If n is a prime number greater than 2, is 1/x 1?

Question

If n is a prime number greater than 2, is 1/x > 1? (1) x^n < x < x^{\frac{1}{n}} (2) x^{n-1} > x^{2n-2}

mau5 · Accepted Answer

The question is asking, is \frac{1}{x} >1 --> Is 0x^3, we have x(1-x^2)>0 --> x(x+1)(x-1)<0 . Hence 0 x^4 --> x^2<1 --> -1

Vetrik · Answer

Statement 1 -> will be true for negative integers and positive fractions. So X can either be a negative integer or positive fraction. If X is a negative integer 1/X will not be greater than 1 but if X is a positive fraction 1/X will be greater than 1. So we have two possibilities.

Hence Statement 1 alone is insufficient.

Statement 2 -> Will be true for negative and positive fractions. If X is a negative fraction 1/X will not be greater than 1 but if X is a positive fraction 1/X will be greater than 1. So we have two possibilities.

Hence Statement 2 alone is insufficient.

But for both statement 1 and 2 to be true, X can only be a positive fraction. Here 1/X will be greater than 1.

Hence both statements together are sufficient.

Answer -> C

thefibonacci · Answer

I got A for this one. 1) x^n < x < x^(1/n) n = odd (since it is prime and > 2) so the inequality will hold true only for 0 < x < 1 it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number. so this is sufficient. 2) x^(n-1) > x^(2n-2) => x^(n-1) > x^2(n-1) n=odd, so n-1=even=2k(say) x^2k > x^4k this inequality holds true for -1

Bunuel · Answer

Odd root from negative number is negative, not imaginary. For, example,  \sqrt {-8}=-2 .

The question asks whether 0 < x < 1. For (1) check x = -8 and n = 3 prime.

mattyahn · Answer

I believe the answer should be B

Statement 1: Insufficient   or   (refer to others)

Statement 2: Fractions get closer to zero when raised to a power greater than 1. I think of it like this - when fractions are raised to a power greater than 1, the fraction gets closer to zero. A positive fraction would decrease, while a negative fraction would increase.
When plugging in x= (1/2) and n=3, statement 2 is correct (  is greater than  ) 
When plugging in x=(-1/2) and n=3, statement 2 is incorrect (  is NOT greater than  )
Therefore:   only ;  >1 in all cases ; Statement 2 Sufficient

Answer is B.

Bunuel · Answer

The OA is given under the spoiler and it's C, not B.

If x=-1/2 and n=3,  x^{n-1} > x^{2n-2}  holds true:    >

exc4libur · Answer

n= 1/x>1…when:x=positive.proper.fraction…0 x^{2n-2} insufic. n=3:x^{n-1} > x^{2n-2}…x^2>x^4…x=|proper.fraction|…0

GMATStudyStudent · Answer

Statement 1: x^n < x < x^\frac{1}{n} This is only valid for 0 < x <1, since if x<0 then x^\frac{1}{n} is not defined. Thus \frac{1}{x} >1 Statement 1 is Sufficient. Statement 2: x^{n-1} > x^{2n-2} x^{n-1}(1 - x^{n-1})>0 x^{n-1}>0 and (1 - x^{n-1})>0 or x^{n-1}<0 and (1 - x^{n-1})<0 -1 < x < 1, x eq{0} or Not Possible, since n is a prime number greater than 2 and thus n is odd, and n-1 is even Thus x^{n-1} cannot be negative. Statement 2 is Not Sufficient Answer A @Bunel please check

Bunuel · Answer

The red part is not correct. For example,  \sqrt {-8}=-2 . BTW, this is explained a above.

sidship21 · Answer

hi  Bunuel ,  VeritasKarishma   mikemcgarry

I generally understand the solutions above, except, I don't understand the following: we use n=3 to work out the individual statements. However, how are we sure that the statements work out the same way for every single odd prime?

Thank you in advance

KarishmaB · Answer

You are given that statement 1 holds. This means that for whatever values n can take,

x^n < x < x^(1/n)

So this will be true for n = 3, 5, 7, 11 etc all values of n.

You can plug in any one of these values and it will hold for that.

amoghhlgr · Answer

One off topic question related to this concept....

If x^(n-1) > x^2(n-1), then let's say by plugging n=3, is the below solution acceptable?

x^2 > x^4
x^2 > x^(2+2)
x^2 > x^2 * x^2
1 > x^2

But the final solution doesn't hold good for the equation we started with in the first place. I don't think x can be equal to 0. Can you please help me understand this better?

Bunuel   VeritasKarishma   mikemcgarry

KarishmaB · Answer

This solution is incorrect. You cannot cancel off x^2 until and unless you know that it is not 0. x^2 > x^4 x^4 - x^2 < 0 x^2 ( x^2 - 1) < 0 x^2 (x + 1)(x - 1) < 0 Since x^2 cannot be negative, it will not change the sign of the expression so ignore x^2 in transition points (-1 and 1) but it will need to be considered while writing the answer later. -1 < x < 1 and x cannot be 0. Check here: https://youtu.be/PWsUOe77__E

Archit3110 · Answer

value of n can be 3,5,7.. let n = 3 #1 x^n < x < x^{\frac{1}{n}} this is possible when 0>x<1 and x<-1 insufficient #2 x^{n-1} > x^{2n-2} possible at fraction +ve and -ve -1x<1 insufficient from 1 &2 0>x<1 sufficient option C

bumpbot · Answer

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If n is a prime number greater than 2, is 1/x > 1?

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