answer:To solve the given system of inequalities for x and determine the range of a that allows for 5 integer solutions, we proceed as follows: 1. **Solve the first inequality:** From the first inequality 5-2x geq -1, we isolate x: begin{align*} 5 - 2x &geq -1 -2x &geq -6 x &leq 3 end{align*} 2. **Solve the second inequality:** From the second inequality x - a > 0, we isolate x: begin{align*} x &> a end{align*} 3. **Combine the solutions:** Combining the solutions from the two inequalities, we get a < x leqslant 3. 4. **Determine the integer solutions:** Given that there are 5 integer solutions, we list the integers that satisfy x leqslant 3: -1, 0, 1, 2, 3. These are the 5 integer values that x can take. 5. **Find the range of a:** For x to take the values -1, 0, 1, 2, 3, a must be less than -1 (to allow x = -1 as a solution) but cannot be equal to or less than -2 (as that would allow x = -2 as a solution, increasing the total number of integer solutions to 6). Therefore, the range of a is -2 leqslant a < -1. Hence, the final answer, encapsulating the range of a, is boxed{-2 leqslant a < -1}.

answer:We have g'(x) = 4x^2 + 4x - 3 = (2x - 1)(2x + 3). Let g'(x) = 0, we get x = frac{1}{2} or x = -frac{3}{2}. According to the problem, we know a = frac{1}{2}. Hence, the function f(x) can be written as: f(x) = begin{cases} (frac{1}{2})^x, & text{if } x < 0 log_{frac{1}{2}}{x}, & text{if } x geq 0 end{cases} So, f(frac{1}{4}) + f(log_2{frac{1}{6}}) = log_{frac{1}{2}}{frac{1}{4}} + (frac{1}{2})^{log_2{frac{1}{6}}} = 2 + 2^{log_2{6}} = 2 + 6 = 8. Therefore, the answer is boxed{8}. Let g'(x) = 0 to find the extreme value point, and according to the problem, we can find the value of a, and then get the analytical expression of the function f(x). Finally, use the properties of logarithms to find the answer. This problem tests the basic skills of finding extreme values of functions using derivatives, evaluating piecewise functions, and the properties of logarithms.

answer:Since y=f(x+1) is an even function, the graph of y=f(x+1) is symmetric about the y-axis. By shifting the graph of y=f(x+1) to the right by 1 unit, we obtain the graph of y=f(x). Therefore, the graph of y=f(x) is symmetric about x=1, hence the correct answer is boxed{C}.

answer:1. **Assumption:** Suppose that the triangle A_1 B_1 C_1 is equilateral. 2. **Analysis of Point A:** First, let's consider if point A lies on the segment B_1 C_1. - Given that angle C_1 B_1 A_1 = angle A B_1 C = 60^circ, this implies that triangle A_1 B_1 C_1 is not equilateral since one side of an equilateral triangle cannot lie on any segment formed by the vertices of a properly formed triangle. - Hence, if angle C_1 B_1 A_1 = 60^circ, then angle B A C = 180^circ - angle B A C_1 - angle C A B_1 = 60^circ. - By this method, we also get angle A C B = 60^circ. - Thus, triangle A B C should be equilateral which is a contradiction to the given condition that its sides are pairwise different. 3. **Conclusion:** Therefore, point A does not lie on the segment B_1 C_1. 4. **Investigation with External and Internal Triangles:** We need to consider the triangles A_1 B C_1, B_1 C A_1, and C_1 A B_1. - We define an external triangle as any triangle intersecting with triangle ABC solely at the corresponding vertex. 5. **Detailed Description:** - If we denote a triangle as external if it intersects triangle ABC in the way depicted in the left illustration (as shown, external triangles are A_1BC_1 and C_1AB_1). - Alternatively, an internal triangle is defined when it does not meet the prior condition. 6. **Critical Observation:** - At least one vertex of triangle A_1, B_1, C_1 will be adjacent to either two external triangles (left illustration) or two internal triangles (right illustration). - In such a scenario, the corresponding angle for triangle A_1 B_1 C_1 in the external case will be greater than 60^circ while in the internal case it will be less than 60^circ. 7. **Conclusion and Contradiction:** - Setting this up for each vertex, an angle mismatch for an equilateral triangle structure leads to a contradiction in each instance. Hence, the triangle A_1 B_1 C_1 cannot be equilateral. Therefore, our initial assumption is incorrect. [ boxed{} ]