answer:(I) From f(-2)=f(0)=-3, and for any real number x, f(x) geqslant -4, we can deduce that the axis of symmetry is x=-1, and the minimum value is -4. Let's assume f(x)=a(x+1)^{2}-4. From f(0)=a-4=-3, we solve for a=1. Therefore, f(x)=(x+1)^{2}-4=x^{2}+2x-3. (II) 1. From the problem, we have g(x)=m(x+1)^{2}-4m+1. Given m < 0, the axis of symmetry is x=-1 < 1. Hence, g(x) is monotonically increasing in (-infty, -1] and monotonically decreasing in (-1, 1]. Since g(1)=1 > 0 and g(-1)=1-4m > 0, g(x) has no zeros in (-1, 1] and exactly one zero in (-infty, -1]. Therefore, g(x) has exactly one zero in (-infty, 1]. 2. We have g(-1)=1-4m, g(-3)=1, and g(frac{3}{2}) = frac{9}{4}m+1. Given m > 0, we have g(frac{3}{2}) > g(3). - When 1-4m geqslant 0, i.e., m leqslant frac{1}{4}, y_{max}=|g(x)|_{max}=g(frac{3}{2}) = frac{9}{4}m + 1. - When 1-4m < 0, i.e., m > frac{1}{4}, there are two cases: a. If frac{1}{4} < m leqslant frac{8}{7}, then y_{max}=|g(x)|_{max}=g(frac{3}{2}) = frac{9}{4}m + 1. b. If m > frac{8}{7}, then y_{max}=|g(x)|_{max}=|g(-1)|=4m-1. In summary, when 0 < m leqslant frac{8}{7}, boxed{y_{max} = frac{9}{4}m + 1}, and when m > frac{8}{7}, boxed{y_{max} = 4m - 1}.

answer:To find the mode and median of the number of students in each study group in Class 9-1, we first need to determine the value of x. We know that the average number of students per group is 6, and there are seven groups in total. The numbers of students in the groups, including x, are 5, 5, 6, x, 6, 7, 8. The formula for the average is the sum of the values divided by the number of values. Given that the average number of students per group is 6, we can set up the equation as follows: [ frac{5 + 5 + 6 + x + 6 + 7 + 8}{7} = 6 ] Multiplying both sides by 7 to clear the denominator, we get: [ 5 + 5 + 6 + x + 6 + 7 + 8 = 6 times 7 ] Simplifying the right side of the equation: [ 5 + 5 + 6 + x + 6 + 7 + 8 = 42 ] Adding the known numbers: [ 37 + x = 42 ] Solving for x: [ x = 42 - 37 = 5 ] Now that we know x = 5, we can list all the numbers of students in the groups in ascending order: 5, 5, 5, 6, 6, 7, 8. The mode is the number that appears most frequently, which in this case is 5. The median is the middle number when the numbers are listed in order. With seven numbers, the fourth number is the median, which is 6. Therefore, the mode is 5 and the median is 6. Hence, the correct answer is boxed{text{A: } 5, 6}.

answer:(1) Proof: Since f(x) = x^2 + bx is an even function, we have f(-x) = f(x), thus b = 0. Given a_{n+1} = 2f(a_n - 1) + 1, we have a_{n+1} - 1 = 2(a_n - 1)^2. Since b_n = log_2(a_n - 1), we get b_{n+1} = 1 + 2b_n, thus b_{n+1} + 1 = 2(b_n + 1). Therefore, the sequence {b_{n}+1} is a geometric sequence with the first term as 2 and the common ratio as 2. (2) Solution: From (1), we have b_{n+1} = 2^n, thus b_n = 2^n - 1. Therefore, c_n = nb_n = n(2^n - 1), thus S_n = 1cdot2 + 2cdot2^2 + ldots + ncdot2^n - frac{n(n+1)}{2}. Let T = 1cdot2 + 2cdot2^2 + ldots + ncdot2^n, 2T = 1cdot2^2 + 2cdot2^3 + ldots + (n-1)cdot2^n + ncdot2^{n+1}. Subtracting the two equations, we get -T = 2 + 2^2 + 2^3 + ldots + 2^n - ncdot2^{n+1} = (1-n)cdot2^{n+1} - 2. Therefore, T = (n-1)cdot2^{n+1} + 2, thus S_n = (n-1)cdot2^{n+1} + 2 - frac{n(n+1)}{2}. So, the final answer is boxed{S_n = (n-1)cdot2^{n+1} + 2 - frac{n(n+1)}{2}}.

answer:: We are given the following conditions and need to find the times at which the motorcyclist and the bicyclist are at specific distances from the intersection. 1. **Intersection Angle and Speeds:** - The paths of the motorcyclist and the bicyclist intersect at an angle of 60^circ. - The bicyclist is traveling at a speed of 36 text{ km/h}, which is half the speed of the motorcyclist. Therefore, the motorcyclist's speed is 72 text{ km/h}. 2. **Time Variables:** - Let's begin the time count t from 13:00. Negative values of t would correspond to times before 13:00. 3. **Position Equations:** - The motorcyclist's distance from the intersection at time t is given by |72t|. - The bicyclist's distance from the intersection at time t is given by |36(t-1)|. 4. **Applying the Law of Cosines:** Using the Law of Cosines for the triangle formed by their distances from the intersection and the included angle of 60^circ, we write: [ |AB|^2 = 72^2 t^2 + 36^2 (t-1)^2 - 2 cdot 72 cdot 36 |t| |t-1| cos(60^circ) ] Since cos(60^circ) = frac{1}{2}, this simplifies to: [ 72^2 t^2 + 36^2 (t-1)^2 - 2 cdot 72 cdot 36 |t| |t-1| cdot frac{1}{2} = 252^2 ] 5. **Simplifying the Equation:** Simplify each term and factor out common terms: [ 72^2 t^2 + 36^2(t-1)^2 - 72 cdot 36 |t| |t-1| = 252^2 ] [ 5184 t^2 + 1296 (t^2 - 2t +1) - 2592 |t| |t-1| = 63504 ] 6. **Dividing by 5184 to simplify:** Notice that the common factor across all terms is manageable by dividing: [ t^2 + frac{1}{4}(t^2 - 2t + 1) - frac{1}{2} |t| |t-1| = 12.25 ] Simplifying further: [ t^2 + frac{1}{4}t^2 - frac{1}{2}t + frac{1}{4} - frac{1}{2} |t| |t-1| = 12.25 ] Multiplying through by 4 to clear the fraction: [ 4t^2 + t^2 - 2t + 1 - 2 |t| |t-1| = 49 ] 7. **Isolating ( t^2 ):** The simplified form is: [ 5t^2 - 2t + 1 - 2 |t| |t-1| = 49 ] Given that this simplifies into a workable form: [ 4t^2 + t^2 - 2t + 1 - 2t^2 + 2t = 49 ] Solving for ( t ) we get: [ 4t^2 + t^2 - 2t + 1 = 49 ] [ t^2 = 16 implies t = pm 4 ] 8. **Conclusion:** - Calculating the times using these ( t ) values: - When ( t = 4 ): [ 13:00 + 4text{ hours} = 17:00 ] - When ( t = -4 ): [ 13:00 - 4text{ hours} = 09:00 ] Hence, the correct times are: (boxed{09:00 text{ and } 17:00})