The question tests the conceptual understanding of the cases where \(base^{exponent} = 1\). A number raised to a power( say \(p^q)\) is equal to 1 in the following cases:

Case-I: \(p = 1\) and \(q\) is any number1 raised to any power will be equal to 1.

Case-II: \(p = -1\) and \(q\) is any even number-1 raised to an even power will be equal to 1

Case-III: \(p\) is any non-zero number and \(q = 0\)Any non-zero number raised to power zero will be equal to 1

With this understanding let's analyze the statements.

Analyze Statement-I independentlySt-I tells us that \(a^{x+y} = 1\). So, there are three possible cases:

i. \(a = 1\) and \(x+y\) is either even/odd

ii. \(a = -1\) and \(x+y\) is even

iii. \(a\) is any number and \(x+y =0\) which is even

Thus, statement-I does note tell us for sure if \(x+y\) is even or odd

Analyze Statement-II independentlySt-II tells us that \(b^{x+y} = 1\). So, there are three possible cases:

i. \(b = 1\) and \(x+y\) is either even/odd

ii. \(b = -1\) and \(x+y\) is even

iii. \(b\) is any number and \(x+y =0\) which is even

Thus, statement-II does note tell us for sure if \(x+y\) is even or odd

Combining statements-I & IICombining st-I & II gives us two scenarios:

i. \(a,b\) are any numbers and \(x+y = 0\) which is an even

ii. \(a = 1\) and \(b = -1\) or vice versa. For both cases \(x+y\) will have to be an even for \(-1^{x+y} =1\). Since a, b are distinct numbers we can say that \(a\) and \(b\) both can't be equal to 1 at the same time.

Both the scenarios tell us that \(x+y\) will be an even. Thus, combination of both the statements is sufficient to give us an unique answer.

Answer: Option C

Takeaway\(p^q = 1\) is one of the special cases in algebra. It's important to consider all its possible scenarios to arrive at the right answer.Regards

Harsh

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