(Please refer to the book "Advanced Problems in Core Mathematics" by Stephen Siklos for a better understanding of the question)
Find all the solutions of the equation
Let f(x) = x + 1 − x + 3x − 1 − 2x − 2 − (x + 2).
Solution:
In the separate regions, we have f(x) =
I don't understand the rationale behind this approach. If we notice the signs of the respective modulus terms together with the respective limits given, the "objective" of this method would be as such:
If we notice the sign change in the above equations, the bottom 4 equations will always give the same sign as the top row (with the exception of the final column, as it is not a modulus function). Notice the first term of each equation. The "negative" modulus is only introduced in the fifth row only to produce a positive term, but not introduced in the rest of the equations to bring up the possibility of having the first term as negative.
(Question 1) What is the purpose of keeping the terms in the same sign as that in the first row (i.e. the terms when the limits are )
(Question 2) What happens if the terms are not kept in the same sign as the first row?
(Question 3) Is the role of the nonmodulus function (i.e. the last term) significant? Will its sign change bring about a different solution for the question apart from the ones above?
Could someone let me know how you analyse the question and how would you justify the validity of the method proposed by Siklos? Why is that the 5 possibilities listed above is sufficient without listing out the 64 (am I correct?) possible combinations?
Any help is greatly appreciated. thank you!
Modulus Question  STEP Siklos Booklet


The objective of the approach is to remove the modulus signs and reduce the problem to simple linear equations.
There are relatively few cases to consider since, for example, if x<0 then clearly x1 and x2 are also both less than zero.
They are not independent of each other. 
(Original post by BabyMaths)
The objective of the approach is to remove the modulus signs and reduce the problem to simple linear equations.
There are relatively few cases to consider since, for example, if x<0 then clearly x1 and x2 are also both less than zero.
They are not independent of each other. 
Here is a simpler example.
x+1 = x2
Case 1
If x<1 then x+1<0 and x2 <0 and so x+1=x1 and x2=2x.
The equation becomes x1=2x.
Case 2
If 1 < x < 2 then x+1>0 and x2<0 and so x+1 = x+1 and x2=2x.
The equation becomes x+1=2x and so x=1/2.
Case 3
If x>2 then x+1>0 and x2>0 and x+1=x+1 and x2=x2.
The equation becomes x+1=x2.
Of course you would not use this method in this case. 
(Original post by BabyMaths)
Of course you would not use this method in this case.
I think I'm not getting the concept of modulus right. I shall google more on that. Thank you.
Any other advice would be greatly appreciated! 
(Original post by BabyMaths)
Here is a simpler example.
x+1 = x2
Case 1
If x<1 then x+1<0 and x2 <0 and so x+1=x1 and x2=2x.
The equation becomes x1=2x.
Case 2
If 1 < x < 2 then x+1>0 and x2<0 and so x+1 = x+1 and x2=2x.
The equation becomes x+1=2x and so x=1/2.
Case 3
If x>2 then x+1>0 and x2>0 and x+1=x+1 and x2=x2.
The equation becomes x+1=x2.
Of course you would not use this method in this case.
So how do I justify cases 1 and 3 in your method? "no 'x' exists, and so... what?"
So from my understanding, we only employ the negative sign by the modulus when the expression inside the modulus gives a negative term (in order to make the term positive, otherwise it wouldn't satisfy the conditions of a modulus function). Am I right to say so? 
(Original post by johnconnor92)
So how do I justify cases 1 and 3 in your method? "no 'x' exists, and so... what?"
(Original post by johnconnor92)
So from my understanding, we only employ the negative sign by the modulus when the expression inside the modulus gives a negative term (in order to make the term positive, otherwise it wouldn't satisfy the conditions of a modulus function). Am I right to say so?
x=x if .
x=x if x<0.
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: