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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: vailkar on November 04, 2011, 08:22:31 pm

Title: Removal of Absolute Value Signs
Post by: vailkar on November 04, 2011, 08:22:31 pm: Greetings all,

I have a query in regards to when integrating fractions which result in the form of $\log|x|$ or similar. Whilst I understand that usually, initial conditions will be given in order to resolve these, I have noticed, particularly in many trial exam solutions, that the suggested solutions will simply integrate and use brackets instead of absolute value signs, showing no indication of why they were eradicated.

I was pondering what the best way of removing these absolute value signs are.

Thank you.
Title: Re: Removal of Absolute Value Signs
Post by: dc302 on November 04, 2011, 08:31:56 pm: Could you post a specific question/example please?
Title: Re: Removal of Absolute Value Signs
Post by: luken93 on November 04, 2011, 08:36:23 pm: If a domain isn't specified, then you keep the signs.

However, if you have a salt solution question such as loge |80 - x|, since you know that x>0 because salt can't be negative weight, then you can remove the modulus...
Title: Re: Removal of Absolute Value Signs
Post by: dc302 on November 04, 2011, 08:38:45 pm: Quote from: luken93 on November 04, 2011, 08:36:23 pm
If a domain isn't specified, then you keep the signs.

However, if you have a salt solution question such as loge |80 - x|, since you know that x>0 because salt can't be negative weight, then you can remove the modulus...

Even if you know x>0, you can't remove the mod, unless you know that x<80. Is that what you meant?
Title: Re: Removal of Absolute Value Signs
Post by: dc302 on November 04, 2011, 08:56:02 pm: Quote from: vailkar on November 04, 2011, 08:42:07 pm
In Neap 2011 Examination 2, Question 5e, it essentially asks for the integral of $-1 + \frac{5}{10-u} + \frac{5}{10+u}$ . In the solutions it gives the integral as equaling $-u - 5\log{(10-u)}+5\log{(10+u)}$ stating no reason for why the expressions inside the logs have the absolute value signs removed.

In the TSSM 2011 Examination 1, Question 10a,
Quote
A particle moving in a straight line slows down under the influence of an acceleration a ms−2
such that a = −kv , where v ms−1 is the velocity of the particle at any instant. The initial velocity is
u. After 2 seconds the speed of the particle has decreased by a factor of e , where e is Euler’s
number. Find the constant k and the expression for the velocity v at any time t.

The solution involves once again involves the emission of the absolute value signs with no explanation.

The first example: Seems to me like they were just lazy--I believe you should have the mod signs (given that the domain of u is supposed to be all of R (reals)).

2nd example: This is because v is always positive anyway.
Title: Re: Removal of Absolute Value Signs
Post by: dc302 on November 04, 2011, 10:18:19 pm: I don't really know what VCAA standards/expectations are so you're probably better off asking your teacher or waiting for someone else to answer. For the 2nd example I would probably just state that v is always positive by assumption (you could work with v negative, but then that just means the particle has been moving backwards, which wouldn't really change anything so it would be useless to do that).
Title: Re: Removal of Absolute Value Signs
Post by: abeybaby on November 04, 2011, 10:42:08 pm: This is one way that you can do it. If it isnt easy to prove that whatever is in your modulus is ALWAYS positive or ALWAYS negative, then you can use this approach :)