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❓:Problem 10.1. A bus leaves city A for city B, which is 240 km away from A, at a speed of 40 km/h. At the same time, a car leaves B for A at a speed of v km/h. Half an hour after meeting the bus, the car, without reaching city A, turns around and heads back towards B at the same speed. Determine all values of v for which the car arrives in B before the bus.
🧠:[<think>Okay, let me try to figure this problem out. So, there's a bus going from city A to city B, which is 240 km apart. The bus speed is 40 km/h. At the same time, a car leaves city B going towards city A at speed v km/h. Then, half an hour after they meet, the car turns around and goes back to B at the same speed. We need to find all values of v so that the car arrives back in B before the bus arrives in B.First, let me visualize the scenario. Both the bus and the car start at the same time, moving towards each other. They meet somewhere in between. After meeting, the car continues towards A for half an hour, then turns around and goes back to B. We need to find the speed v such that the total time the car takes to go back to B is less than the time the bus takes to reach B from A.Let me break it down step by step.1. Time until meeting point: Let's denote the time taken until they meet as t hours. In this time, the bus would have traveled 40t km, and the car would have traveled v*t km. Since they are moving towards each other, the sum of these distances should be equal to the total distance between A and B, which is 240 km. So,40t + vt = 240Therefore, solving for t:t(40 + v) = 240t = 240 / (40 + v)So, they meet after t = 240/(40 + v) hours.2. Distance covered by each until meeting: The bus has traveled 40t km, which is 40*(240/(40 + v)) = 9600/(40 + v) km.Similarly, the car has traveled v*t = v*(240/(40 + v)) = 240v/(40 + v) km.So, at the meeting point, the bus is 9600/(40 + v) km away from A, and the car is 240v/(40 + v) km away from B.Wait, actually, since the car started from B, the distance from the meeting point to B is 240v/(40 + v) km, right? Because the car traveled towards A for t hours, so the remaining distance to A would be 240 - (bus's distance) = 240 - 9600/(40 + v) km. But maybe I need to confirm that.Wait, the total distance is 240 km. The bus has covered 40t km, so the remaining distance from the meeting point to B is 240 - 40t = 240 - 40*(240/(40 + v)) = 240 - 9600/(40 + v) km. Alternatively, since the car has traveled vt km from B towards A, the distance from the meeting point to B is vt = 240v/(40 + v). Hmm, but 240 - 40t should be equal to vt.Let me check:From the equation 40t + vt = 240, so vt = 240 - 40t. Therefore, the distance from meeting point to B is vt = 240 - 40t. Which is correct. So, since the car is at the meeting point after time t, it has to go back to B, which is vt km away. Wait, no. Wait, when the car meets the bus, it is at the meeting point. From that point, if the car continues towards A, then half an hour later, it turns around. So, first, after meeting, the car continues towards A for half an hour (0.5 hours) at speed v, so it covers an additional distance of 0.5v km. Then, it turns around and goes back to B. So, the total distance the car needs to cover after turning around is the distance from the point where it turned around to B.Wait, no. Let me think again.After meeting, the car continues towards A for half an hour. Then, turns around and heads back to B. So, during that half hour, the car was moving towards A, so it increased its distance from B. Wait, no, the meeting point is somewhere between A and B. If the car continues towards A after the meeting, then it is moving away from B. So, when it turns around after half an hour, it has to go back towards B, covering the distance it just traveled towards A plus the original distance from the meeting point to B. Wait, no. Wait, let me clarify.At the time of meeting, the car is at the meeting point, which is a certain distance from B. Let's denote the distance from meeting point to B as D. Then, after meeting, the car continues towards A for 0.5 hours, so it moves further away from B. The distance covered in that 0.5 hours is 0.5v km. So, the new position of the car when it turns around is (D + 0.5v) km away from B. Then, the car has to travel back to B, which is (D + 0.5v) km, at speed v. So, the time taken to return to B would be (D + 0.5v)/v hours.But wait, no. Wait, if the car is moving towards A for half an hour after meeting, then from the meeting point, it's moving towards A, so the distance from B is increasing. Wait, no, actually, the meeting point is between A and B. If the car is moving towards A, then it's moving away from B. So, the distance from B when it turns around is D + 0.5v. Therefore, to return to B, it has to cover that distance at speed v, which takes (D + 0.5v)/v hours. So, total time after meeting until arriving in B is 0.5 hours + (D + 0.5v)/v hours.But D is the distance from meeting point to B, which is equal to vt, as we had earlier. Wait, D is vt. Because in time t, the car traveled vt km from B to the meeting point. So D = vt.Wait, but earlier, we had D = 240v/(40 + v). Since t = 240/(40 + v), so vt = v*(240/(40 + v)) = 240v/(40 + v). Therefore, D = 240v/(40 + v).Therefore, when the car turns around, it is D + 0.5v km away from B. Therefore, the time to return to B is (D + 0.5v)/v = (240v/(40 + v) + 0.5v)/v = [240v/(40 + v) + 0.5v]/v = 240/(40 + v) + 0.5.So total time after meeting is 0.5 hours (time moving towards A) plus [240/(40 + v) + 0.5] hours (time moving back to B). Wait, that would be 0.5 + 240/(40 + v) + 0.5 = 240/(40 + v) + 1 hours.Wait, hold on, let's double-check. The time after meeting is divided into two parts: 0.5 hours going towards A, then (distance)/(speed) going back to B. The distance to go back is D + 0.5v, where D = vt = 240v/(40 + v). So, total distance is 240v/(40 + v) + 0.5v. Then, time is (240v/(40 + v) + 0.5v)/v = 240/(40 + v) + 0.5.Therefore, total time after meeting until arrival in B is 0.5 + (240/(40 + v) + 0.5) = 240/(40 + v) + 1 hours.But wait, 0.5 hours moving towards A, then (D + 0.5v)/v hours moving back. Let's compute that:Time after meeting = 0.5 + (D + 0.5v)/v = 0.5 + (240v/(40 + v) + 0.5v)/vSimplify the second term:(240v/(40 + v) + 0.5v)/v = [240v/(40 + v)]/v + (0.5v)/v = 240/(40 + v) + 0.5So total time after meeting is 0.5 + 240/(40 + v) + 0.5 = 1 + 240/(40 + v) hours.So the total time for the car to reach B after the meeting is 1 + 240/(40 + v) hours.But we need to compare this with the time the bus takes to reach B after the meeting.The bus, after the meeting, still has to cover the remaining distance to B, which is D_bus_remaining = 240 - 40t.But t was the time until meeting, which is 240/(40 + v). So, D_bus_remaining = 240 - 40*(240/(40 + v)) = 240 - 9600/(40 + v)Therefore, the time the bus takes to reach B after the meeting is D_bus_remaining / 40 = [240 - 9600/(40 + v)] / 40Let me compute that:First, 240 = 240*(40 + v)/(40 + v) = (9600 + 240v)/(40 + v)Therefore, D_bus_remaining = (9600 + 240v)/(40 + v) - 9600/(40 + v) = (240v)/(40 + v)Therefore, time for bus after meeting is (240v)/(40 + v) / 40 = (240v)/(40*(40 + v)) = (6v)/(40 + v) hours.So, the bus takes (6v)/(40 + v) hours after meeting to reach B.We need the car's total time after meeting (which is 1 + 240/(40 + v) hours) to be less than the bus's time after meeting (which is 6v/(40 + v) hours).So, set up the inequality:1 + 240/(40 + v) < 6v/(40 + v)Multiply both sides by (40 + v) to eliminate the denominator (since 40 + v is positive, the inequality direction remains the same):1*(40 + v) + 240 < 6vSimplify left side:40 + v + 240 = v + 280So:v + 280 < 6vSubtract v from both sides:280 < 5vDivide both sides by 5:56 < vSo, v > 56 km/h.Therefore, the car needs to travel faster than 56 km/h to arrive in B before the bus.Wait, but let me check if I did the time calculations correctly.The car's total time after meeting is 0.5 hours going towards A, then turning around and going back to B. The distance from the turn-around point to B is D + 0.5v, where D was the original distance from meeting point to B (D = vt = 240v/(40 + v)). So, total distance back is 240v/(40 + v) + 0.5v. Then, time is (240v/(40 + v) + 0.5v)/v = 240/(40 + v) + 0.5. Then total time after meeting is 0.5 + 240/(40 + v) + 0.5 = 1 + 240/(40 + v). Wait, that seems correct.But when the car turns around, it has to cover the distance it went past the meeting point towards A. Wait, no. Wait, at the meeting point, the car is D = 240v/(40 + v) km away from B. Then, it goes towards A for half an hour at speed v, so covers 0.5v km. So, the distance from B becomes D + 0.5v. Therefore, when it turns around, it has to cover D + 0.5v km to get back to B, which takes (D + 0.5v)/v hours. That seems correct.So, total time after meeting is 0.5 + (D + 0.5v)/v = 0.5 + (240v/(40 + v) + 0.5v)/v = 0.5 + 240/(40 + v) + 0.5 = 1 + 240/(40 + v). Correct.On the other hand, the bus's remaining distance is 240 - 40t = 240 - 40*(240/(40 + v)) = 240 - 9600/(40 + v) = (240*(40 + v) - 9600)/(40 + v) = (9600 + 240v - 9600)/(40 + v) = 240v/(40 + v). Then, time is 240v/(40*(40 + v)) = 6v/(40 + v). Correct.Therefore, the inequality is 1 + 240/(40 + v) < 6v/(40 + v)Multiply both sides by (40 + v):40 + v + 240 < 6vWhich simplifies to 280 + v < 6vSubtract v:280 < 5v => v > 56So, v must be greater than 56 km/h. Therefore, all values of v greater than 56 km/h satisfy the condition that the car arrives in B before the bus.Wait, but let me check with a sample value. Let's take v = 60 km/h.First, compute t = 240/(40 + 60) = 240/100 = 2.4 hours. So, meeting occurs after 2.4 hours.At that time, the bus has traveled 40*2.4 = 96 km, so remaining distance is 240 - 96 = 144 km. The bus's time to reach B after meeting is 144/40 = 3.6 hours.The car, after meeting, goes towards A for 0.5 hours at 60 km/h, covering 30 km. So, the car is now 96 km (distance from A to meeting point) + 30 km = 126 km from A, which is 240 - 126 = 114 km from B. Then, the car turns around and goes back to B at 60 km/h. The time to cover 114 km is 114/60 = 1.9 hours. So total time after meeting is 0.5 + 1.9 = 2.4 hours.But according to the formula, total time after meeting for the car should be 1 + 240/(40 + 60) = 1 + 240/100 = 1 + 2.4 = 3.4 hours. Wait, that contradicts the manual calculation. Wait, wait, something's wrong here.Wait, when I calculated manually, the car's time after meeting was 0.5 hours going towards A, then 1.9 hours going back, total 2.4 hours. But according to the formula, it was 1 + 240/(40 + v) = 1 + 2.4 = 3.4 hours. There's a discrepancy here.Wait, perhaps my formula is wrong. Let me check.The formula says total time after meeting is 1 + 240/(40 + v). But according to the manual calculation, it's 0.5 + (D + 0.5v)/v, where D = 240v/(40 + v). Let's compute that.For v=60, D = 240*60/(40 + 60) = (240*60)/100 = 144 km. So, D = 144 km. Then, after moving towards A for 0.5 hours at 60 km/h, the car covers 30 km, so distance from B becomes 144 + 30 = 174 km. Then, time to return to B is 174/60 = 2.9 hours. Wait, wait, that contradicts the previous manual calculation. Wait, hold on.Wait, the meeting point is 96 km from A (bus traveled 96 km in 2.4 hours), so the meeting point is 240 - 96 = 144 km from B. So, D = 144 km. Then, the car goes towards A for half an hour (0.5 hours) at 60 km/h, so it travels 30 km towards A. Therefore, from the meeting point, which is 144 km from B, moving 30 km towards A would put the car 144 + 30 = 174 km from B. Therefore, the distance to B is 174 km. Then, turning around, the car has to go 174 km back to B at 60 km/h, which takes 174/60 = 2.9 hours. Therefore, total time after meeting is 0.5 + 2.9 = 3.4 hours. Which matches the formula. But in my initial manual calculation, I thought the distance from B after moving towards A was 114 km, but that was incorrect. I subtracted 126 km from 240 km, but the meeting point is 144 km from B, moving 30 km towards A adds 30 km to the distance from B, making it 174 km. So, total distance to B is 174 km, not 114. That was my mistake.Therefore, the formula is correct. The car's total time after meeting is 3.4 hours when v=60. The bus's time after meeting is 144 km /40 km/h = 3.6 hours. Therefore, 3.4 < 3.6, which is correct. So, the car arrives earlier.Therefore, the formula gives correct result here.Similarly, if v=56 km/h, let's check.t = 240/(40 + 56) = 240/96 = 2.5 hours. So, meeting occurs after 2.5 hours.Bus's remaining distance: 240 - 40*2.5 = 240 - 100 = 140 km. Time for bus: 140/40 = 3.5 hours.Car's time after meeting: 1 + 240/(40 + 56) = 1 + 240/96 = 1 + 2.5 = 3.5 hours. So, equal. Therefore, at v=56, they arrive at the same time.Therefore, the inequality v>56 is correct.Therefore, the answer is v > 56 km/h.But wait, the problem says "the car arrives in B before the bus". So, strictly greater than 56. So, all values of v greater than 56 km/h. Therefore, the answer is v > 56, which in boxed form is boxed{v > 56}Wait, but the problem says "determine all values of v", so we need to present the answer as an inequality. Since the problem might expect the answer in boxed notation as boxed{v > 56} or maybe using interval notation, but probably as an inequality.Alternatively, check if there's a lower or upper limit. For example, can v be any speed greater than 56, or is there a maximum possible speed? The problem doesn't specify any constraints on v other than it being positive, I think. So, since the car is traveling towards A initially, but turns around half an hour after meeting, so even if v is extremely high, say 1000 km/h, would that make sense? Let's check.Suppose v is 1000 km/h.t = 240/(40 + 1000) ≈ 240/1040 ≈ 0.2308 hours ≈ 13.85 minutes.Meeting occurs very close to city A. Then, the car goes towards A for half an hour (30 minutes) at 1000 km/h, which would be 500 km. But since the meeting point was only 0.2308 hours * 1000 km/h ≈ 230.8 km from B, so the distance from meeting point to A is 240 - 230.8 ≈ 9.2 km. But the car is going towards A for half an hour at 1000 km/h, which would take it 500 km, but there's only 9.2 km left to A. Therefore, the car would reach A in 9.2/1000 = 0.0092 hours ≈ 0.55 minutes, which is much less than half an hour. Therefore, in reality, the car would reach A before the half hour is up, and then perhaps turn around immediately? But the problem states that "half an hour after meeting the bus, the car, without reaching city A, turns around..."Ah! The problem says "without reaching city A", so the car turns around before getting to A. Therefore, the half hour journey towards A must not reach A. Therefore, there's a constraint that the distance the car travels towards A after the meeting must be less than the remaining distance to A at the meeting point.So, at the meeting point, the distance from meeting point to A is 240 - D, where D is the distance from meeting point to B. Wait, the distance from meeting point to A is the distance the bus has traveled, which is 40t. So, 40t = 40*(240/(40 + v)) = 9600/(40 + v) km.Therefore, the car, after meeting, travels towards A for 0.5 hours at speed v, covering 0.5v km. To not reach A, this distance must be less than the remaining distance to A, which is 9600/(40 + v) km.So, 0.5v < 9600/(40 + v)Multiply both sides by (40 + v):0.5v(40 + v) < 9600Which is 0.5v*40 + 0.5v^2 < 960020v + 0.5v^2 < 9600Multiply both sides by 2:40v + v^2 < 19200v^2 + 40v - 19200 < 0Solve the quadratic inequality v^2 + 40v - 19200 < 0Find roots:v = [-40 ± sqrt(1600 + 76800)]/2 = [-40 ± sqrt(78400)]/2 = [-40 ± 280]/2Positive root: (-40 + 280)/2 = 240/2 = 120Negative root: (-40 - 280)/2 = -320/2 = -160Since speed can't be negative, the relevant root is 120. The quadratic opens upwards (coefficient of v^2 is positive), so the inequality v^2 + 40v - 19200 < 0 holds between -160 and 120. Since v is positive, the solution is 0 < v < 120.Therefore, the car cannot have a speed of 120 km/h or higher, because at v=120, the car would just reach A in half an hour. Wait, let's check v=120:At v=120, t = 240/(40 + 120) = 240/160 = 1.5 hours. The distance from meeting point to A is 40*1.5 = 60 km. The car travels towards A at 120 km/h for 0.5 hours, covering 60 km, which is exactly the remaining distance to A. Therefore, the car would reach A exactly when it turns around. But the problem states "without reaching city A", so the car must turn around before reaching A. Therefore, v must be less than 120 km/h.Therefore, there's an upper bound on v: v < 120 km/h.But in our previous analysis, we found that v must be greater than 56 km/h. So, combining both constraints, we have 56 < v < 120.But wait, is this correct? Let's check with v=100 km/h.Check if the car doesn't reach A:t = 240/(40 + 100) = 240/140 ≈ 1.714 hours.Distance from meeting point to A is 40t ≈ 40*1.714 ≈ 68.57 km.After meeting, the car travels towards A at 100 km/h for 0.5 hours, covering 50 km. Since 50 < 68.57, it does not reach A. Then, the distance from turn-around point to B is D + 0.5v = 100t + 0.5*100 = 100*(240/140) + 50 ≈ 171.43 + 50 = 221.43 km.Wait, no. Wait, D was the distance from meeting point to B, which is vt = 100*1.714 ≈ 171.43 km. After moving towards A for 0.5 hours at 100 km/h, the car is 171.43 + 50 = 221.43 km from B. Then, time to return to B is 221.43/100 ≈ 2.214 hours. Total time after meeting is 0.5 + 2.214 ≈ 2.714 hours.Bus's time after meeting is 6v/(40 + v) = 6*100/140 ≈ 4.285 hours.So, 2.714 < 4.285, so car arrives earlier. So, v=100 is acceptable and satisfies both 56 < v < 120.But if we take v=120:t = 240/(40 + 120) = 240/160 = 1.5 hours.Distance from meeting point to A is 40*1.5 = 60 km.Car travels towards A at 120 km/h for 0.5 hours: 120*0.5 = 60 km. So, it reaches A exactly at the time of turning around. But the problem says "without reaching city A", so v must be less than 120. Therefore, the upper limit is v < 120.Similarly, at v approaching 120 from below, the time the car takes would approach 0.5 + (D + 0.5v)/v. When v=120, D = 120*1.5 = 180 km. So, D + 0.5v = 180 + 60 = 240 km. So, time to return is 240/120 = 2 hours. Total time after meeting is 0.5 + 2 = 2.5 hours. The bus's time after meeting is 6*120/(40 + 120) = 720/160 = 4.5 hours. So, even if the car reaches A and turns around immediately (though the problem says it must turn around before reaching A), but in that edge case, it would still arrive earlier. However, according to the problem statement, the car must not reach A, so v must be less than 120.Therefore, combining both constraints:The car must have a speed v such that 56 < v < 120.But wait, let's check with v approaching 120 from below, say v=119 km/h.t = 240/(40 + 119) ≈ 240/159 ≈ 1.509 hours.Distance from meeting point to A is 40*1.509 ≈ 60.36 km.Car travels towards A at 119 km/h for 0.5 hours: 119*0.5 = 59.5 km, which is less than 60.36 km. So, doesn't reach A. Then, distance from turn-around point to B is D + 0.5v = 119*1.509 + 59.5 ≈ 179.57 + 59.5 ≈ 239.07 km. Time to return to B is 239.07/119 ≈ 2.009 hours. Total time after meeting is 0.5 + 2.009 ≈ 2.509 hours.Bus's time after meeting is 6v/(40 + v) = 6*119/159 ≈ 4.49 hours. So, car arrives earlier. So even at v=119, which is less than 120, it works. Therefore, the upper limit is v < 120, but the lower limit is v > 56.Thus, the answer is all v such that 56 < v < 120.But in the original problem statement, there is no mention of an upper limit, but our analysis shows that there is a constraint due to the car not reaching A. Therefore, the correct answer should be 56 < v < 120.Wait, but let me confirm with v=120, even though the problem says "without reaching city A". If the car reaches A exactly when it turns around, is that allowed? The problem says "without reaching city A", so turning around at the city would not satisfy the condition. Therefore, v must be strictly less than 120.Therefore, the values of v are 56 < v < 120.But wait, in our initial inequality, we found that v > 56, but we need to also ensure that the car doesn't reach A, leading to v < 120. Therefore, combining both, the solution is 56 < v < 120.Therefore, the answer is all v between 56 and 120 km/h, not inclusive.But let me check if there's another constraint that we might have missed.For example, the total time the car takes to go back to B must be less than the bus's total time. Wait, the bus's total time to go from A to B is 240/40 = 6 hours. The car's total time is t (time until meeting) + 0.5 hours (towards A) + time to go back to B.But the problem states that the car arrives in B before the bus. The bus arrives in B at 6 hours after departure. So, the car's total time must be less than 6 hours.Wait, let's check that.The car's total time is t + 0.5 + (D + 0.5v)/v.Which is t + 0.5 + (vt + 0.5v)/v = t + 0.5 + t + 0.5 = 2t + 1.So, the car's total time is 2t + 1.But t = 240/(40 + v). Therefore, total time is 2*(240/(40 + v)) + 1 = 480/(40 + v) + 1.The bus's total time is 6 hours.We need 480/(40 + v) + 1 < 6.Solve this inequality:480/(40 + v) < 5Multiply both sides by (40 + v):480 < 5*(40 + v)480 < 200 + 5v280 < 5v56 < vSo, again, we get v > 56. But this seems to ignore the upper limit. Wait, but in this case, the car's total time is calculated as 2t + 1, but if the car turns around before reaching A, then the total time is indeed 2t + 1. However, if the car were to reach A before the half hour is up, then the total time would be different. But since the problem states that the car turns around half an hour after meeting without reaching A, the total time must be 2t + 1, regardless of the value of v, provided that 0.5v < distance from meeting point to A.Wait, but actually, even if the car didn't reach A, the total time is t + 0.5 + (D + 0.5v)/v = t + 0.5 + vt/v + 0.5v/v = t + 0.5 + t + 0.5 = 2t + 1. Therefore, the total time is always 2t + 1, given that the car turns around after exactly half an hour, even if it didn't reach A.But in order for the car to turn around after half an hour without reaching A, it must be that 0.5v < distance from meeting point to A, which is 40t.So, 0.5v < 40t.But t = 240/(40 + v). Therefore:0.5v < 40*(240/(40 + v))Multiply both sides by (40 + v):0.5v*(40 + v) < 9600Which is the same as before, leading to v < 120.Therefore, the total time for the car is 2t + 1, and we have to ensure two things:1. The car arrives before the bus: 2t + 1 < 6.2. The car does not reach A before turning around: 0.5v < 40t.From 1, we get 2*(240/(40 + v)) + 1 < 6, which simplifies to 480/(40 + v) < 5, leading to v > 56.From 2, we get 0.5v < 40*(240/(40 + v)), leading to v < 120.Therefore, combining both conditions, the solution is 56 < v < 120.Therefore, the answer should be all speeds v greater than 56 km/h and less than 120 km/h.But the original problem says "the car arrives in B before the bus". So, if the car's speed is between 56 and 120, it arrives before the bus. If the speed is 56 or 120, what happens?At v=56, car's total time is 2*(240/(40 + 56)) + 1 = 2*(240/96) + 1 = 2*(2.5) + 1 = 5 + 1 = 6 hours, same as the bus. So, arrives at the same time. But the problem says "before", so v must be greater than 56.At v=120, car would reach A while turning around, which is not allowed. So, v must be less than 120.Therefore, the final answer is 56 < v < 120.But the problem says "determine all values of v", so we need to present this as 56 < v < 120. In boxed form, maybe using interval notation: boxed{(56, 120)}. Or as an inequality: boxed{56 < v < 120}.Given that the problem is from a math competition or textbook, likely expects the inequality form. So, I think the answer is boxed{56 < v < 120}Final AnswerThe values of ( v ) are all real numbers greater than 56 and less than 120. Thus, the solution is boxed{56 < v < 120}.</think>To determine the values of ( v ) for which the car arrives in city ( B ) before the bus, we start by analyzing the meeting point of the bus and the car. 1. Time until meeting point: The bus and car meet after time ( t ) hours, where: [ t = frac{240}{40 + v} ] At this meeting point, the bus has traveled ( 40t ) km and the car has traveled ( vt ) km.2. Distance covered by each until meeting: - The bus is ( 40t = frac{9600}{40 + v} ) km away from city ( A ). - The car is ( vt = frac{240v}{40 + v} ) km away from city ( B ).3. Car's journey after meeting: - The car continues towards city ( A ) for 0.5 hours, covering ( 0.5v ) km. - The car then turns around and heads back to city ( B ). The distance from the turn-around point to city ( B ) is ( frac{240v}{40 + v} + 0.5v ). - The time taken to return to city ( B ) is: [ frac{frac{240v}{40 + v} + 0.5v}{v} = frac{240}{40 + v} + 0.5 ] - Total time after meeting for the car is: [ 0.5 + left( frac{240}{40 + v} + 0.5 right) = 1 + frac{240}{40 + v} ]4. Bus's journey after meeting: - The remaining distance for the bus is ( 240 - 40t ). - Time taken by the bus to reach city ( B ) after meeting is: [ frac{240 - 40t}{40} = frac{240 - frac{9600}{40 + v}}{40} = frac{6v}{40 + v} ]5. Inequality for arrival time comparison: - We need the car's total time after meeting to be less than the bus's time: [ 1 + frac{240}{40 + v} < frac{6v}{40 + v} ] - Solving this inequality: [ 40 + v + 240 < 6v implies 280 < 5v implies v > 56 ]6. Constraint for not reaching city ( A ): - The car must not reach city ( A ) before turning around: [ 0.5v < frac{9600}{40 + v} ] - Solving this inequality: [ 0.5v(40 + v) < 9600 implies v^2 + 40v - 19200 < 0 implies v < 120 ]Combining both conditions, the values of ( v ) must satisfy ( 56 < v < 120 ).Thus, the solution is boxed{56 < v < 120}.,<think>Okay, let's try to tackle this problem step by step. So, there's a bus going from city A to city B, which is 240 km apart. The bus's speed is 40 km/h. At the same time, a car leaves city B going towards A at a speed of v km/h. Then, half an hour after meeting the bus, the car turns around without reaching city A and goes back to B at the same speed. We need to find all values of v so that the car arrives in B before the bus does. Hmm, okay.First, let me visualize the scenario. Both the bus and the car start at the same time, moving towards each other. They meet somewhere in between. After the meeting point, the car continues towards A for half an hour, then turns around and heads back to B. We need to make sure that the total time the car takes to go from B to the meeting point, then half an hour towards A, then back to B is less than the time the bus takes to go from A to B.Wait, but the bus is going straight from A to B at 40 km/h. The total time for the bus would just be distance divided by speed: 240 km / 40 km/h = 6 hours. So the bus arrives in B after 6 hours. The car needs to arrive in B before 6 hours. Got it.But the car's journey is more complicated. It starts at B, moves towards A, meets the bus somewhere, continues towards A for half an hour, then turns around and goes back to B. So the car's total time is the time until the meeting point, plus half an hour, plus the time from turning around back to B. This total time has to be less than 6 hours.First, let's find the meeting time and meeting point. Let me denote the time until they meet as t hours. In that time, the bus would have traveled 40t km, and the car would have traveled vt km. Since they are moving towards each other, the sum of the distances they cover until meeting should be equal to the total distance between A and B, which is 240 km. So:40t + vt = 240Therefore, t = 240 / (40 + v). That seems right.So the meeting point is at 40t km from A, which is 40*(240/(40 + v)) = 9600/(40 + v) km from A. Similarly, from B, the meeting point is vt = v*(240/(40 + v)) = 240v/(40 + v) km.After meeting, the car continues towards A for half an hour (0.5 hours) at speed v. So in that half hour, the car would cover 0.5v km. But wait, the car hasn't reached A yet, so the distance from the meeting point to A must be more than 0.5v km. Let me check:The distance from the meeting point to A is 9600/(40 + v) km. So we must have:9600/(40 + v) > 0.5vBecause the car didn't reach A. So that inequality should hold. Let's note that:9600/(40 + v) > 0.5vMultiply both sides by (40 + v):9600 > 0.5v(40 + v)Multiply both sides by 2:19200 > v(40 + v)Which simplifies to:v^2 + 40v - 19200 < 0Hmm, solving this quadratic inequality. Let's find the roots first:v^2 + 40v - 19200 = 0Using quadratic formula:v = [-40 ± sqrt(1600 + 76800)] / 2sqrt(78400) = 280So v = [-40 ± 280]/2Positive root: (240)/2 = 120Negative root: (-320)/2 = -160Since speed can't be negative, the relevant root is 120. So the inequality v^2 + 40v - 19200 < 0 holds when v is between -160 and 120. Since v is positive, we have 0 < v < 120. So this tells us that v must be less than 120 km/h for the car not to reach A before turning around. But we need to keep this in mind when considering possible values of v. So v has to be less than 120 km/h. Okay, that's a constraint.Now, after the car meets the bus, it travels towards A for 0.5 hours, covering 0.5v km, and then turns around. At that point, the car is 0.5v km away from the meeting point towards A. So its distance from A would be the distance from meeting point to A minus 0.5v, which is (9600/(40 + v) - 0.5v) km. But since the car didn't reach A, that distance is positive. But we already considered that in the previous inequality.Alternatively, perhaps it's better to think in terms of position. Let's set up a coordinate system where city A is at position 0 and city B is at position 240 km. Let's assume the meeting happens at position x from A, so the bus has traveled x km in time t, and the car has traveled (240 - x) km in the same time t.Since the bus's speed is 40 km/h, x = 40t. The car's speed is v, so 240 - x = vt. So again, x = 40t and 240 - x = vt. Then, combining these equations, 40t + vt = 240 => t = 240/(40 + v). So that's the same as before. So x = 40t = 9600/(40 + v) km from A.After meeting, the car continues towards A for half an hour. So in half an hour, the car's position would be x - 0.5v (since it's moving towards A). Then, the car turns around and starts moving back towards B at speed v. So the car's new position is x - 0.5v, and needs to go back to B, which is at 240 km. The distance from the turnaround point to B is 240 - (x - 0.5v) = 240 - x + 0.5v. The time it takes for the car to go back to B from there is (240 - x + 0.5v)/v hours.Therefore, the total time for the car is t + 0.5 + (240 - x + 0.5v)/v.But t is 240/(40 + v), and x is 9600/(40 + v). Let's substitute these values:Total car time = 240/(40 + v) + 0.5 + [240 - 9600/(40 + v) + 0.5v]/vLet me compute each term:First term: 240/(40 + v)Second term: 0.5Third term: [240 - 9600/(40 + v) + 0.5v]/vLet's simplify the third term:First, compute 240 - 9600/(40 + v):240 can be written as 240(40 + v)/(40 + v) - 9600/(40 + v) ?Wait, maybe:240 - 9600/(40 + v) = [240(40 + v) - 9600]/(40 + v)Compute numerator:240*40 + 240v - 9600 = 9600 + 240v - 9600 = 240vTherefore, 240 - 9600/(40 + v) = 240v/(40 + v)So then the third term becomes:[240v/(40 + v) + 0.5v]/v = [v*(240/(40 + v) + 0.5)]/v = 240/(40 + v) + 0.5So the total car time is:240/(40 + v) + 0.5 + 240/(40 + v) + 0.5 = 2*(240/(40 + v)) + 1Simplify:480/(40 + v) + 1So total car time is 480/(40 + v) + 1 hours.The bus's total time is 6 hours. So we need:480/(40 + v) + 1 < 6Subtract 1:480/(40 + v) < 5Multiply both sides by (40 + v):480 < 5*(40 + v)Divide both sides by 5:96 < 40 + vSubtract 40:56 < vSo v must be greater than 56 km/h.But earlier, we had the constraint that v must be less than 120 km/h to ensure the car doesn't reach city A before turning around. So combining both inequalities, v must be between 56 km/h and 120 km/h. However, since the car turns around half an hour after meeting the bus, which is before reaching A. Wait, but we already considered that v must be less than 120 km/h for the car not to reach A. So the final answer should be all v such that 56 < v < 120. Therefore, the values of v are 56 < v < 120.But let me double-check the steps to make sure.First, we found the meeting time t = 240/(40 + v). Then, after meeting, the car goes towards A for 0.5 hours, so covers 0.5v km. Then turns around. At that point, the distance from the car to B is 240 - (x - 0.5v) where x is the meeting point from A. Wait, x = 40t = 9600/(40 + v). So distance from car to B after moving towards A for 0.5h is 240 - (x - 0.5v) = 240 - x + 0.5v. Then, the time to go back is (240 - x + 0.5v)/v.But when we calculated that, substituting x = 9600/(40 + v), we had:240 - 9600/(40 + v) + 0.5v, which simplified to 240v/(40 + v) + 0.5v, then dividing by v gives 240/(40 + v) + 0.5, which seems correct.Therefore, total time for car is t + 0.5 + (240/(40 + v) + 0.5). Which is 240/(40 + v) + 0.5 + 240/(40 + v) + 0.5 = 480/(40 + v) + 1. Then setting that less than 6:480/(40 + v) + 1 < 6 => 480/(40 + v) < 5 => 480 < 5*(40 + v) => 96 < 40 + v => v > 56.And from the earlier constraint, v < 120. So yes, 56 < v < 120.Wait, but let me check with some sample values. If v = 60, then total car time would be 480/(100) + 1 = 4.8 + 1 = 5.8 hours, which is less than 6. So that works. If v = 120, the car would reach A at the meeting point? Wait, if v = 120, then the meeting time is t = 240/(40 + 120) = 240/160 = 1.5 hours. Then, the car goes towards A for half an hour. In half an hour at 120 km/h, it would go 60 km. But the distance from meeting point to A is x = 40*1.5 = 60 km. So 60 km - 60 km would be zero. So the car would reach A exactly when it turns around, but the problem states "without reaching city A", so v must be less than 120. So v = 120 is invalid.Similarly, if v = 56, then total car time is 480/(96) +1 = 5 +1 = 6 hours, which is equal to the bus's time. But we need the car to arrive before the bus, so v must be greater than 56. Hence, the interval is (56, 120). Therefore, the answer is all real numbers v such that 56 < v < 120. So in boxed form: boxed{(56, 120)}Wait, but the problem says "determine all values of v", so maybe in interval notation, but since it's a competition problem, perhaps it's expressed as v > 56 and v < 120. So combining those, 56 < v < 120.But let me check once more. Suppose v = 56, then 480/(40 + 56) + 1 = 480/96 +1 = 5 +1 = 6, which is equal. So the car arrives at the same time as the bus, so we need strictly less than 6, so v must be greater than 56. Similarly, at v approaching 120, the total time would be 480/(160) +1 = 3 +1 = 4 hours, which is way less than 6. So as v increases beyond 56, the total time for the car decreases. So indeed, the car arrives earlier for higher v, but v can't be 120 or more. So the answer is all v greater than 56 and less than 120. Therefore, the final answer is boxed{(56, 120)}.Final Answerboxed{(56, 120)}</think>The bus travels from city A to city B at 40 km/h, a distance of 240 km, taking 6 hours. The car leaves city B for city A at speed v km/h. They meet after time t hours, where t = frac{240}{40 + v}. The meeting point is 40t km from A and vt km from B. After meeting, the car travels towards A for 0.5 hours, covering 0.5v km, then turns around and returns to B. The distance from the meeting point to the turnaround point is 0.5v km. The distance from the turnaround point to B is 240 - (vt - 0.5v) km. The total time for the car is the sum of the time to the meeting point, 0.5 hours, and the return time to B. This total time must be less than 6 hours. The total time for the car is calculated as:[frac{240}{40 + v} + 0.5 + frac{240 - vt + 0.5v}{v}]Substituting vt = frac{240v}{40 + v}, we simplify the expression:[frac{240}{40 + v} + 0.5 + frac{240 - frac{240v}{40 + v} + 0.5v}{v}]Simplifying further, we find the total car time is:[frac{480}{40 + v} + 1]Setting this less than 6 hours:[frac{480}{40 + v} + 1 < 6 implies frac{480}{40 + v} < 5 implies 40 + v > 96 implies v > 56]Additionally, the car must not reach A before turning around, leading to:[frac{9600}{40 + v} > 0.5v implies v < 120]Combining both conditions, the values of v are:[boxed{(56, 120)}]]
🔑:Answer: v in(56 ; 120).Solution. Let t be the time of the meeting, measured in hours. Then, by the condition, t=frac{240}{v+40}. Further, the condition states that the car, having driven for half an hour, did not reach point A, which leads to the inequality v(t+0.5) < 240 - 0.5v.Considering the positivity of the denominators, we obtain the equivalent quadratic inequality v^{2}+40 v-19200 < 0.## CriteriaOne of the largest suitable criteria is used:76 . Any correct solution to the problem.## 2 6. Only the lower estimate is obtained.## 2 6. Only the upper estimate is obtained.Points are deducted for the following error in the solution, which is otherwise correct:-1 6. One or both ends of the interval are incorrectly included in the answer.
❓:For any function f(x) defined on mathbb{R}, if a real number x satisfies f(x) = x, then x is called a fixed point of the function f(x). If the function f(x) = ax^2 + (2a-3)x + 1 has exactly one fixed point, then the set of values for the real number a is.
🧠:[<think>Okay, so I need to find all real numbers a such that the function f(x) = ax² + (2a - 3)x + 1 has exactly one fixed point. A fixed point is where f(x) = x, right? So basically, I need to solve the equation ax² + (2a - 3)x + 1 = x and find the values of a for which this equation has exactly one solution.First, let me set up the equation. If f(x) = x, then:ax² + (2a - 3)x + 1 = xI can subtract x from both sides to get everything on one side:ax² + (2a - 3)x + 1 - x = 0Simplify the x terms:(2a - 3)x - x = (2a - 3 - 1)x = (2a - 4)xSo the equation becomes:ax² + (2a - 4)x + 1 = 0Now, this is a quadratic equation in terms of x. For a quadratic equation ax² + bx + c = 0, the discriminant is b² - 4ac. The equation has exactly one real solution when the discriminant is zero. So, in this case, the discriminant of ax² + (2a - 4)x + 1 should be zero.Let me compute the discriminant:D = (2a - 4)² - 4 * a * 1Expand (2a - 4)²:(2a)² - 2 * 2a * 4 + 4² = 4a² - 16a + 16So D = 4a² - 16a + 16 - 4aWait, wait. Wait, the discriminant is (2a - 4)^2 - 4 * a * 1. Let me calculate that again:First term: (2a - 4)^2 = 4a² - 16a + 16Second term: 4 * a * 1 = 4aSo D = 4a² - 16a + 16 - 4a = 4a² - 20a + 16Set discriminant equal to zero for exactly one solution:4a² - 20a + 16 = 0Now, let's solve this quadratic equation for a. First, divide all terms by 4 to simplify:a² - 5a + 4 = 0Factor the quadratic:Looking for two numbers that multiply to 4 and add up to -5. The factors of 4 are 1 & 4, 2 & 2. Since the middle term is -5a, the factors are -1 and -4. So:(a - 1)(a - 4) = 0Therefore, the solutions are a = 1 and a = 4.But wait, hold on. Let me check my steps again to make sure I didn't make a mistake.Original equation: f(x) = ax² + (2a - 3)x + 1. Fixed point when f(x) = x. So subtract x:ax² + (2a - 3)x + 1 - x = ax² + (2a - 4)x + 1 = 0. Correct.Discriminant: (2a - 4)^2 - 4*a*1. Correct.Expanding (2a - 4)^2: 4a² -16a +16. Then subtract 4a: 4a² -16a +16 -4a = 4a² -20a +16. Correct.Set equal to zero: 4a² -20a +16 =0.Divide by 4: a² -5a +4=0.Factor: (a -1)(a -4)=0. So a=1 and a=4. So that's two solutions.But the problem says "the set of values for the real number a" when the function has exactly one fixed point. So the answer should be a=1 and a=4? So the set is {1,4}.Wait, but wait a second. Is there a case where the equation might not be quadratic? Because if a=0, then the original function becomes f(x) = (0)x² + (0 -3)x +1 = -3x +1. Then, the fixed point equation would be -3x +1 = x. Solving that gives -4x = -1 => x=1/4. So even when a=0, there's one fixed point. But according to the previous solution, a=1 and a=4. But when a=0, it's a linear function, which can have exactly one fixed point unless the coefficient of x is 1, which would make it either no solutions or infinitely many. But in this case, when a=0, the equation f(x)=x reduces to -3x +1 =x, which is a linear equation with exactly one solution. So why isn't a=0 included in the solution set?Wait, but according to the problem statement, the function is given as f(x) = ax² + (2a -3)x +1. If a=0, then it's a linear function. However, the problem says "for any function f(x) defined on R, if a real number x satisfies f(x)=x, then x is called a fixed point." So linear functions can have fixed points too. But in the previous calculation, when we set a=0, the equation f(x)=x becomes linear, which has exactly one solution. However, according to our discriminant calculation, we got a=1 and a=4. But when a=0, the quadratic equation becomes a linear equation, which also has one solution. So this seems contradictory.Wait, so maybe the discriminant approach only works when a ≠ 0. Because if a=0, the equation is not quadratic, so we need to check separately.So in the problem statement, the function f(x) is quadratic if a ≠0, but linear if a=0. So to have exactly one fixed point, when a ≠0, the quadratic equation must have exactly one solution (discriminant zero). But when a=0, the linear equation will have exactly one solution (since the slope isn't 1). Let's check when a=0, the equation f(x)=x becomes -3x +1 =x, which gives x=1/4. So that's one fixed point. Therefore, a=0 also gives exactly one fixed point. Therefore, the set of values for a should include a=0, a=1, and a=4? But according to our discriminant calculation, when a=0, the equation is linear and still has one fixed point. So why didn't the discriminant approach account for that?Ah, because when a=0, the original equation is not quadratic but linear, so the discriminant method (which assumes a quadratic) doesn't apply. Therefore, we need to check two cases: when a=0 (linear function) and when a≠0 (quadratic function).Case 1: a=0.Then f(x) = (0)x² + (0 -3)x +1 = -3x +1. Setting f(x)=x:-3x +1 = x → -4x = -1 → x=1/4. So exactly one fixed point. So a=0 is valid.Case 2: a≠0.Then the equation is quadratic: ax² + (2a -4)x +1=0. For this quadratic to have exactly one solution, discriminant must be zero. As before, discriminant D = (2a -4)^2 -4*a*1 = 4a² -16a +16 -4a =4a² -20a +16. Set to zero:4a² -20a +16=0 → a² -5a +4=0 → (a-1)(a-4)=0 → a=1 or a=4.Therefore, combining both cases: a=0, a=1, a=4.But wait, wait. But when a=1 or a=4, the quadratic equation has exactly one solution. But when a=0, the equation is linear and also has exactly one solution. Therefore, the set of values is {0,1,4}.But this contradicts my previous conclusion. So why did the problem not include a=0?Wait, let me check again. The problem says "the function f(x) = ax² + (2a-3)x +1". So if a=0, then f(x) is indeed a linear function. However, in the original problem statement, it's defined for "any function f(x) defined on R", but here, the function is given as quadratic, but if a=0, it's linear. So does the problem consider linear functions as part of the consideration? The problem says "for any function f(x) defined on R", but here we're given a specific function f(x) = ax² + (2a-3)x +1. So regardless of the value of a, even if it's linear (a=0), we have to consider that as a possible function. So yes, a=0 is allowed, and in that case, the function is linear with exactly one fixed point. Therefore, the answer should include a=0 as well as a=1 and a=4.But wait, no. Wait, let's check when a=0: f(x) = -3x +1. Then f(x) = x gives x=1/4. So exactly one fixed point. So why in the discriminant approach, when a=0, which would lead to 0x² + (-4)x +1=0. That equation is -4x +1=0, which is linear and has one solution. But in the discriminant approach for quadratics, if a=0, then the coefficient of x² is zero, so the equation is not quadratic. Therefore, when we set discriminant to zero, we are assuming that the equation is quadratic (i.e., a≠0). Therefore, to get all possible a where the equation f(x)=x has exactly one solution, we need to consider both possibilities: when the equation is linear (a=0) and when it's quadratic (a≠0) with discriminant zero.So in total, the possible a's are a=0, a=1, a=4. Therefore, the set of values is {0,1,4}. However, the original solution I got was a=1 and a=4. So I must have missed a=0.But let me check again. Let's think about the problem again. The problem says "the function f(x) = ax² + (2a-3)x +1 has exactly one fixed point". So fixed points are the solutions to f(x)=x. So regardless of the nature of f(x), whether quadratic or linear, as long as the equation f(x)=x has exactly one solution. Therefore, a=0 should be included if it gives exactly one solution. So when a=0, f(x)=x is a linear equation with exactly one solution, so a=0 is acceptable.Therefore, the answer should be a=0, a=1, a=4.But in the original discriminant approach, we found a=1 and a=4. But why didn't a=0 come up? Because when a=0, the equation becomes linear, so discriminant is not applicable.Therefore, the correct answer is a=0, a=1, a=4.Wait, but let's verify this.For a=0: function is linear, fixed point exists and is unique.For a=1: quadratic equation becomes x² + (2*1 -4)x +1= x² -2x +1=0. Which factors to (x-1)^2=0, so one solution.For a=4: quadratic equation becomes 4x² + (8 -4)x +1=4x² +4x +1=0. Which factors to (2x+1)^2=0, so x=-1/2, one solution.Therefore, all three a=0,1,4 lead to exactly one fixed point. Therefore, the set is {0,1,4}.But wait, but the problem says "the set of values for the real number a". So the answer is a=0,1,4. However, the original calculation missed a=0 because we considered only the quadratic case.Therefore, the correct answer includes a=0 in addition to a=1 and a=4.But the problem is from a source that might expect only the quadratic cases. Let me check again.Wait, the function is given as f(x) = ax² + (2a -3)x +1. If a=0, then it's a linear function, which has exactly one fixed point. So in that case, the answer is 0,1,4.But perhaps the problem assumes that a≠0? The problem states "the function f(x) = ax² + (2a-3)x +1". It doesn't specify that a≠0. Therefore, unless restricted, a can be any real number, including zero. Therefore, a=0 must be considered.Therefore, the set is {0,1,4}.But in the original discriminant approach, we had a quadratic equation ax² + (2a -4)x +1=0. If a=0, this equation becomes (2*0 -4)x +1=0 → -4x +1=0, which has one solution. Therefore, even if a=0, the equation f(x)=x has exactly one solution. Therefore, a=0 is a valid solution.But when we calculated the discriminant, we considered a≠0. So in conclusion, the set of a's is all a where either a=0 or the quadratic equation has discriminant zero. So discriminant zero gives a=1,4 and a=0 gives another case. Therefore, total solutions: a=0,1,4.Therefore, the answer is {0,1,4}.But let me check once more.For a=0: f(x) = -3x +1. Equation f(x)=x gives x=1/4. Exactly one fixed point. So a=0 is valid.For a=1: quadratic equation x² -2x +1=0. Solution x=1 (double root). Exactly one fixed point.For a=4: quadratic equation 4x² +4x +1=0. Solution x=(-4 ± sqrt(16 -16))/8 = (-4)/8 = -0.5. Exactly one fixed point.For other values of a:If a≠0,1,4, then the equation ax² + (2a -4)x +1=0 would have two distinct solutions or no real solutions. For example, take a=2:Equation 2x² + (4 -4)x +1 = 2x² +1=0. No real solutions. So f(x)=x has no fixed points.Take a= -1:Equation -1x² + (-2 -4)x +1= -x² -6x +1=0. Discriminant: 36 +4=40>0. Two solutions. So two fixed points.Therefore, only when a=0,1,4 does f(x)=x have exactly one fixed point.Therefore, the correct answer is {0,1,4}. But the original discriminant approach missed a=0 because we only considered quadratics. Therefore, the answer is {0,1,4}.However, let's check if the problem allows a=0. The problem says "the function f(x) = ax² + (2a-3)x +1". If a=0, it's allowed as per the problem statement. Therefore, the answer should be {0,1,4}.But in many math problems, when a quadratic is considered, a≠0 is implicitly assumed. However, in this problem, since it's defined for any real number a, including zero, we must include a=0 as well.Therefore, the set is {0,1,4}.But let's check once more with the problem statement:"If the function f(x) = ax² + (2a-3)x +1 has exactly one fixed point, then the set of values for the real number a is."It doesn't specify that the function has to be quadratic, so a can be zero. Therefore, the answer is {0,1,4}.But initially, I didn't consider a=0. So this shows the importance of checking edge cases where coefficients might be zero. Hence, the correct answer includes a=0,1,4.Final AnswerThe set of values for ( a ) is (boxed{{0, 1, 4}}).</think>To find the set of values for the real number ( a ) such that the function ( f(x) = ax^2 + (2a - 3)x + 1 ) has exactly one fixed point, we start by solving the equation ( f(x) = x ). First, we set up the equation:[ax^2 + (2a - 3)x + 1 = x]Subtracting ( x ) from both sides, we get:[ax^2 + (2a - 4)x + 1 = 0]For this quadratic equation to have exactly one solution, its discriminant must be zero. The discriminant ( D ) of the quadratic equation ( ax^2 + (2a - 4)x + 1 = 0 ) is:[D = (2a - 4)^2 - 4 cdot a cdot 1]Expanding and simplifying:[D = 4a^2 - 16a + 16 - 4a = 4a^2 - 20a + 16]Setting the discriminant to zero:[4a^2 - 20a + 16 = 0]Dividing by 4:[a^2 - 5a + 4 = 0]Factoring:[(a - 1)(a - 4) = 0]Thus, the solutions are ( a = 1 ) and ( a = 4 ).However, we must also consider the case when ( a = 0 ). If ( a = 0 ), the function becomes linear:[f(x) = -3x + 1]Setting ( f(x) = x ):[-3x + 1 = x implies -4x = -1 implies x = frac{1}{4}]This linear equation has exactly one solution, so ( a = 0 ) is also valid.Therefore, combining both cases, the set of values for ( a ) is ( {0, 1, 4} ).[boxed{{0, 1, 4}}],<think>Okay, let's try to figure this problem out. So, we have a function f(x) = ax² + (2a - 3)x + 1, and we need to find the values of a for which this function has exactly one fixed point. A fixed point is where f(x) = x, right? So, basically, we need to solve the equation f(x) = x and make sure there's exactly one solution. First, let's recall that solving f(x) = x is equivalent to finding the roots of the equation f(x) - x = 0. So, let's subtract x from both sides to set that up. Starting with f(x) = ax² + (2a - 3)x + 1. Subtracting x from both sides gives:ax² + (2a - 3)x + 1 - x = 0Simplify the terms with x:ax² + (2a - 3 - 1)x + 1 = 0Wait, let me check that step again. The original equation is f(x) - x = 0. So:ax² + (2a - 3)x + 1 - x = 0Combine the x terms: (2a - 3)x - x = (2a - 3 - 1)x = (2a - 4)xSo, the quadratic equation becomes:ax² + (2a - 4)x + 1 = 0So, this is a quadratic in x: ax² + (2a - 4)x + 1 = 0Now, for a quadratic equation ax² + bx + c = 0, the discriminant is b² - 4ac. The equation has exactly one real root when the discriminant is zero. So, in our case, we need the discriminant of this quadratic to be zero. So, let's compute the discriminant:Discriminant D = [2a - 4]^2 - 4 * a * 1Let me compute that step by step:First, expand [2a - 4]^2:(2a - 4)^2 = (2a)^2 - 2*2a*4 + 4^2 = 4a² - 16a + 16Then, compute 4*a*1 = 4aSo, D = (4a² - 16a + 16) - 4a = 4a² - 16a + 16 - 4a = 4a² - 20a + 16We need D = 0 for exactly one fixed point. So, set the discriminant equal to zero:4a² - 20a + 16 = 0Hmm, let's solve this quadratic equation for a. Maybe we can divide all terms by 4 to simplify:a² - 5a + 4 = 0Now, factor this quadratic:Looking for two numbers that multiply to 4 and add up to -5. Wait, factors of 4 are 1 and 4, 2 and 2. Let's see:(a - 1)(a - 4) = a² - 5a + 4. Yes, that works because 1*4=4 and 1+4=5, but since the middle term is -5a, it should be (a -1)(a -4) = a² -5a +4. So, the equation factors to:(a - 1)(a - 4) = 0Therefore, the solutions are a = 1 and a = 4.But wait, hold on. The original problem says the function f(x) is defined on ℝ. But if a = 0, then the function f(x) would be linear. However, in our case, the equation f(x) - x = 0 becomes a quadratic. But if a = 0, then the quadratic becomes a linear equation. So, maybe we need to check if a = 0 is a possible case here. Wait, but in the problem statement, the function is given as f(x) = ax² + (2a -3)x + 1. So, if a = 0, then f(x) = (2*0 -3)x +1 = -3x +1, which is a linear function. A linear function can have either one fixed point, unless it's of the form f(x) = x + c, which would have no fixed points if c ≠0. But in this case, if a=0, then f(x) = -3x +1. So, solving -3x +1 = x gives -4x +1=0, so x=1/4. So, even when a=0, there is exactly one fixed point. But wait, in our discriminant calculation, we assumed a quadratic equation. But if a=0, then the equation f(x)-x=0 becomes linear. So, if a=0, then f(x) is linear, and the equation f(x)=x will have exactly one solution unless the coefficient of x is 1. Let's check that.Wait, when a=0, f(x) - x = (-3x +1) -x = -4x +1. So, solving -4x +1=0 gives x=1/4, so one solution. So, even though a=0 makes the function linear, it still has one fixed point. But in our discriminant approach, we assumed that a ≠0 because we divided by a when solving the quadratic? Wait, no. Wait, the original quadratic equation is ax² + (2a -4)x +1=0. If a=0, then this equation becomes 0x² + (0 -4)x +1=0, which is -4x +1=0, which is linear, as we saw. So, even when a=0, the equation is linear and has one solution. So, in that case, we need to check if a=0 is included in the solutions. But in our discriminant calculation, we had D=4a² -20a +16, and we set that to zero, giving a=1 and a=4. But when a=0, D=0 -0 +16=16>0, so the discriminant is 16, which is positive, meaning two real roots. Wait, but when a=0, the equation f(x)-x=0 is linear, so it can only have one root. So, there's a contradiction here. Therefore, maybe our discriminant approach is missing something. Because when a=0, even though the discriminant D=16, which would suggest two roots, but in reality, the equation becomes linear and can only have one root. Therefore, we need to handle the case when a=0 separately. So, perhaps the discriminant approach is only valid when a≠0. So, we need to check two cases: when a=0 and when a≠0. Case 1: a=0. Then f(x) = (2*0 -3)x +1 = -3x +1. Solving -3x +1 =x gives -4x +1=0 => x=1/4. So, there is exactly one fixed point. Therefore, a=0 is a valid solution. Case 2: a≠0. Then the equation f(x)-x=0 is quadratic: ax² + (2a -4)x +1=0. For this quadratic equation to have exactly one real root, discriminant must be zero. As we found earlier, discriminant D=4a² -20a +16. Setting D=0 gives a=1 and a=4. However, when a=1 and a=4, the quadratic equation becomes a perfect square, hence one real root (a double root). So, these are valid solutions. But wait, the problem states "the function f(x) has exactly one fixed point". So, if a=0, we get one fixed point. If a=1 or a=4, we get one fixed point. So, the set of values for a is {0,1,4}? But wait, let me check if when a=0, the equation is linear with one solution. When a=1, the quadratic equation becomes x² + (2*1 -4)x +1 = x² -2x +1 =0, which is (x-1)^2=0, so x=1. Similarly, when a=4, the equation becomes 4x² + (8 -4)x +1=4x² +4x +1=0, which is (2x+1)^2=0, so x=-1/2. So, indeed, those a=1 and a=4 give one fixed point each. But in our discriminant calculation, when we set D=0, we found a=1 and a=4, but a=0 is another case where the equation becomes linear. Therefore, the answer should include a=0,1,4. But wait, the problem says "the set of values for the real number a is". So, is a=0 included? Let me verify. Wait, when a=0, the equation f(x)-x=0 is linear, so it has exactly one solution. Thus, the function f(x) has exactly one fixed point. Therefore, a=0 must be included. However, in the discriminant approach, we considered the quadratic equation and found a=1 and a=4. But when a=0, the equation is linear, which automatically has one solution unless the coefficient of x is zero. Wait, but in the linear case, if the coefficient of x is zero, then it would either have no solutions or infinitely many. But in our case, when a=0, the equation becomes -4x +1=0, which has exactly one solution. Therefore, a=0 is a valid solution. But in our discriminant calculation, we had discriminant D=4a² -20a +16. For a quadratic equation, when a≠0, D=0 gives a=1 and a=4. When a=0, even though D=16≠0, the equation is not quadratic, so the discriminant is not applicable. So, in that case, we need to separately consider a=0. Therefore, the total solutions are a=0,1,4. But wait, let's verify if when a=0, the problem's condition is satisfied. The problem states "the function f(x) has exactly one fixed point". When a=0, the function is linear, f(x)=-3x+1. Solving f(x)=x gives x=1/4, so exactly one fixed point. So yes, a=0 is valid. But why didn't the discriminant approach catch a=0? Because when a=0, the equation is not quadratic, so the discriminant approach (which is for quadratic equations) doesn't apply. Therefore, we need to check separately when the equation becomes linear. Thus, the answer should be a=0,1,4. However, the original problem says "the set of values for the real number a is". So, the answer is {0,1,4}? Let me check once again. Wait, but in the problem statement, the function is given as f(x)=ax² + (2a-3)x +1. If a=0, then the function is linear. If a≠0, it's quadratic. Therefore, in order for the equation f(x)=x to have exactly one solution, either:1. The function is linear (a=0) and the equation f(x)=x has exactly one solution (which it does, as we saw).2. The function is quadratic (a≠0) and the quadratic equation f(x)-x=0 has exactly one solution, i.e., discriminant zero. Therefore, combining these two cases, the possible values for a are a=0,1,4. But wait, hold on. When a=0, even though the equation is linear, the problem didn't specify that the function has to be quadratic. It just says "for any function f(x) defined on ℝ". Therefore, linear functions are allowed. So, a=0 is acceptable. But maybe the answer is supposed to be only a=1 and a=4? Wait, but according to the logic above, a=0 is also a solution. Let me check again. Wait, when a=0, f(x)= -3x +1. Fixed point is x where -3x +1 =x => -4x +1=0 => x=1/4. So, exactly one fixed point. So, according to the problem's condition, a=0 is acceptable. So, the answer should include 0,1,4. But let me check the quadratic equation when a=0. The equation f(x)-x=0 is linear, which is a degenerate case of quadratic (coefficient of x² is zero). So, in the problem statement, they just say "the function f(x)" and it's defined as quadratic, but actually, when a=0, it's linear. However, the problem says "for any function f(x) defined on ℝ", so they probably include linear functions. Therefore, a=0 is a valid solution. But wait, maybe I made a mistake here. Let me check the problem statement again: "For any function f(x) defined on ℝ, if a real number x satisfies f(x) = x, then x is called a fixed point of the function f(x). If the function f(x) = ax² + (2a-3)x + 1 has exactly one fixed point, then the set of values for the real number a is."Wait, the function is given as f(x) = ax² + (2a-3)x +1. So, even when a=0, it's still a valid function, just linear. Therefore, the problem allows a=0, so in that case, the equation f(x)=x is linear, which as we saw, gives one solution. Therefore, a=0 must be included. However, in the discriminant approach, when a=0, the equation f(x)-x=0 is linear, so it's not quadratic, and the discriminant approach for quadratics doesn't apply. Therefore, in addition to the discriminant solutions a=1 and a=4, we need to include a=0. But wait, wait, let me think again. If we consider the equation f(x)-x=0, regardless of a, it can be written as ax² + (2a -4)x +1=0. This is a quadratic equation only if a≠0. If a=0, then it's linear. So, for a=0, the equation is linear and has exactly one solution. For a≠0, it's quadratic and will have exactly one solution when discriminant is zero, which gives a=1 and a=4. Therefore, combining both cases, the possible a values are a=0,1,4. Therefore, the set of values is {0,1,4}. However, let me check again the original problem. The problem says "the function f(x) = ax² + (2a-3)x + 1 has exactly one fixed point". So, even if a=0, the function is still f(x)= -3x +1, which is linear and has one fixed point. Therefore, the answer should include a=0. But wait, in the initial steps, when solving for discriminant, I considered the quadratic equation and found a=1 and a=4. However, the problem is about the function f(x) which can be linear or quadratic depending on a. Therefore, the answer is three values: 0,1,4. But let me check with the problem statement again. Let me see, in the problem statement, they say "for any function f(x) defined on ℝ". Wait, no, the problem is not saying "for any function", but rather "If the function f(x) = ax² + (2a-3)x + 1 has exactly one fixed point, then the set of values for the real number a is." So, the function is given as f(x)=ax² + (2a-3)x +1, and we need to find the real numbers a such that this function has exactly one fixed point. Therefore, the function is allowed to be quadratic or linear (if a=0). So, the answer includes a=0,1,4. But let me check once again if when a=0, the equation has exactly one fixed point. Yes, as we saw, x=1/4. But now, I have conflicting thoughts. In some textbooks, when dealing with quadratic equations, they might consider a quadratic coefficient being zero as a separate case. So, in this problem, since the function is given as quadratic (even though the coefficient is a parameter), perhaps they expect to consider a≠0 and a=0 separately. Therefore, including a=0 as a valid solution. However, let's test a=0. If a=0, the equation becomes linear, which must have exactly one fixed point. Therefore, a=0 is valid. But in the discriminant calculation, when we considered the quadratic equation, we found a=1 and a=4. So, combining both cases, the set of a is {0,1,4}. Wait, but maybe the original problem expects only the quadratic case, but that seems unlikely because the problem says "for any function f(x) defined on ℝ", and then specifies f(x) = ax² + (2a-3)x +1. So, the function can be linear if a=0, so a=0 is allowed, and in that case, it still has one fixed point. Therefore, the answer should be a=0,1,4. But let me check some example. If a=0: f(x) = -3x +1. Fixed point at x=1/4. So, one fixed point. Correct. If a=1: f(x)=x² + (2 -3)x +1 =x² -x +1. Set f(x)=x: x² -x +1 =x →x² -2x +1=0 →(x-1)^2=0. So, x=1. Exactly one fixed point. If a=4: f(x)=4x² + (8 -3)x +1=4x² +5x +1. Wait, wait, hold on. Wait, when a=4, f(x)=4x² + (2*4 -3)x +1=4x² + (8 -3)x +1=4x² +5x +1. Wait, but when we compute f(x)-x=0, that's 4x² +5x +1 -x=4x² +4x +1=0. Which factors as (2x +1)^2=0. So, x=-1/2. Exactly one fixed point. So, yes, a=4 works. Therefore, all three values a=0,1,4 result in the function having exactly one fixed point. But now, I need to check my discriminant calculation again. When I set up the equation f(x)-x=0, which is ax² + (2a -4)x +1=0. The discriminant is [2a -4]^2 -4*a*1. Let's compute that again. [2a -4]^2 =4a² -16a +164*a*1=4aTherefore, discriminant D=4a² -16a +16 -4a=4a² -20a +16. Set to zero:4a² -20a +16=0. Divide by 4: a² -5a +4=0. Factor: (a -1)(a -4)=0. So, a=1,4. So, the discriminant approach gives a=1 and a=4. But when a=0, the equation is not quadratic, so discriminant approach isn't applicable. Therefore, the solutions are a=0,1,4. Therefore, the answer is {0,1,4}. But wait, let me check online if similar problems include the linear case. For example, if a problem asks for the number of fixed points of a quadratic function, they usually consider the quadratic case. But here, the problem specifies the function f(x) = ax² + (2a -3)x +1. If a=0, then it's linear, but since the problem didn't specify that it's quadratic, it's allowed. Therefore, the answer should include a=0. Therefore, the set of values for a is {0,1,4}. So, the final answer is boxed{{0,1,4}}.Wait, but let me check again when a=0. Let me plug a=0 into the original function: f(x)=0x² + (0 -3)x +1=-3x +1. Fixed point at x=1/4. So, exactly one. Correct. Similarly, for a=1 and a=4, as above. So, three values. But I have a feeling that maybe the problem expects only two values. Maybe I made a mistake in considering a=0. Let me check the problem again. Problem says: "If the function f(x) = ax² + (2a-3)x + 1 has exactly one fixed point, then the set of values for the real number a is."The function is defined as f(x)=ax² + (2a-3)x +1. So, if a=0, the quadratic term disappears, but it's still a valid function. Therefore, the answer includes a=0. Alternatively, maybe the problem is written in a way where they assume a≠0, but since it's not specified, we can't assume that. Therefore, the answer is {0,1,4}. But I want to verify with the equation f(x)-x=0. When a=0, equation is linear, one solution. When a≠0, quadratic equation with D=0 gives a=1 and a=4. Therefore, three solutions. Hence, the correct answer is boxed{{0, 1, 4}}.Final AnswerThe set of values for the real number ( a ) is boxed{{0, 1, 4}}.</think>To determine the set of values for the real number ( a ) such that the function ( f(x) = ax^2 + (2a - 3)x + 1 ) has exactly one fixed point, we start by solving the equation ( f(x) = x ). This is equivalent to finding the roots of the equation ( f(x) - x = 0 ).Subtracting ( x ) from both sides of the equation ( f(x) = x ), we get:[ax^2 + (2a - 3)x + 1 - x = 0]Simplifying the terms, we have:[ax^2 + (2a - 4)x + 1 = 0]This is a quadratic equation in ( x ). For a quadratic equation ( ax^2 + bx + c = 0 ) to have exactly one real root, its discriminant must be zero. The discriminant ( D ) of the quadratic equation ( ax^2 + (2a - 4)x + 1 = 0 ) is given by:[D = (2a - 4)^2 - 4 cdot a cdot 1]Expanding and simplifying the discriminant:[D = 4a^2 - 16a + 16 - 4a = 4a^2 - 20a + 16]Setting the discriminant to zero for the equation to have exactly one real root:[4a^2 - 20a + 16 = 0]Dividing by 4:[a^2 - 5a + 4 = 0]Factoring the quadratic equation:[(a - 1)(a - 4) = 0]Thus, the solutions are:[a = 1 quad text{or} quad a = 4]We must also consider the case where ( a = 0 ). If ( a = 0 ), the function ( f(x) ) becomes linear:[f(x) = -3x + 1]Solving ( f(x) = x ) gives:[-3x + 1 = x implies -4x + 1 = 0 implies x = frac{1}{4}]This linear equation has exactly one fixed point, so ( a = 0 ) is also a valid solution.Combining all cases, the set of values for ( a ) is:[boxed{{0, 1, 4}}]]
🔑:According to the problem, we have x = ax^2 + (2a-3)x + 1 has exactly two equal real roots, which means the equation ax^2 + (2a-4)x + 1 = 0 has exactly two equal real roots. Therefore, when a neq 0, Delta = (2a-4)^2 - 4a = 0, solving this yields: a = 1 or a = 4; when a = 0, it obviously also meets the condition. In summary, the set of values for the real number a is boxed{{0, 1, 4}}. Thus, the answer is: boxed{{0, 1, 4}}.
❓:Each of A, B, C, and D is a positive two-digit integer. These integers satisfy each of the equationsbegin{aligned}B & =3 C D & =2 B-C A & =B+Dend{aligned}What is the largest possible value of A+B+C+D ?
🧠:[<think>Okay, let's see. So, we have four positive two-digit integers: A, B, C, and D. They satisfy three equations: B is three times C, D is two times B minus C, and A is B plus D. The goal is to find the largest possible value of the sum A + B + C + D. Hmm, okay. Let's break this down step by step.First, let's write down the equations again to make sure I have them right:1. B = 3C2. D = 2B - C3. A = B + DAnd all of them are two-digit positive integers. So, each of A, B, C, D must be between 10 and 99 inclusive. But since they are defined in terms of each other, maybe some of them have tighter constraints. Let's see.Starting with the first equation: B = 3C. Since B is a two-digit integer, 3C must also be a two-digit integer. Therefore, C must be such that 3C is between 10 and 99. So, C has to be at least 4 (since 3*4=12) and at most 33 (since 3*33=99). But wait, C itself is a two-digit number, so actually, C must be between 10 and 33. Because if C were a single-digit number, like 4, then B would be 12, which is two digits, but C is specified to be a two-digit integer. Wait, the problem says each of A, B, C, D is a positive two-digit integer. So C has to be a two-digit number. Therefore, C must be between 10 and 33, inclusive, since 3*10=30 and 3*33=99. So, C is from 10 to 33. Then B is 3C, so B ranges from 30 to 99. But B has to be a two-digit number, so 30 to 99 inclusive. So, okay.Then the next equation is D = 2B - C. Let's substitute B with 3C here. So D = 2*(3C) - C = 6C - C = 5C. So D is 5C. So D must also be a two-digit number. Since D = 5C, and C is from 10 to 33, 5C would be from 50 to 165. But D has to be a two-digit number, so 5C must be less than or equal to 99. Therefore, 5C ≤ 99 => C ≤ 19.8. Since C is an integer, C ≤ 19. But C is a two-digit number, so C ranges from 10 to 19 inclusive. So now we have a tighter constraint on C: instead of up to 33, it's up to 19. Because if C were 20, then D would be 5*20=100, which is a three-digit number, which is invalid. So C is from 10 to 19. Then B = 3C is from 30 to 57 (since 3*19=57). Then D = 5C is from 50 to 95 (since 5*19=95). So that makes sense.Now, the third equation is A = B + D. Let's substitute B and D here. Since B = 3C and D = 5C, then A = 3C + 5C = 8C. So A is 8C. Now, A must be a two-digit number as well. So 8C must be between 10 and 99. Since C is from 10 to 19, 8*10=80 and 8*19=152. But 152 is a three-digit number, which is invalid. Therefore, A must be 8C ≤ 99, so C ≤ 12.375. Since C is an integer, C ≤ 12. But C is a two-digit number, so C is from 10 to 12 inclusive. So now, C can only be 10, 11, or 12. Let's check that.If C = 10:B = 3*10 = 30D = 5*10 = 50A = 8*10 = 80So A = 80, B = 30, C = 10, D = 50. All two-digit numbers. Good.If C = 11:B = 3*11 = 33D = 5*11 = 55A = 8*11 = 88All two-digit numbers. Good.If C = 12:B = 3*12 = 36D = 5*12 = 60A = 8*12 = 96All two-digit numbers. Good.If C = 13:A = 8*13 = 104, which is three digits. So invalid. So C cannot be 13 or higher. So C is only 10, 11, 12.So our possible values of C are 10, 11, 12. Then A, B, D are determined as above. Let's tabulate them:Case 1: C = 10A = 80B = 30D = 50Sum: 80 + 30 + 10 + 50 = 170Case 2: C = 11A = 88B = 33D = 55Sum: 88 + 33 + 11 + 55 = 187Case 3: C = 12A = 96B = 36D = 60Sum: 96 + 36 + 12 + 60 = 204So among these, 204 is the largest sum. Therefore, the answer should be 204. Wait, but let's check if there's any other possible C values. Wait, we thought C can only be 10, 11, 12. Because higher C would make A three-digit. But just to confirm, let's check C=13:B=3*13=39, D=5*13=65, A=8*13=104. So A=104 is invalid, so C=13 is invalid. Similarly, higher C would make A even larger. So yes, only C=10,11,12 are valid. Therefore, the maximum sum is 204.But wait, let me double-check the math for each case.Case 1:A=80, B=30, C=10, D=50Sum: 80+30=110, 10+50=60, total 170. Correct.Case 2:A=88, B=33, C=11, D=55Sum: 88+33=121, 11+55=66, total 187. Correct.Case 3:A=96, B=36, C=12, D=60Sum: 96+36=132, 12+60=72, total 204. Correct.So yes, 204 is the maximum. Therefore, the answer is 204.Wait, but hold on. Let me check if there's another way to approach this problem where maybe we don't have such tight constraints. Like, could there be cases where even if C is higher, but D is adjusted to be a two-digit number? Wait, but according to the equations given, D is determined by B and C. Since D=2B - C. But since B=3C, D=6C - C=5C. So D is always 5C. Therefore, if C is higher than 12, D would be 5*13=65, which is still two-digit. Wait, wait, 5*13=65, which is two-digit. Then why do we say that A=8C=104, which is three-digit. So even if D is two-digit, A would be three-digit. So even if C=13, D=65 is okay, but A=104 is not. So C is limited by A's two-digit requirement, not D's. Therefore, yes, C must be at most 12. So the three cases are the only possible ones.Therefore, 204 is indeed the maximum sum. So the answer should be boxed{204}.Final AnswerThe largest possible value of ( A + B + C + D ) is boxed{204}.</think>Given the equations:[begin{aligned}B & = 3C D & = 2B - C A & = B + Dend{aligned}]where (A), (B), (C), and (D) are positive two-digit integers, we need to find the largest possible value of (A + B + C + D).1. From (B = 3C), (C) must be a two-digit integer such that (3C) is also a two-digit integer. Therefore, (C) ranges from 10 to 33. However, since (D = 5C) must also be a two-digit integer, (C) must be such that (5C) is a two-digit number, limiting (C) to 10 through 19. Further, since (A = 8C) must be a two-digit integer, (C) must be limited to 10 through 12.2. Checking the valid values of (C): - If (C = 10): [ B = 3 times 10 = 30, quad D = 5 times 10 = 50, quad A = 8 times 10 = 80 ] Sum: (80 + 30 + 10 + 50 = 170) - If (C = 11): [ B = 3 times 11 = 33, quad D = 5 times 11 = 55, quad A = 8 times 11 = 88 ] Sum: (88 + 33 + 11 + 55 = 187) - If (C = 12): [ B = 3 times 12 = 36, quad D = 5 times 12 = 60, quad A = 8 times 12 = 96 ] Sum: (96 + 36 + 12 + 60 = 204)The largest possible value of (A + B + C + D) is (boxed{204}).,<think>Okay, let's see. I need to find the largest possible value of A + B + C + D given these equations. Let me start by writing down the equations again to make sure I have them right:B = 3C D = 2B - C A = B + D And all of A, B, C, D are positive two-digit integers. So each of them is between 10 and 99 inclusive. First, let's express all variables in terms of C. That might help me see the relationships better. Since B is 3C, and B has to be a two-digit number, then 3C must be between 10 and 99. So C must satisfy 10 ≤ 3C ≤ 99. Therefore, C must be at least 4 (since 3*4=12, which is two digits) and at most 33 (since 3*33=99). But wait, C is a two-digit integer, so C must be between 10 and 33. Because if C were a single-digit number, like 4, then it's not a two-digit integer. Wait, the problem says each of A, B, C, D is a positive two-digit integer. So C must be a two-digit number. Therefore, C is between 10 and 33, but since 3*33=99, which is the maximum two-digit number. So C can be from 10 up to 33, but 3*33=99, which is allowed. So C ∈ [10, 33], and C must be an integer. So C is 10 to 33 inclusive. Then B = 3C, which would be 30 to 99 inclusive. But since B has to be a two-digit integer, that's okay because 30 is two-digit and 99 is the maximum. So B is 30 to 99, but in increments such that B is divisible by 3 because B = 3C. So B must be a multiple of 3. Next, D = 2B - C. Let's substitute B with 3C here. So D = 2*(3C) - C = 6C - C = 5C. Therefore, D = 5C. But D also has to be a two-digit number, so 5C must be between 10 and 99. So 10 ≤ 5C ≤ 99. Therefore, C must be at least 2 (since 5*2=10) and at most 19 (since 5*19=95). But wait, C has to be a two-digit integer, so the minimum C is 10. Therefore, 5C would start at 50 (5*10=50) up to 5*19=95. Wait, but earlier we had C up to 33. But here, because D = 5C must be two-digit, C can only go up to 19. Wait, this is a conflict. Let me check.Wait, C has to be a two-digit integer, so C is 10 to 33 as per B = 3C (since B has to be two-digit, up to 99). But then D = 5C must also be two-digit. Therefore, 5C ≤ 99, so C ≤ 19.8. Since C is an integer, C ≤19. But C is a two-digit integer, so 10 ≤ C ≤19. Therefore, that reduces the possible range of C from 10 to 19. That's important. So even though B = 3C could go up to 33 (if C=33, but since C must be ≤19, then B can only go up to 3*19=57. Therefore, B would be from 30 (3*10) to 57 (3*19). So now, C is between 10 and 19 inclusive. Then B is 3*10=30 up to 3*19=57. Then D is 5*10=50 up to 5*19=95. All two-digit numbers. Then A = B + D. Let's see: B is between 30 and 57, D between 50 and 95. So A = B + D would be between 30 +50=80 and 57 +95=152. But A has to be a two-digit integer. Wait, 152 is three digits. So A must be ≤99. Therefore, A = B + D must be ≤99. So we have another constraint here. So A must be two-digit, so B + D ≤99. Since D =5C and B=3C, then A =3C +5C=8C. Therefore, A=8C. But A must be a two-digit number, so 8C ≤99. Thus, C ≤12.375. Since C is an integer, C ≤12. But C is a two-digit integer, so C can be 10, 11, or 12. Wait, 12.375, so integer part is 12. So C must be 10,11,12. Because if C=13, then 8*13=104, which is three digits. Therefore, C can only be 10,11,12.Wait, hold on. Let's verify this. Let me check:If C=10: A=8*10=80 (two-digit)If C=11: A=88 (two-digit)If C=12: A=96 (two-digit)If C=13: A=104 (three-digit) which is invalid.Therefore, the possible values for C are 10,11,12. Therefore, C can only be 10,11,12. Then B=3C would be 30,33,36. D=5C would be 50,55,60. Then A=8C would be 80,88,96. So let's check each possible C:Case 1: C=10Then B=30, D=50, A=30+50=80Sum A+B+C+D=80+30+10+50=170Case 2: C=11B=33, D=55, A=33+55=88Sum=88+33+11+55=88+33=121, 121+11=132, 132+55=187Case 3: C=12B=36, D=60, A=36+60=96Sum=96+36+12+60=96+36=132, 132+12=144, 144+60=204Wait, that's 204. But wait, let's confirm that all numbers are two-digit:For C=12: C=12, B=36, D=60, A=96. All are two-digit numbers. So that works.But wait, in the case of C=12, D=5*12=60, which is two-digit. A=8*12=96, which is two-digit. So that's okay.So the sums are 170, 187, and 204. Therefore, the largest possible value is 204. Wait, but hold on. Wait, I thought earlier that C can be up to 19, but with the constraint that A=8C must be two-digit, so only up to 12. But what if there's a case where even if A is two-digit, but B or D might not be? But in the equations above, since B=3C and C is 10-12, then B is 30-36, which are all two-digit. D=5C, so 50-60, which are two-digit. So all cases are valid. So the maximum sum is 204 when C=12. But let me check again. Maybe I missed some values. Wait, but according to the constraints, C can only be 10,11,12 because higher C would make A three-digit. For example, C=13 would give A=104, which is invalid. Therefore, C can only be 10,11,12. So the maximum sum is 204. Therefore, the answer is 204. But let me check once more.Alternatively, perhaps there's a way to get higher sums by not expressing all variables in terms of C. Wait, but given the equations, all variables are dependent on C. Let me check the equations again. Given:B = 3C D = 2B - C = 2*(3C) - C = 6C - C =5C A = B + D =3C +5C=8C Therefore, all variables are directly proportional to C. So A=8C, B=3C, C=C, D=5C. Therefore, the total sum is A+B+C+D=8C+3C+C+5C=17C. Therefore, the total sum is 17C. Since C is a two-digit integer, and A=8C must also be a two-digit integer, which constrains C to 10,11,12. Therefore, the maximum sum is 17*12=204. Therefore, the answer is 204. Therefore, 204 is the maximum possible sum. But let me just check if there's any other interpretation of the problem. For example, could D be calculated as 2B - C, but if there's some other relation? Wait, no, the equations are given. So D is defined as 2B - C. But since B is 3C, substituting that in gives D=5C. So unless there's a miscalculation here. Wait:D=2B - C. B=3C. So D=2*(3C) - C=6C - C=5C. Yes, that's correct. So D=5C. Then A=B + D=3C +5C=8C. Therefore, all variables are directly proportional to C. So the total sum is 17C, which is maximized when C is as large as possible. But since A must be two-digit, 8C ≤99. So C ≤12.375, so C=12. Therefore, 17*12=204. That's the largest possible value. So unless there's an error in my reasoning here, 204 is the answer. Wait, but wait a second. The problem says "Each of A, B, C, and D is a positive two-digit integer." So C must be two-digit. So if C=10,11,12, those are two-digit. If I take C=13, which is two-digit, but then A=8*13=104, which is three-digit, so that's invalid. Therefore, C can't be 13. So my original conclusion holds. Therefore, the largest possible value of A+B+C+D is 204. Final Answerboxed{204}</think>Given the equations:[begin{aligned}B & = 3C D & = 2B - C A & = B + Dend{aligned}]where (A, B, C,) and (D) are positive two-digit integers, we need to find the largest possible value of (A + B + C + D).First, express all variables in terms of (C):1. (B = 3C)2. Substitute (B = 3C) into (D = 2B - C): (D = 2(3C) - C = 5C)3. Substitute (B) and (D) into (A = B + D): (A = 3C + 5C = 8C)Next, ensure all variables are two-digit integers:- (C) must be a two-digit integer, so (10 leq C leq 99).- (B = 3C) must be a two-digit integer, so (10 leq 3C leq 99). This gives (10 leq C leq 33).- (D = 5C) must be a two-digit integer, so (10 leq 5C leq 99). This gives (10 leq C leq 19).- (A = 8C) must be a two-digit integer, so (10 leq 8C leq 99). This gives (10 leq C leq 12).Thus, (C) can be 10, 11, or 12. Calculate the sum (A + B + C + D) for each possible (C):- For (C = 10): (A = 80), (B = 30), (D = 50), sum = (80 + 30 + 10 + 50 = 170)- For (C = 11): (A = 88), (B = 33), (D = 55), sum = (88 + 33 + 11 + 55 = 187)- For (C = 12): (A = 96), (B = 36), (D = 60), sum = (96 + 36 + 12 + 60 = 204)The largest possible value is when (C = 12), resulting in a sum of (boxed{204}).]
🔑:Substituting B=3 C into D=2 B-C gives D=2(3 C)-C=5 C.Substituting D=5 C and B=3 C into A=B+D gives A=3 C+5 C=8 C.Since A=8 C is a two-digit integer, we must have 10 leq 8 C leq 99, so frac{10}{8} leq C leq frac{99}{8}.Since C is an integer, this implies 2 leq C leq 12.We also know that C is a two-digit integer, so 10 leq C, which means that C=10, C=11, or C=12.Using the equations from earlier, we have A+B+C+D=8 C+3 C+C+5 C=17 C, so the larger C is, the larger A+B+C+D is.With C=12, we get A=8(12)=96, B=3(12)=36, and D=5(12)=60, which are all two-digit integers.Therefore, the conditions are satisfied when C=12, its largest possible value, which means C=12 corresponds to the largest possible value of A+B+C+D.Therefore, the answer is A+B+C+D=17 C=17(12)=204.ANSWER: 204
❓:In the geometric progression {a_n}, S_n denotes the sum of the first n terms, a_3=2S_2+1, and a_4=2S_3+1. Find the common ratio q.
🧠:[<think>Okay, so I need to find the common ratio q of a geometric progression where the third term a₃ is equal to 2 times the sum of the first two terms S₂ plus 1, and the fourth term a₄ is equal to 2 times the sum of the first three terms S₃ plus 1. Hmm, let me think step by step.First, let me recall what a geometric progression is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio, which is q here. So, if the first term is a₁ = a, then the terms go like a, aq, aq², aq³, and so on. Therefore, the nth term aₙ is a * q^(n-1).The sum of the first n terms Sₙ of a geometric progression is given by the formula Sₙ = a₁(1 - q^n)/(1 - q) when q ≠ 1. I think that's right. Let me verify. Yeah, when q is not 1, that's the formula. If q is 1, the sum is just n times the first term, but in this problem, since we're talking about a common ratio, q might not be 1, but maybe we need to check that.But given the problem mentions a geometric progression, and since they have a common ratio, it's possible that q could be 1, but in that case, all terms would be equal, which might not satisfy the given equations. Let me see later.So, given that a₃ = 2S₂ + 1 and a₄ = 2S₃ + 1, I need to find q.First, let's write expressions for a₃, a₄, S₂, and S₃ in terms of a and q.Starting with the terms:- a₁ = a- a₂ = a*q- a₃ = a*q²- a₄ = a*q³Now the sums:- S₂ = a₁ + a₂ = a + a*q = a(1 + q)- S₃ = S₂ + a₃ = a(1 + q) + a*q² = a(1 + q + q²)So, substituting into the given equations:First equation: a₃ = 2S₂ + 1So, a*q² = 2*a*(1 + q) + 1Second equation: a₄ = 2S₃ + 1So, a*q³ = 2*a*(1 + q + q²) + 1Now, we have two equations with two variables: a and q. Let's write them down:1) a*q² = 2a(1 + q) + 12) a*q³ = 2a(1 + q + q²) + 1Our goal is to solve for q. So perhaps we can solve for a from the first equation and substitute into the second equation.From equation 1: Let's rearrange for a.a*q² - 2a(1 + q) = 1a(q² - 2(1 + q)) = 1a(q² - 2 - 2q) = 1Therefore, a = 1 / (q² - 2q - 2)Okay, so now substitute this expression for a into equation 2.Equation 2: a*q³ = 2a(1 + q + q²) + 1Substitute a = 1 / (q² - 2q - 2):[1 / (q² - 2q - 2)] * q³ = 2*[1 / (q² - 2q - 2)]*(1 + q + q²) + 1Multiply both sides by (q² - 2q - 2) to eliminate the denominator:q³ = 2(1 + q + q²) + 1*(q² - 2q - 2)Simplify the right side:First, expand the 2*(1 + q + q²) term: 2 + 2q + 2q²Then add the (q² - 2q - 2) term:2 + 2q + 2q² + q² - 2q - 2Combine like terms:2 - 2 = 02q - 2q = 02q² + q² = 3q²So the right side simplifies to 3q²Therefore, the equation becomes:q³ = 3q²Bring all terms to one side:q³ - 3q² = 0Factor out q²:q²(q - 3) = 0So, either q² = 0 or q - 3 = 0If q² = 0, then q = 0. But if q is 0, then the terms of the geometric sequence would be a, 0, 0, 0,... which would make S₂ = a + 0 = a. Then a₃ = 0 = 2*a + 1. But from the first equation, a₃ = 0 = 2a + 1 => 2a + 1 = 0 => a = -1/2. Then check if this works with the second equation. a₄ = 0, and S₃ = a + 0 + 0 = a = -1/2. Then 2S₃ + 1 = 2*(-1/2) + 1 = -1 + 1 = 0, which matches a₄ = 0. Wait, so q = 0 actually satisfies both equations? But let's check if q = 0 is allowed in a geometric progression. Typically, the common ratio can be any real number except where division by zero occurs, but here q = 0 would make all terms after the first term zero. But in the problem statement, it's just called a geometric progression, not specifying that the common ratio can't be zero. So, perhaps q = 0 is a possible solution?But wait, let's think again. If q = 0, then the sequence is a, 0, 0, 0, ... Then S₂ = a + 0 = a, and a₃ = 0. Then according to the first equation, 0 = 2a + 1 => a = -1/2. Then S₃ = a + 0 + 0 = a = -1/2. Then the second equation says a₄ = 0 = 2*(-1/2) + 1 = -1 + 1 = 0. So that works. But in this case, the common ratio q is 0. However, typically, in some definitions, the common ratio is required to be non-zero to avoid triviality, but the problem didn't specify that. So maybe q = 0 is a valid solution here?But the other solution is q = 3. Let's check that.If q = 3, then from equation 1, a = 1 / (q² - 2q - 2) = 1/(9 - 6 - 2) = 1/(1) = 1. So a = 1.Then check the terms:a₁ = 1a₂ = 3a₃ = 9a₄ = 27S₂ = 1 + 3 = 4Check first equation: a₃ = 9 = 2*4 + 1 = 8 + 1 = 9. Correct.S₃ = 1 + 3 + 9 = 13Second equation: a₄ = 27 = 2*13 + 1 = 26 + 1 = 27. Correct.So q = 3 is a valid solution.So here we have two possible solutions: q = 0 and q = 3.But wait, earlier when we considered q = 0, the problem didn't specify that the common ratio can't be zero, but in standard geometric progressions, the common ratio is allowed to be zero, though such sequences are trivial after the first term. However, let's check the problem statement again. It says "geometric progression {a_n}", and gives equations involving S₂ and S₃. If q = 0, then S₂ = a + 0 = a, S₃ = a + 0 + 0 = a. Then the equations given are a₃ = 0 = 2a + 1 and a₄ = 0 = 2a + 1. Which gives a = -1/2. So that works, but if we accept q = 0, then both equations are satisfied. However, let's check if the problem allows q = 0. If the problem had mentioned that all terms are non-zero, then q = 0 would be invalid, but it didn't. So strictly speaking, both q = 0 and q = 3 are solutions. But maybe the problem expects only q = 3, as the meaningful solution. Let me see.But let's check the problem again. It says "Find the common ratio q." Without any restrictions. So technically, both are solutions. However, perhaps the problem is designed with a non-zero ratio in mind. Let me check.Wait, when q = 0, the terms after the first are all zero. So the third term is zero, but the sum S₂ is a + 0 = a. The equation a₃ = 2S₂ + 1 becomes 0 = 2a + 1, so a = -1/2. Then the fourth term is zero, and S₃ = a + 0 + 0 = a = -1/2. Then 2S₃ + 1 = 2*(-1/2) + 1 = -1 + 1 = 0, which matches a₄ = 0. So technically, this works.But maybe the problem considers q = 0 as invalid. In some contexts, a geometric progression is defined with a non-zero common ratio. For example, the common ratio is a real number not equal to zero. Let me check the definition.According to standard definitions, a geometric progression is a sequence where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. So, in that case, q = 0 would not be allowed because multiplying by zero would not maintain a non-zero ratio. However, different sources might have different definitions. If the problem follows the definition where the common ratio must be non-zero, then q = 0 is invalid, and only q = 3 is the answer. But if the problem allows q = 0, then there are two solutions.But since the problem is given in a mathematical context, likely expecting a non-zero ratio. Let's check if there's a mistake in my calculation.Wait, when I solved the equation, I had q³ = 3q² => q²(q - 3) = 0. So roots at q = 0 and q = 3. But if the definition requires q ≠ 0, then q = 3 is the only solution.Alternatively, maybe even with q = 0, the problem is considered valid, so the answer could be 0 or 3. But let's see. If I put both answers, but the problem says "Find the common ratio q." So maybe expecting one answer. Wait, perhaps there is a unique solution. Let me check my steps again.Wait, when q = 0, the first term is -1/2. Then a₃ = 0, a₄ = 0. The sums S₂ = -1/2, S₃ = -1/2, so 2S₂ + 1 = 2*(-1/2) + 1 = -1 + 1 = 0, which equals a₃. Similarly, 2S₃ +1 = 0, which equals a₄. So q=0 is a valid solution.But if the problem assumes the progression is non-trivial, then q=3 is the answer. But unless specified, both are possible. However, maybe in the problem's context, since the progression is written as {a_n}, and with the conditions given, the ratio is unique. Wait, but according to the equations, there are two solutions.But let me check with the equations again. If q=0, then a is -1/2. But the sequence is -1/2, 0, 0, 0,... which is a valid geometric progression with ratio 0, even if trivial. So unless specified, both are possible. But perhaps the problem expects q=3 as the answer.Wait, maybe there's a mistake in the calculation. Let me check again.From equation 1: a*q² = 2a(1 + q) + 1If q=0, then left side is 0, right side is 2a +1. So 0 = 2a +1 => a= -1/2. Then equation 2: a*q³ = 2a(1 + q + q²) +1. If q=0, then left side is 0, right side is 2a(1 +0 +0) +1 = 2a +1. Again, 0 = 2*(-1/2) +1 => 0 = -1 +1 => 0=0. So yes, both equations are satisfied. Therefore, q=0 is a solution.But maybe the problem expects the non-trivial solution. Let me check if the problem has any hidden constraints.Wait, if q=0, then all terms after the first are zero. Then S₂ = a +0 = a, S₃ = a +0 +0 =a. Then the equations given are:a₃ = 2S₂ +1 => 0 = 2a +1 => a= -1/2.a₄ = 2S₃ +1 => 0 = 2a +1 => same as above.So this works. But if the problem wants a geometric progression where all terms are non-zero, then q=0 would be invalid. But the problem didn't specify that. So in a strict sense, both q=0 and q=3 are solutions.Wait, but in the problem statement, they use S_n as the sum of the first n terms. If q=0, then the sum S_n = a +0 +0 +... So S_n = a for all n >=1, which is possible, but maybe the problem expects the sum to be different when n increases, but since in the equations given, S₂ and S₃ are used, but with q=0, S₂ = S₃ = a. Then 2S₂ +1 and 2S₃ +1 would both be 2a +1, which would equal a₃ and a₄. So yes, that works. But since a₃ and a₄ are both zero, which equals 2a +1. So that's valid.But given that the problem is likely designed for a non-zero ratio, since otherwise it's a trivial case, the answer is q=3. But since the problem didn't specify, perhaps both are acceptable.But in the problem's context, maybe the answer is 3. Let me check if my calculation was correct.Wait, another way to check: if we assume q ≠0, then from equation q²(q -3)=0, the solution is q=3. So if we discard q=0, then q=3 is the solution. But in the problem's case, perhaps they want the non-zero solution, so the answer is 3.Alternatively, the problem might have intended for q to be 3, so that's the answer. Let me check if any steps have an error.Starting from the two equations:1) a*q² = 2a(1 + q) +12) a*q³ = 2a(1 + q + q²) +1Expressed in terms of a:From 1) a = 1/(q² - 2q - 2)Plug into 2):[1/(q² - 2q -2)] * q³ = 2*[1/(q² - 2q -2)]*(1 + q + q²) +1Multiply both sides by (q² - 2q -2):q³ = 2(1 + q + q²) + (q² -2q -2)Simplify RHS:2 + 2q + 2q² + q² -2q -2Combine terms:(2 -2) + (2q -2q) + (2q² + q²) = 0 + 0 + 3q² = 3q²Thus:q³ = 3q²Which gives q²(q -3)=0, so q=0 or q=3.Therefore, the solutions are q=0 or q=3.So the problem is designed to have two possible solutions? But since the problem says "Find the common ratio q", maybe both are acceptable, but in most cases, unless specified, the non-zero solution is expected.Alternatively, perhaps the problem is from a source where the answer is 3, so that's intended. But given that both solutions satisfy, I need to check again.Wait, let's verify with q=0. If q=0, then the sequence is -1/2, 0, 0, 0,... So the third term is 0, which is 2*S₂ +1 = 2*(-1/2) +1 = -1 +1 =0, which works. The fourth term is 0, which is 2*S₃ +1 =2*(-1/2) +1 =0, which works. So yes, q=0 is correct.But perhaps the problem's answer is 3. Maybe when writing the problem, the author considered only the non-trivial solution. Alternatively, in some textbooks, geometric sequences are defined with a non-zero common ratio, so q ≠0. Thus, q=3 is the answer.Since the problem is in Chinese maybe (given the user's Chinese username?), but regardless, in international mathematics, a geometric progression can have q=0, but sometimes restricted to q ≠0. The problem didn't specify, so both solutions are possible. However, since the problem asks for "the common ratio q", implying a singular answer, maybe there's a unique solution. Let me check again.Wait, when we solved the equations, we found two solutions. So unless there's a mistake in my calculations, both q=0 and q=3 are valid. But if the problem is designed to have a unique solution, maybe I made a mistake. Wait, let me check my substitution again.From equation 1: a = 1 / (q² -2q -2)Substituted into equation 2:a*q³ = 2a*(1 + q + q²) +1Yes, then multiplied both sides by denominator:q³ = 2(1 + q + q²) + (q² -2q -2)Which simplifies to 3q², so q³=3q², leading to q²(q -3)=0. Correct.Thus, the solutions are indeed q=0 and q=3.Therefore, unless there's a constraint in the problem I missed, both are acceptable.But given that in the original problem statement, the progression is written as {a_n}, without any restrictions, the answer could be both. But likely, the problem expects q=3 as the answer.Alternatively, maybe q=0 is rejected because when substituting back, the denominator in the expression for a becomes q² -2q -2. If q=0, then denominator is 0² -2*0 -2 = -2, which is okay. So a=1/(-2). Wait, wait, no: when q=0, denominator is (0 -0 -2)= -2, so a=1/(-2)= -1/2. Which is valid, no division by zero.So q=0 is acceptable. Therefore, both solutions are valid. Hmm.But since the problem says "Find the common ratio q", and it's possible to have two answers. However, the problem might expect both, but since the answer is to be put in a box, maybe two answers. But the original problem is in Chinese, maybe expecting only the non-zero solution. Alternatively, maybe there's an error in my thinking.Wait, let's check q=0 again. If q=0, then the sequence is -1/2, 0, 0, 0,... Then S₁ = -1/2, S₂ = -1/2, S₃ = -1/2, etc. Then a₃ = 0, which is 2*S₂ +1 =2*(-1/2) +1= -1+1=0, correct. Similarly, a₄ =0=2*S₃ +1= same result. So that's valid.But in the progression, after the first term, all terms are zero, so it's a valid geometric progression with ratio zero. However, some might argue that a geometric progression requires the common ratio to be non-zero to have a non-degenerate sequence, but I don't recall a strict mathematical definition that enforces this. It's more of a convention in some contexts.Therefore, in conclusion, the problem has two solutions: q=0 and q=3. However, if the problem expects a single answer, likely q=3, as the non-trivial solution. But since the problem didn't specify, I should present both solutions. However, the problem might be designed such that only q=3 is the answer, so maybe I made a mistake in considering q=0. Let me check once more.Wait, the problem says "geometric progression {a_n}", and "S_n denotes the sum of the first n terms". If q=0, then all terms after the first are zero, so S_n = a₁ for all n >=1. Then, the equations given are:a₃ = 2S₂ +1 => 0 = 2a₁ +1 => a₁ = -1/2a₄ = 2S₃ +1 => 0 = 2a₁ +1 => a₁ = -1/2Which is consistent, so the sequence is valid. Therefore, q=0 and a₁=-1/2 is a valid solution. Thus, the common ratio can be 0 or 3.But given that the problem is asking for the common ratio, and it's possible that multiple ratios satisfy the conditions, I should check the answer. However, given the problem's phrasing, perhaps the answer is 3. Alternatively, maybe I missed something.Wait, let's check with q=0, is the progression really considered a geometric progression? According to the definition, yes, as long as each term is multiplied by the common ratio, even if it's zero. So yes, q=0 is allowed.But in many mathematical problems, especially in exams or textbooks, unless specified, sometimes degenerate cases like q=0 are disregarded. So, perhaps the answer is 3.Alternatively, the problem might require the ratio to be positive, but since the problem didn't specify, we can't assume that.Therefore, based on strict mathematical analysis, the possible common ratios are 0 and 3. However, if the problem expects a single answer, likely 3. But since the problem didn't specify, both are possible.But since the user hasn't specified, and given that in most cases, problems like this expect the non-trivial solution, I think the answer is 3. So, final answer is q=3.Final Answerboxed{3}</think>Given a geometric progression ({a_n}) where (S_n) denotes the sum of the first (n) terms, we know that (a_3 = 2S_2 + 1) and (a_4 = 2S_3 + 1). We need to find the common ratio (q).First, express the terms and sums in terms of the first term (a) and the common ratio (q):- The terms are: (a_1 = a), (a_2 = aq), (a_3 = aq^2), (a_4 = aq^3).- The sums are: (S_2 = a + aq = a(1 + q)), (S_3 = a + aq + aq^2 = a(1 + q + q^2)).Using the given conditions:1. (a_3 = 2S_2 + 1) [ aq^2 = 2a(1 + q) + 1 ]2. (a_4 = 2S_3 + 1) [ aq^3 = 2a(1 + q + q^2) + 1 ]From the first equation, solve for (a):[aq^2 - 2a(1 + q) = 1 implies a(q^2 - 2q - 2) = 1 implies a = frac{1}{q^2 - 2q - 2}]Substitute (a) into the second equation:[left(frac{1}{q^2 - 2q - 2}right)q^3 = 2left(frac{1}{q^2 - 2q - 2}right)(1 + q + q^2) + 1]Multiply both sides by (q^2 - 2q - 2):[q^3 = 2(1 + q + q^2) + (q^2 - 2q - 2)]Simplify the right-hand side:[q^3 = 2 + 2q + 2q^2 + q^2 - 2q - 2 implies q^3 = 3q^2]Factor out (q^2):[q^2(q - 3) = 0]This gives the solutions (q = 0) or (q = 3). - For (q = 0), the sequence becomes (-1/2, 0, 0, 0, ldots), which is a valid geometric progression but trivial after the first term.- For (q = 3), the sequence is (1, 3, 9, 27, ldots), which satisfies both given conditions.Given the problem context and common expectations for non-trivial solutions, the common ratio is (boxed{3}).,<think>Alright, let's tackle this problem step by step. It's about a geometric progression, so I know that each term is multiplied by a common ratio q to get the next term. The problem states that the third term, a₃, is equal to twice the sum of the first two terms plus one, and similarly, the fourth term, a₄, is equal to twice the sum of the first three terms plus one. We need to find the common ratio q.First, let me recall the formula for the nth term of a geometric progression. The nth term is a_n = a₁ * q^(n-1), where a₁ is the first term and q is the common ratio. Also, the sum of the first n terms, S_n, is given by S_n = a₁*(1 - q^n)/(1 - q) when q ≠ 1. Since this is a geometric progression and we're talking about sums, I can use these formulas.Given the problem, there are two equations provided:1. a₃ = 2S₂ + 12. a₄ = 2S₃ + 1I need to translate these into equations involving a₁ and q, then solve for q. Let's start with the first equation.Starting with a₃. Since a₃ is the third term, using the formula for the nth term:a₃ = a₁ * q².Now, S₂ is the sum of the first two terms. Using the sum formula:S₂ = a₁ + a₂ = a₁ + a₁*q = a₁(1 + q).So the first equation becomes:a₁*q² = 2*a₁*(1 + q) + 1. ...(1)Similarly, for the second equation:a₄ = a₁ * q³.S₃ is the sum of the first three terms. Using the sum formula:S₃ = a₁ + a₁*q + a₁*q² = a₁*(1 + q + q²).So the second equation is:a₁*q³ = 2*a₁*(1 + q + q²) + 1. ...(2)Now, we have two equations (1) and (2) with two variables: a₁ and q. The goal is to solve for q. Let's see how to eliminate a₁.From equation (1):a₁*q² = 2*a₁*(1 + q) + 1.Let's rearrange this equation to solve for a₁. Bring all terms involving a₁ to the left and constants to the right:a₁*q² - 2*a₁*(1 + q) = 1.Factor out a₁:a₁*(q² - 2*(1 + q)) = 1.Similarly, let's compute the coefficient:q² - 2*(1 + q) = q² - 2q - 2.So:a₁*(q² - 2q - 2) = 1. ...(1a)Similarly, from equation (2):a₁*q³ = 2*a₁*(1 + q + q²) + 1.Rearranging:a₁*q³ - 2*a₁*(1 + q + q²) = 1.Factor out a₁:a₁*(q³ - 2*(1 + q + q²)) = 1.Compute the coefficient:q³ - 2*(1 + q + q²) = q³ - 2 - 2q - 2q².So:a₁*(q³ - 2q² - 2q - 2) = 1. ...(2a)Now, from equations (1a) and (2a), we have:From (1a): a₁ = 1 / (q² - 2q - 2)From (2a): a₁ = 1 / (q³ - 2q² - 2q - 2)Since both equal a₁, we can set them equal to each other:1 / (q² - 2q - 2) = 1 / (q³ - 2q² - 2q - 2)Therefore, cross-multiplying:q³ - 2q² - 2q - 2 = q² - 2q - 2.Bring all terms to one side:q³ - 2q² - 2q - 2 - q² + 2q + 2 = 0Simplify term by term:q³ - 3q² + 0q + 0 = 0So:q³ - 3q² = 0Factor out q²:q²(q - 3) = 0Therefore, possible solutions are q² = 0 => q = 0, or q - 3 = 0 => q = 3.But in a geometric progression, the common ratio q can't be 0, because then all terms after the first would be zero, which would lead to a trivial and non-sensical progression in most cases. Let's check if q = 3 is valid.So q = 3. Let's verify this solution.First, compute a₁ from equation (1a):a₁ = 1 / (q² - 2q - 2) = 1 / (9 - 6 - 2) = 1 / (1) = 1.Therefore, a₁ = 1. Let's check if this works in equation (2a):a₁ = 1 / (q³ - 2q² - 2q - 2) = 1 / (27 - 18 - 6 - 2) = 1 / (1) = 1.Consistent. So both equations give a₁ = 1 when q = 3.Now, let's verify the original equations to ensure there are no mistakes.First equation: a₃ = 2S₂ + 1.Compute a₃: a₁*q² = 1*9 = 9.Compute S₂: a₁ + a₂ = 1 + 3 = 4.Then 2S₂ + 1 = 8 + 1 = 9, which matches a₃.Second equation: a₄ = 2S₃ + 1.Compute a₄: a₁*q³ = 1*27 = 27.Compute S₃: 1 + 3 + 9 = 13.Then 2S₃ + 1 = 26 + 1 = 27, which matches a₄.Therefore, q = 3 is indeed the solution.But wait, let's check if q = 0 is a possible solution, even though it seems invalid. If q = 0, then:a₁ would be from equation (1a):a₁ = 1 / (0 - 0 - 2) = -1/2.But then:a₁ = -1/2, q = 0.Check equation (2a):a₁ = 1 / (0 - 0 - 0 - 2) = -1/2, which matches.But let's check the original equations.First equation: a₃ = 2S₂ + 1.If q = 0, then the sequence is:a₁ = -1/2,a₂ = a₁*0 = 0,a₃ = 0,a₄ = 0.Compute S₂ = a₁ + a₂ = -1/2 + 0 = -1/2.Then 2S₂ + 1 = 2*(-1/2) + 1 = -1 + 1 = 0. Which matches a₃ = 0.Similarly, a₄ = 0.Compute S₃ = a₁ + a₂ + a₃ = -1/2 + 0 + 0 = -1/2.Then 2S₃ + 1 = 2*(-1/2) + 1 = -1 + 1 = 0, which matches a₄.So technically, q = 0 also satisfies the equations. But is q = 0 considered a valid common ratio?In a geometric progression, the common ratio can technically be any real number, including 0. However, if q = 0, then all terms after the first are zero, which might be considered a degenerate case. But depending on the problem's constraints, it might be allowed. However, let's check the problem statement. It says "geometric progression {a_n}" without any restrictions. So mathematically, q = 0 is a valid solution here.But wait, in the problem statement, they mention S_n denotes the sum of the first n terms. If q = 0, then the progression is -1/2, 0, 0, 0, ..., so S₂ = -1/2, S₃ = -1/2, etc. But when they write a₃ = 2S₂ + 1, which would be 0 = 2*(-1/2) +1 => 0 = -1 +1 => 0=0, which holds. Similarly for a₄. So it's mathematically valid.But the problem is asking for the common ratio. If both q = 0 and q = 3 are solutions, then there's an ambiguity. However, in the equations, when we solved, we got q²(q - 3) = 0, so q = 0 or q = 3. But maybe the problem expects us to consider the non-trivial solution, or there's an implicit assumption that the progression is non-degenerate. Let's check.If we take q = 0, then all terms after the first are zero, which might not be considered a "proper" geometric progression in some contexts. For example, in some definitions, the common ratio must be non-zero. But in standard mathematical terms, q = 0 is allowed. However, since the problem mentions S_n, which would be constant from the second term onward if q = 0, but in the problem, the equations still hold. So why does the problem specify "the common ratio q" as if it's unique? Therefore, maybe q = 0 is invalid here, perhaps due to division by zero in the sum formula? Wait, when q = 0, the sum formula S_n = a₁*(1 - q^n)/(1 - q) is still valid, since q ≠ 1. For q = 0, S_n = a₁*(1 - 0)/(1 - 0) = a₁. But in reality, when q = 0, the sum S_n is a₁ + 0 + 0 + ... + 0 = a₁, which matches. So even the sum formula works here.But the problem here is that if q = 0, then S₂ = a₁ + a₂ = a₁ + 0 = a₁. So in equation (1):a₃ = 2S₂ + 1 => 0 = 2a₁ + 1. So a₁ = -1/2, which is consistent. Similarly, equation (2):a₄ = 0 = 2S₃ + 1. But S₃ = a₁ + 0 + 0 = a₁ = -1/2. Therefore, 2*(-1/2) +1 = -1 +1 = 0, which is correct.So both q = 3 and q = 0 satisfy the equations. However, in the context of a problem asking for "the common ratio", implying a unique answer, and given that the user is likely expecting the non-zero ratio, but we need to check.But wait, when we solved the equations, we found q²(q - 3) = 0, so q = 0 or q = 3. However, perhaps there's an error in the algebra steps. Let's verify the step where we subtracted the equations.From equations (1a) and (2a):(1a): a₁ = 1 / (q² - 2q - 2)(2a): a₁ = 1 / (q³ - 2q² - 2q - 2)Therefore, equate them:1 / (q² - 2q - 2) = 1 / (q³ - 2q² - 2q - 2)Cross-multiplying:q³ - 2q² - 2q - 2 = q² - 2q - 2Subtract q² - 2q - 2 from both sides:q³ - 2q² - 2q - 2 - q² + 2q + 2 = 0Simplify:q³ - 3q² = 0Yes, that's correct. So q²(q - 3) = 0, so q = 0 or q = 3.Therefore, both are solutions. However, the problem might have a unique solution, so perhaps q = 0 is invalid for some reason.Wait, if we take q = 0, then the term a₃ = 0, which is equal to 2S₂ +1. But S₂ = a₁ + a₂ = (-1/2) + 0 = -1/2. So 2*(-1/2) +1 = -1 +1 = 0, which is correct. Similarly, a₄ = 0, which equals 2S₃ +1, and S₃ = (-1/2) +0 +0 = -1/2, so 2*(-1/2) +1 = 0. So all equations hold. Therefore, mathematically, both solutions are valid.But the problem is presented as a problem to find the common ratio, which is likely unique. Therefore, perhaps the problem assumes that the common ratio is not zero. Maybe in the context of the problem, they consider only q ≠ 0. Alternatively, perhaps the problem is designed such that q is positive, or an integer. But the problem statement doesn't specify.Alternatively, maybe there's a miscalculation in assuming q = 0 is valid. Let's think again. If we take q = 0, then the progression is -1/2, 0, 0, 0,... Then, S₂ = -1/2, which is 2S₂ +1 = 2*(-1/2)+1= -1 +1=0 = a₃. Correct. Similarly, S₃ = -1/2, 2*S₃ +1=0=a₄. So it's correct. Therefore, q=0 is a valid solution. But perhaps the problem expects us to find the non-trivial solution. However, since both are mathematically valid, perhaps both answers should be considered.But let's check the problem statement again: "Find the common ratio q." Without any restrictions. So technically, there are two solutions. However, in the given equations, when we consider q = 0, the sum S_n is not a geometric series in the traditional sense because after the first term, all terms are zero, but the problem doesn't specify any restrictions. Therefore, both q = 0 and q =3 are solutions. But the answer provided by the system is likely expecting q =3. Maybe the problem is from a source where q=3 is the answer, and q=0 is considered trivial. Alternatively, perhaps when setting up the equations, there's an implicit assumption that q ≠0.Wait, let's think again. When we wrote the sum S_n, we used the formula S_n = a₁(1 - q^n)/(1 - q), which is valid when q ≠1. But when q=0, the formula still holds because (1 - 0)/(1 -0 )=1, so S_n =a₁*1/1= a₁, which is correct, as the sum of the first n terms when q=0 is just a₁ followed by zeros, so the sum is a₁. Therefore, the equations hold. Therefore, mathematically, both solutions are correct. But since the problem asks for "the" common ratio, implying uniqueness, maybe there's a mistake here.Alternatively, perhaps in the original problem, there are constraints like all terms are positive or integers. But the problem as stated doesn't mention that. Hmm.Wait, let's check the problem again: "In the geometric progression {a_n}, S_n denotes the sum of the first n terms, a₃=2S₂+1, and a₄=2S₃+1. Find the common ratio q."Given that the problem is presented in Chinese (from the original user's request), but translated here, and the standard geometric progression problems usually expect a non-trivial ratio unless specified otherwise. Moreover, if both answers are possible, the problem would mention "find all possible common ratios". Since it's asking for "the common ratio", it's likely that the answer is unique, so q=3. Therefore, perhaps q=0 is rejected for some reason. Maybe because if q=0, then S₂ and S₃ are negative (since a₁=-1/2), but the problem doesn't specify positivity. Alternatively, maybe when q=0, the terms beyond the first are zero, but the problem says "geometric progression", and in some contexts, a geometric progression is considered with a non-zero ratio. For example, in some textbooks, the common ratio is required to be non-zero to avoid trivialities. So perhaps in this problem's context, q=0 is invalid, and only q=3 is acceptable.Alternatively, maybe the problem is designed such that a₁ must be positive? The problem doesn't specify, but if we take a₁=1, which is positive, when q=3, then the progression is 1,3,9,27,... which is all positive terms. Whereas when q=0, a₁=-1/2, which is negative, but the problem doesn't state that the terms must be positive. So that's not a valid reason.Alternatively, maybe when q=0, the progression is not considered a "geometric progression" in the problem's context. However, according to the definition, a geometric progression is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio. So even if the common ratio is zero, it's still a geometric progression, albeit a trivial one after the first term. But perhaps the problem expects a non-zero ratio.Given that, and given that in school problems, often the common ratio is expected to be a non-zero real number unless specified otherwise, perhaps the answer is q=3.Moreover, in the problem's equations, if we take q=0, then the first term a₁=-1/2, which is a fraction. The problem doesn't specify whether the terms should be integers or not, but in competition problems, often integer solutions are preferred. However, since the problem doesn't specify, that might not be a valid consideration.But, given that the problem provides two equations and we have two solutions, but the problem asks for "the" common ratio, implying a unique answer, perhaps q=3 is the expected answer. Also, considering that in the case of q=0, the progression is not really a standard geometric progression in an intuitive sense beyond the first term, since all subsequent terms are zero, which might be considered degenerate.Therefore, taking all this into account, the answer is q=3.But just to make sure, let's verify once again with q=3 and a₁=1:The sequence is 1, 3, 9, 27,...a₃=9, S₂=1+3=4, 2*S₂ +1=8+1=9, which matches.a₄=27, S₃=1+3+9=13, 2*S₃ +1=26+1=27, which matches.Perfect. So q=3 is correct.For q=0, the sequence is -1/2, 0, 0, 0,...a₃=0, S₂=-1/2, 2*(-1/2)+1=0, which matches.a₄=0, S₃=-1/2, 2*(-1/2)+1=0, which matches.So mathematically, both are solutions, but since the problem asks for "the" common ratio, likely expecting q=3.Therefore, the common ratio is 3.Final AnswerThe common ratio ( q ) is (boxed{3}).</think>Given the geometric progression ({a_n}) with (S_n) denoting the sum of the first (n) terms, we know the following conditions: (a_3 = 2S_2 + 1) and (a_4 = 2S_3 + 1). We need to find the common ratio (q).First, recall the formulas for the (n)-th term and the sum of the first (n) terms of a geometric progression:- The (n)-th term: (a_n = a_1 cdot q^{n-1})- The sum of the first (n) terms: (S_n = a_1 frac{1 - q^n}{1 - q}) for (q neq 1)Using these formulas, we translate the given conditions into equations involving (a_1) and (q):1. For (a_3 = 2S_2 + 1): [ a_1 cdot q^2 = 2 left( a_1 frac{1 - q^2}{1 - q} right) + 1 ] Simplifying (S_2): [ S_2 = a_1 + a_1 q = a_1 (1 + q) ] Therefore: [ a_1 q^2 = 2 a_1 (1 + q) + 1 ]2. For (a_4 = 2S_3 + 1): [ a_1 cdot q^3 = 2 left( a_1 frac{1 - q^3}{1 - q} right) + 1 ] Simplifying (S_3): [ S_3 = a_1 + a_1 q + a_1 q^2 = a_1 (1 + q + q^2) ] Therefore: [ a_1 q^3 = 2 a_1 (1 + q + q^2) + 1 ]We now have two equations:1. (a_1 q^2 = 2 a_1 (1 + q) + 1)2. (a_1 q^3 = 2 a_1 (1 + q + q^2) + 1)Rearranging these equations to solve for (a_1):1. (a_1 (q^2 - 2q - 2) = 1)2. (a_1 (q^3 - 2q^2 - 2q - 2) = 1)Setting the expressions for (a_1) equal to each other:[frac{1}{q^2 - 2q - 2} = frac{1}{q^3 - 2q^2 - 2q - 2}]Cross-multiplying and simplifying:[q^3 - 2q^2 - 2q - 2 = q^2 - 2q - 2][q^3 - 3q^2 = 0][q^2 (q - 3) = 0]This gives (q = 0) or (q = 3). Since (q = 0) leads to a degenerate geometric progression, we discard it. Thus, the valid common ratio is (q = 3).Verifying (q = 3) with (a_1 = 1):- The sequence is (1, 3, 9, 27, ldots)- (a_3 = 9) and (2S_2 + 1 = 2(1 + 3) + 1 = 9)- (a_4 = 27) and (2S_3 + 1 = 2(1 + 3 + 9) + 1 = 27)Both conditions are satisfied, confirming the common ratio is indeed (3).[boxed{3}]]
🔑:Given that a_3 = 2S_2 + 1 quad text{(1)}a_4 = 2S_3 + 1 quad text{(2)}Subtracting equation (1) from equation (2), we get:a_4 - a_3 = 2(S_3 - S_2) = 2a_3Rearranging the equation, we find that:a_4 = 3a_3 quad text{(3)}Using the formula for the nth term of the geometric progression, where a_n = a_1 cdot q^{n-1}, we can substitute n with 4 and 3 to get:a_1q^3 = 3a_1q^2Since a_1 neq 0 and q neq 0, we can divide both sides by a_1q^2 to isolate q. Doing that, we get:q = 3Thus, the common ratio of the geometric progression is boxed{3}.
❓:Convert 750^{circ} into radians.A: dfrac{28}{3}pitext{rad}B: dfrac{25}{6}pitext{rad}C: dfrac{23}{6}pitext{rad}D: dfrac{23}{3}pitext{rad}
🧠:[<think>Okay, so I need to convert 750 degrees into radians. Hmm, right. I remember that converting between degrees and radians involves a conversion factor. Let me recall... I think the key is that 180 degrees is equal to π radians. So, the formula should be something like multiplying the number of degrees by π over 180 to get radians. Let me check that again.Yes, if you have an angle in degrees, you can convert it to radians by multiplying by π/180. So, the formula is: radians = degrees × (π/180). Let me write that down to make sure. So, for 750 degrees, it would be 750 × (π/180). Then, I need to simplify that fraction.First, let's compute 750 divided by 180. Let me see. Both 750 and 180 can be divided by 10, so that becomes 75/18. Hmm, can that be simplified further? Let's check. 75 divided by 3 is 25, and 18 divided by 3 is 6. So, 75/18 simplifies to 25/6. Therefore, 750/180 reduces to 25/6. So, 750 degrees is equal to (25/6)π radians. Let me confirm that.Wait, 180 degrees is π radians, so 360 degrees is 2π radians, 540 degrees is 3π radians, 720 degrees is 4π radians, and 750 degrees is 720 + 30, so 4π + (30 degrees). Then, converting 30 degrees to radians would be 30 × π/180 = π/6. So, adding that to 4π gives 4π + π/6. Let me convert 4π to sixths: 4π is 24/6 π, so 24/6 + 1/6 is 25/6 π. So that matches the previous result. So, 25/6 π radians.Looking at the options given: A is 28/3 π, B is 25/6 π, C is 23/6 π, D is 23/3 π. So, the correct answer should be B. Let me just cross-verify once more to make sure I didn't make a mistake.Another way: Let's compute 750 divided by 180. 750 divided by 180. Let's do this division step by step. 180 times 4 is 720. 750 minus 720 is 30. So, 750 is 4 and 30/180, which simplifies to 4 and 1/6, which is 25/6 when converted to an improper fraction. Therefore, 25/6 π radians. Yep, that's definitely B. So, the answer should be option B.I think that's solid. I converted using two different methods: simplifying the fraction directly and breaking it down into known angles. Both led to the same answer, so I feel confident that B is the correct choice.Final Answerboxed{B}</think>To convert (750^circ) into radians, we use the conversion factor (pi) radians = (180^circ). The formula to convert degrees to radians is:[text{radians} = text{degrees} times left(frac{pi}{180}right)]Applying this to (750^circ):[750^circ times left(frac{pi}{180}right)]First, simplify the fraction (frac{750}{180}):1. Divide both numerator and denominator by 10: (frac{75}{18})2. Simplify further by dividing both by 3: (frac{25}{6})Thus, we have:[750^circ times left(frac{pi}{180}right) = frac{25}{6}pi]To verify, we can break down (750^circ) as (720^circ + 30^circ). Converting each part:- (720^circ = 4pi) radians (since (360^circ = 2pi) radians)- (30^circ = frac{pi}{6}) radians (since (30^circ times frac{pi}{180} = frac{pi}{6}))Adding these together:[4pi + frac{pi}{6} = frac{24pi}{6} + frac{pi}{6} = frac{25pi}{6}]Both methods confirm the result is (frac{25}{6}pi) radians. Therefore, the correct answer is:[boxed{B}],<think>Okay, so I need to convert 750 degrees into radians. Hmm, let me remember how to do that. I think the conversion between degrees and radians is done using a simple formula. Right, to convert degrees to radians, you multiply the number of degrees by π over 180. So the formula should be: radians = degrees × (π/180). Let me write that down.So, applying that to 750 degrees, it would be 750 multiplied by π divided by 180. Let me compute that. First, 750 divided by 180. Hmm, both of these numbers can be divided by 10 to simplify. That gives me 75/18. Wait, can that be simplified more? Let's check.75 divided by 18. Both numbers are divisible by 3. 75 divided by 3 is 25, and 18 divided by 3 is 6. So, that simplifies to 25/6. So, 750 degrees is 25/6 π radians. Let me check the answer choices.Looking at the options: A is 28/3 π, B is 25/6 π, C is 23/6 π, and D is 23/3 π. So, 25/6 π is option B. That seems straightforward, but wait, let me verify again to make sure I didn't make a mistake in the simplification.Starting over: 750 degrees times (π/180). So 750/180. Let me do this division step by step. 180 goes into 750 how many times? 180 times 4 is 720, which is 4 times 180. 750 minus 720 is 30. So that's 4 and 30/180. 30/180 simplifies to 1/6. So total is 4 + 1/6 = 25/6. Wait, 4 is 24/6, right? So 24/6 + 1/6 is 25/6. Yeah, that's correct. So 750 degrees is indeed 25/6 π radians, which is option B.But hold on, let me think again. Sometimes angles can be more than 360 degrees, but when converting to radians, the multiple doesn't matter, right? Because radians are just a different unit, regardless of how many full circles you've made. So even though 750 degrees is more than 360, converting it to radians is just a direct conversion without subtracting full circles. So 750 degrees is just 25/6 π radians. Hmm, that's about 4.1667 π radians. Let me check with another method.Alternatively, I know that 180 degrees is π radians, so 360 degrees is 2π radians. 750 divided by 360 is equal to... Let me calculate that. 750 divided by 360. Dividing numerator and denominator by 30, that's 25/12. So 750 degrees is 25/12 times 2π radians, which is 25/6 π radians. Yep, same answer. So that confirms it. So option B must be correct.But just to be thorough, let me check if any of the other options could be equivalent. For example, 25/6 π is approximately 4.1667 π. Let's see, 28/3 π is roughly 9.333 π, which is way more. 23/6 π is approximately 3.833 π, and 23/3 π is around 7.666 π. So none of these are the same as 25/6 π, so B is definitely the right answer. I think that's solid. No mistakes in the calculation steps, and cross-verified using a different approach. So the answer should be B.Final Answerboxed{B}</think>To convert (750^circ) into radians, we use the conversion formula:[text{radians} = text{degrees} times left( frac{pi}{180} right)]Applying this to (750^circ):[750 times left( frac{pi}{180} right)]First, simplify the fraction (frac{750}{180}):[frac{750}{180} = frac{75}{18} = frac{25}{6}]Thus, (750^circ) in radians is:[frac{25}{6} pi]Checking the answer choices, we find that (frac{25}{6} pi) corresponds to option B. [boxed{B}]]
🔑:Since 180^{circ}=pi, it follows that 1^{circ} = dfrac{pi}{180}. Therefore, 750^{circ}=750 times dfrac{pi}{180}= dfrac{25}{6}pitext{rad}. Hence, the correct choice is boxed{B}. From 180^{circ}=pi, we get 1^{circ} = dfrac{pi}{180}, substituting into 750^{circ}=750 times dfrac{pi}{180} gives the answer. This question tests the conversion between degree measure and radian measure, which is a basic topic.
❓:For a non negative integer n, set t I_n=int_0^{frac{pi}{4}} tan ^ n x dx to answer the following questions:(1) Calculate I_{n+2}+I_n.(2) Evaluate the values of I_1, I_2 and I_3.1978 Niigata university entrance exam
🧠:[<think>Okay, so I need to solve these two questions about the integrals ( I_n = int_0^{frac{pi}{4}} tan^n x , dx ). Let's start with the first one: Calculate ( I_{n+2} + I_n ).Hmm, integrating powers of tangent functions. I remember that sometimes you can use reduction formulas or recursive relationships for integrals involving powers of trigonometric functions. Maybe I can find a relationship between ( I_{n+2} ) and ( I_n ).Let me write out ( I_{n+2} + I_n ):[I_{n+2} + I_n = int_0^{frac{pi}{4}} tan^{n+2} x , dx + int_0^{frac{pi}{4}} tan^n x , dx]Wait, perhaps I can combine these two integrals into one:[int_0^{frac{pi}{4}} left( tan^{n+2} x + tan^n x right) dx = int_0^{frac{pi}{4}} tan^n x left( tan^2 x + 1 right) dx]Ah! Because ( tan^{n+2} x = tan^n x cdot tan^2 x ). Then, factoring out ( tan^n x ), we get ( tan^n x (tan^2 x + 1) ). And I remember that ( tan^2 x + 1 = sec^2 x ). That's a trigonometric identity. So this simplifies to:[int_0^{frac{pi}{4}} tan^n x cdot sec^2 x , dx]Okay, so now the integral becomes ( int tan^n x cdot sec^2 x , dx ). Let me think about substitution here. If I let ( u = tan x ), then ( du = sec^2 x , dx ). Perfect, that substitution would work.Let's do the substitution:When ( x = 0 ), ( u = tan 0 = 0 ).When ( x = frac{pi}{4} ), ( u = tan frac{pi}{4} = 1 ).So changing variables:[int_{0}^{1} u^n , du]Which is straightforward to integrate:[left[ frac{u^{n+1}}{n+1} right]_0^1 = frac{1^{n+1} - 0^{n+1}}{n+1} = frac{1}{n+1}]Therefore, ( I_{n+2} + I_n = frac{1}{n+1} ).So that's the answer to part (1). Now moving on to part (2): Evaluate ( I_1 ), ( I_2 ), and ( I_3 ).First, let's recall that ( I_n = int_0^{pi/4} tan^n x , dx ). Let's handle each integral one by one.Starting with ( I_1 = int_0^{pi/4} tan x , dx ).The integral of ( tan x ) is a standard one. Remember that ( int tan x , dx = -ln |cos x| + C ). Let's verify that:( frac{d}{dx} (-ln cos x) = - ( frac{-sin x}{cos x} ) = tan x ). Yup, that's correct.So:[I_1 = left[ -ln |cos x| right]_0^{pi/4} = -ln cos left( frac{pi}{4} right) + ln cos 0]Since ( cos frac{pi}{4} = frac{sqrt{2}}{2} ) and ( cos 0 = 1 ):[I_1 = -ln left( frac{sqrt{2}}{2} right) + ln 1 = -ln left( frac{sqrt{2}}{2} right)]Simplify the logarithm:( frac{sqrt{2}}{2} = 2^{-1/2} ), so:[-ln left( 2^{-1/2} right) = frac{1}{2} ln 2]Because ( ln a^b = b ln a ), so the negative exponent comes down as a negative coefficient, and the negative sign in front makes it positive. So ( I_1 = frac{1}{2} ln 2 ).Okay, that's ( I_1 ). Now ( I_2 ). Let's compute ( I_2 = int_0^{pi/4} tan^2 x , dx ).We know that ( tan^2 x = sec^2 x - 1 ), from the identity ( tan^2 x + 1 = sec^2 x ). So substituting:[I_2 = int_0^{pi/4} ( sec^2 x - 1 ) , dx = int_0^{pi/4} sec^2 x , dx - int_0^{pi/4} 1 , dx]Integrate term by term:First integral: ( int sec^2 x , dx = tan x ). Evaluated from 0 to ( pi/4 ):[tan frac{pi}{4} - tan 0 = 1 - 0 = 1]Second integral: ( int_0^{pi/4} 1 , dx = left. x right|_0^{pi/4} = frac{pi}{4} - 0 = frac{pi}{4} )Therefore, ( I_2 = 1 - frac{pi}{4} ).Alright, that's ( I_2 = 1 - frac{pi}{4} ).Now ( I_3 ). Let's compute ( I_3 = int_0^{pi/4} tan^3 x , dx ).Hmm, integrating ( tan^3 x ). Maybe we can use a substitution or perhaps split the integral. Remember that for odd powers of tangent, we can use substitution by splitting off a ( tan^2 x ):( tan^3 x = tan x cdot tan^2 x = tan x ( sec^2 x - 1 ) )So:[I_3 = int_0^{pi/4} tan x ( sec^2 x - 1 ) , dx = int_0^{pi/4} tan x sec^2 x , dx - int_0^{pi/4} tan x , dx]Let's handle the first integral: ( int tan x sec^2 x , dx ). Let me set substitution ( u = tan x ), then ( du = sec^2 x , dx ). Then the integral becomes:[int u , du = frac{u^2}{2} + C = frac{tan^2 x}{2} + C]Therefore, evaluating from 0 to ( pi/4 ):[frac{tan^2 frac{pi}{4}}{2} - frac{tan^2 0}{2} = frac{1^2}{2} - frac{0}{2} = frac{1}{2}]So the first integral is ( frac{1}{2} ).The second integral is ( int_0^{pi/4} tan x , dx ), which we already computed as ( I_1 = frac{1}{2} ln 2 ).Therefore, ( I_3 = frac{1}{2} - frac{1}{2} ln 2 ).Alternatively, factoring out ( frac{1}{2} ):( I_3 = frac{1}{2} (1 - ln 2) ).But maybe we can also use the recursive formula from part (1). Wait, part (1) gives ( I_{n+2} + I_n = frac{1}{n+1} ). Let's check if that works here.For example, if we set ( n = 1 ), then ( I_{3} + I_1 = frac{1}{1 + 1} = frac{1}{2} ). Then ( I_3 = frac{1}{2} - I_1 ).Since ( I_1 = frac{1}{2} ln 2 ), then:( I_3 = frac{1}{2} - frac{1}{2} ln 2 = frac{1}{2}(1 - ln 2) ).Which matches what we calculated earlier. So that's good, consistency check.Similarly, maybe we can use the recursive formula for ( I_2 ) as well. Let's check:If ( n = 0 ), then ( I_2 + I_0 = frac{1}{0 + 1} = 1 ). So ( I_2 = 1 - I_0 ).But ( I_0 = int_0^{pi/4} tan^0 x , dx = int_0^{pi/4} 1 , dx = frac{pi}{4} ). Therefore, ( I_2 = 1 - frac{pi}{4} ), which matches our previous result. Good.So the recursive formula works here as well.Therefore, in general, we can use ( I_{n+2} = frac{1}{n+1} - I_n ).But since we had to compute ( I_1, I_2, I_3 ), we can compute ( I_1 ) directly, then ( I_2 ) via the identity, and ( I_3 ) via the identity using ( I_1 ).But in any case, our computations for ( I_1 ), ( I_2 ), ( I_3 ) seem consistent.Let me just recap:- ( I_1 = frac{1}{2} ln 2 )- ( I_2 = 1 - frac{pi}{4} )- ( I_3 = frac{1}{2}(1 - ln 2) )I think those are the values. Let me check once more.For ( I_1 ):Yes, integral of tan x is -ln |cos x|, evaluated from 0 to π/4. That gives -ln(√2/2) + ln 1. Since ln 1 is 0, and -ln(√2/2) = -ln(2^{-1/2}) = (1/2) ln 2. Correct.For ( I_2 ):Used identity tan²x = sec²x -1, integrated to get 1 - π/4. Correct.For ( I_3 ):Split into two integrals, first integral via substitution gave 1/2, second integral was I1 = (1/2) ln 2, so subtract to get 1/2 - (1/2) ln 2. Correct.Alternatively, using the recursive formula with n=1: I3 + I1 = 1/(1+1)=1/2, so I3 =1/2 - I1 =1/2 - (1/2) ln2. Same result.Good. So all values check out.Therefore, the answers are:(1) ( I_{n+2} + I_n = frac{1}{n+1} )(2) ( I_1 = frac{1}{2} ln 2 ), ( I_2 = 1 - frac{pi}{4} ), ( I_3 = frac{1}{2}(1 - ln 2) )Final Answer(1) (boxed{dfrac{1}{n+1}})(2) (I_1 = boxed{dfrac{ln 2}{2}}), (I_2 = boxed{1 - dfrac{pi}{4}}), and (I_3 = boxed{dfrac{1 - ln 2}{2}})</think>To solve the given problem, we start by analyzing the integrals ( I_n = int_0^{frac{pi}{4}} tan^n x , dx ).(1) We need to calculate ( I_{n+2} + I_n ):[I_{n+2} + I_n = int_0^{frac{pi}{4}} left( tan^{n+2} x + tan^n x right) dx = int_0^{frac{pi}{4}} tan^n x left( tan^2 x + 1 right) dx]Using the trigonometric identity ( tan^2 x + 1 = sec^2 x ), we get:[int_0^{frac{pi}{4}} tan^n x cdot sec^2 x , dx]Substituting ( u = tan x ) (hence ( du = sec^2 x , dx )) with limits from 0 to 1, we obtain:[int_{0}^{1} u^n , du = left[ frac{u^{n+1}}{n+1} right]_0^1 = frac{1}{n+1}]Thus, the result is:[I_{n+2} + I_n = boxed{dfrac{1}{n+1}}](2) We evaluate ( I_1 ), ( I_2 ), and ( I_3 ):- For ( I_1 ):[I_1 = int_0^{frac{pi}{4}} tan x , dx = left[ -ln |cos x| right]_0^{frac{pi}{4}} = -ln left( frac{sqrt{2}}{2} right) = frac{1}{2} ln 2]Result: ( I_1 = boxed{dfrac{ln 2}{2}} )- For ( I_2 ):Using ( tan^2 x = sec^2 x - 1 ):[I_2 = int_0^{frac{pi}{4}} (sec^2 x - 1) , dx = left[ tan x - x right]_0^{frac{pi}{4}} = 1 - frac{pi}{4}]Result: ( I_2 = boxed{1 - dfrac{pi}{4}} )- For ( I_3 ):Using the recursive formula from part (1) with ( n = 1 ):[I_3 + I_1 = frac{1}{2} implies I_3 = frac{1}{2} - I_1 = frac{1}{2} - frac{1}{2} ln 2 = frac{1 - ln 2}{2}]Result: ( I_3 = boxed{dfrac{1 - ln 2}{2}} ) Final Answers(1) (boxed{dfrac{1}{n+1}})(2) (I_1 = boxed{dfrac{ln 2}{2}}), (I_2 = boxed{1 - dfrac{pi}{4}}), and (I_3 = boxed{dfrac{1 - ln 2}{2}}),<think>Okay, so I have this integral problem from an old entrance exam. Let me try to work through it step by step. The problem is about integrals involving powers of tangent from 0 to π/4. Specifically, they define Iₙ = ∫₀^{π/4} tanⁿx dx for non-negative integers n. Then there are two parts: (1) Calculate I_{n+2} + I_n, and (2) Evaluate I₁, I₂, and I₃. Starting with part (1). They want me to find an expression for I_{n+2} + I_n. Hmm. So I need to relate the integral of tan^{n+2}x dx and tan^n x dx. Maybe there's a reduction formula here. Let me recall that tan^{k}x can be expressed as tan^{k-2}x * tan²x, and since tan²x = sec²x - 1, that might help. Let me try that.So, I_{n+2} = ∫₀^{π/4} tan^{n+2}x dx = ∫₀^{π/4} tan^n x * tan²x dx. Since tan²x = sec²x - 1, substitute that in:I_{n+2} = ∫₀^{π/4} tan^n x (sec²x - 1) dx = ∫₀^{π/4} tan^n x sec²x dx - ∫₀^{π/4} tan^n x dx.So, I_{n+2} = ∫₀^{π/4} tan^n x sec²x dx - I_n.Now, the first integral here is ∫ tan^n x sec²x dx. Let me see if I can integrate this. Let’s set u = tan x. Then du/dx = sec²x, so du = sec²x dx. Therefore, ∫ tan^n x sec²x dx = ∫ u^n du. That’s straightforward. The integral of u^n is u^{n+1}/(n+1) + C. So, evaluating from 0 to π/4:When x = π/4, u = tan(π/4) = 1. When x = 0, u = tan 0 = 0. So the integral becomes [u^{n+1}/(n+1)] from 0 to 1, which is 1^{n+1}/(n+1) - 0 = 1/(n+1).Therefore, I_{n+2} = 1/(n+1) - I_n.Therefore, rearranged, I_{n+2} + I_n = 1/(n+1). So the answer to part (1) is 1/(n+1). That seems neat. Let me check if this makes sense. For example, if n = 0, then I_2 + I_0 = 1/(0+1) = 1. Let me compute I_0 and I_2 manually to verify.I_0 is ∫₀^{π/4} tan^0 x dx = ∫₀^{π/4} 1 dx = π/4 ≈ 0.7854.I_2 is ∫₀^{π/4} tan²x dx. As tan²x = sec²x - 1, integrating gives tan x - x from 0 to π/4. So, (1 - π/4) - (0 - 0) = 1 - π/4 ≈ 1 - 0.7854 ≈ 0.2146. Then I_2 + I_0 ≈ 0.2146 + 0.7854 ≈ 1, which matches 1/(0+1) = 1. So that works. Therefore, part (1) seems correct.Now moving on to part (2): Evaluate I₁, I₂, and I₃.Let me start with I₁ = ∫₀^{π/4} tan x dx. The integral of tan x is -ln |cos x| + C. So evaluate from 0 to π/4:At π/4: -ln(cos(π/4)) = -ln(√2/2) = -ln(2^{-1/2}) = (1/2) ln 2.At 0: -ln(cos 0) = -ln(1) = 0.So I₁ = (1/2) ln 2 - 0 = (1/2) ln 2. That's straightforward.Next, I₂. As computed earlier, I₂ = 1 - π/4. Wait, when we checked part (1) earlier, we saw that I_2 + I_0 = 1, and since I_0 = π/4, then I_2 = 1 - π/4. So that's correct.Alternatively, integrating tan²x dx: ∫ tan²x dx = ∫ (sec²x - 1) dx = tan x - x + C. Evaluated from 0 to π/4: [1 - π/4] - [0 - 0] = 1 - π/4. So I₂ = 1 - π/4.Now, I₃. Let's use the recursion formula from part (1). For I_{n+2} + I_n = 1/(n+1). Let's set n = 1. Then I_3 + I₁ = 1/(1+1) = 1/2. Therefore, I₃ = 1/2 - I₁. We already found that I₁ = (1/2) ln 2. So I₃ = 1/2 - (1/2) ln 2 = (1/2)(1 - ln 2).Alternatively, we can compute I₃ directly. Let's do that to check. I₃ = ∫₀^{π/4} tan³x dx.Hmm. Let me recall that integrating tan³x. Let's write tan³x as tan x * tan²x = tan x (sec²x - 1) = tan x sec²x - tan x. So,∫ tan³x dx = ∫ tan x sec²x dx - ∫ tan x dx.For the first integral, let u = tan x, du = sec²x dx. So ∫ tan x sec²x dx = ∫ u du = (1/2) u² + C = (1/2) tan²x + C.The second integral is ∫ tan x dx = -ln |cos x| + C.So combining, ∫ tan³x dx = (1/2) tan²x + ln |cos x| + C.Evaluated from 0 to π/4:At π/4: (1/2)(1)^2 + ln(√2/2) = 1/2 + ln(2^{-1/2}) = 1/2 - (1/2) ln 2.At 0: (1/2)(0)^2 + ln(1) = 0 + 0 = 0.Therefore, I₃ = [1/2 - (1/2) ln 2] - 0 = 1/2 - (1/2) ln 2, which matches the earlier result from the recursion. So that's consistent.Therefore, summarizing:I₁ = (1/2) ln 2I₂ = 1 - π/4I₃ = 1/2 - (1/2) ln 2Alternatively, maybe we can write I₃ as (1 - ln 2)/2. Either way is correct.So I think that's all. Let me check again if the recursion was applied correctly for I₃.Using the recursion formula: I_{n+2} = 1/(n+1) - I_n. So for n =1, I₃ = 1/(1+1) - I₁ = 1/2 - (1/2) ln 2. Yep, that's exactly what we did. So that's correct.Therefore, the answers are:(1) I_{n+2} + I_n = 1/(n+1)(2) I₁ = (1/2) ln 2, I₂ = 1 - π/4, I₃ = (1 - ln 2)/2But maybe they want the answers in a specific form. Let me see:For I₁: (1/2) ln 2 is acceptable. Alternatively, sometimes written as (ln 2)/2.For I₂: 1 - π/4 is straightforward.For I₃: (1 - ln 2)/2 or 1/2 - (ln 2)/2. Both are equivalent. Let me check if the direct computation gives the same. Earlier, I got 1/2 - (1/2) ln 2, which is the same as (1 - ln 2)/2. So either form is okay. Maybe prefer (1 - ln 2)/2 for symmetry with I₁.So I think that's the answer.Final Answer(1) boxed{dfrac{1}{n+1}}(2) ( I_1 = boxed{dfrac{ln 2}{2}} ), ( I_2 = boxed{1 - dfrac{pi}{4}} ), ( I_3 = boxed{dfrac{1 - ln 2}{2}} )</think>To solve the given problem, we start by analyzing the integral ( I_n = int_0^{frac{pi}{4}} tan^n x , dx ). (1) Calculate ( I_{n+2} + I_n )We use the identity ( tan^2 x = sec^2 x - 1 ):[I_{n+2} = int_0^{frac{pi}{4}} tan^{n+2} x , dx = int_0^{frac{pi}{4}} tan^n x cdot tan^2 x , dx]Substituting ( tan^2 x = sec^2 x - 1 ):[I_{n+2} = int_0^{frac{pi}{4}} tan^n x (sec^2 x - 1) , dx = int_0^{frac{pi}{4}} tan^n x sec^2 x , dx - int_0^{frac{pi}{4}} tan^n x , dx]The first integral can be solved by substitution ( u = tan x ), ( du = sec^2 x , dx ):[int_0^{frac{pi}{4}} tan^n x sec^2 x , dx = int_0^1 u^n , du = left[ frac{u^{n+1}}{n+1} right]_0^1 = frac{1}{n+1}]Thus,[I_{n+2} = frac{1}{n+1} - I_n]Therefore,[I_{n+2} + I_n = frac{1}{n+1}] (2) Evaluate ( I_1 ), ( I_2 ), and ( I_3 )For ( I_1 ):[I_1 = int_0^{frac{pi}{4}} tan x , dx = left[ -ln |cos x| right]_0^{frac{pi}{4}} = -ln left( cos frac{pi}{4} right) + ln (cos 0) = -ln left( frac{sqrt{2}}{2} right) = frac{ln 2}{2}]For ( I_2 ):Using the identity ( tan^2 x = sec^2 x - 1 ):[I_2 = int_0^{frac{pi}{4}} tan^2 x , dx = int_0^{frac{pi}{4}} (sec^2 x - 1) , dx = left[ tan x - x right]_0^{frac{pi}{4}} = 1 - frac{pi}{4}]For ( I_3 ):Using the recursion formula from part (1) with ( n = 1 ):[I_3 + I_1 = frac{1}{2} implies I_3 = frac{1}{2} - I_1 = frac{1}{2} - frac{ln 2}{2} = frac{1 - ln 2}{2}] Final Answers(1) boxed{dfrac{1}{n+1}}(2) ( I_1 = boxed{dfrac{ln 2}{2}} ), ( I_2 = boxed{1 - dfrac{pi}{4}} ), ( I_3 = boxed{dfrac{1 - ln 2}{2}} )]
🔑:1. Expressing ( I_n ) in terms of ( I_{n+2} ) and ( I_n ): We start with the integral ( I_n = int_0^{frac{pi}{4}} tan^n x , dx ). To find a relationship between ( I_{n+2} ) and ( I_n ), we use the identity ( tan^2 x = sec^2 x - 1 ): [ I_n = int_0^{frac{pi}{4}} tan^n x , dx = int_0^{frac{pi}{4}} tan^{n-2} x cdot tan^2 x , dx ] Substituting ( tan^2 x = sec^2 x - 1 ): [ I_n = int_0^{frac{pi}{4}} tan^{n-2} x (sec^2 x - 1) , dx ] This can be split into two integrals: [ I_n = int_0^{frac{pi}{4}} tan^{n-2} x sec^2 x , dx - int_0^{frac{pi}{4}} tan^{n-2} x , dx ] The first integral can be simplified using the substitution ( u = tan x ), ( du = sec^2 x , dx ): [ int_0^{frac{pi}{4}} tan^{n-2} x sec^2 x , dx = int_0^1 u^{n-2} , du = left[ frac{u^{n-1}}{n-1} right]_0^1 = frac{1}{n-1} ] Therefore, we have: [ I_n = frac{1}{n-1} - I_{n-2} ]2. Finding ( I_{n+2} + I_n ): Using the relationship derived above, we set ( n = n+2 ): [ I_{n+2} = frac{1}{(n+2)-1} - I_n = frac{1}{n+1} - I_n ] Adding ( I_n ) to both sides: [ I_{n+2} + I_n = frac{1}{n+1} ]3. Evaluating ( I_1, I_2, ) and ( I_3 ): - For ( I_1 ): [ I_1 = int_0^{frac{pi}{4}} tan x , dx ] Using the substitution ( u = tan x ), ( du = sec^2 x , dx ): [ I_1 = int_0^1 u , du = left[ frac{u^2}{2} right]_0^1 = frac{1}{2} ] - For ( I_2 ): [ I_2 = int_0^{frac{pi}{4}} tan^2 x , dx = int_0^{frac{pi}{4}} (sec^2 x - 1) , dx ] [ I_2 = left[ tan x right]_0^{frac{pi}{4}} - int_0^{frac{pi}{4}} 1 , dx = 1 - frac{pi}{4} ] - For ( I_3 ): Using the relationship ( I_{n+2} + I_n = frac{1}{n+1} ): [ I_3 + I_1 = frac{1}{2} ] Since ( I_1 = frac{1}{2} ): [ I_3 + frac{1}{2} = frac{1}{2} implies I_3 = 0 ]The final answer is ( boxed{ I_{n+2} + I_n = frac{1}{n+1} } ), ( I_1 = frac{1}{2} ), ( I_2 = 1 - frac{pi}{4} ), ( I_3 = 0 )