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Higher Maths 2002 Paper 1 - HSN forum

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Higher Maths 2002 Paper 1


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#1 frazboy

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Posted 04 April 2006 - 04:38 PM

Hi was wondering if anyone could aid me with these questions. obviously I know the answers from checking the answer sheet, but i dont know the working.

just like to say what a great site this is. Ive already printed off the revision notes. keep up the good work!

#2 Allana

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Posted 04 April 2006 - 05:47 PM

Ok question 9 is your standard wave function question.

9a) sin(x)-cos(x)

ksin(x-a)

since we have to put the first equation in that form, you should know your sin expansion :

ksinxcosa-kcosxsina

Now rearrange the above to get the x's on the outside, so it corresponds to our initial equation, and generally makes it easier to work with.

therefore,

(kcosa)sinx-(ksina)cosx Note: I have only put brackets around these parts to clearly show the sinx
and cosx

Now match up the values of kcosa and ksina which correspond to your initial equation:

kcosa = 1
ksina = -1

we know that tana=sina/cosa

therefore tana= -1/1

i'm sure you can work out the value of a from here:)


Now for k

k = sqrt.gif (-1) power2.gif +(1) power2.gif and so on.


b) your above equation will be in the form:
ksin(x-a)

Sketch the normal sin graph
K is your amplitude
Move the whole graph 'a' degrees to the right.

Voila.
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#3 SncZ

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Posted 04 April 2006 - 06:48 PM

you dont need a calculator for tan -1 / 1, as you should know tan45 = 1 , you then set up your cast diagram and identify the quadrent(s) where tanx is negative(because it's -1 and not 1 ). you then apply 45 ( pi.gif powerslash.gif 4 in this case ) to the quadrant adding/subtracting your angle from 180 depending on the quadrant chosen.

In this case :

sin is -ve
cos is +ve
Tan is -ve so you set up your cast diagram and tick the quadrants where sin is -ve, cos is +ve and where tan is -ve. If you have done this correctly you should have 6 ticks; 3 of them being in the same quadrant, and the others 3 boxes should have 1 tick each. The box with 3 ticks is the quadrant you use when finding the angle.
(its the cos box )


go on to solve the angle, then find K by the way Allana has shown.

10a)

f(x) = (8-x3)1/2

f(x) =1/2 (8-x3)-1/2 X -3x2 implies.gif (power down to the front, leave the bracket along, subtract 1 from the power and then times the whole thing by the differential of the bracket)

f(x) =-3x2 /2 X (8-x3)-1/2

f(x) -3x2 / 2((8-x3)1/2)


and im not sure about part b, i hate integration but i will try give me 10 mins or so. - even tho its only worth 1 mark

#4 Allana

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Posted 04 April 2006 - 07:09 PM

oops. that's true smile.gif

I also didn't even notice it was paper 1 tongue.gif darn.
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#5 frazboy

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Posted 05 April 2006 - 03:25 PM

Is CAST the same as the Sex And The City quadrant?

#6 The Wedge Effect

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Posted 05 April 2006 - 03:31 PM

Yeah, except it's not as cool to call it CAST. rolleyes.gif

#7 John

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Posted 05 April 2006 - 04:38 PM

I must be really uncool

I just call it All, Sin, Tan and Cos, no fancy names or ways of remembering it sad.gif

#8 frazboy

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Posted 05 April 2006 - 04:42 PM

Sncz answer to Q10 appears to be wrong.

The answer should be

-2/3 (8-x3) -1/2

Or can it be written either way?

#9 ad absurdum

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Posted 05 April 2006 - 06:08 PM

QUOTE(frazboy @ Apr 5 2006, 05:42 PM) View Post

Sncz answer to Q10 appears to be wrong.

The answer should be

-2/3 (8-x3) -1/2

Or can it be written either way?
That's the answer to part (b). His answer to part (a) isn't quite right too though, I'll just go through it again because I can't figure out what he's done wrong by looking at text versions of the workings.

f(x) = (8 - x^3)^1/2
if g(x) = 8- x^3
f(x) = g(x)^1/2
d/dx f(x) = f'(x) * d/dx g(x)
d/dx f(x) = 1/2 g(x)^-1/2 * -3x^2
d/dx f(x) = (-3x^2/2) (8 - x^3)^-1/2

so the differential of f(x) is that, f'(x) = (-3x^2/2) (8 - x^3)^-1/2

I can't seem to get the answer they have for part (b), I'll need to think about this a bit more.
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#10 SncZ

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Posted 05 April 2006 - 06:22 PM

I think your looking at the wrong answers then as i checked my answers and it has what i got ^^

the only difference is that i tidied it up:

by the way the answer is -3/2x2 (8-x3)-1/2

which is the same as -3/2x2 X 1/(8-x3)1/2

= -3x2 / 2(8-x3)1/2

and im sorry if i am not cool enough for you lot, 'cast' diagram is the most recognised in my opinion. If you had told me to set up my 'sex and the city diagram' 2 days ago, i would of slowly backed off.....

Yea i mistyped the last line, gets confusing after a while, but i have fixed it now. i just forgot an x2

and i dont understand part (b) seems a bit silly for 1 mark aswell


#11 ad absurdum

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Posted 06 April 2006 - 08:20 PM

QUOTE(SncZ @ Apr 5 2006, 07:22 PM) View Post

I think your looking at the wrong answers then as i checked my answers and it has what i got ^^

the only difference is that i tidied it up:

by the way the answer is -3/2x2 (8-x3)-1/2

which is the same as -3/2x2 X 1/(8-x3)1/2

= -3x2 / 2(8-x3)1/2

and im sorry if i am not cool enough for you lot, 'cast' diagram is the most recognised in my opinion. If you had told me to set up my 'sex and the city diagram' 2 days ago, i would of slowly backed off.....

Yea i mistyped the last line, gets confusing after a while, but i have fixed it now. i just forgot an x2

and i dont understand part (b) seems a bit silly for 1 mark aswell
Oh, my bad, I misread your capital X as a lower case x, as in the variable. Apologies.

Incidentally, I always remember it as "after sex throw condom". When I'm helping out the younger kids at our school though I always get told to teach it as "cast" sad.gif
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