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# MATHS EXAM 2005

### #81

Posted 20 May 2005 - 03:41 PM

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

Posted 20 May 2005 - 03:42 PM

can any one help?

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.

WWE RAW IS WAR

### #83

Posted 20 May 2005 - 03:49 PM

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

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

Posted 20 May 2005 - 03:49 PM

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.

### #85

Posted 20 May 2005 - 04:16 PM

### #86

Posted 20 May 2005 - 04:19 PM

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

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

WWE RAW IS WAR

### #87

Posted 20 May 2005 - 04:21 PM

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

dave

If i am not here i am somewhere else

### #88

Posted 20 May 2005 - 04:22 PM

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

Posted 20 May 2005 - 04:23 PM

First, get the collision points K and L.

2x

^{2}-9=x

^{2}

x

^{2}-9=0

x

^{2}=9

x= +/- 3

x value of K = -3

x value of L = 3

There are the limits.

Next, integrate 2x

^{2}-9-x

^{2}.

Which is the same as integrating x

^{2}-9

After that, you have the formula:

x

^{3}/3 - 9x

Just put in your limits:

(3

^{3}/3 - 9*3) - ((-3)

^{3}/3 - 9*-3)

= (27/3 - 27) - (-27/3 + 27)

= 9 - 27 + 9 - 27

=-36, which translates to 36 units

^{2}.

<@X-Factor> Wouldnt you be breaking the first 2 rules?

### #90

Posted 20 May 2005 - 04:24 PM

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

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

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

<@X-Factor> Wouldnt you be breaking the first 2 rules?

### #91

Posted 20 May 2005 - 04:26 PM

### #92

Posted 20 May 2005 - 04:26 PM

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

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

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

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

WWE RAW IS WAR

### #94

### #95

Posted 20 May 2005 - 04:29 PM

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!

WWE RAW IS WAR

### #96

Posted 20 May 2005 - 04:31 PM

How did you do it then, out of interest?

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

Posted 20 May 2005 - 04:31 PM

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

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

Posted 20 May 2005 - 04:34 PM

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!

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

Posted 20 May 2005 - 04:35 PM

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

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?

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

Posted 20 May 2005 - 04:37 PM

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

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?

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...

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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