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For all integers x, the function f is defined as follows

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For all integers x, the function f is defined as follows [#permalink] New post 02 Mar 2018, 16:18
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80% (01:49) correct 20% (00:00) wrong based on 5 sessions
For all integers \(x\), the function \(f\) is defined as follows

\(\mathcal{f} (x) =\) \(\left\{ \begin{array}{rcl}{x - 1} & \mbox{if x is even}\\{x + 1} & \mbox{if x is odd}\end{array}\right\)

If \(a\) and \(b\) are integers and \(f (a) + f (b) = a + b\), which of the following statements must be true?

A a = b

B a = -b

C a + b is odd.

D Both a and b are even.

E Both a and b are odd.
[Reveal] Spoiler: OA

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Re: For all integers x, the function f is defined as follows [#permalink] New post 03 Mar 2018, 04:17
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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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Re: For all integers x, the function f is defined as follows [#permalink] New post 09 May 2018, 07:00
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basically the question is testing us to check if we can see that function is converting an even number to odd and an odd number to even and then the summation of the function output.

which is odd + even = Odd hence the answer is C - a+b is odd.
Re: For all integers x, the function f is defined as follows   [#permalink] 09 May 2018, 07:00
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