Original post by leoishush

I think I lost 2 marks even though I was so close to the answer. I just multiplied my answer by -1 at the end but Im pretty sure thats incorrect maths

Your answer is correct and is the same as the mark scheme.

(edited 1 year ago)

Original post by leoishush

You've multiplied the top and the bottom by "-1", which is the same thing as multplying the whole thing by "1". it's fine.

ahhh thats great thanks @ghostwalker @Notnek

Original post by leoishush

. I canâ€™t seem to find where I went wrong with this Intergration from the same paper as well

From your working

du = 1/(2u) dx

I.e. dx = 2u du.

Your replacing the dx in the integral, so that will become the 2u du, not 1/(2u) du.

Original post by ghostwalker

From your working

du = 1/(2u) dx

I.e. dx = 2u du.

Your replacing the dx in the integral, so that will become the 2u du, not 1/(2u) du.

du = 1/(2u) dx

I.e. dx = 2u du.

Your replacing the dx in the integral, so that will become the 2u du, not 1/(2u) du.

ugh I would of gotten the answer right. Thank you very much.

Original post by leoishush

The way to get the answer in the form that the question wants is to take the 2x to the RHS and take the 2x(dy/dx) to the LHS. Then do the same steps as u did and you will get (y-x)/(3y-x)

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can someone please explain what principle domain is and why the answer is a not c?Maths

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