answer:(1) Since f(x) = x^2 - (2t + 1)x + t ln x where x > 0 and t = 1, we have f'(x) = 2x - 3 + frac{1}{x} = frac{(2x^2 - 3x + 1)}{x} = frac{(2x - 1)(x - 1)}{x}. If we set f'(x) > 0, we solve to find 0 < x < frac{1}{2} or x > 1. If we set f'(x) < 0, we solve to find frac{1}{2} < x < 1. Therefore, f(x) is increasing in (0, frac{1}{2}) and (1, +infty), and decreasing in (frac{1}{2}, 1). Thus, the maximum value of f(x) is f(frac{1}{2}) = - frac{5}{4} - ln 2, and the minimum value is f(1) = -2. (2) For g(x) = (1 - t)x, there exists an x_0 in [1, e] such that f(x_0) geq g(x_0) holds if and only if there exists an x in [1, e] such that x^2 - (2t + 1)x + tln x geq (1 - t)x. Since x in [1, e], we have x - ln x > 0. Thus, the original problem is equivalent to finding t leq frac{x^2 - 2x}{x - ln x} for some x in [1, e]. Let h(x) = frac{x^2 - 2x}{x - ln x}, h'(x) = frac{(x - 1)(x + 2 - 2ln x)}{(x - ln x)^2}. As x in [1, e], we have x + 2 - ln x > 0, and therefore h'(x) geq 0 for all x in [1, e], meaning that h(x) is increasing on the interval [1, e]. The maximum value of h(x) is then h(e) = frac{e^2 - 2e}{e - 1}. Hence, the maximum value of t is boxed{frac{e(e - 2)}{e - 1}}.

answer:Solution: From a grade of 1,000 students, 125 students were selected for a statistical analysis of their weight. In option (A), the weight of the 1,000 students is the population, so A is incorrect; In option (B), the weight of each surveyed student is an individual, so B is incorrect; In option (C), the weight of the 125 surveyed students is a sample, so C is correct; In option (D), 125 is the sample size, so D is incorrect. Therefore, the correct answer is boxed{C}. This problem is solved directly using the definitions of population, individual, sample, and sample size. It tests the ability to judge the truthfulness of statements, understanding of the definitions of population, individual, sample, and sample size, as well as computational, data processing abilities, and the concept of functions and equations. It is a basic question.

answer:Let's denote the sum of money as S, which is Rs. 246. Let's assume A gets x paisa, B gets 65 paisa, and C gets 40 paisa for each rupee. Since the total sum of money is Rs. 246, we can write the following equation based on the distribution: A's share + B's share + C's share = Total sum (x paisa + 65 paisa + 40 paisa) * 246 = 246 * 100 paisa (since 1 rupee = 100 paisa) Now, we can simplify the equation: (x + 65 + 40) * 246 = 24600 paisa (x + 105) * 246 = 24600 paisa We need to find C's share, which is 40 paisa for each rupee. So, C's share in paisa is: C's share = 40 paisa * 246 Now, we can convert paisa to rupees by dividing by 100 (since 1 rupee = 100 paisa): C's share in rupees = (40 * 246) / 100 C's share in rupees = 9840 / 100 C's share in rupees = Rs. 98.40 Therefore, C's share is Rs. boxed{98.40} .

answer:Given a positive integer ( n ), we wish to find the smallest number ( k ) such that the masses ( 1, 2, ldots, n ) grams can be measured using ( k ) weights and a balance scale. The goal is to determine the minimum value of ( k ), denoted as ( f(n) ). 1. **Introduction of Variables**: Let the masses of the ( k ) weights be ( a_1, a_2, ldots, a_k ) grams, where ( 1 leq a_1 leq a_2 leq cdots leq a_k ) and ( a_i in mathbf{N}_+ ) (i.e., ( a_i ) are positive integers). 2. **Balance Scale Measurement**: Using the balance scale, we can write any measured mass as: [ sum_{i=1}^{k} x_i a_i, ] where each ( x_i ) can take the value ( -1 ), ( 0 ), or ( 1 ), indicating that the weight ( a_i ) is on the left side, not used, or on the right side of the scale, respectively. 3. **Coverage of Weights**: To measure masses ( 1, 2, ldots, n ) grams, the set of possible sums must cover all ( {-n, ldots, -1, 0, 1, ldots, n} ): [ left{ sum_{i=1}^{k} x_i a_i mid x_i in {-1, 0, 1} right} supseteq {0, pm 1, pm 2, ldots, pm n}. ] 4. **Size of the Set**: The size of this set is: [ 2n + 1 quad text{which is at most} quad 3^k, ] since there are ( 3^k ) possible combinations of ( x_i ). Thus, we get the inequality: [ 3^k > 2n. ] 5. **Minimum ( k )**: Solving for ( k ), we get: [ k > log_3 (2n). ] Therefore: [ k geq lceil log_3 (2n) rceil. ] We can refine this to: [ k geq lfloor log_3 (2n) rfloor + 1. ] 6. **Upper Bound Verification**: For any positive integer ( n ), there exists ( k ) such that: [ 3^k > 2n geq 3^{k-1}, ] giving: [ log_3 (2n) < k < log_3 (2n) + 1 implies k = lfloor log_3 (2n) rfloor + 1. ] Let ( a_1 = 1 ), ( a_2 = 3 ), (ldots), ( a_k = 3^{k-1} ). By representing numbers in base-3, any integer ( p ) in ( [0, 3^k - 1] ) can be written as: [ p = sum_{i=1}^{k} y_i a_i quad text{where} quad y_i in {0, 1, 2}. ] By adjusting ( y_i ) to ( x_i = y_i - 1 , (text{with} , x_i in {-1, 0, 1}) ), we map: [ -frac{3^k - 1}{2} leq p - frac{3^k - 1}{2} leq frac{3^k - 1}{2}. ] For measuring ( n ) grams: [ 2n < 3^k implies 2n leq 3^k - 1 implies n leq frac{3^k - 1}{2}. ] Hence, we can measure any mass ( l in {-n, ldots, n} ) using the weights ( a_1, a_2, cdots, a_k ). # Conclusion: Therefore, the smallest value of ( k ) is: [ f(n) = lfloor log_3 (2n) rfloor + 1. ] boxed{lfloor log_3 (2n) rfloor + 1}