Since \(x^2\) is always non negative, we can cross multiply without changing the inequality sign.

\(2x < \frac{1}{x^2}\)? Becomes Is \(2x^3<1......x^3<\frac{1}{2}\)

1) \(x^2<1\)
x could be \(\frac{9}{10}\), then \(x^3=\frac{729}{1000}>\frac{1}{2}\)
x could be \(\frac{1}{2}\), then \(x^3=\frac{1}{8}<\frac{1}{2}\)
Insufficient

2) \(|x|<\frac{1}{x}\)
Since 1/x is greater than a modulus, c is positive
We can cross multiply without any changes in inequality sign.
\(x*x<1\)
Same as statement I
Insufficient

E

given Is \(2x < \frac{1}{x^2}\)?
simplify
2x-1/x^2<0
2x^3-1<0
2x^3<1
x^3<1/2
this is our target value now

#1
\(x^2 < 1\)
this relation would be valid for all values of -.9 to +.9
at x=0 we get yes to our target
at x= +0.8 we get no
insufficient
#2
\(|x| < \frac{1}{x}\)
this be valid for values of x from + .1 to .9
again we get yes and no for our target
yes till .1 to .5
and from .6 to .9
insufficient

from 1 &2
no 1 value of x satisfies given target
option E is correct