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OCR MEI C3 Maths June 2015 watch

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    For question 4, I had my calculator in radian mode for tan(45), and hence got a different value. However, if I didn't have my calculator in radian mode, I would have got the final answer. I got to a stage where... 5/(100 x pi x tan(45)) though, which would be correct if i put in tan(45)=1. 4 marks? :L

    Also, 3ln3-4 is a negative value so shouldn't the answer be 4-3ln3 for 9)iii)?

    Does anyone think grade boundaries would be below 58? :l
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    (Original post by Another_Brummie)
    For question 4, I had my calculator in radian mode for tan(45), and hence got a different value. However, if I didn't have my calculator in radian mode, I would have got the final answer. I got to a stage where... 5/(100 x pi x tan(45)) though, which would be correct if i put in tan(45)=1. 4 marks? :L

    Also, 3ln3-4 is a negative value so shouldn't the answer be 4-3ln3 for 9)iii)?

    Does anyone think grade boundaries would be below 58? :l
    To me it certainly felt harder than average, although opinions seem to be mixed.
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    (Original post by Another_Brummie)
    For question 4, I had my calculator in radian mode for tan(45), and hence got a different value. However, if I didn't have my calculator in radian mode, I would have got the final answer. I got to a stage where... 5/(100 x pi x tan(45)) though, which would be correct if i put in tan(45)=1. 4 marks? :L

    Also, 3ln3-4 is a negative value so shouldn't the answer be 4-3ln3 for 9)iii)?

    Does anyone think grade boundaries would be below 58? :l
    Yeah, I put 4-3ln3 but lots of people seem to have forgotten this step so I don't know what they'll do about it
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    I don't know if I'm correct in working this out for UMS, but I think I got 36/72 in that paper, so for harshness's sake let's say that's 47% in harsh boundaries. So would I be right in saying that's 38 UMS out of 80 UMS? And I got 16/18 UMS in the coursework giving me 18 UMS. So would that be 56 UMS in total?
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    (Original post by JamboGooner)
    I don't know if I'm correct in working this out for UMS, but I think I got 36/72 in that paper, so for harshness's sake let's say that's 47% in harsh boundaries. So would I be right in saying that's 38 UMS out of 80 UMS? And I got 16/18 UMS in the coursework giving me 18 UMS. So would that be 56 UMS in total?
    Yeah that would work out to be 56 UMS in total haha

    (Original post by a96clark)
    Yeah, I put 4-3ln3 but lots of people seem to have forgotten this step so I don't know what they'll do about it
    I think it will only be worth a mark, if anything.

    (Original post by HenryHein)
    To me it certainly felt harder than average, although opinions seem to be mixed.
    I didn't find it too hard, it was just the questions were so time-consuming D: I ran out of time :/
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    (Original post by JamboGooner)
    I don't know if I'm correct in working this out for UMS, but I think I got 36/72 in that paper, so for harshness's sake let's say that's 47% in harsh boundaries. So would I be right in saying that's 38 UMS out of 80 UMS? And I got 16/18 UMS in the coursework giving me 18 UMS. So would that be 56 UMS in total?
    No
    Assuming you got 36 in paper and 16 in cw,
    UMS score = (36+16)/90 * 100 = 57.8 UMS

    EDIT: Never mind I messed up

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    For integration by parts, if I messed up the u and dv/dx values (got them wrong way round) will i get carried forward marks
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    So, this is what I did for the arccosx=arcsinx question:

    

arcsin(x) = arccos(x)

x = sin(arccos(x))

    But we know that:

    

sin^2x+cos^2x=1 \Rightarrow

sinx = \pm\sqrt {1-cos^2x}

    Therefore:

    

x = \pm\sqrt {1-cos^2 {(arccosx)}}

x = \pm\sqrt {1-x^2}

x^2 = 1-x^2

2x^2 = 1

x^2 = \frac{1}{2}

x = \pm \frac{1}{\sqrt 2}

    I now realise this isn't entirely correct, as the negative root has to be discarded due to the restricted domains of the inverse functions. However, how many marks would I get for this?
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    (Original post by Norbo11)
    So, this is what I did for the arccosx=arcsinx question:

    

arcsin(x) = arccos(x)

x = sin(arccos(x))

    But we know that:

    

sin^2x+cos^2x=1 \Rightarrow

sinx = \pm\sqrt {1-cos^2x}

    Therefore:

    

x = \pm\sqrt {1-cos^2 {(arccosx)}}

x = \pm\sqrt {1-x^2}

x^2 = 1-x^2

2x^2 = 1

x^2 = \frac{1}{2}

x = \pm \frac{1}{\sqrt 2}

    I now realise this isn't entirely correct, as the negative root has to be discarded due to the restricted domains of the inverse functions. However, how many marks would I get for this?
    It was only a two mark question, so you would only drop one mark. You might even not drop any depending on how lenient the mark scheme is. For future reference, the result follows quickly after noting that:

    arccosx = \frac{\pi}{2} - arcsinx
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    (Original post by lizard54142)
    It was only a two mark question, so you would only drop one mark. You might even not drop any depending on how lenient the mark scheme is. For future reference, the result follows quickly after noting that:

    arccosx = \frac{\pi}{2} - arcsinx
    I see, I wasn't aware of this result. Do you have a proof?
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    (Original post by Norbo11)
    I see, I wasn't aware of this result. Do you have a proof?
    There are some lengthy algebraic proofs which are tedious to type up, but I'm sure you can find some online.

    A non-rigorous proof would be to consider the graph of y=arcsinx. If you reflect the graph in the y axis, and translate it by \begin{pmatrix} 0 \\ \pi /2 \end{pmatrix} you will see that you obtain the graph of y=arccosx
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    (Original post by JamboGooner)
    I don't know if I'm correct in working this out for UMS, but I think I got 36/72 in that paper, so for harshness's sake let's say that's 47% in harsh boundaries. So would I be right in saying that's 38 UMS out of 80 UMS? And I got 16/18 UMS in the coursework giving me 18 UMS. So would that be 56 UMS in total?
    Lol well done that's sick
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    (Original post by lizard54142)
    It was only a two mark question, so you would only drop one mark. You might even not drop any depending on how lenient the mark scheme is. For future reference, the result follows quickly after noting that:

    arccosx = \frac{\pi}{2} - arcsinx
    Hope you don't mind me asking, do you think I would lose any if I didn't show any working other than:
    arccos\frac{1}{\sqrt{2}} = \frac{\pi}{4}, arcsin\frac{1}{\sqrt{2}} = \frac{\pi}{4} \therefore x = \frac{\pi}{4}?I couldn't figure out the working and I was worried for time so I just drew graphs on some scrap and figured out roughly where they collide and then just rounded to the nearest significant value I knew and plugged it into my calculator (1/rt2)
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    (Original post by anunoriginaluser)
    Hope you don't mind me asking, do you think I would lose any if I didn't show any working other than:
    arccos\frac{1}{\sqrt{2}} = \frac{\pi}{4}, arcsin\frac{1}{\sqrt{2}} = \frac{\pi}{4} \therefore x = \frac{\pi}{4}?I couldn't figure out the working and I was worried for time so I just drew graphs on some scrap and figured out roughly where they collide and then just rounded to the nearest significant value I knew and plugged it into my calculator (1/rt2)
    You'll probably get one mark for that, possibly two I'm not sure! You probably had to show a bit more working for the method mark.
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    (Original post by lizard54142)
    You'll probably get one mark for that, possibly two I'm not sure! You probably had to show a bit more working for the method mark.
    Awesome, better than nothing I suppose thanks!
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    (Original post by anunoriginaluser)
    Hope you don't mind me asking, do you think I would lose any if I didn't show any working other than:
    arccos\frac{1}{\sqrt{2}} = \frac{\pi}{4}, arcsin\frac{1}{\sqrt{2}} = \frac{\pi}{4} \therefore x = \frac{\pi}{4}?I couldn't figure out the working and I was worried for time so I just drew graphs on some scrap and figured out roughly where they collide and then just rounded to the nearest significant value I knew and plugged it into my calculator (1/rt2)
    That shows that X=1/sqrt2, not x=1/4pi. X=1/sqrt2 will be an answer mark so you'll probs lose one, but you should get the other
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    (Original post by a96clark)
    That shows that X=1/sqrt2, not x=1/4pi. X=1/sqrt2 will be an answer mark so you'll probs lose one, but you should get the other
    Oops yeah, I typed my response wrong - I did write x = 1/rt2 in the actual exam. Thanks!
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    I think I got ~60 in C3, and 14 in the coursework, and in C4 I think I got 80/90, do you think that's enough for an A*?
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    (Original post by anunoriginaluser)
    Oops yeah, I typed my response wrong - I did write x = 1/rt2 in the actual exam. Thanks!
    I think the mark scheme should be pretty lenient so I'd bet you'd get both marks there
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    what happens if in the exam the answer is '4 - 3 ln 3' and you write 4 - ln (3^3)? Technically they are the same, but from their mark schemes they dont shed any light on further unnecessary simplification. I hope I haven't lost stupid marks doing this, as i did it throughout c3 and c4
 
 
 
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