Author Topic: 4U Maths Question Thread (Read 665367 times) Tweet Share

Kle123 · « **Reply #870 on:** February 22, 2017, 07:01:57 pm »

Quote from: RuiAce on February 22, 2017, 06:39:01 pm

$\text{In a similar way, the other one shows our S.P. to be at }y=-2$

THANK YOU RUI. Your explanations are soooo easy to understand. You're the greatest!

Kle123 · « **Reply #871 on:** February 24, 2017, 12:04:35 am »

This questions seems easy enough but i just don't trust my proof (it's kinda dodgy). Could someone show me how it's done. Thank you.

RuiAce · « **Reply #872 on:** February 24, 2017, 12:36:25 am »

Quote from: Kle123 on February 24, 2017, 12:04:35 am

This questions seems easy enough but i just don't trust my proof (it's kinda dodgy). Could someone show me how it's done. Thank you.

$\text{Suppose, w.l.o.g. that }a>b$
$\text{The standard ellipse obviously has eccentricity}\\ e_1=\sqrt{1-\frac{b^2}{a^2}}\\ \text{and thus foci at }\left(\pm a\sqrt{1-\frac{b^2}{a^2}},0\right)\\ \text{Expanding the }a\text{ into that square root tidies things up a bit:}\\ (\pm \sqrt{a^2-b^2},0)$
$\text{The other two ellipses have eccentricity respectively:}\\ e=\sqrt{1-\frac{b^2\pm \lambda^2}{a^2\pm \lambda^2}}$
$\text{So the foci of these ellipses are}\\ \begin{align*}&\quad \left(\pm (a^2\pm \lambda^2) \sqrt{1-\frac{b^2\pm \lambda^2}{a^2\pm \lambda^2}},0\right)\\&=(\pm \sqrt{(a^2\pm \lambda^2)-(b^2 \pm \lambda^2)},0)\\ &= (\sqrt{a^2-b^2},0)\end{align*}\\ \text{Matching the focus found earlier.}$

Kle123 · « **Reply #873 on:** February 24, 2017, 07:31:57 am »

Quote from: RuiAce on February 24, 2017, 12:36:25 am

$\text{Suppose, w.l.o.g. that }a>b$
$\text{The standard ellipse obviously has eccentricity}\\ e_1=\sqrt{1-\frac{b^2}{a^2}}\\ \text{and thus foci at }\left(\pm a\sqrt{1-\frac{b^2}{a^2}},0\right)\\ \text{Expanding the }a\text{ into that square root tidies things up a bit:}\\ (\pm \sqrt{a^2-b^2},0)$
$\text{The other two ellipses have eccentricity respectively:}\\ e=\sqrt{1-\frac{b^2\pm \lambda^2}{a^2\pm \lambda^2}}$
$\text{So the foci of these ellipses are}\\ \begin{align*}&\quad \left(\pm (a^2\pm \lambda^2) \sqrt{1-\frac{b^2\pm \lambda^2}{a^2\pm \lambda^2}},0\right)\\&=(\pm \sqrt{(a^2\pm \lambda^2)-(b^2 \pm \lambda^2)},0)\\ &= (\sqrt{a^2-b^2},0)\end{align*}\\ \text{Matching the focus found earlier.}$

Omg... I just realised this morning when I was thinking about it nonchalantly. Sleeep really does help! I used b^2=a^2(e^2-1) and didnt bother checking. Sorry for wasting your time rui.

RuiAce · « **Reply #874 on:** February 24, 2017, 11:10:09 am »

Quote from: Kle123 on February 24, 2017, 07:31:57 am

Omg... I just realised this morning when I was thinking about it nonchalantly. Sleeep really does help! I used b^2=a^2(e^2-1) and didnt bother checking. Sorry for wasting your time rui.

Sleep is a beautiful thing. I had a lot of those moments over the last two years of my maths life.

VydekiE · « **Reply #875 on:** February 26, 2017, 08:30:26 am »

Hi,
I'm having trouble with this polynomial question. It would be wonderful if I could get some help on it.
1. Find the value of k such that y=x^2-4x+6 and y=k-6x-x^2 are tangential to each other at some point.
Thank you!!

RuiAce · « **Reply #876 on:** February 26, 2017, 09:07:11 am »

Quote from: VydekiE on February 26, 2017, 08:30:26 am

Hi,
I'm having trouble with this polynomial question. It would be wonderful if I could get some help on it.
1. Find the value of k such that y=x^2-4x+6 and y=k-6x-x^2 are tangential to each other at some point.
Thank you!!

$\text{If the two parabolas are tangential to each other}\\ \text{Then there should be one and only one unique}\\ \text{point of intersection}\\ \text{i.e. the point of contact}$
$\text{Suppose we wish to find this point of intersection}\\ \text{Then solving simultaneously}\\ \begin{align*}x^2-4x+6&=k-6x-x^2\\2x^2 + 2x + (6-k)&=0 \end{align*}$
$\text{We've stated that there is only one point of intersection}\\ \text{The above quadratic allows us to find that point of intersection}\\\text{This means the quadratic equation only has 1 solution for }x$
$\text{If a quadratic has only one solution for }x\\ \text{then the discriminant satisfies }\Delta = 0\\ \begin{align*}2^2-4(2)(6-k)&=0\\ k&=\frac{11}2\end{align*}$
The question is of a 3U calibre and applies a common 2U trick, but it's likely that it builds into something meant for only 4U students.

VydekiE · « **Reply #877 on:** February 26, 2017, 09:16:21 am »

Thank you so much!!

VydekiE · « **Reply #878 on:** February 26, 2017, 09:24:17 am »

Hi, I have another polynomial question and it would be great if I could get some help on this
1. Find the values of k for which the polynomial p(x)=2x^3-9x^2+12x-k has
a) one distinct real root
b) repeated roots
Thank you!!

RuiAce · « **Reply #879 on:** February 26, 2017, 10:05:24 am »

Quote from: VydekiE on February 26, 2017, 09:24:17 am

Hi, I have another polynomial question and it would be great if I could get some help on this
1. Find the values of k for which the polynomial p(x)=2x^3-9x^2+12x-k has
a) one distinct real root
b) repeated roots
Thank you!!

$\text{Note that }p(x)\text{ is a cubic}\\ \text{with }p^\prime(x)=6x^2-18x+12$
$\text{To find the stationary points of }p(x)\\ \text{Setting }p^\prime(x)=0\text{ gives }6(x-1)(x-2)=0\\ \text{So the stationary points are at }x=(1,5-k),(2,4-k)\\ \text{We now turn to the graph}$
GeoGebra simulation attached
$\text{Recall that }k\text{ is just a constant}\\ \text{and that if we minus off a constant}\\ \textit{we are translating the graph DOWNWARDS}$
$\text{The graph will inevitably have the stationary point.}\\ \text{Since the stationary points are obviously turning points}\\ \text{depending on how much we }\textit{translate it}\\ \text{we may have 1, 2 or 3 roots.}$
$\text{The repeated root scenario in (b) obviously occurs}\\ \text{when the stationary point IS a root}\\ \text{this is what causes a double root, and clearly occurs when }k=4\text{ or }k=5$
$\text{If we take }k>5\\\text{then we'll bring the stationary point low enough}\\ \text{to only have 1 root as per the graph}\\ \text{and similarly for }k<4\\ \text{So for part (a) we have }k<4 \text{ or }k>5$

michaelalt · « **Reply #880 on:** February 26, 2017, 04:39:10 pm »

Does anyone know how to represent z^6=1 in mod arg form?

Sine · « **Reply #881 on:** February 26, 2017, 04:43:33 pm »

Quote from: michaelalt on February 26, 2017, 04:39:10 pm

Does anyone know how to represent z^6=1 in mod arg form?

$z^6=1cis(0)$
$z^6=cis(0)$

michaelalt · « **Reply #882 on:** February 26, 2017, 04:48:48 pm »

Quote from: Sine on February 26, 2017, 04:43:33 pm

$z^6=1cis(0)$
$z^6=cis(0)$

I did get that one, thanks!
However the answers also had cis (+/- pi/3), cis (+/- 2pi/3) and cis pi, so I didn't understand how they got those values.

jakesilove · « **Reply #883 on:** February 26, 2017, 04:56:51 pm »

Quote from: michaelalt on February 26, 2017, 04:39:10 pm

Does anyone know how to represent z^6=1 in mod arg form?

Hey! So, let's try to do this comprehensively. We're expecting six roots; called the six roots of unity

Let's create some complex number, z,

$z=rcis(\theta)$

Now,

$z^6=r^6cis(6\theta)=1$

Clearly, r=1. So, we're left with

$cis(6\theta)=1$
$cis(6\theta)=cis(0+2\pi n)$

For all real, integer values of n. Now, we're expecting six roots, so we need six ns. It's easiest to use the plus/minus version of the lowest integers.

$n=0, \pm1, \pm2, 3$

This gets us

$cis(6\theta)=cis(0), cis(2\pi), cis(-2\pi), cis(4\pi), cis(-4\pi), cis(6\pi)$

So,

$6\theta=0, 2\pi....$

Divide through all theta values by 6, then then add/subtract 2*pi so that all thetas are within -pi and pi (the required range for arguments in 4U). That should get you the answers you're expecting!

VydekiE · « **Reply #884 on:** February 26, 2017, 06:39:56 pm »

Hi, it would be great if I could get some help on this question.
1.
a) If w is a seventh root of 1, w does not equal to 1, show that w^3+w^2+w+1+1/w+1/w^2+1/w^3=0
b) By letting z=w +1/w reduce this equation into a cubic equation in z.
Thank you!!

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