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Old 02-02-12, 12:11   #1
moodydog
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integrate ((4x-1) / (2x))

Hmm... the boundaries are between 8 and 2

If we split this up into (2x - 1/2x) and differentiate I get 2x - 1/2 ln(2x).

8
[2x -1/2 ln(2x)]
2

[16 - 1/2 ln(16)] - [4 - 1/2 ln(4)]

16 - 1/2 ln(16) - 4 + 1/2 ln(4)

12 + 1/2 ln(1/4)

The answer book gives 12 ln2. My answer is the negative of this. Also wolfram gives:

[IMG]http://i40.************/2cervk8.gif[/IMG]

What has it done?


_____________________________________

sorry new question, the book I am using is actually swimming with errors int he answers.

If you have sin2x it equals out to 2sinxcosx

But if you have sin(x/2) it equals out to 2sin(x/4)cos(x/4)??? Why?

Thinking about it

Sin(x+x) = sinxcosx + sinxcosx hence 2sinxcosx
but then

Sin(x/2+x/2) should I imagine = 2sin(x/2)cos(x/2) and not 2sin(x/4)cos(x/4). Why is the two doubled? Same with anying else, sinx/6 = 2sin(x/12)cos(x/12)
Why is it doubled?


Secondly

Solving tan2x + tanx =0

I get X as 71.56, 251.56, 0 and 180

the way I did that:

2tanx/(1-tan^x) + tanx = 0

2tanx + tanx(1-tan^2x) = 0

2tanx + tanx - tan^3x = 0

tan^3x - 3tanx = 0

tan^2x - 3 = 0

tanx(tanx-3) = 0

tanx = 0 or tanx = 3

x = 0 or x= 71.56

Between 0 and 360

I get X as 71.56, 251.56, 0 and 180

I got two of those values right, but apparently tanx does not equal three? What is wrong with my method? Or like I said, the book is filled with errors.


Hey again, further down the page, there was this.

It uses the same principles as before, I have tried changing the cosec into (1 + cot) with ^2 theta, and tan into sec -1 and sin into 1 - cos, and then expanded everything out, but just came into a big mess.
and also tried the right hand side, and couldn't get further then sec -1...

Sorry this is a newer question I am stuck on;


The question wants me to rearrange the left hand side so it looks identical to the right, using trig identities.

hey, I am stuck on this question;

Solve to find x

(x-2)^2 = (root x) +2

I know one of the answers is 4, but I don't know how to get it.

Last edited by moodydog; 09-03-12 at 14:38.
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Old 02-02-12, 12:18   #2
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So you know that the answer is 4, but you don't know why? Is that what you're saying?
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Old 02-02-12, 12:51   #3
moodydog
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Quote:
Originally Posted by Mystery-King View Post
So you know that the answer is 4, but you don't know why? Is that what you're saying?
Exactly.

I can expand the (x-2)^2 to get:

x^2 - 4x + 4 = (root x) + 2

But to solve that to find x on its on is really hard. I have no idea what I am supposed to do with the (root x)?

I looked at the answers, and it gives it as 4. However, there is no worked solutions in the answer.

I also know there must be two solutions, as it is x^2, however, the question states that x>2. So the other answer must be <2.

Can anyone please help?

Last edited by moodydog; 02-02-12 at 12:52.
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Old 02-02-12, 15:57   #4
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I think this is it. I did it in as many steps as possible:

(x-2)^2 = (root x) + 2
(x-2)(x-2) = (root x) + 2
x^2 - 2x - 2x + 4 = (root x) + 2
x^2 - 4x + 4 = (root x) + 2
x^2 - 4x + 4 - 2 = (root x)
x^2 - 4x + 2 = (root x)
(x^2 - 4x + 2)^2 = x
x^4 - 8x^3 - 20x^2 - 16x + 4 = x

And follow Cochrane's steps below...

Last edited by jonathanrij; 02-02-12 at 17:29. Reason: Made stupid mistakes and fixed them.
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Old 02-02-12, 16:09   #5
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Are you supposed to know that one of the answers is four? If yes, then take jonathanrij (which looks correct, but better check first), bring it in … = 0 form, and divide the polynomial by (x-4). This gives you a standard quadratic formula of the x^2 + px + q = 0 form, and solving that is trivial.
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Old 02-02-12, 17:08   #6
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I'm so sorry, my answer was insufficient . For some reason I stopped and didn't think it through. So it's only part of the whole answer. I wanted to edit it after I noticed it, but I had no computer near me and my cell phone wouldn't let me log in.

So go ahead with what Cochrane said .
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Old 02-02-12, 17:09   #7
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Quote:
Originally Posted by jonathanrij View Post
I think this is it. I did it in as many steps as possible:

(x-2)^2 = (root x) + 2
(x-2)(x-2) = (root x) + 2
x^2 - 2x - 2x + 4 = (root x) + 2
x^2 - 4x + 4 = (root x) + 2
x^2 - 4x + 4 - 2 = (root x)
x^2 - 4x + 2 = (root x)
(x^2)^2 - (4x)^2 + 2^2 = x
x^4 - 16x^2 + 4 = x
Shouldn't this be (x^2 - 4x + 2)^2 = x ? Which would be x^4 - 8x^3 - 20x^2 - 16x + 4 = x ? (I think )
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Old 02-02-12, 17:25   #8
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^ Damn, you're right. I guess I just wasn't thinking clearly. In my defence, I was in a hurry... xD
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Old 02-02-12, 18:30   #9
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OK, this solution sets out to find all solutions of x using the rational roots theorem...hopefully you should be able to work out how to work out the specific answer to your question.





Sorry about the messy handwriting, it's really hard to write equations on plain paper.
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Old 02-02-12, 22:26   #10
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Thank you all for your help.

@ tomekkobialka - Thank you for taking time to solve it, however, being only an A-level question, Rational Roots Theorem is not learnt, or expected to be used. And also, I don't think the question is looking for that much working out from the nature of the question.

@Cochrane - No the question does not give the answer, I just looked at the mark scheme. But thanks anyway.

I should explain the question a bit better.

f(x) = (X-2)^2

1) Find f-1(x):

f-1(x) = (root) x + 2

2) Solve f(x) = f-1(x)

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