Band 4 – Page 101

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)

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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)

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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)

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`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)

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`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)

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`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)

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`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)

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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)

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`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)

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`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)

Show Answers Only

`- 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)

Show Answers Only

`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)

--- 3 WORK AREA LINES (style=lined) ---
Show that `| z + z^2 | = 2 cos theta/2`. (2 marks)

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

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

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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)

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

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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) ---

Show Answers Only

`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) ---

Show Answers Only

`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.)

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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)

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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)

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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)

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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`

Proof, EXT2 P1 2012 HSC 15a

Prove that `sqrt(ab) ≤ (a + b)/2`, where `a ≥ 0` and `b ≥ 0`. (1 mark)

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If `1 ≤ x ≤ y`, show that `x(y − x + 1) ≥ y`. (2 marks)

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Let `n` and `j` be positive integers with `1 ≤ j ≤ n`.

Prove that `sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2.` (2 marks)

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For integers `n ≥ 1`, prove that

`(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`. (1 mark)

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

Show Answers Only

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

Show Worked Solution

i.	`(sqrta − sqrtb)^2`	`≥ 0`
	`a − 2sqrt(ab) + b`	`≥ 0`
	`a + b`	`≥ 2sqrt(ab)`
	`sqrt(ab)`	`≤ (a + b)/2`

ii. `text(Solution 1)`

♦♦ Mean mark part (ii) 28%.

`text(S)text(ince)\ \ 1 ≤ x ≤ y`

`y-x`	`>=0`
`y(x-1)-x(x-1)`	`>=0,\ \ \ \ (x-1>=0)`
`xy-x^2+x-y`	`>=0`
`:.xy-x^2+x`	`>=y`

`text(Solution 2)`

`x( y − x + 1)`	`= xy − x^2 + x`
	`= -y + xy − x^2 + x + y`
	`= y(x − 1) − x(x − 1)+ y`
	`= (x − 1)( y − x) + y`

`text(S)text(ince)\ \ x − 1 ≥ 0\ \ text(and)\ \ y − x ≥ 0`

`=>(x − 1)( y − x) + y`	`>=y`
`:.x(y − x + 1)`	`>=y`

iii. `text(Let)\ c = n − j + 1, \ c > 0\ text(as)\ n ≥ j.`

♦ Mean mark part (iii) 39%.

`=> sqrt(j(n − j + 1))\ \ text(can be written as)\ \ sqrt(jc).`

`sqrt(jc)`	`≤ (j + c)/2\ \ \ \ text{(part (i))}`
`sqrt(j(n − j + 1))`	`≤ (j + n-j+1)/2`
`=>sqrt(j(n − j + 1))`	`≤ (n+1)/2`
`j(n − j +1)`	`≥ n\ \ \ \ text{(part (ii))}`
`=>sqrt(j(n − j + 1))`	`≥ sqrt n`

`:.sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2`

iv. `text{From (iii)}\ n ≤ j(n− j +1) ≤ (n + 1)/2`

♦♦♦ Mean mark part (iv) just 2%!

`text(Let)\ j\ text(take on the values from 1 to)\ n.`

`j = 1:`	`sqrtn ≤ sqrt(1(n)) ≤ (n + 1)/2`
`j = 2:`	`sqrtn ≤ sqrt(2(n−1)) ≤ (n + 1)/2`
	`vdots`
`j = n:`	`sqrtn ≤ sqrt(n(1)) ≤ (n + 1)/2`

`text{Multiply the corresponding parts of each line}`

`(sqrtn)^n ≤ sqrt(n! xx n!) ≤ ((n + 1)/2)^n`

`(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`

Harder Ext1 Topics, EXT2 2012 HSC 14d

The diagram shows points `A` and `B` on a circle. The tangents to the circle at `A` and `B` meet at the point `C`. The point `P` is on the circle inside `ΔABC`. The point `E` lies on `AB` so that `AB ⊥ EP`. The points `F` and `G` lie on `BC` and `AC` respectively so that `FP ⊥ BC` and `GP ⊥ AC`.

Copy or trace the diagram into your writing booklet.

Show that `ΔAPG` and `ΔBPE` are similar. (2 marks)
Show that `EP^2 = FP xx GP`. (2 marks)

Show Answers Only

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

Show Worked Solution

(i)

`text(Join)\ AP\ text(and)\ BP.`

`text(In)\ ΔAPG\ text(and)\ ΔBPE`

`∠GAP`	`= ∠ABP\ \ text{(angle in alternate segment)}`
`∠AGP`	`= ∠BEP = 90^@\ \ text{(given)}`
`:.ΔAPG\ text(\|\|\|)\ ΔBPE\ \ text{(equiagular)}`

(ii) `text(Similarly, ΔBPF ||| ΔAPE)`

`:. (FP)/(EP)`	`= (BP)/(AP)\ \ `	`text{(corresponding sides in}`
		`Delta BPF and Delta APEtext{)}`
`(AP)/(BP) `	`= (GP)/(EP)`	`text{(corresponding sides in}`
		`Delta APG and Delta BPEtext{)}`
`=>(EP)/(GP)`	`=(FP)/(EP)`
`:.EP^2`	`=FP xx GP`

Functions, EXT1′ F1 2012 HSC 14b

The diagram shows the graph `y = (x(2x − 3))/(x − 1)`. The line `l` is an asymptote.

Use the above graph to draw a sketch of the graph

`y = (x − 1)/(x(2x − 3))`

indicating all asymptotes and all `x`- and `y`-intercepts. (2 marks)

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Write `(x(2x − 3))/(x − 1)` in the form `mx + b + a/(x − 1)`. (1 mark)

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

Show Answers Only

`text(See Worked Solutions.)`
`y = 2x − 1`

Show Worked Solution

ii.	`(x(2x− 3))/(x − 1)`	`= (2x^2 − 3x)/(x − 1)`
		`= (2x^2 − 2x − x + 1 − 1)/(x − 1)`
		`= (2x(x −1) − (x − 1) − 1)/(x − 1)`
		`= 2x − 1 − 1/(x − 1)`

Conics, EXT2 2012 HSC 13c

Let `P` be a point on the hyperbola given parametrically by `x = a\ sec\ theta` and `y = b\ tan\ theta`, where `a` and `b` are positive. The foci of the hyperbola are `S(ae,0)` and `S′(–ae,0)` where `e` is the eccentricity. The point `Q` is on the `x`-axis so that `PQ` bisects `∠SPS′`.

Show that `SP = a(e\ sec\ theta\ – 1)`. (1 mark)
It is given that
`S′P = a(e\ sec\ theta + 1),\ \ and\ \ (PS)/(QS) = (PS′)/(QS′)`.
Using this, or otherwise, show that the `x`-coordinate of `Q` is
1. `a/(sec\ theta)`. (2 marks)
The slope of the tangent to the hyperbola at `P` is
1. `(b\ sec\ theta)/(a\ tan\ theta)`. (Do NOT prove this.)
Show that the tangent at `P` is the line `PQ`. (1 mark)

Show Answers Only

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

Show Worked Solution

(i) `text(Solution 1)`

♦ Mean mark 50%.

`text(Let)\ M\ text(be on the directrix at)\ x = a/e`

`M (a/e, b\ tan theta)`

`PM`	`= a\ sec\ theta − a/e`
	`= a/e(e\ sec\ theta − 1)`

`text(S)text(ince)\ PS = ePM\ \ \ text{(by definition)}`

`:.PS`	`= e xx a/e(e\ sec\ theta − 1) `
	`=a(e\ sec\ theta − 1)`

`text(Solution 2)`

`PS`	`= sqrt((a\ sec\ theta − ae)^2 + b^2\ tan^2\ theta)`
	`= sqrt(a^2\ sec^2\ theta − 2a^2e\ sec\ theta + a^2e^2 + b^2\ tan^2\ theta)`
	`= sqrt(a^2\ sec^2\ theta − 2a^2e\ sec\ theta + a^2e^2 + (a^2e^2 − a^2)tan^2\ theta)`
	`= sqrt(a^2\ sec^2\ theta − a^2\ tan^2\ theta − 2a^2e\ sec\ theta + a^2e^2 + a^2e^2\ tan^2\ theta)`
	`= sqrt(a^2 (sec^2\ theta − tan^2\ theta) − 2a^2e\ sec\ theta + a^2e^2 + a^2e^2\ tan^2\ theta)`
`text{(Using}\ \ sec^2 theta – tan^2 theta = 1 text{)}`
	`= sqrt(a^2 − 2a^2e\ sec\ theta + a^2e^2 + a^2e^2 (sec^2\ theta −1))`
	`= a sqrt(1− 2e\ sec\ theta + e^2 + e^2\ sec^2\ theta − e^2)`
	`= a sqrt(e^2\ sec^2\ theta − 2e\ sec\ theta +1)`
	`= a sqrt((e\ sec\ theta −1)^2)`
	`= a(e\ sec\ theta −1)`

(ii) `text(Let)\ Q\ text(be)\ (x, 0)`

`text{Using}\ \ (PS)/(QS)“= (PS′)/(QS′)`

`(a(e\ sec theta − 1))/(ae-x)`	`=(a(e\ sec\ theta + 1))/(x+ae)`
`a(x + ae)(e\ sec\ theta − 1)`	`= a(ae − x)(e\ sec\ theta + 1)`
`xe\ sec\ theta − x + ae^2\ sec\ theta − ae`	`= ae^2\ sec\ theta + ae − xe\ sec\ theta − x`
`2xe\ sec\ theta`	`= 2ae`
`:.x`	`= a/(sec\ theta)\ \ \ text(… as required)`

(iii) `text(Solution 1)`

`m_text(tan)=(b\ sec theta)/(a\ tan theta)`

`P(a\ sec theta, b\ tan\ theta),\ \ Q(a/(sec\ theta), 0)`

`m_(PQ)`	`= (b tan theta)/(a sec theta – a/sec theta)`
	`=(b tan theta sec theta)/(a(sec^2 theta-1))`
	`=(b tan theta sec theta)/(a tan^2 theta)`
	`=(b sec theta)/(a tan theta)`

`:. PQ\ text(and the tangent at)\ P\ text(are the same line.)`

`text(Alternative Solution)`

`text(The equation of the tangent at)\ P\ text(is)`

`y − b\ tan\ theta = (b\ sec\ theta)/(a\ tan\ theta)(x − a\ sec\ theta)`

`text(Check if)\ Q(a/(sec\ theta),0)\ text(satisfies)`

`-b\ tan\ theta`	`= (b\ sec\ theta)/(a\ tan\ theta)(x − a\ sec\ theta)`
`-ab\ tan^2\ theta`	`= bx\ sec\ theta − ab\ sec^2\ theta`
`x\ sec\ theta`	`= a(sec^2\ theta − tan^2\ theta)`
`x\ sec\ theta`	`= a`
`x`	`=a/ sec theta\ \ \ \ text{(proven true in part (ii))}`

`:.\ text(This line passes through)\ Q\ text(so the tangent at)\ P\ text(is the line)\ PQ.`

Mechanics, EXT2 M1 2012 HSC 13a

An object on the surface of a liquid is released at time `t = 0` and immediately sinks. Let `x` be its displacement in metres in a downward direction from the surface at time `t` seconds.

The equation of motion is given by

`(dv)/(dt) = 10 − (v^2)/40`,

where `v` is the velocity of the object.

Show that `v = (20(e^t − 1))/(e^t + 1)`. (4 marks)

--- 10 WORK AREA LINES (style=lined) ---
Use `(dv)/(dt) = v (dv)/(dx)` to show that

`x = 20\ log_e(400/(400 − v^2))` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
How far does the object sink in the first 4 seconds? (2 marks)

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

Show Answers Only

`text(See Worked Solutions.)`
`text(See Worked Solutions.)`
`40log_e((e^4 + 1)/(2e^2))\ text(m)`

Show Worked Solution

i.	`(dv)/(dt)`	`= 10 − (v^2)/40`
	`(dv)/(dt)`	`= (400 − v^2)/40`
	`dt`	`= 40/(400 −v^2)\ dv`
	`int dt`	`=int 40/(400 −v^2)\ dv`
	`t`	`= int (1/(20 + v) + 1/(20 − v))\ dv`
		`= log_e(20 + v) − log_e(20 − v) + c`

`text(When)\ \ t = 0, v = 0\ \ => \ c = 0`

`t=`	` log_e((20 + v)/(20 − v))`
`e^t=`	` (20 + v)/(20 − v)`
`20e^t-ve^t=`	` 20 + v`
`v+ ve^t=`	` 20e^t − 20`
`v(1+e^t)=`	`20(e^t − 1)`
`v=`	` (20(e^t − 1))/(e^t + 1)\ \ \ \ text(… as required)`

ii.	`v (dv)/(dx)`	`= 10 − (v^2)/40`
	`(40v\ dv)/(400 − v^2)`	`= dx`

`int dx`	`= int (40v)/(400 − v^2)\ dv`
`x`	`= -20log_e(400 − v^2) + c`

`text(When)\ \ x = 0, v = 0\ \ \ => c = 20log_e 400`

`:.x`	`= 20log_e400 − 20log_e(400 − v^2)`
	`= 20log_e((400)/(400 − v^2))\ \ \ \ text(… as required)`

iii. `text(When)\ \ t = 4,\ v = (20(e^4 − 1))/(e^4 + 1)`

`x`	`= 20log_e[400/(400 −((20(e^4 − 1))/(e^4 + 1))^2)]`
	`= 20log_e[((e^4 + 1)^2)/((e^4 + 1)^2 − (e^4 − 1)^2)]`
	`= 20log_e(((e^4 + 1)^2)/((e^4+1 + e^4 − 1)(e^4 + 1 − e^4 + 1)))`
	`= 20log_e(((e^4 + 1)^2)/(4e^4))`
	`= 40log_e\ (e^4 + 1)/(2e^2)\ \ text(metres)`

Complex Numbers, EXT2 N2 2012 HSC 12d

On the Argand diagram the points `A_1` and `A_2` correspond to the distinct complex numbers `u_1` and `u_2` respectively. Let `P` be a point corresponding to a third complex number `z`.

Points `B_1` and `B_2` are positioned so that `ΔA_1PB_1` and `ΔA_2B_2P`, labelled in an anti-clockwise direction, are right-angled and isosceles with right angles at `A_1` and `A_2`, respectively. The complex numbers `w_1` and `w_2` correspond to `B_1` and `B_2`, respectively.

Explain why `w_1 = u_1 + i(z − u_1)`. (1 mark)

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Find the locus of the midpoint of `B_1B_2` as `P` varies. (2 marks)

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

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`text(See Worked Solutions.)`
`(u_1 + u_2)/2 + (u_2 − u_1)/2 i`

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i.	`vec (A_1P)`	`= z − u_1`
	`vec (A_1B_1)`	`= w_1 − u_1`

`B_1A_1 ⊥ A_1P\ text(and)\ |vec (A_1P)| = |vec (A_1B_1)|`

`vec (A_1B_1)\ text(is an anticlockwise rotation of)\ vec (A_1P)\ text(through)\ 90^@`

`:.w_1 − u_1 = i(z −u_1)`

`:.w_1 = u_1+ i(z −u_1)`

ii.	`vec (A_2B_2)`	`= w_2 − u_2`
	`vec (A_2P)`	`= z − u_2`

`A_2B_2 ⊥ A_2P\ text(and)\ |vec (A_2B_2)| = |vec (A_2P)|`

`vec (A_2P)\ text(is an anticlockwise rotation of)\ vec (A_2B_2)\ text(through)\ 90^@`

`z − u_2`	`= i(w_2 −u_2)`
`iw_2`	`= z − u_2 + iu_2`
`−w_2`	`= iz − iu_2 − u_2`
`:. w_2`	`= u_2 + i(u_2 − z)`

`:.\ text(The midpoint of)\ B_1B_2\ text(is)\ (w_1 + w_2)/2`

`= 1/2[u_1 + i(z − u_1) + u_2 + i(u_2 − z)]`

`= 1/2[u_1 + u_2 + i(u_2 − u_1)]`

`= (u_1 + u_2)/2 + (u_2 − u_1)/2 i\ \ \ \ text{(which is a fixed point)}`

Conics, EXT2 2012 HSC 12b

The diagram shows the ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1` with `a > b`. The ellipse has focus `S` and eccentricity `e`. The tangent to the ellipse at `P(x_0, y_0)` meets the `x`-axis at `T`. The normal at `P` meets the `x`-axis at `N`.

Show that the tangent to the ellipse at `P` is given by the equation
1. `y − y_0 = -(b^2x_0)/(a^2y_0)(x − x_0)`. (2 marks)
Show that the `x`-coordinate of `N` is `x_0e^2`. (2 marks)
Show that `ON xx OT = OS^2` (2 marks)

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

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

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

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(i) `(x^2)/(a^2) + (y^2)/(b^2) = 1`

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

`text(At)\ P(x_0, y_0),\ \ m_(tan)= -(b^2x_0)/(a^2y_0)`

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

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

(ii) `text(At)\ P(x_0, y_0),\ \ m_(norm) = (a^2y_0)/(b^2x_0)`

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

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

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

`0-y_0=`	` (a^2y_0)/(b^2x_0)(x − x_0)`
`-b^2x_0y_0=`	` a^2y_0x − a^2x_0y_0`
`a^2y_0x=`	` a^2x_0y_0 − b^2x_0y_0`
`a^2x=`	` x_0(a^2 − b^2)`
`x=`	` (x_0(a^2 − b^2))/(a^2)\ \ \ \ \ text{(using}\ \ a^2e^2=a^2-b^2 text{)}`
`=`	`x_0e^2`

(iii) `ON = x_0e^2`

`OS=ae\ \ \ \ text{(given}\ \ S(ae,0) text{)}`

`T\ text(is the)\ x text(-axis intercept of the tangent)\ PT`

`0-y_0`	`= -(b^2x_0)/(a^2y_0)(x − x_0)`
`a^2y_0^2`	`=b^2x_0x-b^2x_0^2`
`b^2x_0x`	`=b^2x_0^2+a^2y_0^2`
`x`	`=(b^2x_0^2+a^2y_0^2)/(b^2x_0)`

`text(S)text(ince)\ P(x_0,y_0)\ text(lies on the ellipse,)`

`=> (x_0^2)/(a^2) + (y_0^2)/(b^2)`	` = 1`
`b^2 x_0^2+a^2y_0^2`	`=a^2b^2`

`:.x=OT`

`=(a^2b^2)/(b^2x_0)=a^2/x_0`

`:.ON xx OT`	`= x_0e^2 xx a^2/x_0`
	`=a^2e^2`
	`=OS^2`

Functions, EXT1′ F1 2012 HSC 11f

Sketch the following graphs, showing the `x`- and `y`-intercepts

`y = |\ x\ |- 1` (1 mark)
`y = x(|\ x\ | - 1)` (2 marks)

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

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

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i. & ii.

`text{Note for part (ii)}`

`=>y = x(|\ x\ | – 1)\ \ text(is an ODD function)`

Functions, EXT1′ F2 2013 HSC 15b

The polynomial `P(x) = ax^4 + bx^3 + cx^2 + e` has remainder `-3` when divided by `x - 1`. The polynomial has a double root at `x = -1.`

Show that `4a + 2c = -9/2.` (2 marks)

--- 7 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find the slope of the tangent to the graph `y = P(x)` when `x = 1.` (1 mark)

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

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`text(Proof)\ \ text{(See Worked Solutions)}`
`-9`

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i.	`P(x)`	`= ax^4 + bx^3 + cx^2 + e`
	`P prime (x)`	`=4ax^3 + 3bx^2 + 2cx`

`P(1)=-3`
`a+b+c+e`	`=-3\ \ \ \ …\ (1)`
`P(-1)=0`
`a-b+c+e`	`=0\ \ \ \ …\ (2)`
`P′(-1)=0`
`-4a+3b-2c`	`=0\ \ \ \ …\ (3)`

`(1)-(2)`

`2b`	`=-3`
`b`	`=-3/2`

`text(Substitute into)\ \ (3)`

`:.4a+2c`	`=3b`
	`=-9/2\ \ \ \ text(… as required)`

ii. `P prime (1)`	`= 4a + 3b + 2c`
	`= -9/2 – 9/2`
	`=-9`

Harder Ext1 Topics, EXT2 2013 HSC 14d

A triangle has vertices `A, B` and `C`. The point `D` lies on the interval `AB` such that `AD = 3` and `DB = 5`. The point `E` lies on the interval `AC` such that `AE = 4`, `DE = 3` and `EC = 2`.

Prove that `Delta ABC` and `Delta AED` are similar. (1 mark)
Prove that `BCED` is a cyclic quadrilateral. (1 mark)
Show that `CD = sqrt 21`. (2 marks)
Find the exact value of the radius of the circle passing through the points `B, C, E and D`. (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`(3 sqrt 105)/10`

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(i) `text(In)\ \ Delta ABC and Delta AED`

`/_ BAC = /_ EAD\ \ \ \ text{(common angle)}`

`(AB)/(AE)`	`= 8/4 = 2/1`
`(AC)/(AD)`	`= 6/3 = 2/1`
`(AB)/(AE)`	`= (AC)/(AD) = 2/1`

`:.\ Delta ABC\ \ text(\|\|\|)\ \ Delta AED`	`\ \ \ \ \ text{(sides about equal angles}`
	`\ \ \ \ \ text{are in the same ratio)}`

(ii) `/_ ABC = /_ AED\ \ \ \ \ text{(corresponding angles of similar triangles)}`

`/_ AED\ \ text(is an exterior angle of quadrilateral)\ \ BCED`

`:.\ BCED\ \ text(is a cyclic quadrilateral as an exterior)`

`text(angle equals the interior opposite angle.)`

(iii) `cos /_ AED = (3^2 + 4^2 – 3^2)/(2 xx 3 xx 4) = 2/3`

♦♦ Mean mark 33%. 

`cos /_ CED`	`= cos(pi – /_AED)`
	`=-cos/_AED`
	`=-2/3`

`text(In)\ \ Delta CDE`

`CD^2`	`= 2^2 + 3^2 – 2 xx 2 xx 3 xx (-2/3)`
	`= 13 + 8`
	`=21`
`:.CD`	`= sqrt 21`

(iv)

`text(Mark the centre)\ \ O,\ \ text(draw the radii)\ \ OC, OD`

`text(Let)\ \ /_ CBD`	`= alpha`
`:. /_ COD`	`= 2 alpha`	` \ \ \ \ text{(angles at circumference and}`
		`\ \ \ \ text{centre on arc}\ \ CD text{)}`
`/_ DEC`	`= pi – alpha`	` \ \ \ \ text{(opposite angles of cyclic}`
		`\ \ \ \ text{quadrilateral}\ \ BCED text{)}`

♦♦♦ Mean mark 7%. 
COMMENT: This part had the lowest mean mark in the entire 2013 exam.

`cos alpha = cos /_ AED = 2/3\ \ \ \ text{(part (iii))}`

`text(Let)\ \ r= text(circle radius)`

`text(In)\ \ Delta DOC`

`CD^2`	`=r^2 + r^2 – 2r xx r xx cos 2 alpha`
`21`	`=2r^2 – 2r^2 (2 cos^2 alpha – 1)`
	`=2r^2 – 2r^2 (8/9 – 1)`
	`=2r^2 + (2r^2)/9`
	`=(20r^2)/9`
`r^2`	`=(9 xx 21)/20`
`:.r`	`=(3 sqrt 21)/(2 sqrt 5)`
	`=(3 sqrt 105)/10`

Calculus, EXT2 C1 2013 HSC 14a

The diagram shows the graph `y = ln x.`

By comparing relevant areas in the diagram, or otherwise, show that

`ln t > 2 ((t - 1)/(t + 1))`, for `t > 1.` (3 marks)

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

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

Show Worked Solution

`text(Area under curve)`	`>\ text(Area of triangle)`
`int_1^t ln x\ dx`	`> 1/2 xx (t – 1)ln t,\ \ \ t > 1`
`underbrace{int_1^t 1\ln x\ dx}_text(integration by parts)`	`> 1/2 xx (t – 1)ln t`
`[x ln x]_1^t – int_1^t x * 1/x\ dx`	`> ((t – 1)ln t)/2`
`(tlnt – ln 1) – [x]_1^t`	`> ((t – 1)ln t)/2`
`t ln t – (t – 1)`	`> ((t – 1) ln t)/2`
`2t ln t – 2t + 2`	`> t ln t – ln t`
`t ln t + ln t`	`> 2(t – 1)`
`(t + 1) ln t`	`> 2 (t – 1)`
`ln t`	`> 2 ((t – 1)/(t + 1)),\ \ \ t > 1`

Harder Ext1 Topics, EXT2 2013 HSC 13c

The points `A, B, C` and `D` lie on a circle of radius `r`, forming a cyclic quadrilateral. The side `AB` is a diameter of the circle. The point `E` is chosen on the diagonal `AC` so that `DE _|_ AC`. Let `alpha = /_DAC and beta = /_ACD.`

Show that `AC = 2r sin (alpha + beta).` (2 marks)
By considering `Delta ABD`, or otherwise, show that `AE = 2r cos alpha sin beta.` (2 marks)
Hence, show that `sin (alpha + beta) = sin alpha cos beta + sin beta cos alpha.` (1 mark)

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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) `/_ ADC = (180 – (alpha + beta))`

`/_ ABC = alpha + beta\ \ \ \ \ `	`text{(opposite angles of cyclic}`
	`text(quadrilateral)\ \ ABCD)`

`/_ ACB = 90^@\ \ \ text{(angle in a semi-circle)}`

`(AC)/(AB)`	`= sin /_ ABC`
`:. AC`	`= 2 r sin (alpha + beta)`

(ii)

`text(In)\ \ Delta ABD`

♦ Mean mark 49%.

`/_ ADB = 90^@\ \ \ text{(angle in a semi-circle)}`

`(AD)/(AB)`	`= cos /_ DAB`
`AD`	`= AB cos /_ DAB`

`text(In)\ \ Delta ABC`

`/_CAB`	`= 180^@ – (90^@ +alpha + beta)\ \ \ \ text{(angle sum of}\ Delta ABC text{)}`
	`= 90^@- (alpha + beta)`

`=>/_ DAB`	`= alpha + 90^@ – (alpha + beta)`
	`= 90^@ – beta`

`:.AD`	`= AB cos ( 90^@ – beta)`
	`=2r sin beta`

`text(In)\ \ Delta ADE`
`(AE)/(AD)`	`= cos alpha`
`:.AE`	`= 2 r cos alpha sin beta\ \ text(… as required)`

(iii) `text(In)\ \ Delta ADE`

♦ Mean mark 35%.

`sin alpha`	`=(DE)/(2r sin beta)`
`DE`	`=2r sin alpha sin beta`

`text(In)\ \ Delta CDE`

`tan beta`	`=(DE)/(EC)`
`EC`	`=(2r sin alpha sin beta)/tan beta`
	`=2r sin alpha cos beta`

`text(S)text(ince)\ \ AC`	`=AE + ED`
`2 r sin (alpha + beta)`	`= 2 r sin beta cos alpha + 2 r sin alpha cos beta`
`:.sin (alpha + beta)`	`= sin alpha cos beta + sin beta cos alpha`

Functions, EXT1′ F1 2013 HSC 13b

The diagram shows the graph of a function `f(x).`

Sketch the following curves on separate half-page diagrams.

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

--- 8 WORK AREA LINES (style=lined) ---
`y = 1/(1 - f(x)).` (3 marks)

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

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

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

`text(Only exist for)\ \ f(x) >= 0`

`y^2 = 1,\ \ \ y = +- 1`

ii. `y = 1/(1 – f(x))`

MARKER’S COMMENT: Correct working sketches such as `y=-f(x)` and `y=1-f(x)` meant that students could obtain some marks, even if their final sketch was wrong. Note this important advice.

`f(x) = 1,\ \ \ y\ text(undefined.)`

`f(x) > 1,\ \ \ y < 0`

`f(x) <= 0, \ \ \ y <= 1`

Calculus, EXT2 C1 2013 HSC 13a

Let `I_n = int_0^1 (1 - x^2)^(n/2)\ dx`, where `n >= 0` is an integer.

Show that

`I_n = n/(n + 1) I_(n-2)` for every integer `n>= 2.` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Evaluate `I_5.` (2 marks)

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

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`text(Proof)\ \ text{(See Worked Solutions)}`
`(5 pi)/32`

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i. `I_n = int_0^1 (1 – x^2)^(n/2)\ dx`

♦ Mean mark 46%.
STRATEGY TIP: Attempting to prove this relationship using induction proved unsuccessful and time consuming for some students.

`text(Let)\ \ u`	`= (1 – x^2)^(n/2)`
`u′`	`= n/2 (1 – x^2)^(n/2 – 1) (-2x)`
	`= -nx (1 – x^2)^(n/2 -1)`
`v`	`=x`
`v′`	`=1`

`I_n`	`= [x (1 – x^2)^(n/2)]_0^1 – int_0^1 – nx (1 – x^2)^(n/2 – 1)x\ dx`
	`= 0 + n int_0^1 x^2 (1 – x^2)^(n/2 – 1)\ dx`
	`= n int_0^1 (x^2 – 1 + 1) (1 – x^2)^(n/2 – 1)\ dx`
	`= n int_0^1 1 (1 – x^2)^(n/2 -1)\ dx – n int_0^1 (1 – x^2) (1 – x^2)^(n/2 – 1)\ dx`
	`= n int_0^1 (1 – x^2)^((n-2)/2)\ dx – n int_0^1 (1 – x^2)^(n/2)\ dx`
	`= nI_(n – 2) – nI_n`

`:.(n + 1) I_n`	`= nI_(n-2)`
`I_n`	`= n/(n + 1) I_(n-2)\ \ \ \ text(where)\ \ n >= 2`

STRATEGY TIP: Realising that the integral `I_1` is equal to the area of a quarter unit circle saved valuable time.

ii. `I_5 = 5/6 I_3 = 5/6 xx 3/4\ I_1`

`I_1`	`= int_0^1 (1 – x^2)^(1/2)\ dx`
	`= 1/4 xx pi xx 1^2\ \ \ \ text{(1st quadrant of a circle, radius 1)}`
	`= pi/4`

`I_5`	`= 5/6 xx 3/4 xx pi/4`
	`= (5 pi)/32`

Conics, EXT2 2013 HSC 12d

The points `P (cp, c/p)` and `Q (cq, c/q)`, where `|\ p\ | ≠ |\ q\ |`, lie on the rectangular hyperbola with equation `xy = c^2.`

The tangent to the hyperbola at `P` intersects the `x`-axis at `A` and the `y`-axis at `B`. Similarly, the tangent to the hyperbola at `Q` intersects the `x`-axis at `C` and the `y`- axis at `D`.

Show that the equation of the tangent at `P` is `x + p^2 y = 2cp.` (2 marks)
Show that `A, B and O` are on a circle with centre `P.` (2 marks)
Prove that `BC` is parallel to `PQ.` (1 mark)

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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) `P (cp, c/p),\ \ Q (cq, c/q),\ \ xy = c^2,\ \ |\ p\ | ≠ |\ q\ |`

`xy = c^2`

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

`text(At)\ \ (cp, c/p)`

`(dy)/(dx) = (-c/p)/(cp) = -1/p^2`

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

`y – c/p`	`= -1/p^2 (x – cp)`
`p^2 y – cp`	`= -x + cp`
`:. x + p^2 y`	`= 2cp\ \ \ text(… as required)`

(ii) `text(Solution 1)`

`text(At)\ \ A, y = 0,\ \ \ x = 2cp`

`text(At)\ \ B, x = 0,\ \ \ y = (2c)/p`

`OP`	`= sqrt (c^2 p^2 + c^2/p^2) = c/p sqrt(p^4 + 1)`
`PA`	`= sqrt {(2cp – cp)^2 + c^2/p^2} = c/p sqrt (p^4 + 1)`
`PB`	`= sqrt{c^2 p^2 + ((2c)/p – c/p)^2} = c/p sqrt (p^4 + 1)`

`OP = PA = PB`

`:. A, B and O\ \ text(are on the circle centre)\ \ P.`

`text(Alternate Solution)`

`text(Mid-point of)\ \ AB`

`=((2cp+0)/2,\ (0+(2c)/p)/2)`

`=(cp,\ c/p)`

`=>P\ \ text(is the midpoint of)\ \ AB`

`text(S)text(ince)\ \ ∠AOB=90^@`

`:. A, B and O\ \ text(are on the circle centre)\ \ P.`

(iii) `text(Equation of tangent at)\ \ Q\ \ \ text{(using part (i))}`

`x + q^2y = 2cq`

`=> C\ \ text(is)\ \ (2cq, 0)`

`m_(BC)`	`= {(2c)/p – 0}/(0 – 2cq)`
	`= – 1/(pq)`
`m_(PQ)`	`= (c/p – c/q)/(cp – cq)`
	`=((cq-cp)/(pq))/(c(p-q))`
	`= {-c(p – q)}/{cpq (p – q)}`
	`= -1/(pq)`

`m_(BC) = m_(PQ)`

`:. BC\ text(||)\ PQ`

Volumes, EXT2 2013 HSC 12c

The diagram shows the region bounded by the graph `y = e^x`, the `x`-axis and the lines `x = 1` and `x = 3`. The region is rotated about the line `x = 4` to form a solid.

Find the volume of the solid. (4 marks)

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`4 pi e (e^2 – 2)`

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`text(Using cylindrical shells)`

`δV`	`~~2 pi r h\ δx`
`V`	`= int_1^3 2 pi (4 – x) y\ dx`
	`= 2 pi int _1^3 (4 – x) e^x\ dx`

`text(Using integration by parts,)`

`text(Let)\ \ u`	`= 4 – x\ ,\ \ `	`du`	`= -dx`
`dv`	`= e^x\ ,\ \ `	`v`	`= e^x`

`V`	`= 2 pi [[(4 – x) e^x]_1^3 – int_1^3 -e^x\ dx]`
	`= 2 pi [e^3 – 3e + [e^x]_1^3]`
	`= 2 pi (e^3 – 3e + e^3 – e)`
	`= 2 pi (2e^3 – 4e)`
	`= 4 pi e (e^2 – 2)`

Complex Numbers, EXT2 N2 2013 HSC 11e

Sketch the region on the Argand diagram defined by `z^2 + bar z^2 <= 8.` (3 marks)

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Show Worked Solution

`z^2 + bar z^2`	`<= 8`
`(x + iy)^2 + (x − iy)^2`	`<= 8`
`x^2 + 2ixy − y^2 + x^2 − 2ixy − y^2`	`<= 8`
`2x^2 − 2y^2`	`<= 8`
`x^2 – y^2`	`<= 4`

Calculus, EXT2 C1 2013 HSC 11d

Evaluate `int_0^1 x^3 sqrt(1 - x^2)\ dx.` (3 marks)

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

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`2/15`

Show Worked Solution

STRATEGY: The choice of `u=1-x^2` or `u^2=1-x^2` provided much less calculation than `u=x^2`. Take note!

`text(Let)\ \ u = 1 – x^2\ ,\ \ du = -2x\ dx`

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

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

`int_0^1 x^3 sqrt(1 – x^2)\ dx`	`= int_0^1 x^2 sqrt (1 – x^2) \ x\ dx`
	`= int_1^0 (1 – u) sqrt u xx ((-du)/2)`
	`= 1/2 int_0^1 (u^(1/2) – u^(3/2))\du`
	`= 1/2 [2/3 u^(3/2) – 2/5 u^(5/2)]_1^0`
	`= 1/2 [(2/3 – 2/5)-0]`
	`= 2/15`

Mechanics, EXT2 2014 HSC 15c

A toy aeroplane `P` of mass `m` is attached to a fixed point `O` by a string of length `l`. The string makes an angle `ø` with the horizontal. The aeroplane moves in uniform circular motion with velocity `v` in a circle of radius `r` in a horizontal plane.

The forces acting on the aeroplane are the gravitational force `mg`, the tension force `T` in the string and a vertical lifting force `kv^2`, where `k` is a positive constant.

By resolving the forces on the aeroplane in the horizontal and the vertical directions, show that
`(sin\ ø)/(cos^2\ ø) = (lk)/m − (lg)/(v^2)`. (3 marks)
Part (i) implies that
`(sin\ ø)/(cos^2\ ø) < (lk)/m`. (Do NOT prove this.)
Use this to show that
1. `sin\ ø < (sqrt(m^2 + 4l^2k^2) − m)/(2lk).` (2 marks)
Show that
`(sin\ ø)/(cos^2\ ø)` is an increasing function of `ø` for `-pi/2 < ø < pi/2`. (2 marks)
Explain why `ø` increases as `v` increases. (1 mark)

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

Show Worked Solution

(i) `text(Vertically)`

`kv^2 = mg + T\ sin\ ø\ \ \ \ \ …\ (1)`

`text(Horizontally)`

`T cos\ ø`	`=(mv^2)/r`
	`=(mv^2)/(l cos\ ø)\ \ \ \ \ \ (r=lcos\ ø)`
`T`	`=(mv^2)/(l cos^2\ ø)\ \ \ \ …\ (2)`

`text{Substitute (2) into (1)}`

`mg + (mv^2)/(l cos^2\ ø)\ sin\ ø`	`=kv^2`
`lmg + (mv^2)/(cos^2\ ø)\ sin\ ø`	`=lkv^2`
`(mv^2)/(cos^2\ ø)\ sin\ ø`	`=lkv^2 – lmg`
`(sin\ ø)/(cos^2\ ø)`	`=(lk)/m − (lg)/(v^2)\ \ \ text(… as required)`

♦♦♦ Mean mark 14%.
STRATEGY: The form of the required proof strongly hints that the quadratic formula is relevant – a clue that was ignored by many students.

(ii)	`(sin\ ø)/(cos^2\ ø)`	`< (lk)/m`
	`sin\ ø`	`< (lk)/m(1 − sin^2\ ø)`
	`m sin\ ø`	`< lk – lksin^2\ ø`
	`lk\ sin^2\ ø + m\ sin\ ø − lk< 0`

`text(Using the quadratic formula)`

`sin\ ø = (-m ± sqrt(m^2 + 4l^2k^2))/(2lk)`

`text(S)text(ince)\ \ 0<ø<pi/2\ \ \ \ =>\ 0 < sin\ ø < 1`

`:. sin\ ø = (sqrt(m^2 + 4l^2k^2)-m)/(2lk)`

`=> lk\ sin^2\ ø + m\ sin\ ø − lk< 0\ \ text(is true)`

`text(when)\ \ sin\ ø = 0`

`=>sqrt(m^2 + 4l^2k^2)>m`

`:.sin\ ø < (sqrt(m^2 + 4l^2k^2) − m)/(2lk)\ \ \ \ text(… as required)`

(iii)	`f(x)`	`= (sin\ ø)/(cos^2\ ø)`
	`f′(x)`	`= (cos\ ø\ cos^2\ ø − sin\ ø xx 2\ cos\ ø(-sin\ ø))/(cos^4\ ø)`
		`= (cos^2\ ø + 2sin^2\ ø)/(cos^3\ ø)`

♦♦ Mean mark 25%.

`text(Consider)\ \ -pi/2 < ø < pi/2`

`cos\ ø>0, \ \ cos^3\ ø > 0, \ \ 2sin^2\ ø>=0`

`=> f′(x) > 0`

`:. f(x)=(sin\ ø)/(cos^2\ ø)\ text(is an increasing function for)\ \ \ -pi/2 < ø < pi/2.`

(iv) `text(Consider)\ \ (sin\ ø)/(cos^2\ ø) = (lk)/m − (lg)/(v^2)`

♦♦♦ Mean mark 8%.
COMMENT: This question had the lowest mean mark in the entire 2014 exam.

`text(As)\ v\ text(increases)`	`=>(lg)/(v^2)\ text(decreases)`
	`=>(lk)/m – (lg)/(v^2)\ text(increases)`
	`=>(sin\ ø)/(cos^2\ ø)\ text(increases.)`

`text{From (iii)},\ (sin\ ø)/(cos^2\ ø)\ text(is an increasing function as)\ ø\ text(increases.)`

`:.ø\ text(increases as)\ \ v\ \ text(increases.)`

Complex Numbers, EXT2 N2 2014 HSC 15b

Using de Moivre’s theorem, or otherwise, show that for every positive integer `n`,

`(1 + i)^n + (1 − i)^n = 2(sqrt2)^n\ cos\ (npi)/4`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, show that for every positive integer `n` divisible by 4,

`((n),(0)) − ((n),(2)) + ((n),(4)) − ((n),(6)) + … + ((n),(n)) = (-1)^(n/4)(sqrt2)^n` (3 marks)

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

Show Answers Only

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

Show Worked Solution

i.	`1+i`	`=sqrt2(1/sqrt2 + 1/sqrt2 i)`
		`=sqrt2(cos\ pi/4 + i sin\ pi/4)`
	`(1+i)^n`	`=(sqrt2)^n (cos\ (n pi)/4 + i sin\ (n pi)/4)`

	`text(Similarly,)`
	`1-i`	`=sqrt2(cos (-pi/4) + i sin (-pi/4))`
		`=sqrt2(cos\ pi/4 – i sin\ pi/4)`
	`(1-i)^n`	`=(sqrt2)^n (cos\ (n pi)/4 – i sin\ (n pi)/4)`

`:.(1 + i)^n + (1 − i)^n`

`= (sqrt2)^n (cos\ (n pi)/4 + i sin\ (n pi)/4) +(sqrt2)^n (cos\ (n pi)/4 – i sin\ (n pi)/4)`

`= (sqrt2)^n(cos\ (npi)/4 + i\ sin\ (npi)/4 + cos\ (npi)/4 − i\ sin\ (npi)/4) `

`= 2(sqrt2)^n\ cos\ (npi)/4\ \ \ text(… as required)`

ii. `(1 + i)^n= ((n), (0)) + i((n),(1)) − ((n),(2)) − i((n), (3)) + ((n),(4)) + … + i^n((n),(n))`

`(1 − i)^n=((n),(0)) − i((n),(1)) − ((n),(2)) + i((n),(3)) + … + (-1)^ni^n((n),(n))`

`(1 + i)^n + (1 − i)^n`

♦♦ Mean mark 22%.
MARKER’S COMMENT: Few students could link part (i) to solving part (ii).

`= 2((n),(0)) − 2((n),(2)) + 2((n),(4)) + … + 2((n),(n))`

`text(Given)\ n\ text(is divisible by 4, the last term)=+((n),(n))`

`text{Using part (i):}`

`2((n),(0)) − 2((n),(2)) + 2((n),(4)) + … + 2((n),(n))`	`=2(sqrt2)^n cos\ (npi)/4`
`((n),(0)) − ((n),(2)) + ((n),(4)) + … + ((n),(n))`	`=(sqrt2)^n cos\ (npi)/4`

`text(If)\ n\ text(is a positive integer divisible by 4 then)\ cos\ (npi)/4 = ±1,`

`text(depending on whether)\ n\ text(is an even or odd multiple of 4.)`

`:. ((n),(0)) − ((n),(2)) + ((n),(4)) + … + ((n),(n)) = (-1)^(n/4)(sqrt2)^n`

Mechanics, EXT2 M1 2014 HSC 14c

A high speed train of mass `m` starts from rest and moves along a straight track. At time `t` hours, the distance travelled by the train from its starting point is `x` km, and its velocity is `v` km/h.

The train is driven by a constant force `F` in the forward direction. The resistive force in the opposite direction is `Kv^2`, where `K` is a positive constant. The terminal velocity of the train is 300 km/h.

Show that the equation of motion for the train is

`m ddot x = F[1 − (v/300)^2]`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Find, in terms of `F` and `m`, the time it takes the train to reach a velocity of 200 km/h. (4 marks)

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

Show Answers Only

`text{Proof (See Worked Solutions)}`
`(150 m ln5)/F`

Show Worked Solution

i. `m ddot x = F – Kv^2,\ \ v_T = 300`

`text(At)\ \ v_T,\ \ ddotx=0`

`m(0)`	`=F-K(300^2)`
`=>K`	`=F/300^2`

`:. m ddot x`	`= F – F/300^2 v^2`
	`=F[1 − (v/300)^2]\ \ \ text(… as required)`

ii.	`m*(dv)/(dt)`	`= F[1 − (v/(300))^2]`
	`(dv)/(1-(v/300)^2)`	`=F/m\ dt`
	`(dv)/(300^2-v^2)`	`=F/(300^2 m)\ dt`
	`dt`	`=(300^2 m)/F xx (dv)/(300^2-v^2)`

`int_0^t dt`	`=(300 m)/F int_0^200 300/((300+v)(300-v))\ dv`
`:. t`	`=(300 m)/F int_0^200 (1/2)/(300+v) + (1/2)/(300-v)\ dv`
	`=(150 m)/F [ln(300+v) – ln(300-v)]_0^200`
	`=(150 m)/F [ln ((300+v)/(300-v))]_0^200`
	`=(150 m)/F (ln5 – ln1)`
	`=(150 m ln5)/F`

Volumes, EXT2 2014 HSC 13b

The base of a solid is the region bounded by `y = x^2`, `y = –x^2` and `x = 2`. Each cross-section perpendicular to the `x`-axis is a trapezium, as shown in the diagram. The trapezium has three equal sides and its base is twice the length of any one of the equal sides.

Find the volume of the solid. (4 marks)

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

Show Answers Only

`(24sqrt3)/5\ \ text(u³)`

Show Worked Solution

`text(Trapezium cross section)`

`h^2`	`=a^2-(a/2)^2`
	`=a^2-a^2/4`
	`=(3a^2)/4`
`:.h`	`=(sqrt3 a)/2`

`text(S)text(ince)\ \ a=x^2`

`=>text(Base length)\ = 2x^2`

`=>text(Height)\ =(sqrt3 x^2)/2`

`:.\ text(Area of trapezium)`

`= 1/2h(a+b)`

`=1/2 xx (sqrt3 x^2)/2 xx(2x^2 + x^2)`

`= (3sqrt3 x^4)/4`

`:.V`	`= int_0^2 (3sqrt3 x^4)/4 dx`
	`= (3sqrt3)/4[(x^5)/5]_0^2`
	`= (3sqrt3)/4 xx 32/5`
	`= (24sqrt3)/5\ \ text(u³)`

Calculus, EXT2 C1 2014 HSC 13a

Using the substitution `t = tan\ x/2`, or otherwise, evaluate

`int_(pi/3)^(pi/2) 1/(3sinx - 4cosx + 5)\ dx`. (3 marks)

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

Show Answers Only

`(2sqrt3 − 3)/6`

Show Worked Solution

`t = tan\ x/2,\ \ \ sinx=(2t)/(1+t^2)`

`cos x = (1-t^2)/(1+t^2),\ \ \ dx=(2 dt)/(1+t^2`

`text(When)\ \ \ x = pi/3,\ \ t = tan\ pi/6 =1/sqrt3`

`text(When)\ \ \ x = pi/2, \ \ t = tan\ pi/4=1`.

`int_(pi/3)^(pi/2) 1/(3\ sin\ x − 4\ cos\ x + 5)`

`= int_(1/sqrt3)^1 1/((6t)/(1 + t^2) − (4(1 − t^2))/(1 + t^2) + 5) xx (2dt)/(1 + t^2)`

`= int_(1/sqrt3)^1 2/(6t − 4 + 4t^2 + 5 + 5t^2)dt`

`= int_(1/sqrt3)^1 2/(9t^2 + 6t + 1)dt`

`= 2 int_(1/sqrt3)^1 (dt)/((3t + 1)^2)`

`= -2/3[1/(3t + 1)]_(1/sqrt3)^1`

`= -2/3(1/4 − 1/(sqrt3 + 1))`

`= -2/3(1/4 − (sqrt3 − 1)/2)`

`= -1/6+sqrt3/3-1/3`

`= (2sqrt3 − 3)/6`

Calculus, EXT2 C1 2014 HSC 12d

Let `I_n = int_0^1 (x^(2n))/(x^2 + 1)\ dx`, where `n` is an integer and `n ≥ 0`.

Show that `I_0 = pi/4`. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Show that

`I_n + I_(n − 1) = 1/(2n − 1)`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find

`int_0^1 (x^4)/(x^2 + 1)\ dx`. (2 marks)

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

Show Answers Only

`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`pi/4 − 2/3`

Show Worked Solution

i. `I_n = int_0^1 (x^(2n))/(x^2 + 1)\ dx, \ \ n ≥ 0`

`I_0`	`= int_0^1 (x^0)/(x^2 + 1)\ dx`
	`= [tan^(−1) x]_0^1`
	` = tan^(−1) 1 − tan^(−1)0`
	`= pi/4`

♦ Mean mark 41%.
STRATEGY: The denominator of the proof, `(2n-1)`, should flag the potential existence of `x^(2n-2)` in the integral.

ii.	`I_(n − 1)`	`= int_0^1 x^(2n-2)/(x^2 + 1)\ dx`
	`I_n`	`= int_0^1 x^(2n)/(x^2 + 1)\ dx`

`I_n + I_(n − 1)`	`= int_0^1(x^(2n))/(x^2 + 1)\ dx + int_0^1(x^(2n − 2))/(x^2 + 1)\ dx`
	`= int_0^1(x^(2n) + x^(2n − 2))/(x^2 + 1)\ dx`
	`= int_0^1(x^(2n − 2)(x^2 + 1))/(x^2 + 1)\ dx`
	`= int_0^1x^(2n − 2)\ dx`
	`= [(x^(2n − 1))/(2n − 1)]_0^1`
	`= 1/(2n − 1)`

iii.	`I_2`	`= int_0^1(x^4)/(x^2 + 1)\ dx`
	`I_1`	`= int_0^1(x^2)/(x^2 + 1)\ dx`

`I_n + I_(n − 1)`	`= 1/(2n − 1)`
`I_2 + I_1`	`= 1/(2(2) − 1) = 1/3\ \ …\ (1)`
`I_1 + I_0`	`= 1/(2(1)-1)=1`
`I_1+pi/4`	`= 1`
`:.I_1`	`=1- pi/4`
`text(Substitute into (1))`
`I_2+1- pi/4`	`= 1/3`
`:.I_2`	`= pi/4 − 2/3`

Harder Ext1 Topics, EXT2 2014 HSC 12b

It can be shown that `4\ cos^3 theta - 3 cos theta = cos 3theta`. (Do NOT prove this.)

Assume that `x = 2cos theta` is a solution of `x^3 − 3x= sqrt3`.

Show that
1. `cos 3theta = sqrt3/2`. (1 mark)
Hence, or otherwise, find the three real solutions of `x^3 - 3x = sqrt3`. (2 marks)

Show Answers Only

`text{Proof (See Worked Solutions)`
`2cos\ pi/18, 2cos\ (11pi)/18, 2cos\ (13pi)/18`

Show Worked Solution

(i) `4\ cos^3 theta − 3\ cos theta = cos 3theta`

`text(Substitute)\ \ x = 2 cos theta\ \ text(into)`

`x^3 − 3x`	`= sqrt3`
`(2\ cos\ theta)^3 − 6\ cos\ theta`	`= sqrt3`
`8\ cos^3\ theta − 6\ cos\ theta`	`= sqrt3`
`4\ cos^3\ theta − 3\ cos\ theta`	`= sqrt3/2`
`:.cos\ 3theta`	`= sqrt3/2\ \ \ text{ … as required}`

MARKER’S COMMENT: Many students failed to complete the final step of part (ii) and lost an easy mark!

(ii)	`3theta`	`= pi/6, 2pi − pi/6, 2pi + pi/6`
	`3theta`	`= pi/6, (11pi)/6, (13pi)/6`
	`theta`	`= pi/18, (11pi)/18, (13pi)/18`
	`:.x`	`= 2cos\ pi/18, \ 2cos\ (11pi)/18, \ 2 cos\ (13pi)/18`

Functions, EXT1′ F1 2014 HSC 11d

Without the use of calculus, sketch the graph `y = x^2 - 1/(x^2)`, showing all intercepts. (2 marks)

Show Answers Only

`text(See Worked Solutions.)`

Show Worked Solution

`y = x^2 − 1/(x^2)\ \ =>text(even function)`

MARKER’S COMMENT: The best solutions showed a curve approaching the asymptote `y=x^2`, as shown in the solution.

`text(Domain: All)\ x, \ x≠0`

`y=x^2\ \ text(is an asymptote)`

`x text(-intercepts at)\ \ x=+-1`

`lim_(x→0^+) (x^2 − 1/(x^2))=-oo`

Complex Numbers, EXT2 N2 2014 HSC 11c

Sketch the region in the Argand diagram where `|\ z\ | ≤ |\ z − 2\ |` and `−pi/4 ≤ text(arg)\ z ≤ pi/4`. (3 marks)

Show Answers Only

`text(See Worked Solutions.)`

Show Worked Solution

`text(Let)\ \ z=x+iy`

`\|\ x + iy\ \|`	`≤ \|\ x − 2 + iy\ \|`
`sqrt(x^2 + y^2)`	`≤ sqrt((x − 2)^2 + y^2)`
`x^2 + y^2`	`≤ x^2 − 4x + 4 + y^2`
`4x − 4`	`≤ 0`
`x`	`≤ 1`

`−pi/4 ≤ text(arg)\ z ≤ pi/4`

Mechanics, EXT2* M1 2004 HSC 7a

The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours. A ship is to sail from a wharf to the harbour entrance and then out to sea. On the morning the ship is to sail, high tide at the wharf occurs at 2 am. The water depths at the wharf at high tide and low tide are 10 metres and 4 metres respectively.

Show that the water depth, `y` metres, at the wharf is given by

`y = 7 + 3 cos\ ((4pit)/(25))`, where `t` is the number of hours after high tide. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
An overhead power cable obstructs the ship’s exit from the wharf. The ship can only leave if the water depth at the wharf is 8.5 metres or less.

Show that the earliest possible time that the ship can leave the wharf is 4:05 am. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
At the harbour entrance, the difference between the water level at high tide and low tide is also 6 metres. However, tides at the harbour entrance occur 1 hour earlier than at the wharf. In order for the ship to be able to sail through the shallow harbour entrance, the water level must be at least 2 metres above the low tide level.

The ship takes 20 minutes to sail from the wharf to the harbour entrance and it must be out to sea by 7 am. What is the latest time the ship can leave the wharf? (2 marks)

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

Show Answers Only

`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`4:28\ text(am)`

Show Worked Solution

i. `text(Period)`

`(2pi)/n =`	` 12.5`
`12.5n =`	` 2pi`
`25n =`	` 4pi`
`n =`	` (4pi)/25`

`text(Amplitude)`

`a`	`= 1/2(10 − 4)`
	`= 3`

`=>\ text(Motion centres around)\ \ x = 7\ \ text(with)`

`y = 10\ \ text(when)\ \ t= 0`

`:.\ text(Water depth is given by)`

`y`	`= 7+a cos nt`
	`= 7 + 3 cos\ ((4pit)/(25))\ \ …\ text(as required.)`

ii.

`text(Find)\ \ t\ \ text(when)\ \ y = 8.5:`

`8.5`	`= 7 + 3\ cos\ ((4pit)/25)`
`3\ cos\ ((4pit)/25)`	`= 1.5`
`cos\ ((4pit)/25)`	`= 1/2`
`(4pit)/25`	`=pi/3`
`t`	`=(25pi)/(3 xx 4pi)`
	`=2 1/12`
	`= 2\ text(hrs 5 mins)`

`:.\ text(The earliest time the ship can leave)`

`text(is 4:05 am … as required.)`

iii. `text(2 metres above low tide = 6 m)`

`text(Find)\ t\ text(when)\ y = 6:`

`6 = 7 + 3\ cos\ ((4pit)/(25))`

`3\ cos\ ((4pit)/(25))`	`= -1`
`cos\ ((4pit)/(25))`	`= -1/3`
`(4pit)/25`	`= 1.9106…`
`:.t`	`= (25 xx 1.9106…)/(4pi)`
	`= 3.801…`
	`= 3\ text{hr 48 min (nearest min)}`

`:.\ text(At the harbour entrance, water depth)`

`text(of 6 m occurs when)\ \ t = 2\ text(hr 48 min.)`

`:.\ text(Given 20 mins sailing time, the latest the ship can)`

`text(leave the wharf is at)\ \ t = 2\ text(hr 28 min, or 4:28 am.)`

Mechanics, EXT2* M1 2007 HSC 7b

A small paintball is fired from the origin with initial velocity `14` metres per second towards an eight-metre high barrier. The origin is at ground level, `10` metres from the base of the barrier.

The equations of motion are

`x = 14t\ cos\ theta`

`y = 14t\ sin\ theta – 4.9t^2`

where `theta` is the angle to the horizontal at which the paintball is fired and `t` is the time in seconds. (Do NOT prove these equations of motion)

Show that the equation of trajectory of the paintball is

`y = mx − ((1 + m^2)/40)x^2`, where `m = tan\ theta`. (2 mark)

--- 6 WORK AREA LINES (style=lined) ---
Show that the paintball hits the barrier at height `h` metres when

`m = 2 ± sqrt(3 − 0.4h)`.

Hence determine the maximum value of `h`. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
There is a large hole in the barrier. The bottom of the hole is `3.9` metres above the ground and the top of the hole is `5.9` metres above the ground. The paintball passes through the hole if `m` is in one of two intervals. One interval is `2.8 ≤ m ≤ 3.2`.

Find the other interval. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Show that, if the paintball passes through the hole, the range is

`(40m)/(1 + m^2)\ \ text(metres.)`

Hence find the widths of the two intervals in which the paintball can land at ground level on the other side of the barrier. (3 marks)

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

Show Answers Only

`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`0.8 ≤ m ≤ 1.2`
`text{1.3 m (to 1 d.p.) and 0.5 m (to 1 d.p.)}`

Show Worked Solution

i.	`x`	`= 14t\ cos\ theta`	`\ \ …\ (1)`
	`y`	`= 14t\ sin\ theta-4.9t^2`	`\ \ …\ (2)`

`text(Substitute)\ \ t = x/(14\ cos\ theta)\ \ text{from (1) into (2)}`

`y`	`= 14(x/(14\ cos\ theta))\ sin\ theta − 4.9(x/(14\ cos\ theta))^2`
	`= x\ tan\ theta − (4.9/(14^2))((x^2)/(cos^2\ theta))`
	`= x\ tan\ theta − (x^2)/40\ sec^2\ theta`
	`= x\ tan\ theta − (x^2)/40(1 + tan^2\ theta)`
	`= mx − ((1 + m^2)/40)x^2\ \ \ \ \ (text(Given)\ \ m = tan\ theta)`

ii. `text(Show paintball hits at)\ \ h\ \ text(when)`

`m = 2 ± sqrt(3 − 0.4h)`

`text(i.e.)\ \ y = h\ \ text(when)\ \ x = 10`

`10m − ((1 + m^2)/40) · 10^2`	`= h`
`10m − 5/2(1 + m^2)`	`= h`
`20m − 5 − 5m^2`	`= 2h`
`5m^2 − 20m + 2h + 5`	`= 0`

`text(Using the quadratic formula)`

`m`	`=(20 ± sqrt((-20)^2 − 4 · 5 · (2h + 5)))/(2 · 5)`
	`= (20 ± sqrt(400 − 40h −100))/10`
	`= (20 ± sqrt(300 − 40h))/10`
	`= (20 ± 10sqrt(3 − 0.4h))/10`
	`= 2 ± sqrt(3 − 0.4h)\ \ \ text(… as required)`

`text(Find maximum)\ \ h`

`sqrt(3 − 0.4h)`	`≥ 0`
`3 − 0.4h`	`≥ 0`
`0.4h`	`≤ 3`
`h`	`≤ 7.5`

`:.\ text(Maximum)\ \ h = 7.5\ text(m)`

iii.

`text{Using part (ii)}`

`text(When)\ \ h = 3.9`

`m`	`= 2 ± sqrt(3 − 0.4(3.9))`
	`= 2 ± sqrt(1.44)`
	`= 2 ± 1.2`
	`= 3.2\ \ text(or)\ \ 0.8`

`text(When)\ \ h = 5.9`

`m`	`= 2 ± sqrt(3 − 0.4(5.9))`
	`= 2 ± sqrt(0.64)`
	`= 2 ± 0.8`
	`= 2.8\ \ text(or)\ \ 1.2`

`:.\ text(The other interval is)\ \ \ 0.8 ≤ m ≤ 1.2`

iv. `text(Find)\ \ x\ \ text(when)\ \ y = 0`

`mx − ((1 + m^2)/40)x^2`	`= 0`
`x[m − ((1 + m^2)/40)x]`	`= 0`
`((1 + m^2)/40)x`	`= m,\ \ \ \ x ≠ 0`
`:. x`	`= (40m)/(1 + m^2)\ \ …\ text(as required)`

`text(Consider the interval)\ \ \ 2.8 ≤ m ≤ 3.2`

`text(When)\ \ m = 2.8`

`=> x = (40(2.8))/(1 + 2.8^2) = 12.669…\ text(m)`

`text(When)\ \ m = 3.2`

`=>x = (40(3.2))/(1 + 3.2^2) = 11.387…\ text(m)`

`:.\ text(Landing width interval)`

`= 12.669… − 11.387…`

`= 1.281…`

`= 1.3\ text(m)\ \ text{to 1 d.p.}`

`text(Consider the interval)\ \ \ 0.8 ≤ m ≤ 1.2`

`text(When)\ \ m = 0.8`

`=>x = (40(0.8))/(1 + 0.8^2) = 19.512…\ text(m)`

`text(When)\ \ m = 1.2`

`=>x = (40(1.2))/(1 + 1.2^2) = 19.672…`

`text(S)text(ince interval includes)\ \ m = 1\ \ text(where the)`

`text(paintball has maximum range.)`

`x_(text(max)) = (40(1))/(1 + 1^2) = 20\ text(m)`

`:.\ text(Landing width interval)`

`= 20 − 19.512…`

`= 0.487…`

`= 0.5\ text(m)\ \ \ text{(to 1 d.p.)}`

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