Edexcel C3,C4 June 2013 Thread

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Alright, then subbing in

tan^2x=sec^2x-1

the integral now becomes

\displaystyle \int \dfrac{4}{1- (sec^2x-1) }

\displaystyle \int \dfrac{4}{2-sec^2x}

\displaystyle \int-2cos^2x

\displaystyle \int -2(\frac{1}{2}cos2x+\frac{1}{2})

which then equals to

\displaystyle \int -cos2x - 1

and i get

\frac{1}{2}sin2x -x +C

Okay, if you go back to the original expression that you had. On the bottom you had 1-{Sin^4(x)/Cos^4(x)} so what can you make this equal?

Spoiler

Reply 1761

nm786

Original post by MathsNerd1

Okay, if you go back to the original expression that you had. On the bottom you had 1-{Sin^4(x)/Cos^4(x)} so what can you make this equal?

Spoiler

do you mean Cos^4(x)-Sin^4(x)/cos^4(x) be expressed as cos(4x)/cos^4(x)? I'm not sure why i have to flip this and multiply by the numerator sorry. :redface:

(edited 10 years ago)

Reply 1762

MathsNerd1

Original post by nm786

do you mean Cos^4(x)-Sin^4(x)/cos^4(x) be expressed as cos(4x)/cos^4(x)? I'm not sure why i have to flip this and multiply by the numerator sorry. :redface:

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Reply 1763

ACBLISS

Original post by MedMed12

good good! how are they are going? I know you'll be OK :smile:

Its been fine, just need to get more confidence with C3 again and build it up with the past papers for Maths. Been doing Chem and Bio questions alot this week.

They are going fine dw lol :smile:

I need to crack on with eco :/

Reply 1764

tsr1

how do you know whether to apply u-substitution or by parts when the question doesn't tell u anything?
e.g. sind exact value of: 1/2 + 7/2 cos2θ

Reply 1765

Namod

Can anyone give me C4 and C3 Jan 2013 please.

Reply 1766

usycool1

Original post by tsr1

how do you know whether to apply u-substitution or by parts when the question doesn't tell u anything?
e.g. sind exact value of: 1/2 + 7/2 cos2θ

I'm guessing this is an integration question? If so, what are the limits? And you should be able to integrate:

Unparseable latex formula:

\dfrac{1}{2} + \dfrac{7}{2} \cos \2 \theta

w.r.t. x

without the need for substitution (it can be done very quickly in your head :tongue:

Reply 1767

pepeeglesfield

if someone could integrate 1/sin^2x cos^2x id be grateful!! :smile:

edit: scratch that, I've done it, was simpler than I thought...but feel free to give it a crack guys and girls

(edited 10 years ago)

Reply 1768

Tikara

anyone know all the volume and surface area formulas we should know?
thanks : )

(I'm just doing connected rates of change C4 thats all)

Reply 1769

Tikara

Original post by Namod

Can anyone give me C4 and C3 Jan 2013 please.

they're both here with mark schemes :smile:

http://mathspapers.co.uk/edexcel.html

Reply 1770

tsr1

Original post by usycool1

I'm guessing this is an integration question? If so, what are the limits? And you should be able to integrate:

Unparseable latex formula:

\dfrac{1}{2} + \dfrac{7}{2} \cos \2 \theta

w.r.t. x

without the need for substitution (it can be done very quickly in your head :tongue:

Yes it is an integration question, it's easy that way but the question says use calculus to find the solution, also they have give 5 marks for using by-parts in the mark scheme.. It's june 2010 paper c4 , question 6 but I wouldn't have know what integration method to use...

Reply 1771

physicso

Can't wait to start revising maths... ****ing Economics is painful.

Reply 1772

justinawe

Original post by tsr1

They were asking for,

\displaystyle \int^{\pi /2}_0 \theta \left( \frac{1}{2} + \frac{7}{2} \cos 2\theta \right) d\theta

NOT,

\displaystyle \int^{\pi /2}_0 \left( \frac{1}{2} + \frac{7}{2} \cos 2\theta \right) d\theta

Reply 1773

Story

Hey guys do you have any tips on doing hard differentiation questions?

Also, the C3 June 11 paper, q)5c) I am trying to differentiate M= 7.5e^-1/4ln3t to get dm/dt... and then make it = to -0.6ln3.

I used the chain rule and did m=e^u and u=1/4ln3t and got dm/du=e^u and dm/dt= -1/12t.

But I dont think im going along the right lines as I did this before and got the wrong answer.

Any help is appreciated!

(edited 10 years ago)

Reply 1774

DELETED ACCOUNT

Original post by pepeeglesfield

if someone could integrate 1/sin^2x cos^2x id be grateful!! :smile:

edit: scratch that, I've done it, was simpler than I thought...but feel free to give it a crack guys and girls

I got to the part where we had to integrate the integral of cosec2x dx.

Do we have to know that the integral of cosec x = ln |cosecx - cotx| + c

EDIT: Just found out that the integral is provided in the formula book

(edited 10 years ago)

Reply 1775

usycool1

Original post by frogs r everywhere

I got to the part where we had to integrate the integral of cosec2x dx.

Do we have to know that the integral of cosec x = ln |cosecx - cotx| + c

Yeah, but it's on the formula booklet.

Reply 1776

usycool1

Original post by Story

I presume you mean du/dt.

Remember:

u=-\dfrac{1}{4} \ln(3)t

; not

u = -\dfrac{1}{4} \ln(3t)

(edited 10 years ago)

Reply 1777

Story

Original post by usycool1

I presume you mean du/dt.

Remember:

u=-\dfrac{1}{4} \ln(3)t

; not

u = -\dfrac{1}{4} \ln(3t)

So du/dt would be - 1/12 t with the t in the middle of the fraction?

Reply 1778

Better

Hey guys! Very sunny day out there! Gonna take a break in a few hours

BUT Does anyone know how to do Solomon Press C3 Paper K Question 4 b
Markscheme here: http://www.tomred.org/uploads/7/7/8/3/778329/c3_qp_solomon_k.pdf

Can someone please explain the justifications?

(b) Use proof by contradiction to prove that log2 3 is irrational. (6)

Reply 1779

MathsNerd1

Original post by Better

Remember you can express a rational number by a fraction, so you'll have a/b if it was rational and you've just got to go about proving that it isn't rational, therefore it's irrational.

I wouldn't worry too much about this as I doubt you'd ever get asked this in the exam :smile:

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