OCR MEI C3 Maths June 2015

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

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

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, 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

To me it certainly felt harder than average, although opinions seem to be mixed.

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?

So, this is what I did for the arccosx=arcsinx question:

$[br]arcsin(x) = arccos(x)[br]x = sin(arccos(x))[br]$

But we know that:

$[br]sin^2x+cos^2x=1 \Rightarrow[br]sinx = \pm\sqrt {1-cos^2x}[br]$

Therefore:

$[br]x = \pm\sqrt {1-cos^2 {(arccosx)}}[br]x = \pm\sqrt {1-x^2}[br]x^2 = 1-x^2[br]2x^2 = 1[br]x^2 = \frac{1}{2}[br]x = \pm \frac{1}{\sqrt 2}[br]$

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:

$arccosx = \frac{\pi}{2} - arcsinx$

I see, I wasn't aware of this result. Do you have a proof?

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?

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)

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.

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

Oops yeah, I typed my response wrong - I did write x = 1/rt2 in the actual exam. Thanks!