Which of the following is always correct

IMO Option D is correct.
In option D x is always positive and graeter than1.
Say x=2
8 is greater than 4.So 16 is greater than 8.
Statement D is true for all values of x greater than1(even fractions).
Please give me kudos. I need them badly to unlock gmatclub tests.

Why is the answer not A? by what is given, x must be negative, so x squared will always be larger, no?

A. If \(\frac{1}{x}\) is greater than \(x\), \(x^2\) is greater than \(x\)

1/x > x holds true for x < -1 and 0 < x < 1. If x < -1, then yes, x^2 > x but if 0 < x < 1, then x^2 < x.
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A. 1/x > x
So, 0<x<1 (x^2<x)or x<-1(x^2>x)
NOT CORRECT

B. 1/x > x
So, 0<x<1 (x>x^2)or x<-1(x<x^2)
NOT CORRECT

C. x^2 > x
So, x>1(x^3 >X^2) or x<0 (x^3 < x^2)
NOT CORRECT

D. x^3 > x^2
So, x>1 (x^4>x^3)
CORRECT

E. x^4 > x^3 ~ x^2 > x (Same as C)
NOT CORRECT

Which of the following is always correct?

A. If \(\frac{1}{x}\) is greater than \(x\), \(x^2\) is greater than \(x\)

\(\frac{1}{x}\) is greater than \(x\) => Either x = fraction ( between 0 and 1) or negative (less than -1)

if x = fraction => \(\frac{1}{(1/2)}\) > \(\frac{1}{2}\)
=> \(x^2\) is greater than \(x\) False

if x = negative => \(\frac{1}{(-2)}\) > \(-2\)
=> \(x^2\) is greater than \(x\) True

2 different solutions. So equation not always true

B. If \(\frac{1}{x}\) is greater than \(x\), \(x\) is greater than \(x^2\)

Same as above
\(\frac{1}{x}\) is greater than \(x\) => Either x = fraction ( between 0 and 1) or negative (less than -1)

if x = fraction => \(\frac{1}{(1/2)}\) > \(\frac{1}{2}\)
=> \(x\) is greater than \(x^2\) True

if x = negative => \(\frac{1}{(-2)}\) > \(-2\)
=>\(x\) is greater than \(x^2\) False

2 different solutions. So equation not always true


C. If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\)

\(x^2\) is greater than \(x\) => either x =-ve or x=+ve
if x =-ve => \(x^3\) is greater than \(x^2\) False
if x =+ve => \(x^3\) is greater than \(x^2\) True

2 different solutions. So equation not always true

D. If \(x^3\) is greater than \(x^2\), \(x^4\) is greater than \(x^3\)

\(x^3\) is greater than \(x^2\) => For this condition to be true. x need to be +ve and greater than 1.

And for above value of x => \(x^4\) will always be greater than \(x^3\)

1 solution. Always true condition
Answer

E. If \(x^4\) is greater than \(x^3\), \(x^5\) is greater than \(x^4\)

\(x^4\) is greater than \(x^3\) => either x =-ve or x=+ve
if x =-ve => \(x^5\) is greater than \(x^4\) False
if x =+ve => \(x^5\) is greater than \(x^4\) True

Answer:D

hi
Can someone please provide a strategy to deal with these types of questions generally I face problems in considering different numbers for these types of questions or sometimes forget to consider all the cases so someone can provide an easy way of solving them?

Thanks