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(A) Given the equation $f(x) \cdot f(f(x)) = x^2 + 1$. Substitute $x = 1$ into the equation: $f(1) \cdot f(f(1)) = 1^2 + 1$ Since $f(1) = 2$,we have $

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$\frac{1}{k} - \frac{1}{2}$

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(A) Given the equation $f(x) \cdot f(f(x)) = x^2 + 1$. Substitute $x = 1$ into the equation: $f(1) \cdot f(f(1)) = 1^2 + 1$ Since $f(1) = 2$,we have $2 \cdot f(2) = 2$,which implies $f(2) = 1$. Differentiate both sides of the original equation with respect to $x$ using the product rule and chain rule: $f'(x) \cdot f(f(x)) + f(x) \cdot f'(f(x)) \cdot f'(x) = 2x$. Substitute $x = 1$ into the derivative equation: $f'(1) \cdot f(f(1)) + f(1) \cdot f'(f(1)) \cdot f'(1) = 2(1)$. Using $f(1) = 2$,$f'(1) = k$,and $f(2) = 1$: $k \cdot f(2) + 2 \cdot f'(2) \cdot k = 2$. Substitute $f(2) = 1$: $k(1) + 2k \cdot f'(2) = 2$. Solve for $f'(2)$: $2k \cdot f'(2) = 2 - k$ $f'(2) = \frac{2 - k}{2k} = \frac{2}{2k} - \frac{k}{2k} = \frac{1}{k} - \frac{1}{2}$.

If $f(x)$ is a continuous and differentiable function satisfying $f(x) \cdot f(f(x)) = x^2 + 1$,$f(1) = 2$,and $f'(1) = k$,then the value of $f'(2)$ is

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