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04 May 2026, 19:00
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If x is positive, which of the following may be the correct ordering of \(\frac{1}{x}\), \(x\), \(x^2\), \(x^3\) ?
I. \(\frac{1}{x} < x < x^2 < x^3\)
II. \(\frac{1}{x} < x^3 < x^2 < x\)
III. \(x^3 < x^2 < x < \frac{1}{x}\)
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
I. \(\frac{1}{x} < x < x^2 < x^3\)
II. \(\frac{1}{x} < x^3 < x^2 < x\)
III. \(x^3 < x^2 < x < \frac{1}{x}\)
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
04 May 2026, 19:57
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Good question to solve by substitution:
Given x>0:
Let x = 2
So: I. 1/x<x<x2<x3
is true, 1/2 < 2<4<8
So, the answer is A?
NO, we have not taken fraction:
It still satisfies the condition x>0
0<x<1
x= 1/2
So, he opposite of I will also be true.
III. x3<x2<x<1/x
1/8<1/4<1/2<2
So, I and III
Option D
Given x>0:
Let x = 2
So: I. 1/x<x<x2<x3
is true, 1/2 < 2<4<8
So, the answer is A?
NO, we have not taken fraction:
It still satisfies the condition x>0
0<x<1
x= 1/2
So, he opposite of I will also be true.
III. x3<x2<x<1/x
1/8<1/4<1/2<2
So, I and III
Option D
04 May 2026, 23:08
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here X>0
but x can be either integer or fraction.
lets consider both case.
case-1. x is integer. say x=3
1/x = 1/3
x=3
x^2= 9
x^3= 27
from this we can say, 1/x<x<x^2<x^3
case-2. x is fraction. say x=1/3
1/x= 3
x=1/3
x^2= 1/9
x^3= 1/27
here x^3 is smallest and 1/x is biggest.
x^3<x^2<x<1/x
only option D states this.
but x can be either integer or fraction.
lets consider both case.
case-1. x is integer. say x=3
1/x = 1/3
x=3
x^2= 9
x^3= 27
from this we can say, 1/x<x<x^2<x^3
case-2. x is fraction. say x=1/3
1/x= 3
x=1/3
x^2= 1/9
x^3= 1/27
here x^3 is smallest and 1/x is biggest.
x^3<x^2<x<1/x
only option D states this.
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