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So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Hope it's clear.[/quote][/quote]

Not sure understand your question. Can you please elaborate?

P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...[/quote]

Hello Bunnel, I have one small doubt. When you say |X| = +X, when Xgeq{0} and -X when xleq{0}, then in this question how could you compare |y-4| to be 4-y when yleq{0}, dont you think it should be when y<0. ?

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Hope it's clear.[/quote][/quote]

Not sure understand your question. Can you please elaborate?

P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...[/quote]

Hello Bunnel, I have one small doubt. When you say |X| = +X, when X<eq0 and -X when <0 , then in this question how could you compare |y-4| to be 4-y when comparing y<eq4 ? Shouldnt it be y<4 only? On x=0 how does the mod function behave, its neither postive nor negative then dont you think at x= 0, we cant be sure wether F(X) is positive or negative ?

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Hope it's clear

Not sure understand your question. Can you please elaborate?

P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...

Hello Bunnel, I have one small doubt. When you say |X| = +X, when X<eq0 and -X when <0 , then in this question how could you compare |y-4| to be 4-y when comparing y<eq4 ? Shouldnt it be y<4 only? On x=0 how does the mod function behave, its neither postive nor negative then dont you think at x= 0, we cant be sure wether F(X) is positive or negative ?

Check your approach with numbers: if \(y=4\) --> \(\sqrt{(4-4)^2}=0\) and \(4-4=0\) --> \(0=0\).

Is \(\sqrt{(y-4)^2}=4-y\)? --> is \(|y-4|=4-y\)? is \(y\leq{4}\)?

(1) |y-3| less than or equal to 1 --> \(|y-3|\) is just the distance between 3 and \(y\) on the number line. We are told that this distance is less than or equal to 1: --2----3----4-- so, \(2\leq{y}\leq{4}\). Sufficient.

(2) y*|y|>0 --> just says that \(y>0\). Not sufficient.

Answer: A.

Bunuel,

\(|y-3| \leq1\) means \(-(y-3) \leq 1\) also. But could you please clarify why we're not considering the same?
_________________

Is \(\sqrt{(y-4)^2}=4-y\)? --> is \(|y-4|=4-y\)? is \(y\leq{4}\)?

(1) |y-3| less than or equal to 1 --> \(|y-3|\) is just the distance between 3 and \(y\) on the number line. We are told that this distance is less than or equal to 1: --2----3----4-- so, \(2\leq{y}\leq{4}\). Sufficient.

(2) y*|y|>0 --> just says that \(y>0\). Not sufficient.

Answer: A.

Bunuel,

\(|y-3| \leq1\) means \(-(y-3) \leq 1\) also. But could you please clarify why we're not considering the same?

\(|y-3|\leq{1}\) means that \(-1\leq{y-3}\leq{1}\) --> \(2\leq{y}\leq{4}\).
_________________

\(|y-3|\leq{1}\) means that \(-1\leq{y-3}\leq{1}\) --> \(2\leq{y}\leq{4}\).

There I'm getting bit confused...For any such inequality(\(|x|\leq{1}\)) it can have two values i.e. either \(x\leq{1}\) or \(-x\leq{1}\) dependning upon whether \(x>=0\) or \(x<0\). Simultaneously both can't hold good...!

Now my question is in case of 'must be true' problems should we consider the range \(-1\leq{x}\leq{1}\) or two individual possibilities separately?
_________________

\(|y-3|\leq{1}\) means that \(-1\leq{y-3}\leq{1}\) --> \(2\leq{y}\leq{4}\).

There I'm getting bit confused...For any such inequality(\(|x|\leq{1}\)) it can have two values i.e. either \(x\leq{1}\) or \(-x\leq{1}\) dependning upon whether \(x>=0\) or \(x<0\). Simultaneously both can't hold good...!

Now my question is in case of 'must be true' problems should we consider the range \(-1\leq{x}\leq{1}\) or two individual possibilities separately?

You are overthinking it: \(|x|\leq{1}\) means \(-1\leq{x}\leq{1}\).
_________________

\(|y-3|\leq{1}\) means that \(-1\leq{y-3}\leq{1}\) --> \(2\leq{y}\leq{4}\).

There I'm getting bit confused...For any such inequality(\(|x|\leq{1}\)) it can have two values i.e. either \(x\leq{1}\) or \(-x\leq{1}\) depending upon whether \(x>=0\) or \(x<0\). Simultaneously both can't hold good...!

Now my question is in case of 'must be true' problems should we consider the range \(-1\leq{x}\leq{1}\) or two individual possibilities separately?

You are overthinking it: \(|x|\leq{1}\) means \(-1\leq{x}\leq{1}\).

Okay...So.even in case of 'must be true' questions we'll consider that \(|x|\leq{1}\) means \(-1\leq{x}\leq{1}\) always, unless otherwise stated explicitly that \(x>=0\) or \(x<0\). Right?
_________________

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Hope it's clear

Not sure understand your question. Can you please elaborate?

P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...

Hello Bunnel, I have one small doubt. When you say |X| = +X, when X<eq0 and -X when <0 , then in this question how could you compare |y-4| to be 4-y when comparing y<eq4 ? Shouldnt it be y<4 only? On x=0 how does the mod function behave, its neither postive nor negative then dont you think at x= 0, we cant be sure wether F(X) is positive or negative ?

Check your approach with numbers: if \(y=4\) --> \(\sqrt{(4-4)^2}=0\) and \(4-4=0\) --> \(0=0\).

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Hope it's clear

Not sure understand your question. Can you please elaborate?

P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...

Hello Bunnel, I have one small doubt. When you say |X| = +X, when X<eq0 and -X when <0 , then in this question how could you compare |y-4| to be 4-y when comparing y<eq4 ? Shouldnt it be y<4 only? On x=0 how does the mod function behave, its neither postive nor negative then dont you think at x= 0, we cant be sure wether F(X) is positive or negative ?

Check your approach with numbers: if \(y=4\) --> \(\sqrt{(4-4)^2}=0\) and \(4-4=0\) --> \(0=0\).

It doesn't matter where you take EQUAL sign, with less than or greater than.. only point is it should be catered in any one..... You are having TWO ranges.. 1) y is LESS than or EQUAL to 4 and y is more than 4.... so this completes the range 2) y is less than 4 and y is GREATER than or EQUAL to 4... this again completes the range

But you cannot have EQUAL to in both as it is duplicated while you combine the two to find the different ranges..

The MOD in my signature below may be of some use here
_________________

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