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MATHS EXAM 2005 - HSN forum - Page 5

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MATHS EXAM 2005


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#81 zaim187

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Posted 20 May 2005 - 03:41 PM

8c= I didnt diffrentiate just got the values for F by substiuting X values in. think i'll get any marks?

9= cosx=4/5

10B didny dow any thing to the graph but move it to the right by pie/3 . cood i mayb get a mark.

11B no idea just subsitutues the 2x and applied the discriminant

can any one help?

#82 tupacshakur

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Posted 20 May 2005 - 03:42 PM

QUOTE(zaim187 @ May 20 2005, 03:41 PM)
8c= I didnt diffrentiate just got the values for F by substiuting X values in. think i'll get any marks?


can any one help?

View Post



thats what i did. I was thinking of differentiation but i didnt recognise the question before. I think it was wrong though - apprently u HAD to differentiate it then get max and min turning point. I f*** that up.
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#83 david__taylor

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Posted 20 May 2005 - 03:49 PM

QUOTE(e-unit @ May 20 2005, 03:35 PM)
QUOTE(david__taylor @ May 20 2005, 03:56 PM)
Q2(b)(i). cos(p+q) = -13/85.
Q2(b)(ii). tan(p+q) = -85/13 = -6.46

View Post


Where do these negatives come from?

I drew out a triangle with (p+q) as the angle and 84 as the opposite and 85 as the hypotenuse

View Post




The negatives:... hmm... for (a) sin(p+q) = 84/85 so in (b)...

cos(p+q)
= cos(p)cos(q) - sin(p)sin(q)
= (8/17) * (8/10) - (15/17) * (8/10)
= (64/170) - (120/170)
= (-56/170)
That's how I got it to a minus and the tan(p+q) is a minus because you divide sin(p+q) by cos(p+q) (which is negative from my previous answer) and so I got a negative. Did I go wrong somewhere?



#84 Scott_Cameron

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Posted 20 May 2005 - 03:49 PM

QUOTE(Raeg @ May 20 2005, 03:39 PM)
Can anyone go over how they got the final answer in Q5, I keep trying it and am still ending up with 18 and I know I'm just forgetting something.


That would be most helpful for myself aswell. ^^ For some reason, I keep ending up with 12? Maybe I'm forgetting to multiply by 3 somewhere. wink.gif

#85 gary

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Posted 20 May 2005 - 04:16 PM

How many marks would you get for a vector question but getting the wrong angle.

#86 tupacshakur

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Posted 20 May 2005 - 04:19 PM

QUOTE(gary @ May 20 2005, 04:16 PM)
How many marks would you get for a vector question but getting the wrong angle.

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yeah can someone advise us because i got 36.9 for the angle and i think its wrong sad.gif

gary, atleast we have done a lot of the working so i wont be surprised if we get atleast 3 marks for it.
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#87 Dave

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Posted 20 May 2005 - 04:21 PM

i went into school today just as the higher lot were going in and i did a few of the question in each paper

I think the papers were very fair and the paper 2 was def more of the more simple ones

Hope you all do well then

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#88 gary

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Posted 20 May 2005 - 04:22 PM

Yeah I was thinking that

There will be one mark for using the scalar product
one for being able to carry through a-b maybe one for knowing how to re-arrange the scalar product.

I still dont know where I went wrong on that question.

#89 coca

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Posted 20 May 2005 - 04:23 PM

Q5:

First, get the collision points K and L.

2x2-9=x2
x2-9=0
x2=9
x= +/- 3

x value of K = -3
x value of L = 3

There are the limits.

Next, integrate 2x2-9-x2.

Which is the same as integrating x2-9

After that, you have the formula:

x3/3 - 9x

Just put in your limits:

(33/3 - 9*3) - ((-3)3/3 - 9*-3)
= (27/3 - 27) - (-27/3 + 27)
= 9 - 27 + 9 - 27
=-36, which translates to 36 units2.
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#90 coca

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Posted 20 May 2005 - 04:24 PM

QUOTE(tupacshakur @ May 20 2005, 05:19 PM)
QUOTE(gary @ May 20 2005, 04:16 PM)
How many marks would you get for a vector question but getting the wrong angle.

View Post



yeah can someone advise us because i got 36.9 for the angle and i think its wrong sad.gif

gary, atleast we have done a lot of the working so i wont be surprised if we get atleast 3 marks for it.

View Post



I messed it up too. It was a real gift of a question, but I fumbled it. Damn 9 A.M. logic!
<MrBob> I hate Uni. At least in film studies we get to talk about Fight Club.
<@X-Factor> Wouldnt you be breaking the first 2 rules?


#91 gary

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Posted 20 May 2005 - 04:26 PM

I didnt do the 8 mark integration that way but I got 36, weird.

#92 tupacshakur

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Posted 20 May 2005 - 04:26 PM

QUOTE(pseudotoxic @ May 20 2005, 04:24 PM)
QUOTE(tupacshakur @ May 20 2005, 05:19 PM)
QUOTE(gary @ May 20 2005, 04:16 PM)
How many marks would you get for a vector question but getting the wrong angle.

View Post



yeah can someone advise us because i got 36.9 for the angle and i think its wrong sad.gif

gary, atleast we have done a lot of the working so i wont be surprised if we get atleast 3 marks for it.

View Post



I messed it up too. It was a real gift of a question, but I fumbled it. Damn 9 A.M. logic!

View Post



u'll prob get atleast 2-3 marks worth of working though.
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#93 coca

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Posted 20 May 2005 - 04:27 PM

QUOTE(tupacshakur @ May 20 2005, 05:26 PM)
u'll prob get atleast 2-3 marks worth of working though.

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Hope so smile.gif
<MrBob> I hate Uni. At least in film studies we get to talk about Fight Club.
<@X-Factor> Wouldnt you be breaking the first 2 rules?


#94 coca

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Posted 20 May 2005 - 04:29 PM

QUOTE(gary @ May 20 2005, 05:26 PM)
I didnt do the 8 mark integration that way but I got 36, weird.

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How did you do it then, out of interest?

<MrBob> I hate Uni. At least in film studies we get to talk about Fight Club.
<@X-Factor> Wouldnt you be breaking the first 2 rules?


#95 tupacshakur

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Posted 20 May 2005 - 04:29 PM

so how many times have u added up the marks trying to estimate what mark u might get.

i have done it 10 times since i came back from the exam and i have come to the conclusion that i think around the early 80s is my best bet

hopefully thats a B!
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#96 gary

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Posted 20 May 2005 - 04:31 PM

QUOTE(pseudotoxic @ May 20 2005, 04:29 PM)
QUOTE(gary @ May 20 2005, 05:26 PM)
I didnt do the 8 mark integration that way but I got 36, weird.

View Post



How did you do it then, out of interest?

View Post



I am crapping myself now.

I split the graph into two because it was symetrical then i did:

area between 0,-3
area between 3,0 then the total area to get 36.

Is this wrong.


#97 david__taylor

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Posted 20 May 2005 - 04:31 PM

or you could do it so you get the positive answer by doing:

integral.gif between 3 and -3 of; (x[^2])- (2x^2 - 9) dx

integral.gif between 3 and -3 of; (-x[^2] + 9 ) dx

[ (-x[^3] / 3) +9x ] between 3 and -3.

which is [(-27/3) + 27] - [(27/3) - 27].

= (-9 + 27) - (-18)
= 18+18
=36 units[^2]

#98 gary

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Posted 20 May 2005 - 04:34 PM

QUOTE(tupacshakur @ May 20 2005, 04:29 PM)
so how many times have u added up the marks trying to estimate what mark u might get.

i have done it 10 times since i came back from the exam and i have come to the conclusion that i think around the early 80s is my best bet

hopefully thats a B!

View Post



I have worked out I have got 60 markjs from questions I got all the marks.

I should get another 12 for working here and there.

#99 e-unit

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Posted 20 May 2005 - 04:35 PM

QUOTE(david__taylor @ May 20 2005, 04:49 PM)
QUOTE(e-unit @ May 20 2005, 03:35 PM)
QUOTE(david__taylor @ May 20 2005, 03:56 PM)
Q2(b)(i). cos(p+q) = -13/85.
Q2(b)(ii). tan(p+q) = -85/13 = -6.46

View Post


Where do these negatives come from?

I drew out a triangle with (p+q) as the angle and 84 as the opposite and 85 as the hypotenuse

View Post




The negatives:... hmm... for (a) sin(p+q) = 84/85 so in (b)...

cos(p+q)
= cos(p)cos(q) - sin(p)sin(q)
= (8/17) * (8/10) - (15/17) * (8/10)
= (64/170) - (120/170)
= (-56/170)
That's how I got it to a minus and the tan(p+q) is a minus because you divide sin(p+q) by cos(p+q) (which is negative from my previous answer) and so I got a negative. Did I go wrong somewhere?

View Post



I didn't think you were supposed to use the same method as in part (a) cos thats worth 4 marks and parts b1 and b2 are worth 3 marks combined.
If sin(p+q)=84/85 then the opposite =84 and the hyp=85. u then work out the adjacent by pythagoras (=13).
cos(p+q)=adj/hyp=13/85
tan(p+q)=opp/adj=84/13
I still cant see how urs is wrong either. Hmm...
You've said sinq=8/10 when it should be 6/10
That gives you the negative of my answer Hmmm....

#100 gary

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Posted 20 May 2005 - 04:37 PM

QUOTE(e-unit @ May 20 2005, 04:35 PM)
QUOTE(david__taylor @ May 20 2005, 04:49 PM)
QUOTE(e-unit @ May 20 2005, 03:35 PM)
QUOTE(david__taylor @ May 20 2005, 03:56 PM)
Q2(b)(i). cos(p+q) = -13/85.
Q2(b)(ii). tan(p+q) = -85/13 = -6.46

View Post


Where do these negatives come from?

I drew out a triangle with (p+q) as the angle and 84 as the opposite and 85 as the hypotenuse

View Post




The negatives:... hmm... for (a) sin(p+q) = 84/85 so in (b)...

cos(p+q)
= cos(p)cos(q) - sin(p)sin(q)
= (8/17) * (8/10) - (15/17) * (8/10)
= (64/170) - (120/170)
= (-56/170)
That's how I got it to a minus and the tan(p+q) is a minus because you divide sin(p+q) by cos(p+q) (which is negative from my previous answer) and so I got a negative. Did I go wrong somewhere?

View Post



I didn't think you were supposed to use the same method as in part (a) cos thats worth 4 marks and parts b1 and b2 are worth 3 marks combined.
If sin(p+q)=84/85 then the opposite =84 and the hyp=85. u then work out the adjacent by pythagoras (=13).
cos(p+q)=adj/hyp=13/85
tan(p+q)=opp/adj=84/13
I still cant see how urs is wrong either. Hmm...

View Post



I think the method with the negatives is correct, I can't see them giving 3 marks for just using SOHCAHTOA.






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