HSC 2015 Maths Marathon (archive) (3 Viewers)

Trebla · Dec 7, 2014

Post any questions within the scope and level of Mathematics (2 unit). Once a question is posted, it needs to be answered before the next question is raised.

I encourage all current students in particular to participate in this marathon.

To get the ball rolling:

$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).$

Sy123 · Dec 8, 2014

Re: HSC 2015 2U Marathon

For those who haven't started series yet, and are still doing differentiation and integration

TOPIC: Integration

$\\ $The area enclosed between the curves$ \ y = x \ $and$ \ y=x^3 \ $is rotated about the$ \ x \ $axis, find the volume of the figure generated$$

jkerr138 · Dec 9, 2014

Re: HSC 2015 2U Marathon

Is the volume solution 20pi/21 units ^3

FrankXie · Dec 9, 2014

Re: HSC 2015 2U Marathon

jkerr138 said:
Is the volume solution 20pi/21 units ^3

sorry but it is 8pi/21. u must have used plus sign instead of minus sign between two pieces?

jkerr138 · Dec 10, 2014

Re: HSC 2015 2U Marathon

Yeah, I did, I just used the minus, and multiplied by the area by 2 for it being symmetrical, and got 8pi/21. Thanks.

Sy123 · Dec 10, 2014

Re: HSC 2015 2U Marathon

TOPIC: Maxima and Minima

$\\ $I have a constant amount of fencing that I want to turn into a rectangular paddock, and I want to maximise the area of the paddock. Show that no matter how much (constant) material I have, I should always make a square shaped paddock$$

InteGrand · Dec 11, 2014

Re: HSC 2015 2U Marathon

Sy123 said:
TOPIC: Maxima and Minima

$\\ $I have a constant amount of fencing that I want to turn into a rectangular paddock, and I want to maximise the area of the paddock. Show that no matter how much (constant) material I have, I should always make a square shaped paddock$$

Let the total available length of fencing be L. Let the dimensions of a rectangular paddock built from this fencing be x and y, so that 2x+2y = L (perimeter), i.e. x+y = L/2, or y = L/2 - x.

Now, the area of the paddock is A = xy = x(L/2 - x).

This represents a concave down parabola, so its turning point will give a max., and the turning point occurs midway between the two roots, which is when x = (L/2)/2 = L/4, so y = L/4 too (as y = L/2 - x). Hence for max. area, dimensions are equal, i.e. the rectangle is a square.

Axio · Dec 11, 2014

Re: HSC 2015 2U Marathon

$Let \ $\alpha \ $and $ \beta \ $be \ the \ roots \ of \ 5x^2+10x+\frac{1}{2}=0, \ find:$

$\frac{\alpha ^{2}+\beta ^{2}}{\alpha }+\frac{\alpha ^{2}+\beta ^{2}}{\beta }$

LostInHSC · Dec 14, 2014

Re: HSC 2015 2U Marathon

Axio said:
$Let \ $\alpha \ $and $ \beta \ $be \ the \ roots \ of \ 5x^2+10x+\frac{1}{2}=0, \ find:$

$\frac{\alpha ^{2}+\beta ^{2}}{\alpha }+\frac{\alpha ^{2}+\beta ^{2}}{\beta }$

$\frac{\alpha^{2}+\beta ^{2}}{\alpha} + \frac{\alpha ^{2}+\beta ^{2}}{\beta } = \frac{\alpha ^{3}+\beta ^{3} +\alpha \beta \left ( \alpha +\beta \right )}{\alpha \beta } \\ \\ $Now, $ \\ \begin{align*}\left ( \alpha +\beta \right )^{3} &= \left ( \alpha +\beta \right )\left ( \alpha ^{2} + 2\alpha \beta +\beta ^{2}\right ) \\ &= \alpha ^{3} + 2\alpha ^{2}\beta +\alpha \beta ^{2} +\alpha ^{2}\beta +2\alpha \beta ^{2} + \beta ^{3} && \text{(Now Rearranging)} \\ & \end{align*} \\ \alpha ^{3} +\beta ^{3} = \left ( \alpha +\beta \right )^{3} -3\alpha \beta \left ( \alpha +\beta \right ) \\ \\ $Therefore, $ \frac{\alpha^{2}+\beta ^{2}}{\alpha} + \frac{\alpha ^{2}+\beta ^{2}}{\beta } = \frac{\left ( \alpha +\beta \right )^{3} -3\alpha \beta \left ( \alpha +\beta \right ) +\alpha \beta \left ( \alpha +\beta \right )}{\alpha \beta } = \frac{\left ( \alpha +\beta \right )^{3} -2\alpha \beta \left ( \alpha +\beta \right ) }{\alpha \beta } \\ \\$

LostInHSC · Dec 14, 2014

Re: HSC 2015 2U Marathon

LostInHSC said:
$\frac{\alpha^{2}+\beta ^{2}}{\alpha} + \frac{\alpha ^{2}+\beta ^{2}}{\beta } = \frac{\alpha ^{3}+\beta ^{3} +\alpha \beta \left ( \alpha +\beta \right )}{\alpha \beta } \\ \\ $Now, $ \\ \begin{align*}\left ( \alpha +\beta \right )^{3} &= \left ( \alpha +\beta \right )\left ( \alpha ^{2} + 2\alpha \beta +\beta ^{2}\right ) \\ &= \alpha ^{3} + 2\alpha ^{2}\beta +\alpha \beta ^{2} +\alpha ^{2}\beta +2\alpha \beta ^{2} + \beta ^{3} && \text{(Now Rearranging)} \\ & \end{align*} \\ \alpha ^{3} +\beta ^{3} = \left ( \alpha +\beta \right )^{3} -3\alpha \beta \left ( \alpha +\beta \right ) \\ \\ $Therefore, $ \frac{\alpha^{2}+\beta ^{2}}{\alpha} + \frac{\alpha ^{2}+\beta ^{2}}{\beta } = \frac{\left ( \alpha +\beta \right )^{3} -3\alpha \beta \left ( \alpha +\beta \right ) +\alpha \beta \left ( \alpha +\beta \right )}{\alpha \beta } \\ \\$

$5x^{2} + 10x +\frac{1}{2} =0 \\ \\ \alpha +\beta =\frac{-10}{5} = -2 \\ \alpha \beta =\frac{\frac{1}{2}}{5} = \frac{1}{10} \\ $Now, sub in for result.$$

RealiseNothing · Dec 14, 2014

Re: HSC 2015 2U Marathon

Axio said:
$Let \ $\alpha \ $and $ \beta \ $be \ the \ roots \ of \ 5x^2+10x+\frac{1}{2}=0, \ find:$

$\frac{\alpha ^{2}+\beta ^{2}}{\alpha }+\frac{\alpha ^{2}+\beta ^{2}}{\beta }$

Just a quicker method that isn't very obvious:

Cancel out the denominators where we can and then add/subtract 2:

$\alpha + \beta + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2 - 2$

$\alpha + \beta + (\frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}})^2 - 2$

$\alpha + \beta + (\frac{\alpha + \beta}{\sqrt{\alpha \beta}})^2-2$

$\alpha + \beta + \frac{(\alpha+\beta)^2}{\alpha \beta} - 2$

Sub in our known values now.

Fiction · Dec 14, 2014

Re: HSC 2015 2U Marathon

Sy123 said:
For those who haven't started series yet, and are still doing differentiation and integration

TOPIC: Integration

$\\ $The area enclosed between the curves$ \ y = x \ $and$ \ y=x^3 \ $is rotated about the$ \ x \ $axis, find the volume of the figure generated$$

Can someone walk me through this question? I'm getting either pie/4 or pie/6 lol
//crappy maths

peri24 · Dec 14, 2014

Re: HSC 2015 2U Marathon

Fiction said:
Can someone walk me through this question? I'm getting either pie/4 or pie/6 lol
//crappy maths

It's very straight forward..

LostInHSC · Dec 14, 2014

Re: HSC 2015 2U Marathon

Fiction said:
Can someone walk me through this question? I'm getting either pie/4 or pie/6 lol
//crappy maths

$$Make a quick sketch of y=x and y=x^3. $ \\ $Now, first find x-coordinates of intercepts$ \\ x^{3} = x \\ x^{3} - x =0 \\ x(x^{2}-1)=0 \\ \therefore x=0, \pm 1 \\ \\ $Volume Generated is found by: \pi\p\int_{a}^{b}f(x)^{2} -g(x)^{2} \, dx \\$

LostInHSC · Dec 14, 2014

Re: HSC 2015 2U Marathon

LostInHSC said:
$$Make a quick sketch of y=x and y=x^3. $ \\ $Now, first find x-coordinates of intercepts$ \\ x^{3} = x \\ x^{3} - x =0 \\ x(x^{2}-1)=0 \\ \therefore x=0, \pm 1 \\ \\ $Volume Generated is found by: \pi\p\int_{a}^{b}f(x)^{2} -g(x)^{2} \, dx \\$

$\begin{align*}V &= 2\pi\int_{0}^{1} (x)^{2} -(x^{3})^{2} \,dx \\&= 2\pi\int_{0}^{1}x^{2}-x^{6} \,dx\\ &= 2\pi\left[\frac{x^{3}}{3} -\frac{x^{7}}{7} \right]_{0}^{1}\\&= 2\pi(\frac{1}{3} -\frac{1}{7}) \\&= \frac{8\pi}{21} units^{3} \end{align*} \\$

Axio · Dec 14, 2014

Re: HSC 2015 2U Marathon

RealiseNothing said:
Just a quicker method that isn't very obvious:

Cancel out the denominators where we can and then add/subtract 2:

$\alpha + \beta + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2 - 2$

$\alpha + \beta + (\frac{\sqrt{\alpha}}{\sqrt{\beta}} + \frac{\sqrt{\beta}}{\sqrt{\alpha}})^2 - 2$

$\alpha + \beta + (\frac{\alpha + \beta}{\sqrt{\alpha \beta}})^2-2$

$\alpha + \beta + \frac{(\alpha+\beta)^2}{\alpha \beta} - 2$

Sub in our known values now.

Good method. But shouldn't the fractions in the first line be a^2/b and b^2/a?

I just went:

$\frac{\alpha ^{2}+\beta ^{2}}{\alpha }+\frac{\alpha ^{2}+\beta ^{2}}{\beta }$

$=\frac{(\alpha +\beta )(\alpha ^{2}+\beta ^{2})}{\alpha \beta }=\frac{(\alpha +\beta )((\alpha +\beta )^{2}-2\alpha \beta )}{\alpha \beta }$

And then subbing in values from there I got =-76.

FrankXie · Dec 14, 2014

Re: HSC 2015 2U Marathon

Axio said:
Good method. But shouldn't the fractions in the first line be a^2/b and b^2/a?

I just went:

$\frac{\alpha ^{2}+\beta ^{2}}{\alpha }+\frac{\alpha ^{2}+\beta ^{2}}{\beta }$

$=\frac{(\alpha +\beta )(\alpha ^{2}+\beta ^{2})}{\alpha \beta }=\frac{(\alpha +\beta )((\alpha +\beta )^{2}-2\alpha \beta )}{\alpha \beta }$

And then subbing in values from there I got =-76.

This is pretty much what I would do.

Fizzy_Cyst · Dec 15, 2014

Re: HSC 2015 2U Marathon

Solve for x:

sin^2(x) - 3sin(x)cos(x) + 2cos^2(x) = 0 where 0<=x<=2(Pi)

RealiseNothing · Dec 15, 2014

Re: HSC 2015 2U Marathon

Axio said:
Good method. But shouldn't the fractions in the first line be a^2/b and b^2/a?

I cant even maths

Fiction · Dec 15, 2014

Re: HSC 2015 2U Marathon

Fiction said:
Can someone walk me through this question? I'm getting either pie/4 or pie/6 lol
//crappy maths

peri24 said:
It's very straight forward..

.

LostInHSC said:
$$Make a quick sketch of y=x and y=x^3. $ \\ $Now, first find x-coordinates of intercepts$ \\ x^{3} = x \\ x^{3} - x =0 \\ x(x^{2}-1)=0 \\ \therefore x=0, \pm 1 \\ \\ $Volume Generated is found by: \pi\p\int_{a}^{b}f(x)^{2} -g(x)^{2} \, dx \\$

LostInHSC said:
$\begin{align*}V &= 2\pi\int_{0}^{1} (x)^{2} -(x^{3})^{2} \,dx \\&= 2\pi\int_{0}^{1}x^{2}-x^{6} \,dx\\ &= 2\pi\left[\frac{x^{3}}{3} -\frac{x^{7}}{7} \right]_{0}^{1}\\&= 2\pi(\frac{1}{3} -\frac{1}{7}) \\&= \frac{8\pi}{21} units^{3} \end{align*} \\$

ooo thanks bb C:
I did the intergration thing wrong -> made x^2 into x^3/3 but forgot about making x^6 into x^7/7

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