smc-1006-95-X-topic

Financial Maths, 2ADV M1 2024 HSC 30

Suppose the geometric series \(x+x^2+x^3+\ \cdots\) has a limiting sum.

By considering the graph \(y=-1-\dfrac{1}{x-1}\), or otherwise, find the range of possible values of \(S\). (3 marks)

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\(S \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

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\(x + x^2+x^3\ \cdots \Rightarrow a=x, r=x\)

\(S=\dfrac{a}{1-r} = \dfrac{x}{1-x}=\dfrac{-(1-x)+1}{1-x} = -1+\dfrac{1}{1-x}=-1-\dfrac{1}{x-1}\)

♦♦ Mean mark 31%.

\(\text{If limiting sum}\ \Rightarrow \abs{r} \lt 1\ \Rightarrow \ \abs{x} \lt 1\)

\(\Rightarrow \text{Possible values of}\ S =\ \text{range of}\ \ y=-1-\dfrac{1}{x-1}\ \text{in domain}\ \abs{x} \lt 1\)

\(\text{Consider}\ \ y=-1-\dfrac{1}{x-1}:\)

\(\text{Asymptote at}\ \ x=1.\)

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -\dfrac{2}{3} & -\dfrac{1}{2} & \ \ 0\ \ & \infty & -2 \\
\hline
\end{array}

\(\text{As}\ x \rightarrow \infty, \ y \rightarrow -1\)

\(\text{As}\ x \rightarrow -\infty, \ y \rightarrow -1\)

\(\abs{x} \lt 1\ \Rightarrow \ y \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

\(\therefore\ \text{Possible values of}\ S \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

Trigonometry, 2ADV T2 2004 HSC 9a

Consider the geometric series `1 − tan^2 theta + tan^4 theta − …`

When the limiting sum exists, find its value in simplest form. (2 marks)

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For what values of `theta` in the interval

`−pi/2 < theta < pi/2` does the limiting sum of the series exist? (2 marks)

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`cos^2 theta`
`− pi/4 < theta < pi/4`

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i. `1 − tan^2 theta + tan^4 theta − …`

`=>\ text(GP where)\ \ a=1,\ \ r=T_2/T_1= − tan^2 theta`

`:. S_∞`	`= 1/(1 − (−tan^2 theta))`
	`= 1/(1 + tan^2 theta)`
	`= 1/(sec^2 theta)`
	`= cos^2 theta`

ii.	`text(Find)\ \ theta\ \ text(such that)\ \ \ \|\ r\ \|`	`< 1`
	`\|−tan^2 theta\ \|`	`< 1`
	` tan^2 theta`	`< 1`
	`−1 < tan theta`	`< 1`
	`:. − pi/4 < theta`	`< pi/4`

Trigonometry, 2ADV T1 2005 HSC 9b

The triangle `ABC` has a right angle at `B, \ ∠BAC = theta` and `AB = 6`. The line `BD` is drawn perpendicular to `AC`. The line `DE` is then drawn perpendicular to `BC`. This process continues indefinitely as shown in the diagram.

Find the length of the interval `BD`, and hence show that the length of the interval `EF` is `6 sin^3\ theta`. (2 marks)

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Show that the limiting sum

`qquad BD + EF + GH + ···`

is given by `6 sec\ theta tan\ theta`. (3 marks)

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

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`text(Show)\ EF = 6\ sin^3\ theta`

`text(In)\ ΔADB`

`sin\ theta`	`= (DB)/6`
`DB`	`= 6\ sin\ theta`

`∠ABD`	`= 90 − theta\ \ \ text{(angle sum of}\ ΔADB)`
`:.∠DBE`	`= theta\ \ \ (∠ABE\ text{is a right angle)}`

`text(In)\ ΔBDE:`

`sin\ theta`	`= (DE)/(DB)`
	`= (DE)/(6\ sin\ theta)`

`DE`	`= 6\ sin^2\ theta`
`∠BDE`	`= 90 − theta\ \ \ text{(angle sum of}\ ΔDBE)`
`∠EDF`	`= theta\ \ \ (∠FDB\ text{is a right angle)}`

`text(In)\ ΔDEF:`

`sin\ theta`	`= (EF)/(DE)`
	`= (EF)/(6\ sin^2\ theta)`
`:.EF`	`= 6\ sin^3\ theta\ \ …text(as required)`

ii. `text(Show)\ \ BD + EF + GH\ …`

`text(has limiting sum)\ =6 sec theta tan theta`

`underbrace{6\ sin\ theta + 6\ sin^3\ theta +\ …}_{text(GP where)\ \ a = 6 sin theta, \ \ r = sin^2 theta}`

`text(S)text(ince)\ \ 0 < theta < 90^@`

`−1`	`< sin\ theta`	`< 1`
`0`	`< sin^2\ theta`	`< 1`

`:. |\ r\ | < 1`

`:.S_∞`	`= a/(1 − r)`
	`= (6\ sin\ theta)/(1 − sin^2\ theta)`
	`= (6\ sin\ theta)/(cos^2\ theta)`
	`= 6 xx 1/(cos\ theta) xx (sin\ theta)/(cos\ theta)`
	`= 6 sec\ theta\ tan\ theta\ \ …text(as required.)`

Probability, 2ADV S1 2013 HSC 15d

Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

Find the probability that Pat wins the game on the first throw. (1 mark)

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What is the probability that Pat wins the game on the first or on the second throw? (2 marks)

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Find the probability that Pat eventually wins the game. (2 marks)

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`1/36`
`(2521)/(46\ 656)\ \ text(or)\ \ 0.054`
`36/71`

Show Worked Solutions

i. `P\ text{(Pat wins on 1st throw)}=P(W)`

`P(W)`	`=P\ text{(Pat throws 2 sixes)}`
	`=1/6 xx 1/6`
	`=1/36`

ii. `text(Let)\ P(L)=P text{(loss for either player on a throw)}=35/36`

`P text{(Pat wins on 1st or 2nd throw)}`

♦♦ Mean mark 33%
MARKER’S COMMENT: Many students did not account for Chandra having to lose when Pat wins on the 2nd attempt.

`=P(W) + P(LL W)`

`=1/36\ + \ (35/36)xx(35/36)xx(1/36)`

`=(2521)/(46\ 656)`

`=0.054\ \ \ text{(to 3 d.p.)}`

iii. `P\ text{(Pat wins eventually)}`

`=P(W) + P(LL\ W)+P(LL\ LL\ W)+ … `

`=1/36\ +\ (35/36)^2 (1/36)\ +\ (35/36)^2 (35/36)^2 (1/36)\ +…`

`=>\ text(GP where)\ \ a=1/36,\ \ r=(35/36)^2=(1225)/(1296)`

♦♦♦ Mean mark 8%!
COMMENT: Be aware that diminishing probabilities and `S_oo` within the Series and Applications are a natural cross-topic combination.

`text(S)text(ince)\ |\ r\ |<\ 1:`

`S_oo`	`=a/(1-r)`
	`=(1/36)/(1-(1225/1296))`
	`=1/36 xx 1296/71`
	`=36/71`

`:.\ text(Pat’s chances to win eventually are)\ 36/71`.