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!

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# Higher Maths 2002 Paper 1

Started by frazboy, Apr 04 2006 04:38 PM

10 replies to this topic

### #1

Posted 04 April 2006 - 04:38 PM

### #2

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 = (-1) +(1) 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.

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 = (-1) +(1) 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

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 ( 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-x

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

f(x) =-3x

f(x) -3x

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

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-x

^{3})^{1/2}f(x) =1/2 (8-x

^{3})^{-1/2}X -3x^{2}(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) =-3x

^{2}/2 X (8-x^{3})^{-1/2}f(x) -3x

^{2}/ 2((8-x^{3})^{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

Posted 04 April 2006 - 07:09 PM

oops. that's true

I also didn't even notice it was paper 1 darn.

I also didn't even notice it was paper 1 darn.

### #5

Posted 05 April 2006 - 03:25 PM

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

### #6

Posted 05 April 2006 - 03:31 PM

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

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### #8

Posted 05 April 2006 - 04:42 PM

Sncz answer to Q10 appears to be wrong.

The answer should be

-2/3 (8-x

Or can it be written either way?

The answer should be

-2/3 (8-x

^{3})^{ -1/2 }Or can it be written either way?

### #9

Posted 05 April 2006 - 06:08 PM

Sncz answer to Q10 appears to be wrong.

The answer should be

-2/3 (8-x

^{3})

^{ -1/2 }

Or can it be written either way?

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

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 -3/2x

which is the same as -3/2x

= -3x

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 x

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

the only difference is that i tidied it up:

by the way the answer is -3/2x

^{2}(8-x^{3})^{-1/2}which is the same as -3/2x

^{2}X 1/(8-x^{3})^{1/2}= -3x

^{2}/ 2(8-x^{3})^{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 x

^{2}and i dont understand part (b) seems a bit silly for 1 mark aswell

### #11

Posted 06 April 2006 - 08:20 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 -3/2x

^{2}(8-x

^{3})

^{-1/2}

which is the same as -3/2x

^{2}X 1/(8-x

^{3})

^{1/2}

= -3x

^{2}/ 2(8-x

^{3})

^{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 x

^{2}

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

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"

HMFC - Founded 1874, beefing the Cabbage since 1875

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