Aussie Maths & Science Teachers: Save your time with SmarterEd
An access code consists of 4 digits chosen from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The code will only work if the digits are entered in the correct order.
Some access codes contain exactly two different digits, for example 3377 or 5155.
How many such access codes can be made using exactly two different digits?
The diagram shows the two curves `y = sin x` and `y = sin(x-alpha) + k`, where `0 < alpha < pi` and `k > 0`. The two curves have a common tangent at `x_0` where `0 < x_0 < pi/2`.
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| i. | `y_1` | `= sin x` |
| `(dy_1)/(dx)` | `= cos x` | |
| `y_2` | `= sin(x-alpha) + k` | |
| `(dy_2)/(dx)` | `= cos (x-alpha)` |
`text(At)\ \ x = x_0,\ \ text(tangent is common)`♦ Mean mark part (i) 47%.
`:. cos x_0 = cos(x_0-alpha)`
ii. `x_0\ text{is in 1st quadrant (given).}`
`text{Using part (i):}`
`cos\ x_0 = cos(x_0-alpha) >0`♦♦♦ Mean mark part (ii) 19%.
`=> x_0-alpha\ text(is in 4th quadrant)\ (0 < alpha < pi)`
`text(S)text(ince sin is positive in 1st quadrant and)`
`text(negative in 4th quadrant)`
`=> sin x_0 = -sin(x_0-alpha)`
| iii. |  |
`text(When)\ \ x = x_0:`
| `y_1` | `=sin x_0` | |
| `y_2` | `=sin(x_0-alpha) + k` | |
| `sin x_0` | `=sin (x_0-alpha) + k` | |
| `sin x_0` | `= -sin x_0 + k` | |
| `k` | `== 2\ sin x_0` |
♦♦ Mean mark part (iii) 21%.
| `text(S)text(ince)\ \ cos x_0` | `= cos(x_0-alpha)` |
| `x_0` | `= -(x_0-alpha)` |
| `2x_0` | `= alpha` |
| `x_0` | `= alpha/2` |
`:. k = 2 sin\ alpha/2`
Use mathematical induction to prove that
`n^5 + n^3 + 2n`
is divisible by 4 for integers `n >= 1.` (4 marks)
smc-6514
The relationship between British pounds `(p)` and Australian dollars `(d)` on a particular day is shown in the graph.
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Convert 93 100 Japanese yen to British pounds. (2 marks)
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Ashley has a credit card with the following conditions:
Ashley's credit card statement for April is shown, with some figures missing.
The minimum payment is calculated as 2% of the closing balance on 30 April.
Calculate the minimum payment. (3 marks)
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Two right-angled triangles, `ABC` and `ADC`, are shown.
Calculate the size of angle `theta`, correct to the nearest minute. (3 marks)
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The container shown is initially full of water.
Water leaks out of the bottom of the container at a constant rate.
Which graph best shows the depth of water in the container as time varies?
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A school collected data related to the reasons given by students for arriving late. The Pareto chart shows the data collected.
What percentage of students gave the reason 'Train or bus delay'?
The function `f(x) = -sqrt(1 + sqrt(1 + x))` has inverse `f^(-1) (x)`.
The graph of `y = f^(-1) (x)` forms part of the curve `y = x^4-2x^2`.
The diagram shows the curve `y = x^4-2x^2`.
How many points do the graphs of `y = f(x)` and `y = f^(-1) (x)` have in common?
The diagram shows the region `R`, bounded by the curve `y = x^r`, where `r >= 1`, the `x`-axis and the tangent to the curve at the point `(1, 1)`.
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A person wins $1 000 000 in a competition and decides to invest this money in an account that earns interest at 6% per annum compounded quarterly. The person decides to withdraw $80 000 from this account at the end of every fourth quarter. Let `A_n` be the amount remaining in the account after the `n`th withdrawal.
`qquad A_2 = 1\ 000\ 000 xx 1.015^8-80\ 000(1 + 1.015^4)`. (2 marks)
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A museum is planning an exhibition using five rooms.
The museum manager draws a network to help plan the exhibition. The vertices `A`, `B`, `C`, `D` and `E` represent the five rooms. The number on the edges represent the maximum number of people per hour who can pass through the security checkpoints between the rooms.
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Draw the minimum cut onto the network below and recommend a change that the manager could make to one or more security checkpoints to increase the flow capacity to 240 visitors per hour. (2 marks)
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| a. | `text(Capacity)` | `= 130 + 90 + 70` |
| `= 290` |
♦♦ Mean mark (a) 32%.
COMMENT: In part (a), edge BC flows from the exit to the entry and is therefore not counted.
b. `text(Maximum flow capacity:)`
`text(Minimum cut = 80 + 40 + 65 + 45 = 230)`♦♦♦ Mean mark (b) 19%.
COMMENT: In part (b), edge BC now flows from entry to exit in the new “minimum” cut and is counted.
`text(If security is improved to increase the flow)`
`text(between Room C and Room B by 10 visitors)`
`text(per hour, the network’s flow capacity increases)`
`text(to 240.)`
The scores on an examination are normally distributed with a mean of 70 and a standard deviation of 6. Michael received a score on the examination between the lower quartile and the upper quartile of the scores.
Which shaded region most accurately represents where Michael's score lies?
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A school collected data related to the reasons given by students for arriving late. The Pareto chart shows the data collected.
What percentage of students gave the reason 'Train or bus delay'?
What is the interest earned, in dollars, if $800 is invested for `x` months at a simple interest rate of 3% per annum?
Steve and Jess are two students who have agreed to take part in a psychology experiment. Each has to answer several sets of multiple-choice questions. Each set has the same number of questions, `n`, where `n` is a number greater than 20. For each question there are four possible options A, B, C or D, of which only one is correct.
Let the random variable `X` be the number of questions that Steve answers correctly in a particular set.
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If `P(Y > 23) = 6 xx P(Y = 25)`, show that the value of `p=5/6`. (2 marks)
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If `|underset ~a + underset ~b| = |underset ~a| + |underset ~b|` and `underset ~a, underset ~b != underset ~0`, which one of the following is necessarily true?
A. `underset ~a\ text(is parallel to)\ underset ~b`
B. `|underset ~a| = |underset ~b|`
C. `underset ~a = underset ~b`
D. `underset ~a\ text(is perpendicular to)\ underset ~b`
Consider a part of the graph of `y = xsin(x)`, as shown below.
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ii. Given that `int(xsin(x))\ dx = sin(x)-xcos(x) + c`, evaluate `int_(npi)^((n + 1)pi)(xsin(x))\ dx` when `n` is a positive odd integer.
Give your answer in simplest form. (1 mark)
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The continuous random variable \(X\) has a probability density function given by
\(f(x)= \begin{cases}
\cos(2x)& \text {if}\quad \dfrac{3 \pi}{4}<x<\dfrac{5 \pi}{4} \\
\ \\
0 & \text{elsewhere}
\end{cases}\)
Find the value of \(a\) such that \(P(X < a) = 0.25\). Give your answer correct to 2 decimal places. (3 marks)
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The gradient of the the perpendicular line to a curve at any point `P(x,y)` is twice the gradient of the line joining `P` and the point `Q(1,1)`.
The coordinate of points on the curve satisfy the differential equation
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).`
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Give your answer in the form `(a sqrt b)/c, \ a, b, c in Z.` (3 marks)
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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` |
The diagram above shows a mass suspended in equilibrium by two light strings that make angles of `60^@` and `30^@` with a ceiling. The tensions in the strings are `T_1` and `T_2`, and the weight force acting on the mass is `underset~W`. The correct statement relating the given forces is
A. `underset~T_1 + underset~T_2 + underset~W = underset~0`
B. `underset~T_1 + underset~T_2 - underset~W = underset~0`
C. `underset~T_1 xx 1/2 + underset~T_2 xx sqrt3/2 = underset~0`
D. `underset~T_1 xx sqrt3/2 + underset~T_2 xx 1/2 = underset~W`
E. `underset~T_1 xx 1/2 + underset~T_2 xx sqrt3/2 = underset~W`
On a particular morning, the position vectors of a boat and a jet ski on a lake `t` minutes after they have started moving are given by `underset~r_B(t) = (1-2cos(t)) underset~i + (3 + sin(t))underset~j` and `underset~r_J(t) = (1-sin(t)) underset~i + (2-cos(t))underset~j` respectively for `t >= 0`, where distances are measured in kilometres. The boat and the jet ski start moving at the same time. The graphs of their paths are shown below.
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| a. | |
`text(Initial positions occur at)\ \ t=0.`
`text(Consider the graph)\ \ underset~r_B(t)\ \ text(at)\ \ t=pi/4:`
`(1-2cos0) < (1-2cos(pi/4)) =>\ text(moves higher)`
`text(Consider the graph)\ \ underset~r_J(t)\ \ text(at)\ \ t=pi/4:`
`(1-sin(0)) > (1-sin(pi/4)) =>\ text(moves left)`
| b.i. | `underset~dotr_B(t)` | `= 2sin(t)underset~i + cos(t)underset~j` |
| `|underset~dotr_B(t)|` | `= sqrt(4sin^2(t) + cos^2(t))` |
| `underset~dotr_J(t)` | `= −cos(t)underset~i + sin(t)underset~j` |
| `|underset~dotr_J(t)|` | `= sqrt(cos^2(t) + sin(t))=1` |
`text(Find)\ \ t\ \ text(when)\ \ sqrt(4sin^2(t) + cos^2(t))=1:`
`t = pi\ text(seconds)\ \ \ (t!=0)`
♦ Mean mark (b)(ii) 48%.
| b.ii. | `(underset~r)_B(pi)` | `= (1-2cos(pi))underset~i + (3 + sin(pi))underset~j` |
| `= 3underset~i + 3underset~j` |
`:.\ text(Boat coordinates):\ (3,3)`
| c.i. | `underset~r_B-underset~r_J` | `=(sin(t)-2 cos(t))i +(1+sin(t) + cos(t))` |
| `:. d` | `= |underset~r_B-underset~r_J|` | |
| `= sqrt((sin(t)-2cos(t))^2 + (1 + sin(t) + cos (t))^2)` |
c.ii. `d_text(min) ~~ 0.33\ \ \ text{(by CAS)}`♦♦♦ Mean mark part (c)(ii) 15%.
d. `text(Equating coefficients for collision:)`
| `x:\ \ \ 1-sin(t)` | `= 1-2cos(t)\ …\ (1)` | |
| `sin(t)` | `= 2cos(t)` | |
| `tan(t)` | `= 2` | |
| `t` | `=tan^(−1)(2)` |
`y:\ \ \ a-cos(t) = 3 + sin(t)\ …\ (2)`♦ Mean mark part (d) 41%.
`text(Using)\ \ tan(t)=2,`
`=> sin(t) = (2sqrt5)/5,\ \ cos(t) = sqrt5/5`
`text{Substitute into (1):}`
| `a-1/sqrt5` | `= 3 + 2/sqrt5` | |
| `:. a` | `= 3 + 3/sqrt5` |
`:.\ text(Collision occurs when)\ \ t=tan^(-1)2\ \ text(and)\ \ a=3 + 3/sqrt5`
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Find `b` in terms of `a`. (1 mark)
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| a. | `r` | `= sqrt((−2)^2 + (−2sqrt3)^2)=4` |
| `theta` | `= −pi + tan^(−1)((2sqrt3)/2)` |
| `= −pi + pi/3` | |
| `=(-2pi)/3` |
`:. −2-2sqrt3 i = 4text(cis)((−2pi)/3)`
| b. `z^2 + 4z + z^2-4 + 16` | `=0` |
| `(z + 2)^2 + 12` | `= 0` |
| `(z + 2)^2` | `= -12` |
| `(z + 2)^2` | `= 12i^2` |
| `z + 2` | `= ±sqrt12 i` |
| `z + 2` | `= ±2sqrt3 i` |
| `:. z` | `= -2 ± 2sqrt3 i` |
♦♦ Mean mark part (c) 31%.
| c. | `z_1` | `= -2 + 2sqrt3 i = -(2-2sqrt3 i)` |
| `z_2` | `=-2-2sqrt3 i =-bar((2-2sqrt3 i))` |
| d. | `|z|` | `= |z-(2-2sqrt3 i)|` |
| `x^2 + y^2` | `= (x-2)^2 + (y + 2sqrt3)^2` | |
| `x^2 + y^2` | `= x^2-4x + 4+ y^2 + 4sqrt3 y + 12` | |
| `0` | `= −4x + 4sqrt3 y + 16` | |
| `0` | `= −x + sqrt3 y + 4` |
`:. x-sqrt3 y-4=0`
| e. | |
f. `x = − 2\ \ text(is equidistant from)\ \ z_1 = a\ \ text(and)\ \ z_2 = b`♦♦♦ Mean mark 1%!
`=> text(Im)(a) = text(Im)(a)`
`text(Let)\ \ a = alpha + betai, \ b = gamma + betaj`
| `(alpha + gamma)/2` | `= −2` |
| `alpha + gamma` | `= −4` |
| `gamma` | `= -4-alpha` |
| `:. b` | `= -4-alpha + betaj` |
| `= -4-(alpha + betaj)` | |
| `= -4-bara` |
♦♦ Mean mark 31%.
| g. | `text(Area)\ DeltaOAB` | `= 1/2 xx (4sqrt3 xx 2)` |
| `= 4sqrt3` |
| `text(Area of sector)\ AOB` | `= pi xx 4^2 xx (2 xx pi/3)/(2pi)` |
| `= (16pi)/3` |
`:.\ text(Area of major segment area)`
`=pi(4)^2-((16pi)/3-4sqrt3)`
`= 4sqrt3 + (32pi)/3`
One root of a quadratic equation with real coefficients is `sqrt 3 + i`.
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Find the equation of this circle, expressing it in the form `|z-alpha| = beta`, where `alpha, beta in R`. (3 marks)
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a.i. `z_1 = sqrt 3 + i`
`z_2 = bar z_1 = sqrt 3-i\ \ \ text{(conjugate root)}`
| a.ii. | `(z-(sqrt 3 + i))(z-(sqrt 3-i))` | `= 0` |
| `((z-sqrt 3)-i)((z-sqrt 3) + i)` | `= 0` | |
| `(z-sqrt 3)^2-i^2` | `= 0` | |
| `z^2-2 sqrt 3 z + 3 + 1` | `= 0` | |
| `z^2-2 sqrt 3 z + 4` | `= 0` |
| b. | `z(z^2-2 sqrt 3 z + 4)` | `= 0` |
| `z(z-(sqrt 3 + i))(z-(sqrt 3-i))` | `= 0` |
`text(Convert to polar form:)`
| `sqrt 3 + i` | `= sqrt((sqrt 3)^2 + 1^2) * text(cis) (tan^(-1)(1/sqrt 3))` | |
| `= 2 text(cis) (pi/6)` | ||
| `=> sqrt 3-i` | `= 2 text(cis) (-pi/6)` |
c. `text(Equidistant from)\ (0, 0) and (sqrt 3, 1)`
| `text(Midpoint)\ (x_1, y_1)` | `= (sqrt 3/2, 1/2)` |
`m = (1-0)/(sqrt 3-0) = 1/sqrt 3`
`m_(_|_) = (-1)/m = -sqrt 3`
`:.\ text(Equation of ⊥ bisector:)`
| `y-1/2` | `=-sqrt3(x-sqrt3/2)` | |
| `y` | `=-sqrt3 x +3/2+1/2` | |
| `:.y` | `=-sqrt3 x +2` |
d. `text(Let)\ O = (0, 0),\ P = (sqrt 3, 1),\ Q = (sqrt 3, -1)`
`text(⊥ bisector of two points on arc of a circle passes)`
`text(through the centre of the circle.)`
`OP = OQ = PQ = 2`
`=> Delta OPQ\ text(is equilateral)`
`text(⊥ bisector of)\ PQ\ text(is)\ y=0.`
`text(Centre of circle occurs when:)`
`0 = -sqrt 3 x + 2\ \ text{(using part c)`
`x=2/sqrt3`
`=>\ text(Radius)\ = 2/sqrt3`
`:. |z-2/sqrt 3| = 2/sqrt 3`
A horizontal beam is supported at its endpoints, which are 2 m apart. The deflection `y` metres of the beam measured downwards at a distance `x` metres from the support at the origin `O` is given by the differential equation `80 (d^2y)/(dx^2) = 3x-4`.
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The heights of mature water buffaloes in northern Australia are known to be normally distributed with a standard deviation of 15 cm. It is claimed that the mean height of the water buffaloes is 150 cm.
To decide whether the claim about the mean height is true, rangers selected a random sample of 50 mature water buffaloes. The mean height of this sample was found to be 145 cm.
A one-tailed statistical test is to be carried out to see if the sample mean height of 145 cm differs significantly from the claimed population mean of 150 cm.
Let `bar X` denote the mean height of a random sample of 50 mature water buffaloes.
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Consider the curve represented by `x^2 - xy + 3/2 y^2 = 9.`
Write each equation in the form `y = ax + b.` (2 marks)
Give your answer in the form `k pi`, where `k` is a real constant (2 marks)
Consider `f(x) = 3x arctan (2x)`. --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) ---
a. `f(x) = 3x arctan (2x)` `f(0) = 0` `text(When)\ \ x < 0, \ \ 3x<0,\ \ text(and)\ \ y = tan^(-1) (2x) < 0` `text(When)\ \ x>0,\ \ 3x>0,\ \ text(and)\ \ y = tan^(-1) (2x) > 0` b. `u = 3x, qquad v = tan^(-1)(2x)` `u prime = 3, qquad v prime = 2/(1 + 4x^2)` Show Worked Solution
♦♦♦ Mean mark 4%.
`:. f(x) in [0, oo)`
`:. f prime (x)`
`= 3(arctan (2x)) + (2/(1 + 4x^2))(3x)`
`= 3 arctan (2x) + (6x)/(1 + 4x^2)`
c.
`A`
`= int_(1/2)^(sqrt 3/2) arctan (2x)\ dx`
`= 1/3 int_(1/2)^(sqrt 3/2) 3arctan (2x)\ dx`
`=1/3 int_(1/2)^(sqrt 3/2) (3 arctan (2x) + (6x)/(1 + 4x^2)-(6x)/(1 + 4x^2))\ dx`
`= 1/3 int_(1/2)^(sqrt 3/2) 3 arctan (2x) + (6x)/(1 + 4x^2) dx-int_(1/2)^(sqrt 3/2) (2x)/(1 + 4x^2)\ dx`
`= 1/3 [3x arctan (2x)]_(1/2)^(sqrt 3/2)-1/4 int_(1/2)^(sqrt 3/2) (8x)/(1 + 4x^2)\ dx`
`= [sqrt 3/2 arctan (sqrt 3)-1/2 arctan (1)]-1/4[ln (1 + 4x^2)]_(1/2)^(sqrt 3/2)`
`= sqrt 3/2 (pi/3)-1/2 (pi/4)-1/4 [ln (1 + 4(3/4))-ln(1 + 4 (1/4)]`
`= (pi sqrt 3)/6-pi/8-1/4 (ln4-ln 2)`
`= (pi sqrt 3)/6-pi/8-1/4 ln2`
Let `f: R -> R, \ f(x) = x^2e^(kx)`, where `k` is a positive real constant.
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Let `g(x) = −(2xe^(kx))/k`. The diagram below shows sections of the graphs of `f` and `g` for `x >= 0`.
Let `A` be the area of the region bounded by the curves `y = f(x), \ y = g(x)` and the line `x = 2`.
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The differentiable function `f : R -> R` is a probability density function. It is known that the median of the probability density function `f` is at `x = 0` and `f^{′} (0) = 4`.
The transformation `T : R^2 -> R^2` maps the graph of `f` to the graph of `g`, where `g : R -> R` is a probability density function with a median at `x = 0` and `g^{′} (0) = -1`.
The transformation `T` could be given by
Consider the functions `f:R^+ -> R,\ f(x) = x^(p/q)` and `g: R^+ -> R,\ g(x) = x^(m/n)`, where `p, q, m` and `n` are positive integers, and `p/q` and `m/n` are fractions in simplest form.
If `{x: f(x) > g(x)} = (0, 1)` and `{x: g(x) > f(x)} = (1, oo)`, which of the following must be false?
A. `q > n and p = m`
B. `m > p and q = n`
C. `pn < qm`
D. `f prime (c) = g prime(c)\ \ text(for some)\ \ c in (0, 1)`
E. `f prime (d) = g prime(d)\ \ text(for some)\ \ d in (1, oo)`
Alex is buying a used car which has a sale price of $13 380. In addition to the sale price there are the following costs:
The graph of `y = tan(ax)`, where `a ∈ R^+`, has a vertical asymptote `x = 3 pi` and has exactly one `x`-intercept in the region `(0, 3 pi)`.
The value of `a` is
Frank owns a tennis court.
A diagram of his tennis court is shown below
Assume that all intersecting lines meet at right angles.
Frank stands at point `A`. Another point on the court is labelled point `B`.
Round your answer to one decimal place. (1 mark)
Assume that the ball travels in a straight line to the ground at point `B`.
What is the straight-line distance, in metres, that the ball travels?
Round your answer to the nearest whole number. (1 mark)
Frank hits two balls from point `A`.
For Frank’s first hit, the ball strikes the ground at point `P`, 20.7 m from point `A`.
For Frank’s second hit, the ball strikes the ground at point `Q`.
Point `Q` is `x` metres from point `A`.
Point `Q` is 10.4 m from point `P`.
The angle, `PAQ`, formed is 23.5°.
Round your answers to one decimal place. (1 mark)
Round your answer to the nearest metre. (1 mark)
| a. | `text(Using Pythagoras,)` | |
| `AB` | `= sqrt(4.1^2 + (6.4 + 6.4 + 5.5)^2` | |
| `= 18.75…` | ||
| `= 18.8\ text{m (to 1 d.p.)}` | ||
b. `text(Let)\ \ d = text(distance travelled)`
| `d` | `= sqrt(2.5^2 + 18.8^2)` |
| `= 18.96…` | |
| `= 19\ text{m (nearest m)}` |
| c.i. |  |
`/_AQP ->\ text(2 possibilities)`♦♦ Mean mark 25%.
`text(Using Sine rule,)`
| `(sin/_AQP)/20.7` | `= (sin 23.5^@)/10.4` |
| `sin /_AQP` | `= (20.7 xx sin 23.5^@)/10.4` |
| `= 0.7936…` | |
| `:. /_AQP` | `= 52.5^@ or 127.5^@\ \ \ text{(to 1 d.p.)}` |
♦♦♦ Mean mark 18%.
| c.ii. | `Q\ text(is within court when)\ \ /_AQP = 127.5^@` | |
| `/_APQ = 180 – (127.5 + 23.5) = 29^@` | ||
| `text(Using sine rule,)` | ||
| `x/(sin 29^@)` | `= 10.4/(sin 23.5^@)` | |
| `:. x` | `= (10.4 xx sin 29^@)/(sin 23.5^@)` | |
| `= 12.64…` | ||
| `= 13\ text{m (nearest m)}` | ||
Annie, Buddhi, Chuck and Dorothy work in a factory.
Today each worker will complete one of four tasks, 1, 2, 3 and 4.
The usual completion times for Annie, Chuck and Dorothy are shown in the table below.
Buddhi takes 3 minutes for Task 3.
He takes `k` minutes for each other task.
Today the factory supervisor allocates the tasks as follows
This allocation will achieve the minimum total completion time if the value of `k` is at least
Parcel deliveries are made between five nearby towns, `P` to `T`.
The roads connecting these five towns are shown on the graph below. The distances, in kilometres, are also shown.
A road inspector will leave from town `P` to check all the roads and return to town `P` when the inspection is complete. He will travel the minimum distance possible.
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Julie has retired from work and has received a superannuation payment of $492 800. She has two options for investing her money. Option 1 Julie could invest the $492 800 in a perpetuity. She would then receive $887.04 each fortnight for the rest of her life. --- 4 WORK AREA LINES (style=lined) --- Option 2 Julie could invest the $492 800 in an annuity, instead of a perpetuity. The annuity earns interest at the rate of 4.32% per annum, compounding monthly. The balance of Julie’s annuity at the end of the first year of investment would be $480 242.25 --- 1 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) ---
How much interest would Julie’s annuity earn in the second year of investment?
Round your answer to the nearest cent. (1 mark)
Table 3 shows the yearly average traffic congestion levels in two cities, Melbourne and Sydney, during the period 2008 to 2016. Also shown is a time series plot of the same data.
The time series plot for Melbourne is incomplete.
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`text(congestion level = −2280 + 1.15 × year)`
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iii. Use the least squares line to predict when the percentage congestion level in Sydney will be 43%. (1 mark)
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The yearly average traffic congestion level data for Melbourne is repeated in Table 4 below.
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| congestion level = |
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× year |
Explain why, quoting the values of appropriate statistics. (2 marks)
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| a. |  |
| b.i. |  |
b.ii. `text(The least squares line is 1.15% higher each year.)`♦ Mean mark (b)(ii) 36%.
COMMENT: Major problems caused by part (b)(ii). Review!
` :.\ text(Average rate of increase) = 1.15 text(%)`
| b.iii. | `text(Find year when:)` | |
| `43` | `= -2280 + 1.15 xx text(year)` | |
| `text(year)` | `= 2323/1.15` | |
| `= 2020` | ||
c. `-1515`
d. `text(congestion level) = -1515 + 0.7667 xx text(year)`
e. `text(Melbourne congestion level in 2008)`♦♦♦ Mean mark 18%.
`= -1515 + 0.7667 xx 2008`
`= 24.5 text(%)`
`text{In 2008 Sydney has higher congestion (29.2 > 24.5)}`
`text(After 2008, Sydney congestion grows at 1.15% per)`
`text(year and Melbourne grows at 0.7667% per year.)`
`:.\ text(Sydney predicted to always exceed Melbourne.)`
A tower is built on flat ground.
Three tourists, `A`, `B` and `C` are observing the tower from ground level.
`A` is due north of the tower, `C` is due east and `B` is on the line of sight from `A` and `C` and between them.
The angles of elevation to the top of the tower from `A`, `B` and `C` are 26°, 28° and 30°, respectively.
What is the bearing of `B` from the tower? (4 marks)
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`text(Let)\ \ h =\ text(height of tower)`
`text(In)\ DeltaOAT:`
`tan26^@= h/(d_A)\ \ =>\ \ d_A= h/(tan26^@)`
`text(Similarly,)`
`d_B= h/(tan28^@)\ \ =>\ \ d_C= h/(tan30^@)`
`text(In)\ DeltaOAC:`
`tan angleOAC= (d_C)/(d_A)= (h/(tan30^@))/(h/(tan26^@))= (tan26^@)/(tan30^@)= 0.8447…`
`angleOAC=40.19^@`
`text(Using sine rule in)\ DeltaOAB:`
| `(sinangleABO)/(d_A)` | `= (sinangleOAC)/(d_B)` |
| `sinangleABO` | `= sin40.2^@ xx (tan28^@)/(tan26^@)= 0.7035…` |
| `angleABO` | `= 44.71^@\ text(or)\ 135.29^@` |
`text(S)text(ince)\ \ angleOCA= tan^(−1)((tan30)/(tan26))= 49.8^@`
`=> angleOBC = 44.71°`
`(text(otherwise angle sum)\ DeltaOBC > 180°)`
`angleAOB= 180-(40.19 + 135.29)= 4.52`
`:.\ text(Bearing of)\ B\ text(from tower is)\ 005°.`
A function `f`, its derivative `fprime` and its second derivative `f″` are defined for `x ∈ R` with the following properties.
`f(a) = 1, f(−a) = −1`
`f(b) = −1, f(−b) = 1`
and `f″(x) = ((x + a)^2(x - b))/(g(x))`, where `g(x) < 0`
The coordinates of any points of inflection of `|\ f(x)\ |` are
Given `f(x) = sqrt (x^2 - 9)` and `g(x) = x + 5`
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| a. | `f(g(x))` | `= sqrt {(x + 5)^2 – 9}` |
| `= sqrt (x^2 + 10x + 16)` | ||
| `= sqrt {(x + 2) (x + 8)}` |
`:. c = 2, d = 8 or c = 8, d = 2`
b. `text(Find)\ x\ text(such that:)`
`(x+2)(x+8) >= 0`
`(x + 2) (x + 8) >= 0\ \ text(when)`
`x <= -8 or x >= -2`
`:.\ text(Domain:)\ \ x<=-8\ \ and\ \ x>=-2`
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`.
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a. `t > 5\ \ text(seconds)`
b. `7.4\ \ text(seconds)`
c.
a. `text(Displacement is reducing when the velocity is negative.)`
`:. t > 5\ \ text(seconds)`
b. `text(At)\ B,\ text(the displacement) = 7\ text(units)`
`text(Consider the displacement from)\ B\ text(to)\ D:`
`text(Since the area below the graph from)\ B\ text(to)\ C\ text(equals )`
`text(the area above the graph from)\ C\ text(to)\ D,\ text(there is no )`
`text(change in displacement from)\ B\ text(to)\ D.`
`text(Consider)\ t >= 6,`
`text(Time required to return to origin:)`
`t=d/v= 7/5= 1.4\ \ text(seconds)`
`:.\ text(The particle returns to the origin after 7.4 seconds.)`
c.
The diagram shows the front of a tent supported by three vertical poles. The poles are 1.2 m apart. The height of each outer pole is 1.5 m, and the height of the middle pole is 1.8 m. The roof hangs between the poles.
The front of the tent has area `A\ text(m)^2`.
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| a. | `A` | `~~ h/2 [y_0 + 2y_1 + y_2]` |
| `~~ 1.2/2 [1.5 + (2 xx 1.8) + 1.5]` | ||
| `~~ 0.6 [6.6]~~ 3.96\ text(m)^2` |
| b. |  |
`text(The tent roof is concave up. Since the Trapezoidal rule uses)`
`text(straight lines, it will estimate a higher area.)`
Two events, `A` and `B`, from a given event space, are such that `P(A) = 1/5` and `P(B) = 1/3`.
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a. `text(Sketch Venn diagram:)`
`:.P(A^{′} ∩ B)=P(B)-P(A ∩B)=1/3-1/8=5/24`
| b. |  |
`text(Mutually exclusive means)\ \ P(A ∩ B)=0.`
`:. P(A^{′} ∩ B) = 1/3`
Consider the discrete probability distribution with random variable `X` shown in the table below.
\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ -1\ \ \ & \ \ \ \ 0\ \ \ \ & \ \ \ \ b\ \ \ \ & \ \ \ \ 2b\ \ \ \ &\ \ \ \ 4\ \ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & a & b & b & 2b & 0.2 \\
\hline
\end{array}
The smallest and largest possible values of `text(E)(X)` are respectively
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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| a. |  |
`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^@`
b. `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)}` |
c. `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` | `=cos^(-1)0.9018…` |
| `=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 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. `\text{Solution 1}`
`h=200/4=50`
| `A` | `~~50/2(360+300)+50/2(300+270)+50/2(270+140)+50/2(140+0)` |
| `~~25(660+570+410+140)~~44\ 500\ \text{m}^3` |
`\text{Solution 2}`
`V~~50/2[360 + 2(300 + 270 + 140) + 0]~~ 44\ 500\ text(m)^3`
b. `text(Convert 2 km)^2\ \text{to m}^2:`
`text(2 km)^2= 2000\ text(m) xx 1000\ \text{m} = 2\ 000\ 000\ text(m)^2`
`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)}` |
Mick has a bag of marbles. His marbles are orange, white, blue and green.
Mick picks one marble from his bag.
Which of the following could be the probability that the marble he picks is green.
| `13/3` | `1.52` | `3/13` | `6` |
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Kim made these solid prisms out of identical cubes.
Which prism has the largest volume?
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`text{The prism with the largest volume (36 cubes) is:}`
Madison uses the number sentence 15 × 12 = 180 to solve a problem.
Which of the following could be the problem?
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Madison buys 15 showbags. How much does each showbag cost? |
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Madison spends $15 on 180 showbags. How much does she spend? |
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Madison buys 15 showbags that cost $12 each. How many showbags does she buy? |
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Madison buys 12 showbags that cost $15 each. How much does she spend? |
Caroline has 56 match sticks.
She uses 6 match sticks to make two small triangles.
What is the largest number of small triangles like this that Caroline can make with her 56 match sticks?
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A bottle of milk can fill 4 glasses exactly.
Which of the following holds the most milk?
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`text(Converting each option to glasses)`
`text(Option 1: 3 × 4 = 12 glasses)`
`text(Option 2: 1 × 4 + 8 = 12 glasses)`
`text(Option 3: 2 × 4 + 1 = 9 glasses)`
`text(Option 4: 2 × 4 + 5 = 13 glasses)`
Groucho, Harpo and Zeppo are weighing their grapes.
Harpo's grapes weigh more than Zeppo's, but less than Groucho's grapes.
Which of these could be the weight of Harpo's grapes?
| 0.11 kg | 0.32 kg | 0.6 kg | 0.7 kg |
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Conrad followed these instructions to create a drawing:
Which of these did he draw?
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Cleopatra uses balls and sticks to make a model of a triangular pyramid.
Each stick is the same length.
She decides to change her model into a cube.
How many more balls and sticks will Cleopatra need to make the cube?
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At lunch time, a hamburger shop sold 260 hamburgers.
At dinner time on the same day, the shop sold 487 hamburgers.
Which number sentence shows how many more hamburgers were sold at dinnertime than at lunchtime?
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`+\ 487 = 260` |
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`487 - 260 =` | |
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`- 487 = 260` |
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`260 + 487 = ` | |
| `487 – 260 =` | |
| `487 – 260 =` | |
Let `alpha, beta, gamma` be the zeros of `p(x) = x^3 + px + q`, where `p` and `q` are real and `q != 0`.
In `DeltaABC`, point `D` is chosen on side `AB` and point `E` is chosen on side `AC` so that `DE` is parallel to `BC` and `(BC)/(DE) = sqrt2`.
The process is repeated two more times. Point `F` is chosen on `BC` and point `G` on `BA` so that `FG` is parallel to `CA` and `(CA)/(FG) = sqrt2`.
Point `H` is chosen on side `CB` and `I` on `CA` so that `HI` is parallel to `BA` and `(BA)/(HI) = sqrt2`.
The segments `FG` and `HI` intersect at `X`, `DE` and `HI` intersect at `Y`, and `DE` and `FG` intersect at `Z`.
Let `n` be a positive integer and let `x` be a positive real number.
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