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
AnswerE. 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