Band 4 – Page 99

Graphs, MET2 2015 VCAA 5 MC

Part of the graph of `y = f(x)` is shown below.

The corresponding part of the graph of the inverse function `y = f^-1(x)` is best represented by

Show Answers Only

`E`

Show Worked Solution

`f^(-1)\ text(is a reflection in)\ \ y=x,`

`=> E`

Calculus, MET2 2014 VCAA 19 MC

Jake and Anita are calculating the area between the graph of `y = sqrt x` and the `y`-axis between `y = 0` and `y = 4.`

Jake uses a partitioning, shown in the diagram below, while Anita uses a definite integral to find the exact area.

The difference between the results obtained by Jake and Anita is

A. `0`

B. `22/3`

C. `26/3`

D. `14`

E. `35`

Show Answers Only

`C`

Show Worked Solution

`text(Jake:)`

`A_J`	`~~ 1[1 + 4 + 9 + 16]`
	`~~ 30`

`text(Anita:)`

`A_A = 16 xx 4 – int_0^16 sqrtx\ dx`

`text(or)`

`A_A`	`= int_0^4 y^2\ dy`
	`= 64/3`

`:.\ text(Difference)`	`= 30 – 64/3`
	`= 26/3`

`=> C`

Probability, MET2 2014 VCAA 11 MC

A bag contains five red marbles and four blue marbles. Two marbles are drawn from the bag, without replacement, and the results are recorded.

The probability that the marbles are different colours is

`20/81`
`5/18`
`4/9`
`40/81`
`5/9`

Show Answers Only

`E`

Show Worked Solution

`text(Pr)(RB) + text(Pr)(BR)`	`= 5/9 xx 4/8 + 4/9 xx 5/8`
	`= 5/9`

`=> E`

Calculus, MET2 2014 VCAA 8 MC

If `int_1^4 f(x)\ dx = 6`, then `int_1^4 (5 - 2\ f(x))\ dx` is equal to

A. `3`

B. `4`

C. `5`

D. `6`

E. `16`

Show Answers Only

`A`

Show Worked Solution

`int_1^4 5 – 2\ f(x)\ dx`

`= int_1^4 5\ dx – 2 int_1^4 f(x)\ dx`

`= 15 – 2(6)`

`= 3`

`=> A`

Algebra, MET2 2014 VCAA 6 MC

The function `f: D -> R` with rule `f(x) = 2x^3 - 9x^2 - 168x` will have an inverse function for

`D = R`
`D = (7, oo)`
`D = text{(−4, 8)}`
`D = text{(−∞, 0)}`
`D = [text(−)1/2, oo)`

Show Answers Only

`B`

Show Worked Solution

`text(Inverse exists if)\ f(x)\ text(is)`

`:. D = (7, ∞)\ \ text(gives)\ \ f(x)\ text(one-to-one.)`

`=> B`

Probability, MET2 2014 VCAA 5 MC

The random variable `X` has a normal distribution with mean 12 and standard deviation 0.5.

If `Z` has the standard normal distribution, then the probability that `X` is less than 11.5 is equal to

`text(Pr)(Z > – 1)`
`text(Pr)(Z < – 0.5)`
`text(Pr)(Z > 1)`
`text(Pr)(Z >= 0.5)`
`text(Pr)(Z < 1)`

Show Answers Only

`C`

Show Worked Solution

`text(Pr)(X < 11.5)`	`= text(Pr)(Z < (11.5 – 12)/0.5)`
	`= text(Pr)(Z < −1)`
	`= text(Pr)(Z > 1)`

`=> C`

Calculus, MET2 2014 VCAA 4 MC

Let `f` be a function with domain `R` such that `f (5) = 0` and `f prime (x) < 0` when `x != 5.`

At `x = 5`, the graph of `f` has a

local minimum.
local maximum
gradient of 5
gradient of – 5
stationary point of inflection.

Show Answers Only

`E`

Show Worked Solution

`:.\ text(S)text(ince concavity doesn’t change,)`

`text(point of inflection at)\ \ x = 5.`

`=> E`

Mechanics, EXT2* M1 2008 HSC 7

A projectile is fired from `O` with velocity `V` at an angle of inclination `theta` across level ground. The projectile passes through the points `L` and `M`, which are both `h` metres above the ground, at times `t_1` and `t_2` respectively. The projectile returns to the ground at `N`.

The equations of motion of the projectile are

`x = Vtcos theta` and `y = Vtsin theta − 1/2 g t^2`. (Do NOT prove this.)

Show that `t_1 + t_2 = (2V)/g sin theta` AND `t_1t_2 = (2h)/g`. (2 marks)

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Let `∠LON = α` and `∠LNO = β`. It can be shown that

`tan alpha = h/(Vt_1 cos theta)` and `tan beta = h/(Vt_2 cos theta)`. (Do NOT prove this.)

Show that `tan alpha + tan beta = tan theta`. (2 marks)

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Show that `tan alpha tan beta = (gh)/(2V^2cos^2theta)`. (1 mark)

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Let `ON = r` and `LM = w`.

Show that `r = h(cot alpha + cot beta)` and `w = h(cot beta - cot alpha)`. (2 marks)

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Let the gradient of the parabola at `L` be `tan phi`.

Show that `tan phi = tan alpha - tan beta`. (3 marks)

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Show that `w/(tan phi) = r/(tan theta)`. (2 marks)

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Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `text(At)\ \ y = h,`

`-1/2g t^2 + Vtsin theta`	`=h`
`g t^2 − 2Vtsin theta + 2h`	`=0`

`text(Roots are)\ \ t_1\ text(and)\ t_2`

`:. t_1 + t_2`	`= -b/a`
	`= (2V)/g sin theta …\ \ text(as required)`

`:. t_1t_2`	`= c/a`
	`= (2h)/g …\ \ text(as required)`

ii. `text(Show)\ \ tan alpha + tan beta = tan theta`

`text(LHS)`	`= h/(Vt_1cos theta) + h/(Vt_2cos theta)`
	`= (h(t_1 + t_2))/(t_1t_2Vcos theta)`
	`= (h((2V)/g sin theta))/(((2h)/g)Vcos theta),\ \ \ \ text{(using part (i))}`
	`= ((2Vh)/g)/((2Vh)/g) xx (sin theta)/(cos theta)`
	`= tan theta …\ \ text(as required.)`

iii. `text(Show)\ \ tan alpha tan beta = (gh)/(2V^2cos^2 theta)`

`text(LHS)`	`= h/(Vt_1cos theta) xx h/(Vt_2cos theta)`
	`= (h^2)/(t_1t_2V^2cos^2 theta)`
	`= (h^2)/((2h)/g · V^2cos^2 theta)`
	`= (gh)/(2V^2cos^2 theta) …\ \ text(as required.)`

iv. `text(From the diagram,)`

`r = ON = OP + PN`

`text(In)\ DeltaOLP,`

`tan alpha`	`= h/(OP)`
`OP`	`= hcot alpha`
`text(In)\ DeltaNLP,`
`tan beta`	`= h/(PN)`
`PN`	`= hcot beta`

`:.r`	`= hcot alpha + hcot beta`
	`= h(cot alpha + cot beta) …\ \ text(as required.)`

`w`	`= r − (OP + QN)`
	`= r − 2 xx OP,\ \ \ text{(symmetry of parabola)}`
	`= h cot alpha + hcot beta − 2hcot alpha`
	`= h (cot beta − cot alpha) …\ \ text(as required.)`

v. `text(Show)\ \ tan phi = tan alpha − tan beta`

`text(At)\ L, \ t = t_1\ \ text(and)`

`tan phi = (dot y)/(dot x)`

`dot y`	`= Vsin theta − g t`
`dot x`	`= Vcos theta`

`:. tan phi`	`= (Vsin theta – g t_1)/(Vcos theta)`
	`= tan theta – (g t_1)/(Vcos theta)`
	`= tan theta – g/(Vcos theta) xx h/(Vcos theta tan alpha)`
	`= tan theta – (gh)/(V^2cos^2 theta tan alpha)`
	`= tan theta – 2 tan beta,\ \ \ text{(from part (iii))}`
	`= tan alpha + tan beta − 2tan beta,\ \ \ text{(from part (ii))}`
	`= tan alpha – tan beta …\ \ text(as required.)`

vi. `text(Show)\ \ w/(tan phi) = r/(tan theta)`

`w/r`	`= (h(cot beta − cot alpha))/(h(cot alpha + cot beta))`
	`= (1/(tan beta) − 1/(tan alpha))/(1/(tan alpha) + 1/(tan beta)) xx (tan alpha tan beta)/(tan alpha tan beta)`
	`= (tan alpha − tan beta)/(tan beta + tan alpha)`
	`= (tan phi)/(tan theta)`

`:. w/(tan phi) = r/(tan theta) …\ \ text(as required.)`

MATRICES, FUR1 2014 VCAA 3 MC

Regular customers at a hairdressing salon can choose to have their hair cut by Shirley, Jen or Narj.

The salon has 600 regular customers who get their hair cut each month.

In June, 200 customers chose Shirley (S) to cut their hair, 200 chose Jen (J) to cut their hair and 200 chose Narj (N) to cut their hair.

The regular customers’ choice of hairdresser is expected to change from month to month as shown in the transition matrix, `T`, below.

`{:(qquad qquad qquad qquad text(this month)), (qquad qquad qquad quad {:(S, quad quad J, qquad N):}), (T = [(0.75, 0.10, 0.10), (0.10, 0.75, 0.15), (0.15, 0.15, 0.75)] {:(S), (J), (N):} qquad text(next month)):}`

In the long term, the number of regular customers who are expected to choose Shirley is closest to

A. `150`

B. `170`

C. `185`

D. `195`

E. `200`

Show Answers Only

`B`

Show Worked Solution

`text(Consider)\ \ n =50,`

`[(0.75, 0.10, 0.10), (0.10, 0.75, 0.15), (0.15, 0.15, 0.75)]^50 [(200), (200), (200)] = [(171.4), (203.6), (225)]`

`=>B`

MATRICES, FUR1 2014 VCAA 4 MC

Two hundred and fifty people buy bread each day from a corner store. They have a choice of two brands of bread: Megaslice (M) and Superloaf (S).

The customers’ choice of brand changes daily according to the transition diagram below.

On a given day, 100 of these people bought Megaslice bread while the remaining 150 people bought Superloaf bread.

The number of people who are expected to buy each brand of bread the next day is found by evaluating the matrix product

Show Answers Only

`C`

Show Worked Solution

`text(The)\ 2 xx 2\ text(matrix columns need)`

`text(to add up to 100%)`

`:.\ text(Eliminate)\ A and B.`

`text(The transition diagram corresponds)`

`text(to the matrix equation in)\ C.`

`=>C`

Calculus, EXT1 C1 2008 HSC 4a

A turkey is taken from the refrigerator. Its temperature is 5°C when it is placed in an oven preheated to 190°C.

Its temperature, `T`° C, after `t` hours in the oven satisfies the equation

`(dT)/(dt) = -k(T − 190)`.

Show that `T = 190 - 185e^(-kt)` satisfies both this equation and the initial condition. (2 marks)

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The turkey is placed into the oven at 9 am. At 10 am the turkey reaches a temperature of 29°C. The turkey will be cooked when it reaches a temperature of 80°C.

At what time (to the nearest minute) will it be cooked? (3 marks)

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Show Answers Only

`text(See Worked Solutions)`
`12:44\ text(pm)`

Show Worked Solution

i.	`T`	`= 190 − 185e^(-kt)`
	`(dT)/(dt)`	`= -k xx -185e^(-kt)`
		`= -k(190 − 185e^(-kt) − 190)`
		`= -k(T − 190)`

`:.T = 190 − 185e^(-kt)\ text(satisfies equation.)`

`text(When)\ \ t = 0,`

`T`	`= 190 − 185e^0`
	`= 5°`

`:.\ text(Initial conditions are satisfied.)`

ii. `text(When)\ \ t = 1,\ T = 29`

`29`	`= 190 − 185e^(-k)`
`185e^(-k)`	`= 161`
`e^(-k)`	`= 161/185`
`-k`	`= ln\ 161/185`
`:.k`	`= -ln\ 161/185`
	`= 0.1389…`

`text(Find)\ \ t\ \ text(when)\ \ T = 80 :`

`80`	`= 190 − 185e^(-kt)`
`185e^(-kt)`	`= 110`
`e^(-kt)`	`= 110/185`
`-kt`	`= ln\ 110/185`
`:.t=`	`= ln\ 110/185 -: -0.1389…`
	`= 3.741…`
	`= 3\ text(hours)\ 44\ text{mins (nearest minute)}`

`:.\ text(The turkey will be cooked at 12:44 pm.)`

Calculus, 2ADV C3 2004 HSC 10b

The diagram shows a triangular piece of land `ABC` with dimensions `AB = c` metres, `AC = b` metres and `BC = a` metres, where `a ≤ b ≤ c`.

The owner of the land wants to build a straight fence to divide the land into two pieces of equal area. Let `S` and `T` be points on `AB` and `AC` respectively so that `ST` divides the land into two pieces of equal area.

Let `AS = x` metres, `AT = y` metres and `ST = z` metres.

Show that `xy = 1/2 bc`. (1 mark)

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Use the cosine rule in triangle `AST` to show that

`z^2 = x^2 + (b^2c^2)/(4x^2) − bc cos A.` (2 marks)

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Show that the value of `z^2` in the equation in part (ii) is a minimum when

`x = sqrt((bc)/2)`. (4 marks)

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Show that the minimum length of the fence is `sqrt(((P − 2b)(P − 2c))/2)` metres, where `P = a + b + c`.

(You may assume that the value of `x` given in part (iii) is feasible.) (2 marks)

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Show Answers Only

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

Show Worked Solution

i.	`text(Area)\ \ ΔABC`	`= 1/2\ bc sin A`
	`text(Area)\ \ ΔAST`	`= 1/2\ xy sin A`

`text(Area)\ \ ΔAST`	`= 1/2 xx text(Area)\ \ ΔABC\ \ \ text{(given)}`
`1/2\ xy sin A`	`= 1/2 xx 1/2\ bc sin A`
`xy sin A`	`= 1/2\ bc sin A`
`:. xy`	`= 1/2\ bc\ \ …\ text(as required.)`

ii. `text(Using the cosine rule in)\ \ ΔAST,`

`z^2 = x^2 + y^2 − 2xy cos A`

`text(S)text(ince)\ \ xy`	`= 1/2\ bc`
`:. y`	`= (bc)/(2x)`

`:. z^2`	`= x^2 + ((bc)/(2x))^2 − 2x((bc)/(2x)) cos A`
	`= x^2 + (b^2c^2)/(4x^2) − bc cos A`

iii.	`z^2`	`= x^2 + ((b^2c^2)/4)x^(−2) − bc cos A`
	`(d(z^2))/(dx)`	`= 2x − (2b^2c^2)/4 x^(−3)= 2x − (b^2c^2)/(2x^3)`
	`(d^2(z^2))/(dx^2)`	`= 2 + (3b^2c^2)/2 x^(−4)= 2 + (3b^2c^2)/(2x^4)`

`text(SP’s occur when)\ \ (d(z^2))/(dx)=0`

`2x − (b^2c^2)/(2x^3)`	`= 0`
`4x^4 − b^2c^2`	`= 0`
`4x^4`	`= b^2c^2`
`x^4`	`= (b^2c^2)/4`
`x^2`	`= (bc)/2`
`x`	`= sqrt((bc)/2),\ \ \ (x > 0)`

` (d^2(z^2))/(dx^2)=2 + (3b^2c^2)/(2x^4)>0\ \ \ text{(for all}\ xtext{)}`

`:.\ text(A minimum value of)\ z^2\ text(when)\ x = sqrt((bc)/2).`

iv. `cos A = (b^2 + c^2 − a^2)/(2bc)`

`text(When)\ \ x = sqrt((bc)/2),`

`:. z^2`	`= x^2 + (b^2c^2)/(4x^2) − bc cos A`
	`= (bc)/2 + (b^2c^2)/(4((bc)/2)) − bc((b^2 + c^2 − a^2)/(2bc))`
	`= (bc)/2 + (bc)/2 − ((b^2 + c^2 − a^2)/2)`
	`= (2bc – b^2 − c^2 +a^2)/2`
	`= (a^2 − (b^2 − 2bc + c^2))/2`
	`= 1/2[a^2 − (b − c)^2]`
	`= 1/2[(a − (b − c))(a + (b − c))]`
	`= 1/2[(a − b + c)(a + b − c)]`
	` = 1/2[(a + b + c − 2b)(a + b + c − 2c)]`
	`= ((P − 2b)(P − 2c))/2\ \ \ text{(using}\ \ P = a + b + c text{)}`

`:.z = sqrt(((P − 2b)(P − 2c))/2)\ \ text(metres)\ \ …\ text(as required.)`

Calculus, 2ADV C3 2004 HSC 9c

Consider the function `f(x) = (log_e x)/x`, for `x > 0`.

Show that the graph of `y = f(x)` has a stationary point at `x = e`. (2 marks)

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By considering the gradient on either side of `x = e`, or otherwise, show that the stationary point is a maximum. (1 mark)

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Use the fact that the maximum value of `f(x)` occurs at `x = e` to deduce that `e^x ≥ x^e` for all `x > 0`. (2 marks)

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Show Answers Only

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

Show Worked Solution

i. `f(x) = (log_e x)/x, text(for)\ x > 0`

`text(Using the quotient rule,)`

`f′(u/v)`	`=(u′v-uv′)/v^2`
`f′(x)`	`= (x · d/(dx) (log_e x) − log_e x · d/(dx) (x))/(x^2)`
	`= (x(1/x) − log_e x · 1)/(x^2)`
	`= (1 − log_e x)/(x^2)`

`text(SP’s occur when)\ \ f′(x)=0`

`(1 − log_e x)/(x^2)`	`= 0`
`1 − log_e x`	`= 0`
`log_e x`	`= 1`
`:. x`	`= e`

`:. text(There is a stationary point at)\ \ x = e.`

ii.

`text(When)\ \ x < e, log_e x<1\ \ \ text{(from graph)}`

`1 − log_e x`	`> 0`
`(1 − log_e x)/(x^2)`	`> 0`
`:. f′(x)`	`> 0`

`text(When)\ x > e, log_e x>1\ \ \ text{(from graph)}`

`1 − log_e x`	`< 0`
`(1 − log_e x)/(x^2)`	`< 0`
`:. f′(x)`	`< 0`

`:.\ text(The stationary point at)\ \ x = e\ \ text(is a maximum.)`

iii.	`f(e)`	`= (log_e x)/e`
		`= 1/e`

`:. f(x)\ \ text(has a maximum value at)\ \ (e,1/e)`

MARKER’S COMMENT: Very challenging for most students. Successful students recognised the link to part (ii).

`:. (log_e x)/x`	`≤ 1/e`
`log_e x`	`≤ x/e`
`e log_e x`	`≤ x`
`log_e x^e`	`≤ x`
`e^(log_e x^e)`	`≤ e^x`
`x^e`	`≤ e^x\ \ \ text{(using}\ e^(log x) = x)`
`:. e^x`	`≥ x^e \ \ \ text{(for}\ \ x > 0text{)}`

Calculus, 2ADV C4 2004 HSC 8b

The diagram shows the graph of the parabola `x^2 = 16y`. The points `A (4, 1)` and `B (−8, 4)` are on the parabola, and `C` is the point where the tangent to the parabola at `A` intersects the directrix.

Write down the equation of the directrix of the parabola `x^2 = 16y`. (1 mark)
Find the equation of the tangent to the parabola at the point `A`. (2 marks)
Show that `C` is the point `(−6, −4)`. (1 mark)
Given that the equation of the line `AB` is `y = 2 − x/4`, find the area bounded by the line `AB` and the parabola. (2 marks)
Hence, or otherwise, find the shaded area bounded by the parabola, the tangent at `A` and the line `BC`. (3 marks)

Show Answers Only

`y = −4`
`y = x/2 − 1`
`text(Proof)\ \ text{(See Worked Solutions)}`
`18\ \ text(u²)`
`27\ \ text(u²)`

Show Worked Solution

(i)	`x^2`	`= 4ay`
		`= 16y`
	`:.4a`	`= 16`
	`a`	`= 4`

`:.\ text(Equation of the directrix is)\ \ y = −4.`

(ii)	`x^2`	`= 16y`
	`y`	`= (x^2)/(16)`
	`:.(dy)/(dx)`	`= (2x)/(16)=x/8`
	`text(At)\ x = 4`
	`dy/dx`	`= 1/2`

`:.\ text(Equation of the tangent passes through)\ \ A(4, 1)`

`\ text(with)\ \ m = 1/2`

`y − 1`	`= 1/2(x − 4)`
`y − 1`	`= x/2 − 2`
`y`	`= x/2 − 1`

(iii) `C\ text(lies on)\ \ y=-4\ \ text{(directrix)}`

`text(S)text(ince)\ \ C\ \ text(also lies on the tangent,)`

`y`	`= x/2 − 1`
`−4`	`= x/2 − 1`
`−3`	`= x/2`
`:. x`	`= −6`

`:.C\ \ text{is the point (–6, – 4) … as required}`

(iv)	`text(Equation of parabola is)\ \ x^2`	`= 16y`
	`:. y`	`= (x^2)/(16)`

`text(Required area)`

`= int_(−8)^4 (2 − x/4)\ dx − int_(−8)^4 (x^2)/16\ dx`

`= int_(−8)^4 (2 − x/4 − (x^2)/16)\ dx`

`= [2x − (x^2)/8 − (x^3)/48]_(−8)^(\ \ \ 4)`

`= (2(4) − (4^2)/8 − (4^3)/48) − (2(−8) − ((−8)^2)/8 − ((−8)^3)/48)`

`= (8 − 16/8 − 64/48) − (−16 − 64/8 − ((−512)/48))`

`= (6 − 4/3) − (−16 − 8 + 32/3)`

`= 18`

`:.\ text(Area between the parabola and)\ \ AB`

`= 18\ \ text(u²).`

(v) `text(Shaded area = Area of)\ ΔABC − 18\ \ \ text{(from part (iv))}`

`text(Area of)\ \ ΔABC = 1/2 xx b × h`

`h=\ text(⊥ distance of)\ \ C text{(−6, − 4)}\ \ text(from)\ \ AB,`

`text(where AB has equation)`	`4y`	`= 8 − x`
	`x + 4y − 8`	`= 0`

`:.h`	`= \|(ax_1 + by_1 + c)/(sqrt(a^2 + b^2))\|`
	`= \|(1(−6) + 4(−4) + (−8))/(sqrt((1)^2 + (4)^2))\|`
	`= \|(−6 − 16 − 8)/(sqrt(1 + 16))\|`
	`= \|(−30)/sqrt(17)\|`
	`= 30/sqrt17`
`AB`	`= sqrt((−8 − 4)^2 + (4 − 1)^2)`
	`= sqrt(144 + 9)`
	`= sqrt(153)`
	`= 3sqrt17`

`:.\ text(Area of)\ ΔABC`

`= 1/2 × 3sqrt17 × 30/sqrt17`

`= 45\ \ text(u²).`

`:.\ text(Shaded area)`	`= 45 − 18`
	`= 27\ \ text(u²).`

Trigonometry, 2ADV T2 2004 HSC 8a

Show that `cos theta tan theta = sin theta`. (1 mark)

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Hence solve `8 sin theta cos theta tan theta = text(cosec)\ theta` for `0 ≤ theta ≤ 2pi`. (2 marks)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`pi/6, (5pi)/6`

Show Worked Solution

i. `text(Prove)\ \ cos theta tan theta = sin theta`

`text(LHS)`	`= cos theta tan theta`
	`= cos theta ((sin theta)/(cos theta))`
	`= sin theta`
	`=\ text{RHS}`

ii.	`8 sin theta cos theta tan theta`	`= text(cosec)\ theta`
	`:. 8 sin theta(sin theta)`	`= text(cosec)\ theta,\ \ \ \ text{(part (i))}`
	`8 sin^2 theta`	`= 1/(sin theta)`
	`8 sin^3 theta`	`= 1`
	`sin^3 theta`	`= 1/8`
	`sin theta`	`= 1/2`
	`:. theta`	`= pi/6, (5pi)/6\ \ \ \ text{(for}\ \ 0 ≤ theta ≤ 2pi text{)}`

Financial Maths, 2ADV M1 2004 HSC 7c

Betty decides to set up a trust fund for her grandson, Luis. She invests $80 at the beginning of each month. The money is invested at 6% per annum, compounded monthly.

The trust fund matures at the end of the month of her final investment, 25 years after her first investment. This means that Betty makes 300 monthly investments.

After 25 years, what will be the value of the first $80 invested? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
By writing a geometric series for the value of all Betty’s investments, calculate the final value of Luis’ trust fund. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`$357.20\ \ text(to the nearest cent)`
`$55\ 716.71\ \ text(to the nearest cent)`

Show Worked Solution

i. `A = P(1 + r/(100))^n`

`P`	`= $80`
`n`	`= 300\ \ text(months)`
`r`	`=\ text(6% per annum)`
	`=\ text(0.5% per month)`

`:.A`	`= 80(1 + (0.5)/(100))^300`
	`= 80(1.005)^300`
	`= 357.1975…`
	`= $357.20\ \ text{(nearest cent)}`

ii.	`A_1`	`= 80(1.005)^300`
	`A_2`	`= 80(1.005)^299`
	`A_3`	`= 80(1.005)^298`
		` vdots`
	`A_300`	`= 80(1.005)^1`

`:.\ text(Final value of fund)`

`= 80(1.005)^300 + 80(1.005)^299 + … + 80(1.005)^1`

`= 80[1.005^300 + 1.005^299 + … + 1.005^1]`

`= 80[1.005^1 + 1.005^2 + … + 1.005^300]`

`=>\ text(GP where)\ \ \ a = 1.005, \ r = 1.005, \ n = 300`

`= 80[(a(r^n − 1))/(r − 1)]`

`= 80[(1.005(1.005^300 − 1))/(1.005 − 1)]`

`= 55\ 716.714…`

`= $55\ 716.71\ \ text(to the nearest cent.)`

Calculus, EXT1* C1 2004 HSC 7b

At the beginning of 1991 Australia’s population was 17 million. At the beginning of 2004 the population was 20 million.

Assume that the population `P` is increasing exponentially and satisfies an equation of the form `P = Ae^(kt)`, where `A` and `k` are constants, and `t` is measured in years from the beginning of 1991.

Show that `P = Ae^(kt)` satisfies `(dP)/(dt) =kP`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
What is the value of `A`? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the value of `k`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Predict the year during which Australia’s population will reach 30 million. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`kP`
`1.7 × 10^7`
`0.013\ \ text(to 3 decimal places.)`
`2036`

Show Worked Solution

i.	`P`	`= Ae^(kt)`
	`(dP)/(dt)`	`= kAe^(kt)`
		`= kP`

ii. `P = Ae^(kt)`

`text(When)\ \ t = 0, \ P = 1.7 × 10^7`

`1.7 × 10^7`	`= Ae^0`
`1.7 × 10^7`	`= A xx 1`
`:. A`	`= 1.7 × 10^7`

iii. `P = 1.7 × 10^7e^(kt)`

`text(When)\ t = 13, \ P = 2 × 10^7`

`2 × 10^7`	`= 1.7 × 10^7e^(13k)`
`(2 × 10^7)/(1.7 × 10^7)`	`= e^(13k)`
`ln (2/(1.7))`	`= ln e^(13k)`
	`= 13k`
`:.k`	`= 1/13 ln (2/(1.7))`
	`= 0.0125…`
	`= 0.013\ \ \ text{(to 3 d.p.)}`

iv. `P`

`= 1.7 × 10^7e^(kt)`

`text(Find)\ \ t\ \ text(when)\ \ P = 3 × 10^7,`

MARKER’S COMMENT: Many students had the correct calculations but didn’t answer the question by identifying the exact year and lost a valuable mark.

`3 × 10^7`	`= 1.7 × 10^7e^(kt)`
`(3 × 10^7)/(1.7 × 10^7)`	`= e^(kt)`
`ln (3/(1.7))`	`= ln e^(kt)`
`ln (3/(1.7))`	`= kt ln e`
	`= kt`
`:.t`	`= (ln (3/(1.7)))/k`
	`= (ln(3/(1.7)))/(0.0125…)`
	`= 45.433…`
	`= 45.4\ \ text{years (to 1 d.p.)}`

`:.\ text(The population will reach 30 million in 2036.)`

Financial Maths, 2ADV M1 2004 HSC 7a

Evaluate `sum_(n = 2)^4 n^2`. (1 mark)

Show Answers Only

`29`

Show Worked Solution

`sum_(n = 2)^4 n^2`	`= 2^2 + 3^2 + 4^2`
	`= 4 + 9 + 16`
	`= 29`

Probability, 2ADV S1 2004 HSC 6c

In a game, a turn involves rolling two dice, each with faces marked 0, 1, 2, 3, 4 and 5. The score for each turn is calculated by multiplying the two numbers uppermost on the dice.

What is the probability of scoring zero on the first turn? (2 marks)

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What is the probability of scoring `16` or more on the first turn? (1 mark)

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What is the probability that the sum of the scores in the first two turns is less than 45? (2 marks)

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Show Answers Only

`(11)/(36)`
`1/9`
`1291/1296`

Show Worked Solution

MARKER’S COMMENT: Students who drew up the table for the sample space were “overwhelmingly” more successful in all parts of this question.

`:. P(0) = (11)/(36)`

ii. `P(≥ 16)= 4/36=1/9`

iii. `Ptext{(Sum} < 45) = 1 − Ptext{(Sum} ≥ 45)`

`Ptext{(Sum} ≥ 45)`	`=P(20,25)+P(25,20)+P(25,25)`
	`=(2/36 xx 1/36) + (2/36 xx 1/36)+(1/36 xx 1/36)`
	`=2/1296 + 2/1296+ 1/1296`
	`=5/1296`

`:.Ptext{(Sum} < 45)`	`= 1 − 5/1296`
	`= 1291/1296`

Plane Geometry, 2UA 2004 HSC 6b

The diagram shows a right-angled triangle `ABC` with `∠ABC = 90^@`. The point `M` is the midpoint of `AC`, and `Y` is the point where the perpendicular to `AC` at `M` meets `BC`.

Show that `ΔAYM ≡ ΔCYM`. (2 marks)
Suppose that it is also given that `AY` bisects `∠BAC`. Find the size of `∠YCM` and hence find the exact ratio `MY : AC`. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`1 : 2sqrt3`

Show Worked Solution

(i)

`text(In)\ ΔAYM\ text(and)\ ΔCYM`

MARKER’S COMMENT: Markers strongly recommend that students copy the diagram in geometry questions and label it along with the workings of their solution.

`∠AMY`	`= ∠CMY = 90^@\ \ \ (MY ⊥ AC)`
`AM`	`=CM\ \ \ text{(given)}`
`YM\ text(is common)`

`:.ΔAYM ≡ ΔCYM\ \ text{(SAS)}`

(ii) `text(Let)\ \ /_BAC = 2 theta`

`∠YAB = ∠YAM = theta\ \ \ \ (AY\ text(bisects)\ ∠BAC)`

`∠YAM`	`= ∠YCM=theta`	`\ \ \ text{(corresponding angles of}`
		`\ \ \ text{congruent triangles)}`
`:.∠YAB= ∠ YAM = ∠YCM=theta`

♦♦ Very few students answered this part correctly.
MARKER’S COMMENT: Students who began by making `∠BAC=2theta` were the most successful.

`text(In)\ \ ΔABC,`

`∠ABC + ∠BAC + ∠YCM`	`= 180^@`
`:.90^@ +2theta+theta`	`= 180^@`
`3theta`	`= 90^@`
`:.theta`	`= 30^@`

`text(In)\ \ Delta MYC,`

`:.tan 30^@`	`= (MY)/(MC)`
`(MY)/(MC)`	`= 1/sqrt3`
`(MY)/(2MC)`	`=1/(2 sqrt3)`
`(MY)/(AC)`	`= 1/(2sqrt3)\ \ \ text{(given}\ \ AC=2MC text{)}`
`:.MY : AC`	`= 1 : 2sqrt3.`

L&E, 2ADV E1 2004 HSC 6a

Solve the following equation for `x`:

`e^(2x) + 3e^x − 10 = 0`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`x = ln 2`

Show Worked Solution

`e^(2x) + 3e^x − 10`	`= 0`
`:. (e^x)^2 + 3e^x − 10`	`= 0`

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

`:. X^2 + 3X – 10`	`= 0`
`(X + 5)(X − 2)`	`= 0`
`:. X =2 or -5`

`text(If)\ \ X`	`=2`
`e^x`	`=2`
`ln e^x`	`=ln 2`
`x`	`=ln 2`

`text(If)\ \ X`	`=-5`
`e^x`	`=-5\ \ \ text{(no solution)}`

`:. x=ln 2`

Calculus, 2ADV C3 2004 HSC 5b

A particle moves along a straight line so that its displacement, `x` metres, from a fixed point `O` is given by `x = 1 + 3 cos 2t`, where `t` is measured in seconds.

What is the initial displacement of the particle? (1 mark)

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Sketch the graph of `x` as a function of `t` for `0 ≤ t ≤ pi`. (2 marks)

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Hence, or otherwise, find when AND where the particle first comes to rest after `t = 0`. (2 marks)

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Find a time when the particle reaches its greatest magnitude of velocity. What is this velocity? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(4 m to the right of)\ O.`
`text(See Worked Solutions)`
`t = pi/2\ text(seconds, 2 m to the left of)\ O.`
`text(6 m s)^(−1)`

Show Worked Solution

i. `x = 1 + 3 cos 2t`

`text(When)\ \ t = 0,`

`x`	`= 1 + 3 cos 0`
	`= 1 + 3`
	`= 4`

`:.\ text(Initial displacement is 4 m to the right of)\ O.`

ii. `text(Period)\ = (2pi)/n = (2pi)/2 = pi`

`text(Considering the range)`

`-1`	`<=cos 2t<=1`
`-3`	`<=3cos 2t<=3`
`-2`	`<=1 + 3 cos 2t<=4`

iii.	`x`	`= 1 + 3 cos 2t`
	`:.v`	`= −6 sin 2t`

`text(The particle comes to rest when)\ \ v=0`

`-6 sin 2t`	`= 0`
`sin 2t`	`= 0`
`2t`	`= 0, pi, 2pi…`
`t`	`= 0, pi/2, pi…`

`:.\ text(After)\ \ t=0, text(particle first comes to rest when)`

`t = pi/2\ text(seconds.)`

`text(When)\ t = pi/2,`

`x`	`= 1 + 3 cos 2(pi/2)`
	`= 1 + 3 cos pi`
	`= 1 + 3(−1)`
	`= −2`

`:.\ text(Particle first comes to rest at 2 m to the left of)\ O.`

iv. `x = 1 + 3 cos 2t`

`v`	`= -6 sin 2t`
`a`	`= -12 cos 2t`

`text(MAX occurs when)\ \ a=0`

`−12 cos 2t`	`= 0`
`cos 2t`	`= 0`
`2t`	`= pi/2, (3pi)/2, …`
`t`	`= pi/4, (3pi)/4, …`

`:.\ text(Maximum at)\ \ t=pi/4,\ \ (3pi)/4, …\ text(seconds,)`

`text(When)\ \ t = pi/4\ text(seconds,)`

`v`	`= -6 sin 2(pi/4)`
	`= -6 sin(pi/2)`
	`= −6`

`:.\ text(Maximum is 6 m s)^(−1).`

Financial Maths, 2ADV M1 2004 HSC 5a

Clare is learning to drive. Her first lesson is 30 minutes long. Her second lesson is 35 minutes long. Each subsequent lesson is 5 minutes longer than the lesson before.

How long will Clare’s twenty-first lesson be? (1 mark)

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How many hours of lessons will Clare have completed after her twenty-first lesson? (2 marks)

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During which lesson will Clare have completed a total of 50 hours of driving lessons? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(130 minutes)`
`28\ text(hours)`
`text(30th lesson)`

Show Worked Solution

i. `30, 35, 40, …`

`=>\ text(AP where)\ \ a = 30,\ \ d = 5`

`T_n`	`= a + (n − 1) d`
`T_(21)`	`= 30 + 20(5)`
	`= 30 + 100`
	`= 130`

`:.\ text(Clare’s twenty-first lesson will be 130)`

`text(minutes long.)`

ii. `text(Find)\ \ S_21\ \ text(given)\ \ a = 30, \ d = 5`

`S_n`	`= n/2[2a + (n − 1)d]`
`:.S_21`	`= 21/2[2(30) + 20 × 5]`
	`= 21/2[60 + 100]`
	`= 21/2 × 160`
	`= 1680\ text(minutes)`
	`= 28\ text(hours.)`

(iii) `text(50 hours)`	`= 50 × 60`
	`= 3000\ text(minutes)`

`text(Find)\ \ n,\ \ text(given)\ \ a = 30, \ d = 5, \ S_n = 3000`

`S_n`	`= n/2[2a + (n − 1)d]`
`:. 3000`	`= n/2[2(30) + (n − 1)5]`
`6000`	`= n[60 + 5n − 5]`
`6000`	`= n[55 + 5n]`
`6000`	`= 55n + 5n^2`

`5n^2 + 55n − 6000`	`= 0`
`n^2 + 11n − 1200`	`= 0`

`:.n`	`= (−11 ± sqrt((11)^2 − 4(1)(−1200)))/(2(1))`
	`= (−11 ± sqrt(121 + 4800))/(2)`
	`= (−11 ± sqrt4921)/2`
	`= 29.5749…\ \ \ (n> 0)`

`:.\ text(Clare completes 50 hours of lessons during)`

`text(her 30th lesson.)`

Calculus, EXT1* C3 2004 HSC 4c

In the diagram, the shaded region is bounded by the curve `y = 2 sec x`, the coordinate axes and the line `x = pi/3`. The shaded region is rotated about the `x`-axis.

Calculate the exact volume of the solid of revolution formed. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`4sqrt3\ pi\ \ text(u³)`

Show Worked Solution

`y`	`= 2 sec x`
`:. y^2`	`= 4 sec^2 x`

`V`	`= pi int_0^(pi/3) y^2\ dx`
	`= pi int_0^(pi/3) 4 sec^2 x\ dx`
	`= 4pi int_0^(pi/3) sec^2 x\ dx`
	`= 4pi[tan x]_0^(pi/3)`
	`= 4pi(tan\ pi/3 − tan 0)`
	`= 4pi(sqrt3 − 0)`
	`= 4sqrt3\ pi\ \ text(u³)`

Calculus, MET2 2012 VCAA 1

A solid block in the shape of a rectangular prism has a base of width `x` cm. The length of the base is two-and-a-half times the width of the base.

The block has a total surface area of 6480 sq cm.

Show that if the height of the block is `h` cm, `h = (6480-5x^2)/(7x).` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
The volume, `V` cm³, of the block is given by `V(x) = (5x(6480-5x^2))/14.`
Given that `V(x) > 0` and `x > 0`, find the possible values of `x`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Find `(dV)/(dx)`, expressing your answer in the form `(dV)/(dx) = ax^2 + b`, where `a` and `b` are real numbers. (3 marks)

--- 4 WORK AREA LINES (style=lined) ---
Find the exact values of `x` and `h` if the block is to have maximum volume. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`x ∈ (0,36)`
`(dV)/(dx) = −75/14 x^2 + (16\ 200)/7`
`(120sqrt3)/7`

Show Worked Solution

a. `text(Total surface area) = 6480`

`6480`	`= 2 xx (5/2 x xx h) + 2xx(5/2x^2) + 2 xx (xh)`
`6480`	`= 5xh + 5x^2 + 2xh`
`7xh`	`= 6480-5x^2`
`:. h`	`= (6480-5x^2)/(7x)\ …\ text(as required)`

b. `V (x) > 0quadtext(for)quadx > 0,`

`=>(6480-5x^2)`	`>0`
`5x^2`	`<6480`
`x`	`<36`

`:. x ∈ (0,36)`

c.	`V(x)`	`=(5x(6480-5x^2))/14`
		`=(-25x^3)/14 + (16\ 200x)/7`
	`(dV)/(dx)`	`=- 75/14 x^2 + (16\ 200)/7`

d. `text(S.P. when)\ (dV)/(dx) = 0quadtext(for)quad x > 0`

`text(Solve:)\ \-75/14 x^2 + (16\ 200)/7=0\ \ text(for)\ x,`

`=> x = 12sqrt3`

`text(Substitute)\ \ x = 12sqrt3quadtext(for)quadh,`

`:. h`	`= (6480-5(12sqrt3)^2)/(7(12sqrt3))`
	`= (120sqrt3)/7`

Calculus, MET2 2012 VCAA 21 MC

The volume, `V` cm³, of water in a container is given by `V = 1/3 pi h^3` where `h` cm is the depth of water in the container at time `t` minutes. Water is draining from the container at a constant rate of 300 cm³/min. The rate of decrease of `h`, in cm/min, when `h = 5` is

A. `12/pi`

B. `4/pi`

C. `25 pi`

D. `60/pi`

E. `30 pi`

Show Answers Only

`A`

Show Worked Solution

`V = 1/3 pi h^3`

`(dV)/(dh) = pi h^2`

`(dV)/(dt) = (dV)/(dh) xx (dh)/(dt),`

`(dV)/(dt) = pi h^2 xx (dh)/(dt)`

`text(When)\ \ h = 5,`

`-300`	`= 25 pi xx (dh)/(dt)`
`(dh)/(dt)`	`= -12/pi`

`:.\ text(Rate of decrease is)\ \12/pi\ \ text(cm/min.)`

`=> A`

Calculus, MET2 2012 VCAA 14 MC

The graph of `f: R^+ uu {0} -> R,\ f(x) = sqrt x` is shown below.

In order to find an approximation to the area of the region bounded by the graph of `f`, the `y`-axis and the line `y = 4`, Zoe draws four rectangles, as shown, and calculates their total area.

Zoe's approximation to the area of the region is

A. `14`

B. `21`

C. `29`

D. `30`

E. `64/3`

Show Answers Only

`D`

Show Worked Solution

`text(Width of each rectangle)\ = 1`

`text(Height of each rectangle:)`

`text(When)\ \ y=1 => 1=sqrt x, \ x=1`

`text(When)\ \ y=2\ => 2=sqrt x, \ x=4`

`text(When)\ \ y=3 => x=9`

`text(When)\ \ y=2 => x=16`

`:. A`	`~~ 1[1 + 4 + 9 + 16]`
	`~~ 30\ \ text(u²)`

`=> D`

Probability, MET2 2012 VCAA 13 MC

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

`text(Pr)(A nn B) = 2/5` and `text(Pr)(A nn B prime) = 3/7`

`text(Pr)(B prime | A)` is equal to

`6/35`
`15/29`
`14/35`
`29/35`
`2/3`

Show Answers Only

`B`

Show Worked Solution

`text(Pr)(B prime \| A)`	`= (text(Pr)(B prime nn A))/(text(Pr)(A))`
	`= (text(Pr)(B prime nn A))/(text(Pr)(B prime nn A)+ text(Pr)(A nn B))`
	`= (3/7)/(3/7 + 2/5)`
	`= 15/29`

`=> B`

Statistics, MET2 2012 VCAA 11 MC

The weights of bags of flour are normally distributed with mean 252 g and standard deviation 12 g. The manufacturer says that 40% of bags weigh more than `x` g.

The maximum possible value of `x` is closest to

`249.0`
`251.5`
`253.5`
`254.5`
`255.0`

Show Answers Only

`E`

Show Worked Solution

`X = text(weight), X ∼ N(252, 12^2)`

`Pr(X < x)`	`= 0.6`
`:. x`	`= 255.04`

`[text(CAS: invNorm) (0.6,252,12)]`

`=> E`

Calculus, MET2 2012 VCAA 10 MC

The average value of the function `f: [0, 2 pi] -> R,\ f(x) = sin^2(x)` over the interval `[0, a]` is 0.4.

The value of `a`, to three decimal places, is

`0.850`
`1.164`
`1.298`
`1.339`
`4.046`

Show Answers Only

`C`

Show Worked Solution

`1/(a – 0) int_0^a(sin^2x)\ dx`	`= 0.4`
`[x- (sin 2x)/2]_0^a`	`=0.4a`
`a-(sin 2a)/2`	`=0.4a`
`:. a`	`= 1.298, quad a∈ (0, 2pi)`

`=> C`

Calculus, MET2 2012 VCAA 9 MC

The normal to the graph of `y = sqrt (b - x^2)` has a gradient of 3 when `x = 1.`

The value of `b` is

A. `-10/9`

B. `10/9`

C. `4`

D. `10`

E. `11`

Show Answers Only

`D`

Show Worked Solution

`y`	`=sqrt (b – x^2)`
`dy/dx`	`=(-2x)/(2 sqrt(b-x^2))`

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

`dy/dx`	`=(-1)/sqrt(b-1)`

`text(S)text(ince)\ \ m_(norm) xx m_(tan) = -1,`

`dy/dx=- 1/3`

`(-1)/sqrt(b-1)`	`=- 1/3`
`sqrt(b-1)`	`=3`
`b-1`	`=9`
`b`	`=10`

`=> D`

CORE*, FUR2 2014 VCAA 1

The adult membership fee for a cricket club is $150.

Junior members are offered a discount of $30 off the adult membership fee.

Write down the discount for junior members as a percentage of the adult membership fee. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Adult members of the cricket club pay $15 per match in addition to the membership fee of $150.

If an adult member played 12 matches, what is the total this member would pay to the cricket club? (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

If a member does not pay the membership fee by the due date, the club will charge simple interest at the rate of 5% per month until the fee is paid.

Michael paid the $150 membership fee exactly two months after the due date.

Calculate, in dollars, the interest that Michael will be charged. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

The cricket club received a statement of the transactions in its savings account for the month of January 2014.

The statement is shown below.

1. Calculate the amount of the withdrawal on 17 January 2014. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
2. Interest for this account is calculated on the minimum balance for the month and added to the account on the last day of the month.
3. What is the annual rate of interest for this account? Write your answer, correct to one decimal place. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(20%)`
`$330`
`$15`
1. `$17\ 000`
2. `3.5`

Show Worked Solution

a. `text(Discount for junior members)`

`= 30/150 xx 100text(%)`

`= 20text(%)`

b. `text(Match Payments)= 12 xx 15=$180`

`:.\ text(Total paid to the club)`	`= 150 + 180`
	`= $330`

c.	`I`	`= (PrT)/100`
		`= (150 xx 5 xx 2)/100`
		`= $15`

d.i. `text(Withdrawal on 17 Jan)`

`= 59\ 700-42\ 700`

`= $17\ 000`

d.ii. `text(Minimum Jan balance) = $42\ 700`

`47\ 200 xx r xx 1/12`	`= 125.12`
`:. r`	`= (125.12 xx 12)/(42\ 700)`
	`= 0.0351…`

`:.\ text(Annual interest rate) = 3.5text(%)`

Calculus, MET2 2012 VCAA 7 MC

The temperature, `T^@C`, inside a building `t` hours after midnight is given by the function

`f: [0, 24] -> R,\ T(t) = 22 - 10\ cos (pi/12 (t - 2))`

The average temperature inside the building between 2 am and 2 pm is

`10°text(C)`
`12°text(C)`
`20°text(C)`
`22°text(C)`
`32°text(C)`

Show Answers Only

`D`

Show Worked Solution

`text(Period) = (2pi)/n = (2pi)/(pi/12) = 24`

`text(At 2 am,)\ \ t=2,`

`T(2) = 22 – 10\ cos (0) = 12`

`text(At 2 pm,)\ \ t=14,`

`T(14) = 22 – 10\ cos (pi) = 32`

`text(Symmetry of graph means the average)`

`text(temperature occurs at)\ \ t=8:`

`T(8) = 22 – 10\ cos ((pi)/2) = 22`

`=> D`

Graphs, MET2 2012 VCAA 3 MC

The range of the function `f: text{[−2, 3)} -> R,\ f(x) = x^2 - 2x - 8` is

A. `R`

B. `text{(−9, −5]}`

C. `text{(−5, 0)}`

D. `text{[−9, 0]}`

E. `text{[−9, −5)}`

Show Answers Only

`D`

Show Worked Solution

`text(Sketching the quadratic:)`

`:.\ text(Range) = [−9,0]`

`=> D`

Graphs, MET2 2012 VCAA 5 MC

Let the rule for a function `g` be `g (x) = log_e ((x - 2)^2).` For the function `g`, the

maximal domain `= R^+` and range `= R`
maximal domain `= R text(\{2})` and range `= R`
maximal domain `= R text(\{2})` and range `=\ text{(−2, ∞)`
maximal domain `= [2, oo)` and range `= (0, oo)`
maximal domain `= [2, oo)` and range `= [0, oo)`

Show Answers Only

`B`

Show Worked Solution

`text(Domain:)\ (x – 2)^2 > 0`

`:. x != 2`

`text(Range: sketch graph)`

`:. text(Range) = R`

`=> B`

Algebra, STD2 A1 SM-Bank 9

The volume of a sphere is given by `V = 4/3 pi r^3` where `r` is the radius of the sphere.

If the volume of a sphere is `\text{220 cm}^3`, find the radius, to 1 decimal place. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`3.7\ \ text{cm (to 1 d.p.)}`

Show Worked Solution

`V`	`= 4/3 pi r^3`
`3V`	`= 4 pi r^3`
`r^3`	`= (3V)/(4 pi)`

`text(When)\ \ V = 220`

`r^3`	`= (3 xx 220)/(4 pi)`
	`= 52.521…`
`:. r`	`=root3 (52.521…)`
	`= 3.744…\ \ \ text{(by calc)}`
	`= 3.7\ \ text{cm (to 1 d.p.)}`

Algebra, MET2 2013 VCAA 13 MC

If the equation `f(2x) - 2f(x) = 0` is true for all real values of `x`, then the rule for `f` could be

`x^2/2`
`sqrt (2x)`
`2x`
`log_e (x/2)`
`x - 2`

Show Answers Only

`C`

Show Worked Solution

`text(We need)\ \ f(2x)=2\ f(x),`

`text(Consider)\ C,`

`f(x)`	`=2x,`
`f(2x)`	`= 2(2x)`
	`= 2\ f(x)`

`text(CAS can be used to quickly prove other answers)`

`text(do not satisfy equation.)`

`=> C`

Probability, MET2 2013 VCAA 9 MC

Harry is a soccer player who practises penalty kicks many times each day.

Each time Harry takes a penalty kick, the probability that he scores a goal is 0.7, independent of any other penalty kick.

One day Harry took 20 penalty kicks.

Given that he scored at least 12 goals, the probability that Harry scored exactly 15 goals is closest to

A. `0.1789`

B. `0.8867`

C. `0.8`

D. `0.6396`

E. `0.2017`

Show Answers Only

`E`

Show Worked Solution

`text(Let)\ \ X =\ text(number of goals,)`

`X ∼\ text(Bi)(20, 0.7)`

`text(Pr) (X = 15 | X >= 12)`

`= (text(Pr) (X = 15))/(text(Pr) (X >= 12))`

`~~ 0.2017`

`=> E`

Calculus, MET2 2013 VCAA 6 MC

For the function `f(x) = sin (2 pi x) + 2x,` the average rate of change for `f(x)` with respect to `x` over the interval `[1/4, 5]` is

A. `0`

B. `34/19`

C. `7/2`

D. `(2 pi + 10)/4`

E. `23/4`

Show Answers Only

`B`

Show Worked Solution

`text(Average rate of change for)\ \f(x)\ \ text(over)\ \ [1/4, 5]`

`= (f(5) – f(1/4))/(5 – 1/4)`

`= (sin (10 pi) + 10 – (sin\ pi/2 +1/2))/(19/4)`

`= 34/19`

`=> B`

Algebra, MET2 2013 VCAA 3 MC

If `x + a` is a factor of `7x^3 + 9x^2 - 5ax`, where `a in R text(\){0}`, then the value of `a` is

A. `-4`

B. `-2`

C. `-1`

D. `1`

E. `2`

Show Answers Only

`E`

Show Worked Solution

`text(S)text(ince)\ \ x+a\ \ text(is a factor),\ \ f(-a)=0`

`7(-a)^3 + 9(-a)^2 – 5a(-a) = 0,`

`-7a^3 + 14a^2`	`=0`
`-7a^2 (a – 2)`	`=0`
`:. a = 2,`	`\ \ \ a!=0`

`=> E`

Algebra, MET2 2013 VCAA 2 MC

The midpoint of the line segment that joins `text{(1, −5)}` to `(d, 2)` is

`((d + 1)/2, -3/2)`
`((1 - d)/2, -7/2)`
`((d - 4)/2, 0)`
`(0, (1 - d)/3)`
`((5 + d)/2, 2)`

Show Answers Only

`A`

Show Worked Solution

`text(Midpoint)`	`= ((d + 1)/2, (2 + (-5))/2)`
	`= ((d + 1)/2, -3/2)`

`=> A`

Polynomials, EXT2 2015 HSC 16b

Let `n` be a positive integer.

By considering `(cos alpha + i sin alpha)^(2n)`, show that
`cos(2n alpha) = cos^(2n) alpha - ((2n), (2)) cos^(2n - 2) alpha sin^2 alpha + ((2n), (4)) cos^(2n - 4) alpha sin^4 alpha - …`
1. `+ … + (-1)^(n - 1) ((2n), (2n - 2)) cos^2 alpha sin^(2n - 2) alpha + (-1)^n sin^(2n) alpha.`
Let `T_(2n) (x) = cos(2n cos^-1 x)`, for `-1 <= x <= 1.` (2 marks)
Show that
`T_(2n)(x) = x^(2n) - ((2n), (2)) x^(2n - 2)(1 - x^2) + ((2n), (4)) x^(2n - 4) (1 - x^2)^2 +`
1. `… + (-1)^n (1 - x^2)^n.` (2 marks)
By considering the roots of `T_(2n) (x)`, find the value of
1. `cos(pi/(4n)) cos((3 pi)/(4n)) …\ cos (((4n - 1) pi)/(4n)).` (3 marks)
Prove that
1. `1 - ((2n), (2)) + ((2n), (4)) - ((2n), (6)) + … + (-1)^n ((2n), (2n)) = 2^n cos ((n pi)/2).` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(-1)^n/2^(2n – 1)`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `text(By De Moivre)`

`(cos alpha + i sin alpha)^(2n)=cos (2n alpha) + i sin (2n alpha)`

`text(By the Binomial Theorem)`

`(cos alpha + i sin alpha)^(2n)`

`=cos^(2n) alpha + ((2n), (1)) cos^(2n – 1) alpha (i sin alpha) + ((2n), (2)) cos^(2n – 2) alpha (isin alpha)^2 +`

`… ((2n), (2n – 2)) cos^2 alpha (isin alpha)^(2n-2) + ((2n), (2n – 1)) cos alpha (i sin alpha)^(2n-1) + (i sin alpha)^(2n)`

`text(Equating real parts,)`

`cos(2n alpha) = cos^(2n) alpha – ((2n), (2)) cos^(2n – 2) alpha sin^2 alpha + … `

`+ (-1)^(n – 1) ((2n), (2n – 2)) cos^2 alpha sin^(2n – 2) alpha + (-1)^n sin^(2n) alpha`

(ii) `T_(2n) (x) = cos (2n cos^-1 x)\ \ text(for)\ \ -1 <= x <= 1`

`text(Given)\ \ alpha = cos^-1 x,\ \ =>x = cos alpha`

`text(Substituting)\ \ x = cos alpha\ \ text{into part (i),}`

`cos (2n cos^-1 x)`

`= x^(2n) – ((2n), (2)) x^(2n – 2) (1 – cos^2 alpha) + ((2n), (4)) x^(2n – 4) (1 – cos^2 alpha)^2 +`

`… + (-1)^n (1 – cos^2 alpha)^n`

`:. T_(2n) (x) = x^(2n) – ((2n), (2)) x^(2n – 2) (1 – x^2) + ((2n), (4)) x^(2n – 4)(1 – x^2)^2 +`

`… + (-1)^n (1 – x^2)^n`

♦♦♦ Mean mark 8%!
STRATEGY: Note that finding the correct roots gained 2 full marks in this part.

(iii) `T_(2n) (x) = 0\ \ text(when)`

`cos (2n cos^-1 x)`	`=0`
`2n cos^-1 x`	`=pi/2, (3 pi)/2, (5 pi)/2, … , ((2k + 1) pi)/2,\ \ k = 0, 1, 2 … , (2n – 1)`
`cos^-1 x`	`= ((2k + 1) pi)/(4n),\ \ k = 0, 1, 2 …, (2n – 1)`
`:. x`	`= cos(pi/(4n)), cos((3 pi)/(4n)), … , cos((4n – 1) pi)/(4n)`

`=>T_(2n) (x) = 0\ \ text(has degree)\ \ 2n`

`=>T_(2n) (x) = 0\ \ text(has)\ \ 2n\ \ text(distinct roots)`

`:.\ text(Product of roots of)\ \ T_(2n) (x) = 0\ \ text(is the constant term)`

`text(divided by the coefficient of)\ \ x^(2n).`

`text(Constant term is)\ \ T_(2n) (0)=(-1)^n`

`text(Coefficient of)\ \ x^(2n)\ \ text(is)\ \ (1 + ((2n), (2)) + ((2n), (4)) + … + 1)`

`:.cos(pi/(4n)) cos((3 pi)/(4n)) cos ((5 pi)/(4n)) …\ cos(((4n – 1) pi)/(4n))`

`=((-1)^n)/((1 + ((2n), (2)) + ((2n), (4)) + … + 1))`

`text(*The denominator can be further simplified to)\ \ 2^(2n-1)\ \ text(by)`

`text(using the binomial expansion of)\ \ (1+x)^(2n)`

♦♦ Mean mark 14%.

(iv) `cos(2n cos^-1 x)= cos ((n pi)/2)\ \ text(when)`

`cos^-1 x=pi/4\ \ \ =>x=1/sqrt2`

`text{Using part (ii)}`

`cos((n pi)/2)`	`=(1/sqrt 2)^(2n) – ((2n), (2)) (1/sqrt 2)^(2n – 2) (1/2) +`
	`((2n), (4)) (1/sqrt 2)^(2n – 4) (1/2)^2 – … + (-1)^n (1/2)^n`
	` = 1/2^n – ((2n), (2)) 1/2^(n-2) 1/2^2 + ((2n), (4)) 1/2^(n-4) 1/2^4 – … + (-1)^n 1/2^n`
	`=1/2^n – ((2n), (2)) 1/2^n + ((2n), (4)) 1/2^n – … + (-1)^n 1/2^n`
`:.2^n cos ((n pi)/2)`	`=1 – ((2n), (2)) + ((2n), (4)) – ((2n), (6)) + … + (-1)^n`

Harder Ext 1 Topics, EXT2 2015 HSC 16a

A table has `3` rows and `5` columns, creating `15` cells as shown.
Counters are to be placed randomly on the table so that there is one counter in each cell. There are `5` identical black counters and `10` identical white counters.
Show that the probability that there is exactly one black counter in each column is `81/1001.` (2 marks)
The table is extended to have `n` rows and `q` columns. There are `nq` counters, where `q` are identical black counters and the remainder are identical white counters. The counters are placed randomly on the table with one counter in each cell.
Let `P_n` be the probability that each column contains exactly one black counter.
Show that `P_n = n^q/(((nq), (q))).` (2 marks)
Find `lim_(n -> oo) P_n.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(q!)/q^q`

Show Worked Solution

(i) `text(Arrangements of 5 black counters in 15 cells)`

`=\ ^15C_5`

`text(In each column, there are 3 cells to place a black counter.)`

`:.\ text(Ways to place a black counter in each column)`

`\ =3^5=243`

`:.P text{(1 black counter in each column)`

`=243/(\ ^15C_5`

`=81/1001`

(ii) `text{Similarly to part (i),}`

♦♦ Mean mark 16%.

`text(Arrangements of)\ \ q\ \ text(black counters in)\ \ nq\ \ text(cells)`

`=\ ^(nq)C_q`

`text(In each column, there are)\ \ n\ \ text(cells to place a black counter.)`

`:.\ text(Ways to place a black counter in each of)\ \ q\ \ text(columns)`

`\ =n^q`

`:.P text{(1 black counter in each column)`

`=n^q/(\ ^(nq)C_q)`

♦♦♦ Mean mark 5%.
COMMENT: Lowest State mean mark on the 2015 exam.

(iii)	`lim_(n -> oo) P_n`	`= lim_(n -> oo) n^q/(\ ^(nq)C_q)`
		`= lim_(n -> oo)[n^q xx (q!)/((nq)(nq – 1)…(nq – q + 1))]`
		`= lim_(n -> oo)[(n^q q!)/(n^q q(q – 1/n)(q – 2/n) … (q – (q – 1)/n))]`
		`= lim_(n -> oo)[(q!)/(q(q – 1/n)(q – 2/n) … (q – (q – 1)/n))]`
		`= (q!)/q^q`

Proof, EXT2 P1 2015 HSC 15c

For positive real numbers `x` and `y`, `sqrt (xy) <= (x + y)/2`. (Do NOT prove this.)

Prove `sqrt (xy) <= sqrt ((x^2 + y^2)/2)`, for positive real numbers `x` and `y.` (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Prove `root4(abcd) <= sqrt ((a^2 + b^2 + c^2 + d^2)/4)`, for positive real numbers `a, b, c` and `d.` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i. `text(S)text(ince)\ \ (x – y)^2 = x^2 + y^2 – 2xy`

`and \ (x – y)^2 >= 0`

`0`	`≤x^2 + y^2 – 2xy`
`2xy`	`≤x^2 + y^2`
`:.sqrt (xy)`	`≤sqrt ((x^2 + y^2)/2)`

ii. `sqrt(ab) <= sqrt((a^2 + b^2)/2),\ \ \ \ text{(part (i))}`

♦♦ Mean mark 29%.

`sqrt(cd) <= sqrt((c^2 + d^2)/2),\ \ \ \ text{(part (i))}`

`sqrt(ab) sqrt(cd)`	`<=sqrt((a^2 + b^2)/2)*sqrt((c^2 + d^2)/2)`
	`<=sqrt(((a^2 + b^2)/2) * ((c^2 + d^2)/2))`

`sqrt (xy) <= (x + y)/2\ \ \ \ text{(given):}`
`sqrt(ab) sqrt(cd)`	`<=1/2((a^2 + b^2+c^2+d^2)/2)`
`sqrt(abcd)`	`<=(a^2 + b^2+c^2+d^2)/4`
`:.root4(abcd)`	`<=sqrt((a^2 + b^2+c^2+d^2)/4)`

Proof, EXT2 P1 2015 HSC 15b

Suppose that `x >= 0` and `n` is a positive integer.

Show that `1 - x <= 1/(1 + x) <= 1.` (2 marks)

--- 2 WORK AREA LINES (style=lined) ---
Hence, or otherwise, show that

`1 - 1/(2n) <= n ln (1 + 1/n) <= 1.` (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Hence, explain why

`lim_(n -> oo) (1 + 1/n)^n = e.` (1 mark)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i.	`1-x^2`	`<=1\ \ \ text(for)\ x>=0`
	`(1-x)(1+x)`	`<=1`
	`(1-x)`	`<=1/(1+x)`

`text(S)text(ince)\ \ 1 + x >= 1\ \ text(when)\ \ x >= 0`

`=>1/(1 + x) <= 1`

`:. 1 – x <= 1/(1 + x) <= 1.`

♦♦ Mean mark 21%.
STRATEGY: The conversion of the middle term from a fraction into a logarithm should flag the need for integration of each term.

ii.	`int_0^(1/n) (1 – x)\ dx`	`<= int_0^(1/n) (dx)/(1 + x) <= int_0^(1/n) 1\ dx`
	`[x – x^2/2]_0^(1/n)`	`<= [ln (1 + x)]_0^(1/n) <= [x]_0^(1/n)`
	`1/n-1/(2n^2)`	`<= ln (1 + 1/n) <= 1/n`
	`1 – 1/(2n) `	`<= n ln (1 + 1/n) <= 1,\ \ \ \ \ (n>=1)`

iii.	`lim_(n -> oo) (1 – 1/(2n))`	`<= lim_(n -> oo){n ln (1 + 1/n)} <= lim_(n -> oo) (1)`
	`1`	`<= lim_(n -> oo) {l ln (1 + 1/n)} <= 1`

♦♦ Mean mark 29%.

`:. lim_(n -> oo) (n ln (1 + 1/n))`	`=1`
`lim_(n -> oo) ln (1 + 1/n)^n`	`=1`
`:.lim_(n -> oo) (1 + 1/n)^n`	`=e`

Mechanics, EXT2 M1 2015 HSC 15a

A particle `A` of unit mass travels horizontally through a viscous medium. When `t = 0`, the particle is at point `O` with initial speed `u`. The resistance on particle `A` due to the medium is `kv^2`, where `v` is the velocity of the particle at time `t` and `k` is a positive constant.

When `t = 0`, a second particle `B` of equal mass is projected vertically upwards from `O` with the same initial speed `u` through the same medium. It experiences both a gravitational force and a resistance due to the medium. The resistance on particle `B` is `kw^2`, where `w` is the velocity of the particle `B` at time `t`. The acceleration due to gravity is `g`.

Show that the velocity `v` of particle `A` is given by

`1/v = kt + 1/u.` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
By considering the velocity `w` of particle `B`, show that

`t = 1/sqrt(gk) (tan^-1(u sqrt(k/g)) - tan^-1 (w sqrt(k/g))).` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that the velocity `V` of particle `A` when particle `B` is at rest is given by

`1/V = 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g)).` (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Hence, if `u` is very large, explain why

`V ~~ 2/pi sqrt(g/k).` (1 mark)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

i. `text(Particle)\ A:`

`ddot x`	`= -kv^2`
`(dv)/(dt)`	`= -kv^2`
`(dt)/(dv)`	`=- 1/(kv^2)`
`t`	`= -1/k int 1/v^2\ dv`
`-kt`	`= -1/v + c`
`text(When)\ \ t=0,\ \ v=u\ \ \ \ =>c=1/u`
`-kt`	`= -1/v + 1/u`
`:.1/v`	`= kt + 1/u`

ii. `text(Particle)\ B:`

`ddot x`	`= -g – kw^2`
`(dw)/(dt)`	`= -g – kw^2`
`(dt)/(dw)`	`=-1/(g + kw^2)`
`t`	`= – int (dw)/(g + kw^2)`
	`= -1/k int (dw)/(g/k + w^2)`
	`= -1/k xx 1/sqrt (g/k) tan^-1(w/sqrt (g/k)) + c`
	`= -1/sqrt (gk)\ tan^-1 ((sqrt k w)/sqrt g) + c`

`text(When)\ \ t = 0,\ \ w = u`

`=>c= 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`

`:. t`	`= -1/sqrt (gk) tan^-1 ((sqrt k w)/sqrt g) + 1/sqrt (gk) tan^-1 ((sqrt k u)/sqrt g)`
	`=1/sqrt (gk) (tan^-1 (u sqrt(k/g)) – 1/sqrt (gk) tan^-1 (w sqrt (k/g)))`

iii. `B\ \ text(at rest when)\ \ w = 0`

`t = 1/sqrt (gk) (tan^-1 (u sqrt (k/g)))`

`:.1/V`	`= k xx 1/sqrt(gk) tan^-1 (u sqrt (k/g)) + 1/u,\ \ \ \ \ text{(part (i))}`
	`= 1/u + sqrt(k/g) tan^-1 (u sqrt (k/g))`

iv. `1/V = 1/u + sqrt (k/g) tan^-1 (u sqrt (k/g))`

`text(As)\ \ u -> oo,\ \ tan^-1 (u sqrt (k/g)) -> pi/2`

`:.\ text(If)\ \ u\ \ text(is very large,)`

`1/V`	`~~ 0 + sqrt (k/g) xx pi/2`
`:.V`	`~~ 2/pi sqrt (g/k)`

Mechanics, EXT2 2015 HSC 14c

A car of mass `m` is driven at speed `v` around a circular track of radius `r`. The track is banked at a constant angle `theta` to the horizontal, where `0 < theta < pi/2`. At the speed `v` there is a tendency for the car to slide up the track. This is opposed by a frictional force `mu N`, where `N` is the normal reaction between the car and the track, and `mu > 0`. The acceleration due to gravity is `g`.

Show that
`v^2 = rg((tan theta + mu)/(1 - mu tan theta)).` (3 mark)
At the particular speed `V`, where `V^2 = rg`, there is still a tendency for the car to slide up the track.
Using the result from part (i), or otherwise, show that `mu < 1.` (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

`text(Resolving the forces vertically)`

`N cos theta`	`= mg + mu N sin theta`
`N (cos theta – mu sin theta)`	`= mg\ \ \ …\ (1)`

`text(Resolving the forces horizontally)`

`N sin theta + mu N cos theta`	`=(m v^2)/r`
`N (sin theta + mu cos theta)`	`=(m v^2)/r\ \ \ …\ (2)`

`text(Divide)\ \ (2)÷(1)`

`((m v^2)/r)/(mg)`	`=(sin theta + mu cos theta)/(cos theta – mu sin theta)`
`v^2/(rg)`	`=(sin theta + mu cos theta)/(cos theta – mu sin theta)`
`v^2`	`=rg ((sin theta + mu cos theta)/(cos theta – mu sin theta))`
	`= rg ((tan theta + mu)/(1 – mu tan theta))`

(ii) `text(Given that)\ \ V^2=rg`

♦ Mean mark 46%.

`=> (tan theta + mu)/(1 – mu tan theta)=1`

`(tan theta + mu)`	`=(1 – mu tan theta)`
`mu(1+ tan theta)`	`=1-tan theta`
`mu`	`=(1-tan theta)/(1+ tan theta)`
	`=1- (2tan theta)/(1+ tan theta)`

`text(S)text(ince)\ \ tan theta>0\ \ text(for)\ \ 0<theta<pi/2`

`=> (2tan theta)/(1+ tan theta) >0`

`:. mu<1`

Functions, EXT1′ F2 2015 HSC 14b

The cubic equation `x^3 – px + q = 0` has roots `alpha, beta` and `gamma`.

It is given that `alpha^2 + beta^2 + gamma^2 = 16` and `a^3 + beta^3 + gamma^3 = -9`.

Show that `p = 8.` (1 mark)
Find the value of `q.` (2 marks)
Find the value of `alpha^4 + beta^4 + gamma^4.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`3`
`128`

Show Worked Solution

(i) `alpha + beta + gamma = – b/a=0`

`alpha beta + beta gamma + gamma alpha = c/a= -p`

`(alpha + beta + gamma)^2`	`= alpha^2 + beta^2 + gamma^2 + 2(alpha beta + beta gamma + gamma alpha)`
`0`	`= 16 – 2p`
`:. p`	`= 8`

(ii)	`alpha^3 – 8 alpha + q`	`=0\ \ \ …\ (1)`
	`beta^3 – 8 beta + q`	`=0\ \ \ …\ (2)`
	`gamma^3 – 8 gamma + q`	`=0\ \ \ …\ (3)`

`(1)+(2)+(3)`

`(alpha^3 + beta^3 + gamma^3) – 8(alpha + beta + gamma) + 3q = 0`

`-9 – 0 + 3q`	`=0`
`q`	`=3`

COMMENT: Part (iii) proved challenging for many students with the State mean mark just over 50%.

(iii)	`alpha(alpha^3 – 8 alpha + 3)`	`=0\ \ \ …\ (1)`
	`beta(beta^3 – 8 beta + 3)`	`=0\ \ \ …\ (2)`
	`gamma(gamma^3 – 8 gamma + 3)`	`=0\ \ \ …\ (3)`

`(1)+(2)+(3)`

`alpha^4 + beta^4 + gamma^4 – 8(alpha^2 + beta^2 + gamma^2) + 3(alpha + beta + gamma) = 0`

`alpha^4 + beta^4 + gamma^4`	`= 8 xx 16 – 0`
	`= 128`

Calculus, EXT2 C1 2015 HSC 14a

Differentiate `sin^(n - 1) theta cos theta`, expressing the result in terms of `sin theta` only. (2 marks)

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Hence, or otherwise, deduce that

`int_0^(pi/2) sin^n theta\ d theta = ((n-1))/n int_0^(pi/2) sin^(n - 2) theta\ d theta`, for `n>1.` (2 marks)

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Find `int_0^(pi/2) sin^4 theta\ d theta.` (1 mark)

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`text(See Worked Solutions)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(3 pi)/16`

Show Worked Solution

i. `d/(d theta) (sin^(n – 1) theta cos theta)`

`=(n – 1) sin^(n – 2) theta cos theta cos theta + sin^(n – 1) theta xx (-sin theta)`

`=(n – 1) sin^(n – 2) theta cos^2 theta – sin^n theta`

`=(n – 1) sin^(n – 2) theta (1 – sin^2 theta) – sin^n theta`

`=(n – 1) sin^(n – 2) theta – (n – 1) sin^n theta – sin^n theta`

`=(n – 1) sin^(n – 2) theta – n sin^n theta`

ii. `text{From part (i)}`

`n sin^n theta = (n – 1) sin^(n – 2) theta – d/(d theta) (sin^(n – 1) theta cos theta)`

`:. int_0^(pi/2) sin^n theta\ d theta`

`=1/n int_0^(pi/2) ((n – 1) sin^(n – 2) theta – d/(d theta) (sin^(n – 1) theta cos theta)) d theta`

`=1/n int_0^(pi/2) (n – 1) sin^(n – 2) theta\ d theta – 1/n int_0^(pi/2) d/(d theta) (sin^(n – 1) theta cos theta)\ d theta`

`= 1/n int_0^(pi/2) (n – 1) sin^(n – 2) theta\ d theta – 1/n int_0^(pi/2) d/(d theta) (sin^(n – 1) theta cos theta)\ d theta`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta – 1/n [sin^(n – 1) theta cos theta]_0^(pi/2)`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta – 1/n (0 – 0)`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta,\ \ \ \ (n>1)`

iii.	`int_0^(pi/2) sin^4 theta\ d theta`	`= 3/4 int_0^(pi/2) sin^2 theta\ d theta`
		`= 3/4 xx [(2-1)/2 int_0^(pi/2) d theta]`
		`= 3/8 xx [theta]_0^(pi/2)`
		`= (3 pi)/16`

Harder Ext1 Topics, EXT2 2015 HSC 13c

A small spherical balloon is released and rises into the air. At time `t` seconds, it has radius `r` cm, surface area `S = 4 pi r^2` and volume `V = 4/3 pi r^3`.

As the balloon rises it expands, causing its surface area to increase at a rate of `((4 pi)/3)^(1/3)\ \text(cm)^2 text(s)^-1`. As the balloon expands it maintains a spherical shape.

By considering the surface area, show that
1. `(dr)/(dt) = 1/(8 pi r) (4/3 pi)^(1/3).` (2 marks)
Show that
1. `(dV)/(dt) = 1/2 V^(1/3).` (2 marks)
When the balloon is released its volume is `8000\ text(cm³)`. When the volume of the balloon reaches `64000\ text(cm³)` it will burst.
How long after it is released will the balloon burst? (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`1\ \ text(hour)`

Show Worked Solution

(i)	`(dS)/(dt)`	`= (dS)/(dr) xx (dr)/(dt)`
	`((4 pi)/3)^(1/3)`	`=8 pi r xx (dr)/(dt)`
	`(dr)/(dt)`	`=((4 pi)/3)^(1/3) xx 1/(8 pi r)`
		`=1/(8 pi r) (4/3 pi)^(1/3)`

(ii) `(dV)/(dt) = (dV)/(dr) xx (dr)/(dt)`

`(dV)/(dt) =`	`4 pi r^2 xx ((4 pi)/3)^(1/3) xx 1/(8 pi r)`
`=`	`r/2 xx ((4 pi)/3)^(1/3)`
`=`	`1/2 (4/3 pi r^3)^(1/3)`
`=`	`1/2 V^(1/3)`

(iii) `text(Find)\ \ t\ \ text(when)\ \ V=64\ 000`

`text(When)\ \ t = 0,\ \ V = 8000`

`(dV)/(dt)`	`=1/2 V^(1/3)`
`(dt)/(dV)`	`=2/V^(1/3)`
`int_0^t\ dt`	`= int_8000^64000 2/V^(1/3)\ dV`
`:.t`	`= [3V^(2/3)]_8000^64000`
	`= 3(1600 – 400)`
	`= 3600\ \ text(seconds)`
	`= 1\ \ text(hour)`

Volumes, EXT2 2015 HSC 13b

Two quarter cylinders, each of radius `a`, intersect at right angles to form the shaded solid.

A horizontal slice `ABCD` of the solid is taken at height `h` from the base. You may assume that `ABCD` is a square, and is parallel to the base.

Show that `AB = sqrt(a^2 - h^2).` (1 mark)
Find the volume of the solid. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`(2a^3)/3\ \ text(u³)`

Show Worked Solution

(i)

`text(The cylinder has equation)\ \ x^2 + y^2 = a^2`

`text(When)\ \ y = h,`

`x`	`= sqrt (a^2 – h^2)`
`:. AB`	`= sqrt (a^2 – h^2)`

(ii) `BC = sqrt (a^2 – h^2)`

`text(Area of)\ \ ABCD = (a^2 – h^2)`

`:.text(Volume of solid) =`	`int_0^a (a^2 – h^2) dh`
`=`	`[a^2 h – h^3/3]_0^a`
`=`	`[(a^3 – a^3/3) – 0]`
`=`	`(2a^3)/3\ \ text(u³)`

Volumes, EXT2 2015 HSC 12d

The diagram shows the graph `y = sqrt (x + 1)` for `0 <= x <= 3`. The shaded region is rotated about the line `x = 3` to form a solid.

Use the method of cylindrical shells to find the volume of the solid. (4 marks)

Show Answers Only

`(188 pi)/15\ \ text(u³)`

Show Worked Solution

`text(Solution 1)`

`δV=2pi r h\ δx,\ \ \ text(where)`

`r=3-x,\ \ and\ \ h=y=sqrt (x+1)`

`:.V`	`=2 pi int_0^3 (3 – x) sqrt (x + 1)\ dx`
	`=2 pi int_0^3 (3 sqrt (x + 1) – x sqrt (x + 1))\ dx`
	`=2 pi [2 (x + 1)^(3/2)]_0^3- 2 pi int_0^3 x sqrt (x + 1))\ dx`

`text(Integrating by parts)`

`u`	`=x`	`u′`	`=1`
`v′`	`=(x+1)^(1/2)`	`v`	`=2/3 (x+1)^(3/2)`

`:.V`	`=2 pi [2 (x + 1)^(3/2)]_0^3 – 2 pi ([2/3 x (x + 1)^(3/2)]_0^3 – int_0^3 2/3 (x + 1)^(3/2)\ dx)`
	`=4 pi (8 – 1) – (4 pi)/3 (24 – 0 – [2/5 (x + 1)^(5/2)]_0^3)`
	`=28 pi – (4 pi)/3 (24 – [2/5 xx 32 – 2/5])`
	`=28 pi – (4 pi)/3 xx 58/5`
	`=(188 pi)/15\ \ text(u³)`

`text{Solution 2 (Substitution)}`

`int_0^3 x sqrt(x + 1)\ dx`

`text(Let)\ \ u = x + 1,\ \ du = dx,\ \ x = u – 1`

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

`text(When)\ \ x = 3,\ \ u = 4`

`int_0^3 x sqrt (x + 1)\ dx =`	`int_1^4 (u – 1) sqrt u\ du`
`=`	`int_1^4 (u^(3/2) – u^(1/2)) du`
`=`	`[2/5 u^(5/2) – 2/3 u^(3/2)]_1^4`
`=`	`(64/5 – 16/3) – (2/5 – 2/3)`
`=`	`62/5 – 14/3`
`=`	`116/15`

`:.V =`	`2 pi [2 (x + 1)^(3/2)]_0^3 – 2 pi xx 116/15`
`=`	`28 pi – (232 pi)/15`
`=`	`(188 pi)/15\ \ text(u³)`

Polynomials, EXT2 2015 HSC 12b

The polynomial `P(x) = x^4 - 4x^3 + 11x^2 - 14x + 10` has roots `a + ib` and `a + 2ib` where `a` and `b` are real and `b != 0.`

By evaluating `a` and `b`, find all the roots of `P(x).` (3 marks)
Hence, or otherwise, find one quadratic polynomial with real coefficients that is a factor of `P(x).` (1 mark)

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`1 +- i,\ \ 1 +- 2i`
`x^2 − 2x + 2\ \ or\ \ x^2 − 2x + 5`

Show Worked Solution

(i) `text(S)text(ince coefficients of)\ \ P(x)\ \ text(are real,)`

`=>\ text(Complex roots occur in conjugate pairs)`

`=>\ text(Roots are)\ \ a +- ib\ \ and\ \ a +- 2ib`

`text(Sum of roots) = -b/a=4`

`4`	`=a + ib + a – ib + a + 2ib + a – 2ib`
`4a`	`=4`
`:.a`	`=1`

`text(Products of roots)`

`(a + ib) (a – ib) (a – 2ib) (a – 2ib)`	`= 10`
`(a^2 + b^2) (a^2 + 4b^2)`	`= 10`
`(1 + b^2) (1 + 4b^2)`	`= 10`
`4b^4 + 5b^2 + 1`	`= 10`
`4b^4 + 5b^2 – 9`	`= 0`
`(4b^2 + 9) (b^2 – 1)`	`= 0`

`:.b^2 = 1,\ \ \ \ (b\ \ text{is real})`

`:.b = +- 1`

`:.P(x)\ text(has roots)\ \ \ 1 +- i,\ 1 +- 2i.`

(ii)	`P(x)`	`=(x – 1 – i) (x – 1 + i)(x-1-2i)(x-1+2i)`
		`=(x^2 – 2x + 2)(x^2 – 2x + 5)`

Complex Numbers, EXT2 N1 2015 HSC 12a

The complex number `z` is such that `|\ z\ |=2` and `text(arg)(z) = pi/4.`

Plot each of the following complex numbers on the same half-page Argand diagram.

`z` (1 mark)

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`u = z^2` (1 mark)

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`v = z^2 - bar z` (1 mark)

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`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

i. `z =`	`2 text(cis) pi/4`
`=`	`sqrt 2 (1 + i)`

ii. `u =`	`z^2`
`=`	`4 text(cis)\ pi/2`
`=`	`4i`

COMMENT: 12a(iii) had the lowest mean mark (55%) of any part within Q12 and deserves attention.

iii. `v =`	`z^2 – bar z`
`=`	`4i – sqrt 2 (1 – i)`
`=`	`- sqrt 2 + (4 + sqrt 2) i`

Calculus, EXT2 C1 2015 HSC 11f

Show that

`cot theta + text(cosec)\ theta = cot(theta/2).` (2 marks)

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Hence, or otherwise, find

`int (cot theta + text(cosec)\ theta)\ d theta.` (1 mark)

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Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`2 ln\ |sin\ theta/2| + c`

Show Worked Solution

i. `cot theta + text(cosec)\ theta =`	`(cos theta)/(sin theta) + 1/(sin theta)`
`=`	`(1 + cos theta)/(sin theta)`
`=`	`(1 + 2 cos^2 (theta/2) – 1)/(2 sin (theta/2) cos (theta/2))`
`=`	`(2 cos^2(theta/2))/(2 sin (theta/2) cos (theta/2))`
`=`	`(cos (theta/2))/(sin (theta/2))`
`=`	`cot (theta/2)`

COMMENT: Part (ii) mean mark 51%.

ii. `int (cot theta + text(cosec)\ theta)\ d theta =`	`int cot (theta/2)\ d theta`
`=`	`int (cos (theta/2))/(sin (theta/2))\ d theta`
`=`	`2 ln\ \|sin\ theta/2\| + c`

Complex Numbers, EXT2 N1 2015 HSC 11b

Consider the complex numbers `z = -sqrt 3 + i` and `w = 3 (cos\ pi/7 + i sin\ pi/7).`

Evaluate `|\ z\ |.` (1 mark)

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Evaluate `text(arg)(z).` (1 mark)

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Find the argument of `z/w.` (1 mark)

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`2`
`(5 pi)/6`
`(29 pi)/42`

Show Worked Solution

i. `\|\ z\ \|`	`= sqrt ((-sqrt3)^2 + 1^2)`
	`= 2`

ii. `text(arg)\ (z) =`	`tan^-1 (1- sqrt 3)`
`=`	`pi – pi/6`
`=`	`(5 pi)/6`

iii. `text(arg) (z/w) =`	`text(arg)\ z – text(arg)\ w`
`=`	`(5 pi)/6 – pi/7`
`=`	`(29 pi)/42`

`xprime`	`= −x + 1`
`1 – xprime`	`= x`

`yprime`	`= 2y – 2`
`(yprime + 2)/2`	`= y`

`(1 – xprime) – 2((yprime + 2)/2)`	`= 3`
`−xprime – yprime`	`= 4`

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

`text(Area)`	`= int_-2^0 x(x + 2)(x – 4)\ dx – int_0^4 x(x + 2)(x – 4)\ dx`
	`= 148/3`