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The Edexcel C3 (14/06/12 - AM) Revision Thread

GAHHHHH. I thought I understood range and domain. :frown:

With y= e^(x-4) -2

I thought the domain of x could be any number since when I plug it in, it's all fine. Why is it x had to be less than or equal to 4?
The numbers work. :s-smilie:

I realise the range of it's inverse is x has to be less than or equal to four
BUT if x is greater than 4 in the e^(x-4) -2 it still works... I am so confused. :frown:

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Reply 781

MuffinMonster

Original post by LifeIsGood

Yes.

Do we need to know those graphs off by heart - sin, cos, tan?

How do I differentiate : y = 2lnx/3 ?

It's the same thing as 2/3 X (ln x )

Reply 782

otrivine

14

hi guys so what formulas do we need to know to prove????/

Reply 783

Huda134

Original post by Souriretoujours

anyone else feel physically sick thinking about tomorrow? :/

YES. but at least it'll be over tomorrow.

This was posted from The Student Room's iPhone/iPad App

Reply 784

Pinkhead

14

Original post by Jorgeyy

If you get time I would love to see the solution :smile:

I think I've worked out how to do question 5... not sure if it's correct though. Want me to post a working out and you can check if I've gone wrong somewhere?

Reply 785

Arva

14

Original post by LifeIsGood

Yes.

Do we need to know those graphs off by heart - sin, cos, tan?

How do I differentiate : y = 2lnx/3 ?

Is that

\frac {2lnx}{3}

or

2ln \frac{x}{3}

?
If the former, you end up with

\frac {2}{3} lnx

which of course differentiates to

\frac {2}{3x}

If the latter you end up with

\frac {2}{x}

Reply 786

F1's Finest

21

Original post by otrivine

hi guys so what formulas do we need to know to prove????/

tan^2(x) + 1 = sec^2(x)

cot^2(x) + 1 = cosec^2(x)

cos2x = 1 - 2sin^2(x)

cos2x = 2cos^2(x) - 1

there are some more, if someone could go over then please

Reply 787

otrivine

14

Original post by James A

tan^2(x) + 1 = sec^2(x)

cot^2(x) + 1 = cosec^2(x)

cos2x = 1 - 2sin^2(x)

cos2x = 2cos^2(x) - 1

there are some more, if someone could go over then please

hi how are u :wink:

those two last ones the book makes it confusing or can u show me those last 2 thanks :wink:

Reply 788

LifeIsGood

12

Original post by Arva

Is that

\frac {2lnx}{3}

or

2ln \frac{x}{3}

?
If the former, you end up with

\frac {2}{3} lnx

which of course differentiates to

\frac {2}{3x}

If the latter you end up with

\frac {2}{x}

It's the second one but I don't get how to do it :s-smilie:

Picked it from a Solomon Paper

Reply 789

hellie7

2

are the review exercise in the book designed to be harder than the past papers? I could complete all the past papers fine, but the old P3 papers and review exercises are really difficult!

Reply 790

Vividness

1

Just a quick question from my textbook:

Using the half angle formulae or otherwise show that:

\frac{1-\cos \theta}{\sin \theta} = \tan \frac{\theta}{2}

Dont really know where to start :s-smilie:

thanks

Reply 791

Mathematix

2

Original post by Pinkhead

I think I've worked out how to do question 5... not sure if it's correct though. Want me to post a working out and you can check if I've gone wrong somewhere?

Yes <3

Reply 792

raheem94

13

Original post by LifeIsGood

It's the second one but I don't get how to do it :s-smilie:

Picked it from a Solomon Paper

y = 2 \ln \left( \dfrac{x}3 \right)

The general rule is,

y = \ln f(x) \\ \dfrac{dy}{dx} = \dfrac{f'(x)}{f(x)}

In your question

f(x) = \dfrac{x}3

hence

f'(x) = \dfrac13

Does it makes sense?

Reply 793

Arva

14

Original post by LifeIsGood

It's the second one but I don't get how to do it :s-smilie:

Picked it from a Solomon Paper

Use the chain rule. Let us make a substitution:

u = \frac{x}{3}[br][br]du/dx = 1/3

Also, if

y = \ln u[br][br]dy/du = 1/u

We know from the chain rule that:

dy/dx = (dy/du)(du/dx)

Therefore

[br]dy/dx = 1/u x 1/3 = 1/(x/3) x 1/3 = 1/x

EDIT: forgot to times by two but I'm sure you can work that part out from the fact 2 is a constant.

(edited 11 years ago)

Reply 794

Arva

14

Original post by raheem94

y = 2 \ln \left( \dfrac{x}3 \right)

The general rule is,

y = \ln f(x) \\ \dfrac{dy}{dx} = \dfrac{f'(x)}{f(x)}

In your question

f(x) = \dfrac{x}3

hence

f'(x) = \dfrac13

Does it makes sense?

Damn it Raheem I can never answer before you. :colonhash:

This one I can blame on my latexing troubles - I can never get fractions to work.

Reply 795

Pinkhead

14

Original post by Jorgeyy

Yes <3

sorry, it's a bit hard to read. I'm not too sure on some of the working out though.

Reply 796

simransanghera

hey has anyone done solomon paper H? Im stuck on q 4b.

Thanks

Reply 797

gadzoinks

2

Original post by LifeIsGood

It's the second one but I don't get how to do it :s-smilie:

Picked it from a Solomon Paper

Looks like its been worked out above,
but heres my workings :smile:

quickworking.jpg37.3KB

Reply 798

Dark Lord of Mordor

1

Original post by Vividness

Just a quick question from my textbook:

Using the half angle formulae or otherwise show that:

\frac{1-\cos \theta}{\sin \theta} = \tan \frac{\theta}{2}

Dont really know where to start :s-smilie:

thanks

I'm pretty sure half angle formulae aren't on the Edexcel C3 specification

Reply 799

raheem94

13

Original post by Arva

Damn it Raheem I can never answer before you. :colonhash:

This one I can blame on my latexing troubles - I can never get fractions to work.

I started answering it very late :tongue:

i was giving you chance to answer.

I am much more experienced member of this forum than you, so i am quite fast at LaTexing and also my LaTex comes out much nicer than your's :biggrin:

May be you need to practice some LaTex and read the guide again :tongue:

One day you will become as fast as me but till that time i will be even faster :tongue:

The Edexcel C3 (14/06/12 - AM) Revision Thread

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