# PC2 Maths, AQA 24/05/10Watch

8 years ago
#321
For the binomial question, consdiering that it said it should be in the form with only one negative at the end, would you get marked down for putting p as -3, but leaving it like that in the binomial expansion?
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#322
(Original post by Doughnuts!!)
For the binomial question, consdiering that it said it should be in the form with only one negative at the end, would you get marked down for putting p as -3, but leaving it like that in the binomial expansion?
I'm not entirely sure I know what you mean by that, but it you were to put + (-3/x^2) it's exatcly the same as - (3/x^2) so you'd get the makrs for either. I expect one'd even get the marks if you just worte down the values of p & q qithout any working at all, also.
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8 years ago
#323
(Original post by rob...)
I'm not entirely sure I know what you mean by that, but it you were to put + (-3/x^2) it's exatcly the same as - (3/x^2) so you'd get the makrs for either. I expect one'd even get the marks if you just worte down the values of p & q qithout any working at all, also.
Yh sorry, I realise now that the post was a bit muddled

I meant if you put - (3/x^2), because even though it's exactly the same, it wasn't in the form shown in the question.

But you're probably right about just writing down p and q!
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#324
(Original post by Doughnuts!!)
Yh sorry, I realise now that the post was a bit muddled

I meant if you put - (3/x^2), because even though it's exactly the same, it wasn't in the form shown in the question.

But you're probably right about just writing down p and q!

Well I didn't write it in the form of the question, I just wrote - (3/x^2), but also stated p & q too. I think if the examiner has any sense then they'll accept any form... But we all know how picky examiners can be; any excuse to dock a mark!
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8 years ago
#325
Jesus christ you scared me with that Q9 jpg. lol

I worked out im pretty sure i'll get roughly around 63-65.

A bit dissapointing as I was getting 90% on past papers, but its allright as I have 95UMS on C1 and stats tomorrow :O

Anyone have an idea what UMS score 63-65 out of 75 would be on Core 2 Paper.

Last year you needed 58/75 for an A?
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8 years ago
#326
I figured that if last year 58/75 was 80UMS then:

(80/58) x 63 = 87ums
(80/58) x 65 = 90ums

So I hopefully should get around there
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#327
(Original post by Ally451)
Jesus christ you scared me with that Q9 jpg. lol

I worked out im pretty sure i'll get roughly around 63-65.

A bit dissapointing as I was getting 90% on past papers, but its allright as I have 95UMS on C1 and stats tomorrow :O

Anyone have an idea what UMS score 63-65 out of 75 would be on Core 2 Paper.

Last year you needed 58/75 for an A?
Someone asked me what I got for Question 9 as a joke after the exam, and I felt my heart sinking

I couldn't tell you what UMS mark that'd translate to though unfortunately I guess you'll just have to wait and find out on the 19th August
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#328
(Original post by Ally451)
I figured that if last year 58/75 was 80UMS then:

(80/58) x 63 = 87ums
(80/58) x 65 = 90ums

So I hopefully should get around there
Good thinking!
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8 years ago
#329
(Original post by Ally451)
Jesus christ you scared me with that Q9 jpg. lol

I worked out im pretty sure i'll get roughly around 63-65.

A bit dissapointing as I was getting 90% on past papers, but its allright as I have 95UMS on C1 and stats tomorrow :O

Anyone have an idea what UMS score 63-65 out of 75 would be on Core 2 Paper.

Last year you needed 58/75 for an A?
You've also got to take into account the B grade boundary to determine where the full UMS 'cap' lies.

For January 2010, MPC2 was 58/75 for an A and 50/75 for a B. Full UMS would be 58+8+8=74 (8 being the difference between B and A).

For Jan 10:
63/75 = 10[(63-58)/8]+80 = 86.25 = 86 UMS
65/75 = 10[(65-58)/8]+80 = 88.75 = 89 UMS

For June 09 the A boundary was 60/75 and the B was 52.

63/75 = 10[(63-60)/8]+80 = 83.75 = 84 UMS
65/75 = 10[(65-60)/8]+80 = 86.25 = 86 UMS

60/75 is a very common A boundary for summer sessions for MPC2.

For Jan 09 the A boundary was 65/75 and the B was 57 and the C was 49.

63/75 = 10[(63-57)/8]+70 = 77.5 = 78 UMS
65/75 = 80 UMS (A boundary)

June 08 - identical to June 09.
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8 years ago
#330
(Original post by Ollie F)
-.- , I'm still adament i'm right on this, consider this.

y=x^2

F(X) =X^2 agreed?

F(X-2) = (X+2)^2

Now by your argument, you would argue that it aught to be the root of 2, am I correct? because the square of 2 is 4, meaning a translation of 4 not too. But this is wrong. I wil explain.

Say we have the graph (X+2)^2, now w/e this you want this graph to look like, imagine now we do a horizontal stretch paralell to the x axis of 1/2. Now this function should look like, [2(x+2)]^2. However if this is so, surely you would say, the graph would be translated more so, as the +2 is now multiplied. But this is silly.

Now if your still not convinced, I have scoured youtube for a decent video which supports my point, (there were alot).

Are you now happy that the translation of the graph IS [3,0]. If not PLEASE explain because I'd like to know if you have anything more to contribute.

All the best, Ollie.

The curve with equation y= root(x^3 +1) is stretched parallel to the x-axis with scale
factor 0.5 to give the curve with equation y=f(x) .

F(x) would = to root(2x^3+1) where as the correct answer is root((2x)^3+1) = root(8x^3+1)
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8 years ago
#331
(Original post by rob...)
And the answers: to which I hope you're all very grateful as it's taken me ages to find all the right symbols on microsoft word, and re-do the paper haha

There is the document, in the formats:
word 2009;
word 97-03;
PDF.
i dont understand your answer for 8 e ii ,(4x-3)log102 = log10(5 ÷ 4) (4x)log102 = log10(23 x 5 ÷ 4) how come did u 4x ignoring -3?
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8 years ago
#332
Here is my proof for the translation question. -.- final, surely there is nothing which can contradict this. I mean its maths for ffs, its not made to be opinionated. The translation IS [3,0] if the graph shifts 3 places to the right. and here is why.

Y= 2^4x = F(x)

Translation of 3 to the right

= F(x-3)

Therefore

2^4(x-3)

4 is outside brackets because it is part of the function F (x) and therefore does not act as horizontal translation as would be identified by F (4x-3), with additional translation of 3 to the right.

Logarithm proof.

Let Y = K.

K = 2 ^ 4x
Log K = Log 2^4x
Log K / 4Log 2 = x. This bit was simple to identify and we are both agreed.

Now.

K = 2^4(x-3) with the 4 being outside the brackets as explained above. This is the significant change.

Log K = Log 2^4(x-3)
Log K = 4(x-3)Log 2
Log K / 4Log 2 = X – 3

Therefore

(Log K / 4Log 2) + 3 = X.
Log K / 4Log 2 = X

This demonstrates a translation of 3 places to the right and NOT that of ¾.
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8 years ago
#333
(Original post by chronicl3)
The curve with equation y= root(x^3 +1) is stretched parallel to the x-axis with scale
factor 0.5 to give the curve with equation y=f(x) .

F(x) would = to root(2x^3+1) where as the correct answer is root((2x)^3+1) = root(8x^3+1)
*blank expression* nope i'm lost to many formulas and words. Look above for my proof of the initial problem.
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8 years ago
#334
Oh well, lets wait for the mark scheme .....

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8 years ago
#335
(Original post by frh_xo)
HI LEILAAAAAAAAAAA
I'm still amazed you knew it was me. Of all the million of people. Obviously made a big impact in your life
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8 years ago
#336
(Original post by Weeman5872)
I put p = -3, all they said was p and q were integers. Just because there is a plus sign in front of p, doesn't make it positive.
p HAS to be a minus otherwise the expansion wouldn't work. So it was -3 and +3
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8 years ago
#337
(Original post by L_93)
I'm still amazed you knew it was me. Of all the million of people. Obviously made a big impact in your life
you actuallllyy have leilaaa, you are one of a kind ! dontt ever changeeeeee
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8 years ago
#338
(Original post by Averagegirl:))
p HAS to be a minus otherwise the expansion wouldn't work. So it was -3 and +3
That too.

Though, this is why I put -3. My previous post was to prove a point that people couldn't say -3 was wrong just because the question said + p/x^2
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8 years ago
#339
kwik question! for the binomial question i panicked and left it out for me to return to at the end of the paper and finish! i managed to complete it but got 3 and 3 instead of 3 and -3! I gt part c) right except i got positive 1.7 instead of -1.7. will i still get majority of the method marks and if so how much will this be?
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8 years ago
#340
For 6bii) I have a feeling I did something very stupid. Even though I worked out f'(x)= 2x-0.5x^-1.5 I think for some reason unknown I worked out the value of f'(1) as 2 . Therefore getting the normal equation wrong. However I worked out correctly the Y coordinate of P being 2. This gave me y-2=-2(x-1). y=-2x+4. Do you think I would get any method marks for this?

For the integration question I also put the 1/5x^5 value first as 0.2x^-5 and then as (1/5)(1/x^5) would I lose a mark for not simplifying? Or would I get the mark as I put it as 0.2x^-5 to start with anyway?
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