Aussie Maths & Science Teachers: Save your time with SmarterEd
The cricket club borrowed $400 000 to build a clubhouse.
Interest is calculated at the rate of 4.5% per annum, compounding monthly.
The cricket club will make monthly repayments of $2500.
After a number of monthly repayments, the balance of the loan will be reduced to $143 585.33.
What percentage of the next monthly repayment will reduce the balance of the loan?
Write your answer, correct to the nearest percentage. (2 marks)
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A shop owner bought 100 kg of Arthur’s tomatoes to sell in her shop.
She bought the tomatoes for $3.50 per kilogram.
The shop owner will offer a discount to her customers based on the number of kilograms of tomatoes they buy in one bag.
The revenue, in dollars, that the shop owner receives from selling the tomatoes is given by
`text(revenue)\ {{:(qquadqquad5.4n),(10.8 + 4(n - 2)),(a + 2(n - 10)):}qquadqquad{:(0 < n <= 2),(2 < n <= 10),(10 < n < 100):}`
where `n` is the number of kilograms of tomatoes that a customer buys in one bag.
Jane and Michael borrow $50 000 to expand their business.
Interest on the unpaid balance is charged to the loan account monthly.
The $50 000 is to be fully repaid in equal monthly repayments of $485.60 for 12 years.
Write your answer correct to two decimal places. (1 mark)
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Write your answer correct to the nearest dollar. (1 mark)
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Assume no other changes are made to their loan conditions.
Determine how much time Jane and Michael will save in repaying their loan.
Give your answer correct to the nearest number of months. (2 marks)
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Let `f: R -> R, \ \ f (x) = (x-3)(x-1)(x^2 + 3) and g: R-> R, \ \ g (x) = x^4-8x.`
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a. `text(Solution 1)`
`g(x) = x^4-8x`
`= x(x-2)(x^2 + 2x + 4)\ \ \ text([by CAS])`
`= x(x-2)((x + 1)^2 + 3)`
`text(Solution 2)`
| `x^4-8x` | `=x(x^3-2^3)` |
| `=x(x-2)(x^2 +2x+4)` | |
| `=x(x-2)(x^2 +2x+1+3)` | |
| `=x(x-2)((x+1)^2+3)` |
| b. | `f(x + 1)` | `= ((x + 1)-3)((x + 1)-1)((x + 1)^2 + 3)` |
| `= x(x-2)((x + 1)^2 + 3)` | ||
| `= g(x)` |
`:.\ text(Horizontal translation of 1 unit to the left.)`
c.i. `text(Consider part of the)\ \ f(x)\ text(graph below:)`
`text(For one positive)\ xtext(-axis intercept, translate at least)`
`text(1 unit left, but not more than 3 units left.)`
`:. d ∈ [1,3)`
c.ii. `text(Translate less than 1 unit left, or translate)`
`text(right.)`
`:. d < 1`
d. `text(If)\ \ g(x)=n\ \ text(has one solution, then it)`
`text(will occur when)\ \ g^{′}(x)=0 and x>0.`
| `g^{′}(x)` | `=4x^3-8` |
| `4x^3` | `=8` |
| `x` | `=2^(1/3)` |
| `n` | `=g(2^(1/3))` |
| `=2^(4/3)-8xx2^(1/3)` | |
| `=-6 xx 2^(1/3)` |
e.i. `gprime(u) = mqquadgprime(v) = −m`
| `g^{prime}(u)` | `= -g^{prime}(v)` |
| `4u^3-8` | `= -(4v^3-8)` |
| `4u^3 + 4v^3` | `= 16` |
| `:. u^3 + v^3` | `= 4` |
e.ii. `u^3 + v^3 = 4\ …\ (1)`
`u + v = 1\ …\ (2)`
`text(Solve simultaneous equation for)\ \ u > 0:`
`:. u = (sqrt5 + 1)/2,quad v = (-sqrt5 + 1)/2`
f.i. `text(Solution 1)`
`text(Using the point-gradient formula,)`
| `y-g(p)` | `=g^{prime}(p)(x-p)` |
| `y-(p^4-8p)` | `=(4p^3-8)(x-p)` |
| `y` | `=(4p^3-8)x -3p^4` |
`text(Solution 2)`
`y = 4(p^3-2)x-3p^4`
`text([CAS: tangentLine)\ (g(u),x,p)]`
f.ii. `text(Sub)\ (3/2,-12)\ text(into tangent equation,)`
`text(Solve:)\ \ -12 = 4(p^3-2)(3/2)-3p^4\ \ text(for)\ p,`
`p = 0\ \ text(or)\ \ p = 2`
| `text(When)\ \ p = 0:` | `y` | `= 4(-2)x` |
| `= -8x` |
| `text(When)\ \ p = 2:` | `y` | `= 4(2^3-2)x-3(2)^4` |
| `= 24x-48` |
`:. text(Equations are:)\ \ y =-8x, \ y = 24x-48`
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If the minimum value of the function occurs when `t = log_e (9)`, find the value of `d`. (2 marks)
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| a.i. | `S(0)` | `=2e^0+8e^0=10` |
| `S(5)` | `=2e^(5/3)+8e^(-10/3)` |
a.ii. `text(When)\ \ S^{′}(t)=0:`
| `t` | `= 3ln(2)` |
| `= log_e(2^3)` |
`:. c = 8`
`:. S_text(min) = S(log_e(8)) = 6`
| a.iii. | |
| a.iv. | `text(Average ROC)` | `= (S(log_e(8))-S(0))/(log_e(8)-0)` |
| `=(6-10)/(log_(8))` | ||
| `= (−4)/(log_e(8))` |
b. `V(t) = de^(t/3) + (10-d)e^(-(2t)/3)`
`text(Solve:)\ \ V^{′}(log_e(9)) = 0\ text(for)\ d,`
`:. d = 20/11`
c.i. `text(Solve:)\ \ V^{′}(0) = 0\ \ text(for)\ \ d,`
`d=20/3,`
`:. 20/3 <= d <10`
c.ii. `text(Solve:)\ \ V^{′}(5) = 0\ \ text(for)\ \ d,`
`d=20/(2+e^5),`
`:. 0< d <=20/(2 + e^5)`
| d. | `text(Local min when)\ V^{′}(a)` | `= 0` |
`text(Solve)\ \ V^{′}(a)=0\ \ text(for)\ a:`
`a= log_e(20/d-2)`
`text(Solve)\ \ V(log_e(20/d-2))=k/2 d^(3/2)(10-d)^(1/3)\ \ text(for)\ k,`
`k= 3 xx 2^(1/3)`
An electronics company is designing a new logo, based initially on the graphs of the functions
`f(x) = 2 sin (x) and g(x) = 1/2 sin (2x),\ text(for)\ 0 <= x <= 2 pi`
These graphs are shown in the diagram below, in which the measurements in the `x` and `y` directions are in metres.
The logo is to be painted onto a large sign, with the area enclosed by the graphs of the two functions (shaded in the diagram) to be painted red.
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The electronics company considers changing the circular functions used in the design of the logo.
Its next attempt uses the graphs of the functions `f(x) = 2 sin(x) and h(x) = 1/3 sin (3x),\ text(for)\ \ 0 <= x <= 2 pi`.
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The electronics company now considers using the graphs of the functions `k(x) = m sin(x) and q (x) = 1/n sin (nx)`, where `m` and `n` are positive integers with `m >= 2` and `0<= x <= 2pi`.
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| a. | `text(Area)` | `= 2 xx int_0^pi (2 sin (x))\ dx` |
| `= 4 xx int_0^pi sin (x)\ dx` |
`:. a = 4`
| b. | |
c. `text(Find sequence that takes)\ f(x) -> h(x)`
`text(A dilation by factor of)\ 1/6\ text(from the)\ xtext(-axis.)`
`text(A dilation by factor of)\ 1/3\ text(from the)\ ytext(-axis.)`
d.i. `text(Area when)\ n\ text(is even)`
| `= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx` |
| `=2 [-mcos x + 1/n^2 cos (nx)]_0^pi` |
| `=2[m+1/n^2 cos (npi) – (-m+1/n^2)]` |
| `= 2(2m+(cos(npi)-1)/n^2)` |
| `= 4m + 0/(n^2)` |
d.ii. `text(Area when)\ n\ text(is even)`
| `= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx` |
| `=2 [-mcos x + 1/n^2 cos (nx)]_0^pi` |
| `=2[m+1/n^2 cos (npi) – (-m+1/n^2)]` |
| `= 2(2m+(cos(npi)-1)/n^2)` |
| `= 4m + (-4)/(n^2)` |
Harry offers dog washing and dog clipping services.
Let `x` be the number of dogs washed in one day
`y` be the number of dogs clipped in one day.
It takes 20 minutes to wash a dog and 25 minutes to clip a dog.
There are 200 minutes available each day to wash and clip dogs.
This information can be written as Inequalities 1 to 3.
Inequality 1: `x ≥ 0`
Inequality 2: `y ≥ 0`
Inequality 3: `20x + 25y ≤ 200`
In any one day the number of dogs clipped is at least twice the number of dogs washed.
The profit from washing one dog is $40 and the profit from clipping one dog is $30.
Let `P` be the total profit obtained in one day from washing and clipping dogs.
| a. | |
b. `text(Inequality 4:)\ \ y >= 2x`
| c.i. | |
c.ii. `text(When)\ y =5,\ text(the maximum value of)\ x`
`text{(whole number) in the feasible region is 2.)`
`:. 2\ text(dogs can be washed.)`
d. `P = 40x + 30y`
e.i. `text(Test points for maximum profit)`
`text(At)\ (0, 8),`
`text(Profit) = 40 xx 0 + 30 xx 8 = $240`
`text(At)\ (1, 7),`
`text(Profit) = 40 xx 1 + 30 xx 7 = $250`
`text(At)\ (2, 6),`
`text(Profit) = 40 xx 2 + 30 xx 6 = $260`
`:.\ text(2 washes and 6 clips produce maximum profit.)`
e.ii. `text(Maximum profit)`
`= 40 xx 2 + 30 xx 6`
`= $260`
A farmer owns a flat allotment of land in the shape of triangle `ABC` shown below.
Boundary `AB` is 251 metres.
Boundary `AC` is 142 metres.
Angle `BAC` is 45°.
A straight track, `XY`, runs perpendicular to the boundary `AC`.
Point `Y` is 55 m from `A` along the boundary `AC`.
Write your answer, in metres, correct to one decimal place. (1 mark)
Determine the bearing of `B` from `A`. (1 mark)
Write your answer, in metres, correct to one decimal place. (2 marks)
The length of the boundary `BC` is 181 metres (correct to the nearest metre).
A farmer plans to build a fence, `MN`, perpendicular to the boundary `AC`.
The land enclosed by triangle `AMN` will have an area of 3200 m².
| a. | `/_ AXY` | `= 180 – (90 + 45)` |
| `= 45^@` |
b. `text(In)\ Delta AXY,`
| `cos 45^@` | `= 55/(AX)` |
| `:. AX` | `= 55/(cos 45^@` |
| `= 77.78…` | |
| `= 77.8\ text{m (1 d.p.)}` |
| c. |  |
`text(Bearing of)\ B\ text(from)\ A`
`= 78 – 45`
`= 033^@`
| d. |  |
`text(Using the cosine rule,)`
| `XC^2` | `= 77.8^2 + 142^2 – 2 xx 77.8 xx 142 xx cos 45^@` |
| `= 10\ 593.17…` | |
| `:. XC` | `=sqrt (10\ 593.17…)` |
| `= 102.92…` | |
| `= 102.9\ text{m (1 d.p.)}` |
e. `text(Using sine rule,)`
| `text(Area)\ Delta ABC` | `= 1/2 bc sin A` |
| `= 1/2 xx 142 xx 251 xx sin 45^@` | |
| `= 12601.34…` | |
| `= 12\ 601\ text{m² (nearest m²)}` |
f.i. `text(Using the cosine rule,)`
| `BC^2` | `= 142^2 + 251^2 – 2 xx 142 xx 251 xx cos 45^@` |
| `= 32\ 759.60…` | |
| `:. BC` | `= 180.99…` |
| `= 181\ text{m (nearest m) … as required}` |
f.ii. `text(Using the cosine rule,)`
| `cos /_ ABC` | `= (251^2 + 181^2 – 142^2)/(2 xx 251 xx 181)` |
| `= 0.832…` | |
| `:. /_ ABC` | `= 33.69…` |
| `= 33.7^@\ text{(1 d.p.)}` |
g. `/_ AMN = 180 – (90 + 45) = 45^@`
`:. Delta AMN\ text(is isosceles.)`
| `text(Area)\ Delta AMN` | `= 1/2 xx AN xx MN` |
| `3200` | `= 1/2 xx MN xx MN\ (AN = MN)` |
| `MN^2` | `= 6400` |
| `:. MN` | `= 80\ text(m)` |
Tessa has a task that involves removing the top 24 cm of the height of her right regular pyramid below.
The shape remaining is shown in Figure 5 below. The top surface, `JKLM`, is parallel to the base, `ABCD`.
By the end of each academic year, students at the university will have either passed, failed or deferred the year. Experience has shown that
Twelve hundred and thirty students began a business degree in 2007.
By the end of the 2007 academic year, 880 students had passed, 230 had failed, while 120 had deferred the year.
No students have dropped out of the business degree permanently.
Use this information to predict the number of business students who will defer the 2009 academic year. (2 marks)
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A tree, 12 m tall, is growing at point `T` near a shed.
The distance, `CT`, from corner `C` of the shed to the centre base of the tree is 13 m.
Write your answer, in degrees, correct to one decimal place. (1 mark)
`N` and `C` are two corners at the base of the shed. `N` is due north of `C`.
The angle, `TCN`, is 65°.
Explain your answer showing appropriate calculations. (2 marks)
| a. |  |
| `text(Let)\ \ theta` | `=\ text(angle of elevation)` |
| `= 12/13` | |
| `:. theta` | `= 42.709…` |
| `= 42.7^@\ text{(1 d.p.)}` |
| b. |  |
`text(Using the cosine rule:)`
| `NT^2` | `= 10^2 + 13^2 – 2 xx 10 xx 13 xx cos65^@` |
| `= 159.11…` | |
| `:. NT` | `= sqrt(159.11…)` |
| `= 12.61…` | |
| `= 12.6\ text{m (1 d.p.) … as required}` |
c. `text(Using the sine rule:)`
| `(sin /_ CNT)/13` | `= (sin 65^@)/12.6` |
| `sin /_ CNT` | `= (13 xx sin 65^@)/12.6` |
| `= 0.935…` | |
| `:. /_ CNT` | `= 69.24…` |
| `= 69^@\ text{(nearest degree)}` |
| d. |  |
`text(Bearing of)\ T\ text(from)\ N`
`= 180 – 69`
`= 111^@`
| e. |  |
`ST\ text(is the shortest distance between the)`
`text(tree and the shed.)`
`text(In)\ \ Delta NST,`
| `sin 69^@` | `= (ST)/12.6` |
| `:. ST` | `= 12.6 xx sin 69^@` |
| `= 11.76…\ text(m)` |
`:.\ text(S) text(ince the tree is 12 m, and 12 m) > 11.76\ text(m),`
`text(it could hit the shed if it falls.)`
The ferry has two fuel filters, `A` and `B`.
Filter `A` has a hemispherical base with radius 12 cm.
A cylinder of height 30 cm sits on top of this base.
Filter `B` is a right cone with height 50 cm.
If the height of the oil in the cone is now 20 cm, what percentage of the original volume of oil was removed? (2 marks)
Part of the graph of a function `g: R -> R, \ g (x) = (16-x^2)/4` is shown below.
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The tangent to the graph of `g` at a point `P` has a negative gradient and intersects the `y`-axis at point `D(0, k)`, where `5 <= k <= 8.`
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a.i. `B(4,0), C(0,4), A(a,(16-a^2)/4)`
`m_(BC) = m_text(tan) = (4-0)/(0-4) = -1`
| `g(x)` | `= (16-x^2)/4` |
| `g^{prime}(x)` | `=-x/2` |
`text(S)text(ince)\ \ g^{prime}(a) = -1,`
`=>a = 2`
`text(T)text(angent passes through)\ \ (2,3),`
| `y-3` | `=-(x-2)` |
| `y` | `=-x + 5` |
`text(Alternatively, using technology:)`
`[text(CAS: tangent Line)]\ (g(x),x,2)`
a.ii. `text(Solution 1)`
`D(5,0), E(0,5)`
| `text(Area)` | `= DeltaEOD-int_0^4 g(x)\ dx` |
| `= 1/2 xx 5 xx 5-32/3` | |
| `= 11/6\ text(u²)` |
`text(Solution 2)`
| `text(Area)` | `= int_0^5 (-x+5)\ dx-int_0^4 g(x)\ dx` |
| `= 11/6\ text(u²)` |
b. `Q(x, (16-x^2)/4), O(0,0)`
| `z` | `= OQ` |
| `= sqrt(x^2 + ((16-x^2)/4)^2), x > 0` |
`text(Max or min when)\ \ (dz)/(dx)=0,`
`text(Solve:)\ (dz)/dx = 0quadtext(for)quadx > 0`
`=> x = 2sqrt2`
`=>z(2sqrt2) = 2sqrt3`
`:. text(Min distance of)\ \ 2sqrt3\ \ text(units when)\ \ x = 2sqrt2`
c. `text(Let)\ \ P(p, (16-p^2)/4)`
`m_text(tan)\ text(at)\ P =-p/2`
`m_(PD) = ((16-p^2)/4-k)/(p-0)`
`text(Equating gradients,)`
`text(Solve:)\ \ ((16-p^2)/4-k)/p=-p/2\ \ text(for)\ p,`
`=> p=2sqrt(k-4)`
`:.\ text(Gradient of tangent:)`
`g^{prime}(2sqrt(k-4)) =-sqrt(k-4)`
d.i. `text(Equation of tangent:)`
`y = -sqrt(k-4)x + k`
`text(CAS: tangent Line)\ (g(x),x,2sqrt(k-4))`
`xtext(-int of tangent:)\ x = k/(sqrt(k-4))`
| `:. A(k)` | `= 1/2 xx (k/(sqrt(k-4))) xx k-int_0^4 g(x)\ dx` |
| `= (k^2)/(2sqrt(k-4))-32/3, \ \ k ∈ [5,8]` |
d.ii. `text(Solve:)\ \ A(k)=0quadtext(for)quadk ∈ [5,8]`
`=> k=16/3`
`text(Sketch the graph of)\ \ A(k)\ \ text(for)\ \ 5<=k<=8`
`:. A_text(max) = 16/3quadtext(when)quadk = 8`
d.iii. `A_text(min)\ text(occurs at the turning point)\ (k=16/3).`
| `A_text(min)` | `=A(16/3)` |
| `=(64sqrt3)/9-32/3` |
Tasmania Jones is in Switzerland. He is working as a construction engineer and he is developing a thrilling train ride in the mountains. He chooses a region of a mountain landscape, the cross-section of which is shown in the diagram below.
The cross-section of the mountain and the valley shown in the diagram (including a lake bed) is modelled by the function with rule
`f(x) = (3x^3)/64-(7x^2)/32 + 1/2.`
Tasmania knows that `A (0, 1/2)` is the highest point on the mountain and that `C(2, 0)` and `B(4, 0)` are the points at the edge of the lake, situated in the valley. All distances are measured in kilometres.
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Tasmania’s train ride is made by constructing a straight railway line `AB` from the top of the mountain, `A`, to the edge of the lake, `B`. The section of the railway line from `A` to `D` passes through a tunnel in the mountain.
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In order to ensure that the section of the railway line from `D` to `B` remains stable, Tasmania constructs vertical columns from the lake bed to the railway line. The column `EF` is the longest of all possible columns. (Refer to the diagram above.)
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Tasmania’s train travels down the railway line from `A` to `B`. The speed, in km/h, of the train as it moves down the railway line is described by the function.
`V: [0, 4] -> R, \ V(x) = k sqrt x-mx^2,`
where `x` is the `x`-coordinate of a point on the front of the train as it moves down the railway line, and `k` and `m` are positive real constants.
The train begins its journey at `A (0, 1/2)`. It increases its speed as it travels down the railway line.
The train then slows to a stop at `B(4, 0)`, that is `V(4) = 0.`
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Tasmania is able to change the value of `m` on any particular day. As `m` changes, the relationship between `k` and `m` remains the same.
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Trigg the gardener is working in a temperature-controlled greenhouse. During a particular 24-hour time interval, the temperature `(Ttext{°C})` is given by `T(t) = 25 + 2 cos ((pi t)/8), \ 0 <= t <= 24`, where `t` is the time in hours from the beginning of the 24-hour time interval.
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Trigg is designing a garden that is to be built on flat ground. In his initial plans, he draws the graph of `y = sin(x)` for `0 <= x <= 2 pi` and decides that the garden beds will have the shape of the shaded regions shown in the diagram below. He includes a garden path, which is shown as line segment `PC.`
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In further planning for the garden, Trigg uses a transformation of the plane defined as a dilation of factor `k` from the `x`-axis and a dilation of factor `m` from the `y`-axis, where `k` and `m` are positive real numbers.
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a. `T_text(max)\ text(occurs when)\ \ cos((pit)/8) = 1,`
`T_text(max)= 25 + 2 = 27^@C`
`text(Max occurs when)\ \ t = 0, or 16\ text(h)`
| b. | `text(Period)` | `= (2pi)/(pi/8)` |
| `= 16\ text(hours)` |
c. `text(Solve:)\ \ 25 + 2 cos ((pi t)/8)=26\ \ text(for)\ t,`
| `t` | `= 8/3,40/3,56/3\ \ text(for)\ t ∈ [0,24]` |
| `t_text(min)` | `= 8/3` |
d. `text(Consider the graph:)`
| `text(Time above)\ 26 text(°C)` | `= 8/3 + (56/3-40/3)` |
| `= 8\ text(hours)` |
e.i. `(dy)/(dx) = cos(x)`
`text(At)\ x = (2pi)/3,`
| `(dy)/(dx)` | `= cos((2pi)/3)=-1/2` |
e.ii. `text(Solution 1)`
`text(Equation of)\ \ PC,`
| `y-sqrt3/2` | `=-1/2(x-(2pi)/3)` |
| `y` | `=-1/2 x +pi/3 +sqrt3/2` |
`PC\ \ text(passes through)\ \ (c,0),`
| `0` | `=-1/2 c +pi/3 + sqrt3/2` |
| `c` | `=sqrt3 + (2 pi)/3\ …\ text(as required)` |
`text(Solution 2)`
`text(Equating gradients:)`
| `- 1/2` | `= (sqrt3/2-0)/((2pi)/3-c)` |
| `-1` | `= sqrt3/((2pi-3c)/3)` |
| `3c-2pi` | `= 3sqrt3` |
| `3c` | `= 3 sqrt3 + 2pi` |
| `:. c` | `= sqrt3 + (2pi)/3\ …\ text(as required)` |
f.i. `X^{′} ((2pi)/3 m,0)qquadP^{′}((2pi)/3 m, sqrt3/2 k)qquadC^{′} ((sqrt3 + (2pi)/3)m, 0)`
| `X^{′}P^{′}` | `= 10` |
| `sqrt3/2 k` | `= 10` |
| `:. k` | `= 20/sqrt3` |
| `=(20sqrt3)/3` |
`X^{′}C^{′}=30`
| `((sqrt3 + (2pi)/3)m)-(2pi)/3 m` | `= 30` |
| `:. m` | `= 30/sqrt3` |
| `=10sqrt3` |
| f.ii. | `P^{′}((2pi)/3 m, sqrt3/2 k)` | `= P^{′}((2pi)/3 xx 10sqrt3, sqrt3/2 xx 20/sqrt3)` |
| `= P^{′}((20pisqrt3)/3,10)` |
The shaded region in the diagram below is the plan of a mine site for the Black Possum mining company.
All distances are in kilometres.
Two of the boundaries of the mine site are in the shape of the graphs of the functions
`f: R -> R,\ f(x) = e^x and g: R^+ -> R,\ g(x) = log_e (x).`
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Tasmania Jones is in the jungle, searching for the Quetzalotl tribe’s valuable emerald that has been stolen and hidden by a neighbouring tribe. Tasmania has heard that the emerald has been hidden in a tank shaped like an inverted cone, with a height of 10 metres and a diameter of 4 metres (as shown below).
The emerald is on a shelf. The tank has a poisonous liquid in it.
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The tank has a tap at its base that allows the liquid to run out of it. The tank is initially full. When the tap is turned on, the liquid flows out of the tank at such a rate that the depth, `h` metres, of the liquid in the tank is given by
`h = 10 + 1/1600 (t^3 - 1200t)`,
where `t` minutes is the length of time after the tap is turned on until the tank is empty.
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From the moment the liquid starts to flow from the tank, find how long, in minutes, it takes until `h = 2`.
(Give your answer correct to one decimal place.) (2 marks)
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Steve, Katerina and Jess are three students who have agreed to take part in a psychology experiment. Each student is 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. --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) ---
Let `f: R text(\{2}) -> R,\ f(x) = 1/(2x-4) + 3.`
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a. `text(Asymptotes:)`
`x = 2`
`y = 3`
b.i. `f^{′}(x) = (−2)/((2x-4)^2)`
b.ii. `text(Range) = (−∞,0), or R^-`
| b.iii. | `text(As)\ \ ` | `f^{′}(x) < 0quadtext(for)quadx ∈ R text(\{2})` |
| `f^{′}(x) != 0` |
`:. f\ text(has no stationary points.)`
c. `text(Point of tangency) = P(p,1/(2p-4) + 3)`
| `m_text(tang)` | `= f^{′}(p)` |
| `= (-2)/((2p-4)^2)` |
`text(Equation of tangent using:)`
| `y-y_1` | `= m(x-x_1)` |
| `y-(1/(2p-4) + 3)` | `= (-2)/((2p-4)^2)(x-p)` |
| `y-3` | `= (-2(x-p))/((2p-4)^2) + (2p-4)/((2p-4)^2)` |
| `(2p-4)^2(y-3)` | `=-2x + 2p + 2p-4` |
| `:. (2p-4)^2(y-3)` | `=-2x + 4p-4\ \ text(… as required)` |
d. `text(Substitute)\ \ (−1,7/2)\ text{into tangent (part c),}`
`text(Solve)\ \ (2p-4)^2(7/2-3) = −2(-1) + 4p-4\ \ text(for)\ p:`
`:. p = 1,\ text(or)\ 5`
`text(Substitute)\ \ p = 1\ text(and)\ p = 5\ text(into)\ \ P(p,1/(2p-4) + 3)`
`:. text(Coordinates:)\ (1,5/2)\ text(or)\ (5,19/6)`
e. `text(Determine transformations that that take)\ f -> g:`
`text(Dilate the graph of)\ \ f(x) = 1/(2x-4) + 3\ \ text(by a)`
`text(factor of 2 from the)\ \ ytext(-axis).`
`y = 1/(2(x/2)-4) + 3= 1/(x-4) + 3`
`text(Translate the graph 4 units to the left and 3)`
`text(units down to obtain)\ \ g(x).`
`text(Using the transformation matrix,)`
| `x^{′}` | `=ax+c` |
| `y^{′}` | `=y+d` |
`f -> g:\ \ 1/(2x-4) -> 1/(x^{′})`
`x^{′}=2x-4`
`=> a=2,\ \ c=-4`
`f -> g:\ \ y -> y^{′} + 3`
`y^{′}=y -3`
`=>\ \ d=-3`
Michael began his hike at the national park office and followed a track towards a camping ground, 16 km away.
The distance-time graph below shows Michael's distance from the national park office, `d` kilometres, after `t` hours of hiking.
It took Michael seven hours to complete this hike.
Write your answer correct to one decimal place. (1 mark)
The equation of Michael's distance-time graph from `t = 3` to `t = 7` is
`d = at + b`
Katie hiked along the same track as Michael, but hiked in the opposite direction.
She begun at the camping ground and hiked towards the national park office.
Katie's distance from the national park office, `d` kilometres, after `t` hours of hiking, can be determined from the equation
`d = -3t + 16`
Katie and Michael both starter hiking at the same time.
Katie and Michael both carry radio transmitters that allow them to talk to each other while hiking.
The transmitters will not work if Katie and Michael are more than three kilometres apart.
Write your answer in hours correct to two decimal places. (2 marks)
| a. | `text(Average speed)` | `= text(total kms)/text(total time)` |
| `= 16/7` | ||
| `= 2.285…` | ||
| `= 2.3\ text{km/hr (1 d.p.)}` |
b. `text(When)\ \ t = 3\ \ text{(from graph)},`
`10 = 3a + b\ \ text{… (1)}`
`text(When)\ \ t = 7`
`16 = 7a + b\ \ text{… (2)}`
`text(Substitute)\ \ b = 10 – 3a\ \ text{from (1) into (2)}`
| `16` | `= 7a + 10 – 3a` |
| `6` | `= 4a` |
| `:. a` | `= 1.5` |
`text(Substitute)\ a = 1.5\ \ text{into (1)}`
| `10` | `= 3 xx 1.5 + b` |
| `:. b` | `= 5.5` |
| c. |  |
`text(From the graph, Katie’s distance from)`
`text(the office is the same as Michael’s at)`
`text(the intersection of)`
| `d` | `= -3t + 16\ \ text{… (1)}` |
| `d` | `= 5t\ \ text{… (2)}` |
`text(Intersection when)`
| `5t` | `= -3t + 16` |
| `8t` | `= 16` |
| `t` | `= 2` |
`:.\ text(Katie passes Michael after 2 hours.)`
d. `text(Find the times when they are 3km apart.)`
| `-3t + 16 – 5t` | `= 3` |
| `8t` | `= 13` |
| `t` | `= 1.625\ text(hours)` |
`text(After passing each other, the graph)`
`text(shows the next time they are 3km)`
`text(apart is when)\ \ t = 3\ text(hours.)`
`:.\ text(Time the radios will work)`
`= 3 – 1.625`
`= 1.375`
`= 1.38\ text{hours (2 d.p.)}`
A lighthouse tower, shaded on Figure 5 below, is in the shape of a truncated cone.
It has circular cross-sections that decrease uniformly from a radius of 3.5 metres at ground level to a radius of 2 metres at the walkway.
The height of the lighthouse tower is 18 metres.
The angle marked as `alpha` is the angle that the outer wall of the lighthouse tower makes with the horizontal at ground level.
Write your answer in degrees correct to one decimal place. (1 mark)
The lighthouse tower is part of a cone. The height of this cone is `h` metres and the base radius is 3.5 metres, as shown in Figure 6.
Write your answer to the nearest cubic metre. (1 mark)
| a. |  |
| `tan alpha` | `= 18/1.5` |
| `:. alpha` | `= tan^-1 12` |
| `= 85.23…` | |
| `= 85.2^@\ text{(1 d.p.)}` |
| b.i. |  |
| `tan 85.2^@` | `= h/3.5` |
| `:. h` | `= 3.5 xx tan 85.2^@` |
| `= 41.68…` | |
| `= 42\ text{m (nearest m)}` |
| ii. |  |
`text(Volume of large cone)`
`= 1/3 pi r^2 h`
`= 1/3 pi xx (3.5)^2 xx 42`
`= 538.78…\ text(m³)`
`text(Volume of Upper Cone)`
`= 1/3 xx pi xx 2^2 xx h_(uc)`
`= 4/3 pi xx (42 – 18)`
`= 100.53…\ text(m³)`
`:.\ text(Volume of lighthouse tower)`
`= 538.78… – 100.53…`
`= 438.2…`
`= 438\ text{m³ (nearest m³)}`
There are three candidates in the election: Ms Aboud (`A`), Mr Broad (`B`) and Mr Choi (`C`).
Mr Choi may need to withdraw from the election at the end of May.
Matrix `T`, shown below, shows the percentage of voters who change their preferred candidate, from month to month, before Mr Choi would withdraw from the election.
`{:(qquadqquadqquadqquadtext(this month)),(qquadqquadqquad\ {:(A,qquadB,qquadC):}),(T = [(0.75,0.10,0.20),(0.05,0.80,0.40), (0.20,0.10,0.40)]{:(A),(B),(C):}qquadtext(next month)):}`
Matrix `T_1`, shown below, shows the percentage of voters who change their preferred candidate, from May to June, after Mr Choi would withdraw from the election.
`{:(qquadqquadqquadqquadqquadquadtext(May)),(qquadqquadqquadquad{:(A,qquadB,qquadC):}),(T_1 = [(0.75,0.15,0.6),(0.25,0.85,0.4),(0,0,0)]{:(A),(B),(C):}qquadtext(June)):}`
Consider the voters who preferred Mr Broad in May and who were expected to prefer Mr Choi in June.
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The state matrix that indicates the number of voters who are expected to have a preference for each candidate in January, `S_1`, is given below.
`S_1 = [(6000), (3840), (2160)]{:(A), (B), (C):}`
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The heights (in cm) and ages (in months) of the 15 boys are shown in the scatterplot below.
| a. | |
b. `text{Using the points (23,86) and (35, 94),}`
| `text(Slope)` | `= (94 – 86)/(35 – 23)` |
| `= 0.666…` | |
| `=0.7\ \ text{(1 d.p.)}` |
`:.\ text(height =)\ c + 0.7 xx text(age)`
`text{Substituting (35,94) into the equation,}`
| `94` | `=c + 0.66… xx 35` |
| `:.c` | `=94-23.33…` |
| `=70.66…` | |
| `=70.7\ \ text{(1 d.p.)}` |
`:.text(height)\ = 70.7 + 0.7 xx text(age)`
c. `text(The three median line is not as influenced)`
`text{by extreme values such as (20, 93).}`
The number of hours spent doing homework each week (homework hours) and the number of hours spent watching television each week (television hours) were recorded for a group of 20 Year 12 students.
The results are displayed in the table below and a scatterplot constructed as shown.
The relationship between homework hours per week and television hours per week is clearly nonlinear.
A reciprocal transformation applied to the variable, television hours, can be used to linearise the scatterplot.
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The average age of women at first marriage in years (average age) and average yearly income in dollars per person (income) were recorded for a group of 17 countries.
The results are displayed in Table 2. A scatterplot of the data is also shown.
The relationship between average age and income is nonlinear.
A log transformation can be applied to the variable income and used to linearise the scatterplot.
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The maximum temperature and the minimum temperature at this weather station on each of the 30 days in November 2011 are displayed in the scatterplot below.
The correlation coefficient for this data set is `r = 0.630`.
The equation of the least squares regression line for this data set is
maximum temperature = `13 + 0.67` × minimum temperature
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On the day that the minimum temperature was 11.1 °C, the actual maximum temperature was 12.2 °C.
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a. `text(The two widest points in this data range are,)`
`text{(0, 13) and (20, 26.4).}`
b. `text(On average, when the minimum temperature is)`
`text(0 °C, the maximum temperature is 13 °C.)`
c. `text(Given)\ r = 0.630,`
`text(Strength: moderate)`
`text(Direction: positive)`
d. `text(On average, it is predicted that \the maximum)`
`text(temperature increases by 0.67 °C for every 1 °C)`
`text(increase in the minimum temperature.)`
| e. | `r^2` | `= 0.630^2` |
| `=0.3969` | ||
| `=40text{% (nearest %)}` |
f. `text(When the minimum temperature was)\ 11.1 text(°C),`
| `text(Predicted Value)` | `= 13 + 0.67 xx 11.1` |
| `=20.437…` |
| `:.\ text(Residual)` | `= 12.2 − 20.437…` |
| `= – 8.237…` | |
| `= – 8\ text{°C (nearest degree)}` |
The time series plot below shows the average pay rate, in dollars per hour, for workers in a particular country for the years 1991 to 2005.
A three median line will be used to model the increasing trend in the average pay rate shown in this time series.
The explanatory variable to be used is year.
Write your answer, correct to one decimal place. (1 mark)
a.i. & ii. `text(The three medians points are)`
`(1993,13.7), (1998,16.7), (2003,21.8)`
| b. | `text(Slope)` | `=(y_2-y_1)/(x_2-x_1)` |
| `=(21.8 − 13.7)/(2003 − 1993)` | ||
| `= 0.81` | ||
| `=0.8\ \ text{(to 1 d.p.)}` |
c. `text(The average pay rate increases, on average,)`
`text(by $0.80 each year between 1991 and 2005.)`
The scatterplot below shows the population density, in people per square kilometre, and the area, in square kilometres, of 38 inner suburbs of the same city.
For this scatterplot, `r^2 = 0.141`
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Assume the areas of these inner suburbs are approximately normally distributed.
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The shaded region in the graph below represents the feasible region for a linear programming problem.
Which objective function, `Z`, has its maximum value at the point `M`?
A. `Z = x + y`
B. `Z = x - y`
C. `Z = 3x + y`
D. `Z = 3x - 2y`
E. `Z = x + 4y`
The cost of posting a parcel that weighs 500 grams or less depends on its weight, as shown on the graph below.
The total cost of posting three separate parcels weighing respectively 100 grams, 150 grams and 210 grams is $11.
The total cost of posting three separate parcels weighing respectively 60 grams, 110 grams and 350 grams is $8.
The total cost of posting two separate parcels weighing respectively 170 grams and 500 grams is
A. `$5`
B. `$7`
C. `$8`
D. `$10`
E. `$12`
The shaded region in the graph below represents the feasible region for a linear programming problem.
An objective function `Z = ax + by` has its value maximised at both vertex `M` and vertex `N`.
The values of `a` and `b` could be
A. `a = 15\ and\ b = – 15`
B. `a = 15\ and\ b = 15`
C. `a = 15\ and\ b = 25`
D. `a = 25\ and\ b = 50`
E. `a = 50\ and\ b = – 25`
In the diagram below, `/_ ABD = /_ ACB = theta^@`.
`BD = 24\ text(cm)` and `BC = 40\ text(cm.)`
The area of triangle `ABD` is `100\ text(cm².)`
The area of triangle `ABC`, in cm² is closest to
A. `167`
B. `178`
C. `267`
D. `278`
E. `378`
`/_ ABD = /_ BCA = theta^@\ \ \ text{(given)}`
`/_ BAD\ text(is common)`
`:. Delta BAD\ text(|||)\ Delta CAB\ \ text{(equiangular)}`
`text(Linear scale factor)`
`k= (BC)/(DB) = 40/24 = 5/3`
| `:.\ text(Area)\ Delta CAB` | `= k^2 xx text(Area)\ Delta BAD` |
| `= (5/3)^2 xx 100` | |
| `= 277.77…\ \ text(cm²)` |
`=> D`
A composite shape is made up of a parallelogram and a triangle, as shown below.
Which one of the following is always true?
A. `b = 2a`
B. `a = 2b`
C. `a + b = 180`
D. `a + 2b = 180`
E. `2a - b = 180`
`text(Base angles of isosceles triangle)`
`=(180 – b)°`
`text(Summing up the angles of the isosceles triangle,)`
| `b + 2(180^@ – a)` | `=180°` |
| `b-2a+360°` | `=180°` |
| `:. 2a – b` | `=180°` |
`=> E`
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.)
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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.)
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Let `ON = r` and `LM = w`.
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Let the gradient of the parabola at `L` be `tan phi`.
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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`.
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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.
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(You may assume that the value of `x` given in part (iii) is feasible.) (2 marks)
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Consider the function `f(x) = (log_e x)/x`, for `x > 0`.
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a. `f(x) = (log_e x)/x, text(for)\ x > 0`
`text(Using the quotient rule,)`
| `f^{prime}(u/v)` | `=(u′v-uv′)/v^2` |
| `f^{prime}(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^{prime}(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.`
| b. |
|
`text(When)\ \ x < e, log_e x<1\ \ \ text{(from graph)}`
| `1-log_e x` | `> 0` |
| `(1-log_e x)/(x^2)` | `> 0` |
| `:. f^{prime}(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^{prime}(x)` | `< 0` |
`:.\ text(The stationary point at)\ \ x = e\ \ text(is a maximum.)`
| c. | `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`.
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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.
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For `x > 0`, let `f(x) = x^n e^-x`, where `n` is an integer and `n >= 2.`
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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.`
