i guess we cannot take 1 because it is a square but we can use 2. if so,

if x = 2, n = 2 (1 and 2) and 2*sqrt(2)-1 = 1.83 approx.

so, n>2*sqrt(x)-1.

Professor wrote:

suppose, n = no of divisors

(I) x could be 1 or 3 or 6 or so on.

=> if x = 1, n = 1 and 2*sqrt(x)-1 = 1.

so n = 2*sqrt(x)-1.

=> if x = 3, n = 2 (1 and 3) and 2*sqrt(3)-1 = 2.46 approx. so, n>2*sqrt(x)-1.

=> if x = 6, n = 4 (1, 2, 3 and 6) and 2*sqrt(6)-1 = 3.9 approx. divisors<2*sqrt(x)-1.

so not sufficient.

(II) if x is a prime, n is always 2 but 2*sqrt(x)-1 greater or smaller than n. so not sufficient.

from i and ii, togather, also not sufficient. so E.

from statement i, [the following table (calculated in XL format) would be more useful].

if x = 2, n = 2, 2sqrt(x)-1 = 1.83

if x = 3, n = 2, 2sqrt(x)-1 = 2.46

if x = 8, n = 4, 2sqrt(x)-1 = 4.66

if x = 11, n = 2, 2sqrt(x)-1 = 5.63

if x = 12, n = 6, 2sqrt(x)-1 = 5.93

from above when x = 2, and 12, n> 2sqrt(x)-1 and when x = 3, 8, and 11, n<2sqrt(x)-1. this proovs that A is out.

from st ii,

if x = 8, n = 4, 2sqrt(x)-1 = 4.66

if x = 12, n = 6, 2sqrt(x)-1 = 5.93

from above when x = 8, n< 2sqrt(x)-1 and when x = 12, n>2sqrt(x)-1. this proovs that A is out. this also rules out B.

togather also we cannot say which one is greater. so E.