answer:1. **Define the Rational Number Representation in Continued Fractions:** Any rational number ( x ) can be uniquely written as a finite continued fraction: [ x = a_{0} + frac{1}{a_{1} + frac{1}{a_{2} + ldots + frac{1}{a_{n}}}} ] where ( a_{0} ) is an integer, and ( a_{1}, a_{2}, ldots, a_{n} ) are positive integers with ( a_{n} > 1 ). This representation is obtained through a simple greedy algorithm where ( a_0 ) is the integer part of ( x ), and the continued fraction for (frac{1}{x - a_0} ) is further decomposed. 2. **Define the Function (ell(x)):** Let (ell(x)) be the number of fraction bars in the continued fraction representation of (x). According to the rules of continued fraction formation: - (ell(x) = 0) if (x) is an integer. - (ell(x) = ell(frac{1}{x}) + 1) if (0 < x < 1). 3. **Definition of the Function (f):** Define (f) on the set of rational numbers as follows: - ( f(0) = -1 ) - ( f(x) = (-1)^{ell(x)} ) if (x > 0) - ( f(x) = -(-1)^{ell(|x|)} ) if (x < 0 ) 4. **Verifying the Property (fleft(frac{1}{x}right) = -f(x)):** - Consider positive (x) such that (x neq 0, pm 1). - For (0 < x < 1), we use the relationship (ell(x) = ellleft(frac{1}{x}right) + 1): [ fleft(frac{1}{x}right) = (-1)^{ell(frac{1}{x})} = (-1)^{ell(x) - 1} = -(-1)^{ell(x)} = -f(x) ] - Since (f(-x) = -f(x)) and (fleft(-frac{1}{x}right) = -fleft(frac{1}{x}right)), it suffices to verify for positive (x). 5. **Verifying the Property (f(x + k) = f(x)):** If (x) is positive and (k) is a positive integer: - Since adding an integer to (x) does not change the fraction part: [ ell(x + k) = ell(x) quad text{and hence} quad f(x + k) = f(x) ] 6. **Verifying (f(1 - x) = -f(x)) for (x neq frac{1}{2}):** Assume symmetry and consider only (x > frac{1}{2}): - For (x > 1): [ f(x) = f(x - 1) = -f(1 - x) ] - For (x = 1): [ f(1) = 1 quad text{and} quad f(1 - 1) = f(0) = -1 ] - For (frac{1}{2} < x < 1), assume (frac{1}{x} = y) where (y > 1): [ f(x) = -fleft( frac{1}{x} right) = -f(y) = -(-f(1 - y)) = f(1 - x) ] 7. **Conclusion:** The function (f) fulfills the given conditions, ensuring that (f(x) = -f(y)) whenever (x) and (y) are distinct rational numbers with (x y = 1) or (x + y in {0, 1}). Therefore, such a function does exist and the construction of (f) meets all the criteria. (blacksquare)

answer:Solution: Given f(x) = 2cos^2x + 2sqrt{3}sin xcos x + m, we have f(x) = cos 2x + sqrt{3}sin 2x + m + 1, which simplifies to f(x) = 2sin(2x + frac{pi}{6}) + m + 1. By the period formula T = frac{2pi}{omega} = pi, let frac{pi}{2} + 2kpi leq 2x + frac{pi}{6} leq frac{3pi}{2} + 2kpi, (k in mathbb{Z}), we get frac{pi}{6} + kpi leq x leq frac{2pi}{3} + kpi, (k in mathbb{Z}), the intervals of monotonic decrease for the function f(x) are left[frac{pi}{6} + kpi, frac{2pi}{3} + kpiright], (k in mathbb{Z}); (2) For x in left[0, frac{pi}{2}right], then 2x + frac{pi}{6} in left[frac{pi}{6}, frac{7pi}{6}right], sin(2x + frac{pi}{6}) in left[frac{1}{2}, 1right], thus f(x) in [m, m+3], therefore, there exists m satisfying the condition, which is begin{cases} m= frac{1}{2} m+3= frac{7}{2}end{cases}, solving this gives: m= frac{1}{2}, hence, there exists m= frac{1}{2}, making the range of the function f(x) exactly left[frac{1}{2}, frac{7}{2}right]. - The smallest positive period of the function f(x) is boxed{pi}. - The intervals of monotonic decrease for the function f(x) are boxed{left[frac{pi}{6} + kpi, frac{2pi}{3} + kpiright]}, (k in mathbb{Z}). - There exists m= boxed{frac{1}{2}}, making the range of the function f(x) exactly left[frac{1}{2}, frac{7}{2}right].

answer:The biologist captures 90 dolphins on June 1st, 50% of which were not present on January 1st, leaving 45 dolphins that were present originally. Thus, the proportion of tagged dolphins in June should be equal to the proportion of tagged dolphins from January, resulting in the equation (frac{4}{45} = frac{100}{x}). Solving for (x), the number of dolphins on January 1st is calculated as: (frac{4}{45} = frac{100}{x}) (4x = 4500) (x = 1125) So the biologist estimates there were (boxed{1125}) dolphins initially.

answer:The cosine function equals 0 at certain standard angles in the unit circle. Specifically, cos frac{pi}{2} = 0. Therefore, when seeking the angle whose cosine is 0 within the range of the arccos function (which is from 0 to pi radians), we find that: [ arccos 0 = frac{pi}{2} ] Thus, the solution to arccos 0 is boxed{frac{pi}{2}}.