OCR (Not MEI) C2 January 2013 Discussion

It makes 80(9y^2+6y^3+9y^4)

Thats what i did

log₂(x-3) - log₂x = 4
log₂(x-3)/x = 4
(x-3)/x = 2^4
(x-3)/x = 16
(x-3) = 16x
-1/5 = x

Reckon I'd get a few method marks even though the final answer is most likely wrong?

Reply 143

Original post by olivers16

Can someone please explian how on Q9 i) the 4/x^2 integrated and put the 2a and a into becomes 2? Thanks.

I'll try my best to help!

Basically:

You need to integrate between 2a and a: (2x³-5x²- 4)x^-2
Which equals 2x - 5 - 4x^-2

Integrating it makes it become [x²-5x + 4x^-1] between 2a and a

So therefore (4a² - 10a + 4/2a) - (a²- 5a + 4/a)

does that help?

Reply 144

For 6:
just sub in values to U1 U2 and U3 to get x=15/2
put x in to find 8 works
ar/a=ar^2/ar=r
So (x+4)/2x=(2x-7)/(x+4)
goes to x^2+8x+16=4x^2-14x
3x^2-22x-16=(3x+2)(x-8)

Reply 145

9 has been answered by someone else :smile:

Reply 146

I could really do with someone going through the very last parts of 1 and 9, for both of those I used the quadratic formula and ended up with two answers.

Reply 147

Original post by cleverslacker

I could really do with someone going through the very last parts of 1 and 9, for both of those I used the quadratic formula and ended up with two answers.

a=1 is a root, so therefore f(-1) must equal zero.

Then divide the cubic by (a - 1) to get a quadratic

By using the quadratic formula you can then find the roots of the quadratic :smile:

You should get two answers from the quadratic formula for 9 ii) but one of them was negative - remember in the question it says that a is positive.

(edited 11 years ago)

Reply 148

olivers16

Original post by stirkee

Ahh, yes it does :smile:

Thank you. I found I got the first two values but not the +2

Reply 149

Original post by stirkee

a=1 is a root, so therefore f(-1) must equal zero.

Then divide the cubic by (a - 1) to get a quadratic

By using the quadratic formula you can then find the roots of the quadratic :smile:

Makes more sense to use completing the square as they want exact answer and is quicker tbh, only (1-sqrt7)/3=a as it said a was a positive constant

Reply 150

mossss

what did everyone get for 6iib) ?

Reply 151

Original post by stirkee

a=1 is a root, so therefore f(-1) must equal zero.

Then divide the cubic by (a - 1) to get a quadratic

By using the quadratic formula you can then find the roots of the quadratic :smile:

thanks but it was when trying to find the roots of the quadratic that I had the problem. it says "other possible value" not "values", but I got two: (1/3)+(1/3)sqrt(7), (1/3)-(1/3)sqrt(7)
Anyone else get this?

Reply 152

Original post by cleverslacker

See my edit :smile:

it said in the question that a is a positive constant.

Reply 153

Selvey

Original post by mossss

what did everyone get for 6iib) ?

My earlier post:

Original post by Selvey

Not sure if the below 'x' values are right (I can't remember them exactly), but this is the method I used.

a=2x
ar = x+4
ar²= 2x-7

Therefore, as 'ar²'÷'ar' = r, that can be set equal to 'ar'÷'a'.Therefore (2x-7)÷(x+4)=(x+4)÷(2x). From there I cross multiplied and solved; and as one of the solutions was 8, I knew the other one was correct.(If I remember correctly x was -2/3).

Reply 154

Original post by Lstigant

Makes more sense to use completing the square as they want exact answer and is quicker tbh, only (1-sqrt7)/3=a as it said a was a positive constant

Oh bugger I didn't realise it said "positive" constant. Dya guys think I'll get like 5/6 marks?
Thanks for all your help in this thread :smile:

btw. I think you mean a=(1+sqrt7)/3, that would be the positive one

(edited 11 years ago)

Reply 155

Original post by mossss

what did everyone get for 6iib) ?

ar/a=ar^2/ar=r
So (x+4)/2x=(2x-7)/(x+4)
goes to x^2+8x+16=4x^2-14x
3x^2-22x-16=(3x+2)(x-8)
x=8 or x=-2/3

Reply 156

Original post by cleverslacker

Oh bugger I didn't realise it said "positive" constant. Dya guys think I'll get like 5/6 marks?

I'd imagine that putting both roots from the quadratic would only lose you 1 mark yeah, with the rest of the working being correct!

Reply 157

Original post by stirkee

See my edit :smile:

it said in the question that a is a positive constant.

Original post by stirkee

I'd imagine that putting both roots from the quadratic would only lose you 1 mark yeah, with the rest of the working being correct!

Cool, thanks.
Not looking tooo bad for me, will have to check Mr. M's for silly mistakes though.

Reply 158