💡Alright, so I have this problem about a ninth-grade math interest group that did a market survey on a product's price and sales volume over x days, where x is between 1 and 70. They gave me a table with two intervals for x: 1 to less than 40, and 40 to 70. For each interval, they provided the price per item and the daily sales volume. The cost price is 30 yuan per item, and I need to find the daily profit y in terms of x, find when the profit is maximized, and determine how many days the profit doesn't fall below 3250 yuan.Okay, let's start with part (1): finding the functional relationship between y and x. So, profit is generally calculated as (price - cost) multiplied by the number of items sold. That makes sense because for each item, you make a certain profit, and then you multiply by how many you sold to get the total profit.Looking at the table, for 1 ≤ x < 40, the price is x + 45 yuan per item, and the daily sales are 150 - 2x items. For 40 ≤ x ≤ 70, the price is fixed at 85 yuan, and the daily sales are still 150 - 2x items. The cost price is 30 yuan per item, so the profit per item is price minus cost.So, for the first interval, 1 ≤ x < 40, the profit per item would be (x + 45) - 30, which simplifies to x + 15. Then, the daily profit y would be this profit per item multiplied by the number of items sold, which is 150 - 2x. So, y = (x + 15)(150 - 2x). I should probably expand this to get a quadratic function.Let me do that: (x + 15)(150 - 2x) = x*150 + x*(-2x) + 15*150 + 15*(-2x) = 150x - 2x² + 2250 - 30x. Combining like terms: 150x - 30x is 120x, so y = -2x² + 120x + 2250.Okay, that seems right. Now, for the second interval, 40 ≤ x ≤ 70, the price is fixed at 85 yuan, so the profit per item is 85 - 30 = 55 yuan. The daily sales are still 150 - 2x, so the daily profit y is 55*(150 - 2x). Let me calculate that: 55*150 is 8250, and 55*(-2x) is -110x. So, y = -110x + 8250.Alright, so I have two functions for y depending on the interval of x. That answers part (1). I think I did that correctly, but let me double-check.For 1 ≤ x < 40: y = (x + 45 - 30)*(150 - 2x) = (x + 15)*(150 - 2x). Expanding gives -2x² + 120x + 2250. That seems correct.For 40 ≤ x ≤ 70: y = (85 - 30)*(150 - 2x) = 55*(150 - 2x) = 8250 - 110x. That also seems correct.Okay, moving on to part (2): finding the day when the daily profit is maximized and the maximum profit. So, I need to find the maximum value of y in both intervals and see which one is larger.Starting with the first interval, 1 ≤ x < 40, where y = -2x² + 120x + 2250. This is a quadratic function, and since the coefficient of x² is negative, it opens downward, meaning the vertex is the maximum point.The vertex of a quadratic function ax² + bx + c is at x = -b/(2a). So here, a = -2, b = 120. Plugging in, x = -120/(2*(-2)) = -120/(-4) = 30. So, the maximum profit in this interval occurs at x = 30.Now, let's calculate y at x = 30: y = -2*(30)^2 + 120*30 + 2250. Calculating each term:-2*(900) = -1800120*30 = 3600So, y = -1800 + 3600 + 2250 = (3600 - 1800) + 2250 = 1800 + 2250 = 4050.So, the maximum profit in the first interval is 4050 yuan on day 30.Now, for the second interval, 40 ≤ x ≤ 70, y = -110x + 8250. This is a linear function with a negative slope, meaning it decreases as x increases. Therefore, the maximum profit in this interval occurs at the smallest x, which is x = 40.Calculating y at x = 40: y = -110*40 + 8250 = -4400 + 8250 = 3850.So, the maximum profit in the second interval is 3850 yuan on day 40.Comparing the two maximums: 4050 yuan at day 30 and 3850 yuan at day 40. Clearly, 4050 is larger. Therefore, the maximum daily profit occurs on day 30 with a profit of 4050 yuan.Wait a second, but the first interval is up to x < 40, so x = 30 is within that interval. So, that's correct.Just to be thorough, let me check the profit at x = 39, which is the last day of the first interval. y = -2*(39)^2 + 120*39 + 2250.Calculating:39 squared is 1521.-2*1521 = -3042120*39 = 4680So, y = -3042 + 4680 + 2250 = (4680 - 3042) + 2250 = 1638 + 2250 = 3888.So, at x = 39, y = 3888, which is less than 4050. So, yes, the maximum is indeed at x = 30.Alright, moving on to part (3): determining how many days the daily profit does not fall below 3250 yuan. So, I need to find all x in 1 ≤ x ≤ 70 where y ≥ 3250.Since y is defined differently in two intervals, I need to solve y ≥ 3250 for both intervals and then combine the results.First, for 1 ≤ x < 40: y = -2x² + 120x + 2250 ≥ 3250.Let's set up the inequality: -2x² + 120x + 2250 ≥ 3250.Subtract 3250 from both sides: -2x² + 120x + 2250 - 3250 ≥ 0 → -2x² + 120x - 1000 ≥ 0.Let me write it as: -2x² + 120x - 1000 ≥ 0.I can multiply both sides by -1 to make it easier, but remember to reverse the inequality sign: 2x² - 120x + 1000 ≤ 0.Now, let's solve 2x² - 120x + 1000 ≤ 0.First, find the roots of the quadratic equation 2x² - 120x + 1000 = 0.Using the quadratic formula: x = [120 ± sqrt(120² - 4*2*1000)] / (2*2).Calculating discriminant: 120² = 14400; 4*2*1000 = 8000; so sqrt(14400 - 8000) = sqrt(6400) = 80.So, x = [120 ± 80] / 4.Calculating the two roots:x = (120 + 80)/4 = 200/4 = 50x = (120 - 80)/4 = 40/4 = 10So, the quadratic equals zero at x = 10 and x = 50. Since the coefficient of x² is positive, the parabola opens upward, meaning the quadratic is ≤ 0 between its roots. So, 10 ≤ x ≤ 50.But in this interval, x is from 1 to less than 40. So, the overlap is 10 ≤ x < 40.Therefore, in the first interval, the profit is ≥ 3250 yuan from day 10 to day 39.Now, for the second interval, 40 ≤ x ≤ 70: y = -110x + 8250 ≥ 3250.Let's solve for x:-110x + 8250 ≥ 3250Subtract 8250 from both sides: -110x ≥ 3250 - 8250 → -110x ≥ -5000Divide both sides by -110, remembering to reverse the inequality: x ≤ (-5000)/(-110) → x ≤ 5000/110 ≈ 45.4545.Since x must be an integer, x ≤ 45.But in this interval, x is from 40 to 70. So, the overlap is 40 ≤ x ≤ 45.Therefore, in the second interval, the profit is ≥ 3250 yuan from day 40 to day 45.Now, combining both intervals:From day 10 to day 39: that's 39 - 10 + 1 = 30 days.From day 40 to day 45: that's 45 - 40 + 1 = 6 days.Total days: 30 + 6 = 36 days.Wait, let me verify that calculation.From day 10 to day 39 inclusive: 39 - 10 = 29, plus 1 is 30 days.From day 40 to day 45 inclusive: 45 - 40 = 5, plus 1 is 6 days.Total: 30 + 6 = 36 days.Yes, that seems correct.Just to double-check, let's pick a day in each interval and ensure y is ≥ 3250.For example, day 10: y = -2*(10)^2 + 120*10 + 2250 = -200 + 1200 + 2250 = 3250. So, exactly 3250.Day 39: y = -2*(39)^2 + 120*39 + 2250 = -2*1521 + 4680 + 2250 = -3042 + 4680 + 2250 = 3888, which is above 3250.Day 40: y = -110*40 + 8250 = -4400 + 8250 = 3850, which is above 3250.Day 45: y = -110*45 + 8250 = -4950 + 8250 = 3300, which is above 3250.Day 46: y = -110*46 + 8250 = -5060 + 8250 = 3190, which is below 3250. So, day 46 is excluded, which aligns with our previous result.Therefore, the total number of days where the daily profit is not below 3250 yuan is 36 days.I think that covers all parts of the problem. I should summarize my findings:1. The functional relationship between y and x is: - For 1 ≤ x < 40: y = -2x² + 120x + 2250 - For 40 ≤ x ≤ 70: y = -110x + 82502. The maximum daily profit occurs on day 30 with a profit of 4050 yuan.3. There are 36 days where the daily profit is not below 3250 yuan.

💡First, I need to determine the amount invested by B. I'll denote this amount as Rs. ( x ).Next, I recognize that the profit shares of A, B, and C are directly proportional to their respective investments. Therefore, the ratio of their profits is the same as the ratio of their investments:[text{Profit Ratio} = 8000 : x : 12000]I know that the difference between the profit shares of A and C is Rs. 680. Let ( P_A ) be the profit share of A and ( P_C ) be the profit share of C. Then:[P_C - P_A = 680]I also know that the profit share of B is Rs. 1700. Let ( P_B = 1700 ).Using the profit ratio, I can express ( P_A ) and ( P_C ) in terms of ( P_B ):[P_A = left( frac{8000}{x} right) P_B][P_C = left( frac{12000}{x} right) P_B]Substituting these into the equation for the difference in profits:[left( frac{12000}{x} right) P_B - left( frac{8000}{x} right) P_B = 680][frac{4000}{x} P_B = 680]Now, substituting ( P_B = 1700 ):[frac{4000}{x} times 1700 = 680][frac{6800000}{x} = 680][x = frac{6800000}{680}][x = 10000]Therefore, B invested Rs. 10,000 in the business.

💡Okay, so I have this geometry problem here. Let me try to understand it step by step. We have a right angle with vertex O. On one leg of this right angle, there are two points, A and B, such that OA is greater than OB. That means A is farther from O than B is on the same leg. Now, I need to construct a point C on the other leg such that the angle BCA is twice the angle BAC. Hmm, okay. So, angle at C is twice the angle at A. First, I should probably draw a diagram to visualize this. Let me imagine a right angle with O at the origin, one leg along the x-axis, and the other along the y-axis. Points A and B are on the x-axis, with A further out than B. So, O is at (0,0), B is at some point (b,0) where b > 0, and A is at (a,0) where a > b. Now, I need to find a point C on the y-axis such that when I connect points B, C, and A, the angle at C (∠BCA) is twice the angle at A (∠BAC). So, ∠BCA = 2∠BAC. I think I should start by considering triangle ABC. Since C is on the y-axis, its coordinates will be (0,c) for some c > 0. So, points are A(a,0), B(b,0), and C(0,c). Let me denote ∠BAC as α. Then, according to the problem, ∠BCA should be 2α. So, in triangle ABC, we have angles at A, B, and C. The sum of angles in a triangle is 180°, so ∠ABC = 180° - α - 2α = 180° - 3α. Hmm, maybe I can use the Law of Sines here. In triangle ABC, the Law of Sines states that:AB / sin(∠BCA) = BC / sin(∠BAC) = AC / sin(∠ABC)So, plugging in the known angles:AB / sin(2α) = BC / sin(α) = AC / sin(180° - 3α)But sin(180° - 3α) is sin(3α), so:AB / sin(2α) = BC / sin(α) = AC / sin(3α)Let me write down the ratios:AB / sin(2α) = BC / sin(α)AndAB / sin(2α) = AC / sin(3α)So, from the first ratio:AB / sin(2α) = BC / sin(α)Which implies:BC = AB * sin(α) / sin(2α)Similarly, from the second ratio:AC = AB * sin(3α) / sin(2α)Hmm, okay. Let me compute these expressions.First, sin(2α) is 2 sin α cos α, so:BC = AB * sin α / (2 sin α cos α) = AB / (2 cos α)Similarly, sin(3α) is 3 sin α - 4 sin^3 α, but maybe it's better to express it in terms of sin and cos:sin(3α) = sin(2α + α) = sin 2α cos α + cos 2α sin α = 2 sin α cos^2 α + (1 - 2 sin^2 α) sin αSimplify that:= 2 sin α cos^2 α + sin α - 2 sin^3 α= sin α (2 cos^2 α + 1 - 2 sin^2 α)But since cos^2 α = 1 - sin^2 α, substitute:= sin α [2(1 - sin^2 α) + 1 - 2 sin^2 α]= sin α [2 - 2 sin^2 α + 1 - 2 sin^2 α]= sin α [3 - 4 sin^2 α]Hmm, so sin(3α) = 3 sin α - 4 sin^3 α, which is another identity. So, AC = AB * (3 sin α - 4 sin^3 α) / (2 sin α cos α)Simplify:AC = AB * [3 - 4 sin^2 α] / (2 cos α)Hmm, okay. So, now, let's think about the coordinates. Points A(a,0), B(b,0), C(0,c). Let me compute the lengths AB, BC, and AC.AB is the distance between A and B, which is |a - b|, since they are on the x-axis.BC is the distance between B(b,0) and C(0,c). So, BC = sqrt(b^2 + c^2)Similarly, AC is the distance between A(a,0) and C(0,c). So, AC = sqrt(a^2 + c^2)So, AB = a - b (since a > b)BC = sqrt(b^2 + c^2)AC = sqrt(a^2 + c^2)So, from earlier, we have:BC = AB / (2 cos α)So,sqrt(b^2 + c^2) = (a - b) / (2 cos α)Similarly,AC = AB * [3 - 4 sin^2 α] / (2 cos α)So,sqrt(a^2 + c^2) = (a - b) * [3 - 4 sin^2 α] / (2 cos α)Hmm, okay. So, now, I have two equations:1. sqrt(b^2 + c^2) = (a - b) / (2 cos α)2. sqrt(a^2 + c^2) = (a - b) * [3 - 4 sin^2 α] / (2 cos α)Let me denote (a - b) as d for simplicity. So, d = a - b.Then, equation 1 becomes:sqrt(b^2 + c^2) = d / (2 cos α) => sqrt(b^2 + c^2) = d / (2 cos α)Equation 2 becomes:sqrt(a^2 + c^2) = d * [3 - 4 sin^2 α] / (2 cos α)Hmm, okay. Let me square both sides of equation 1:b^2 + c^2 = d^2 / (4 cos^2 α)Similarly, square equation 2:a^2 + c^2 = [d^2 * (3 - 4 sin^2 α)^2] / (4 cos^2 α)Now, subtract equation 1 squared from equation 2 squared:(a^2 + c^2) - (b^2 + c^2) = [d^2 * (3 - 4 sin^2 α)^2 / (4 cos^2 α)] - [d^2 / (4 cos^2 α)]Simplify:a^2 - b^2 = [d^2 / (4 cos^2 α)] * [(3 - 4 sin^2 α)^2 - 1]Factor the right-hand side:First, compute (3 - 4 sin^2 α)^2 - 1:= 9 - 24 sin^2 α + 16 sin^4 α - 1= 8 - 24 sin^2 α + 16 sin^4 αFactor:= 8(1 - 3 sin^2 α + 2 sin^4 α)Wait, let me check:Wait, 8 - 24 sin^2 α + 16 sin^4 α = 8(1 - 3 sin^2 α + 2 sin^4 α). Hmm, is that factorable?Let me see: 1 - 3 sin^2 α + 2 sin^4 αLet me set x = sin^2 α:= 1 - 3x + 2x^2 = 2x^2 - 3x + 1Factor:= (2x - 1)(x - 1)So, 2x^2 - 3x + 1 = (2x - 1)(x - 1)So, substituting back:= (2 sin^2 α - 1)(sin^2 α - 1)But sin^2 α - 1 = -cos^2 αSo,= (2 sin^2 α - 1)(-cos^2 α) = -cos^2 α (2 sin^2 α - 1)So, putting it all together:a^2 - b^2 = [d^2 / (4 cos^2 α)] * [8 * (-cos^2 α)(2 sin^2 α - 1)]Simplify:= [d^2 / (4 cos^2 α)] * [ -8 cos^2 α (2 sin^2 α - 1) ]= [d^2 / (4 cos^2 α)] * [ -8 cos^2 α (2 sin^2 α - 1) ]The cos^2 α cancels out:= [d^2 / 4] * [ -8 (2 sin^2 α - 1) ]= [d^2 / 4] * [ -16 sin^2 α + 8 ]= [d^2 / 4] * [8 - 16 sin^2 α]= [d^2 / 4] * 8 (1 - 2 sin^2 α)= 2 d^2 (1 - 2 sin^2 α)But 1 - 2 sin^2 α is cos(2α), so:a^2 - b^2 = 2 d^2 cos(2α)But d = a - b, so:a^2 - b^2 = 2 (a - b)^2 cos(2α)Let me compute a^2 - b^2:a^2 - b^2 = (a - b)(a + b) = d (a + b)So,d (a + b) = 2 d^2 cos(2α)Divide both sides by d (assuming d ≠ 0, which it is since a > b):a + b = 2 d cos(2α)But d = a - b, so:a + b = 2 (a - b) cos(2α)Let me solve for cos(2α):cos(2α) = (a + b) / [2 (a - b)]Hmm, interesting. So, cos(2α) is expressed in terms of a and b.But we also know that in triangle ABC, angle at A is α, so tan α = opposite / adjacent = BC / AB? Wait, no. Wait, in triangle ABC, angle at A is α, so tan α = opposite / adjacent. Wait, let me think.In triangle ABC, angle at A is α. The sides adjacent to angle A are AB and AC. Wait, no. Let me clarify.Wait, in triangle ABC, angle at A is α, so the sides are:- Opposite to α: BC- Adjacent to α: ABWait, no. Wait, in triangle ABC, angle at A is α. So, the sides are:- Opposite to α: BC- Adjacent to α: AB and AC.Wait, no, actually, in triangle ABC, angle at A is between sides AB and AC. So, the sides adjacent to angle A are AB and AC, and the side opposite is BC.So, tan α = opposite / adjacent = BC / AB?Wait, no, tan α is opposite over adjacent, but in triangle ABC, angle at A is α, so the opposite side is BC, and the adjacent side is AB? Wait, no, adjacent side would be the side adjacent to angle A, which is AB, but actually, in triangle ABC, the sides adjacent to angle A are AB and AC, but AB is adjacent in the sense of the triangle, but in terms of the coordinate system, it's along the x-axis.Wait, maybe it's better to use coordinates to find tan α.In triangle ABC, angle at A is α. So, tan α is the ratio of the opposite side to the adjacent side. The opposite side to angle A is BC, and the adjacent side is AB. Wait, but BC is not directly opposite to angle A in the coordinate system.Wait, perhaps I should think in terms of coordinates. The angle at A is between the lines AB and AC. So, AB is along the x-axis from A to B, and AC goes from A to C on the y-axis. So, the angle at A is between AB (along negative x-axis from A) and AC (going up to C). So, tan α would be the slope of AC.Wait, AC goes from A(a,0) to C(0,c). So, the slope is (c - 0)/(0 - a) = -c/a. So, tan α = |slope| = c/a. Wait, but since it's going from A to C, the angle α is the angle between AB (which is along the negative x-axis from A) and AC. So, the angle α is measured from the negative x-axis up to AC. So, tan α = (c)/a, because the rise is c and the run is a.Wait, actually, if you consider point A at (a,0), then AC goes to (0,c). So, the vector from A to C is (-a, c). So, the angle α is the angle between the negative x-axis and the vector AC. So, tan α = c / a.So, tan α = c / a.Similarly, in triangle ABC, angle at C is 2α. So, tan(2α) would be related to the sides as well.But maybe instead of tan, I can use the Law of Sines or Cosines.Wait, earlier, I had expressions for BC and AC in terms of α. Maybe I can relate them.From equation 1:sqrt(b^2 + c^2) = (a - b) / (2 cos α)From equation 2:sqrt(a^2 + c^2) = (a - b) * [3 - 4 sin^2 α] / (2 cos α)Let me denote sqrt(b^2 + c^2) as BC and sqrt(a^2 + c^2) as AC.So, from equation 1:BC = (a - b) / (2 cos α)From equation 2:AC = (a - b) * [3 - 4 sin^2 α] / (2 cos α)But from the coordinates, AC = sqrt(a^2 + c^2) and BC = sqrt(b^2 + c^2). So, maybe I can express AC in terms of BC.Let me compute AC / BC:AC / BC = [sqrt(a^2 + c^2)] / [sqrt(b^2 + c^2)] = [ (a - b) * (3 - 4 sin^2 α) / (2 cos α) ] / [ (a - b) / (2 cos α) ] = (3 - 4 sin^2 α)So,sqrt(a^2 + c^2) / sqrt(b^2 + c^2) = 3 - 4 sin^2 αLet me square both sides:(a^2 + c^2) / (b^2 + c^2) = (3 - 4 sin^2 α)^2But earlier, I found that (3 - 4 sin^2 α)^2 = 9 - 24 sin^2 α + 16 sin^4 αSo,(a^2 + c^2) / (b^2 + c^2) = 9 - 24 sin^2 α + 16 sin^4 αHmm, this seems complicated. Maybe I can express sin^2 α in terms of c and a, since tan α = c/a, so sin α = c / sqrt(a^2 + c^2). So, sin^2 α = c^2 / (a^2 + c^2)Similarly, cos^2 α = a^2 / (a^2 + c^2)So, let me substitute sin^2 α = c^2 / (a^2 + c^2) into the equation:(a^2 + c^2) / (b^2 + c^2) = 9 - 24*(c^2 / (a^2 + c^2)) + 16*(c^4 / (a^2 + c^2)^2)Let me multiply both sides by (b^2 + c^2)*(a^2 + c^2)^2 to eliminate denominators:(a^2 + c^2)^3 = [9 - 24*(c^2 / (a^2 + c^2)) + 16*(c^4 / (a^2 + c^2)^2)] * (b^2 + c^2)*(a^2 + c^2)^2Wait, this seems too messy. Maybe there's a better approach.Wait, earlier, I had:cos(2α) = (a + b) / [2 (a - b)]But cos(2α) can also be expressed in terms of sin^2 α or cos^2 α. Since I have tan α = c/a, I can express cos(2α) as (1 - tan^2 α) / (1 + tan^2 α) = (1 - (c^2/a^2)) / (1 + (c^2/a^2)) = (a^2 - c^2)/(a^2 + c^2)So,cos(2α) = (a^2 - c^2)/(a^2 + c^2) = (a + b)/(2(a - b))Therefore,(a^2 - c^2)/(a^2 + c^2) = (a + b)/(2(a - b))Cross-multiplying:2(a - b)(a^2 - c^2) = (a + b)(a^2 + c^2)Let me expand both sides:Left side: 2(a - b)(a^2 - c^2) = 2a(a^2 - c^2) - 2b(a^2 - c^2) = 2a^3 - 2a c^2 - 2b a^2 + 2b c^2Right side: (a + b)(a^2 + c^2) = a(a^2 + c^2) + b(a^2 + c^2) = a^3 + a c^2 + b a^2 + b c^2So, set left side equal to right side:2a^3 - 2a c^2 - 2b a^2 + 2b c^2 = a^3 + a c^2 + b a^2 + b c^2Bring all terms to left side:2a^3 - 2a c^2 - 2b a^2 + 2b c^2 - a^3 - a c^2 - b a^2 - b c^2 = 0Simplify:(2a^3 - a^3) + (-2a c^2 - a c^2) + (-2b a^2 - b a^2) + (2b c^2 - b c^2) = 0So,a^3 - 3a c^2 - 3b a^2 + b c^2 = 0Factor:a^3 - 3b a^2 - 3a c^2 + b c^2 = 0Let me factor terms:a^2(a - 3b) - c^2(3a - b) = 0So,a^2(a - 3b) = c^2(3a - b)Therefore,c^2 = [a^2(a - 3b)] / (3a - b)So,c = a sqrt[(a - 3b)/(3a - b)]Hmm, interesting. So, c is expressed in terms of a and b. But we need to ensure that the expression inside the square root is positive, because c is a real number.So,(a - 3b)/(3a - b) > 0This inequality holds when both numerator and denominator are positive or both are negative.Case 1: Both positive.a - 3b > 0 => a > 3b3a - b > 0 => 3a > b, which is always true since a > b (given OA > OB, so a > b). So, if a > 3b, then both numerator and denominator are positive.Case 2: Both negative.a - 3b < 0 => a < 3b3a - b < 0 => 3a < b => a < b/3But since a > b (given OA > OB), a < b/3 is impossible because b/3 < b < a. So, Case 2 is impossible.Therefore, the expression inside the square root is positive only when a > 3b.So, c = a sqrt[(a - 3b)/(3a - b)] is real only when a > 3b.Wait, but in the problem statement, OA > OB, so a > b, but not necessarily a > 3b. So, does this mean that point C exists only when a > 3b?Hmm, that seems to be the case from this derivation. So, the necessary condition for the existence of point C is that OA > 3 OB, i.e., a > 3b.Wait, but in the problem statement, it's given that OA > OB, which is a > b, but not necessarily a > 3b. So, does this mean that the construction is only possible when a > 3b?Alternatively, maybe I made a mistake in the derivation. Let me check.Wait, when I set up the equation:(a^2 - c^2)/(a^2 + c^2) = (a + b)/(2(a - b))And then cross-multiplied to get:2(a - b)(a^2 - c^2) = (a + b)(a^2 + c^2)Then expanded both sides:Left: 2a^3 - 2a c^2 - 2b a^2 + 2b c^2Right: a^3 + a c^2 + b a^2 + b c^2Subtracting right from left:2a^3 - 2a c^2 - 2b a^2 + 2b c^2 - a^3 - a c^2 - b a^2 - b c^2 = 0Simplify:a^3 - 3a c^2 - 3b a^2 + b c^2 = 0Factor:a^2(a - 3b) - c^2(3a - b) = 0So,c^2 = [a^2(a - 3b)] / (3a - b)Yes, that seems correct.So, c^2 must be positive, so (a - 3b)/(3a - b) > 0As before, since a > b, denominator 3a - b is always positive because 3a > a > b, so 3a - b > 0.Therefore, numerator a - 3b must be positive for c^2 to be positive, so a > 3b.Therefore, point C exists only when OA > 3 OB, i.e., a > 3b.Wait, but in the problem statement, it's given that OA > OB, but not necessarily OA > 3 OB. So, does this mean that the construction is only possible when OA > 3 OB?Alternatively, maybe I missed something in the construction.Wait, perhaps there's another way to approach this problem without coordinates.Let me think about it geometrically.We have a right angle at O, with A and B on one leg (x-axis), and we need to find C on the other leg (y-axis) such that ∠BCA = 2∠BAC.Let me denote ∠BAC = α, so ∠BCA = 2α.In triangle ABC, the sum of angles is 180°, so ∠ABC = 180° - α - 2α = 180° - 3α.Hmm, okay.Alternatively, maybe I can use the concept of circle of Apollonius or some other geometric construction.Wait, another approach: Let me consider the locus of points C such that ∠BCA = 2∠BAC.This might be a circle or some conic section.Alternatively, maybe I can use trigonometric identities.Wait, earlier, I found that c = a sqrt[(a - 3b)/(3a - b)].So, if a > 3b, then c is real, otherwise, it's imaginary, meaning no such point C exists.Therefore, the necessary and sufficient condition for the existence of point C is that OA > 3 OB, i.e., a > 3b.So, in terms of the problem, the construction is possible only when OA > 3 OB.Therefore, the answer is that point C exists if and only if OA > 3 OB, i.e., a > 3b.So, the condition is OA > 3 OB.But let me verify this with an example.Suppose OA = 4, OB = 1, so a = 4, b = 1.Then, c = 4 sqrt[(4 - 3*1)/(3*4 - 1)] = 4 sqrt[(1)/(11)] = 4/sqrt(11) ≈ 1.206.So, point C is at (0, 4/sqrt(11)).Let me check the angles.Compute tan α = c/a = (4/sqrt(11))/4 = 1/sqrt(11) ≈ 0.3015.So, α ≈ arctan(0.3015) ≈ 16.8°.Then, 2α ≈ 33.6°.Now, compute ∠BCA.Point C is at (0, 4/sqrt(11)), B is at (1,0), A is at (4,0).Compute vectors CB and CA.Vector CB = B - C = (1,0) - (0,4/sqrt(11)) = (1, -4/sqrt(11))Vector CA = A - C = (4,0) - (0,4/sqrt(11)) = (4, -4/sqrt(11))Compute the angle between CB and CA, which is ∠BCA.The angle between vectors CB and CA can be found using the dot product:cos(theta) = (CB · CA) / (|CB| |CA|)Compute CB · CA = (1)(4) + (-4/sqrt(11))(-4/sqrt(11)) = 4 + (16/11) = (44 + 16)/11 = 60/11|CB| = sqrt(1^2 + (4/sqrt(11))^2) = sqrt(1 + 16/11) = sqrt(27/11) = 3 sqrt(3/11)|CA| = sqrt(4^2 + (4/sqrt(11))^2) = sqrt(16 + 16/11) = sqrt(192/11) = 4 sqrt(12/11)So,cos(theta) = (60/11) / [3 sqrt(3/11) * 4 sqrt(12/11)] = (60/11) / [12 sqrt(36/121)] = (60/11) / [12 * 6/11] = (60/11) / (72/11) = 60/72 = 5/6So, theta = arccos(5/6) ≈ 33.56°, which is approximately 2α (since α ≈ 16.8°, 2α ≈ 33.6°). So, it checks out.Therefore, when OA = 4, OB = 1, point C exists at (0, 4/sqrt(11)).Now, let me test when OA = 2, OB = 1, so a = 2, b = 1.Then, c = 2 sqrt[(2 - 3*1)/(3*2 - 1)] = 2 sqrt[(-1)/(5)] which is imaginary. So, no such point C exists.Therefore, the condition OA > 3 OB is necessary and sufficient.So, the answer is that point C exists if and only if OA > 3 OB, i.e., OA > 3 OB.Therefore, the condition is OA > 3 OB.So, the final answer is OA > 3 OB, which can be written as OA > 3 OB.But in terms of the problem statement, it's OA > OB, but for the construction to be possible, OA must be greater than 3 OB.Therefore, the necessary and sufficient condition is OA > 3 OB.So, the answer is OA > 3 OB.**Final Answer**The necessary and sufficient condition for the existence of point ( C ) is boxed{OA > 3OB}.

💡First, I identify the known quantities: 15 kg of carrots and 8 kg of broccoli. Let Z represent the unknown weight of zucchini.Next, I calculate the total weight of all vegetables installed by adding the known quantities and the unknown weight:15 kg (carrots) + Z kg (zucchini) + 8 kg (broccoli) = 23 + Z kg.The merchant sold half of the total vegetables, which equals 18 kg. Therefore, I set up the equation:(23 + Z) / 2 = 18.To solve for Z, I multiply both sides of the equation by 2:23 + Z = 36.Finally, I subtract 23 from both sides to find the value of Z:Z = 13 kg.

💡Okay, so I need to figure out where the vertical asymptotes are for this function: ( f(x) = frac{x^3 + 3x^2 + 2x + 12}{x^2 - 5x + 6} ). I remember that vertical asymptotes happen where the denominator is zero, but the numerator isn't zero at those points. So, first things first, I should factor the denominator to find those x-values.Let me look at the denominator: ( x^2 - 5x + 6 ). I need to factor this quadratic. Hmm, what two numbers multiply to 6 and add up to -5? Well, -2 and -3 do that because (-2) * (-3) = 6 and (-2) + (-3) = -5. So, the denominator factors to ( (x - 2)(x - 3) ). That means the denominator is zero when x is 2 or 3. So, potential vertical asymptotes at x = 2 and x = 3.But before I conclude that, I need to check if the numerator is also zero at these points. If the numerator is zero at x = 2 or x = 3, then those would be holes instead of vertical asymptotes. So, let's plug x = 2 and x = 3 into the numerator: ( x^3 + 3x^2 + 2x + 12 ).First, x = 2:( 2^3 + 3(2)^2 + 2(2) + 12 = 8 + 12 + 4 + 12 = 36 ). That's not zero.Next, x = 3:( 3^3 + 3(3)^2 + 2(3) + 12 = 27 + 27 + 6 + 12 = 72 ). Also not zero.So, since the numerator isn't zero at x = 2 or x = 3, those points are indeed vertical asymptotes. I don't think I made any mistakes here, but just to be sure, maybe I should double-check my factoring of the denominator. Let me expand ( (x - 2)(x - 3) ) to see if it matches the original denominator.Multiplying out: ( x*x = x^2 ), ( x*(-3) = -3x ), ( (-2)*x = -2x ), and ( (-2)*(-3) = 6 ). Combining like terms: ( x^2 - 5x + 6 ). Yep, that's correct.Alright, so I think I've got it. The vertical asymptotes are at x = 2 and x = 3.

💡Okay, so I have this problem here: I need to find the constant term in the expansion of (x² - 1/x)^n, where n is given as the integral from 1 to e of 6/x dx. Hmm, let me break this down step by step.First, I need to figure out what n is. It says n equals the integral from 1 to e of 6/x dx. I remember that the integral of 1/x dx is ln|x| + C, so integrating 6/x should just be 6 times ln|x|, right? So, evaluating from 1 to e, that would be 6 times (ln e - ln 1). I know that ln e is 1 because the natural logarithm of e is 1, and ln 1 is 0 because e^0 is 1. So, n should be 6*(1 - 0) which is 6. Okay, so n is 6. Got that down.Now, the main part is finding the constant term in the expansion of (x² - 1/x)^6. I remember that when expanding a binomial like (a + b)^n, each term is given by the binomial theorem: C(n, k) * a^(n-k) * b^k, where C(n, k) is the combination of n things taken k at a time. So, in this case, a is x² and b is -1/x.So, the general term in the expansion would be C(6, k) * (x²)^(6 - k) * (-1/x)^k. Let me write that out: C(6, k) * x^(12 - 2k) * (-1)^k * x^(-k). Combining the exponents of x, that would be x^(12 - 2k - k) which simplifies to x^(12 - 3k). So, each term has an exponent of x equal to 12 - 3k.I need to find the constant term, which means the exponent of x should be zero. So, I set 12 - 3k equal to zero and solve for k: 12 - 3k = 0. Subtracting 12 from both sides gives -3k = -12, and dividing both sides by -3 gives k = 4. So, the constant term corresponds to k = 4.Now, plugging k = 4 back into the general term: C(6, 4) * (-1)^4 * x^(12 - 12). The x term becomes x^0, which is 1, so we can ignore that. Now, C(6, 4) is the combination of 6 things taken 4 at a time. I remember that C(n, k) is equal to n! / (k!(n - k)!). So, C(6, 4) is 6! / (4! * 2!) which is (720) / (24 * 2) = 720 / 48 = 15.Also, (-1)^4 is 1 because any negative number raised to an even power is positive. So, putting it all together, the constant term is 15 * 1 = 15.Wait, let me double-check that. So, n = 6, k = 4. C(6, 4) is indeed 15, and (-1)^4 is 1. So, yes, the constant term is 15. I think that's correct.Just to make sure I didn't make a mistake earlier, let me recap: n was found by integrating 6/x from 1 to e, which gave me 6. Then, using the binomial theorem, I found that the exponent of x in each term is 12 - 3k, set that equal to zero for the constant term, solved for k = 4, and then calculated the coefficient as C(6, 4) * (-1)^4 = 15 * 1 = 15. Yep, that seems solid.I don't think I made any calculation errors. Maybe I should verify the combination calculation. C(6, 4) is the same as C(6, 2) because C(n, k) = C(n, n - k). C(6, 2) is 15 as well, since 6*5 / 2*1 = 15. So, that checks out.Alright, I'm pretty confident now that the constant term is 15.