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\(\text{, }\!\!~\!\!\text{ }\frac{dy}{dx}=f\left( x \right)\text{ }\!\!~\!\!\)with y(0) = 1, then \(y\left( \frac{1}{4} \right)+y\left( \frac{3}{4} \right)\) is equal to:

Let f: [0, 1] → R be such that f(xy) = f(x)∙f(y) for all x, y ∈

1.	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\(\text{, }\!\!~\!\!\text{ }\frac{dy}{dx}=f\left( x \right)\text{ }\!\!~\!\!\)with y(0) = 1, then \(y\left( \frac{1}{4} \right)+y\left( \frac{3}{4} \right)\) is equal to:
A.	3
B.	4
C.	2
D.	5
Answer» B. 4

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