\(m < +-3\)

\(-3<m<3\)

\(m<-1\) is redundant Is there to trick you

Only A must be true

30 seconds approach

Sir, I am confused in this highlighted part. Would you mind helping me?

IMO answer must be E

Thank you very much sir for your swift response.

The answer is A cannot be E

if answer is A and \(m>-3\) must be true, then \(m^2<9\) must not be true, for all cases of \(m>-3\)
\(m>-3\) implies all values of m to the right of the negative 3

-3<m<3

Thi sis a must be true question and not could

C is m>3 so it is out immediately

B is m > 1 but we do know that m could be also 0 so is a could be true

D is m<1 and could be true but rules out the possibility to take into account values such as 2

A is m >-3 cover all our entire range.

My reasoning is that

If we select option A which is m > -3 then it also counts the possibility of m being equal to 3 or any other integer greater than that. Then this would contradict the statement mentioned in the question (i.e. m^2 < 9).

That's is also true. clever observation. from -3 to infinite

However, the meaning of the question is : which one among the answer choices is " amust be true" regardless

\(m^2<9\) and \(m>-1\) imply \(-3<m<3\) and \(m>-1\). This is only possible, if one takes on both intervals.

answer A \(m>-3\) is true for \(m>-1\), and it's true on the interval \(-3<m<3\), but it's not true for the interval as defined in the answer choice A itself \(m>-3\). It must be true implies true at all times.

ok ok guys

sorry. It is not the end.

It is just a discussion around a question


Simple is that

Hi,

\(m^2<9\) & \(m>-1\)

So,
\(-3<m<3\) & \(m>-1\)

When we combine both these inequalities, we will get

\(-1 < m < 3\)

So in every case \(m\) will be greater than \(-3\)

Hence, answer A

Here we go. Join the army

IF it is true that -1<m<3, THEN which of the following must be true

A. m>−3, IF A is FALSE, THEN m is not greater than -3. A is NOT FALSE, because m is greater than -3

B. m>1, IF B is FALSE, THEN m is not greater than 1. B is FALSE, because m can be -0.5 or 0; m is not greater than 1.

C. m>3, IF C is FALSE, THEN m is not greater than 3. C is FALSE, because m cannot be 3 and it's not greater than 3.

D. m<1, IF D is FALSE, THEN m is not less than 1. D is FALSE, because m can be 2 or 2.5; m is not less than 1.

E. ..

this is called proof by contradiction, and answer is indeed A

sorry if I am wrong but the post with the paper attached was not E...minutes ago ???

It was repeated from the above solution lol. I managed to check If/Then statement for this question. Thanks for your original post. This question is quite didactic ☺️ another vocab entry popes out of my mind


Posted from my mobile device

m^2 < 9
divided by 3 ==> m < 3
divided by (-3) ==> m > -3