Could the answer be an integer if x is an integer greater than 1
a) X^10 + X^(-10)
b) X^(1/6) + X^(1/2)
a) X^10 + X^(-10)
b) X^(1/6) + X^(1/2)
ST1: X^10 + X^(-10) or {x^20 + 1}/x^10
given x is an integer greater than 1, this will never be an integer.
Hence Sufficient.
St2: X^(1/6) + X^(1/2)
if x = 64, the above becomes 2 + 8 = 10 (INTEGER)
if x = 2, the above becomes 2^1/6 + root2 (NOT an INTEGER)
Hence Insufficient
IMO A
