AQA Core 4 Exam Discussion 14/06/2012 !Poll, paper and unofficial MS (first post)!

Anyone care to help me with June 09 question 3(c)? Its just finding the range of values for a binomial expansion but there's a subtle difference I haven't seen before, I just assumed it was mod(x)<1/3 but it isn't so can anyone explain it to me?

When you split it in to partial fractions and worked out the expansion from there, you should have used

Unparseable latex formula:

3(\dfrac{1}{2}\ (1 - \dfrac{3}{2}\ x)^-^1)

as part of it right? Because it is

\dfrac{3}{2}

in the bracket not just

3x

then the range is now this:

-1<

\dfrac{3}{2} \ x

< 1

Which you can then change (multiply by 2 and then divide by 3) to give

\dfrac{-2}{3}\ < x < \dfrac{2}{3}\

(or mod x is less than 2/3)

Hope this helps! :smile:

Reply 181

lee_vassallo

Thank you both, great answers :smile:

Reply 182

lara_ox

10

Original post by hash007

Anyone do maths with statistics?

I really dont understand continous probability distributions/cumulative distribution functions. Wasn't really taught it as college and I don't have a clue what to do in the past papers, looking at the mark scheme doesn't make much sense either. Anyone know any sites that would help explain it, or some worked examples. I'm on track for A* in C3/C4 but I hope S2 doesn't drag me down so that I don't get an overall D:

Ive looked on examsolutions, but it's a real brief section on there.

I feel the exactly same S2 is really confusing i havent started past papers yet as im just reading through the aqa stats 2 book. I wish i done decision maths instead everyone said that paper was really easy :frown:

Reply 183

hash007

15

January 2010, 8bii)

h = 60 ( 1 - e^-t/4 )

show that dh/dt = 15 - h/4

Dont understand how to do this :confused:

Reply 184

Oromis263

2

Original post by hash007

January 2010, 8bii)

h = 60 ( 1 - e^-t/4 )

show that dh/dt = 15 - h/4

Dont understand how to do this :confused:

Just did this paper, that bit was a little sneaky, I'll spoiler my method:

Spoiler

Reply 185

Thechillyone

4

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC4-W-QP-JAN08.PDF

Can anyone help with question 2b) ? I always get confused when i do this type of question. And gooood luccck guys :smile:

Reply 186

AlexBrown1994

Have a look at the actual grade boundaries for the papers you do, because often teachers mark them as standard 70% B 80% A and so on. But they are often slightly lower that that on core 4, for instance jan 12 76% for and A and 68% for a B!

Reply 187

AlexBrown1994

2b) Use long division to divide 2x squared after expanding the brackets, then A is the number of top and Bx+C is the remainder you get from dividing.

Reply 188

mojopin1

Original post by Thechillyone

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC4-W-QP-JAN08.PDF

Can anyone help with question 2b) ? I always get confused when i do this type of question. And gooood luccck guys :smile:

That's an improper fraction so you do long division and then you have quotient + remainder/divisor.

Reply 189

sydney07

hey(: Do we need to learn the triple angle equations? ...cos I remember doing them at school, but I haven't seen any in past papers(:

Reply 190

hash007

15

Original post by Oromis263

Just did this paper, that bit was a little sneaky, I'll spoiler my method:

Spoiler

Thought it would be something like that, just could figure out the steps :tongue:

What did you get in the paper?

Reply 191

Cath-ay

OP

Original post by sydney07

hey(: Do we need to learn the triple angle equations? ...cos I remember doing them at school, but I haven't seen any in past papers(:

I've seen it in the june 2005 paper, it wasn't too bad. You can just use what you know about double angles to work it out.

(edited 11 years ago)

Reply 192

ao_mango

Can someone help me with this vector question please. It seem pretty straight forward, dunno if i'm missing something here but i keep getting the wrong answer :s-smilie:

Triangle PQR is defined by points P (5, -3, 1), Q (-2, 1, 5) and R (9,5,0). Find the angles of the triangle.

Reply 193

Iepnauy

9

Question guys.

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC4-W-QP-JUN08.PDF
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC4-W-MS-JUN08.PDF

Question 8.a.i
I don't understand the method mark?
I just write down the answer as dy/dx = -kx
What does the mark scheme mean by p dy/dx = q
Where p and q are functions?

Thanks :smile:

Reply 194

Iepnauy

9

Original post by ao_mango

Can someone help me with this vector question please. It seem pretty straight forward, dunno if i'm missing something here but i keep getting the wrong answer :s-smilie:

Triangle PQR is defined by points P (5, -3, 1), Q (-2, 1, 5) and R (9,5,0). Find the angles of the triangle.

45 degrees?

Find QP. Then QR

(QP)(QR) = Distance of QP x Distance of QR x cos (theta)
I got
81 = √81 √162 Cos(theta)

Did it really quickly, so probably a mistake or two in there somewhere.

Reply 195

JulietheCat

16

Original post by ao_mango

Can someone help me with this vector question please. It seem pretty straight forward, dunno if i'm missing something here but i keep getting the wrong answer :s-smilie:

Triangle PQR is defined by points P (5, -3, 1), Q (-2, 1, 5) and R (9,5,0). Find the angles of the triangle.

Find QP and QR and use the a.b formula to calculate the angle. :smile:

Reply 196

sydney07

Original post by Cath-ay

I've seen it in the june 2005 paper, it wasn't too bad. You can just use what you know about double angles to work it out.

Ahh right okay, thank you(:

Reply 197

jammysmt

5

Sorry if someone has already explained, but could somebody help me with question 7c, June 2010.

I don't know how you can work out both solutions, I only come out with one and the mark scheme seems to be confusing. I understand that you need to manipulate the info given, that AC=PB but HOW do you know to use BOTH AC=PB AND AP=BC. Surely they should given the same result? Help appreciated.

http://store.aqa.org.uk/qual/gce/pdf/AQA-MPC4-W-QP-JUN10.PDF

Reply 198

mojopin1

Original post by jammysmt

Sorry if someone has already explained, but could somebody help me with question 7c, June 2010.

I don't know how you can work out both solutions, I only come out with one and the mark scheme seems to be confusing. I understand that you need to manipulate the info given, that AC=PB but HOW do you know to use BOTH AC=PB AND AP=BC. Surely they should given the same result? Help appreciated.

http://store.aqa.org.uk/qual/gce/pdf/AQA-MPC4-W-QP-JUN10.PDF

I'm really struggling with this area of vectors too. I don't understand how adding or subtracting vectors gives coordinates. There seems to be no where that covers this topic.

Reply 199

username896004

Original post by jammysmt

Sorry if someone has already explained, but could somebody help me with question 7c, June 2010.

I don't know how you can work out both solutions, I only come out with one and the mark scheme seems to be confusing. I understand that you need to manipulate the info given, that AC=PB but HOW do you know to use BOTH AC=PB AND AP=BC. Surely they should given the same result? Help appreciated.

http://store.aqa.org.uk/qual/gce/pdf/AQA-MPC4-W-QP-JUN10.PDF

I'll do the question and post a picture

AQA Core 4 Exam Discussion 14/06/2012 !Poll, paper and unofficial MS (first post)!

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