y(0)` `= -3`

 

`0` `= x^2 – 3`
`x^2` `= 3`
`x` `= +- sqrt 3`

 

c.    `underset ~ dot r (t)` `= underset ~ dot i + (2t – 4) underset ~j`
  `underset ~ dot r (1)` `= underset ~ dot i + (2 – 4) underset ~j`
    `= underset ~ dot i – 2 underset ~j`
  `|underset ~ dot r(1)|` `= sqrt (1^2 + (-2)^2)`
    `= sqrt 5`

Vectors, SPEC1 2014 VCAA 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
a.    `|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%.

b.    `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^@`

 

c.   `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`  

Statistics, SPEC2 2017 VCAA 20 MC

In a one-sided statistical test at the 5% level of significance, it would be concluded that

  1. `H_0` should not be rejected if  `p = 0.04`
  2. `H_0` should be rejected if  `p = 0.06`
  3. `H_0` should be rejected if  `p = 0.03`
  4. `H_0` should not be rejected if  `p != 0.05`
  5. `H_0` should not be rejected if  `p = 0.01`
Show Answers Only

`C`

Show Worked Solution

`p < 0.05:\ text(Reject)\ H_0`

`p > 0.05:\ text(Do not reject)\ H_0`

`=> C`

Vectors, SPEC2-NHT 2018 VCAA 11 MC

Let  `underset ~a = 2 underset ~i - 2 underset ~j + underset ~k`  and  `underset ~b = 2 underset ~i + 3 underset ~j + 6 underset ~k`.

The acute angle between  `underset ~a`  and  `underset ~b`  is closest to

  1. `11º`
  2. `75º`
  3. `79º`
  4. `86º`
  5. `88º`
Show Answers Only

`C`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= 4 – 6 + 6 = 4`
`4` `=sqrt(4 + 4 + 1) sqrt(4 + 9 + 36) cos theta`
`4` `= sqrt 9 xx sqrt 49 cos theta`
`4` `= 21 cos theta`
`theta` `= cos^(-1) (4/21)`
`:. theta` `= 79.0194…`

 
`=>  C`

Calculus, SPEC2-NHT 2018 VCAA 9 MC

`int (1-cos(10x))\ dx`  is equivalent to

  1. `int(sin^2 (5x))\ dx`
  2. `1/2 int(sin^2 (20x))\ dx`
  3. `int(cos^2 (5x))\ dx`
  4. `2 int (cos^2 (10x))\ dx`
  5. `2 int(sin^2 (5x))\ dx`
Show Answers Only

`E`

Show Worked Solution

`int (1-cos(10x))\ dx`

 `= int 1-cos (2xx (5x))\ dx`

`= int 1-(2 cos^2 (5x)-1)\ dx`

`= int 2-2 cos^2 (5x)\ dx`

`= int 2 (1-cos^2 (5x))\ dx`

`= 2 int sin^2 (5x)\ dx`

 
`=>  E`

Calculus, SPEC2-NHT 2018 VCAA 7 MC

The gradient of the line that is perpendicular to the graph of the relation  `3y^2 - 5xy - x^2 = 1`  at the point  `(1, 2)`  is

A.  `-1/12`

B.      `12/7`

C.       `21`

D.   `-7/12`

E.   `-7/13`

Show Answers Only

`D`

Show Worked Solution
`6y *(dy)/(dx) – (5y + 5x* (dy)/(dx)) – 2x` `= 0`
`6y *(dy)/(dx) – 5x *(dy)/(dx)` `= 2x + 5y`
`6 xx 2 xx m_T – 5 xx 1 xx m_T` `= 2 xx 1 + 5 xx 2`
`(12 – 5)m_T` `= 12`
`7 m_T` `= 12`
`m_T` `= 12/7`

 `:.m_N = -7/12`

`=>  D`

Complex Numbers, SPEC2-NHT 2018 VCAA 4 MC

On the Argand diagram shown above, `4 text(cis)(-(2pi)/3)`  is represented by the point

  1. `P`
  2. `Q`
  3. `R`
  4. `S`
  5. `T`
Show Answers Only

`D`

Show Worked Solution

`text(Point is in 3rd quadrant, and)\ \ r = 4.`

`text(Consider:)\  R, S, T`

`r_R` `~~ sqrt((-1)^2 + (-1.75)^2)`
  `~~ 2.02 < 4`

 
`r_T > sqrt((-2)^2 + (-4)^2)`

`r_T > sqrt 20> 4`

 
`text(By elimination, point must be)\ \ S.` 

`=>   D`

Calculus, SPEC1-NHT 2018 VCAA 9

  1. i.   Given that  `cot(2 theta) = a`, show that  `tan^2(theta) + 2a tan(theta) - 1 = 0`.  (2 marks)
  2. ii.  Show that  `tan(theta) = -a +- sqrt(a^2 + 1)`.  (1 mark)
  3. iii. Hence, show that  `tan(pi/12) = 2 - sqrt 3`, given that  `cot(2 theta) = sqrt 3`, where  `theta in (0, pi)`.  (1 mark)
  4.  Find the gradient of the tangent to the curve  `y = tan (theta)`  at  `theta = pi/12`.  (2 marks)
  5.  A solid of revolution is formed by rotating the region between the graph of  `y = tan(theta)`, the horizontal axis, and the lines  `theta = pi/12`  and  `theta = pi/3`  about the horizontal axis.

     

    Find the volume of the solid of revolution.  (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)}`
  2. `8 – 4 sqrt 3`
  3. `pi (2 sqrt 3 – 2 – pi/4)`
Show Worked Solution
a.i.   `1/(tan 2 theta)` `= a`
  `1` `= a tan (2 theta)`
  `1` `= a ((2 tan (theta))/(1 – tan^2(theta)))`
  `1/a (1 – tan^2 (theta))` `= 2 tan (theta)`
  `1 – tan^2 (theta)` `= 2a tan (theta)`
  `:. tan^2 (theta) + 2 a tan (theta) – 1` `=0\ \ text(… as required)`

 

a.ii.   `[tan^2 (theta) + 2a tan (theta) + a^2] – a^2 – 1` `= 0`
  `(tan (theta) + a)^2` `= a^2 + 1`
  `tan (theta) + a` `= +- sqrt(a^2 + 1)`
  `:. tan (theta)` `= -a +- sqrt(a^2 + 1)`

 

a.iii.   `theta in (0, pi) \ => \ 2 theta in (0, 2 pi)`

`text(S)text(ince)\ \ cot(2theta)=sqrt3\ \ \ =>\ \ \ tan(2theta)=1/sqrt3`

  `:. 2 theta` `= pi/6, pi + pi/6`
  `theta` `= pi/12, (7 pi)/12`
  `tan (theta)` `= – sqrt 3 +- sqrt((sqrt 3)^2 + 1)`
    `= – sqrt 3 +- sqrt(3 + 1)`
    `= – sqrt 3 +- 2`

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

 

b.   `y` `=tan(theta)`
  `y prime` `= sec^2 (theta)`
    `= 1 + tan^2 (theta)`

 

`y prime (pi/12)` `= 1 + tan^2 (pi/12)`
  `= 1 + (2 – sqrt 3)^2`
  `= 1 + 4 – 4 sqrt 3 + 3`
  `= 8 – 4 sqrt 3`

 

c.   `V` `= pi int_(pi/12)^(pi/3) y^2\ d theta`
    `= pi int_(pi/12)^(pi/3) tan^2 (theta)\ d theta`
    `= pi int_(pi/12)^(pi/3) (1 + tan^2 (theta) – 1)\ d theta`
    `= pi int_(pi/12)^(pi/3) (sec^2 (theta) – 1)\ d theta`
    `= pi [tan (theta) – theta]_(pi/12)^(pi/3)`
    `= pi (tan (pi/3) – pi/3 – (tan (pi/12) – pi/12))`
    `= pi (sqrt 3 – (4 pi)/12 – (2 – sqrt 3) + pi/12)`
    `= pi (2 sqrt 3 – 2 – pi/4)\ \ text(u³)`

Complex Numbers, SPEC1-NHT 2018 VCAA 8

A circle in the complex plane is given by the relation  `|z - 1 - i| = 2, \ z in C`.

  1. Sketch the circle on the Argand diagram below.  (1 mark)

 

  1. i.  Write the equation of the circle in the form  `(x - a)^2 + (y - b)^2 = c`  and show that the gradient of a tangent to the circle can be expressed as  `(dy)/(dx) = (1 - x)/(y - 1)`.  (2 marks)
  2. ii. Find the gradient of the tangent to the circle where  `x = 2`  in the first quadrant of the complex plane.  (1 mark)
  3. Find the equations of all rays that are perpendicular to the circle in the form  `text(Arg) (z) = alpha`.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
    1.  `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `(-1)/sqrt 3`
  3. `text(Arg) (z) = pi/4; quad text(Arg) (z) = (-3 pi)/4`
Show Worked Solution
a.  

 

b.i.   `(x – 1)^2 + (y – 1)^2` `= 4`
  `2(x – 1) + d/(dx) ((y – 1)^2)` `= 0`
  `2(x – 1) + 2(y-1)*(dy)/(dx)` `= 0`
  `2 (y-1)*(dy)/(dx)` `= -2(x – 1)`
  `(dy)/(dx)` `= (-(x – 1))/(y-1)`
    `= (1 – x)/(y – 1)`

 

b.ii.   `(2 – 1)^2 + (y – 1)^2` `= 4`
  `1 + (y – 1)^2` `= 4`
  `(y – 1)^2` `= 3`
  `y` `= 1 + sqrt 3\ \ \ (y > 0)`
     
  `(dy)/(dx)|_{(2, 1 + sqrt 3)}` `= (1 – 2)/(1 + sqrt 3 – 1`
    `= (-1)/sqrt 3`

 

c.   `P (0, 0) quad C(1, 1)`
  `-> y = x`
  `:. alpha = pi/4, quad (-3 pi)/4` 

 
`text(Arg) (z) = pi/4`

`text(Arg) (z) = (-3 pi)/4`

Calculus, SPEC1-NHT 2018 VCAA 7

  1. Find  `d/(dx) ((1 - x^2)^(1/2))`.  (2 marks)
  2. Hence, find the length of the curve specified by  `y = sqrt (1 - x^2)`  from  `x = 1/2`  to  `x = sqrt 3/2`.

     

    Give your answer in the form  `k pi, k in R`.  (2 marks)

Show Answers Only
  1. `(-x)/sqrt(1 – x^2)`
  2. `pi/6`
Show Worked Solution

a.  `d/(dx) ((1 – x^2)^(1/2))`

`= (1/2 (-2x) (1 – x^2)^(-1/2))`

`= -x(1 – x^2)^(-1/2)`

`= (-x)/sqrt (1 – x^2)`

 

b.   `l` `= int_(1/2)^(sqrt 3/2) sqrt(1 + ((-x)/sqrt(1 – x^2))^2)\ dx`
    `= int_(1/2)^(sqrt 3/2) sqrt((1 – x^2 + x^2)/(1 – x^2))\ dx`
    `= int_(1/2)^(sqrt 3/2) sqrt(1/(1 – x^2))\ dx`
    `= int_(1/2)^(sqrt 3/2) 1/sqrt(1 – x^2)\ dx`
    `= [sin^(-1) (x)]_(1/2)^(sqrt 3/2)`
    `= sin^(-1) (sqrt 3/2) – sin^(-1) (1/2)`
    `= pi/3 – pi/6`
    `= pi/6`

Vectors, SPEC2 2018 VCAA 11 MC

Consider the vectors given by  `underset ~a = m underset ~i + underset ~j`  and  `underset ~b = underset ~i + m underset ~j`, where  `m in R`.

If the acute angle between  `underset ~a`  and  `underset ~b`  is 30°, then `m` equals

  1. `sqrt 2 +- 1`
  2. `2 +- sqrt 3`
  3. `sqrt 3, 1/sqrt 3`
  4. `sqrt 3/(4 - sqrt 3)`
  5. `sqrt 39/13` 
Show Answers Only

`C`

Show Worked Solution
`underset ~a *underset ~b` `= m + m = 2m`
`underset ~a *underset ~b` `= |underset ~a||underset ~b| cos 30^@`
  `= sqrt(m^2 + 1) *sqrt(1 + m^2) *cos 30^@`
  `= {(m^2 + 1) sqrt 3}/2`

 

`{(m^2 + 1) sqrt 3}/2` `=2m`  
`m^2 sqrt 3 + sqrt 3` `=4m`  
`m^2 sqrt 3 – 4m + sqrt 3` `=0`  
`(sqrt 3 m)^2 – 4(sqrt 3 m) + 3` `=0`  
`(sqrt 3 m)^2 – 4(sqrt 3 m) + 2^2 – 1` `=0`  
`(sqrt 3 m – 2)^2 – 1` `=0`  
`sqrt 3 m – 2` `= +-1`  
`sqrt 3 m` `= 2 +- 1`  

 
`:. m = (2 +- 1)/sqrt 3 = 3/sqrt 3 or 1/sqrt 3`

`= sqrt 3, 1/sqrt 3`

`=>  C`

Calculus, SPEC2 2018 VCAA 7 MC

A curve is described parametrically by  `x = sin(2t), y = 2 cos (t)`  for  `0 <= t <= 2pi`.

The length of the curve is closest to

A.    9.2

B.    9.5

C.  12.2

D.  12.5

E.  38.3 

Show Answers Only

`C`

Show Worked Solution

`x prime (t) = 2 cos (2t)`

`y prime (t) = -2 sin(t)`

`ℓ` `= int_0^(2 pi) sqrt{(2 cos(2t))^2 + (-2 sin (t))^2 dt}`
  `~~ 12.2`

 
`=>  C`

Calculus, SPEC1 2018 VCAA 9

A curve is specified parametrically by  `underset ~r(t) = sec(t) underset ~i + sqrt 2/2 tan(t) underset ~j, \ t in R`.

  1.  Show that the cartesian equation of the curve is  `x^2 - 2y^2 = 1`.  (2 marks)
  2.  Find the `x`-coordinates of the points of intersection of the curve  `x^2 - 2y^2 = 1`  and the line  `y = x - 1`.  (1 mark)
  3.  Find the volume of the solid of revolution formed when the region bounded by the curve and the line is rotated about the `x`-axis.  (2 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = 1 or x = 3`
  3. `(2 pi)/3`
Show Worked Solution

a.     `x = sec(t), qquad y = sqrt 2/2 tan(t)`

`x^2 = sec^2(t), qquad y^2 = 1/2 tan^2(t)`

`x^2 = sec^2(t), qquad 2y^2 = tan^2(t)`

`1 + tan^2(t)` `= sec^2(t)`
`1 + 2y^2` `= x^2`
`:.x^2 – 2y^2` `=1\ \ text(.. as required)`

 

b.    `x^2 – 2(x – 1)^2` `= 1`
  `x^2 – 2(x^2 – 2x + 1)` `=1`
  `x^2 – 2x^2 + 4x – 2` `=1`
  `-x^2 + 4x – 2 – 1` `=0`
  `x^2 – 4x + 3` `=0`
  `(x – 3) (x – 1)` `=0`

 
`:. x = 1 or x = 3`

♦♦ Mean mark 30%.

 

c.   `x^2 – {:2y_1:}^2` `=1`
  `{:2y_1:}^2` `=x^2 – 1`
  `{:y_1:}^2` `= (x^2 – 1)/2`
  `{:y_2:}^2` `= (x – 1)^2`

 

`V` `=pi int_1^3 {:y_1:}^2 – {:y_2:}^2 \ dx`
  `= pi int_1^3 (x^2 – 1)/2 – (x-1)^2\ dx`
  `= pi [x^3/6 – x/2 – (x – 1)^3/3]_1^3`
  `= pi [(3^3/6 – 3/2 – 2^3/3) – (1^3/6 – 1/2 – 0)]`
  `= pi (9/2 – 3/2 – 8/3 – 1/6 + 1/2)`
  `= pi (7/2 – 8/3 – 1/6)`
  `= pi ((21 – 16 – 1)/6)`
  `= (2 pi)/3`

Trigonometry, SPEC1 2018 VCAA 7

Given that  \(\cot (2 x)+\dfrac{1}{2}\, \tan (x)=a \cot (x)\), use a suitable double angle formula to find the value of \(a , a\) in RR. (3 marks)

Show Answers Only

\(a=\dfrac{1}{2}\)

Show Worked Solution

\(\dfrac{1}{\tan (2 x)}+\dfrac{\tan (x)}{2}=\dfrac{a}{\tan (x)}, \quad \tan (x) \neq 0\)

\(\dfrac{1-\tan ^2(x)}{2 \tan (x)}+\dfrac{\tan (x)}{2}=\dfrac{a}{\tan (x)}\)

 
\(\text{If \(\tan(x) \neq 0\):}\)

\(\dfrac{1-\tan ^2(x)+\tan ^2(x)}{2 \tan (x)}\) \(=\dfrac{a}{\tan (x)}\)
\(1\) \(=2 a\)
\(\therefore a\) \(=\dfrac{1}{2}\)

Statistics, SPEC1 2018 VCAA 4

`X` and `Y` are independent random variables. The mean and the variance of `X` are both 2, while the mean and the variance of `Y` are 2 and 4 respectively.

Given that `a` and `b` are integers, find the values of `a` and `b` if the mean and the variance of  `aX + bY`  are 10 and 44 respectively.  (4 marks)

Show Answers Only

`a = 2, qquad b = 3`

Show Worked Solution

`E(aX + bY) = aE(X) + bE(Y) = 10`

`10` `= 2a + 2b`
`5` `= a + b`
`b` `= 5-a \ \ …\ (1)`

 
`text(Var) (aX + bY) = a^2 text(Var) (X) + b^2 text(Var) (Y) = 44`

`44` `= 2a^2 + 4b^2`
`22` `= a^2 + 2b^2 \ \ …\ (2)`

 
`text{Substitute (1) into (2)}`

`22 = a^2 + 2 (5-a)^2`

`22 = a^2 + 2(25-10 a + a^2)`

`22 = 3a^2-20a + 50`

`0 = 3a^2-20a + 28`

`0 = (3a-14)(a-2)`

`a = 14/3 or a = 2`

`:. a = 2, \ b = 3,\ \ \ (a ∈ ZZ)`

Calculus, SPEC1 2018 VCAA 3

Find the gradient of the curve with equation  `2x^2 sin(y) + xy = pi^2/18`  at the point  `(pi/6, pi/6)`.

Give your answer in the form  `a/(pi sqrt b + c)`, where `a, b` and `c` are integers.  (4 marks)

Show Answers Only

`m = (-18)/(pi sqrt 3 + 6)`

Show Worked Solution

`d/(dx) (2x^2 sin (y)) + d/(dx) (xy) = d/(dx) (pi^2/18)`

`4x sin(y) + 2x^2 cos(y) (dy)/(dx) + y + x (dy)/(dx)` `=0`  
`(4 pi)/6 sin (pi/6) + (2 pi^2)/36 cos (pi/6) m + pi/6 + pi/6 m` `=0`  
`(2 pi)/3 xx 1/2 + pi^2/18 xx sqrt 3/2 m + pi/6 + pi/6 m` `=0`  
`pi/3 + pi/6 + ((pi^2 sqrt 3)/36 + pi/6) m` `=0`  
`pi/2 + pi/36 (pi sqrt 3 + 6) m` `=0`  
`(pi m)/36 (pi sqrt 3 + 6)` `= (-pi)/2`
`m (pi sqrt 3 + 6)` `= -18`
`:. m` `= (-18)/(pi sqrt 3 + 6)`

Complex Numbers, SPEC1 2018 VCAA 2

  1. Show that  `1 + i = sqrt 2\ text(cis)(pi/4)`.  (1 mark)
  2. Evaluate  `(sqrt 3 - i)^10/(1 + i)^12`, giving your answer in the form  `a + bi`, where  `a, b ∈ R`.  (3 marks)
Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-8 – 8 sqrt 3 i`
Show Worked Solution
a.   `r` `= sqrt(1^2 + 1^2)`
    `= sqrt 2`
  `theta` `= tan^(-1) (1/1) = pi/4`
     
  `:. 1 + i` `= sqrt 2\ text(cis)(pi/4)`

 

b.   
 `r_2` `= sqrt((sqrt 3)^2 + (-1)^2)`
    `= sqrt (3 + 1)`
    `= 2`

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

`sqrt 3 – i` `= 2\ text(cis)(- pi/6)`
`(sqrt 3 – i)^10` `= 2^10\ text(cis) ((-10 pi)/6)`
   
`=>(1 + i)^12` `= (sqrt 2)^12text(cis)((12pi)/4)`
  `=2^6 text(cis)(3pi)`

 

`:. (sqrt 3 – i)^10/(1 + i)^12` `= (2^10 text(cis)((-5pi)/3))/(2^6\text(cis)(3pi))`
  `= 2^4 text(cis)((-5pi)/3 – (9pi)/3)`
  `= 16 text(cis)((-14pi)/3)`
  `= 16 text(cis) ((-2pi)/3)`
  `= 16(-1/2 + (-sqrt3)/2 i)`
  `= -8 – 8 sqrt 3 i`