If f (x) = K-2 ^ X / 1 + K2 ^ x (k is a constant) is an odd function in the domain of definition, then k =? The answer is 1 or - 1, | Math enthusiast | Mathematical art

If f (x) = K-2 ^ X / 1 + K2 ^ x (k is a constant) is an odd function in the domain of definition, then k =? The answer is 1 or - 1,

The solution is an odd function in the domain of definition by the function f (x) = K-2 ^ X / 1 + K2 ^ x (k is a constant),
Then f (- x) = - f (x)
That is, (K-2 ^ (- x)) / (1 + K2 ^ (- x)) = - (K-2 ^ x) / (1 + K2 ^ x)
So (K-2 ^ (- x)) (1 + K2 ^ x) = - (K-2 ^ x) (1 + K2 ^ (- x))
That is, K + K ^ 2 × 2 ^ X-2 ^ (- x) - K2 ^ x × 2 ^ (- x) = - (K + K ^ 2 × 2 ^ (- x) - 2 ^ x-k2 ^ x × 2 ^ (- x))
That is, K + K ^ 2 × 2 ^ X-2 ^ (- x) - k = - (K + K ^ 2 × 2 ^ (- x) - 2 ^ x-k)
That is, K ^ 2 × 2 ^ X-2 ^ (- x) = - (k ^ 2 × 2 ^ (- x) - 2 ^ x)
That is, K ^ 2 × 2 ^ X-2 ^ (- x) = - K ^ 2 × 2 ^ (- x) + 2 ^ X
That is, K ^ 2 (2 ^ x + 2 ^ (- x)) = 2 ^ (- x) + 2 ^ X
That is, K ^ 2 = 1
So k = ± 1

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