FP1 Edexcel (New Spec) - 30 January 2009

the_neg_master_ · 29-01-2009: 29th January 2009 23:15

Hi guys! How's revision going? Are you finding FP1 easy or difficult? Tbh, I personally find it quite hard especially the induction bit. I always find myself stuck with divisibility question (hope we get matrices/series tomorrow on induction). Oh, I also want to inform you that there is a new formulae book for the new FP1 specification (orange book instead of the old green book).

So to start with, how do you do this divisibilty induction part:
" Use the method of mathematical induction to prove that $3^{2n+2} + 8n - 9$ is divisible by 8 for all n is a member of all positive integers.

So.. n= 1.. tsk tsk.. 3^4 + 8 - 9

= 80 which is divisible by 8..

Assume n= k : $3^{2k+2} + 8k - 9$
therefore, n= k+1 : $3^{2(k+1)+2} + 8(k+1) - 9$
then.. f(k+1)-f(k) = $3^{2(k+1)+2} + 8(k+1) -9 - f(k)$

$3^{2(k+1)}.3^{2} + 8(k+1) -9 - f(k)$

$9(3^{2(k+1)}) + 8k+8-9 -( 3^{2k+2} + 8k - 9 )$

$9(3^{2k+2})-3^{2k+2} +8$

$8(3^{2k+2}) +8$

$8[(3^{2k+2}) +1]$

Therefore, f(n) is divisble by 8 when n=k+1, where n is a subset of all the positive integers.

ok.. so did i do it right? i'm struggling with the conclusion bit and i'm not sure if my conclusion is right.. help please..

cheemaj187 · 29-01-2009: 29th January 2009 23:23

f(k) = 3^(2k+2) + 8k -9

f(k+1) = 3^(2k+4) +8k +8 -9 = 3^(2k+4) +8k -1

= 3^2. 3^(2k+2) +8k -1

take out the 9 and put the f(k) in:

f(k+1) =3^2. 3^(2k+2) +8k -1

= 9(3^(2k+2) +8k -9) -72k +80

=9(3^(2k+2) +8k -9) +8(-9k +10)

sorry my latex is appalling, but the brackets after 3 is to the power of...if you see what I mean

sohanshah · 29-01-2009: 29th January 2009 23:26

Originally Posted by cheemaj187

take out the 9 and put the f(k) in:

f(k+1) =3^2. 3^(2k+2) +8k -1

= 9(3^(2k+2) +8k -9) -72k +80

=9(3^(2k+2) +8k -9) +8(-9k +10)

sorry my latex is appalling, but the brackets after 3 is to the power of...if you see what I mean

You need to use {} brackets around your powers.

linus_k · 29-01-2009: 29th January 2009 23:28

what you've done is fine, you've shown that the difference f(k+1) - f(k) is divisible by 8, from which it follows that f(k+1) is divisible by eight, so since [8 divides f(k)] => [8 divides f(k+1)] and 8 divides f(1) then 8 divides f(k) for all positive integers k, basically just state the PMI for the conclusion

the_neg_master_ · 29-01-2009: 29th January 2009 23:30

Originally Posted by cheemaj187

take out the 9 and put the f(k) in:

f(k+1) =3^2. 3^(2k+2) +8k -1

= 9(3^(2k+2) +8k -9) -72k +80

=9(3^(2k+2) +8k -9) +8(-9k +10)

sorry my latex is appalling, but the brackets after 3 is to the power of...if you see what I mean

that's ok.. thanks.. so i have to expand it first then simplify? FP1 is hard.. i'm still in Year 12 (haven''t done C2) and i have to learn how to turn degrees to radians.. hm..
how bout my conclusion? is it.. good?

Yiikes010 · 29-01-2009: 29th January 2009 23:31

i think the conclusion shud b lke: if summation is true for n=k, its shown 2 b tru 4 n=k+1....summation true 4 n=1, then by induction tru 4 n subset of positive integers

the_neg_master_ · 29-01-2009: 29th January 2009 23:32

Originally Posted by linus_k

what you've done is fine, you've shown that the difference f(k+1) - f(k) is divisible by 8, from which it follows that f(k+1) is divisible by eight, so since [8 divides f(k)] => [8 divides f(k+1)] and 8 divides f(1) then 8 divides f(k) for all positive integers k, basically just state the PMI for the conclusion

ah. i see. you have to say everything like n=1 is divisble by 8 then n= k etc.... also, what's PMI?

linus_k · 29-01-2009: 29th January 2009 23:34

Originally Posted by the_neg_master_

ah. i see. you have to say everything like n=1 is divisble by 8 then n= k etc.... also, what's PMI?

formally yes, although its pretty well known, depends on what the spec for the exam asks
PMI = Principle of Mathematical Induction

moleymus · 29-01-2009: 29th January 2009 23:34

I am finding FP1 really easy, it just like going back to the days of C1, FP2 is way harder for anyone out there thinking its much the same

the_neg_master_ · 29-01-2009: 29th January 2009 23:35

oh.. okay.. 'cause i wasn't in lessons when we did induction (got exams) which is why i don't know how to conclude.. oh, can anyone explain the difference between the interval bisection and interpolation, coz i always mix them up.. Is interval bisection the same with trial and error, right? and interpolation is the one where you use similar triangles, right? or is it the other way round?

Tyrotoxism · 29-01-2009: 29th January 2009 23:36

I can definitely imagine this exam being hard for those in year 12 taking it tomorrow.

the_neg_master_ · 29-01-2009: 29th January 2009 23:36

Originally Posted by moleymus

I am finding FP1 really easy, it just like going back to the days of C1, FP2 is way harder for anyone out there thinking its much the same

i just did my C1 exam (still in year 12) and it was ok.. but i think FP1 is harder compared to C1 and probz C2.

the_neg_master_ · 29-01-2009: 29th January 2009 23:38

Originally Posted by Tyrotoxism

I can definitely imagine this exam being hard for those in year 12 taking it tomorrow.

cheers bud.. pheww, someone out there can emphatise with my feelings.

Yiikes010 · 29-01-2009: 29th January 2009 23:41

has any1 gt the mark schemes 4 the 5 practice papers..it wud b reeeeeellly helpful if u cud post them...thnx!!!

Tyrotoxism · 29-01-2009: 29th January 2009 23:43

Originally Posted by the_neg_master_

cheers bud.. pheww, someone out there can emphatise with my feelings.

This exam would be so easy if you've already covered C3/C4. Our teacher has a hunch that tomorrows paper is going to be heavily based on the specimen paper.

Dom K. Jordan · 29-01-2009: 29th January 2009 23:46

All you need for the induction conclusion is to write "true for n=k+1 if true for n=k". You'll already have stated true for n=1 when you showed that earlier so then close with "true for all n E Z+ by mathematical induction".

moleymus · 29-01-2009: 29th January 2009 23:51

It is like all maths really, you learn a set way of doing something and apply it to everything you need it for until you learn and even better new way of doing things.

For all of you in year 12 who just sat C1, integration is just starting for you wait till FP3 oh my days!!!!!!

the_neg_master_ · 29-01-2009: 29th January 2009 23:53

Originally Posted by Tyrotoxism

This exam would be so easy if you've already covered C3/C4. Our teacher has a hunch that tomorrows paper is going to be heavily based on the specimen paper.

oh.. can you help me with the specimen paper question 2? how do u find a if the area of the image R is 18?
i found det(A) = a+4
and the coordinates of the image: (0,0), (a,-1), (a+8,1) (8,2)

Tyrotoxism · 29-01-2009: 29th January 2009 23:53

Originally Posted by moleymus

It is like all maths really, you learn a set way of doing something and apply it to everything you need it for until you learn and even better new way of doing things.

For all of you in year 12 who just sat C1, integration is just starting for you wait till FP3 oh my days!!!!!!

Haha, it's not that bad

Tyrotoxism · 29-01-2009: 29th January 2009 23:55

Originally Posted by the_neg_master_

oh.. can you help me with the specimen paper question 2? how do u find a if the area of the image R is 18?
i found det(A) = a+4
and the coordinates of the image: (0,0), (a,-1), (a+8,1) (8,2)

The area of R is 2

The area of the image is 18

Therefore the scale factor is 9

set Det(A) to equal nine and hey presto.

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FP1 Edexcel (New Spec) - 30 January 2009