The graphs of the functions `y = kx^n` and `y = log_e x` have a common tangent at `x = a`, as shown in the diagram.
- By considering gradients, show that
- `a^n = 1/(nk)`. (1 mark)
- Express `k` as a function of `n` by eliminating `a`. (2 marks)
Functions, EXT1 F1 2007 HSC 6b
Consider the function `f(x) = e^x − e^(-x)`.
- Show that `f(x)` is increasing for all values of `x`. (1 mark)
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- Show that the inverse function is given by
`qquad qquad f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)` (3 marks)
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- Hence, or otherwise, solve `e^x - e^(-x) = 5`. Give your answer correct to two decimal places. (1 mark)
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Mechanics, EXT2* M1 2007 HSC 6a
A particle moves in a straight line. Its displacement, `x` metres, after `t` seconds is given by
`x = sqrt3\ sin\ 2t − cos\ 2t + 3`.
- Prove that the particle is moving in simple harmonic motion about `x = 3` by showing that `ddot x = -4(x − 3)`. (2 marks)
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- What is the period of the motion? (1 mark)
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- Express the velocity of the particle in the form `dotx = A\ cos\ (2t − α)`, where `α` is in radians. (2 marks)
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- Hence, or otherwise, find all times within the first `pi` seconds when the particle is moving at `2` metres per second in either direction. (2 marks)
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Quadratic, EXT1 2007 HSC 5d
The diagram shows a point `P(2ap, ap^2)` on the parabola `x^2= 4ay`. The normal to the parabola at `P` intersects the parabola again at `Q(2aq,aq^2)`.
The equation of `PQ` is `x + py − 2ap − ap^3 = 0`. (Do NOT prove this.)
- Prove that `p^2+ pq + 2 = 0`. (1 mark)
- If the chords `OP` and `OQ` are perpendicular, show that `p^2 = 2`. (2 marks)
Trigonometry, EXT1 T1 2007 HSC 5c
Find the exact values of `x` and `y` which satisfy the simultaneous equations
`sin^(-1)\ x + 1/2\ cos^(-1)\ y = pi/3` and
`3\ sin^(-1)\ x-1/2\ cos^(-1)\ y = (2pi)/3`. (3 marks)
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Combinatorics, EXT1 A1 2007 HSC 5b
Mr and Mrs Roberts and their four children go to the theatre. They are randomly allocated six adjacent seats in a single row.
What is the probability that the four children are allocated seats next to each other? (2 marks)
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Trig Calculus, EXT1 2006 HSC 7
A gutter is to be formed by bending a long rectangular metal strip of width `w` so that the cross-section is an arc of a circle.
Let `r` be the radius of the arc and `2 theta` the angle at the centre, `O`, so that the cross-sectional area, `A`, of the gutter is the area of the shaded region in the diagram on the right.
- Show that, when `0 < theta <= pi/2`, the cross-sectional area is
- `A = r^2 (theta - sin theta cos theta).` (2 marks)
- `A = r^2 (theta - sin theta cos theta).` (2 marks)
- The formula in part (i) for `A` is true for `0 < theta < pi.` (Do NOT prove this.)
- By first expressing `r` in terms of `w` and `theta`, and then differentiating, show that
- `(dA)/(d theta) = (w^2 cos theta (sin theta - theta cos theta))/(2 theta^3).`
- for `0 < theta < pi.` (3 marks)
- Let `g(theta) = sin theta - theta cos theta.`
- By considering `g prime(theta)`, show that `g(theta) > 0` for `0 < theta < pi.` (3 marks)
- Show that there is exactly one value of `theta` in the interval `0 < theta < pi` for which
- `(dA)/(d theta) = 0.` (2 marks)
- `(dA)/(d theta) = 0.` (2 marks)
- Show that the value of `theta` for which `(dA)/(d theta) = 0` gives the maximum cross-sectional area. Find this area in terms of `w.` (2 marks)
Show Worked Solution
(i) `text(Show)\ \ A = r^2(theta – sin theta cos theta)`
`text(Area of segment)\ \ OBC`
`= (2 theta)/(2 pi) xx pi r^2`
`= r^2 theta`
| `text(Area of)\ \ Delta OBC` | `= 1/2 ab sin C` |
| `= 1/2 r * r * sin 2 theta` | |
| `= 1/2 r^2 * 2 sin theta cos theta` | |
| `= r^2 sin theta cos theta` |
`:.\ text(Shaded Area (A))`
`= text(Area of segment)\ \ OBC – text(Area of)\ \ Delta OBC`
`= r^2 theta – r^2 sin theta cos theta`
`= r^2 (theta – sin theta cos theta)\ \ text(… as required.)`
(ii) `text(Consider Arc length)\ \ BC`
| `w` | `= (2 theta)/(2 pi) xx 2 pi r` |
| `= 2 theta r` | |
| `:.\ r` | `= w/(2 theta)` |
| `:. A` | `= (w/(2 theta))^2 (theta – sin theta cos theta)` |
| `= w^2/(4 theta) – (w^2 sin theta cos theta)/(4 theta^2)` |
| `:. (dA)/(d theta)` | `= (-w^2)/(4 theta^2) – (w^2/4) [((sin theta xx -sin theta + cos theta * cos theta) theta^2 – 2 theta sin theta cos theta)/theta^4]` |
| `= (-w^2)/(4 theta^2) – (w^2/(4 theta^4))[(cos^2 theta – sin^2 theta) theta^2 – 2 theta sin theta cos theta]` | |
| `= (-w^2)/(4 theta^2) – (w^2/(4 theta^3))[(2 cos^2 theta – 1) theta – 2 sin theta cos theta]` | |
| `= (-w^2)/(4 theta^2) – (w^2/(4 theta^3))[(2 cos^2 theta – 1) theta – 2 sin theta cos theta]` | |
| `= w^2/(4 theta^3) [-theta – 2 theta cos^2 theta + theta + 2 sin theta cos theta]` | |
| `= w^2/(4 theta^3) [2 sin theta cos theta – 2 theta cos^2 theta]` | |
| `= w^2/(2 theta^3) (sin theta cos theta – theta cos^2 theta)` | |
| `= (w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3)\ \ text(… as required)` |
| (iii) | `g(theta)` | `= sin theta- theta cos theta` |
| `g prime (theta)` | `= cos theta – (theta xx -sin theta + cos theta * 1)` | |
| `= cos theta + theta sin theta – cos theta` | ||
| `= theta sin theta` |
`text(S)text(ince)\ \ 0 < theta < pi,`
`=> sin theta > 0`
`:.\ g prime (theta) > 0`
`g(0) = sin 0 – 0 * cos 0 = 0`
`:.\ text(S)text(ince)\ \ g(0) = 0 and g(theta)\ \ text(is an increasing function)`
`(g prime (theta) > 0)\ \ text(for)\ \ 0 < theta < pi , text(then)\ \ g(theta) > 0.`
(iv) `text(If)\ \ (dA)/(d theta) = 0\ ,\ \ \ 0 < theta < pi`
`(w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3) = 0`
`text(Consider)`
`(w^2 cos theta)/(2 theta^3) = 0\ ,\ \ theta != 0`
`=>theta = pi/2`
`text(Consider)`
| `sin theta – theta cos theta=` | ` 0` |
| `text(i.e.)\ \ g(theta)=` | ` 0` |
`=>\ text(No solution for)\ \ 0 < theta < pi\ \ text{(using part (iii))}`
`:.\ text(There is only one value of)\ \ theta.`
| (v) `(dA)/(d theta)` | `= (w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3)` |
| `= (w^2 cos theta * g(theta))/(2 theta^3)` |
`text(If)\ \ theta = pi/4\ ,\ \ g(pi/4) > 0\ ,\ \ cos (pi/4) > 0`
`:.\ (dA)/(d theta) > 0`
`text(If)\ \ theta = (3 pi)/4\ ,\ \ g((3 pi)/4) > 0\ ,\ \ cos ((3 pi)/4) < 0`
`:.\ (dA)/(d theta) < 0`
`:.\ text(Maximum when)\ \ theta = pi/2`
`text(When)\ \ theta = pi/2`
| `A` | `= r^2 (theta – sin theta cos theta)` |
| `= (w/(2 theta))^2 (theta – sin theta cos theta)` | |
| `= w^2/(2^2 xx (pi/2)^2) (pi/2 – sin\ pi/2 cos\ pi/2)` | |
| `= w^2/pi^2 (pi/2)` | |
| `= w^2/(2 pi)\ \ text(u²)` |
Statistics, EXT1 S1 2006 HSC 6b
In an endurance event, the probability that a competitor will complete the course is `p` and the probability that a competitor will not complete the course is `q = 1 - p.` Teams consist of either two or four competitors. A team scores points if at least half its members complete the course.
- Show that the probability that a four-member team will have at least three of its members not complete the course is `4pq^3 + q^4.` (1 mark)
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- Hence, or otherwise, find an expression in terms of `q` only for the probability that a four-member team will score points. (2 marks)
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- Find an expression in terms of `q` only for the probability that a two-member team will score points. (1 mark)
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- Hence, or otherwise, find the range of values of `q` for which a two-member team is more likely than a four-member team to score points. (2 marks)
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Show Worked Solution
i. `P text{(at least 3 don’t complete)}`
`= P\ text{(3 don’t complete)} + P\ text{(4 don’t complete)}`
`= \ ^4C_3 q^3 p + \ ^4C_4 q^4`
`= 4pq^3 + q^4`
ii. `P text{(4-member team scores)}`
`= 1 – P\ text{(at least 3 don’t complete)}`
`= 1 – 4pq^3 + q^4`
`= 1 – [4 (1 – q) q^3 + q^4]`
`= 1 – (4q^3 – 4q^4 + q^4)`
`= 1 – 4q^3 + 3q^4`
iii. `P text{(2-member team scores)}`
`= 1 – P\ text{(both don’t complete)}`
`= 1 – q*q`
`= 1 – q^2`
iv. `text(A 2-member team is more likely to score when)`
| `1 – q^2` | `> 1 -4q^3 + 3q^4` |
| `3q^4 – 4q^3 + q^2` | `< 0` |
| `q^2 (3q^2 – 4q + 1)` | `< 0` |
| `q^2 (3q – 1) (q – 1)` | `< 0` |
`text(Consider)\ \ (3q – 1) (q – 1) < 0`
`:.\ text(S)text(ince)\ \ q\ \ text(is positive and)\ != 1`
`q^2 (3q – 1) (q – 1) < 0\ \ text(when)`
`1/3 < q < 1.`
Mechanics, EXT2* M1 2006 HSC 6a
Two particles are fired simultaneously from the ground at time `t = 0.`
Particle 1 is projected from the origin at an angle `theta, \ \ 0 < theta < pi/2`, with an initial velocity `V.`
Particle 2 is projected vertically upward from the point `A`, at a distance `a` to the right of the origin, also with an initial velocity of `V.`
It can be shown that while both particles are in flight, Particle 1 has equations of motion:
`x = Vt cos theta`
`y = Vt sin theta -1/2 g t^2,`
and Particle `2` has equations of motion:
`x = a`
`y = Vt -1/2 g t^2.` Do NOT prove these equations of motion.
Let `L` be the distance between the particles at time `t.`
- Show that, while both particles are in flight,
`L^2 = 2V^2t^2 (1 - sin theta) - 2aVt cos theta + a^2.` (2 marks)
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- An observer notices that the distance between the particles in flight first decreases, then increases.
Show that the distance between the particles in flight is smallest when
`t = (a cos theta)/(2V(1 - sin theta))` and that this smallest distance is `a sqrt ((1 - sin theta)/2).` (3 marks)
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- Show that the smallest distance between the two particles in flight occurs while Particle 1 is ascending if
`V > sqrt((a g cos theta)/(2 sin theta \ (1 - sin theta))).` (1 mark)
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Show Worked Solution
| i. |  |
`text(Show)\ \ L^2 = 2V^2t^2 (1 – sin theta) – 2aVt cos theta + a^2`
`text(Consider)\ \ P_1`
`x_1 = Vt cos theta`
`y_1 = Vt sin theta – 1/2 g t^2`
`text(Consider)\ \ P_2`
`x_2 = a`
`y_2 = Vt -1/2 g t^2`
`text(Let)\ \ d=\ text(Vertical distance between particles)`
`d= Vt -1/2 g t^2 – (Vt sin theta – 1/2 g t^2)`
`d= Vt (1 – sin theta)`
`text(Using Pythagoras:)`
| `L^2` | `= (a – x_1)^2 + d^2` |
| `= (a – Vt cos theta)^2 + V^2t^2 (1 – sin theta)^2` | |
| `= a^2 – 2aVt cos theta + V^2t^2 cos ^2 theta + V^2t^2 (1 – 2 sin theta + sin^2 theta)` | |
| `= a^2 – 2aVt cos theta + V^2 t^2 (cos^2 theta + sin^2 theta + 1 – 2 sin theta)` | |
| `= a^2 – 2aVt cos theta + V^2 t^2 (2 – 2 sin theta)` | |
| `= 2 V^2 t^2 (1 – sin theta) – 2aVt cos theta + a^2\ \ text(… as required.)` |
ii. `L^2 = 2V^2 t^2 (1 – sin theta) – 2a Vt cos theta + a^2`
`(d(L^2))/(dt) = 4 V^2 t\ (1 – sin theta) – 2aV cos theta`
`text(Max or min when)\ \ (d(L^2))/(dt) = 0`
| `4V^2t\ (1 – sin theta)` | `= 2aV cos theta` |
| `t` | `= (2a V cos theta)/(4V^2 (1 – sin theta))` |
| `= (a cos theta)/(2V(1 – sin theta)` |
`(d^2(L^2))/(dt^2) = 4V^2(1 – sin theta) > 0\ \ text(for)\ \ V > 0,\ \ 0 < theta < pi/2`
`:.\ L^2\ \ text(is a minimum)`
`:.\ L\ \ text(is a minimum when)\ \ t = (a cos theta)/(2V (1 – sin theta)`
`text(Show minimum distance is)\ \ a sqrt {(1 – sin theta)/2}`
`text(When)\ \ t = (a cos theta)/(2V(1 – sin theta))`
| `L^2` | `= 2V^2 ((a^2 cos ^2 theta)/(4V^2 (1 – sin theta)^2)) (1 – sin theta)` |
| `\ \ – 2aV ((a cos theta)/(2V (1 – sin theta))) cos theta + a^2` | |
| `= (a^2 cos^2 theta)/(2 (1 – sin theta)) – (a^2 cos^2 theta)/((1 – sin theta)) + a^2` | |
| `= a^2[(cos^2 theta – 2 cos^2 theta + 2 (1 – sin theta))/(2(1 – sin theta))]` | |
| `= a^2[(-cos^2 theta + 2 – 2 sin theta)/(2(1 – sin theta))]` | |
| `= a^2[(-(1 – sin^2 theta) + 2 – 2 sin theta)/(2 (1 – sin theta))]` | |
| `= a^2 [(sin^2 theta – 2 sin theta + 1)/(2(1 – sin theta))]` | |
| `= a^2 [(1 – sin theta)^2/(2 (1 – sin theta))]` | |
| `= a^2 [((1 – sin theta))/2]` | |
| `:.\ L` | `= sqrt ((a^2(1 – sin theta))/2)` |
| `= a sqrt ((1 – sin theta)/2)\ \ text(… as required.)` |
iii. `text(Smallest distance occurs when)`
`t = (a cos theta)/(2V (1 – sin theta)`
`text(If)\ \ P_1\ \ text(is ascending,)\ \ dot y_1 > 0`
`y_1 = Vt sin theta – 1/2 g t^2`
`dot y_1 = V sin theta – g t`
| `:.\ V sin theta – g ((a cos theta)/(2V (1 – sin theta)))` | `> 0` |
| `2V^2 sin theta\ (1 – sin theta) – a g cos theta` | `> 0` |
| `2V^2 sin theta\ (1 – sin theta)` | `> ag cos theta` |
| `V^2` | `> (a g cos theta)/(2 sin theta\ (1 – sin theta))` |
| `V` | `> sqrt ((a g cos theta)/(2 sin theta\ (1 – sin theta)))\ \ text(… as required.)` |
Plane Geometry, EXT1 2007 HSC 4c
The diagram shows points `A`, `B`, `C` and `D` on a circle. The lines `AC` and `BD` are perpendicular and intersect at `X`. The perpendicular to `AD` through `X` meets `AD` at `P` and `BC` at `Q`.
Copy or trace this diagram into your writing booklet.
- Prove that `∠QXB =∠QBX`. (3 marks)
- Prove that `Q` bisects `BC`. (2 marks)
Show Worked Solution
| (i) |
|
`text(Prove)\ \ ∠QXB =∠QBX`
MARKER’S COMMENT: Many students did not copy the diagram into their answer! Efficient solutions labelled angles with a symbol.
`∠ADX = ∠ACB = theta`
`text{(angles in the same segment on arc}\ AB)`
`text(In)\ ΔDPX`
| `∠DPX` | `= 90^@\ \ (PQ ⊥ AD)` |
| `∠PXD` | `= (90 – theta)^@\ \ text{(angle sum of}\ Δ DPX)` |
| `∠QXB` | `= (90 – theta)^@\ \ text{(vertically opposite angle)}` |
`text(In)\ Δ BXC`
| `∠BXC` | `= 90^@\ \ (∠AXC\ text{is a straight angle)}` |
| `∠QBX` | `= (90 – theta)^@\ \ text{(angle sum of}\ ΔBXC)` |
| `:.∠QXB` | `= ∠QBX` |
(ii) `text(Prove)\ \ Q\ \ text(bisects)\ \ BC`
| `BQ = QX\ \ ` | `text{(sides opposite equal angles}` |
| `text{of isosceles}\ Delta BXQ text{)}` |
| `∠QXC` | `= 180 − 90 − (90 − theta)\ \ (∠AXC\ text{is a straight angle)}` |
| `= theta` | |
| `∠XCB` | `=theta\ \ \ text{(from part (i))}` |
| `:. ΔXQC\ text(is isosceles)` | |
| `QX = QC\ \ ` | `text{(sides opposite base angles}` |
| `text{of isosceles}\ ΔQXC)` |
`:. BQ = QC`
`:. Q\ text(bisects)\ BC`
Proof, EXT1 P1 2007 HSC 4b
Use mathematical induction to prove that `7^(2n – 1) + 5` is divisible by 12, for all integers `n ≥ 1`. (3 marks)
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Statistics, EXT1 S1 2007 HSC 4a
In a large city, 10% of the population has green eyes.
- What is the probability that two randomly chosen people both have green eyes? (1 mark)
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- What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal places. (1 mark)
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- What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places. (2 marks)
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Mechanics, EXT2* M1 2007 HSC 3c
A particle is moving in a straight line with its acceleration as a function of `x` given by `ddot x = -e^(-2x)`. It is initially at the origin and is travelling with a velocity of 1 metre per second.
- Show that `dot x = e^(-x)`. (2 marks)
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- Hence show that `x = log_e(t + 1)`. (2 marks)
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Functions, EXT1 F1 2007 HSC 3b
- Find the vertical and horizontal asymptotes of the hyperbola `y = (x − 2)/(x − 4)`
and hence sketch the graph of `y = (x − 2)/(x − 4)`. (3 marks)
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- Hence, or otherwise, find the values of `x` for which
`qquad qquad (x − 2)/(x − 4) ≤ 3`. (2 marks)
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Show Worked Solution
i. `y = (x − 2)/(x − 4)`
`text(Vertical asymptote at)\ x = 4`
| `lim_(x → ∞) (x − 2)/(x − 4)` | `= lim_(x → ∞) (1 − 2/x)/(1 − 4/x)=1` |
`ytext(–intercept)\ = 1/2`
`xtext(–intercept)\ = 2`
ii. `text(Find)\ \ x\ \ text(so that)\ \ (x − 2)/(x − 4) ≤ 3`
| `text(When)\ \ (x − 2)/(x − 4)` | `= 3` |
| `x − 2` | `= 3x − 12` |
| `2x` | `= 10` |
| `x` | `= 5` |
`=>(5, 3)\ \ text(is the intersection of)\ \ y = 3\ and\ y = (x − 2)/(x − 4)`
`:. (x − 2)/(x − 4) ≤ 3\ \ text(when)\ \ x < 4\ \ text(and)\ \ x ≥ 5.`
Calculus, EXT1* C3 2007 HSC 3a
Find the volume of the solid of revolution formed when the region bounded by the curve `y = 1/(sqrt(9 + x^2))`, the `x`-axis, the `y`-axis and the line `x = 3`, is rotated about the `x`-axis. (3 marks)
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Mechanics, EXT2* M1 2007 HSC 2d
A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, `v` metres per second, at which she is falling `t` seconds after jumping is given by `v =50(1 - e^(-0.2t))`.
- Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place. (2 marks)
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- Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre. (2 marks)
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Trigonometry, EXT1 T1 2007 HSC 2b
Let `f(x) = 2 cos^(-1)x`.
- Sketch the graph of `y = f(x)`, indicating clearly the coordinates of the endpoints of the graph. (2 marks)
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- State the range of `f(x)`. (1 mark)
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Show Worked Solution
i. `y= 2\ cos^(-1)x`
`text(Domain:)\ -1 ≤ x ≤ 1`
| `text{Range:}\ \0 ≤` | `y/2 ≤ pi` | |
| `0 ≤` | `y ≤ 2pi` |
ii. `text(Range:)\ \ 0 ≤ y ≤ 2pi`
Trigonometry, EXT1 T3 2007 HSC 2a
By using the substitution `t = tan\ theta/2`, or otherwise, show that `(1 − cos\ theta)/(sin\ theta) = tan\ theta/2`. (2 marks)
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Show Worked Solution
COMMENT: The values of `sin theta` and `cos theta` can be committed to memory or quickly derived using double angle formulae (as shown in the worked solution).
`=>tan\ theta/2 = t`
| `=>sin theta` | `= 2 · t/(sqrt(1 + t^2)) · 1/(sqrt(1 + t^2)) = (2t)/(1 + t^2)` |
| `=>cos theta` | `= 1/(1 + t^2) − t^2/(1 + t^2) = (1 − t^2)/(1 + t^2)` |
`text(Show)\ \ (1 − cos theta)/(sin theta) = tan\ theta/2 :`
| `(1 − cos\ theta)/(sin theta)` | `= (1 − ((1 − t^2)/(1 + t^2)))/((2t)/(1 + t^2)) xx (1+t^2)/(1+t^2)` |
| `= (1 + t^2 − (1 − t^2))/(2t)` | |
| `= (2t^2)/(2t)` | |
| `= t` | |
| `= tan\ theta/2\ \ …text(as required)` |
Calculus, EXT1 C2 2007 HSC 1e
Use the substitution `u = 25 - x^2` to evaluate `int_3^4 (2x)/(sqrt(25 - x^2))\ dx`. (3 marks)
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Mechanics, EXT2* M1 2004 HSC 6b
A fire hose is at ground level on a horizontal plane. Water is projected from the hose. The angle of projection, `theta`, is allowed to vary. The speed of the water as it leaves the hose, `v` metres per second, remains constant. You may assume that if the origin is taken to be the point of projection, the path of the water is given by the parametric equations
`x = vt\ cos\ theta`
`y = vt\ sin\ theta − 1/2 g t^2`
where `g\ text(ms)^(−2)` is the acceleration due to gravity. (Do NOT prove this.)
- Show that the water returns to ground level at a distance`(v^2\ sin\ 2theta)/g` metres from the point of projection. (2 marks)
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This fire hose is now aimed at a 20 metre high thin wall from a point of projection at ground level 40 metres from the base of the wall. It is known that when the angle `theta` is 15°, the water just reaches the base of the wall.
- Show that `v^2 = 80g`. (1 mark)
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- Show that the cartesian equation of the path of the water is given by
`y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`. (2 marks)
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- Show that the water just clears the top of the wall if
`tan^2\ theta − 4\ tan\ theta + 3 = 0`. (2 marks)
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- Find all values of `theta` for which the water hits the front of the wall. (2 marks)
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Show Worked Solution
| i. | `x` | `= vt\ cos\ theta` |
| `y` | `= vt\ sin\ theta − 1/2 g t^2` |
`text(Find)\ \ t\ \ text(when)\ \ y = 0`
| `vt\ sin\ theta − 1/2 g t^2` | `= 0` |
| `t(v\ sin\ theta − 1/2 g t)` | `= 0` |
| `v\ sin\ theta − 1/2 g t` | `= 0, \ \ t ≠ 0` |
| `1/2 g t` | `= v\ sin\ theta` |
| `t` | `= (2v\ sin\ theta)/g` |
`text(Find)\ \ x\ \ text(when)\ \ t = (2v\ sin\ theta)/g`
| `x` | `= v · (2v\ sin\ theta)/g\ cos\ theta` |
| `= (v^2 · \ 2\ sin\ theta\ cos\ theta)/g` | |
| `= (v^2\ sin\ 2theta)/g` |
`:.\ text(When)\ \ x = (v^2\ sin\ 2theta)/g,\ \ \text(the water returns)`
`text(to ground level … as required.)`
ii. `text(Show)\ \ v^2 = 80\ text(g)`
`text(When)\ \ theta = 15^@, \ x = 40`
`text{From part (i)}`
| `40` | `= (v^2\ sin\ 30^@)/g` |
| `v^2 xx 1/2` | `= 40g` |
| `v^2` | `= 80g\ \ \ …text(as required)` |
iii. `text(Show)\ \ y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`
| `x` | `= vt\ cos\ theta` |
| `:.t` | `= x/(v\ cos\ theta)` |
`text(Substitute into)`
| `y` | `= vt\ sin\ theta − 1/2 g t^2` |
| `= v · x/(v\ cos\ theta) · sin\ theta −1/2 g (x/(v\ cos\ theta))^2` | |
| `= x\ tan\ theta − 1/2 g (x^2/(v^2\ cos^2\ theta))` | |
| `= x\ tan\ theta − 1/2 g · x^2/(80g\ cos^2\ theta)` | |
| `= x\ tan\ theta − (x^2\ sec^2\ theta)/160\ \ \ …text(as required.)` |
iv. `text(Water clears the top if)\ \ y = 20\ \ text(when)\ \ x = 40`
`text{Substitute into equation from (iii)}`
| `40\ tan\ theta − (40^2\ sec^2\ theta)/160` | `= 20` |
| `40\ tan\ theta − 10\ sec^2\ theta` | `= 20` |
| `40\ tan\ theta − 10(1 + tan^2\ theta)` | `= 20` |
| `40\ tan\ theta − 10 − 10\ tan^2\ theta` | `= 20` |
| `10\ tan^2\ theta − 40\ tan\ theta\ + 30` | `= 0` |
| `tan^2\ theta − 4\ tan\ theta + 3` | `= 0\ \ \ text(… as required)` |
| v. |
|
`text(Water hits the bottom of the wall when)`
`x = 40\ \ text(and)\ \ y = 0`
| `40\ tan\ theta − (40^2\ sec^2\ theta)/160` | `= 0` |
| `40\ tan\ theta − 10\ sec^2\ theta` | `= 0` |
| `4\ tan\ theta − (1 + tan^2\ theta)` | `= 0` |
| `tan^2\ theta − 4\ tan\ theta + 1` | `= 0` |
`text(Using the quadratic formula)`
| `tan\ theta` | `= (+4 ± sqrt(16 − 4 · 1 · 1))/ (2 xx 1)` |
| `= (4 ± sqrt12)/2` | |
| `= 2 ± sqrt3` | |
| `theta` | `= 15^@\ \ text(or)\ \ 75^@` |
`text(Water hits the top of the wall when)`
`x = 40\ text(and)\ \ y = 20`
| `tan^2\ theta − 4\ tan\ theta + 3` | `= 0\ \ \ text{from (iv)}` |
| `(tan\ theta − 1)(tan\ theta − 3)` | `= 0` |
| `tan\ theta` | `= 1` | `text(or)` | `tan\ theta` | `= 3` |
| `theta` | `= 45^@` | `theta` | `= tan^(−1)\ 3` | |
| `= 71.565…` | ||||
| `= 71.6^@\ \ \text{(to 1 d.p.)}` |
`:.\ text(Following the motion of the water as)\ \ theta\ \ text(increases,)`
`text(water hits the wall when)`
| `15^@ ≤ theta ≤ 45^@` | `\ text(and)` |
| `71.6^@ ≤ theta ≤ 75^@` |
Plane Geometry, EXT1 2004 HSC 6a
The points `A, \ B, \ C` and `D` are placed on a circle of radius `r` such that `AC` and `BD` meet at `E`. The lines `AB` and `DC` are produced to meet at `F`, and `BECF` is a cyclic quadrilateral.
Copy or trace this diagram into your writing booklet.
- Find the size of `∠DBF`, giving reasons for your answer. (2 marks)
- Find an expression for the length of `AD` in terms of `r`. (1 mark)
Show Worked Solution
| (i) |
|
`∠ACD = ∠ABD = theta\ \ \ text{(angles in the same segment on arc}\ AD)`
`text(Consider cyclic quad)\ \ BECF`
| `∠DBF = ∠ECD = theta` | `\ \ \ text{(exterior angle of cycle quad}` |
| `\ \ \ text{equals its interior opposite)}` |
| `2theta` | `= 180^@\ \ \ (∠ABF\ text{is a straight angle)}` |
| `theta` | `= 90^@` |
| `:.∠DBF` | `= 90^@` |
(ii) `text(S)text(ince)\ AD\ text(subtends right angles)\ \ ∠ABD\ \ text(and)\ \ ∠ACD`
`text{(from part (i))}`
`⇒ AD\ \ text(is a diameter)`
`:.AD = 2r`
Inverse Functions, EXT1 2004 HSC 5b
The diagram below shows a sketch of the graph of `y = f(x)`, where `f(x) = 1/(1 + x^2)` for `x ≥ 0`.
- Copy or trace this diagram into your writing booklet.
On the same set of axes, sketch the graph of the inverse function, `y = f^(−1)(x)`. (1 mark) - State the domain of `f^(−1)(x)`. (1 mark)
- Find an expression for `y = f^(−1)(x)` in terms of `x`. (2 marks)
- The graphs of `y = f(x)` and `y = f^(−1)(x)` meet at exactly one point `P`.
- Let `α` be the `x`-coordinate of `P`. Explain why `α` is a root of the equation
- `x^3 + x − 1 = 0`. (1 mark)
- Take 0.5 as a first approximation for `α`. Use one application of Newton’s method to find a second approximation for `α`. (2 marks)
Show Worked Solution
| (i) |
|
(ii) `text(Domain of)\ \ f^(−1)(x)\ text(is)`
`0 < x ≤ 1`
(iii) `f(x) = 1/(1 + x^2)`
`text(Inverse: swap)\ \ x↔y`
| `x` | `= 1/(1 + y^2)` |
| `x(1 + y^2)` | `= 1` |
| `1 + y^2` | `= 1/x` |
| `y^2` | `= 1/x − 1` |
| `= (1 − x)/x` | |
| `y` | `= ± sqrt((1 − x)/x)` |
`:.y = sqrt((1 − x)/x), \ \ y >= 0`
(iv) `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`
`text(i.e. where)`
| `1/(1 + x^2)` | `= x` |
| `1` | `= x(1 + x^2)` |
| `1` | `= x + x^3` |
| `x^3 + x − 1` | `= 0` |
`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`
`text(it is a root of)\ \ \ x^3 + x − 1 = 0`
| (v) | `f(x)` | `= x^3 + x − 1` |
| `f′(x)` | `= 3x^2 + 1` | |
| `f(0.5)` | `= 0.5^3 + 0.5 − 1` | |
| `= −0.375` | ||
| `f′(0.5)` | `= 3 xx 0.5^2 + 1` | |
| `= 1.75` |
| `α_2` | `= α_1 − (f(0.5))/(f′(0.5))` |
| `= 0.5 − ((−0.375))/1.75` | |
| `= 0.5 − (−0.2142…)` | |
| `= 0.7142…` | |
| `= 0.714\ \ \ text{(to 3 d.p.)}` |
Mechanics, EXT2* M1 2004 HSC 5a
A particle is moving along the `x`-axis, starting from a position `2` metres to the right of the origin (that is, `x = 2` when `t = 0`) with an initial velocity of `5\ text(ms)^(−1)` and an acceleration given by
`ddot x = 2x^3 + 2x`.
- Show that `dot x = x^2 + 1`. (2 marks)
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- Hence find an expression for `x` in terms of `t`. (3 marks)
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Combinatorics, EXT1 A1 2004 HSC 4c
Katie is one of ten members of a social club. Each week one member is selected at random to win a prize.
- What is the probability that in the first 7 weeks Katie will win at least 1 prize? (1 mark)
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- Show that in the first 20 weeks Katie has a greater chance of winning exactly 2 prizes than of winning exactly 1 prize. (2 marks)
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- For how many weeks must Katie participate in the prize drawing so that she has a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes? (2 marks)
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Induction, EXT1 2004 HSC 4a
Use mathematical induction to prove that for all integers `n ≥ 3`,
`(1 − 2/3)(1 − 2/4)(1 − 2/5)…(1 − 2/n) = 2/(n(n − 1))`. (3 marks)
Trig Ratios, EXT1 2004 HSC 3d
The length of each edge of the cube `ABCDEFGH` is 2 metres. A circle is drawn on the face `ABCD` so that it touches all four edges of the face. The centre of the circle is `O` and the diagonal `AC` meets the circle at `X` and `Y`.
- Explain why `∠FAC = 60^@`. (1 mark)
- Show that `FO = sqrt6` metres. (1 mark)
- Calculate the size of `∠XFY` to the nearest degree. (1 mark)
Show Worked Solution
| (i) |
|
`text(S)text(ince)\ \ FA, \ AC\ \ text(and)\ \ FC\ \ text(are all)`
`text(diagonals of sides of a cube,)`
`FA = AC = FC`
`:.ΔFAC\ \ text(is equilateral)`
`:.∠FAC = 60^@`
| (ii) |
|
`text(In)\ \ ΔAEF`
| `AF^2` | `= EF^2 + EA^2` |
| `= 2^2 + 2^2` | |
| `= 8` | |
| `AF` | `= sqrt8` |
| `= 2sqrt2` |
`text(In)\ \ ΔAFO`
| `sin\ 60^@` | `= (FO)/(AF)` |
| `sqrt3/2` | `= (FO)/(2sqrt2)` |
| `FO` | `= sqrt3/2 xx 2sqrt2` |
| `= sqrt6\ text(metres … as required.)` |
| (iii) |
|
`XY\ \ text(is the diameter of a circle AND the width)`
`text(of the cube.)`
| `:.XY` | `= 2` |
| `:.OX` | `= OY = 1` |
| `tan\ ∠OFX` | `=1 /sqrt6` |
| `∠OFX` | `= 22.207…^@` |
| `:.∠XFY` | `= 2 xx 22.407…` |
| `= 44.415…` | |
| `= 44^@\ text{(nearest degree)}` |
Calculus, EXT1 C1 2004 HSC 3c
A ferry wharf consists of a floating pontoon linked to a jetty by a 4 metre long walkway. Let `h` metres be the difference in height between the top of the pontoon and the top of the jetty and let `x` metres be the horizontal distance between the pontoon and the jetty.
- Find an expression for `x` in terms of `h`. (1 mark)
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When the top of the pontoon is 1 metre lower than the top of the jetty, the tide is rising at a rate of 0.3 metres per hour.
- At what rate is the pontoon moving away from the jetty? (3 marks)
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Functions, EXT1 F2 2004 HSC 3b
Let `P(x) = (x + 1) (x − 3)Q(x) + a(x + 1) + b`, where `Q(x)` is a polynomial and `a` and `b` are real numbers.
When `P(x)` is divided by `(x + 1)` the remainder is `−11`.
When `P(x)` is divided by `(x − 3)` the remainder is `1`.
- What is the value of `b`? (1 mark)
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- What is the remainder when `P(x)` is divided by `(x + 1)(x − 3)`? (2 marks)
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Trig Calculus, EXT1 2004 HSC 3a
Find `int cos^2 4x\ dx.` (2 marks)
Combinatorics, EXT1 A1 2004 HSC 2e
A four-person team is to be chosen at random from nine women and seven men.
- In how many ways can this team be chosen? (1 mark)
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- What is the probability that the team will consist of four women? (1 mark)
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Trigonometry, EXT1 T3 2004 HSC 2d
- Write `8 cos x + 6 sin x` in the form `A cos(x −α)`, where `A > 0` and `0 ≤α ≤ π/2`. (2 marks)
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- Hence, or otherwise, solve the equation `8cos x + 6 sin x = 5` for `0 ≤ x ≤ 2π`.
Give your answers correct to three decimal places. (2 marks)
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Calculus, EXT1 C2 2004 HSC 2b
Find `d/(dx)\ cos^(−1)\ (3x^2).` (2 marks)
Trig Calculus, EXT1 2004 HSC 2a
Evaluate `lim_(x → 0)(sin (x/5))/(2x).` (2 marks)
Functions, EXT1 F1 2006 HSC 5b
Let `f(x) = log_e (1 + e^x)` for all `x`.
Show that `f(x)` has an inverse. (2 marks)
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Mechanics, EXT2* M1 2006 HSC 4c
A particle is moving so that `ddot x = 18x^3 + 27x^2 + 9x.`
Initially `x = – 2` and the velocity, `v`, is `– 6.`
- Show that `v^2 = 9x^2 (1 + x)^2.` (2 marks)
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- Hence, or otherwise, show that
`int 1/(x(1 + x)) \ dx = -3t.` (2 marks)
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- It can be shown that for some constant `c`,
`log_e (1 + 1/x) = 3t + c.` (Do NOT prove this.)
Using this equation and the initial conditions, find `x` as a function of `t.` (2 marks)
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Functions, EXT1 F2 2006 HSC 4a
The cubic polynomial `P(x) = x^3 + rx^2 + sx + t`. where `r, \ s` and `t` are real numbers, has three real zeros, `1, alpha` and `-alpha.`
- Find the value of `r.` (1 mark)
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- Find the value of `s + t.` (2 marks)
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Plane Geometry, EXT1 2006 HSC 3d
The points `P, Q` and `T` lie on a circle. The line `MN` is tangent to the circle at `T` with `M` chosen so that `QM` is perpendicular to `MN`. The point `K` on `PQ` is chosen so that `TK` is perpendicular to `PQ` as shown in the diagram.
- Show that `QKTM` is a cyclic quadrilateral. (1 mark)
- Show that `/_KMT = /_KQT.` (1 mark)
- Hence, or otherwise, show that `MK` is parallel to `TP.` (2 marks)
Show Worked Solution
| (i) |  |
`/_ QMT = 90^@\ \ \ (QM _|_ MN)`
`/_ QKT = 90^@\ \ \ (PQ _|_ TK)`
`:.\ /_ QMT + /_ QKT = 180^@`
`:.\ QKTM\ \ text(is cyclic)\ \ text{(opposite angles are supplementary)}`
`text(… as required.)`
(ii) `text(Show)\ \ /_ KMT = /_ KQT`
`/_ KMQ = /_ KTQ = theta`
`text{(angles in the same segment on arc}\ \ KQ text{)}`
| `/_ KQT` | `= 90 – theta\ \ text{(angle sum of}\ \ Delta KQT text{)}` |
| `/_ KMT` | `= /_ QMT – /_ KMQ` |
| `= 90 – theta` |
`:.\ /_ KMT = /_ KQT\ \ text(… as required.)`
(iii) `text(Show)\ \ MK\ text(||)\ TP`
| `/_ NTP` | `= /_ KQT\ \ text{(angle in alternate segment)` |
| `= 90 – theta` |
`:.\ /_ NTP = /_ KMT\ \ text{(from part (ii))}`
`:.\ MK\ text(||)\ TP\ \ text{(corresponding angles are equal)}`
Combinatorics, EXT1 A1 2006 HSC 3c
Sophie has five coloured blocks: one red, one blue, one green, one yellow and one white. She stacks two, three, four or five blocks on top of one another to form a vertical tower.
- How many different towers are there that she could form that are three blocks high? (1 mark)
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- How many different towers can she form in total? (2 marks)
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Polynomials, EXT1 2006 HSC 3b
- By considering `f(x) = 3log_e x - x`, show that the curve `y = 3 log_e x` and the line `y = x` meet at a point `P` whose `x`-coordinate is between `1.5` and `2`. (1 mark)
- Use one application of Newton’s method, starting at `x = 1.5`, to find an approximation to the `x`-coordinate of `P`. Give your answer correct to two decimal places. (2 marks)
Trig Calculus, EXT1 2006 HSC 3a
Find `int_0^(pi/4) sin^2 x\ dx.` (2 marks)
Quadratic, EXT1 2006 HSC 2c
The points `P(2ap, ap^2), Q(2aq, aq^2)` and `R(2ar, ar^2)` lie on the parabola `x^2 = 4ay`. The chord `QR` is perpendicular to the axis of the parabola. The chord `PR` meets the axis of the parabola at `U`.
The equation of the chord `PR` is `y = 1/2(p + r)x - apr.` (Do NOT prove this.)
The equation of the tangent at `P` is `y = px - ap^2.` (Do NOT prove this.)
- Find the coordinates of `U.` (1 mark)
- The tangents at `P` and `Q` meet at the point `T`. Show that the coordinates of `T` are `(a(p + q), apq).` (2 marks)
- Show that `TU` is perpendicular to the axis of the parabola. (1 mark)
Binomial, EXT1 2006 HSC 2b
- By applying the binomial theorem to `(1 + x)^n` and differentiating, show that
- `n(1 + x)^(n - 1) = ((n), (1)) + 2((n), (2)) x + … + r((n), (r)) x^(r - 1) + … + n((n), (n)) x^(n - 1).` (1 mark)
- Hence deduce that
- `n3^(n - 1) = ((n), (1)) + … + r((n), (r)) 2^(r - 1) + … + n((n), (n)) 2^(n - 1).` (1 mark)
Calculus, EXT1 C2 2006 HSC 2a
Let `f(x) = sin^-1 (x + 5).`
- State the domain and range of the function `f(x).` (2 marks)
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- Find the gradient of the graph of `y = f(x)` at the point where `x = -5.` (2 marks)
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- Sketch the graph of `y = f(x).` (2 marks)
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Show Answers Only
- `text(Domain):\ -6 <= x <= -4, \ \ \ text(Range):\ -pi/2 <= y <= pi/2`
- `1`
-
Show Worked Solution
i. `f(x) = sin^-1 (x + 5)`
`text(Domain)`
`-1 <= x + 5 <= 1`
`-6 <= x <= -4`
`text(Range)`
`-pi/2 <= y <= pi/2`
ii. `y = sin^-1 (x + 5)`
`(dy)/(dx) = 1/sqrt(1-(x + 5)^2)`
`text(When)\ \ x = -5`
| `(dy)/(dx)` | `= 1/sqrt(1-(-5 + 5)^2)` |
| `= 1/sqrt(1-0)` | |
| `= 1` |
`:.\ text(Gradient of)\ \ y = f(x)\ \ text(at)\ \ x = -5\ \ text(is)\ \ 1.`
|
iii. |
 |
Differentiation, EXT1 2006 HSC 1e
For what values of `b` is the line `y =12x + b` tangent to `y =x^3`? (3 marks)
Trig Calculus, EXT1 2006 HSC 1c
Evaluate `lim_(x -> 0) (sin 5x)/(3x).` (2 marks)
Mechanics, EXT2* M1 2005 6b
An experimental rocket is at a height of 5000 m, ascending with a velocity of ` 200 sqrt 2\ text(m s)^-1` at an angle of 45° to the horizontal, when its engine stops.
After this time, the equations of motion of the rocket are:
`x = 200t`
`y = -4.9t^2 + 200t + 5000,`
where `t` is measured in seconds after the engine stops. (Do NOT show this.)
- What is the maximum height the rocket will reach, and when will it reach this height? (2 marks)
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- The pilot can only operate the ejection seat when the rocket is descending at an angle between 45° and 60° to the horizontal. What are the earliest and latest times that the pilot can operate the ejection seat? (3 marks)
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- For the parachute to open safely, the pilot must eject when the speed of the rocket is no more than `350\ text(m s)^-1`. What is the latest time at which the pilot can eject safely? (2 marks)
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Show Worked Solution
i.
`y = -4.9t^2 + 200t + 5000`
`dot y = -9.8t + 200`
`text(Max height occurs when)\ \ dot y = 0`
| `9.8t` | `= 200` |
| `t` | `= 20.408…` |
| `= 20.4\ text{seconds (to 1 d.p.)}` |
`text(When)\ t = 20.408…`
| `y` | `= -4.9 (20.408…)^2 + 200 (20.408…) + 5000` |
| `= 7040.816…` | |
| `= 7041\ text{m (to nearest metre)}` |
`:.\ text(The rocket will reach a maximum height of)`
`text(7041 metres when)\ \ t = 20.4\ text(seconds.)`
ii. `text(The rocket is descending at 45° at point)\ A\ text(on the graph.)`
`text(The symmetry of the parabolic motion means that)\ \ A`
`text(occurs when)\ \ t = 2 xx text(time to reach max height, or)`
`40.8\ text(seconds.)`
`text(Point)\ B\ text(occurs when the rocket is descending at 60°)`
`text(to the horizontal.)`
`text(At)\ \ B,`
`-tan 60° = (dot y)/(dot x)`
`text{(The gradient of the projectile becomes}`
`text{negative after reaching its max height.)}`
`dot y = -9.8t + 200`
`dot x = d/(dt) (200t) = 200`
| `:.\ – sqrt 3` | `= (-9.8t + 200)/200` |
| `-200 sqrt 3` | `= -9.8t + 200` |
| `9.8t` | `= 200 + 200 sqrt 3` |
| `t` | `= (200 + 200 sqrt 3)/9.8` |
| `= (546.410…)/9.8` | |
| `= 55.756…` | |
| `= 55.8\ text{seconds (nearest second)}` |
`:.\ text(The pilot can operate the ejection seat)`
`text(between)\ \ t = 40.8 and t = 55.8\ text(seconds.)`
|
iii. |
 |
`v^2 = (dot x)^2 + (dot y)^2`
`text(When)\ \ v = 350`
| `350^2` | `= 200^2 + (-9.8t + 200)^2` |
| `122\ 500` | `= 40\ 000 + (-9.8t + 200)^2` |
| `(-9.8t + 200)^2` | `= 82\ 500` |
| `-9.8t + 200` | `= +- sqrt (82\ 500)` |
| `9.8t` | `= 200 +- sqrt (82\ 500)` |
| `t` | `= (200 +- sqrt (82\ 500))/9.8` |
| `= 49.717…\ \ \ (t > 0)` | |
| `= 49.7\ text{seconds (to 1 d.p.)}` |
`:.\ text(The latest time the pilot can eject safely)`
`text(is when)\ \ t = 49.7\ text(seconds.)`
Statistics, EXT1 S1 2005 HSC 6a
There are five matches on each weekend of a football season. Megan takes part in a competition in which she earns one point if she picks more than half of the winning teams for a weekend, and zero points otherwise. The probability that Megan correctly picks the team that wins any given match is `2/3`.
- Show that the probability that Megan earns one point for a given weekend is 0.7901, correct to four decimal places. (2 marks)
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- Hence find the probability that Megan earns one point every week of the eighteen-week season. Give your answer correct to two decimal places. (1 mark)
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- Find the probability that Megan earns at most 16 points during the eighteen-week season. Give your answer correct to two decimal places. (2 marks)
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Mechanics, EXT2* M1 2005 HSC 5c
A particle moves in a straight line and its position at time `t` is given by
`x = 5 + sqrt 3 sin3t - cos 3t.`
- Express `sqrt 3 sin3t − cos 3t` in the form `R sin(3t - alpha)` where `alpha` is in radians. (2 marks)
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- The particle is undergoing simple harmonic motion. Find the amplitude and the centre of the motion. (2 marks)
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- When does the particle first reach its maximum speed after time `t = 0`? (1 mark)
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Plane Geometry, EXT1 2005 HSC 5b
Two chords of a circle, `AB` and `CD`, intersect at `E`. The perpendiculars to `AB` at `A` and `CD` at `D` intersect at `P`. The line `PE` meets `BC` at `Q`, as shown in the diagram.
- Explain why `DPAE` is a cyclic quadrilateral. (1 mark)
- Prove that `/_ APE = /_ ABC.` (2 marks)
- Deduce that `PQ` is perpendicular to `BC.` (1 mark)
Show Worked Solution
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(i)
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`/_ PAE = /_ PDE = 90°\ \ text{(given)}`
`:. DPAE\ \ text(is a cyclic quadrilateral)`
`text{(opposite angles are supplementary)}`
(ii) `text(Prove)\ \ /_ APE = /_ ABC`
`/_ ABC = /_ ADE`
`text{(angles in the same segment on arc}\ AC text{)}`
`text(S) text(ince)\ \ DPAE\ \ text(is a cyclic quad)\ \ text{(from (i))}`
`/_ APE = /_ ADE`
`text{(angles in the same segment on arc}\ AE text{)}`
`:.\ /_ APE = /_ ABC\ \ text(… as required.)`
(iii) `/_ APE = /_ ABC\ \ \ text{(from part (ii))}`
`/_ AEP = /_ BEQ\ \ \ text{(vertically opposite angles)}`
`:.\ Delta APE\ \ text(|||)\ \ Delta QBE\ \ text{(equiangular)}`
`:.\ /_ EQB = /_ EAP = 90°`
`text{(corresponding angles of similar triangles)}`
`:.\ PQ _|_ BC\ \ text(… as required.)`
Calculus, EXT1 C3 2005 HSC 5a
Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve `y = sin 2x`, the `x`-axis and the line `x = pi/8` is rotated about the `x`-axis. (3 marks)
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Quadratic, EXT1 2005 HSC 4c
The points `P (2ap, ap^2)` and `Q (2aq, aq^2)` lie on the parabola `x^2 = 4ay`.
The equation of the normal to the parabola at `P` is `x + py = 2ap + ap^3` and the equation of the normal at `Q` is similarly given by `x + qy = 2aq + aq^3.`
- Show that the normals at `P` and `Q` intersect at the point `R` whose coordinates are
- `(–apq[p + q], a[p^2 + pq + q^2 + 2]).` (2 marks)
- `(–apq[p + q], a[p^2 + pq + q^2 + 2]).` (2 marks)
- The equation of the chord `PQ` is
- `y = 1/2 (p + q) x - apq.` (Do NOT show this.)
- If the chord `PQ` passes through `(0, a)`, show that `pq = –1.` (1 mark)
- Find the equation of the locus of `R` if the chord `PQ` passes through `(0, a).` (2 marks)
Show Worked Solution
(i) `text(Show)\ \ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`
`text(Equations of normals through)\ \ P and Q`
| `x + py` | `= 2ap + ap^3` | `\ \ text{… (1)}` |
| `x + qy` | `= 2aq + aq^3` | `\ \ text{… (2)}` |
`text{Multiply (1)} xx q\ ,\ \ text{(2)} xx p`
| `qx + pqy` | `= 2apq + ap^3q` | `\ \ text{… (3)}` |
| `px + pqy` | `= 2apq + apq^3` | `\ \ text{… (4)}` |
`text{Subtract (4) – (3)}`
| `x (p – q)` | `= apq^3 – ap^3q` |
| `= apq (q^2 – p^2)` | |
| `= apq (q – p) (q + p)` | |
| `= -apq (p -q) (p + q)` | |
| `x` | `= -apq [p + q]` |
`text(Substitute)\ \ x = -apq [p + q]\ \ text{into (1)}`
| `py – apq[p + q]` | `= 2ap + ap^3` |
| `py` | `= 2ap + ap^3 + apq (p + q)` |
| `y` | `= 2a + ap^2 + aq (p + q)` |
| `= 2a + ap^2 + apq + aq^2` | |
| `= a[p^2 + pq + q^2 + 2]` |
`:.\ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`
`text(… as required.)`
(ii) `PQ\ \ text(is)\ \ y = 1/2 (p + q)x – apq`
`text(Passes)\ \ (0, a)`
| `a` | `= 1/2 (p + q)0 – apq` |
| `a` | `= -apq` |
| `:. pq` | `= -1\ \ text(… as required)` |
(iii) `R\ \ text(has coordinates)`
| `x` | `= -apq [p + q]` |
| `x` | `= a(p + q)\ \ \ text{(using part (ii))}` |
| `x/a` | `=(p + q)` |
| `y` | `= a [p^2 + pq + q^2 + 2]` |
| `= a (p^2 – 1 + q^2 + 2)` | |
| `= a (p^2 + q^2 +1)` | |
| `= a [(p + q)^2 – 2pq + 1]` | |
| `= a [(p + q)^2 + 3]` | |
| `= a [(x/a)^2 + 3]` | |
| `= x^2/a + 3a` |
`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = x^2/a + 3a`
Plane Geometry, EXT1 2005 HSC 3d
In the circle centred at `O` the chord `AB` has length `7`. The point `E` lies on `AB` and `AE` has length `4`. The chord `CD` passes through `E`.
Let the length of `CD` be `l` and the length of `DE` be `x`.
- Show that `x^2 - lx + 12 = 0.` (2 marks)
- Find the length of the shortest chord that passes through `E.` (2 marks)
Show Worked Solution
(i) `text(Show)\ \ x^2 – lx + 12 = 0`
| `AB` | `= 7` |
| `:.\ EB` | `= 7 – 4 = 3` |
`AE xx EB = ED xx CE`
`text{(intercepts of intersecting chords)}`
| `4 xx 3` | `= x (l – x)` |
| `12` | `= xl – x^2` |
`:.\ x^2 – lx + 12 = 0\ \ text(… as required.)`
(ii) `x^2 – lx + 12 = 0\ \ text(has a solution)`
MARKER’S COMMENT: A majority of students tried to differentiate this equation, not realising that `l` is a variable, NOT a constant.
`text(When)\ \ Delta >= 0`
| `b^2 – 4ac` | `>= 0` |
| `(-l)^2 – 4 * 1 * 12` | `>= 0` |
| `l^2 – 48` | `>= 0` |
| `l^2` | `>= 48` |
| `l` | `>= sqrt 48` |
| `>= 4 sqrt 3` |
`:.\ text(The shortest chord that could pass through)`
`E\ \ text(is)\ \ 4 sqrt 3\ text(units.)`
Calculus, EXT1 C2 2005 HSC 3b
- By expanding the left-hand side, show that
- `qquad sin(5x + 4x) + sin(5x-4x) = 2 sin (5x) cos(4x)` (1 mark)
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- Hence find `int sin(5x) cos (4x)\ dx.` (2 marks)
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Polynomials, EXT1 2005 HSC 3a
- Show that the function `g(x) = x^2 - log_e (x + 1)` has a zero between `0.7` and `0.9.` (1 mark)
- Use the method of halving the interval to find an approximation to this zero of `g(x)`, correct to one decimal place. (2 marks)
Calculus, EXT1 C2 2005 HSC 2a
Find `d/(dx) (2 sin^-1 5x).` (2 marks)
Linear Functions, EXT1 2005 HSC 1f
The acute angle between the lines `y = 3x + 5` and `y = mx + 4` is `45°`. Find the two possible values of `m`. (2 marks)
Calculus, EXT1 C2 2005 HSC 1d
Using the substitution `u = 2x^2 + 1`, or otherwise, find `int x (2x^2 + 1)^(5/4)\ dx.` (3 marks)
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Linear Functions, EXT1 2005 HSC 1b
Sketch the region in the plane defined by `y <= |\ 2x + 3\ |.` (2 marks)
Show Answers Only
Show Worked Solution
`y <= |2x + 3|`
♦ This question surprised the markers by being poorly done.
`text(Consider)\ \ (0, 0)`
`0 <= |\ 0 + 3\ |\ \ =>\ \ text(correct.)`
`:.\ text(Shaded area defines)\ \ y <= |2x + 3|`
Calculus, 2ADV C3 2007 HSC 10b
The noise level, `N`, at a distance `d` metres from a single sound source of loudness `L` is given by the formula
`N = L/d^2.`
Two sound sources, of loudness `L_1` and `L_2` are placed `m` metres apart.
The point `P` lies on the line between the sound sources and is `x` metres from the sound source with loudness `L_1.`
- Write down a formula for the sum of the noise levels at `P` in terms of `x`. (1 mark)
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- There is a point on the line between the sound sources where the sum of the noise levels is a minimum.
Find an expression for `x` in terms of `m`, `L_1` and `L_2` if `P` is chosen to be this point. (4 marks)
--- 8 WORK AREA LINES (style=lined) ---
Show Worked Solution
| i. |  |
`N = L/d^2`
| `text(Noise from)\ L_1` | `= L_1/x^2` |
| `text(Noise from)\ L_2` | `= L_2/(m-x)^2` |
| `:. N` | `= L_1/x^2 + L_2/(m-x)^2` |
ii. `N = L_1\ x^-2 + L_2 (m – x)^-2`
| `(dN)/(dx)` | `= -2 L_1 x^-3 + -2 L_2 (m – x)^-3 xx d/(dx) (m – x)` |
| `= (-2 L_1)/x^3 + (2 L_2)/(m – x)^3` |
`text(Max or min when)\ (dN)/(dx) = 0`
| `(2 L_1)/x^3` | `= (2 L_2)/(m – x)^3` |
| `2 L_1 (m – x)^3` | `= 2 L_2\ x^3` |
| `L_1 (m – x)^3` | `= L_2\ x^3` |
| `root 3 L_1 (m – x)` | `= root 3 L_2\ x` |
| `root 3 L_1\ m – root 3 L_1\ x` | `= root 3 L_2\ x` |
| `root 3 L_2\ x + root 3 L_1\ x` | `= root 3 L_1\ m` |
| `x (root 3 L_2 + root 3 L_1)` | `= root 3 L_1\ m` |
| `x` | `= (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}` |
| `(dN)/(dx)` | `= -2 L_1\ x^-3 + 2 L_2 (m – x)^-3` |
| `(d^2N)/(dx^2)` | `= 6 L_1\ x^-4 – 6 L_2 (m – x)^-4 xx -1` |
| `= (6 L_1)/x^4 + (6 L_2)/(m – x)^4 > 0` |
`:.\ text(A minimum occurs when)`
`x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`
Calculus in the Physical World, 2UA 2007 HSC 10a
An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for `t >= 6`.
- Using Simpson’s rule, estimate the distance travelled between `t = 0` and `t = 4`. (2 marks)
- The object is initially at the origin. During which time(s) is the displacement of the object decreasing? (1 mark)
- Estimate the time at which the object returns to the origin. Justify your answer. (2 marks)
- Sketch the displacement, `x`, as a function of time. (2 marks)
Show Answers Only
- `~~ 6\ \ text(units)`
- `t > 5\ \ text(seconds)`
- `7.2\ \ text(seconds)`
Show Worked Solution
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(i) |
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`text(Distance travelled)`
`= int_0^4 (dx)/(dt)\ dt`
`~~ h/3 [y_0 + 4y_1 + y_2]`
`~~ 2/3 [0 + 4 (1) + 5]`
`~~ 2/3 [9]`
`~~ 6\ \ text(units)`
(ii) `text(Displacement is reducing when the velocity is negative.)`
`:. t > 5\ \ text(seconds)`
(iii) `text(At)\ B,\ text(the displacement) = 6\ text(units)`
`text(Considering displacement from)\ B\ text(to)\ D.`
`text(S)text(ince the area below the graph from)`
`B\ text(to)\ C\ text(equals the area above the)`
`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`
`text(in displacement from)\ B\ text(to)\ D.`
`text(Considering)\ t >= 6`
`text(Time required to return to origin)`
| `t` | `= d/v` |
| `= 6/5` | |
| `= 1.2\ \ text(seconds)` |
`:.\ text(The particle returns to the origin after 7.2 seconds.)`
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(iv) |
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