## 2004 - Paper 2 - Question 1 and 5

### JiLL

Posted 14 April 2005 - 05:13 PM

See for question 1 a) am i right in thinking that you re-arrange the equation as y=mx+c? but if i do this then it all goes a bit wonky! So how do u find the value of 'a'??

and for qustion b) would you just use M=tan
tan 30
and so on.........??
Also how on earth do u do question 5 in this paper please??!?

Help is much appreciated !!

Ta !

### Shaun

Posted 14 April 2005 - 05:29 PM

Ok, for 1a,

You have the equation x-2y = 0.

Rewrite this as y = -x/-2

Which is y = x / 2

Pick a value - i.e. when x = 1, and this gives you y is a half. This allows you to draw a triangle and use Tan a = opposite/adjacent.

For b) all you do is add 30 degrees to what you worked out as 'a' and take the tan of it.

For 5, you take the derivative and let it equal 12.

So you have 12x - 3x2 = 12

Bring everything to one side and you get:

-3x2 + 12x - 12 = 0

Multiply by minus 1 to get rid of the -3x2

You get:

3x2 - 12x + 12 = 0

Take out common factor of 3.

3(x2 - 4x + 4) = 0

Factorise:

3(x - 2)(x - 2)

Value of x = 2

### Shaun

Posted 14 April 2005 - 05:35 PM

To do part b. You know x = 2. So stick this in your original equation (6x2 - x3 and you get 16.

Coordinates are (2, 16). You know gradient is 12.

Your y - b = m(x - a)

a = 2, b = 16 and m = 12.

You end up with equation: y = 12x - 8.

### JiLL

Posted 14 April 2005 - 05:35 PM

Sorry, but for 1a)

i dont really understand what the opposite and adjacent values would be in this case???

arrh!

Ta 4 question 5, i now understand it!

### Shaun

Posted 14 April 2005 - 05:46 PM

Ok, I'll try to clarify. You have an equation of a line. You can pick any x coordinate and and stick it in your equation and get an answer for y.

For example, say you choose x = 1. You stick it in and you get y = 0.5

This tells you that you have a line from the origin, 1 along and 0.5 up.

It's a triangle.

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### JiLL

Posted 14 April 2005 - 05:54 PM

woo hoo i sooooooooo get it now!

I'm kicking myself for how silly i'm beng!!

Sorry for any inconveinence that this may have caused

### Shaun

Posted 14 April 2005 - 06:12 PM

Don't be daft you never caused any inconvienance, I'm here to help - and even want to be a maths teacher .

### JiLL

Posted 14 April 2005 - 06:55 PM

o thats good!

Well rather u than me!!

lol

### werlop

Posted 14 April 2005 - 08:28 PM

If you need any more help with the 2004 paper I put some answers together last year before the official marking schemes came out.