\(a-b\) even --> either both even or both odd

\(\frac{a}{b}\) even --> either both even or \(a\) is even and \(b\) is odd.

As both statements are true --> \(a\) and \(b\) must be even.

As \(\frac{a}{b}\) is an even integer --> \(a\) must be multiple of 4.


Options A is always even.

Options B can be even or odd.

Options C can be even or odd.

Options D: \(\frac{a+2}{2}=\frac{a}{2}+1\), as \(a\) is multiple of \(4\), \(\frac{a}{2}\) is even integer --> even+1=odd. Hence option D is always odd.

Options E can be even, odd.


Answer: D.
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I didnt get why a has to be multiple of 4

Now if u take a as 4 and b as 2 then a-b is even a/b is even and a+b/2 is odd which is option C.

Please explain where i am going wrong?

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 always

E) (b+2)/2
==> \(\frac{b+2}{2} = 1 + \frac{b}{2}\)
==> Can be even or odd
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The video solution is explained here.

Answer: Option D


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