AQA Core 3 - June 11th 2010

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Reply 40

KitCat93

Jeester

Core 4.

(But you do need to know how to differentiate Cos2X etc.

Cheers

Reply 41

Jeester

KitCat93

Cheers

Looks like you've been doing tonnes of past papers.

Reply 42

Elimelim

anyone got jan 05 paper?

Reply 43

c0nfus3d

Jeester

Integration by parts is by far, a mon avis, the hardest part of Core 3.
I always make silly slips.
But well done mucker!

"ln x" should always be the "u" for integration by parts I've been told!

I've just gone through 6 past papers...marking them as I type...

:work:

Reply 44

KitCat93

Jeester

Looks like you've been doing tonnes of past papers.

Haha, i've actually done loads but i've been doing C4 ones as well, so all the stuff is just jumbled up

Reply 45

Jeester

c0nfus3d

"ln x" should always be the "u" for integration by parts I've been told!

I've just gone through 6 past papers...marking them as I type...

:work:

6?!
I might do one more today which will take me to three.

You're crazy.

Reply 46

KitCat93

c0nfus3d

"ln x" should always be the "u" for integration by parts I've been told!

There's the LAET rule.
Logs, Algebra, Exponentials and Trig. You only need the first two for C3 though :smile:

Reply 47

c0nfus3d

KitCat93

There's the LAET rule.
Logs, Algebra, Exponentials and Trig. You only need the first two for C3 though :smile:

I don't get that bit?! :frown:

Reply 48

Va_Va_ Voom

Am strugling big time.

June 2005 q3c

Integration by parts question can any1 do that and explain please

Reply 49

KitCat93

c0nfus3d

I don't get that bit?! :frown:

Your x terms :smile:

So if it was xsinx
x would be U and sinx would be dv/dx

Reply 50

busdepp

Is there a jan 05 paper because I can't seem to find one for that year!

Reply 51

Elimelim

does anyone have the january 05 aqa paper?

ive done all the papers since jan 10

Reply 52

thewait

Feeling quite confident about this exam now (& thats when it generally goes bad for me :rolleyes:

). Start making silly mistakes like forgetting the minus sign :o:

.

Couple of thing keeps dragging my marks down, like finding the range (no clue how to do this & just rely on luck) and finding inequalities in modulus functions. All fairly easy but I just can't get the hang of them, whilst the integration & differentation I find them straightforward :mad:

Any ideas on a method I can follow for finding the inequalities in modulus functions? Especially when they have two mod functions = to each other e.g. [x+3]<[2x+4] ( this [] means modulus).

Reply 53

c0nfus3d

KitCat93

Your x terms :smile:

So if it was xsinx
x would be U and sinx would be dv/dx

Thankyou!

+rep :biggrin:

Reply 54

steph_v

carlylwright

Excuse me, are you me?
Because everything you wrote is exactly the same with me!
I always get the range/domain bit wrong - I've had it explained to me a thousand times, but I still get it wrong!

If you get someone to explain it, let me know!

I've done all the past papers from Jan 05 to Jan 10, most about twice each. I'm not sure what good it's done though, I'm not particularly good at maths.

Reply 55

c0nfus3d

Jeester

6?!
I might do one more today which will take me to three.

You're crazy.

Aim is to reach 10 before 10 o clock! :biggrin:

Really wanna ace C3 :frown:

Reply 56

username261841

You guys should do FP2/3... you apply everything in C3 about a million times it becomes second nature, haha.

Reply 57

student92

From the January 2010 paper, Question 7b i still don't get, even after looking at the mark scheme, could someone plz explain to me how to differentiate: 4(1 + tan^2 4x)

Reply 58

username261841

student92

From the January 2010 paper, Question 7b i still don't get, even after looking at the mark scheme, could someone plz explain to me how to differentiate: 4(1 + tan^2 4x)

I'm assuming it's the tan^2 that is throwing you.

Tan^2(4x) is the same as [tan(4x)]^2, so you can use the chain rule, or product rule if you wish, on it.

Chain rule is simpler, so, multiply by the power, take one from it, and multiply by the differencial of what's in the brackets.

2(tan(4x) is the first bit, then to multiply by the differencial of tan4x.

From the Formulae book, tan(4x) diffs to 4sec^2(4x).

So

2(tan(4x)4sec^2(4x))
= 8tan(4x)sec^2(4x)

The actual answer to the Q after you do the rest is 4x that.

32tan(4x)sec^2(4x)

Sec^2=1+tan^2
32tan(4x)(1+tan^2(4x))
32y(1+y^2)

Reply 59

Jeester