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Edexcel FP1 Thread - 20th May, 2016

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What is the general consensus about the hard FP1 papers (which ones are hard).

I've done June 2013 and that was fine.
Original post by Glavien
For the June 2015 IAL paper Q5 (b), will my answer get full marks? As the mark scheme says the 'use of negative lengths scores M0'. What does that mean?

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I can't see your answer.
Original post by Zacken
I can't see your answer.


Sorry, can you see this post?

ImageUploadedByStudent Room1463661467.114392.jpg


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Original post by thesmallman
For the proof of divisibility qs- I always use the following method- can someone pls check for me if it is the right way to do it/would get me full marks:

E.g. prove that f(n)=7^(2n)-48n-1 is divisible by 2304
sub in 1, make assumption n=k etc... then

f(k+1)-f(k)= 7^(2k+2) - 48(k+1) -1 - 7^(2k) +48k + 1
= 48(7^(2k)) - 48

then I rearrange f(k) to make 7^(2k) subject of the equation
i.e. 7^(2k)= f(k) + 48k + 1
Then sub it in to f(k+1)-f(k)

Is this a correct way of doing it??

yeah that's correct but it's abit convoluted because:
f(k+1)=7^(2k+2) - 48(k+1) -1=49.7^(2k)-48k-49 then you could have just taken 69f(k) away from f(k+1) instead of taking f(k) away from f(k+1) then using a substitution: f(k+1)-49f(k)=2304k therefore f(k+1)=2304k+49f(k)
if you want to use substitution approach that's less convoluted then in my opinion:
you know that 49f(k)-49.7^(2k)+2352k+49=0 & f(k+1)=7^(2k+2) - 48(k+1) -1=49.7^(2k)-48k-49+0 then sub for zero and you get the same result f(k+1)=2304k+49f(k)
Original post by Glavien
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Nope, you're not allowed to use negative lengths. You're implicitly using similar triangles which have positive lengths.
Original post by Zacken
Nope, you're not allowed to use negative lengths. You're implicitly using similar triangles which have positive lengths.


So, I wont get full marks? Can I use similar triangles?
Original post by Glavien
So, I wont get full marks? Can I use similar triangles?


Nopes. All you have to do is write down what you did but just change all the negative lengths to positive.
Original post by NotNotBatman
Is ZZ∗=∣Z∣2 ZZ^* =\left |Z \right |^2 For all complex numbers?


Yeah :smile:

∣z∣2=a2+b2 |z|^2 = a^2 + b^2
zz∗=(a+bi)(a−bi)=a2+b2 z z^{*} = (a + bi)(a - bi) = a^2 + b^2
(edited 7 years ago)
Original post by NotNotBatman
Is ZZ∗=∣Z∣2 ZZ^* =\left |Z \right |^2 For all complex numbers?

yes because if you let z=a+ib then z*=a-ib and z z*=a2+b2=mod(z)2 mod(z)=(a2+b2)0.5
Original post by Zacken
Nopes. All you have to do is write down what you did but just change all the negative lengths to positive.


Ohh ok, thanks!! So, I should factorize out negative 1 from the denominator of both sides of the equation, so the negative signs cancel to give positive lengths?
Original post by Glavien
Ohh ok, thanks!! So, I should factorize out negative 1 from the denominator of both sides of the equation, so the negative signs cancel to give positive lengths?


Yes, but do it in your head, don't do it on paper or you'll score M0. You might want to look up the examsolutions video about this, he explains it well.
Original post by Zacken
Yes, but do it in your head, don't do it on paper or you'll score M0. You might want to look up the examsolutions video about this, he explains it well.


Ok, thanks for the help and advice!!!! :smile:
Are their any changes in specification to the IAL F1 papers? Compared to the standard Edexcel past papers for FP1
(edited 7 years ago)
Original post by Zacken
Nopes. All you have to do is write down what you did but just change all the negative lengths to positive.


WOW, you just made me very worried because I don't use similar triangles or the method at uses the gradient of the line. I just use this formula as it is extremely quick,easy to remember and always works. Let us say your root is in interval [a,b] then x=(a|f (b) | +b|f (a) |)/(|f (b) | +|f (a) |). I was wondering would they give me the marks on a linear interpolation question if I just used this formula.
Original post by circuits4life
WOW, you just made me very worried because I don't use similar triangles or the method at uses the gradient of the line. I just use this formula as it is extremely quick,easy to remember and always works. Let us say your root is in interval [a,b] then x=(a|f (b) | +b|f (a) |)/(|f (b) | +|f (a) |). I was wondering would they give me the marks on a linear interpolation question if I just used this formula.


Yes. That's basically similar triangles.
Original post by Zacken
Yes. That's basically similar triangles.


thanks so much man
Original post by Louisb19
What is the general consensus about the hard FP1 papers (which ones are hard).

I've done June 2013 and that was fine.


full marks - it wasn't hard at all - just felt tedious.
Original post by Louisb19
What is the general consensus about the hard FP1 papers (which ones are hard).

I've done June 2013 and that was fine.


Try IAL papers or gold papers - they're usually better to do.
Original post by Marxist
Try IAL papers or gold papers - they're usually better to do.


Doesn't IAL have a different spec?
Original post by JustDynamite
Are their any changes in specification to the IAL F1 papers? Compared to the standard Edexcel past papers for FP1


Those roots of polynomials and alpha, beta stuff you see in IAL F1 papers aren't on the normal FP1 edexcel papers, that's about it.

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