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Let's check in which ranges of \(x\) expression \(|x|+|x-1|=1\) holds true:

Two check points for \(x\): 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check: A. \(x<0\) --> \(-x-x+1=1\) --> \(x=0\), not good as we are checking for \(x<0\); B. \(0\leq{x}\leq{1}\) --> \(x-x+1=1\) --> \(1=1\), which is true. This means that in this range given equation holds true for all x-es; C. \(x>1\) --> \(x+x-1=1\) --> \(x=1\), not good as we are checking for \(x>1\).

So we got that the equation \(|x|+|x-1|=1\) holds true only in the range \(0\leq{x}\leq{1}\).

(1) \(x \geq 0\). Not sufficient. (2) \(x \leq 1\). Not sufficient.

(1)+(2) gives us the range \(0\leq{x}\leq{1}\), which is exactly the range for which given equation holds true. Sufficient.

stmt1: x >= 0 => x is positive, if we apply this to eq 1, we cannot say that 2x-1 = 1 cos x can be >= 0 but less than 1 so insuff stmt2: x<=1 => if we apply this to eq 1, we cannot say that 2x-1 = 1 cos x<=1 can lead to negative numbers also.

both stmt taken together 0<= x <= 1 x is a fraction between 0 to 1 so always x + 1-x = 1 in our case |x| will be x and |x - 1| is actually value of 1-x so it will always be 1 this is why C both the stmts are reqd.
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Let's check in which ranges of \(x\) expression \(|x|+|x-1|=1\) holds true:

Two check points for \(x\): 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check: A. \(x<0\) --> \(-x-x+1=1\) --> \(x=0\), not good as we are checking for \(x<0\); B. \(0\leq{x}\leq{1}\) --> \(x-x+1=1\) --> \(1=1\), which is true. This means that in this range given equation holds true for all x-es; C. \(x>1\) --> \(x+x-1=1\) --> \(x=1\), not good as we are checking for \(x>1\).

So we got that the equation \(|x|+|x-1|=1\) holds true only in the range \(0\leq{x}\leq{1}\).

(1) \(x \geq 0\). Not sufficient. (2) \(x \leq 1\). Not sufficient.

(1)+(2) gives us the range \(0\leq{x}\leq{1}\), which is exactly the range for which given equation holds true. Sufficient.

Answer: C.

Hope it helps.

Can you please clarify as to how is \(x-x+1=1\) derived inB when it is provided that \(0\leq{x}\leq{1}\) that means X is +ve.

Let's check in which ranges of \(x\) expression \(|x|+|x-1|=1\) holds true:

Two check points for \(x\): 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check: A. \(x<0\) --> \(-x-x+1=1\) --> \(x=0\), not good as we are checking for \(x<0\); B. \(0\leq{x}\leq{1}\) --> \(x-x+1=1\) --> \(1=1\), which is true. This means that in this range given equation holds true for all x-es; C. \(x>1\) --> \(x+x-1=1\) --> \(x=1\), not good as we are checking for \(x>1\).

So we got that the equation \(|x|+|x-1|=1\) holds true only in the range \(0\leq{x}\leq{1}\).

(1) \(x \geq 0\). Not sufficient. (2) \(x \leq 1\). Not sufficient.

(1)+(2) gives us the range \(0\leq{x}\leq{1}\), which is exactly the range for which given equation holds true. Sufficient.

Answer: C.

Hope it helps.

Can you please clarify as to how is \(x-x+1=1\) derived inB when it is provided that \(0\leq{x}\leq{1}\) that means X is +ve.

Thanks !!!

When \(x\) is in the range \(0\leq{x}\leq{1}\) then \(|x|=x\) and \(|x-1|=-(x-1)=-x+1\), thus \(|x|+|x-1|=1\) in this range becomes \(x-x+1=1\).

I have got the answer as E. Can someone please clarify, what is wrong in my approach ?

Statement 1: X>=0

I assume X as 0 |0|+|0-1|= |0|+|1|= 1 ------------> Equation satisfied

I assume X as 3/4 |3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 1 is not sufficient

Statement 2: X<=1

I assume X as 1 |1|+|1-1|= |1|+|0|= 1 --------------> Equation satisfied

I assume X as 3/4 again |3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 2 is not sufficient

Taking together both the equation, I assume X as 0, 3/4 and 1

When X=0, the answer is 1--------> Equation satisfied When X=3/4, the answer is 2---------> Equation unsatisfied When X=1, the answer is 1----------> Equation satisfied

Hence answer choice E is correct.

Clarifications needed, please.
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A Ship in port is safe but that is not what Ships are built for !

l x l + lx-1l= 1 implies that sum of distance of x from 0 and from 1 should be 1. We can easily see that the distance between 0 and 1 is 1. So, x can lie on 0, on 1 or anywhere in between 0 and 1. It cannot go beyond 1 or before 0 because then the sum of the distance will become larger than 1. Both statements together tell you that 0 <= x <= 1 and hence, the equation will hold for all such values of x. Answer (C)
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I have got the answer as E. Can someone please clarify, what is wrong in my approach ?

Statement 1: X>=0

I assume X as 0 |0|+|0-1|= |0|+|1|= 1 ------------> Equation satisfied

I assume X as 3/4 |3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 1 is not sufficient

Statement 2: X<=1

I assume X as 1 |1|+|1-1|= |1|+|0|= 1 --------------> Equation satisfied

I assume X as 3/4 again |3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 2 is not sufficient

Taking together both the equation, I assume X as 0, 3/4 and 1

When X=0, the answer is 1--------> Equation satisfied When X=3/4, the answer is 2---------> Equation unsatisfied When X=1, the answer is 1----------> Equation satisfied

Hence answer choice E is correct.

Clarifications needed, please.

The correct answer is C, not E.

If x = 3/4, then Is |x| + |x-1| = |3/4| + |3/4 - 1| = 3/4 + 1/4 = 1.

Wouldn't that imply that for both cases of x (where we have positive and negative values inside the | | signs) the equation is true?

Thanks!

No. In the first case for an \(x=\frac{1}{2}\) (\(0\leq{x}\leq{1}\)) you get that the equation is true \(1=1\) it's like saying ALWAYS TRUE. But in the second case for an \(x>1\) you get x=1, but since x>1 that is not true Statement 1 is not sufficient
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It is beyond a doubt that all our knowledge that begins with experience.

x>1 |x| + |x-1| = 1 x + (x-1) = 1 2x = 2 x = 1 x = 1 which ISN'T greater than 1. Insufficient

0≤x<1 x = 1/2 |x| + |x-1| = 1 x + -(x-1) =1 x+ -x +1 = 1 0=0 I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!

Zarrolou wrote:

WholeLottaLove wrote:

Is |x| + |x-1| = 1?

For #1 (x≥0) I see two possible cases:

x=1/2 x+ -(x-1)=1 x+ -x+1=1 1=1

OR

x=2 x+x-1=1 2x=2 x=1

Wouldn't that imply that for both cases of x (where we have positive and negative values inside the | | signs) the equation is true?

Thanks!

No. In the first case for an \(x=\frac{1}{2}\) (\(0\leq{x}\leq{1}\)) you get that the equation is true \(1=1\) it's like saying ALWAYS TRUE. But in the second case for an \(x>1\) you get x=1, but since x>1 that is not true Statement 1 is not sufficient

x>1 |x| + |x-1| = 1 x + (x-1) = 1 2x = 2 x = 1 x = 1 which ISN'T greater than 1. Insufficient

0≤x<1 x = 1/2 |x| + |x-1| = 1 x + -(x-1) =1 x+ -x +1 = 1 0=0 I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!

0=0, and 1=1,... means that that equation is always true. You should read the results as follow: x>1 x = 1 which ISN'T greater than 1. So in this case the equation if FALSE (as 1 does not equal a value greater than 1) 0≤x<1 0=0. So in this case the equation in TRUE( because 0=0 always)

So from statement 1 you get two different results (F/T)=> not sufficient
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It is beyond a doubt that all our knowledge that begins with experience.

x>1 |x| + |x-1| = 1 x + (x-1) = 1 2x = 2 x = 1 x = 1 which ISN'T greater than 1. Insufficient

0≤x<1 x = 1/2 |x| + |x-1| = 1 x + -(x-1) =1 x+ -x +1 = 1 0=0 I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!

0=0, and 1=1,... means that that equation is always true. You should read the results as follow: x>1 x = 1 which ISN'T greater than 1. So in this case the equation if FALSE (as 1 does not equal a value greater than 1) 0≤x<1 0=0. So in this case the equation in TRUE( because 0=0 always)

So from statement 1 you get two different results (F/T)=> not sufficient

So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say, x < -8 ------ x is less than -8 -8 <= x < -3 ------- x is greater than or equal to -8 but less than -3 -3 <= x < 4 ------- x is greater than or equal to -3 but less than 4 x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as x <= -8 -8 < x <= -3 -3 < x <= 4 x > 4

The above explanation is easily understandable but it creates a small doubt if i compare it with this question Is |x| + |x-1| = 1? (1) \(x\geq{0}\) (2) \(x\leq{1}\)

Quote:

Two check points for \(x\): 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check: A. \(x<0\) --> \(-x-x+1=1\) --> \(x=0\), not good as we are checking for \(x<0\); B. \(0\leq{x}\leq{1}\) --> \(x-x+1=1\) --> \(1=1\), which is true. This means that in this range given equation holds true for all x-es; C. \(x>1\) --> \(x+x-1=1\) --> \(x=1\), not good as we are checking for \(x>1\).

So we got that the equation \(|x|+|x-1|=1\) holds true only in the range \(0\leq{x}\leq{1}\)

Here, we have to include the transition points (0 and 1) but only once. Are these okay, for B. \(0\leq{x}<1\) --> 1=1 not in the range. for C. \(x\geq{1}\) --> x = 1. It is in the range.

Or

A. \(x\leq{0}\) --> x= 0, It is in the range. B. \(0<{x}\leq{1}\) --> 1=1 It is also in the range.

Are my ranges correct ? if yes, then i still don't feel close to the answer.

1. From 1st equation , if x = 1 , then |x| + |x-1| = 1 , then its yes if x = 5 , 9 # 1 , so No so Equation 1 is No 2. From 2nd equation , of x = -1 , then its No x = 1 its yes so equation is No

3. using both , x = 0 then its Definite yes So C..

1. From 1st equation , if x = 1 , then |x| + |x-1| = 1 , then its yes if x = 5 , 9 # 1 , so No so Equation 1 is No 2. From 2nd equation , of x = -1 , then its No x = 1 its yes so equation is No

3. using both , x = 0 then its Definite yes So C..

Is my approach is Right

Using both, you get 0 <= x <= 1 x can lie anywhere from 0 to 1. Just trying the extreme values is not a good idea. You must try some values from the middle too i.e. x = 1/2, 1/4 etc.

Though establishing something by plugging in values is always tricky. You can try to negate a universal statement by looking for one value for which the statement doesn't hold but since you cannot try every value for which a statement must hold, its treacherous to depend on number plugging in this case. You should try to use logic.
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