AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme]

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It's fine. That's equivalent to 1/2 x y^-1/2, so you just integrate from there (it becomes 1/4 x y^1/2).

don't you need to divide by 1/2 not mulitply? so 1/2 * 1/2 =1/4 is wrong, should be 1/2 divided by 1/2 giving 1. Idk if I've just got that very wrong tho

Reply 1241

datpr0

Original post by 2014_GCSE

You've done 1/2 x 1/2. It's 1/2 / 1/2

so it's y^1/2

:smile:

Ay, yes, haha :P... I just realised and edited it. Don't worry, though, I will go to sleep earlier tonight so I am actually awake in the exam ;D!

Reply 1242

gloria0816

Original post by moinoi

1/2rootY is equal to 1/2(y^-1/2)
So integrates to y^1/2 or root Y

Thanks

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Reply 1243

datpr0

Original post by moinoi

don't you need to divide by 1/2 not mulitply? so 1/2 * 1/2 =1/4 is wrong, should be 1/2 divided by 1/2 giving 1. Idk if I've just got that very wrong tho

Damn, everyone was too fast for my editing skills :biggrin:

! Yes, you're correct in the sense that it was 1/2 divided by 1/2 (i.e. 1/2 x 2) :smile:

Reply 1244

yasmin221b

Original post by student0042

I commented earlier on this on this post. It is a trapezium, therefore you need to smaller value because if you use the larger value, the shape will be a parallelogram. (As the lengths will be the same CD, i believe, but the shape wrong). If you still don't understand let me know and I will draw a diagram.

A diagram would be very useful, thank you!

Reply 1245

saeed97

Are you guys ready for this? I cant wait to get it overw with.

Reply 1246

Tom^

Original post by saeed97

Are you guys ready for this? I cant wait to get it overw with.

Im scared they will make it different to a degree of difficulty

Reply 1247

amyrah

8a) is ln(y) = -cos(t) + c but in the mark scheme they changed the -cos(t) to A and this gave the final mark. Why did they do this? And would we have to do it in all questions of this type?

Thanks!

Screen Shot 2015-06-08 at 14.59.43.png25.1KB

Reply 1248

Tiwa

Does anybody know if we need to know these formulae?:

sinA + sinB = 2sin(\frac{A +B}{2})cos(\frac{A - B}{2})
sinA - sinB = 2cos(\frac{A + B}{2})sin(\frac{A - B}{2})
cosA + cosB = 2cos(\frac{A + B}{2})cos(\frac{A - B}{2})
cosA - cosB = 2sin(\frac{A + B}{2})sin(\frac{A - B}{2})

Reply 1249

2014_GCSE

Original post by Tiwa

Does anybody know if we need to know these formulae?:

sinA + sinB = 2sin(\frac{A +B}{2})cos(\frac{A - B}{2})

sinA - sinB = 2cos(\frac{A + B}{2})sin(\frac{A - B}{2})
cosA + cosB = 2cos(\frac{A + B}{2})cos(\frac{A - B}{2})
cosA - cosB = 2sin(\frac{A + B}{2})sin(\frac{A - B}{2})

Nope we don't need to know them

Reply 1250

Tom^

Original post by CD223

Plot twist when vectors doesn't appear because AQA forgot to put them in.

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AKA The dream

Reply 1251

Tiwa

Original post by 2014_GCSE

Nope we don't need to know them

Thanks!

Reply 1252

Jankie

With vectors, how do you use the AB = OA+OB? Like how does that work? Thanks!

Reply 1253

Haza100

Original post by 2014_GCSE

How do you manipulate

3sin4x

My teacher swears that we'll get 4x's and stuff where we have to manipulate the double angle formulae to do this.
How does it work?

Does it become 3Sin2xCos2x?

Would also like to know this.

Would something like Cos(4x) be equivalent to
2Cos^2(2x) - 1
1 - 2Sin^2(2x)
cos^2(2x) - sin^2(2x)

Reply 1254

CD223

Original post by Tiwa

Does anybody know if we need to know these formulae?:

sinA + sinB = 2sin(\frac{A +B}{2})cos(\frac{A - B}{2})

sinA - sinB = 2cos(\frac{A + B}{2})sin(\frac{A - B}{2})
cosA + cosB = 2cos(\frac{A + B}{2})cos(\frac{A - B}{2})
cosA - cosB = 2sin(\frac{A + B}{2})sin(\frac{A - B}{2})

That's Edexcel.

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Reply 1255

datpr0

Original post by Jankie

With vectors, how do you use the AB = OA+OB? Like how does that work? Thanks!

The rule is AB = OB - OA.

This is because vector AB is unknown. However, travelling from vector a to vector b is the same as travelling from vector a to the origin and the from the origin to vector b (i.e. AO + OB). You then proceed to write AO as -OA (AO = -OA), giving AB = OB - OA.