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2016 May 25th Edexcel Core 2 Questions and answers. [Unofficial mark scheme] 2016

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Original post by medapplicant2016
That's what I thought, I did get 16.something though, and did work out f"(x) > 0, which usually gives you 2 marks (you don't need to sub)

If you got 16.6 then i think you may only lose 1 mark then just for not subbing for the minimum value
Original post by dididid
the answer was b= (3a+5)/9


I did this

log 3 (3b+1/a-2) = -1
3^-1 = 3b+1/a-2
1/3= 3b+1/a-2
a-2=9b+3
a-5=9b
b=a-5/9
Original post by tory88
Is question 6 written out correctly? I get 11π30\frac{11\pi}{30}or 31π30\frac{31\pi}{30}

EDIT: Seen that there have been amendments since then. Shouldn't 8i) be a−59\frac{a-5}{9}?



I think you're correct, this was my workings.

log3(3b+1)−log3(a−2)=−1log_3(3b+1)-log_3(a-2)=-1

log3(3b+1)+1=log3(a−2)log_3(3b+1)+1=log_3(a-2)

log3(3b+1)+log33=log3(a−2)log_3(3b+1)+log_33=log_3(a-2)

log33(3b+1)=log3(a−2)log_33(3b+1)=log_3(a-2)

3(3b+1)=a−23(3b+1)=a-2

9b+3=a−29b+3=a-2

9b=a−59b=a-5

b=a−59b=\frac{a-5}{9}
(edited 7 years ago)
Original post by bulletman54
If you got 16.6 then i think you may only lose 1 mark then just for not subbing for the minimum value


Cool, thanks so much!
Original post by UKoE Luna
I think you're correct, this was my workings.

log3(3b+1)−log3(a−2)=−1log_3(3b+1)-log_3(a-2)=-1

log3(3b+1)+1=log3(a−2)log_3(3b+1)+1=log_3(a-2)

log3(3b+1)+log33=log3(a−2)log_3(3b+1)+log_33=log_3(a-2)

log33(3b+1)=log3(a−2)log_33(3b+1)=log_3(a-2)

3(3b+1)=a−23(3b+1)=a-2

9b+3=a−29b+3=a-2

9b=a−59b=a-5

b=a−59b=\frac{a-5}{9}


At this point I give up, Yes you are right however the question in the exam may have been log⁡3(3b+1)−log⁡3(2a−2)=−1\log_3 (3b+1)-\log_3 (2a-2)=-1. It has been too long, I forget. I'll wait till I can see the paper again tomorow
Original post by X_IDE_sidf
we are on to V4 now http://docdro.id/18BTG2C

Thanks


Shouldn't question 9c) be to prove that

P=1000x+x12(4π+36−33)P=\frac{1000}{x}+\frac{x}{12}(4\pi+36-3\sqrt{3})

I forget how to LATEX... Shouldn't the fraction outside the brackets be x12\frac{x}{12}
(edited 7 years ago)
Original post by X_IDE_sidf
At this point I give up, Yes you are right however the question in the exam may have been log⁡3(3b+1)−log⁡3(2a−2)=−1\log_3 (3b+1)-\log_3 (2a-2)=-1. It has been too long, I forget. I'll wait till I can see the paper again tomorow

Fair enough, probably the best shout, you can only get so far with memory, but I must thank you for the awesome mark scheme!
Original post by X_IDE_sidf
At this point I give up, Yes you are right however the question in the exam may have been log⁡3(3b+1)−log⁡3(2a−2)=−1\log_3 (3b+1)-\log_3 (2a-2)=-1. It has been too long, I forget. I'll wait till I can see the paper again tomorow


Please quote me in if you give any corrections to the questions.
how many marks would b 90 UMS
for question 6 b, are we supposed tto give answrs to 1 decimal place?
Original post by tory88
Shouldn't question 9c) be to prove that

P=1000x+x12(4π+36−33)P=\frac{1000}{x}+\frac{x}{12}(4\pi+36-3\sqrt{3})

I forget how to LATEX... Shouldn't the fraction outside the brackets be x12\frac{x}{12}


yes, you're right
Original post by dadawdawd
for question 6 b, are we supposed tto give answrs to 1 decimal place?


Pretty sure it said to give answers to 1dp yes.
Original post by X_IDE_sidf
The question didn't have limits at that point, part (a) was simply ∫(3x−x32)dx\int (3x-x^{\frac{3}{2}})dx


Well then it's just +c
Can somebody type out the actual question 7b? I cant remember what it was and may help me to remember if i got 24.3
Original post by X_IDE_sidf
un=arn−1u_n=ar^{n-1}
∴∣u9−u10∣=∣(64)(34)9−1−(64)(34)10−1∣=1.602\therefore |u_9-u_{10}| = |(64)(\frac{3}{4})^{9-1}-(64)(\frac{3}{4})^{10-1}| = 1.602 (3 d.p.)


Hey guys I did exactly this in my exam, I remember writing it identically however I got the wrong answer (calculator error maybe?) Would I get the method marks but not the answer mark?
Original post by Dieselblue
Hey guys I did exactly this in my exam, I remember writing it identically however I got the wrong answer (calculator error maybe?) Would I get the method marks but not the answer mark?


I did u10 - u9 or is that wrong? You should get the method mark if it was only a calculating error.
Original post by Brydon hurst
I did u10 - u9 or is that wrong? You should get the method mark if it was only a calculating error.


I did ar^(9-1)-ar^(10-1) I got -0.025 though
Original post by Dieselblue
I did ar^(9-1)-ar^(10-1) I got -0.025 though


Yeah you will definitely get method marks.
Original post by Brydon hurst
Yeah you will definitely get method marks.


Ok thanks thats good to know
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obeyyy
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any ideas for grade boundaries? roughly how many marks for an a?

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