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FP2 Revision Thread

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Reply 60

jakezg

Can someone help me with this question from solomon paper D.

Show that INT of (a^2+x^2)^1/2= x/2(a^2+x^2)^1/2 + a^2/2arsinh(x/a)+c

I must be missing a trick here..

Is it just me or does every1 else find the solomon papers quite hard...? Im getting really annoyed with my poor performances.. my best is 51/75!!

INT(a^2+x^2)^1/2
INT(a^2(1+(x/a)^2))^1/2
a INT((1+(x/a)^2))^1/2, Let x/a = sinhu, x = asinhu, dx/du = acoshu

a^2 INT cosh^2u.du
a^2/2 INT 1+cosh(2u).du
a^2/2(u+(1/2)sinh(2u)) + c
a^2/2(u+sinh(u)cosh(u)) + c
a^2/2(u+sinh(u)(1+sinh^2(u))^(1/2)) + c
a^2/2(arcsinh(x/a)+(x/a)(1+(x/a)^2)^(1/2)) + c
(a^2/2)arcsinh(x/a)+(x/2)(a^2+x^2)^(1/2) + c

Reply 61

Eau

I am finding FP2 past papers okay. I get an overall high mark, but some questions are just nasty (compensated by relatively easy peesy questions).

Nasties include:

- the use of half-angle formulae (wtf - we rarely used it in C1 to FP1 and now you want us to use it so frequently?!)

- the use of substitutions to solve integrals - sometimes I just can't think of the substitution - it would be nice to provide them!!

- +/- signs in many hyperbolics questions- hard to decide whether there is only one or two solutions

Reply 62

Eau

So anyone taking FP2 Edexcel this year?

Reply 63

aeiou81

:O So incredibly useful, thanks!

You basically just taught me all the areas I have trouble with... Golly my teacher will be pleased with me tomorrow!

Reply 64

insparato

Mauve

So anyone taking FP2 Edexcel this year?

Ofcourse there is.

Btw you know that Mauve was one of the first synthetically made dyes :smile:

. It was made when they were trying to create modified quinine and then surprisingly they made an aniline dye called Mauveine.

Reply 65

Eau

insparato

Ofcourse there is.

Btw you know that Mauve was one of the first synthetically made dyes :smile:

. It was made when they were trying to create modified quinine and then surprisingly they made an aniline dye called Mauveine.

Lol, that's why I chose Mauve, and not Red.

Reply 66

datr

BTW Mauve, here's the solution for the integral in your blog written up. :smile:

$I = \int\frac{1 + 2x}{\sqrt{1 - 3x - 2x^2}} dx = \int \frac{\frac{3}{2} + 2x}{\sqrt{1-3x-2x^2}} dx - \frac{1}{2} \int \frac{1}{\sqrt{1-3x-2x^2}} dx$

$I = -\sqrt{1 - 3x - 2x^2} - \frac{1}{2\sqrt{2}} \int \frac{1}{\sqrt{\frac{17}{16} - (x + \frac{3}{4})^2}} dx$

$I = -\sqrt{1-3x-2x^2} - \frac{1}{2\sqrt{2}}\arcsin\left(\frac{4x + 3}{\sqrt{17}}\right) + C$

Reply 67

Eau

datr

BTW Mauve, here's the solution for the integral in your blog written up. :smile:

$I = \int\frac{1 + 2x}{\sqrt{1 - 3x - 2x^2}} dx = - \frac{1}{2} \int \frac{-3 - 4x}{\sqrt{1-3x-2x^2}} dx - \frac{1}{2} \int \frac{1}{\sqrt{1-3x-2x^2}} dx$

$I = -\sqrt{1 - 3x - 2x^2} - \frac{1}{2\sqrt{2}} \int \frac{1}{\sqrt{\frac{17}{16} - (x + \frac{3}{4})^2}} dx$

$I = -\sqrt{1-3x-2x^2} - \frac{1}{2\sqrt{2}}\arcsin\left(\frac{4x + 3}{\sqrt{17}}\right) + C$

Exactly what I got! :smile:

Changed the first line to make it more understandable:

$I = \int\frac{1 + 2x}{\sqrt{1 - 3x - 2x^2}} dx = - \frac{1}{2} \int \frac{-3 - 4x}{\sqrt{1-3x-2x^2}} dx - \frac{1}{2} \int \frac{1}{\sqrt{1-3x-2x^2}} dx$

Reply 68

Eau

Anyone got the recent papers for FP2? Jan 07?

Reply 69

Nadeshka

Eau

Anyone got the recent papers for FP2? June 2006/Jan 07?

Which board? I've got the January 06 and 07 papers for OCR MEI if you want them, though I don't know how their syllabus compares to other boards.

Reply 70

Eau

Nadeshka

Which board? I've got the January 06 and 07 papers for OCR MEI if you want them, though I don't know how their syllabus compares to other boards.

For Edexcel.

Reply 71

Breadster™

I'm starting learning this module from scratch tomorrow, got it on the 7th! Should be fun!

Reply 72

babydoll_89

this question has been really annoying me:
I have no idea how 2 use superscript so t1'2 is t1 to the power of 2 and t2'2 is t2 to the power of 2. I am new on this forum
The variable chord PQ on the parabola with equation y2 = 4x subtends a right angle at the origin O. By taking P as (t1'2 , 2t1) and Q as (t2'2 , 2t2) , find a relation between t1 and t2 . and hence show that PQ passes through a fixed point on the x-axis

The answer is :
t1 t2 -4 = 0
but I have no idea how to get it . Please Help!!!

Reply 73

insparato

The variable chord PQ on the parabola with equation y2 = 4x subtends a right angle at the origin O. By taking P as (t1'2 , 2t1) and Q as (t2'2 , 2t2) , find a relation between t1 and t2 . and hence show that PQ passes through a fixed point on the x-axis

You need to find
distance OP using pythagoras
distance OQ using pythagoras

Then using these distances, do pythagoras again to find to distance PQ. Everything eventually cancels out.

Reply 74

Rabite

Hmm...
I would have found the gradients, and set their product equal to -1.

Reply 75

insparato

That would be nicer :smile:

Reply 76

username80328

hmm can't find a general formula for the differential of arccos with the constant a in it... is it just the same as the sin one but negative? thats what i got when i tried to derive it, but i could be wrong, and better check before learning it!

Reply 77

zrancis

$y=cos^{-1}ax$

$cosy=ax$

$-siny\frac{dy}{dx}=a$

$\frac{dy}{dx}=-\frac{a}{siny}$

$\frac{dy}{dx}=-\frac{a}{\sqrt{(1-a^2x^2)}}$

Reply 78

datr

If you know the derivative of arccosx you can find the derivative of arccosax by chain rule immediately.

Reply 79

username80328

ooo thanks, but i actually meant $\frac{\mathrm{d}}{\mathrm{d}x} \cos^{-1} \frac{x}{a}$

woop i just taught myself latex, about time i learnt really, wudda stopped that confusion before having sed that i always cudve sed arcos (x/a) sorry bout that ppl

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