Given:

a > 0, b > 0 and a, b are Integers

a - b and a/b are both even intergers

==> \(a - b = 2m_1\)

\(\frac{a}{b} = 2m_2\) ==> \(a = (b)(2m_2)\)

==> a & b are both even, otherwise the above conditions won't be satisfied.

Now lets go with options one by one ...

A) a/2

==> can be even or odd ( a = 6 or a = 8)

B) b/2

==> Can be either even or odd ( b = 6 or b = 8)

C) (a+b) /2

==> Since, \(a = (b)(2m_2)\), \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2) + b}{2}\)

==> \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2 + 1)}{2}\)

==> Can be even or odd

D) (a+2)/2

==> \(\frac{a + b}{2}\) =\(\frac{2bm_2 + 2}{2}\)

==> \(\frac{a + b}{2}\) =\(bm_2 + 1\)

==> Since, b is even, \(bm_2\) is even and \(bm_2 + 1\) is odd

alwaysE) (b+2)/2

==> \(\frac{b+2}{2} = 1 + \frac{b}{2}\)

==> Can be even or odd

_________________

Cheers!

Ravi

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