Author Topic: 3U Maths Question Thread (Read 1235475 times) Tweet Share

f_tan · « **Reply #2640 on:** August 08, 2017, 09:30:46 pm »

How do you graph:

RuiAce · « **Reply #2641 on:** August 08, 2017, 09:35:15 pm »

Quote from: f_tan on August 08, 2017, 09:30:46 pm

How do you graph:
(Image removed from quote.)

Please find the posts on similar graphs, and then show any progress on how you applied this to $\sin^{-1}(\sin x)$ if you need further help.

winstondarmawan · « **Reply #2642 on:** August 08, 2017, 10:36:18 pm »

Would appreciate help for the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20707680_1294490204009833_905080762_n.png?oh=2d3d1518479730bd35a1a7e57c561304&oe=598BDB8F
TIA

RuiAce · « **Reply #2643 on:** August 08, 2017, 10:50:36 pm »

Quote from: winstondarmawan on August 08, 2017, 10:36:18 pm

Would appreciate help for the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20707680_1294490204009833_905080762_n.png?oh=2d3d1518479730bd35a1a7e57c561304&oe=598BDB8F
TIA

$\text{Wouldn't this just be }\binom{n}{2}\binom{n}{2}\text{ i.e. D?}$

stephjones · « **Reply #2644 on:** August 09, 2017, 01:03:53 pm »

hey guys, how would you answer this question?

RuiAce · « **Reply #2645 on:** August 09, 2017, 01:51:56 pm »

Quote from: stephjones on August 09, 2017, 01:03:53 pm

hey guys, how would you answer this question?

$\text{The trick relies on drawing your graph to scale}\\ \text{and identifying the line through the endpoints of motion}\\ \text{That, and another extremity.}$
$\text{Recall that the number of solutions to }\sin^{-1}x-mx=0\\ \text{is the number of points of intersection}\\\text{between }y=\sin^{-1}x\text{ and }y=mx$
GeoGebra simulation attached. Note that for the GeoGebra simulation's purpose, $m=\frac{\pi}2 M $ just to make the slider easier to use.
$\text{If your line passes through the endpoints, i.e. }\left(\pm 1, \pm\frac\pi2\right)\\ \text{Then the equation of the line is }y=\frac{\pi}2x\\ \text{This line, of course, intersects three times, because it also passes through }(0,0)$
$\text{You can use the GeoGebra simulation to verify that for any }0 < m \le \frac{\pi}{2}\\ \text{there will be three points of intersection.}$
$\text{When }m=0\text{, you will realise you only have one point of intersection}\\ \text{On the GeoGebra simulation, if you bring the slider to }0\\ \text{you will see how the two other points of intersection basically 'collapse'.}$
$\text{Also, if }m < 0\text{ or }m > \frac{\pi}2\\ \text{the line basically 'falls off' the sine curve}\\ \text{So }0 < m \le \frac{\pi}2\text{ is the domain of choices.}\\ \text{and hence, reading off the domain, the max is }m=\frac{\pi}2$

winstondarmawan · « **Reply #2646 on:** August 10, 2017, 07:05:49 am »

Answer is C but I'm not sure how. Would appreciate an explanation.

https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20786429_1295637207228466_1938049319_n.png?oh=8a3b574e546d86461738f7011515e009&oe=598E3EAC

Shadowxo · « **Reply #2647 on:** August 10, 2017, 08:50:57 am »

Quote from: winstondarmawan on August 10, 2017, 07:05:49 am

Answer is C but I'm not sure how. Would appreciate an explanation.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20786429_1295637207228466_1938049319_n.png?oh=8a3b574e546d86461738f7011515e009&oe=598E3EAC

A and B start from the same vertical distance from the ground. The only force acting on each of them is gravity which changes their vertical speed

a) since they start from the same distance vertically, their initial vertical velocity is the same (0) and the acceleration vertically is the same, this means they are in the air for the same length of time. The horizontal component is irrelevant, only the vertical component affects how long they're in the air. So, True

b) Their initial vertical speed is 0, they are accelerating vertically at the same acceleration (9.

so after the same length of time they will have covered the same distance and have the same vertical velocity. So, True

c) Vertically, A and B are travelling at the same speed on impact. Horizontally, B is travelling twice the speed of A (horizontal component of velocity is unchanged, no horizontal force). Overall velocity takes into account both horizontal and vertical speeds (note: horizontal and vertical speed don't just add to give the final speed, you'd use pythagoras as they're perpendicular). So, False

d) They are in the air the same length of time, and B's horizontal speed is twice that of A. The horizontal speeds stay constant, so we can use distance = velocity * time, therefore B travels twice the distance of A horizontally. So, true

Hope this helps

12carpim · « **Reply #2648 on:** August 10, 2017, 11:51:42 am »

Hey guys,

I am struggling to understand this circle geometry concept as I don't understand what they exactly mean by the same arc?

Thanks

chelseam · « **Reply #2649 on:** August 10, 2017, 12:05:46 pm »

Quote from: 12carpim on August 10, 2017, 11:51:42 am

Hey guys,

I am struggling to understand this circle geometry concept as I don't understand what they exactly mean by the same arc?

Thanks

Hey! It means that they're both standing on the exact same section of the circumference of the circle

Does that help to clear it up?

RuiAce · « **Reply #2650 on:** August 10, 2017, 12:06:26 pm »

Quote from: 12carpim on August 10, 2017, 11:51:42 am

Hey guys,

I am struggling to understand this circle geometry concept as I don't understand what they exactly mean by the same arc?

Thanks

In every circle, the arc has been emphasised in red and labelled $AB$.

In each circle, the angles being subtended come from the arc $AB$, to either the centre of the circle. Equivalently, the angles are always standing on the same arc, just that one is being subtended to the centre, whilst the other is being subtended to the circumference.

The important thing to realise that both angles stand on the same arc, and not say another arc $CD$ or something.

Piccolo · « **Reply #2651 on:** August 10, 2017, 12:32:14 pm »

Hey guys,

So I effectively missed the auxiliary method stuff in year 11, and I never got around to catching up with it! So here's a question from the 2014 CSSA trial, if someone could guide me through it that would be fab

Thanks!

RuiAce · « **Reply #2652 on:** August 10, 2017, 12:40:18 pm »

Quote from: Piccolo on August 10, 2017, 12:32:14 pm

Hey guys,

So I effectively missed the auxiliary method stuff in year 11, and I never got around to catching up with it! So here's a question from the 2014 CSSA trial, if someone could guide me through it that would be fab

Thanks!

$\text{Any equation of the form }a\sin x + b\cos x\\ \text{can be expressed as }R\sin (x+\alpha)\\ \text{and any equation of the form }a\cos x + b\sin x\\ \text{can be expressed as }R\cos (x-\alpha)$
$\text{where in both cases, }R=\sqrt{a^2+b^2}\\ \text{and }\tan \alpha = \frac{b}{a}$
_____________________________
$\text{Here, we have }\cos x -\sin x\\ \text{so matching, we have }a=1, b=1\\ \text{Hence, }R=\sqrt{1^2+1^2}=2\\ \text{and }\tan \alpha = 1 \implies \alpha = \frac{\pi}{4}$
$\therefore \cos x - \sin x = \sqrt2\cos \left(x-\frac{\pi}4\right)$

12carpim · « **Reply #2653 on:** August 10, 2017, 01:14:53 pm »

Thank you very much RuiAce for you help

bronwoolbank · « **Reply #2654 on:** August 10, 2017, 01:53:39 pm »

Hallo, I am having trouble with this

Search the forums now!

We have moved!

Author Topic: 3U Maths Question Thread (Read 1235475 times) Tweet Share

f_tan

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

winstondarmawan

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

stephjones

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

winstondarmawan

Re: 3U Maths Question Thread

Shadowxo

Re: 3U Maths Question Thread

12carpim

Re: 3U Maths Question Thread

chelseam

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Piccolo

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

12carpim

Re: 3U Maths Question Thread

bronwoolbank

Re: 3U Maths Question Thread

Recent Posts

User:
Password: