Bunuel wrote:

samark wrote:

Bunuel,

I am confused here..

"B. 0<=x<=1 --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0<=x<=1, equation |x| + |x -1| = 1 holds true."

I am confused that how first x is +ive and second one -ve...after we take condition 0<=x<=1?

Pls, explain.

Thanks!

We know that for

|x|:

When x\leq{0}, then |x|=-x;When

x\geq{0}, then

|x|=x.

We have

|x| + |x -1| = 1.

Now for the range:

0\leq{x}\leq{1} -->

|x|=x (as

x in given range is positive) and

|x-1|=-(x-1)=-x+1 (as expression

x-1 in the given range is negative, to check this try some

x from this range, let

x=-0.5 then

x-1=0.5-1=-0.5=negative). So

|x| + |x -1| = 1 in this range becomes:

x-x+1=1 -->

1=1, which is true. That means that for ANY value from the range

0\leq{x}\leq{1}, equation

|x| + |x -1| = 1 holds true.

Hope it's clear.

Bunuel, I have a question on the part in red. Shouldn't it actually be:

We know that for

|x|:

When x<{0}, then |x|=-x; (I have changed the "less than or equal to" to only "less than")

When

x\geq{0}, then

|x|=x.

Because we should consider 2 cases -

a) greater than or equal to zero

AND

b) less than zero. [Not less than or equal to zero]

In the part B of your solution we are also considering the case where

x=1, right? If this is the case, how can

|x-1| be

-x +1? At

x=1, I am guessing

|x-1| =

x-1.

No, it works with = sign as well: |0|=0=-0.