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Given that \(A \text{@}B = A\) if A is greater than B else \(A \text{@}B= B\) and \(A \% B = AB\) if A x B is positive else \(A \% B = A\) and we need to find the value of \(\frac{1\text{@}-1}{(-K)\text{@}(-K)}\%K\)

Lets start by finding the value of \(1\text{@}-1\)

To find \(1\text{@}-1\) we need to compare what is before and after \(\text{@}\) in \(1\text{@}-1\) and \(A\text{@}B\)

=> We need to substitute A with 1 and B with -1 in \(A\text{@}B\) to get the value of \(1\text{@}-1\)
Since A(1) > B(-1)
=> \(1\text{@}-1\) = 1

=> In case of \((-K)\text{@}(-K)\) A = B = -K => A is NOT greater than B
=> \((-K)\text{@}(-K)\) = -K

=> \(\frac{1\text{@}-1}{(-K)\text{@}(-K)}\%K\) = \(\frac{1}{-K}\%K\)

Now, \(\frac{-1}{K}\) * K = -1

=> \(\frac{1}{-K}\%K\) = \(\frac{-1}{K}\)

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

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Bunuel : I think the question needs to be changed to find the value of \(\frac{1\text{@}-1}{(-K)\text{@}(-K)}\%K\) and NOT find the value of K.

Please check, confirm and update if required.
Thank you.

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