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(A) Let $f(x) = \frac{(1 + x)^{3/5}}{1 + x^{3/5}}$ for $x \in [0, 1]$.Taking the derivative $f'(x)$:$f'(x) = \frac{(1 + x^{3/5}) \cdot \frac{3}{5}(1 +

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$(2^{-0.4}, 1)$

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(A) Let $f(x) = \frac{(1 + x)^{3/5}}{1 + x^{3/5}}$ for $x \in [0, 1]$.Taking the derivative $f'(x)$:$f'(x) = \frac{(1 + x^{3/5}) \cdot \frac{3}{5}(1 + x)^{-2/5} - (1 + x)^{3/5} \cdot \frac{3}{5}x^{-2/5}}{(1 + x^{3/5})^2}$$f'(x) = \frac{3}{5} \left[ \frac{1 + x^{3/5}}{(1 + x)^{2/5}} - \frac{(1 + x)^{3/5}}{x^{2/5}} \right]$$f'(x) = \frac{3}{5} \left[ \frac{x^{2/5}(1 + x^{3/5}) - (1 + x)^{3/5}(1 + x)^{2/5}}{x^{2/5}(1 + x)^{2/5}} \right]$$f'(x) = \frac{3}{5} \left[ \frac{x^{2/5} + x - (1 + x)}{x^{2/5}(1 + x)^{2/5}} \right] = \frac{3}{5} \left[ \frac{x^{2/5} - 1}{x^{2/5}(1 + x)^{2/5}} \right]$Since $x \in [0, 1]$,$x^{2/5} \le 1$,so $f'(x) \le 0$. The function is monotonically decreasing.Thus,the maximum value $K = f(0) = \frac{(1+0)^{0.6}}{1+0^{0.6}} = 1$.The minimum value $k = f(1) = \frac{(1+1)^{0.6}}{1+1^{0.6}} = \frac{2^{0.6}}{2} = 2^{0.6 - 1} = 2^{-0.4}$.Therefore,the ordered pair $(k, K)$ is $(2^{-0.4}, 1)$.

Let $k$ and $K$ be the minimum and the maximum values of the function $f(x) = \frac{(1 + x)^{0.6}}{1 + x^{0.6}}$ in $[0, 1]$ respectively,then the ordered pair $(k, K)$ is equal to

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