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?
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    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!!!!!!!!!!!!!!!!!!!!
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    (Original post by Mr M)
    Yes but don't worry. Full marks.
    Wicked thanks
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    (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
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    (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.
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    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?
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    (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.
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    (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.
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    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
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    (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.
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    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!?
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    (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.
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    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!
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    (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.
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    Question 2 asked for the answer in the form a+b root5, but that's just a small nitpick
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    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
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    (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.
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    (Original post by liziepie)
    Question 2 asked for the answer in the form a+b root5, but that's just a small nitpick
    The answer is in that form.
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    (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.
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    (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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