[JulietheCat]

(Original post by sunnybacon)
I was hoping to get full marks on Core 3 so I could take this a bit more relaxed and still get A*, but I know for a fact I've dropped 3-4 raw marks on Core 3, dunno how that will affect my UMS score though

Quick question, you know how when you integrate stuff like 1/(x+1) you end up with ln|x+1| , and for Core 3 we always got taught to use modulus signs for the function within ln.
But in Core 4 mark schemes, I always see it written as ln(x+1), haven't seen any modulus signs yet...any reason for this? And will it matter whether I use brackets or modulus?

Thanks

Usually I write modulus if I know that the inside of the bracket is going to be negative. But it shouldn't really matter. The examiners will mark you for your method and accuracy, not for different (but still correct) notation.

[Lollyage]

Guys I'm stuck on question 5 on June 2008, it says "The angle a is acute and sin a = 4/5. Find the value of cos a". According to the mark scheme, cos a = 3/5 but I have NO idea why! I hate this topic, please help

[24DJF]

(Original post by Lollyage)
Guys I'm stuck on question 5 on June 2008, it says "The angle a is acute and sin a = 4/5. Find the value of cos a". According to the mark scheme, cos a = 3/5 but I have NO idea why! I hate this topic, please help

I don't have latex software, but I just did this question.

You know that sin a = 4/5 so the opp is 4 and the hyp is 5.
Use pythagoras to find out the adjacent.
4^2 + x^2 = 5^2
So x, or the adj, is 3. Therefore as cos a = adj/hyp it's 3/5

Then the 2nd part, use the identity and sub in. :-)

[24DJF]

Quick message, good luck everyone!! :-D

[Xaerion]

What's the hardest Core 4 paper you guys have come across? - I haven't been able to revise much for this exam, but feel quite confident on most of it, however, I want to try something difficult to test topic areas Anyone have suggestions for a paper to do? ^_^

[24DJF]

(Original post by Xaerion)
What's the hardest Core 4 paper you guys have come across? - I haven't been able to revise much for this exam, but feel quite confident on most of it, however, I want to try something difficult to test topic areas Anyone have suggestions for a paper to do? ^_^

Difficulty is subjective, however I would recommend the Jan 2007 and then June 2011 paper.

[Xaerion]

(Original post by 24DJF)
Difficulty is subjective, however I would recommend the Jan 2007 and then June 2011 paper.

thanks I'll give both of them a go tomorrow ^_^

[Lollyage]

(Original post by 24DJF)
I don't have latex software, but I just did this question.

You know that sin a = 4/5 so the opp is 4 and the hyp is 5.
Use pythagoras to find out the adjacent.
4^2 + x^2 = 5^2
So x, or the adj, is 3. Therefore as cos a = adj/hyp it's 3/5

Then the 2nd part, use the identity and sub in. :-)

Ah thanks! Yeah I thought it would be something like that, but it seemed like a lot of work for 1 mark so I assumed the value would be a lot more obvious I get it now

[Pol]

Hi,

June 2011 7bi) Differential equation question,
I roughly understand it, but not fully as I think I may be going the long way around to get the answer. So would someone please give their solution please?

Thanks

[lee_vassallo]

Was there an implicit differentiation question in Jan 2012 and if not, why not? :s

[GY11]

(Original post by Pol)
Hi,

June 2011 7bi) Differential equation question,
I roughly understand it, but not fully as I think I may be going the long way around to get the answer. So would someone please give their solution please?

Thanks

Im not sure, but i think you need to work out c and k first by using the 4(pi)r^2=A
We know from the q that at t=0 then r=60 and when t=9 then r=30
we know from part a) that dA/dt= -k so we need to work out what A= its done by dA = -k dt then you integrate to get A=-kt +c therefore A= -kt +c = 4(pi)r^2 so we sub in.. t=0 and r= 60...to give 4pi(60)^2 = -k(0)+c therefore c=14400pi. now we know what c equal we can work out k...by using t=9 and r =30 so A= 4pi(30)^2=3600pi
so A=3600pi ... therefore -k(9)+14400pi =3600pi than you rearrange to get k=1200pi..
than using the equation A= -kt +c we sub in k and c values... so its A= -1200(pi)(t)+14400pi which can be simplified as A= -1200(pi)(12-t)

[sunnybacon]

(Original post by JulietheCat)
Usually I write modulus if I know that the inside of the bracket is going to be negative. But it shouldn't really matter. The examiners will mark you for your method and accuracy, not for different (but still correct) notation.

Ah yeah, that's true, thanks

[Lollyage]

I hate vectors.

[24DJF]

(Original post by lee_vassallo)
Was there an implicit differentiation question in Jan 2012 and if not, why not? :s

It may imply there will most probably be one on this paper then!

[lee_vassallo]

(Original post by 24DJF)
It may imply there will most probably be one on this paper then!

Good omen then, at least we can predict what one question will be!

[Sagacious]

3 past papers to go! I shall complete them all tomorrow

Who's feeling confident for Thursday??

[Rachael1993]

Hi,

June 2011 Q8bii), How do you integrate 1/2(squareroot(y)) dy

I've managed to complete the whole paper and It's probably too late in the evening to finish a past paper, buts its bugging me...

Apologies if it's really simple!

[PhysicsGirl]

Hey, struggling with the final part of q7 on the June 2010 C4 paper (link: http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN10.PDF)
I ended up getting c as either (1, -1, 3) or (-3, -1, 1), whereas in the mark scheme, c is given as either (3, -1, 3) or (-1, -1, 1). The similarity in numbers would suggest a sign error somewhere, maybe?
The method I used was:
C lies on a line parallel to l2 that runs through B, so C= K(2, 0, 1) + B
Because it's a parallelogram, magnitude of AP = BC (I think this is wrong. Maybe I drew the diagram wrong?)
Therefore, found AP to be (-2, 0, -1) and since BC is K(2, 0, 1), sqrt(5)=sqrt(5k^2), therefore K=+or-1
I put that back into initial statement, and that gave me those values of C; either (1, -1, 3) or (-3, -1, 1), which is wrong.
If anyone wants to point out where I'm going wrong and how to put it right, I'd really appreciate it! Thanks :-)
EDIT: DAMN. Found out where I went wrong; part a of q7, worked out the coordinates of B wrongly, because apparently I can't do -1+2. Oh dear :'(

[f1mad]

(Original post by Sagacious)
3 past papers to go! I shall complete them all tomorrow

Who's feeling confident for Thursday??

Me, and I've hardly done any papers .

[Cath-ay]

(Original post by Rachael1993)
Hi,

June 2011 Q8bii), How do you integrate 1/2(squareroot(y)) dy

I've managed to complete the whole paper and It's probably too late in the evening to finish a past paper, buts its bugging me...

Apologies if it's really simple!

Seeing as it's bugging you... You need to change it into indices form I.e. (2y^0.5)^-1 or 1/2 x y^-0.5
then you can intergrate from there.