OCR C4 (not mei) 18th June 2013 revision

Great, I agree with you on all of them :-) Although, I couldn't do 10i or 10ii (the proof part), but I spotted what they were getting at for 10iii so grabbed the marks there :P The volume of the pyramid was 9.33 I believe (to 3sf)

Did you take -x out as a factor, to prove it was the same as the binomial function? :smile:

Reply 521

AlesanaWill

16

Original post by Chrissii

I only paid £25 for my graphical, but it forgot to even use it to help me with that parametric question! :angry:

That sucks. I'll freely admit my graphical calculator saved me on that question as it made me realise I'd made an error (initially said it was a max point).

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Reply 522

RandomMedic

11

Original post by AlesanaWill

It's the 171 I think for volume to use the 1/2 ab sin C formula. It helped to draw a diagram.

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I stayed with 8. whatever, because on the question, you worked out what AC and AB were, then it asked for BAC, so you'd do dot product of BA and AC, which means you use the negative of what you have previously worked out, giving you a positive number to do cos^-1, giving 8. whatever. Then, this could be used directly in the 1/2absinC. That was my thinking, anyway

Reply 523

Lindseyjm

Original post by SJ12345

OMG! I've done 1/2lnsecx instead of lnsec2x. Now I'm depressed

So did I! I just read that and strung out swear words. How stupid.

Reply 524

Jack976

Sin 171 = sin 9 doesn't it? so it won't mess up the volume question

Reply 525

benex381

I think it was a decent paper, only tough one was the last question and the volume of the pyramid. The highest grade boundaries have been 61 for an A and the lowest have been 51 so i would guess at about 56/57 for an A

Reply 526

ActaNonVerba

Original post by Maid Marian

I don't understand that? :frown:

I don't understand why you have to divide by 2...

It's just the same as intergrating cos2x.. You get 1/2sin2x ..

Reply 527

JASApplications

Tough paper! I completely forgot how to integrate tan2x.

I ended up integrating to 1/2tan2x (I know that's completely wrong) and then I applied the limits and got a random answer! - I should get 1 mark for applying the limits?

I think the grade boundaries will be around 52 for an A.

Reply 528

Benjy100

5

Original post by Theo115

Ben, I do not agree with your answer for the angle BAC because you had to find the angle between the vector BA and the vector AC. I believe what you did was find the angle between AB and AC which is incorrect. I got the same answer as you but did 180-ans, to achieve an answer of 8.68 degrees. I am so annoyed though because I messed up on the differential equation and consequently the question after about the volume. Hopefully I will get error carried forward. Hopefully.

It's odd because although I don't know whether AB dot AC (what I did) or BA dot AC was required I got 171.3 degrees but then went on to say - thus acute angle is 8.68 degrees - does it even matter which one you do? I'm not too sure about this one really - I mean it makes sense to think that AB dot AC leads you to the angle BAC just by drawing the diagram - so now I thinking I am thinking that I should have just stayed with 171.3 degrees. Then again I may have even taken the wrong dot product so who knows. I guess we'll have to wait for Mr M :smile:

(edited 10 years ago)

Reply 529

94Singh

Grade boundaries for June are always lower than Jan .. so some hope - perhaps??

Reply 530

ActaNonVerba

Original post by Lindseyjm

So did I! I just read that and strung out swear words. How stupid.

It was 1/2ln(sec2x) anyway..

Reply 531

RandomMedic

11

Original post by Kirity

Also Guys the Angel was 171. It wasn't asking for the acute angle.

The triangle can have an angle of 171 since the total degrees for a triangle is 180 that leave 9 degrees to be split between the other two.

no where did it state it was a regular pyramid.

When two lines meet, there is an acute and obtuse angle, so both are acceptable, as it didn't state which it wanted. They always allow both in the mark scheme, unless they've asked for acute

Reply 532

Kenna

The problem I had with the int(tan2x) was that in the formula book it said the int(tanx) is lnIsecxl so, does that mean the int(tan2x) is lnlsec2xl? (it obviously does not now) . But what I'm trying to say is that the "k" coefficient of the tankx is not in the formula book. Unlike the other inegration formulas. So in my opinion it should say int(tankx) is 1/klnlseckxl.

Anyone know what I'm trying to say? lol :biggrin:

Reply 533

a10

19

Original post by Kirity

Also Guys the Angel was 171. It wasn't asking for the acute angle.

The triangle can have an angle of 171 since the total degrees for a triangle is 180 that leave 9 degrees to be split between the other two.

no where did it state it was a regular pyramid.

is that even possible?

Reply 534

Vip3rgt9

10

Original post by Holz888

When two lines meet, there is an acute and obtuse angle, so both are acceptable, as it didn't state which it wanted. They always allow both in the mark scheme, unless they've asked for acute

I put 8.something as well but they were actually vectors, they have length, they weren't lines that continued forever.

EDIT: By the way I don't know what the correct answer is.

Reply 535

benex381

Original post by 94Singh

Grade boundaries for June are always lower than Jan .. so some hope - perhaps??

No point putting your hopes on this, last year it was higher grade boundaries in June

Reply 536

tooambitious

16

Original post by samthegooner

It's so unfair, I'd much rather have an easier paper and actually give students the chance to show their talent.
That paper was so bad, the difference between the top and bottom achievers won't be very much because there were so 'few' standard questions. I don't mind them being hard, just that they are fairly standard. Question 10 WTF?

I'm sorry, but I completely disagree. What you've said actually makes no sense. When there is an easier paper, it's difficult to differentiate between the top and bottom students becaus even th bottom students can do all the questions, consequently, the results are dependent on who makes the fewest silly mistakes.
Certainly your claim that there will be no difference between top and bottom is rubbish, considering many people at my centre thought the paper was fine and some thought it was awful.
The paper was hard but the questions really weren't as 'left field' as you're making out :confused:

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Reply 537

RandomMedic

11

Original post by Vip3rgt9

I put 8.something as well but they were actually vectors, they have length, they weren't lines that continued forever.

EDIT: By the way I don't know what the correct answer is.

Yeah I understand that, but what I said still stands. If you draw any two lines that meet, whether they carry on or don't there will always be two angles to measure

Reply 538

Theo115

Original post by Benjy100

It's odd because although I don't know whether AB dot AC (what I did) or BA dot AC was required I got 171.3 degrees but then went on to say - thus acute angle is 8.68 degrees - does it even matter which one you do? I'm not too sure about this one really - I mean it makes sense to think that AB dot AC leads you to the angle BAC just by drawing the diagram - so now I thinking I am thinking that I should have just stayed with 171.3 degrees. Then again I may have even taken the wrong dot product so who knows. I guess we'll have to wait for Mr M :smile:

The answer you got will not affect the answer to the second part of the question, but it surely the diagram that you drew would explain how the angle had to be 8.68 degrees. The only reason is because I draw the points on the Cartesian graph and it was very small indeed, so I went for 8.68 degrees. Who knows :P