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 A Further Function Puzzle (Posted on 2007-01-19)
Define f(x+y) = f(x) + f(y) - 1 for all real x and y such that f(x) is differentiable for all values of x with f ’(0) = cos A.

Determine whether or not the value of f(1) – cos A is equal to 2.

 Submitted by K Sengupta Rating: 2.0000 (1 votes) Solution: (Hide) For x=y=0; we obtain: f(0) = 2f(0) – 1; giving f(0) = 1……(i) Now: f’(x) = Limit (h->0) [f(x+h) – f(x)]/h = Limit (h->0) [f(x) + f(h) +2xh-1 – f(x)]/h = 2x + Limit (h->0) [f(h) – f(0)]/h = f’(0); giving: f(x) = x cosA + C; where C is a constant. Substituting x=0, we observe from (i) that: C = f(0), giving: f(x) = x cosA + f(0) Accordingly, f(1) = 1 + cosA, so that: f(1) – cos A = 1. Consequently, the value of f(1) – cos A cannot be equal to 2.----------------------------------------------------------- Also refer to the solution posted by Larry in this location.

 Subject Author Date A solution Larry 2007-01-19 22:53:11

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