HSC 2015 MX2 Marathon (archive) (3 Viewers)

InteGrand · Apr 8, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ P(x) \ $and$ \ Q(x) \ $be polynomials of degree$ \ n \ $and are such that$ \\ P(x) = Q(x) \ $for some$ \ n + 1 \ $distinct values of$ \ x \\\\ $Show that$ \ P(x) = Q(x) \ $for all$ \ x$

$$Suppose $P(x)$ and $Q(x)$ are not identical. Then consider the polynomial $f(x):=P(x)-Q(x)$. As $P(x)$ and $Q(x)$ are degree $n$ polynomials and non-identical, $f(x)$ is of degree $d\in \mathbb{Z}:d\in[0,n]$, which means it has exactly $d\leq n$ zeros (this follows from the Fundamental Theorem of Algebra). This contradicts the fact that $P(x)=Q(x)\Rightarrow f(x)=0$ for some $n+1 >n$ distinct values of $x$. Hence $P(x)$ and $Q(x)$ must be identical.$

InteGrand · Apr 8, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let there be a triangle$ \ ABC \ $and points$ \ P,Q,R \ $on$ \ BC, AC, AB \ $respectively, such that$ \ P,Q,R \ $are midpoints on the sides$ \ BC,AC,AB \ $respectively$ \\ \\ $Show that$ \\ 4(AP^2 + BQ^2 + CR^2) = 3(AB^2 + AC^2 + BC^2)$

$Use Apollonius' theorem (probably prove it first for HSC purposes, or it would be a sub-part of the question) to get the relationship between a squared median and the triangle's three sides, and then add up for all the medians and manipulate and simplify.$

InteGrand · Apr 8, 2015

Re: HSC 2015 4U Marathon

$\underline{NEW QUESTION}$

$Consider a projectile thrown up with initial speed $u$ that is under the influence of gravity and a drag (resistive) force $F_\text{drag} (v) = -mkv^2$, where $v$ is the velocity of the projectile at time $t$, $k>0$ is a constant, and $m>0$ is the mass of the projectile. Does it take longer for the projectile to reach its maximum height, or to come back down to its original height from the maximum height?$
$Extra challenge (most probably not required for HSC-level): Can you show your result for a general resistive drag force $F_\text{drag}(v)$?$

Drsoccerball · Apr 8, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$\underline{NEW QUESTION}$

$Consider a projectile thrown up with initial speed $u$ that is under the influence of gravity and a drag (resistive) force $F_\text{drag} (v) = -mkv^2$, where $v$ is the velocity of the projectile at time $t$, $k>0$ is a constant, and $m>0$ is the mass of the projectile. Does it take longer for the projectile to reach its maximum height, or to come back down to its original height from the maximum height?$
$Extra challenge (most probably not required for HSC-level): Can you show your result for a general resistive drag force $F_\text{drag}(v)$?$

havnt done mechanics

we've finished:
-complex numbers
-polynomials
-graphs
-conics
-integration
Havnt finished:
-volumes
-mechanics

braintic · Apr 8, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$\underline{NEW QUESTION}$

$Consider a projectile thrown up with initial speed $u$ that is under the influence of gravity and a drag (resistive) force $F_\text{drag} (v) = -mkv^2$, where $v$ is the velocity of the projectile at time $t$, $k>0$ is a constant, and $m>0$ is the mass of the projectile. Does it take longer for the projectile to reach its maximum height, or to come back down to its original height from the maximum height?$
$Extra challenge (most probably not required for HSC-level): Can you show your result for a general resistive drag force $F_\text{drag}(v)$?$

Extra challenge: As resistance is a non-conservative force (since it is dependent on velocity and not position), only reducing the total energy of the body, at any point in its downward path it must have less kinetic energy and hence less speed than at that point in its upward path (because potential energy is the same at both times). Since it is slower all the way down, it takes longer on the way down.

RealiseNothing · Apr 11, 2015

Re: HSC 2015 4U Marathon

Show that $x^3+x+1=0$ has no rational roots.

VBN2470 · Apr 11, 2015

Re: HSC 2015 4U Marathon

RealiseNothing said:
Show that $x^3+x+1=0$ has no rational roots.

$Since $P(x)=x^3+x+1$ is an increasing function over the reals, this means that only one root of $P(x)$ exists. Note that $P(-1)\,<\,0\,<\,P(0)$, so our root must lie in the open interval $(-1, 0)$. But by the rational root theorem, there exists no such ordered pair $(p, q)$ such that $x=\frac{p}{q}$, since the only candidate roots for $P(x)$ are $\pm1$, which contradicts the fact that our root must lie in $(-1, 0)$ $\Rightarrow$ root of $P(x)$ must be irrational.$

InteGrand · Apr 11, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Since $P(x)=x^3+x+1$ is an increasing function over the reals, this means that only one root of $P(x)$ exists. Note that $P(-1)\,<\,0\,<\,P(0)$, so our root must lie in the open interval $(-1, 0)$. But by the rational root theorem, there exists no such ordered pair $(p, q)$ such that $x=\frac{p}{q}$, since the only candidate roots for $P(x)$ are $\pm1$, which contradicts the fact that our root must lie in $(-1, 0)$ $\Rightarrow$ root of $P(x)$ must be irrational.$

Or, just sub. 1 and -1 into the LHS of the equation to show they're not roots.

InteGrand · Apr 11, 2015

Re: HSC 2015 4U Marathon

NEW QUESTION:

$Explain the flaw (if any) in this argument:$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)= \frac{\mathrm{d}}{\mathrm{d}x} \left(\underbrace{x+x+ \cdots+x}_{x\text{ times}}\right)$

$=\underbrace{\frac{\mathrm{d}}{\mathrm{d}x}(x)+ \frac{\mathrm{d}}{\mathrm{d}x}(x)+\cdots + \frac{\mathrm{d}}{\mathrm{d}x}(x)}_{x\text{ times}}$

$= \underbrace{1+1+\cdots + 1}_{x\text{ times}}$

$=x.$

Sy123 · Apr 11, 2015

Re: HSC 2015 4U Marathon

RealiseNothing said:
Show that $x^3+x+1=0$ has no rational roots.

Without the rational root theorem:

$\\ $If there is a rational root of the polynomial, then the root is in the form$ \ \frac{p}{q} \ $where$ \ p,q \ $are coprime integers$ \\ $So,$ \ \frac{p^3}{q^3} + \frac{p}{q} + 1 = 0 \Rightarrow \ p^3+q^3 = -pq^2 \\ \Rightarrow \ (p+q)^3 -3pq(p+q) = -pq^2 \\ \Rightarrow \ (p+q)^3 = pq(3p + 2q) \\ \therefore \ \frac{(p+q)^3}{p} \ $is an integer $ \ (*) \\ \\ $However, if$ \ p,q \ $are coprime, then$ \ (p+q) \ $and$ \ p \ $are coprime (1)$ \\ $If,$ \ (p+q) \ $and$ \ p $are coprime, then$ \ (p+q)^3 \ $and$ \ p \ $are coprime (2)$ \\ \\ $So$ \ \frac{(p+q)^3}{p} \ $cannot be an integer$ \ (**) \\ \\ $Contradiction from$ \ (*) \ $and$ \ (**) $!, so there is no rational root of the polynomial$$

------

$\\ (1) \ $Assume for reductio that$ \ (p+q) \ $and$ \ p \ $have a common divisor, then$ \\ p+q = kA \ , \ p = kB \ $for integers$ \ k, A, B \\ \\ $So,$ \ q = k(A-B) \ $so,$ \ p,q \ $have common divisor$ \ k \\ \\ $So$ \ p,q \ $are not coprime$ \\ \\ $Therefore, if$ \ (p+q) \ $and$ \ p \ $are not coprime, then$ \ p,q \ $are not coprime. Since$ \ p,q \ $are coprime, then it is not the case that$ \ (p+q) \ $and$ \ p \ $are not coprime, meaning thus they are coprime$$

$\\ (2) \ $If$ \ a,b \ $are coprime, then their prime factorisation contains no common divisor, the prime factorisation of$ \ b^3 \ $is simply the cube of each prime in the factorisation of$ \ b \ $So,$ \ a,b^3 \ $also do not have any prime in common, and thus do not have any common divisor$$

FrankXie · Apr 11, 2015

Re: HSC 2015 4U Marathon

new question

$$ Prove that $ P(x)=x^5-ax+1=0 $ has three distinct real roots if and only if $ a>5\left(\frac{1}{2}\right)^{8/5}.$

braintic · Apr 11, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
NEW QUESTION:

$Explain the flaw (if any) in this argument:$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)= \frac{\mathrm{d}}{\mathrm{d}x} \left(\underbrace{x+x+ \cdots+x}_{x\text{ times}}\right)$

$=\underbrace{\frac{\mathrm{d}}{\mathrm{d}x}(x)+ \frac{\mathrm{d}}{\mathrm{d}x}(x)+\cdots + \frac{\mathrm{d}}{\mathrm{d}x}(x)}_{x\text{ times}}$

$= \underbrace{1+1+\cdots + 1}_{x\text{ times}}$

$=x.$

We could also try to differentiate x² by writing it as 1+1+1+...+1 (x² times), giving an answer of zero. When we do this, we are using x² only as a counter of terms, and not a contributor in its own right to the rate of change.

So the error is in splitting the derivative into parts. This can only be done when the number of parts is not dependent on the variable of differentiation.

Edit:
And a somewhat more obvious reason:
Since x is being used as a counter, that implies it is an integer.
Which takes away the continuity required for differentiation to make sense.

simpleetal · Apr 11, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
new question

$$ Prove that $ P(x)=x^5-ax+1=0 $ has three distinct real roots if and only if $ a>5\left(\frac{1}{2}\right)^{8/5}.$

$$ Okie. When you differentiate the function, you get $ P'(x)=5x^4 - a $ Suppose alpha is a root to this function, then you find that $ alpha = (\frac{a}{5})^{1/4} $. Since there's only two turning points (at x= plus minus alpha), for there to be three real roots there should be only one turning point below and one above the x axis. so $ P(-alpha) * P(alpha)< 0 $. Then you get $ 1<16\left(\frac{a}{5}\right)^{5/2} $. From this, the desired condition follows.$

simpleetal · Apr 12, 2015

Re: HSC 2015 4U Marathon

I think I saw a good question earlier on this forum like 3 days ago or something, but for some reason I think it's gone? And I don't remember who the original poster was. But this was the question and it required some thinking!

$\text{Given that} x^{5x} = y^{y} \text{find the largest possible x, given x and y are both positive integers}$

InteGrand · Apr 12, 2015

Re: HSC 2015 4U Marathon

braintic said:
Edit:
And a somewhat more obvious reason:
Since x is being used as a counter, that implies it is an integer.
Which takes away the continuity required for differentiation to make sense.

$We don't need the counter to be an integer. For example, the following is valid:$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\pi x\right)= \frac{\mathrm{d}}{\mathrm{d}x} \left(\underbrace{x+x+ \cdots+x}_{\pi\text{ times}}\right)$

$=\underbrace{\frac{\mathrm{d}}{\mathrm{d}x}(x)+ \frac{\mathrm{d}}{\mathrm{d}x}(x)+\cdots + \frac{\mathrm{d}}{\mathrm{d}x}(x)}_{\pi\text{ times}}$

$= \underbrace{1+1+\cdots + 1}_{\pi\text{ times}}$

$=\pi.$

$(By ``$\pi$ times'', I mean there are $\left \lceil \pi \right \rceil$ $x$'s written down, with the first $\left \lfloor \pi \right \rfloor$ $x$'s having coefficient 1, and the last $x$ having coefficient $\{\pi\}$, where $\{\pi\}$ is the fractional part of $\pi$. In other words, $\pi$ lots of $x$.)$

braintic · Apr 13, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$We don't need the counter to be an integer. For example, the following is valid:$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\pi x\right)= \frac{\mathrm{d}}{\mathrm{d}x} \left(\underbrace{x+x+ \cdots+x}_{\pi\text{ times}}\right)$

$=\underbrace{\frac{\mathrm{d}}{\mathrm{d}x}(x)+ \frac{\mathrm{d}}{\mathrm{d}x}(x)+\cdots + \frac{\mathrm{d}}{\mathrm{d}x}(x)}_{\pi\text{ times}}$

$= \underbrace{1+1+\cdots + 1}_{\pi\text{ times}}$

$=\pi.$

$(By ``$\pi$ times'', I mean there are $\left \lceil \pi \right \rceil$ $x$'s written down, with the first $\left \lfloor \pi \right \rfloor$ $x$'s having coefficient 1, and the last $x$ having coefficient $\{\pi\}$, where $\{\pi\}$ is the fractional part of $\pi$. In other words, $\pi$ lots of $x$.)$

Nevertheless, that is the invalid step for the other reason I stated. The number of terms cannot depend on x.

InteGrand · Apr 13, 2015

Re: HSC 2015 4U Marathon

braintic said:
Nevertheless, that is the invalid step for the other reason I stated. The number of terms cannot depend on x.

Yes, that's true.

braintic · Apr 16, 2015

Re: HSC 2015 4U Marathon

This question from the 1950 Leaving Certificate.

View attachment 1950 Maths 1 Leaving Certificate.pdf

simpleetal · Apr 16, 2015

Re: HSC 2015 4U Marathon

braintic said:
This question from the 1950 Leaving Certificate.

View attachment 32020

don't you have to answer the previous question before posting a new one?
anyways
$i) $ (x+y+z)((x+y+z)^2 -3(xy + xz + zy)) $ \\ $ ii) Just expand $ \\ $ iii) let \alpha = (b-c)(x-a), \beta = (c-a)(x-b), \gamma = (a-b)(x-c). $ $Then, f(x) = (\alpha + \beta + \gamma)((\alpha + \beta + \gamma)^2 - 3((\alpha)(\beta) + (\alpha)(\gamma)+ (\beta)(\gamma))) + 3(\alpha)(\beta)(\gamma) \text{from part i)} $. $ The first factor is zero, from part ii), and thus f(x) = 3(b-c)(c-a)(a-b)(x-a)(x-b)(x-c) $ \\ $ iv) Then, the roots of f(x) are x= a, b or c$

Sy123 · Apr 19, 2015

Re: HSC 2015 4U Marathon

$\\ $Prove that the only way an ellipse and a circle with the same centre, can be tangent to each other at their intersection, is if the radius of the circle is the same in length as the semi-minor or semi-major axis of the ellipse$$

HSC 2015 MX2 Marathon (archive) (3 Viewers)

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