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OCR MEI AS Mathematics Core 1 13/05/2013

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Original post by Deceived
(r^2)(a+b) = 3V/pi

(r^2) = 3V/pi/(a+b)

^ This is the wrong step.

r2(a+b)=3V/π

r2=(3V/π)/(a+b)

r2=(3V/π)*(1/(a+b))

r2=3V/(π*(a+b))

I hope this makes sense. Basically, dividing by something is the same as multiplying by the reciprocal.



yeah thanks, should get a mark. silly mistake. dont think i messed up on the rest
one last time guys hahah
is this ok?
i did it a long way kind of


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Deceived
Original post by Rubyturner94
one last time guys hahah
is this ok?
i did it a long way kind of


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Yeah that's correct.

I'm so mad at myself for putting ± before the square root. Next time I'm going to read the damn question. Hopefully, that was my only silly mistake.
Reply 263
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LRJ
Original post by Badman1231
Are well done mate. I did this method but instead of putting (2+k)(2-k) I put (2-k)(2-k) which got my final awnser to
K^2-12k
K(K-12) therfore K=0 or 12


I think I meant to write (2+k)(2+k), I went back to correct what I wrote when I typed it up but forgot that second bracket, oops :redface:
Yes! haha
Forget about C1 now lets focus on C2!
Original post by Rubyturner94
one last time guys hahah
is this ok?
i did it a long way kind of


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Reply 265
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LRJ
Original post by Deceived
Damn. That actually works. Spent ~5 minutes trying to do that but for some reason I couldn't get it to work so I used the differentiation method. Way to go for using that method. I assume that's the method OCR want their candidates to use as chain rule isn't on the C1 syllabus. Do you think I'll still receive full marks for this question even though I used differentiation? I got the same answer (k=0 or 4).

Thanks


I confused myself trying to work it out again!

I think it should be fully credited, as you ended up with the right answer - I mean the Jan11 question was a show that question so it should have been marked differently :smile:
It's a bit like for the question where you had to find the y coordinate for the minimum point(?) in Section A, I read -7 off the completing the square intercept, then checked it by differentiating the original expression to find the y value from the x value when dy/dx =0 if that makes sense :smile:
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Deceived
Original post by LRJ
I confused myself trying to work it out again!

I think it should be fully credited, as you ended up with the right answer - I mean the Jan11 question was a show that question so it should have been marked differently :smile:
It's a bit like for the question where you had to find the y coordinate for the minimum point(?) in Section A, I read -7 off the completing the square intercept, then checked it by differentiating the original expression to find the y value from the x value when dy/dx =0 if that makes sense :smile:



Thanks. I pray it does end up getting full credit. Yes, I did the exact same thing for the minimum question. Differentiation is so useful haha. Best of luck with C2 :smile:
Reply 267
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LRJ
Original post by KanKan
Omg, I actually got all this, untill this point
then somehow ended up with x = +-(-1)^1/2
.. which can't be right because you cant root a negative (in real terms) :frown:

How many marks do you reckon I picked up then?


I'm not sure, how many marks was it out of? :s-smilie:

I'm guessing here, but there should have been a mark for eliminating y, finding the quadratic expression involving k's, a mark for showing that b^2 -4ac = 0 for a tangent, a mark for the right answer, and perhaps another method mark somewhere in there?
-and for the differentiation method I'm not too sure xD
Reply 268
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vincentky
Hi can someone confirm the workings for question 2.) section A i seem to have got Integers and I'm not sure how :/
Original post by vincentky
Hi can someone confirm the workings for question 2.) section A i seem to have got Integers and I'm not sure how :/

same!
I think I need to stop reading this thread and calculating how many marks I lost. I've lost about 8-10 marks on stupid human errors and probably more. Hopefully I'll make up for the lost marks in C2.

Can anyone comment on how hard this paper was to previous past papers? I hope the grade boundaries for an A is around 57.
Reply 271
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Deceived
Original post by vincentky
Hi can someone confirm the workings for question 2.) section A i seem to have got Integers and I'm not sure how :/


What was question 2 again?
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vincentky
Original post by Deceived
What was question 2 again?

Something to do with the intersection between two lines?
I think I got one of them to be... y=-1/3x+1/3 when I rearranged for Y. then equated it to the other lines y=3x+2 and attempted to solve..
Reply 273
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Deceived
Original post by vincentky
Something to do with the intersection between two lines?
I think I got one of them to be... y=-1/3x+1/3 when I rearranged for Y. then equated it to the other lines y=3x+2 and attempted to solve..


Oh yeah, that's right.

One equation was x+3y=1 and the other was y=3x-2 I think

3y=-x+1
y=-(1/3)x+(1/3)

3x-2=-(1/3)x+(1/3)
(10/3)x-(7/3)=0
(10/3)x=(7/3)
10x=7
x=7/10

y=3x-2
y=(3*(7/10))-2
y=(21/10)-2
y=1/10

Point of intersection = (7/10,1/10)

I'm fairly sure that was the answer I got.

I hope this helps. :smile:
Reply 274
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vincentky
Original post by Deceived
Oh yeah, that's right.

One equation was x+3y=1 and the other was y=3x-2 I think

3y=-x+1
y=-(1/3)x+(1/3)

3x-2=-(1/3)x+(1/3)
(10/3)x-(7/3)=0
(10/3)x=(7/3)
10x=7
x=7/10

y=3x-2
y=(3*(7/10))-2
y=(21/10)-2
y=1/10

Point of intersection = (7/10,1/10)

I'm fairly sure that was the answer I got.

I hope this helps. :smile:


I did everything correct up until the part 10/3x =7/3 i couldn't remember the equations of the lines exact. but i didn't multiply the numerators by 3 i just multiplied 7/3 by the inverse of 10/3 and got 21/30 but I did not simplify and used the x val to find y. do you think I would still get credibility?
and thanks btw :biggrin:
(edited 10 years ago)
Original post by krishparscrotam
My answers were as follows:
8. 3(x-2)^2 -7 and minimum y value was -7


Hi, did they ask for the minimum y value, or the turning point. I put (2,-7). Am i wrong?
Reply 276
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Deceived
Original post by vincentky
I did everything correct up until the part 10/3x =7/3 i couldn't remember the equations of the lines exact. but i didn't multiply the numerators by 3 i just multiplied 7/3 by the inverse of 10/3 and got 21/30 but I did not simplify and used the x val to find y. do you think I would still get credibility?


You'd definitely get a mark for substituting out y. You may receive an additional mark but I guess it really depends on how many marks it was out of. I assume 1 mark is rewarded for finding the y coordinate at the end. So if it was a 3 mark question, you're probably looking at receiving 2 marks because 21/30=7/10 so you shouldn't lose an A1 mark. So 2/3 marks probably.

Hope the exam went well for you and best of luck with C2. :smile:
Reply 277
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Deceived
Original post by Studenthsr
Hi, did they ask for the minimum y value, or the turning point. I put (2,-7). Am i wrong?


They asked for the minimum y value. I assume you'll receive full credit because you put (2,-7).
Reply 278
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cloudhang
How did everyone find the Paper?
I personally found is okay - easy, compared to the Jan one
60 for an A? or lower?
That was a really nice exam. I mucked up a little on 12)iii but got 4 and 0 out graphically, so should have picked up a couple of marks. Also had a stupid moment on the surds rearrangement question in section A, but other than that I'm reasonably confident of a high A :smile:

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