Aussie Maths & Science Teachers: Save your time with SmarterEd
The volume of the solid of revolution formed by rotating the graph of `y = sqrt (9-(x-1)^2)` above the `x`-axis is given by
`y = sqrt (9-(x-1)^2)\ \ => text(Semi-circle of circle)`♦ Mean mark 48%.
`text{with centre (1,0), radius = 3}`
| `:. V` | `= 4/3 pi r^3` |
| `= 4/3 pi (3)^3` | |
| `=4 pi (3)^2` |
`=> A`
On the argand diagram below, the twelve points `P_1, P_2, P_3 …, P_12` are evenly spaced around the circle of radius 3.
The points which represent complex numbers such that `z^3 = -27i` are
A. `P_10` only
B. `P_4` only
C. `P_2, P_6, P_10`
D. `P_3, P_7, P_11`
E. `P_4, P_8, P_12`
The number of straight line asymptotes of the graph of `y = (2x^3 + x^2 - 1)/(x^2 - x - 2)` is
A. 0
B. 1
C. 2
D. 3
E. 4
Find the coordinates of the points of intersection of the graph of the relation
`y = text(cosec)^2 ((pi x)/6)` with the line `y = 4/3`, for `0 < x < 12.` (3 marks)
For any complex number `z`, the location on an Argand diagram of the complex number `u = i^3 bar z` can be found by
A. rotating `z` through `(3 pi)/2` in an anticlockwise direction about the origin
B. reflecting `z` about the `x`-axis and then reflecting about the `y`-axis
C. reflecting `z` about the `y`-axis and then rotating anticlockwise through `pi/2` about the origin
D. reflecting `z` about the `x`-axis and then rotating anticlockwise through `pi/2` about the origin
E. rotating `z` through `(3 pi)/2` in a clockwise direction about the origin
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).`
Give your answer in the form `(a sqrt b)/c, \ a, b, c in Z.` (3 marks)
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 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.`
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` |
Consider the equation `z^3 - z^2 - 2z - 12 = 0, \ z in C.`
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. |  |
Find all real solutions of the equation `2 cos(x) = sqrt 3 cot (x).` (3 marks)
A parachutist of mass `m` kg falls towards Earth and is slowed by air resistance of `kv^2` newtons, where `v` is the velocity of the parachutist `t` seconds after commencing the fall.
The equation of motion for the parachutist is
A. `(dv)/(dt) = g - kv^2`
B. `m(d^2x)/(dt^2) = g - kv^2`
C. `m(d(v^2))/(dx) = (mg - kv^2)/2`
D. `v(dv)/(dx) = 1 - (kv^2)/(mg)`
E. `(dv)/(dx) = g/v - (kv)/m`
The graphs of `y = ax` and `y = arctan(bx)` intersect exactly three times if
A. `0 < b < a`
B. `a < b < 0`
C. `a = b`
D. `b < a < 0`
E. `0 < b^2 < a^2`
The graph of `y = 1/(ax^2 + bx + c)` has asymptotes at `x = − 5`, `x = 3` and `y = 0`.
Given that the graph has one stationary point with a `y`-coordinate of `−1/8`, it follows that
A. `a = 1`, `b = 2`, `c = −15`
B. `a = 1/2`, `b = −1`, `c = −15/2`
C. `a = −1/2`, `b = −1`, `c = 15`
D. `a = −1`, `b = −2`, `c = −15`
E. `a = 1/2`, `b = 1`, `c = −15/2`
Find all solutions of `z^4 - 2z^2 + 4 = 0,\ \ z in C` in cartesian form. (4 marks)
The position vector `underset ~r (t)` of a particle moving relative to an origin `O` at time `t` seconds is given by
`underset ~r(t) = 4 sec (t) underset ~i + 2 tan (t) underset ~j,\ t in [0, pi/2)`
where the components are measured in metres.
| a. | `x` | `= 4sec(t)` |
| `x/4` | `= sec(t)` | |
| `y` | `= 2tan(t)` | |
| `y/2` | `= tan(t)` |
`text(Using)\ \ tan^2 (t) + 1 = sec^2(t),`
| `(x^2)/16 +1 ` | `= y^2/4` |
| `:. (x^2)/16 – (y^2)/4` | `=1` |
b. `y^2/4 = x^2/16 -1\ \ =>\ \ y=+- sqrt(x^2/4 -4)`♦ Mean mark part (b) 45%.
`lim_(x->oo) y = +- x/2`
♦ Mean mark part (c) 39%.
| c. | `overset·underset~r(t)` | `= d/(dt)(4(cos(t))^(−1))underset~i + d/(dt)(2tan(t)) underset~j` |
| `= 4(−1)(−sin(t))(cos(t))^(−2)underset~i + 2sec^2(t)underset~j` | ||
| `= 4sin(t)sec^2(t)underset~i + 2sec^2(t)underset~i` |
| `|overset·underset~r(pi/4)|` | `= sqrt(16sin^2(pi/4)sec^4(pi/4) + 4sec^4(pi/4))` |
| `= sqrt(16(1/sqrt2)^2(sqrt2)^4 + 4(sqrt2)^4)` | |
| `= sqrt(16(1/2)(4) + 4(4))` | |
| `= sqrt(48)` | |
| `= 4sqrt3\ \ text(ms)^(-1)` |
The velocity–time graph below shows the motion of a body travelling in a straight line, where `v\ text(ms)^(−1)` is its velocity after `t` seconds.
The velocity of the body over the time interval `t in [4,9]` is given by `v(t) = −9/16(t - 4)^2 + 9`.
The distance, in metres, travelled by the body over nine seconds is closest to
A. 45.6
B. 47.5
C. 48.6
D. 51.0
E. 53.4
The velocity vector of a 5 kg mass moving in the cartesian plane is given by `underset ~v(t) = 3 sin(2t) underset ~i + 4 cos(2t) underset ~j`, where velocity components are measured in m/s.
During its motion, the maximum magnitude of the net force, in newtons, acting on the mass is
A body on a horizontal smooth plane is acted upon by four forces, `underset ~F_1`, `underset ~F_2`, `underset ~F_3` and `underset ~F_4` as shown.
The force `underset ~F_1` acts in a northerly direction and the force `underset ~F_4` acts in a westerly direction.
Given that `|\ underset ~F_1\ | = 1`, `|\ underset ~F_2\ | = 2`, `|\ underset ~F_3\ | = 4` and `|\ underset ~F_4\ | = 5`, the motion of the body is such that it
A. is in equilibrium.
B. moves to the west.
C. moves to the north.
D. moves in the direction 30° south of west.
E. moves to the east.
`text(Resolving forces horizontally:)`♦ Mean mark 45%.
| `sum F` | `= underset ~F_2 cos30 + underset ~F_3 sin60 – underset ~F_4` |
| `= 2 xx sqrt3/2 + 4 xx sqrt3/2 -5` | |
| `~~0.2\ \ text(East)` |
`text(Resolving forces vertically:)`
| `sum F` | `= underset ~F_1 + underset ~F_2 sin30 – underset ~F_3 cos 60` |
| `= 1 + 2 xx 1/2 -4 xx 1/2` | |
| `= 0` |
`:.\ text(Body moves to the east.)`
`=> E`
A ball is thrown vertically up with an initial velocity of `7sqrt6` ms`\ ^(–1)`, and is subject to gravity and air resistance.
The acceleration of the ball is given by `ddotx = −(9.8 + 0.1v^2)`, where `x` metres is its vertical displacement, and `v` ms`\ ^(–1)` is its velocity at time `t` seconds.
The time taken for the ball to reach its maximum height is
A. `pi/3`
B. `(5pi)/(21sqrt2)`
C. `log_e(4)`
D. `(10pi)/(21sqrt2)`
E. `10log_e(4)`
Points `A`, `B` and `C` have position vectors `underset~a = 2underset~i + underset~j`, `underset~b = 3underset~i - underset~j + underset~k` and `underset~c = -3underset~j + underset~k` respectively.
The cosine of angle `ABC` is equal to
The component of the force `underset~F = aunderset~i + bunderset~j`, where `a` and `b` are non-zero real constants, in the direction of the vector `underset~w = underset~i + underset~j`, is
A. `((a + b)/2)underset~w`
B. `underset~F/(a + b)`
C. `((a + b)/(a^2 + b^2))underset~F`
D. `(a + b)underset~w`
E. `((a + b)/sqrt2)underset~w`
The direction field for a certain differential equation is shown above.
The solution curve to the differential equation that passes through the point `(–2.5, 1.5)` could also pass through
A. `(0, 2)`
B. `(1, 2)`
C. `(3, 1)`
D. `(3, –0.5)`
E. `(–0.5, 2)`
`text{Draw a graph that goes through (–2.5, 1.5) such that all}`♦ Mean mark 47%.
`text{gradient curve lines are tangential:}`
`(3, –0.5)`
`=> D`
A cricket ball is hit from the ground at an angle of 30° to the horizontal with a velocity of 20 ms¯¹. The ball is subject only to gravity and air resistance is negligible.
Given that the field is level, the horizontal distance travelled by the ball, in metres, to the point of impact is
A. `(10 sqrt 3)/g`
B. `20/g`
C. `(100 sqrt 3)/g`
D. `(200 sqrt 3)/g`
E. `400/g`
| `u_x` | `= 20 cos 30^@` |
| `= 10 sqrt 3` |
| `u_y` | `= 20 sin 30^@` |
| `= 10` |
| `h` | `= u_y t + 1/2 a_yt^2` |
| `= 10t – (g t^2)/2` |
`text(Find)\ \ t\ \ text(when)\ \ h=0:`
`0 = t(10 – (g t)/2)`♦ Mean mark 46%.
`t = 0, or t=20/g`
`text(Horizontal distance travelled:)`
| `x` | `= u_xt + 1/2 a_xt^2` |
| `= 10 sqrt 3 xx 20/g +0` | |
| `= (200 sqrt 3)/g` |
`=> D`
Two light strings of length 4 m and 3 m connect a mass to a horizontal bar, as shown below
The strings are attached to the horizontal bar 5 m apart.
Given the tension in the longer string is `T_1` and the tension in the shorter string is `T_2`, the ratio of the tensions `T_1/T_2` is
A. `3/5`
B. `3/4`
C. `4/5`
D. `5/4`
E. `4/3`
`text{Strings make up 3-4-5 (right-angled) triangle:}`♦ Mean mark 41%.
`text(Let)\ \ theta =\ text(angle between 3m string and horizontal bar)`
| `T_1/T_2` | `= cot theta` |
| `= 3/4` |
`=> B`
Given that `x = sin(t) - cos(t)` and `y = 1/2 sin(2t)`, then `(dy)/(dx)` in terms of `t` is
A. `cos(t) - sin(t)`
B. `cos(t) + sin(t)`
C. `sec(t) + text(cosec)(t)`
D. `sec(t) - text(cosec)(t)`
E. `(cos (2t))/(cos(t) - sin(t))`
Solve the differential equation `sqrt(2-x^2) (dy)/(dx) = 1/(2-y)`, given that `y(1) = 0`. Express `y` as a function of `x`. (5 marks)
--- 9 WORK AREA LINES (style=lined) ---
A farmer grows peaches, which are sold at a local market. The mass, in grams, of peaches produced on this farm is known to be normally distributed with a variance of 16. A bag of 25 peaches is found to have a total mass of 2625 g.
Based on this sample of 25 peaches, calculate an approximate 95% confidence interval for the mean mass of all peaches produced on this farm. Use an integer multiple of the standard deviation in your calculations. (3 marks)
A bank claims that the amount it lends for housing is normally distributed with a mean of $400 000 and a standard deviation of $30 000.
A consumer organisation believes that the average loan amount is higher than the bank claims.
To check this, the consumer organisation examines a random sample of 25 loans and finds the sample mean to be $412 000.
A 5 kg mass is initially held at rest on a smooth plane that is inclined at 30° to the horizontal. The mass is connected by a light inextensible string passing over a smooth pulley to a 3 kg mass, which in turn is connected to a 2 kg mass.
The 5 kg mass is released from rest and allowed to accelerate up the plane.
Take acceleration to be positive in the directions indicated.
`qquad` 
Find the magnitude of `R` in terms of `g`. (2 marks)
| a. | `2\ text(kg): \ 2a` | `= 2g – T_2` |
| `3\ text(kg): \ 3a` | `= 3g + T_2 – T_1` | |
| `5\ text(kg): \ 5a` | `= T_1 – 5g sin (30^@)` | |
| `5a` | `= T_1 – (5g)/2` |
b. `sum F = 3g + 2g – 5g sin 30^@ = (5 + 3 + 2)a`
| `5g – (5g)/2` | `= 10a` |
| `a` | `= (5g)/(2 xx 10)` |
| `:. a` | `= g/4\ text(ms)^(-2)` |
| c. ` T_1 – (5g)/2` | `= 5a` |
| `:. T_1` | `= (5g)/2 + 5(g/4)` |
| `= (15g)/4` |
| `2g – T_2` | `= 2a` |
| `:. T_2` | `= 2g – 2(g/4)` |
| `= (3g)/2` |
d. `u = 0,\ \ a = g/4,\ \ s = 2`
`text(Find)\ \ v\ \ text(when)\ \ s=2:`
| `v^2` | `= u^2 + 2as` | |
| `=0 + 2 (g/4) xx 2` | ||
| `=g` | ||
| `v` | `=sqrtg\ \ \ (v>0)` |
`:. p = 5 sqrt g`
e. `sum F = 2g + 3g – 5g sin 30^@ – R = 0`
| `:. R` | `= 2g + 3g – 5g sin 30^@` |
| `= 5g – (5g)/2` | |
| `= (5g)/2` |
A cricketer hits a ball at time `t = 0` seconds from an origin `O` at ground level across a level playing field.
The position vector `underset ~r(t)`, from `O`, of the ball after `t` seconds is given by
`qquad underset ~r(t) = 15t underset ~i + (15 sqrt 3 t - 4.9t^2)underset ~j`,
where, `underset ~i` is a unit vector in the forward direction, `underset ~j` is a unit vector vertically up and displacement components are measured in metres.
How far horizontally from `O` is the fielder when the ball is caught? Give your answer in metres, correct to one decimal place. (2 marks)
Bacteria are spreading over a Petri dish at a rate modelled by the differential equation
`(dP)/(dt) = P/2 (1 - P),\ 0 < P < 1`
where `P` is the proportion of the dish covered after `t` hours.
After one hour, a toxin is added to the Petri dish, which harms the bacteria and reduces their rate of growth. The differential equation that models the rate of growth is now
`(dP)/(dt) = P/2 (1 - P) - sqrt P/20` for `t >= 1`
`qquad qquad T = int_q^r (1/(P/2(1 - P) - sqrt P/20)) dP + s`
where `q, r and s in R`.
Find the values of `q, r` and `s`, giving the value of `q` correct to two decimal places. (2 marks)
A dairy factory produces milk in bottles with a nominal volume of 2 L per bottle. To ensure most bottles contain at least the nominal volume, the machine that fills the bottles dispenses volumes that are normally distributed with a mean of 2005 mL and a standard deviation of 6 mL.
Bottles of milk are packed in crates of 10 bottles, where the nominal total volume per crate is 20 L.
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.
| 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`
Find `b` in terms of `a`. (1 mark)
| 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`
A brooch is designed using inverse circular functions to make the shape shown in the diagram below.
The edges of the brooch in the first quadrant are described by the piecewise function
`f(x){(3text(arcsin)(x/2)text(,), 0 <= x <= sqrt2),(3text(arccos)(x/2)text(,), sqrt2 < x <= 2):}`
Give your answer in square centimetres, correct to one decimal place. (3 marks)
a. `text(Corner point occurs when)\ \ x=sqrt2.`
`y=3 sin^(-1) (sqrt2/2) = (3pi)/4`
`:.\ text(Coordinates are:)\ \ (sqrt2, (3pi)/4)`
b. `text(Reflect in the)\ xtext(-axis and then the)\ ytext(-axis:)`♦ Mean mark 49%.
`g(x){(-3text(arccos)(- x/2)text(,), -2 <= x < -sqrt2),(-3text(arcsin)(- x/2)text(,), -sqrt2 <= x <= 0):}`
| c. | `A` | `= 4 xx (int_0^sqrt2 3sin^(−1)(x/2)dx + int_sqrt2^2 3cos^(−1)(x/2)dx)` |
| `~~ 9.9\ text(cm²)` |
d. `text(Find the gradient of graph at)\ \ x=0:`
`f′(0) = 3/2`
`alpha = tan^(−1)(3/2) = 56.31…°`
`beta = pi/2 – tan^(−1)(1.5) = 33.69°`
`:.\ text(Acute angle between the edges)`
`=2 xx 33.69`
`=67.4°`
e. `f′(x)\ {(3/sqrt(4-x^2)text(,), 0<= x <= sqrt2),((-3)/sqrt(4-x^2)text(,), sqrt2 < x <= 2):}`
`:.\ text(Length of border)`♦♦ Mean mark 27%.
`= 4 int_0^sqrt2 sqrt(1 + (3/sqrt(4 – x^2))^2)\ dx + 4 int_sqrt2^2 sqrt(1 + ((-3)/sqrt(4 – x^2))^2)\ dx`
`= 4 int_0^2 sqrt(1 + (3/sqrt(4 – x^2))^2)\ dx`
`= int_0^2 sqrt(16 + 144/(4 – x^2)\ dx`
`:. a=16, \ b=144`
A helicopter is hovering at a constant height above a fixed location. A skydiver falls from rest for two seconds from the helicopter. The skydiver is subject only to gravitational acceleration and air resistance is negligible for the first two seconds. Let downward displacement be positive.
After two seconds, air resistance is significant and the acceleration of the skydiver is given by `a = g -0.01v^2`.
One root of a quadratic equation with real coefficients is `sqrt 3 + i`.
Find the equation of this circle, expressing it in the form `|z - alpha| = beta`, where `alpha, beta in R`. (3 marks)
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`
Consider `f: [sqrt 3, 6 + 3 sqrt 3] -> R,\ \ f(x) = arctan (x/3) - pi/6`.
`qquad V = tan(h + pi/6) - h - sqrt 3/3`
If water is flowing into the pond at a rate of 0.03 m³ per minute, find the rate at which the depth is increasing when the depth is 0.6 m. Give your answer in metres per minute, correct to three decimal places. (3 marks)
a.i. `tan (theta/2) = tan ((5pi)/12) = t`
| `tan (theta)` | `=(2 tan (theta/2))/(1 – tan^2(theta/2)` |
| `tan((5 pi)/6)` | `= (2tan ((5pi)/12))/(1 – tan^2((5pi)/2)` |
| `-1/sqrt 3` | `= (2t)/(1 – t^2), quad (t != +- 1)` |
| `-(1 – t^2)` | `= 2t sqrt 3` |
| `-1 + t^2` | `= 2t sqrt 3` |
`:. t^2 – 2 sqrt 3 t – 1 = 0`
| a.ii. | `t^2 – 2 sqrt 3 t + (-sqrt 3)^2 +3 -4 = 0` | |
| `(t – sqrt 3)^2` | `= 4` | |
| `t – sqrt 3` | `= +- 2` | |
| `t` | `= sqrt 3 +-2` | |
`=> tan ((5 pi)/12)\ \ text(is in 1st quadrant,)`
| `:. t` | `= tan ((5 pi)/12) = 2 + sqrt 3` |
| b. | `f(sqrt 3)` | `= 0` |
| `f(6 + 3 sqrt 3)` | `=pi/4` |
`text(Graphing)\ \ f(x) = arctan (x/3) – pi/6\ \ text(on CAS will)`
`text(show the shape of the graph.)`
| c.i. | `y` | `= tan^(-1)(x/3) – pi/6` |
| `y + pi/6` | `= tan^(-1)(x/3)` | |
| `x/3` | `= tan(y + pi/6)` | |
| `x` | `= 3 tan (y + pi/6)` | |
| `x^2` | `= 9 tan^2 (y + pi/6)` |
| `:. V` | `= pi int_0^(pi/4) x^2\ dy` | |
| `=pi int_0^(pi/4) 9 tan^2 (y + pi/6)\ dy` |
c.ii. `pi int_0^(pi/4) 9 tan^2 (y + pi/6)\ dy`
`=66.99…`
`=67\ text(u³)`
| d. | `(dV)/(dt)` | `= 0.03` |
| `V` | `= tan(h + pi/6) – h – sqrt 3/3` | |
| `(dV)/(dh)` | `= tan^2 (h+pi/6)\ \ \ text{(by CAS)}` | |
| `(dh)/(dt)` | `= (dh)/(dV) * (dV)/(dt)` | |
| `= 1/(tan^2 (h+pi/6)) xx 3/100` |
`(dh)/(dt)|_(h = 0.6)~~0.007\ text(m/min)`
According to medical records, the blood pressure of the general population of males aged 35 to 45 years is normally distributed with a mean of 128 and a standard deviation of 14. Researchers suggested that male teachers had higher blood pressures than the general population of males.
To investigate this, a random sample of 49 male teachers from this age group was obtained and found to have a mean blood pressure of 133.
A coffee machine dispenses coffee concentrate and hot water into a 200 mL cup to produce a long black coffee. The volume of coffee concentrate dispensed varies normally with a mean of 40 mL and a standard deviation of 1.6 mL.
Independent of the volume of coffee concentrate, the volume of water dispensed varies normally with a mean of 150 mL and a standard deviation of 6.3 mL.
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`.
A basketball player aims to throw a basketball through a ring, the centre of which is at a horizontal distance of 4.5 m from the point of release of the ball and 3 m above floor level. The ball is released at a height of 1.75 m above floor level, at an angle of projection `alpha` to the horizontal and at a speed of `V\ text(ms)^(-1)`. Air resistance is assumed to be negligible.
The position vector of the centre of the ball at any time, `t` seconds, for `t >= 0`, relative to the point of release is given by
`qquad underset ~r(t) = Vt cos (alpha) underset ~i + (Vt sin(alpha) - 4.9t^2) underset ~j`,
where `underset ~i` is a unit vector in the horizontal direction of motion of the ball and `underset ~j` is a unit vector vertically up. Displacement components are measured in metres.
A 200 kg crate rests on a smooth plane inclined at `theta` to the horizontal. An external force of `F` newtons acts up the plane, parallel to the plane, to keep the crate in equilibrium.
The magnitude of the external force `F` is changed to 780 N and the plane is inclined at `theta = 30^@`.
Starting from rest, the crate slides down a smooth plane inclined at `alpha` degrees to the horizontal.
A force of `295 cos(alpha)` newtons, up the plane and parallel to the plane, acts on the crate.
| a. |  |
b. `F – 200g sin (theta) = 0`
`:. F = 200g sin (theta)`
| c.i. | `sum F` | `= -F + 200g sin (30^@)` |
| `200a` | `= -780 + 200g xx 1/2` | |
| `=- 780 + 980` | ||
| `:.a` | `=1\ text(ms)^(-1)` |
| c.ii. |  |
c.iii. `text(Distance travelled in 4 seconds)`
`=\ text(Area under graph between)\ \ t=0 and t=4`
`=1/2 xx 4 xx 4`
`= 8\ text(m)`
| d. | `p` | `=mv` |
| `800` | `=200v` | |
| `:.v` | `=4` |
`text(Find)\ \a\ \ text(given)\ \ x = 10,\ \ v=4:`
| `v^2` | `= u^2 + 2ax` | |
| `16` | `=20a` | |
| `:.a` | `=0.8\ \ text(ms)^(-2) quad text(down the incline)` |
e. `200g sin(alpha) – 295 cos (alpha) = 200 xx 0.75`
`alpha ~~ 12.9^@\ \ \ text{(by CAS)}`
In the complex plane, `L` is the line given by `|z + 1| = |z + 1/2 - sqrt 3/2 i|`.
The part of the line `L` in the fourth quadrant can be expressed in the form `text(Arg)(z) = a`.
Find `k` in the form `r text(cis)(theta)`, where `theta` is the principal argument of `k`. (2 marks)
a. `(x + 1)^2 + y^2 = (x + 1/2)^2 + (y – sqrt 3/2)^2`
`x^2 + 2x + 1 + y^2 = x^2 + x + 1/4 + y^2 – y sqrt 3 + 3/4`
| `0` | `=-x – y sqrt 3` |
| `y sqrt 3` | `= -x` |
| `:. y` | `= -1/sqrt 3 x` |
| b. | `(x + iy) (x – iy)` | `=4` |
| `x^2 – i^2 y^2` | `=4` | |
| `x^2 + y^2` | `=4` |
`text(Substitute)\ \ y = -1/sqrt 3 x\ \ text(into)\ \ x^2 + y^2 = 4:`
| `x^2 + 1/3 x ^2` | `=4` |
| `4/3 x^2` | `=4` |
| `x^2` | `=3` |
| `x` | `=+- sqrt3` |
| `=>y` | `= +- 1` |
`:.\ text(Intersection at):\ (sqrt 3, -1), quad (-sqrt 3, 1)`
| c. |  |
d. `alpha = tan^(-1) (-1/sqrt 3), quad alpha in (-pi/2, 0)`
`alpha = -pi/6`
e.
`text(Total Area)`
`=\ text(Area of sector + Area of triangle)`
`=pi xx 2^2 xx ((pi/3 – pi/6)/(2 pi)) + 1/2 xx sqrt 3 xx 2`
`=4 pi (1/12) + sqrt3`
`=pi/3 + sqrt3\ \ text(u²)`
| f. | `r\ text(cis)(theta)` | `=k (r\ text(cis)(-theta))` |
| `:. k` | `=(r\ text(cis)(theta))/(r\ text(cis)(-theta))` | |
| `= text(cis)(2 xx ((-pi)/6))` | ||
| `= text(cis)(- pi/3)` |
Consider the function `f` with rule `f(x) = 10 arccos (2 - 2x)`.
A vase is to be modelled by rotating the graph of `f` about the `y`-axis to form a solid of revolution, where units of measurement are in centimetres.
What is the minimum distance the bee will need to travel? Give your answer in centimetres, correct to one decimal place. (1 mark)
| a. |  |
`2 – 2x in [-1, 1]`
`-2x in [-3, -1]`
`:. x in [1/2, 3/2]`
| `f(1/2)` | `= 10 cos^(-1) (1)=0` |
| `f(3/2)` | `= 10 cos^(-1) (1)=10pi` |
| b.i. | `y` | `= 10 cos^(-1) (2 – 2x)` |
| `y/10` | `= cos^(-1) (2 – 2x)` | |
| `cos (y/10)` | `= 2 – 2x` | |
| `2x` | `= 2 – cos (y/10)` | |
| `x` | `= 1 – 1/2 cos (y/10)` |
| `:. V` | `= pi int_0^(10 pi) x^2\ dy` | |
| `= pi int_0^(10 pi) (1 – 1/2 cos (y/10))^2\ dy` |
| b.ii. | `V = (45 pi^2)/4 text(cm)^3` |
c. `(dV)/(dt) = 20\ text(cm³/s)\ \ \ text{(given)}`
`(dV)/(dh) = pi (1 – 1/2 cos (y/10))^2\ \ text(when)\ \ y=5pi`
`=> (dV)/(dh) = pi`
| `:. (dh)/(dt)` | `= (dh)/(dV)*(dV)/(dt)` |
| `= 1/pi * 20` | |
| `= 20/pi\ text(cm s)^(-1)` |
d. `f(x) = 10cos^(-1)(2-2x)`
`l=int_(1/2)^(3/2) sqrt(1 + (f′(x))^2)\ dx`
`~~31.4\ text(cm)\ \ \ text{(by CAS)}`
Let `f:D ->R, \ f(x) = x/(1 + x^3)`, where `D` is the maximal domain of `f`.
a.i. `text(Graphing the function on CAS:`♦♦ Mean mark 36%.
`text(Vertical asymptote:)\ x = −1`
`text(Horizontal asymptote:)\ \ y = 0`
a.ii. `u = x, \ u′ = 1,\ \ v = 1 + x^3, \ v′ = 3x^2\ \ \ text{(manual or by CAS)}`
| `f′(x)` | `= (1(1 + x^3) – x(3x^2))/((1 + x^3)^2)` |
| `= (1 – 2x^3)/((1 + x^3)^2)` |
`text(S.P. when)\ \ f′(x)=0:\ `
`=> (0.79, 0.53)\ \ \ text{(by CAS)}`
a.iii. `text(When)\ \ f″(x)=0,\ \ \ text{(by CAS)}`
`=> x = 0, \ x = -1, \ x = 2`
`x != -1`
`text(Check concavity changes:)`
`f″(−1/2) = −1632/343`
`f″(1) = −3/4`
`f″(3) = 675/(10\ 976)`
`text(P.O.I. at)\ \ x = sqrt2 ~~ 1.26\ \ text{(concavity changes)}`
`=> f(sqrt2) ~~ 0.42`
`:. text(P.O.I.)\ (1.26, 0.42)`
b.
| c.i. | `V_1` | `= pi int_0^a y^2\ dx` |
| `V_2` | `= pi int_a^3 y^2\ dx` |
`:.\ text(Equation to solve for)\ a:`
`int_0^a x^2/(1 + x^3)^2\ dx = int_a^3 x^2/(1 + x^3)^2\ dx`
c.ii. `a=0.98\ \ \ text{(by CAS)}`
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.
Luggage at an airport is delivered to its owners via a 15 m ramp that is inclined at 30° to the horizontal. A 20 kg suitcase, initially at rest at the top of the ramp, slides down the ramp against a resistance of `v` newtons per kilogram, where `v\ text(ms)^(-1)` is the speed of the suitcase.
| a. | |
♦ Mean mark part (a) 44%.
| b.i. | `20g sin 30^@ – 20v` | `= sumF` |
| `10g – 20v` | `= 20a` |
| b.ii. | `(10g – 20v)/20` | `= a` |
| `g/2 – v` | `= a` | |
| `:.a` | `= (g – 2v)/2` |
♦ Mean mark part (c) 48%.
| c. | `v *(dv)/(dx)` | `= (g – 2v)/2` |
| `(dv)/(dx)` | `= (g – 2v)/(2v)` | |
| `(dx)/(dv)` | `= (2v)/(g – 2v)` | |
| `(dx)/(dv)` | `= -(2v)/(2v – g)` | |
| `= – ((2v – g + g))/(2v – g)` | ||
| `= -1 – g/(2v – g)` |
| `x` | `= int_0^v – 1 – g/(2v – g)\ dv` |
| `= int_0^v – 1 – g/2 (2/(2v – g))\ dv` | |
| `= [-v – 4.9ln\ |2v – g|]_0^v` |
`text(When)\ \ x=0, v=0:`
`2v – g < 0\ \ =>\ \ |2v – g| = g – 2v`
| `x` | `= [-v – 4.9 ln (g – 2v)]_0^v` |
| `= -v – 4.9 ln (g – 2v) – (0 – 4.9 ln (g))` | |
| `= -v – 4.9 ln (g – 2v) + 4.9 ln (g)` | |
| `= -v + 4.9 ln (g/(g – 2v))` | |
| `=-v + 4.9 ln(9.8/(9.8-2v))` | |
| `:. x` | `= -v + 4.9 ln ((4.9)/(4.9 – v))` |
d. `text(Find)\ \ v\ \ text(when)\ \ x=15:`
`15 = -v + 4.9 ln (4.9/(4.9 – v))`♦♦ Mean mark part (d) 26%.
`:. v ~~ 4.81\ text(ms)^(-1)\ \ \ (v>0)`
♦ Mean mark 41%.
| e.i | `(dv)/(dt)` | `= (g – 2v)/2` |
| `(dt)/(dv)` | `= 2/(g – 2v)` | |
| `:.t` | `= int_0^4.5 2/(g – 2v)\ dv` |
♦ Mean mark 41%.
e.ii. `t=[-ln (9.8 -2v)]_0^4.5`
`=> t ~~ 2.51 text(s)`
Two yachts, `A` and `B`, are competing in a race and their position vectors on a certain section of the race after time `t` hours are given by
`underset ~ r_A (t) = (t + 1) underset ~i + (t^2 + 2t) underset ~j \ and \ underset ~r_B (t) = t^2 underset ~i + (t^2 + 3) underset ~j, \ t >= 0`
where displacement components are measured in kilometres from a given reference buoy at origin `O`.
One of the rules for the race is that the yachts are not allowed to be within 0.2 km of each other. If this occurs there is a time penalty for the yacht that is travelling faster.
The mass of suspended matter in the air in a particular locality is normally distributed with a mean of `mu` micrograms per cubic metre and a standard deviation of `sigma = 8` micrograms per cubic metre. The mean of 100 randomly selected air samples was found to be 40 micrograms per cubic metre.
Based on this, a 90% confidence interval for `mu`, correct to two decimal places, is
A. `(38.68, 41.32)`
B. `(26.84, 53.16)`
C. `(38.43, 41.57)`
D. `(24.32, 55.68)`
E. `(37.93, 42.06)`
The acceleration, `a\ text(ms)^(-2)`, of a particle moving in a straight line is given by `a = v^2 + 1`, where `v` is the velocity of the particle at any time `t`. The initial velocity of the particle when at origin `O` is `2\ text(ms)^(-1)`.
The displacement of the particle from `O` when its velocity is `3\ text(ms)^(-1)` is
Let `OABCD` be a right square pyramid where `underset ~a = vec(OA),\ underset ~b = vec(OB),\ underset ~c = vec(OC)` and `underset ~d = vec(OD)`.
An equation correctly relating these vectors is
A. `underset ~a + underset ~c = underset ~b + underset ~d`
B. `(underset ~a - underset ~c) ⋅ (underset ~d - underset ~c) = 0`
C. `underset ~a + underset ~d = underset ~b + underset ~c`
D. `(underset ~a - underset ~d) ⋅ (underset ~c - underset ~b) = 0`
E. `underset ~a + underset ~b = underset ~c + underset ~d`
`|underset ~a| = |underset ~b| = |underset ~c| = |underset ~d|`
| `text(Let)\ \ underset ~a` | `= alpha underset ~i + beta underset ~j + gamma underset ~k` |
| `underset ~b` | `= -alpha underset ~i + beta underset ~j + gamma underset ~k` |
| `underset ~c` | `= -alpha underset ~i + beta underset ~j – gamma underset ~k` |
| `underset ~d` | `= alpha underset ~i + beta underset ~j – gamma underset ~k` |
| `A:\ \ underset ~a + underset ~ c` | `= 2 beta underset ~ j` |
| `underset ~b + underset ~d` | `= 2 beta underset ~j` |
`=> A`
If `underset ~u = 3 underset ~i + 6 underset ~j - 2 underset ~k` and `underset ~v = 2 underset ~i + 2 underset ~j - underset ~k`, then the vector resolute of `underset ~u` in the direction of `underset ~v` is
A. `20/7 (3 underset ~i + 6 underset ~j - 2 underset ~k)`
B. `20/9 (2 underset ~i + 2 underset ~j - underset ~k)`
C. `20/49 (3 underset ~i + 6 underset ~j - 2 underset ~k)`
D. `20/3 (2 underset ~i + 2 underset ~j - underset ~k)`
E. `3/7 (3 underset ~i + 6 underset ~j - 2 underset ~k)`
Let `z_1 = r_1 text(cis)(theta_1)` and `z_2 = r_2 text(cis)(theta_2)`, where `z_1` and `z_1z_2` are shown in the Argand diagram below; `theta_1` and `theta_2` are acute angles.
A statement that is necessarily true is
A. `r_2 > 1`
B. `theta_1 < theta_2`
C. `|\ (z_1)/(z_2)\ | > r_1`
D. `theta_1 = theta_2`
E. `r_1 > 1`
`text(Consider each option:)`♦ Mean mark 47%.
`A:\ \ r_1r_2 < r_1\ \ =>\ \ r_2 < 1`
`B:`
`alpha + pi/2 – theta_1 = theta_2 ~~ theta_1\ \ text(not necessarily true)`
`C:\ \ (r_1)/(r_2) = (|\ z_1\ |)/(|\ z_2\ |) = |(z_1)/(z_2)|`
`text(S)text(ince)\ \ r_2<1\ \ =>\ \ |(z_1)/(z_2)| > r_1`
`D:\ \ theta_1 = theta_2\ \ text(possible but no scale given)`
`E:\ \ r_1>1\ \ text(possible but no scale given.)`
`=> C`
Find the length of the curve specified parametrically by `x = a theta - a sin (theta), \ y = a - a cos (theta)` from `theta = (2 pi)/3` to `theta = 2 pi`, where `a in R^+`. Give your answer in terms of `a`. (4 marks)
Part of the graph of `y = 1/2 sqrt(4x^2 - 1)` is shown below.
The curve shown is rotated about the `y`-axis to form a volume of revolution that is to model a fountain, where length units are in metres.
The fountain is initially empty. A vertical jet of water in the centre fills the fountain at a rate of 0.04 cubic metres per second and, at the same time, water flows out from the bottom of the fountain at a rate of `0.05 sqrt h` cubic metres per second when the depth is `h` metres.
The line given by `|z - 1| = |z - 3|` intersects the circle given by `|z + 1| = sqrt 2 |z - i|` in two places.
--- 6 WORK AREA LINES (style=lined) ---
a. `(x – 1)^2 + (y – 2)^2 = 2^2`
`:.\ text(Centre)\ (1, 2), quad text(radius) = 2`
| b. | `y^2 + (x + 1)^2` | `= (sqrt 2)^2 (x^2 + (y – 1)^2)` |
| `y^2 + x^2 + 2x + 1` | `= 2(x^2 + y^2 – 2y + 1)` | |
| `y^2 + x^2 + 2x + 1` | `= 2x^2 + 2y^2 – 4y + 2` |
| `0` | `= 2x^2 + 2y^2 – 4y + 2 – y^2 – x^2 – 2x – 1` |
| `0` | `= x^2 + y^2 – 4y – 2x + 1` |
| `0` | `= (x^2 – 2x + 1) + (y^2 – 4y)` |
| `0` | `= (x – 1)^2 + (y^2 – 4y + 2^2) – 4` |
| `0` | `= (x – 1)^2 + (y – 2)^2 – 4` |
| `4` | `= (x – 1)^2 + (y – 2)^2` |
`:.\ text(Centre)\ (1, 2), quad text(radius) = 2`
c. `ytext(-axis intercepts occur when)\ \ x=0:`
| `(0 – 1)^2 + (y – 2)^2` | `= 4` |
| `1 + (y – 2)^2` | `= 4` |
| `(y – 2)^2` | `= 3` |
| `y – 2` | `= +- sqrt 3` |
| `y` | `= 2 +- sqrt 3` |
d. `|z – 1| = |z – 3|`
`=>\ text(Graph is a line equidistant from:)\ \ 1 + 0i and 3 + 0i`
`text(Circle intercepts occur when)\ \ x=2:`
| `(2- 1)^2 + (y – 2)^2` | `= 4` |
| `(y – 2)^2` | `= 3` |
| `y – 2` | `= +- sqrt 3` |
| `y` | `= 2 +- sqrt 3` |
| e. |  |
`sin theta = sqrt3/2\ \ =>\ \ theta = pi/3`♦ Mean mark 39%.
MARKER’S COMMENT: Students who used standard formulae rather than definite integrals were more successful.
`text(Angle at centre of segment) = (2pi)/3`
`text(Height of triangle)\ (h) = 2^2 – (sqrt3)^2 =1`
`:.\ text(Shaded Area)`
`=\ text(Area of segment – Area of triangle)`
`=((2pi)/3)/(2pi) xx pi xx 2^2 – 1/2 xx 2sqrt3 xx 1`
`=(4pi)/3 – sqrt3`
Consider the function `f: D -> R`, where `f(x) = 2 text(arcsin)(x^2 - 1)`.
a. `-1 <= x^2 – 1 <= 1`
`=>\ 0 <= x^2<= 1`
`=>\ -sqrt2 <= x <=sqrt2`
`:.\ text(Domain:)\ [- sqrt 2 , sqrt 2]`
`sin^(-1)(x) in [-pi/2, pi/2]`
`=> 2 sin^(-1)(x) in [-2xx pi/2, 2 xx pi/2]`
`:.\ text(Range:)\ [-pi, pi]`
| b. |  |
c. `f(x) = 2 sin^(-1)(x^2 – 1)`
| `f(x)` | `=2 sin^(-1)(x^2 – 1)` |
`:. f prime(x)= 4/sqrt(2 – x^2)\ \ \ (x > 0,\ \ text(by CAS))`
| d. | `f prime(x)` | `= (4x)/sqrt(x^2(2 – x^2))` |
| `= (-4)/sqrt(2 – x^2)\ \ \ (x < 0)` |
e.i. `f prime(x)\ \ text(is defined when:)`♦♦ Mean mark part (e)(i) 21%.
`2 – x^2 > 0\ \ and\ \ x!=0`
`:. x in (-sqrt2, 0) ∪ (0, sqrt 2)`
e.ii. `g(x) = +- 4`♦ Mean mark part (e)(ii) 49%, part (e)(iii) 48%.
`:. g(x) = {(-4text(,), x < 0), (\ \ \ 4text(,), x > 0):}`
| e.iii. |  |
Let `(dy)/(dx) = (4-y)^2`.
--- 7 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
Find all real solutions of `tan(2x) = -tan(x)`. (3 marks)