Author Topic: 4U Maths Question Thread (Read 665231 times) Tweet Share

RuiAce · « **Reply #375 on:** August 10, 2016, 12:54:42 pm »

Quote from: Neutron on August 10, 2016, 11:38:58 am

Ahh yeah it came up in half yearlies, fingers crossed that it won't come up again :/ But anyhow, with complex loci such as:

arg(z+1)-arg(z-1)=pi/3

It looks like an arc of a circle, but how do you know whether to draw this arc above or below the x axis? Thanks ^^

For the locus arg(z-z₁)-arg(z-z₂)=α

I've found from experimentation that you always go in an anti-clockwise direction from z₁ to z₂

GeoGebra simulation to experiment with:
https://www.geogebra.org/m/AgXWc6mD

massive · « **Reply #376 on:** August 13, 2016, 11:29:39 am »

Hey guys for part iii of this question how do you get what b equals to. I've attached the worked solution as well but i don't understand where the third line comes from (how the p^4 just factorises out? :S). Thanks!

RuiAce · « **Reply #377 on:** August 13, 2016, 12:28:45 pm »

Quote from: massive on August 13, 2016, 11:29:39 am

Hey guys for part iii of this question how do you get what b equals to. I've attached the worked solution as well but i don't understand where the third line comes from (how the p^4 just factorises out? :S). Thanks!

b is the product of the roots. Note that the general quadratic Ax²+Bx+C has sum of roots α+β=-B/A, αβ=C/A

The p⁴ is pulled out by simply pulling p from the first bracket and p³ from the second.

massive · « **Reply #378 on:** August 13, 2016, 04:17:44 pm »

how do you do this guys?

RuiAce · « **Reply #379 on:** August 13, 2016, 04:21:15 pm »

Quote from: massive on August 13, 2016, 04:17:44 pm

how do you do this guys?

$\text{Without the degree of the denominator being higher than that of the numerator}\\ \text{We must perform polynomial long division (or something similar).}\\ \begin{align*}\frac{x^2}{(x+1)(x+2)}&=\frac{x^2+3x+2-(3x+2)}{x^2+3x+2}\\ &= 1 - \frac{3x+2}{(x+1)(x+2)}\end{align*}\\ \text{This is now something we can just do partial fractions with.}$
$\int \frac{x^2}{(x+1)(x+2)}dx=\int \left(1- \frac{4}{x+2}+\frac{1}{x+1} \right)dx$

massive · « **Reply #380 on:** August 13, 2016, 05:20:31 pm »

after proving that lhs=rhs do you use t formula? because it says "hence" :S

RuiAce · « **Reply #381 on:** August 13, 2016, 05:22:39 pm »

Quote from: massive on August 13, 2016, 05:20:31 pm

after proving that lhs=rhs do you use t formula? because it says "hence" :S

$\text{Observe that since }\\ I=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin x + \cos x}dx\\ I=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sin x + \cos x}dx$
$\text{Upon addition:} \\ 2I=\int_0^{\frac{\pi}{2}}\frac{\sin x+\cos x}{\sin x + \cos x}dx\\ \text{so }I=\frac{1}{2}\int_0^{\frac{\pi}{2}}dx=\frac{\pi}{4}$

massive · « **Reply #382 on:** August 13, 2016, 05:30:28 pm »

Quote from: RuiAce on August 13, 2016, 05:22:39 pm

$\text{Observe that since }\\ I=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin x + \cos x}dx\\ I=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sin x + \cos x}dx$
$\text{Upon addition:} \\ 2I=\int_0^{\frac{\pi}{2}}\frac{\sin x+\cos x}{\sin x + \cos x}dx\\ \text{so }I=\frac{1}{2}\int_0^{\frac{\pi}{2}}dx=\frac{\pi}{4}$

WOAAAH that's so smart lol thnx!

massive · « **Reply #383 on:** August 13, 2016, 07:28:51 pm »

how do you get this guys??

RuiAce · « **Reply #384 on:** August 13, 2016, 07:48:13 pm »

Quote from: massive on August 13, 2016, 07:28:51 pm

how do you get this guys??

$\text{If }x=4\sin^2\theta \implies dx=8\sin \theta \cos \theta \, d\theta$
$\begin{align*}\int_0^2\sqrt{x(4-x)}dx&=\int_0^{\frac{\pi}{4}}\sqrt{4\sin^2\theta(4\cos^2\theta)}(8\sin \theta \cos \theta\, d\theta)\\ &= 8\int_0^{\frac{\pi}{4}}4\sin^2\theta\cos^2\theta \, d\theta\\ &= 8\int_0^{\frac{\pi}{4}}\sin^2 2\theta \, d\theta \\ &= 4\int_0^{\frac{\pi}{4}}(1-\cos 4\theta)d\theta\end{align*}\\ \text{to which after evaluating equals }\pi$

massive · « **Reply #385 on:** August 13, 2016, 09:14:41 pm »

how do you do this?

jakesilove · « **Reply #386 on:** August 13, 2016, 09:37:00 pm »

Quote from: massive on August 13, 2016, 09:14:41 pm

how do you do this?

This was just a brute-force question, hope the steps make sense!

RuiAce · « **Reply #387 on:** August 13, 2016, 09:49:32 pm »

Quote from: jakesilove on August 13, 2016, 09:37:00 pm

This was just a brute-force question, hope the steps make sense!

(Image removed from quote.)

Woooaaaahhh Jake got a 4U question before me

Actually I was just studying for my discrete maths test.

jakesilove · « **Reply #388 on:** August 13, 2016, 10:18:11 pm »

Quote from: RuiAce on August 13, 2016, 09:49:32 pm

Woooaaaahhh Jake got a 4U question before me

Actually I was just studying for my discrete maths test.

You're just too good at LaTeX mate, I don't stand a chance

massive · « **Reply #389 on:** August 13, 2016, 10:24:44 pm »

guys how do you do this?

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