Aussie Maths & Science Teachers: Save your time with SmarterEd
Cindy took out a reducing balance loan of $8400 to finance an overseas holiday.
Interest was charged at a rate of 9% per annum, compounding quarterly.
Her loan is to be fully repaid in six years, with equal quarterly payments.
After three years, Cindy will have reduced the balance of her loan by approximately
A. 9%
B. 35%
C. 43%
D. 50%
E. 57%
`A_n` is the `n`th term in a sequence.
Which one of the following expressions does not define a geometric sequence?
| A. `A_(n + 1) = n` | `\ \ \ \ A_0 = 1` |
| B. `A_(n + 1) = 4` | `\ \ \ \ A_0 = 4` |
| C. `A_(n + 1) = A_n + A_n` | `\ \ \ \ A_0 = 3` |
| D. `A_(n + 1) = –A_n` | `\ \ \ \ A_0 = 5` |
| E. `A_(n + 1) = 4A_n` | `\ \ \ \ A_0 = 2` |
A student uses the following data to construct the scatterplot shown below.
A reciprocal transformation is applied to the `x`-axis and is used to linearise the scatterplot.
With `y` as the dependent variable, the slope of the least squares regression line that is fitted to the linearised plot is closest to
A. `–249`
B. `–25`
C. `0.004`
D. `25`
E. `249`
`text(By calculator,)`
`text(Gradient of)\ y\ text(versus)\ 1/x\ text(is closest to 249.)`
`=> E`
The following information relates to the repayment of a home loan of $300 000.
• The loan is to be repaid fully with monthly payments of $2500.
• Interest compounds monthly.
• After the first monthly payment has been made, the amount owing on the loan is $299 000.
Which one of the following statements is true?
A healthy eating and gym program is designed to help football recruits build body weight over an extended period of time.
Roh, a new recruit who initially weighs 73.4 kg, decides to follow the program.
In the first week he gains 400 g in body weight.
In the second week he gains 380 g in body weight.
In the third week he gains 361 g in body weight.
If Roh continues to follow this program indefinitely, and this pattern of weight gain remains the same, his eventual body weight will be closest to
A. `74.5\ text(kg)`
B. `77.1\ text(kg)`
C. `77.3\ text(kg)`
D. `80.0\ text(kg)`
E. `81.4\ text(kg)`
The following information relates to Parts 1, 2 and 3.
The table shows the seasonal indices for the monthly unemployment numbers for workers in a regional town.
Part 1
The seasonal index for October is missing from the table.
The value of the missing seasonal index for October is
A. `0.93`
B. `0.95`
C. `0.96`
D. `0.98`
E. `1.03`
Part 2
The actual number of unemployed in the regional town in September is 330.
The deseasonalised number of unemployed in September is closest to
A. `310`
B. `344`
C. `351`
D. `371`
E. `640`
Part 3
A trend line that can be used to forecast the deseasonalised number of unemployed workers in the regional town for the first nine months of the year is given by
deseasonalised number of unemployed = 373.3 – 3.38 × month number
where month 1 is January, month 2 is February, and so on.
The actual number of unemployed for June is predicted to be closest to
A. `304`
B. `353`
C. `376`
D. `393`
E. `410`
Eleven speed bumps are placed on a road.
The speed bumps are placed so that the distance between consecutive speed bumps decreases according to an arithmetic sequence.
The distance between the first and last speed bumps is exactly 100 m.
The smallest distance between consecutive speed bumps is 2 m.
The largest distance, in m, between two consecutive speed bumps is
A. `16`
B. `18`
C. `20`
D. `22`
E. `24`
Let `h` be a function with an average value of 2 over the interval `[0, 6].`
The graph of `h` over this interval could be
`text(Draw the line)\ \ y=2\ \ text(on each graph.)`
`=>\ text(An average value of 2 occurs when the same area)`
`text(is enclosed above and below the graph.)`
`text(Consider)\ C,`
`text(Area)\ A =\ text(Area)\ B`
`=> C`
Let `f(x) = ax^m` and `g(x) = bx^n`, where `a, b, m` and `n` are positive integers. The domain of `f = text(domain of)\ g = R.`
If `f prime (x)` is an antiderivative of `g(x)`, then which one of the following must be true?
A. `m/n` is an integer
B. `n/m` is an integer
C. `a/b` is an integer
D. `b/a` is an integer
E. `n - m = 2`
The transformation that maps the graph of `y = sqrt(8x^3 + 1)` onto the graph of `y = sqrt(x^3 + 1)` is a
The rule for a function with the graph above could be
The trapezium `ABCD` is shown below. The sides `AB, BC` and `DA` are of equal length, `p.` The size of the acute angle `BCD` is `x` radians
The area of the trapezium is a maximum when the value of `x` is
| `sin x` | `= h/p` |
| `:. h` | `= p sinx` |
| `cosx` | `= w/p` |
| `:.w` | `= pcos x` |
| `A_text(trap)` | `= 1/2h xx [p + (p + 2w)]` |
| `= (p sin x)/2 [2p + 2pcosx]` | |
| `= p^2sin x (1 + cosx), \ \ x ∈ (0,pi/2)` |
`text(Find when)\ \ (dA)/(dp)=0,\ \ x ∈ (0,pi/2)`
`:. x = pi/3`
`=> D`
A projectile is fired from `O` with velocity `V` at an angle of inclination `theta` across level ground. The projectile passes through the points `L` and `M`, which are both `h` metres above the ground, at times `t_1` and `t_2` respectively. The projectile returns to the ground at `N`.
The equations of motion of the projectile are
`x = Vtcos theta` and `y = Vtsin theta − 1/2 g t^2`. (Do NOT prove this.)
--- 5 WORK AREA LINES (style=lined) ---
Let `∠LON = α` and `∠LNO = β`. It can be shown that
`tan alpha = h/(Vt_1 cos theta)` and `tan beta = h/(Vt_2 cos theta)`. (Do NOT prove this.)
--- 5 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
Let `ON = r` and `LM = w`.
--- 4 WORK AREA LINES (style=lined) ---
Let the gradient of the parabola at `L` be `tan phi`.
--- 8 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
The graph of `f: [1, 5] -> R,\ f(x) = sqrt (x - 1)` is shown below.
Which one of the following definite integrals could be used to find the area of the shaded region?
A. `int_1^5 (sqrt (x - 1))\ dx`
B. `int_0^2 (sqrt (x - 1))\ dx`
C. `int_0^5 (2 - sqrt (x - 1))\ dx`
D. `int_0^2 (x^2 + 1)\ dx`
E. `int_0^2 (x^2)\ dx`
A transformation `T: R^2 -> R^2, T ([(x), (y)]) = [(1, 0), (0, -1)] [(x), (y)] + [(5), (0)]` maps the graph of a function `f` to the graph of `y = x^2, x in R.`
The rule of `f` is
Let `p` and `q` be positive integers with `p ≤ q`.
The diagram shows a triangular piece of land `ABC` with dimensions `AB = c` metres, `AC = b` metres and `BC = a` metres, where `a ≤ b ≤ c`.
The owner of the land wants to build a straight fence to divide the land into two pieces of equal area. Let `S` and `T` be points on `AB` and `AC` respectively so that `ST` divides the land into two pieces of equal area.
Let `AS = x` metres, `AT = y` metres and `ST = z` metres.
--- 2 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
(You may assume that the value of `x` given in part (iii) is feasible.) (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
Consider the function `f(x) = (log_e x)/x`, for `x > 0`.
--- 6 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. `f(x) = (log_e x)/x, text(for)\ x > 0`
`text(Using the quotient rule,)`
| `f′(u/v)` | `=(u′v-uv′)/v^2` |
| `f′(x)` | `= (x · d/(dx) (log_e x) − log_e x · d/(dx) (x))/(x^2)` |
| `= (x(1/x) − log_e x · 1)/(x^2)` | |
| `= (1 − log_e x)/(x^2)` |
`text(SP’s occur when)\ \ f′(x)=0`
| `(1 − log_e x)/(x^2)` | `= 0` |
| `1 − log_e x` | `= 0` |
| `log_e x` | `= 1` |
| `:. x` | `= e` |
`:. text(There is a stationary point at)\ \ x = e.`
| ii. |
|
`text(When)\ \ x < e, log_e x<1\ \ \ text{(from graph)}`
| `1 − log_e x` | `> 0` |
| `(1 − log_e x)/(x^2)` | `> 0` |
| `:. f′(x)` | `> 0` |
`text(When)\ x > e, log_e x>1\ \ \ text{(from graph)}`
| `1 − log_e x` | `< 0` |
| `(1 − log_e x)/(x^2)` | `< 0` |
| `:. f′(x)` | `< 0` |
`:.\ text(The stationary point at)\ \ x = e\ \ text(is a maximum.)`
| iii. | `f(e)` | `= (log_e x)/e` |
| `= 1/e` |
`:. f(x)\ \ text(has a maximum value at)\ \ (e,1/e)`
| `:. (log_e x)/x` | `≤ 1/e` |
| `log_e x` | `≤ x/e` |
| `e log_e x` | `≤ x` |
| `log_e x^e` | `≤ x` |
| `e^(log_e x^e)` | `≤ e^x` |
| `x^e` | `≤ e^x\ \ \ text{(using}\ e^(log x) = x)` |
| `:. e^x` | `≥ x^e \ \ \ text{(for}\ \ x > 0text{)}` |
A particle moves along the `x`-axis. Initially it is at rest at the origin. The graph shows the acceleration, `a`, of the particle as a function of time `t` for `0 ≤ t ≤ 5`.
--- 2 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
The diagram shows the graph of the parabola `x^2 = 16y`. The points `A (4, 1)` and `B (−8, 4)` are on the parabola, and `C` is the point where the tangent to the parabola at `A` intersects the directrix.
A discrete random variable `X` has the probability function `text(Pr)(X = k) = (1 -p)^k p`, where `k` is a non-negative integer.
`text(Pr)(X > 1)` is equal to
Let `n` be a positive integer.
Suppose that `x >= 0` and `n` is a positive integer.
--- 2 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
For `x > 0`, let `f(x) = x^n e^-x`, where `n` is an integer and `n >= 2.`
--- 8 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
In the acute-angled triangle `ABC, \ K` is the midpoint of `AB, \ L` is the midpoint of `BC` and `M` is the midpoint of `CA`. The circle through `K, L` and `M` also cuts `BC` at `P` as shown in the diagram.
Copy or trace the diagram into your writing booklet.
| (i) |
 |
`(AK)/(KB) = (AM)/(MC) = 1/1`
| `:. KM\ text(||)\ BC` | `\ \ \ text{(parallel lines cut in the}` `\ \ \ \ text{same proportion)` |
`text(Similarly,)\ \ (CL)/(LB) = (CM)/(MA) = 1/1`
`:. ML\ text(||)\ AB`
`:. KMLB\ \ text(is a parallelogram)`
| (ii) | `/_ BPK` | `= /_ KML` | `text{(exterior angle of a cyclic}` `text{quadrilateral}\ \ KMLP text{)}` |
| (iii) | `/_ KBP` | `= /_ KML\ \ \ text{(opposite angles of a parallelogram)}` |
| `:. /_ KBP` | `= /_ KPB\ \ \ text{(both equal}\ \ /_ KML text{)}` |
`:. Delta BKP\ \ text(is isosceles)`
`text(S)text(ince)\ \ BK = KP=KA\ \ \ text{(given}\ K\ \ text(is the midpoint of)\ \ ABtext{)}`
`=>K\ \ text(is the centre of a circle, diameter)\ \ AB,`
`text(that passes through)\ \ A, B and P`
`/_ APB = 90^@\ \ \ \ text{(angle in semi-circle)}`
`:. AP _|_ BC.`
The diagram shows a regular `n`-sided polygon with vertices `X_1, X_2, …, X_n`. Each side has unit length. The length `d_k` of the ‘diagonal’ `X_n X_k` where `k = 1, 2, …, n - 1` is given by
`d_k = (sin\ (k pi)/n)/(sin\ pi/n).` (Do NOT prove this.)
--- 5 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
Using part (i), or otherwise, show that
`int_0^a f(x)\ dx = a/2\ f(a).` (2 marks)
--- 8 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
Show that the graph of `y = f(x)` is concave up for `x > 0.` (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
Without evaluating the integrals, which one of the following integrals is greater than zero?
Which integral is necessarily equal to `int_(−a)^a f(x)\ dx`?
A hostel has four vacant rooms. Each room can accommodate a maximum of four people.
In how many different ways can six people be accommodated in the four rooms?
A game is being played by `n` people, `A_1, A_2, ..., A_n`, sitting around a table. Each person has a card with their own name on it and all the cards are placed in a box in the middle of the table. Each person in turn, starting with `A_1`, draws a card at random from the box. If the person draws their own card, they win the game and the game ends. Otherwise, the card is returned to the box and the next person draws a card at random. The game continues until someone wins.
Let `W` be the probability that `A_1` wins the game.
Let `p = 1/n and q = 1 - 1/n`.
--- 5 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
`qquad tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + ….` (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
Let `z = cos theta + i sin theta.`
--- 6 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
The diagram shows a circle of radius `r`, centred at the origin, `O`. The line `PQ` is tangent to the circle at `Q`, the line `PR` is horizontal, and `R` lies on the line `x = c`.
Let
`A_n = int_0^(pi/2) cos^(2n) x\ dx` and `B_n = int_0^(pi/2) x^2cos^(2n)x\ dx`,
where `n` is an integer, `n ≥ 0`. (Note that `A_n > 0`, `B_n > 0`.)
Let `P(x) = (n − 1)x^n − nx^(n − 1) + 1`, where `n` is an odd integer, `n ≥ 3`.
| (i) | `P′(x)` | `= n(n − 1)x^(n − 1) − n(n − 1)x^(n − 2)` |
| `= n(n − 1)x^(n − 2)(x − 1)` |
`P′(x) = 0\ \ text(when)\ \ x = 0\ \ text(or)\ \ x = 1.`
`:.P(x)\ text(has exactly two stationary points.)`
| (ii) | `P(1)` | `= (n − 1) xx 1 − n xx 1 + 1 = 0, and` |
| `P′(1) ` | `= 0` |
`:. P(x)\ text(has a double zero at)\ \ x=1`
(iii) `P(x)\ text{has exactly two stationary points at}\ \ x=0 and 1`
`text(S)text(ince)\ \ P(0) = 1 and P(1)=0`
`=>\ text{A local maximum occurs at (0,1)}`
| `text(Noting that)\ \ P(x)` | `-> oo\ \ text(as)\ \ x-> oo, and` |
| `P(x)` | `-> -oo\ \ text(as)\ \ x-> -oo` |
`:.\ text(The double zero at)\ \ x=1,\ text{means that}\ \ P(x)`
`text(has exactly one real zero other than)\ \ x=1.`
| (iv) | `P(-1)` | `=(n-1)(-1)^n-n xx (-1)^(n-1) + 1` |
| `=(n-1)(-1) – nxx1 +1\ \ \ \ text{(given}\ n\ text{is odd)}` | ||
| `=-2n+2` | ||
| `<0\ \ \ text{(for odd}\ n>=3text{)}` |
| `P(-1/2)` | `= (n − 1)(-1/2)^n − n(-1/2)^(n − 1)+1` |
| `= -(n − 1) xx 1/(2^n) − n/(2^(n − 1)) + 1\ \ \ \ text{(given}\ n\ text{is odd)}` | |
| `= (-n + 1 − 2n + 2^n)/(2^n)` | |
| `= (2^n − (3n − 1))/(2^n)` | |
| `>=0\ \ \ \ \ text{(given}\ \ 2^n>=3n-1\ \ text{for}\ \ n>=3, and 2^n > 0)` |
`text(S)text(ince)\ \ P(x)\ \ text(is continuous, when it changes sign, it cuts the)\ x text(-axis)`
`:. -1 < α ≤ -1/2`
(v) `P(x) = 4x^5 − 5x^4 + 1\ \ text(is of the form)`
`P(x) = (n − 1)x^n − nx^(n − 1) + 1`
`:.P(x)\ \ text(has 5 zeros)\ \=> 1, 1, alpha, beta, bar beta\ \ \ \ (text(where)\ beta\ text{is not real})`
`text(The zeros 1 and α have a modulus ≤ 1.)`
`text(Consider the product of the roots)`
| `1*1*alpha*beta*bar beta` | `=- 1/4` |
| `alpha*beta*bar beta` | `=- 1/4` |
| `|\ alpha*beta*bar beta\ |` | `=|\ – 1/4\ |` |
| `|\ alpha\ |*|\ beta\ |^2` | `= 1/4` |
`text(S)text(ince)\ \ |\ α\ | > 1/2,\ \ |\ beta\ |<1`
`:.text(Each of the zeros of)\ x^5 − 5x^4 + 1\ text(has a modulus)\ <=1.`
In the diagram `ABCD` is a cyclic quadrilateral. The point `K` is on `AC` such that `∠ADK = ∠CDB`, and hence `ΔADK` is similar to `ΔBDC`.
Copy or trace the diagram into your writing booklet.
| (i) |  |
| `text(S)text(ince)\ \ ∠ADB` | `=∠ADK +∠KDB, and` |
| `∠KDC` | `=∠CDB +∠KDB` |
| `=>∠ADB` | `= ∠KDC` |
| `∠ABD` | `= ∠ACD\ \ \ text{(angles in the same segment on arc}\ ADtext{)}` |
| `∠ADK` | `= ∠CDB\ \ \ ` | `text{(corresponding angles of similar}` |
| `text{triangles,}\ ΔADK\ text(|||)\ ΔBDC text{)}` |
`:.ΔADB\ text(|||)\ ΔKDC\ \ text{(equiangular)}`
(ii) `text(S)text(ince)\ ΔADB\ text(|||)\ ΔKDC`
`(BD)/(DC) = (AD)/(DK) = (AB)/(KC)\ \ text{(corresponding sides of similar triangles)}`
`:.AB xx DC = KC xx BD\ \ \ \ …\ (1)`
`text(Similarly, using)\ \ ΔADK\ text(|||)\ ΔBDC\ \ text{(given)}`
`(AD)/(BD)=(AK)/(BC)`
`:. BC xx AD = AK xx BD\ \ \ \ …\ (2)`
`text{Add (1) + (2)}`
| `AB xx DC + AD xx BC` | `=KC xx BD+AK xx BD` |
| `=BD(AK+KC)` | |
| `:.BD xx AC` | `=AD xx BC+AB xx DC` |
(iii) `text(Each diagonal of the pentagon is)\ x.`
`text{Hence in applying the result in (i) you have}`
`BD = AC = x, AD = DC = CB = 1, BA = x`
| `x xx x` | `= 1 xx 1 + x xx 1` |
| `x^2 − x − 1` | `= 0` |
| `x` | `= (1 ± sqrt(1 + 4))/2` |
| `= (1 ± sqrt(5))/2` | |
| `:.x` | `= (1 + sqrt(5))/2,\ \ \ (x>0)` |
A TV channel has estimated that if it spends `$x` on advertising a particular program it will attract a proportion `y(x)` of the potential audience for the program, where
`(dy)/(dx) = ay(1 − y)`
and `a > 0` is a given constant.
(i) `(dy)/(dx) = ay(1 − y)`
`=>text(Given)\ \ a>0,\ \ ay(1 − y)\ text(is an inverted parabola)`
`text(with zeros at)\ \ y = 0\ \ text(and)\ \ y = 1`
`=>\ text(Parabola symmetry means that a maximum occurs when)`
`y = (0+1)/2=1/2`
`:.(dy)/(dx)\ \ text(has its maximum value when)\ y = 1/2.`
(ii) `dy/dx=ay(1 − y)`
| `int (dy)/(y(1 − y))` | `=int a\ dx` |
| `ax` | `= ln (y/(1 − y))+c` |
| `e^(ax − c)` | `= y/(1 − y)` |
| `e^(ax − c) − ye^(ax − c)` | `= y` |
| `y(1 + e^(ax − c))` | `= e^(ax − c)` |
| `y` | `= (e^(ax − c))/(1 + e^(ax − c))` |
| `y` | `= 1/(e^c e^(− ax) + 1)` |
`text(Let)\ k = e^c`
`:.y(x) = 1/(ke^(-ax) + 1)\ text(for some constant)\ \ k > 0.`
(iii) `text(When)\ \ x = 0, y = 0.1`
| `0.1` | `= 1/(ke^0 + 1)` |
| `k + 1` | `= 10` |
| `k` | `= 9` |
(iv) `text(The gradient the curve is greatest at)\ y = 1/2\ \ \ text{from part (i)}`
`:. text(A point of inflection occurs at)\ y =1/2.`
(v) `text(As)\ \ x->oo,\ \ y->1/(0+1)=1`
Let `k` be a real number, `k ≥ 4`.
Show that, for every positive real number `b`, there is a positive real number `a` such that `1/a + 1/b = k/(a + b)`. (3 marks)
The diagram shows the rectangular hyperbola `xy = c^2`, with `c > 0`.
The points `A(c, c)`, `R(ct, c/t)` and `Q(-ct, -c/t)` are points on the hyperbola,
with `t ≠ ±1`.
| (i) |
|
| `m_(QA)` | `= (c + c/t)/(c + ct)` |
| `= (1 + 1/t)/(1 + t)` | |
| `= (t + 1)/(t(1 + t))` | |
| `= 1/t` |
`:.\ text(Gradient of perpendicular to)\ QA\ \ (l_1) = -t`
`:.text(Equation of)\ \ l_1`
| `y − c/t` | `= -t(x − ct)` |
| `y` | `= -tx + ct^2 + c/t` |
| `= -tx + c(t^2 + 1/t)\ \ \ …\ (1)` |
| (ii) | `m_(RA)` | `= (c − c/t)/(c − ct)` |
| `= (1 − 1/t)/(1 − t)` | ||
| `= (t − 1)/(t(1 − t))` | ||
| `= -1/t` |
`:.m\ text(of)\ l_2 = t`
`:.\ text(Equation of)\ l_2`
| `y + c/t` | `= t(x + ct)` |
| `y` | `= tx + ct^2 – c/t` |
| `= tx + c(t^2 – 1/t)\ \ \ …\ (2)` |
(iii) `text{Solving (1) and (2) simultaneously}`
| `-tx + c(t^2 + 1/t)` | `= tx + c(t^2 − 1/t)` |
| `ct^2+c/t-ct^2+c/t` | `=2tx` |
| `(2c)/t` | `= 2tx` |
| `x` | `= c/(t^2)` |
`text{Substitute into (2)}`
| `y` | `= (ct)/(t^2) + ct^2 − c/t` |
| ` = ct^2` |
`:.P\ text(has coordinates)\ (c/(t^2), ct^2)`
(iv) `text(Using the coordinates of)\ P`
`=>t^2 = c/x,\ \ t^2 = y/c`
| `y/c` | `= c/x` |
| `:.xy` | `=c^2` |
`text(S)text(ince the parameter of)\ P\ text(is)\ t^2`
`=>\ text(the locus is in the first quadrant.)`
`:.\ text(The locus of)\ P\ text(is the branch of the rectangular hyperbola)`
`xy = c^2\ text{in the first quadrant (excluding}\ A, t≠±1 text{)}`
Two identical biased coins are each more likely to land showing heads than showing tails.
The two coins are tossed together, and the outcome is recorded. After a large number of trials it is observed that the probability that the two coins land showing a head and a tail is `0.48`.
What is the probability that both coins land showing heads? (2 marks)
Let `beta` be a root of the complex monic polynomial
`P(z) = z^n + a_(n - 1) z^(n - 1) + … + a_1z + a_0.`
Let `M` be the maximum value of `|\ a_(n - 1)\ |,\ |\ a_(n - 2)\ |,\ …\ ,\ |\ a_0\ |.`
(i) `P(z) = z^n + a_(n – 1) z^(n – 1) + … + a_1z + a_0.`
`M = text(Max)\ (\ |\ a_(n – 1)\ |, |\ a_(n – 2)\ |, … , |\ a_0|\ )`
`P(beta) = 0`
| `0` | `=beta^n + a_(n – 1) beta^(n – 1) + … + a_1 beta + a_0` |
| `beta^n` | `= -(a_(n – 1) beta^(n – 1) + … + a_1 beta + a_0)` |
| `|\ beta^n\ |` | `=|\ a_(n – 1) beta^(n – 1) + … + a_1 beta + a_0\ |` |
| `<= |\ a_(n – 1) beta^(n – 1)\ | + … + |\ a_1 beta\ | + |\ a_0\ |` | |
| `|\ beta\ |^n` | `<= |\ a_(n – 1)\ ||\ beta\ |^(n – 1) + … + |\ a_1\ ||\ beta\ | + |\ a_0\ |` |
| `<= M (|\ beta\ |^(n – 1) + … + |\ beta\ | + 1)` |
(ii) `(1 + |\ beta\ | + … + |\ beta\ |^(n – 1))`
`=>text(GP where)\ \ a = 1,\ \ r = |\ beta\ |, and n\ text(terms)`
`text(Consider)\ |\ beta\ | > 1`
`1 + |\ beta\ | + … + |\ beta\ |^(n – 1) = (1(|\ beta\ |^n – 1))/(|\ beta\ | – 1)`
`text(Using)\ \ |\ beta\ |^n <= M(|\ beta\ |^(n – 1) + … + |\ beta\ | + 1)\ \ \ text{(part (i))}`
| `|\ beta\ |^n` | `<= M((|\ beta\ |^n – 1))/(|\ beta\ | – 1)` |
| `|\ beta\ |^n (|\ beta\ | – 1)` | `<= M(|\ beta\ |^n – 1)` |
| `|\ beta\ | – 1` | `<= M ((|\ beta\ |^n – 1))/|\ beta\ |^n` |
| `|\ beta\ |` | `<= 1 + M (1 – 1/|\ beta\ |^n)` |
| `|\ beta\ |` | `< 1 + M` |
`text(Consider)\ |\ beta\ | <= 1`
`|\ beta\ | < 1 + M,\ \ text(as)\ \ M > 0.`
`:.|\ beta\ | < 1 + M\ \ text(for any root)\ \ beta`
| (iii) `S(x)` | `=sum_(k=0)^n c_k(x + 1/x)^k` |
| `=c_n(x + 1/x)^n + c_(n – 1)(x + 1/x)^(n – 1) + … + c_1(x + 1/x) + c_0` | |
| `(S(x))/c_n` | `=(x + 1/x)^n + (c_(n-1))/c_n (x + 1/x)^(n – 1) + … + c_1/c_n(x + 1/x) + c_0/c_n` |
`text(This is a monic polynomial of the form in part (i))`
`text(Let)\ \ z = x + 1/x`
`P(z) = z^n + a_(n – 1) z^(n – 1) + … + a_1 z + a_0`
`text(Consider the coefficients)\ \ a_k = c_k/c_n`
`text(S)text(ince)\ \ |\ c_k\ |<=|\ c_n\ |,\ \ \ text{(given)}`
| `=>|\ a_k\ | <=` | `(|\ c_k\ |)/|\ c_n\ | <= 1\ \ text(from above)` |
| `=>M <=` | `1\ \ \ \ text{(since}\ M\ text(is the max of)\ |\ a_k\ | text{)}` |
`text(If)\ \ beta\ \ text(is a real root of)\ \ P(z) and |\ beta\ | < 1 + M,\ \ text(then)`
`=>|\ beta\ | < 2`
| `text(However,)\ \ beta` | `= x + 1/x` |
| `:. |\ beta\ |` | `= |\ x + 1/x\ |` |
| `= |\ x\ | + 1/|\ x\ |\ \ \ \ text{(for}\ x\ text{real)}` | |
| `>= 2` |
`:.text(We have)\ \ |\ beta\ | < 2 and |\ beta\ | = |\ x + 1/x\ | >= 2`
`text(This is a contradiction and hence our assumption that)`
`beta\ \ text(is a real root of)\ \ S(x)\ \ text(is incorrect.)`
`text(Alternative strategy to show a contradiction)`
`text(Graphing)\ \ y = x + 1/x`
`text(S)text(ince the turning points occur at)\ \ y=+- 2`
`=>|\ x + 1/x\ | >= 2.`
`:.S(x)\ \ text(has no real solutions.)`
A bag contains seven balls numbered from `1` to `7`. A ball is chosen at random and its number is noted. The ball is then returned to the bag. This is done a total of seven times.
For every integer `m >= 0` let
`I_m = int_0^1 x^m (x^2 - 1)^5\ dx.`
Prove that for `m >= 2`
`I_m = (m - 1)/(m + 11) \ I_(m - 2).` (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
On an Argand diagram, sketch the region described by the inequality
`|\ 1 + 1/z\ | <= 1.` (2 marks)
| `|\ 1 + 1/z\ |<= ` | `1` |
| `|\ (z + 1)/z\ |<= ` | `1` |
| `|\ z + 1\ |/|\ z\ |<= ` | `1` |
| `|\ z + 1\ |<= ` | `|\ z\ |` |
| `sqrt ((x + 1)^2 + y^2) <=` | `sqrt (x^2 + y^2)` |
| `(x + 1)^2 + y^2 <=` | `x^2 + y^2` |
| `x^2 + 2x + 1 <=` | ` x^2` |
| `:.x <=` | `-1/2` |
Let `f (x)` be a function with a continuous derivative.
(i) `text(If)\ \ f(x)\ \ text(is a function with a continuous derivative,)`
`=> f prime (x)\ \ text(exists for all)\ \ x.`
| `y` | `= (f(x))^3` |
| `(dy)/(dx)` | `= 3 xx (f(x))^2 xx f prime(x)` |
`:.\ (dy)/(dx) = 0\ \ text(when)\ \ f(x) = 0 or f prime (x) = 0`
`:.\ x = a\ \ text(is a stationary point if)\ \ f(a) = 0 or f prime(a) = 0`
(ii) `text(If)\ \ f prime (a) != 0\ \ text(then either)\ \ f prime (a) > 0 or f prime (a) < 0,`
`and f prime (x)\ \ text(keeps that sign either side of)\ \ x = a.`
`:. text(If)\ \ f(a) = 0 and f prime(a) != 0, text(there is a stationary point at)\ \ x = a`
`text(where the slope of the curve does not change either side of)\ \ x = a.`
`text(i.e. a horizontal point of inflection occurs at)\ \ x=a`
(iii) `y = (f(x))^3`
`text(When)\ \ f(x) = 0,\ \ (f(x))^3 = 0\ \ text{(horizontal P.I. from part (ii))}`
`text(When)\ \ f(x) = 1,\ \ (f(x))^3 = 1`
`text(If)\ \ 0 < f(x) < 1,\ \ 0 < (f(x))^3 < f(x)`
`text(If)\ \ f(x) > 1,\ \ (f(x))^3 > f(x)`
`text(When)\ f prime (a) = 0,\ \ (f(x))^3\ \ text(has a maximum turning point)`
`f(x) < 0,\ \ (f(x))^3 < 0`
Jac jumps out of an aeroplane and falls vertically. His velocity at time `t` after his parachute is opened is given by `v(t)`, where `v(0) = v_0` and `v(t)` is positive in the downwards direction. The magnitude of the resistive force provided by the parachute is `kv^2`, where `k` is a positive constant. Let `m` be Jac’s mass and `g` the acceleration due to gravity. Jac’s terminal velocity with the parachute open is `v_T.`
Jac’s equation of motion with the parachute open is
`m (dv)/(dt) = mg - kv^2.` (Do NOT prove this.)
--- 3 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
Jac opens his parachute when his speed is `1/3 v_T.` Gil opens her parachute when her speed is `3v_T.` Jac’s speed increases and Gil’s speed decreases, both towards `v_T.`
Show that in the time taken for Jac's speed to double, Gil's speed has halved. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
The diagram shows the ellipse `x^2/a^2 + y^2/b^2 = 1`, where `a > b`. The line `l` is the tangent to the ellipse at the point `P`. The foci of the ellipse are `S` and `S prime`. The perpendicular to `l` through `S` meets `l` at the point `Q`. The lines `SQ` and `S prime P` meet at the point `R`.
Copy or trace the diagram into your writing booklet.
| (i) |
 |
`text(Let)\ \ l\ \ text(cut the)\ \ y text(-axis at)\ \ M`
| `/_ MPS prime` | `= /_ QPS\ \ \ \ \ text{(reflection property of an ellipse)}` |
| `/_ RPQ` | `=/_ MPS prime\ \ \ \ \ text{(vertically opposite angles)}` |
`text(In)\ \ Delta SPQ and Delta RPQ`
`/_ SPQ = /_ RPQ\ \ \ text{(both equal}\ \ /_ MPS prime text{)}`
`/_ SQP = /_ RQP=90^@\ \ \ text{(given that}\ \ PQ⊥SRtext{)}`
`PQ\ \ text(is a common side)`
`:.\ Delta SPQ -= Delta RPQ\ \ \ \ text{(AAS)}`
`:. SQ = RQ\ \ \ text{(corresponding sides of congruent triangles)}`
(ii) `SP = PR\ \ \ text{(corresponding sides of congruent triangles)}`
| `S prime P+ PS` | `= 2a\ \ \ \ text{(locus property of an ellipse)` |
| `:.S prime P+PR` | `= 2a` |
| `:.S prime R` | `= 2a` |
(iii) `text(Join)\ \ QO`
`text(Consider)\ \ Delta SS prime R and Delta SOQ`
`(SO)/(SS prime) = 1/2\ \ \ \ text{(}O\ \ text(is the midpoint of)\ \ SS prime text{)}`
`(SQ)/(SR) = 1/2\ \ \ \ text{(}Q\ \ text(is the midpoint of)\ \ SR prime text{)}`
`/_ S prime SR = /_ OSQ\ \ text{(common angle)}`
| `:.\ Delta SS prime R\ text(|||)\ Delta SOQ` | `\ \ \ text{(AAS)}` |
| `(OQ)/(S prime R)` | `= 1/2` | `\ \ \ text{(corresponding sides of}` |
| `\ \ \ text{similar triangles)}` | ||
| `(OQ)/(2a)` | `=1/2` | |
| `:.\ OQ` | `=a` |
`:.\ Q\ \ text(lies on the circle)\ \ x^2 + y^2 = a^2`
A mass is attached to a spring and moves in a resistive medium. The motion of the mass satisfies the differential equation
`(d^2y)/(dt^2) + 3 (dy)/(dt) + 2y = 0,`
where `y` is the displacement of the mass at time `t.`
The equation `x^2/16 - y^2/9 = 1` represents a hyperbola.
(i) `x^2/16 – y^2/9 = 1,\ \ \ =>a^2 = 16,\ \ \ b^2 = 9`
| `b^2` | `= a^2 (e^2 – 1)` |
| `9` | `= 16 (e^2 – 1)` |
| `e^2` | `= 1 + 9/16 ` |
| `= 25/16` | |
| `:. e` | `= 5/4` |
(ii) `S (ae, 0),\ \ S prime (-ae, 0)`
`S(5, 0),\ \ S prime (–5, 0)`
(iii) `y = +- b/a x`
`y = +- 3/4 x`
| (iv) | |
(v) `text(As)\ \ e -> oo,\ \ b -> oo\ \ text(and the asymptotes approach the)`
`y text{-axis from both sides (i.e. the gradients of the asymptotes →∞)}`
` text(Thus the hyperbola approaches the)\ \ y text(-axis from both sides.)`
Let `n` be an integer where `n > 1`. Integers from `1` to `n` inclusive are selected randomly one by one with repetition being possible. Let `P(k)` be the probability that exactly `k` different integers are selected before one of them is selected for the second time, where `1 ≤ k ≤ n`.
--- 6 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
You may use part (iii) and also that `k^2 − k − n >0` if `P(k)< P(k − 1)`. (2 marks)
--- 6 WORK AREA LINES (style=lined) ---
(i) `((m + n)!)/(m!n!)\ text{ways (By definition)}`
(ii) `text{Consider this as being “arrange 10 coins in 4 boxes}`
`text{with 3 separators between the boxes, making a total}`
`text{of 13 items” (as per the diagram).}`
`:.\ text(10 identical coins and 3 identical separators to arrange.)`
`:.\ text(Number of ways) = (13!)/(10!3!)\ \ \ \ text{(from part (i))}`
Let `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`, where `k` is real.
Let `α = x + iy`, where `x` and `y` are real.
Suppose that `α` and `iα` are zeros of `P(z)`, where `bar α ≠ iα`.
--- 3 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
--- 3 WORK AREA LINES (style=lined) ---
The solid `ABCD` is cut from a quarter cylinder of radius `r` as shown. Its base is an isosceles triangle `ABC` with `AB = AC`. The length of `BC` is `a` and the midpoint of `BC` is `X`.
The cross-sections perpendicular to `AX` are rectangles. A typical cross-section is shown shaded in the diagram.
Find the volume of the solid `ABCD`. (4 marks)
`text(Need to find the width of the rectangular cross-section)`
`text(Using similar triangles)`
| `(AF)/(AX)` | `= (DF)/(BX)` |
| `h/r` | `= (DF)/(a/2)` |
| `DF` | `= (ah)/(2r)` |
| `:.DE` | `= (ah)/r` |
`text(Need to find the height of the cross-section)`
| `JF` | `=sqrt((AJ)^2-(AF)^2)` |
| ` = sqrt(r^2 − h^2)` |
`=>text(Rectangle has base length)\ (ah)/r\ text(and height)\ sqrt(r^2 − h^2)`
| `text(Volume of solid)` | `=lim_(δh→0) ∑_(h=0)^r (ah)/r sqrt(r^2 − h^2)\ δh` |
| `= int_0^r (ah)/r sqrt(r^2 − h^2)\ dh` | |
| `= a/r int_0^r h sqrt(r^2 − h^2)\ dh` |
| `text(Let)\ \ u^2` | `=r^2-h^2` |
| `2u\ du` | `=-2h\ dh` |
| `-u\ du` | `=h\ dh` |
`text(If)\ \ h=r,\ \ u^2=0,\ \ u=0`
`text(If)\ \ h=0,\ \ u^2=r^2,\ \ u=r\ \ (r>0)`
| `:.\ text(Volume of solid)` | `=a/r int_r^0 u xx -u\ du` |
| `=-a/r [u^3/3]_r^0` | |
| `=-a/r((-r^3)/3)` | |
| `=(ar^2)/3\ \ text(u³)` |
A small bead `P` of mass `m` can freely move along a string. The ends of the string are attached to fixed points `S` and `S′`, where `S′` lies vertically above `S`. The bead undergoes uniform circular motion with radius `r` and constant angular velocity `omega` in a horizontal plane.
The forces acting on the bead are the gravitational force and the tension forces along the string. The tension forces along `PS` and `PS′` have the same magnitude `T`.
The length of the string is `2a` and `SS′= 2ae`, where `0 < e < 1`. The horizontal plane through `P` meets `SS′` at `Q`. The midpoint of `SS′` is `O` and `beta = /_S′PQ`. The parameter `theta` is chosen so that `OQ =a cos theta.`
(i) `SP + S′P = 2a and SS prime = 2ae,\ \ 0 < e < 1`
`text(This is the condition for an ellipse with foci)`
`text(at)\ \S and S′ text(and eccentricity)\ e.`
| (ii) | `SP` | `= ePM\ \ \ \ \ text{(where}\ M\ text{is on the directrix)}` |
| `=e(OM-OQ)` | ||
| `= e (a/e – a cos theta)` | ||
| `= a – a e cos theta` | ||
| `= a(1 – e cos theta)` |
| (iii) `S prime P` | `= 2a – (a – a e cos theta)` |
| `= a + a e cos theta` |
| `sin beta` | `= (QS prime)/(S prime P)` |
| `= (ae + a cos theta)/(a + a e cos theta)` | |
| `= (e + cos theta)/(1 + e cos theta)` |
| (iv) |  |
`T sin beta = T ((e + cos theta)/(1 + e cos theta))`
| `sin /_ QPS` | `= (QS)/(SP)` |
| `= (ae – a cos theta)/(a (1 – e cos theta)` | |
| `= (e – cos theta)/(1 – e cos theta)` |
`text(Resolving the forces vertically)`
| `mg` | `= T sin beta – T sin /_ QPS` |
| `mg` | `= T ((e + cos theta)/(1 + e cos theta) – (e – cos theta)/(1 – e cos theta))` |
| `mg` | `= T ((e-e^2 cos theta + cos theta – e cos^2 theta-(e + e^2 cos theta – cos theta – e cos^2 theta))/(1 – e^2 cos^2 theta))` |
| `mg` | `= T((-2e^2 cos theta + 2 cos theta)/(1 – e^2 cos^2 theta))` |
| `mg` | `= (2T(1 – e^2) cos theta)/(1 – e^2 cos^2 theta)` |
(v) `text(Resolving the forces horizontally)`
`mr omega^2= T cos beta + T cos /_ QPS`
| `cos beta` | `= r/(S prime P)` |
| `= r/(2a – SP)` | |
| `= r/(2a – (a – a e cos theta))` | |
| `= r/(a (1 + e cos theta))` |
| `cos /_ QPS` | `= r/(SP)` |
| `= r/(a (1 – e cos theta))` |
| `r` | `= sqrt (SP^2 – SQ^2)` |
| `= sqrt (a^2 (1 – e cos theta)^2 – a^2 (e – cos theta)^2)` | |
| `= a sqrt (1 – 2e cos theta + e^2 cos ^2 theta – (e^2 – 2 e cos theta + cos^2 theta))` | |
| `= a sqrt (1 – e^2 – (1 – e^2) cos^2 theta)` | |
| `= a sqrt ((1 – e^2) (1 – cos^2 theta))` | |
| `= a sin theta sqrt (1 – e^2)` |
| `:. mr omega^2` | `= T ((a sin theta sqrt(1 – e^2))/(a(1 + e cos theta)) + (a sin theta sqrt(1 – e^2))/(a(1 – e cos theta)))` |
| `= T sin theta sqrt (1 – e^2) ((1 – e cos theta + 1 + e cos theta)/(1 – e^2 cos^2 theta))` | |
| `= (2T sqrt (1 – e^2) sin theta)/(1 – e^2 cos^2 theta)` |
| (vi) `cos theta` | `= (mg (1 – e^2 cos^2 theta))/(2T (1 – e^2))` |
| `sin theta` | `= (mr omega^2(1 – e^2 cos^2 theta))/(2T sqrt (1 – e^2))` |
| `tan theta` | `= (mr omega^2(1 – e^2 cos^2 theta))/(2T sqrt(1 – e^2)) xx (2T(1 – e^2))/(mg(1 – e^2 cos^2 theta))` |
| `= (r omega^2 sqrt (1 – e^2))/g` |
