Is |x| + |x -1| = 1? (1) x ≥ 0 (2) x ≤ 1

When we have equation of a type: \(|x|=3x-2\), we should consider two cases:

1. \(x<0\) --> \(|x|=-x\) --> \(-x=3x-2\) --> \(x=\frac{1}{2}\), not a vaild solution for this range as we are considering \(x<0\). Which means that when \(x<0\) equation \(|x|=3x-2\) has no real roots.
2. \(x\geq{0}\) --> \(|x|=x\) --> \(x=3x-2\) --> \(x=1\), good as \(x>0\).
Final answer: equation \(|x|=3x-2\) has one root \(x=1\).

BUT the problem above can be solved in another way:
You should notice following: \(3x-2\) is equal to "something" and that "something", as it is an absolute value, cannot be negative. So, we conclude that \(3x-2\) cannot be negative. \(3x-2\geq{0}\) --> \(x\geq{\frac{2}{3}}\). After that when we definitely know that \(x\geq{\frac{2}{3}}>0\), we can check only ONE range for \(x\), which is \(x\geq{\frac{2}{3}}>0\) and write \(|x|=x\). So, equation will become \(x=3x-2\) --> \(x=1\).

But if it were: \(x=|3x-2|\), we can definitely say that \(x\geq{0}\), but we cannot write \(x=3x-2\). We should still check two ranges:
1. \(0\leq{x}\leq{\frac{2}{3}}\) --> \(x=-3x+2\) --> \(x=\frac{1}{2}\);
2. \(\frac{2}{3}<x\) -->\(x=3x-2\) --> \(x=1\).
So, this equation has two roots 1/2 and 1.

Third case: \(|x|=|3x-2|\). In this case we should consider 3 ranges:
1. \(x<0\) --> \(-x=-3x+2\) --> \(x=1\), not good as \(x<0\);
2. \(0\leq{x}\leq{\frac{2}{3}}\) --> \(x=-3x+2\) --> \(x=\frac{1}{2}\);
3. \(\frac{2}{3}<x\) --> \(x=3x-2\) --> \(x=1\).
Again two roots 1/2 and 1.


Back to our original question:
Is \(|x|+|x-1|=1\)?
(1) x ≥ 0
(2) x ≤ 1

Absolute value, \(|x|\), cannot be negative, BUT \(x\) in it can be. \(|x|\geq{0}\) but \(x\) can be \(-3<0\) --> \(|-3|=3>0\).

Basically here we have the sum of two absolute values, or the sum of two non negative values totaling 1. But again \(x\) in it can take negative values and still these two can give us 1 as their sum.

Q: is \(|x|+|x-1|=1\)?
This equation can be true in some ranges of \(x\) and false in another, or can be true/false in all of the ranges. Thus we should check ALL ranges for \(x\) to answer the question. How to do this? There are two crucial points when absolute values \(|x|\) and \(|x -1|\) flip signs, two check points 0 (\(x=0\)) and 1 \((x-1=0\), \(x=1\)). Thus three ranges must be checked:

1. \(x<0\) --> \(-x-x+1=1\) --> \(x=0\). Not good as \(x<0\);

2. \(0\leq{x}\leq{1}\) --> \(x-x+1=1\) --> \(1=1\). Which means that for ANY value from the range \(0\leq{x}\leq{1}\), equation \(|x|+|x-1|=1\) holds true.

3. \(x>1\) --> \(x+x-1=1\) --> \(x=1\). Not good as \(x>1\).

Note that we aren't considering the statements (1) and (2) yet. We are just checking in which range the equation \(|x|+|x-1|=1\) holds true. And from 1. 2. and 3. we get that it's true ONLY in the range \(0\leq{x}\leq{1}\) and not true in all other ranges. But at this stage we still don't know in which range \(x\) is.

Moving to the statements:
(1) x ≥ 0. Not sufficient, as x must be also <=1
(2) x ≤ 1. Not sufficient, as x must be also >=0.

(1)+(2) \(0\leq{x}\leq{1}\), exactly the range we needed. Sufficient.

Answer: C.

Hope it's clear.