If y=−m^2, which of the following must be true?

I say that from y=−m^2 we don't have ANY restriction on the value of m, so it can take ANY value: positive, negative or zero. So, option II which says that "m is non-negative" (read zero or positive) is not always true since m can also be negative.

Condition I
If m=0 then y=0, hence not always true

COndition II
m can be anything, +ve, 0 or -ve. m^2 will be either 0 (for m=0) or +ve

Condition III
if m is -ve (or to say m is less than 0) than m^2 is +ve.
Hence -m^2 is always -ve.
if m is -ve than y IS -ve
Always True

If m is positive then too y is negative so statement iii is contradicted ?

I think you don't understand the question.

Option III says: If m is negative then y is negative --> m=negative --> y=-(negative^2)=-positive=negative. So, you can see that this option is true.

Why do you even bother to consider positive m for this statement? Anyway, even if m=positive then we would have: y=-(positive^2)=-positive=negative.

In response to the doubt (also on PM)

Joy the question is asking which must always be true

I will go thru the conditions 1 and 2 once again to help you understand the condition 3

Condition 1 says "y is -ve"
That means it is saying that "y=-m^2" is always -ve
But if m=0 then y becomes 0.
So condition 1 is not true

We rule out choices A and D

Condition 2 says "m is non-negative"
That means it is saying that in "y=-m^2" m is either 0 or +ve
But it is not true, m can be -ve also
From the question stem, there is no restriction on value of m, and hence m can be any number

We rule out choices B and E

solved
Answer is C

Pursuing further
Codition 3: "If m is negative then y is negative."
This condition limits the values of m
It is saying that IF m<0 then y is always -ve
The only time when y is not -ve is when m=0
as the condition says that m is less than 0 (that is to say that m is not 0)

Hence when m is -ve (less than 0) y is -ve
Always True

I think both of you are correct, I was thinking the question to be " Which of the following is ALWAYS true "

Please correct me if I am wrong .

I) which of the following MUST be true

II) which of the following is ALWAYS true ,

I think for both of these the answer to this question will be different ?

No, they are the same. Option III must be true because it's always true.

hm,

I ignored m=0, since 0 cant have a sign. is it ok to write y=-0!?

y=-0 simply means that y=0, so there is nothing wrong with that.

aha, I found in wiki - https://en.wikipedia.org/wiki/Sign_%28mathematics%29

The number zero is neither positive nor negative, and therefore has no sign. In arithmetic, +0 and −0 both denote the same number 0, and the negation of zero is zero itself.

I thought it is not allowed to write -0 ,since 0 is neither negative, nor positive. but wiki says -0 is ok and it means just 0.
ok, Bunuel, now I agree with u

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Given that \(y = -m^2\)

If m=0, then y =0 --> This eliminates option I. Again, m can be a non-negative value as there is no restriction on that, thus statement II gets eliminated.

Multiply both sides by m assuming \(m\neq{0}\). Thus, \(my = -m^3\). Now, if m<0,then \(m^3<0\)and \(-m^3>0\). Hence, as my>0, m and y must have the same sign.

C.

The biggest issue here is obviously noticing that m could be 0, thus making y=0 as well. The first option is a tricky one, but normally the other options will at least offer a hint and/or insight into the trick. In this case, option II mentions "non-negative," which suggests that the reader considers that m could be 0.

Summary: Read all of the option choices! They're trying to give you hints

If y=−m^2, which of the following must be true?

I) y is negative.

II) m is non-negative.

III) If m is negative then y is negative.


I only

II only

III only

I and II only

II and III only

Please help with this question. Thanks

Please refer to the discussion above. Also, please search before posting. Thank you.

yep, that m=0 does the trick. tricky one!

\(y=-m^2; let \ m=0\)
\(y=0\)

\(y=-m^2; let \ m=1\)
\(y=-1\)

\(y=-m^2; let \ m=-1\)
\(y=-1\)

So, y can be negative if \(m>0 \ or \ M<0\)

The answer is C