Let's analyze the given property. We have a function \(f(x)\) that is positive, differentiable, and decreasing. When we draw a tangent to the graph of \(f(x)\) at a point \((x, f(x))\), the tangent intersects the x-axis at \((x - f(x), 0)\) and the y-axis at \((0, f(x))\). This forms a right triangle with the x-axis and y-axis, and the tangent line acts as the hypotenuse of the triangle.
Given that the length of chord \(MN\) (which is the hypotenuse of the right triangle) is constant, it means that the triangle's hypotenuse has a fixed length. Let's denote this constant length as \(k\).
Using the Pythagorean theorem for the right triangle, we have:
\[(x - f(x))^2 + f(x)^2 = k^2.\]
Simplify:
\[x^2 - 2xf(x) + f(x)^2 + f(x)^2 = k^2.\]
Combine like terms:
\[x^2 - 2xf(x) + 2f(x)^2 = k^2.\]
Now, we need to find a function \(f(x)\) that satisfies this equation for all \(x\) where the function is defined.
Notice that \(k^2\) is a constant, so the equation implies that \(x^2 - 2xf(x) + 2f(x)^2\) is also a constant. This means that the derivative of this expression with respect to \(x\) must be \(0\):
\[\frac{d}{dx} (x^2 - 2xf(x) + 2f(x)^2) = 0.\]
Simplify and differentiate:
\[2x - 2f(x) - 2xf'(x) + 4f(x)f'(x) = 0.\]
Factor out \(2\) and \(f'(x)\):
\[2(f(x) - x) + 2f'(x)(2f(x) - x) = 0.\]
Since \(f(x)\) is decreasing and positive, \(f'(x) < 0\), which means that the equation above holds if and only if both terms in the equation are \(0\):
\[f(x) - x = 0\]
\[2f(x) - x = 0.\]
Solve these two equations for \(f(x)\):
\[f(x) = x\]
\[f(x) = \frac{x}{2}.\]
Now, let's check if these solutions satisfy the given conditions.
For \(f(x) = x\), it's a linear function, which is both decreasing and increasing. This doesn't satisfy the requirement of being only decreasing.
For \(f(x) = \frac{x}{2}\), it is a decreasing function, and the length of the chord formed by the tangent is indeed constant (equal to \(\frac{k}{2}\)).
So, the only function that satisfies the given property is \(f(x) = \frac{x}{2}\).
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