HSC 2016 MX1 Marathon (archive) (4 Viewers)

leehuan · Feb 4, 2016

Re: Shuuya's endless maths questions SOS thread

Tbh I was more or less influenced by Flop.

The idea was already floating around with appleibeats using the Yr 11 Cambridge thread, but I never cared to make a personalised one until now cause

a) Didn't even need it
b) Yeah, didn't want to be the odd one out either

Flop21 · Feb 4, 2016

Re: Flop math question thread

Solve the equation sin(x) + cos(x) = 1/(root 2) 0<x<2pi

I'm not getting the correct answer. I put it in the aux form, (root 2)sin(x - pi/4) = 1/(root 2)

InteGrand · Feb 4, 2016

Re: Flop math question thread

Flop21 said:
Solve the equation sin(x) + cos(x) = 1/(root 2) 0<x<2pi

I'm not getting the correct answer. I put it in the aux form, (root 2)sin(x - pi/4) = 1/(root 2)

$\noindent It should become $\sqrt{2} \cos \left(x-\frac{\pi}{4} \right)$. If you want to use $\sin$, it's $\sqrt{2} \sin \left(x+\frac{\pi}{4} \right)$.$

leehuan · Feb 4, 2016

Re: Flop math question thread

$\\ $Suppose $ \sin { x } +\cos { x } \equiv R\sin { \left( x+\alpha \right) } \\ \sin { x } +\cos { x } \equiv R\sin { x } \cos { \alpha } +R\cos { x } \sin { \alpha } \\ \\ \therefore $ Equating: $ \\ 1=R\cos { \alpha } \quad ---(1)\\ 1=R\sin { \alpha } \quad ---(2)\\ \\ \left( 2 \right) \div \left( 1 \right) :\quad \frac { R\sin { \alpha } }{ R\cos { \alpha } } =\frac { 1 }{ 1 } \\ \\ \tan { \alpha } =1\Rightarrow \alpha =\frac { \pi }{ 4 }$

$\\ $Sub this into 1:$ \\ 1=R\cos { \frac { \pi }{ 4 } } \quad \\ R=\sqrt { 2 }$

Flop21 · Feb 4, 2016

Re: Flop math question thread

InteGrand said:
$\noindent It should become $\sqrt{2} \cos \left(x-\frac{\pi}{4} \right)$. If you want to use $\sin$, it's $\sqrt{2} \sin \left(x+\frac{\pi}{4} \right)$.$

Oh right thanks, dunno why I put a negative sign.

Flop21 · Feb 4, 2016

Re: Flop math question thread

Okay yeah still not getting the correct answer.

So I've got it in aux form, (root 2)sin(x+pi/4) = 1(root 2)

Now make x+pi/4 = theta

Can someone work it out after here, please.

InteGrand · Feb 4, 2016

Re: Flop math question thread

$\noindent For interests's sake, in general, $a\cos \theta + b\sin \theta = R\cos \left(\theta - \varphi \right)$ for all $\theta \in \mathbb{R}$, where $R=\sqrt{a^2 + b^2}$ and $\varphi$ is the angle to the positive $x$-axis made by the vector $(a,b)$ (with $a,b \in \mathbb{R}$ being constants). For those who know complex numbers, they will see that $R=|z|$ and $\varphi = \mathrm{Arg}\left(z\right)$, where $z=a+bi$. This fact can be proved using complex numbers as follows:$

$\noindent Consider $\zeta:=(a-bi)(\cos \theta + i\sin \theta)$, where $a,b,\theta \in \mathbb{R}$. Let $a+bi = R\mathrm{e}^{i\varphi}$ (for MX2-friendly notation, replace all $\mathrm{e}^{it}$'s with $\mathrm{c}i\mathrm{s}(t)$, so we're just saying the polar form of $a+bi$ is $R\, \mathrm{c}i\mathrm{s}\left(\varphi \right)$. Also replace any $\mathrm{e}^{-it}$'s you see with $\mathrm{c}i\mathrm{s}(-t)$). Here, $R=|a+bi| = \sqrt{a^2 + b^2}$ and $\varphi = \mathrm{Arg}\left(a+bi\right)$. On the one hand, $\Re \left(\zeta \right) = a\cos \theta + b\sin \theta$ (just by expanding and only considering the real part).$

$\noindent On the other hand, $a-bi = \overline{a+bi}= \overline{R\mathrm{e}^{i\varphi}}=R\mathrm{e}^{-i\varphi}$, and $\cos \theta + i\sin \theta=\mathrm{e}^{i\theta}$, so $\zeta = R\mathrm{e}^{-i\varphi}\cdot \mathrm{e}^{i\theta} = R\mathrm{e}^{i\left(\theta - \varphi\right)}$. So using the fact that $\mathrm{e}^{i\left(\theta - \varphi\right)}$ just means $\cos \left(\theta - \varphi\right) + i\sin \left(\theta - \varphi\right)$, we see that $\Re \left(\zeta\right) = R \cos \left(\theta -\varphi \right)$.$

$\noindent Equating the two expressions for the real part of $\zeta$, we have $a\cos \theta + b\sin \theta = R\cos \left(\theta - \varphi \right)$.$

InteGrand · Feb 4, 2016

Re: Flop math question thread

Flop21 said:
Okay yeah still not getting the correct answer.

So I've got it in aux form, (root 2)sin(x+pi/4) = 1(root 2)

Now make x+pi/4 = theta

Can someone work it out after here, please.

$\noindent We have$

$\begin{align*}\sqrt{2}\sin \left(x+\frac{\pi}{4}\right) &= \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\ \Longleftrightarrow \sin \left(x+\frac{\pi}{4}\right) &= \frac{1}{2}.\end{align*}$

$\noindent Assuming you're after general solutions, the general solution is $x+\frac{\pi}{4} = (-1)^m \sin^{-1}\frac{1}{2} + m\pi$. Since $\sin^{-1}\frac{1}{2}=\frac{\pi}{6}$, this becomes $x=-\frac{\pi}{4}+(-1)^m \frac{\pi}{6} + m\pi$, where $m\in \mathbb{Z}$.$

Flop21 · Feb 4, 2016

Re: Flop math question thread

InteGrand said:
$\noindent We have$

$\begin{align*}\sqrt{2}\sin \left(x+\frac{\pi}{4}\right) &= \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\ \Longleftrightarrow \sin \left(x+\frac{\pi}{4}\right) &= \frac{1}{2}.\end{align*}$

$\noindent Assuming you're after general solutions, the general solution is $x+\frac{\pi}{4} = (-1)^m \sin^{-1}\frac{1}{2} + m\pi$, i.e. $x=-\frac{\pi}{4}+(-1)^m \sin^{-1}\frac{1}{2} + m\pi$, where $m\in \mathbb{Z}$.$

Answers say 7pi/12, 23pi/12

The original domain was 0<x<2pi

InteGrand · Feb 4, 2016

Re: Flop math question thread

Flop21 said:
Answers say 7pi/12, 23pi/12

The original domain was 0<x<2pi

$\noindent Oh yeah, sorry, forgot to simplify in my original answer that $\sin^{-1} \frac{1}{2}=\frac{\pi}{6}$.$

InteGrand · Feb 4, 2016

Re: Flop math question thread

Flop21 said:
Answers say 7pi/12, 23pi/12

The original domain was 0<x<2pi

$\noindent So the solutions are of the form $x=-\frac{\pi}{4} + (-1)^m \frac{\pi}{6}+m\pi$ (and any number of this form is a solution). If we only want solutions in the range $\left(0,2\pi\right)$, we just take the integers $m$ that give solutions in this range. Note that there will be two such solutions, since our equation was a trig. function with period $2\pi$ being equated to a number with absolute value strictly less than the amplitude but greater than 0. It is easy to see then that in any period, there are exactly two solutions (just draw a horizontal line in any period to see this). So we just try integers $m$ until we get the two solutions in the right range.$

$\noindent Taking $m=1$, $x=-\frac{\pi}{4}-\frac{\pi}{6}+\pi = \frac{7\pi}{12}$, so this is one of the solutions as it is in the right range.$

$\noindent Taking $m=2$, $x=-\frac{\pi}{4}+\frac{\pi}{6}+2\pi = \frac{23\pi}{12}$, so this is the other solution as it is in the right range.$

drsabz101 · Feb 4, 2016

Re: Shuuya's endless maths questions SOS thread

Help ASAP

Solve these 3 equations to find real number p,q,r

P+2q-r=5
2p-q+r=3
P+3q-2r=8

InteGrand · Feb 4, 2016

Re: Shuuya's endless maths questions SOS thread

drsabz101 said:
Help ASAP

Solve these 3 equations to find real number p,q,r

P+2q-r=5
2p-q+r=3
P+3q-2r=8

$\noindent We have the system$

$\left\{\begin{matrix} p+2q-r=5& (1) & \\2p-q+r=3 & (2)& \\ p+3q-2r = 8& (3) & \end{matrix}\right.$

$\noindent The goal is to first eliminate one of the variables completely, so we are left with only two variables.$

$(1) + (2): 3p + q = 8\quad (4)$

$2\times (2)+ (3): 5p +q = 14 \quad (5)$

$\noindent This is a system of equations with two unknowns and two equations, which you should know how to go about. From (4), we have $q = -3p + 8$. Now, sub. this into (5): $5p +(-3p+8)=14 \Rightarrow 2p = 6 \Rightarrow p = 3$.$

$\noindent So $q = -3p + 8\Rightarrow q = -3\cdot 3 + 8=-1$. Now, from (2), $r= 3-2p + q = 3 -2\cdot 3 +\left(-1\right)\Rightarrow r=-4$. So the solution is $(p,q,r) = \left(3,-1,-4\right)$.$

drsabz101 · Feb 4, 2016

Re: Shuuya's endless maths questions SOS thread

Thank you !

Shuuya · Feb 4, 2016

Re: Shuuya's endless maths questions SOS thread

PLEASE SAVE ME T_T

I got an answer for (i) but not sure

Paradoxica · Feb 4, 2016

Re: Shuuya's endless maths questions SOS thread

Shuuya said:
PLEASE SAVE ME T_T

I got an answer for (i) but not sure

Clearly, since it is a positive quartic, the convexity of the quartic term will eventually dominate for sufficiently large values of x.

This means that there is an infinite domain where the gradient is decreasing along the positive x axis, as well as an infinite domain where it is increasing.

Is your answer an infinite domain?

Shuuya · Feb 4, 2016

Re: Shuuya's endless maths questions SOS thread

Paradoxica said:
Clearly, since it is a positive quartic, the convexity of the quartic term will eventually dominate for sufficiently large values of x.

This means that there is an infinite domain where the gradient is decreasing along the positive x axis, as well as an infinite domain where it is increasing.

Is your answer an infinite domain?

$I\quad got\quad x\le -\frac { \sqrt { 3 } }{ 2 } \quad for\quad the\quad domain\quad$

InteGrand · Feb 4, 2016

Re: Shuuya's endless maths questions SOS thread

Shuuya said:
PLEASE SAVE ME T_T

I got an answer for (i) but not sure

$\noindent (i) Since $f(x) = 2x^4 - 3x^2 +1$, we have $f^\prime (x) = 8x^3 - 6x = 2x(4x^3 -3) = 2x\cdot 4\left(x^2 - \frac{3}{4}\right)$. So to analyse the sign of $f^\prime (x)$, we need to analyse the sign of the factors $x$ and $x^2 - \frac{3}{4}$. Clearly $x^2 - \frac{3}{4} \geq 0$ if and only if $x\geq \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$ or $x\leq -\frac{\sqrt{3}}{2} $, so $x^2 - \frac{3}{4} \leq 0$ if and only if $-\frac{\sqrt{3}}{2} \leq x \leq \frac{\sqrt{3}}{2}$. Since a product of two numbers is non-positive if and only if the two numbers don't have the same sign, we see that $f^\prime (x) \leq 0$ if either $0\leq x \leq \frac{\sqrt{3}}{2}$ or $x\leq -\frac{\sqrt{3}}{2}$ (since it is in these intervals and these intervals only where those two factors have opposing signs or are 0). So the largest interval of decrease is $\left(-\infty, -\frac{\sqrt{3}}{2}\right]$.$

$\noindent (ii) We want the inverse to the function $y= 2x^4 - 3x^2 +1$, $x\leq -\frac{\sqrt{3}}{2}$. Since the function was decreasing on this interval and at $x=-\frac{\sqrt{3}}{2}$, $y = 2 \times \left(\frac{3}{4}\right)^2 - 3 \times \frac{3}{4}+1 = -\frac{1}{8}$, we have $y\geq -\frac{1}{8}$ here (also $y\to +\infty$ as $x\to -\infty$). For the inverse function $y=f^{-1}(x)$, we have $x=2y^4 - 3y^2 +1$, with $y \leq -\frac{\sqrt{3}}{2}$ and $x\geq -\frac{1}{8}$. We now for $y$:$

$\begin{align*}2y^4 - 3y^2 +1&= x \\ \Rightarrow y^4 - \frac{3}{2}y^2 +\frac{1}{2} &= \frac{x}{2} \\ \Rightarrow \left(y^2 - \frac{3}{4} \right)^2 - \frac{9}{16} +\frac{1}{2} &=\frac{x}{2}\quad (\text{completing the square}) \\ \Rightarrow \left(y^2 - \frac{3}{4} \right)^2 &=\frac{x}{2}+\frac{1}{16} \\ \Rightarrow y^2 - \frac{3}{4} &= \pm \sqrt{\frac{x}{2}+\frac{1}{16}} \\ \Rightarrow y^2 &= \frac{3}{4}\pm \sqrt{\frac{x}{2}+\frac{1}{16}} \\ \Rightarrow y = -\sqrt{\frac{3}{4}\pm \sqrt{\frac{x}{2}+\frac{1}{16}} }\quad (\text{as }y\text{ is negative}). \\ \end{align*}$

$\noindent Now for this inverse, $x$ is allowed to be arbitrarily large positive, as $y\to +\infty$ in the original function. So we need to take the positive square root in the expression inside the outer radical so that we have a real value for all large $x$. So the answer is that $f^{-1} (x) = -\sqrt{\frac{3}{4}+\sqrt{\frac{x}{2}+\frac{1}{16}} }, \, x \geq -\frac{1}{8}$. This can be simplified using index laws etc. to: $f^{-1} (x) = -\frac{1}{2}\sqrt{3+\sqrt{8x+1}}, \, x \geq -\frac{1}{8}$.$

$\noindent $\Big{(}$For interest's sake: if we just obtained the full solution for $y$ with no restrictions, it would be $y = \pm\frac{1}{2}\sqrt{3\pm\sqrt{8x+1}}$. Here we mean that the two $\pm$'s are independent, so there are four functions here, as there are two choices for the first $\pm$, and two for the second. Each of these four choices corresponds to one of the four monotonic ``pieces'' of the original graph, i.e. the parts of the graphs in each of the following domains (each of which the function is monotonic in): $\left(-\infty, -\frac{\sqrt{3}}{2}\right],\, \left[-\frac{\sqrt{3}}{2}, 0\right], \, \left[0, \frac{\sqrt{3}}{2}\right], \,\left[\frac{\sqrt{3}}{2}, +\infty \right) $.$\Big{)}$$

InteGrand · Feb 4, 2016

Re: HSC 2016 3U Marathon

Also, realised that Shuuya's SOS thread got merged with the HSC 3U Marathon.

Shuuya · Feb 4, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
Also, realised that Shuuya's SOS thread got merged with the HSC 3U Marathon.

Naww I'm not special enough to have a personalised thread :'(

HSC 2016 MX1 Marathon (archive) (4 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)