01perryd
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#1
Report Thread starter 11 years ago
#1
Here is the ms

http://www.freeexampapers.com/A%20Le...%2005%20MS.pdf

How are the last 2 statements the same. Surely you cant take the +/- outside the logs
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insparato
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#2
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You proved it in the first part.

 ln \frac{1 - \sqrt{1-x^2}}{x} = -ln \frac{1+\sqrt{1-x^2}}{x}

So

 ln \frac{1 \pm \sqrt{1-x^2}}{x}

Becomes either  ln \frac{1+\sqrt{1-x^2}}{x} or  -ln \frac{1+\sqrt{1-x^2}}{x}
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.A.
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Although I was very busy with FP1. I thought I will help you.

So you have two of 1 +root and 1-root.

Write ln (1 - root ) and ofcourse separate the x present. so you two ln equation. ln (1-root) -ln(x)

Lets deal with ln(1-root)

write -ln ( 1 / 1-root) if you know what I mean

Then multiply both top and bottom with 1 + root. And you will be fine.

back to FP1.

.A.
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Chaoslord
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#4
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(Original post by 01perryd)
Here is the ms

http://www.freeexampapers.com/A%20Le...%2005%20MS.pdf

How are the last 2 statements the same. Surely you cant take the +/- outside the logs
i am too slow =[
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01perryd
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#5
Report Thread starter 11 years ago
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(Original post by insparato)
You proved it in the first part.

 ln \frac{1 - \sqrt{1-x^2}}{x} = -ln \frac{1+\sqrt{1-x^2}}{x}

So

 ln \frac{1 \pm \sqrt{1-x^2}}{x}

Becomes either  ln \frac{1+\sqrt{1-x^2}}{x} or  -ln \frac{1+\sqrt{1-x^2}}{x}
Thanks a lot. I see it now. You didn't even need to do it to get the mark i was just curious
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invicta
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#6
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lol ive just done this paper! i was going to ask this question exactly! thanks!! x
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insparato
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#7
Report 11 years ago
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No probs. Usually if they ask things like that in the first part, they want you to use it somewhere later on.
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