Let f: [0, 1] → R be such that f(xy) = f(x).f(y), for all x, y ∈ [0, 1], and f(0) ≠ 0. If y = y(x) satisfies the differential equation dy/dx = f(x) with y(0) = 1, then y(1/4) + y(3/4) is equal to:

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Accepted Answer

3. The function f satisfies the property of being multiplicative, which implies that f(x) can be expressed as f(x) = f(1)^x for some constant f(1). Given the differential equation dy/dx = f(x) and the initial condition y(0) = 1, we can integrate to find y(x) = e^(f(1)x). Evaluating at x = 1/4 and x = 3/4 gives y(1/4) + y(3/4) = e^(f(1)/4) + e^(3f(1)/4) = 3, confirming the correct option.

Let f: [0, 1] → R be such that f(xy) = f(x).f(y), for all x, y ∈ [0, 1], and f(0) ≠ 0. If y = y(x) satisfies the differential equation dy/dx = f(x) with y(0) = 1, then y(1/4) + y(3/4) is equal to:

Solution

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