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Mr M's OCR Core 1 (not OCR MEI) May 2016 Answers

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For Question 6 I got -2(x+3)^2 +22 and for 6ii I ended up getting (-3,22).

Also, for question 9 i did all the steps required however at the start i ended up putting 2kx + k instead of 2kx - k, which gave me a different answer.

How many marks will i lose on each of these questions?
does it matter for question 2 if I put -3/4 root 5 rather than -3root 5 all over 4?
also for last question I attempted to rearrange so it was equal to x and then subbed it back in to original eqn, will I get marks for this?

also I didn't draw a graph for q 9?

pleaseeeeeeeeee help!!!!!!!!!!!!!!!!!!!!
Original post by Mr M
Yes but don't worry. Full marks.


Wicked :biggrin: thanks
Original post by Mr M
As that question is worth 8 marks, I think you will lose 1 for failing to simplify.

And you are welcome.


Okay thanks very much I think I got between 64-68 marks which is better than I thought considering how badly I thought it went hopefully I will still get an a:smile:
Original post by NotGambit
For Question 6 I got -2(x+3)^2 +22 and for 6ii I ended up getting (-3,22).

Also, for question 9 i did all the steps required however at the start i ended up putting 2kx + k instead of 2kx - k, which gave me a different answer.

How many marks will i lose on each of these questions?


Drop 1 and drop 2.
For question 9, I recognised that the discriminant would have to be >0, and I substituted the correct values of a, b, c into b^2 - 4ac. However, I expanded the quadratic incorrectly, and therefore was left with an unsolvable quadratic, for which I tried (and failed) to use the quadratic formula. How many marks would I lose for this?

Also, any grade boundary predictions?
Original post by mathsgeek5454
does it matter for question 2 if I put -3/4 root 5 rather than -3root 5 all over 4?
also for last question I attempted to rearrange so it was equal to x and then subbed it back in to original eqn, will I get marks for this?

also I didn't draw a graph for q 9?

pleaseeeeeeeeee help!!!!!!!!!!!!!!!!!!!!


For Q2 it doesn't matter.

Drop 2 marks for Q9.

Get some credit for the last question.
Reply 27
Avatar for S.G.
S.G.
22
Original post by Mr M
The answer I provided is correct here.


I think I have got around 58-60 marks. Based on your experience of teaching, what do you think the grade boundary could be.

By the way, thanks for doing this and sorry for bothering you.
Reply 28
Avatar for Devkj
Devkj
3
Sir how many marks would I get for the last question if I found that a=8x^3 and then subbed in 32 into the formula but then rearranging it to make a the subject. Subbing a back in as 8x^3 coming to an answer of somehow 16root2
Original post by theycallmejamo
For question 9, I recognised that the discriminant would have to be >0, and I substituted the correct values of a, b, c into b^2 - 4ac. However, I expanded the quadratic incorrectly, and therefore was left with an unsolvable quadratic, for which I tried (and failed) to use the quadratic formula. How many marks would I lose for this?

Also, any grade boundary predictions?


You'll probably get 3 or 4 marks.

I don't guess boundaries but I expect they will be lower than they have been for the last few years.
Reply 30
Avatar for fjfb1
fjfb1
I have a quick question about question 10 iii? Is the tangent not y=2x-5 not y=2x+10? Since
2 = (y-3)/(x-4)
2(x-4) = y-3
2x-8 = y-3
Hence y=2x-5
If not, would you be able to just mention what I have done wrong so I know before the next exam!?
Original post by Devkj
Sir how many marks would I get for the last question if I found that a=8x^3 and then subbed in 32 into the formula but then rearranging it to make a the subject. Subbing a back in as 8x^3 coming to an answer of somehow 16root2


Definitely worth some marks. 4 minimum and probably more.
For 10iii I found the y intercept as (0,10) and input it to get the correct answer, but I'm concerned this wasn't actually the right working- will I not get marks even if i used the correct point and gradient to get the right answer?

Also for 8ii I differentiated to find the gradient at A as 20 and said that this showed that its gradient was lower than at B as it was 20 + 2h, will that get zero?

Thanks for your help!
Original post by fjfb1
I have a quick question about question 10 iii? Is the tangent not y=2x-5 not y=2x+10? Since
2 = (y-3)/(x-4)
2(x-4) = y-3
2x-8 = y-3
Hence y=2x-5
If not, would you be able to just mention what I have done wrong so I know before the next exam!?


That's the equation of the parallel line that passes through the centre not the other tangent.
Question 2 asked for the answer in the form a+b root5, but that's just a small nitpick :tongue:
Reply 35
Avatar for blinin7
blinin7
1
Hi, would appreciate some help.

For question 6 i) I got
2(x+3)-22
instead of,

as for some reason I re-arranged the equation given so that it began with 2x^2...

Question 7 iii)
I put stretch of scale factor 2 in x direction instead of Stretch parallel to the y axis scale factor 0.5

Question 9
I got k>5 and K<-4
instead of
or
however i got the factors (2k-10) (2k+4), I was just so rushed i put K as -4 instead of K as -2

Any ideas how many Marks I would have dropped from those 3.

Many thanks :smile:
(edited 7 years ago)
Original post by Nomes24
For 10iii I found the y intercept as (0,10) and input it to get the correct answer, but I'm concerned this wasn't actually the right working- will I not get marks even if i used the correct point and gradient to get the right answer?

Also for 8ii I differentiated to find the gradient at A as 20 and said that this showed that its gradient was lower than at B as it was 20 + 2h, will that get zero?

Thanks for your help!


The point you wanted to use was (-2, 6) so I'm not sure I know what you did. You'll get one mark for knowing parallel lines have the same gradient.

8ii) will get the mark I think.
Original post by liziepie
Question 2 asked for the answer in the form a+b root5, but that's just a small nitpick :tongue:


The answer is in that form.
Original post by SGHD26716
I think I have got around 58-60 marks. Based on your experience of teaching, what do you think the grade boundary could be.

By the way, thanks for doing this and sorry for bothering you.


I don't like to guess boundaries but it was more difficult than usual.
Original post by Mr M
The point you wanted to use was (-2, 6) so I'm not sure I know what you did. You'll get one mark for knowing parallel lines have the same gradient.

8ii) will get the mark I think.


Thanks for your reply! I thought the second tangent met the circle (according to the diagram) at the y intercept, I know that isn't the most mathematical way to do it but I was desperate, which is why I think I might have lost marks. I got the gradient as 2 and (0,10) then did y-10=2(x-0) to get y=2x+10?

I found the y intercept by setting x to zero and then factorising.

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