HSC 2015 MX2 Marathon (archive) (7 Viewers)

kawaiipotato · May 19, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
You're right, that's what the working should be (and there should be no mg).

Unless the question says that it's a vertical motion question right?

InteGrand · May 19, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
Unless the question says that it's a vertical motion question right?

Well it doesn't have to be completely vertical, is long as there is gravity influencing it somehow (e.g. for conical pendulums, the motion is in a horizontal circle, but gravity still needs to be taken into account, or for a ball rolling down an inclined plane, there is a vertical component to motion, so gravity needs to be considered).

In the case with the boat in that question, gravity didn't affect its motion in the horizontal direction, so we didn't need to consider it. And I think the HSC will usually tell you all the relevant forces acting on the particle in question.

Drsoccerball · May 23, 2015

Re: HSC 2015 4U Marathon

My test for :
integration
Volumes
Mechanics
is on friday can we get some hard volumes and mechanics question

?

InteGrand · May 31, 2015

Re: HSC 2015 4U Marathon

kz4x said:
Hi if a particle is moving in a circular path of radius 'r' and VARIABLE angular velocity (θ') what would be the tangential and normal components of the acceleration in terms of θ' and θ"
Thanks!

$If we denote the angle as $\theta \equiv \theta (t)$ and $\omega \equiv \omega (t) = \frac{\mathrm{d}\theta}{\mathrm{d}t}$ and $\alpha \equiv \alpha (t) = \frac{\mathrm{d}^2 \theta}{ \text{d} t^2}=\frac{\mathrm{d}\omega}{\mathrm{d} t}$, we can use vector analysis to find that the acceleration vector for the particle as a function of time is: $\bold{a}(t) = -r\left(\alpha \sin \theta +\omega ^2 \cos \theta\right)\bold{i} + r\left(\alpha \cos \theta - \omega ^2 \sin \theta \right)\bold{j}$.$

$The projection of this acceleration vector in the normal direction is then: $\bold{a}_{\perp} = \mathrm{proj} _{\bold{r}}{\bold{a}}$, where $\bold{r}\equiv \bold{r}(t)$ is the position vector of the particle at time $t$: $\bold{r} = (r\cos \theta)\bold{i} + (r\sin \theta)\bold{j}$. The projection of the acceleration vector in the tangential direction is $\bold{a}_{\text{tg}} = \mathrm{proj}_{\bold{r}_{\perp}} \bold{a}$, where $\bold{r}_{\perp}$ is a vector perpendicular to $\bold{r}$ (the vector $\bold{a}-\bold{a}_{\perp}$ is such a vector).$

InteGrand · May 31, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$If we denote the angle as $\theta \equiv \theta (t)$ and $\omega \equiv \omega (t) = \frac{\mathrm{d}\theta}{\mathrm{d}t}$ and $\alpha \equiv \alpha (t) = \frac{\mathrm{d}^2 \theta}{ \text{d} t^2}=\frac{\mathrm{d}\omega}{\mathrm{d} t}$, we can use vector analysis to find that the acceleration vector for the particle as a function of time is: $\bold{a}(t) = -r\left(\alpha \sin \theta +\omega ^2 \cos \theta\right)\bold{i} + r\left(\alpha \cos \theta - \omega ^2 \sin \theta \right)\bold{j}$.$

$The projection of this acceleration vector in the normal direction is then: $\bold{a}_{\perp} = \mathrm{proj} _{\bold{r}}{\bold{a}}$, where $\bold{r}\equiv \bold{r}(t)$ is the position vector of the particle at time $t$: $\bold{r} = (r\cos \theta)\bold{i} + (r\sin \theta)\bold{j}$. The projection of the acceleration vector in the tangential direction is $\bold{a}_{\text{tg}} = \mathrm{proj}_{\bold{r}_{\perp}} \bold{a}$, where $\bold{r}_{\perp}$ is a vector perpendicular to $\bold{r}$.$

$Performing some computations, we can show that $\bold{a}_{\perp} = (-r\omega ^2 \cos \theta)\bold{i} + (-r\omega ^2 \sin \theta)\bold{j}$ and $\bold{a}_{\text{tg}} = \bold{a} - \bold{a}_{\perp} = (-r\alpha \sin \theta)\bold{i}+(r\alpha \cos \theta)\bold{j}$.$ .

$The required \emph{components} are then $\left \| \bold{a}_{\perp} \right \| = r\omega ^2$ and $\left \| \bold{a}_{\text{tg}} \right \|=r|\alpha|$.$

InteGrand · May 31, 2015

Re: HSC 2015 4U Marathon

$Also, here were the position and velocity vectors respectively: $\bold{r}(t) = (r\cos \theta)\bold{i} + (r\sin \theta)\bold{j}$ and $\bold{v}(t) = \frac{\mathrm{d}\bold{r}}{\mathrm{d}t}=(-r\omega \sin \theta)\bold{i} + (r\omega \cos \theta)\bold{j}$. Also note that the radial acceleration points in the opposite direction to the position vector at all times (since we can see from our vector expressions that $\bold{a}(t)=-\omega ^2 (t) \bold{r}(t)$ (the minus sign means the vectors point in opposite ways), which means the radial acceleration is pointing towards the centre, as you're probably aware of.).$

seanieg89 · May 31, 2015

Re: HSC 2015 4U Marathon

Given that every degree n polynomial with real coefficients has exactly n complex roots (counting multiplicity), show that the same is true of every degree n polynomial with complex coefficients.

(Obviously, you may not assume the fundamental theorem of algebra for this question.)

seanieg89 · Jun 1, 2015

Re: HSC 2015 4U Marathon

Two more polynomial questions:

Let p be a degree n polynomial with complex coefficients.

1. If the roots of p form an equilateral n-gon, show that p' has a single root of multiplicity (n-1), located at the centroid of this n-gon.

2. Given that p(ix) is real for every real x, show that p(-x) is the conjugate of p(x) for every real x.

(1 and 2 are completely unrelated to each other.)

Sy123 · Jun 3, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Given that every degree n polynomial with real coefficients has exactly n complex roots (counting multiplicity), show that the same is true of every degree n polynomial with complex coefficients.

(Obviously, you may not assume the fundamental theorem of algebra for this question.)

I've proven it for when the polynomial's domain is restricted among the reals

$\\ $Let$ \ P \ $be a complex polynomial among the reals with complex co-efficients$ \\ c_j = r_j + is_j \ $for$ \ j= 0, 1, 2, \dots, n \ $and$ \ r_j, s_j \in \mathbb{R}$

$\\ $So,$ \ P(x) = \sum_{j=0}^n c_jx^j = \sum_{j=0}^n r_jx^j + i \sum_{j=0}^n s_jx^j \ (*)$

$\\ $Let$ \ R(x) = \sum_{j=0}^n r_jx^j \ $and$ \ S(x) = \sum_{j=0}^n s_jx^j$

$\\ $So,$ \ P(x) = R(x) + i S(x) \ $by$ \ (*)$

$\\ P(x) + \overline{P(x)} = R(x) + i S(x) + \overline{R(x) + iS(x)} \\ \\ = R(x) + \overline{R(x)} + i (S(x) - \overline{S(x)}) \\ \\= 2R(x) \ $since$ \ R \ $is real$$

$\\ $Suppose$ \ P(x) \ $has$ \ k \ $roots, then$ \ \overline{P(x)} \ $has the exact same$ \ k \ $roots, for if there is$ \ P(w) = 0 \ $then$ \ \overline{P(w)} = \overline{0} = 0 \\ \\ $Let the roots be$ \ \alpha_1, \alpha_2,\dots, \alpha_k$

--------

$\\ $Then the following cases are exhaustive$$

$\\ (1): \ k > n$

$\\ (2): \ k < n$

$\\ (3): \ k = n$

----------

$\\ $If$ \ (1) \ $then$ \ \frac{1}{2} (P(\alpha_j) + \overline{P(\alpha_j)}) = 0 = R(\alpha_j) \ $for$ \ j = 1,1,\dots,n+1$

$\\ $Therefore,$ \ R(x) \ $has at least$ \ n+1 \ $complex roots$$

$\\ R(x) \ $is real however, and so should have only$ \ n \ $complex roots, therefore contradiciton$$

---------

$\\ $If$ \ (2) \ $then there is some$ \ \beta \ $where$ \ R(\beta) = 0 \ $and$ \ P(\beta) \neq 0$

$\\ R(\beta) = 0 \Rightarrow P(\beta) + \overline{P(\beta)} = 0 \Rightarrow \ \frac{P(\beta)^2 }{|P(\beta)|^2 } = - 1 \ $since$ \ P(\beta) \neq 0$

$\\ (2.1): \ P(\beta) \in \mathbb{R}$

$\\ (2.2): \ P(\beta) \in \mathbb{C}$

$\\ $If$ \ (2.1) \ $then$ \ \frac{P(\beta)}{|P(\beta)|} = A \in \mathbb{R} \Rightarrow \ A^2 = -1 \Rightarrow \ A \not \in \mathbb{R} \ $, contradiction$$

$\\ $If$ \ (2.2) \ $then$ \ P(\beta)^2 \in \mathbb{C} \Rightarrow \ - |P(\beta)|^2 \in \mathbb{C} \ $but$ \ - |P(\beta)|^2 \not \in \mathbb{C} \ $, contradiction$$

$\\ $Therefore$ \ (2) \ $yields a contradiction$$

------

$\\ $Therefore$ \ k = n$

$\\ $Therefore$ \ P(x) \ $has$ \ n \ $complex roots$$

InteGrand · Jun 4, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Two more polynomial questions:

Let p be a degree n polynomial with complex coefficients.

1. If the roots of p form an equilateral n-gon, show that p' has a single root of multiplicity (n-1), located at the centroid of this n-gon.

2. Given that p(ix) is real for every real x, show that p(-x) is the conjugate of p(x) for every real x.

(1 and 2 are completely unrelated to each other.)

$2. Let $p(z) = \alpha _0 + \sum_{k=1}^{n}\alpha_k z^k$, where $\alpha_k = a_k + ib_k, a_k,b_k\in \mathbb{R}$. Then we can rewrite this as $p(z) = A(z)+iB(z)$, where $A(z) = a_0 + \sum_{k=1}^{n}a_k z^k$ and $B(z) = b_0 + \sum_{k=1}^{n}b_k z^k$ (note that $A$ and $B$ are polynomials that have only real coefficients). We can write $A(z) = o_A (z) + e_A (z)$, where $o_A$ contains all the terms in $A$ with odd powers, and $e_A$ contains all terms in $A$ with even powers. Similarly, we can write $B(z) = o_B (z) + e_B (z)$ (each $o$ and $e$ is a polynomial with real coefficients). Then for both $o$'s and $e$'s, $o(-z)=-o(z),e(z)=e(-z),e(ix)\in \mathbb{R} \text{ } \forall x \in \mathbb{R}$. Also, for real $x$, $o(ix)$ is always of the form $iy$, where $y\in\mathbb{R}$.$

$Now, since we are given that $p(ix)= \underbrace{o_A (ix)}_{=iy_1} + \underbrace{e_A (ix)}_{\text{real}} + i(\underbrace{o_B (ix)}_{=iy_2} + \underbrace{e_B (ix)}_{\text{real}})$ is real for all real $x$, we must have $q(ix)\equiv o_A (ix) + ie_B (ix)$ be real (and hence 0) for all real $x$ (since 0 is the only real value it can attain, as it is always of the form $iy$, where $y\in\mathbb{R}$.). If we take out a factor of $i$, we see that $q(ix)$ is $i$ times a polynomial in $x$ (call it $r(x)$) with real coefficients, whose coefficients are those of $o_A$ for the odd powers and those of $e_B$ for the even powers, except with some sign changes. Since $q(ix)=ir(x)$ must be identically 0, $r(x)$ must be identically 0, and since all its coefficients are equal in absolute value to those of $o_A$ and $e_B$, it follows that $o_A$ and $e_B$'s coefficients are all 0 too (i.e. $o_A$ and $e_B$ must be zero polynomials). So $p(z) = 0 + e_A (z) + i(o_B (z) + 0)$.$

$\therefore p(-x) = \underbrace{e_A (-x)}_{\text{real as }x\text{ is real}} + i\underbrace{o_B (-x)}_{\text{real as }x\text{ is real}}$

$$=e_A (x) -io_B (x)$, as $e_A$ is an even function, and $o_B$ is an odd function$

$$=\overline{e_A (x) + io_B (x)}$, as $e_A (x)$ and $o_B(x)$ are real as $x$ is real$

$$=\overline{p(x)}.$ \blacksquare$

seanieg89 · Jun 5, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
I've proven it for when the polynomial's domain is restricted among the reals

What do you mean by this? Are you then only saying something about the real roots of an n'th degree polynomial? You definitely shouldn't be able to obtain an "exactly n" statement in this case!

$\\ P(x) + \overline{P(x)} = R(x) + i S(x) + \overline{R(x) + iS(x)} \\ \\ = R(x) + \overline{R(x)} + i (S(x) - \overline{S(x)}) \\ \\= 2R(x) \ $since$ \ R \ $is real$$

This identity holds for all real x at least.

$\\ $Suppose$ \ P(x) \ $has$ \ k \ $roots, then$ \ \overline{P(x)} \ $has the exact same$ \ k \ $roots, for if there is$ \ P(w) = 0 \ $then$ \ \overline{P(w)} = \overline{0} = 0 \\ \\ $Let the roots be$ \ \alpha_1, \alpha_2,\dots, \alpha_k$

Real roots or complex roots? Given that you use the previous identity I can only assume real?
--------

$\\ $Then the following cases are exhaustive$$

$\\ (1): \ k > n$

$\\ (2): \ k < n$

$\\ (3): \ k = n$

----------

$\\ $If$ \ (1) \ $then$ \ \frac{1}{2} (P(\alpha_j) + \overline{P(\alpha_j)}) = 0 = R(\alpha_j) \ $for$ \ j = 1,1,\dots,n+1$

$\\ $Therefore,$ \ R(x) \ $has at least$ \ n+1 \ $complex roots$$

Should "complex" be replaced by real here? This is fair enough then, and says that any complex deg n poly has at most n real roots. (The "at most" statement in solving polynomials is the easy part though, we can get it straight from the factor theorem and the notion of degree. In fact this would tell us that any complex deg n poly has at most n COMPLEX roots, which is closer to what we need overall in this problem.)

$\\ R(x) \ $is real however, and so should have only$ \ n \ $complex roots, therefore contradiciton$$

---------

$\\ $If$ \ (2) \ $then there is some$ \ \beta \ $where$ \ R(\beta) = 0 \ $and$ \ P(\beta) \neq 0$

$\\ R(\beta) = 0 \Rightarrow P(\beta) + \overline{P(\beta)} = 0 \Rightarrow \ \frac{P(\beta)^2 }{|P(\beta)|^2 } = - 1 \ $since$ \ P(\beta) \neq 0$

$\\ (2.1): \ P(\beta) \in \mathbb{R}$

$\\ (2.2): \ P(\beta) \in \mathbb{C}$

Remember that R is a subset of C! By C I will assume you mean C-R.

$\\ $If$ \ (2.1) \ $then$ \ \frac{P(\beta)}{|P(\beta)|} = A \in \mathbb{R} \Rightarrow \ A^2 = -1 \Rightarrow \ A \not \in \mathbb{R} \ $, contradiction$$

$\\ $If$ \ (2.2) \ $then$ \ P(\beta)^2 \in \mathbb{C} \Rightarrow \ - |P(\beta)|^2 \in \mathbb{C} \ $but$ \ - |P(\beta)|^2 \not \in \mathbb{C} \ $, contradiction$$

Why should the square of a nonreal number be nonreal? P(\beta)=i is a counterexample to your claim. (2) certainly shouldn't be a contradiction given that you are just talking about real numbers...it is very possible for a complex deg n poly to have < n real roots.

$\\ $Therefore$ \ (2) \ $yields a contradiction$$

------

$\\ $Therefore$ \ k = n$

$\\ $Therefore$ \ P(x) \ $has$ \ n \ $complex roots$$
complex or real? I don't see how you have proven the complex statement.

.

seanieg89 · Jun 5, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$2. Let $p(z) = \alpha _0 + \sum_{k=1}^{n}\alpha_k z^k$, where $\alpha_k = a_k + ib_k, a_k,b_k\in \mathbb{R}$. Then we can rewrite this as $p(z) = A(z)+iB(z)$, where $A(z) = a_0 + \sum_{k=1}^{n}a_k z^k$ and $B(z) = b_0 + \sum_{k=1}^{n}b_k z^k$ (note that $A$ and $B$ are polynomials that have only real coefficients). We can write $A(z) = o_A (z) + e_A (z)$, where $o_A$ contains all the terms in $A$ with odd powers, and $e_A$ contains all terms in $A$ with even powers. Similarly, we can write $B(z) = o_B (z) + e_B (z)$ (each $o$ and $e$ is a polynomial with real coefficients). Then for both $o$'s and $e$'s, $o(-z)=-o(z),e(z)=e(-z),e(ix)\in \mathbb{R} \text{ } \forall x \in \mathbb{R}$. Also, for real $x$, $o(ix)$ is always of the form $iy$, where $y\in\mathbb{R}$.$

$Now, since we are given that $p(ix)= \underbrace{o_A (ix)}_{=iy_1} + \underbrace{e_A (ix)}_{\text{real}} + i(\underbrace{o_B (ix)}_{=iy_2} + \underbrace{e_B (ix)}_{\text{real}})$ is real for all real $x$, we must have $q(ix)\equiv o_A (ix) + ie_B (ix)$ be real (and hence 0) for all real $x$ (since 0 is the only real value it can attain, as it is always of the form $iy$, where $y\in\mathbb{R}$.). If we take out a factor of $i$, we see that $q(ix)$ is $i$ times a polynomial in $x$ (call it $r(x)$) with real coefficients, whose coefficients are those of $o_A$ for the odd powers and those of $e_B$ for the even powers, except with some sign changes. Since $q(ix)=ir(x)$ must be identically 0, $r(x)$ must be identically 0, and since all its coefficients are equal in absolute value to those of $o_A$ and $e_B$, it follows that $o_A$ and $e_B$'s coefficients are all 0 too (i.e. $o_A$ and $e_B$ must be zero polynomials). So $p(z) = 0 + e_A (z) + i(o_B (z) + 0)$.$

$\therefore p(-x) = \underbrace{e_A (-x)}_{\text{real as }x\text{ is real}} + i\underbrace{o_B (-x)}_{\text{real as }x\text{ is real}}$

$$=e_A (x) -io_B (x)$, as $e_A$ is an even function, and $o_B$ is an odd function$

$$=\overline{e_A (x) + io_B (x)}$, as $e_A (x)$ and $o_B(x)$ are real as $x$ is real$

$$=\overline{p(x)}.$ \blacksquare$

Nice

.

My way was essentially the same:

$For $P(z)=\sum_{k=0}^na_k z^k$, we define the real polynomials $\Re(P)(z)=\sum_{k=0}^n \Re(a_k)z^k$, $\Im(P)(z)=\sum_{k=0}^n \Im(a_k)z^k$, and the conjugate polynomial $\overline{P}(z)=\sum_{k=0}^n \overline{a_k}z^k.$ \\ \\ Then $Q(x):=P(ix)\in \mathbb{R}$ for all $x\in\mathbb{R}$.\\ So for real $x$ we have $\Im(Q(x))=\Im(Q)(x)=0$.\\ \\ As the only polynomial which can vanish on the entire real line is the zero polynomial, we have that $Q=\Re(Q)$ has all coefficients real. \\ \\ But $Q(z)=\sum_{k=0}^n i^ka_kz^k$, so each $i^ka_k$ is real.\\ \\ That means that for odd $k$, $a_k$ lies on the imaginary axis, and for even $k$, $a_k$ is real. \\ \\ So $P(z)=E(z)+iO(z)$, from which the conjugate property is clear.$

seanieg89 · Jun 6, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Given that every degree n polynomial with real coefficients has exactly n complex roots (counting multiplicity), show that the same is true of every degree n polynomial with complex coefficients.

(Obviously, you may not assume the fundamental theorem of algebra for this question.)

So an approach similar to Sy's in spirit works here.

Recall from my previous post that a complex polynomial p such that p(x) is real for all real x must in fact be a real polynomial.

$q(x):=p(x)\overline{p}(x)=p(x)\overline{p(x)}\in \mathbb{R} \quad \forall x\in\mathbb{R}.$

As a real degree 2n polynomial, q has exactly 2n roots counting multiplicity by the problem assumption.

But the roots of $\overline{p}(z)$ are just the conjugates of those of $p(z)$ , counting multiplicity.

So p(z) must have n roots, counting multiplicity.

simpleetal · Jun 7, 2015

Re: HSC 2015 4U Marathon

Hey Seanieg, or Sy or whoever sees this, my friend claims that if sin(2a)*sin(2b)*cosa + sin(2a)sin(a)cos(2b)= sin(2b)sin(2a)cos(b)+sin(2b)sin(b)cos(2a) because the lhs is basically the rhs except with the a and b switched, this automatically means that a must equal b. That was his reasoning, given that a and b are actue angles of a triangle. Is his reasoning valid?

seanieg89 · Jun 7, 2015

Re: HSC 2015 4U Marathon

simpleetal said:
Hey Seanieg, or Sy or whoever sees this, my friend claims that if sin(2a)*sin(2b)*cosa + sin(2a)sin(a)cos(2b)= sin(2b)sin(2a)cos(b)+sin(2b)sin(b)cos(2a) because the lhs is basically the rhs except with the a and b switched, this automatically means that a must equal b. That was his reasoning, given that a and b are actue angles of a triangle. Is his reasoning valid?

a might be forced to equal b (not going to check this myself right now), but the reasoning is not.

Eg x+y = y+x for all x and y whatsoever, but this does not tell us that x=y.

simpleetal · Jun 7, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
a might be forced to equal b (not going to check this myself right now), but the reasoning is not.

Eg x+y = y+x for all x and y whatsoever, but this does not tell us that x=y.

Thank you, for some reason he isn't convinced though, apparently his tutor said otherwise

seanieg89 · Jun 7, 2015

Re: HSC 2015 4U Marathon

simpleetal said:
Thank you, for some reason he isn't convinced though, apparently his tutor said otherwise

Lol, his tutor really shouldn't be tutoring then.

Is he really not convinced? Ask him how the situation in his question differs from that in my example.

It seems his claim is that for an arbitrary function f(x,y), that if f(x,y)=f(y,x) then x=y. Ask him if he thinks this is true.

Nearly any function you can think of violates this. (Any symmetric function violates this really badly!)

InteGrand · Jun 7, 2015

Re: HSC 2015 4U Marathon

I'm guessing the tutor was including the fact that a and b were acute angles, although it still doesn't seem obvious that a must equal b.

seanieg89 · Jun 7, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
I'm guessing the tutor was including the fact that a and b were acute angles, although it still doesn't seem obvious that a must equal b.

Of course. I suspect it could well be true for this function with that condition, but you certainly can't phrase it as an injectivity type argument as he/the tutor did. Injectivity is precisely what you need to prove to draw that conclusion!

Sy123 · Jun 13, 2015

Re: HSC 2015 4U Marathon

$\\ $Let the polynomial$ \ P(x) \ $be a monic cubic polynomial with non-negative real roots$ \ a,b,c$

$\\ $i) Find$ \ P(a) + P(b) + P(c) \ $in terms of$ \ a,b,c$

$\\ $ii) Prove for non-negative reals$ \ x,y,z \ $that$ \ x^2 + y^2 + z^2 \geq xy + xz + zx$

$\\ $iii) Hence prove$ \ \frac{a^3 + b^3 + c^3}{3} \geq abc \ $for$ \ a,b,c \geq 0$

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