The graph shown could have equation
- `y = 2cos(x + pi/6) + 1`
- `y = 2cos4(x - pi/6) + 1`
- `y = 4sin2(x - pi/12) - 1`
- `y = 3cos(2x + pi/6) - 1`
Trigonometry, 2ADV T3 SM-Bank 6 MC
The function with equation `f(x) = 4 tan (x/3)` has period
- `(2pi)/3`
- `6 pi`
- `3`
- `3 pi`
Trigonometry, 2ADV T3 SM-Bank 5 MC
The UV index, `y`, for a summer day in Newcastle East is illustrated in the graph below, where `t` is the number of hours after 6 am.
The graph is most likely to be the graph of
- `y = 5 + 5 cos ((pi t)/7)`
- `y = 5 - 5 cos ((pi t)/7)`
- `y = 5 + 5 cos ((pi t)/14)`
- `y = 5 - 5 cos ((pi t)/14)`
Trigonometry, 2ADV T3 SM-Bank 4 MC
A section of the graph of `f(x)` is shown below.
The equation of `f(x)` could be
- `f (x) = tan (x)`
- `f (x) = tan (x - pi/4)`
- `f (x) = tan(2 (x - pi/4))`
- `f (x) = tan(2 (x - pi/2))`
Measurement, STD2 M7 SM-Bank 12
Francis is buying a fridge. The energy usage of his two choices are below:
Fridge 1 (5 star rating): 227 kWh per year
Fridge 2 (2 star rating): 492 kWh per year
Each fridge has an expected life of 8 years.
If electricity costs 32.4 cents per kWh, how much will Francis save in electricity over the expected life of the more energy efficient fridge, to the nearest dollar? (2 marks)
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Measurement, STD2 M7 SM-Bank 10
A potting mix is made up of two ingredients, potting soil and manure, in the ratio 5:2.
21 kilograms of potting mix are used to fertilise the rectangular vegetable garden pictured below.
- What rate must the whole 21 kilograms of potting mix be applied to be spread evenly over the garden. Give your answer in grams per square metre. (2 marks)
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- How much manure is needed to make the required potting mix? (1 mark)
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Networks, STD2 N3 SM-Bank 42 MC
The network diagram below flows from the source (S) to sink (T).
Which of the edges is not at maximum capacity?
- `AC`
- `BC`
- `CT`
- `DT`
Show Worked Solution
`text(Consider the minimum cut line: all edges are saturated.)`
`:.\ text(Edge)\ CT\ text(is not at maximum capacity.)`
`=>\ C`
Networks, STD2 N3 SM-Bank 46
A network diagram is drawn below.
- Calculate the maximum flow through this network. (2 marks)
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- Copy the network above and illustrate the maximum flow capacity. (2 marks)
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Show Answers Only
- `28`
-
Show Worked Solution
| i. |  |
| `text(Maximum flow)` | `=\ text(minimum cut)` |
| `= 8 + 8 + 7 + 5` | |
| `= 28` |
| ii. |
|
Networks, STD2 N3 SM-Bank 39
Bianca is designing a project for producing an advertising brochure. It involves activities A-M.
The network below shows these activities and their completion time in hours.
- What is the earliest starting time of activity J? (1 mark)
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- What is the latest starting time of activity H? (1 mark)
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- What is the float time of activity I? (1 mark)
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- What is the minimum time required to produce the brochure? (2 marks)
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Show Worked Solution
i. `text(Scanning forwards and backwards)`
`text{EST (activity}\ J) = 11\ text(hours)`
| ii. | `text{LST (activity}\ H)` | `=\ text{LST}\ (H)-text(weight)\ (H)` |
| `= 20-2` | ||
| `=18\ text(hours)` |
| iii. | `text{Float time}\ (I)` | `=\ text(LST)\ (I)-text(EST)\ (I)` |
| `=12-10` | ||
| `= 2\ text(hours)` |
iv. `text(Critical Path is)\ CGJLM`
| `:.\ text(Minimum time)` | `= 2 + 9 + 6 + 7 + 10` |
| `= 34\ text(hours)` |
Algebra, STD2 A4 SM-Bank 1 MC
The graph below represents `y = a^x`.
What is the approximate value of `2(a^5)`?
- 4
- 8
- 32
- 64
Calculus, SPEC2 2012 VCAA 5
At her favourite fun park, Katherine’s first activity is to slide down a 10 m long straight slide. She starts from rest at the top and accelerates at a constant rate, until she reaches the end of the slide with a velocity of `6\ text(ms)^(-1)`.
- How long, in seconds, does it take Katherine to travel down the slide? (1 mark)
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When at the top of the slide, which is 6 m above the ground, Katherine throws a chocolate vertically upwards. The chocolate travels up and then descends past the top of the slide to land on the ground below. Assume that the chocolate is subject only to gravitational acceleration and that air resistance is negligible.
- If the initial speed of the chocolate is 10 m/s, how long, correct to the nearest tenth of a second, does it take the chocolate to reach the ground? (2 marks)
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- Assume that it takes Katherine four seconds to run from the end of the slide to where the chocolate lands.
At what velocity would the chocolate need to be propelled upwards, if Katherine were to immediately slide down the slide and run to reach the chocolate just as it hits the ground?
Give your answer in `text(ms)^(-1)`, correct to one decimal place. (2 marks)
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Katherine’s next activity is to ride a mini speedboat. To stop at the correct boat dock, she needs to stop the engine and allow the boat to be slowed by air and water resistance.
At time `t` seconds after the engine has been stopped, the acceleration of the boat, `a\ text(ms)^(-2)`, is related to its velocity, `v\ text(ms)^(-1)`, by
`a = -1/10 sqrt(196-v^2)`.
Katherine stops the engine when the speedboat is travelling at `7\ text(m/s)`.
- i. Find an expression for `v` in terms of `t`. (3 marks)
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- ii. Find the time it takes the speedboat to come to rest.Give your answer in seconds in terms of `pi`. (2 marks)
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- iii. Find the distance it takes the speedboat to come to rest, from when the engine is stopped.Give your answer in metres, correct to one decimal place. (3 marks)
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Vectors, SPEC2 2012 VCAA 4
The position vector of the International Space Station `text{(S)}`, when visible above the horizon from a radar tracking location `text{(O)}` on the surface of Earth, is modelled by
`underset ~r(t) = 6800 sin(pi(1.3t-0.1))underset~i + (6800 cos(pi(1.3t-0.1))-6400)underset ~j`,
for `t in [0, 0.154]`,
where `underset ~i` is a unit vector relative to `text(O)` as shown and `underset ~j` is a unit vector vertically up from point `text(O)`. Time `t` is measured in hours and displacement components are measured in kilometres.
- Find the height, `h` km, of the space station above the surface of Earth when it is at point `text(P)`, directly above point `text(O)`.
- Give your answer correct to the nearest km. (1 mark)
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- Find the acceleration of the space station, and show that its acceleration is perpendicular to its velocity. (3 marks)
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- Find the speed of the space station in km/h.
- Give your answer correct to the nearest integer. (2 marks)
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- Find the equation of the path followed by the space station in cartesian form. (2 marks)
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- Find the times when the space station is at a distance of 1000 km from the radar tracking location `text(O)`.
- Give your answers in hours, correct to two decimal places. (3 marks)
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Calculus, SPEC2 2012 VCAA 3
A car accelerates from rest. Its speed after `T` seconds is `V\ text(ms)^(−1)`, where
`V = 17 tan^(−1)((pi T)/6), T >= 0`
- Write down the limiting speed of the car as `T -> oo`. (1 mark)
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- Calculate, correct to the nearest `0.1\ text(ms)^(−2)`, the acceleration of the car when `T = 10`. (1 mark)
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- Calculate, correct to the nearest second, the time it takes for the car to accelerate from rest to `25\ text(ms)^(−1)`. (2 marks)
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After accelerating to `25\text(ms)^(−1)`, the car stays at this speed for 120 seconds and then begins to decelerate while braking. The speed of the car `t` seconds after the brakes are first applied is `v\ text(ms)^(−1)` where
`(dv)/(dt) =-1/100 (145-2t),`
until the car comes to rest.
- i. Find `v` in terms of `t`. (2 marks)
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- ii. Find the time, in seconds, taken for the car to come to rest while braking. (2 marks)
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- i. Write down the expressions for the distance travelled by the car during each of the three stages of its motion. (2 marks)
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- ii. Find the total distance travelled from when the car starts to accelerate to when it comes to rest.
Give your answer in metres correct to the nearest metre. (1 mark)
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Complex Numbers, SPEC2 2012 VCAA 2
- Given that `cos(pi/12) = (sqrt (sqrt 3 + 2))/2`, show that `sin(pi/12) = (sqrt (2-sqrt 3))/2`. (2 marks)
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- Express `z_1 = (sqrt(sqrt3 + 2))/2 + i(sqrt(2-sqrt3))/2` in polar form. (1 mark)
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- i. Write down `z_1^4` in polar form. (1 mark)
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- ii. On the Argand diagram below, shade the region defined by
`{z: text(Arg)(z_1) <= text(Arg)(z) <= text(Arg)(z_1^4)} ∩ {z: 1 <= |\ z\ | <= 2}, z ∈ C`. (2 marks)
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- Find the area of the shaded region in part c. (2 marks)
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- i. Find the value(s) of `n` such that `text(Re)(z_1^n) = 0`, where `z_1 = (sqrt(sqrt3 + 2))/2 + i(sqrt(2-sqrt3))/2`. (3 marks)
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- ii. Find `z_1^n` for the value(s) of `n` found in part i. (1 mark)
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Show Worked Solution
| a. | `cos^2(pi/12) + sin^2(pi/12)` | `= 1` |
| `sin^2(pi/12)` | `= 1-(sqrt 3 + 2)/4` | |
| `= (2-sqrt 3)/4` |
`:. sin (pi/12) = sqrt(2-sqrt 3)/2,\ \ \ (sin (pi/12) > 0)`
| b.i. | `z_1` | `= cos(pi/12) + i sin (pi/12)` |
| `= text(cis)(pi/12)` |
| b.ii. | `z_1^4` | `= 1^4 text(cis) ((4 pi)/12)` |
| `= text(cis)(pi/3)` |
| c. |  |
d. `text(Area of large sector)`Mean mark 51%.
`=theta/(2pi) xx pi r^2`
`=(pi/3-pi/12)/(2pi) xx pi xx 2^2`
`=pi/2`
`text(Area of small sector)`
`=1/2 xx pi/4 xx 1`
`=pi/8`
`:.\ text(Area shaded)`
`= pi/2-pi/8`
`= (3 pi)/8`
♦ Mean mark (e)(i) 34%.
| e.i. | `z_1^n` | `= 1^n text(cis) ((n pi)/12)` |
| `text(Re)(z_1^n)` | `= cos((n pi)/12) = 0` | |
| `(n pi)/12` | `=cos^(-1)0 +2pik,\ \ \ k in ZZ` | |
| `(n pi)/12` | `= pi/2 + k pi` | |
| `n` | `= 6 + 12k,\ \ \ k in ZZ` |
e.ii. `text(When)\ \ n=(6 + 12k):`♦ Mean mark (e)(ii) 36%.
| `z_1^n` | `= 1^(6 + 12k) text(cis) (((6 + 12k)pi)/12), quad k in ZZ` | |
| `= i xx sin (((6 + 12k)pi)/12)` | ||
| `= i xx sin (pi/2 + k pi)` |
`text(If)\ \ k=0 or text(even,)\ \ z_1^n=i`
`text(If)\ \ k\ text(is odd,)\ \ z_1^n=-i`
`:. z_1^n = +-i,\ \ \ k in ZZ`
Mechanics, SPEC2 2012 VCAA 22 MC
A 12 kg mass is suspended in equilibrium from a horizontal ceiling by two identical light strings. Each string makes an angle of 60° with the ceiling, as shown.
The magnitude, in newtons, of the tension in each string is equal to
A. `6 g`
B. `12 g`
C. `24 g`
D. `4 sqrt 3 g`
E. `8 sqrt 3 g`
Show Worked Solution
| `T/(sin 30^@)` | `= (12 g)/(sin 120^@)` |
| `2T` | `= (24 g)/sqrt3` |
| `T` | `= (12 g)/sqrt3` |
| `= 4g sqrt 3` |
`=> D`
Vectors, SPEC2 2012 VCAA 17 MC
If `underset ~u = 2 underset ~i - 2 underset ~j + underset ~k` and `underset ~v = 3 underset ~i - 6 underset ~j + 2 underset ~k`, the vector resolute of `underset ~v` in the direction of `underset ~u` is
A. `20/49(3 underset ~i - 6 underset ~j + 2 underset ~k)`
B. `20/3(2 underset ~i - 2underset ~j + underset ~k)`
C. `20/7(3 underset ~i - 6 underset ~j + 2 underset ~k)`
D. `20/9(2 underset ~i - 2 underset ~j + underset ~k)`
E. `1/9(−2 underset ~i + 2 underset ~j - underset ~k)`
Mechanics, SPEC2 2012 VCAA 14 MC
A particle is acted on by two forces, one of 6 newtons acting due south, the other of 4 newtons acting in the direction N60° W.
The magnitude of the resultant force, in newtons, acting on the particle is
- `10`
- `2sqrt7`
- `2sqrt19`
- `sqrt(52 - 24sqrt3)`
- `sqrt(52 + 24sqrt3)`
Show Worked Solution
Mean mark 51%.
| `x^2` | `= 6^2 + 4^2 – 2(6)(4) cos 60^@` |
| `= 28` | |
| `:. x` | `= sqrt 28` |
| `= 2 sqrt 7` |
`=> B`
Calculus, SPEC2 2012 VCAA 13 MC
Using a suitable substitution, `int_0^(pi/3) sin^3 (x) cos^4 (x)\ dx` can be expressed in terms of `u` as
A. `int_0^(pi/3) (u^6 - u^4)\ du`
B. `int_1^(1/2) (u^6 - u^4)\ du`
C. `int_(1/2)^1 (u^6 - u^4)\ du`
D. `int_0^(sqrt3/2) (u^6 - u^4)\ du`
E. `int_0^(sqrt3/2) (u^4 - u^6)\ du`
Calculus, SPEC2 2012 VCAA 11 MC
If `(d^2y)/(dx^2) = x^2 - x` and `(dy)/(dx) = 0` at `x = 0`, then the graph of `y` will have
- a local minimum at `x = 1/2`
- a local maximum at `x = 0` and a local minimum at `x = 1`
- stationary points of inflection at `x = 0` and `x = 1`, and a local minimum at `x = 3/2`
- a stationary points of inflection at `x = 0`, no other points of inflection and a local minimum at `x = 3/2`
- a stationary point of inflection at `x = 0`, a non-stationary point of inflection at `x = 1` and a local minimum at `x = 3/2`
Calculus, SPEC2 2012 VCAA 10 MC
The diagram that best represents the direction field of the differential equation `(dy)/(dx) = xy` is
| A. |  |
B. |  |
| C. |  |
D. |  |
Calculus, SPEC2 2012 VCAA 9 MC
Euler's formula is used to find `y_2`, where `(dy)/(dx) = cos(x),\ x_0 = 0, \ y_0 = 1` and `h = 0.1`.
The value of `y_2` correct to four decimal places is
A. 1.1000 and this is an underestimate of `y(0.2)`
B. 1.1995 and this is an overestimate of `y(0.2)`
C. 1.1995 and this is an underestimate of `y(0.2)`
D. 1.2975 and this is an underestimate of `y(0.2)`
E. 1.2975 and this is an overestimate of `y(0.2)`
Calculus, SPEC2 2011 VCAA 22 MC
A particle moves in a straight line. At time `t` seconds, where `t >= 0`, its displacement `x` metres from the origin and its velocity `v` metres per second are such that `v = 25 + x^2`.
If `x = 5` initially, then `t` is equal to
- `25x + (x^3)/3`
- `25x + (x^3)/3 - 500/3`
- `1/5tan^(−1)(x/5) + 5`
- `tan^(−1)(x/5) - pi/4`
- `1/5tan^(−1)(x/5) - pi/20`
Calculus, SPEC2 2011 VCAA 20 MC
A body moves in a straight line such that its velocity `v\ text(ms)^(-1)` is given by `v = 2sqrt(1 - x^2)`, where `x` metres is its displacement from the origin.
The acceleration of the body in `text(ms)^(-2)` is given by
- `(−2x)/(sqrt(1 - x^2))`
- `−2x`
- `2/(sqrt(1 - x^2))`
- `2(1 - 2x)`
- `−4x`
Calculus, SPEC2 2011 VCAA 19 MC
The motion of a lift (elevator) in a shopping centre is given by the velocity-time graph below, where time `t` is in seconds, and the velocity of the lift is `v` metres per second. For `v > 0` the lift is moving upwards.
The graph shows that at the end of 30 seconds, the position of the lift is
A. 17.5 metres above its starting level.
B. 5 metres above its starting level.
C. at the same position as its starting level.
D. 5 metres below its starting level.
E. 17.5 metres below its starting level.
Calculus, SPEC2 2011 VCAA 18 MC
The amount of chemical `x` in a tank at time `t` is given by the differential equation `dx/dt = -10/(10 - t)` and when `t = 0`, `\ x_0 = 5`. Euler's method is used with a step size of 0.5 in the values of `t`.
The value of `x` correct to two decimal places when `t = 1` is found to be
A. 3.95
B. 3.97
C. 4.50
D. 5.50
E. 6.03
Calculus, SPEC2 2011 VCAA 17 MC
The differential equation which best represents the above direction field is
A. `(dy)/(dx) = (y - 2x)/(2y + x)`
B. `(dy)/(dx) = (2x - y)/(y - 2x)`
C. `(dy)/(dx) = (2y - x)/(y + 2x)`
D. `(dy)/(dx) = (y - 2x)/(2y - x)`
E. `(dy)/(dx) = (x - 2y)/(2y + x)`
Calculus, SPEC2 2011 VCAA 15 MC
Using a suitable substitution, the definite integral `int_0^(pi/24)tan(2x)sec^2(2x)\ dx` is equivalent to
A. `1/2int_0^(pi/24)(u)\ du`
B. `2int_0^(pi/24)(u)\ du`
C. `int_0^(2 - sqrt3)(u)\ du`
D. `1/2int_0^(2 - sqrt3)(u)\ du`
E. `2int_0^(2 - sqrt3)(u)\ du`
Calculus, SPEC2 2011 VCAA 14 MC
If `f″(x) = 2e^xsin(x)`, `f′(0) = 0` and `f(0) = 0`, the `f(x)` equals
A. `−e^x(cos(x) + sin(x))`
B. `−e^x(cos(x) - sin(x)) + 1`
C. `−e^xcos(x)`
D. `x - e^xcos(x) + 1`
E. `x - e^xcos(x)`
Vectors, SPEC2 2011 VCAA 13 MC
The position of a particle at time `t` is given by `underset~r(t) = (sqrt(t - 2))underset~i + (2t)underset~j` for `t >= 2`.
The cartesian equation of the path of the particle is
| A. `y = 2x^2 + 4`, | `x >= 2` |
| B. `y = 2x^2 + 2`, | `x >= 2` |
| C. `y = 2x^2 + 4`, | `x >= 0` |
| D. `y = sqrt((x - 4)/2)`, | `x >= 4` |
| E. `y = 2x^2 + 2`, | `x >= 0` |
Vectors, SPEC2 2011 VCAA 11 MC
Consider the three forces
`underset~F_1 = −sqrt3underset~j`, `underset~F_2 = underset~i + sqrt3underset~j` and `underset~F_3 = −1/2underset~i + sqrt3/2underset~j`.
The magnitude of the sum of these three forces is equal to
- the magnitude of `(underset~F_3 - underset~F_1)`
- the magnitude of `(underset~F_2 - underset~F_1)`
- the magnitude of `underset~F_1`
- the magnitude of `underset~F_2`
- the magnitude of `underset~F_3`
Vectors, SPEC2 2011 VCAA 10 MC
The diagram below shows a rhombus, spanned by the two vectors `underset~a` and `underset~b`.
It follows that
A. `underset~a.underset~b = 0`
B. `underset~a = underset~b`
C. `(underset~a + underset~b).(underset~a - underset~b) = 0`
D. `|\ underset~a + underset~b\ | = |\ underset~a - underset~b|`
E. `2underset~a + 2underset~b = underset~0`
Show Worked Solution
`text(Consider A:)`
`text(If)\ \ underset~a · underset~b = 0\ \ =>\ \ underset~a ⊥ underset~b\ \ text{(incorrect)}`
`text(Consider B:)`
`underset~a != underset~b\ \ text{(incorrect)}`
`text(Consider C:)`
| `underset~a + underset~b` | `= overset(->)(OC)` |
| `underset~a – underset~b` | `= overset(->)(BA)` |
`overset(->)(OC) · overset(->)(BA)=0`
`text{The diagonals of a rhombus are perpendicular (correct)}`
`=> C`
Complex Numbers, SPEC2 2011 VCAA 7 MC
In the complex plane, the circle with equation `|\ z - (2 + 3i)\ | = 1` is intersected exactly twice by the curve with equation
A. `|\ z - 3i\ | = 1`
B. `|\ z + 3\ | = |\ z - 3i\ |`
C. `|\ z - 3\ | = |\ z - 3i\ |`
D. `text(Im)(z) = 4`
E. `text(Re)(z) = 3`
Show Worked Solution
`text(Consider option C:)`
`|z – 3| = |z – 3i|`
`=> y = x\ text{(perpendicular bisector}`
`text{between (3, 0) and (0, 3)}`
`=> C`
Complex Numbers, SPEC2 2011 VCAA 6 MC
The polynomial `P(z)` has real coefficients. Four of the roots of the equation `P(z) = 0` are `z = 0`, `z = 1 - 2i`, `z = 1 + 2i` and `z = 3i`.
The minimum number of roots that the equation `P(z) = 0` could have is
A. 4
B. 5
C. 6
D. 7
E. 8
Complex Numbers, SPEC2 2011 VCAA 5 MC
A certain complex number `z`, where `|\ z\ | > 1`, is represented by the point on the following argand diagram.
All axes below have the same scale as those in the diagram above.
The complex number `1/barz` is best represented by
Calculus, SPEC1 2011 VCAA 11
The region in the first quadrant enclosed by the curve `y = sin(x)`, the line `y = 0` and the line `x = pi/6` is rotated about the `x`-axis. Find the volume of the resulting solid of revolution. (3 marks) --- 8 WORK AREA LINES (style=lined) ---
Show Worked Solution
`V`
`= pi int_(0)^(pi/6) y^2\ dx`
`= pi int_0^(pi/6) sin^2(x)\ dx`
`= pi/2 int_0^(pi/6) (1-cos 2x)\ dx`
`= pi/2 [x-1/2 sin (2x)]_0^(pi/6)`
`=pi/2 [(pi/6-1/2 sin (pi/3))-0]`
`=pi/2 (pi/6-(sqrt 3)/4)`
`=pi^2/12-(sqrt3 pi)/8\ \ text(u³)`
Calculus, SPEC2 2012 VCAA 18 MC
The velocity–time graph for the first 2 seconds of the motion of a particle that is moving in a straight line with respect to a fixed point is shown below.
The particle’s velocity `v` is measured in cm/s. Initially the particle is `x_0` cm from the fixed point.
The distance travelled by the particle in the first 2 seconds of its motion is given by
A. `int_0^2 v\ dt`
B. `int_0^2 v\ dt + x_0`
C. `int_1^2 v\ dt - int_0^1 v\ dt`
D. `|\ int_0^2 v\ dt\ |`
E. `int_1^2 v\ dt - int_0^1 v\ dt + x_0`
Vectors, SPEC1 2011 VCAA 9
Consider the three vectors `underset ~a = underset ~i - underset ~j + 2underset ~k,\ underset ~b = underset ~i + 2 underset ~j + m underset ~k` and `underset ~c = underset ~i + underset ~j - underset ~k`, where `m in R.` --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
Mechanics, SPEC1 2011 VCAA 7
A flowerpot of mass `m` kg is held in equilibrium by two light ropes, both of which are connected to a ceiling. The first rope makes an angle of 30° to the vertical and has tension `T_1` newtons. The second makes an angle of 60° to the vertical and has tension `T_2` newtons.
- Show that `T_2 = T_1/sqrt 3.` (1 mark)
- The first rope is strong, but the second rope will break if the tension in it exceeds 98 newtons.
Find the maximum value of `m` for which the flowerpot will remain in equilibrium. (3 marks)
Show Worked Solution
a. `text(Forces in equilibrium:)`
| `tan 30^@` | `=T_2/T_1` |
| `T_2` | `= T_1 xx tan30^@` |
| `T_2` | `= T_1/sqrt3\ \ text(.. as required)` |
| b. | `sin30^@` | `= (T_2)/(mg)` |
| `T_2` | `= (mg)/2` | |
| `98` | `=(m_text(max) xx 9.8)/2` | |
| `m_text(max)` | `=(2 xx 98)/9.8` | |
| `=20\ text(kg)` |
Calculus, SPEC1 2011 VCAA 6
Evaluate `int_0^1 e^x cos (e^x)\ dx.` (2 marks)
Calculus, SPEC1 2011 VCAA 5
For the curve with parametric equations
`x = 4 sin (t) - 1`
`y = 2 cos (t) + 3`
Find `(dy)/(dx)` at the point `(1, sqrt 3 + 3).` (3 marks)
Complex Numbers, SPEC2 2012 VCAA 8 MC
If `z = a + bi`, where both `a` and `b` are non-zero real numbers and `z in C`, which of the following does not represent a real number?
- `z + barz`
- `|\ z\ |`
- `zbarz`
- `z^2-2abi`
- `(z-bar z) (z + bar z)`
Complex Numbers, SPEC2 2012 VCAA 7 MC
The set of points in the complex plane defined by `|\ z + 2i\ | = |\ z\ |` corresponds to
A. the point given by `z = −i`
B. the line `text(Im)(z) = −1`
C. the line `text(Im)(z) = −i`
D. the line `text(Re)(z) = −1`
E. the circle with centre `−2i` and radius `1`
Complex Numbers, SPEC2 2012 VCAA 5 MC
If `z = sqrt 2 text(cis)(-(4 pi)/5)` and `w = z^9`, then
A. `w = 16 sqrt 2 text(cis)((36 pi)/5)`
B. `w = 16 sqrt 2 text(cis)(−pi/5)`
C. `w = 16 sqrt 2 text(cis)((4pi)/5)`
D. `w = 9 sqrt 2 text(cis)(-pi/5)`
E. `w = 9 sqrt 2 text(cis)((4pi)/5)`
Calculus, SPEC1 2011 VCAA 3
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Show Answers Only
b. `underset (x→oo) (lim y) = 2` `text(S)text(ince)\ \ x^2>0\ \ text(for all)\ x,` `=> f(x)_text(max)\ \ text(occurs when)\ \ x=0\ \ text(at)\ \ (0,3)` c. `f(x)\ \ text(is an even function.)`Show Worked Solution
a.
`f(x)`
`= (2x^2 + 3)/(x^2 + 1)`
`= (2(x^2 + 1) + 1)/(x^2 + 1)`
`= 2 + 1/(x^2 + 1)`
`:.\ text(Area)`
`= 2 int_0^1 2 + 1/(x^2 + 1)\ dx`
`= 2[2x + tan^(−1)(x)]_0^1`
`=2[(2+pi/4)-0]`
`= 4 + pi/2`
Calculus, SPEC1 2011 VCAA 2
Find the value of the real constant `k` given that `kxe^(2x)` is a solution of the differential equation
`(d^2y)/(dx^2)-2 (dy)/(dx) + 5y = e^(2x) (15x + 6).` (3 marks)
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Calculus, SPEC1 2011 VCAA 1
Find an antiderivative of `(1 + x)/(9-x^2),\ \ x in R \ text(\{– 3, 3}).` (3 marks)
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Graphs, SPEC2 2012 VCAA 3 MC
The graph of `y = f(x)` is shown below.
All the axes below have the same scale as the axes in the diagram above.
The graph of `y = 1/(f(x))` is best represented by
| A. |  |
B. |  |
| C. |  |
D. |  |
| E. |  |
Graphs, SPEC1 2012 VCAA 10
Consider the functions with rules `f(x) = arcsin (x/2) + 3/sqrt (25 x^2-1)` and `g(x) = arcsin (3x)-3/sqrt (25x^2-1).`
-
- Find the maximal domain of `f_1(x) = arcsin (x/2).` (1 mark)
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- Find the maximal domain of `f_2(x) = 3/sqrt (25x^2-1).` (1 mark)
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- Find the largest set of values of `x in R` for which `f(x)` is defined. (1 mark)
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- Find the maximal domain of `f_1(x) = arcsin (x/2).` (1 mark)
- Given that `h(x) = f(x) + g(x)` and that `theta = h(1/4)`, evaluate `sin (theta).`
Give your answer in the form `(a sqrt b)/c, \ a, b, c in Z.` (3 marks)
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Show Worked Solution
a.i. `text(Maximal Domain occurs when:)`
| `-1 <= x/2 <= 1` | |
| `{x: -2 <= x <= 2}` |
a.ii. `text(Maximal Domain occurs when:)`♦♦ Mean mark part (a)(ii) 33%.
| `25x^2-1 > 0` | |
| `x^2 > 1/25` | |
| `{x: x < -1/5 uu x > 1/5}` |
a.iii. `text(Max domain for which)\ \ f(x)\ \ text(is defined:)`♦♦ Mean mark part (a)(iii) 26%.
| `(-2 <= x <= 2) nn (x < -1/5 uu x > 1/5)` | |
| `{x: -2 <= x < -1/5 \ uu \ 1/5 < x <= 2}` |
| b. | `h(x)` | `= sin^(-1) (x/2) + 3/sqrt(25x^2-1) + sin^(-1)(3x)-3/sqrt(25x^2-1)` |
| `= sin^(-1)(x/2) + sin^(-1)(3x)` |
`text(When)\ \ x=1/4,\ \ h(x)=theta`♦♦♦ Mean mark part (b) 20%.
`theta=sin^(-1) (1/8) + sin^(-1) (3/4)`
`text(Let)\ \ theta_1 = sin^(-1) (1/8),\ \ theta_2 = sin^(-1) (3/4)`
`text(Using)\ \ sin(theta_1 + theta_2)=sin theta_1 cos theta_2 + cos theta_1 sin theta_2:`
| `sin(theta)` | `= 1/8 * sqrt7/4 + sqrt63/8 * 3/4` |
| `=sqrt7/32 + (9sqrt7)/32` | |
| `=(5 sqrt7)/16` |
Vectors, SPEC1 2012 VCAA 9
The position of a particle at time `t` is given by
`underset ~r (t) = (2 sqrt (t^2 + 2) - t^2) underset ~i + (2 sqrt (t^2 + 2) + 2t) underset ~j,\ \ t >= 0.`
- Find the velocity of the particle at time `t.` (1 mark)
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- Find the speed of the particle at time `t = 1` in the form `(a sqrt b)/c`, where `a, b` and `c` are positive integers. (2 marks)
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- Show that at time `t = 1,\ \ (dy)/(dx) = (1 + sqrt 3)/(1 - sqrt 3).` (2 marks)
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- Find the angle in terms of `pi`, between the vector `-sqrt 3 underset ~i + underset ~j` and the vector `underset ~r (t)` at time `t = 0.` (2 marks)
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Show Worked Solution
a. `underset ~r (t) = (2 sqrt (t^2 + 2) – t^2) underset ~i + (2 sqrt (t^2 + 2) +2t) underset ~j`
| `underset ~v(t)` | `=dot underset ~r(t)` | |
| `= (2(2t)(1/2)(t^2 + 2)^(-1/2) – 2t) underset ~i + (2(2t)(1/2)(t^2 + 2)^(-1/2) + 2)underset ~j` | ||
| `= ((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j` |
| b. | `underset ~v(1)` | `= (2/sqrt(1 + 2) – 2)underset ~i + (2/sqrt(1 + 2) + 2) underset ~j` |
| `= (2/sqrt 3 – 2) underset ~i + (2/sqrt 3 + 2)underset ~j` | ||
| `|\ underset ~v(1)\ |` | `= sqrt((2/sqrt 3 – 2)^2 + (2/sqrt 3 + 2)^2)` | |
| `= sqrt(4/3 – 8/sqrt 3 + 4 + 4/3 + 8/sqrt 3 + 4)` | ||
| `= sqrt(8/3 + 8)` | ||
| `= sqrt(32/3)` | ||
| `= (4sqrt2)/sqrt3` | ||
| `= (4 sqrt 6)/3` |
♦ Mean mark part (c) 41%.
| c. | `(dy)/(dt)` | `= (dy)/(dx) xx (dx)/(dt)` |
| `(dy)/(dx)` | `=((dy)/(dt))/((dx)/(dt))` | |
| `:. (dy)/(dx)|_(t=1)` | `= (2/sqrt 3 + 2)/(2/sqrt 3 – 2)` | |
| `= ((2 + 2 sqrt 3)/sqrt 3) xx (sqrt 3/(2 – 2 sqrt 3))` | ||
| `= (2 + 2 sqrt 3)/(2 – 2 sqrt 3)` | ||
| `= (1 + sqrt 3)/(1 – sqrt 3)` |
d. `text(At)\ \ t=0:` ♦♦ Mean mark part (d) 32%.
`underset ~r(0) = 2 sqrt 2 underset ~i + 2 sqrt 2 underset ~j`
| `theta_1` | `= tan^(-1)((2 sqrt 2)/(2 sqrt 2)) = pi/4` |
| `theta_2` | `= tan^(-1)(1/sqrt 3)=pi/6` |
`text(Let)\ \ theta=\ text(angle between the vectors:)`
| `theta` | `= pi – theta_1 – theta_2` |
| `= pi – pi/4 – pi/6` | |
| `= (12 pi – 3 pi – 2 pi)/12` | |
| `= (7 pi)/12` |
Calculus, SPEC1 2012 VCAA 8
The velocity, `v` m/s, of a body when it is `x` metres from a fixed point `O` is given by
`v = (2x)/sqrt(1 + x^2).`
Find an expression for the acceleration of the body in terms of `x` in simplest form. (3 marks)
Calculus, SPEC1 2012 VCAA 7
Consider the curve with equation `y = (x - 1) sqrt (2 - x),\ \ 1 <= x <= 2.`
Calculate the area of the region enclosed by the curve and the `x`-axis. (3 marks)
Calculus, SPEC1 2012 VCAA 6
Find the gradient of the tangent to the curve `xy^2 + y + (log_e (x - 2))^2 = 14` at the point `(3, 2).` (3 marks)
Calculus, SPEC1 2012 VCAA 5
Let `y = arctan (2x).`
Find the value of `a` given that `(d^2y)/(dx^2) = ax ((dy)/(dx))^2`, where `a` is a real constant. (3 marks)
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Calculus, SPEC1 VCE SM-Bank 5
Find `int(3x^2 + 8)/(x(x^2 +4))\ dx`. (3 marks)
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Complex Numbers, SPEC1 2012 VCAA 3
Consider the equation `z^3-z^2-2z-12 = 0, \ z in C.`
- Given that `z = 2 text(cis) ((2pi)/3)` is a root of the equation, find the other two roots in the form `a + ib`, where `a, b in R.` (3 marks)
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- Plot all of the roots clearly on the Argand diagram below. (1 mark)
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Show Worked Solution
a. `text(Given)\ \ z = 2 text(cis)((2 pi)/3)\ \ text(is a root, and)`
`z^3-z^2-2z-12 = 0\ \ text(has real coefficients,)`
`=> z= 2 text(cis)(-(2 pi)/3)\ \ text{is a root (conjugate).}`
`=> text(Roots:)\ \ -1+sqrt3i,\ \ -1-sqrt3i`
`(z-alpha) (z-beta) (z-gamma) = z^3-z^2-2z-12`
`text(Using)\ \ alpha beta gamma = 12:`
| `(-1 + i sqrt 3) (-1-i sqrt 3) gamma` | `=12` | |
| `((-1)^2-(sqrt3i)^2)gamma` | `=12` | |
| `4gamma` | `=12` | |
| `gamma` | `=3` |
`:.\ text(Other two roots:)`♦♦ Mean mark part (b) 33%.
| `z_2` | `= -1-i sqrt 3` |
| `z_3` | `= 3 + 0i` |
| b. |  |
Calculus, SPEC1 2012 VCAA 1
Find an antiderivative of `(6 + x)/(x^2 + 4).` (2 marks)
Mechanics, SPEC2 2013 VCAA 20 MC
A 5 kg parcel is on the floor of a lift that is accelerating downwards at 3 m/s².
The reaction, in newtons, of the floor of the lift on the parcel is
A. `−15 + 5g`
B. `15 + 5g`
C. `−15 + 3g`
D. `−15 − 5g`
E. `15 + 3g`
Calculus, SPEC2 2013 VCAA 19 MC
A tourist in a hot air balloon, which is rising at 2 m/s, accidentally drops a camera over the side and it falls 100 m to the ground.
Neglecting the effect of air resistance on the camera, the time taken for the camera to hit the ground, correct to the nearest tenth of a second, is
A. 4.3 s
B. 4.5 s
C. 4.7 s
D. 4.9 s
E. 5.0 s
Calculus, SPEC2 2013 VCAA 18 MC
A particle moves in a straight line such that its acceleration is given by `a = sqrt(v^2 - 1)` , where `v` is its velocity and `x` is its displacement from a fixed point.
Given that `v = sqrt2` when `x = 0`, the velocity `v` in terms of `x` is
A. `v = sqrt(2 + x)`
B. `v = 1 + |\ x + 1\ |`
C. `v = sqrt(2 + x^2)`
D. `v = sqrt(1 + (1 + x)^2)`
E. `v = sqrt(1 + (x - 1)^2)`
Vectors, SPEC2 2013 VCAA 17 MC
Consider the four vectors `underset~a = underset~j + 3underset~k`, `underset~b = underset~i - 4underset~k`, `underset~c = 3underset~j - k` and `underset~d = 2underset~j + underset~k`.
Which one of the following is a linearly dependent set of vectors?
- `{underset~a, underset~b, underset~c}`
- `{underset~a, underset~c, underset~d}`
- `{underset~a, underset~b, underset~d}`
- `{underset~b, underset~c, underset~d}`
- `{underset~a, underset~b}`
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