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# OCR (not MEI) Core 3 - 19/01/2011 Watch

1. (Original post by ElMoro)
I got b = 3pi - a
And I got WTF. Seriously how was that like anything we've ever done?!!?
2. Ahhhhh why didn't I recognise (ln(x))^2 + 8 as a chain rule question?

Ahhh no facepalm can take away the shame I'm feeling after doing that exam.
3. Surely for 4 i) A is inverse tan of 7/25, not inverse sine?

(Original post by yourworstnightmare)
Just saw that the paper was uploaded, so had a crack at the first page. I'm not sure how correct it is since I did it in 5 minutes and have no calculator (being borrowed by current students ) so answers are either done in my head or in exact form:

1) x = 3a, x = a/3

2) sketch; move 3 to left and multiply entire function by -4 (by stretching and reflecting)

3) 14400pi = 45000cm^2 (2 sf)

4i) R = 25, a = sin-1(7/25) (no calc sorry )

ii) theta = sin-1 (12/15) - a (alpha from part i)

5) a = 9

volume = pi x (12ln25) = 12pi ln25

Shall I do the rest of them for you too?
4. (Original post by ElMoro)
I got b = 3pi - a
i got sin^-1(sin theta)
5. (Original post by a.nk)
Surely for 4 i) A is inverse tan of 7/25, not inverse sine?
Nope tis sine inverse, where did you get tan from?
6. (Original post by a.nk)
Surely for 4 i) A is inverse tan of 7/25, not inverse sine?
Rsina = 7

R = 25 (worked out)

25sina = 7

sina = 7/25

a = sin-1 (7/25)

where a = alpha
7. (Original post by nikita_atikin)
Yep

What did people get for that question that said cosec(a)=cosec(b)?

I said b= 2pi + a

I didn't get what it was going on about.
I got 3pi - a. Since, if you had a value that was between 0.5pi and pi, then, to get the other angle that gives the same sine (and thus cosec) between zero and pi, you'd do pi- a. Then, that'd give you one between 0 and 0.5pi, and to get the next answer you'd add 2pi, to give an answer in the 2pi to 2.5pi range, and so overall it'd be (pi-a) + 2pi = 3pi-a, which also sorta made sense from looking at the graph and doing some rough measurements.
8. for question 1 i got x=a/3 and x= -3a. did we have to find out what a was?
9. How come people only got a/3 for the first question, don't you also get -3a?
10. (Original post by ElMoro)
I got b = 3pi - a
Fudge. Yeah, my sister said she got that so she's obv right.

(Original post by Revolution is my Name)
I got 3pi - a. Since, if you had a value that was between 0.5pi and pi, then, to get the other angle that gives the same sine (and thus cosec) between zero and pi, you'd do pi- a. Then, that'd give you one between 0 and 0.5pi, and to get the next answer you'd add 2pi, to give an answer in the 2pi to 2.5pi range, and so overall it'd be (pi-a) + 2pi = 3pi-a, which also sorta made sense from looking at the graph and doing some rough measurements.
Aaaaooooohhhhh, I understand. Oh, I didn't know that was the method haha
11. what did every1 put for 9ii?
12. (Original post by gildartz)
How come people only got a/3 for the first question, don't you also get -3a?
I'll have another look in a second but i'm pretty sure it said a is a positive integer and the modulus sign sort of gives away the fact that it must be positive anyway
13. (Original post by a.nk)
Surely for 4 i) A is inverse tan of 7/25, not inverse sine?
There are two potential ways of finding alpha, one involving inverse tan of 7/24 (by dividing the two expressions involving alpha), the other inverse sin of 7/25 (by using the R value found), both of which give the same alpha value
14. (Original post by Princess_perfect786)
for question 1 i got x=a/3 and x= -3a. did we have to find out what a was?
When there's a modulus sign: |-3a| = +3a
15. Woah, ok. That exam was tricky, and since it's pretty likely that my uni place will be riding on C3 and C4 it's pretty stressful. BUT, I think I cracked most of the questions (touch wood) and am seeing a lot of answers that I agree with on here. I think my favourite question was sketching cosecX. My graph looked so pretty! The next question was the one that stumped me most I think, but it's not too many marks so it's not that bad. Hopefully.

For the one where you had to find the exact value to TanXCot2XTan4X, I spent agggggeeeeeeeessssssss trying to expand it all and simplify everything, before realising I was doing it a stupid way and it was much much simpler just to work out the individual parts separately with TanX=1/4. I got something that I've seen on here so HOPEFULLY, it's right. Although I hope they can see my answer as I used up like 2 sheets before getting it! Also I found that bloody symbol really annoying to draw!

Other than that, for the translation of the graph one did anyone else notice how they were sneakily trying to make you mark off the x-axis intercept as well as the stationary points?

16. (Original post by yourworstnightmare)
I'll have another look in a second but i'm pretty sure it said a is a positive integer and the modulus sign sort of gives away the fact that it must be positive anyway
but if you sub -3a into x you still end up with 5a I think.
17. On question 9 how was f(x)greater than 0 for all values???
18. (Original post by yourworstnightmare)
Rsina = 7

R = 25 (worked out)

25sina = 7

sina = 7/25

a = sin-1 (7/25)

where a = alpha
Oh ok, I've been taught a different method.... I think it gives the same value of a though.
19. (Original post by yourworstnightmare)
When there's a modulus sign: |-3a| = +3a
the retake in june is not going to be any easier, is it?
20. (Original post by Rokaf0)
On question 9 how was f(x)greater than 0 for all values???
f(x) was e^something i think. Well you can't get an answer below 0 for any power to e. I think anyway, sure ill be corrected

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