HSC 2016 MX2 Marathon (archive) (1 Viewer)

Paradoxica · Dec 12, 2015

Re: HSC 2016 4U Marathon

braintic said:
For anyone who wants to give this a go, the answers are 2 and 22. It's up to you to explain those answers.

yeah, but I cbbs explaining

PPPPPPPP Let's just say by inspection.

leehuan · Dec 12, 2015

Re: HSC 2016 4U Marathon

KingOfActing said:
Note $|z-w|$ is the distance between a point on the circle centered on the origin with radius 12 and the circle with center $(3,4)$ and radius 5. Looking at this diagram:

We can see that $AB$ is the shortest distance whilst $BC$ is the longest.

$AB = OA - OB = 12 - 2\times5 = 12 - 10 = 2\\BC = OB + OC = 12 + 2\times5 = 12 + 10 = 22$

Don't you have to justify something before using the diameter/radius despite it's obviousness?

InteGrand · Dec 12, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Don't you have to justify something before using the diameter/radius despite it's obviousness?

It seems to take too long to prove to make it part of that question. So we can make it the next question.

$\noindent $\textbf{NEW QUESTION}$$

$\noindent Let $\mathcal{C}_1$ and $\mathcal{C}_2$ be two circles lying in the plane without any intersection. Let $X$ be a variable point on $\mathcal{C}_1$ and similarly $Y$ on $\mathcal{C}_2$. Show that when the minimum and maximum distances for $XY$ are attained, $X$ and $Y$ lie on the line joining the two circles' centres.$

Chensanity · Dec 13, 2015

Re: HSC 2016 4U Marathon

Help

Find the roots of the following equation: (hint divide by x^2 and reduce to quadratic)
3x^4-x^3+4x^2-x+3=0

InteGrand · Dec 13, 2015

Re: HSC 2016 4U Marathon

Chensanity said:
Help

Find the roots of the following equation: (hint divide by x^2 and reduce to quadratic)
3x^4-x^3+4x^2-x+3=0

$Note that $0$ is not a root. So we can divide through by $x^2$ to obtain$

$$3x^2 - x + 4 - \frac{1}{x}+\frac{3}{x^2} = 0$.$

$This can be written as $3\left(x^2+\frac{1}{x^2} \right)-\left(x+\frac{1}{x}\right) + 4 = 0$.$

$Now, note that $x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right) ^2 - 2$. So letting $u\equiv x+\frac{1}{x}$, our equation becomes$

$3\left(u^2 - 2\right) - u+4 = 0 \Longleftrightarrow 3u^2 -u -2 = 0.$

$The discriminant is $\Delta = 1 + 4 \times 3 \times 2 = 25 \Rightarrow \sqrt{\Delta} = 5$. So by the quadratic formula, the solutions for $u$ are $u = \frac{1 \pm 5}{6}$, i.e. $u=1$ or $u=-\frac{2}{3}$.$

$For the case where $u=1$, we have $x+\frac{1}{x}=u\Rightarrow x^2 -ux+ 1 =0$, so $x^2 - x +1 = 0.$ \

$We can complete the square to solve this to obtain$

$\begin{align*} \left(x-\frac{1}{2} \right)^2 + \frac{3}{4}&= 0 \\ \Rightarrow \left(x-\frac{1}{2} \right)^2 &= -\frac{3}{4} \\ x-\frac{1}{2} &= \pm \frac{\sqrt{3}}{2}i \\ \Rightarrow x&= \frac{1}{2} \pm \frac{\sqrt{3}}{2}i .\end{align*}$

$For the case where $u=-\frac{2}{3}$, we have$

$\begin{align*}x^2 -ux+ 1 &=0 \\ \Rightarrow x^2 + \frac{2}{3}x + 1&= 0 \\ \Rightarrow \left(x+\frac{1}{3} \right)^2 +\frac{8}{9} &= 0 \\ \Rightarrow \left(x+\frac{1}{3} \right)^2 &= -\frac{8}{9} \\ \Rightarrow x + \frac{1}{3} &= \pm \frac{2\sqrt{2}}{3}i \\ \Rightarrow x &= - \frac{1}{3}\pm \frac{2\sqrt{2}}{3}i. \end{align*}$

$So the solutions are $x= \frac{1}{2} \pm \frac{\sqrt{3}}{2}i, - \frac{1}{3}\pm \frac{2\sqrt{2}}{3}i$.$

omegadot · Dec 13, 2015

Re: HSC 2016 4U Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent For all complex numbers $z,w$ satisfying $|z| = 12$ and $|w - 3 - 4i| = 5$, find the minimum and maximum value of $|z - w|.$

Paradoxica said:
yeah, but I cbbs explaining PPPPPPPP Let's just say by inspection.

leehuan said:
Don't you have to justify something before using the diameter/radius despite it's obviousness?

$$\noindent Here is an algebraic approach which makes use of a number of inequalities stemming from the triangle inequality.$

$$\noindent For the \textit{mimimum} value, using $ |z_1 - z_2| \geqslant \Big{|}|z_1| - |z_2| \Big{|}. $\\Clearly $|z_1 - z_2| \geqslant |z_1| - |z_2|.$

$$\noindent Let $z_1 = w$ and $z_2 = 3 + 4i.$ So$\\\begin{align*}|w - 3 - 4i| &\geqslant |w| - |3 + 4i| = |w| - 5.\end{align*}$

$$\noindent Given $|w - 3 - 4i| = 5$ the above inequality reduces to $|w| \leqslant 10.$

$$\noindent Also, if we let $z_1 = z$ and $z_2 = w$ we have$\\\begin{align*}|z - w| &\geqslant |z| - |w| = 12 - |w|,\end{align*}$\\since $|z| = 12.$ On rearranging we have$\\\begin{align*}|z - w| -12 &\geqslant -|w|\\\Rightarrow 12 - |z - w| &\leqslant |w|\\\Rightarrow 12 - |z - w| &\leqslant |w| \leqslant 10 \quad \mbox{as} \quad |w| \leqslant 10\\\Rightarrow |z - w| &\geqslant 2\end{align*}$

$$\noindent Thus $ |z - w|_{\rm min} = 2.$

$$\noindent And for the \textit{maximum} value, using $|z_1 - z_2| \leqslant |z_1| + |z_2|.$

$$\noindent Let $z_1 = z$ and $z_2 = w$ so$\\\begin{align*}|z - w| \leqslant |z| + |w| = 12 + |w|,\end{align*}\\$since $|z| = 12.$ Thus$\\\begin{align*}|z - w| &\leqslant 12 + |w|\\\Rightarrow |z - w| - 12 &\leqslant |w|\\\Rightarrow |z - w| - 12 &\leqslant |w| \leqslant 10 \quad \mbox{as} \quad |w| \leqslant 10\\\Rightarrow |z - w| \leqslant 22.\end{align*}$

$$\noindent Thus $|z - w|_{\rm max} = 22.$

internalsarekey · Dec 14, 2015

Re: HSC 2016 4U Marathon

b) The polynomial equationx4+ 4x3−2x2−12x−3 = 0 has rootsα,β,γandδ.Find the polynomial equation with roots (α+ 1), (β+ 1), (γ+ 1) and (δ+ 1).(c) Hence, or otherwise, solve the equationx4+ 4x3−2x2−12x−3 = 0.

leehuan · Dec 14, 2015

Re: HSC 2016 4U Marathon

omegadot said:
$$\noindent Here is an algebraic approach which makes use of a number of inequalities stemming from the triangle inequality.$

$$\noindent For the \textit{mimimum} value, using $ |z_1 - z_2| \geqslant \Big{|}|z_1| - |z_2| \Big{|}. $\\Clearly $|z_1 - z_2| \geqslant |z_1| - |z_2|.$

$$\noindent Let $z_1 = w$ and $z_2 = 3 + 4i.$ So$\\\begin{align*}|w - 3 - 4i| &\geqslant |w| - |3 + 4i| = |w| - 5.\end{align*}$

$$\noindent Given $|w - 3 - 4i| = 5$ the above inequality reduces to $|w| \leqslant 10.$

$$\noindent Also, if we let $z_1 = z$ and $z_2 = w$ we have$\\\begin{align*}|z - w| &\geqslant |z| - |w| = 12 - |w|,\end{align*}$\\since $|z| = 12.$ On rearranging we have$\\\begin{align*}|z - w| -12 &\geqslant -|w|\\\Rightarrow 12 - |z - w| &\leqslant |w|\\\Rightarrow 12 - |z - w| &\leqslant |w| \leqslant 10 \quad \mbox{as} \quad |w| \leqslant 10\\\Rightarrow |z - w| &\geqslant 2\end{align*}$

$$\noindent Thus $ |z - w|_{\rm min} = 2.$

$$\noindent And for the \textit{maximum} value, using $|z_1 - z_2| \leqslant |z_1| + |z_2|.$

$$\noindent Let $z_1 = z$ and $z_2 = w$ so$\\\begin{align*}|z - w| \leqslant |z| + |w| = 12 + |w|,\end{align*}\\$since $|z| = 12.$ Thus$\\\begin{align*}|z - w| &\leqslant 12 + |w|\\\Rightarrow |z - w| - 12 &\leqslant |w|\\\Rightarrow |z - w| - 12 &\leqslant |w| \leqslant 10 \quad \mbox{as} \quad |w| \leqslant 10\\\Rightarrow |z - w| \leqslant 22.\end{align*}$

$$\noindent Thus $|z - w|_{\rm max} = 22.$

Nice. I was trying to do something with the triangular inequality but I couldn't be bothered putting enough effort in to go about it.

leehuan · Dec 14, 2015

Re: HSC 2016 4U Marathon

internalsarekey said:
b) The polynomial equationx4+ 4x3−2x2−12x−3 = 0 has rootsα,β,γandδ.Find the polynomial equation with roots (α+ 1), (β+ 1), (γ+ 1) and (δ+ 1).(c) Hence, or otherwise, solve the equationx4+ 4x3−2x2−12x−3 = 0.

Use _^_ notation to denote a power.

Replace x with x-1 for the equation with roots (α+ 1) etc.

(x-1)^4 + 4(x-1)^3 - 2(x-1)^2 - 12(x-1) - 3 = 0
With the help of the binomial theorem this is simplified to
x^4-8x^2+4=0
Use completing the square to factorise said expression:
x^4-8x^2+16=12
(x^2-4)^2=12
x^2-4=+-2sqrt(3)
x^2=4+-2sqrt(3)
Hence:
x=sqrt(4+2sqrt3)
x=sqrt(4-2sqrt3)
x=-sqrt(4+2sqrt3)
x=-sqrt(4-2sqrt3)

If one can see ahead and identify that:
(1+sqrt3)^2 = 1 + 2sqrt3 + 3 = 4 + 2sqrt(3)
and similarly (1-sqrt3)^2 = 4 - 2sqrt(3)

The roots can be rewritten as:
x=1+sqrt3
x=1-sqrt3
x=-1+sqrt3
x=-1-sqrt3

Now, note that these are the roots (α+ 1) etc. because we already established these are the roots to the polynomial equation which we just solved.

Hence:
α=sqrt3
β=-sqrt3
γ=-2+sqrt3
δ=-2-sqrt3

hedgehog_7 · Dec 17, 2015

Re: HSC 2016 4U Marathon

The equation x^3 + 2x + 1 = 0 has the roots a,b,c

Find the monic cubic equation with roots (b+c)/a^2 , (a+c)/b^2 and (a+b)/c^2

leehuan · Dec 17, 2015

Re: HSC 2016 4U Marathon

Not sure if this is a next question or SOS so, one way of approaching this:

a+b+c=0. So b+c=-a. Hence a root (b+c)/a^2 just turns into -a/a^2 --> -1/a. And similarly other roots -1/b and -1/c. So we're finding the polynomial with THOSE roots.

glittergal96 · Dec 22, 2015

Re: HSC 2016 4U Marathon

Here is a good exercise to test understanding of the connection between integration and differentiation:

$Let $F(x):=\int_{a(x)}^{b(x)} f(x,t)\, dt$\\ \\where $a$, $b$, and $f$ are smooth (infinitely differentiable) functions.\\ \\ Prove that $F$ is differentiable, and find an expression for $F'(x)$ (as simple an expression as you can.)\\ \\ You may find the notion of a partial derivative useful:\\ \\ $\frac{\partial f}{\partial x}(x,t):= \lim_{h\rightarrow 0} \frac{f(x+h,t)-f(x,t)}{h}, \frac{\partial f}{\partial t}(x,t):= \lim_{h\rightarrow 0} \frac{f(x,t+h)-f(x,t)}{h}.$\\ \\ (So if a function is dependent on several variables, its partial derivative with respect to one of them comes from treating the other variables as constants and differentiating as normal for functions of a single variable.)\\ \\ For the university students who frequent this forum that know some multivariable calculus, try to formulate and prove an analogous result in higher dimensions.$

Ambility · Dec 22, 2015

Re: HSC 2016 4U Marathon

glittergal96 said:
Here is a good exercise to test understanding of the connection between integration and differentiation:

$Let $F(x):=\int_{a(x)}^{b(x)} f(x,t)\, dt$\\ \\where $a$, $b$, and $f$ are smooth (infinitely differentiable) functions.\\ \\ Prove that $F$ is differentiable, and find an expression for $F'(x)$ (as simple an expression as you can.)\\ \\ You may find the notion of a partial derivative useful:\\ \\ $\frac{\partial f}{\partial x}(x,t):= \lim_{h\rightarrow 0} \frac{f(x+h,t)-f(x,t)}{h}, \frac{\partial f}{\partial t}(x,t):= \lim_{h\rightarrow 0} \frac{f(x,t+h)-f(x,t)}{h}.$\\ \\ (So if a function is dependent on several variables, its partial derivative with respect to one of them comes from treating the other variables as constants and differentiating as normal for functions of a single variable.)\\ \\ For the university students who frequent this forum that know some multivariable calculus, try to formulate and prove an analogous result in higher dimensions.$

The 2016 cohort hasn't supposed to have done integration at this level, has it?

Paradoxica · Dec 22, 2015

Re: HSC 2016 4U Marathon

Ambility said:
The 2016 cohort hasn't supposed to have done integration at this level, has it?

"supposed to"

There are some far and few in between...

leehuan · Dec 22, 2015

Re: HSC 2016 4U Marathon

Lol multidimensional calculus was never a part of something like 4u. That's why glittergal gave reference to the partial derivative

glittergal96 · Dec 22, 2015

Re: HSC 2016 4U Marathon

Ambility said:
The 2016 cohort hasn't supposed to have done integration at this level, has it?

Having thought about this a little more, indeed the question requires a little more than can be expected of HS students.

If any want to still give it a crack, the following tools from outside curriculum are the basic things you might need:

-The epsilon-delta definition of limits.
-The mean value theorem.
-The extreme value theorem.
(there are several alternate routes though, some of which use different things)

It is first year calculus basically.

The problem in higher dimensions is a bit harder.

P.s. There really should be a university marathon, it is a shame the actual tertiary maths subforums are so dead.

InteGrand · Dec 22, 2015

Re: HSC 2016 4U Marathon

glittergal96 said:
Having thought about this a little more, indeed the question requires a little more than can be expected of HS students.

If any want to still give it a crack, the following tools from outside curriculum are the basic things you will need:

-The epsilon-delta definition of limits.
-The mean value theorem.
-The extreme value theorem.
(there are several alternate routes though, some of which use different things)

It is first year calculus basically.

The problem in several variables is a bit harder.

Can't it also be done using the Leibniz rule for differentiating under the integral sign and also multivariable chain rule? Or is this assuming too many rules for what you had in mind?

InteGrand · Dec 22, 2015

Re: HSC 2016 4U Marathon

There is a "first year university maths thread", which seems to be a bit like these marathons, but yeah, it's a dead thread. Here it is for anyone curious: http://community.boredofstudies.org/1003/maths/321004/1st-year-university-mathematics-thread.html

glittergal96 · Dec 22, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
Can't it also be done using the Leibniz rule for differentiating under the integral sign and also multivariable chain rule? Or is this assuming too many rules for what you had in mind?

Finding an expression for this quantity is precisely a derivation of one form of the Leibniz rule. (And so assuming a weaker version of the Leibniz rule as a starting point is cheating a little.)

What I was looking for was such a derivation based on first principles. (And the fundamental theorem of calculus of course).

Paradoxica · Dec 22, 2015

Re: HSC 2016 4U Marathon

glittergal96 said:
Here is a good exercise to test understanding of the connection between integration and differentiation:

$Let $F(x):=\int_{a(x)}^{b(x)} f(x,t)\, dt$\\ \\where $a$, $b$, and $f$ are smooth (infinitely differentiable) functions.\\ \\ Prove that $F$ is differentiable, and find an expression for $F'(x)$ (as simple an expression as you can.)\\ \\ You may find the notion of a partial derivative useful:\\ \\ $\frac{\partial f}{\partial x}(x,t):= \lim_{h\rightarrow 0} \frac{f(x+h,t)-f(x,t)}{h}, \frac{\partial f}{\partial t}(x,t):= \lim_{h\rightarrow 0} \frac{f(x,t+h)-f(x,t)}{h}.$\\ \\ (So if a function is dependent on several variables, its partial derivative with respect to one of them comes from treating the other variables as constants and differentiating as normal for functions of a single variable.)\\ \\ For the university students who frequent this forum that know some multivariable calculus, try to formulate and prove an analogous result in higher dimensions.$

my guts tell me to taylor the bounds but my mind says no

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