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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, dx dy = f(x) with y(0) = 1, then y ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 4 1 + y ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 4 3 is equal to :

1. 2
2. 3
3. 5
4. 4

Correct answer: 3

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