Removal of Absolute Value Signs

[vailkar]

Greetings all,

I have a query in regards to when integrating fractions which result in the form of 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.

[dc302]

Could you post a specific question/example please?

[luken93]

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...

[dc302]

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?

[dc302]

In Neap 2011 Examination 2, Question 5e, it essentially asks for the integral of . In the solutions it gives the integral as equaling stating no reason for why the expressions inside the logs have the absolute value signs removed.



In the TSSM 2011 Examination 1, Question 10a,

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.

[dc302]

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).

[abeybaby]

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