Now there are 4 case possible:
Case 1: a is even b is even
Case 2: a is even b is odd
Case 3: a is odd b is even
Case 4: a is odd b is odd
rewriting the f(a) and f(b) in each case:
Case 1: \(f(a)= a-1\) and \(f(b)= b-1\) so .... \(f(a)+f(b)= a+b-2\)
Case 2: \(f(a)= a-1\) and \(f(b)= b+1\) so .... \(f(a)+f(b)= a+b\)
Case 3: \(f(a)= a+1\) and \(f(b)= b-1\) so .... \(f(a)+f(b)= a+b\)
Case 4: \(f(a)= a+1\) and \(f(b)= b+1\) so .... \(f(a)+f(b)= a+b+2\)
\(f(a)+f(b)= a+b\) holds only when either a is odd and b is even or vice versa.
Hence option C is correct!
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