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Reply 220
Steff21
hmm, well after reading this, i havnt done 2 bad. i got A=1 and B=-1 on the partial fractions qu, i put B on the last qu on the comprehension, and 100 and the speed with the least emissions. also got 0 and 62ish for 1 of the qu's, cant remember which. urm... on the graph on qu 3, did any1 get it crossing the axis at 1,-1 and 1/2,-1/2 (cant remember which was y and which was x)? thanks


Yeah, I got cut the x axis at ±1\pm1 cut y axis at ±12\pm\frac{1}{2}
Reply 221
work now, enjoy later
for the volume of revolution, what did people get?
i got 4/3pi


oh craaap
forgot the pi
oh well
Reply 222
Yay :smile: thats a few more marks then, thanks :biggrin:
Tallon
Think I did quite well. Hoping for the A*. I don't want to sound too big headed but if anybody has any questions about the exam quote me and ask and I'll try and remember.

What did you put for the last question on the comprehension?
matt2k8
There's a link to the "curriculum 2000" papers on the MEI site somewhere (on the page of the current papers I think) - you'd be wanting to look at P2 and P3


P2 is a bit Core 2, with things like sequences in :/ is there much C3 in? Don't know if it's worth doing.

DeanT
For the comprehension graph, were you supposed to just draw 5 horizontal lines for the correct boundaries, or a curve/line?

I got 10cm for Q5 in the comprehension, I think I messed up 2 parts of the vector question so unless the markers are kind there goes my A* :frown:.


I assumed we had to draw that 'step function' graph thing.

Tbh I thought a bar chart would've been more appropriate:p:
Reply 225
DeanT
4+x2+x4x264\sqrt{4 + x} \approx 2 + \frac{x}{4} - \frac{x^2}{64}

-_- how did you get this :frown:
Reply 226
JoshC
-_- how did you get this :frown:

(4+x)12=412(1+x4)12=2(1+x4)12(4 + x)^{\frac{1}{2}} = 4^{\frac{1}{2}}(1 + \frac{x}{4})^{\frac{1}{2}} = 2(1 + \frac{x}{4})^{\frac{1}{2}} and then just use the general expansion.
JoshC
-_- how did you get this :frown:

By binomial
(4+x)12=2(1+x4)12(4+x)^{\frac{1}{2}}=2(1+\frac{x}{4})^{\frac{1}{2}}
which by the binomial is approx2[1+(12)(x4)+(12)(12)(12)(x4)2+...]2[1+(\frac{1}{2})(\frac{x}{4})+(\frac{1}{2})(\frac{-1}{2})(\frac{1}{2})(\frac{x}{4})^{2}+...]
=2[1+x8x2128+...]=2[1+\frac{x}{8}-\frac{x^2}{128}+...]
=2+x4x264+...=2+\frac{x}{4}-\frac{x^2}{64}+...

Hope this helps
Reply 228
DeanT
(4+x)12=412(1+x4)12=2(1+x4)12(4 + x)^{\frac{1}{2}} = 4^{\frac{1}{2}}(1 + \frac{x}{4})^{\frac{1}{2}} = 2(1 + \frac{x}{4})^{\frac{1}{2}} and then just use the general expansion.

-_- i obviously never came across one of these questions before.. (taking the 4 out that is)

looks like ive lost a few there then, i just did:

1 + x/2 + (1/2*-1/2)/2! x^2 :l
Reply 229
Oh My God -.- I thought the vertical was the x-axis, even though it said z was "vertically" upwards - I thought they were trying to catch you out, they should REALLY clear that up, I used (1, 0, 0) as my direction vector instead of the (0, 0, 1). Ffs I could have got the marks for that =.=. Oh well hopefully ecf.

The Comprehension was so easy but while reading it, I was like..."Oh my god...WHO GIVES A **** ABOUT THIS?" I felt like George Bush#.

I put C for the last question....but I don't think I justified it correctly. I basically stated that since they used 7 significant figures, when calculating, you knock off one significant figure, hence you use 6 significant figures, therefore it was to the nearest metre.

Lol :colondollar:.

C4 was fine apart from that vertical thing, I might have also got some of the angles wrong, I got 89.5 Degrees for one & 48.5 Degrees for another one. Apart from that & the first parametric equations x^2 + 4y^2 = 1 or something similiar, everything went smoothly.
foolsihboy
By binomial
(4+x)12=2(1+x4)12(4+x)^{\frac{1}{2}}=2(1+\frac{x}{4})^{\frac{1}{2}}
which by the binomial is approx2[1+(12)(x4)+(12)(12)(12)(x4)2+...]2[1+(\frac{1}{2})(\frac{x}{4})+(\frac{1}{2})(\frac{-1}{2})(\frac{1}{2})(\frac{x}{4})^{2}+...]
=2[1+x8x2128+...]=2[1+\frac{x}{8}-\frac{x^2}{128}+...]
=2+x4x264+...=2+\frac{x}{4}-\frac{x^2}{64}+...

Hope this helps

Ugh just realised I made an arithmetic error on the 3rd term and put /128 lol. Ah well, only 1 mark, should still get an A* :smile:
matt2k8
Ugh just realised I made an arithmetic error on the 3rd term and put /128 lol. Ah well, only 1 mark, should still get an A* :smile:

Immediately after my paper had been collected my brain went
"IDIOT! TAKE THE ANGLE OFF 90 TO GET THE INTERSECTION WITH PLANE"

and likewise for the comp I put "to the nearest millimetre" which again is quite clearly wrong. Urgh. If only I'd checked thoroughly.

By the way, did you find that, after STEP/AEA preperations, C4 seemed sooo much simpler than it used to?
Reply 232
treemantris
What did you put for the last question on the comprehension?



B.
Reply 233
Steff21
hmm, well after reading this, i havnt done 2 bad. i got A=1 and B=-1 on the partial fractions qu, i put B on the last qu on the comprehension, and 100 and the speed with the least emissions. also got 0 and 62ish for 1 of the qu's, cant remember which. urm... on the graph on qu 3, did any1 get it crossing the axis at 1,-1 and 1/2,-1/2 (cant remember which was y and which was x)? thanks


Yeah I got some ovally shaped graph crossing at 1, -1, -1/2, 1/2

Instead of getting 2tanT(tanT+2) i seemed to get 2tanT(tanT+1) so I got T= 0 and 45 which is obviously wrong

I put B on the last question but I forgot to write down 110 or whatever the y coord was for Q2 on the comp and only wrote 100. :frown:
Reply 234
I don't see why anybody would even need to talk about the partial fractions questions. You can just chuck in x =7.34343 or whatever and see minus your answer from the starting fraction in the book and it should equal 0? But yeah, I got A =1 and B =-1, assuming we're using the same denominators for A and B.
DAMNIT - for the third term of the binomial expansion I think I wrote /32 instead of /64.
I seem to have got the same answers for paper A and am confident on that.

However, comprehension the last two questions I'm not too sure.

For the last one I put C because with 16264440cm it could be 16264449cm or something with a more accurate value of the distance so would be 16264450 to the nearest ten centimetre, but I think that's wrong.
foolsihboy
Immediately after my paper had been collected my brain went
"IDIOT! TAKE THE ANGLE OFF 90 TO GET THE INTERSECTION WITH PLANE"

and likewise for the comp I put "to the nearest millimetre" which again is quite clearly wrong. Urgh. If only I'd checked thoroughly.

By the way, did you find that, after STEP/AEA preperations, C4 seemed sooo much simpler than it used to?

I find C3/4 really easy because of STEP/AEA prep :smile:

I realised as soon as I went through it with people who also sat it that I'd made a mistake on the last bit of the comprehension, I remember at first thinking its not even to the nearest metre for some reason or some other thing :\. Hopefully I'll still get 1 or 2 of the marks though for having the right idea.
DOUBLE DAMNIT - I think I labelled the x-intercepts and y-intercepts of the graph of the ellipse the wrong way round!

Easy marks lost!



Don't you just hate when you don't find out about these mistakes until after the exam?
Jk.d
Hmm.. I think I put x2128-\frac{x^2}{128}

That should only cost about 1 or 2 marks?


I did that too:confused:

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