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Edexcel C3,C4 June 2013 Thread

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Reply 700
Original post by Better
Dan helped me on the second part.

To solve an Equation. You can either.
1. Equate and Solve Algebraically. You will then get your solution.
2. Draw the graph for the both and see where the intersect. Intersection point is the solution.

Hence they drew a graph

The other bit I can't remember but Post a link so other people can see it.


I get it! They manipulated f(x) = 0 to get two equal graphs. Ahh.
Reply 701
Does anyone know how to work out Q8c on the C3 Solomon Paper A?
Original post by sounique
Does anyone know how to work out Q8c on the C3 Solomon Paper A?


The root is where the lines intersect...

x^2 + 1 = 1/x-2

(x-2)(x^2+1)-1=0

x^3 - 2x^2 + x - 3 = 0

f(x) = x^3 - 2x^2 + x - 3

Test out a few integers... when the sign changes, you know to start of with one of the 2 integers :smile: then do the usual process....
Reply 703
Hey guys, can someone help me please :smile:

1366136050303.jpg

Integration by parts question from the edexcel book... Answer is meant to be -xcotx + ln(sinx) + C. Where have I gone wrong?

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Reply 704
Original post by posthumus
The root is where the lines intersect...

x^2 + 1 = 1/x-2

(x-2)(x^2+1)-1=0

x^3 - 2x^2 + x - 3 = 0

f(x) = x^3 - 2x^2 + x - 3

Test out a few integers... when the sign changes, you know to start of with one of the 2 integers :smile: then do the usual process....


Thanks! So, can we literally choose any integers for x0, and then sub into the iterative formula and we should get the same answer?
Original post by sounique
Thanks! So, can we literally choose any integers for x0, and then sub into the iterative formula and we should get the same answer?


I would have thought so yes... I just checked the mark scheme and they used Xo= 3 , although personally I would have thought 2 would be better since it gave answer closer to 0 (sorry if I ruined any surprises). but it does say "e.g." :smile:
Reply 706
Original post by suncake
Hey guys, can someone help me please :smile:

1366136050303.jpg

Integration by parts question from the edexcel book... Answer is meant to be -xcotx + ln(sinx) + C. Where have I gone wrong?

Posted from TSR Mobile


I think when you integrated sin(x)^-2 you never used the reverse chain rule.
for example if v = -cosecx then dv/dx = cosecxcotx not (sinx)^-2
(edited 11 years ago)
Reply 707
Original post by suncake
Hey guys, can someone help me please :smile:

1366136050303.jpg

Integration by parts question from the edexcel book... Answer is meant to be -xcotx + ln(sinx) + C. Where have I gone wrong?

Posted from TSR Mobile


This is how I would work it out.
Reply 708
Original post by posthumus
I would have thought so yes... I just checked the mark scheme and they used Xo= 3 , although personally I would have thought 2 would be better since it gave answer closer to 0 (sorry if I ruined any surprises). but it does say "e.g." :smile:


I did notice the e.g, I just assumed there was something lying deeper than that, something I didn't yet fully understand. Thanks mate
Original post by suncake
Hey guys, can someone help me please :smile:

1366136050303.jpg

Integration by parts question from the edexcel book... Answer is meant to be -xcotx + ln(sinx) + C. Where have I gone wrong?

Posted from TSR Mobile

third line - you integrated
Unparseable latex formula:

sinx^-^2

wrong, you need to use:12(cos2x1)\frac{-1}{2}(cos2x-1) since
Unparseable latex formula:

sinx^-^2 =\frac{1}{(sinx)^2}

(edited 11 years ago)
Reply 710
Original post by sounique
This is how I would work it out.


Yeah I realised that way was easier afterwards :redface: Thanks!

Original post by nmanvi
I think when you integrated sin(x)^-2 you never used the reverse chain rule.
for example if v = -cosecx then dv/dx = cosecxcotx not (sinx)^-2


How would you reverse chain rule sinx^-2 when it's just a single x? I don't get what you mean :colondollar:

Original post by gaffer dean
third line - you integrated
Unparseable latex formula:

sinx^-^2

wrong, you need to use the identities since:
Unparseable latex formula:

sinx^-^2 =\frac{1}{(sinx)^2}



Oh I see haha. Thanks! :smile:

(This is what happens when you neglect maths for a whole two weeks :rolleyes:)

Posted from TSR Mobile
(edited 11 years ago)
Reply 711
Original post by suncake
Yeah I realised that way was easier afterwards :redface: Thanks!



How would you reverse chain rule sinx^-2 when it's just a single x? I don't get what you mean :colondollar:



Oh I see haha. Thanks! :smile:

(This is what happens when you neglect maths for a whole two weeks :rolleyes:)

Posted from TSR Mobile


ignore me hun was chatting nonsense :colondollar:
Reply 712
Can someone please explain where they got the 3 pi /2 on Q7 in the C3 Solomon C Paper. Shouldn't they equate it to pi where the minimum occurs?
Reply 713
c4 solomon paper aren't to different to the normal exam papers
Original post by yaboy
c4 solomon paper aren't to different to the normal exam papers


I agree, I haven't really seen any big difference between the two myself...
Reply 715
just came across a very odd question in the c3 solomon C on question 4d. have no idea why they used OD
Reply 716
Vector diagram.jpg
Original post by yaboy
just came across a very odd question in the c3 solomon C on question 4d. have no idea why they used OD


Umm, does the following diagram explain what they did? Basically AB is the base of the triangle, OD is the height, they've used the standard way of calculating modulus then used the basic triangle area formula (1/2 x base x height).

Let me know if you want a more detailed solution :smile:
Reply 717
Original post by Lilmzbest
Vector diagram.jpg

Umm, does the following diagram explain what they did? Basically AB is the base of the triangle, OD is the height, they've used the standard way of calculating modulus then used the basic triangle area formula (1/2 x base x height).

Let me know if you want a more detailed solution :smile:


but that would only work if D is the midpoint of A and B and I dont know how you would know that.
Came across an unfamiliar question earlier from the Edexcel C3 January 2007 paper: question 8ii. The question is on inverse trigonometric functions (arcsin x, arcos x, arctan x), and whilst it isn't hugely difficult once you see the method, the second part was only answered correctly by 10% of candidates according to the examiner's report (I think this is simply due to unfamiliarity - it's one of those areas of the spec which doesn't appear very often in the exam.) It's covered briefly in the blue and green Edexcel C3 textbook but the examples given aren't that helpful. A useful source of help is the "PatrickJMT" Youtube account: http://www.youtube.com/user/patrickJMT?feature=watch
Reply 719
Original post by Vistaprint
Came across an unfamiliar question earlier from the Edexcel C3 January 2007 paper: question 8ii. The question is on inverse trigonometric functions (arcsin x, arcos x, arctan x), and whilst it isn't hugely difficult once you see the method, the second part was only answered correctly by 10% of candidates according to the examiner's report (I think this is simply due to unfamiliarity - it's one of those areas of the spec which doesn't appear very often in the exam.) It's covered briefly in the blue and green Edexcel C3 textbook but the examples given aren't that helpful. A useful source of help is the "PatrickJMT" Youtube account: http://www.youtube.com/user/patrickJMT?feature=watch


that video left me knowing no more on how to do that question than before I watched it. which was nothing

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