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# OCR C4 (not mei) 18th June 2013 revision watch

1. (Original post by Holz888)
I got 8.something.

You did it again and got the negative. Both are right, because if you think, when two lines meet there is always an obtuse and acute angle. They didn't specify which they wanted, so mark schemes always allow both.

The only problem is which one is used in the volume equation, and I'm not sure about that one
Aaaah. So they were both right. I used the 171 for the volume, but I think that's wrong.
2. (Original post by Holz888)
I got 8.something.

You did it again and got the negative. Both are right, because if you think, when two lines meet there is always an obtuse and acute angle. They didn't specify which they wanted, so mark schemes always allow both.

The only problem is which one is used in the volume equation, and I'm not sure about that one
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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3. (Original post by Vip3rgt9)
I got something with r^(3/2) in it.
You're right
4. (Original post by Benjy100)
This is effectively a preliminary markscheme until Mr M posts his full definitive solutions. Some of my answers may be wrong. Spoilered incase you don't want to see them

Spoiler:
Show
1.

2.

3. Skew. Solved the first two equations for and and then substituted into the third equation for 18=-38 which is clearly untrue, thus the equations are inconsistent and the lines are skew.

4.

Stationary points are

5.

Integral was

6.

7.

Angle BAC = 171.3 degrees (1 d.p.)

I messed up on the volume of the pyramid

8.

Letting t=0, finding r and substituting gives V = 30.5 cm^3 (3 s.f.)

9.

Solving dy/dx yields t=1 which corresponds to the point (0,3)

When t>1 x<0 and dy/dx is negative
When t<1 x>0 and dy/dx is positive

Hence it is a minimum

10. Show.

Let x=0.1 for 0.136

Same substitution would be unsuitable as the expansion here is valid for which is the same as
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)
5. (Original post by SJ12345)
I got 8.68
That was the first answer I got but I scribbled it out because I thought it was too low
6. (Original post by ActaNonVerba)
I can see grade boundaries dropping to about 50 for an A (52 is the lowest it's ever dropped)
I agree and I thought that this paper was as hard or harder than that paper.
7. THAT EXAM COULDN'T HAVE GONE ANY WORSE
WHY OCR? WHY
the exam was so long I couldn't even think for long enough without thinking I'm not going to have time to do the other questions
8. (Original post by eggfriedrice)
Isnt it slightly unfair people can plot the graphs using theirs calculators during the exam? :| "pay £80 and get a slight advantage. " Meh.
I totally agree with this....
9. I got 180-171.3= 8.67 degrees for angle.
Mucked up on tan2x.
Didn't have a clue on later parts of Q10.
Can't certainly say any answer looks convincing.
10. can someone confirm this:

for the tan question my final integral was

and I subbed in values wrong for the first part so I ended up with

can someone confirm whether they got the -6+4 root 3 part?
11. (Original post by Maid Marian)
That's what I did, but apparently it's not right and I didn't think it could be right either because it was worth 5 marks
you can change it to sin2x/cos2x
12. Just fml, its all over.
13. Did anyone else not have a clue what q10 was trying to get at?

I got the first part, but the rest of it
14. Did anyone get t= -1? Fml :/
15. I think the first half of the exam was lovely. The second half though *kill me*
16. (Original post by JoshL123)
Did anyone get t= -1? Fml :/
i did

but i can tell you now we're wrong
17. (Original post by louise_234)
you can change it to sin2x/cos2x
Would that have worked? Worst thing is, I'd written some identities on the post it note with my desk number on it, and just as I gave the lady the paper I noticed them and I realised that's what I should have done
18. (Original post by Benjy100)
This is effectively a preliminary markscheme until Mr M posts his full definitive solutions. Some of my answers may be wrong. Spoilered incase you don't want to see them

Spoiler:
Show
1.

2.

3. Skew. Solved the first two equations for and and then substituted into the third equation for 18=-38 which is clearly untrue, thus the equations are inconsistent and the lines are skew.

4.

Stationary points are

5.

Integral was

6.

7.

Angle BAC = 171.3 degrees (1 d.p.)

I messed up on the volume of the pyramid

8.

Letting t=0, finding r and substituting gives V = 30.5 cm^3 (3 s.f.)

9.

Solving dy/dx yields t=1 which corresponds to the point (0,3)

When t>1 x<0 and dy/dx is negative
When t<1 x>0 and dy/dx is positive

Hence it is a minimum

10. Show.

Let x=0.1 for 0.136

Same substitution would be unsuitable as the expansion here is valid for which is the same as
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.
19. Did anyone get 0.136 on any part of question 10?
20. (Original post by printergirl)
i did

but i can tell you now we're wrong
Oh man :/

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