HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

Paradoxica · Nov 9, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$$Prove $\int_0^1 (1-\sqrt[p]{x})^q dx = \int_0^1 (1-\sqrt[q]{x})^p dx$

Sy123 · Nov 9, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$Prove $\int_0^1 (1-\sqrt[p]{x})^q dx = \int_0^1 (1-\sqrt[q]{x})^p dx$

$\\ L = \int_0^1 (1-\sqrt[p]{x})^q \ dx$

$\\ x = u^p \ \Rightarrow \ L = \int_0^1 pu^{p-1} (1-u)^q \ du$

$\\ $By integrating by parts$ \ L = u^p(1-u)^q|_0^1 + \int_0^1 q(1-u)^{q-1} u^p \ du$

$\\ \Rightarrow \ L = \int_0^1 q(1-u)^{q-1} u^p \ du$

$\\ v = 1 - u \ \Rightarrow \ L = \int_0^1 qv^{q-1} (1-v)^p \ du$

$\\ z = v^q \ \Rightarrow \ L = \int_0^1 (1-\sqrt[q]{z})^p \ dz = \int_0^1 (1-\sqrt[q]{x})^p \ dx = R \ \square$

Sy123 · Nov 9, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $Let there be coins$ \ C_k \ $for each positive integer$ \ k \ $where the probability$ \ C_k \ $returns heads is$ \ \frac{1}{2k+1}$

$\\ $Let the probability that flipping the first$ \ n \ $coins returns an odd number of heads in total be$ \ p_n$

$\\ $Show that$ \ p_n = \frac{n}{2n+1}$

simpleetal · Nov 9, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let there be coins$ \ C_k \ $for each positive integer$ \ k \ $where the probability$ \ C_k \ $returns heads is$ \ \frac{1}{2k+1}$

$\\ $Let the probability that flipping the first$ \ n \ $coins returns an odd number of heads in total be$ \ p_n$

$\\ $Show that$ \ p_n = \frac{n}{2n+1}$

$$The base cases are trivial. Assume statement is true for n=k. Suppose we now have k+1 coins. To get an odd number of heads, we can either have an odd number of heads in the first k coins and toss a tails for the k+1th coin, or get an even number of heads in the first k coins and toss a head for the k+1th coin. So the probability of getting an odd number of heads in the k+1 coins is (k/(2k+1))*((2k+2)/(2k+3))+((k+1)/(2k+1))*(1/(2k+3))=(2k+1)(k+1)/((2k+1)(2k+3))=(k+1)/(2k+3). Hence, by induction true for n$\geq $1. \\\\New Question: \\\\i) By considering $(z-1/z)^5$ find sin5x in terms of $sinx, sin^5x, $and $sin3x.$\\ii) Hence find all solutions to sin5x=5cos3x in radians in the range $ 0\leq x\leq 4\pi$

Sy123 · Nov 9, 2015

Re: HSC 2016 4U Marathon - Advanced Level

simpleetal said:
$$The base cases are trivial. Assume statement is true for n=k. Suppose we now have k+1 coins. To get an odd number of heads, we can either have an odd number of heads in the first k coins and toss a tails for the k+1th coin, or get an even number of heads in the first k coins and toss a head for the k+1th coin. So the probability of getting an odd number of heads in the k+1 coins is (k/(2k+1))*((2k+2)/(2k+3))+((k+1)/(2k+1))*(1/(2k+3))=(2k+1)(k+1)/((2k+1)(2k+3))=(k+1)/(2k+3). Hence, by induction true for n$\geq $1. \\\\New Question: \\\\i) By considering $(z-1/z)^5$ find sin5x in terms of $sinx, sin^5x, $and $sin3x.$\\ii) Hence find all solutions to sin5x=5cos3x in radians in the range $ 0\leq x\leq 4\pi$

Yep well done

My approach was

$\\ p_{n} = \frac{1}{2n+1} (1-p_{n-1}) + \frac{2n}{2n+1}p_{n-1} \ $(same reasoning as you use)$$

$\\ (2n+1)p_n - (2n-1)p_n = 1$

Then the sum telescopes and we get n/(2n+1)

It's pretty much induction in disguise

dan964 · Nov 9, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
$i) x= p+\sqrt{q}$

$x^2-2px+p^2-q=0$

$x=p-\sqrt{q}$

$x^2-2px+p^2-q=0$

$$ Since Both equations are equal if $ p +\sqrt{q} $ is a root, then so is $ p=\sqrt{q}$

The problem with your solution is also the second line, you seem to introduce the root that you are looking for by squaring it.
If I am misguided, then you would be better just using quadratic formula on second line, producing the two desired roots.

dan964 · Nov 14, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$$New Question: \\\\i) By considering $(z-1/z)^5$ find sin5x in terms of $sinx, sin^5x, $and $sin3x.$\\ii) Hence find all solutions to sin5x=5cos3x in radians in the range $ 0\leq x\leq 4\pi$

FYI: this is the currently posted question

glittergal96 · Nov 17, 2015

Re: HSC 2016 4U Marathon - Advanced Level

The following question is as much a test of choosing good notation in possibly unfamiliar situations as it is of your problem solving abilities.

$A polynomial in multiple variables $P(x_1,\ldots,x_n)$ is a formal expression formed by combining sums and products of the monomials $x_i$ together with the operation of multiplying by a complex number. Examples in three variables include: $2x_1^2x_2x_3, ix_2^4, (x_1^2+x_2x_3)/5,x_1^2+x_2^2+x_3^2.$\\ \\ A polynomial is said to be symmetric if permuting some of it's variables leaves it unchanged. Only the last of the original examples is symmetric.\\ \\ The elementary symmetric polynomial $S_k$ of degree $k$ in $n$ variables is given by the sum of all $k$-fold products of the monomials $x_i$.\\ \\ Eg. $S_2(x,y,z)=xy+xz+yz$ \\ \\ Prove that a polynomial $P$ in several variables is symmetric if and only if it can be written as $Q(S_1,\ldots,S_n).$\\ \\Remark: This theorem is exactly the reason why we can evaluate symmetric polynomials in the roots of a polyomial in terms of the coefficients of the polynomial, as is very commonly asked in MX2 exams.$

If you cannot provide a complete proof, but think you have made progress or have a good understanding of why this theorem is true, you are welcome to post as well. The problem itself is of course not one an average MX2 student could be expected to solve.

Paradoxica · Nov 21, 2015

Re: HSC 2016 4U Marathon - Advanced Level

So how would this apply to special cases, such as the AM-GM symmetric polynomials?

glittergal96 · Nov 22, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
So how would this apply to special cases, such as the AM-GM symmetric polynomials?

Not too sure what you are asking here?

I assume by AM-GM polynomials, you mean:

$P_n(x_1,\ldots,x_n)=\frac{1}{n}\sum_{k=1}^n x_k^n-\prod_{k=1}^n x_k.$

(The AM-GM inequality then being the statement that P_n is non-negative if each x_k is non-negative).

Indeed P_n is symmetric and so this theorem applies, that is we can express the P_n in terms of the elementary symmetric polynomials.

Eg, for n=2,3:

$P_2=\frac{1}{2}(x_1^2+x_2^2)-x_1x_2=\frac{1}{2}(x_1+x_2)^2-x_1x_2=\frac{1}{2}S_1(x_1,x_2)^2-S_2(x_1,x_2).$

$P_3=\frac{1}{3}(x_1^3+x_2^3+x_3^3)-x_1x_2x_3=\frac{1}{3}(x_1+x_2+x_3)^3-(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)=\frac{1}{3}S_1(x_1,x_2,x_3)^3-S_1(x_1,x_2,x_3)S_2(x_1,x_2,x_3).$

Sy123 · Nov 23, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $The proposition in question can be split into two individual propositions$$

$\\ \textbf{(1)} $ Prove that polynomials of the form$ \ Q(S_1,S_2,\dots, S_n) \ $are symmetric$$

$\\ \textbf{(2)} $ Prove that for polynomials$ \ P \ $that are symmetric, there exists some$ \ Q \ $so that$ \ P(x_1, \dots , x_n) = Q(S_1, S_2, \dots , S_n)$

--------------------------

$\textbf{Proof of proposition (1)}$

$\\ $The proof will start with a proof of (1), consider the polynomial$ \ Q(S_1, S_2, \dots , S_n) = P(x_1, x_2, \dots , x_n) $. We mean by the operation$ \ \sigma \ $a permutation of a given sequence, that is$ \ \sigma (x) \ $is the permutation of$ \ x_1, x_2, \dots , x_n \ $and$ \ \sigma(R) \ $is a permutation of the variables of a multivariable polynomial$ \ R$

$\\ $Then,$ \ P(\sigma(x)) = Q(\sigma(S_1), \sigma(S_2), \dots , \sigma(S_n)) = Q(S_1, S_2, \dots , S_n) = P(x_1, x_2, \dots, x_n)$

$\\ $Thus,$ \ P \ $is equal to any permutation of its variables, and thus is symmetric$$

--------------------------

$\textbf{Proof of proposition (2)}$

$\\ $Now, to move onto proposition (2), the following will be given:$ \\\\ $Let the proposition that, if polynomial$ \ P(x_1, \dots , x_n) \ $is symmetric, with number of variables$ \ n \ $and maximum degree$ \ d \ $then, it can be written in the form$ \ Q(S(n,1), S(n,2), \dots , S(n,n))$, be$ \ \rho (n,d)$

$\\ $Note: The maximum degree of the polynomial is simply the maximum of the degree of each individual variable$ \\ $The function$ \ S(n,k) \ $is simply$ \ S_k \ $on the variables$ \ x_1, x_2, \dots , x_n$

$\\ $Now, it will be proven that:$$

$\\ \textbf{(2.1)} \ \rho(n,1) \ $is true for all integers$ \ n\geq 1$

$\\ \textbf{(2.2)} \ \rho(1,d) \ $is true for all integers$ \ d \geq 1$

$\\ \textbf{(2.3)} \ \rho(n+1,d-1) \wedge \rho(n,d) \Rightarrow \rho(n+1,d)$

$\\ \textbf{(2.4)} $ Therefore, we conclude by induction,$ \ \rho(n,d) \ $is true for all$ \ n,d$

$\\ $Note: The symbol$ \ \wedge \ $simply means 'AND'$$

$\\ $The validity of the inductive conclusion will be assumed, it's truth is clear to see by drawing a table of truth values for each integer$ \ n,d \ $and seeing that each cell in the table has a 'chain' back to the base cases$$

--------------------------

$\textbf{Proof of proposition (2.1)}$

$\\ $Proof of (2.1) will be split into smaller steps to make the proof clearer$$

$\\ \textbf{(2.1.1)} $ Every symmetric polynomial can be written as the sum of polynomials of the form$ \ C_\alpha (x_1, x_2, \dots , x_n) = \sum_{\beta \in \sigma (\alpha)} x_1^{\beta_1} x_2^{\beta_2} \dots x_n^{\beta_n} \ $where$ \ \beta \ $is some element of the set of permutations for some sequence$ \ \alpha \ $specific to polynomial$ \ C_\alpha$

$\\ \textbf{(2.1.2)} $ Polynomials of the form$ \ C_\alpha \ $given above for some sequence$ \ \alpha \ $, when they have degree 1, then$ \ C_\alpha (x_1, x_2, \dots , x_n) = S_k \ $for some integer$ \ k$

$\\ \textbf{(2.1.3)} $ Therefore, from (2.1.1) and (2.1.2), we conclude that the degree 1 polynomial,$ \ P(x_1,x_2, \dots , x_n) = \sum_{i \in X} S_i \ $where$ \ X \ $is an appropriate set of integers, and hence$ \ P(x_1,x_2, \dots , x_n) = Q(S_1, S_2, \dots , S_n) \ $for degree 1, and hence$ \ \rho(n,1) \ $is true$$

--------------------------

$\textbf{Proof of proposition (2.1.1)}$

$\\ $Starting with a proof of (2.1.1), consider that there exists a term in the polynomial$ \ P(x_1, x_2, \dots ,x_n) \ $that is$ \ c\cdot x_1^{\alpha_1} x_2^{\alpha_2} \dots x_n^{\alpha_n}$

$\\ $Since the polynomial is identical given the permutation of variables (as it is symmetric), then since the term$ \ c\cdot x_1^{\alpha_{\sigma(1)}} x_2^{\alpha_{\sigma(2)}} \dots x_n^{\alpha_{\sigma(n)}} \ $exists in$ \ P(\sigma(x)) \ $and$ \ P(\sigma(x)) = P(x_1, x_2, \dots, x_n) \ $then the permuted term also exists in$ \ P(x_1, x_2, \dots x_n) \ $and since multiple kinds of terms can exist in$ \ P \ $then$ \ P \ $can be written in the form:$$

$\\ P(x_1, x_2, \dots x_n) = \sum_{\alpha \in S_P} c_\alpha \sum_{\beta \in \sigma(S)} x_1^{\beta_1} x_2^{\beta_2} \dots x_n^{\beta_n}$

$\\ $Where$ \ S \ $are all the associated sequences for the polynomial$ \ P$

$\\ $Then$ \ P \ $can be written as a linear combination of polynomials of the form$ \ C = \sum_{\beta \in \sigma(S)} x_1^{\beta_1} x_2^{\beta_2} \dots x_n^{\beta_n}$

--------------------------

$\textbf{Proof of proposition (2.1.2)}$

$\\ $The proof of (2.1.2) is simple, since$ \ C(x_1,x_2, \dots x_n) = \sum_{\beta \in \sigma(\alpha)} x_1^{\beta_1} x_2^{\beta_2} \dots x_n^{\beta_n}$

$\\ $Now, we are free to choose$ \ \alpha \ $as long as the maximum number in this sequence (the degree) is 1$$

$\\ $Then, if$ \ \alpha \ $has$ \ k \ $1s and$ \ n-k \ $0s, then since every permutation is exhausted in$ \ C \ $then$ \ C = S_k$

--------------------------

$\textbf{Proof of proposition (2.2)}$

$\\ $The proof of (2.2) is similar, but shorter, we simply see that a multivariate polynomial of 1 variable, is simply$ \ P(x) = \sum_{i=0}^d c_i x^i = \sum_{i=0}^d c_i S_1^i = Q(S_1) \ $thus, proposition$ \ \rho(1,d) \ $is true$$

--------------------------

$\textbf{Proof of proposition (2.3)}$

$\\ $The proof of proposition (2.3) follows:$$

$\\ $Since, by$ \ (2.1.1) \ $then all symmetric polynomials$ \ P \ $can be written in the form of$ \ C_\alpha \ $for sequences$ \ \alpha \ $associated with$ \ P \ $then, denote$ \\\\ P(x_1,x_2, \dots , x_n) = \sum_{\alpha \in S(P)}c_\alpha \sum_{\beta \in \sigma{\alpha}} x_1^{\beta_1}x_2^{\beta_2} \dots x_n^{\beta_n}$

$\\ $Here,$ \ S(P) \ $simply denotes all the associated sequences with polynomial$ \ P \ $and associated constant$ \ c_\alpha$

$\\ $Then,$ \ P(x_1,x_2,\dots , x_n) - P(x_1, x_2, \dots , x_{n-1}, 0) = \sum_{\alpha \in \chi}c_\alpha \sum_{\beta \in \sigma{\alpha}} x_1^{\beta_1}x_2^{\beta_2} \dots x_n^{\beta_n}$

$\\ $Here,$ \ \chi \ $denotes the subset of$ \ S(P) \ $in which the sequence does not contain a zero, i.e. those sequences in which every variable is present in every sum$$

$\\ $Therefore, by the definition, we see that$ \ x_1x_2\dots x_n \ $is a factor of$ \ P(x_1,x_2,\dots , x_n) - P(x_1, x_2, \dots , x_{n-1}, 0)$

$\\ $So,$ \ P(x_1,x_2, \dots , x_{n+1}) = P(x_1, x_2, \dots , x_{n}, 0) + S(n+1,n+1) R(x_1, x_2, \dots x_{n+1}), \ (*)\ $for some polynomial$ \ R$

$\\ $Then, we see that$ \ P \ $is a polynomial, being of degree$ \ d \ $it is clear that$ \ R \ $has degree$ \ d-1$

$\\ $Suppose$ \ \rho(n,d) \ $is true, then$ \ P(x_1, x_2, \dots , x_n, 0) \ $can be written in the form$ \ Q(S(n,1), \dots , S(n,n))$

$\\ $Suppose also that$ \ \rho(n+1,d-1) \ $is true, then$ \ R \ $can be written in the form$ \ Q^{*}(S(n+1,1) \dots , S(n+1,n+1))$

$\\ $Since, any polynomial of the form$ \ Q(S(n,1), \dots , S(n,n)) \ $is a polynomial of the form$ \ Q^{*}(S(n+1,1) \dots , S(n+1,n+1)) \ $by simply setting$ \ x_{n+1} = 0$

$\\ $Then, by equation$ \ (*) \ $the supposition of$ \ \rho(n+1,d-1) \ $and$ \ \rho(n,d) \ $implies that$ \ P \ $can be written in the form$ \ Q^{*}(S(n+1,1) \dots , S(n+1,n+1)) \ $implying the proposition$ \ \rho(n+1,d)$

$\\ $Therefore,$ \ \rho(n+1,d-1) \wedge \rho(n,d) \Rightarrow \rho(n+1,d)$

--------------------------

$\\ $Then, by conclusion (2.4) the proposition$ \ (2) \ $has been proven, and thus the proposition in question has been proven$$

glittergal96 · Nov 23, 2015

Re: HSC 2016 4U Marathon - Advanced Level

This seems close to the right idea, but not quite I think.

To start with, $P(x_1,\ldots,x_n)-P(x_1,\ldots,x_{n-1,0})$ is NOT necessarily divisible by $x_1\ldots x_n$ (the error in your argument being that we cannot write this difference using the set $\chi$ you introduced. A simple counterexample is $P(x,y)=x^2+y^2$ ).

It is however true that $x_n$ must divide this difference, so you might be able to do something with that.

Also a quick note that the degree of a monomial in several variables normally refers to the sum of the indices, rather than the maximum of them. (This notion of degree interacts with multiplication etc in a nicer way.)

So, assuming I am correct in interpreting your overall proof strategy as:

I) Write $p(x,y,z)=p(x,y,0)+q(x,y,z)$
II) deduce q is a symmetric polynomial of lesser degree than p (your definition of degree), and $p(x,y,0)$ is a symmetric polynomial of the same degree in fewer variables than $p(x,y,z)$
III) use the inductive hypotheses to write the two summands as polynomials in $x+y+z, xy+xz+yz, xyz$

My objections are:
-By my initial paragraph, q is not necessarily of lesser degree. (Nor is it necessarily symmetric in all three variables).
-p(x,y,0) is a symmetric polynomial in the two variables x and y, and so by hypothesis can be written as a polynomial in x+y and xy, but this does not mean that we can write it is a polynomial in x+y+z, xy+xz+yz, xyz. In fact such a thing is impossible unless p is constant.

Both of these objections are evident in the example $P(x,y)=x^2+y^2.$

---------------------------------------------------------------------------------------------

Such an induction (which could alternatively be viewed as in nested induction to prove a statement about the cells in a countable square array by first proving the statement for the first row by using induction, and then proving it over the whole array row-by-row by using another induction) is definitely the idea behind the kind of proof I have in mind here though.

dan964 · Nov 23, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Since P(x_1...x_n) is a permutation of different variables added or multiplied together
Q(S1...Sn) implies that permutation of different Sn added or multipled together.

It surfices to show that only permutations of the form Q(S1...Sn) are symmetric.
I don't honestly think I would be able to do it.

dan964 · Nov 23, 2015

Re: HSC 2016 4U Marathon - Advanced Level

An even simpler counter-example is $S_1(x_1 ... x_n)$ itself.

I came across this
that $S_{k}(x_{1}, x_{2}....x_{n})=S_{n}(x_{1}, x_{2}....x_{n}) S_{k-n}(x_{1}, x_{2}....x_{n})$
Note that for k>n it works nicely. For k-n
You have to define $S_{-k}(x_{1}, x_{2}....x_{n}) = (\frac{1}{x_{1}}, \frac{1}{x_{2}}....\frac{1}{x_{n}})$
and $S_{0}(x_{1}, x_{2}....x_{n}) = 1$

Here is the related result (I looked up the topic on the web). If anyone is able to derive this, apparently from here is should be able to done easily.
$\sum_{i=0}^k(-1)^iS_i(X_1,\ldots,X_n)P_{k-i}(X_1,\ldots,X_n)=0,$
Doesn't apply obviously to those counter examples
$x^{2}+y^{2}$ which is $S_{2}(x^{2}, y^{2})$

Not a fool-proof way, but I can now understand what Sy123 was trying to do.

note: this is not a question, just me trying to understand a previous question.

Paradoxica · Nov 24, 2015

Re: HSC 2016 4U Marathon - Advanced Level

omegadot said:
$Let $ u, v \in \mathbb C $ be the roots of the equation$

$z + \frac{1}{z} = 2(\cos{\alpha} + i\sin{\alpha}), 0 < \alpha< \pi$

$$(a) Show that $ u+i $ and $ v+i $ have the same argument and that$

$u-i $ and $ v-i $ have the same modulus.$$

$$(b) Find the locus of the roots $ u, v $ when $ \alpha $ varies from $ 0 $ to $ \pi.$

Nobody ever answered this in the 2015 reg marathon and I have trouble finding someone who can do it, so I'll just put it here.

glittergal96 · Nov 24, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
Nobody ever answered this in the 2015 reg marathon and I have trouble finding someone who can do it, so I'll just put it here.

ai) We are solving the quadratic p(z)=z^2-2wz+1=0, where w is in the open upper half unit circle.

The complex numbers u+i and v+i then must be roots of q(z)=p(z-i)=z^2-2(i+w)z+2wi.

It suffices then to prove that the roots of q lie on the same open ray from the origin.

A useful substitution here is z=2(i+w)y. If you cannot see from the form of q why this should be useful, one can observe that if the roots of q both lie on the same open ray, then their sum 2(i+w) must lie on this same ray.

Our new polynomial is r(y)=q(z)=q(2(i+w)y)=4y(y-1)(w+i)^2+2wi.

Dividing by (w+i)^2, we see that the roots of r satisfy 4y^2-4y+2wi/(w+i)^2=0.

But 2wi/(w+i)^2=((w/i+i/w)/2+1)^(-1). Since w is of unit modulus and lies in the upper half plane, we can conclude that 2wi/(w+i)^2 is a real number between 0 and 1.

The quadratic formula then implies that the roots of r are both positive real numbers. Since our change of variables from z to y just corresponded to a rotation about zero, we have proven that the roots of q lie on the same open ray from the origin, and hence that arg(u+i)=arg(v+i).

ii) We proceed similarly. Our definition of q is now:
q(z)=p(z+i)=z^2+2(i-w)z-2wi.

Our substitution is z=2(i-w)y, so

q(z)=4y(y+1)(i-w)^2-2wi.

Dividing the equation be (i-w)^2, we are left with

4y^2+4y-2wi/(w-i)^2.

Since w has unit modulus and is in the upper half plane, we have 2wi/(w-i)^2=((w/i+i/w)/2-1)^(-1) which is real and strictly less than -1.

Hence our equation in y has negative discriminant. As it's coefficients are real, we must conclude that its roots are a pair of conjugate complex numbers, and hence of equal modulus. (Again, this property is invariant under the rotational change of variables.)

b) It doesn't quite make sense to talk individually about the locii of u and v, as for each the alpha there is an arbitrary choice in which to label u. We can however talk about the locus of z satisfying the original equation.

We have |z+1/z|=2 so

|(x+iy)+(x-iy)/(x^2+y^2)|^2=4

=> |x(1+1/(x^2+y^2))+iy(1-1/(x^2+y^2))|^2=4

=> x^2(x^2+y^2+1)^2+y^2(x^2+y^2-1)^2=4(x^2+y^2)^2

=> x^4+y^4+2x^2y^2-2x^2-6y^2+1=0.

=> (x^2+(y-1)^2-2)(x^2+(y+1)^2-2)=0.

That is, the locus lies on the union of the two circles centred at i and -i respectively with radius sqrt(2).

However, note that as alpha varies, we do not hit this full union of two circles, as we have the additional constraint that z+1/z has positive imaginary part, that is y(x^2+y^2-1)>0. So we have to exclude the region outside the open unit disk and below the real axis, as well as the region inside the unit disk and above the real axis.

This leaves us with the circle centred at i with radius sqrt(2), excluding the points +-1.

omegadot · Nov 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$$\noindent Let $P(x)$ be a polynomial with positive coefficients. Show that if$\\\begin{align*}P \left (\frac{1}{x} \right ) &\geqslant \frac{1}{P(x)}\end{align*}\\$ holds for $x = 1$, then it holds for all $x > 0$.$

Paradoxica · Nov 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

omegadot said:
$$\noindent Let $P(x)$ be a polynomial with positive coefficients. Show that if$\\\begin{align*}P \left (\frac{1}{x} \right ) &\geqslant \frac{1}{P(x)}\end{align*}\\$ holds for $x = 1$, then it holds for all $x > 0$.$

The Cauchy-Schwarz Inequality is not something that can be cited in the exam.

Drsoccerball · Nov 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
The Cauchy-Schwarz Inequality is not something that can be cited in the exam.

Which is why this is known as the "Advanced Level" thread.

lita1000 · Nov 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
The Cauchy-Schwarz Inequality is not something that can be cited in the exam.

uhhhh Cauchy Schwarz isn't needed to solve that question mate.

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