though i wouldnt bet my last tenner on it
1
MATHS EXAM 2005
Started by milnedaniel, May 19 2005 09:50 PM
212 replies to this topic
#201
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Posted 20 May 2005 - 09:35 PM
If i am not here i am somewhere else
#202
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Posted 21 May 2005 - 04:27 PM
oh crap, im getting nervous now cos i think i MITE have forgotton to put the "+ c" after i integrated in question 1 in paper 2
Maybe its just after exam nerves but im thinking bak and i still cant remember if i did put the "+ c" in.
how many marks do u lose for not putting plus c in?
Maybe its just after exam nerves but im thinking bak and i still cant remember if i did put the "+ c" in.
how many marks do u lose for not putting plus c in?
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Posted 21 May 2005 - 04:57 PM
QUOTE(gary @ May 20 2005, 12:01 PM)
Yes I got 36 units wasn't sure about that question paper 1 sucked.
I agree! I thought that that was the hardest paper 1 i had ever done (and people are saying it was easy!!!! AHHHHHHHHHHH!)
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Posted 21 May 2005 - 05:17 PM
QUOTE(tupacshakur @ May 21 2005, 05:27 PM)
oh crap, im getting nervous now cos i think i MITE have forgotton to put the "+ c" after i integrated in question 1 in paper 2
Maybe its just after exam nerves but im thinking bak and i still cant remember if i did put the "+ c" in.
how many marks do u lose for not putting plus c in?
Maybe its just after exam nerves but im thinking bak and i still cant remember if i did put the "+ c" in.
how many marks do u lose for not putting plus c in?
I can't imagine them taking any more than one mark off
#205
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Posted 21 May 2005 - 06:53 PM
Yeah, they only deduct one mark if you miss something like "+ C" or "dx", something like that. That what I used to always forget and I lost only one mark by doing that...lucky. However, I'm panicking about the function combination question "h(x)=g(f(x))" as I always make mistakes in those...
Even more panicky about the differentiation question. How many marks do you reckon they will take off for having 7³=313 instead of 7³=343?
Even more panicky about the differentiation question. How many marks do you reckon they will take off for having 7³=313 instead of 7³=343?
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#206
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Posted 21 May 2005 - 11:14 PM
surley u cant lose a mark for not writing dx!
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Posted 22 May 2005 - 01:04 AM
I noticed the discussion about P2, Q4b earlier in the paper. I got 50.9 for the angle also. Those who thought it was 36 something? The angle doesn't look like it could be any less than 45degrees but I could be wrong.
#208
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Posted 22 May 2005 - 09:04 AM
QUOTE(tupacshakur @ May 22 2005, 12:14 AM)
surley u cant lose a mark for not writing dx!
Yeah, you can lose marks for not writing "dx" or "+ C". Don't believe me? Ask your Maths teacher next time you see him/her. Anyway, about the angle question, I got 80 something degree. I think my answer is correct however...
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#209
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Posted 22 May 2005 - 09:05 AM
QUOTE(Mr H @ May 20 2005, 08:44 PM)
Paper 2 Q8
I am writing out solutions which I will soon put on the web.
Just done P2 Q8 and proved the answer but can't see how it is worth 5 marks.
Can anyone check this and see if I missed something?
Cheers!
ksin2x = sinx
2ksinxcosx = sinx (double angle substitution)
2kcosx = 1 (divide by sinx)
cosx = 1/2k
Proved!
Generous at 4 marks I thought but 5?!*?
I am writing out solutions which I will soon put on the web.
Just done P2 Q8 and proved the answer but can't see how it is worth 5 marks.
Can anyone check this and see if I missed something?
Cheers!
ksin2x = sinx
2ksinxcosx = sinx (double angle substitution)
2kcosx = 1 (divide by sinx)
cosx = 1/2k
Proved!
Generous at 4 marks I thought but 5?!*?
Or you could do:
2ksinxcosx = sinx, subtract:
2ksinxcosx - sinx = 0
sinx(2kcosx - 1) = 0
so sinx = 0 and 2kcosx - 1 = 0
therfore cosx = 1/2k.
Harder but also correct
Ignore this post as people have already talked about it.
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Posted 22 May 2005 - 09:32 AM
OK, I worked out the solutions for question 4 in 2nd paper as so many people had been asking for them, and the angle came out to 50.9 
so I realised I got the wrong answer. >=( Solution is below. Vector names have to have arrows on the top but crappy keyboard lacks arrows (or huge brackets for component form)
4. a)TA=a-t
= (-5, 15, 1)
TB=b-t
=(-40, 15, 2)
b) TA.TB=ta
tb + ta 
tb +ta 
tb
=((-5)x(-40))+(15²)+(1x2)
= 427
|TA|=
251
|TB|= 1829
Cos
= (TA.TB)/(|TA|x|TB|)
= 427/( 251x 1829)
50.9
Some lines of working were missed out as I couldn't be bothered to type them.
If you notice any mistakes, please do not hesitate to let me know and I'll edit it.
so I realised I got the wrong answer. >=( Solution is below. Vector names have to have arrows on the top but crappy keyboard lacks arrows (or huge brackets for component form)4. a)TA=a-t
= (-5, 15, 1)
TB=b-t
=(-40, 15, 2)
b) TA.TB=ta
tb + ta tb +ta tb =((-5)x(-40))+(15²)+(1x2)
= 427
|TA|=
251|TB|=
1829Cos
= (TA.TB)/(|TA|x|TB|)= 427/(
251x 1829) 50.9 Some lines of working were missed out as I couldn't be bothered to type them.
If you notice any mistakes, please do not hesitate to let me know and I'll edit it.
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#211
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Posted 22 May 2005 - 10:19 AM
QUOTE(Wedge37 @ May 22 2005, 09:32 AM)
OK, I worked out the solutions for question 4 in 2nd paper as so many people had been asking for them, and the angle came out to 50.9 so I realised I got the wrong answer. >=( Solution is below. Vector names have to have arrows on the top but crappy keyboard lacks arrows (or huge brackets for component form)
4. a)TA=a-t
= (-5, 15, 1)
TB=b-t
=(-40, 15, 2)
b) TA.TB=ta tb + ta tb +ta tb
=((-5)x(-40))+(15²)+(1x2)
= 427
|TA|= 251
|TB|= 1829
Cos = (TA.TB)/(|TA|x|TB|)
= 427/( 251x 1829)
50.9
Some lines of working were missed out as I couldn't be bothered to type them.
If you notice any mistakes, please do not hesitate to let me know and I'll edit it.
so I realised I got the wrong answer. >=( Solution is below. Vector names have to have arrows on the top but crappy keyboard lacks arrows (or huge brackets for component form)4. a)TA=a-t
= (-5, 15, 1)
TB=b-t
=(-40, 15, 2)
b) TA.TB=ta
tb + ta tb +ta tb =((-5)x(-40))+(15²)+(1x2)
= 427
|TA|=
251|TB|=
1829Cos
= (TA.TB)/(|TA|x|TB|)= 427/(
251x 1829) 50.9 Some lines of working were missed out as I couldn't be bothered to type them.
If you notice any mistakes, please do not hesitate to let me know and I'll edit it.
oh crap, i dont know where i went wrong. I got 36.9 degrees but im hopeful of atleast 3 marks.
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Posted 22 May 2005 - 10:30 AM
QUOTE(tupacshakur @ May 22 2005, 10:19 AM)
QUOTE(Wedge37 @ May 22 2005, 09:32 AM)
OK, I worked out the solutions for question 4 in 2nd paper as so many people had been asking for them, and the angle came out to 50.9 so I realised I got the wrong answer. >=( Solution is below. Vector names have to have arrows on the top but crappy keyboard lacks arrows (or huge brackets for component form)
4. a)TA=a-t
= (-5, 15, 1)
TB=b-t
=(-40, 15, 2)
b) TA.TB=ta tb + ta tb +ta tb
=((-5)x(-40))+(15²)+(1x2)
= 427
|TA|= 251
|TB|= 1829
Cos = (TA.TB)/(|TA|x|TB|)
= 427/( 251x 1829)
50.9
Some lines of working were missed out as I couldn't be bothered to type them.
If you notice any mistakes, please do not hesitate to let me know and I'll edit it.
so I realised I got the wrong answer. >=( Solution is below. Vector names have to have arrows on the top but crappy keyboard lacks arrows (or huge brackets for component form)4. a)TA=a-t
= (-5, 15, 1)
TB=b-t
=(-40, 15, 2)
b) TA.TB=ta
tb + ta tb +ta tb =((-5)x(-40))+(15²)+(1x2)
= 427
|TA|=
251|TB|=
1829Cos
= (TA.TB)/(|TA|x|TB|)= 427/(
251x 1829) 50.9 Some lines of working were missed out as I couldn't be bothered to type them.
If you notice any mistakes, please do not hesitate to let me know and I'll edit it.
oh crap, i dont know where i went wrong. I got 36.9 degrees but im hopeful of atleast 3 marks.
I also followed all those steps and got something like 70 degrees. Hopefully will still get 3 marks though
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Posted 22 May 2005 - 10:39 AM
well, at least I done the method correctly but I still got the wrong answer. I am quite confident I got the coordinates as they look familiar and I must have checked it a few times. Hopefully I got 5/7 PLEAAAASSSEEEE!!! GOD
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