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Modulus Question - STEP Siklos Booklet

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    (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|x+1| - |x| +3|x-1| -2|x-2| = x +2
    Let f(x) = |x + 1| − |x| + 3|x − 1| − 2|x − 2| − (x + 2).

    Solution:
    In the separate regions, we have f(x) =

    (x + 1) - x + 3(x - 1) - 2(x - 2) - (x + 2) = 0 \mathrm{for} 2 \leq x < \infty



(x + 1) - x + 3(x - 1) + 2(x - 2) - (x + 2) = 4x - 8 for 1\leq x \leq 2



(x + 1) - x - 3(x - 1) + 2(x - 2) - (x + 2) = -2x - 2 for 0 \leq x \leq 1



(x + 1) + x - 3(x - 1) + 2(x - 2) - (x + 2) = -2 for -1 \leq x \leq 0



-(x + 1) + x - 3(x - 1) + 2(x - 2) - (x + 2) = -2x - 4 for -\infty < x \leq -1

    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:

     (+ve) - (+ve) + (+ve) - (0 / +ve) - (+ve) for 2 \leq x < \infty

(+ve) - (+ve) + (+ve) + (0 / -ve) - (+ve) for 1\leq x \leq 2

(+ve) - (+ve) - (0 / -ve) + (-ve) - (+ve) for 0\leq x \leq 1

(0 / +ve) + (0 / -ve) - (-ve) + (-ve) - (+ve) for -1 \leq x \leq 0

- (0 / -ve) + (-ve) - (-ve) + (-ve) - (-ve / +ve) for -\infty < x \leq -1

    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 2 \leq x < \infty)

    (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 non-modulus 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!
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    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 x-1 and x-2 are also both less than zero.

    They are not independent of each other.
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    (Original post by BabyMaths)
    The objective of the approach is to remove the modulus signs and reduce the problem to simple linear equations.
    Yes the purpose is to remove modulus signs, but why is that the "negative sign" is introduced ONLY in the case when the first term of negative value (thereby producing a positive term)? Why are we not introducing the negative sign at cases when the first term gives a positive value (thereby producing a negative term)? Similarly, there will be many more possibilities if we take into consideration the second, third and fourth terms too. Why aren't they taken into consideration?

    There are relatively few cases to consider since, for example, if x<0 then clearly x-1 and x-2 are also both less than zero.

    They are not independent of each other.
    Could you please elaborate further? Thank you.
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    Here is a simpler example.

    |x+1| = |x-2|

    Case 1

    If x<-1 then x+1<0 and x-2 <0 and so |x+1|=-x-1 and |x-2|=2-x.

    The equation becomes -x-1=2-x.

    Case 2

    If -1 < x < 2 then x+1>0 and x-2<0 and so |x+1| = x+1 and |x-2|=2-x.

    The equation becomes x+1=2-x and so x=1/2.

    Case 3

    If x>2 then x+1>0 and x-2>0 and |x+1|=x+1 and |x-2|=x-2.

    The equation becomes x+1=x-2.

    Of course you would not use this method in this case.
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    (Original post by BabyMaths)
    Of course you would not use this method in this case.
    What you meant was: if I had the determination to use this method in this case, the question would still be solved, albeit the working will be far longer. Yes?

    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!
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    (Original post by BabyMaths)
    Here is a simpler example.

    |x+1| = |x-2|

    Case 1

    If x<-1 then x+1<0 and x-2 <0 and so |x+1|=-x-1 and |x-2|=2-x.

    The equation becomes -x-1=2-x.

    Case 2

    If -1 < x < 2 then x+1>0 and x-2<0 and so |x+1| = x+1 and |x-2|=2-x.

    The equation becomes x+1=2-x and so x=1/2.

    Case 3

    If x>2 then x+1>0 and x-2>0 and |x+1|=x+1 and |x-2|=x-2.

    The equation becomes x+1=x-2.

    Of course you would not use this method in this case.
    From what I learned (my A Level dealt very little with modulus functions), I'd square both sides of the equation and obtain the solution x=1/2. Should the question be an inequality, I'd draw a number line and determine which region satisfies the inequality. It's pretty much like solving locus questions in complex numbers too (am I wrong in saying so?).

    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?
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    (Original post by johnconnor92)
    So how do I justify cases 1 and 3 in your method? "no 'x' exists, and so... what?"
    Cases 1 and 3. Yes, you could just say no such x exists.

    (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?
    Put simply.

    |x|=x if x \ge 0.

    |x|=-x if x<0.

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Updated: May 21, 2012
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