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Measurement, STD2 M1 SM-Bank 26 MC

During a flood, 12.5 hectares of land was covered by water to a depth of 30 cm.

How many kilolitres of water covered the land?

(1 hectare = 10 000 m² and 1 m³ = 1000 L)

  1. 3.75 kL
  2. 37.5 kL
  3. 37 500 kL
  4. 37 500 000 kL
Show Answers Only

`=>\ text(C)`

Show Worked Solution

`text(Volume of water in m³:)`

`text(Vol)` `= Ah`
  `= 12.5 xx 10\ 000 xx 30\ text(cm)`
  `= 12.5 xx 10\ 000 xx 0.3`
  `= 37\ 500 \ text(m³)`
  `= 37\ 500\ text(kL)`

 
`=>\ text(C)`

Algebra, SPEC2 2018 VCAA 2 MC

Consider the function  `f` with rule  `f(x) = 1/sqrt(sin^(-1)(cx + d))`, where  `c, d in R`  and  `c > 0`.

The domain of  `f` is

A.  `x > -d/c`

B.  `-d/c < x <= (1 - d)/c`

C.  `(-1 - d)/c <= x <= (1 - d)/c`

D. `x in R\ text(\) {-d/c}`

E.  `x in R` 

Show Answers Only

`B`

Show Worked Solution
`cx + d` `in [-1, 1]`
`cx` `in [-1-d, 1-d]`
`x` `in [(-1-d)/c, (1 – d)/c]\ \ …\ (1)`

 

`sin^(-1) (cx + d)` `> 0`
`cx + d` `>0`
`cx` `> -d`
`x` `> -d/c\ \ …\ (2)`

 

`(1) nn (2):`

`x in (-d/c, (1 – d)/c]`

 
`=>  B`

Algebra, STD2 A4 EQ-Bank 8

Two friends, Sequoia and Raven, sold organic chapsticks at the the local market.

Sequoia sold her chapsticks for $4 and Raven sold hers for $3 each. In the first hour, their total combined sales were $20.

If Sequoia sold `x` chapsticks and Raven sold `y` chapsticks, then the following equation can be formed:

`4x + 3y = 20`

In the first hour, the friends sold a total of 6 chapsticks between them.

Find the number of chapsticks each of the friends sold during this time by forming a second equation and solving the simultaneous equations graphically.  (5 marks)

Show Answers Only

`text(Sequoia sold 2 and Raven sold 4.)`

Show Worked Solution
`text(Graphing)\ \ 4x + 3y` `= 20`
`y` `= −4/3x + 20/3`

 
`ytext(-intercept) = (0, 20/3)`

`xtext(-intercept)= (5, 0)`

`text(Gradient)\ = -4/3`
 

`text(S)text(econd equation:)`

`x + y = 6`

`text(From the graph,)`

`text(Sequoia sold 2 and Raven sold 4.)`

Algebra, STD2 A4 SM-Bank 7

The graph of the line  `y = 4 - x`  is shown.
 


 

By graphing  `y = 2x + 1`  on the grid provided, find the point of intersection of  `y = 4 - x`  and  `y = 2x + 1`.  (3 marks)

Show Answers Only

`(1, 3)`

Show Worked Solution

`text(Graphing)\ y = 2x + 1 :`

`ytext(-intercept = 1)`

`text(Gradient = 2)`
 

 
`:.\ text{Point of intersection is (1, 3).}`

Algebra, STD2 A4 SM-Bank 6

A student was asked to solve the following simultaneous equations.

`y = 2x - 8`

`x - 4y + 3 = 0`

After graphing the equations, the student found the point of intersection to be `(5,2)`?

Is the student correct? Support your answer with calculations.  (2 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text{Substitute (5, 2) into}\ \ y = 2x – 8`

`text(LHS) = y = 2`

`text(RHS) = 2(5) – 8 = 2`

`:.\ text(LHS = RHS)`
 

`text{Substitute (5, 2) into}\ \ x – 4y + 3 = 0`

`text(LHS)` `= 5 – 4(2) + 3`
  `= 0`
  `=\ text(RHS)`

 
`=> (5,2)\ text(satisfies both equations.)`

`:.\ text(Student is correct.)`

Probability, STD2 S2 SM-Bank 1

A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 red tokens and 3 black tokens. The player keeps the two tokens drawn.

  1.  Complete the probability tree by writing the missing probabilities in the boxes.  (2 marks)
     
     
       
     
  2.  What is the probability that a player draws at least one red token. Give your answer in exact form.  (2 marks)

Show Answers Only
  1.  
  2. `187/190`
Show Worked Solution
i.   

 

ii.   `P(text(at least one red))`

`= 1 – P(BB)`

`= 1 – 3/20 · 2/19`

`= 187/190`

Financial Maths, STD2 F4 SM-Bank 1

An investment fund purchases 4500 shares of Bank ABC for a total cost of $274 500 (ignore any transaction costs).

The investment fund is paid a divided of $3.66 per share in the first year.

  1. What was the purchase price of 1 share?  (1 mark)
  2. Calculate the divided yield.  (1 mark)
Show Answers Only
  1. `$61`
  2. `text(6%)`
Show Worked Solution

i.   `text(Price per share)`

`= (274\ 500)/4500`

`= $61`
 

ii.   `text(Dividend Yield)`

`= text(Dividend)/text(Share Price) xx 100`

`= 3.66/61 xx 100`

`= 6text(%)`

Functions, 2ADV F2 SM-Bank 1

  1.  Draw the graph  `y = ln x`.  (1 mark)

  2.  Explain how the above graph can be transformed to produce the graph
     
             `y = 3ln(x + 2)`
     
    and sketch the graph, clearly identifying all intercepts.  (3 marks)

Show Answers Only

  1.  

  2.  
Show Worked Solution

i.

 

ii.   `text(Transforming)\ \ y = ln x => \ y = ln(x + 2)`

`y = ln x\ \ =>\ text(shift 2 units to left.)`
 

`text(Transforming)\ \ y = ln(x + 2)\ \ text(to)\ \ y = 3ln(x + 2)`

`=>\ text(increase each)\ y\ text(value by a factor of 3)`
 

Calculus, 2ADV C1 SM-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = 2t^3 - t^2 - 3t + 11`     when   `t >= 0`

  1.  Calculate the velocity when  `t = 2`.  (1 mark)

  2.  When is the particle stationary?  (2 marks)

Show Answers Only
  1.  `17\ text(ms)^(−1)`
  2.  `(1 + sqrt19)/6`
Show Worked Solution

i.   `x =2t^3 – t^2 – 3t + 11` 

`v = (dx)/(dt) = 6t^2 – 2t – 3`

 
`text(When)\ t = 2,`

`v` `= 6 xx 2^2 – 2 · 2 – 3`
  `= 17\ text(ms)^(−1)`

 

ii.   `text(Particle is stationary when)\ \ v = 0`

`6t^2 – 2t – 3` `= 0`
`:. t` `= (2 ±sqrt((−2)^2 – 4 · 6 · (−3)))/12`
  `= (2 ± sqrt76)/12`
  `= (1 ± sqrt19)/6`
  `= (1 + sqrt19)/6 qquad(t >= 0)`

Calculus, 2ADV C1 SM-Bank 2

  1.  Find the equations of the tangents to the curve  `y = x^2 - 3x`  at the points where the curve cuts the `x`-axis.  (2 marks)

  2.  Where do the tangents intersect?  (2 marks)

Show Answers Only
  1. `y = −3x`
    `y = 3x – 9`
  2. `(3/2, −9/4)`
Show Worked Solution
i.   `y` `= x^2 – 3x`
  `= x(x – 3)`

 
`text(Cuts)\ xtext(-axis at)\ \ x = 0\ \ text(or)\ \ x = 3`
 

`(dy)/(dx) = 2x – 3`

 
`text(At)\ \ x = 0 \ => \ (dy)/(dx) = -3`

`T_1\ text(has)\ \ m = −3,\ text{through (0, 0)}`

`y – 0` `= -3(x – 0)`
`y` `= -3x`

  

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 3`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y – 0` `= 3(x – 3)`
`y` `= 3x – 9`

 

ii.   `text(Intersection occurs when:)`

`3x – 9` `= -3x`
`6x` `= 9`
`x` `= 3/2`

  

`y = -3 xx 3/2 = −9/2`

`:.\ text(Intersection at)\ \ (3/2, −9/2)`

Trigonometry, 2ADV T2 SM-Bank 33

Given  `sectheta = −37/12`  for  `0 < theta < pi`,

find the exact value of  `text(cosec)\ theta`.   (2 marks)

Show Answers Only

`37/35`

Show Worked Solution

`sec\ theta = -37/12 => cos\ theta = -12/37`

`text(Graphically:)`

`x=sqrt(37^2-12^2)=35`
 

`:.\ text(cosec)\ theta` `= 1/(sintheta)`
  `= 1/(35/37)`
  `= 37/35`

Trigonometry, 2ADV T2 SM-Bank 32

Express  `5cot^2 x - 2text(cosec)\ x + 2`  in terms of  `text(cosec)\ x`  and hence solve

`5cot^2 x - 2text(cosec)\ x + 2 = 0`  for  `0 < x < 2pi`.  (3 marks)

Show Answers Only

`x = pi/2`

Show Worked Solution
`cot^2 x` `= (cos^2 x)/(sin^2 x)`
  `= (1 – sin^2 x)/(sin^2 x)`
  `= text(cosec)^2 x – 1`

 

`5cot^2 x – 2text(cosec)\ x + 2` `= 0`
`5(text(cosec)^2 x – 1) – 2text(cosec)\ x + 2` `= 0`
`5text(cosec)^2 x – 2text(cosec)\ x – 3` `= 0`
`(5text(cosec)\ x + 3)(text(cosec)\ x – 1)` `= 0`
`text(cosec)\ x` `= −3/5` `text(cosec)\ x` `= 1`
`sinx` `= −5/3` `sinx` `= 1`
`(text(no solution))` `x` `= pi/2`

 
`:. x = pi/2`

Trigonometry, 2ADV T2 SM-Bank 31

Given  `cottheta = −24/7`  for  `−pi/2 < theta < pi/2`, find the exact value of

  1.  `sectheta`   (2 marks)

  2.  `sintheta`   (1 mark)

Show Answers Only
  1. `25/24`
  2. `-7/25`
Show Worked Solution

i.   `cot theta = −24/7\ \ =>\ \ tan theta=– 7/24`

`text(Graphically, given)\ −pi/2 < theta < pi/2:`

`x= sqrt(24^2 + 7^2)=25`
 

`sectheta` `= 1/(costheta)`
  `= 1/(24/25)`
  `= 25/24`

 

ii.   `sintheta = −7/25`

Functions, 2ADV F1 EQ-Bank 27

The stopping distance of a car on a certain road, once the brakes are applied, is directly proportional to the square of the speed of the car when the brakes are first applied.

A car travelling at 70 km/h takes 58.8 metres to stop.

How far does it take to stop if it is travelling at 105 km/h?  (3 marks)

Show Answers Only

`132.3\ text(metres)`

Show Worked Solution

`text(Let)\ \ d\ text(= stopping distance)`

`d \prop s^2\ \ =>\ \ d = ks^2`
 

`text(Find)\ k,`

`58.8` `= k xx 70^2`
`k` `= 58.8/(70^2)= 0.012`

 
`text(Find)\ \ d\ \ text(when)\ \ s = 105:`

`d` `= 0.012 xx 105^2`
  `= 132.3\ text(metres)`

Functions, 2ADV F1 EQ-Bank 26

Fuifui finds that for Giant moray eels, the mass of an eel is directly proportional to the cube of its length.

An eel of this species has a length of 25 cm and a mass of 4350 grams.

What is the expected length of a Giant moray eel with a mass of 6.2 kg? Give your answer to one decimal place.  (3 marks)

Show Answers Only

`28.1\ text{cm}`

Show Worked Solution

`text(Mass) prop text(length)^3`

`m = kl^3`
 

`text(Find)\ k:`

`4350` `= k xx 25^3`
`k` `= 4350/25^3`
  `= 0.2784`

 
`text(Find)\ \ l\ \ text(when)\ \ m = 6200:`

`6200` `= 0.2784 xx l^3`
`l^3` `= 6200/0.2784`
`:. l` `= 28.13…`
  `= 28.1\ text{cm  (to 1 d.p.)}`

Functions, 2ADV F1 SM-Bank 25

Damon owns a swim school and purchased a new pool pump for $3250.

He writes down the value of the pool pump by 8% of the original price each year.

  1.  Construct a function to represent the value of the pool pump after `t` years.   (1 mark)

  2.  Draw the graph of the function and state its domain and range.   (2 marks)

Show Answers Only
  1. `text(Value) = 3259 – 260t`
  2. `text(Domain)\ {t: 0 <= t <= 12.5}`
    `text(Range)\ {y: 0 <= y <= 3250}`
Show Worked Solution
i.   `text(Depreciation each year)` `= 8text(%) xx 3250`
  `= $260`

 
`:.\ text(Value) = 3250 – 260t`
 

ii.   

`text(Find)\ \ t\ \ text(when value = 0)`

`3250 – 260t` `= 0`
`t` `= 3250/260`
  `= 12.5\ text(years)`

 
`text(Domain)\ \ {t: 0 <= t <= 12.5}`

`text(Range)\ \ {y: 0 <= y <= 3250}`

Functions, 2ADV F1 SM-Bank 23

Find the values of `k` for which the expression  `x^2-3x + (4-2k)`  is always positive.  (3 marks)

Show Answers Only

`k < 7/8`

Show Worked Solution

`x^2-3x + (4-2k) > 0`

`x^2-3x + (4-2k) = 0\ \ text(is a concave up parabola)`

`=>\ text{Always positive (no roots) if}\ \ Delta < 0`
 

`b^2-4ac < 0`

`(−3)^2-4 · 1 · (4-2k)` `< 0`
`9-16 + 8k` `< 0`
`8k` `< 7`
`k` `< 7/8`

Functions, 2ADV F1 EQ-Bank 22

Worker A picks a bucket of blueberries in `a` hours. Worker B picks a bucket of blueberries in `b` hours.

  1.  Write an algebraic expression for the fraction of a bucket of blueberries that could be picked in one hour if A and B worked together.  (2 marks)

  2.  What does the reciprocal of this fraction represent?  (1 mark)

Show Answers Only
  1. `(a + b)/(ab)`
  2. `text(The reciprocal represents the number of hours it would)`
  3. `text(take to fill one bucket, with A and B working together.)`
Show Worked Solution

i.    `text(In one hour:)`

COMMENT: Note that the question asks for “a fraction”.

`text(Worker A picks)\ 1/a\ text(bucket.)`

`text(Worker B picks)\ 1/b\ text(bucket.)`
 

`:.\ text(Fraction picked in 1 hour working together)`

`= 1/a + 1/b`

`= (a + b)/(ab)`
 

ii.   `text(The reciprocal represents the number of hours it would)`

`text(take to fill one bucket, with A and B working together.)`

Calculus, 2ADV C1 SM-Bank 11

A particle is moving along the `x`-axis. Its velocity `v` at time `t` is given by

`v = sqrt(20t - 2t^2)`  metres per second

Find the acceleration of the particle when  `t = 4`.

Express your answer as an exact value in its simplest form.  (3 marks)

Show Answers Only

`sqrt3/6\ \ text(ms)^(−2)`

Show Worked Solution

`v = sqrt(20t – 2t^2)`

`alpha` `= (dv)/(dt)`
  `= 1/2 · (20t – 2t^2)^(−1/2) · (20 – 4t)`

 
`text(When)\ \ t = 4,`

`alpha` `= 1/2(20 · 4 – 2 · 4^2)^(−1/2)(20 – 16)`
  `= 2/(sqrt48)`
  `= 2/(4sqrt3) xx sqrt3/sqrt3`
  `= sqrt3/6\ \ text(ms)^(−2)`

Trigonometry, EXT1 T2 EQ-Bank 7

Find the exact value of  `cos((11pi)/12)`.  (2 marks)

Show Answers Only

`−((sqrt2 + sqrt6))/4`

Show Worked Solution
`cos((11pi)/12)` `= cos((2pi)/3 + pi/4)`
  `= cos((2pi)/3) · cos(pi/4) – sin((2pi)/3) · sin(pi/4)`
  `= cos(pi – pi/3) · 1/sqrt2 – sin (pi – pi/3) · 1/sqrt2`
  `= −1/2 · 1/sqrt2 – sqrt3/2 · 1/sqrt2`
  `= −((1 + sqrt3))/(2sqrt2)`
  `= −((sqrt2 + sqrt6))/4`

Trigonometry, EXT1 T2 EQ-Bank 5

If  `sintheta = −4/6`  and  `−pi/2 < theta < pi/2`,

determine the exact value of  `costheta`  in its simplest form.  (2 marks)

Show Answers Only

`sqrt5/3`

Show Worked Solution

`text(Consider the angle graphically:)`

COMMENT: Pay careful attention to the range of  `theta`.

`text(S)text(ince)\ sintheta\ text(is negative) => underbrace(text(4th quadrant))_(−pi/2 <\ theta\ < pi/2)`

`text(Using Pythagoras:)`

`x^2 = 6^2-4^2`

`x = sqrt20 = 2sqrt5`

`:. costheta= (2sqrt5)/6= sqrt5/3`

Trigonometry, EXT1 T2 SM-Bank 2

Find the exact value of  `cos\ pi/8`.  (2 marks)

Show Answers Only

`(sqrt(sqrt2 + 2))/2`

Show Worked Solution

`text(Using:)\ \ cos2A = 2cos^2A – 1`

`2cos^2\ pi/8 – 1` `= cos\ pi/4`
`2cos^2\ pi/8` `= 1/sqrt2 + 1`
`cos^2\ pi/8` `= (1 + sqrt2)/(2sqrt2) xx sqrt2/sqrt2`
  `= (sqrt2 + 2)/4`
`:. cos\ pi/8` `= sqrt((sqrt2 + 2)/4)`
  `= (sqrt(sqrt2 + 2))/2`

Functions, EXT1 F1 SM-Bank 3

A circle has centre `(5,3)` and radius 3.

  1.  Describe, with inequalities, the region that consists of the interior of the circle and more than 2 units above the `x`-axis.  (2 marks)

  2.  Sketch the region.  (1 mark)

Show Answers Only
  1. `(x-5)^2 + (y-3)^2 < 9\ ∩\ y > 2`

  2.  

Show Worked Solution

i.   `text(Equation of circle:)`

`(x-5)^2 + (y-3)^2 = 3^2`
 

`:.\ text(Region is:)`

`(x-5)^2 + (y- 3)^2 < 9\ ∩\ y > 2`

COMMENT: The broken line on the graph represents an excluded boundary.

 

ii.   

Combinatorics, EXT1 A1 EQ-Bank 3

Find the coefficient of  `x^4`  in the expansion of  `(x^2 - 3/x)^5`.  (2 marks)

Show Answers Only

`90`

Show Worked Solution

`text(General term:)`

`T_k` `= \ ^5C_k(x^2)^(5 – k) · (−3/x)^k`
  `= \ ^5C_k · x^(10 – 2k)(−3)^k · x^(−k)`
  `= \ ^5C_k · x^(10 – 3k) · (−3)^k`

 
`text(Coefficient of)\ \ x^4\ \ text(occurs when)`

`10 – 3k` `= 4`
`3k` `= 6`
`k` `= 2`

 
`:.\ text(Coefficient of)\ \ x^4`

`= \ ^5C_2·(−3)^2`

`= 90`

Combinatorics, EXT1 A1 SM-Bank 1

`(1-2sqrt2)^6 = x + ysqrt2`

Evaluate the value of `x` and `y`.  (2 marks)

Show Answers Only

`x = 1593, \ y =-1100`

Show Worked Solution

`text(Using the binomial expansion:)`

`(1-2sqrt2)^6`

`= \ ^6C_0 + \ ^6C_1(-2sqrt2) + \ ^6C_2(-2sqrt2)^2 + \ ^6C_3(-2sqrt2)^3 + \ ^6C_4(-2sqrt2)^4`

`+ \ ^6C_5(-2sqrt2)^5 + \ ^6C_6(-2sqrt2)^6`
 

`= 1-12sqrt2 + 120 -320sqrt2 + 960-768sqrt2 + 512`

`= 1593-1100sqrt2`
 

`:.x = 1593, \ y =-1100`

Trigonometry, 2ADV T1 EQ-Bank 2

Determine all possible dimensions for triangle  `ABC`  given  `AB = 6.2\ text(cm)`, `angleABC = 35°`  and  `AC = 4.1`.

Give all dimensions correct to one decimal place.  (3 marks)

Show Answers Only

`text(7.1 cm, 6.2 cm, 4.1 cm or)`

`text(3.0 cm, 6.2 cm, 4.1 cm.)`

Show Worked Solution

`text(Using the sine rule:)`

`(sinangleACB)/6.2` `= (sin35^@)/4.1`
`sinangleACB` `= (6.2 xx sin35^@)/4.1`
  `= 0.8673…`
`angleACB` `= 60.15…^@\ text(or)\ 119.84…^@`

  
`text(If)\ \ angleACB = 60.15^@,`

`angleBAC = 180 – (35 + 60.15) = 84.85^@`
 

`(BC)/(sin84.85)` `= 4.1/(sin35^@)`
`BC` `= 7.11…`
  `= 7.1\ text(cm)`

 
`text(If)\ \ angleACB = 119.85^@,`

`angleBAC = 180 – (35 + 119.85) = 25.15^@`
 

`(BC)/(sin25.15)` `= 4.1/(sin35^@)`
`BC` `= 3.03…`
  `= 3.0\ text(cm)`

 
`:.\ text(Possible dimensions are:)`

`text(7.1 cm, 6.2 cm, 4.1 cm or)`

`text(3.0 cm, 6.2 cm, 4.1 cm.)`

Calculus, EXT1 C3 SM-Bank 1

The region enclosed by the semicircle  `y = sqrt(1 - x^2)`  and the `x`-axis is to be divided into two pieces by the line  `x = h`, when  `0 <= h <1`.
 


 

The two pieces are rotated about the `x`-axis to form solids of revolution. The value of `h` is chosen so that the volumes of the solids are in the ratio `2 : 1`.

Show that `h` satisfies the equation  `3h^3 - 9h + 2 = 0`.  (3 marks)

Show Answers Only

`text(Show Worked Solution)`

Show Worked Solution

(i)   `text(Volume of smaller solid)`

`= pi int_h^1 (sqrt(1 – x^2))^2\ dx`

`= pi int_h^1 1 – x^2\ dx`

`= pi[x – (x^3)/3]_h^1`

`= pi[(1 – 1/3) – (h – (h^3)/3)]`

`= pi(2/3 – h + (h^3)/3)`

 
`text(S)text(ince smaller solid is)\ 1/3\ text(volume of sphere,)`

`pi(2/3 – h + (h^3)/3)` `= 1/3 xx 4/3 · pi · 1^3`
`(h^3)/3 – h + 2/3` `= 4/9`
`3h^3 – 9h + 6` `= 4`
`:. 3h^3 – 9h + 2` `= 0\ \ text(… as required)`

Calculus, SPEC2 2017 VCAA 7 MC

With a suitable substitution `int_1^2 x^2 sqrt(2 - x)\ dx` can be expressed as

  1. `−int_1^2(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
  2. `int_1^2(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
  3. `int_0^1(−4u^(1/2) + 4u^(3/2) - u^(5/2)) du`
  4. `−int_1^0(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
  5. `int_1^0(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
Show Answers Only

`D`

Show Worked Solution
`u` `= 2 – x`
`x` `= 2 – u\ \ =>\ \ x^2 = (2 – u)^2`

 
`(du)/(dx) = −1\ \ =>\ \ du = – dx`

`u(2)` `= 0`
`u(1)` `= 2 – 1=1`

 
`int_1^2 – (2 – u)^2sqrtu\ du`

`= −int_1^0(2 – u)^2 u^(1/2)\ du`

`= −int_1^0(4 – 4u + u^2) u^(1/2)\ du`

`= −int_1^0 (4u^(1/2) – 4u^(3/2) + u^(5/2))\ du`

 
`=>D`

Complex Numbers, SPEC2 2017 VCAA 4 MC

The solutions to  `z^n = 1 + i, \ n ∈ Z^+`  are given by

  1. `2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ R`
  2. `2^(1/n)text(cis)(pi/(4n) + 2pik), k ∈ Z`
  3. `2^(1/(2n))text(cis)(pi/4 + (2pik)/n), k ∈ R`
  4. `2^(1/n)text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
  5. `2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
Show Answers Only

`E`

Show Worked Solution
`z^n` `= sqrt(1^2 + 1^2)text(cis)(tan^(−1)(1/1))`
  `= sqrt2text(cis)(pi/4)`
`z_1` `= (2^(1/2))^(1/n)text(cis)(pi/(4n))`
`z_k` `= (2^(1/2))^(1/n)text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
  `= 2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`

 
`=>E`

Functions, EXT1 F1 2010 HSC 3b*

Let  `f(x) = e^(-x^2)`.  The diagram shows the graph  `y = f(x)`.
 

 Inverse Functions, EXT1 2010 HSC 3b

  1. The graph has two points of inflection.  

     

    Find the  `x`  coordinates of these points.   (3 marks)

  2. Explain why the domain of  `f(x)`  must be restricted if  `f(x)`  is to have an inverse function.    (1 mark)

  3. Find a formula for  `f^(-1) (x)`  if the domain of  `f(x)`  is restricted to  `x ≥ 0`.   (2 marks)

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

  5. Sketch the curve  `y = f^(-1) (x)`.    (1 mark)

Show Answers Only
  1. `x = +- 1/sqrt2` 
  2. `text(There can only be 1 value of)\ y\ text(for each value of)\ x.`
  3. `f^(-1)x = sqrt(ln(1/x))`
  4. `0 <= x <= 1`
  5.  
  6. Inverse Functions, EXT1 2010 HSC 3b Answer
Show Worked Solution
i.    `y` `= e^(-x^2)`
  `dy/dx` `= -2x * e^(-x^2)`
  `(d^2y)/(dx^2)` `= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
    `= 4x^2 e^(-x^2)\ – 2e^(-x^2)`
    `= 2e^(-x^2) (2x^2\ – 1)`

 

`text(P.I. when)\ \ (d^2y)/(dx^2) = 0`

`2e^(-x^2) (2x^2\ – 1)` `= 0` 
 `2x^2\ – 1` `= 0` 
 `x^2` `= 1/2`
 `x` `= +- 1/sqrt2` 
COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.
`text(When)\ \ ` `x < 1/sqrt2,` `\ (d^2y)/(dx^2) < 0`
  `x > 1/sqrt2,` `\ (d^2y)/(dx^2) > 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = 1/sqrt2`

`text(When)\ \ ` `x < – 1/sqrt2,` `\ (d^2y)/(dx^2) > 0`
  `x > – 1/sqrt2,` `\ (d^2y)/(dx^2) < 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = – 1/sqrt2`

 

ii.   `text(In)\ f(x), text(there are 2 values of)\ y\ text(for)`
  `text(each value of)\ x.`
  `:.\ text(The domain of)\ f(x)\ text(must be restricted)`
  `text(for)\ \ f^(-1) (x)\ text(to exist).`

 

iii.   `y = e^(-x^2)`

`text(Inverse function can be written)` 

`x` `= e^(-y^2),\ \ \ x >= 0`
`lnx` `= ln e^(-y^2)`
`-y^2` `= lnx`
`y^2` `= -lnx`
  `=ln(1/x)`
`y` `= +- sqrt(ln (1/x))`

 

`text(Restricting)\ \ x>=0,\ \ =>y>=0`

`:.  f^(-1) (x)=sqrt(ln (1/x))`
 

♦ Parts (iv) and (v) were poorly answered with mean marks of 39% and 49% respectively.
iv.   `f(0) = e^0 = 1`

`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`

`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`

 

v. 

Inverse Functions, EXT1 2010 HSC 3b Answer

Functions, EXT1 F1 2004 HSC 5b*

The diagram below shows a sketch of the graph of  `y = f(x)`, where  `f(x) = 1/(1 + x^2)`  for  `x ≥ 0`.
 

Inverse Functions, EXT1 2004 HSC 5b

  1. On the same set of axes, sketch the graph of the inverse function,  `y = f^(−1)(x)`.  (1 mark)
  2. State the domain of  `f^(−1)(x)`.  (1 mark)

  3. Find an expression for  `y = f^(−1)(x)`  in terms of  `x`.  (2 marks)

  4. The graphs of  `y = f(x)`  and  `y = f^(−1)(x)`  meet at exactly one point  `P`.

     

    Let  `α`  be the `x`-coordinate of  `P`. Explain why  `α`  is a root of the equation  `x^3 + x − 1 = 0`.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0 < x ≤ 1`
  3. `y = sqrt((1 − x)/x), \ y > 0`
  4. `text(See Worked Solutions)`
Show Worked Solution
i.  

Inverse Functions, EXT1 2004 HSC 5b Answer

ii.   `text(Domain of)\ \ f^(−1)(x)\ text(is)`

`0 < x ≤ 1`
 

iii.  `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\ \ x↔y`

`x` `= 1/(1 + y^2)`
`x(1 + y^2)` `= 1`
`1 + y^2` `= 1/x`
`y^2` `= 1/x − 1`
  `= (1 − x)/x`
`y` `= ± sqrt((1 − x)/x)`

 

`:.y = sqrt((1 − x)/x), \ \ y >= 0`

 

iv.   `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`

`text(i.e. where)`

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

 

`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`

 `text(it is a root of)\ \ \ x^3 + x − 1 = 0`

Functions, 2ADV F1 SM-Bank 8 MC

Let  `f (x) = x^2`

Which one of the following is not true?

A.   `f(xy) = f (x) f (y)`

B.   `f(x) - f(-x) = 0`

C.   `f (2x) = 4 f (x)`

D.   `f (x - y) = f(x) - f(y)`

Show Answers Only

`D`

Show Worked Solution

`text(By trial and error,)`

`text(Consider option)\ D:`

`f(x-y)` `=(x-y)^2`
  `=x^2 -2xy+y^2`
`f(x)-f(y)` `= x^2-y^2`
   
`:.f(x-y)` `!=f(x)-f(y)`

 
`=>D`

Functions, 2ADV F1 SM-Bank 7

Let  `f(x) = log_e(x)`  for  `x>0,`  and  `g (x) = x^2 + 1`  for  all `x`.

  1. Find  `h(x)`, where  `h(x) = f (g(x))`.  (1 mark)

  2. State the domain and range of  `h(x)`.  (2 marks)

  3. Show that  `h(x) + h(−x) = f ((g(x))^2 )`.  (2 marks)
Show Answers Only
  1. `log_e(x^2 + 1)`
  2. `text(Domain)\ (h):\ text(all)\ x`

     

    `text(Range)\ h(x):\ \  h>=0`

  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `h(x)` `= f(x^2 + 1)`
    `= log_e(x^2 + 1)`

 

ii.   `text(Domain)\ (h) =\ text(Domain)\ (g):\ text(all)\ x`

♦♦ Mean mark part (a)(ii) 30%.
`=> x^2 + 1 >= 1`
`=> log_e(x^2 + 1) >= 0`

 
`:.\ text(Range)\ h(x):\ \  h>=0`

 

MARKER’S COMMENT: Many students were unsure of how to present their working in this question. Note the layout in the solution.
iii.   `text(LHS)` `= h(x) + h(−x)`
    `= log_e(x^2 _ 1) + log_e((−x)^2 + 1)`
    `= log_e(x^2 + 1) + log_e(x^2 + 1)`
    `= 2log_e(x^2 + 1)`

 

`text(RHS)` `= f((x^2 + 1)^2)`
  `= 2log_e(x^2 + 1)`

 
`:. h(x) + h(−x) = f((g(x))^2)\ \ text(… as required)`

Functions, 2ADV F1 SM-Bank 3

Let  `f(x) = sqrt(x + 1)`   for   `x>=0`

  1. State the range of  `f(x)`.   (1 mark)

  2. Let  `g(x)=x^2+4x+3`,  where  `x<=c`  and  `c<=0`.
  3. Find the largest possible value of `c` such that the range of `g(x)` is a subset of the domain of `f(x)`.   (2 marks)

Show Answers Only
  1.   `y>= 1`
  2.  `-3`
Show Worked Solution

i.  `text(Sketch of)\ \ f(x):`
 

`:.\ text(Range:)\ \ y>= 1`

 

ii.  `text(Sketch)\ \ g(x) = (x + 1) (x + 3)`

 

 

 

 

 

 

`text(Domain of)\ \ f(x):\ \ x>=0`

`text(Find domain of)\ g(x)\ text(such that range)\ g(x):\ y>=0`

`text(Graphically, this occurs when)\ \ g(x)\ \ text(has domain:)`

`x <= –3\ \ text(and)\ \ x>=-1`
 

`:. c = -3`

Calculus, 2ADV C4 2007* HSC 10a

An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are  `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for  `t >= 6`.
 


 

  1. The object is initially at the origin. During which time(s) is the displacement of the object decreasing?  (1 mark)

  2. If the object travels 7 units in the first 4 seconds, estimate the time at which the object returns to the origin. Justify your answer.  (2 marks)

  3. Sketch the displacement, `x`, as a function of time.  (2 marks)

Show Answers Only
  1. `t > 5\ \ text(seconds)`
  2. `7.2\ \ text(seconds)`
  3.    
Show Worked Solution

i.  `text(Displacement is reducing when the velocity is negative.)`

`:. t > 5\ \ text(seconds)`

 

ii.  `text(At)\ B,\ text(the displacement) = 7\ text(units)`

`text(Considering displacement from)\ B\ text(to)\ D:`

`text(S)text(ince the area below the graph from)`

`B\ text(to)\ C\ text(equals the area above the)`

`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`

`text(in displacement from)\ B\ text(to)\ D.`

 

`text(Considering)\ t >= 6`

`text(Time required to return to origin)`

`t` `= d/v`
  `= 7/5`
  `= 1.4\ \ text(seconds)`

 

`:.\ text(The particle returns to the origin after 7.4 seconds.)`
   

iii.   

Statistics, NAPX-K2-04 SA

Sharnie records the number of items she recycles in one week.

The tally table shows Sharnie's results.
 

 
How many items does Sharnie recycle in total?
Show Answers Only

`25`

Show Worked Solution
`text(Total items)` `= 9 + 12 + 3 + 1`
  `= 25`

Probability, 2ADV S1 2017 MET1 8

For events `A` and `B` from a sample space, `P(A text(|)B) = 1/5`  and  `P(B text(|)A) = 1/4`.  Let  `P(A nn B) = p`.

  1. Find  `P(A)` in terms of `p`.  (1 mark)

  2. Find  `P(A nn B^{′})` in terms of `p`.  (2 marks)

  3. Given that  `P(A uu B) <= 1/5`, state the largest possible interval for `p`.  (2 marks)

Show Answers Only
  1. `P (A) = 4p,\ \ p > 0`
  2. `1-8p`
  3. `0 < p <= 1/40`
Show Worked Solution
i.    `P\ (A)` `=(P\ (A nn B))/(P\ (B text(|) A))`
    `=p/(1/4)`
    `=4p`

 

ii.  `text(Consider the Venn diagram:)`

♦ Mean mark 40%.
MARKER’S COMMENT: The most successful answers used a Venn diagram or table.
 

`P\ (A^{′} nn B^{′}) = 1-8p`

 

iii.  `text(Given)\ P(A uu B) = 8p`

♦ Mean mark 37%.

`=> 0 < 8p <= 1/5`

`:. 0 < p <= 1/40`

Probability, 2ADV S1 2014 MET1 9

Sally aims to walk her dog, Mack, most mornings. If the weather is pleasant, the probability that she will walk Mack is `3/4`, and if the weather is unpleasant, the probability that she will walk Mack is `1/3`.

Assume that pleasant weather on any morning is independent of pleasant weather on any other morning.

  1. In a particular week, the weather was pleasant on Monday morning and unpleasant on Tuesday morning.

     

     

    Find the probability that Sally walked Mack on at least one of these two mornings.  (2 marks)

  2. In the month of April, the probability of pleasant weather in the morning was `5/8`.
    1. Find the probability that on a particular morning in April, Sally walked Mack.  (2 marks)

    2. Using your answer from part b.i., or otherwise, find the probability that on a particular morning in April, the weather was pleasant, given that Sally walked Mack that morning.  (2 marks)

Show Answers Only
  1. `5/6`
  2. i. `19/32`
     
    ii. `15/19`
Show Worked Solution
a.    `Ptext{(at least 1 walk)}` `= 1 – Ptext{(no walk)}`
    `= 1 – 1/4 xx 2/3`
    `= 5/6`

 

b.i.   `text(Construct tree diagram:)`
 

met1-2014-vcaa-q9-answer1 
 

`P(PW) + P(P′W)` `= 5/8 xx 3/4 + 3/8 xx 1/3`
  `= 19/32`

 

♦ Part (b)(ii) mean mark 38%.
b.ii.    `P(P | W)` `= (P(P ∩ W))/(P(W))`
    `= (5/8 xx 3/4)/(19/32)`
    `= 15/32 xx 32/19`
    `= 15/19`

Probability, 2ADV S1 2011 MET1 8

Two events, `A` and `B`, are such that  `P(A) = 3/5`  and  `P(B) = 1/4.`

If  `A^{′}` denotes the compliment of `A`, calculate  `P (A^{′} nn B)` when

  1. `P (A uu B) = 3/4`  (2 marks)

  2. `A` and `B` are mutually exclusive.  (1 mark)

Show Answers Only
  1. `3/20`
  2. `1/4`
Show Worked Solution

i.   `text(Sketch Venn Diagram)`

vcaa-2011-meth-8i

`P (A uu B)` `= P (A) + P (B)-P (A nn B)`
`3/4` `= 3/5 + 1/4-P (A nn B)`
`P (A nn B)` `= 1/10`

 

 `:.\ P(A^{′} nn B) = 1/4-1/10 = 3/20`

 

ii.   vcaa-2011-meth-8ii

`P (A∩ B)=0\ \ text{(mutually exclusive)},`

`:.\ P (A^{′} nn B) = P (B) = 1/4`

Probability, 2ADV S1 2007 MET1 11

There is a daily flight from Paradise Island to Melbourne. The probability of the flight departing on time, given that there is fine weather on the island, is 0.8, and the probability of the flight departing on time, given that the weather on the island is not fine, is 0.6.

In March the probability of a day being fine is 0.4.

Find the probability that on a particular day in March

  1. the flight from Paradise Island departs on time  (2 marks)

  2. the weather is fine on Paradise Island, given that the flight departs on time.  (2 marks)

Show Answers Only
  1. `0.68`
  2. `8/17`
Show Worked Solution
i.   

 
`P(FT) +\ P(F^{′}T)`

`= 0.4 xx 0.8 + 0.6 xx 0.6`

`= 0.32 + 0.36`

`= 0.68`

 

ii.   `text(Conditional probability:)`

♦♦ Mean mark 29%.
MARKER’S COMMENT: Students continue to struggle with conditional probability. Attention required here.
`P(F\ text(|)\ T)` `= (P(F ∩ T))/(P(T))`
  `= 0.32/0.68`

 

`:. P(F\ text(|)\ T) = 8/17`

Probability, 2ADV S1 2015 MET1 8

For events `A` and `B` from a sample space, `P(A | B) = 3/4`  and  `P(B) = 1/3`.

  1.  Calculate  `P(A ∩ B)`.  (1 mark)

  2.  Calculate  `P(A^{′} ∩ B)`, where `A^{′}` denotes the complement of `A`.  (1 mark)

  3.  If events `A` and `B` are independent, calculate  `P(A ∪ B)`.  (1 mark)

Show Answers Only
  1. `1/4`
  2. `1/12`
  3. `5/6`
Show Worked Solution

i.   `text(Using Conditional Probability:)`

`P(A | B)` `= (P(A ∩ B))/(P(B))`
`3/4` `= (P(A ∩ B))/(1/3)`
`:. P(A ∩ B)` `= 1/4`

 

ii.    met1-2015-vcaa-q8-answer
`P(A^{′} ∩ B)` `= P(B)-P(A ∩B)`
  `= 1/3-1/4`
  `= 1/12`

 

iii.   `text(If)\ A, B\ text(independent)`

♦♦ Mean mark 28%.
MARKER’S COMMENT: A lack of understanding of independent events was clearly evident.
`P(A ∩ B)` `= P(A) xx P(B)`
`1/4` `= P(A) xx 1/3`
`:. P(A)` `= 3/4`

 

`P(A ∪ B)` `= P(A) + P(B)-P(A ∩ B)`
  `= 3/4 + 1/3-1/4`
`:. P(A ∪ B)` `= 5/6`

Probability, 2ADV S1 2009 MET1 5

Four identical balls are numbered 1, 2, 3 and 4 and put into a box. A ball is randomly drawn from the box, and not returned to the box. A second ball is then randomly drawn from the box.

  1. What is the probability that the first ball drawn is numbered 4 and the second ball drawn is numbered 1?  (1 mark)

  2. What is the probability that the sum of the numbers on the two balls is 5?  (1 mark)

  3. Given that the sum of the numbers on the two balls is 5, what is the probability that the second ball drawn is numbered 1?  (2 marks)

Show Answers Only
  1. `1/12`
  2. `1/3`
  3. `1/4`
Show Worked Solution
i.    `P (4, 1)` `= 1/4 xx 1/3`
    `= 1/12`

 

ii.   `P (text(Sum) = 5)`

`= P (1, 4) + P (2, 3) + P (3, 2) + P (4, 1)`

`= 4 xx (1/4 xx 1/3)`

`= 1/3`

 

iii.   `text(Conditional Probability)`

♦ Mean mark 46%.

`P (2^{text(nd)} = 1\ |\ text(Sum) = 5)`

`= (P (4, 1))/(P (text(Sum) = 5))`

`= (1/12)/(1/3)`

`= 1/4`

Probability, 2ADV S1 2013 MET2 10 MC

For events `A` and `B,\ P(A ∩ B) = p,\ P(A^{′}∩ B) = p -1/8`  and  `P(A ∩ B^{′}) = (3p)/5.`

If `A` and `B` are independent, then the value of  `p`  is

  1. `0`
  2. `1/4`
  3. `3/8`
  4. `1/2`
Show Answers Only

`C`

Show Worked Solution
`P(A)` `= P(A ∩ B) + P(A ∩ B^{′})`
  `= p + (3p)/5`
  `= (8p)/5`

 

`P(B)` `= P(B ∩ A) + P(B ∩ A^{′})`
  `= p + p-1/8`
  `= 2p – 1/8`

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

`P(A ∩ B)` ` = P(A) xx P(B)`
`p` `=(8p)/5 (2p-1/8)`
`5p` `=16p^2-p`
`16p^2-6p` `=0`
`2p(8p-3)` `=0`
`:.p` `=3/8,\ \ \ p!=0`

 
`=>   C`

Probability, 2ADV S1 2012 MET2 13 MC

`A` and `B` are events of a sample space `S.`

`P(A nn B) = 2/5`  and  `P(A nn B^(′)) = 3/7`

`P(B^(′) | A)`  is equal to

  1. `6/35`
  2. `15/29`
  3. `14/35`
  4. `29/35`
Show Answers Only

`B`

Show Worked Solution

met2-2012-vcaa-13-mc-answer

`P(B ^(′) | A)` `= (P(B^(′) nn A))/(P(A))`
  `= (P(B^(′) nn A))/(P(B^(′) nn A)+ P(A nn B))`
  `= (3/7)/(3/7 + 2/5)`
  `= 15/29`

`=>   B`

Probability, 2ADV S1 2013 MET1 7

The probability distribution of a discrete random variable, `X`, is given by the table below.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \ &\ \ \ 4\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.6p^{2} & 0.1 & 1-p & 0.1 \\
\hline
\end{array}

  1. Show that  `p = 2/3 or p = 1`.  (3 marks)

  2. Let  `p = 2/3`.

    1. Calculate  `E(X)`.  Answer in exact form.  (2 marks)

    2. Find  `P(X >= E(X))`.  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
    1. `28/15`
    2. `8/15`
Show Worked Solution

a.   `text(S)text(ince probabilities must sum to 1:)`

`0.2 + 0.6p^2 + 0.1 + 1-p + 0.1` `= 1`
`0.6p^2-p + 0.4` `= 0`
`6p^2-10p + 4` `= 0`
`3p^2-5p + 2` `= 0`
`(p-1) (3p-2)` `= 0`

`:. p = 1 or p = 2/3`

 

b.i.   `E(X)` `= sum x * P (X = x)`
    `= 1 xx (3/5 xx 2^2/3^2) + 2 (1/10) + 3 (1-2/3) + 4 (1/10)`
    `= 4/15 + 1/5 + 1 + 2/5`
    `= 28/15`

 

♦♦ Part (b)(ii) mean mark 32%.
  ii.   `P(X >= 28/15)` `=P(X = 2) + P(X = 3) + P(X = 4)`
    `= 1/10 + 1/3 + 1/10`
    `= 8/15`

Probability, 2ADV S1 2012 MET1 4

On any given day, the number `X` of telephone calls that Daniel receives is a random variable with probability distribution given by

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.2 & 0.5 & 0.1 \\
\hline
\end{array}

  1. Find the mean of  `X`.   (2 marks)

  2. What is the probability that Daniel receives only one telephone call on each of three consecutive days?   (1 mark)

  3. Daniel receives telephone calls on both Monday and Tuesday.

     

    What is the probability that Daniel receives a total of four calls over these two days?   (3 marks)

Show Answers Only
  1. `1.5`
  2. `0.008`
  3. `29/64`
Show Worked Solution
i.    `E(X)` `= 0 xx 0.2 + 1 xx 0.2 + 2 xx 0.5 + 3 xx 0.1`
    `= 0 + .2 + 1 + 0.3`
    `= 1.5`

 

ii.   `P(1, 1, 1)` `= 0.2 xx 0.2 xx 0.2`
    `= 0.008`

 

iii.   `text(Conditional Probability:)`

♦ Mean mark 36%.

`P(x = 4 | x >= 1\ text{both days})`

`= (P(1, 3) + P(2, 2) + P(3, 1))/(P(x>=1\ text{both days}))`

`= (0.2 xx 0.1 + 0.5 xx 0.5 + 0.1 xx 0.2)/(0.8 xx 0.8)`

`= (0.02 + 0.25 + 0.02)/0.64`

`= 0.29/0.64`

`= 29/64`

Probability, 2ADV S1 2010 MET1 8

The discrete random variable `X` has the probability distribution

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ \ -1\ \ \ \  & \ \ \ \ \ 0\ \ \ \ \  & \ \ \ \ \ 1\ \ \ \ \  & \ \ \ \ \ 2\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p^{2} & p^{2} & \dfrac{p}{4} & \dfrac{4p+1}{8} \\
\hline
\end{array}

Find the value of `p.`  (3 marks)

Show Answers Only

`1/2`

Show Worked Solution
`text(Sum of probabilities)` `= 1`
`p^2 + p^2 + p/4 + (4p + 1)/8` `= 1`
`16p^2 + 2p + 4p + 1` `= 8`
`16p^2 + 6p – 7` `= 0`
`(2p – 1) (8p + 7)` `= 0`

 

`:. p = 1/2,\ \ \ (p>0)`

Probability, 2ADV S1 2009 MET1 7

The random variable `X` has this probability distribution.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  &\ \ \ 4\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \\
\hline
\end{array}

Find

  1.  `P (X > 1 | X <= 3)`  (2 marks)

  2.  `P (X),` the variance of  `X.`  (3 marks)
Show Answers Only
  1. `2/3`
  2. `1.2`
Show Worked Solution

i.   `P(X > 1 | X <= 3)`

`= (P(X = 2) + P(X = 3))/(1-P(X = 4))`

`= (0.4 + 0.2)/(1-0.1)`

`= 0.6/0.9`

`= 2/3`

 

ii.   `E(X)` `= 0.1 (0) + 1 (0.2) + 2 (0.4) + 3 (0.2) + 4 (0.1)`
  `= 0 + 0.2 + 0.8 + 0.6 + 0.4`
  `= 2`

 

`E(X^2)` `= 0^2 (0.1) + 1^2 (0.2) + 2^2 (0.4) + 3^2 (0.2) + 4^2 (0.1)`
  `= 0 + 0.2 + 1.6 + 1.8 + 1.6`
  `= 5.2`

 

`:.\ text(Var) (X)` `= E(X^2)-[E(X)]^2`
  `= 5.2-(2)^2`
  `= 1.2`

Probability, 2ADV S1 2008 MET1 7

Jane drives to work each morning and passes through three intersections with traffic lights. The number `X` of traffic lights that are red when Jane is driving to work is a random variable with probability distribution given by

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.1 & 0.2 & 0.3 & 0.4 \\
\hline
\end{array}

  1. What is the mode of  `X`?  (1 mark)

  2. Jane drives to work on two consecutive days. What is the probability that the number of traffic lights that are red is the same on both days?  (2 marks)

Show Answers Only
  1. `3`
  2. `0.3`
Show Worked Solution

i.   `3`

 

ii.   `P(0,0) + P(1,1) + P(2,2) + P(3,3)`

`= 0.1^2 + 0.2^2 + 0.3^2 + 0.4^2`

`= 0.3`

Probability, 2ADV S1 2009 MET2 10 MC

The discrete random variable `X` has a probability distribution as shown.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.4 & 0.2 & 0.3 & 0.1 \\
\hline
\end{array}

The median of `X` is

  1. `0`
  2. `1`
  3. `1.1`
  4. `2`
Show Answers Only

`B`

Show Worked Solution
`P(X <= 0)` `= 0.4`
`P(X <= 1)` `= 0.6`

 
`:.\ text(Median) = 1`

`=>   B`

Probability, 2ADV S1 2017 MET2 14 MC

The random variable `X` has the following probability distribution, where  `0 < p < 1/3`.

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

The variance of `X` is

  1. `2p(1-3p)`
  2. `p(5-9p)`
  3. `(1-3p)^2`
  4. `6p-16p^2`
Show Answers Only

`D`

Show Worked Solution
`text(Var)(X)` `= E(X^2)-[E(X)]^2`
  `= [(-1)^2p + 0^2 xx 2p + 1^2(1-3p)]-[-p + 0 + 1-3p]^2`
  `= 6p-16p^2`

 
`=> D`

Calculus, 2ADV C4 2018* HSC 15c

The shaded region is enclosed by the curve  `y = x^3 - 7x`  and the line  `y = 2x`, as shown in the diagram. The line  `y = 2x`  meets the curve  `y = x^3 - 7x`  at  `O(0, 0)`  and  `A(3, 6)`. Do NOT prove this.
 


 

  1.  Use integration to find the area of the shaded region.  (2 marks)

  2. Use the Trapezoidal rule and four function values to approximate the area of the shaded region.  (2 marks)

The point `P` is chosen on the curve  `y = x^3 − 7x`  so that the tangent at `P` is parallel to the line  `y = 2x`  and the `x`-coordinate of `P` is positive

  1.  Show that the coordinates of `P` are  `(sqrt 3, -4 sqrt 3)`.  (2 marks)

  2.  Using the perpendicular distance formula  `|ax_1 + by_1 + c|/sqrt(a^2 + b^2)`,  find the area of  `Delta OAP`.  (2 marks)

Show Answers Only
  1. `81/4\ text(units²)`
  2. `18\ text(u²)`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `9 sqrt 3\ text(units²)`
Show Worked Solution
i.   `text(Area)` `= int_0^3 2x – (x^3 – 7x)\ dx`
    `= int_0^3 9x – x^3\ dx`
    `= [9/2 x^2 – 1/4 x^4]_0^3`
    `= [(9/2 xx 3^2 – 1/4 xx 3^4) – 0]`
    `= 81/2 – 81/4`
    `= 81/4\ text(units²)`

 

ii.  `f(x) = 9x – x^3`

`text(Area)` `~~ 1/2[0 + 2(8 + 10) + 0]`
  `~~ 1/2(36)`
  `~~ 18\ text(u²)`

 

iii.   `y = x^3 – 7x`

`(dy)/(dx) = 3x^2 – 7`

`text(Find)\ \ x\ \ text(such that)\ \ (dy)/(dx) = 2`

`3x^2 – 7` `= 2`
`3x^2` `= 9`
`x^2` `= 3`
`x` `= sqrt 3 qquad (x > 0)`

 

`y` `= (sqrt 3)^3 – 7 sqrt 3`
  `= 3 sqrt 3 – 7 sqrt 3`
  `= -4 sqrt 3`

 
`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`

 

iv.  

 

`text(dist)\ OA` `= sqrt((3 – 0)^2 + (6 – 0)^2)`
  `= sqrt 45`
  `= 3 sqrt 5`

 
`text(Find)\ _|_\ text(distance of)\ \ P\ \ text(from)\ \ OA`

`P(sqrt 3, -4 sqrt 3),\ \ 2x – y=0`

`_|_\ text(dist)` `= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|`
  `= (6 sqrt 3)/sqrt 5`
   
`:.\ text(Area)` `= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5`
  `= 9 sqrt 3\ text(units²)`

Calculus, 2ADV C4 2017* HSC 14b

  1. Find the exact value of  `int_0^(pi/3) cos x\ dx`.  (1 mark)

  2. Using the Trapezoidal rule with three function values, find an approximation to the integral  `int_0^(pi/3) cos x\ dx,` leaving your answer in terms of `pi` and `sqrt 3`.  (2 marks)

  3. Using parts (i) and (ii), show that  `pi ~~ (12 sqrt 3)/(3 + 2 sqrt 3)`.  (1 mark)

Show Answers Only
  1. `sqrt 3/2`
  2. `((2sqrt3 + 3)pi)/24`
  3. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i.   `int_0^(pi/3) cos x\ dx` `= [sin x]_0^(pi/3)`
    `= sin\ pi/3-0`
    `= sqrt 3/2`

 

ii.  

\begin{array} {|l|c|c|c|}\hline
x & \ \ \ 0\ \ \  & \ \ \ \dfrac{\pi}{6}\ \ \  & \ \ \ \dfrac{\pi}{3}\ \ \ \\ \hline
\text{height} & 1 & \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ \hline
\text{weight} & 1 & 2 & 1  \\ \hline \end{array}

`int_0^(pi/3) cos x\ dx` `~~ 1/2 xx pi/6[1 + 2(sqrt3/2) + 1/2]`
  `~~ pi/12((3 + 2sqrt3)/2)`
  `~~ ((3+2sqrt3)pi)/24`

 

♦ Mean mark part (iii) 49%.

(iii)    `((3+2sqrt3)pi)/24` `~~ sqrt3/2`
  `:. pi` `~~ (24sqrt3)/(2(3+2sqrt3))`
    `~~ (12sqrt3)/(3 + 2sqrt3)`

Calculus, 2ADV C4 2011* HSC 5c

The table gives the speed `v` of a jogger at time `t` in minutes over a  20-minute period. The speed `v` is measured in metres per minute, in intervals of 5 minutes.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ t\ \ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ &\ \ \ 5\ \ \ &\ \ \ 10\ \ \ &\ \ \ 15\ \ \ &\ \ \ 20\ \ \  \\
\hline
\rule{0pt}{2.5ex} \ \ \ v\ \ \  \rule[-1ex]{0pt}{0pt} & 173 & 81 & 127 & 195 & 168 \\
\hline
\end{array}

The distance covered by the jogger over the 20-minute period is given by  `int_0^20 v\ dt`.

Use the Trapezoidal rule and the speed at each of the five time values to find the approximate distance the jogger covers in the 20-minute period.   (3 marks)

Show Answers Only

 `text(2867.5 metres)`

Show Worked Solution

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ t\ \ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ &\ \ \ 5\ \ \ &\ \ \ 10\ \ \ &\ \ \ 15\ \ \ &\ \ \ 20\ \ \  \\
\hline
\rule{0pt}{2.5ex} \ \ \ v\ \ \  \rule[-1ex]{0pt}{0pt} & 173 & 81 & 127 & 195 & 168 \\
\hline
\rule{0pt}{2.5ex} \text{weight} \rule[-1ex]{0pt}{0pt} & 1 & 2 & 2 & 2 & 1 \\
\hline
\end{array}

`int_0^20 v\ dt`

 `~~ 5/2[173 + 2(81 + 127 + 195) + 168]`
  `~~ 5/2(1147)`
  `~~ 2867.5\ text(metres)`

Calculus, 2ADV C4 SM-Bank 07

The diagram shows a block of land and its dimensions, in metres. The block of land is bounded on one side by a river. Measurements are taken perpendicular to the line  `AB`, from  `AB`  to the river, at equal intervals of 50 m.
 

Integration, 2UA 2009 HSC 3d

 
Use the Trapezoidal rule with six subintervals to find an approximation to the area of the block of land.   (3 marks)

Show Answers Only

 `text(64 250 m²)`

Show Worked Solution

 

`A` `~~ 50/2[210 + 2(220 + 200 + 190 + 210 + 240) + 240]`
  `~~ 25(2570)`
  `~~ 64\ 250\ text(m²)`

 
`:.\ text(The block of land)\ ~~64 250 m²`