shahrukh wrote:

There are three possibilities for value of x here,

i) 0<x>1 & x=1

ii)1<x>2 & x=2

iii)x>2

Now for i) taking from previous examples x=0.5

then the ordering is x^2<2x<1/x (this is I) & for x=1, 1/x=x^2<2x

For ii) lets say x=1.5

the ordering is 1/x<x^2<2x & for x=2, 1/x< 2x=x^2

for iii) lets say x=3

the ordering is 1/x<2x<x^2

SO in my opinion, the answer is "b"

Tell me plz how its d and not b

From my post above.

II. x^2 < 1/x < 2x

.8^2=.64 < 1/.8=1.25 < 2*.8=1.6 so II. is possible

You only have to prove it can work for a single case not that it is always true.