answer:To determine the periodicity of the function ( y = f_1(x) + f_2(x) ), we start from the given properties of ( f_1(x) ) and ( f_2(x) ): 1. Given that ( f_1(x) ) and ( f_2(x) ) are periodic functions with the same smallest positive period ( T ), we have: [ f_1(x+T) = f_1(x) quad text{and} quad f_2(x+T) = f_2(x) ] 2. For the function ( y = f_1(x) + f_2(x) ), we check its periodicity: [ y(x+T) = f_1(x+T) + f_2(x+T) ] Since both ( f_1 ) and ( f_2 ) have period ( T ), we substitute: [ y(x+T) = f_1(x) + f_2(x) = y(x) ] Therefore, ( y = f_1(x) + f_2(x) ) is a periodic function with period ( T ). 3. Next, we must consider whether ( y ) has a minimum period ( T ), or could it be that ( y ) might not have a smallest positive period at all. 4. To illustrate this, let's consider specific examples: - Let ( f_1(x) = sin x ) and ( f_2(x) = -sin x ): [ y = sin x + (-sin x) = 0 ] The function ( y(x) = 0 ) is periodic for any non-zero period, meaning there is no smallest positive period. Hence, ( y ) does not have a smallest positive period, demonstrating it could be a periodic function but without a smallest positive period. 5. Let's also analyze other scenarios to understand further. Consider the case where ( f(x) = sin x ) and ( g(x) = sin(pi x) ): - The function ( sin x + 1 ) is a periodic function, while ( sin x + sin(pi x) ) is not periodic due to the irrational period ratio. Therefore, the function ( y = f_1(x) + f_2(x) ) is concluded to be periodic but might not necessarily have a smallest positive period, based on the given examples and reasoning. **Conclusion:** [ boxed{text{C}} ]

answer:1. Consider the sequence of numbers ( a_1, a_2, ldots, a_n ). 2. Each number ( a_i ) will be underlined if: - ( a_i ) is positive, or - The sum of ( a_i ) and the next ( k ) consecutive numbers ( a_{i + 1}, a_{i + 2}, ldots, a_{i + k} ) is positive for the smallest such ( k ). 3. Let's denote by ( k_i ) the smallest non-negative integer such that [ a_i + a_{i+1} + cdots + a_{i+k_i} > 0 ] Then, ( a_i ) is underlined ( k_i + 1 ) times. 4. If ( a_i ) is underlined ( k_i + 1 ) times, then the numbers ( a_{i+1}, a_{i+2}, ldots, a_{i+k_i} ) are also underlined, but less frequently: - ( a_{i+1} ) is underlined ( k_i ) times, - ( a_{i+2} ) is underlined ( k_i - 1 ) times, and so on. 5. To ensure clarity, group the numbers that have the same underlining condition (i.e., grouped by the maximum value ( K ) of their underlines): - Begin with numbers underlined by ( K ). - Follow with those underlined by ( K-1 ), and so on. 6. By grouping these numbers and noting that the sum of the numbers in each group is guaranteed to be positive because: [ a_i + a_{i+1} + cdots + a_{i+k_i} > 0 ] for each group, the overall sum of all underlined numbers is positive. This fulfills the problem's conditions. 7. Conclude that the sum of all underlined numbers is positive. [ boxed{text{The sum of all underlined numbers is positive.}} ]

answer:To identify the real and imaginary parts of the complex number (1+sqrt{3})i, we can express it in the standard form of a complex number, which is a + bi, where a is the real part and b is the imaginary part. Given: (1+sqrt{3})i Since there is no real number outside the imaginary unit i, the real part, a, is 0. The coefficient of i gives us the imaginary part, b, which is 1+sqrt{3}. Therefore, the real part is 0 and the imaginary part is 1+sqrt{3}. This corresponds to option C: 0, 1+sqrt{3}. Hence, the correct answer is boxed{C}.

answer:We rewrite the function g(x) as follows: begin{align*} g(x) &= x + frac{2x}{x^2 + 1} + frac{x(x + 3)}{x^2 + 3} + frac{3(x + 1)}{x(x^2 + 3)} &= frac{x(x^2 + 1) + 2x}{x^2 + 1} + frac{x^2(x + 3)}{x(x^2 + 3)} + frac{3(x + 1)}{x(x^2 + 3)} &= frac{x^3 + 3x}{x^2 + 1} + frac{x^3 + 3x^2 + 3x + 3}{x(x^2 + 3)} &= frac{x(x^2 + 3)}{x^2 + 1} + frac{3x^2 + 3}{x(x^2 + 3)} + frac{x(x^2 + 3)}{x(x^2 + 3)} &= frac{x(x^2 + 3)}{x^2 + 1} + 3 cdot frac{x^2 + 1}{x(x^2 + 3)} + 1. end{align*} Apply AM-GM inequality: [frac{x(x^2 + 3)}{x^2 + 1} + 3 cdot frac{x^2 + 1}{x(x^2 + 3)} geq 2 sqrt{frac{x(x^2 + 3)}{x^2 + 1} cdot 3 cdot frac{x^2 + 1}{x(x^2 + 3)}} = 6,] Therefore, g(x) geq 7. Equality occurs when [frac{x(x^2 + 3)}{x^2 + 1} = 3,] or x(x^2 + 3) = 3x^2 + 3. Simplifying, x^3 - 3x^2 + 3x - 3 = 0. Let h(x) = x^3 - 3x^2 + 3x - 3. By the Intermediate Value Theorem, since h(1) < 0 and h(2) > 0, there exists a root between 1 and 2. Thus, the minimum value of g(x) for x > 0 is boxed{7}.