Band 4 – Page 72

Harder Ext1 Topics, EXT2 2007 HSC 4a

Two circles intersect at `A` and `B`.

The lines `LM` and `PQ` pass through `B`, with `L` and `P` on one circle and `M` and `Q` on the other circle, as shown in the diagram.

Copy or trace this diagram into your writing booklet.

Show that `/_ LAM = /_ PAQ.` (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`text(Join)\ \ LA, PA, MA, QA`

`text(Let)\ \ /_ ALM = x^@,\ \ /_AML = y^@`

`/_ APB`	`= /_ ALB = x^@\ \ text{(angles in the same segment on arc}\ AB text{)}`
`text(Similarly, but in the circle through)\ \ ABMQ`
`/_ AQB`	`= /_ AMB = y^@\ \ text{(angles in the same segment on arc}\ AB text{)}`

`/_ LAM`	`= 180^@ – (x + y)^@\ \ \ text{(angle sum of}\ \ Delta LAMtext{)}`
`/_ PAQ`	`= 180^@ – (x + y)^@\ \ text{(angle sum of}\ \ Delta PAQtext{)}`
`:.\ /_ LAM = /_ PAQ`

Mechanics, EXT2 2007 HSC 3d

A particle `P` of mass `m` undergoes uniform circular motion with angular velocity `omega` in a horizontal circle of radius `r` about `O`. It is acted on by the force due to gravity, `mg`, a force `F` directed at an angle `theta` above the horizontal and a force `N` which is perpendicular to `F`, as shown in the diagram.

By resolving forces horizontally and vertically, show that
1. `N = mg cos theta - m r omega^2 sin theta.` (3 marks)
For what values of `omega` is `N > 0?` (1 mark)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`g/r cot theta`

Show Worked Solution

(i)

`text(Resolving forces vertically)`

`F sin theta + N cos theta`	`= mg`
`N cos theta`	`=mg-F sin theta\ \ \ …\ (1)`
`text(Resolving forces horizontally)`
`F cos theta – N sin theta`	`= m r omega^2`
`N sin theta`	`=F cos theta-m r omega^2\ \ \ …\ (2)`

`text{Multiply (1)}\ xx cos theta and text{(2)}\ xx sin theta`

`N cos^2 theta`	`=mg cos theta -F sin theta cos theta\ \ \ …\ (3)`
`N sin^2 theta`	`=F cos theta sin theta-m r omega^2 sin theta\ \ \ …\ (4)`

`text{Add (3) + (4)}`

`:.N`	`=mg cos theta-F sin theta cos theta + F cos theta sin theta-m r omega^2 sin theta`
	`=mg cos theta – m r omega^2 sin theta`

(ii) `text(When)\ \ N > 0,`

`mg cos theta`	`>m r omega^2 sin theta`
`omega^2`	`<(g cos theta)/(r sin theta)`
	`<g/r cot theta`

Volumes, EXT2 2007 HSC 3c

Use the method of cylindrical shells to find the volume of the solid formed when the shaded region bounded by

`y = 0,\ \ y = (log_e x)/x,\ \ x = 1 and x = e`

is rotated about the `y`-axis. (4 marks)

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`2 pi\ \ text(u³)`

Show Worked Solution

(i) `y = 0,\ \ y = (log_e x)/x,\ \ x = 1 and x = e`

MARKER’S COMMENT: A “significant number” of students were confused about the radius of the cylindrical shell.

`δV`	`=2 pi x y\ δx`
`V`	`=2 pi int_1^e x*(log_e x)/x\ dx`
	`= 2pi int_1^e log_e x\ dx`

`text(Integrating by parts,)`

`u`	`=log_e x,`	`u′`	`=1/x`
`v′`	`=1,`	`v`	`=x`

`:.V`	`=2pi [[x log_e x]_1^e – int_1^e x * 1/x\ dx]`
	`=2pi [(e – log_e 1) – [x]_1^e]`
	`=2pi (e – (e – 1))`
	`=2pi\ \ text(u³)`

Functions, EXT1′ F2 2007 HSC 3b

The zeros of `x^3 - 5x + 3` are `alpha, beta` and `gamma.`

Find a cubic polynomial with integer coefficients whose zeros are `2 alpha, 2 beta` and `2 gamma.` (2 marks)

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`x^3 – 20x + 24`

Show Worked Solution

`text(Solution 1)`

`α+β+γ`	`=0`
`:.2α+2β+2γ`	`=2(α+β+γ)`
	`=0`
`αβ+βγ+γα`	`=-5`
`:.2α2β+2β2γ+2γ2α`	`=4(αβ+βγ+γα)`
	`=-20`
`αβγ`	`=-3`
`:.2α2β2γ`	`=8αβγ`
	`=-24`

`:.\ text(Polynomial is)\ \ x^3 – 20x + 24`

`text(Solution 2)`

`P(x) = x^3 – 5x + 3`

`text(New zeros are)\ \ 2 alpha, 2 beta and 2 gamma`

`:.H(x)`	`=(x/2)^3 – 5(x/2) + 3`
	`=x^3/8-(5x)/2+3`

`:.\ text(New Polynomial with integer coefficients is)`

`x^3 – 20x + 24`

Functions, EXT1′ F1 2007 HSC 3a

The diagram shows the graph of `y = f(x)`. The line `y = x` is an asymptote.

Draw separate one-third page sketches of the graphs of the following:

`f(-x).` (1 mark)

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`f(|\ x\ |).` (2 marks)

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`f(x) - x.` (2 marks)

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Show Worked Solution

MARKER’S COMMENT: In part (ii), a significant number of students graphed `y=|f(x)|`.

ii.

iii.

Complex Numbers, EXT2 N2 2007 HSC 2d

The points `P,Q` and `R` on the Argand diagram represent the complex numbers `z_1, z_2` and `a` respectively.

The triangles `OPR` and `OQR` are equilateral with unit sides, so `|\ z_1\ | = |\ z_2\ | = |\ a\ | = 1.`

Let `omega = cos pi/3 + i sin pi/3.`

Explain why `z_2 = omega a.` (1 mark)

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Show that `z_1 z_2 = a^2.` (1 mark)

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Show that `z_1` and `z_2` are the roots of `z^2 - az + a^2 = 0.` (2 marks)

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`text(See Worked Solutions)`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

i. `text(S) text(ince)\ \ Delta ORQ\ \ text(is equilateral, each angle is)\ \ pi/3\ \ text(radians.)`

`text(From)\ \ R(a):`

`Q(z_2)\ \ text(is an anticlockwise rotation through)\ \ pi/3.`

`:. z_2 = a(cos\ pi/3+i sin\ pi/3)=omega a.`

ii. `text(Solution 1)`

`text(Similarly,)\ \ a`	`=z_1 omega`
`z_1`	`=a/omega`
`:z_1z_2`	`=a/omega xx omega a`
	`=a^2`

`text(Solution 2)`

`P(z_1)\ \ text(is a clockwise rotation of)\ \ R(a)\ \ text(through)\ \ pi/3.`

`:.z_1`	`= bar omega a.`
`:. z_1 z_2`	`= bar omega a xx omega a`
	`=a^2(cos pi/3 – i sin pi/3) xx (cos pi/3 + i sin pi/3)`
	`=a^2(cos^2 pi/3 + sin^2 pi/3)`
	`= a^2`

iii. `z^2-az + a^2 = 0`

`text(Let the roots be)\ \ alpha and beta.`

`alpha + beta`	`=-b/a=a`
`alpha beta`	`=c/a=a^2`

`z_1 z_2`	`= a^2\ \ \ \ \ text{(part (ii))}`
`z_1 + z_2`	`=bar omega a + omega a`
	`=(cos pi/3 + i sin pi/3 + cos pi/3-i sin pi/3) a`
	`=2 cos pi/3 xx a`
	`=2 xx 1/2 xx a`
	`=a`

`:.\ z_1 and z_2\ \ text(are the roots of)\ \ z^2-az + a^2 = 0.`

Complex Numbers, EXT2 N2 2007 HSC 2c

The point `P` on the Argand diagram represents the complex number `z`, where `z` satisfies

`1/z + 1/bar z = 1.`

Give a geometrical description of the locus of `P` as `z` varies. (3 marks)

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`text(Locus is a circle, centre)\ (1, 0), text(radius 1,)`

`text(excluding the point)\ (0, 0).`

Show Worked Solution

`1/z + 1/bar z`	`=1/(x + iy) +1/(x – iy)`
	`=(x – iy + x+iy)/(x^2+y^2)`
	`=(2x)/(x^2+y^2)`

`(2x)/(x^2+y^2)`	`=1`
`x^2 + y^2`	`=2x`
`x^2 – 2x + 1 + y^2`	`=1`
`(x – 1)^2 + y^2`	`=1`

`:.\ text(Locus is a circle, centre)\ (1, 0), text(radius 1,)`

`text(excluding the point)\ (0, 0).`

Calculus, EXT2 C1 2007 HSC 1e

It can be shown that

`2/(x^3 + x^2 + x + 1) = 1/(x + 1) - x/(x^2 + 1) + 1/(x^2 + 1).` (Do NOT prove this.)

Use this result to evaluate `int_(1/2)^2 2/(x^3 + x^2 + x + 1)\ dx.` (4 marks)

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`tan^-1 2 – tan^-1 1/2`

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`int_(1/2)^2 2/(x^3 + x^2 + x + 1)\ dx`

`=int_(1/2)^2 (1/(x + 1) – x/(x^2 + 1) + 1/(x^2 + 1)) dx`

`=[log_e(x + 1) – 1/2 log_e (x^2 + 1) + tan^-1 x]_(1/2)^2`

`=[(log_e 3 – 1/2 log_e 5 + tan^-1 2) – (log_e 3/2 – 1/2 log_e 5/4 + tan^-1 1/2)]`

`=log_e\ 3/sqrt5 -log_e (3/2 xx 2/sqrt5) + tan^-1 2 – tan^-1 1/2`

`=log_e (3/sqrt(5) xx sqrt (5)/3) + tan^-1 2 – tan^-1 1/2`

`=tan^-1 2 – tan^-1 1/2`

Calculus, EXT2 C1 2007 HSC 1d

Evaluate `int_0^(3/4) x/sqrt (1 - x)\ dx.` (4 marks)

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`5/12`

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`text(Let)\ \ u = 1 – x,\ \ du = -dx,\ \ x = 1 – u`

`text(When)\ \ x = 0,\ \ u = 1`

`text(When)\ \ x = 3/4,\ \ u = 1/4`

`int_0^(3/4) (x\ dx)/sqrt (1 – x)`	`=int_1^(1/4) ((1 – u)(-du))/sqrt u`
	`=int_(1/4)^1 (u^(-1/2) – u^(1/2)) du`
	`=[2 u^(1/2) – 2/3 u^(3/2)]_(1/4)^1`
	`=[(2 – 2/3) – (1 – 1/12)]`
	`=5/12`

Harder Ext1 Topics, EXT2 2015 HSC 10 MC

Consider the expansion of

`(1 + x + x^2 + … + x^n) (1 + 2x + 3x^2 + … + (n + 1) x^n).`

What is the coefficient of `x^n` when `n = 100?`

`4950`
`5050`
`5151`
`5253`

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`C`

Show Worked Solution

`text(By expanding, the co-efficients of)\ \ x^n\ \ text(will be,)`

`1+2+3+…+100+101`

`=> text(A.P. with)\ \ a=1,\ \ d=1,\ \ n=101`

`:.S_n`	`=n/2(a + l)`
	`=101/2(1+101)`
	`=5151`

`=> C`

Complex Numbers, EXT2 N2 2015 HSC 9 MC

The complex number `z` satisfies `| z - 1 | = 1.`

What is the greatest distance that `z` can be from the point `i` on the Argand diagram?

`1`
`sqrt 5`
`2 sqrt 2`
`sqrt 2 + 1`

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`D`

Show Worked Solution

`| z – 1 | = 1\ \ text{is a circle, centre (1,0), radius 1.}`

`text(Consider the graph:)`

`text(Distance from)\ \(0, i)\ \ text(to the centre)=sqrt2`

`:.\ text(Greatest distance)=sqrt2 + text(radius)=sqrt2+1`

`=> D`

Complex Numbers, EXT2 N1 2014 HSC 8 MC

The Argand diagram shows the complex numbers `w`, `z` and `u`, where `w` lies in the first quadrant, `z` lies in the second quadrant and `u` lies on the negative real axis.

Which statement could be true?

`u = zw` and `u = z + w`
`u = zw` and `u = z − w`
`z = uw` and `u = z + w`
`z = uw` and `u = z − w`

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`B`

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`text(Using the parallelogram method, it could be true that)`

`u = z − w`

`=>\ text{Eliminate (A) and (C)}`

`text{arg}(u)`	`=\ text{arg}(w) + text{arg}(z)`
`u`	` = zw`

`=> B`

Functions, EXT1′ F1 2013 HSC 9 MC

Which diagram best represents the graph `y = (sin x)/x?`

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`B`

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`f(x)`	`=(sin x)/x`
`f(−x)`	`=(sin (−x))/-x`
	`=(sin (x))/x`
	`=f(x)`

`=>f(x)\ \ text(is even)`

`lim_(x -> 0) (sin(x))/x =1`

`=> B`

Mechanics, EXT2 2013 HSC 7 MC

The angular speed of a disc of radius `5` cm is `10` revolutions per minute.

What is the speed of a mark on the circumference of the disc?

`50\ \ text(cm min)^-1`
`1/2\ \ text(cm min)^-1`
`100 pi\ \ text(cm min)^-1`
`1/(4 pi)\ \ text(cm min)^-1`

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`C`

Show Worked Solution

`omega`	`= 2 pi f`
	`= 2 pi xx 10`
	`= 20 pi`

` v`	`=r omega`
	`= 5 xx 20 pi`
	`= 100 pi\ \ text(cm min)^-1`

`=> C`

Complex Numbers, EXT2 N2 2013 HSC 5 MC

Which region on the Argand diagram is defined by `pi/4 <= | z - 1 | <= pi/3?`

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`B`

Show Worked Solution

`| z-1 |=r\ \ text{is a circle with centre (1, 0), and}`

`text(radius)\ \ r.`

`=> B`

Complex Numbers, EXT2 N1 2013 HSC 3 MC

The Argand diagram below shows the complex number `z.`

Which diagram best represents `z^2?`

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`D`

Show Worked Solution

`text(Consider)\ \ z\ \ text(in polar form:)`

`\|\ z\ \|`	`< 1`
`:.\|\ z^2\ \|`	`< \|\ z\ \|`

`text(arg) (z^2) = 2text(arg) (z)`

`=> D`

Proof, EXT2 P2 2009 HSC 8a

Using the substitution `t = tan\ theta/2`, or otherwise, show that

`qquad cot theta + 1/2 tan\ theta/2 = 1/2 cot\ theta/2.` (2 marks)

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Use mathematical induction to prove that, for integers `n >= 1`,

`qquad sum_(r = 1)^n 1/2^(r - 1) tan x/2^r = 1/2^(n - 1) cot\ x/2^n - 2 cot x.` (3 marks)

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Show that `lim_(n -> oo) sum_(r = 1)^n 1/2^(r - 1) tan\ x/2^r = 2/x - 2 cot x.` (2 marks)

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Hence find the exact value of
`qquad tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + ….` (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`4/pi`

Show Worked Solution

i. `t=tan\ theta/2, \ \ \ sin theta=(2t)/(1+t^2),\ \ \ cos theta=(1-t^2)/(1+t^2)`

`text(Prove)\ \ cot theta + 1/2 tan\ theta/2 = 1/2 cot\ theta/2`

`text(LHS)`	`=cot theta + 1/2 tan\ theta/2`
	`=cos theta/sin theta+ 1/2 tan\ theta/2`
	`=(1-t^2)/(2t)+t/2`
	`=(1-t^2+t^2)/(2t)`
	`=1/(2t)`
	`=1/2 cot\ theta/2`
	`=\ text(RHS)`

ii. `text(If)\ \ n = 1`

`text(LHS)`	`=1/2^0 tan\ x/2^1=tan\ x/2`
`text(RHS)`	`=1/2^0 cot\ x/2 – 2 cot x`
	`=cot\ x/2 – 2 cot x`

`text{Using part (i)},\ \ 1/2 tan\ theta/2`	`= 1/2 cot\ theta/2 – cot theta,`
`:.tan\ theta/2`	`= cot\ theta/2 – 2 cot theta`

`text(RHS)`	`=tan\ x/2`
	`=\ text(LHS)`
`:.\ text(True for)\ \ n=1`

`text(Assume true for)\ \ n = k`

`text(i.e.)\ \ sum_(r = 1)^k 1/2^(r – 1) tan\ x/2^r = 1/2^(k – 1) cot\ x/2^k – 2 cot x.`

`text(Prove the result true for)\ \ n = k+1`

`text(i.e.)\ \sum_(r = 1)^(k + 1) 1/2^(r – 1) tan x/2^r = 1/2^k cot x/2^(k + 1) – 2 cot x`

`text(LHS)`	`=sum_(r = 1)^(k) 1/2^(r – 1) tan x/2^r + 1/2^k tan x/2^(k + 1)`
	`=1/2^(k – 1) cot x/2^k – 2 cot x + 1/2^k tan x/2^(k + 1)`
	`=1/2^(k – 1) (cot x/2^k + 1/2 tan x/2^(k +1)) – 2 cot x,\ \ \ \ text{(Let}\ \ theta=x/2^ktext{)}`
	`=1/2^(k – 1)(cot\ theta + 1/2 tan\ theta/2) – 2 cot x`
	`=1/2^(k – 1)(1/2 cot\ theta/2) – 2 cot x,\ \ \ \ text{(from part (i))}`
	`=1/2^k cot\ theta/2 – 2 cot x`
	`=1/2^k cot x/2^(k + 1) – 2 cot x`
	`=\ text(RHS)`

`=>text(True for)\ \ n = k + 1\ \ text(if it is true for)\ \ n = k.`

`:.text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ \ n >= 1.`

iii. `lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan x/2^r `

`=lim_(n -> oo) (1/2^(n-1) cot x/2^n – 2 cot x)`

`=lim_(n -> oo) (2/x * x/2^n * 1/(tan x/2^n) – 2 cot x)`

`=2/x xx lim_(n -> oo) ((x/2^n)/(tan x/2^n)) – 2 cot x`

`=>text(As)\ \ n -> oo,\ \ x/2^n=theta->0, and`

`=>lim_(theta-> 0) (theta)/(tan theta) =1`

`:.lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan x/2^r =2/x – 2 cot x`

iv. `tan\ pi/4 + 1/2 tan\ pi/8 + 1/4 tan\ pi/16 + …`

`=lim_(n -> oo) sum_(r = 1)^n 1/2^(r – 1) tan (pi/2)/2^r`

`=(2/(pi/2)) – 2 cot pi/2`

`=4/pi`

Complex Numbers, EXT2 N2 2009 HSC 7b

Let `z = cos theta + i sin theta.`

Show that `z^n + z^-n = 2 cos n theta`, where `n` is a positive integer. (2 marks)

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Let `m` be a positive integer. Show that

`(2 cos theta)^(2m) = 2 [cos 2 m theta + ((2m), (1)) cos (2m - 2) theta + ((2m), (2)) cos (2m - 4) theta`

`+ … + ((2m), (m - 1)) cos 2 theta] + ((2m), (m)).` (3 marks)

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Hence, or otherwise, prove that

`int_0^(pi/2) cos^(2m) theta\ d theta = pi/(2^(2m + 1)) ((2m), (m))`

where `m` is a positive integer. (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

i.	`z`	`= cos theta + i sin theta`
	`z^n`	`= cos n theta + i sin n theta\ \ \ \ text{(De Moivre)}`
	`z^-n`	`= cos (-n theta) + i sin (-n theta)\ \ \ \ text{(De Moivre)}`
		`= cos n theta – i sin n theta`
	`z^n + z^-n`	`= cos n theta + i sin n theta + cos n theta – i sin n theta`
		`= 2 cos n theta,\ \ \ \ n > 0`

ii. `z + z^-1 = 2 cos theta`

`:.(2 cos theta)^(2m)`

`=(z + z^-1)^(2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 1) z^-1 + ((2m), (2)) z^(2m – 2) z^-2+`

` … + ((2m), (2m – 1)) z^1 z^-(2m – 1) + z^-(2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 2) + ((2m), (2)) z^(2m – 4)+`

` … + ((2m), (2m – 1)) z^-(2m – 2) + z^(-2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 2) + ((2m), (2)) z^(2m – 4) + … + ((2m), (m)) z^(2m-2m) …`

`+ ((2m), (2)) z^-(2m – 4) + ((2m), (1)) z^-(2m – 2) + z^(-2m)`

`=(z^(2m) + z^(-2m)) + ((2m), (1)) (z^(2m – 2) + z^-(2m – 2)) + ((2m), (2))`

`(z^(2m – 4) + z^-(2m – 4)) + … + ((2m), (m – 1)) (z + z^-1) + ((2m), (m))`

`=2 [cos 2 m theta + ((2m), (1)) cos (2m – 2) theta + ((2m), (2)) cos (2m – 4) theta`

`+ … + ((2m), (m – 1)) cos 2 theta] + ((2m), (m))`

iii. `int_0^(pi/2) cos^(2m) d theta`

`=1/(2^(2m)) int_0^(pi/2) (2 cos theta)^(2m)`

`=1/(2^(2m)) int_0^(pi/2)[2(cos 2 m theta + ((2m), (1)) cos (2m – 2) theta + ((2m), (2))`

`cos (2m – 4) theta + … + ((2m), (m – 1)) cos 2 theta) + ((2m), (m))] d theta`

`=1/(2^(2m)) [2((sin 2 m theta)/(2m) + ((2m), (1)) (sin (2m – 2) theta)/(2m – 2)`

`+ … + ((2m), (m – 1)) (sin 2 theta)/2) + ((2m), (m)) theta]_0^(pi/2)`

`=1/(2^(2m)) [2(0 + 0 + … + 0) + ((2m), (m)) pi/2 – (0)]`

`=pi/(2^(2m + 1)) ((2m), (m))`

Mechanics, EXT2 M1 2009 HSC 7a

A bungee jumper of height 2 m falls from a bridge which is 125 m above the surface of the water, as shown in the diagram. The jumper’s feet are tied to an elastic cord of length `L` m. The displacement of the jumper’s feet, measured downwards from the bridge, is `x` m.

The jumper’s fall can be examined in two stages. In the first stage of the fall, where `0 <= x <= L`, the jumper falls a distance of `L` m subject to air resistance, and the cord does not provide resistance to the motion. In the second stage of the fall, where `x > L`, the cord stretches and provides additional resistance to the downward motion.

The equation of motion for the jumper in the first stage of the fall is

`ddot x = g - rv`

where `g` is the acceleration due to gravity, `r` is a positive constant, and `v` is the velocity of the jumper.

(1) Given that `x = 0` and `v = 0` initially, show that

`qquad x = g/r^2 ln (g/(g - rv)) - v/r.` (3 marks)

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(2) Given that `g = 9.8\ text(ms)^-2` and `r = 0.2\ text(s)^-1`, find the length, `L`, of the cord such that the jumper’s velocity is `30\ text(ms)^-1` when `x = L`. Give your answer to two significant figures. (1 mark)

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In the second stage of the fall, where `x > L`, the displacement `x` is given by

`x = e^(-t/10)(29 sin t - 10 cos t) + 92`

where `t` is the time in seconds after the jumper’s feet pass `x = L`.

Determine whether or not the jumper’s head stays out of the water. (4 marks)

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`(1)\ \ text{Proof}\ \ text{(See Worked Solutions)}`
`(2)\ \ 82\ \ text(m)`
`text(The jumper’s head will stay out of the water)`

Show Worked Solution

i. (1)	`ddot x`	`=g-rv`
	`v (dv)/(dx)`	`= g – rv`
	`(dx)/(dv)`	`=v/(g-rv)`

`:. int dx`	`= int v/(g-rv)\ dv`
`x`	`=-1/r int ((g-rv-g)/(g-rv))\ dv`
	`=-1/r int (1-g/(g-rv))\ dv`
	`=-1/r (v + g/r ln(g-rv)) +c`

`text(When)\ \ x=0,\ \ v=0`

`:.c=1/r(g/r lng)=g/r^2 ln g`

`:.x`	`=-1/r (v + g/r ln(g-rv))+g/r^2 ln g`
	`=-v/r- g/r^2 ln(g-rv) +g/r^2 ln g`
	`=g/r^2 ln (g/(g – rv)) – v/r`

i. (2) `g = 9.8\ \ text(ms)^-1,\ \ r = 0.2\ \ text(s)^-1`

`text(If)\ \ x = L,\ \ v = 30\ \ text(ms)^-1`

`:.L`	`=9.8/0.2^2 log_e (9.8/(9.8 – 0.2 xx 30)) – 30/0.2`
	`=82\ \ text(m)\ \ \ \ text{(2 sig.)`

ii.	`x`	`= e^(-t/10) (29 sin t – 10 cos t) + 92`
	`dx/dt`	`=e^(-t/10) (29cos t + 10 sin t)`
		`+(-1/10 e^(-t/10) )(29 sin t – 10 cos t)`
		`=e^(-t/10)(30 cos t+7.1 sin t)`

`text(When)\ \ dx/dt=0\ \ => text(maximum occurs)`

`30 cos t+7.1 sin t`	`=0`
`tan t`	`=-30/7.1`
`:.t`	`=tan^-1 (-30/7.1)`
	`=pi-1.338…`
	`~~1.8\ \ text(s)`

`text(When)\ \ t=1.8`

`x`	`=e^-0.18 (29 sin 1.8 – 10 cos 1.8) + 92`
	`~~117.5\ \ text(m)`

`:.\ text(Distance from the bridge to the jumper’s head) = 119.5\ \ text(m)`

`:.\ text(The jumper’s head will not enter the water.)`

Harder Ext1 Topics, EXT2 2009 HSC 6c

The diagram shows a circle of radius `r`, centred at the origin, `O`. The line `PQ` is tangent to the circle at `Q`, the line `PR` is horizontal, and `R` lies on the line `x = c`.

Find the length of `PQ` in terms of `x, y and r.` (1 mark)
The point `P` moves such that `PQ = PR`.
Show that the equation of the locus of `P` is
1. `y^2 = r^2 + c^2 - 2cx.` (2 marks)
Find the focus, `S`, of the parabola in part (ii). (2 marks)
Show that the difference between the length `PS` and the length `PQ` is independent of `x.` (2 marks)

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`PQ = sqrt (x^2 + y^2 – r^2)`
`text(Proof)\ \text{(See Worked Solutions)}`
`(r^2/(2c), 0)`
`text(Proof)\ \text{(See Worked Solutions)}`

Show Worked Solution

(i)	`OP`	`= sqrt (x^2 + y^2)`
	`PQ^2`	`= OP^2 – OQ^2`
	`PQ^2`	`= x^2 + y^2 – r^2`
	`:.\ PQ`	`= sqrt (x^2 + y^2 – r^2)`

(ii) `text(When)\ \ PQ = PR`

`sqrt (x^2 + y^2 – r^2)`	`=c-x`
`x^2 + y^2 – r^2`	`= (c – x)^2`
`x^2 + y^2 – r^2`	`= c^2 – 2cx + x^2`
`y^2 – r^2`	`= c^2 – 2cx`

`:.\ y^2 = r^2 + c^2 – 2cx\ \ text(is the locus of)\ \ P`

(iii) `text(Rearranging the locus of)\ \ P\ \ text(in the form)`

♦♦♦ “Few” students answered part (iii) correctly (exact data unavailable).

`(y-y_0)^2`	`=4a(x-x_0)`
`y^2`	`= r^2 + c^2 – 2cx`
	`= -2c(x – (r^2 + c^2)/(2c))`

`=>\ text(The parabola is lying on its side and opening to the left.)`

`text(Vertex) = ((r^2 + c^2)/(2c),0)`

`text(Focal length) \ \ \ \ 4a`	`=-2c`
`a`	`=- c/2`

`:.S\ text(has coordinates)\ \ ((r^2 + c^2)/(2c) – c/2, 0) -=(r^2/(2c), 0)`

(iv) `text(The directrix of the parabola has the equation)`

♦♦♦ “Few” students answered part (iv) correctly (exact data unavailable).
MARKER’S COMMENT: Very few used the efficient focus-directrix definition here.

`x`	`=(r^2+c^2)/(2c) + c/2`
	`=(r^2+2c^2)/(2c)`

`text(The definition of a parabola requires that)`

`PS`	`=PM`
	`=(r^2+2c^2)/(2c)-x`

`text(S)text(ince)\ \ \ PQ`	`=PR\ \ \ \ \ \ text{(from part (ii))}`
	`=c-x`

`=> PS-PQ`	`=(r^2+2c^2)/(2c)-x-(c-x)`
	`=r^2/(2c)`

`:. PS-PQ\ \ text(is independent of)\ x.`

Polynomials, EXT2 2009 HSC 6b

Let `P(x) = x^3 + qx^2 + qx + 1`, where `q` is real. One zero of `P(x)` is `-1`.

Show that if `alpha` is a zero of `P(x)` then `1/alpha` is a zero of `P(x).` (1 mark)
Suppose that `alpha` is a zero of `P(x)` and `alpha` is not real.
(1) Show that `|\ alpha\ | = 1.` (2 marks)
(2) Show that `text(Re)(alpha) = (1 - q)/2.` (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i) `P(x) = x^3 + qx^2 + qx + 1`

`text(Let roots be)\ \ -1, alpha, beta`

`text(Product of roots)`	`=d/a`
` -1 xx alpha xx beta`	`=-1`
`:. beta`	`=1/alpha`

(ii)(1) `text(If)\ \ alpha\ \ text(is a zero of)\ \ P(x)`

`=>1/alpha\ \ text{is also a zero}\ \ \ \ text{(part (i))}`

`text(S)text(ince all coefficients are real)`

`=>bar alpha\ \ text(is also a zero)`

`:.bar alpha`	`=1/alpha`
`alpha bar alpha`	`=1`
`\|\ alpha\ \|^2`	`=1`
`:.\|\ alpha\ \|`	`=1`

(ii)(2) `text{Sum of roots} = -b/a`

`:.\ -1 + alpha + bar alpha =`	`-q`
`alpha + bar alpha =`	`1 – q`
`2 text(Re)(alpha) =`	`1 – q`
`:.text(Re)(alpha) =`	`(1 – q)/2`

Volumes, EXT2 2009 HSC 6a

The base of a solid is the region enclosed by the parabola `x =4 – y^2` and the `y`-axis. The top of the solid is formed by a plane inclined at `45^@` to the `xy`-plane. Each vertical cross-section of the solid parallel to the `y`-axis is a rectangle. A typical cross-section is shown shaded in the diagram.

Find the volume of the solid. (3 marks)

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`128/5\ \ text(u³)`

Show Worked Solution

`text(Shaded cross section)`

`text(Height)`	`= 4 – x,\ \ \ \ 0 <= x <= 4`
`text(Length)`	`=2y\ \ \ \ \ \ (text{using}\ \ x=4-y^2)`
	`=2 sqrt(4-x)`
`:. text(Area)\ \ `	`=2 sqrt(4-x) (4-x)`
	`=2 (4-x)^(3/2),\ \ \ \ 0 <= x <= 4`

`delta V`	` = 2 (4 – x)^(3/2)\ delta x,\ \ 0 <= x <= 4`
`:.V `	`=2 int_0^4 (4 – x)^(3/2)\ dx`
	`=2[-2/5 (4 – x)^(5/2)]_0^4`
	`=- 4/5[(4 – x)^(5/2)]_0^4`
	`=- 4/5 (0 – 2^5)`
	`=128/5\ \ text(u³)`

Harder Ext1 Topics, EXT2 2009 HSC 5c

Let `f(x) = (e^x - e^-x)/2 - x.`

Show that `f″(x) > 0` for all `x > 0.` (2 marks)
Hence, or otherwise, show that
1. `f prime (x) > 0` for all `x > 0.` (2 marks)
Hence, or otherwise, show that
1. `(e^x - e^-x)/2 > x` for all `x > 0.` (1 mark)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)	`f(x) =`	`(e^x – e^-x)/2 – x`
	`f prime (x) =`	`(e^x + e^-x)/2 – 1`
	`f″(x) =`	`(e^x – e^-x)/2`

`text(When)\ \ x > 0, \ \ e^x > e^-x`

`=>e^x – e^-x > 0,`

`:.f ″(x) > 0\ \ text(for)\ \ x > 0`

(ii) `text(Solution 1)`

`f prime (0) = (e^0 – e^0)/2 – 1 = 0`

`text(S)text(ince)\ \ f″(x) > 0\ \ text(for)\ \ x > 0,`

`=>f prime(x)\ \ text(is increasing.)`

`:.f prime (x) > 0\ \ \ \ (text{for}\ \ x > 0)`

`text(Solution 2)`

`f prime (x)`	`=(e^x + e^-x)/2 – 1`
	`=(e^x + e^-x -2)/2`
	`=(e^(2x) – 2e^x + 1)/(2e^x)`
	`=(e^x – 1)^2/(2e^x),\ \ \ \ (e^x > 0)`

`:.f prime (x) > 0\ \ \ \ (text{for}\ \ x > 0)`

(iii) `f(0) = (e^0 – e^0)/2 – 0 = 0`

`text(S)text(ince)\ \ f′(x) > 0,\ \ \ \ (x > 0)`

`=>f(x)\ \ text(is increasing.)`

`:.f (x)> 0`

`(e^x – e^-x)/2 – x`	`> 0`
`:.(e^x – e^-x)/2`	`>x\ \ \ \ (text{for}\ \ x > 0)`

Calculus, EXT2 C1 2009 HSC 5b

For each integer `n >= 0`, let

`I_n = int_0^1 x^(2n + 1) e^(x^2)\ dx.`

Show that for `n >= 1,`

`I_n = e/2 - nI_(n-1).` (2 marks)

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Hence, or otherwise, calculate `I_2.` (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`(e – 2)/2`

Show Worked Solution

i. `text(Integrating by parts:)`

`u`	`=x^(2n),`	`u′`	`=2nx^(2n-1)`
`v′`	`=xe^(x^2),`	`v`	`=1/2 e^(x^2)`

`:.I_n`	`= int_0^1 x^(2n + 1) e^(x^2)\ dx`
	`= int_0^1 x^(2n) x e^(x^2)\ dx`
	`= [(x^(2n) e^(x^2))/2]_0^1 – int_0^1 (2n x^(2n) e^(x^2))/2 \ dx`
	`= e/2 – n int_0^1 x^(2n) e^(x^2)\ dx`
	`= e/2 – nI_(n-1)`

ii. `I_2 = e/2 – 2I_1`

`I_1 = e/2 – I_0`

`I_0`	`= int_0^1 xe^(x^2)\ dx`
	`= [(e^(x^2))/2]_0^1`
	`= (e-1)/2`

`I_1`	`= e/2 – (e – 1)/2=1/2`
`:.I_2`	`= e/2 – 2 xx 1/2`
	`= (e – 2)/2`

Harder Ext1 Topics, EXT2 2009 HSC 5a

In the diagram `AB` is the diameter of the circle. The chords `AC` and `BD` intersect at `X`. The point `Y` lies on `AB` such that `XY` is perpendicular to `AB`. The point `K` is the intersection of `AD` produced and `YX` produced.

Copy or trace the diagram into your writing booklet.

Show that `/_ AKY = /_ ABD.` (2 marks)
Show that `CKDX` is a cyclic quadrilateral. (2 marks)
Show that `B, C and K` are collinear. (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`/_ ADB = 90^@\ \ \ text{(angle in a semicircle on diameter}\ \ AB text{)}`

`text(In)\ \ Delta BDA and Delta KYA`

`/_ BAD`	`= /_ KAY\ \ text{(common angle)}`
`/_ ADB`	`= /_ KYA = 90^@`
`:./_ AKY`	`= /_ ABD\ \ text{(angle sum of triangle)}`

(ii) `text(Join)\ \ DC`

`/_ DCA`	`= /_ DBA\ \ text{(angles in the same segment on chord}\ DA text{)}`
`/_ DCA`	`= /_ XKD\ \ text{(both equal to}\ /_ DBA text{)}`

`=>text(S)text(ince)\ \ /_ DKX = /_ DCX\ \ text(are a pair of equal angles)`

`text{standing on arc}\ DX\ \ text{(in the same segment)}.`

`:.CKDX\ \ text(is a cyclic quadrilateral.)`

(iii) `/_ KDX`

`= 90^@\ \ text{(} /_ ADK\ text{is a straight angle)}`

`/_ KCX`

`= 90^@\ \ text{(opposite angles of a cyclic}`

`text{quadrilateral are supplementary)}`

`/_ ACB`

`= 90^@\ \ text{(angle in a semicircle)}`

`/_ KCX + /_ ACB = 180^@`

`:./_ BCK\ \ text(is a straight angle.)`

`:.B, C and K\ \ text(are collinear.)`

Mechanics, EXT2 2009 HSC 4b

A light string is attached to the vertex of a smooth vertical cone. A particle `P` of mass `m` is attached to the string as shown in the diagram. The particle remains in contact with the cone and rotates with constant angular velocity `omega` on a circle of radius `r`. The string and the surface of the cone make an angle of `alpha` with the vertical, as shown.

The forces acting on the particle are the tension, `T`, in the string, the normal reaction, `N`, to the cone and the gravitational force `mg`.

Resolve the forces on `P` in the horizontal and vertical directions. (2 marks)
Show that `T = m (g cos alpha + r omega^2 sin alpha)` and find a similar expression for `N.` (2 marks)
Show that if `T = N` then
1. `omega^2 = g/r ((tan alpha - 1)/(tan alpha + 1)).` (2 marks)
For which values of `alpha` can the particle rotate so that `T = N`? (1 mark)

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`text(Horizontal:)\ \ \ \ \ T sin alpha – N cos alpha = mr omega^2`
`text(Vertical:)\ \ \ \ \ \ T cos alpha + N sin alpha = mg`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`pi/4 < alpha < pi/2`

Show Worked Solution

(i)

`text(Resolving the forces at)\ \ P`

`text(Horizontal)`

`T sin alpha – N cos alpha = mrω^2\ \ \ \ …\ (1)`

`text(Vertical)`

`T cos alpha + N sin alpha = mg\ \ \ \ …\ (2)`

(ii) `text{Multiply}\ \ (1) xx sin alpha\ \ and\ \ (2) xx cos alpha`

`T sin^2 alpha – N cos alpha sin alpha`	`= mr omega^2 sin alpha\ \ \ \ …\ (3)`
`T cos^2 alpha + N sin alpha cos alpha`	`= mg cos alpha\ \ \ \ …\ (4)`
`text{Add (3) + (4)}`
`T (sin^2 alpha + cos^2 alpha)`	`= mr omega^2 sin alpha + mg cos alpha`
`:.T`	`= m(g cos alpha + r omega^2 sin alpha)`

`text{Multiply}\ \ (1) xx cos\ α\ \ and\ \ (2) xx sin\ α`

`T sin alpha cos alpha – N cos^2 alpha`	`= mr omega^2 cos alpha\ \ \ \ …\ (5)`
`T cos alpha sin alpha + N sin^2 alpha`	`= mg sin alpha\ \ \ \ …\ (6)`
`text{Subtract (6) – (5)}`
`N (sin^2 alpha + cos^2 alpha)`	`= mg sin alpha – mr omega^2 cos alpha`
`:.N`	`= m(g sin alpha – r omega^2 cos alpha)`

(iii) `text(When) \ \ T = N`

`m(g cos alpha + r omega^2 sin alpha)`	`= m(g sin alpha – r omega^2 cos alpha)`
`g cos alpha + r omega^2 sin alpha`	`= g sin alpha – r omega^2 cos alpha`
`r omega^2 sin alpha + r omega^2 cos alpha`	`= g sin alpha – g cos alpha`
`r omega^2(sin alpha + cos alpha)`	`= g(sin alpha – cos alpha)`

`:.omega^2`	`=g/r((sin alpha – cos alpha)/(sin alpha + cos alpha))`
	`=g/r((tan alpha – 1)/(tan alpha + 1))`

♦ Part (iv) proved challenging (exact data not available).

MARKER’S COMMENT: Most students did not realise `ω² >0`.

(iv) `text(S) text(ince)\ \ omega^2 > 0`

`=>(tan alpha-1) > 0\ \ \ \ text{(using part (iii))}`

`:.\ \ pi/4 < alpha < pi/2.`

Conics, EXT2 2009 HSC 4a

The ellipse `x^2/a^2 + y^2/b^2 = 1` has foci `S(ae, 0)` and `S prime (– ae, 0)` where `e` is the eccentricity, with corresponding directrices `x = a/e` and `x = -a/e`. The point `P(x_0, y_0)` is on the ellipse. The points where the horizontal line through `P` meets the directrices are `M` and `M prime`, as shown in the diagram.

Show that the equation of the normal to the ellipse at the point `P` is
1. `y - y_0 = (a^2y_0)/(b^2x_0) (x - x_0).` (2 marks)
The normal at `P` meets the `x`-axis at `N`. Show that `N` has coordinates `(e^2x_0, 0).` (2 marks)
Using the focus-directrix definition of an ellipse, or otherwise, show that
1. `(PS)/(PS prime) = (NS)/(NS prime).` (2 marks)
Let `alpha = /_ S prime PN` and `beta = /_ NPS.`
By applying the sine rule to `Delta S prime PN` and to `Delta NPS`, show that `alpha = beta.` (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)	`x^2/a^2 + y^2/b^2`	`= 1`
	`(2x)/a^2 + (2y)/b^2 * (dy)/(dx)`	`= 0`
	`(dy)/(dx)`	`= (-b^2x)/(a^2y)`

`text(At)\ \ P(x_0, y_0)`

`(dy)/(dx) = (-b^2 x_0)/(a^2 y_0)`

`:.m\ text{of normal at}\ \ P = (a^2 y_0)/(b^2 x_0)`

`:.\ text(Equation of normal at)\ \ P\ \ text(is)`

`y – y_0 = (a^2 y_0)/(b^2 x_0) (x – x_0)`

(ii) `N\ \ text(occurs when)\ \ y = 0`

`0-y_0`	`= (a^2 y_0)/(b^2 x_0) (x – x_0)`
`-b^2 x_0`	`= a^2 x – a^2 x_0`
`a^2x`	`=x_0(a^2-b^2)`
`x`	`= (a^2 – b^2)/a^2 x_0`
	`= (a^2-a^2(1-e^2))/a^2 x_0`
	`=e^2 x_0`

`:.N\ \ text(has coordinates)\ \ (e^2 x_0, 0)`

(iii) `(PS)/(PM)`	`= (P S′)/(PM′) = e`
`:.(PS)/(PS prime)`	`= (PM)/(PM prime)`

`PM`	`=a/e-x_0,`	`\ \ PM′`	`=a/e +x_0`
`text(S)text(ince)\ \ N(e^2 x_0, 0)`
`NS`	`=ae-e^2x_0,`	`NS′`	`=ae+e^2 x_0`

`(PM)/(PM prime)`	`= (a/e – x_0)/(a/e+x_0) xx e^2/e^2`
	`= (ae – e^2x_0)/(ae+e^2 x_0)`
`:.(PS)/(PS prime)`	`= (NS)/(NS prime)`

(iv) `text(In)\ \ Delta S prime PN,`

`(P S′)/(sin /_ S′NP)`	`= (NS′)/(sin alpha)`
`=>P S′`	`= (NS′sin /_ S′NP)/(sin alpha)`

`text(In)\ \ Delta NPS,`

`(PS)/(sin /_ SNP)`	`= (NS)/(sin beta)`
`=>PS`	`= (NS sin /_ SNP)/(sin beta)`

`(PS)/(PS prime)`	`=(NS sin /_ SNP)/(sin beta) xx (sin alpha)/(NS′sin /_ S′NP)`
	`=(NS)/(NS prime) xx (sin /_ SNP sin alpha)/(sin beta sin /_ S′NP)`
`1`	`=sin alpha/sin beta xx (sin /_ SNP)/(sin /_ S′NP)\ \ \ \ text{(using part (iii))}`

`text(S)text(ince)\ \ /_ SNP + /_ S′NP = 180^@`

`=> sin/_ SNP`	`= sin /_ S′NP`
`:. sin alpha`	`= sin beta`
`:. alpha`	`=beta\ \ \ \ \ (text{since}\ \ alpha + beta<180^@)`

Volumes, EXT2 2009 HSC 3d

The diagram shows the region enclosed by the curves `y = x + 1` and `y = (x - 1)^2.`

The region is rotated about the `y`-axis.

Use the method of cylindrical shells to find the volume of the solid formed. (3 marks)

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`(27 pi)/2\ \ text(u³)`

Show Worked Solution

`text(Intersection occurs when)`

`x + 1`	`= x^2 – 2x + 1`
`x^2 – 3x`	`= 0`
`:.x = 0, 3`

`delta V`	`=2pi xy\ δx`
	`=2pi x[x+1 -(x-1)^2]\ δx`
	`=2pi x(3x-x^2)\ δx`
`:.V`	`=2 pi int_0^3 x (3x – x^2)\ dx`
	`=2 pi int_0^3 (3x^2 – x^3)\ dx`
	`=2 pi [x^3 – x^4/4]_0^3`
	`=2 pi [(27 – 81/4)-0]`
	`=(27 pi)/2\ \ text(u³)`

Functions, EXT1′ F2 2009 HSC 3c

Let `P(x) = x^3 + ax^2 + bx + 5`, where `a` and `b` are real numbers.

Find the values of `a` and `b` given that `(x - 1)^2` is a factor of `P(x).` (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

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`a = 3,\ \ \ b = -9`

Show Worked Solution

`P(x)`	`= x^3 + ax^2 + bx + 5`
`P(1)`	`= 1 + a + b + 5=0`
`:.b`	`= -a -6\ \ \ \ …\ (1)`

`P prime (x)`	`= 3x^2 + 2ax + b`
`P prime (1)`	`= 3 + 2a + b=0`
`2a+b`	`= -3\ \ \ \ …\ (2)`

`text(Substitute)\ \ b=-a-6\ \ text{from (1) into (2)}`

`2a+(-a-6)=-3`

`:.a = 3,\ \ \ b = -9`

Functions, EXT1′ F1 2009 HSC 3a

The diagram shows the graph `y = f(x).`

Draw separate one-third page sketches of the graphs of the following:

`y = 1/(f(x)) .` (2 marks)

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`y = f(x)\ f(x)` (2 marks)

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`y = f(x^2).` (2 marks)

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ii.

iii.

Show Worked Solution

i. `text(Vertical asymptotes at)\ \ x=0 and 4`

`text(Horizontal asymptote at)\ \ y=-1/3`

ii. `y=f(x)\ f(x) = [f(x)]^2`

iii. `y=f(x^2) =>text(even function)`

`text(When)\ \ x=±2,\ \ y=f(4)=0`

Complex Numbers, EXT2 N2 2009 HSC 2e

Find all the 5th roots of `–1` in modulus-argument form. (2 marks)

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Sketch the 5th roots of `–1` on an Argand diagram. (1 mark)

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`z_1 = text(cis)\ pi/5,\ \ \ z_2 = text(cis) (3 pi)/5,\ \ \ z_3 = text(cis) pi=-1,``z_4 = text(cis) (7 pi)/5,\ \ z_5 = text(cis) (9 pi)/5`

Show Worked Solution

i.	`z`	`=cos theta+i sin theta`
	`z^5`	`=cos\ 5 theta+i sin\ 5 theta=-1,\ \ \ \ text{(De Moivre)}`
	`:. cos\ 5 theta`	`=-1`
	`5 theta`	`=pi,\ 3pi,\ 5pi,\ 7pi,\ 9pi`
	`theta`	`=pi/5,\ (3pi)/5,\ pi,\ (7pi)/5,\ (9pi)/5`

`:.\ text(The roots are)`

`z_1 = text(cis)\ pi/5,\ \ \ z_2 = text(cis)\ (3 pi)/5,\ \ \ z_3 = text(cis) pi=-1,`

`z_4 = text(cis)\ (7 pi)/5,\ \ z_5 = text(cis)\ (9 pi)/5`

ii.

Calculus, EXT2 C1 2009 HSC 1e

Evaluate `int_1^sqrt 3 1/(x^2 sqrt (1 + x^2))\ dx.` (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`sqrt 2 – (2 sqrt 3)/3`

Show Worked Solution

`text(Let)\ \ x = tan theta,\ \ \ \ dx = sec^2 theta\ d theta`

`text(When)\ \ x = 1,\ \ theta = pi/4`

`text(When)\ \ x = sqrt 3,\ \ theta = pi/3`

`int_1^sqrt 3 1/(x^2 sqrt (1 + x^2))\ dx`	`= int_(pi/4)^(pi/3) (sec^2 theta)/(tan^2 theta sqrt (1 + tan^2 theta))\ d theta`
	`= int_(pi/4)^(pi/3) (sec^2 theta)/(tan^2 theta sec theta)\ d theta`
	`= int_(pi/4)^(pi/3) (sec theta)/(tan^2 theta)\ d theta`
	`= int_(pi/4)^(pi/3) (cos^2 theta)/(cos theta sin^2 theta)\ d theta`
	`= int_(pi/4)^(pi/3) (cos theta\ d theta)/(sin^2 theta)`
	`= [-1/(sin theta)]_(pi/4)^(pi/3)`
	`= (-2/sqrt 3 + sqrt 2)`
	`= sqrt 2 – (2 sqrt 3)/3`

Calculus, EXT2 C1 2009 HSC 1d

Evaluate `int_2^5 (x-6)/(x^2 + 3x-4)\ dx.` (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`2 ln (3/4)`

Show Worked Solution

`text(Using partial fractions:)`

`(x-6)/(x^2 + 3x-4)`	`=(x-6)/((x+4)(x-1))`
	`= A/(x+4) + B/(x-1)`
`:. x-6`	`=A(x-1)+B(x+4)`

`text(When)\ \ x=1,\ \ 5B=-5\ =>B=-1`

`text(When)\ \ x=-4,\ \ -5A=-10\ =>A=2`

`:. int_2^5 (x-6)/(x^2 + 3x-4)\ dx`	`= int_2^5 2/(x+4)\ dx-int_2^5 (dx)/(x-1)`
	`= [2 ln (x+4)]_2^5 -[ln(x-1)]_2^5`
	`= 2(ln 9-ln 6)-(ln4-ln1)`
	`= 2ln(3/2)-ln4`
	`= 2ln(3/2)-2ln2`
	`= 2ln(3/4)`

Polynomials, EXT2 2010 HSC 7c

Let `P(x) = (n − 1)x^n − nx^(n − 1) + 1`, where `n` is an odd integer, `n ≥ 3`.

Show that `P(x)` has exactly two stationary points. (1 mark)
Show that `P(x)` has a double zero at `x = 1`. (1 mark)
Use the graph `y = P(x)` to explain why `P(x)` has exactly one real zero other than `1`. (2 marks)
Let `α` be the real zero of `P(x)` other than `1`.
Given that `2^x>=3x-1` for `x>=3`, or otherwise, show that `-1 < α ≤ -1/2`. (2 marks)
Deduce that each of the zeros of `4x^5 − 5x^4 + 1` has modulus less than or equal to `1`. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

(i)	`P′(x)`	`= n(n − 1)x^(n − 1) − n(n − 1)x^(n − 2)`
		`= n(n − 1)x^(n − 2)(x − 1)`

`P′(x) = 0\ \ text(when)\ \ x = 0\ \ text(or)\ \ x = 1.`

`:.P(x)\ text(has exactly two stationary points.)`

(ii)	`P(1)`	`= (n − 1) xx 1 − n xx 1 + 1 = 0, and`
	`P′(1) `	`= 0`

`:. P(x)\ text(has a double zero at)\ \ x=1`

(iii) `P(x)\ text{has exactly two stationary points at}\ \ x=0 and 1`

♦♦ Mean mark part (iii) 26%.

`text(S)text(ince)\ \ P(0) = 1 and P(1)=0`

`=>\ text{A local maximum occurs at (0,1)}`

`text(Noting that)\ \ P(x)`	`-> oo\ \ text(as)\ \ x-> oo, and`
`P(x)`	`-> -oo\ \ text(as)\ \ x-> -oo`

`:.\ text(The double zero at)\ \ x=1,\ text{means that}\ \ P(x)`

`text(has exactly one real zero other than)\ \ x=1.`

♦♦♦ Mean mark part (iv) 3%.

(iv)	`P(-1)`	`=(n-1)(-1)^n-n xx (-1)^(n-1) + 1`
		`=(n-1)(-1) – nxx1 +1\ \ \ \ text{(given}\ n\ text{is odd)}`
		`=-2n+2`
		`<0\ \ \ text{(for odd}\ n>=3text{)}`

`P(-1/2)`	`= (n − 1)(-1/2)^n − n(-1/2)^(n − 1)+1`
	`= -(n − 1) xx 1/(2^n) − n/(2^(n − 1)) + 1\ \ \ \ text{(given}\ n\ text{is odd)}`
	`= (-n + 1 − 2n + 2^n)/(2^n)`
	`= (2^n − (3n − 1))/(2^n)`
	`>=0\ \ \ \ \ text{(given}\ \ 2^n>=3n-1\ \ text{for}\ \ n>=3, and 2^n > 0)`

`text(S)text(ince)\ \ P(x)\ \ text(is continuous, when it changes sign, it cuts the)\ x text(-axis)`

`:. -1 < α ≤ -1/2`

(v) `P(x) = 4x^5 − 5x^4 + 1\ \ text(is of the form)`

♦♦♦ Mean mark part (v) 2%.

`P(x) = (n − 1)x^n − nx^(n − 1) + 1`

`:.P(x)\ \ text(has 5 zeros)\ \=> 1, 1, alpha, beta, bar beta\ \ \ \ (text(where)\ beta\ text{is not real})`

`text(The zeros 1 and α have a modulus ≤ 1.)`

`text(Consider the product of the roots)`

`11alphabetabar beta`	`=- 1/4`
`alphabetabar beta`	`=- 1/4`
`\|\ alphabetabar beta\ \|`	`=\|\ – 1/4\ \|`
`\|\ alpha\ \|*\|\ beta\ \|^2`	`= 1/4`

`text(S)text(ince)\ \ |\ α\ | > 1/2,\ \ |\ beta\ |<1`

`:.text(Each of the zeros of)\ x^5 − 5x^4 + 1\ text(has a modulus)\ <=1.`

Graphs, EXT2 2010 HSC 7b

The graphs of `y = 3x − 1` and `y = 2^x` intersect at `(1, 2)` and at `(3, 8)`.

Using these graphs, or otherwise, show that `2^x ≥ 3x − 1` for `x ≥ 3`. (1 mark)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(When)\ x > 3, y = 2^x\ text(is above the graph of)\ y = 3x − 1.`

`text(S)text(ince the graphs intersect when)\ x = 3,`

`2^x ≥ 3x − 1\ \ text(for)\ x ≥ 3.`

Harder Ext1 Topics, EXT2 2010 HSC 7a

In the diagram `ABCD` is a cyclic quadrilateral. The point `K` is on `AC` such that `∠ADK = ∠CDB`, and hence `ΔADK` is similar to `ΔBDC`.

Copy or trace the diagram into your writing booklet.

Show that `ΔADB` is similar to `ΔKDC`. (2 marks)
Using the fact that `AC = AK + KC`,
show that `BD xx AC = AD xx BC + AB xx DC`. (2 marks)
A regular pentagon of side length `1` is inscribed in a circle, as shown in the diagram.

Let `x` be the length of a chord in the pentagon.
Use the result in part (ii) to show that `x = (1 + sqrt5)/2`. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`

Show Worked Solution

(i)

Harder Ext1 Topics, EXT2 2010 HSC 7a Answer

`text(S)text(ince)\ \ ∠ADB`	`=∠ADK +∠KDB, and`
`∠KDC`	`=∠CDB +∠KDB`
`=>∠ADB`	`= ∠KDC`
`∠ABD`	`= ∠ACD\ \ \ text{(angles in the same segment on arc}\ ADtext{)}`

`∠ADK`	`= ∠CDB\ \ \ `	`text{(corresponding angles of similar}`
		`text{triangles,}\ ΔADK\ text(\|\|\|)\ ΔBDC text{)}`

`:.ΔADB\ text(|||)\ ΔKDC\ \ text{(equiangular)}`

(ii) `text(S)text(ince)\ ΔADB\ text(|||)\ ΔKDC`

`(BD)/(DC) = (AD)/(DK) = (AB)/(KC)\ \ text{(corresponding sides of similar triangles)}`

`:.AB xx DC = KC xx BD\ \ \ \ …\ (1)`

♦♦ Mean mark part (ii) 31%.

`text(Similarly, using)\ \ ΔADK\ text(|||)\ ΔBDC\ \ text{(given)}`

`(AD)/(BD)=(AK)/(BC)`

`:. BC xx AD = AK xx BD\ \ \ \ …\ (2)`

`text{Add (1) + (2)}`

`AB xx DC + AD xx BC`	`=KC xx BD+AK xx BD`
	`=BD(AK+KC)`
`:.BD xx AC`	`=AD xx BC+AB xx DC`

(iii) `text(Each diagonal of the pentagon is)\ x.`

♦♦ Mean mark part (iii) 24%.

`text{Hence in applying the result in (i) you have}`

`BD = AC = x, AD = DC = CB = 1, BA = x`

`x xx x`	`= 1 xx 1 + x xx 1`
`x^2 − x − 1`	`= 0`
`x`	`= (1 ± sqrt(1 + 4))/2`
	`= (1 ± sqrt(5))/2`
`:.x`	`= (1 + sqrt(5))/2,\ \ \ (x>0)`

Polynomials, EXT2 2010 HSC 6c

Expand `(cos theta + i sin theta)^5` using the binomial theorem. (1 mark)
Expand `(cos theta + i sin theta)^5` using de Moivre’s theorem, and hence show that
1. `sin 5theta = 16 sin^5 theta − 20sin^3 theta + 5 sin theta`. (3 marks)
Deduce that
`x = sin (pi/10)` is one of the solutions to
1. `16x^5 − 20x^3 + 5x − 1 = 0`. (1 mark)
Find the polynomial `p(x)` such that `(x − 1) p(x) = 16x^5 − 20x^3 + 5x − 1`. (1 mark)
Find the value of `a` such that `p(x) = (4x^2 + ax − 1)^2`. (1 mark)
Hence find an exact value for
1. `sin (pi/10)`. (1 mark)

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`cos^5 theta + 5i cos^4 theta sin theta − 10 cos^3 theta sin^2 theta − 10i cos^2 theta sin^3 theta`
`+ 5 cos theta sin^4 theta + i sin^5 theta`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`16x^4 + 16x^3 − 4x^2 − 4x + 1`
`a = 2`
`(-1 + sqrt5)/4`

Show Worked Solution

(i) `(cos theta + i sin theta)^5`

`=cos^5 theta + 5cos^4 theta (i sin theta) + 10 cos^3 theta (i sin theta)^2 + `

`10 cos^2 theta (isin theta)^3 + 5 cos theta (i sin theta)^4 + (i sin theta)^5`

`= cos^5 theta + 5i cos^4 theta sin theta − 10 cos^3 theta sin^2 theta − 10i cos^2 theta sin^3 theta +`

`5 cos theta sin^4 theta + i sin^5 theta`

(ii) `text(Using De Moivre)`

`(cos theta + i sin theta)^5 = cos 5theta + i sin 5theta`

`text(Equating imaginary parts)`

`sin 5theta`	`= 5 cos^4theta sin theta − 10cos^2 theta sin^3 theta + sin^5 theta`
	`= 5 sin theta (1 − sin^2 theta)^2 − 10 sin^3 theta (1 − sin^2 theta) + sin^5 theta`
	`= 5 sin theta (1 − 2 sin^2 theta + sin^4 theta) − 10 sin^3 theta + 10 sin^5 theta + sin^5 theta`
	`= 5 sin theta − 10 sin^3 theta + 5 sin^5 theta − 10 sin^3 theta + 11 sin^5 theta`
	`= 16 sin^5 theta − 20 sin^3 theta + 5 sin theta`

(iii) `text(If)\ x = sin (pi/10)`

`16 sin^5 (pi/10) − 20 sin^3 (pi/10) + 5 sin (pi/10)`	`=sin (5 xx pi/10)`
`16x^5 − 20x^3 + 5x`	`= sin\ pi/2`
`16x^5 − 20x^3 + 5x`	`=1`

`:. sin\ pi/10\ \ text(is one solution to)\ \ 16x^5 − 20x^3 + 5x − 1=0`

(iv) `16x^5 − 20x^3 + 5x − 1`

`=(x-1)(16x^4 + 16x^3 − 4x^2 − 4x + 1)`

`:.p(x) = 16x^4 + 16x^3 − 4x^2 − 4x + 1`

(v) `(4x^2 + ax − 1)^2`

`= 16x^4 + 4ax^3 − 4x^2 + 4ax^3 + a^2x^2 − ax − 4x^2 − ax + 1`

`= 16x^4 + 8ax^3 − 8x^2 + a^2x^2 − 2ax +1`

`text(By equating coefficients of)\ \ x^3`

`:.a = 2`

(vi) `4 x^2 + 2 x − 1 = 0`

♦♦ Mean mark part (vi) 27%.

`x`	`=(-2 ± sqrt(4 + 16))/8`
	`=(-1 ± sqrt5)/4`

`:. sin\ pi/10 = (-1 + sqrt5)/4\ \ \ \ (sin\ pi/10\ > 0)`

Harder Ext1 Topics, EXT2 2010 HSC 5c

A TV channel has estimated that if it spends `$x` on advertising a particular program it will attract a proportion `y(x)` of the potential audience for the program, where

`(dy)/(dx) = ay(1 − y)`

and `a > 0` is a given constant.

Explain why
`(dy)/(dx)` has its maximum value when `y = 1/2`. (1 mark)
Using
`int (dy)/(y(1 − y)) = ln (y/(1 − y)) + c`, or otherwise, deduce that
`y(x) = 1/(ke^(-ax) + 1)` for some constant `k > 0`. (3 marks)
The TV channel knows that if it spends no money on advertising the program then the audience will be one-tenth of the potential audience.
Find the value of the constant `k` referred to in part (c)(ii). (1 mark)
What feature of the graph
`y = 1/(ke^(-ax) + 1)` is determined by the result in part (c)(i)? (1 mark)
Sketch the graph
`y(x) = 1/(ke^(-ax) + 1)` (1 mark)

Show Answers Only

`text(See Worked Solutions.)`
`text{Proof (See Worked Solutions)}`
`9`
`y = 1/2`
`text(See Worked Solutions.)`

Show Worked Solution

(i) `(dy)/(dx) = ay(1 − y)`

♦♦ Mean mark part (i) 32%.

`=>text(Given)\ \ a>0,\ \ ay(1 − y)\ text(is an inverted parabola)`

`text(with zeros at)\ \ y = 0\ \ text(and)\ \ y = 1`

`=>\ text(Parabola symmetry means that a maximum occurs when)`

`y = (0+1)/2=1/2`

`:.(dy)/(dx)\ \ text(has its maximum value when)\ y = 1/2.`

♦ Mean mark part (ii) 41%.

(ii) `dy/dx=ay(1 − y)`

`int (dy)/(y(1 − y))`	`=int a\ dx`
`ax`	`= ln (y/(1 − y))+c`
`e^(ax − c)`	`= y/(1 − y)`
`e^(ax − c) − ye^(ax − c)`	`= y`
`y(1 + e^(ax − c))`	`= e^(ax − c)`
`y`	`= (e^(ax − c))/(1 + e^(ax − c))`
`y`	`= 1/(e^c e^(− ax) + 1)`

`text(Let)\ k = e^c`

`:.y(x) = 1/(ke^(-ax) + 1)\ text(for some constant)\ \ k > 0.`

(iii) `text(When)\ \ x = 0, y = 0.1`

`0.1`	`= 1/(ke^0 + 1)`
`k + 1`	`= 10`
`k`	`= 9`

♦♦ Mean mark part (iv) 25%.

(iv) `text(The gradient the curve is greatest at)\ y = 1/2\ \ \ text{from part (i)}`

`:. text(A point of inflection occurs at)\ y =1/2.`

(v) `text(As)\ \ x->oo,\ \ y->1/(0+1)=1`

♦♦ Mean mark part (v) 12%.

Calculus, EXT2 C1 2010 HSC 5b

Show that `int (dy)/(y(1 − y)) = ln (y/(1 − y)) + c`

for some constant `c`, where `0 < y < 1`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Solution 1)`

`d/(dy)[ln(y/(1 − y)) + c]`	`= d/(dy)[ln y − ln(1 − y) + c]`
	`= 1/y − (-1)/(1 − y) + 0`
	`= (1 − y + y)/(y(1 − y))`
	`= 1/(y(1 − y))`

`text(Solution 2)`

`text(Using partial fractions:)`

`1/(y(1 − y))=`	`A/y+B/(1-y)`

`=> A(y-1)+By=1`

`text(When)\ \ y=1,\ B=1`

`text(When)\ \ y=0,\ A=1`

`:.int (dy)/(y(1 − y))`	`= int (1/y + 1/(1 − y))\ dy`
	`= ln y − ln(1 − y) + c`
	`= ln(y/(1 − y)) + c,\ \ \ \ \ y/(1 − y) > 0\ \ text(for)\ \ 0 < y < 1.`

Conics, EXT2 2010 HSC 5a

The diagram shows two circles, `C_1` and `C_2`, centred at the origin with radii `a` and `b`, where `a > b`.

The point `A` lies on `C_1` and has coordinates `(a cos theta, a sin theta)`.

The point `B` is the intersection of `OA` and `C_2`.

The point `P` is the intersection of the horizontal line through `B` and the vertical line through `A`.

Write down the coordinates of `B`. (1 mark)
Show that `P` lies on the ellipse
`(x^2)/(a^2) + (y^2)/(b^2) = 1`. (1 mark)
Find the equation of the tangent to the ellipse
`(x^2)/(a^2) + (y^2)/(b^2) = 1` at `P`. (2 marks)
Assume that `A` is not on the `y`-axis.
Show that the tangent to the circle `C_1` at `A`, and the tangent to the ellipse
`(x^2)/(a^2) + (y^2)/(b^2) = 1` at `P`, intersect at a point on the `x`-axis. (2 marks)

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`B(b cos theta, b sin theta)`
`text{Proof (See Worked Solutions)}`
`b cos theta x + a sin theta y = ab, or`
`(x cos theta)/a + (y sin theta)/b = 1`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

(i) `B(b cos theta, b sin theta)`

(ii) `text(Substitue)\ \ P(a cos theta, b sin theta)\ \ text(into)\ \ (x^2)/(a^2) + (y^2)/(b^2) = 1`

`text(LHS)`	`= (a^2 cos^2 theta)/(a^2) + (b^2 sin^2 theta)/(b^2)`
	`= cos^2 theta + sin^2 theta`
	`= 1`
	`=\ text(RHS)`

`:.P\ text(lies on the ellipse.)`

(iii) `text(Solution 1)`

`(x^2)/(a^2) + (y^2)/(b^2)`	`= 1`
`(2x)/(a^2) + (2y)/(b^2) xx (dy)/(dx)`	`= 0`
`(dy)/(dx)`	`= -(b^2x)/(a^2y)`

`text(At)\ \ P(a cos theta, b sin theta)`

`(dy)/(dx)`	`= (b^2a cos theta)/(a^2b sin theta)`
	`= -(b cos theta)/(a sin theta)`

`:.\ text(Equation of tangent at)\ P`

`y − b sin theta=`	` -(b cos theta)/(a sin theta)(x − a cos theta)`
`a sin theta y − ab sin^2 theta=`	` −b cos theta x + ab cos^2 theta`
`b cos theta x + a sin theta y=`	` ab(sin^2 theta + cos^2 theta)`
`:.b cos theta x + a sin theta y=`	`ab,\ \ \ text(or)`
`(x cos theta)/a + (y sin theta)/b=`	`1`

`text(Alternative Solution)`

`dy/dx`	`=(dy)/(d theta) xx (d theta)/(dx)`
	`=b cos theta xx 1/(-a sin theta)`
	`= -(bcos theta)/(a sin theta)`

`text{(then use the point-gradient formula as shown above)}`

(iv) `text(Finding the tangent to)\ x^2 + y^2 = 1\ text(at)\ \ A(a cos theta, a sin theta)`

`2x + 2y* (dy)/(dx)`	`=0`
`(dy)/(dx)`	`= (-x)/y`

`:.\ text(Equation of tangent)`

`y – a sin theta`	`=-(a cos theta)/(a sin theta)(x − a cos theta)`
`y sin theta – a sin^2 theta`	`=-x cos theta + a cos^2 theta`
`x cos theta + y sin theta`	`=a(sin^2 theta + cos^2 theta)`
`x cos theta + y sin theta`	`=a`
`:.y sin theta`	`=a- x cos theta`

`text(Substitute into the equation of the tangent at)\ \ P`

`b cos theta x + a sin theta y`	`=ab`
`b cos theta x+a(a- x cos theta)`	`=ab`
`bx cos theta + a^2 – ax cos theta`	`=ab`
`(b – a)x cos theta`	`=a(b – a)`
`x cos theta`	`= a`

`text(When)\ x cos theta = a,\ \ y sin theta = a − a = 0\ \ =>y=0,\ \ theta≠0`

`:.\ text(Intersection occurs on the)\ x text(-axis.)`

Harder Ext1 Topics, EXT2 2010 HSC 4d

A group of `12` people is to be divided into discussion groups.

In how many ways can the discussion groups be formed if there are `8` people in one group, and `4` people in another? (1 mark)
In how many ways can the discussion groups be formed if there are `3` groups containing `4` people each? (2 marks)

Show Answers Only

`495`
`5775`

Show Worked Solution

(i)	`text(Number of combinations)`	`=\ ^(12)C_4`
		`= 495`

(ii) `text(Number of discussion groups)`

♦ Mean mark part (ii) 35%.

`=(\ ^12C_4 xx\ ^8C_4 xx\ ^4C_4) / (3!)`

`= (34\ 650) / (3!)`

`= 5775`

Mechanics, EXT2 2010 HSC 4b

A bend in a highway is part of a circle of radius `r`, centre `O`. Around the bend the highway is banked at an angle `α` to the horizontal.

A car is travelling around the bend at a constant speed `v`. Assume that the car is represented by a point `P` of mass `m`. The forces acting on the car are a lateral force `F`, the gravitational force `mg` and a normal reaction `N` to the road, as shown in the diagram.

By resolving forces, show that
`F = mg sin α − (mv^2)/r cos α`. (3 marks)
Find an expression for `v` such that the lateral force `F` is zero. (1 mark)

Show Answers Only

`text{Proof (See Worked Solutions)}`
`sqrt(rg tan α)`

Show Worked Solution

(i)

`text(Resolving forces vertically)`
`N cos α + F sin α=`	`mg\ \ \ \ …\ (1)`
`text(Resolving forces horizontally)`
`N sin α − F cos α=`	`(mv^2)/r\ \ \ \ …\ (2)`
`text{Multiply (1) by sin α and (2) by cos α}`
`N sin α\ cos α + F sin^2 α=`	`mg sin α\ \ \ \ …\ (3)`
`N sin α\ cos α − F cos^2 α=`	`(mv^2)/r cos α\ \ \ \ …\ (4)`
`text{Subtract (3) – (4)}`
`F sin^2 α + F cos^2 α=`	`mg sin α − (mv^2)/r cos α`
`:.F=`	`mg sin α − (mv^2)/r cos α`

(ii) `F = mg sin α − (mv^2)/r cos α`

`mg sin α − (mv^2)/r cos α`	`=0`
`(v^2)/r cos α`	`=g sin α`
`v^2`	`=rg tan α`
`v`	`=sqrt(rg tan α)`

Graphs, EXT2 2010 HSC 4a

A curve is defined implicitly by `sqrtx + sqrty = 1`.
Use implicit differentiation to find `(dy)/(dx)`. (2 marks)
Sketch the curve `sqrtx + sqrty = 1`. (2 marks)
Sketch the curve `sqrt(|\ x\ |) + sqrt(|\ y\ |) = 1` (1 mark)

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`- sqrty/sqrtx`
`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

Show Worked Solution

(i)	`sqrtx + sqrt y`	`= 1`
	`1/(2sqrtx) + 1/(2sqrty)*(dy)/(dx)`	`= 0`
	`:.(dy)/(dx)`	`= – sqrty/sqrtx`

(ii)

♦♦ Mean mark part (iii) 46%.

(iii)

Volumes, EXT2 2010 HSC 3b

The region shaded in the diagram is bounded by the `x`-axis and the curve `y = 2x − x^2`.

The shaded region is rotated about the line `x = 4`.

Find the volume generated. (4 marks)

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`8pi\ \ text(u³)`

Show Worked Solution

`text(Using cylindrical shells)`

`delta V`	`=2 pi y (4-x)\ delta x`
`:.V`	`=lim_(delta x->0) sum_(x=0)^2 2 pi y (4-x)\ delta x`
	`= int_0^2 2pi y (4 − x)\ dx`
	`= 2pi int_0^2 (2x − x^2)(4 − x)\ dx`
	`= 2pi int_0^2(8x − 6x^2 + x^3)\ dx`
	`= 2pi[4x^2 − 2x^3 + (x^4)/4]_0^2`
	`= 2pi[(16 − 16 + 4) − 0]`
	`= 8pi\ \ text(u³)`

Complex Numbers, EXT2 N2 2010 HSC 2d

Let `z = cos theta + i sin theta` where `0 < theta < pi/2`.

On the Argand diagram the point `A` represents `z`, the point `B` represents `z^2` and the point `C` represents `z + z^2`.

Copy or trace the diagram into your writing booklet.

Explain why the parallelogram `OACB` is a rhombus. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Show that `text(arg)\ (z + z^2) = (3theta)/2`. (1 mark)

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Show that `| z + z^2 | = 2 cos theta/2`. (2 marks)

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By considering the real part of `z + z^2`, or otherwise deduce that
`cos theta + cos 2theta = 2 cos theta/2 cos (3theta)/2`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(See Worked Solutions.)`
`text(See Worked Solutions.)`
`text(See Worked Solutions.)`
`text(See Worked Solutions.)`

Show Worked Solution

i. `z = cos theta + i sin theta`

`|\ OA\ |=|\ z\ |=1`

`|\ OB\ | =|\ z^2\ |=|\ z\ |^2 = 1`

`:.\ OACB\ text(is a parallelogram with a pair of adjacent sides equal.)`

`:.\ OACB\ text(is a rhombus.)`

ii. `text(arg)\ z^2 = 2\ text(arg)\ z = 2 theta`

`∠BOA = 2 theta − theta = theta`

`text(S)text(ince)\ OACB\ text(is a rhombus then)\ CO\ text(bisects)\ ∠BOA`

`:.∠COA = theta/2`

`:.\ text(arg)(z + z^2) = theta + theta/2 = (3theta)/2`

iii. `OC = |\ z + z^2\ |`

`text(Join)\ \ AB\ \ text(so that it meets)\ \ OC\ \ text(at)\ \ M`

`AB ⊥ OC, and OM=OC\ \ \ text{(diagonals of a rhombus)}`

♦♦ Mean mark part (iii) 26%.

`text(In)\ \ Delta OAM:`

`cos\ theta/2`	`=(OM)/(OA)`
	`=OM`
`:.OC`	`=2 xx OM`
	`=2 cos\ theta/2`

iv.	`z + z^2`	`= cos theta + i sin theta + (cos theta + i sin theta)^2`
		`= cos theta + cos 2theta + i(sin theta + sin 2theta)\ \ \ \ text{(De Moivre)}`
`:.\ text(Re)(z+z^2)=cos theta + cos 2theta`

♦♦ Mean mark part (iv) 44%.

`text{Using parts (ii) and (iii),}`

`z + z^2=2 cos\ theta/2(cos (3theta)/2+ i sin (3theta)/2)`

`text(Equating real parts:)`

`cos theta + cos 2theta = 2 cos theta/2 cos\ (3theta)/2`

Calculus, EXT2 C1 2010 HSC 1e

Find `int (dx)/(1 + sqrtx)`. (3 marks)

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`2sqrtx − 2log_e\ (1 + sqrtx) + c_1\ \ text(or)`

`2+2sqrtx − 2log_e\ (1 + sqrtx) + c_2`

Show Worked Solution

`text(Solution 1)`

`text(Let)\ \ u = sqrtx,\ \ du = 1/(2sqrtx)\ dx,\ \ dx = 2u\ du`
`int (dx)/(1 + sqrtx)`	`=int (2u\ du)/(1 + u)`
	`=int((2+2u-2)/(1+u))\ du`
	`=int (2 − 2/(1 + u))\ du`
	`=2u − 2log_e(1 + u) + c_1`
	`=2sqrtx − 2log_e\ (1 + sqrtx) + c_1`

`text(Alternative Solution)`

`text(Let)\ \ u = 1+sqrtx,\ \ du = 1/(2sqrtx)\ dx,\ \ dx = 2(u-1)\ du`
`int (dx)/(1 + sqrtx)`	`=2int (u-1)/u\ du`
	`=2 int(1 − 1/u)du`
	`=2u − 2log_e\ \|\ u\ \| + c_2`
	`=2+2sqrtx − 2log_e\ (1 + sqrtx) + c_2`

`text(NB. These solutions are equivalent by making)\ \ c_1=c_2+2`

Calculus, EXT2 C1 2010 HSC 1c

Find `int 1/(x(x^2 + 1))\ dx`. (3 marks)

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`ln\ |\ x\ | + 1/2ln\ (x^2 + 1) + c`

Show Worked Solution

`text(Using partial fractions:)`

`1/(x(x^2 + 1)) =`	`a/x + (bx + c)/(x^2 + 1)`
`1=`	`a(x^2+1)+x(bx+c)`
`1=`	`(a + b)x^2 + cx + a`
`:.c = 0, \ \ a = 1,\ \ b = -1`

`:.int 1/(x(x^2 + 1))\ dx`	`=int(1/x − x/(x^2 + 1))\ dx`
	`=ln\ \|\ x\ \|-1/2ln\ (x^2 + 1) + c`

Harder Ext1 Topics, EXT2 2011 HSC 8b

A bag contains seven balls numbered from `1` to `7`. A ball is chosen at random and its number is noted. The ball is then returned to the bag. This is done a total of seven times.

What is the probability that each ball is selected exactly once? (1 mark)
What is the probability that at least one ball is not selected? (1 mark)
What is the probability that exactly one of the balls is not selected? (2 marks)

Show Answers Only

`(6!)/7^6=720/(117\ 649)`
`1 – (6!)/7^6=(116\ 929)/(117\ 649)`
`(3 xx 6!)/7^5=2160/(16\ 807)`

Show Worked Solution

(i) `P text{(each ball is selected once)}`

♦♦ Mean mark part (i) 32%.

`=7/7 xx 6/7 xx 5/7 xx 4/7 xx 3/7 xx 2/7 xx 1/7`

`=(6!)/7^6`

`=720/(117\ 649)`

(ii) `P text{(at least one ball is not selected)}`

`=1 – P text{(each ball once)}`

`=1 – (6!)/7^6`

`=(116\ 929)/(117\ 649)`

(iii) `text{Consider when 1 of the balls is not selected (say #1).}`

♦♦♦ Mean mark part (iii) 2%.

`=>\ text(One of the other 6 balls is selected twice.)`

`text(e.g. 2, 2, 3, 4, 5, 6, 7).`

`text(This can be done in)\ \ (7!)/(2!)\ \ text(ways.)`

`text(With 6 differently numbered pairs possible,)`

`text(This can be done in)\ \ 6 xx (7!)/(2!)\ \ text(ways)`

`text(S)text(ince there are 7 numbers that can be left out,)`

`text(Total number of ways to leave 1 number out)`

`=7 xx 6 xx (7!)/(2!)`

`=21 xx 7!`

`:.P text{(exactly one ball not selected)}`	`=(21 xx 7!)/7^7`
	`=(3 xx 6!)/7^5`
	`=2160/(16\ 807)`

Volumes, EXT2 2011 HSC 7a

The diagram shows the graph of `f(x) = x/(1 + x^2)` for `0 <= x <= 1.`

The area bounded by `y = f(x)`, the line `x = 1` and the `x`-axis is rotated about the line `x = 1` to form a solid.

Use the method of cylindrical shells to find the volume of the solid. (4 marks)

Show Answers Only

`pi(ln 2 + pi/2 – 2)\ \ text(u³)`

Show Worked Solution

`delta V`	`= 2 pi R h delta x`
	`= 2 pi (1 – x)* f (x) delta x`

`text(where)\ \ 2 pi(1 – x) delta x\ \ text(is the approximate area)`

`text(of the base and)\ \ f(x)\ \ text(is the height.)`

`text(Summing over the shells and letting)\ \ delta x -> 0,`

`V`	`=2 pi int_0^1 (1 – x) f(x)\ dx`
	`=2 pi int_0^1 (1 – x) xx x/(1 + x^2)\ dx`
	`=2 pi int_0^1 (x-x^2)/(1+x^2)\ dx`

`text(Using partial fractions)`

`(x – x^2)/(1 + x^2)`	`=A+(Bx+C)/(1+x^2)`
`x-x^2`	`=A(1+x^2)+Bx+C`
	`=Ax^2+Bx+(A+C)`
`:. A=-1,\ \ B=1,\ \ C=1`

`V`	`=2 pi int_0^1 (-1+(x+1)/(1+x^2))\ dx`
	`=2 pi int_0^1 (x/(1 + x^2) – 1 + 1/(1 + x^2))\ dx`
	`=2 pi [1/2 ln (1 + x^2) – x + tan^-1 x]_0^1`
	`=2 pi [1/2 ln 2 – 1 + tan^-1 1 – (1/2 ln 1 + tan^-1 0)]`
	`=2 pi (1/2 ln 2 – 1 + pi/4)`
	`=pi (ln 2 + pi/2 – 2)\ \ text(u³)`

Graphs, EXT2 2011 HSC 6b

Let `f (x)` be a function with a continuous derivative.

Prove that `y = (f(x))^3` has a stationary point at `x = a` if `f(a) = 0` or `f prime(a) = 0.` (2 marks)
Without finding `f″(x)`, explain why `y = (f(x))^3` has a horizontal point of inflection at `x = a` if `f(a) = 0` and `f prime (a) != 0.` (1 mark)
The diagram shows the graph `y = f(x).`
Copy or trace the diagram into your writing booklet.
On the diagram in your writing booklet, sketch the graph `y = (f(x))^3`, clearly distinguishing it from the graph `y = f(x).` (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(See Worked Solutions)`

Show Worked Solution

(i) `text(If)\ \ f(x)\ \ text(is a function with a continuous derivative,)`

`=> f prime (x)\ \ text(exists for all)\ \ x.`

`y`	`= (f(x))^3`
`(dy)/(dx)`	`= 3 xx (f(x))^2 xx f prime(x)`

`:.\ (dy)/(dx) = 0\ \ text(when)\ \ f(x) = 0 or f prime (x) = 0`

`:.\ x = a\ \ text(is a stationary point if)\ \ f(a) = 0 or f prime(a) = 0`

♦♦♦ Mean mark part (ii) 0%!
A BEAST!

(ii) `text(If)\ \ f prime (a) != 0\ \ text(then either)\ \ f prime (a) > 0 or f prime (a) < 0,`

`and f prime (x)\ \ text(keeps that sign either side of)\ \ x = a.`

`:. text(If)\ \ f(a) = 0 and f prime(a) != 0, text(there is a stationary point at)\ \ x = a`

`text(where the slope of the curve does not change either side of)\ \ x = a.`

`text(i.e. a horizontal point of inflection occurs at)\ \ x=a`

(iii) `y = (f(x))^3`

`text(When)\ \ f(x) = 0,\ \ (f(x))^3 = 0\ \ text{(horizontal P.I. from part (ii))}`

`text(When)\ \ f(x) = 1,\ \ (f(x))^3 = 1`

`text(If)\ \ 0 < f(x) < 1,\ \ 0 < (f(x))^3 < f(x)`

`text(If)\ \ f(x) > 1,\ \ (f(x))^3 > f(x)`

`text(When)\ f prime (a) = 0,\ \ (f(x))^3\ \ text(has a maximum turning point)`

`f(x) < 0,\ \ (f(x))^3 < 0`

Mechanics, EXT2 2011 HSC 5a

A small bead of mass `m` is attached to one end of a light string of length `R`. The other end of the string is fixed at height `2h` above the centre of a sphere of radius `R`, as shown in the diagram. The bead moves in a circle of radius `r` on the surface of the sphere and has constant angular velocity `omega > 0`. The string makes an angle of `theta` with the vertical.

Three forces act on the bead: the tension force `F` of the string, the normal reaction force `N` to the surface of the sphere, and the gravitational force `mg`.

By resolving the forces horizontally and vertically on a diagram, show that
1. `F sin theta - N sin theta = m omega^2 r`
and
1. `F cos theta + N cos theta = mg.` (2 marks)
Show that
1. `N = 1/2 mg sec theta - 1/2 m omega^2 r\ text(cosec)\ theta.` (2 marks)
Show that the bead remains in contact with the sphere if
1. `omega <= sqrt (g/h).` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`text(Resolving forces horizontally)`

`text(Net force)`	`= m r omega^2\ \ \ text{(towards the centre of the circle)}`
`F sin theta – N sin theta`	`= mr omega^2\ \ \ \ text{… (1)}`

♦ Mean mark part (i) 50%.

`text(Resolving forces vertically)`

`text(Net force)`	`= mg\ \ \ \ text{(gravitational force)}`
`F cos theta + N cos theta`	`= mg\ \ \ \ text{… (2)}`

(ii) `text{Divide (1) by}\ \ sin theta`

`F – N = mr omega^2\ text(cosec)\ theta\ \ \ \ text{… (3)}`

`text{Divide (2) by}\ \ cos theta`

`F+N = mg sec theta\ \ \ \ text{… (4)}`

`text{Subtract (4) – (3)}`

`2N`	`= mg sec theta – mr omega^2\ text(cosec)\ theta`
`:.N`	`= 1/2 mg sec theta – 1/2 mr omega^2\ text(cosec)\ theta`

(iii) `text(When in contact with the sphere,)\ \ N >= 0.`

`1/2 mg sec theta – 1/2 mr omega^2`	`>= 0`
`1/2 mg sec theta`	`>= 1/2 mr omega^2\ text(cosec)\ theta`
`g sec theta`	`>= r omega^2\ text(cosec)\ theta`
`omega^2`	`<= (g sec theta)/(r\ text(cosec)\ theta)`
	`<= g/r tan theta,\ \ \ \ (text{since}\ \ tan theta = r/h)`
	`<= g/r xx r/h`
	`<=g/h`
`:. omega`	`<= sqrt (g/h)`

Harder Ext1 Topics, EXT2 2011 HSC 4c

A mass is attached to a spring and moves in a resistive medium. The motion of the mass satisfies the differential equation

`(d^2y)/(dt^2) + 3 (dy)/(dt) + 2y = 0,`

where `y` is the displacement of the mass at time `t.`

Show that, if `y = f(t)` and `y = g(t)` are both solutions to the differential equation and `A` and `B` are constants, then
1. `y = A f (t) + Bg (t)`
is also a solution. (2 marks)
A solution of the differential equation is given by `y = e^(kt)` for some values of `k`, where `k` is a constant.
Show that the only possible values of `k` are `k = -1` and `k = -2.` (2 marks)
A solution of the differential equation is
1. `y = Ae^(-2t) + Be^-t.`
When `t = 0`, it is given that `y = 0` and `(dy)/(dt) = 1`.
Find the values of `A` and `B.` (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`A = -1,\ \ B = 1`

Show Worked Solution

(i) `(d^2y)/(dt^2) + 3(dy)/(dt) + 2y = 0`

♦♦♦ Mean mark 18%.

`text(Given)\ \ y`	`=Af(t) + Bg (t)\ \ text(then)`
`y′`	`=Af′(t)+ Bg′(t)`
`y″`	`=Af″(t)+Bg″(t)`

`text(LHS)`	`=d^2/(dt^2) (Af (t) + Bg (t)) + 3 d/(dt) (Af (t) + Bg (t)) + 2 (Af (t) + Bg(t))`
	`=Af″ (t) + Bg″ (t) + 3 Af prime (t) + 3 Bg prime (t) + 2 Af (t) + 2 Bf (t)`
	`=A (f″(t) + 3 f prime (t) + 2 f (t)) + B(g″ (t) + 3g prime (t) + 2 f(t))`
	`=A(0) + B(0)`
	`=0`

`:.y = Af(t) + Bg(t)\ \ text(is a solution of)\ \ (d^2y)/(dt^2) + 3 (dy)/(dt) + 2y = 0`

(ii) `y = e^(kt),\ \ (dy)/(dt) = ke^(kt),\ \ (d^2y)/(dt^2) = k^2 e^(kt)`

`text(Substituting into)`

`(d^2y)/(dt^2) + 3 (dy)/(dt) + 2y`	`= 0`
`k^2 e^(kt) + 3 ke^(kt) + 2e^(kt)`	`=0`
`e^(kt) (k^2 + 3k + 2)`	`=0`
`k^2 + 3k + 2`	`=0\ \ \ \ (e^(kt) != 0)`
`(k + 1) (k + 2)`	`=0`

`:.k = -1 or -2`

(iii)	`y`	`= Ae^(-2t) + Be^-t`
	`(dy)/(dt)`	`= -2 Ae^(-2t) – Be^-t`

`text(When)\ \ t = 0,\ \ y = 0`

`=>A + B = 0\ \ \ …\ text{(i)}`

`text(When)\ \ t = 0,\ \ (dy)/(dt) = 1`

`=>-2A – B = 1\ \ \ …\ text{(ii)}`

`text{Add (i) + (ii)}`

`-A = 1 `

`:.A = -1 and B=1`

Harder Ext1 Topics, EXT2 2011 HSC 4b

In the diagram, `ABCD` is a cyclic quadrilateral. The point `E` lies on the circle through the points `A, B, C` and `D` such that `AE\ text(||)\ BC`. The line `ED` meets the line `BA` at the point `F`. The point `G` lies on the line `CD` such that `FG\ text(||)\ BC.`

Copy or trace the diagram into your writing booklet.

Prove that `FADG` is a cyclic quadrilateral. (2 marks)
Explain why `/_ GFD =/_ AED.` (1 mark)
Prove that `GA` is a tangent to the circle through the points `A, B, C` and `D.` (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`/_ GFD = /_ AED\ \ text{(alternate angles,}\ \ FG\ text(||)\ AE text{)}`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

(i)

`text(Let)\ /_ BCD`	`= alpha`
`/_ FAD`	`= alpha\ \ text{(exterior angle of a cyclic quadrilateral}\ ABCD text{)}`
`/_ FGC`	`= pi – alpha\ \ text{(cointerior angles,}\ \ FG\ text(\|\|)\ BC text{)}`

`:.\ \ /_ FAD + /_ FGD = pi`

`:.\ FADG\ \ text{is a cyclic quadrilateral (opposite angles are supplementary)}`

(ii) `/_ GFD = /_ AED\ \ text{(alternate angles,}\ \ FG\ text(||)\ AE text{)}`

(iii) `text(Join)\ GA`

`/_ GAD = /_ GFD\ \ text{(angles in the same segment on arc}\ GDtext{)}`

`text(S)text(ince)\ /_ GFD`	`= /_ AED\ \ \ text{(part (ii))}`
`/_ GAD`	`= /_ AED`

`:.GA\ \ text(is a tangent to the circle through)\ \ A, B, C and D.`

`text{(angle in the alternate segment equals the angle}`

`text(between)\ GA\ text(and chord)\ AD text{).}`

Complex Numbers, EXT2 N2 2011 HSC 4a

Let `a` and `b` be real numbers with `a != b`. Let `z = x + iy` be a complex number such that

`|\ z - a\ |^2 - |\ z - b\ |^2 = 1.`

Prove that `x = (a + b)/2 + 1/(2 (b - a)).` (2 marks)

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Hence, describe the locus of all complex numbers `z` such that `|\ z - a\ |^2 - |\ z - b\ |^2 = 1.` (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

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`text(Proof)\ \ text{(See Worked Solutions)`
`x = (a + b)/2 + 1/(2(b – a))`

Show Worked Solution

i.	`\|\ z – a\ \|^2 – \|\ z – b\ \|^2`	`= 1`
	`\|\ (x-a)+iy\ \|^2-\|\ (x-b)+iy\ \|^2`	`=1`
	`(x – a)^2 + y^2 – ((x – b)^2 + y^2)`	`=1`
	`(x – a)^2 – (x – b)^2`	`=1`
	`(x – a – (x – b)) (x – a + x – b)`	`=1`
	`(b – a) (2x – a – b)`	`=1`

`2x – a – b`	`= 1/(b – a)`
`2x`	`= a + b + 1/(b – a)`
`:. x`	`= (a + b)/2 + 1/(2(b – a))`

♦ Mean mark part (ii) 44%.

ii. `text(The locus is the vertical line)`

`x = (a + b)/2 + 1/(2(b – a)).`

Proof, EXT2 P2 2011 HSC 3c

Use mathematical induction to prove that `(2n)! >= 2^n (n!)^2` for all positive integers `n`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(If)\ \ n=1,`

`text(LHS) = 2! = 2`

`text(RHS) = 2 xx (1!)^2 = 2 = text(LHS)`

`text(Assume true for)\ \ n = k,`

`text(i.e.)\ \ (2k)! >= 2^k (k!)^2\ \ \ …\ text{(i)}`

`text(Prove true for)\ \ n = k +1,`

`text(i.e.)\ \ (2(k +1))! >= 2^(k + 1) ((k +1)!)^2`

`text(LHS)`	`= (2 (k + 1))!`
	`= (2k + 2)(2k +1) xx underbrace{(2k)!}_text{using part (i)}`
	`>= (2k + 2) (2k +1) xx 2^k (k!)^2`
	`>= 2(k + 1) (2k + 1) xx 2^k (k!)^2`
	`>=2^(k+1) (k+1)(k+1) (k!)^2,\ \ \ \ \ (2k+1)>(k+1),\ text(for)\ k>0`
	`>= 2^(k + 1) ((k + 1)!)^2`

`=>\ text(True for)\ \ n = k + 1`

`:.\ text(S)text(ince true for)\ \ n = 1,\ text(by PMI, true for integral)\ n>=1.`

Volumes, EXT2 2011 HSC 3b

The base of a solid is formed by the area bounded by `y = cos x` and `y = -cos x` for `0 <= x <= pi/2.`

Vertical cross-sections of the solid taken parallel to the `y`-axis are in the shape of isosceles triangles with the equal sides of length `1` unit as shown in the diagram.

Find the volume of the solid. (3 marks)

Show Answers Only

`1/2\ \ text(u³)`

Show Worked Solution

`text(Length of base of cross-section) = 2 cos x`

`text(Using Pythagoras to find the height of cross-section)`

`h^2+cos^2 x`	`=1`
`h^2`	`=1-cos^2 x`
`h`	`=sin x\ \ \ \ \ (h>0)`

`text(Volume of vertical slice) = 1/2 xx 2 cos x xx sin x xx delta x`

`V`	`=lim_(delta x -> 0) sum_(x=0)^(pi/2) cos x sin x\ delta x`
	`=int_0^(pi/2) sin x cos x\ dx`
	`=1/2 int_0^(pi/2) sin 2 x\ dx`
	`=-1/4[cos 2x]_0^(pi/2)`
	`=-1/4[(cos pi)-(cos 0)]`
	`=-1/4 [-1 – (1)]`
	`=1/2`

`text(OR)\ \ `	`= [1/2 sin^2 x]_0^(pi/2)`
	`= 1/2`

`:. text(The volume is)\ \ 1/2\ \ text(u³)`

Functions, EXT1′ F1 2011 HSC 3a

Draw a sketch of the graph

`quad y = sin\ pi/2 x` for `0 < x < 4.` (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
Find `lim_(x -> 0) x/(sin\ pi/2 x).` (1 mark)

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Draw a sketch of the graph

`quad y = x/(sin\ pi/2 x)` for `0 < x < 4.` (2 marks)
(Do NOT calculate the coordinates of any turning points.)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

ii. `2/pi`

iii.

Show Worked Solution

ii. `lim_(x->0) x/(sin\ pi/2 x)`	`= 2/pi lim_(x->0) (pi/2 x)/(sin\ pi/2 x)`
	`= 2/pi xx 1`
	`= 2/pi`

iii.

Complex Numbers, EXT2 N2 2011 HSC 2d

Use the binomial theorem to expand `(cos theta + i sin theta)^3.` (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Use de Moivre’s theorem and your result from part (i) to prove that

`cos^3 theta = 1/4 cos 3 theta + 3/4 cos theta.` (3 marks)

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Hence, or otherwise, find the smallest positive solution of

`4 cos^3 theta - 3 cos theta = 1.` (2 marks)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`cos^3 theta + 3 i cos^2 theta sin theta – 3 cos theta sin^2 theta – i sin^3 theta`
`text(Proof)\ \ text{(See Worked Solutions)}`
`theta = (2 pi)/3`

Show Worked Solution

i. `(cos theta + i sin theta)^3`

`=sum_(k=0)^3 \ ^3C_k (cos theta)^(3-k) (i sin theta)^k`

`= cos^3 theta + 3 cos^2 theta (i sin theta)+ 3 cos theta (i sin theta)^2 + (i sin theta)^3`

`= cos^3 theta + 3 i cos^2 theta sin theta- 3 cos theta sin^2 theta – i sin^3 theta`

ii. `text(Using De Moivre’s Theorem)`

`(cos theta + i sin theta)^3 = cos 3 theta + i sin 3 theta`

`text(Equate real parts)`

`cos 3 theta`	`= cos^3 theta – 3 cos theta sin^2 theta`
`cos 3 theta`	`= cos^3 theta – 3 cos theta (1 – cos^2 theta)`
`cos 3 theta`	`= 4 cos^3 theta – 3 cos theta`
`4 cos^3 theta`	`=cos 3 theta+3cos theta`
`:.cos^3 theta`	`= 1/4 cos 3 theta + 3/4 cos theta\ \ \ text(… as required)`

iii. `text(If)\ \ \ 4 cos^3 theta – 3 cos theta = 1`

`=>cos 3 theta = 1\ \ \ \ text{(from part (ii))}`

`3 theta`	`= 2 k pi`
`:. theta`	`= (2 k pi)/3`

`:.\ text(Smallest positive solution occurs when)`

`theta = (2 pi)/3\ \ \ \ text{(i.e. when}\ k = 1 text{)}`

Proof, EXT2 P2 2012 HSC 16b

Show that `tan^(-1)\ x + tan^(-1)\ y = tan^(-1)((x + y)/(1 − xy))` for `|\ x\ | < 1` and `|\ y\ | < 1`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Use mathematical induction to prove

`sum_(j = 1)^n\ tan^(-1)(1/(2j^2)) = tan^(-1)(n/(n + 1))` for all positive integers `n`. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---
Find `lim_(n → ∞) sum_(j = 1)^n\ tan^(-1)(1/(2j^2))`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`pi/4`

Show Worked Solution

i. `text(Let)\ \ A=tan^(-1)\ x, and B=tan^(-1)y`

`=> tan\ A=x, and tan\ B=y`

`tan\ (A+B)`	`=(tan A + tan B)/(1-tanAtanB)`
	`=(x+y)/(1-xy)`
`:. A+B`	`= tan^(-1)\ ((x+y)/(1-xy))\ \ \ text(… as required)`

ii. `sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ (n/(n + 1))`

`text(If)\ \ j=1`

`text(LHS)`	`=tan^(-1)\ (1/2)`
`text(RHS)`	`=tan^(-1)\ (1/(1 + 1))`
	`=tan^(-1)\ (1/2)`
	`=\ text(LHS)`

`=>\ text(True for)\ \ j = 1.`

`text(Assume true)`

`sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ (n/(n + 1))`

`text(Need to prove)`

`sum_(j = 1)^(n + 1)\ tan^(-1)\ (1/(2j^2)) = tan^(-1)\ ((n + 1)/(n + 2))`

`text(LHS)`	`=sum_(j = 1)^(n + 1)\ tan^(-1)\ (1/(2j^2))`
	`=tan^(-1)\ (n/(n + 1)) + tan^(-1)\ (1/(2(n + 1)^2))`
	`=tan^(-1)\ ((n/(n + 1) + 1/(2(n + 1)^2))/(1 − n/(n + 1) xx 1/(2(n + 1)^2))),\ \ \ text{(using part (i))}`
	`=tan^(-1)\ ((2n(n + 1)^2 + n + 1)/(2(n + 1)^3 − n))`
	`=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2(n + 1)^3 − n))`
	`=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2n^3 + 6n^2 + 6n + 2 − n))`
	`=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/(2n^3 + 6n^2 + 5n + 2))`
	`=tan^(-1)\ (((n + 1)(2n^2 + 2n + 1))/((n + 2)(2n^2 + 2n + 1)))`
	`=tan^(-1)\ ((n + 1)/(n + 2))`
	`=\ text(RHS)`

`=> text(True for)\ \ j=n+1`

`:.text(S)text(ince true for)\ \ j = 1,\ text(by PMI, true for integral)\ \ j>=1`

iii.	`lim_(n → ∞) sum_(j = 1)^n\ tan^(-1)\ (1/(2j^2))`	`= lim_(n → ∞)\ tan^(-1)\ (n/(n + 1))`
		`= lim_(n → ∞)\ tan^(-1)\ (1/(1 + 1/n))`
		`= tan^(-1)\ 1`
		` = pi/4`

Polynomials, EXT2 2012 HSC 15b

Let `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`, where `k` is real.

Let `α = x + iy`, where `x` and `y` are real.

Suppose that `α` and `iα` are zeros of `P(z)`, where `bar α ≠ iα`.

Explain why `bar α` and `-i bar α` are zeros of `P(z)`. (1 mark)
Show that `P(z) = z^2(z − k)^2 + (kz − 1)^2`. (1 mark)
Hence show that if `P(z)` has a real zero then
1. `P(z) = (z^2 + 1)(z+ 1)^2` or `P(z) = (z^2 + 1)(z − 1)^2.` (2 marks)
Show that all zeros of `P(z)` have modulus `1`. (2 marks)
Show that `k = x − y`. (1 mark)
Hence show that `-sqrt2 ≤ k ≤ sqrt2`. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`
`text{Proof (See Worked Solutions.)}`

Show Worked Solution

(i) `P(z) = z^4 − 2kz^3 + 2k^2z^2 − 2kz +1`

`P(α) = P(iα) = 0, \ \ bar α ≠ iα.`

`text(S)text(ince)\ \ P(z)\ text(has real coefficients,)`

`=>\ text(Its zeros occur in conjugate pairs.)`

`text(i.e.)\ \ P(α) = 0\ text(and)\ P(bar α) = 0`

`bar α`	`= x − iy`
`bar(i α)`	`=bar (i(x+iy))`
	`=bar (ix-y)`
	`=-y-ix`
	`=-i(x-iy)`
	`=-i barα`

`:. -i bar α\ text(is a zero of)\ P(z)\ text(as it is the conjugate of)\ iα.`

(ii)	`H(z)`	`= z^2(z − k)^2 + (kz − 1)^2`
		`= z^2 (z^2 − 2kz + k^2) + (k^2z^2 − 2kz + 1)`
		`= z^4 − 2kz^3 + k^2z^2 + k^2z^2 − 2kz +1`
		`= z^4 − 2kz^3 + 2k^2z^2 − 2kz + 1`
		`= P(z)`

(iii) `text(If)\ z\ text(is real then)\ z^2(z − k)^2\ text(and)\ (kz −1)^2\ text(are real and)`

`text(either positive or zero.)`

♦♦♦ Mean mark part (iii) 11%.

`:.text(If)\ \ P(z) = z^2(z − k)^2 + (kz − 1)^2 = 0`

`=>(z − k)^2 = 0\ text(and)\ (kz − 1)^2 = 0`

`text(When)\ \ z = k,\ \ \ k^2 − 1 = 0\ \ text(or)\ k = ±1.`

`text(If)\ k = 1`

`P(z)`	`= z^2(z − 1)^2 + (z − 1)^2`
	`= (z^2 + 1)(z − 1)^2.`

`text(If)\ k = -1`

`P(z)`	`= z^2(z + 1)^2 + (-z − 1)^2`
	`= (z^2 + 1)(z + 1)^2`

(iv) `text(Product of the roots) = e/a=1`

♦♦♦ Mean mark part (iv) 20%.

`:. α bar α iα *(-i barα)`	`=1`
`(α bar α)^2`	`=1`
`(\|α\|)^4`	`=1`
`\|α\|`	`=1`

`:.\ text(All zeros have modulus 1.)`

(v) `text(Sum of zeroes)\ = -b/a=-(-2k)/1 = 2k`

♦♦♦ Mean mark part (v) 20%.

`:. 2k`	`=α + bar α + iα + (-i bar α )`
	`=x + iy + x − iy + (-y + ix) − i(x − iy)`
	`=2x − y + ix − ix − y`
	`=2x − 2y`
`:. k`	`=x-y`

(vi) `text(S)text(ince)\ \ |α| = 1\ \ \ text{(part (iv))}`

♦♦♦ Mean mark part (vi) 2%.

`=>x^2 + y^2`	`=1`
`text(Substitute)\ \ y=x-k\ \ \ text{(part (v))}`
`x^2 + (x − k)^2`	`=1`
`2x^2 − 2kx + k^2 − 1`	`=0`

`text(For a real solution to exist), Δ ≥ 0`

`4k^2 − 8(k^2 − 1) `	`≥ 0`
`-4k^2 + 8`	`≥ 0`
`k^2`	`≥ 2`

`:. -sqrt2 ≤ k ≤ sqrt2`

`\|\ z\ \|`	`< 1`
`:.\|\ z^2\ \|`	`< \|\ z\ \|`