HSC 2015 Maths Marathon (archive) (3 Viewers)

davidgoes4wce · Aug 29, 2015

Re: HSC 2015 2U Marathon

part b) When will Paul's annual earnings first exceed Megan's annual earnings, if Megan receives her bonus every year?

davidgoes4wce · Aug 29, 2015

Re: HSC 2015 2U Marathon

My Working out, if someone could confirm that would be good.

keepLooking · Sep 12, 2015

Re: HSC 2015 2U Marathon

Drsoccerball · Sep 12, 2015

Re: HSC 2015 2U Marathon

keepLooking said:

$i) 5(k^2 -4k +\frac{24}{5})$

$5(k-2)^2 +4$

$\therefore $ always positive $$

ii) $$ Discriminant $ \geq 0$

$k^2-4(k-2)(3-k) \geq 0$

$5k^2 -20k +24 \geq 0 $ Proven in (i) $ \therefore 2 real roots$

iii) $$ Since the graph has always two solutions it would be meaningless to prove that discriminant=0$

$\therefore $ Observe the equation of the graph and notice that if $ k=2 $ The squared term becomes 0 leaving $ 2x+1= 0....$

$\therefore k =2$

Flop21 · Sep 13, 2015

Re: HSC 2015 2U Marathon

How does sin^(3)x = 1/8

then turn into

sinx = 1/2

rand_althor · Sep 13, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
How does sin^(3)x = 1/8

then turn into

sinx = 1/2

Take the cube root of both sides, since $\sin^3{x}=\left (\sin{x} \right )^3$ .

Flop21 · Sep 13, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
Take the cube root of both sides, since $\sin^3{x}=\left (\sin{x} \right )^3$ .

Oh right, duh lol. Thanks!

davidgoes4wce · Sep 16, 2015

Re: HSC 2015 2U Marathon

The solution:

Must we show the 2nd derivative that it is a minimum point, since there is only one solution for the first derivative? (0=25800t-18000)

I'm not sure how examiners marking are in terms of penalising missing content.

psyc1011 · Sep 16, 2015

Re: HSC 2015 2U Marathon

Yes, David. Or otherwise, how would you show it's the minimum point continuiing with calculus? (you can't say parabola argument since that disrupts the efficiency of solution)

Drsoccerball · Sep 16, 2015

Re: HSC 2015 2U Marathon

psyc1011 said:
Yes, David. Or otherwise, how would you show it's the minimum point continuiing with calculus? (you can't say parabola argument since that disrupts the efficiency of solution)

Or you can just fudge it and say that you tested positive and negative of the stationary point.

InteGrand · Sep 17, 2015

Re: HSC 2015 2U Marathon

davidgoes4wce said:
The solution:

Must we show the 2nd derivative that it is a minimum point, since there is only one solution for the first derivative? (0=25800t-18000)

I'm not sure how examiners marking are in terms of penalising missing content.

Don't need to bother with second derivative or any other calculation, you can just say that the objective function (the function to be optimised) is a concave up parabola, so its stationary point is indeed the minimum. But you do need to provide some justification.

Trebla · Sep 17, 2015

Re: HSC 2015 2U Marathon

By the way did anyone actually try answering the question in the original post?

flavurr · Sep 18, 2015

Re: HSC 2015 2U Marathon

Trebla said:
By the way did anyone actually try answering the question in the original post?

Whats the question again?

Drsoccerball · Sep 18, 2015

Re: HSC 2015 2U Marathon

flavurr said:
Whats the question again?

Its the first post on this forum

leehuan · Sep 18, 2015

Re: HSC 2015 2U Marathon

Is this how to do it?

Trebla · Sep 19, 2015

Re: HSC 2015 2U Marathon

leehuan said:
Is this how to do it?View attachment 32410

Not quite correct. The scenario is that the person is receiving income of A each year, rather than as a one-off deposit.

leehuan · Sep 19, 2015

Re: HSC 2015 2U Marathon

That explains why I didn't arrive at a progression. But I think my progression is wrong.

$\\ { A }_{ 1 }=A\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ { A }_{ 2 }=\left( { A }_{ 1 }+A \right) \left( 1+{ r }_{ 1 } \right) -\left( { A }_{ 1 }+A \right) \left( 1+{ r }_{ 1 } \right) { r }_{ 2 }\\ =\left( { A }_{ 1 }+A \right) \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ =A\left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +1 \right] \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ { A }_{ 3 }=\left( { A }_{ 2 }+A \right) \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ =\left[ A\left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +1 \right] \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +A \right] \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right)$

$\\ =A\left[ { \left( 1+{ r }_{ 1 } \right) }^{ 2 }{ \left( 1-{ r }_{ 2 } \right) }^{ 2 }+\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +1 \right] \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ \vdots \\ { A }_{ n }=A\left\{ 1+\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +{ \left( 1+{ r }_{ 1 } \right) }^{ 2 }{ \left( 1-{ r }_{ 2 } \right) }^{ 2 }+\dots A{ \left( 1+{ r }_{ 1 } \right) }^{ n-1 }{ \left( 1-{ r }_{ 2 } \right) }^{ n-1 } \right\} \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ =A\left\{ \frac { { \left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \right] }^{ n }-1 }{ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) -1 } \right\} \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right)$

$\\ $We need ${ A }_{ n }>A\\ \therefore \left\{ \frac { { \left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \right] }^{ n }-1 }{ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) -1 } \right\} \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) >1\\ { \left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \right] }^{ n+1 }-\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) >\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) -1$

EDIT 2: Wait, how do we know if |common ratio|<1?

Trebla · Sep 19, 2015

Re: HSC 2015 2U Marathon

leehuan said:
That explains why I didn't arrive at a progression. But I think my progression is wrong.

$\\ { A }_{ 1 }=A\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ { A }_{ 2 }=\left( { A }_{ 1 }+A \right) \left( 1+{ r }_{ 1 } \right) -\left( { A }_{ 1 }+A \right) \left( 1+{ r }_{ 1 } \right) { r }_{ 2 }\\ =\left( { A }_{ 1 }+A \right) \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ =A\left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +1 \right] \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ { A }_{ 3 }=\left( { A }_{ 2 }+A \right) \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ =\left[ A\left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +1 \right] \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +A \right] \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right)$

$\\ =A\left[ { \left( 1+{ r }_{ 1 } \right) }^{ 2 }{ \left( 1-{ r }_{ 2 } \right) }^{ 2 }+\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +1 \right] \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ \vdots \\ { A }_{ n }=A\left\{ 1+\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) +{ \left( 1+{ r }_{ 1 } \right) }^{ 2 }{ \left( 1-{ r }_{ 2 } \right) }^{ 2 }+\dots A{ \left( 1+{ r }_{ 1 } \right) }^{ n-1 }{ \left( 1-{ r }_{ 2 } \right) }^{ n-1 } \right\} \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \\ =A\left\{ \frac { { \left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \right] }^{ n }-1 }{ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) -1 } \right\} \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right)$

$\\ $We need ${ A }_{ n }>A\\ \therefore \left\{ \frac { { \left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \right] }^{ n }-1 }{ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) -1 } \right\} \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) >1\\ { \left[ \left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) \right] }^{ n+1 }-\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) >\left( 1+{ r }_{ 1 } \right) \left( 1-{ r }_{ 2 } \right) -1$

EDIT 2: Wait, how do we know if |common ratio|<1?

Think about what the question is asking. You want your savings to actually grow without bound over time, so there is a certain condition you need to impose.

rand_althor · Sep 19, 2015

Re: HSC 2015 2U Marathon

Trebla said:
$Suppose that a person receives an income of $A$ at the beginning of every year which is deposited into a bank account. The bank account earns an interest rate of $100r_1$ per cent per annum. However, just before he deposits his income into the bank account each year, he withdraws $100r_2$ per cent of his balance for everyday expenses.\\\\Explain why this person must choose the proportion of his withdrawal $r_2$ to be such that $r_2<\dfrac{r_1}{1+r_1}$ in order to ensure that his savings balance grows at a divergent rate over time (otherwise his savings will never exceed a certain value).$

$Let $A_n$ be the amount in the bank account at the end of the $n$th year.$

$\begin{align*}A_1&=A(1+r_1)(1-r_2)+A \\A_2&=\left [A(1+r_1)(1-r_2)+A \right ] \left [(1+r_1)(1-r_2) \right ] + A \\&=A(1+r_1)^2(1-r_2)^2+A(1+r_1)(1-r_2)+A \\A_3&=\left [A(1+r_1)^2(1-r_2)^2+A(1+r_1)(1-r_2)+A \right ] \left [(1+r_1)(1-r_2) \right ] + A \\&=A(1+r_1)^3(1-r_2)^3+A(1+r_1)^2(1-r_2)^2+A(1+r_1)(1-r_2)+A \\\vdots \\A_n&=A(1+r_1)^n(1-r_2)^n + A(1+r_1)^{n-1}(1-r_2)^{n-1}+...+A(1+r_1)(1-r_2)+A \\&=A+A(1+r_1)(1-r_2)+...+A(1+r_1)^{n-1}(1-r_2)^{n-1}+A(1+r_1)^n(1-r_2)^n \\\end{align*}$

$$This is a geometric series with a common ratio of $r=(1+r_1)(1-r_2)$. For this geometric series to diverge, a limiting sum must not exist, i.e. $\left |r \right |>1$.$

$\begin{align*}\left | (1+r_1)(1-r_2) \right |&>1 \\(1+r_1)(1-r_2)&>1 \textup{ Since }(1+r_2)>0, \, 0<(1-r_2)<1 \\1-r_2&>\frac{1}{1+r_1} \textup{ Since } (1+r_1)>0 \\r_2&<1-\frac{1}{1+r_1} \\r_2&<\frac{1+r_1-1}{1+r_1} \\r_2&<\frac{r_1}{1+r_1} \\\end{align*}$

leehuan · Sep 19, 2015

Re: HSC 2015 2U Marathon

Should've trusted my instincts and forced the (absolute value of) the common ratio to be greater than 1. But yep, makes sense.

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