💡Alright, so I have this problem here about finding the range of the inclination angle of the line given by the equation ( x sin alpha + y + 2 = 0 ). The options are A, B, C, and D, each providing different intervals. I need to figure out which one is correct.First, I remember that the inclination angle of a line is the angle measured from the positive direction of the x-axis to the line in the counterclockwise direction. This angle, usually denoted by ( theta ), has a range from 0 to ( pi ) radians, which is 0 to 180 degrees. So, the inclination angle is always between 0 and ( pi ), but depending on the slope of the line, this angle can vary within that interval.The equation given is ( x sin alpha + y + 2 = 0 ). I need to find the slope of this line because the slope is directly related to the inclination angle. To find the slope, I can rewrite the equation in the slope-intercept form, which is ( y = mx + b ), where ( m ) is the slope and ( b ) is the y-intercept.Starting with the given equation:[ x sin alpha + y + 2 = 0 ]Let me solve for ( y ):[ y = -x sin alpha - 2 ]So, the equation in slope-intercept form is:[ y = -sin alpha cdot x - 2 ]From this, I can see that the slope ( m ) is ( -sin alpha ). Now, the slope is related to the inclination angle ( theta ) by the formula:[ m = tan theta ]Therefore, ( tan theta = -sin alpha ). To find the range of ( theta ), I need to understand the possible values of ( tan theta ) based on the values of ( sin alpha ).I know that the sine function ( sin alpha ) has a range of ([-1, 1]). Therefore, ( -sin alpha ) will also have a range of ([-1, 1]). So, the slope ( m = -sin alpha ) can vary between -1 and 1.Now, let's think about the tangent function ( tan theta ). The tangent function is periodic with a period of ( pi ), and it's undefined at ( frac{pi}{2} ) because the cosine of ( frac{pi}{2} ) is zero, leading to division by zero in the definition of tangent.Given that ( tan theta = m ) and ( m in [-1, 1] ), we need to find all angles ( theta ) in the interval ([0, pi)) such that ( tan theta ) is between -1 and 1.Let me consider the unit circle to visualize this. The tangent function is positive in the first and third quadrants and negative in the second and fourth quadrants. However, since the inclination angle ( theta ) is measured from the positive x-axis and goes counterclockwise up to ( pi ), we are only concerned with angles in the first and second quadrants (i.e., between 0 and ( pi )).In the first quadrant, ( theta ) ranges from 0 to ( frac{pi}{2} ), and ( tan theta ) is positive. In the second quadrant, ( theta ) ranges from ( frac{pi}{2} ) to ( pi ), and ( tan theta ) is negative.Given that ( tan theta ) can be both positive and negative, but within the range of -1 to 1, let's break this down:1. **Positive Slope (( m > 0 )):** When ( tan theta ) is positive, ( theta ) is in the first quadrant. The maximum value of ( tan theta ) is 1, which occurs at ( theta = frac{pi}{4} ). So, for positive slopes between 0 and 1, ( theta ) ranges from 0 to ( frac{pi}{4} ).2. **Negative Slope (( m < 0 )):** When ( tan theta ) is negative, ( theta ) is in the second quadrant. The minimum value of ( tan theta ) is -1, which occurs at ( theta = frac{3pi}{4} ). So, for negative slopes between -1 and 0, ( theta ) ranges from ( frac{3pi}{4} ) to ( pi ).However, I need to be careful here. The slope ( m = -sin alpha ) can take any value between -1 and 1, but depending on the value of ( alpha ), ( sin alpha ) can vary, affecting the slope.Wait, actually, ( sin alpha ) is a parameter here, right? The equation is given as ( x sin alpha + y + 2 = 0 ), so ( alpha ) is a parameter that can vary, which affects the slope of the line. So, the slope ( m = -sin alpha ) can vary between -1 and 1 as ( alpha ) changes.Therefore, the inclination angle ( theta ) must satisfy ( tan theta = -sin alpha ), and since ( sin alpha ) can be any value between -1 and 1, ( tan theta ) can be between -1 and 1.But wait, actually, ( sin alpha ) is bounded between -1 and 1, so ( -sin alpha ) is also between -1 and 1. Therefore, ( tan theta ) is between -1 and 1.So, to find the range of ( theta ), we need to find all angles ( theta ) in ([0, pi)) such that ( tan theta ) is between -1 and 1.Let me recall that ( tan theta = 1 ) at ( theta = frac{pi}{4} ) and ( tan theta = -1 ) at ( theta = frac{3pi}{4} ). Between 0 and ( frac{pi}{4} ), ( tan theta ) increases from 0 to 1. Between ( frac{pi}{4} ) and ( frac{pi}{2} ), ( tan theta ) increases beyond 1 to infinity. Between ( frac{pi}{2} ) and ( frac{3pi}{4} ), ( tan theta ) is negative and decreases from negative infinity to -1. Between ( frac{3pi}{4} ) and ( pi ), ( tan theta ) increases from -1 to 0.But since ( tan theta ) must be between -1 and 1, the angles where ( tan theta ) is between -1 and 1 are:- From 0 to ( frac{pi}{4} ) (where ( tan theta ) is between 0 and 1)- From ( frac{3pi}{4} ) to ( pi ) (where ( tan theta ) is between -1 and 0)Therefore, the range of ( theta ) is ( [0, frac{pi}{4}] cup [frac{3pi}{4}, pi) ).Looking back at the options:A: ( [0, pi) ) — This is too broad because it includes angles where the slope would be greater than 1 or less than -1, which isn't the case here.B: ( [0, frac{pi}{4}] cup [frac{3pi}{4}, pi) ) — This matches what I derived.C: ( [0, frac{pi}{4}] ) — This only includes the first part, missing the second quadrant angles.D: ( [0, frac{pi}{4}] cup (frac{pi}{2}, pi) ) — This includes angles between ( frac{pi}{2} ) and ( pi ), but not specifically between ( frac{3pi}{4} ) and ( pi ). It also includes angles where the slope would be less than -1, which isn't allowed.Therefore, the correct answer should be option B.

💡Alright, so I have this problem where I need to prove that two functions, f(n) and g(n), are equal for all positive integers n. The functions are defined as follows:f(n) = 1 - 1/2 + 1/3 - 1/4 + ... + 1/(2n-1) - 1/(2n)g(n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n)Okay, so f(n) is an alternating series that goes up to 1/(2n), and g(n) is a sum of reciprocals from 1/(n+1) to 1/(2n). I need to show that these two expressions are equal for any positive integer n.First, maybe I should try plugging in some small values of n to see if the equality holds. That might give me some intuition.Let's try n = 1:f(1) = 1 - 1/2 = 1/2g(1) = 1/(1+1) = 1/2Okay, so f(1) = g(1). Good.Now n = 2:f(2) = 1 - 1/2 + 1/3 - 1/4Let me calculate that:1 = 1-1/2 = 1 - 1/2 = 1/2+1/3 = 1/2 + 1/3 = 5/6-1/4 = 5/6 - 1/4 = (10/12 - 3/12) = 7/12So f(2) = 7/12g(2) = 1/(2+1) + 1/(2+2) = 1/3 + 1/4 = 7/12Okay, so f(2) = g(2). That's promising.n = 3:f(3) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6Calculating step by step:1 = 1-1/2 = 1/2+1/3 = 1/2 + 1/3 = 5/6-1/4 = 5/6 - 1/4 = 10/12 - 3/12 = 7/12+1/5 = 7/12 + 1/5 = 35/60 + 12/60 = 47/60-1/6 = 47/60 - 10/60 = 37/60So f(3) = 37/60g(3) = 1/(3+1) + 1/(3+2) + 1/(3+3) = 1/4 + 1/5 + 1/6Calculating:1/4 = 15/601/5 = 12/601/6 = 10/60Adding them up: 15 + 12 + 10 = 37/60So g(3) = 37/60, which equals f(3). Nice.So for n=1,2,3, the equality holds. That gives me some confidence, but I need a general proof.I think induction might be a good approach here. Let me recall how mathematical induction works. I need to show that the statement is true for a base case (which I've done for n=1), and then assume it's true for some arbitrary n=k, and then prove it for n=k+1.So, let's set up the induction.**Base Case:**n = 1, as above, f(1) = 1 - 1/2 = 1/2, and g(1) = 1/2. So f(1) = g(1). Base case holds.**Inductive Hypothesis:**Assume that for some integer k ≥ 1, f(k) = g(k). That is,1 - 1/2 + 1/3 - 1/4 + ... + 1/(2k-1) - 1/(2k) = 1/(k+1) + 1/(k+2) + ... + 1/(2k)**Inductive Step:**We need to show that f(k+1) = g(k+1).First, let's write expressions for f(k+1) and g(k+1).f(k+1) = f(k) + 1/(2k+1) - 1/(2k+2)Because f(k+1) is f(k) plus the next two terms in the alternating series.Similarly, g(k+1) = g(k) + 1/(2k+1) + 1/(2k+2) - 1/(k+1)Wait, let me think about that.g(k) is the sum from 1/(k+1) to 1/(2k). So g(k+1) would be the sum from 1/(k+2) to 1/(2k+2). So, to get from g(k) to g(k+1), we remove the first term 1/(k+1) and add the last two terms 1/(2k+1) and 1/(2k+2).So, g(k+1) = g(k) - 1/(k+1) + 1/(2k+1) + 1/(2k+2)Therefore, f(k+1) = f(k) + 1/(2k+1) - 1/(2k+2)And g(k+1) = g(k) - 1/(k+1) + 1/(2k+1) + 1/(2k+2)But from the inductive hypothesis, f(k) = g(k). So, substituting f(k) with g(k) in the expression for f(k+1):f(k+1) = g(k) + 1/(2k+1) - 1/(2k+2)Now, let's compare this with g(k+1):g(k+1) = g(k) - 1/(k+1) + 1/(2k+1) + 1/(2k+2)So, if f(k+1) = g(k+1), then:g(k) + 1/(2k+1) - 1/(2k+2) = g(k) - 1/(k+1) + 1/(2k+1) + 1/(2k+2)Subtracting g(k) from both sides:1/(2k+1) - 1/(2k+2) = -1/(k+1) + 1/(2k+1) + 1/(2k+2)Simplify both sides:Left side: 1/(2k+1) - 1/(2k+2)Right side: -1/(k+1) + 1/(2k+1) + 1/(2k+2)Let me rearrange the right side:= 1/(2k+1) + 1/(2k+2) - 1/(k+1)So, both sides have 1/(2k+1). Let's subtract 1/(2k+1) from both sides:Left side: -1/(2k+2)Right side: 1/(2k+2) - 1/(k+1)So, we have:-1/(2k+2) = 1/(2k+2) - 1/(k+1)Let me solve the right side:1/(2k+2) - 1/(k+1) = [1 - 2]/(2k+2) = (-1)/(2k+2)So, right side is -1/(2k+2), which equals left side.Therefore, the equality holds.Thus, f(k+1) = g(k+1).Since the base case holds and the inductive step is valid, by mathematical induction, f(n) = g(n) for all positive integers n.Wait, that seems a bit too straightforward. Let me double-check my steps.Starting from f(k+1) = f(k) + 1/(2k+1) - 1/(2k+2)And g(k+1) = g(k) - 1/(k+1) + 1/(2k+1) + 1/(2k+2)Since f(k) = g(k), substitute:f(k+1) = g(k) + 1/(2k+1) - 1/(2k+2)g(k+1) = g(k) - 1/(k+1) + 1/(2k+1) + 1/(2k+2)So, to show f(k+1) = g(k+1):g(k) + 1/(2k+1) - 1/(2k+2) = g(k) - 1/(k+1) + 1/(2k+1) + 1/(2k+2)Subtract g(k) and 1/(2k+1) from both sides:-1/(2k+2) = -1/(k+1) + 1/(2k+2)Bring all terms to one side:-1/(2k+2) - 1/(2k+2) + 1/(k+1) = 0Combine like terms:-2/(2k+2) + 1/(k+1) = 0Simplify:-1/(k+1) + 1/(k+1) = 0Which is 0 = 0.Yes, that checks out. So the inductive step is valid.Therefore, the proof by induction is correct.Alternatively, maybe there's another way to see this without induction, perhaps by manipulating the series directly.Let me think about f(n):f(n) = 1 - 1/2 + 1/3 - 1/4 + ... + 1/(2n-1) - 1/(2n)This is the alternating harmonic series up to 2n terms. It's known that the alternating harmonic series converges to ln(2), but here we're dealing with finite sums.I recall that the sum of the first 2n terms of the alternating harmonic series can be expressed as:f(n) = sum_{k=1}^{2n} (-1)^{k+1}/kWhich can also be written as:f(n) = sum_{k=1}^{n} 1/(2k-1) - sum_{k=1}^{n} 1/(2k)So, f(n) = [1 + 1/3 + 1/5 + ... + 1/(2n-1)] - [1/2 + 1/4 + ... + 1/(2n)]Now, let's consider the harmonic series H_{2n} = 1 + 1/2 + 1/3 + ... + 1/(2n)We can write H_{2n} = [1 + 1/3 + 1/5 + ... + 1/(2n-1)] + [1/2 + 1/4 + ... + 1/(2n)]So, the sum of the odd terms is H_{2n} - [1/2 + 1/4 + ... + 1/(2n)]But [1/2 + 1/4 + ... + 1/(2n)] = (1/2)(1 + 1/2 + ... + 1/n) = (1/2)H_nTherefore, the sum of the odd terms is H_{2n} - (1/2)H_nThus, f(n) = [H_{2n} - (1/2)H_n] - (1/2)H_n = H_{2n} - H_nSo, f(n) = H_{2n} - H_nBut H_{2n} - H_n is exactly the sum from 1/(n+1) to 1/(2n), which is g(n).Therefore, f(n) = g(n).That's another way to see it, using properties of harmonic numbers.So, either through induction or through expressing f(n) in terms of harmonic numbers and recognizing it as g(n), we can see that f(n) equals g(n).I think both approaches are valid, but the induction might be more elementary, while the harmonic number approach is perhaps more elegant but requires knowledge of harmonic series properties.In any case, both methods confirm that f(n) = g(n) for all positive integers n.

💡First, I need to determine Alton's total weekly earnings. He earns 8 per day, and there are 7 days in a week.Next, I calculate his weekly earnings by multiplying his daily earnings by 7:8/day × 7 days = 56.Then, I need to account for his weekly expenses, which include a 20 rent for the business space.Finally, I subtract the weekly expenses from the total weekly earnings to find the profit:56 - 20 = 36.Therefore, Alton's total profit every week is 36.

💡Alright, so I have this problem about convex curves and their properties. It mentions three points on a curve with abscissas (which I think are just the x-coordinates) in increasing order. The problem talks about how convex curves are characterized by certain properties related to the chords between these points. Specifically, it says that the chord between the first and second points runs below the chord between the first and third points, and similarly for other combinations of points. It also mentions that these properties can be connected to the two-term weighted Jensen inequality.Okay, first, I need to understand what it means for a curve to be convex. From what I remember, a function is convex if the line segment between any two points on the graph of the function lies above or on the graph. So, if I have three points on a convex curve, the chords between them should have certain relationships.The problem wants me to express these properties in algebraic terms. So, I need to translate the geometric descriptions into mathematical inequalities. Let me think about how to do that.Let's denote the three points as ( P_1(x_1, y_1) ), ( P_2(x_2, y_2) ), and ( P_3(x_3, y_3) ) where ( x_1 < x_2 < x_3 ). The first property is that the chord ( P_1P_2 ) runs below the chord ( P_1P_3 ). How can I express this algebraically?Well, the chord between two points can be represented by the slope of the line connecting them. So, the slope of ( P_1P_2 ) is ( frac{y_2 - y_1}{x_2 - x_1} ), and the slope of ( P_1P_3 ) is ( frac{y_3 - y_1}{x_3 - x_1} ). If ( P_1P_2 ) runs below ( P_1P_3 ), that should mean that the slope of ( P_1P_2 ) is less than the slope of ( P_1P_3 ). So, I can write:[frac{y_2 - y_1}{x_2 - x_1} < frac{y_3 - y_1}{x_3 - x_1}]Similarly, the problem mentions that the chord ( P_2P_3 ) runs below the chord ( P_1P_3 ). Using the same logic, the slope of ( P_2P_3 ) is ( frac{y_3 - y_2}{x_3 - x_2} ), and we want this to be less than the slope of ( P_1P_3 ). So:[frac{y_3 - y_2}{x_3 - x_2} < frac{y_3 - y_1}{x_3 - x_1}]The third property is that the extension of the chord ( P_1P_2 ) runs below the chord ( P_2P_3 ). This is a bit trickier. The extension of ( P_1P_2 ) beyond ( P_2 ) should lie below ( P_2P_3 ). To express this, I think we need to consider the slopes again. If the extension of ( P_1P_2 ) is below ( P_2P_3 ), then the slope of ( P_1P_2 ) should be less than the slope of ( P_2P_3 ). So:[frac{y_2 - y_1}{x_2 - x_1} < frac{y_3 - y_2}{x_3 - x_2}]Okay, so now I have three inequalities:1. ( frac{y_2 - y_1}{x_2 - x_1} < frac{y_3 - y_1}{x_3 - x_1} )2. ( frac{y_3 - y_2}{x_3 - x_2} < frac{y_3 - y_1}{x_3 - x_1} )3. ( frac{y_2 - y_1}{x_2 - x_1} < frac{y_3 - y_2}{x_3 - x_2} )I need to see if these inequalities are connected or if they imply each other. Let me try to manipulate the first inequality to see if I can get the third one.Starting with the first inequality:[frac{y_2 - y_1}{x_2 - x_1} < frac{y_3 - y_1}{x_3 - x_1}]Multiply both sides by ( (x_2 - x_1)(x_3 - x_1) ) which is positive since ( x_1 < x_2 < x_3 ):[(y_2 - y_1)(x_3 - x_1) < (y_3 - y_1)(x_2 - x_1)]Expanding both sides:[y_2x_3 - y_2x_1 - y_1x_3 + y_1x_1 < y_3x_2 - y_3x_1 - y_1x_2 + y_1x_1]Simplify by subtracting ( y_1x_1 ) from both sides:[y_2x_3 - y_2x_1 - y_1x_3 < y_3x_2 - y_3x_1 - y_1x_2]Rearrange terms:[y_2x_3 - y_3x_2 < y_2x_1 - y_3x_1 + y_1x_3 - y_1x_2]Factor terms on the right:[y_2x_3 - y_3x_2 < y_2x_1 - y_3x_1 + y_1x_3 - y_1x_2][y_2x_3 - y_3x_2 < (y_2 - y_3)x_1 + (y_1)x_3 - y_1x_2][y_2x_3 - y_3x_2 < (y_2 - y_3)x_1 + y_1(x_3 - x_2)]Hmm, not sure if this is helpful. Maybe I should try another approach. Let's consider the third inequality:[frac{y_2 - y_1}{x_2 - x_1} < frac{y_3 - y_2}{x_3 - x_2}]Multiply both sides by ( (x_2 - x_1)(x_3 - x_2) ):[(y_2 - y_1)(x_3 - x_2) < (y_3 - y_2)(x_2 - x_1)]Expanding both sides:[y_2x_3 - y_2x_2 - y_1x_3 + y_1x_2 < y_3x_2 - y_3x_1 - y_2x_2 + y_2x_1]Simplify:[y_2x_3 - y_2x_2 - y_1x_3 + y_1x_2 < y_3x_2 - y_3x_1 - y_2x_2 + y_2x_1]Cancel out ( -y_2x_2 ) on both sides:[y_2x_3 - y_1x_3 + y_1x_2 < y_3x_2 - y_3x_1 + y_2x_1]Rearrange terms:[y_2x_3 - y_1x_3 + y_1x_2 - y_3x_2 + y_3x_1 - y_2x_1 < 0]Factor:[(y_2x_3 - y_3x_2) + (-y_1x_3 + y_1x_2) + (y_3x_1 - y_2x_1) < 0][(y_2x_3 - y_3x_2) + y_1(x_2 - x_3) + y_3x_1 - y_2x_1 < 0]This seems complicated. Maybe instead of trying to manipulate the inequalities directly, I should think about the relationship with Jensen's inequality.Jensen's inequality states that for a convex function ( f ) and weights ( lambda ) and ( 1 - lambda ), we have:[f(lambda x + (1 - lambda)y) leq lambda f(x) + (1 - lambda)f(y)]In the two-term weighted case, this is similar to what we have with the slopes. Maybe if I can express the slopes in terms of weighted averages, I can connect them to Jensen's inequality.Let me consider the first inequality:[frac{y_2 - y_1}{x_2 - x_1} < frac{y_3 - y_1}{x_3 - x_1}]Let me denote ( lambda = frac{x_2 - x_1}{x_3 - x_1} ). Then ( 1 - lambda = frac{x_3 - x_2}{x_3 - x_1} ). So, ( x_2 = lambda x_1 + (1 - lambda)x_3 ).If I apply Jensen's inequality to ( x_1 ) and ( x_3 ) with weights ( lambda ) and ( 1 - lambda ), I get:[f(lambda x_1 + (1 - lambda)x_3) leq lambda f(x_1) + (1 - lambda)f(x_3)]But ( lambda x_1 + (1 - lambda)x_3 = x_2 ), so:[f(x_2) leq lambda f(x_1) + (1 - lambda)f(x_3)]Substituting back ( lambda = frac{x_2 - x_1}{x_3 - x_1} ):[f(x_2) leq frac{x_2 - x_1}{x_3 - x_1} f(x_1) + frac{x_3 - x_2}{x_3 - x_1} f(x_3)]Rearranging:[f(x_2) leq frac{(x_2 - x_1)f(x_1) + (x_3 - x_2)f(x_3)}{x_3 - x_1}]Multiplying both sides by ( x_3 - x_1 ):[(x_3 - x_1)f(x_2) leq (x_2 - x_1)f(x_1) + (x_3 - x_2)f(x_3)]Which simplifies to:[x_3 f(x_2) - x_1 f(x_2) leq x_2 f(x_1) - x_1 f(x_1) + x_3 f(x_3) - x_2 f(x_3)]Rearranging terms:[x_1 (f(x_2) - f(x_3)) + x_2 (f(x_3) - f(x_1)) + x_3 (f(x_1) - f(x_2)) geq 0]Wait, this looks familiar. Earlier, when I tried to manipulate the first inequality, I ended up with a similar expression. So, this suggests that the inequality derived from Jensen's is equivalent to the one we get from the slope condition.Similarly, if I apply Jensen's inequality to other pairs of points, I should get the other inequalities. So, all three slope conditions are equivalent to the two-term weighted Jensen inequality.Therefore, the inequalities hold if and only if the two-term weighted Jensen inequality is satisfied. This means that the convexity of the curve is equivalent to satisfying these slope conditions, which in turn is equivalent to satisfying Jensen's inequality.I think I've connected the geometric properties to the algebraic inequalities and linked them to Jensen's inequality. So, the key takeaway is that the convexity of a curve can be characterized by these slope inequalities, and they are all tied together through Jensen's inequality.**Final Answer**The inequalities hold if and only if the two-term weighted Jensen inequality is satisfied, which is expressed as:boxed{x_1 (f(x_2) - f(x_3)) + x_2 (f(x_3) - f(x_1)) + x_3 (f(x_1) - f(x_2)) > 0}

💡Alright, so I have this problem about purchasing sewage treatment equipment. There are two models, A and B. The office wants to buy 10 units in total. Model A costs 12 million yuan and can treat 240 tons of sewage per month. Model B costs 10 million yuan and can treat 200 tons per month.First, I need to figure out how many purchasing plans are possible without exceeding a budget of 105 million yuan. Then, under the same budget, I have to ensure that the monthly sewage treatment volume is at least 2040 tons and choose the most cost-effective plan.Okay, let's start with part (1). I need to find the number of possible purchasing plans. Let me denote the number of model A equipment as x. Then, the number of model B equipment would be 10 - x since the total is 10 units.The total cost should not exceed 105 million yuan. So, the cost equation would be:12x + 10(10 - x) ≤ 105Let me simplify this:12x + 100 - 10x ≤ 105Combine like terms:2x + 100 ≤ 105Subtract 100 from both sides:2x ≤ 5Divide both sides by 2:x ≤ 2.5But x has to be an integer because you can't buy half a unit. So, x can be 0, 1, or 2. That means there are three possible purchasing plans:1. Buy 0 units of A and 10 units of B.2. Buy 1 unit of A and 9 units of B.3. Buy 2 units of A and 8 units of B.Okay, that seems straightforward. Now, moving on to part (2). I need to ensure that the monthly sewage treatment volume is at least 2040 tons. So, I need to set up another inequality based on the treatment capacities.The total treatment capacity would be:240x + 200(10 - x) ≥ 2040Let me simplify this:240x + 2000 - 200x ≥ 2040Combine like terms:40x + 2000 ≥ 2040Subtract 2000 from both sides:40x ≥ 40Divide both sides by 40:x ≥ 1So, x has to be at least 1. That means we can't buy 0 units of A; we need to buy at least 1 unit of A. So, the possible plans now are:1. Buy 1 unit of A and 9 units of B.2. Buy 2 units of A and 8 units of B.Now, I need to choose the most cost-effective plan. Let's calculate the total cost for both options.For 1 unit of A and 9 units of B:12*1 + 10*9 = 12 + 90 = 102 million yuan.For 2 units of A and 8 units of B:12*2 + 10*8 = 24 + 80 = 104 million yuan.Comparing the two, 102 million yuan is cheaper than 104 million yuan. Therefore, buying 1 unit of A and 9 units of B is more cost-effective.Wait, but I should double-check if there are any other constraints or if I missed something. Let me verify the calculations.Total cost for 1A and 9B:12 + 90 = 102. Yes, that's correct.Total cost for 2A and 8B:24 + 80 = 104. Correct.And the treatment capacity for 1A and 9B:240*1 + 200*9 = 240 + 1800 = 2040 tons. That meets the requirement.For 2A and 8B:240*2 + 200*8 = 480 + 1600 = 2080 tons. That also meets the requirement but costs more.So, yes, 1A and 9B is the most cost-effective plan that meets both the budget and treatment capacity requirements.I think that's all. I don't see any other possible purchasing plans that satisfy both conditions, and the calculations seem correct.

💡Okay, so I have this trigonometry problem here. It says that tan(α + π/4) equals 1/2, and α is between -π/2 and 0. I need to find the value of sin2α + 2sin²α. The options are A: -2√5/5, B: -2/5, C: 2/5, and D: 2√5/5.Hmm, let me think about how to approach this. I remember that there are tangent addition formulas. Maybe I can use that to find tanα first. The formula for tan(A + B) is (tanA + tanB)/(1 - tanA tanB). So, in this case, A is α and B is π/4. Since tan(π/4) is 1, the formula becomes (tanα + 1)/(1 - tanα) = 1/2.Wait, let me write that down:tan(α + π/4) = (tanα + tan(π/4)) / (1 - tanα tan(π/4)) = (tanα + 1)/(1 - tanα) = 1/2.So, I can set up the equation:(tanα + 1)/(1 - tanα) = 1/2.Now, I can solve for tanα. Let's cross-multiply:2(tanα + 1) = 1(1 - tanα).Expanding both sides:2 tanα + 2 = 1 - tanα.Now, bring all the tanα terms to one side and constants to the other:2 tanα + tanα = 1 - 2.That simplifies to:3 tanα = -1.So, tanα = -1/3.Alright, so tanα is -1/3. Now, I need to find sin2α + 2sin²α. Let me recall that sin2α is 2 sinα cosα, so the expression becomes 2 sinα cosα + 2 sin²α.I can factor out a 2 sinα:2 sinα (cosα + sinα).Hmm, that might be useful. Alternatively, maybe I can express everything in terms of tanα since I already know tanα. Let me think.I remember that sinα can be expressed as tanα / sqrt(1 + tan²α) and cosα is 1 / sqrt(1 + tan²α). But since α is between -π/2 and 0, which is in the fourth quadrant, sinα will be negative and cosα will be positive.Let me compute sinα and cosα.Given tanα = -1/3, which is opposite over adjacent. So, if we consider a right triangle where the opposite side is -1 (but since lengths are positive, we can think of it as 1 with a negative sign) and the adjacent side is 3. Then, the hypotenuse would be sqrt(1² + 3²) = sqrt(10).So, sinα = opposite / hypotenuse = -1 / sqrt(10), and cosα = adjacent / hypotenuse = 3 / sqrt(10).Wait, but since α is in the fourth quadrant, sinα is negative and cosα is positive, which matches.So, sinα = -1/√10 and cosα = 3/√10.Now, let's compute sin2α + 2 sin²α.First, sin2α = 2 sinα cosα = 2*(-1/√10)*(3/√10) = 2*(-3/10) = -6/10 = -3/5.Next, 2 sin²α = 2*(sinα)^2 = 2*(1/10) = 2/10 = 1/5.So, adding them together: sin2α + 2 sin²α = (-3/5) + (1/5) = (-2)/5.So, the result is -2/5, which is option B.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I used the tangent addition formula correctly: (tanα + 1)/(1 - tanα) = 1/2. Solved for tanα and got -1/3. That seems correct.2. Then, I found sinα and cosα using the right triangle approach. Since tanα = -1/3, and α is in the fourth quadrant, sinα is negative and cosα is positive. So, sinα = -1/√10 and cosα = 3/√10. That makes sense.3. Calculated sin2α: 2 sinα cosα = 2*(-1/√10)*(3/√10) = -6/10 = -3/5. Correct.4. Calculated 2 sin²α: 2*(1/10) = 1/5. Correct.5. Added them: -3/5 + 1/5 = -2/5. Correct.So, yeah, the answer should be B: -2/5.Alternatively, I can think of another method using identities. Maybe express everything in terms of tanα.I know that sin2α = 2 tanα / (1 + tan²α) and sin²α = tan²α / (1 + tan²α). So, let's try that.Given tanα = -1/3, so tan²α = 1/9.Compute sin2α: 2*(-1/3)/(1 + 1/9) = (-2/3)/(10/9) = (-2/3)*(9/10) = -18/30 = -3/5.Compute 2 sin²α: 2*(1/9)/(1 + 1/9) = (2/9)/(10/9) = (2/9)*(9/10) = 18/90 = 1/5.Add them: -3/5 + 1/5 = -2/5. Same result. So, that confirms it.I think that's solid. So, the answer is B.**Final Answer**boxed{B}