Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

Kle123 · « **Reply #1425 on:** September 28, 2017, 07:45:00 pm »

Hi, Could I get help on this proof:
(n-1)Ck+(n-2)Ck+(n-3)Ck+...+kCk=nC(k+1)
Thankss

RuiAce · « **Reply #1426 on:** September 28, 2017, 07:47:23 pm »

Quote from: Kle123 on September 28, 2017, 07:45:00 pm

Hi, Could I get help on this proof:
(n-1)Ck+(n-2)Ck+(n-3)Ck+...+kCk=nC(k+1)
Thankss

This was one of the 3U questions in 2009.

raymatar · « **Reply #1427 on:** September 29, 2017, 10:16:48 am »

Can anyone help with v please?

Thanks

RuiAce · « **Reply #1428 on:** September 29, 2017, 10:50:02 am »

Quote from: raymatar on September 29, 2017, 10:16:48 am

Can anyone help with v please?

Thanks

$\text{Suppose, without loss of generality, that }\\ \alpha\text{ is the right angle.}$
This is because if $\beta$ or $\gamma$ were chosen to be the right angle instead, the same argument would work. This falls from just the fact that $\alpha,\beta,\gamma$ are just the angles of the triangle.
$\text{In a right-angled isosceles triangle, the angles}\\ \textbf{must be }\frac{\pi}{2},\frac{\pi}{4},\frac\pi4\\ \text{so it follows that }\beta=\gamma=\frac\pi4$
_________________________________

Remark: Internationally, cosec is denoted as "csc", and is what LaTeX renders. Hence it will be what I type.
$\text{Upon substitution,}\\ \left(\cot\theta-\cot\frac\pi2\right)\left(\cot\theta-\cot\frac\pi4\right)\left(\cot\theta-\cot\frac\pi4\right) = \csc \frac\pi2\csc \frac\pi4\csc \frac\pi4$
$\text{After evaluating the exact values}\\ \begin{align*}\cot\theta \left(\cot\theta-1\right)^2&=2\\ \cot^3\theta - 2\cot^2\theta + \cot\theta - 2 &=0\\ \cot^2\theta(\cot\theta-2)+(\cot\theta-2)&=0\\ (\cot^2\theta+1)(\cot\theta-2)&=0\end{align*}$
$\cot^2\theta + 1 = 0\implies \cot^2\theta=-1\text{ will have no solution}\\ \text{Hence }\cot\theta-2=0 \iff \tan \theta = \frac12 \iff \theta =\tan^{-1}\frac12$

raymatar · « **Reply #1429 on:** September 29, 2017, 12:12:48 pm »

Quote from: RuiAce on September 29, 2017, 10:50:02 am

$\text{Suppose, without loss of generality, that }\\ \alpha\text{ is the right angle.}$
This is because if $\beta$ or $\gamma$ were chosen to be the right angle instead, the same argument would work. This falls from just the fact that $\alpha,\beta,\gamma$ are just the angles of the triangle.
$\text{In a right-angled isosceles triangle, the angles}\\ \textbf{must be }\frac{\pi}{2},\frac{\pi}{4},\frac\pi4\\ \text{so it follows that }\beta=\gamma=\frac\pi4$
_________________________________

Remark: Internationally, cosec is denoted as "csc", and is what LaTeX renders. Hence it will be what I type.
$\text{Upon substitution,}\\ \left(\cot\theta-\cot\frac\pi2\right)\left(\cot\theta-\cot\frac\pi4\right)\left(\cot\theta-\cot\frac\pi4\right) = \csc \frac\pi2\csc \frac\pi4\csc \frac\pi4$
$\text{After evaluating the exact values}\\ \begin{align*}\cot\theta \left(\cot\theta-1\right)^2&=2\\ \cot^3\theta - 2\cot^2\theta + \cot\theta - 2 &=0\\ \cot^2\theta(\cot\theta-2)+(\cot\theta-2)&=0\\ (\cot^2\theta+1)(\cot\theta-2)&=0\end{align*}$
$\cot^2\theta + 1 = 0\implies \cot^2\theta=-1\text{ will have no solution}\\ \text{Hence }\cot\theta-2=0 \iff \tan \theta = \frac12 \iff \theta =\tan^{-1}\frac12$

How do you find the cot equation?

RuiAce · « **Reply #1430 on:** September 29, 2017, 12:31:16 pm »

Quote from: raymatar on September 29, 2017, 12:12:48 pm

How do you find the cot equation?

That was what you proved in iv.

itssona · « **Reply #1431 on:** September 29, 2017, 04:14:26 pm »

halp pls

RuiAce · « **Reply #1432 on:** September 29, 2017, 07:46:11 pm »

Quote from: sssona09 on September 29, 2017, 04:14:26 pm

halp pls

Just looking at this, part b is clearly wrong, as $ w = z(\cos\theta+i\sin\theta)$ so $ w\overline{z} = z^2(\cos\theta-i\sin\theta)$

itssona · « **Reply #1433 on:** September 30, 2017, 10:17:03 am »

Quote from: RuiAce on September 29, 2017, 07:46:11 pm

Just looking at this, part b is clearly wrong, as $ w = z(\cos\theta+i\sin\theta)$ so $ w\overline{z} = z^2(\cos\theta-i\sin\theta)$

ohh okay thank you

help with this please

Mod edit: Post merge

RuiAce · « **Reply #1434 on:** September 30, 2017, 02:22:23 pm »

Quote from: sssona09 on September 30, 2017, 10:17:03 am

help with this please

$\text{Since }\triangle OAB\text{ is right-angled isosceles at }A\\ \text{ it follows that }\angle OBA = \angle BOA = \frac{\pi}{4}$
$\text{Since a base angle of }\triangle COB\text{ is equal to }\frac{7\pi}{24}\\ \text{it then follows that the other base angle is }\frac{7\pi}{24}\\ \text{so }\angle COB = \frac{5\pi}{12}\text{ using the angle sum of a triangle}$
$\text{Hence }\angle AOC = \frac{5\pi}{12}+\frac\pi4=\frac{2\pi}3$
___________________________________________
$\text{From vector addition,}\\ \begin{align*}\overrightarrow{OB}&=\overrightarrow{OA}+\overrightarrow{AB}\\ &= \alpha + i\alpha = (1+i)\alpha\end{align*}\\ \text{noting that }\overrightarrow{AB}\text{ is a rotation of }\overrightarrow{OA}\text{ by }90^\circ$
$\text{It then follows that }|\beta| = |(1+i)\alpha| = \sqrt{2}|\alpha|\\ \text{and hence, as }\triangle COB\text{ is isosceles, }|\gamma| = \sqrt{2}|\alpha|$
$\text{So since }\alpha\text{ must be dilated by a factor of }\sqrt2\\ \text{and rotated by }\frac{2\pi}{3}\text{ to get }\gamma\\ \text{it follows that }\gamma = \sqrt{2} \left(\cos \frac{2\pi}3+i\sin \frac{2\pi}3\right)\alpha$
___________________________________________
$\text{With the aid of De Moivre's theorem}\\\begin{align*}&\quad 2\alpha^2+\gamma^2+\alpha\gamma\sqrt{2}\\ &= 2\alpha^2+2\left(\cos \frac{4\pi}3+i\sin \frac{4\pi}3\right)\alpha^2 + 2\alpha^2\left(\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)\\ &= 2\left(1 + (\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}) + (\cos \frac{4\pi}{3}+i\sin \frac{4\pi}{3})\right)\\ &=0\end{align*}\\ \text{after explicitly computing the exact values}$
$\text{It may be worth remarking that the expression equating to 0 is NOT a coincidence}\\ \text{as you've just investigated the cube roots of unity here}$

itssona · « **Reply #1435 on:** October 01, 2017, 12:19:26 pm »

hiii if i wanna sketch locus of arg (z+1/z-i)

then do i first see where its undefined and put open circles on the graph at those points? I think i should but i only ever heard of it once so I'm not sure.. thank you

edit: please show how to do this? the solution is a semi circle but how

RuiAce · « **Reply #1436 on:** October 01, 2017, 02:29:53 pm »

Quote from: sssona09 on October 01, 2017, 12:19:26 pm

hiii if i wanna sketch locus of arg (z+1/z-i)

then do i first see where its undefined and put open circles on the graph at those points? I think i should but i only ever heard of it once so I'm not sure.. thank you

edit: please show how to do this? the solution is a semi circle but how

I've never spent the time to fully understand the locus. I just remember how to draw it
$\text{The locus of }\arg(z-z_1)-\arg(z-z_2) = \arg \left( \frac{z-z_1}{z-z_2}\right) = \theta\\ \text{always looks like the arc of a circle, drawn counterclockwise}\\ \textbf{from }z_1\textbf{, to }z_2$

seventeenboi · « **Reply #1437 on:** October 02, 2017, 03:56:42 pm »

hellooo I don't understand this question or the solutions given pls help

thanks in advanced

RuiAce · « **Reply #1438 on:** October 02, 2017, 04:00:22 pm »

Quote from: seventeenboi on October 02, 2017, 03:56:42 pm

hellooo I don't understand this question or the solutions given pls help
thanks in advanced

$\text{A) is possible as }\frac{d^2x}{dt^2} = 2\text{ which is constant}$
$\text{B) is possible as }\frac{d^2x}{dt^2} = \frac{d}{dx} \frac{v^2}{2} = 1\text{ which is constant}$
$\text{C) is possible as }\frac{d^2x}{dt^2} = \frac{dv}{dt} = 4\text{ which is constant}$
$\text{D) fails the initial condition as }\frac{d^2x}{dt^2} = 6t\text{ which now depends on }t,\text{ so is not constant}$

lsong · « **Reply #1439 on:** October 02, 2017, 10:02:34 pm »

Hi all,

Just a general question if anyone knows.
If we copy down a diagram (e.g. circle geometry) and add a new element (e.g. a new point, or line), do we still need to write down what was added?

Thanks

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