What is the solution for this question?
My thought: I think that option E is correct because x^2>1 and x<-1 both satisfy the given equation, but in the forum they say that (A) is the correct answer. I don’t understand why x<-1 is not correct.
Source: GMAT Club - https://gmatclub.com/forum/if-x-x-2-which-of-the-following-must-be-true-99506.html
Thats true, but its asking MUST be true, not CAN be true. What if x=2? Is the third one satisfied?
Okay I get it.
Hey Ganesh,
I have a similar doubt in another problem
In this n would be less than -2, so the answer would be only II. But in the discussion they have mentioned the answer as (D). So my question is, when you consider n<3 and let’s take n=2. At this point n=2 won’t satisfy the equation right? Then how could the answer be option (D).
Source: https://gmatclub.com/forum/if-n-is-an-integer-such-that-3-4n-3-6-n-which-of-the-327384.html
Can you pick n that satisfies the inequality, but is also > 3?
You contradict the “n would be less than -2” part you gathered
I still don’t get it. Let me putdown my thought process clearly.
The question is asking, what are the possible inputs of n such that the inequality remains true.
When solving the inequality, the negative sign will disappear because of the even power.
So, the possible inputs of n for the inequality to remain true would be all integers less than -2
I. n could be -3 which is odd so (I) does not satisfy
II. Negative number are not prime. So this is always true
III. n should always be less than 3 which includes 2,1,0,-1,-2 which are all integers
Lets try substituting few of these values
When n=2,1,0,-1,-2 the inequality does not hold true.
So my question is, when the inequality does not hold true for n<3 how could you choose option (III).
"n is an integer such that (-3)^{-4n} > 3^{6 - n}, so yes the solution n < -2 is correct.
More mathematically, n is any element in the set \{\text{all real numbers less than -2}\}
Option III) is really asking you whether 3 is an upper bound for this set. As you can tell, 3 is greater than any element in the set, so it can definitely be an upper bound. Similarly, 1000 could also be a potential upper bound for the exact same reason.
I think your point of confusion is that x < -2 doesn’t imply x < 3. That is certainly true, but apparently, that is not quite what the question is asking. The difference here is that we aren’t comparing two solution sets, which obviously don’t match. Instead, we’re asked to verify whether a certain number could work as an upper bound. For instance, if I say a set consists of \{1,3\} then you know that any number greater than 3 (e.g., 5 or even something like 100,000) can function as an upper bound.
Tldr; there’s a subtle difference between the two cases. Here, we’re not comparing the solution sets of x < -2 and x < 3. Rather, we’re asking the following question: “is 3 a valid upper bound of a set consisting of all real numbers less than -2?”
Okay, so for my 1st question since x can take values greater than 1 option (III) does not satisfy our inequality is that right?
What is your 1st question?
To supplement what was mentioned above, choose a particular n (such that it is an integer and also less than -2). For all possible n (that satisfies the constraint so you can’t say something like n = 2 because that contradicts the “less than -2” bit), you see that n < 3 is a true statement; hence, the statement is true for all n.
