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| A Further Function Puzzle (Posted on 2007-01-19) |
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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.
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Submitted by K Sengupta
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Rating: 3.5000 (2 votes)
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Solution:
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(Hide)
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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.
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Also refer to the solution posted by Larry in this location.
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Subject |
Author |
Date |
 | A solution | Larry | 2007-01-19 22:53:11 |
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