To find if (x^2+1) * (x+5) is even

If either (x^2+1) or (x+5) is even or both are even then the product will be even. For this to happen x has to be odd

since odd*odd = odd
even*odd or odd*even = even
EVEN*EVEN = EVEN

Stmt I
x is odd
Exactly what we need
SUFF

Stmt II

each prime factor of x^2 greater than 7. This means each prime factor of x is greater than 7 since if x had a prime factor less than 7 then x^2 definitely has that prime factor. So x cannot be even as 2 is a prime factor of every even number.

This means x is odd

SUFF

Choose D)

Both Stmts are sueff.

Stmt1
(X^2 + 1) (X+5)..

to make the product even. either (X^2 +1 ) or (X+5) should be even.. If x is odd, then X+ 5, will always be even Stmt 1 provides that.
hence Sueff.

Stmt2.
the prime factors of X^2 will be greater than 7. there fore 2 cannot be factor of x. Hence again x is odd. same as stmt1
sueff

Poonam,

i guess A is preety clear

for Option B

x^2 is primefactor > 7

now since primecfactor by defination has only two factors 1 and itsefl

so x would also be having prime factor greater than 7. hence it would be a ODD ( 2 prime factor will be missing)

so now proved X is ODD...

Choose D

Takeaway:

2 is the ONLY even prime number!