HSC 2016 Maths Marathon (archive) (1 Viewer)

davidgoes4wce · Nov 8, 2015

Re: HSC 2016 2U Marathon

Glyde said:
Calculate the area between the curve (4-x^2)^(1/2) and the x-axis for the 1st quadrant only.

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here is my answer

davidgoes4wce · Nov 8, 2015

Re: HSC 2016 2U Marathon

wait^^^^ I thought it was a volume of solid question....

Glyde · Nov 8, 2015

Re: HSC 2016 2U Marathon

Why do we solve this differently ? I don't get why we need to take out pi

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davidgoes4wce · Nov 8, 2015

Re: HSC 2016 2U Marathon

The area should just be A=(Pi x r^2)/4 = Pi

davidgoes4wce · Nov 8, 2015

Re: HSC 2016 2U Marathon

Glyde · Nov 8, 2015

Re: HSC 2016 2U Marathon

Yes but we are trying to use integration

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leehuan · Nov 8, 2015

Re: HSC 2016 2U Marathon

Glyde said:
Yes but we are trying to use integration

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An area formulae is integration anyway? Where do you think people got area formulae to begin with? Integration.

Note that in the 2U course, direct integration of sqrt(4-x^2) is not possible. Direct integration will require the trigonometric substitution of x=2sin(theta), which is 4U. Therefore, the only method to approach this question is the area of a circle.

leehuan · Nov 8, 2015

Re: HSC 2016 2U Marathon

kawaiipotato said:
Question :
$\sum_{k=0}^{80} \frac{1}{ \sqrt{k+1} + \sqrt {k}}$

$\\ $By rationalising the denominator: $\frac { 1 }{ \sqrt { k+1 } +\sqrt { k } } =\left( \sqrt { k+1 } -\sqrt { k } \right)$

Now, when we consider our values given, we have:
k=0 -> + sqrt(1) - sqrt(0)
k=1 -> + sqrt(2) - sqrt(1)
k=2 -> +sqrt(3) - sqrt(2)
k=3 -> +sqrt(4) - sqrt(3)
...
k=80 -> +sqrt(81)-sqrt(80)

The fact is, when we add all of these values up (note that our original question has a SIGMA there), all the terms will cancel out EXCEPT
+sqrt(81) - sqrt(0)

Evaluating this and our answer is 9.

leehuan · Nov 14, 2015

Re: HSC 2016 2U Marathon

NEXT QUESTION

$\\ $Write $0.2183\overline { 599 } $ as a fraction$

MilkyCat_ · Nov 14, 2015

Re: HSC 2016 2U Marathon

let x=0.2183599599...
1000x=218.3599599...
999x=218.1416
9990000x=2181416
x=2181416/9990000

leehuan · Nov 14, 2015

Re: HSC 2016 2U Marathon

NEXT QUESTION

$\\ $State the necessary conditions for a point on any arbitrary graph $y=f(x)$ to be a horizontal point of inflexion.$$

sharoooooo · Nov 14, 2015

Re: HSC 2016 2U Marathon

leehuan said:
NEXT QUESTION

$\\ $State the necessary conditions for a point on any arbitrary graph $y=f(x)$ to be a horizontal point of inflexion.$$

y' = 0
y'' = 0

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leehuan · Nov 14, 2015

Re: HSC 2016 2U Marathon

sharoooooo said:
y' = 0
y'' = 0

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Not fully accurate. To verify a point of inflexion, you should recall that you have to test a bit to each side. Because clearly for the case y=x^4 (or x^6 etc.) y' = y'' = 0 at x=0 but that's a local minimum.
----------------
NEXT QUESTION
$\\ $leehuan hopes to make an annual salary of at least 100,000 dollars in the future. He will deposit $90 \%$ of his earnings into an investment fund which will earn $6 \%$ interest p.a. compounded monthly. \\ leehuan hopes to get his career goals going by the time he is 25 years old, and hopefully retire when he turns 65 years old. Assuming lady luck smiled on him, someone be a dear and calculate how much his investment will have reached for his retirement.$$

InteGrand · Nov 14, 2015

Re: HSC 2016 2U Marathon

leehuan said:
Not fully accurate. To verify a point of inflexion, you should recall that you have to test a bit to each side. Because clearly for the case y=x^4 (or x^6 etc.) y' = y'' = 0 at x=0 but that's a local minimum.

$Well you asked for \textsl{necessary} conditions, not \textsl{necessary and sufficient} conditions (there's a difference between \textsl{necessary} and \textsl{sufficient} in mathematics and logic etc.). The conditions given by sharoooooo are indeed necessary conditions, \textit{assuming that the function f is a function that is twice differentiable at the point in question}.$

$A function $f$ can have a stationary point of inflexion at $x=a$ even if its second derivative does not exist at the point $a$. Since you said the function was arbitrary, the necessary and sufficient conditions for a horizontal point of inflexion are just that $f^\prime (a)=0$ and concavity changes about $x=a$ (this is because the definition of horizontal is that the slope is 0, and the definition of inflexion point is that the concavity changes about the point).$

$For example, consider the function $f:\mathbb{R}\to \mathbb{R},f(x) = x^{\frac{5}{3}}$. Then clearly, $f^\prime (x) = \frac{5}{3}x^{\frac{2}{3}}$ and $f''(x) = \frac{5}{3}\cdot \frac{2}{3} x^{-\frac{1}{3}}$. Then $f^\prime (0)=\frac{5}{3}\cdot 0^{\frac{2}{3}}=0$, and $f''$ changes sign about the point $0$. So $x=0$ is a stationary point of inflexion for $f$. However, $f''$ is undefined at $x=0$. This shows that it is possible to have stationary points of inflexion even if $f''$ does not exist at the point in question.$

InteGrand · Nov 14, 2015

Re: HSC 2016 2U Marathon

leehuan said:
Not fully accurate. To verify a point of inflexion, you should recall that you have to test a bit to each side. Because clearly for the case y=x^4 (or x^6 etc.) y' = y'' = 0 at x=0 but that's a local minimum.
----------------
NEXT QUESTION
$\\ $leehuan hopes to make an annual salary of at least 100,000 dollars in the future. He will deposit 90$percent$ of his earnings into an investment fund which will earn 6$percent$ interest p.a. compounded monthly. \\ leehuan hopes to get his career goals going by the time he is 25 years old, and hopefully retire when he turns 65 years old. Assuming lady luck smiled on him, someone be a dear and calculate how much his investment will have reached for his retirement.$$

$To write the percentage sign in LaTeX, do a backslash and then the percentage symbol: $\%$.$

leehuan · Nov 14, 2015

Re: HSC 2016 2U Marathon

InteGrand said:
$Well you asked for \textsl{necessary} conditions, not \textsl{necessary and sufficient} conditions (there's a difference between \textsl{necessary} and \textsl{sufficient} in mathematics and logic etc.). The conditions given by sharoooooo are indeed necessary conditions, \textit{assuming that the function f is a function that is twice differentiable at the point in question}.$

$A function $f$ can have a stationary point of inflexion at $x=a$ even if its second derivative does not exist at the point $a$. Since you said the function was arbitrary, the necessary and sufficient conditions for a horizontal point of inflexion are just that $f^\prime (a)=0$ and concavity changes about $x=a$ (this is because the definition of horizontal is that the slope is 0, and the definition of inflexion point is that the concavity changes about the point).$

$For example, consider the function $f:\mathbb{R}\to \mathbb{R},f(x) = x^{\frac{5}{3}}$. Then clearly, $f^\prime (x) = \frac{5}{3}x^{\frac{2}{3}}$ and $f''(x) = \frac{5}{3}\cdot \frac{3}{3} x^{-\frac{1}{3}}$. Then $f^\prime (0)=\frac{5}{3}\cdot 0^{\frac{2}{3}}=0$, and $f''$ changes sign about the point $0$. So $x=0$ is a stationary point of inflexion for $f$. However, $f''$ is undefined at $x=0$. This shows that it is possible to have stationary points of inflexion even if $f''$ does not exist at the point in question.$

Hm, fair enough. Still have to familiarise myself with proper mathematical terminology.

davidgoes4wce · Nov 14, 2015

Re: HSC 2016 2U Marathon

From the 2011 HSC Exam paper, the answer for part (iii) in the solutions was 7 decibels. (10 ln 3)

I had an answer of 6.93 decibels. Would this be a value that gives me full marks?

InteGrand · Nov 14, 2015

Re: HSC 2016 2U Marathon

davidgoes4wce said:
From the 2011 HSC Exam paper, the answer for part (iii) in the solutions was 7 decibels. (10 ln 3)

I had an answer of 6.93 decibels. Would this be a value that gives me full marks?

You should always write the exact answer unless specified otherwise. Even when rounding the answer because they asked for a rounded value, you should write the answer out to many decimal places and put a "..." at the end of the long numeral, before rounding it, so that if you made a silly mistake in rounding, the marker can see what your value was.

davidgoes4wce · Nov 15, 2015

Re: HSC 2016 2U Marathon

Confirming there is a mistake with the integration from this solution in Coroneos Publications:

leehuan · Nov 15, 2015

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Confirming there is a mistake with the integration from this solution in Coroneos Publications:

That mistake is funny.
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Tbh though, I'm not sure how many of us really care, especially since when people do find the mistake they don't know to come here. This is just a forum which only some of the HSC community use. :x

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