Band 6 – Page 26

Mechanics, EXT2* M1 2006 HSC 4b

A particle is undergoing simple harmonic motion on the `x`-axis about the origin. It is initially at its extreme positive position. The amplitude of the motion is 18 and the particle returns to its initial position every 5 seconds.

Write down an equation for the position of the particle at time `t` seconds. (2 marks)

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How long does the particle take to move from a rest position to the point halfway between that rest position and the equilibrium position? (3 marks)

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Show Answers Only

`x = 18 cos ((2 pi)/5 t)`
`5/6\ \ text(seconds)`

Show Worked Solution

i. `text{Amplitude (A)} = 18`

♦♦♦ “Very poorly” answered (exact data unavailable).

`text(Period) = (2 pi)/n = 5`

`5n`	`= 2 pi`
`n`	`= (2 pi)/5`

`text(Using)\ \ x`	`= A cos n t`
`x`	`= 18 cos ((2 pi)/5 t)`

ii. `text(When)\ \ t= 0,\ \ \ x = 18`

`text(Find)\ \ t\ \ text(when)\ \ x = 9`

`9`	`= 18 cos ((2 pi)/5 t)`
`cos ((2 pi)/5 t)`	`= 1/2`
`(2 pi)/5 t`	`= pi/3`
`t`	`= (5 pi)/(3 xx 2 pi)`
	`= 5/6\ \ text(seconds)`

`:.\ text(It takes the particle)\ \ 5/6\ \ text(seconds to move from)`

`text(rest position and half way to equilibrium.)`

Plane Geometry, EXT1 2006 HSC 3d

The points `P, Q` and `T` lie on a circle. The line `MN` is tangent to the circle at `T` with `M` chosen so that `QM` is perpendicular to `MN`. The point `K` on `PQ` is chosen so that `TK` is perpendicular to `PQ` as shown in the diagram.

Show that `QKTM` is a cyclic quadrilateral. (1 mark)
Show that `/_KMT = /_KQT.` (1 mark)
Hence, or otherwise, show that `MK` is parallel to `TP.` (2 marks)

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)

`/_ QMT = 90^@\ \ \ (QM _|_ MN)`

`/_ QKT = 90^@\ \ \ (PQ _|_ TK)`

`:.\ /_ QMT + /_ QKT = 180^@`

`:.\ QKTM\ \ text(is cyclic)\ \ text{(opposite angles are supplementary)}`

`text(… as required.)`

(ii) `text(Show)\ \ /_ KMT = /_ KQT`

`/_ KMQ = /_ KTQ = theta`

`text{(angles in the same segment on arc}\ \ KQ text{)}`

`/_ KQT`	`= 90 – theta\ \ text{(angle sum of}\ \ Delta KQT text{)}`
`/_ KMT`	`= /_ QMT – /_ KMQ`
	`= 90 – theta`

`:.\ /_ KMT = /_ KQT\ \ text(… as required.)`

(iii) `text(Show)\ \ MK\ text(||)\ TP`

`/_ NTP`	`= /_ KQT\ \ text{(angle in alternate segment)`
	`= 90 – theta`

`:.\ /_ NTP = /_ KMT\ \ text{(from part (ii))}`

`:.\ MK\ text(||)\ TP\ \ text{(corresponding angles are equal)}`

Mechanics, EXT2* M1 2005 6b

An experimental rocket is at a height of 5000 m, ascending with a velocity of ` 200 sqrt 2\ text(m s)^-1` at an angle of 45° to the horizontal, when its engine stops.

After this time, the equations of motion of the rocket are:

`x = 200t`

`y = -4.9t^2 + 200t + 5000,`

where `t` is measured in seconds after the engine stops. (Do NOT show this.)

What is the maximum height the rocket will reach, and when will it reach this height? (2 marks)

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The pilot can only operate the ejection seat when the rocket is descending at an angle between 45° and 60° to the horizontal. What are the earliest and latest times that the pilot can operate the ejection seat? (3 marks)

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For the parachute to open safely, the pilot must eject when the speed of the rocket is no more than `350\ text(m s)^-1`. What is the latest time at which the pilot can eject safely? (2 marks)

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Show Answers Only

`7041\ text(metres)`

`20.4\ text(seconds.)`
`40.8\ text(seconds)\ ;\ 55.8\ text(seconds)`
`49.7\ text(seconds)`

Show Worked Solution

`y = -4.9t^2 + 200t + 5000`

`dot y = -9.8t + 200`

`text(Max height occurs when)\ \ dot y = 0`

`9.8t`	`= 200`
`t`	`= 20.408…`
	`= 20.4\ text{seconds (to 1 d.p.)}`

`text(When)\ t = 20.408…`

`y`	`= -4.9 (20.408…)^2 + 200 (20.408…) + 5000`
	`= 7040.816…`
	`= 7041\ text{m (to nearest metre)}`

`:.\ text(The rocket will reach a maximum height of)`

`text(7041 metres when)\ \ t = 20.4\ text(seconds.)`

ii. `text(The rocket is descending at 45° at point)\ A\ text(on the graph.)`

`text(The symmetry of the parabolic motion means that)\ \ A`

`text(occurs when)\ \ t = 2 xx text(time to reach max height, or)`

`40.8\ text(seconds.)`

`text(Point)\ B\ text(occurs when the rocket is descending at 60°)`

`text(to the horizontal.)`

`text(At)\ \ B,`

`-tan 60° = (dot y)/(dot x)`

`text{(The gradient of the projectile becomes}`

`text{negative after reaching its max height.)}`

`dot y = -9.8t + 200`

`dot x = d/(dt) (200t) = 200`

`:.\ – sqrt 3`	`= (-9.8t + 200)/200`
`-200 sqrt 3`	`= -9.8t + 200`
`9.8t`	`= 200 + 200 sqrt 3`
`t`	`= (200 + 200 sqrt 3)/9.8`
	`= (546.410…)/9.8`
	`= 55.756…`
	`= 55.8\ text{seconds (nearest second)}`

`:.\ text(The pilot can operate the ejection seat)`

`text(between)\ \ t = 40.8 and t = 55.8\ text(seconds.)`

iii.

`v^2 = (dot x)^2 + (dot y)^2`

`text(When)\ \ v = 350`

`350^2`	`= 200^2 + (-9.8t + 200)^2`
`122\ 500`	`= 40\ 000 + (-9.8t + 200)^2`
`(-9.8t + 200)^2`	`= 82\ 500`
`-9.8t + 200`	`= +- sqrt (82\ 500)`
`9.8t`	`= 200 +- sqrt (82\ 500)`
`t`	`= (200 +- sqrt (82\ 500))/9.8`
	`= 49.717…\ \ \ (t > 0)`
	`= 49.7\ text{seconds (to 1 d.p.)}`

`:.\ text(The latest time the pilot can eject safely)`

`text(is when)\ \ t = 49.7\ text(seconds.)`

Polynomials, EXT1 2005 HSC 3a

Show that the function `g(x) = x^2 - log_e (x + 1)` has a zero between `0.7` and `0.9.` (1 mark)
Use the method of halving the interval to find an approximation to this zero of `g(x)`, correct to one decimal place. (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`0.7\ \ text{(to 1 d.p.)}`

Show Worked Solution

(i) `g(x)`	`= x^2 – log_e (x + 1)`
`g(0.7)`	`= 0.7^2 – log_e (0.7 + 1)`
	`= 0.49 – log_e 1.7`
	`= -0.04…`
`g(0.9)`	`= 0.9^2 – log_e (0.9 + 1)`
	`= 0.81 – log_e 1.9`
	`= 0.168…`

`:.\ text(S)text(ince the sign changes, a zero exists)`

`text(between 0.7 and 0.9.)`

(ii) `text(Halving the interval)`

`g(0.8)`	`= 0.8^2 – log_e (0.8 + 1)`
	`= 0.64 – log_e 1.8`
	`= 0.0522…`

`:.\ text(S)text(ince)\ \ g(0.8) > 0,\ \ text(the zero exists)`

`text(between 0.7 and 0.8.)`

♦♦♦ Part (ii) was “very poorly done”.

`g(0.75)`	`= 0.75^2 – log_e 1.75`
	`= 0.00288…`

`=>text(S)text(ince)\ \ g(0.75) > 0,\ \ text(the zero exists)`

`text(between 0.7 and 0.75.)`

`:.\ text{The zero will be 0.7 (to 1 d.p.)}`

Calculus, 2ADV C3 2007 HSC 10b

The noise level, `N`, at a distance `d` metres from a single sound source of loudness `L` is given by the formula

`N = L/d^2.`

Two sound sources, of loudness `L_1` and `L_2` are placed `m` metres apart.

The point `P` lies on the line between the sound sources and is `x` metres from the sound source with loudness `L_1.`

Write down a formula for the sum of the noise levels at `P` in terms of `x`. (1 mark)

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There is a point on the line between the sound sources where the sum of the noise levels is a minimum.

Find an expression for `x` in terms of `m`, `L_1` and `L_2` if `P` is chosen to be this point. (4 marks)

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Show Answers Only

`N = L_1/x^2 + L_2/(m-x)^2`
`x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

Show Worked Solution

`N = L/d^2`

`text(Noise from)\ L_1`	`= L_1/x^2`
`text(Noise from)\ L_2`	`= L_2/(m-x)^2`
`:. N`	`= L_1/x^2 + L_2/(m-x)^2`

ii. `N = L_1\ x^-2 + L_2 (m – x)^-2`

`(dN)/(dx)`	`= -2 L_1 x^-3 + -2 L_2 (m – x)^-3 xx d/(dx) (m – x)`
	`= (-2 L_1)/x^3 + (2 L_2)/(m – x)^3`

`text(Max or min when)\ (dN)/(dx) = 0`

`(2 L_1)/x^3`	`= (2 L_2)/(m – x)^3`
`2 L_1 (m – x)^3`	`= 2 L_2\ x^3`
`L_1 (m – x)^3`	`= L_2\ x^3`
`root 3 L_1 (m – x)`	`= root 3 L_2\ x`

`root 3 L_1\ m – root 3 L_1\ x`	`= root 3 L_2\ x`
`root 3 L_2\ x + root 3 L_1\ x`	`= root 3 L_1\ m`
`x (root 3 L_2 + root 3 L_1)`	`= root 3 L_1\ m`
`x`	`= (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

`(dN)/(dx)`	`= -2 L_1\ x^-3 + 2 L_2 (m – x)^-3`
`(d^2N)/(dx^2)`	`= 6 L_1\ x^-4 – 6 L_2 (m – x)^-4 xx -1`
	`= (6 L_1)/x^4 + (6 L_2)/(m – x)^4 > 0`

`:.\ text(A minimum occurs when)`

`x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

Calculus in the Physical World, 2UA 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`.

Using Simpson’s rule, estimate the distance travelled between `t = 0` and `t = 4`. (2 marks)
The object is initially at the origin. During which time(s) is the displacement of the object decreasing? (1 mark)
Estimate the time at which the object returns to the origin. Justify your answer. (2 marks)
Sketch the displacement, `x`, as a function of time. (2 marks)

Show Answers Only

`~~ 6\ \ text(units)`
`t > 5\ \ text(seconds)`
`7.2\ \ text(seconds)`

Show Worked Solution

(i)

`text(Distance travelled)`

`= int_0^4 (dx)/(dt)\ dt`

`~~ h/3 [y_0 + 4y_1 + y_2]`

`~~ 2/3 [0 + 4 (1) + 5]`

`~~ 2/3 [9]`

`~~ 6\ \ text(units)`

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

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

(iii) `text(At)\ B,\ text(the displacement) = 6\ 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`
	`= 6/5`
	`= 1.2\ \ text(seconds)`

`:.\ text(The particle returns to the origin after 7.2 seconds.)`

(iv)

Financial Maths, 2ADV M1 2007 HSC 9c

Mr and Mrs Caine each decide to invest some money each year to help pay for their son’s university education. The parents choose different investment strategies.

Mr Caine makes 18 yearly contributions of $1000 into an investment fund. He makes his first contribution on the day his son is born, and his final contribution on his son’s seventeenth birthday. His investment earns 6% compound interest per annum.

Find the total value of Mr Caine’s investment on his son’s eighteenth birthday. (3 marks)

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Mrs Caine makes her contributions into another fund. She contributes $1000 on the day of her son’s birth, and increases her annual contribution by 6% each year. Her investment also earns 6% compound interest per annum.

Find the total value of Mrs Caine’s investment on her son’s third birthday (just before she makes her fourth contribution). (2 marks)

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Mrs Caine also makes her final contribution on her son’s seventeenth birthday. Find the total value of Mrs Caine’s investment on her son’s eighteenth birthday. (1 mark)

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Show Answers Only

`$32\ 760\ \ \ text{(nearest dollar)}`
`$3573\ \ \ text{(nearest dollar)}`
`$51\ 378\ \ \ text{(nearest dollar)}`

Show Worked Solution

i. `text(Let)\ A_n = text(value of the investment after)\ n\ text(years)`

`A_1`	`= 1000 (1.06)`
`A_2`	`= A_1 (1.06) + 1000 (1.06)`
	`= [1000 (1.06)] (1.06) + 1000 (1.06)`
	`= 1000 (1.06)^2 + 1000 (1.06)`
`A_3`	`= A_2 (1.06) + 1000 (1.06)`
	`= 1000 (1.06)^3 + 1000 (1.06)^2 + 1000 (1.06)`
	`= 1000 [1.06 + 1.06^2 + 1.06^3]`

`vdots`

`A_n =`	`1000 [1.06 + 1.06^2 + … + 1.06^n]`
	`text(Note that) (1.06 + 1.06^2 + … + 1.06^n)\ text(is a)`
	`text(GP where)\ a = 1.06 and r = 1.06`
`:. A_n =`	`1000 [(a(r^n – 1))/(r – 1)]`

`text(The son’s 18th birthday occurs when)\ n = 18`

`:. A_18`	`= 1000 [(1.06 (1.06^18 -1))/(1.06 – 1)]`
	`= 1000 xx 32.7599…`
	`= 32\ 759.991…`
	`= $32\ 760\ \ \ text{(nearest $)}`

`:.\ text(The value of Mr Caine’s investment is $32 760.)`

ii. `text(Let)\ V_n = text(value of investment after)\ n\ text(years)`

`V_1`	`= 1000 (1.06)`
`V_2`	`= V_1 (1.06) + 1000 (1.06) (1.06)`
	`= 1000 (1.06)^2 + 1000 (1.06)^2`
	`= 2000 (1.06)^2`
`V_3`	`= V_2 (1.06) + 1000 (1.06)^3`
	`= 2000 (1.06)^3 + 1000 (1.06)^3`
	`= 3000 (1.06)^3`
	`= 3573.048…`
	`= $3573\ \ \ text{(nearest dollar)}`

iii. `text(Continuing the pattern)`

`V_4 = 4000 (1.06)^4`

`vdots`

`V_n = n xx 1000 (1.06)^n`

`:. V_18`	`= 18 xx 1000 xx (1.06)^18`
	`= 51\ 378.104…`
	`= $51\ 378\ \ \ text{(nearest dollar)}`

`:.\ text(The value of Mrs Caine’s investment on her)`

`text(son’s 18th birthday will be $51 378.)`

Geometry and Calculus, EXT1 2005 HSC 7b

Let `f(x) = Ax^3 - Ax + 1`, where `A > 0`.

Show that `f(x)` has stationary points at
1. `x = +- (sqrt 3)/3.` (1 mark)
Show that `f(x)` has exactly one zero when
1. `A < (3 sqrt 3)/2.` (2 marks)
By observing that `f(-1) = 1`, deduce that `f(x)` does not have a zero in the interval `-1 <= x <= 1` when
1. `0 < A < (3 sqrt 3)/2.` (1 mark)
Let `g(theta) = 2 cos theta + tan theta`, where `-pi/2 < theta < pi/2.`
By calculating `g prime (theta)` and applying the result in part (iii), or otherwise, show that `g(theta)` does not have any stationary points. (3 marks)
Hence, or otherwise, deduce that `g(theta)` has an inverse function. (1 mark)

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

Show Worked Solution

(i) `f(x) = Ax^3 – Ax + 1\ \ ,\ \ \ A > 0`

`f prime (x) = 3Ax^2 – A`

`text(S.P.’s when)\ \ f prime (x) = 0`

`3Ax^2 – A`	`= 0`
`A (3x^2 – 1)`	`= 0`
`3x^2`	`= 1`
`x^2`	`= 1/3`
`x`	`= +- 1/sqrt3 xx (sqrt 3)/(sqrt 3)`
	`= +- (sqrt 3)/3\ \ text(… as required)`

(ii) `text(Consider)\ \ f(x)\ \ text(with only 1 zero.)`

`f(0) = 1`

`:.\ f(x)\ \ text(passes through)\ \ (0, 1)\ \ text(and has)`

`text(turning points either side at)\ \ x = +- (sqrt 3)/3.`

`text(Given 1 zero)`

`=> f((sqrt 3)/3) > 0`

`A ((sqrt 3)/3)^3 – A ((sqrt 3)/3) + 1`	`> 0`
`A((3 sqrt 3)/27) – A ((9 sqrt 3)/27)`	`> -1`
`A ((-6 sqrt 3)/27)`	`> -1`
`A`	`< 27/(6 sqrt 3)`
`A`	`< 9/(2 sqrt 3) xx (sqrt 3)/(sqrt 3)`
`A`	`< (9 sqrt 3)/6`
`A`	`< (3 sqrt 3)/2\ \ text(… as required.)`

(iii) `text(S)text(ince 1 zero occurs when)\ \ A < (3 sqrt 3)/2\ \ text{(part (ii))}`

`=> text(Minimum TP when)\ \ x = (sqrt3)/3`

`=> text(Maximum TP when)\ \ x = -(sqrt 3)/3\ \ text{(see graph)}`

`text(S)text(ince)\ \ f(-1) = 1`

`:. f(x)\ \ text(cannot have a zero for)\ \ \ -1 <= x <= 1`

(iv) `g(theta) = 2 cos theta + tan theta,\ \ \ \ \ \ -pi/2 < theta < pi/2`

`g prime (theta) = -2 sin theta + sec^2 theta`

`text(S.P.’s when)\ \ g prime (theta) = 0`

`-2 sin theta + sec^2 theta`	`= 0`
`-2 sin theta + 1/(cos^2 theta)`	`= 0`
`-2 sin theta + 1/((1 – sin^2 theta))`	`= 0`
`(-2 sin theta (1 – sin^2 theta) + 1)/((1 – sin^2 theta))`	`= 0`
`-2 sin theta (1 – sin^2 theta) + 1`	`= 0`
`-2 sin theta + 2 sin^3 theta + 1`	`= 0`
`2 sin^3 theta – 2 sin theta + 1`	`= 0\ \ text(… (1))`

`text(Let)\ \ x = sin theta and A = 2`

`text{Equation (1) becomes}`

`Ax^3 – 2x + 1 = 0`

`text(S)text(ince)\ \ A = 2`

`0 < A < (3 sqrt 3)/2`

`text(S)text(ince)\ -pi/2`	`< \ \ \ theta`	`< pi/2`
`-1`	`< sin theta`	`< 1`
`-1`	`< \ \ \ x`	`< 1`

`:.\ text{Using Part (iii)} => g prime (theta)\ \ text(has no zeros and)`

`text(therefore)\ \ g (theta)\ \ text(has no S.P.’s for)\ \ -pi/2 < theta < pi/2.`

(v) `text(Given)\ \ g(theta)\ \ text(has no stationary points, it will be)`

`text(either an increasing or decreasing function in the)`

`text(range)\ \ (-pi)/2 < theta < pi/2\ \ text(and therefore it has an inverse)`

`text(function.)`

Calculus, EXT1 C1 2005 HSC 7a

An oil tanker at `T` is leaking oil which forms a circular oil slick. An observer is measuring the oil slick from a position `P`, 450 metres above sea level and 2 kilometres horizontally from the centre of the oil slick.

At a certain time the observer measures the angle, `alpha`, subtended by the diameter of the oil slick, to be 0.1 radians. What is the radius, `r`, at this time? (2 marks)

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At this time, `(d alpha)/(dt) = 0.02` radians per hour. Find the rate at which the radius of the oil slick is growing. (2 marks)

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Show Answers Only

`102.6\ text{m (to 1 d.p.)}`
`20.6\ text{metres (to 1 d.p.)}`

Show Worked Solution

`text(Let)\ \ Q\ \ text(be the point on the sea such)\ \ /_PQT`

`text(is a right angle.)`

`text(Consider)\ \ Delta PQT,`

`text(Using Pythagoras,)`

`PT^2`	`= PQ^2 + QT^2`
	`= 450^2 + 2000^2`
`PT`	`= sqrt(450^2 + 2000^2)`
	`= 2050\ text(m)`

`text(Consider)\ \ Delta PMT`

`tan\ /_ MPT`	`= r/(PT)`
`tan 0.05`	`= r/2050`
`:. r`	`= 2050 xx tan 0.05`
	`= 102.585…`
	`= 102.6\ text{m (to 1 d.p.)}`

ii. `text(Find)\ \ (dr)/(dt)\ \ text(when)\ \ alpha = 0.1`

`(dr)/(dt)`	`= (dr)/(d alpha) * (d alpha)/(dt)`

`r`	`= 2050 xx tan\ alpha/2`
`:.\ (dr)/(d alpha)`	`= 2050 xx 1/2 xx sec^2\ alpha/2`
	`= 1025 sec^2\ alpha/2`

`text(When)\ \ (d alpha)/(dt)= 0.02\ \ text(radians per hour),`

`alpha = 0.1\ \ \ text{(given)}.`

`:. (dr)/(dt)`	`= 1025 sec^2 0.05 xx 0.02`
	`= 20.5513…`
	`= 20.6\ text{metres per hour (to 1 d.p.)}`

Linear Functions, 2UA 2005 HSC 10b

Xuan and Yvette would like to meet at a cafe on Monday. They each agree to come to the cafe sometime between 12 noon and 1 pm, wait for 15 minutes, and then leave if they have not seen the other person.

Their arrival times can be represented by the point `(x, y)` in the Cartesian plane, where `x` represents the fraction of an hour after 12 noon that Xuan arrives, and `y` represents the fraction of an hour after 12 noon that Yvette arrives.

Thus `(1/3, 2/5)` represents Xuan arriving at 12:20 pm and Yvette arriving at 12:24 pm. Note that the point `(x, y)` lies somewhere in the unit square `0 ≤ x ≤ 1` and `0 ≤ y ≤ 1` as shown in the diagram.

Explain why Xuan and Yvette will meet if
`x - y ≤ 1/4` or `y - x ≤1/4`. (1 mark)
The probability that they will meet is equal to the area of the part of the region given by the inequalities in part (i) that lies within the unit square `0 ≤ x ≤1` and `0≤ y ≤ 1`.
Find the probability that they will meet. (2 marks)
Xuan and Yvette agree to try to meet again on Tuesday. They agree to arrive between 12 noon and 1 pm, but on this occasion they agree to wait for `t` minutes before leaving.
For what value of `t` do they have a 50% chance of meeting? (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`7/16`
`text(17.6 minutes)`

Show Worked Solution

(i) `text(Xuan and Yvette must arrive within)\ 1/4\ text(of an hour)`

`text(of each in order to meet.)`

`text(If Xuan arrives first,)\ \ y − x ≤ 1/4.`

`text(If Yvette arrives first,)\ \ x − y ≤ 1/4.`

(ii) `text{From part (i), the inequalities are}`

`y ≤ x + 1/4`

`y ≥ x − 1/4`

`text(The shaded area satisfies the inequalities)`

`text(and the conditions)`

`0 ≤ x ≤ 1`

`0 ≤ y≤ 1`

`text(Shaded Area)`

`=\ text(Area of square − 2 triangles)`

`= 1^2 −2 xx (1/2 xx b xx h)`

`= 1 − 2 xx (1/2 xx 3/4 xx 3/4)`

`= 1 − 9/16`

`= 7/16\ \ text(u²)`

`:.\ text{P(Xuan and Yvette meet)}`

`=\ text(Shaded Area)/text(Area of square)`

`= 7/16`

(iii) `text(We need the shaded area to be)\ 1/2\ text(u²)`

`text(for a 50% chance.)`

`text(If they wait)\ t\ text{minutes, the inequalities from part (i) are:}`

`y ≤ x + t/(60)`

`y ≥ x − t/(60)`

`text(Shaded Area)\ `	`= 1 − 2 xx (1/2 xx b xx h)`
`1/2`	`= 1 − 2 xx [1/2 xx (1 − t/(60))(1 − t/(60))]`
`1/2`	`= 1 − (1 − t/(60))^2`

`(1 − t/(60))^2`	`= 1/2`
`1 − t/(60)`	`= 1/sqrt2`
`t/(60)`	`= 1 − 1/sqrt2`
`t`	`= 60(1 − 1/sqrt2)`
	`= 17.593…`
	`= text(17.6 minutes (to 1 d.p.))`

`:.\ text(They have 50% chance of meeting if they)`

`text(wait 17.6 minutes.)`

Quadratic, 2UA 2005 HSC 10a

The parabola `y = x^2` and the line `y = mx + b` intersect at the points `A(α,α^2)` and `B(β, β^2)` as shown in the diagram.

Explain why `α + β = m` and `αβ = –b`. (1 mark)
Given that
`(α − β)^2 + (α^2 − β^2)^2 = (α − β)^2[1 + (α + β)^2]`, show that the distance `AB = sqrt((m^2 + 4b)(1 + m^2)).` (2 marks)
The point `P(x, x^2)` lies on the parabola between `A` and `B`. Show that the area of the triangle `ABP` is given by `1/2(mx − x^2 + b)sqrt(m^2 + 4b).` (2 marks)
The point `P` in part (iii) is chosen so that the area of the triangle `ABP` is a maximum.
Find the coordinates of `P` in terms of `m`. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`P(m/2, m^2/4)`

Show Worked Solution

(i) `text(Instersection)`

`y = x^2`	`\ \ …\ (1)`
`y = mx + b`	`\ \ …\ (2)`

`text(Substitute)\ y = x^2\ text(into)\ \ (2)`

`x^2 = mx + b`

`x^2 − mx − b = 0`

`text(We know the intersection occurs when)`

`x = α\ \ text(and)\ \ x = β`

`:.α\ \ text(and)\ \ β\ \ text(are roots of)\ \ x^2 − mx − b = 0`

`:.α + β`	`= (−b)/a`	`= m`
`αβ`	`= c/a`	`= −b`	`\ \ …text(as required)`

(ii) `A(α, α^2),\ \ B(β, β^2)`

`AB`	`= sqrt((α − β)^2 + (α^2 − β^2)^2)`
	`= sqrt((α − β)^2[1 + (α + β)^2])`
	`= sqrt([(α + β)^2 − 4αβ][1 + (α + β)^2])`
	`= sqrt([m^2 − 4(−b)][1 + m^2])`
	`= sqrt((m^2 + 4b)(1 + m^2))\ \ …text(as required)`

(iii) `⊥\ text(distance of)\ \ (x, x^2)\ \ text(from)\ \ \ y=mx+b`

`text(i.e.)\ \ mx – y+b=0`

`= |(ax_1 + by_1 + c)/(sqrt(a^2 + b^2))|`

`= |(mx − 1(x^2) + b)/(sqrt(m^2 + (−1)^2))|`

`= |(mx − x^2 + b)/(sqrt(m^2 + 1))|`

`text(Area of)\ ΔABP`

`= 1/2 xx AB xx h`

`= 1/2 xx sqrt((m^2 + 4b)(1 + m^2)) xx (mx − x^2 + b)/(sqrt(m^2 + 1))`

`= 1/2 (mx − x^2 + b)sqrt(m^2 + 4b)`

(iv)	`A`	`= 1/2 sqrt(m^2 + 4b)(mx − x^2 + b)`
	`(dA)/(dx)`	`= 1/2 sqrt(m^2 + 4b)(m − 2x)`
	`(d^2A)/(dx^2)`	`= 1/2 sqrt(m^2 + 4b)(−2)`
		`= −sqrt(m^2 + 4b)`

`text(Max or min when)\ (dA)/(dx) = 0`

`1/2sqrt(m^2 + 4b)(m − 2x)`	`= 0`
`m-2x`	`=0`
`2x`	`= m`
`x`	` = m/2`

`text(When)\ x = 2`

`(d^2A)/(dx^2) = −sqrt(m^2 + 4b) < 0\ \ \ text{(constant)}`

`:.\ text(Maximum when)\ x = m/2`

`:.P\ text(is)\ (m/2, m^2/4)`

Combinatorics, EXT1 A1 2015 HSC 14c

Two players `A` and `B` play a series of games against each other to get a prize. In any game, either of the players is equally likely to win.

To begin with, the first player who wins a total of 5 games gets the prize.

Explain why the probability of player `A` getting the prize in exactly 7 games is `((6),(4))(1/2)^7`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Write an expression for the probability of player `A` getting the prize in at most 7 games. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Suppose now that the prize is given to the first player to win a total of `(n + 1)` games, where `n` is a positive integer.
By considering the probability that `A` gets the prize, prove that
`((n),(n))2^n + ((n + 1),(n))2^(n-1) + ((n + 2),(n))2^(n-2) + … + ((2n),(n)) = 2^(2n)`. (2 marks)
--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{Proof (See Worked Solutions)}`
`\ ^4C_4(1/2)^5 +\ ^5C_4(1/2)^6 +\ ^6C_4(1/2)^7`
`text{Proof (See Worked Solutions.)}`

Show Worked Solution

i. `text(To win in exactly 7 games, player)\ A\ text(must win the 7th game.)`

`:.P(A\ text{wins in 7 games)}`

`=\ ^6C_4 · (1/2)^4(1/2)^2 xx 1/2`

`=\ ^6C_4(1/2)^7`

ii. `Ptext{(wins in at most 7 games)}`

♦♦♦ Mean mark (i) 19%, (ii) 23%.

`=Ptext{(wins in 5, 6 or 7 games)}`

`=\ ^4C_4(1/2)^4 xx 1/2 +\ ^5C_4(1/2)^4(1/2) xx 1/2 +\ ^6C_4(1/2)^7`

`=\ ^4C_4(1/2)^5 +\ ^5C_4(1/2)^6 +\ ^6C_4(1/2)^7`

♦♦♦ Mean mark (iii) 9%.

iii. `text(Prove that)`

`\ ^nC_n · 2^n +\ ^(n + 1)C_n · 2^(n-1) + … +\ ^(2n)C_n = 2^(2n)`

`P(A\ text(wins in)\ (n + 1)\ text{games)}`

`=\ ^nC_n(1/2)^(n + 1) +\ ^(n + 1)C_n(1/2)^(n + 2) + … +\ ^(2n)C_n(1/2)^(2n + 1)`

`text{One player must have won after (2n + 1) games are played.}`

`text(S)text(ince each player has an equal chance,)`

`\ ^nC_n(1/2)^(n + 1) +\ ^(n + 1)C_n(1/2)^(n + 2) + … +\ ^(2n)C_n(1/2)^(2n + 1) = 1/2`

`text(Multiply both sides by)\ 2^(2n + 1):`

`\ ^nC_n2^(-(n + 1)) · 2^(2n + 1) +\ ^(n + 1)C_n · 2^(-(n + 2)) · 2^(2n + 1) + …`

`… +\ ^(2n)C_n · 2^(-(2n + 1)) · 2^(2n + 1) = 2^(-1) · 2^(2n + 1)`

`:. \ ^nC_n · 2^n +\ ^(n + 1)C_n · 2^(n-1) + … +\ ^(2n)C_n = 2^(2n)`

Measurement, STD2 M6 2015 HSC 30e

From point `S`, which is 1.8 m above the ground, a pulley at `P` is used to lift a flat object `F`. The lengths `SP` and `PF` are 5.4 m and 2.1 m respectively. The angle `PSC` is 108°.

Show that the length `PC` is 6.197 m, correct to 3 decimal places. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Calculate `h`, the height of the object above the ground. (4 marks)

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

Show Answers Only

`6.197\ text{m (to 3 d.p.) … as required}`
`1.37\ text{m (to 2 d.p.)}`

Show Worked Solution

`text(Show)\ PC = 6.197\ text(m)`

♦ Mean mark 41%.

`text(Using the cosine rule in)\ Delta PSC`

`PC^2`	`= PS^2 + SC^2-2 xx PS xx SC xx cos\ 108^@`
	`= 5.4^2 + 1.8^2-2 xx 5.4 xx 1.8 xx cos\ 108^@`
	`= 38.4072…`
`:.PC`	`= 6.19736…`
	`= 6.197\ text{m (to 3 d.p.) …as required}`

ii. `text(Let)\ \ SD⊥PE`

♦♦ Mean mark below 19%.
STRATEGY: Finding `PC` in part (i) and needing `PE` to find `h` should flag the strategy of finding `EC` and using Pythagoras.

`∠DSC\ text(is a right angle)`

`:.∠DSP = 108^@-90^@ = 18^@`

`text(In)\ ΔPDS`

`cos\ 18^@`	`= (DS)/5.4`
`DS`	`= 5.4 xx cos\ 18^@`
	`= 5.1357…\ text(m)`

`EC = DS = 5.1357…\ text{m (opposite sides of rectangle}\ DECS text{)}`

`text(Using Pythagoras in)\ Delta PEC:`

`PE^2 + EC^2 = PC^2`

`PE^2 + 5.1357^2`	`= 6.197^2`
`PE^2`	`= 12.027…`
`PE`	`= 3.468…\ text(m)`

`text(From the diagram,)`

`h`	`= PE-PF`
	`= 3.468…-2.1`
	`= 1.368…`
	`= 1.37\ text{m (to 2 d.p.)}`

Algebra, STD2 A4 2015 HSC 29e

A diver springs upwards from a diving board, then plunges into the water. The diver’s height above the water as it varies with time is modelled by a quadratic function. Graphing software is used to produce the graph of this function.

Explain how the graph could be used to determine how high above the height of the diving board the diver was when he reached the maximum height. (2 marks)

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

Show Answers Only

`1.2\ text(m)`

Show Worked Solution

`text(We can calculate the height of the board by)`

♦♦ Mean mark 19%.

`text(finding the)\ ytext(-value at)\ t = 0,\ text(which is 8 m.)`

`text(The diver’s maximum height occurs at)\ t=0.5,`

`text(which is approximately 9.2 m.)`

`:.\ text(Maximum height above the board)`

`= 9.2-8`

`= 1.2\ text(m)`

Statistics, STD2 S1 2015 HSC 29d

Data from 200 recent house sales are grouped into class intervals and a cumulative frequency histogram is drawn.

Use the graph to estimate the median house price. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
By completing the table, calculate the mean house price. (3 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Class Centre} \rule[-1ex]{0pt}{0pt} & \text{Frequency} \\ \text{(\$'000)} & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

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

Show Answers Only

a. `text(See Worked Solutions.)`

b. `text(See Worked Solutions.)`

Show Worked Solution

`text(From the graph, the estimated median)`

`text(house price = $392 500)`

♦♦♦ Mean marks of 9% for part (i) and 34% for part (ii)!

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Class Centre} \rule[-1ex]{0pt}{0pt} & \text{Frequency} \\ \text{(\$’000)} & \\
\hline
\rule{0pt}{2.5ex} 375 \rule[-1ex]{0pt}{0pt} & 30\\
\hline
\rule{0pt}{2.5ex} 385 \rule[-1ex]{0pt}{0pt} & 50 \\
\hline
\rule{0pt}{2.5ex} 395 \rule[-1ex]{0pt}{0pt} & 70 \\
\hline
\rule{0pt}{2.5ex} 405 \rule[-1ex]{0pt}{0pt} & 50 \\
\hline
\end{array}

`text(Mean house price ($’000))`

`= (375xx30 + 385xx50 + 395xx70 + 405xx50)/200`

`=$392`

`:. text(Mean house price is)\ $392\ 000`

Measurement, STD2 M7 2015 HSC 29c

The image shows a rectangular farm shed with a flat roof.

The real width of the shed indicated by the dotted line was measured using an online ruler tool, and found to be approximately 12 metres.

By measurement and calculation, show that the area of the roof of the shed is approximately 216 m². (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
All the rain that falls onto this roof is diverted into a cylindrical water tank which has a diameter of 3.6 m. During a storm, 5 mm of rain falls onto the roof.

Calculate the increase in the depth of water in the tank due to the rain that falls onto the roof during the storm. (3 marks)

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

Show Answers Only

`text(See Worked Solutions.)`
`10.6\ text{cm (to 1 d.p.)}`

Show Worked Solution

i. `text(By measurement,)`

♦ Mean mark 35%.

`text(Roof length = 1.5 times the roof width)`

`text{Width = 12 m (given)}`

`text(Length)`	`= 1.5 xx 12`
	`= 18\ text(m)`

`:.\ text(Area of roof)`	`= 12 xx 18`
	`= 216\ text(m²)\ \ text(…as required)`

ii. `text(Volume of water)`

♦♦♦ Mean mark 14%.

`= Ah`

`= 216 xx 0.005\ \ \ (5\ text(mm) = 0.005\ text(m))`

`= 1.08\ text(m³)`

`text(Volume of cylinder =)\ pir^2h`

`r = 3.6/2 = 1.8\ text(m)`

`text(Find)\ h\ text(when)\ V= 1.08\ text(m³),`

`pi xx 1.8^2 xx h`	`= 1.08`
`:. h`	`= 1.08/(pi xx 1.8^2)`
	`= 0.1061…\ text(m)`
	`= 10.6\ text{cm (to 1 d.p.)}`

Calculus, 2ADV C3 2015 HSC 16c

The diagram shows a cylinder of radius `x` and height `y` inscribed in a cone of radius `R` and height `H`, where `R` and `H` are constants.

The volume of a cone of radius `r` and height `h` is `1/3 pi r^2 h.`

The volume of a cylinder of radius `r` and height `h` is `pi r^2 h.`

Show that the volume, `V`, of the cylinder can be written as

`V = H/R pi x^2 (R-x).` (3 marks)

--- 6 WORK AREA LINES (style=lined) ---
By considering the inscribed cylinder of maximum volume, show that the volume of any inscribed cylinder does not exceed `4/9` of the volume of the cone. (4 marks)

--- 10 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(Show)\ \ V = H/R pi x^2 (R-x)`

♦♦ Mean mark (i) 16%.

`text(Consider)\ \ Delta ABC and Delta DEC`

`/_ ABC = /_DEC = 90°`

`/_ BCA\ \ text(is common)`

`:. Delta ABC\ \ text(|||)\ \ Delta DEC\ \ text{(equiangular)}`

`:.\ (DE)/(EC)`

`= (AB)/(BC)`

`\ text{(corresponding sides of}`

`\ text{similar triangles)}`

`y/(R-x)`

`= H/R`

`y`

`= H/R (R-x)`

`text(Volume of cylinder)`

`= pi r^2 h`

`= pi x^2 xx H/R (R-x)`

`= H/R pi x^2 (R-x)\ \ text(… as required.)`

♦♦ Mean mark (ii) 21%.

ii. `V= H/R pi x^2 (R-x)`

`(dV)/(dx)`	`= H/R pi [(x^2 xx -1) + 2x (R-x)]`
	`= H/R pi [-x^2 + 2xR-2x^2]`
	`= H/R pi [2xR-3x^2]`
`(dV^2)/(dx^2)`	`= H/R pi [2R-6x]`

`text(Max or min when)\ \ (dV)/(dx) = 0`

`H/R pi [2xR-3x^2]`	`= 0`
`2xR-3x^2`	`= 0`
`x (2R-3x)`	`= 0`

`x = 0\ \ or\ \ 3x`	`= 2R`
`x`	`= (2R)/3`

`text(When)\ \ x = 0:`

`(d^2V)/(dx^2) = H/R pi [2R-0] > 0\ \ =>\ text(MIN)`

`text(When)\ \ x = (2R)/3:`

`(d^2V)/(dx^2)`	`= H/R pi [2R-6 xx (2R)/3]`
	`= H/R pi [-2R] < 0\ \ =>\ \text{MAX}`

`text(Maximum Volume of cylinder)`

`= H/R pi ((2R)/3)^2 (R-(2R)/3)`

`= H/R pi xx (4R^2)/9 xx R/3`

`= (4 pi H R^2)/27`

`text(Volume of cone)\ = 1/3 pi R^2 H`

`:.\ text(Max Volume of Cylinder)/text(Volume of cone)`

`= ((4 pi H R^2)/27)/(1/3 pi R^2 H)`

`= (4 pi H R^2)/27 xx 3/(pi R^2 H)`

`= 4/9`

`:.\ text(The volume of the inscribed cylinder does)`

`text(not exceed)\ 4/9\ text(of the cone volume.)`

Plane Geometry, 2UA 2015 HSC 15b

The diagram shows `Delta ABC` which has a right angle at `C`. The point `D` is the midpoint of the side `AC`. The point `E` is chosen on `AB` such that `AE = ED`. The line segment `ED` is produced to meet the line `BC` at `F`.

Copy or trace the diagram into your writing booklet.

Prove that `Delta ACB` is similar to `Delta DCF.` (2 marks)
Explain why `Delta EFB` is isosceles. (1 mark)
Show that `EB = 3AE.` (2 marks)

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) `text(Prove)\ \ Delta ACB\ \ text(|||)\ \ Delta DCF`

`/_ EAD = /_ ADE = theta\ \ text{(base angles of isosceles}\ \ Delta AED text{)}`

`/_ CDF = /_ ADE = theta\ \ text{(vertically opposite angles)}`

`/_ DCF = /_ ACB = 90°\ \ text{(}/_ FCB\ text{is a straight angle)}`

`:.\ Delta ACB\ \ text(|||)\ \ Delta DCF\ \ text{(equiangular)}`

♦ Mean mark 45%.

(ii) `/_ DFC = 90 – theta\ \ text{(angle sum of}\ \ Delta DCF text{)}`

`/_ ABC = 90 – theta\ \ text{(angle sum of}\ \ Delta ACB text{)}`

`:.\ Delta EFB\ \ text{is isosceles (base angles are equal)}`

(iii) `text(Show)\ \ EB = 3AE`

♦♦ Mean mark 23%.

`(DC)/(AC)`

`= (DF)/(AB)`

`\ \ \ \ \ text{(corresponding sides of}`

`\ \ \ \ \ text{similar triangles)}`

`1/2`

`= (DF)/(AB)`

`2DF`

`= AB`

`2(EF – ED)`	`= AE + EB`
`2(EB – AE)`	`= AE + EB`	`\ \ \ \ \ text{(given}\ EF = EB, ED = AE text{)}`
`2EB – 2AE`	`= AE + EB`

`:. EB = 3AE\ \ text(… as required)`

Calculus, 2ADV C4 2015 HSC 9 MC

A particle is moving along the `x`‑axis. The graph shows its velocity `v` metres per second at time `t` seconds.

When `t = 0` the displacement `x` is equal to `2` metres.

What is the maximum value of the displacement `x`?

`text(8 m)`
`text(14 m)`
`text(16 m)`
`text(18 m)`

Show Answers Only

`D`

Show Worked Solution

`text(Distance travelled)`

♦♦ Mean mark 31%.

`= int_0^4 v\ dt`

`= text(Area under the velocity curve)`

`= 1/2 xx b xx h`

`= 1/2 xx 4 xx 8`

`= 16\ \ text(metres.)`

`text(S)text(ince velocity is always positive between)\ t=0`

`text(and)\ t = 4,\ text(and the original displacement = 2,)`

`text(the maximum displacement) = 16 + 2 = 18\ \ text(metres)`

`=> D`

Plane Geometry, 2UA 2006 HSC 10b

A rectangular piece of paper `PQRS` has sides `PQ = 12` cm and `PS = 13` cm. The point `O` is the midpoint of `PQ`. The points `K` and `M` are to be chosen on `OQ` and `PS` respectively, so that when the paper is folded along `KM`, the corner that was at `P` lands on the edge `QR` at `L`. Let `OK = x` cm and `LM = y` cm.

Copy or trace the diagram into your writing booklet.

Show that `QL^2 = 24x`. (1 mark)
Let `N` be the point on `QR` for which `MN` is perpendicular to `QR`.
By showing that `Delta QKL\ text(|||)\ Delta NLM`, deduce that `y = {sqrt 6 (6 + x)}/sqrt x`. (3 marks)
Show that the area, `A`, of `Delta KLM` is given by
1. `A = {sqrt 6 (6 + x)^2}/(2 sqrt x)` (1 mark)
Use the fact that `12 <= y <= 13` to find the possible values of `x`. (2 marks)
Find the minimum possible area of `Delta KLM`. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`2 2/3 <= x <= 6`
`53.33…\ text(cm²)`

Show Worked Solution

(i)

`text(Show)\ QL^2 = 24x`

`KP = KL = 6 + x`

`KQ = 6 – x`

`text(Using Pythagoras)`

`QL^2`	`= KL^2 – KQ^2`
	`= (6 + x)^2 – (6 – x)^2`
	`= 36 + 12x + x^2 – 36 + 12x – x^2`
	`= 24x\ …\ text(as required)`

(ii) `text(Show)\ y = {sqrt 6 (6 + x)}/sqrt x`

`text(In)\ Delta QKL`

`/_LQK=90°\ \ text{(given)}`

`text(Let)\ /_QKL = theta`

`:. /_QLK = 90 – theta\ \ text{(angle sum of}\ Delta QKL text{)}`

`text(In)\ Delta NLM`

`/_LNM=90°\ \ text{(given)}`

`/_NLM`	`= 180 – (90 + 90 – theta)\ \ (/_QLN\ text(is a straight angle))`
	`= theta`

`:. Delta QKL\ text(|||)\ Delta NLM\ \ \ text{(equiangular)}`

`:. y/(MN)`	`= (KL)/(QL)\ \ text{(corresponding sides of}`
	`text{similar triangles)}`
`y/12`	`= (6 + x)/sqrt(24x)`
`y`	`= (12(6 + x))/(2 sqrt (6x))`
	`= (6 (6 + x))/(sqrt x xx sqrt 6) xx sqrt 6/sqrt 6`
	`= (sqrt 6 (6 + x))/sqrt x\ …\ text(as required)`

(iii) `text(Area)\ DeltaKLM`	`= 1/2 xx y xx KL`
	`= 1/2 xx (sqrt 6 (6 + x))/sqrt x xx (6 + x)`
	`= (sqrt 6 (6 + x)^2)/(2 sqrt x)\ …\ text(as required.)`

(iv) `text(Given that)\ \ \ \ \ \ \ \ 12 <= y <= 13`

`12 <= (sqrt 6 (6 + x))/sqrt x <= 13`

`text(Consider)\ (sqrt 6 ( 6 + x))/sqrt x`	`>= 12`
`(6 (6 + x)^2)/x`	`>= 144`
`(6 + x)^2`	`>= 24x`

`36 + 12x + x^2`	`>= 24x`
`x^2 – 12x + 36`	`>= 0`
`(x – 6)^2`	`>= 0`
`:. x`	`>= 6`

`text(However, we know)\ OP=6,\ text(and)\ x <= 6`.

`:. x = 6\ text(satisfies both conditions)`

`text(Consider)\ (sqrt 6 (6 + x))/sqrt x`	`<= 13`
`(6 (6 + x)^2)/x`	`<= 169`
`6 (6 + x)^2`	`<= 169x`
`6 (36 + 12x + x^2)`	`<= 169x`
`216 + 72x + 6x^2`	`<= 169x`
`6x^2 – 97x + 216`	`<= 0`

`text(Using the quadratic formula)`

`x`	`= (-b +- sqrt (b^2 -4ac))/(2a)`
	`= (97 +- sqrt ((-97)^2 -4 xx 6 xx 216))/(2 xx 6)`
	`= (97 +- sqrt 4225)/12`
	`= (97 +- 65)/12`
	`= 13 1/2 or 2 2/3`

`:. 2 2/3 <= x <= 13 1/2`

`text(However, we know)\ x <= 6,\ text(so)`

`2 2/3 <= x <= 6\ text(satisfies both conditions)`

`:.\ text(All possible values of)\ x\ text(are)`

`2 2/3 <= x <= 6`.

(v) `A = (sqrt 6 (6 + x)^2)/(2 sqrt x)`

`text(Using the quotient rule)`

`(dA)/(dx)`	`= (u prime v – uv prime)/v^2`
	`= {2 sqrt 6 (6 + x) xx 2 sqrt x – sqrt 6 (6 + x)^2 xx 1/2 xx 2 xx x^(-1/2)}/(2 sqrt x)^2`
	`= {4 sqrt x sqrt 6 (6 + x) – sqrt 6 (6 + x)^2 * 1/sqrt x}/(4x)`
	`= {4 sqrt 6 x (6 + x) – sqrt 6 (6 + x)^2}/(4x sqrt x)`
	`= (sqrt 6 (6 + x) (4x – 6 – x))/(4x sqrt x)`
	`= (sqrt 6 (6 + x) (3x – 6))/(4x sqrt x)`

`text(Max or min when)\ (dA)/(dx) = 0`

`sqrt 6 (6 + x) (3x – 6) = 0`

`:. 3x = 6\ \ ,\ \ x ≠ -6`

`x = 2`

`text(However,)\ x = 2\ text(lies outside the range)`

`text(of possible values)\ \ 2 2/3 <= x <= 6`

`:.\ text(Check limits)`

`text(At)\ \ x = 2 2/3`

`A`	`= (sqrt 6 (6 + 2 2/3)^2)/(2 sqrt (2 2/3))`
	`= 56.33…\ text(cm²)`

`text(At)\ \ x = 6`

`A`	`= (sqrt 6 (6 + 6)^2)/(2 sqrt 6)`
	`= 72.0\ text(cm²)`

`:.\ text(Minimum area of)\ Delta KLM\ text(is 56.33… cm²)`

Calculus, 2ADV C3 2006 HSC 9c

A cone is inscribed in a sphere of radius `a`, centred at `O`. The height of the cone is `x` and the radius of the base is `r`, as shown in the diagram.

Show that the volume, `V`, of the cone is given by

`V = 1/3 pi(2ax^2 - x^3)`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Find the value of `x` for which the volume of the cone is a maximum. You must give reasons why your value of `x` gives the maximum volume. (3 marks)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`x = 4/3 a`

Show Worked Solution

i. `text(Show)\ V = 1/3 pi (2ax^2 – x^3)`

`V = 1/3 pi r^2 h`

`text(Using Pythagoras)`

`(x – a)^2 + r^2`	`= a^2`
`r^2`	`= a^2 – (x – a)^2`
	`= a^2 – x^2 + 2ax – a^2`
	`= 2ax – x^2`

`:. V`	`= 1/3 xx pi xx (2ax – x^2) xx x`
	`= 1/3 pi (2ax^2 – x^3)\ …\ text(as required)`

ii. `(dV)/(dx) = 1/3 pi (4ax – 3x^2)`

`(d^2V)/(dx^2) = 1/3 pi (4a – 6x)`

`text(Max or min when)\ (dV)/(dx) = 0`

`1/3 pi (4ax – 3x^2)`	`= 0`
`4ax – 3x^2`	`= 0`
`x(4a – 3x)`	`= 0`

`3x`	`= 4a,`	` \ \ \ \ x ≠ 0`
`x`	` =4/3 a`

`text(When)\ \ x = 4/3 a`

`(d^2V)/(dx^2)`	`= 1/3 pi (4a – 6 xx 4/3 a)`
	`= 1/3 pi (-4a) < 0`
`=>\ text(MAX)`

`:.\ text(Cone volume is a maximum when)\ \ x = 4/3 a.`

Calculus, 2ADV C4 2006 HSC 9b

During a storm, water flows into a 7000-litre tank at a rate of `(dV)/(dt)` litres per minute, where `(dV)/(dt) = 120 + 26t-t^2` and `t` is the time in minutes since the storm began.

At what times is the tank filling at twice the initial rate? (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Find the volume of water that has flowed into the tank since the start of the storm as a function of `t`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Initially, the tank contains 1500 litres of water. When the storm finishes, 30 minutes after it began, the tank is overflowing.

How many litres of water have been lost? (2 marks)

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

Show Answers Only

`t = 6 or 20\ text(minutes)`
`V = 120t + 13t^2 – 1/3t^3`
`800\ text(litres)`

Show Worked Solution

i. `(dV)/(dt) = 120 + 26t-t^2`

`text(When)\ t = 0, (dV)/(dt) = 120`

`text(Find)\ t\ text(when)\ (dV)/(dt) = 240`

`240 = 120 + 26t-t^2`

`t^2-26t + 120 = 0`

`(t-6)(t-20) = 0`

`t = 6 or 20`

`:.\ text(The tank is filling at twice the initial rate)`

`text(when)\ t = 6 and t = 20\ text(minutes)`

ii. `V`	`= int (dV)/(dt)\ dt`
	`= int 120 + 26t-t^2\ dt`
	`= 120t + 13t^2-1/3t^3 + c`

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

`=> c = 0`

`:. V= 120t + 13t^2-1/3t^3`

iii. `text(Storm water volume into the tank when)\ t = 30`

`= 120(30) + 13(30^2)-1/3 xx 30^3`

`= 3600 + 11\ 700-9000`

`= 6300\ text(litres)`

`text(Total volume)`	`= 6300 + 1500`
	`= 7800\ text(litres)`

`:.\ text(Overflow)`	`= 7800-7000`
	`= 800\ text(litres)`

Trigonometry, 2ADV T1 2005 HSC 9b

The triangle `ABC` has a right angle at `B, \ ∠BAC = theta` and `AB = 6`. The line `BD` is drawn perpendicular to `AC`. The line `DE` is then drawn perpendicular to `BC`. This process continues indefinitely as shown in the diagram.

Find the length of the interval `BD`, and hence show that the length of the interval `EF` is `6 sin^3 theta`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Show that the limiting sum
`qquad BD + EF + GH + ···`
is given by `6 sec theta tan theta`. (3 marks)

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

Show Answers Only

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

Show Worked Solution

`text(Show)\ EF = 6 sin^3 theta`

`text(In)\ ΔADB:`

`sin theta`	`= (DB)/6`
`DB`	`= 6 sin theta`

`∠ABD`	`= 90-theta\ \ \ text{(angle sum of}\ ΔADB)`
`:.∠DBE`	`= theta\ \ \ (∠ABE\ text{is a right angle)}`

`text(In)\ ΔBDE:`

`sin theta= (DE)/(DB)= (DE)/(6 sin theta)`

`DE`	`= 6 sin^2 theta`
`∠BDE`	`= 90-theta\ \ \ text{(angle sum of}\ ΔDBE)`
`∠EDF`	`= theta\ \ \ (∠FDB\ text{is a right angle)}`

`text(In)\ ΔDEF:`

`sin theta`	`= (EF)/(DE)= (EF)/(6 sin^2 theta)`
`:.EF`	`= 6 sin^3 theta\ \ …text(as required)`

ii. `text(Show)\ \ BD + EF + GH\ …`

`text(has limiting sum)\ =6 sec theta tan theta`

`underbrace{6 sin theta + 6 sin^3 theta +\ …}_{text(GP where)\ \ a = 6 sin theta, \ \ r = sin^2 theta}`

`text(S)text(ince)\ \ 0 < theta < 90^@`

`-1`	`< sin\ theta`	`< 1`
`0`	`< sin^2\ theta`	`< 1`

`:. |\ r\ | < 1`

`:.S_∞`	`= a/(1-r)`
	`= (6 sin theta)/(1-sin^2 theta)`
	`= (6 sin theta)/(cos^2 theta)`
	`= 6 xx 1/(cos theta) xx (sin theta)/(cos theta)`
	`= 6 sec theta tan theta\ \ …text(as required.)`

Calculus, EXT1* C1 2005 HSC 7b

The graph shows the velocity, `(dx)/(dt)`, of a particle as a function of time. Initially the particle is at the origin.

At what time is the displacement, `x`, from the origin a maximum? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
At what time does the particle return to the origin? Justify your answer. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Draw a sketch of the acceleration, `(d^2x)/(dt^2)`, as a function of time for `0 ≤ t ≤ 6`. (2 marks)

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

Show Answers Only

`2`
`4`
`text(See Worked Solutions.)`

Show Worked Solution

i. `text(Maximum displacement in graph when)`

`(dx)/(dt)`	`= 0`
`:.t`	`= 2`

ii. `text(Particle returns to the origin when)\ t = 4.`

`text(The displacement can be calculated by the)`

`text(net area below the curve and since the)`

`text(area above the curve between)\ t = 0\ text(and)\ t = 2`

`text(is equal to the area below the curve between)`

`t = 2\ text(and)\ t = 4,\ text(the displacement returns to)`

`text{the initial displacement (i.e. the origin).}`

iii.

Calculus in the Physical World, 2UA 2005 HSC 7b Answer

Algebra, STD2 A4 2006 HSC 28b

A new tunnel is built. When there is no toll to use the tunnel, 6000 vehicles use it each day. For each dollar increase in the toll, 500 fewer vehicles use the tunnel.

Find the lowest toll for which no vehicles will use the tunnel. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
For a toll of $5.00, how many vehicles use the tunnel each day and what is the total daily income from tolls? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
If `d` (dollars) represents the value of the toll, find an equation for the number of vehicles `(v)` using the tunnel each day in terms of `d`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Anne says ‘A higher toll always means a higher total daily income’.

Show that Anne is incorrect and find the maximum daily income from tolls. (Use a table of values, or a graph, or suitable calculations.) (3 marks)

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

Show Answers Only

`text($12 toll is the lowest for which)`

`text(no vehicles will use the tunnel.)`
`$17\ 500`
`v = 6000 – 500d`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

i. `text(500 less vehicles per $1 toll)`

`12 xx 500 = 6000`

`:. $12\ text(toll is the lowest for which no)`

`text(vehicles will use the tunnel.)`

ii. `text(If the toll is $5)`

`5 xx 500 = 2500\ text(less vehicles)`

`:.\ text(Vehicles using the tunnel)`

`= 6000 – 2500`

`= 3500`

`:.\ text(Daily toll income)`	`= 3500 xx $5`
	`= $17\ 500`

iii. `d`	`=\ text(toll)`
`v`	`=\ text(Number of vehicles using the tunnel)`
`:. v`	`= 6000 – 500d`

iv. `text(Income from tolls)`

`=\ text(Number of vehicles) xx text(toll)`

`= (6000 – 500d) xx d`

`= 6000d – 500d^2`

`= 500d (12 – d)`

`text(From the graph, the maximum income from tolls)`

`text(occurs when the toll is $6.)`

`:.\ text(Anne is incorrect.)`

`text(Alternate Solution)`

`text{The table of values shows that income (I) increases}`

`text(and peaks when the toll hits $6 before decreasing)`

`text(again as the toll gets more expensive.)`

`:.\ text(Anne is incorrect.)`

Statistics, STD2 S4 2006 HSC 27b

Each member of a group of males had his height and foot length measured and recorded. The results were graphed and a line of fit drawn.

Why does the value of the `y`-intercept have no meaning in this situation? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
George is 10 cm taller than his brother Harry. Use the line of fit to estimate the difference in their foot lengths. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Sam calculated a correlation coefficient of −1.2 for the data. Give TWO reasons why Sam must be incorrect. (2 marks)

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

Show Answers Only

`text(The y-intercept occurs when)\ x = 0.\ text(It has`
`text(no meaning to have a height of 0 cm.)`
`text(A 10 cm height difference means George should)`
`text(have a 3 cm longer foot.)`
`text(A correlation co-efficient must be between –1 and 1.)`
`text(Foot length is positively correlated to a person’s)`
`text(height and therefore can’t be a negative value.)`

Show Worked Solution

i. `text(The y-intercept occurs when)\ x = 0.\ text(It has)`

`text(no meaning to have a height of 0 cm.)`

ii. `text(A 20 cm height difference results in a foot length)`

`text(difference of 6 cm.)`

`:.\ text(A 10 cm height difference means George should)`

`text(have a 3 cm longer foot.)`

iii. `text(A correlation co-efficient must be between –1 and 1.)`

`text(Foot length is positively correlated to a person’s)`

`text(height and therefore isn’t a negative value.)`

Statistics, STD2 S1 2007 HSC 22 MC

A set of examination results is displayed in a cumulative frequency histogram and polygon (ogive).

Sanath knows that his examination mark is in the 4th decile.

Which of the following could have been Sanath’s examination mark?

Show Answers Only

`B`

Show Worked Solution

`text(4th decile occurs when cumulative frequency)`

`text(is between 15 and 20.)`

`:.\ text(Examination mark must be between 55 and 60.)`

`=> B`

Probability, STD2 S2 2006 HSC 28a

On a bridge, the toll of $2.50 is paid in coins collected by a machine. The machine only accepts two-dollar coins, one-dollar coins and fifty-cent coins.

List the different combinations of coins that could be used to pay the $2.50 toll. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Jill has three two-dollar coins, six one-dollar coins and two fifty-cent coins. She selects two coins at random.

What is the probability that she selects exactly $2.50? (3 marks)

--- 6 WORK AREA LINES (style=lined) ---
At the end of a day, the machine contains `x` two-dollar coins, `y` one-dollar coins and `w` fifty-cent coins.

Write an expression for the total value of coins in dollars in the machine. (1 mark)

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

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`6/55`
`$(2x + y + 0.5w)`

Show Worked Solution

i. `text(Combinations for $2.50)`

`$2, 50c`

`$1, $1, 50c`

`$1, 50c, 50c, 50c`

`50c, 50c, 50c, 50c, 50c`

ii. `3 xx $2, 6 xx $1, 2 xx 50c`

`P($2.50)`	`= (3/11 xx 2/10) + (2/11 xx 3/10)`
	`= 6/110 + 6/110`
	`= 6/55`

iii. `text(Total value) = $ (2x + y + 0.5w)`

Measurement, STD2 M1 2005 HSC 28a

The Mitchell family has moved to a new house which has an empty swimming pool. The base of the pool is in the shape of a rectangle, with a semicircle on each end.

Explain why the expression for the area of the base of the pool is `2xy + πy^2`. (1 mark)

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

The pool is 1.1 metres deep.

The sides and base of the pool are covered in tiles. If `x =6` and `y = 2.5`, find the total area covered by tiles. (Give your answer correct to the nearest square metre.) (4 marks)

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

Before filling the pool, the Mitchells need to install a new shower head, which saves 6 litres of water per minute.

The shower is used 5 times every day, for 3 minutes each time.

If the charge for water is $1.013 per kilolitre, how much money would be saved in one year by using this shower head? (Assume there are 365 days in a year.) (2 marks)

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

Show Answers Only

a. `text(See Worked Solutions)`

b. `text(80 m)^2\ text{(nearest m}^2text{)}`

c. `$33.28\ \ text{(nearest cent)}`

Show Worked Solution

a.	`text(Area of base)`	`=\ text(Area of rect.) +2 xx text(Area of semi-circle)`
		`= (x xx 2y) + 2 xx (1/2 xx pi xx y^2)`
		`= 2xy + piy^2`

b.	`text(Area of base)`	`= (2 xx 6 xx 2.5) + (pi xx 2.5^2)`
		`= 49.634…\ text(m²)`

`text(Area of walls) = text(Length) xx text(Height)`

`text(Length)`	`= 2x + 2 xx text(semi-circle perimeter)`
	`= (2 xx 6) + 2 xx (1/2 xx 2 xx pi xx 2.5)`
	`= 12 + 15.707…= 27.707…\ text(m)`

`:.\ text(Area of walls)= 27.707 xx 1.1= 30.478…\ text(m)^2`

`:.\ text(Total Area covered by tiles)`

`= 49.634… + 30.478…= 80.11…`

`= 80\ text{m² (nearest m²)}`

c. `text(Water saved)= 5 xx 3 xx 6= 90\ text(L per day.)`

`text(Water saved per year)`

`= 90 xx 365= 32\ 850\ text(L)`

`= 32.85\ text(kL)`

`:.\ text(Money saved)`	`= 32.85 xx $1.013= $33.277…`
	`= $33.28\ text{(nearest cent)}`

Statistics, STD2 S1 2005 HSC 27d

Nine students were selected at random from a school, and their ages were recorded.

\begin{array} {|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Ages} \rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex} \ \ \ \text{12 11 16} \ \ \ \rule[-1ex]{0pt}{0pt} \\ \rule{0pt}{2.5ex} \text{14 16 15} \rule[-1ex]{0pt}{0pt} \\ \rule{0pt}{2.5ex} \text{14 15 14} \rule[-1ex]{0pt}{0pt} \\
\hline
\end{array}

What is the sample standard deviation, correct to two decimal places? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Briefly explain what is meant by the term standard deviation. (1 mark)

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

Show Answers Only

a. `text{1.69 (to 2 d.p.)}`

b. `text(Standard deviation is a measure of how much)`

`text(members of a data group differ from the mean)`

`text(value of the group.)`

Show Worked Solution

a. `text(Sample standard deviation)`

`= 1.6914…\ text{(by calculator)}`

`= 1.69\ \ \ text{(to 2 d.p.)}`

b. `text(Standard deviation is a measure of how much)`

`text(members of a data group differ from the mean)`

`text(value of the group.)`

Algebra, 2UG 2004 HSC 28b

A set of garden gnomes is made so that the cost (`$C`) varies directly with the cube of the base length (`b` centimetres). A gnome with a base length of `text(10 cm)` has a cost of `$50`.

Write an equation relating the variables `C` and `b`, and a constant `k`. (1 mark)
Find the value of `k`. (1 mark)
Felicity says, ‘If you double the base length, you double the cost.’ Is she correct? Justify your answer with mathematical calculations. (2 marks)

Show Answers Only

`C = kb^3`
`0.05`
`text(Felicity is incorrect, because doubling)`
`text(the base from 10 cm to 20 cm causes the)`
`text(cost to increase 8 times.)`

Show Worked Solution

(i) `C`	`∝ b^3`
`:.C`	`= kb^3`

(ii) `C = 50\ \ text(when)\ \ b = 10`

`50`	`= k xx 10^3`
`:.k`	`= 50/10^3`
	`= 0.05`

(iii) `text(If)\ b = 20`

`C`	`= 0.05 xx 20^3`
	`= $400`

`:.\ text(Felicity is incorrect, because doubling)`

`text(the base from 10 cm to 20 cm causes the)`

`text(cost to increase 8 times.)`

Algebra, STD2 A4 2004 HSC 28a

A health rating, `R`, is calculated by dividing a person’s weight, `w`, in kilograms by the square of the person’s height, `h`, in metres.

Fred is 150 cm and weighs 72 kg. Calculate Fred’s health rating. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Over several years, Fred expects to grow 10 cm taller. By this time he wants his health rating to be 25. How much weight should he gain or lose to achieve his aim? Justify your answer with mathematical calculations. (2 marks)

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

Show Answers Only

i. `32`

ii. `text(8 kg)`

Show Worked Solution

i. `R = w/h^2`

`text(When)\ w = 72 and h = 1.5:`

`R=72/1.5^2= 32`

ii. `text(Find)\ w\ text(if)\ R = 25 and h = 1.6:`

`25`	`= w/1.6^2`
`w`	`= 25 xx 1.6^2`
	`= 64\ text(kg)`

`:.\ text(Weight Fred should lose)`

`= 72-64`

`= 8\ text(kg)`

Financial Maths, STD2 F1 2006 HSC 22 MC

This income tax table is used to calculate Evelyn’s tax payable.

Evelyn’s taxable income increases from `$50\ 000` to `$80\ 000`.

What percentage of her increase will she pay in additional tax?

`15.25%`
`40.7%`
`43.5%`
`52%`

Show Answers Only

`B`

Show Worked Solution

`text(Tax on $50 000)`	`= 2500 + 0.35 xx (50\ 000-45\ 000)`
	`= 2500 + 1750= $4250`
`text(Tax on $80 000)`	`= 11\ 250 + 0.52 xx (80\ 000-70\ 000)`
	`= 11\ 250 + 5200= $16\ 450`

`:.\ text(Extra tax)= 16\ 450-4250= $12\ 200`

`:.\ text(% Increase paid in tax)= (12\ 200) / (30\ 000) xx 100=\ text(40.66… %)`

`=> B`

Statistics, STD2 S1 2005 HSC 22 MC

Two groups of people were surveyed about their weekly wages. The results are shown in the box-and-whisker plots.

Which of the following statements is true for the people surveyed?

The same percentage of people in each group earned more than $325 per week.
Approximately 75% of people under 21 years earned less than $350 per week.
Approximately 75% of people 21 years and older earned more than $350 per week.
Approximately 50% of people in each group earned between $325 and $350 per week.

Show Answers Only

`B`

Show Worked Solution

`text{Option A: 50% of Under 21 group earned over $325 and 75%}`

`text{of Over 21 group did. NOT TRUE.}`

`text{Option B: 75% of Under 21 group earned below $350 is TRUE.}`

`text{Options C and D: can both be proven to be untrue using their}`

`text{median and quartile values.}`

`=> B`

Probability, STD2 S2 2005 HSC 20 MC

Dave’s school has computer security codes made up of four digits (eg 0773). Juanita’s school has computer security codes made up of five digits (eg. 30 568).

How many more codes are available at Juanita’s school than at Dave’s school?

10
50
90 000
100 000

Show Answers Only

`C`

Show Worked Solution

`text(# Codes at Dave’s school)`

`= 10 × 10 × 10 × 10`

`= 10\ 000`

`text(# Codes at Juanita’s school)`

`= 10^5`

`= 100\ 000`

`:.\ text(Extra Codes)`	`= 100\ 000 − 10 \ 000`
	`= 90\ 000`

`=> C`

Financial Maths, 2UG 2006 HSC 16 MC

Two families borrow different amounts of money on the same day.

The Wang family has a flat rate loan. The Salama family has a reducing balance loan and repays the loan earlier than the Wang family.

Which graph best represents this situation?

Show Answers Only

`C`

Show Worked Solution

`text(By elimination)`

`text(The families borrow different amounts)`

`:.\ text(NOT)\ D`

`text{The reducing balance loan (curved graph)}`

`text(is paid off earlier, and therefore intersects the)`

`text(‘time’ axis closer to zero)`

`:.\ text(NOT)\ A\ text(or)\ B`

`=> C`

CORE*, FUR1 2008 VCAA 9 MC

An amount of $8000 is invested for a period of 4 years.

The interest rate for this investment is 7.2% per annum compounding quarterly.

The interest earned by the investment in the fourth year (in dollars) is given by

A. `4 xx (7.2/100 xx 8000)`

B. `8000 xx 1.018^4 - 8000 xx 1.018^3`

C. `8000 xx 1.018^16 - 8000 xx 1.018^12`

D. `8000 xx 1.072^4 - 8000 xx 1.072^3`

E. `8000 xx 1.072^16 - 8000 xx 1.072^12`

Show Answers Only

`C`

Show Worked Solution

`text(4th year interest)`	`=\ text(Value after 4 years) -`
	`text(Value after 3 years.)`

`text(Using)\ \ A = PR^n,`

♦♦ Mean mark 33%.
MARKERS’ COMMENT: The most common errors were not converting and applying the interest rate to a quarterly one.

`text(where)\ R = 1 + 7.2/(100 xx 4) = 1.018`

`text(Value after 4 years:)`

`A_4`	`= 8000 xx 1.018^((4 xx 4))`
	`= 8000 xx 1.018^(16)`

`text(Value after 3 years:)`

`A_3`	`= 8000 xx 1.018^((4 xx 3))`
	`= 8000 xx 1.018^(12)`

`:.\ text(4th year interest) = 8000 xx 1.018^(16) − 8000 xx 1.018^(12)`

`=> C`

CORE*, FUR1 2008 VCAA 7 MC

Ernie took out a reducing balance loan to buy a new family home.

He correctly graphed the amount paid off the principal of his loan each year for the first five years.

The shape of this graph (for the first five years of the loan) is best represented by

Show Answers Only

`B`

Show Worked Solution

`text(A reducing balance means that the amount of)`

♦♦ Mean mark 25%.

`text(interest paid decreases each year, and therefore)`

`text(the amount paid off the principal will not only)`

`text(increase each year, but will do so at an increasing)`

`text(rate.)`

`B\ text(correctly shows this trend.)`

`=> B`

CORE*, FUR1 2007 VCAA 7 MC

At the start of each year Joe's salary increases to take inflation into account.

Inflation averaged 2% per annum last year and 3% per annum the year before that.

Joe's salary this year is $42 000.

Joe's salary two years ago, correct to the nearest dollar, would have been

A. `$39\ 900`

B. `$39\ 925`

C. `$39\ 926`

D. `$39\ 976`

E. `$39\ 977`

Show Answers Only

`E`

Show Worked Solution

`text(Let last year’s salary)\ = x`

♦♦ Mean mark 23%.
MARKERS’ COMMENT: Depreciating this year’s salary of $42 000 by 3% and then 2% to get ($42 000 x 0.97 x 0.98) is an incorrect strategy used by almost half of students. Make sure you understand why!

`x xx 1.03`	`= 42 \ 000`
`x`	`= (42\ 000)/1.03`
	`= $40 \ 776.70`

`text(Let salary 2 years ago)\ = y`

`y xx 1.02`	`= $40 \ 776.70`
`:. y`	`=$39 \ 977.15`

`=> E`

CORE*, FUR1 2006 VCAA 9 MC

Jenny borrowed $18 000. She will fully repay the loan in five years with equal monthly payments.

Interest is charged at the rate of 9.2% per annum, calculated monthly, on the reducing balance.

The amount Jenny has paid off the principal immediately following the tenth repayment is

A. $1876.77

B. $2457.60

C. $3276.00

D. $3600.44

E. $3754.00

Show Answers Only

`B`

Show Worked Solution

`text(Find monthly repayments,)`

♦♦ Mean mark 27%.

`text(By TVM Solver:)`

`N`	`= 5 xx 12 = 60`
`I(%)`	`= 9.2text(%)`
`PV`	`= − 18\ 000`
`PMT`	`= ?`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`:. \ PMT = $375.40`

`text(Find the loan value after the 10th repayment,)`

`text(By TVM Solver:)`

`N`	`= 10`
`I(%)`	`= 9.2text(%)`
`PV`	`= − 18\ 000`
`PMT`	`= 375.40`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`:. \ FV= 15\ 542.40`

`:.\ text(Amount paid off)`	`= 18\ 000 − 15\ 542.40`
	`= $2457.60`

`=>B`

CORE*, FUR1 2006 VCAA 3 MC

Grandpa invested in an ordinary perpetuity from which he receives a monthly pension of $584.

The interest rate for the investment is 6.2% per annum.

The amount Grandpa has invested in the perpetuity is closest to

A. $3600

B. $9420

C. $94 200

D. $43 400

E. $113 000

Show Answers Only

`E`

Show Worked Solution

`I = 584, \ r = 6.2, \ T = 1/12`

♦♦ Mean mark 23%.
MARKERS’ COMMENT: Nearly half of students read the monthly pension amount as a yearly amount. Be careful!

`I`	`= (PrT)/100`
`584`	`=(P xx 6.2 xx 1/12)/100`
`:. P`	`=(584 xx 100)/(6.2 xx 1/12)`
	`=$113\ 032.25…`

`=> E`

CORE*, FUR1 2011 VCAA 7 MC

Anthony invested $15 000 in an account. It earned `rtext(%)` interest per annum, compounding monthly.

The amount of interest that is earned in the third year of the investment is given by

A. `15\ 000 (1 + r / 1200)^3 - 15\ 000 (1 + r / 1200)^2`

B. `15\ 000 (1 + r / 1200)^36 - 15\ 000 (1 + r / 1200)^24`

C. `15\ 000 (1 + r / 100)^3 - 15\ 000 (1 + r / 100)^2`

D. `15\ 000 (1 + r / 100)^36 - 15\ 000 (1 + r / 100)^24`

E. `15\ 000 (1 + r / 1200)^4 - 15\ 000 (1 + r / 1200)^3`

Show Answers Only

`B`

Show Worked Solution

`text(Interest in 3rd year)`

♦♦ Mean mark 33%.

`=\ text(Value after 3 yrs) – text(Value after 2 years)`

`text(Using)\ \ A = PR^n,\ \ text(where)\ \ A = 15\ 000, and`

`R`	`= (1 + r / {12 xx 100})`
	`= (1 + r / 1200)`

`text(In first 3 years:)`

`A_3`	`= 15\ 000 (1 + r / 1200)^(12 xx 3)`
	`= 15\ 000 (1 + r / 1200)^36`

`text(In first 2 years:)`

`A_2`	`= 15\ 000 (1 + r / 1200)^(12 xx 2)`
	`= 15\ 000 (1 + r / 1200)^24`

`:.\ text(Interest in 3rd year)`

`= 15\ 000 (1 + r/1200)^36 – 15\ 000 (1 + r/1200)^24`

`=> B`

CORE*, FUR1 2012 VCAA 9 MC

Peter took out a reducing balance loan where interest was calculated monthly. He planned to repay this loan fully, with eight equal monthly payments of $260.

Peter missed the fourth payment, but made a double payment of $520 in the fifth month. He then continued to make payments of $260 for the remaining three months.

Which graph could show the balance of the loan each month over the eight-month period?

Show Answers Only

`A`

Show Worked Solution

`text(By elimination,)`

`text(As Peter missed the 4th payment, the balance of the)`

♦♦ Mean mark 20%.

`text(loan will increase after 4 months due to interest.)`

`:.\ text(Eliminate B, C, D and E.)`

`=> A`

CORE*, FUR1 2013 VCAA 8 MC

A car is purchased for $25 000. The value of the car is to be depreciated each year by 20% using the reducing balance method.

In the fourth year, the car will depreciate in value by

A. $2048

B. $2560

C. $5000

D. $10 240

E. $14 760

Show Answers Only

`B`

Show Worked Solution

`text(Depreciated value after 3 years)`

♦♦ Mean mark 29%.
MARKERS’ COMMENT: A common mistake was to not read the question carefully and give the depreciated value of the car after 4 years.

`= 25\ 000 xx (1 – 0.2)^3`

`= 12\ 800`

`:.\ text(Depreciation in 4th year)`

`= 12\ 800 xx 0.2`

`= $2560`

`=> B`

CORE*, FUR1 2013 VCAA 6 MC

A worker has received an annual salary increase of 3% for the past two years.

This year, the worker’s annual salary is $46 500.

Two years ago, her salary was closest to

A. `$42\ 315`

B. `$43\ 750`

C. `$43\ 830`

D. `$45\ 140`

E. `$49\ 330`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ x\ text(be the worker’s salary 2 years ago)`

♦♦ Mean mark 30%.

`x xx 1.03 xx 1.03`	`= 46\ 500`
`x (1.03)^2`	`= 46\ 500`
`:. x`	`= (46\ 500) / 1.03^2`
	`= 43\ 830.70…`

`=> C`

CORE*, FUR1 2014 VCAA 9 MC

Leslie borrowed $35 000 from a bank.

Interest is charged at the rate of 4.75% on the reducing monthly balance.

The loan is to be repaid with 47 monthly payments of $802.00 and a final payment that is to be adjusted so that the loan will be fully repaid after exactly 48 monthly payments.

Correct to the nearest cent, the amount of the final payment will be

A. $0.39

B. $3.57

C. $802.00

D. $802.39

E. $805.57

Show Answers Only

`E`

Show Worked Solution

`text(Calculate the FV of the loan after 47 payments)`

♦♦♦ Mean mark 18%.
MARKERS’ COMMENT: A majority of students found the amount still owed after the second last payment but failed to add interest during the last month.

`text(By TVM Solver:)`

`N`	`= 47`
`I(%)`	`= 4.75text(%)`
`PV`	`= -35\ 000`
`PMT`	`= 802`
`FV`	`= text(?)`
`text(P/Y)`	`= text(C/Y) = 12`

`:.\ text(Balance owing after 47 payments)\ =$802.39…`

`text(Final payment)`

`=\ text(Balance after 47 payments + month’s interest)`

`=802.39… + (802.39… xx 4.75/12 xx 100text{%})`

`=$805.57`

`=> E`

GRAPHS, FUR1 2007 VCAA 7 MC

The graph above shows a relationship between `y` and `x^3`.

The graph that shows the same relationship between `y` and `x` is

Show Answers Only

`C`

Show Worked Solution

`text(The gradient of the linear graph of)`

♦♦♦ Mean mark 26%.

`y\ text(versus)\ x^3 = 1/3`

`:.\ y = 1/3x^3`

`text(Test if the coordinates in each graph satisfy)`

`text(the equation)\ \ y=1/3x^3.`

`text(Consider)\ C,`

`text(Substitute)\ (1, 1/3)\ text(into equation,)`

`1/3 = 1/3 xx 1^3\ \ \ text{(true)}`

`text(All other options can be shown to not satisfy equation.)`

`=> C`

GRAPHS, FUR1 2007 VCAA 6 MC

Russell is a wine producer. He makes both red and white wine.

Let `x` represent the number of bottles of red wine he makes and `y` represent the number of bottles of white wine he makes.

This year he plans to make at least twice as many bottles of red wine as white wine.

An inequality representing this situation is

A. `y <= x + 2`

B. `y <= 2x`

C. `y >= 2x`

D. `x <= 2y`

E. `x >= 2y`

Show Answers Only

`E`

Show Worked Solution

`x = text(no. red wine bottles)`

♦♦ Mean mark 34%.

`y = text(no. white wine bottles)`

`text(The constraint states that the number of)`

`text(of bottles of red wine will be at least twice)`

`text(the number of bottles of white wine.)`

`x >= 2y`

`=> E`

GRAPHS, FUR1 2008 VCAA 4 MC

When shopping, Betty can use either Easypark or Safepark to park her car.

At Easypark, cars can be parked for up to 8 hours per day.

The fee structure is as follows.

`text(Fee)={(text($5.00,), 0 < text(hours) <= 2),(text($8.00,), 2 < text(hours) <= 5),(text($11.00,), 5 < text(hours) <= 8) :}`

Safepark charges fees according to the formula

`text(Fee) = $2.50 xx text(hours)`

Betty wants to park her car for 5 hours on Monday and 3 hours on Tuesday.

The minimum total fee that she can pay for parking for the two days is

A. `$7.50`

B. `$11.00`

C. `$15.50`

D. `$16.00`

E. `$20.00`

Show Answers Only

`C`

Show Worked Solution

`text(Fee for 5 hours)`

♦♦ Mean mark 29%.
MARKER’S COMMENT: Almost half of students failed to take into account that the same car park did not have to be used on each day, and incorrectly answered D.

`text(Easypark = $8)`

`text(Safepark = $2.50) xx 5 = $12.50`

`text(Fee for 3 hours)`

`text(Easypark = $8)`

`text(Safepark = $2.50) xx 3 = $7.50`

`:.\ text(The minimum total fee is)`

`$8 + $7.50 = $15.50`

`=> C`

GRAPHS, FUR1 2009 VCAA 9 MC

The graph below shows the cost (in dollars) of producing birthday cards.

If the profit from the sale of 150 birthday cards is $175, the selling price of one card is

A. `$0.30`

B. `$1.60`

C. `$3.10`

D. `$3.50`

E. `$4.40`

Show Answers Only

`D`

Show Worked Solution

`text{Profit = Revenue – Costs = $175 (given)}`

♦ Mean mark 37%.

`text{Revenue}\ (R) = 150 xx S,\ \ \ (S= text{selling price})`

`text(Calculating cost of 150 cards,)\ C,`

`text(Fixed cost) = $50\ \ text{(}y\ text(intercept) text{)}`

`text{40 cards cost $130 (from graph)}`

`:.\ text(C)text(ost per card)= (130 − 50)/40=$2`

`:. C=50 + 150 xx 2=350`

`R-C`	`=175`
`150S-350`	`= 175`
`150S`	`=525`
`:. S`	`= $3.50`

`=> D`

GEOMETRY, FUR1 2006 VCAA 9 MC

Points `M` and `P` are the same distance from a third point `O`.

The bearing of `M` from `O` is 038° and the bearing of `P` from `O` is 152°.

The bearing of `P` from `M` is

A. between 000° and 090°

B. between 090° and 180°

C. exactly 180°

D. between 180° and 270°

E. between 270° and 360°

Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 31%.

`/_ MOP`	`= 152 − 38`
	`= 114^@`

`text(Let)\ R\ text(be directly south of)\ O`

`/_POR`	`= 180 − (38 + 114)\ \ text{(straight line)}`
	`= 28^@`

`text(Given)\ OM = OP`

`text(Diagram shows that)\ M\ text(is further east than)\ P.`

`:.\ text(Bearing of)\ P\ text(from)\ M\ text(just over 180°)`

`=> D`

GEOMETRY, FUR1 2006 VCAA 8 MC

The cross-section of a water pipe is circular with a radius, `r`, of 50 cm, as shown above.

The surface of the water has a width, `w`, of 80 cm.

The depth of water in the pipe, `d`, could be

A. `20\ text(cm)`

B. `25\ text(cm)`

C. `30\ text(cm)`

D. `40\ text(cm)`

E. `50\ text(cm)`

Show Answers Only

`A`

Show Worked Solution

♦♦ Mean mark 22%.
MARKER’S COMMENT: 39% of students calculated `x` in the solution, without then using it to find the depth.

`text(Using Pythagoras,)`

`x^2 + 40^2`	`= 50^2`
`x^2`	`= 2500 − 1600`
	`= 900`
`x`	`= 30\ text(cm)`

`d + x`	`= 50\ text{(radius)}`
`:.\ d`	`= 20\ text(cm)`

`=> A`

Measurement, 2UG 2007 HSC 28c

A piece of plaster has a uniform cross-section, which has been shaded, and has dimensions as shown.

Use two applications of Simpson’s rule to approximate the area of the cross-section. (3 marks)
The total surface area of the piece of plaster is `7480.8\ text(cm²)`.
Calculate the area of the curved surface as shown on the diagram. (2 marks)

Show Answers Only

`50.4\ text(cm²)`
`3500\ text(cm²)`

Show Worked Solution

(i)

`A`	`= h/3 [y_0 +4y_1 + y_2]\ text(… applied twice)`
	`= 3.6/3 (5 + 4 xx 4.6 + 3.7) + 3.6/3 (3.7 + 4 xx 2.8 + 0)`
	`= 32.52 + 17.88`
	`= 50.4\ text(cm²)`

(ii) `text(Total Area) = 7480.8\ text(cm²)`

`text(Area of Base)`	`= 14.4 xx 200`
	`= 2880\ text(cm²)`

`text(Area of End)`	`= 5 xx 200`
	`= 1000\ text(cm²)`

`text(Area of sides)`	`= 2 xx 50.4\ \ \ text{(from (i))}`
	`= 100.8\ text(cm²)`

`:.\ text(Area of curved surface)`

`= 7480.8 – (2880 + 1000 + 100.8)`

`= 3500\ text(cm²)`

Measurement, STD2 M1 2007 HSC 28b

This shape is made up of a right-angled triangle and a regular hexagon.

The area of a regular hexagon can be estimated using the formula `A = 2.598H^2` where `H` is the side-length.

Calculate the total area of the shape using this formula. (3 marks)

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

Show Answers Only

`22.784\ text(cm)^2`

Show Worked Solution

`text(Area) = 2.598H^2`

`text(Using Pythagoras:)`

`H^2`	`= 2^2 + 2^2= 8`
`H`	`= sqrt 8`

`text(Area of hexagon)= 2.598 xx (sqrt 8)^2= 20.784\ text(cm)^2`

`text(Area of triangle)= 1/2 bxx h= 1/2 xx 2 xx 2= 2\ text(cm)^2`

`:.\ text(Total Area)= 20.784 + 2= 22.784\ text(cm)^2`

Probability, 2UG 2007 HSC 28a

Two unbiased dice, `A` and `B`, with faces numbered `1`, `2`, `3`, `4`, `5` and `6` are rolled.

The numbers on the uppermost faces are noted. This table shows all the possible outcomes.

A game is played where the difference between the highest number showing and the lowest number showing on the uppermost faces is calculated.

What is the probability that the difference between the numbers showing on the uppermost faces of the two dice is one? (1 mark)
In the game, the following applies.
What is the financial expectation of the game? (3 marks)
If Jack pays `$1` to play the game, does he expect a gain or a loss, and how much will it be? (1 mark)

Show Answers Only

`5/18`
`$0.75`
`text(If Jack pays $1 to play, he should expect)`
`text(a loss of $0.25.)`

Show Worked Solution

(i) `text(# Outcomes with a difference of 1)`

`= 10`

`:.\ P text{(diff of 1)} = 10/36 = 5/18`

(ii) `P text{(no difference)} = 6/36 = 1/6`

`P text{(2, 3, 4 or 5)}`	`= 1 – [P(0) + P(1)]`
	`= 1 – [5/18 + 1/6]`
	`= 1 – 8/18`
	`= 5/9`

`:.\ text(Financial Expectation)`

`= (1/6 xx 3.50) – (5/18 xx 5) + (5/9 xx 2.80)`

`= $0.75`

(iii) `text(If Jack pays $1 to play, he should expect)`

`text(a loss of $0.25.)`

GEOMETRY, FUR1 2007 VCAA 6 MC

A solid cylinder has a height of 30 cm and a diameter of 40 cm.

A hemisphere is cut out of the top of the cylinder as shown below.

In square centimetres, the total surface area of the remaining solid (including its base) is closest to

A. `1260`

B. `2510`

C. `6280`

D. `7540`

E. `10\ 050`

Show Answers Only

`D`

Show Worked Solution

`text(Total surface area of solid)`

`=\ text{S.A. (base)} + text{S.A. (side)} + text{S.A. (hemisphere)}`

♦♦ Mean mark 28%.

`text{S.A. (base)}`	`= pir^2`
	`= pi xx 20^2`
	`= 1256.63…`

`text{S.A. (side)}`	`= 2 pi r h`
	`= 2 xx pi xx 20 xx 30`
	`= 3769.91…`

`text{S.A. (hemisphere)}`	`= 1/2 xx 4pir^2`
	`= 1/2 xx 4 xx pi xx 20^2`
	`= 2513.27…`

`:.\ text(Total S.A.)`	`= 1256.63… + 3769.91… + 2513.27…`
	`= 7539.81…\ text(cm²)`

`=> D`

GEOMETRY, FUR1 2007 VCAA 5 MC

A block of land has an area of 4000 m².

When represented on a map, this block of land has an area of 10 cm².

On the map 1 cm would represent an actual distance of

A. `10\ text(m)`

B. `20\ text(m)`

C. `40\ text(m)`

D. `400\ text(m)`

E. `4000\ text(m)`

Show Answers Only

`B`

Show Worked Solution

`text(Area scale factor) = k^2`

♦♦♦ Mean mark 18%.
MARKER’S COMMENT: A majority of students obtained the the “area” scale factor (k² = 400) but failed to convert this to the corresponding linear scale factor.

`k^2`	`= 4000/10`
	`= 400`
`:. k`	`= sqrt(400)`
	`= 20`

`:.\ text(Linear scale factor) = 20,`

`text(i.e.)\ \ 1\ text(cm):20\ text(m)`

`=> B`

GEOMETRY, FUR1 2011 VCAA 6 MC

In parallelogram `PQRS`, `/_ QPS = 74°.`

In this parallelogram, `PQ = 18\ text(cm)` and `PS = 25\ text(cm.)`

The length of the longer diagonal of this parallelogram is closest to

A. `26.5\ text(cm)`

B. `30.1\ text(cm)`

C. `30.8\ text(cm)`

D. `34.6\ text(cm)`

E. `39.9\ text(cm)`

Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 34%.

`/_ PQR`	` = 180 – 74\ \ \ text{(co-interior,}\ QR text(\|\|) PS text{)}`
	` = 106^@`

`text(Using cosine rule,)`

`PR^2`	` = 18^2 + 25^2 – 2 xx 18 xx 25 xx cos 106^@`
	` = 324 + 625 + 248.07…`
	` = 1197.07…`
`:. PR`	` = 34.59…\ \ text(cm)`

`=> D`

GEOMETRY, FUR1 2009 VCAA 9 MC

A vertical pole, `TP`, is 4 metres tall and stands on level ground near a vertical wall.

The wall is 6 metres long and 4 metres high.

The base of the pole, `T`, is 5 metres from one end of the wall at `N` and 4 metres from the other end of the wall at `M`.

The pole falls and hits the wall.

The maximum height above ground level at which the pole could hit the wall is closest to

A. `0\ text(m)`

B. `1.5\ text(m)`

C. `2.3\ text(m)`

D. `2.7\ text(m)`

E. `3.3\ text(m)`

Show Answers Only

`C`

Show Worked Solution

♦♦ Mean mark 27%.
MARKER’S COMMENT: The key concept to finding the solution is to realise that max height occurs when the pole falls at right angles to the wall.

`text(Using cosine rule in)\ Delta MTN\ text(to find)\ theta`

`cos theta`	`= (6^2 + 4^2 − 5^2) / (2 × 6 × 4)`
	`= 0.5625`
`theta`	`= 55.77…^@`

`text(In)\ Delta AMT,`
`sin theta`	`= x/4`
`x`	`= 4 xx sin 55.77…^@`
	`= 3.307…`
`text(When the pole falls, max height) = h`

`text(Using Pythagoras,)`

`4^2`	`= h^2 + 3.307^2`
`h^2`	`= 4^2 − 3.307^2`
	`= 5.0625`
`h`	`= 2.25…\ text(m)`
`=> C`

PATTERNS, FUR1 2009 VCAA 9 MC

There are 10 checkpoints in a 4500 metre orienteering course.

Checkpoint 1 is the start and checkpoint 10 is the finish.

The distance between successive checkpoints increases by 50 metres as each checkpoint is passed.

The distance, in metres, between checkpoint 2 and checkpoint 3 is

A. `225`

B. `275`

C. `300`

D. `350`

E. `400`

Show Answers Only

`D`

Show Worked Solution

`text(9 distances exist between 10 checkpoints)`

♦♦ Mean mark 26%.

`text(9 distances form an AP, where)`

`d = 50,\ \ \ S_9 = 4500`

`S_n`	`= n/2[2a + (n − 1)d]`
`4500`	`= 9/2 [2a + (9 − 1) × 50]`
`4500`	`= 9/2[2a + 400]`
`9a`	`= 2700`
`a`	`= 300`

`:.\ text(Distance between checkpoint 1 and 2)`

`= a=300\ text(m)`

`:.\ text(Distance between checkpoint 2 and 3)`

`= a+d=350\ text(m)`

`=> D`