answer:# Solution: Part (1): Given a=1, the function simplifies to f(x)=frac{e^x}{x}. To find if there's a tangent line parallel to the x-axis, we calculate the derivative of f(x): f'(x) = frac{d}{dx}left(frac{e^x}{x}right) = frac{e^x(x-1)}{x^2} For the tangent to be parallel to the x-axis, f'(x) = 0: frac{e^x(x-1)}{x^2} = 0 implies x = 1 At x=1, f(1) = frac{e^1}{1} = e. Therefore, there exists a tangent line at the point (1, e), and the equation of this tangent line is y = e. boxed{text{The equation of the tangent line is } y = e.} Part (2): For g(x) = f(x) - e(x - aln x) to have exactly two zeros in (0, +infty), we consider the equivalent condition for frac{e^x}{x^a} = e(ln e^x - ln x^a). This simplifies to: frac{e^x}{x^a} = elnfrac{e^x}{x^a} Letting t = frac{e^x}{x^a} > 0, we rewrite the equation as t = eln t. Define h(t) = eln t - t and find its derivative: h'(t) = frac{e}{t} - 1 Analyzing h'(t), we find: - For 0 < t < e, h'(t) > 0, indicating h(t) is increasing. - For t > e, h'(t) < 0, indicating h(t) is decreasing. Since h(e) = 0, the equation t = eln t has a unique solution at t = e. This implies that the condition for g(x) to have exactly two zeros translates to the equation aln x = x - 1 having two distinct real solutions. Define varphi(x) = aln x - x + 1 and find its derivative: varphi'(x) = frac{a}{x} - 1 Analyzing varphi'(x), we find: - For 0 < x < a, varphi'(x) > 0, indicating varphi(x) is increasing. - For x > a, varphi'(x) < 0, indicating varphi(x) is decreasing. Given the behavior of varphi(x) and the hints provided, for varphi(a) = aln a - a + 1 > 0 to hold, we analyze the inequality lna + frac{1}{a} - 1 > 0. Define u(a) = lna + frac{1}{a} - 1 and find its derivative: u'(a) = frac{1}{a} - frac{1}{a^2} = frac{a-1}{a^2} Analyzing u'(a), we find: - For 0 < a < 1, u'(a) < 0, indicating u(a) is decreasing. - For a > 1, u'(a) > 0, indicating u(a) is increasing. Since u(1) = 0, for a neq 1, the inequality lna + frac{1}{a} - 1 > 0 always holds. Therefore, the range of values for a is: boxed{(0,1) cup (1, +infty).}

answer:To find the total number of students who voted, we need to add up all the votes for each category: Veggies: 337 votes Meat: 335 votes Dairy: 274 votes Plant-based protein: 212 votes Total votes = 337 + 335 + 274 + 212 Total votes = 1158 So, boxed{1158} students voted in total.

answer:Let's solve this step by step. **Step 1: Evaporation from Solution Y** Original Solution Y (12 kg): - Liquid X: 20% of 12 kg = 2.4 kg - Water: 55% of 12 kg = 6.6 kg - Liquid Z: 25% of 12 kg = 3 kg After 4 kg of water evaporates: - Liquid X: 2.4 kg (unchanged) - Water: 6.6 kg - 4 kg = 2.6 kg - Liquid Z: 3 kg (unchanged) **Step 2: Adding 3 kg of Solution Y** New Solution Y (3 kg): - Liquid X: 20% of 3 kg = 0.6 kg - Water: 55% of 3 kg = 1.65 kg - Liquid Z: 25% of 3 kg = 0.75 kg Adding to the remaining mixture: - Liquid X: 2.4 kg + 0.6 kg = 3 kg - Water: 2.6 kg + 1.65 kg = 4.25 kg - Liquid Z: 3 kg + 0.75 kg = 3.75 kg **Step 3: Adding 2 kg of Solution B** Solution B (2 kg): - Liquid X: 35% of 2 kg = 0.7 kg - Water: 15% of 2 kg = 0.3 kg - Liquid Z: 50% of 2 kg = 1 kg Adding to the mixture: - Liquid X: 3 kg + 0.7 kg = 3.7 kg - Water: 4.25 kg + 0.3 kg = 4.55 kg - Liquid Z: 3.75 kg + 1 kg = 4.75 kg **Step 4: Evaporation to 75%** The total weight before evaporation: 3.7 kg + 4.55 kg + 4.75 kg = 13 kg After evaporation to 75%, the weight becomes 75% of 13 kg = 9.75 kg Since evaporation only affects water, we subtract the evaporated water from the total water: Evaporated water: 25% of 4.55 kg = 1.1375 kg Remaining water: 4.55 kg - 1.1375 kg = 3.4125 kg **Step 5: Adding 6 kg of Solution D** Solution D (6 kg): - Liquid X: 15% of 6 kg = 0.9 kg - Water: 60% of 6 kg = 3.6 kg - Liquid Z: 25% of 6 kg = 1.5 kg Adding to the mixture: - Liquid X: 3.7 kg + 0.9 kg = 4.6 kg - Water: 3.4125 kg + 3.6 kg = 7.0125 kg - Liquid Z: 4.75 kg + 1.5 kg = 6.25 kg **Final Solution:** Total weight of the final solution: 4.6 kg + 7.0125 kg + 6.25 kg = 17.8625 kg Percentage of Liquid X in the final solution: (4.6 kg / 17.8625 kg) * 100% = 25.75% Therefore, the final solution is approximately boxed{25.75%} liquid X.

answer:Let g(x)=x^{2}+2ax+4. Since the inequality x^{2}+2ax+4 > 0 with respect to x holds for all x in R, the graph of the function g(x) opens upwards and has no intersection points with the x-axis, thus triangle = 4a^{2}-16 < 0, -2 < a < 2 ldots (1) If q is a true proposition, then a leqslant x^{2} always holds, i.e., a leqslant 1 ldots (2) Since p or q is true and p and q is false, it is known that p and q are either true or false ldots (3) mathbf{1)} If p is true and q is false, then begin{cases} overset{-2 < a < 2}{a > 1} end{cases} therefore 1 < a < 2; ldots (4) mathbf{2)} If p is false and q is true, then begin{cases} overset{a leqslant -2}{a < 1} end{cases} therefore a leqslant -2; ldots (5) In summary, the range of values for the real number a is boxed{a | 1 < a < 2 text{ or } a leqslant -2} ldots (6)