HSC 2012 MX2 Marathon (archive) (2 Viewers)

httton · May 27, 2012

Re: 2012 HSC MX2 Marathon

Hey guys just two questions

1/ what are some ways to determine whether a question should be done with shells or slices
2/ if I'm given a shape like x^2(6-x^2) find area enclosed in 1st quadrant rotated about the y axis how do I go about it?

bleakarcher · May 27, 2012

Re: 2012 HSC MX2 Marathon

^ cylindrical shells would be easier there

Carrotsticks · May 27, 2012

Re: 2012 HSC MX2 Marathon

httton said:
Hey guys just two questions

1/ what are some ways to determine whether a question should be done with shells or slices
2/ if I'm given a shape like x^2(6-x^2) find area enclosed in 1st quadrant rotated about the y axis how do I go about it?

Generally, rotation around a line parallel to the Y axis is more easily done using Shells.

X axis is more easily done using Slicing.

But of course both will yield correct answers (but one will yield a harder integral than the other).

Example: Rotate the area bounded by the curve y=-x(x-1) and the X axis, around the line x=2.

If you use Slicing, you will need to eventually use the Quadratic formula. and then integrate that expression.

httton · May 27, 2012

Re: 2012 HSC MX2 Marathon

haha i went out to have a good think and came back with the same idea!

for the others, essentially try to have your slice correspond to 1 x to 1 y values each

seanieg89 · May 28, 2012

Re: 2012 HSC MX2 Marathon

When thinking about whether shells or annuli will be more efficient, consider the resulting integrands. There will generally be a sort of inverse relation between our integrands (as we are integrating in perpendicular directions for these two methods), and we prefer to have things like tan(x) in our integrand rather than tan^{-1}(x).

Trebla · May 30, 2012

Re: 2012 HSC MX2 Marathon

This one might look like it is beyond the course but it can be solved using methods within the scope of Ext2 level.

Suppose that f(x) is an even function. Show that

$\displaystyle\int_{-a}^a f(x)(\cos x + i\sin x)\,dx = 2\displaystyle\int_{0}^a f(x)\cos x\,dx$

Hence evaluate

$\displaystyle\int_{0}^{\pi} \left|\dfrac{\pi}{2} - x\right|(\sin x + i\cos x)\,dx$

Fus Ro Dah · May 31, 2012

Re: 2012 HSC MX2 Marathon

Trebla said:
This one might look like it is beyond the course but it can be solved using methods within the scope of Ext2 level.

Suppose that f(x) is an even function. Show that

$\displaystyle\int_{-a}^a f(x)(\cos x + i\sin x)\,dx = 2\displaystyle\int_{0}^a f(x)\cos x\,dx$

Hence evaluate

$\displaystyle\int_{0}^{\pi} \left|\dfrac{\pi}{2} - x\right|(\sin x + i\cos x)\,dx$

First part split the integral as two terms then the integral with the i in it cancels out due to the sine function and we are left with 2x integral of f(x)cos(x). Second part use the property f(x) = f(a-x), add the two integrals, the icos(x) term cancels and the result follows.

I am a bit skeptical regarding this question being within Ext2 level without proper justification because we are taking the Riemann integral of a complex-valued function. It may require justification that it works in this case because $\mathbb{C}$ is a vector space. Other students might get confused about having an i in the integral.

OMGITzJustin · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

Fus Ro Dah said:
First part split the integral as two terms then the integral with the i in it cancels out due to the sine function and we are left with 2x integral of f(x)cos(x). Second part use the property f(x) = f(a-x), add the two integrals, the icos(x) term cancels and the result follows.

I am a bit skeptical regarding this question being within Ext2 level without proper justification because we are taking the Riemann integral of a complex-valued function. It may require justification that it works in this case because $\mathbb{C}$ is a vector space. Other students might get confused about having an i in the integral.

yeah I agree with with you, the question seems like it isnt in the scope of the ext 2 course

Trebla · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

Okay made it heaps easier...

Suppose that f(x) is an even function and k is a constant. Show that

$\displaystyle\int_{-a}^a f(x)(\cos x + k\sin x)\,dx = 2\displaystyle\int_{0}^a f(x)\cos x\,dx$

Hence evaluate

$\displaystyle\int_{0}^{\pi} \left|\dfrac{\pi}{2} - x\right|(\sin x + k\cos x)\,dx$

bleakarcher · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

can you treat 'i' as a constant?

funnytomato · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
can you treat 'i' as a constant?

why not?

bleakarcher · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

wasnt sure..

seanieg89 · Jul 12, 2012

Re: 2012 HSC MX2 Marathon

Might as well bring this back to life.

Prove that the polynomial

$p(x)=\sum_{k=0}^{2n} \frac{x^k}{k!}$

has no real roots. (Where n is a non-negative integer.)

karnbmx · Jul 12, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Might as well bring this back to life.

Prove that the polynomial

$p(x)=\sum_{k=0}^{2n} \frac{x^k}{k!}$

has no real roots. (Where n is a non-negative integer.)

I don't know if this is right, but I'll give it a shot.

$P(x) = p(x)=\sum_{k=0}^{2n} \frac{x^k}{k!}$
Expanding, we get:

$P(x) = 1 + x + \frac{x^2}{2!} + \frac{x^4}{4!} + .... + \frac{x^2^n}{2n!}$

Thus for P(x) to be zero, the terms after 1 must add up to -1. Hence:

$x + \frac{x^2}{2!} + \frac{x^4}{4!} + .... + \frac{x^2^n}{2n!} = -1$

BUT, there is no real value of x for which the above sum can possibly equal to -1. Therefore, there are no real roots for P(x).

EDIT: say. if you put x = -2, the value of the sum will be a limit that approaches -1, but never reaches -1. this would be the same for all x = m, where the sum will approach m + 1.

seanieg89 · Jul 12, 2012

Re: 2012 HSC MX2 Marathon

Not quite, the polynomial does not contain only even powers...some justification would be needed to say that the sum of the nonconstant terms can never be -1. (I also do not understand what you are saying about limits.)

A hint: One way to do it is by induction.

karnbmx · Jul 12, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Not quite, the polynomial does not contain only even powers...some justification would be needed to say that the sum of the nonconstant terms can never be -1. (I also do not understand what you are saying about limits.)

A hint: One way to do it is by induction.

Alright, I'll have to think about it then. :S

Godmode · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Might as well bring this back to life.

Prove that the polynomial

$p(x)=\sum_{k=0}^{2n} \frac{x^k}{k!}$

has no real roots. (Where n is a non-negative integer.)

Let's see if this works...

$p(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^{2n}}{2n!}$

We know that n is an integer, so 2n must be even. Therefore, p(x) is an even polynomial, as it has a degree of 2n.
From the fundamental theorem of algebra, we also know that there are 2n roots of p(x), therefore we have an even number of roots.

Now the product of all roots (which I am going to call: $\alpha_{i}$ ) is

$\prod_{i=1}^{2n} \alpha_{i} = \alpha_{1}\alpha_{2}\alpha_{3} \times... \times \alpha_{2n} = (-1)^{2n} \frac{a_{2n}}{a}$

Where a and a_2n are the coefficients of the first and last term respectively (I flipped p(x) around so the x^2n/2n! is the first term)
We know that the a_2n/a must't have a negative 1 out the front, because p(x) is even. For example, look at a quadratic or quartic. The product of roots being c/a and e/a respectively.

Subbing in values for a and a_2n:

$\prod_{i=1}^{2n} \alpha_{i} = (1) \times\frac{1}{\frac{1}{2n!}} = 2n!$

Now if we take a look at the sum of roots taken 2n-1 at a time (i.e. the product divided by a single root) we find that:

$\sum_{j=1}^{2n} \frac{\prod_{i=1}^{2n} \alpha_{i}}{\alpha_{j}} = -\frac{a_{2n-1}}{a_{1}} =-2n!$

As a_2n = a_(2n-1) = 1.

Now, if we take a closer look at the above sum, we find it can be expressed as:

$\begin{align*} \sum_{j=1}^{2n} \frac{\prod_{i=1}^{2n} \alpha_{i}}{\alpha_{j}} &= \frac{\prod_{i=1}^{2n} \alpha_{i}}{\alpha_{1}} + \frac{\prod_{i=1}^{2n} \alpha_{i}}{\alpha_{2}} + ... + \frac{\prod_{i=1}^{2n} \alpha_{i}}{\alpha_{2n}} \\ &= \frac{2n!}{\alpha_{1}} + \frac{2n!}{\alpha_{2}} + ... + \frac{2n!}{\alpha_{2n}} \\ &= -2n! \end{align*}$

Cancelling the 2n!, we get

$\frac{1}{\alpha_{1}} + \frac{1}{\alpha_{2}} + ... + \frac{1}{\alpha_{2n}} = -1$

now:

$\text{Let} \: \alpha_{i} = \beta_{i}^2 \\ \text{Now, we are going to Substitute} \: \beta_{i} \: \text{for} \: \alpha_{i}} \\ \\ \text{i.e.} \: \frac{1}{\beta_{1}^2} + \frac{1}{\beta_{2}^2} + ... + \frac{1}{\beta_{2n}^2} = -1$

In order for this statement to be true, $\beta_{i}$ must be complex.

i.e. $\beta_{i} = c + di , \text{Where c, d} \in \mathbb{R}$

This means that $\alpha_i = (c + di)^2 = c^2 - d^2 + 2icd$

In order for $\alpha_i$ to be complex, both c and d =! 0.
d =! 0 is obvious, as $\beta_{i}$ would not be complex if d = 0.

Now, we must show that c =! 0

$\text{Assume} \: c = 0 \\ \therefore \: \alpha_{i} = -d^2\\ \text{Now,} \: \frac{1}{-d_{1}^2} + \frac{1}{-d_{2}^2} + ... + \frac{1}{-d_{2n}^2} = -1$
(Where each d_i belongs to its respective $\alpha_i$ )

Cancelling the negative sign,

$\frac{1}{d_{1}^2} + \frac{1}{d_{2}^2} + ... + \frac{1}{d_{2n}^2} = 1$

Although this looks plausible, d^2 > 1 for it to hold true.

i.e. d < -1 & d > 1 , but d is still not equal to 0

Now, as

$\alpha_i = -d_i^2$

We can substitute this back into the original sum.

$\begin{align*} 0 = p(-d_i^2) &= \sum_{k=0}^2n \frac{(-d_i^2)^k}{k!} \\ &= 1 -d_i^2 + \frac{-d_i^4}{2!} + \frac{-d_i^6}{3!} + ... + \frac{-d_i^4n}{2n!} \end{align*} \\ \therefore \: d_i^2 + \frac{d_i^4}{2} + \frac{d_i^6}{6} + ... + \frac{d_i^4n}{2n!} = 1$

Now, in order for this expression to hold true, d^2 < 1 or -1< d<1.

However, we already know that d^2 > 1. This leads us to a contradiction; how can d^2 be both less than and greater than 1? This means that our initial assumption, c = 0 is false.

As c, d =! 0, This means that $\alpha_i$ is complex.

$\therefore \: p(x)$ has no real roots.

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This goddamn better be right, I spent all night on this...

EDIT: Hang on, I don't think I can sub in d^2 into p(x) like that, the even powers will cancel the negative sign

seanieg89 · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

Also, the sum of the squares of the reciprocals of the \beta_i's being negative does not imply that every \beta_i has nonzero imaginary part, it only implies that SOME of the \beta_i's have nonzero imaginary part.

Eg (i*sqrt(2))^2+1^2=-1<0 even though 1 is real.

Godmode · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

Damn, guess I will have to start again lol

Does the fact that the limit of the sum as 2n approaches infinity = e^x have anything to do with the question?

seanieg89 · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

Nope, at least not the way I did it.

(And I don't see how that fact would be helpful. Remember, for a fixed n, this polynomial will resemble e^x less and less the further you get from x=0, so the properties of the function e^x have little bearing here.)

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