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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Thanks for your help everyone with that question I asked! :smile:
Original post by Twiggyrose
I've got both clashes: I'm resitting M1 and I've had a new teacher this year who can teach D1 and D2 so rather than attempt FP3 or S3 I'm doing that. And the worst thing is I've seen some days on Edexcel's exam timetable that have no exams in the morning (or afternoon)!


i know, mI worried that I might mess up on M1 and still be thinking about that during FP2!
Original post by Twiggyrose
I've got both clashes: I'm resitting M1 and I've had a new teacher this year who can teach D1 and D2 so rather than attempt FP3 or S3 I'm doing that. And the worst thing is I've seen some days on Edexcel's exam timetable that have no exams in the morning (or afternoon)!


Ouch that's horrible :/ also i thought that FP3 was compulsory ?
Reply 1183
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oinkk
3
What happens when we're finding a Maclaurin expansion of a function and when it comes to evaluating a particular part of it, we end up doing 1/0?
Reply 1184
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Nirm
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Original post by TheMarshmallows


Now that makes a lot more sense to me. Thanks!
Original post by TheMarshmallows
Ouch that's horrible :/ also i thought that FP3 was compulsory ?


According to the spec you have to have either FP2 or FP3, but most people go for both. I was told I could either do FP3 or M2 so I chose M2 because I thought it might be easier (I'm having second thoughts now though!)
Original post by ImJared
No one else did one, so here is what I got.



Original post by TheMarshmallows
Here's my way worse method using u+iv:
ImageUploadedByStudent Room1465317464.441976.jpg

the key to this question is getting
arg(z) = pi/4
arctan(y/X) = pi/4
y/x = 1
y = X

Then it's just a simple matter of mapping y=x with whatever method you're comfortable with



Posted from TSR Mobile


Thats how i do it but i wanna learn the easier way. How did you guys turn (x+xi+1)/(x+xi+i) into one.
Original post by Nirm
Now that makes a lot more sense to me. Thanks!


No problem, I'd also suggest you check out this video walk-through of it that someone posted earlier: https://youtu.be/hOxtcUJa1J4?t=2138
Original post by fpmaniac
Thats how i do it but i wanna learn the easier way. How did you guys turn (x+xi+1)/(x+xi+i) into one.


they're using modulus(w) = 1 so modulus( (x+xi+1)/(x+xi+i) ) = 1

mod(x +xi +1) = mod(x+xi+i)
square both sides
show they're equal
Original post by TheMarshmallows
they're using modulus(w) = 1 so modulus( (x+xi+1)/(x+xi+i) ) = 1

mod(x +xi +1) = mod(x+xi+i)
square both sides
show they're equal


But if you have to show that |w| = 1 in the end can you use it while solving it. Also im really confused with that now... Ive always used u+vi but took ages on every question but how do you do it the other way can you explain pls.
Original post by thatapanydude
i know, mI worried that I might mess up on M1 and still be thinking about that during FP2!


If you did well last year in normal maths and you're aiming for an A* you shouldn't be worried. I worked out if I get A* on both C3/C4 then I only need 15% ums on M1. So I haven't revised at all for M1. Though I think I'll get a bit more than 15%!
Original post by AmarPatel98
Which paper is this from?

I've only glanced at it, but because it's a modulus, it can either be positive of negative.

Take this example for instance
|x| = 2. Therefore, x can either be +2 or -2.

In this question, you have to possible cases:
arg (z+1) = pi/2
or
arg (z+1) = -pi/2

Hence you get the full line,


Original post by oinkk
My thoughts: a half line at an angle of pi/2 is just a vertical line. We only draw this above or on the real axis, as that's the only place where we can have an angle of pi/2 anticlockwise from the real axis.

Anything below the real axis would be an angle of -pi/2.

However, since it is a modulus, as you quite rightly said, the modulus, that is, the absolute value of the angle, is pi/2. So even points below the real axis (where they'd normally be at an angle of -pi/2) satisfy this.


Gg, thank you both :five:

Do well tomorrow!:smile:
Reply 1192
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wr123
1
Can anyone show a way to get from the top to the bottom? I honestly can't understand the (d^2y/dx^2)(dt/dx) part..



Screen Shot 2016-06-07 at 18.03.54.png
Reply 1193
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Original post by oinkk
What happens when we're finding a Maclaurin expansion of a function and when it comes to evaluating a particular part of it, we end up doing 1/0?


Where did this come up because I've only ever come across one case where it asked to find the value of k where k-1/4 was the denominator and it had to equal 0 so I had to say k=1/4.
Nah, he meant to do you need to show how you can obtain the auxiliary function, which can be done by substituting y=e^mx and differentiating then factorising e^mx
Original post by wr123
Can anyone show a way to get from the top to the bottom? I honestly can't understand the (d^2y/dx^2)(dt/dx) part..



Screen Shot 2016-06-07 at 18.03.54.png


Product rule with u=1xu = \dfrac{1}{x} and v=dydtv = \dfrac{dy}{dt}

d2ydx2=uv′+u′v\dfrac{d^{2}y}{dx^{2}} = uv' + u'v
Original post by mathsmann
Can anyone help me on 7b june 2014 R,,, why do you keep getting same answers for sin(theta) and when do you stop? I dont get it at all . Anyone care to help please?

https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/June%202014%20(R)%20QP%20-%20FP2%20Edexcel.pdf


Keep going til you get 5 distinct values for sin theta
Reply 1197
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Original post by wr123
Can anyone show a way to get from the top to the bottom? I honestly can't understand the (d^2y/dx^2)(dt/dx) part..



Screen Shot 2016-06-07 at 18.03.54.png


http://imgur.com/xKBlDyE

Hope this helps, it really brain teasing I know!
Original post by fpmaniac
But if you have to show that |w| = 1 in the end can you use it while solving it. Also im really confused with that now... Ive always used u+vi but took ages on every question but how do you do it the other way can you explain pls.


if its a "show that" question, then you can use the answer whilst solving it (to show that it is true). Generally using u+iv will work always, doing it with modulus' is usually less algebra/shorter but isn't always possible (and takes some thinking).

For example in the question we were talking about before, arg(z) = pi/4 is the half line y=x, so you can rewrite z = x + iy into z = x + ix (or z = y + iy if you prefer, it doesnt matter).

Then subbing into W = (Z+1)/(z+i) we get W = (x + ix + 1)/(x + ix + i)
grouping real and imaginary gives W = ( (x+1) + ix ) / (x + i(x+1) )

Looking at the answer, they want you to show that mod(W) = 1 so mod(our W) = 1

mod( (x+1) +ix) / (x + i(x+1)) ) = 1

mod((x+1) +ix) / mod(x+ i(x+1)) = 1

Remembering that mod(a + ib) = sqrt(a^2 + b^2) gives us

sqrt( (x+1)^2 + x^2) / sqrt( x^2 + (x+1)^2 ) = 1

which is clearly true because they're the same thing.
(edited 7 years ago)
Original post by oinkk
What happens when we're finding a Maclaurin expansion of a function and when it comes to evaluating a particular part of it, we end up doing 1/0?


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