Aussie Maths & Science Teachers: Save your time with SmarterEd
Use the Trapezoidal rule with three function values to find an approximation to the value of
`int_0.5^1.5 (log_e x )^3\ dx`.
Give your answer correct to three decimal places. (2 marks)
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`text(Let)\ \ f(x) = (log_e x )^3`
| `int_0.5^1.5(log_e x)^3 dx` | `~~ 0.5/2[−0.3330… + 2(0) + 0.0666…]` |
| `~~ 0.25(−0.2663…)` | |
| `~~ −0.06659…` | |
| `~~ −0.067\ \ \ text{(to 3 d.p.)}` |
Five values of the function `f(x)` are shown in the table.
Use the Trapezoidal rule with the five values given in the table to estimate
`int_0^20 f(x)\ dx`. (3 marks)
| `:. int_0^20 f(x)\ dx` | `~~ 5/2[15 + 2(25 + 22 + 18) + 10]` |
| `~~ 5/2(155)` | |
| `~~ 387.5` |
The length of each edge of the cube `ABCDEFGH` is 2 metres. A circle is drawn on the face `ABCD` so that it touches all four edges of the face. The centre of the circle is `O` and the diagonal `AC` meets the circle at `X` and `Y`.
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| i. |
|
`text(S)text(ince)\ \ FA, \ AC\ \ text(and)\ \ FC\ \ text(are all)`
`text(diagonals of sides of a cube,)`
`FA = AC = FC`
`:.ΔFAC\ \ text(is equilateral)`
`:.∠FAC = 60^@`
| ii. |
|
`text(In)\ \ ΔAEF`
| `AF^2` | `= EF^2 + EA^2` |
| `= 2^2 + 2^2` | |
| `= 8` | |
| `AF` | `= sqrt8` |
| `= 2sqrt2` |
`text(In)\ \ ΔAFO`
| `sin\ 60^@` | `= (FO)/(AF)` |
| `sqrt3/2` | `= (FO)/(2sqrt2)` |
| `FO` | `= sqrt3/2 xx 2sqrt2` |
| `= sqrt6\ text(metres … as required.)` |
| iii. |
|
`XY\ \ text(is the diameter of a circle AND the width)`
`text(of the cube.)`
| `:.XY` | `= 2` |
| `:.OX` | `= OY = 1` |
| `tan\ ∠OFX` | `=1 /sqrt6` |
| `∠OFX` | `= 22.207…^@` |
| `:.∠XFY` | `= 2 xx 22.407…` |
| `= 44.415…` | |
| `= 44^@\ text{(nearest degree)}` |
A boat is sailing due north from a point `A` towards a point `P` on the shore line.
The shore line runs from west to east.
In the diagram, `T` represents a tree on a cliff vertically above `P`, and `L` represents a landmark on the shore. The distance `PL` is 1 km.
From `A` the point `L` is on a bearing of 020°, and the angle of elevation to `T` is 3°.
After sailing for some time the boat reaches a point `B`, from which the angle of elevation to `T` is 30°.
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From a point `A` due south of a tower, the angle of elevation of the top of the tower `T`, is 23°. From another point `B`, on a bearing of 120° from the tower, the angle of elevation of `T` is 32°. The distance `AB` is 200 metres.
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i.
ii. `text(Find)\ \ OT = h`
`text(Using the cosine rule in)\ Delta AOB :`
`200^2 = OA^2 + OB^2 – 2 * OA * OB * cos 60\ …\ text{(*)}`
`text(In)\ Delta OAT,\tan 23^@= h/(OA)`
`=> OA= h/(tan 23^@)\ …\ (1)`
`text(In)\ Delta OBT,\ tan 32^@= h/(OB)`
`=> OB= h/(tan 32^@)\ \ \ …\ (2)`
`text(Substitute)\ (1)\ text(and)\ (2)\ text(into)\ text{(*):}`
| `200^2` | `= (h^2)/(tan^2 23^@) + (h^2)/(tan^2 32^@) – 2 * h/(tan 23^@) * h/(tan 32^@) * 1/2` |
| `= h^2 (1/(tan^2 23^@) + 1/(tan^2 32^@) + 1/(tan23^@ * tan32^@) )` | |
| `= h^2 (4.340…)` | |
| `h^2` | `= (40\ 000)/(4.340…)` |
| `= 9214.55…` | |
| `:. h` | `= 95.99…` |
| `= 96\ text(m)\ \ \ text{(to nearest m)}` |
A person walks 2000 metres due north along a road from point `A` to point `B`. The point `A` is due east of a mountain `OM`, where `M` is the top of the mountain. The point `O` is directly below point `M` and is on the same horizontal plane as the road. The height of the mountain above point `O` is `h` metres.
From point `A`, the angle of elevation to the top of the mountain is 15°.
From point `B`, the angle of elevation to the top of the mountain is 13°.
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i. `text(Show)\ \ OA = h\ cot\ 15^@`
`text(In)\ \ Delta MOA,`
| `tan\ 15^@` | `= h/(OA)` |
| `OA` | `= h/(tan\ 15^@)` |
| `= h\ cot\ 15^@\ \ …text(as required)` |
ii. `text(In)\ \ ΔMOB`
| `tan\ 13^@` | `= h/(OB)` |
| `OB` | `= h/(tan\ 13^@)` |
| `= h\ cot\ 13^@` |
`text(In)\ \ ΔAOB:`
| `OA^2 + AB^2` | `= OB^2` |
| `OB^2 − OA^2` | `= AB^2` |
| `(h\ cot\ 13^@)^2 − (h\ cot\ 15^@)^2` | `= 2000^2` |
| `h^2[(cot^2\ 13^@ − cot^2\ 15^@)]` | `= 2000^2` |
| `h^2` | `= (2000^2)/(cot^2\ 13^@ − cot^2\ 15^@)` |
| `:. h` | `= sqrt((2000^2)/(cot^2\ 13^@ − cot^2\ 15^@))` |
| `= 909.704…` | |
| `= 910\ text{m (nearest metre)}` |
A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`. The
size of angle `ABC` is 114°.
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| i. |  |
`text(Let point)\ D\ text(be due North of point)\ B`
`/_ABD=180-121\ text{(cointerior with}\ \ /_A text{)}\ =59^@`
`/_DBC=114-59=55^@`
`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`
ii. `text(Using cosine rule:)`
| `AC^2` | `=AB^2+BC^2-2xxABxxBCxxcos/_ABC` |
| `=6^2+9^2-2xx6xx9xxcos114^@` | |
| `=160.9275…` | |
| `:.AC` | `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}` |
| `=13\ text(km)\ text{(nearest km)}` |
iii. `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`
| `cos/_ACB` | `=(AC^2+BC^2-AB^2)/(2xxACxxBC)` |
| `=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)` | |
| `=0.9018…` | |
| `/_ACB` | `=25.6^@\ text{(to 1 d.p.)}` |
`text(From diagram,)`
`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`
`:.\ text(Bearing of)\ A\ text(from)\ C`
| `=180+55+25.6` | |
| `=260.6` | |
| `=261^@\ text{(nearest degree)}` |
A yacht race follows the triangular course shown in the diagram. The course from `P` to `Q` is 1.8 km on a true bearing of 058°.
At `Q` the course changes direction. The course from `Q` to `R` is 2.7 km and `/_PQR = 74^@`.
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What is the area of this ‘no-go’ zone? (1 mark)
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| i. |  |
`/_ PQS = 58^@ \ \ \ (text(alternate to)\ /_TPQ)`
`text(Bearing of)\ R\ text(from)\ Q`
| `= 180^@ + 58^@ + 74^@` |
| `= 312^@` |
(ii) `text(Using Cosine rule:)`
| `RP^2` | `=RQ^2` + `PQ^2` `- 2` `xx RQ` `xx PQ` `xx cos` `/_RQP` |
| `= 2.7^2` + `1.8^2` `- 2` `xx 2.7` `xx 1.8` `xx cos74^@` | |
| `=7.29 + 3.24\ – 2.679…` | |
| `=7.851…` |
| `:.RP` | `= sqrt(7.851…)` |
| `=2.8019…` | |
| `~~ 2.8\ text(km) (text(1 d.p.) )` |
(iii) `text(Using)\ A = 1/2 ab sinC`
| `A` | `= 1/2` `xx 2.7` `xx 1.8` `xx sin74^@` |
| `= 2.3358…` | |
| `= 2.3\ text(km²)` |
`:.\ text(No-go zone is 2.3 km²)`
A plane flies on a bearing of 150° from `A` to `B`.
What is the bearing of `A` from `B`?
`/_TBA=30^@\ \ \ text{(angle sum of triangle)}`
`:.\ text(Bearing of)\ A\ text{from}\ B`
`=360-30`
`=330^@`
`=> D`
The diagram shows towns `A`, `B` and `C`. Town `B` is 40 km due north of town `A`. The distance from `B` to `C` is 18 km and the bearing of `C` from `A` is 025°. It is known that `∠BCA` is obtuse.
What is the bearing of `C` from `B`?
The following information is given about the locations of three towns `X`, `Y` and `Z`:
• `X` is due east of `Z`
• `X` is on a bearing of `145^@` from `Y`
• `Y` is on a bearing of `060^@` from `Z`.
Which diagram best represents this information?
`text(S)text(ince)\ X\ text(is due east of)\ Z`
`=> text(Cannot be)\ B\ text(or)\ D`
`text(The diagram shows we can find)`
`/_ZYX = 60 + 35^@ = 95^@`
`text(Using alternate angles)\ (60^@)\ text(and)`
`text(the)\ 145^@\ text(bearing of)\ X\ text(from)\ Y`
`=> C`
The diagram shows the position of `Q`, `R` and `T` relative to `P`.
In the diagram,
`Q` is south-west of `P`
`R` is north-west of `P`
`/_QPT` is 165°
What is the bearing of `T` from `P`?
`/_QPS=45^@\ \ \ text{(} Q\ text(is south west of)\ Ptext{)}`
`/_TPS = 165 – 45 = 120^@`
`:.\ /_NPT = 60^@\ \ text{(} 180^@\ text(in straight line) text{)}`
`=> A`
Town `B` is 80 km due north of Town `A` and 59 km from Town `C`.
Town `A` is 31 km from Town `C`.
What is the bearing of Town `C` from Town `B`?
A piece of plaster has a uniform cross-section, which has been shaded, and has dimensions as shown.
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| a. |  |
| `A` | `~~ 3.6/2 [5 + 2(4.6 + 3.7 + 2.8) + 0]` |
| `~~ 1.8(27.2)` | |
| `~~ 48.96\ text(cm²)` |
b. `text(Total Area) = 7480.8\ text{cm² (given)}`
| `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 48.96` |
| `= 97.92\ text(cm²)` |
`:.\ text(Area of curved surface)`
`= 7480.8 – (2880 + 1000 + 97.92)`
`= 3502.88`
`=3503\ text{cm² (nearest cm²)}`
A new 200-metre long dam is to be built.
The plan for the new dam shows evenly spaced cross-sectional areas.
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Assuming no wastage, calculate how much rainfall is needed, to the nearest mm, to fill the dam. (2 marks)
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| a. |  |
| `V` | `~~ 50/2[360 + 2(300 + 270 + 140) + 0]` |
| `~~ 25(1780)` | |
| `~~ 44\ 500\ text(m³)` |
b. `text(Convert 2 km² → m²:)`
| `text(2 km²)` | `= 2000\ text(m × 1000 m)` |
| `= 2\ 000\ 000\ text(m²)` |
`text(Using)\ \ V=Ah\ \ text(where)\ \ h= text(rainfall):`
| `44\ 500` | `= 2\ 000\ 000 xx h` |
| `:.h` | `= (44\ 500)/(2\ 000\ 000)` |
| `= 0.02225…\ text(m)` | |
| `= 22.25…\ text(mm)` | |
| `= 22\ text{mm (nearest mm)}` |
A tunnel is excavated with a cross-section as shown.
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If the value of `a` increases by 2 metres, by how much will `b` change? (2 marks)
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| a. |  |
| `A` | `~~ h/2[0 + 2(a + b + a) + 0]` |
| `~~ h/2(4a + 2b)` | |
| `~~ h(2a + b)` |
b. `A = 600\ text(m²)`
`text(If tunnel is 80 metres wide)`
| `4h` | `= 80` |
| `h` | `= 20` |
`text{Using part (i):}`
`600 = 20(2a + b)`
| `2a + b` | `= 30` |
| `b` | `= 30 – 2a` |
`:.\ text(If)\ a\ text(increases by 2,)\ b\ text(must decrease by 4.)`
An aerial diagram of a swimming pool is shown.
The swimming pool is a standard length of 50 metres but is not in the shape of a rectangle.
In the diagram of the swimming pool, the five widths are measured to be:
`CD = 21.88\ text(m)`
`EF = 25.63\ text(m)`
`GH = 31.88\ text(m)`
`IJ = 36.25\ text(m)`
`KL = 21.88\ text(m)`
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Calculate the approximate volume of the swimming pool, in litres. (1 mark)
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| a. |  |
`text(Surface Area of pool)`
`~~ 12.5/2[21.88 + 2(25.63 + 31.88 + 36.25) + 21.88]`
`~~ 1445.5\ text(m²)`
| b. | `V` | `= Ah` |
| `~~ 1445.5 xx 1.2` | ||
| `~~ 1734.6\ text(m³)` | ||
| `~~ 1\ 734\ 600\ text(L)` |
There is a lake inside the rectangular grass picnic area `ABCD`, as shown in the diagram.
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The lake is 60 cm deep. Bozo the clown thinks he can empty the lake using a four-litre bucket.
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Three equally spaced cross-sectional areas of a vase are shown.
Use the Trapezoidal rule to find the approximate capacity of the vase in litres. (3 marks)
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`text(Solution 1)`
| `V` | `≈ 15/2(45 + 180) + 15/2(180 + 35)` |
| `≈ 15/2(225 + 215)` | |
| `≈ 3300\ text{mL (1 cm³ = 1 mL)}` | |
| `~~3.3\ text(L)` |
`text(Solution 2)`
| `V` | `≈ 15/2(45 + 2 xx 180 + 35)` |
| `≈ 15/2(440)` | |
| `≈ 3300\ text{mL}` | |
| `~~3.3\ text(L)` |
Cassius is a boxer and skips to keep fit.
The table below shows the average energy used, in kilojuoles per kilogram of body mass, by a person skipping for 10 minutes at different speeds.
Cassius eats a hamburger that contains 550 kilocalories.
If Cassuis weighs 72 kilograms, how long must he skip at 150 skips per minute to burn off the energy contained in the hamburger? (1 kilocalorie = 4.184 kJ) Give your answer to 1 decimal place. (3 marks)
The table shows the average energy used, in kilojoules per kilogram of body mass, by a person walking for 30 minutes at different speeds.
\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Walking speed} \rule[-1ex]{0pt}{0pt} & \text{Energy used in 30 minutes} \\
\hline
\rule{0pt}{2.5ex} \text{3 km/h} \rule[-1ex]{0pt}{0pt} & \text{5.53 kJ/kg} \\
\hline
\rule{0pt}{2.5ex} \text{5 km/h} \rule[-1ex]{0pt}{0pt} & \text{7.37 kJ/kg} \\
\hline
\end{array}
Rob, who weighs 90 kg, drinks a large cappuccino made with full cream milk. It contains 146 kilocalories.
For approximately how long must Rob walk at 3 km/hr to burn off the energy contained in the cappuccino? (1 kilocalorie = 4.184 kilojoules)
The scale diagram shows the aerial view of a block of land bounded on one side by a road. The length of the block, `AB`, is known to be 90 metres.
Calculate the approximate area of the block of land, using three applications of the Trapezoidal rule. (3 marks)
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`text(Solution 1)`
`text(6 cm → 90 metres)`
` text(1 cm → 15 metres)`
`text(Height) = 2 xx 15 = 30\ text(metres)`
| `text(Area)` | `~~ 30/2(75 + 60) + 30/2(60 + 45) + 30/2(45 + 60)` |
| `~~ 15(135 + 105 + 105)` | |
| `~~ 5175\ text(m²)` |
`text(Solution 2)`
`text(After converting from scale:)`
| `text(Area)` | `~~ 30/2(75 + 2 xx 60 + 2 xx 45 + 60)` |
| `~~ 5175\ text(m²)` |
Let `n` be a positive integer and let `x` be a positive real number.
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The point `P(acostheta, bsintheta)`, where `0 < theta < pi/2`, lies on the ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1`, where ` a > b`. The point `A(acostheta, asintheta)` lies vertically above `P` on the auxiliary circle `x^2 + y^2 = a^2`. The point `B` lies on the auxiliary circle such that `angleAOB = pi/2` and the point `Q` lies on the ellipse vertically below `B`, as shown.
The line `QO` meets the ellipse again at `Qprime(asintheta, −bcostheta)`. (Do NOT prove this.)
Let `I_n = int_(−3)^0 x^n sqrt(x + 3)\ dx` for `n = 0, 1, 2…`
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A particle of mass `m` is attached to a light inextensible string of length `l`. The string makes an angle `theta` to the vertical.
The particle is moving in a circle of radius `r` on a smooth horizontal surface. The particle is moving with uniform angular velocity `ω`. The forces on the particle are the tension `T` in the string; the normal reaction `N` to the horizontal surface; and the gravitational force `mg`.
The particle remains in contact with the horizontal surface.
By resolving the forces on the particle in the horizontal and vertical directions, show that
`ω^2 <= g/(lcostheta)`. (3 marks)
Let `z = 1 - cos2theta + isin2theta`, where `0 < theta <= pi`.
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i. `z = 1 – cos2theta + isin2theta`
| `|\ z\ |` | `= sqrt((1 – cos2theta)^2 + sin^2 2theta)` |
| `= sqrt(1 – 2cos2theta + cos^2 2theta + sin^2 2theta)` | |
| `= sqrt(2 – 2cos2theta)` | |
| `= sqrt(2(1 – cos2theta))` | |
| `= sqrt(2(2sin^2theta))` | |
| `= 2sintheta\ \ \ text(… as required)` |
ii. `z= 1 – cos2theta + isin2theta`
| `text(arg)(z)` | `= tan^(−1)((sin2theta)/(1 – cos2theta))` |
| `= tan^(−1)((2sinthetacostheta)/(2sin^2theta))` | |
| `= tan^(−1)(cottheta)` | |
| `= tan^(−1)(tan(pi/2 – theta))` | |
| `= pi/2 – theta` |
`(text(S)text(ince)\ \ 0 < theta < pi => \ −pi/2 <= pi/2 – theta < pi/2)`
The graph `y^2 = x(1 - x)^2` is shown.
Use the method of cylindrical shells to find the volume of the solid formed when the shaded region is rotated about the line `x = 1`. (3 marks)
The base of a solid is the region enclosed by the parabola `x = 1 - y^2` and the `y`-axis. Each cross-section perpendicular to the `y`-axis is an equilateral triangle, as shown in the diagram.
Find the volume of the solid. (3 marks)
`text(Cross section of triangle)`
| `sin60°` | `= h/(1 – y^2)` |
| `:.h` | `= sqrt3/2(1 – y^2)` |
| `δV` | `= 1/2(1 – y^2) · sqrt3/2(1 – y^2)\ δy` |
| `= sqrt3/4(1 – y^2)^2\ δy` |
| `:. V` | `= int_(−1)^1 sqrt3/4(1 – y^2)\ dy` |
| `= sqrt3/2 int_0^1 1 – 2y^2 + y^2\ dy` | |
| `= sqrt3/2 [y – 2/3y^3 + 1/5y^5]` | |
| `= sqrt3/2 [(1 – 2/3 + 1/5) – 0]` | |
| `= sqrt3/2(8/15)` | |
| `= (4sqrt3)/15\ text(u³)` |
The circle centred at `O` with radius `r` has a diameter `AB`. Points `C` and `D` are chosen on the circle as shown in the diagram. The chord `AC` has length `d`.
Show that `d = 2r sin D`. (2 marks)
`text(Join)\ CB`
`angleB = angleD\ \ (text{angles on circumference on arc}\ AC)`
`angleACB = 90°\ \ (text{angle in a semi-circle})`
`text(In)\ DeltaABC,`
| `sin B` | `= d/(2r)` |
| `:. d` | `= 2r sin B` |
| `= 2r sin D\ \ \ text(.. as required)` |
The points `A`, `B` and `C` on the Argand diagram represent the complex numbers `u`, `v` and `w` respectively.
The points `O`, `A`, `B` and `C` form a square as shown on the diagram.
It is given that `u = 5 + 2i`.
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Consider the functions `f( x ) = sinx` and `g( x ) = x sinx`.
The `x`-coordinate of each stationary point of `f(x)` is very close to the `x`-coordinate of a stationary point of `g(x)`.
Suppose `f(x)` has a stationary point at `x = a` and the stationary point of `g(x)` with `x`-coordinate closest to `x = a` is at `x = b`.
Which statement is always true?
`f(x) = sin x\ => fprime(x) = cos x`
Mean mark 57%. A challenging question well answered.
`fprime(x) = 0\ \ text(when)\ \ x = pi/2, (3pi)/2, …`
`g(x) = x sin x\ => gprime(x) = sin x – x cos x`
`gprime(x) = 0\ \ text(when)\ \ −x = tan x`
`text(When)\ x > 0,\ text{intersection (where}\ gprime(x) = 0)\ text(is to)`
`text(the right of)\ \ x = pi/2.`
`text(When)\ x < 0,\ text(intersection is to the left of)\ \ x = – pi/2 .`
`:. |\ a\ | < |\ b\ |`
`=>\ text(C)`
It is given that `a`, `b` are real and `p`, `q` are purely imaginary.
Which pair of inequalities must always be true?
Which diagram best represents the solutions to the equation `text(arg)(z) = text(arg)(z + 1 - i)`?
| A. |  |
B. |  |
| C. |  |
D. |  |
Which complex number is a 6th root of `i`?
`text(Consider option A:)`
`|−1/sqrt2 + 1/sqrt2i|= sqrt((−1/sqrt2)^2 + (1/sqrt2)^2) = 1`
| `text(arg)(z)` | `= (3pi)/4` |
| `z` | `= 1(cos\ (3pi)/4 + i sin\ (3pi)/4)` |
| `z^6` | `= cos\ (18pi)/4 + i sin\ (18pi)/4\ \ \ text{(De Moivre)}` |
| `= i` |
`=> A`
Kelly takes off in a helicopter and heads for Cray Island.
In which direction is Kelly flying?
| north-east | north-west | south-east | south-west |
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The table shows the time (in seconds) some children can run 100 metres.
Which column graph correctly shows the data in the table?
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`text(Considering the vertical scale of each option:)`
In a tennis competition a player won 7 games and lost the other games.
Altogether she played 15 games.
Finish the subtraction sentence below to show the number of games she lost.
| `\ - 7 =` |
Cassie records the number of items she recycles in one week.
The tally table shows Cassie's results.
How many items does Cassie recycle in total?
The shaded region is enclosed by the curve `y = x^3 - 7x` and the line `y = 2x`, as shown in the diagram. The line `y = 2x` meets the curve `y = x^3 - 7x` at `O(0, 0)` and `A(3, 6)`. Do NOT prove this.
| (i) | `text(Area)` | `= int_0^3 2x – (x^3 – 7x)\ dx` |
| `= int_0^3 9x – x^3\ dx` | ||
| `= [9/2 x^2 – 1/4 x^4]_0^3` | ||
| `= [(9/2 xx 3^2 – 1/4 xx 3^4) – 0]` | ||
| `= 81/2 – 81/4` | ||
| `= 81/4\ text(units²)` |
(ii) `text(Using Simpson’s rule)`
♦ Mean mark 45%.
`y = 9x – x^3`
| `text(A)` | `~~ h/3 (y_0 + 4y_1 + y_2)` |
| `= 3/6 (0 + 4 xx 81/8 + 0)` | |
| `= 1/2 (81/2)` | |
| `= 81/4\ text( … as required)` |
(iii) `y = x^3 – 7x`
`(dy)/(dx) = 3x^2 – 7`
`text(Find)\ \ x\ \ text(such that)\ \ (dy)/(dx) = 2`
| `3x^2 – 7` | `= 2` |
| `3x^2` | `= 9` |
| `x^2` | `= 3` |
| `x` | `= sqrt 3 qquad (x > 0)` |
| `y` | `= (sqrt 3)^3 – 7 sqrt 3` |
| `= 3 sqrt 3 – 7 sqrt 3` | |
| `= -4 sqrt 3` |
`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`
| (iv) |  |
| `text(dist)\ OA` | `= sqrt((3 – 0)^2 + (6 – 0)^2)` |
| `= sqrt 45` | |
| `= 3 sqrt 5` |
`text(Find)\ _|_\ text(distance of)\ \ P\ \ text(from)\ \ OA`♦♦ Mean mark 27%.
`P(sqrt 3, -4 sqrt 3),\ \ 2x – y=0`
| `_|_\ text(dist)` | `= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|` |
| `= (6 sqrt 3)/sqrt 5` | |
| `:.\ text(Area)` | `= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5` |
| `= 9 sqrt 3\ text(units²)` |
A sector with radius 10 cm and angle `theta` is used to form the curved surface of a cone with base radius `x` cm, as shown in the diagram.
The volume of a cone of radius `r` and height `h` is given by `V = 1/3 pi r^2 h`.
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The diagram shows the region bounded by the curve `y = 1/(x + 3)` and the lines `x = 0`, `x = 45` and `y = 0`. The region is divided into two parts of equal area by the line `x = k`, where `k` is a positive integer.
What is the value of the integer `k`, given that the two parts have equal areas? (3 marks)
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The length of daylight, `L(t)`, is defined as the number of hours from sunrise to sunset, and can be modelled by the equation
`L(t) = 12 + 2 cos ((2 pi t)/366)`,
where `t` is the number of days after 21 December 2015, for `0 ≤ t ≤ 366`.
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Two machines, `A` and `B`, produce pens. It is known that 10% of the pens produced by machine `A` are faulty and that 5% of the pens produced by machine `B` are faulty.
What is the probability that at least one of the pens is faulty? (1 mark)
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What is the probability that neither pen is faulty? (2 marks)
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The shaded region shown in the diagram is bounded by the curve `y = x^4 + 1`, the `y`-axis and the line `y = 10`.
Find the volume of the solid of revolution formed when the shaded region is rotated about the `y`-axis. (3 marks)
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In `Delta KLM, KL` has length 3, `LM` has length 6 and `/_KLM` is 60°. The point `N` is chosen on side `KM` so that `LN` bisects `/_KLM`. The length `LN` is `x`.
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In `Delta ABC`, sides `AB` and `AC` have length 3, and `BC` has length 2. The point `D` is chosen on `AB` so that `DC` has length 2.
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i. `text(Prove)\ \ Delta ABC\ text(|||)\ Delta CBD`
`Delta ABC\ text{is isosceles:}`
`/_ ABC = /_ ACB qquad text{(angles opposite equal sides)}`
`Delta CBD\ text{is isosceles:}`
`/_ CBD = /_ CDB qquad text{(angles opposite equal sides)}`
`text{Since}\ \ /_ ABC = /_ CBD`
`:. Delta ABC\ text(|||)\ Delta CDB qquad text{(equiangular)}`
ii. `text(Using ratios of similar triangles)`
| `(DB)/(CB)` | `= (BC)/(AC)` |
| `{(3-AD)}/2` | `= 2/3` |
| `3-AD` | `= 4/3` |
| `:. AD` | `= 5/3` |
Consider the curve `y = 6x^2 - x^3`.
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i. `y = 6x^2 – x^3`
`(dy)/(dx) = 12x – 3x^2`
`(d^2y)/(dx^2) = 12 – 6x`
`text(S.P.s occur when)\ \ (dy)/(dx) = 0`
`12x – 3x^2 = 0`
`3x(4 – x) = 0`
`x = 0 or 4`
`text(When)\ \ x = 0,\ (d^2y)/(dx^2) > 0`
`:.\ text(MIN at)\ (0, 0)`
`text(When)\ \ x = 4,\ (d^2y)/(dx^2) < 0`
`:.\ text(MAX at)\ (4, 32)`
ii. `text(P.I. occur when)\ \ (d^2y)/(dx^2) = 0,`
| `12 – 6x` | `= 0` |
| `x` | `= 2` |
`text(When)\ \ x = 2,\ y = 16`
`text(S)text(ince the concavity changes)`
`=>\ text(P.I. occurs at)\ \ (2, 16)`
| iii. |  |
The diagram shows the square `ABCD`. The point `E` is chosen on `BC` and the point `F` is chosen on `CD` so that `EC = FC`.
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Find the equation of the tangent to the curve `y = cos 2x` at `x = pi/6`. (3 marks)
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A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 tokens labelled 10 cents and 3 tokens labelled $2. The player wins the total value of the two tokens drawn.
| i. | |
ii. `text(Expectated Gain/Loss)`♦♦ Mean mark part (ii) 27%.
COMMENT: Financial expectation included here (one example only in database) as it is possible to argue that it is an application of expectation.
`= [17/20 xx 16/19 xx 0.20 + 17/20 xx 3/19 xx 2.1`
`+ 3/20 + 17/19 xx 2.1 + 3/20 xx 2/19 xx 4] – 1`
`= 0.77 – 1`
`= −$0.23\ \ (text(or $0.23 loss))`
Last year, Luke’s taxable income was `$87\ 000` and the tax payable on this income was `$19\ 822`. This year, Luke’s taxable income has increased by `$16\ 800`.
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Simplify `(8x^3 - 27y^3)/(2x - 3y)`. (2 marks)
Andrew borrowed $20 000 to be repaid in equal monthly repayments of $243 over 10 years. Having made this monthly repayment for 4 years, he increased his monthly repayment to $281. As a result, Andrew paid off the loan one year earlier.
How much less did he repay altogether by making this change? (2 marks)
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The time in Brisbane is 4.5 hours ahead of the time in New Delhi. John flew from New Delhi to Brisbane via Singapore. His plane left New Delhi at 11.30 am (New Delhi time), stopped for 3 hours in Singapore, and arrived in Brisbane at 9.00 am the following day (Brisbane time).
What was the plane’s total flying time? (3 marks)
Every day, a 1200-watt microwave oven is used for 45 minutes at 40% power. Electricity is charged at $0.25 per kWh.
What is the cost of running this microwave oven for 180 days? (3 marks)
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| `y` | `= x + 5` |
| `y + 2x` | `= 2` |
Draw these two linear graphs on the number plane below and determine their intersection. (3 marks)
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`text(Table of coordinates:)\ \ y = x + 5`
`text(Table of coordinates:)\ \ y + 2x = 2 \ => \ y = −2x + 2`
`text{From graph (and table), intersection occurs}`
`text(at)\ \ (−1,4).`
The line `3x-4y + 3 = 0` is a tangent to a circle with centre (3, – 2).
What is the equation of the circle?
Joanna sits a Physics test and a Biology test.
Calculate the `z`-score for Joanna’s mark in this test. (1 mark)
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Joanna’s `z`-score is the same in both the Physics test and the Biology test.
What is her mark in the Biology test? (2 marks)
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