HSC 2016 MX2 Marathon (archive) (1 Viewer)

leehuan · Jun 25, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
By lateral force, Katebate most likely meant the friction force (which is a lateral force, i.e. up or down the bend).

Ah.

Yes, that has to be included in. Unless an indication is made that the track was banked to allow 0 friction.

Katebate · Jun 25, 2016

Re: HSC 2016 4U Marathon

leehuan said:
Ah.

Yes, that has to be included in. Unless an indication is made that the track was banked to allow 0 friction.

InteGrand said:
By lateral force, Katebate most likely meant the friction force (which is a lateral force, i.e. up or down the bend).

Ok thank you.

Does that mean in this question for example "a car is travelling round a section of race track which is banked at an angle of 15 degrees. The radius of the track is 100 m. What is the speed at which the car can travel without tending to slip" that we set the lateral force to zero and only equate the normal and gravitation forces because we are trying to find when it continues without slipping?

InteGrand · Jun 25, 2016

Re: HSC 2016 4U Marathon

Katebate said:
Ok thank you.

Does that mean in this question for example "a car is travelling round a section of race track which is banked at an angle of 15 degrees. The radius of the track is 100 m. What is the speed at which the car can travel without tending to slip" that we set the lateral force to zero and only use normal and gravity because we are trying to find when it continues without slipping?

Yes.

Katebate · Jun 25, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
Yes.

Thank you

Paradoxica · Jun 25, 2016

HSC 2016 4U Marathon

Find all general solutions to

$\tan{3\theta} = \tan{2\theta} + \tan{\theta}$

RainbowBacon · Jun 26, 2016

Re: HSC 2016 4U Marathon

0+k(pi)
(pi)/3 + k(pi)
-(pi)/3 + k(pi)

maybe?

leehuan · Jun 26, 2016

Re: HSC 2016 4U Marathon

RainbowBacon said:
0+k(pi)
(pi)/3 + k(pi)
-(pi)/3 + k(pi)

maybe?

Yep all good.

May as well just combine the solutions though.
$\theta=k\pi \pm \frac{\pi}{3}\\ \theta=k\pi$

RainbowBacon · Jun 28, 2016

Re: HSC 2016 4U Marathon

Oops, I'm blind. thanks!

hedgehog_7 · Jul 3, 2016

Re: HSC 2016 4U Marathon

A mass of 1kg is fastened by a string of length 1m to a point 0.5m above a smooth horizontal table and is describing a circl o nthe table with uniform angular speed of 1 revolution in 2 seconds.

How do i find the force exerted on the table? Answer is (g - 0.5pi^2)N

InteGrand · Jul 3, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
A mass of 1kg is fastened by a string of length 1m to a point 0.5m above a smooth horizontal table and is describing a circl o nthe table with uniform angular speed of 1 revolution in 2 seconds.

How do i find the force exerted on the table? Answer is (g - 0.5pi^2)N

$\noindent Draw a diagram with the fixed point $P$ $0.5$ m above the table, the mass on the table, and a string connecting the point and the mass. Note that the string is $1$ m in length, so using trigonometry, we can find all angles in the right-angled triangle formed by $P$, the mass, and the point on the table directly below $P$ (which is the circle's centre). (In fact, it'll just be a 30\textendash 60\textendash 90 triangle.) Find the horizontal component of the tension in the string (using $T_{x} = mr\omega^2$, where $T_x$ is the horizontal component of the tension in newtons, $m$ is the object's mass in kilograms, $r$ is the radius of the circle in metres, and $\omega$ is the angular velocity in radians per second. To find $r$, use Pythagoras' Theorem.). Once we have found $T_x$, we can find $T_y$ using trigonometry. The force exerted by the mass on the table is then $mg - T_y$ downwards.$

More working is below (in a spoiler in case you want to do it yourself first).

$\noindent So essentially, we have $r=\frac{\sqrt{3}}{2}$ m using Pythagoras. Then since $\omega = \frac{2\pi \text{ rad}}{2\text{ s}}=\pi \text{ rad s}^{-1}$, we have $T_x = mr\omega^2 = 1 \cdot \frac{\sqrt{3}}{2} \cdot \pi^2 = \frac{\sqrt{3}}{2}\pi^2 \text{ N}$. So $T_y = \frac{1}{2}\pi^2\text{ N}$ by trigonometry. So the force exerted by the mass on the table is $mg - T_y = 1\cdot g - \frac{1}{2}\pi^2 = \left(g-\frac{1}{2}\pi^2 \right)\text{ N}$ downwards.$

Paradoxica · Jul 3, 2016

Re: HSC 2016 4U Marathon

The roots of the equation

$x^5 - x +1 =0$

are

$\alpha, \beta, \gamma, \delta, \epsilon$

Evaluate:

$\alpha^{25}+ \beta^{25}+ \gamma^{25}+ \delta^{25}+ \epsilon^{25}$

KingOfActing · Jul 4, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
The roots of the equation

$x^5 - x +1 =0$

are

$\alpha, \beta, \gamma, \delta, \epsilon$

Evaluate:

$\alpha^{25}+ \beta^{25}+ \gamma^{25}+ \delta^{25}+ \epsilon^{25}$

$$First, notation: $e_n$ will represent the elementary symmetric polynomial of degree $n$ of the roots, and $p_n $ will represent the $n$th degree power sum of the roots. We are looking for$p_25$. It is clear by Vieta's formulas to see that $e_n = 0 $ for any $n \neq 4,\,5 $ and that $e_4 = e_5 = -1$.$\\$By Newton's Identity, and considering the values of $e_n$ that aren't $0$ we arrive the reccurence relation $\\p_n = p_{n-4} - p_{n-5}\\$with $p_1 = p_2 = p_3 = 0,\,p_4 = 4,\, p_5 = -5$.$\\$It is not difficult* to continue this recurrence and obtain\\$0,\,0\,,0\,,4\,,-5\,,0\,,0\,,4\,,-9\,,5\,,0\,,4,\\-13\,,14\,,-5\,,4\,,-17\,,27\,,-19\,,9\,,-21\,,44\,,-46\,,28\,,-30\\$and hence $\alpha^{25} + \beta^{25} + \gamma^{25} + \delta^{25}+\epsilon^{25}=-30$

_{*Apparently it was really difficult for me at 2am}

Paradoxica · Jul 4, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
$$\noindent First, notation: $e_n$ will represent the elementary symmetric polynomial of degree $n$ of the roots, and $p_n $ will represent the $n$th degree power sum of the roots. We are looking for$p_25$. It is clear by Vieta's formulas to see that $e_n = 0 $ for any $n \neq 4,\,5 $ and that $e_4 = e_5 = -1$.$\\$By Newton's Identity, and considering the values of $e_n$ that aren't $0$ we arrive the reccurence relation $\\p_n = p_{n-4} - p_{n-5}\\$with $p_1 = p_2 = p_3 = 0,\,p_4 = 4,\, p_5 = -5$.$\\$It is not difficult* to continue this recurrence and obtain\\$0,\,0\,,0\,,4\,,-5\,,0\,,0\,,4\,,-9\,,5\,,0\,,4,\\-13\,,14\,,-5\,,4\,,-17\,,27\,,-19\,,9\,,-21\,,44\,,-46\,,28\,,-30\\$and hence $\alpha^{25} + \beta^{25} + \gamma^{25} + \delta^{25}+\epsilon^{25}=-30$

_{*Apparently it was really difficult for me at 2am}

Nice. Here's mine:

$$\noindent Since we want $x^{25}$ summed over all the roots, we take the equation, rearrange it, and take the fifth power of both sides. Expanding by inspection, we have:$ \\ x^{25} = (x-1)^5 = x^5 - 5x^4 +10x^3 -10x^2 +5x-1 \\ $Now, note the following:$\\ x^5 = x-1 \\ x^4 = 1-\frac{1}{x} \\ x^3 = \frac{1}{x}-\frac{1}{x^2} \\ $Substituting these into the above gives us: $ \\ x^{25} = 6x+\frac{15}{x} - 10x^2 - \frac{10}{x^2} - 7 \\ $By inspection, the equation with roots $\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma},\frac{1}{\delta},\frac{1}{\epsilon}$, is $x^5-x^4+1=0 \\ \sum \alpha = 0, \sum \alpha \beta = 0, \sum \alpha^2 = \left(\sum \alpha \right)^2 - 2\sum \alpha \beta = 0 \\ \sum \frac{1}{\alpha} = 1, \sum \frac{1}{\alpha \beta} = 0, \sum \frac{1}{\alpha^2} = \left(\sum \frac{1}{\alpha} \right)^2 - 2\sum \frac{1}{\alpha \beta} = 1$

$$\noindent Now we can make our conclusion.$ \\ \sum \alpha^5 = 6\sum \alpha + 15\sum \frac{1}{\alpha} - 10\sum \alpha^2 - 10\sum \frac{1}{\alpha^2} -35 \\ \sum \alpha^5 = 15 - 10 - 35 = -30 \\ $\Huge$ \mathfrak{QED}$

Paradoxica · Jul 4, 2016

Re: HSC 2016 4U Marathon

Evaluate the following:

$\lim_{n \to \infty} \int_0^\frac{\pi}{4} n\tan^{n}{x} $d$x$

KingOfActing · Jul 4, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Evaluate the following:

$\lim_{n \to \infty} \int_0^\frac{\pi}{4} n\tan^{n}{x} $d$x$

Assuming the limit exists:

$Let $ x = \tan^{-1}u \iff dx = \frac{du}{1+u^2}$. Our integral becomes $ n\int_0^1 \frac{u^n}{1+u^2}\,du$. Letting $ I_n = n\int_0^1 \frac{u^n}{1+u^2}\,du, $ we wish to compute $\lim_{n\to\infty} I_n.\\$ Note that:$

$\begin{align*} n\int_0^1 \frac{u^n}{1+u^2}\,du &= n\int_0^1 u^{n-2}-\frac{u^{n-2}}{u^2+1}\,du\\I_n &= \frac{n}{n-1} - \frac{n}{n-2}I_{n-2}\end{align*}$

$$Assuming that $L = \lim_{n\to\infty}I_n $ exists, we get$\\L = 1 - L \iff L = \frac12$

Paradoxica · Jul 4, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
Assuming the limit exists:

$Let $ x = \tan^{-1}u \iff dx = \frac{du}{1+u^2}$. Our integral becomes $ n\int_0^1 \frac{u^n}{1+u^2}\,du$. Letting $ I_n = n\int_0^1 \frac{u^n}{1+u^2}\,du, $ we wish to compute $\lim_{n\to\infty} I_n.\\$ Note that:$

$\begin{align*} n\int_0^1 \frac{u^n}{1+u^2}\,du &= n\int_0^1 u^{n-2}-\frac{u^{n-2}}{u^2+1}\,du\\I_n &= \frac{n}{n-1} - \frac{n}{n-2}I_{n-2}\end{align*}$

$$Assuming that $L = \lim_{n\to\infty}I_n $ exists, we get$\\L = 1 - L \iff L = \frac12$

It's fairly easy to prove the limit does exist by using basic graphical inequalities and integration by parts.

Actually, do that as an exercise.

KingOfActing · Jul 4, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
It's fairly easy to prove the limit does exist by using basic graphical inequalities and integration by parts.

Actually, do that as an exercise.

$0 \leq u \leq 1 \iff 1 \leq u^2 + 1 \leq 2 \iff \frac{n}{2(n+1)} \leq I_n \leq \frac{n}{n+1}$

This doesn't prove convergence, but it does prove boundedness. Ceebs finding good upper limit and using squeeze theorem

hedgehog_7 · Jul 10, 2016

Re: HSC 2016 4U Marathon

Complex number question

I) If w is a non real seventh root of smallest positive argument, show that w = cis 2pi/7
II) Show that 1 + w^2 + w^3 + w^3 + w^4+ w^5 + w^6 = 0
III) Show that w + w^-1 = 2 cos (2pi/7) , hence, find the exact value of cos (pi/7)cos(2pi/7)cos(3pi/7)
IV) Show that z1 = w + w^2 + w^4 and z2 = w^3 + w^5 + w^6 are the roots of the equation z^2+z+2 = 0
V) State the exact value of sin(2pi/7) + sin(3pi/7) - sin(pi/7).

I know how to do parts I,II,III and IV. But im having trouble with (V). I understand that you have to equate Im(z1) for w + w^2 + w^4 and then manipulate the sin to get what the question is asking for. However, the problem is i dont know why it has to be equated with z1 and not z2?. The answer provided is root7/2 which is the Im(z1) value but why is the value taken for Im(z1) and not Im(z2) which provides -root7/2.

Thanks!

Paradoxica · Jul 10, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
Complex number question

I) If w is a non real seventh root of smallest positive argument, show that w = cis 2pi/7
II) Show that 1 + w^2 + w^3 + w^3 + w^4+ w^5 + w^6 = 0
III) Show that w + w^-1 = 2 cos (2pi/7) , hence, find the exact value of cos (pi/7)cos(2pi/7)cos(3pi/7)
IV) Show that z1 = w + w^2 + w^4 and z2 = w^3 + w^5 + w^6 are the roots of the equation z^2+z+2 = 0
V) State the exact value of sin(2pi/7) + sin(3pi/7) - sin(pi/7).

I know how to do parts I,II,III and IV. But im having trouble with (V). I understand that you have to equate Im(z1) for w + w^2 + w^4 and then manipulate the sin to get what the question is asking for. However, the problem is i dont know why it has to be equated with z1 and not z2?. The answer provided is root7/2 which is the Im(z1) value but why is the value taken for Im(z1) and not Im(z2) which provides -root7/2.

Thanks!

Plot the roots on an argand diagram. You will observe (by vector addition) that the real/imaginary parts after addition are in specific portions of the complex plane. This, though informal, should be sufficient (for the purposes of the HSC) to deduce which root is which.

InteGrand · Jul 10, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
Complex number question

I) If w is a non real seventh root of smallest positive argument, show that w = cis 2pi/7
II) Show that 1 + w^2 + w^3 + w^3 + w^4+ w^5 + w^6 = 0
III) Show that w + w^-1 = 2 cos (2pi/7) , hence, find the exact value of cos (pi/7)cos(2pi/7)cos(3pi/7)
IV) Show that z1 = w + w^2 + w^4 and z2 = w^3 + w^5 + w^6 are the roots of the equation z^2+z+2 = 0
V) State the exact value of sin(2pi/7) + sin(3pi/7) - sin(pi/7).

I know how to do parts I,II,III and IV. But im having trouble with (V). I understand that you have to equate Im(z1) for w + w^2 + w^4 and then manipulate the sin to get what the question is asking for. However, the problem is i dont know why it has to be equated with z1 and not z2?. The answer provided is root7/2 which is the Im(z1) value but why is the value taken for Im(z1) and not Im(z2) which provides -root7/2.

Thanks!

$\noindent If we want to do it algebraically, note since $w = \mathrm{cis}\left( \frac{2\pi}{7}\right)$, we have $w^2 = \mathrm{cis}\left(\frac{4\pi}{7}\right)\Rightarrow \mathrm{Im}\left(w^2\right) = \sin \frac{4\pi}{7} = \sin \frac{3\pi}{7}$ (supplementary angles trig. identity for sine).$

$\noindent Also, $w^4 = \mathrm{cis} \left(\frac{8\pi}{7}\right) \Rightarrow \mathrm{Im}\left(w^4\right) = \sin \frac{8\pi}{7} = -\sin \frac{\pi}{7}$.$

$\noindent So the given sum of sines equals $\mathrm{Im}\left( w \right) + \mathrm{Im}\left(w^2\right) +\mathrm{Im} \left(w^4\right) = \mathrm{Im}\left(w + w^2 + w^4 \right) = \mathrm{Im}\left( z_1 \right)$.$

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