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Old 15-06-2008: 15th June 2008 18:50 #1 
01perryd 01perryd is offline Male
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Default FP2 june 05 q8b
 
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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Old 15-06-2008: 15th June 2008 19:02 #2 
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Default Re: FP2 june 05 q8b
 
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}
 
Old 15-06-2008: 15th June 2008 19:02 #3 
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Default Re: FP2 june 05 q8b
 
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.
 
Old 15-06-2008: 15th June 2008 19:05 #4 
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Default Re: FP2 june 05 q8b
 
Originally Posted 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 =[
 
Old 15-06-2008: 15th June 2008 19:05 #5 
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Default Re: FP2 june 05 q8b
 
Originally Posted 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
Old 15-06-2008: 15th June 2008 20:16 #6 
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Default Re: FP2 june 05 q8b
 
lol ive just done this paper! i was going to ask this question exactly! thanks!! x
Old 15-06-2008: 15th June 2008 20:20 #7 
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Default Re: FP2 june 05 q8b
 
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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