HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Oct 3, 2013

Re: HSC 2013 4U Marathon

$\sum_{k=1}^n \frac{\cos (kx)}{\cos^k x}$

JJ345 · Oct 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$\sum_{k=1}^n \frac{\cos (kx)}{\cos^k x}$

Do we have to use Euler's method and compare real part etc. or is there a MX2 way to do it ?

Sy123 · Oct 4, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Do we have to use Euler's method and compare real part etc. or is there a MX2 way to do it ?

What do you mean by euler's method?

there is a MX2 way of doing it

JJ345 · Oct 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
What do you mean by euler's method?

there is a MX2 way of doing it

Euler's Formula sorry, applying
e^{iy}=cos y+isin y
But if you didn't think of it, I probably did something retarded...

Sy123 · Oct 4, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Euler's Formula sorry
e^{iy}=cos y+sin y
But if you didn't think of it, I probably did something retarded...

Well the HSC method would be very similar to it, in your method, just change every e^(iy) into cos y + i sin y, and it will still work out with De Moivere's

seanieg89 · Oct 5, 2013

Re: HSC 2013 4U Marathon

$Suppose $p(x)$ is a polynomial with rational coefficients such that whenever $p(x)$ is rational, $x$ is rational. What possible degrees can $p$ have?$

Sy123 · Oct 6, 2013

Re: HSC 2013 4U Marathon

$f(x) \ \ $is an increasing function over all x reals$ \ \ f(2013) = 2013$

$Describe the graph of$

$g(x) = \frac{af(x) +b}{cf(x) + d} \ \ $for$ \ \ ad = bc$

gustavo28 · Oct 7, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Suppose $p(x)$ is a polynomial with rational coefficients such that whenever $p(x)$ is rational, $x$ is rational. What possible degrees can $p$ have?$

$$Call a polynomial $P(x)$ good if it satisfies the conditions in the question. So, if $P(x)$ is good, then $aP(x)+b$ where $a,c$ is rational will also be good. Moreover, if $P(a)=0$, then a must be rational. So, all of the roots of a good polynomial must also be rational. Now, from what I can see, you probably need to use the rational root theorem or some variant to solve this problem. The theorem states that if an integer polynomial P has a rational root $p/q$ where $gcd(p,q)=1$, then p divides the constant coefficient and q divides the leading coefficient. You can prove this fairly quickly (we did this in class as an example once) but it involves a little bit of number theory. Now, all linear polynomials with rational coefficients are good so we'll move on to polynomials with degree greater than 1. Suppose there is a good polynomial P with degree greater than 1 that is good.$

gustavo28 · Oct 7, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Suppose $p(x)$ is a polynomial with rational coefficients such that whenever $p(x)$ is rational, $x$ is rational. What possible degrees can $p$ have?$

Multiply it by a sufficiently large integer to make it an integer polynomial Q(x) which is also good. Let the leading coefficient of Q be A and the constant be C. Now, considering the graph of Q(x)-C, we can make a prime p large enough so that Q(x)+p-c is above the x-axis for -1<=x<=1. We can also make p a prime that does not divide A. So, let such a prime be called q. Now, Q(x)+q-C where q is prime is also good and hence has only rational roots. Its constant term is q. Consider a root of this polynomial c/d. So, d divides A and, since q does not divide A, q doesn't divide d. However, c divides q. So, c=1 or c=q. If c=1, then -1<=c/d<=1 which is a contradiction since Q(x)+q-c is above the x-axis for such values of x. So, c=q.

Hence, the product of the roots is equal to q^n/stuff where stuff is an integer with no factors of q in it and n is the degree of Q. This is equal to q/A or -q/A. So, q^(n-1)*A=stuff but since n>1, q divides an integer with no factors of q in it.

Contradiction. SO the only good polynomials are linear.

Sy123 · Oct 7, 2013

Re: HSC 2013 4U Marathon

$A function can be expressed as$

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$

$$Where$ \ \ f^{(n)} (x) = \frac{d^n}{dx^n} f(x) \ \ f^{(0)}(x) = f(x)$

$Express$ \ \ e^x \sin x \ \ $in this form$

Sy123 · Oct 8, 2013

Re: HSC 2013 4U Marathon

$$Prove that for positive integers$ \ \ n$

$\left(\frac{n+1}{2} \right)^n \geq n!$

$You may not use the inequality$

$\sum a_k \geq \left( \prod a_k \right)^{1/k}$

Sy123 · Oct 10, 2013

Re: HSC 2013 4U Marathon

$i) A line segment of radius$ \ \ r \ \ $is rotated about the fixed origin by a small angle$

$\delta \theta \ \ $sweeps out an area of$ \ \ \delta A \ \ \ $Show that$$

$\delta A = \frac{1}{2} r^2 \delta \theta$

$ii) Hence show that for some curve with the distance of a variable point on$

$$this curve to the origin is defined by the function$ \ \ r(\theta)$

$The area sweeped out from$ \ \ y=0 \ \ $to$ \ \ y=x\tan \alpha \ \ $is given by$

$A = \int_0^{\alpha} \frac{1}{2}r^2 \ d\theta$

$$iii) For the plane curve$ \ \ x^{2/3} + y^{2/3} = a^{2/3} \ \ $the line$ \ \ y=\frac{x}{\sqrt{3}}$

$is rotated about the origin by$ \ \ \frac{\pi}{6} \ \ $radians$

$Find an expression for the distance from a variable point on the curve$

$to the origin, and using the method above, find the area sweeped out$

---

Not sure if I 'proved' the method correctly lol

Sy123 · Oct 12, 2013

Re: HSC 2013 4U Marathon

Here is a good one:

$Let$ \ \ p(x) \ \ $be an even degree polynomial with a positive leading co-efficient such that$

$p(x) - p''(x) \geq 0 \ \ \ \ $for all real values of$ \ \ x$

$$Prove that$ \ \ p(x) \geq 0$

JJ345 · Oct 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Here is a good one:

$Let$ \ \ p(x) \ \ $be an even degree polynomial with a positive leading co-efficient such that$

$p(x) - p''(x) \geq 0 \ \ \ \ $for all real values of$ \ \ x$

$$Prove that$ \ \ p(x) \geq 0$

Let p(x) have its minimum value point at Q.
So p(x)>= p(Q) (1)
p(Q)>=p''(Q) ...given (2)
p''(Q)>=0 (because its the concavity at the min point of even degree positive coeff. polynomial)...(3)
So combining (1), (2) & (3)
We get that p(x)>=0

I think my (3) is pretty dodgy...

Sy123 · Oct 13, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
Let p(x) have its minimum value point at Q.
So p(x)>= p(Q) (1)
p(Q)>=p''(Q) ...given (2)
p''(Q)>=0 (because its the concavity at the min point of even degree positive coeff. polynomial)...(3)
So combining (1), (2) & (3)
We get that p(x)>=0

I think my (3) is pretty dodgy...

That is the sketch of the proof, but can you explain why we get the conclusion from (1) (2) and (3)?

JJ345 · Oct 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
That is the sketch of the proof, but can you explain why we get the conclusion from (1) (2) and (3)?

I don't get what you want me to explain.
Like actually define a polynomial and then do some random algebra or just explain what I did in more words?

Sy123 · Oct 13, 2013

Re: HSC 2013 4U Marathon

JJ345 said:
I don't get what you want me to explain.
Like actually define a polynomial and then do some random algebra or just explain what I did in more words?

Wait yea nevermind my method was a little different to yours, it involved a proof by contradiction (kinda).
Well done
---

$A function is defined for$ \ \ x,y > 0 \ \ $such that$

$f(xy) = f(x) + f(y)$

$$If$ \ \ f'(1) = 3 \ \ $find$ \ \ f(x)$

Sy123 · Oct 13, 2013

Re: HSC 2013 4U Marathon

$Evaluate (which you may assume converges)$

$\lim_{n \to \infty} \left(\prod_{k=1}^n \left(1+ \frac{k}{n} \right) \right)^{1/n}$

(can definitely be done within syllabus)

RealiseNothing · Oct 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Evaluate (which you may assume converges)$

$\lim_{n \to \infty} \left(\prod_{k=1}^n \left(1+ \frac{k}{n} \right) \right)^{1/n}$

(can definitely be done within syllabus)

You can prove that it converges easily as:

$(1+\frac{k}{n}) \leq 2$

$(2^n)^{\frac{1}{n}} = 2$

$\lim_{n \to \infty} \left(\prod_{k=1}^n \left(1+ \frac{k}{n} \right) \right)^{1/n} < 2$

Sy123 · Oct 13, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
You can prove that it converges easily as:

$(1+\frac{k}{n}) \leq 2$

$(2^n)^{\frac{1}{n}} = 2$

$\lim_{n \to \infty} \left(\prod_{k=1}^n \left(1+ \frac{k}{n} \right) \right)^{1/n} < 2$

True (provided you can prove the product is not periodic) I didn't think about convergence anyway haha

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