# OCR (Not MEI) C2 January 2013 Discussion Watch

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#143

log

log

(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?

_{2}(x-3) - log_{2}x = 4log

_{2}(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?

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#144

(Original post by

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

**olivers16**)Can someone please explian how on Q9 i) the 4/x^2 integrated and put the 2a and a into becomes 2? Thanks.

Basically:

You need to integrate between 2a and a: (2

*x*

^{3 }-5

*x*

^{2 }- 4)

*x*

^{-2 }

Which equals 2

*x*- 5 - 4

*x*

^{-2}Integrating it makes it become [x

*-5x + 4x*

^{2 }*] between 2a and a*

^{-1}So therefore (4a

*- 10a + 4/2a) - (a*

^{2}*5a + 4/a)*

^{2 }-does that help?

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#145

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)

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)

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#147

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.

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#148

(Original post by

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.

**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.

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

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.

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#149

(Original post by

I'll try my best to help!

Basically:

You need to integrate between 2a and a: (2

Which equals 2

Integrating it makes it become [x

So therefore (4a

does that help?

**stirkee**)I'll try my best to help!

Basically:

You need to integrate between 2a and a: (2

*x*^{3 }-5*x*^{2 }- 4)*x*^{-2 }Which equals 2

*x*- 5 - 4*x*^{-2}Integrating it makes it become [x

*-5x + 4x*^{2 }*] between 2a and a*^{-1}So therefore (4a

*- 10a + 4/2a) - (a*^{2}*5a + 4/a)*^{2 }-does that help?

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#150

(Original post by

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

**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

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#152

**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

Anyone else get this?

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#153

(Original post by

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?

**cleverslacker**)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?

it said in the question that a is a

*positive*constant.

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#154

(Original post by

what did everyone get for 6iib) ?

**mossss**)what did everyone get for 6iib) ?

(Original post by

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

ar

Therefore, as 'ar

**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

ar

^{2}= 2x-7Therefore, as 'ar

^{2}'÷'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).
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#155

(Original post by

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

**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

Thanks for all your help in this thread

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

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#156

(Original post by

what did everyone get for 6iib) ?

**mossss**)what did everyone get for 6iib) ?

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

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#157

(Original post by

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

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

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#158

(Original post by

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!

**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!

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

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#159

(Original post by

Cool, thanks.

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

**cleverslacker**)Cool, thanks.

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

I know I've lost 2 marks (maybe 3 depending on how many working marks I get) so hopefully I've made few/no stupid mistakes..

From what I've seen in this thread, a lot of the answers that I got (and can remember..) are being repeated so fingers crossed!!

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#160

(Original post by

What did anyone get for the value of k?

**KRJACK**)What did anyone get for the value of k?

Sorry, im not very helpful

Posted from TSR Mobile

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