🧠:[<think>Okay, so I have two parts to solve here. Let me start with part (1). The problem says that we have the quadratic function y = x² - (m + 2)x + 3. We need to find the range of m such that the maximum value M and minimum value N on the interval 1 ≤ x ≤ 2 satisfy M - N ≤ 2.First, since it's a quadratic function, its graph is a parabola. The coefficient of x² is 1, which is positive, so the parabola opens upwards. That means the function has a minimum at its vertex and no maximum, except at the endpoints of the interval. Wait, but the problem mentions both a maximum and a minimum on the interval. Since the parabola opens upwards, the minimum will be at the vertex if the vertex is within the interval [1, 2], otherwise, the minimum will be at one of the endpoints. The maximum in such a case would be at the other endpoint. So, maybe depending on where the vertex is, the maximum and minimum can be at different points.First, let me find the vertex of the parabola. The x-coordinate of the vertex is given by -b/(2a). Here, a = 1, and b = -(m + 2). So, the x-coordinate is (m + 2)/(2*1) = (m + 2)/2. So, the vertex is at x = (m + 2)/2. Since the parabola opens upwards, this is the point where the function attains its minimum.Now, the problem is about the interval [1, 2]. So, if the vertex x = (m + 2)/2 is inside this interval, then the minimum N is at the vertex, and the maximum M is either at x = 1 or x = 2. If the vertex is to the left of the interval (i.e., (m + 2)/2 < 1), then the minimum is at x = 1 and the maximum is at x = 2. If the vertex is to the right of the interval (i.e., (m + 2)/2 > 2), then the minimum is at x = 2 and the maximum is at x = 1.So, first, let's consider three cases based on the position of the vertex:Case 1: Vertex in [1, 2], i.e., 1 ≤ (m + 2)/2 ≤ 2 ⇒ 2 ≤ m + 2 ≤ 4 ⇒ 0 ≤ m ≤ 2.Case 2: Vertex < 1, i.e., (m + 2)/2 < 1 ⇒ m + 2 < 2 ⇒ m < 0.Case 3: Vertex > 2, i.e., (m + 2)/2 > 2 ⇒ m + 2 > 4 ⇒ m > 2.For each case, we need to compute M and N, then set M - N ≤ 2.Let me start with Case 1: 0 ≤ m ≤ 2.In this case, the minimum N is at the vertex x = (m + 2)/2. Let's compute N:N = y at vertex. The general formula for the vertex y-coordinate is c - b²/(4a). Here, a = 1, b = -(m + 2), c = 3. So,N = 3 - [(-(m + 2))²]/(4*1) = 3 - (m + 2)^2 / 4.Alternatively, since we can compute y at x = (m + 2)/2:y = [(m + 2)/2]^2 - (m + 2)*[(m + 2)/2] + 3.Let me compute that:First term: [(m + 2)/2]^2 = (m + 2)^2 / 4.Second term: -(m + 2)*[(m + 2)/2] = -(m + 2)^2 / 2.Third term: +3.So total N = (m + 2)^2 / 4 - (m + 2)^2 / 2 + 3 = - (m + 2)^2 / 4 + 3. Yep, same as before.So, N = 3 - (m + 2)^2 / 4.Now, the maximum M in this interval will be at either x = 1 or x = 2. Let's compute y at both endpoints.At x = 1:y = 1 - (m + 2)*1 + 3 = 1 - m - 2 + 3 = (1 - 2 + 3) - m = 2 - m.At x = 2:y = 4 - (m + 2)*2 + 3 = 4 - 2m - 4 + 3 = (4 - 4 + 3) - 2m = 3 - 2m.So, to find M, we need to compare y at x=1 and x=2, which are 2 - m and 3 - 2m. Which one is larger?Let's set 2 - m > 3 - 2m ⇒ 2 - m - 3 + 2m > 0 ⇒ (2 - 3) + ( -m + 2m ) ⇒ -1 + m > 0 ⇒ m > 1.So, if m > 1, then y at x=1 is larger; if m < 1, y at x=2 is larger; if m = 1, they are equal.But since in Case 1, m is between 0 and 2, so:- When 0 ≤ m < 1: M = 3 - 2m.- When 1 ≤ m ≤ 2: M = 2 - m.Therefore, for Case 1, we have:If 0 ≤ m < 1: M = 3 - 2m, N = 3 - (m + 2)^2 /4. Then M - N = [3 - 2m] - [3 - (m + 2)^2 /4] = -2m + (m + 2)^2 /4.Set this ≤ 2.Similarly, if 1 ≤ m ≤ 2: M = 2 - m, N = 3 - (m + 2)^2 /4. Then M - N = [2 - m] - [3 - (m + 2)^2 /4] = -1 - m + (m + 2)^2 /4.Set this ≤ 2.Let me compute both scenarios.First, for 0 ≤ m < 1:M - N = (m + 2)^2 /4 - 2m = [ (m² + 4m + 4) /4 ] - 2m = (m² + 4m + 4 - 8m)/4 = (m² - 4m + 4)/4 = (m - 2)^2 /4.So, (m - 2)^2 /4 ≤ 2.Multiply both sides by 4: (m - 2)^2 ≤ 8.Take square roots: |m - 2| ≤ 2√2 ⇒ -2√2 ≤ m - 2 ≤ 2√2 ⇒ 2 - 2√2 ≤ m ≤ 2 + 2√2.But in this case, m is in [0, 1). So, we need the intersection of 2 - 2√2 ≤ m ≤ 2 + 2√2 and 0 ≤ m < 1.Compute 2 - 2√2 ≈ 2 - 2.828 ≈ -0.828. So, the lower bound is less than 0, so the intersection is 0 ≤ m < 1. Therefore, for m in [0,1), the inequality (m - 2)^2 /4 ≤ 2 is automatically satisfied since (m - 2)^2 is between (1 - 2)^2 =1 and (0 - 2)^2 =4, so (m - 2)^2 /4 is between 0.25 and 1, which is always ≤ 2. So actually, in 0 ≤ m <1, M - N is (m -2)^2 /4 ≤ 1, which is already ≤2. So, for all 0 ≤ m <1, M - N ≤1 ≤2, which satisfies the condition. Therefore, in this subcase, all m in [0,1) are acceptable.Now, for 1 ≤ m ≤2:M - N = (m + 2)^2 /4 - m -1.Let me compute that:First, expand (m + 2)^2: m² +4m +4.Divide by 4: (m² +4m +4)/4 - m -1 = m²/4 + m +1 -m -1 = m²/4.So, M - N = m²/4. Wait, that can't be. Let's check the steps again.Wait, original expression:M - N = [2 - m] - [3 - (m + 2)^2 /4] = 2 - m -3 + (m + 2)^2 /4 = -1 - m + (m + 2)^2 /4.Compute (m + 2)^2 /4 - m -1:Let me expand (m + 2)^2:m² +4m +4. So, divided by 4: m²/4 + m +1.Then subtract m and 1: m²/4 + m +1 -m -1 = m²/4.So, indeed, M - N = m²/4.Therefore, we set m² /4 ≤2 ⇒ m² ≤8 ⇒ |m| ≤2√2. Since in this case m is between 1 and 2, which is within -2√2 ≈ -2.828 and 2√2≈2.828. So, the condition is satisfied for all m in [1,2]. But m is only up to 2, which is less than 2√2≈2.828, so m²/4 at m=2 is 4/4=1 ≤2. Therefore, for all 1 ≤m ≤2, M - N = m²/4 ≤1 ≤2. So, all m in [1,2] satisfy the condition.Therefore, in Case 1 (0 ≤m ≤2), all m from 0 to2 are acceptable because in both subcases, M - N is ≤1, which is ≤2. So, Case 1 contributes m ∈ [0,2].Now, Case 2: Vertex <1, i.e., m <0.In this case, the parabola is increasing on the interval [1,2] because the vertex is to the left, so the minimum is at x=1 and maximum at x=2.Therefore, N = y(1) =2 -m (as computed earlier) and M = y(2)=3 -2m.Then, M - N = (3 -2m) - (2 -m) =3 -2m -2 +m =1 -m.We need 1 -m ≤2 ⇒ -m ≤1 ⇒m ≥-1.But in Case 2, m <0. So combining m <0 and m ≥-1 ⇒ -1 ≤m <0.Therefore, in Case 2, m ∈ [-1,0).Case 3: Vertex >2, i.e., m >2.In this case, the parabola is decreasing on the interval [1,2] because the vertex is to the right. So the function is decreasing, hence maximum at x=1 and minimum at x=2.Therefore, M = y(1)=2 -m, N = y(2)=3 -2m.Thus, M - N = (2 -m) - (3 -2m) =2 -m -3 +2m= -1 +m.We need -1 +m ≤2 ⇒m ≤3.But in Case 3, m >2. So, combining m >2 and m ≤3 ⇒2 <m ≤3.So, in Case 3, m ∈ (2,3].Now, combining all cases:Case1: [0,2], Case2: [-1,0), Case3: (2,3]. So total range is [-1,3].But wait, let me check if these are correct.But wait, in Case 2, we had m ∈ [-1,0). In Case1, m ∈ [0,2]. In Case3, m ∈ (2,3]. So the union is [-1,3].But we need to check if there are overlaps or gaps. At m=0: Case2 ends and Case1 starts. At m=2: Case1 ends and Case3 starts. So, the union is continuous from -1 to 3. Therefore, the range of m is -1 ≤m ≤3.But let me verify with some test values.Take m = -1, which is the lower bound. In Case2, m=-1. Then, vertex is (m +2)/2 = (1)/2=0.5 <1. So, the function is increasing on [1,2]. So, N at x=1: y=2 - (-1)=3, M at x=2: y=3 -2*(-1)=5. Then, M -N=5 -3=2, which is equal to 2. So, acceptable.Take m=3, upper bound. Case3, vertex=(3 +2)/2=2.5>2. The function is decreasing on [1,2]. So, M at x=1: y=2 -3= -1, N at x=2: y=3 -2*3= -3. M - N= -1 - (-3)=2, which is equal to 2. So, acceptable.Take m=0, which is in Case1. Then, vertex=(0 +2)/2=1. So, vertex at x=1. So, the minimum is at x=1, which is y=2 -0=2. The maximum is at x=2: y=3 -0=3. So, M -N=1≤2. So, okay.Take m=2: vertex=(2 +2)/2=2. So, minimum at x=2: y=3 - (2 +2)^2 /4=3 - 16/4=3 -4= -1. The maximum is at x=1: y=2 -2=0. So, M - N=0 - (-1)=1≤2. Okay.Take m=1.5: in Case1. M -N= (1.5)^2 /4=2.25/4=0.5625≤2.Take m=-0.5: in Case2. M -N=1 -(-0.5)=1.5≤2.Take m=2.5: in Case3. M -N= -1 +2.5=1.5≤2.So all test points work. Therefore, the range of m is from -1 to3 inclusive. So, the answer for part(1) is m ∈ [-1,3].Now, moving on to part (2): Given the equation x² - (m +2)x +3 = -(2m +1)x +2 has a solution in 0 ≤x ≤2, find the range of m.First, let's write the equation:x² - (m +2)x +3 = -(2m +1)x +2.Bring all terms to left side:x² - (m +2)x +3 + (2m +1)x -2 =0.Simplify:x² + [ - (m +2) + (2m +1) ]x + (3 -2)=0.Compute coefficients:x² + [ (-m -2 +2m +1 ) ]x +1 =0.Which is x² + (m -1)x +1=0.So, the equation reduces to x² + (m -1)x +1=0. We need this quadratic equation to have at least one solution in the interval [0,2].So, we need to find all real numbers m such that the equation x² + (m -1)x +1=0 has at least one real root in [0,2].First, let's note that for the quadratic to have real roots, the discriminant must be non-negative.Discriminant D = (m -1)^2 -4*1*1 = (m -1)^2 -4.So, D ≥0 ⇒ (m -1)^2 -4 ≥0 ⇒ (m -1)^2 ≥4 ⇒ |m -1| ≥2 ⇒ m -1 ≤-2 or m -1 ≥2 ⇒ m ≤-1 or m ≥3.So, the equation has real roots only when m ≤-1 or m ≥3. So, for m between -1 and3, there are no real roots, so no solution. But the problem says "has a solution in 0 ≤x ≤2". So, if m is in (-1,3), the equation has no real roots, so certainly no roots in [0,2]. Therefore, possible m must be in (-∞, -1] ∪ [3, ∞). But even for m in these intervals, we need to check if the roots lie within [0,2].So, the strategy is: for m ≤-1 or m ≥3, find the roots of x² + (m -1)x +1=0 and check if at least one root is in [0,2].Alternatively, since it's a quadratic equation, we can use the following approach:Let f(x) = x² + (m -1)x +1. We need f(x)=0 for some x in [0,2].The function f(x) is a parabola opening upwards. The roots of f(x)=0 are real when m ≤-1 or m ≥3. Let's consider the two cases:Case1: m ≤-1.Case2: m ≥3.In each case, we need to check if there is a root in [0,2].Alternatively, another method is to consider that f(x)=0 has a solution in [0,2] if either:1. f(0)*f(2) ≤0 (by Intermediate Value Theorem, if f(0) and f(2) have opposite signs, there is a root between 0 and2). Note that this applies even if the function doesn't have real roots outside [0,2], but in our case, since the quadratic only has real roots when m ≤-1 or m ≥3, we have to be careful.But perhaps combining both approaches.First, compute f(0)=0 +0 +1=1.f(2)=4 +2(m -1) +1=4 +2m -2 +1=3 +2m.So, f(0)=1, f(2)=3 +2m.If f(0)*f(2) ≤0, then there is a root between 0 and2. But f(0)=1>0, so f(2) must be ≤0. Therefore, 3 +2m ≤0 ⇒m ≤-3/2.But note that when m ≤-3/2, which is ≤-1.5, which is in m ≤-1. So, for m ≤-3/2, f(2) ≤0, so by IVT, there is a root in [0,2].But also, when m ≤-1 or m ≥3, the quadratic has real roots, so even if f(0) and f(2) are both positive, maybe one of the roots is in [0,2]. So, we need to check that.Alternatively, since the parabola opens upwards, and f(0)=1>0, if there is a root in [0,2], then either:- There are two roots, one on each side of [0,2], but since f(0)>0, that's not possible.Wait, actually, since f(0)=1>0, the parabola is upwards. If there is a root in [0,2], then either:1. The quadratic has two real roots, and one is in [0,2], while the other is outside. Since f(0)=1>0 and the parabola opens upwards, to have a root in [0,2], the function must cross the x-axis once between 0 and2. Therefore, f(2) must be ≤0 (since at x=2, the function is below or on the axis), which would mean that the root is between 0 and2. Wait, but since f(0)=1>0 and f(2) ≤0, then there must be a root between 0 and2 by IVT. So, in that case, m ≤-3/2.Alternatively, if both roots are in [0,2], but since f(0)=1>0, this would require that the vertex is in [0,2] and the minimum value ≤0. Let's check that.Wait, but first, let's note that for m ≤-1 or m ≥3, the quadratic has real roots.First, let's handle the case m ≤-1:If m ≤-1, then the roots are:x = [-(m -1) ± sqrt((m -1)^2 -4)] /2.Which is x = [1 - m ± sqrt((m -1)^2 -4)] /2.But m ≤-1, so m -1 ≤-2, so (m -1)^2 ≥4. Therefore, sqrt((m -1)^2 -4) is real.Let me denote sqrt((m -1)^2 -4) as S.Therefore, roots are [1 - m + S]/2 and [1 - m - S]/2.But since m ≤-1, 1 - m ≥2. So, both roots are positive because numerator is positive (since 1 - m ≥2 and S ≥0). So, the roots are positive. But we need to check if they are ≤2.Wait, let's take an example. Suppose m=-2.Then, the equation is x² + (-2 -1)x +1= x² -3x +1=0.The roots are [3 ± sqrt(9 -4)]/2 = [3 ± sqrt(5)]/2 ≈ [3 ±2.24]/2 ≈ 2.62 or 0.38. Both roots are positive. The smaller root ≈0.38 is in [0,2], and the larger ≈2.62 is outside. So, in this case, there is a root in [0,2]. Wait, but when m=-2, f(2)=3 +2*(-2)=3 -4=-1 ≤0, so by IVT, there is a root in [0,2].But according to the roots, the smaller root is 0.38, which is in [0,2], and the larger is 2.62, outside. So, even though m=-2 is ≤-1.5 (since -2 ≤-1.5), so in this case, there is a root in [0,2].Wait, but when m=-1. Let's see m=-1:Equation becomes x² + (-1 -1)x +1= x² -2x +1=0, which factors as (x -1)^2=0. So, double root at x=1. Since 1 is in [0,2], so m=-1 is acceptable.But m=-1. Let's check f(2)=3 +2*(-1)=3 -2=1>0. So, f(2)=1>0, but f(0)=1>0, but the root is at x=1. So, even though f(0)*f(2)=1*1=1>0, there is a root in [0,2]. So, this case is not covered by IVT. Therefore, IVT gives roots when f(0)*f(2) ≤0, but when the root is a double root inside the interval, IVT may not catch it. Therefore, we need to consider two scenarios:1. f(2) ≤0, which gives m ≤-3/2. Then, by IVT, there is a root in [0,2].2. For m >-3/2 but m ≤-1, check if the quadratic has a root in [0,2]. But since f(0)=1>0 and f(2)=3 +2m. If m >-3/2, then f(2)=3 +2m >0. So, f(0) and f(2) both positive. However, since m ≤-1, the quadratic has real roots. The question is whether one of the roots is in [0,2].Wait, but if f(0)=1>0 and f(2)=3 +2m >0 (since m >-3/2), then since the parabola opens upwards, the roots are either both greater than2 or both less than0. Wait, but when m ≤-1, the quadratic has two real roots. If m ≤-1, then the roots are:x = [1 - m ± sqrt((m -1)^2 -4)] /2.Since m ≤-1, 1 - m ≥2. Let's check if the smaller root is less than2.Let me denote the two roots as r1 and r2, with r1 < r2.Since 1 - m is positive and S = sqrt((m -1)^2 -4) is also positive, both roots are positive.So, r1 = [1 - m - S]/2, r2 = [1 - m + S]/2.We need to check if r1 ≤2 or r2 ≤2.But since r2 is the larger root, if r2 ≤2, then both roots are ≤2. If r2 >2, then the smaller root r1 may or may not be ≤2.Alternatively, since f(2)=3 +2m, which when m ≤-1, 3 +2m ≤3 -2=1. Wait, no, if m ≤-1, then 2m ≤-2, so 3 +2m ≤1. But when m >-3/2, i.e., m >-1.5, then 3 +2m >0. So, if m is between -1.5 and -1, then f(2)=3 +2m >0, but m ≤-1.So, for m between -1.5 and -1, f(2) is positive. So, with f(0)=1>0 and f(2)>0, but since the quadratic has real roots (m ≤-1), the roots must be both outside [0,2]. Wait, but if the parabola opens upwards and f(0)>0, f(2)>0, then the roots are either both less than0 or both greater than2. But since m ≤-1, and we saw earlier that roots are positive (since 1 - m ≥2), so the roots are positive. Therefore, if f(0)>0 and f(2)>0, the roots must be both greater than2 or both less than0. But since the roots are positive, they must be both greater than2. Wait, but if m=-1.25, which is in (-1.5, -1):Then, m=-1.25, compute the roots:x = [1 - (-1.25) ± sqrt((-1.25 -1)^2 -4)] /2 = [2.25 ± sqrt(5.0625 -4)] /2 = [2.25 ± sqrt(1.0625)] /2 = [2.25 ±1.0308]/2.So, roots:(2.25 +1.0308)/2 ≈3.2808/2≈1.6404,(2.25 -1.0308)/2≈1.2192/2≈0.6096.Wait, but this contradicts the previous reasoning. Wait, m=-1.25, which is in m ≤-1. So, the roots are approximately 1.64 and0.61. Both are in [0,2]. But f(0)=1>0 and f(2)=3 +2*(-1.25)=3 -2.5=0.5>0. So, even though f(0) and f(2) are positive, the roots are inside [0,2]. How is that possible?Wait, maybe the previous reasoning is flawed. Wait, the parabola opens upwards, f(0)=1>0, f(2)=0.5>0, but there are two roots in [0,2]. But that's impossible because between the roots, the function would be negative. But if both roots are in [0,2], then between them, the function is negative, but f(0) and f(2) are positive. That can't happen. Wait, there must be a mistake.Wait, let's compute f(x)=x² + (m -1)x +1 at m=-1.25:f(x)=x² + (-1.25 -1)x +1= x² -2.25x +1.Compute f(0)=1.f(2)=4 -2.25*2 +1=4 -4.5 +1=0.5.Then, the roots are at x=(2.25 ±sqrt(5.0625 -4))/2=(2.25 ±1.0308)/2≈1.6404 and0.6096.So, between x=0.6096 and1.6404, the function is negative. But f(0)=1>0, f(2)=0.5>0. So, the function is positive at x=0, dips below zero between the roots, and comes back up to 0.5 at x=2. Therefore, there are two roots in [0,2]. But how is that possible if the parabola opens upwards?Wait, that seems contradictory. Wait, no. Wait, if the parabola opens upwards, then it can only have a minimum. If there are two roots, then the function goes from positive infinity to the minimum (below zero) and back to positive infinity. So, if the minimum is below zero and the function is positive at both ends, there must be two roots. But in our case, f(0)=1>0, f(2)=0.5>0, but the function dips below zero in between. So, two roots in [0,2]. But then, how does that reconcile with the previous assumption?Wait, so my earlier reasoning was wrong. If the parabola opens upwards and has two roots in [0,2], then the function is positive at x=0 and x=2, but negative in between. Therefore, even if f(0) and f(2) are both positive, there can be two roots in [0,2]. Therefore, for such cases, we need to check whether the quadratic has roots in [0,2] even if f(0)*f(2) >0.Therefore, the IVT method only catches one root when the function crosses from positive to negative or vice versa. But if the function has two roots within the interval, then f(0) and f(2) can both be positive, but the function dips below zero in between. Therefore, in such cases, even if f(0) and f(2) are positive, there can be two roots in [0,2]. So, my previous reasoning was incorrect.Therefore, we need another approach. The correct way is to consider the following conditions:The quadratic equation has at least one root in [0,2] if either:1. f(0)*f(2) ≤0 (IVT), or2. The quadratic has its vertex within [0,2] and the minimum value ≤0.So, combining these two conditions.First, compute the vertex of f(x)=x² + (m -1)x +1. The x-coordinate of the vertex is -b/(2a)= -(m -1)/(2*1)= (1 -m)/2.The y-coordinate is f((1 -m)/2). Compute it:f((1 -m)/2) = [(1 -m)/2]^2 + (m -1)*[(1 -m)/2] +1.Let me compute:First term: (1 -2m +m²)/4.Second term: (m -1)(1 -m)/2 = -(1 -m)^2 /2.Third term: +1.So total:(1 -2m +m²)/4 - (1 -2m +m²)/2 +1.Convert to common denominator:= [1 -2m +m² -2(1 -2m +m²) +4]/4= [1 -2m +m² -2 +4m -2m² +4]/4= [ (1 -2 +4) + (-2m +4m) + (m² -2m²) ] /4= (3 +2m -m²)/4.So, the minimum value is (3 +2m -m²)/4.For the quadratic to have a root in [0,2], either:- There is a sign change between f(0) and f(2) (i.e., f(0)*f(2) ≤0), or- The vertex is in [0,2] and the minimum value ≤0.So, let's check these conditions.First condition: f(0)*f(2) ≤0.As before, f(0)=1, f(2)=3 +2m. So, 1*(3 +2m) ≤0 ⇒3 +2m ≤0 ⇒m ≤-3/2.Second condition: The vertex is in [0,2], i.e., 0 ≤(1 -m)/2 ≤2 ⇒0 ≤1 -m ≤4 ⇒-3 ≤m ≤1.And the minimum value ≤0: (3 +2m -m²)/4 ≤0 ⇒3 +2m -m² ≤0 ⇒ -m² +2m +3 ≤0 ⇒ multiply by -1 (reverse inequality): m² -2m -3 ≥0 ⇒ (m -3)(m +1) ≥0 ⇒m ≤-1 or m ≥3.But the vertex is in [0,2] only if -3 ≤m ≤1. So, combining with m ≤-1 or m ≥3, we get -3 ≤m ≤-1.Therefore, the second condition is -3 ≤m ≤-1 and the vertex is in [0,2] with the minimum ≤0.Therefore, combining both conditions:1. m ≤-3/2, or2. -3 ≤m ≤-1.But wait, let's clarify.The second condition gives m in [-3,-1]. But within that interval, the vertex is in [0,2], and the minimum is ≤0.So, if m is in [-3,-1], then the quadratic has its vertex in [0,2] and the minimum is ≤0, hence there are two roots in [0,2] (since the parabola opens upwards and the minimum is below zero). So, in this interval, even though f(0) and f(2) are positive, there are roots inside [0,2].Therefore, combining both conditions:- For m ≤-3/2, f(2) ≤0, so by IVT, there is a root in [0,2].- For -3 ≤m ≤-1, the quadratic has its vertex in [0,2] and minimum ≤0, so there are two roots in [0,2].But wait, m ≤-3/2 is m ≤-1.5. So, overlapping with the second condition, which is -3 ≤m ≤-1. So, m in [-3,-1.5] would satisfy both conditions? Wait, no. Let me see:If m is in [-3,-1.5], then m ≤-1.5, so f(2)=3 +2m ≤0, so IVT applies. But also, m is in [-3,-1], so the vertex is in [0,2] and minimum ≤0. So, in this interval, there are two roots in [0,2]. So, overlapping.But the important thing is that:- For m ≤-1.5, there's one root in [0,2] by IVT.- For -1.5 <m ≤-1, there are two roots in [0,2] because the vertex is in [0,2] and the minimum is ≤0.Therefore, the total range of m for which the equation has a solution in [0,2] is m ≤-1.5 or -1.5 <m ≤-1, which combines to m ≤-1. However, wait:Wait, the second condition is -3 ≤m ≤-1. But the discriminant condition requires that m ≤-1 or m ≥3. So, combining the two, for the second condition, m must be in [-3,-1]. But the discriminant requires m ≤-1. Therefore, the allowable m in the second condition is [-3,-1].But m must also satisfy the discriminant condition, which is m ≤-1 or m ≥3. So, the second condition is m in [-3,-1].Therefore, combining the two cases:1. m ≤-3/2 (which is m ≤-1.5) where there is a root by IVT.2. m in [-3,-1] where there are two roots by vertex condition.But wait, m in [-3,-1.5] is already included in case1. So, the total range is m in [-3,-1].Wait, perhaps not. Let me structure this:The quadratic has real roots when m ≤-1 or m ≥3. So, for m in [-3,-1], the quadratic has real roots (since m ≤-1). Also, for m ≥3, but we need to check if those roots are in [0,2].So, the second condition is:Vertex in [0,2] (which is -3 ≤m ≤1) and minimum ≤0 (which is m ≤-1 or m ≥3). Therefore, combining, m in [-3,-1].Therefore, for m in [-3,-1], the quadratic has two real roots in [0,2].For m ≤-3/2 (i.e., m ≤-1.5), even outside [-3,-1], but since m ≤-1.5 is within m ≤-1, so overlapping.Wait, this is getting confusing. Let's take another approach.We can parametrize the problem as follows:We need to find all real numbers m such that the equation x² + (m -1)x +1=0 has at least one solution x in [0,2].The quadratic equation has solutions x = [-(m -1) ± sqrt((m -1)^2 -4)] /2.Given that real solutions exist only when (m -1)^2 -4 ≥0 ⇒m ≤-1 or m ≥3.So, split into two cases:Case1: m ≤-1.Case2: m ≥3.For each case, determine if any root lies in [0,2].Case1: m ≤-1.The roots are:x = [1 -m ± sqrt((m -1)^2 -4)] /2.Given m ≤-1, so 1 -m ≥2.Let’s denote sqrt((m -1)^2 -4) as S, which is real and ≥0.Thus, the two roots are:x1 = [1 -m - S]/2,x2 = [1 -m + S]/2.Since 1 -m ≥2 and S ≥0, x1 and x2 are both positive.We need to check if either x1 ≤2 or x2 ≤2.But since x2 is the larger root, if x2 ≤2, then both roots are ≤2. If x2 >2, then check if x1 ≤2.Alternatively, we can solve for m such that x1 ≤2 or x2 ≤2.But this might be complicated. Let's instead express the problem as:There exists x in [0,2] such that x² + (m -1)x +1=0.Let me solve for m:x² + (m -1)x +1=0 ⇒ (m -1)x = -x² -1 ⇒ m -1 = (-x² -1)/x ⇒ m = 1 - (x² +1)/x = 1 - x - 1/x.Therefore, m = 1 - x - 1/x, where x ∈ (0,2]. (Note: x=0 is not allowed since division by zero occurs, but x=0 gives equation 0 +0 +1=1≠0, so no solution at x=0.)Thus, for each x ∈ (0,2], m =1 -x -1/x. So, the range of m is the image of the function f(x)=1 -x -1/x over x ∈ (0,2].We can analyze this function to find its range.Let’s compute f(x)=1 -x -1/x.First, find its derivative to find extrema:f’(x) = -1 +1/x².Set to zero:-1 +1/x²=0 ⇒1/x²=1 ⇒x²=1 ⇒x=1 (since x ∈(0,2]).At x=1, f(1)=1 -1 -1= -1.Check the limits:As x→0+, f(x)=1 -x -1/x → -∞.At x=2, f(2)=1 -2 -1/2=1 -2.5= -1.5.So, the function f(x) decreases from -∞ to x=1, reaching a minimum of -1 at x=1, then increases from x=1 to x=2, reaching -1.5 at x=2.Wait, compute derivative again:f’(x)= -1 +1/x².For x ∈(0,1):x <1 ⇒x² <1 ⇒1/x² >1 ⇒f’(x)= -1 +1/x² >0. So, f(x) is increasing on (0,1).Wait, but at x approaching0+, f(x) approaches -∞, then increases to x=1 where f(1)=-1. Then, for x >1, x ∈(1,2]:x >1 ⇒x² >1 ⇒1/x² <1 ⇒f’(x)= -1 +1/x² <0. So, f(x) is decreasing on (1,2).Thus, the function f(x) increases from -∞ to x=1, reaching -1, then decreases from x=1 to x=2, reaching -1.5.Therefore, the range of f(x) over x ∈(0,2] is (-∞, -1] ∪ [-1.5, ∞)? Wait, no.Wait, when x approaches0 from the right, f(x)→-∞. From x=0 to x=1, f(x) increases from -∞ to -1. Then, from x=1 to x=2, it decreases from -1 to -1.5. Therefore, the range is (-∞, -1] ∪ [-1.5, -1]? Wait, that can't be.Wait, no. From x ∈(0,1], f(x) increases from -∞ to -1.From x ∈[1,2], f(x) decreases from -1 to -1.5.Therefore, the overall range is (-∞, -1] ∪ [-1.5, -1]? No, because when x ∈[1,2], f(x) goes from -1 to -1.5, which is [-1.5, -1].Therefore, the entire range is (-∞, -1] ∪ [-1.5, -1]?Wait, no. Because for x ∈(0,1], f(x) takes all values from -∞ up to -1. For x ∈[1,2], f(x) takes all values from -1 down to -1.5. So, combining both intervals, the range is (-∞, -1] ∪ [-1.5, -1]. But [-1.5, -1] is a subset of (-∞, -1]. Therefore, the total range is (-∞, -1]. Wait, no. Wait, x ∈(0,1] gives f(x) ∈(-∞, -1]. x ∈[1,2] gives f(x) ∈[-1.5, -1]. So, combining, the total range is (-∞, -1] ∪ [-1.5, -1] = (-∞, -1]. But [-1.5, -1] is included in (-∞, -1]. So, the total range is just (-∞, -1].Wait, no. Wait, for x ∈[1,2], the function decreases from -1 to -1.5, so the range here is [-1.5, -1]. For x ∈(0,1], the function increases from -∞ to -1. So, combined, the total range is (-∞, -1] ∪ [-1.5, -1] = (-∞, -1]. But [-1.5, -1] is within (-∞, -1]. So, the entire range is (-∞, -1]. But that can't be right because when x=2, m= -1.5, which is greater than -∞. Wait, this is confusing.Wait, the function f(x)=1 -x -1/x.For x ∈(0,1):As x approaches0+, f(x)→ -∞.At x=1, f(x)= -1.So, for x ∈(0,1], f(x) ∈ (-∞, -1].For x ∈[1,2]:At x=1, f(x)= -1.At x=2, f(x)= -1.5.So, as x increases from1 to2, f(x) decreases from -1 to -1.5.Thus, for x ∈[1,2], f(x) ∈ [-1.5, -1].Therefore, the total range of f(x) over x ∈(0,2] is (-∞, -1] ∪ [-1.5, -1] = (-∞, -1]. Because [-1.5, -1] is already included in (-∞, -1]. Therefore, the total range is (-∞, -1].But when x ∈[1,2], f(x) is between -1.5 and -1. So, if you take the union, it's all values ≤-1. But the m values that come from x ∈[1,2] are between -1.5 and -1. So, m ∈ (-∞, -1] comes from x ∈(0,1], and m ∈ [-1.5, -1] comes from x ∈[1,2]. Therefore, the total range is m ≤-1. But how?Wait, no, because for each m ≤-1, there exists some x ∈(0,2] such that m=1 -x -1/x. So, the range of m is all real numbers ≤-1. But when x ∈(0,1], m can be any number ≤-1. When x ∈[1,2], m is between -1.5 and -1. Therefore, the total range is m ≤-1. Because even though x ∈[1,2] only gives m down to -1.5, the x ∈(0,1] allows m to go to -∞. Therefore, the equation x² + (m -1)x +1=0 has a solution in [0,2] if and only if m ≤-1.Wait, but when x ∈[1,2], m is between -1.5 and -1. But if we consider m=-2, which is less than -1.5, then according to our previous calculation, there is a solution in [0,2]. Wait, but according to this parameterization, m=1 -x -1/x. So, for x ∈(0,1], m=1 -x -1/x can be less than -1.5. For example, take x=0.5:m=1 -0.5 -1/0.5=1 -0.5 -2= -1.5.Take x=0.4:m=1 -0.4 -1/0.4=0.6 -2.5= -1.9.So, as x approaches0+, m approaches -∞. Therefore, the parameterization shows that m can be any value ≤-1. So, the equation has a solution in [0,2] if and only if m ≤-1.But this contradicts the earlier discriminant condition where real roots exist for m ≤-1 or m ≥3. So, when m ≥3, can there be a solution in [0,2]?Let me check m=3:Equation becomes x² + (3 -1)x +1= x² +2x +1=0 ⇒(x +1)^2=0 ⇒x=-1. But x=-1 is not in [0,2]. So, no solution.Take m=4:Equation is x² +3x +1=0. Roots are [-3 ±sqrt(9 -4)]/2=[-3 ±sqrt(5)]/2≈[-3 ±2.24]/2≈-0.38 or -2.62. Both roots negative, not in [0,2].So, for m ≥3, the quadratic equation has two real roots, both negative (since sum of roots is -(m -1) and product is1. For m ≥3, sum of roots is -(m -1) ≤-2. Since product is1, both roots are negative. Hence, no roots in [0,2].Therefore, the conclusion is that the equation has a solution in [0,2] if and only if m ≤-1.But wait, earlier example with m=-1.25 had roots in [0,2]. According to the parameterization, m ≤-1, which includes m=-1.25. Also, when m=-1, there is a root at x=1. So, the correct range is m ≤-1.But wait, initially, we thought that the IVT condition gives m ≤-1.5 and vertex condition gives m ∈[-3,-1], but according to the parameterization, it's all m ≤-1.This suggests that the correct answer is m ≤-1.But why is there a discrepancy?Because the parameterization approach shows that for any m ≤-1, there exists an x ∈(0,2] such that m=1 -x -1/x, implying that the equation has a solution in (0,2]. Therefore, the answer should be m ≤-1.But the previous analysis with IVT and vertex conditions suggested m ≤-1.5 or m ∈[-3,-1]. But clearly, the parameterization shows that m can be any value ≤-1, including between -1.5 and -1, which is covered by the vertex condition, and less than -1.5 covered by IVT.But how does that align with the discriminant condition?Ah, because when we parameterize m=1 -x -1/x, we are essentially expressing m in terms of x, and for each x ∈(0,2], this gives an m. The range of m obtained is all m ≤-1. And since the quadratic equation only has real roots when m ≤-1 or m ≥3, but when m ≥3, the roots are negative, so no solution in [0,2]. Therefore, the equation has a solution in [0,2] if and only if m ≤-1.Thus, the answer for part (2) is m ≤-1.But let's verify with m=-2 (which is ≤-1):Equation: x² + (-2 -1)x +1= x² -3x +1=0.Roots are [3 ±sqrt(9 -4)]/2= [3 ±sqrt(5)]/2≈ (3 ±2.24)/2≈2.62 and0.38. Both roots are in [0,2] for0.38≈0.38 and2.62≈2.62. Wait, 2.62 is outside [0,2], but0.38 is inside. Therefore, yes, there is a root in [0,2].For m=-1:Equation: x² -2x +1=0 ⇒(x -1)^2=0. Root x=1∈[0,2].For m=-0.5, which is greater than -1, but m must be ≤-1. So, according to the answer, m=-0.5 is invalid. But wait, m=-0.5 is in (-1,0). The equation would have no real roots, so no solution in [0,2]. Correct.For m=-3:Equation: x² + (-3 -1)x +1= x² -4x +1=0. Roots are [4 ±sqrt(16 -4)]/2= [4 ±sqrt(12)]/2= [4 ±2*sqrt(3)]/2=2 ±sqrt(3)≈2 ±1.732≈3.732 and0.267. The root0.267 is in [0,2], so valid.For m=-1.5:Equation: x² + (-1.5 -1)x +1= x² -2.5x +1=0.Roots are [2.5 ±sqrt(6.25 -4)]/2= [2.5 ±sqrt(2.25)]/2= [2.5 ±1.5]/2= (4)/2=2 or (1)/2=0.5. Both roots are in [0,2].Therefore, m=-1.5 is allowed.But according to the parameterization approach, m can be any ≤-1. So, m=-1.5 is allowed.But according to the parameterization, m=1 -x -1/x, when x=2, m=1 -2 -0.5= -1.5. So, x=2 gives m=-1.5. Similarly, when x approaches0+, m approaches-∞. Therefore, the entire range is m ≤-1. Therefore, the answer should be m ≤-1.But why did the earlier analysis with the vertex suggest m ∈[-3,-1]?Because the parameterization shows that when x ∈[1,2], m ranges from -1.5 to -1. However, when x ∈(0,1), m can be any value ≤-1.5. Therefore, combining both intervals of x, m can be any ≤-1.Therefore, the correct answer is m ≤-1.But in the previous analysis, we thought that the vertex condition requires m ∈[-3,-1], but that was incorrect. The correct analysis is via parameterization, which shows that for any m ≤-1, there exists an x ∈(0,2] that solves the equation.Hence, the answer for part (2) is m ≤-1.Final Answer(1) boxed{[-1, 3]}(2) boxed{(-infty, -1]}</think>Given the quadratic function ( y = x^2 - (m + 2)x + 3 ):1. Finding the range of ( m ) such that ( M - N leq 2 ): - The vertex of the quadratic function is at ( x = frac{m + 2}{2} ). - Case 1: Vertex in ([1, 2]) (i.e., ( 0 leq m leq 2 )): - Minimum value ( N = 3 - frac{(m + 2)^2}{4} ). - Maximum value ( M ) at endpoints ( x = 1 ) or ( x = 2 ). - For ( 0 leq m < 1 ), ( M = 3 - 2m ) and ( M - N = frac{(m - 2)^2}{4} leq 2 ). - For ( 1 leq m leq 2 ), ( M = 2 - m ) and ( M - N = frac{m^2}{4} leq 2 ). - Case 2: Vertex < 1 (i.e., ( m < 0 )): - Minimum value ( N = 2 - m ) and maximum value ( M = 3 - 2m ). - ( M - N = 1 - m leq 2 ) gives ( -1 leq m < 0 ). - Case 3: Vertex > 2 (i.e., ( m > 2 )): - Minimum value ( N = 3 - 2m ) and maximum value ( M = 2 - m ). - ( M - N = -1 + m leq 2 ) gives ( 2 < m leq 3 ). Combining all cases, the range of ( m ) is ( [-1, 3] ).2. Finding the range of ( m ) such that the equation ( x^2 - (m + 2)x + 3 = -(2m + 1)x + 2 ) has a solution in ( 0 leq x leq 2 ): - Simplify the equation to ( x^2 + (m - 1)x + 1 = 0 ). - The quadratic equation has real roots when ( m leq -1 ) or ( m geq 3 ). - For ( m leq -1 ), analyze the roots and parameterization: - The equation can be solved for ( m = 1 - x - frac{1}{x} ) where ( x in (0, 2] ). - The range of ( m ) is all real numbers ( leq -1 ).Final answers:(1) boxed{[-1, 3]}(2) boxed{(-infty, -1]},<think>Okay, let's try to tackle these two problems step by step. Both are about quadratic functions and involve finding ranges for the parameter m. Let's start with problem (1).Problem (1): We have the quadratic function y = x² - (m + 2)x + 3. It says that in the interval 1 ≤ x ≤ 2, the maximum value is M and the minimum value is N, and we need to find m such that M - N ≤ 2.First, I remember that for a quadratic function y = ax² + bx + c, the vertex is at x = -b/(2a). Since the coefficient of x² here is 1 (positive), the parabola opens upwards, so the vertex is a minimum point. But since we're looking at the interval [1, 2], the maximum and minimum values could be at the endpoints or at the vertex if the vertex lies within the interval.But wait, the function here is y = x² - (m + 2)x + 3. So the leading coefficient is 1, which is positive, so indeed the parabola opens upwards. Therefore, the vertex is a minimum. So in the interval [1, 2], the minimum N will be either at the vertex (if the vertex is inside the interval) or at one of the endpoints. The maximum M will be at one of the endpoints since the parabola opens upwards. However, maybe the vertex is outside the interval, so we need to check.First, let's find the vertex's x-coordinate. The formula is x = -b/(2a). Here, a = 1, and b = -(m + 2). So x = (m + 2)/(2*1) = (m + 2)/2. So the vertex is at x = (m + 2)/2.Now, we need to check if this vertex lies within the interval [1, 2]. So, (m + 2)/2 should be between 1 and 2. Let's solve for m:1 ≤ (m + 2)/2 ≤ 2Multiply all parts by 2: 2 ≤ m + 2 ≤ 4Subtract 2: 0 ≤ m ≤ 2.So, if m is between 0 and 2, the vertex is inside [1, 2]. Otherwise, if m < 0 or m > 2, the vertex is outside the interval.Therefore, to find the minimum N and maximum M:- If 0 ≤ m ≤ 2, the minimum N is at the vertex x = (m + 2)/2, and the maximum M is either at x = 1 or x = 2.- If m < 0, then the vertex is to the left of the interval [1, 2], so the minimum is at x = 1, and the maximum at x = 2.- If m > 2, the vertex is to the right of the interval, so the minimum is at x = 2, and the maximum at x = 1.Wait, hold on. If the parabola opens upwards, then if the vertex is left of the interval, the function is increasing on the interval, so the minimum is at x=1 and maximum at x=2. Similarly, if the vertex is to the right, the function is decreasing on the interval, so the maximum is at x=1 and the minimum at x=2. That makes sense.So, depending on where the vertex is, the min and max can switch between endpoints or involve the vertex.So, first, let's split into cases:Case 1: m < 0. Then vertex at (m + 2)/2 < (0 + 2)/2 = 1. So vertex is left of interval. Then, function is increasing on [1,2]. Therefore, N = y(1) and M = y(2).Case 2: 0 ≤ m ≤ 2. Then vertex is in [1,2]. So N is at the vertex, and M is the larger of y(1) and y(2).Case 3: m > 2. Then vertex at (m + 2)/2 > (2 + 2)/2 = 2. So vertex is right of interval. Function is decreasing on [1,2], so N = y(2) and M = y(1).So for each case, we need to compute M - N and set that ≤ 2.Let me compute expressions for y(1), y(2), and y at the vertex.First, compute y(1):y(1) = 1² - (m + 2)*1 + 3 = 1 - m - 2 + 3 = (1 - 2 + 3) - m = 2 - m.Similarly, y(2):y(2) = 2² - (m + 2)*2 + 3 = 4 - 2m - 4 + 3 = (4 - 4 + 3) - 2m = 3 - 2m.Now, compute the y-coordinate at the vertex. The vertex x is (m + 2)/2. So plug that into the equation:y( (m + 2)/2 ) = [ (m + 2)/2 ]² - (m + 2)*( (m + 2)/2 ) + 3.Let me compute this step by step.First, expand the square:[ (m + 2)/2 ]² = (m + 2)² / 4.Second term: (m + 2)*( (m + 2)/2 ) = (m + 2)² / 2.So:y(vertex) = (m + 2)² / 4 - (m + 2)² / 2 + 3Combine the terms:= [ (m + 2)² / 4 - 2(m + 2)² / 4 ] + 3= [ - (m + 2)² / 4 ] + 3So y(vertex) = 3 - (m + 2)² / 4.Therefore, the minimum value N is:Case 1 (m < 0): N = y(1) = 2 - m.Case 2 (0 ≤ m ≤ 2): N = 3 - (m + 2)² / 4.Case 3 (m > 2): N = y(2) = 3 - 2m.Similarly, the maximum M is:Case 1: M = y(2) = 3 - 2m.Case 2: M is max{ y(1), y(2) } = max{ 2 - m, 3 - 2m }.Case 3: M = y(1) = 2 - m.Then, M - N must be ≤ 2 in each case.Let me handle each case separately.Case 1: m < 0Then M = 3 - 2m, N = 2 - m.So M - N = (3 - 2m) - (2 - m) = 3 - 2m - 2 + m = 1 - m.Set 1 - m ≤ 2 ⇒ -m ≤ 1 ⇒ m ≥ -1.But in this case, m < 0. So combining m < 0 and m ≥ -1, we get -1 ≤ m < 0.Case 2: 0 ≤ m ≤ 2Here, N = 3 - (m + 2)² / 4.M = max{2 - m, 3 - 2m}.We need to compute max{2 - m, 3 - 2m} and subtract N, then set ≤ 2.First, let's find where 2 - m ≥ 3 - 2m. Solve 2 - m ≥ 3 - 2m ⇒ 2m - m ≥ 3 - 2 ⇒ m ≥ 1.So in 0 ≤ m ≤ 2:- For 0 ≤ m < 1: max{2 - m, 3 - 2m} = 3 - 2m.- For 1 ≤ m ≤ 2: max{2 - m, 3 - 2m} = 2 - m.Therefore, split Case 2 into two subcases:Subcase 2a: 0 ≤ m < 1M = 3 - 2mN = 3 - (m + 2)² / 4M - N = (3 - 2m) - [3 - (m + 2)² / 4] = -2m + (m + 2)² / 4Simplify:= ( (m + 2)² - 8m ) / 4Expand (m + 2)²: m² + 4m + 4So:= (m² + 4m + 4 - 8m) / 4= (m² - 4m + 4) / 4= (m - 2)² / 4Set this ≤ 2:(m - 2)² / 4 ≤ 2 ⇒ (m - 2)² ≤ 8 ⇒ |m - 2| ≤ 2√2 ≈ 2.828But in Subcase 2a: 0 ≤ m < 1. So m is between 0 and 1. Let's check if (m - 2)^2 ≤ 8 for all m in [0,1).Compute (m - 2)^2 when m=0: 4. When m=1: (1 - 2)^2 = 1. So in [0,1), (m - 2)^2 ranges from 1 to 4, which is ≤8. Therefore, for Subcase 2a: 0 ≤ m <1, M - N = (m -2)^2 /4 ≤ 4 /4 = 1 ≤ 2. So automatically satisfies M - N ≤2. Therefore, for 0 ≤ m <1, the inequality holds.Subcase 2b: 1 ≤ m ≤2M = 2 - mN = 3 - (m + 2)^2 /4M - N = (2 - m) - [3 - (m + 2)^2 /4] = 2 - m -3 + (m + 2)^2 /4 = -1 - m + (m² + 4m +4)/4Simplify:= -1 - m + (m²)/4 + m +1Wait:Wait, (m + 2)^2 /4 = (m² +4m +4)/4 = m²/4 + m +1.So:M - N = -1 - m + m²/4 + m +1 = m²/4Therefore, M - N = m² /4.Set this ≤2: m² /4 ≤2 ⇒ m² ≤8 ⇒ |m| ≤2√2 ≈2.828But in Subcase 2b: 1 ≤ m ≤2. Since 2 ≤2.828, then m²/4 ≤4/4=1 when m=2, so 1²/4=0.25 up to 4/4=1. So for 1 ≤m ≤2, m²/4 ≤1 ≤2. So automatically satisfies the inequality.Therefore, in Case 2 (0 ≤ m ≤2), regardless of the subcase, M - N ≤2. So the entire interval 0 ≤m ≤2 is valid.Case 3: m >2Here, M = y(1)=2 - m, N = y(2)=3 - 2mSo M - N = (2 - m) - (3 - 2m) =2 - m -3 +2m= (-1) +mSet this ≤2: -1 +m ≤2 ⇒m ≤3But in this case, m >2. So combining, 2 <m ≤3.Therefore, in Case 3, m is in (2,3]Now, compiling all cases:Case1: -1 ≤m <0Case2:0 ≤m ≤2Case3:2 <m ≤3But notice that Case1 ends at m <0, and Case2 starts at 0, so overlapping at m=0. Similarly, Case2 ends at m=2, and Case3 starts at m>2, but there is a gap at m=2? Wait, no. When m=2, in Case2 (0 ≤m ≤2), so m=2 is included in Case2. Then Case3 starts at m>2. So overall, combining:-1 ≤m <0, 0 ≤m ≤2, and 2 <m ≤3. But overlapping at m=0 and m=2. So combining all, the total range is -1 ≤m ≤3.Wait, but hold on. Wait, let's check:From Case1: -1 ≤m <0From Case2:0 ≤m ≤2From Case3:2 <m ≤3So merging these intervals:-1 ≤m ≤3But wait, when m=0, it's included in Case2. When m=2, included in Case2, and m=3 included in Case3. So overall, the allowable m is from -1 up to 3.But we need to verify if this is correct. Let me check for m=3.At m=3, in Case3: M - N = -1 +3=2, which is exactly equal to 2, so satisfies M -N ≤2.Similarly, for m=-1:In Case1, m=-1. Then M -N =1 -(-1)=2, which also equals 2. So m=-1 is included.So the total range is m ∈ [-1, 3].But wait, let me check m=3 in Case3. Let's compute M and N.y(1) =2 -3= -1y(2)=3 -2*3=3 -6= -3So M = y(1)= -1, N = y(2)= -3, so M -N= -1 - (-3)=2. Correct.Similarly, m=-1:In Case1, y(1)=2 -(-1)=3, y(2)=3 -2*(-1)=3 +2=5. So M -N=5 -3=2. Correct.So the endpoints m=-1 and m=3 are included, giving M -N=2. Hence, the solution is m ∈ [-1, 3].Wait, but wait in the problem statement, part (1) says "the maximum value of M and the minimum value of N in the range 1 ≤x ≤2, and M - N ≤2". So we need to collect all m's where in their respective cases, M - N ≤2. So according to our analysis, the total range is m ∈ [-1,3].But let me check m=4, which is outside the supposed range. If m=4, which is >3. Then in Case3: m >3. Wait, but our Case3 was m>2 and up to m=3. Wait, for m >3, let's check:Wait, when m >3, in the original Case3, m >2. But we concluded that for m >2, the condition M -N ≤2 is equivalent to m ≤3. So if m >3, then M -N = m -1 > 3 -1 =2, which violates the condition. Therefore, m must be ≤3.Hence, the total range is m from -1 to 3, inclusive.Therefore, the answer to part (1) is m ∈ [-1, 3].Problem (2): Given the equation x² - (m + 2)x +3 = -(2m +1)x +2 has a solution in the range 0 ≤x ≤2. Find the range of m.First, let's write the equation:x² - (m +2)x +3 = -(2m +1)x +2Bring all terms to left-hand side:x² - (m +2)x +3 + (2m +1)x -2 =0Simplify:x² + [ - (m +2) + (2m +1) ]x + (3 -2) =0Compute coefficients:The x term: - (m +2) + (2m +1) = -m -2 +2m +1 = m -1Constant term: 3 -2 =1So the equation simplifies to x² + (m -1)x +1 =0So, need to find m such that the equation x² + (m -1)x +1 =0 has at least one solution in 0 ≤x ≤2.So, this is a quadratic equation. Let's denote f(x) =x² + (m -1)x +1. We need f(x)=0 has a solution in [0,2].There are a few possibilities:1. The quadratic has a root in [0,2]. This can happen if either: a. The quadratic has exactly one root in [0,2] (i.e., the quadratic touches the x-axis in this interval). b. The quadratic has two roots, at least one of which is in [0,2].To find the values of m for which this is true, we can use the following approach:First, compute the discriminant to ensure real roots. Then, check if any of the roots lie in [0,2].Alternatively, since the quadratic is continuous, we can use the Intermediate Value Theorem: if f(0) and f(2) have opposite signs, then there's a root between 0 and 2. Also, even if f(0) and f(2) don't have opposite signs, there could be roots within [0,2] if the quadratic has a minimum (or maximum) within [0,2] and the function crosses the x-axis there.But given that the quadratic opens upwards (coefficient of x² is 1 >0), the parabola opens upwards, so if the minimum of the quadratic is below or on the x-axis within [0,2], then there could be roots.Therefore, steps:1. Find discriminant D = (m -1)² -4*1*1 = (m -1)² -4.For real roots, need D ≥0 ⇒(m -1)² ≥4 ⇒ |m -1| ≥2 ⇒ m -1 ≥2 or m -1 ≤-2 ⇒ m ≥3 or m ≤-1.So, real roots exist only when m ≤-1 or m ≥3.If m is between -1 and 3, there are no real roots, so the equation has no solution. But since the problem states "has a solution in 0 ≤x ≤2", so we need to consider only m ≤-1 or m ≥3, but check whether the roots lie in [0,2].So even if the discriminant is non-negative, the roots may be outside [0,2].So, next, when D ≥0, we need to check if any root is in [0,2].The roots are x = [ -(m -1) ±√D ] / 2Let me write them explicitly.Given f(x)=x² + (m -1)x +1=0, the roots are:x = [ -(m -1) ±√( (m -1)^2 -4 ) ] / 2Let me denote:sqrt_term = √( (m -1)^2 -4 )So, roots:x₁ = [ -(m -1) + sqrt_term ] / 2x₂ = [ -(m -1) - sqrt_term ] / 2But since the quadratic is x² + (m -1)x +1, the sum of roots is -(m -1)/1 = 1 -m, and the product is 1.Wait, product of roots is 1. So both roots are positive or both negative. Since product is positive. If both roots are positive, their sum is 1 - m. If both roots are negative, sum is positive? Wait, sum is 1 -m. If both roots are negative, their sum would be negative, but sum is 1 -m. So, if sum is negative, 1 -m <0 ⇒m >1. If sum is positive, 1 -m >0 ⇒m <1. But product is positive. So:If m <1: sum is positive, product positive ⇒both roots positive.If m >1: sum is negative, product positive ⇒both roots negative.Therefore, when m ≤-1 or m ≥3 (the discriminant condition), the roots are real.But for m ≤-1:Since m ≤-1 <1, so sum is 1 -m ≥1 - (-1)=2>0, and product 1>0 ⇒ both roots positive.But need to check if any of them is in [0,2].For m ≥3:Sum is 1 -m ≤1 -3= -2<0, product 1>0 ⇒both roots negative, so not in [0,2].Therefore, only when m ≤-1, both roots are positive, so possible to have roots in [0,2].But we need to check if either of the roots x₁, x₂ is in [0,2].But wait, since m ≤-1, let's compute the roots.Wait, x = [ -(m -1) ±√( (m -1)^2 -4 ) ] /2Let me rewrite this as:x = [ (1 - m) ±√( (m -1)^2 -4 ) ] /2Since m ≤-1, let's substitute m with, say, a variable k where k = m, k ≤-1.Then, x = [1 - k ±√( (k -1)^2 -4 ) ] /2But since k ≤-1, 1 - k ≥1 - (-1)=2, so 1 -k is at least 2.The sqrt term: sqrt( (k -1)^2 -4 ). Let's see:(k -1)^2 -4 = (since k ≤-1, k -1 ≤-2, so squared is ≥4. So (k -1)^2 -4 ≥0, which is consistent with discriminant ≥0.So, sqrt( (k -1)^2 -4 ) = sqrt( ( (k -1) )² -4 )Let me compute (k -1)^2 -4 = k² -2k +1 -4 =k² -2k -3.But since k ≤-1, k is negative. Let's see:But maybe better to compute the roots for m ≤-1.Let me take an example. Let m = -2.Then, equation becomes x² + (-2 -1)x +1 =x² -3x +1=0. Roots:[3 ±√(9 -4)]/2= [3 ±√5]/2≈(3 ±2.236)/2≈2.618 or 0.381. So 0.381 is in [0,2]. So for m=-2, there is a solution in [0,2].Similarly, at m=-1: equation becomes x² + (-1 -1)x +1=x² -2x +1=0. This is (x -1)^2=0. So root x=1, which is in [0,2]. So m=-1 is included.At m approaching -∞:For m →-∞, the equation is x² + (m -1)x +1=0. For very large negative m, the linear term dominates, so approximate roots:x ≈ [ -(m -1) ± |m -1| ] /2. Since m is very negative, m -1 is negative, so -(m -1)=1 -m (positive). sqrt((m -1)^2 -4)≈ |m -1| =1 -m (since m -1 negative). So the roots are:[1 -m ±(1 -m)] /2. So two roots:[1 -m + (1 -m)] /2 = [2 - 2m]/2=1 -m[1 -m - (1 -m)] /2=0/2=0.But as m→-∞, the root 1 -m tends to infinity, and the other root approaches 0. Therefore, for very large negative m, one root approaches 0 and the other goes to infinity. So for any finite m ≤-1, the smaller root x₂ is positive and possibly in [0,2], while the larger root is greater than 1 -m, which is at least 2 when m ≤-1 (since 1 -m ≥2).Wait, but when m ≤-1, 1 -m ≥2. So the larger root is [ (1 -m) + sqrt(...) ] /2. Since sqrt(...) is positive, so larger root is at least (1 -m)/2. Given 1 -m ≥2, then (1 -m)/2 ≥1. So the larger root is at least 1. The smaller root is [ (1 -m) - sqrt(...) ] /2. Let's see if the smaller root can be less than or equal to 2.But given that for m ≤-1, we need to check if either root is in [0,2].But since both roots are positive (as product is 1 and sum is positive), and the smaller root may be less than 2. Let's formalize.Let me denote the two roots as x₁ and x₂, with x₁ ≤x₂.We need to check if x₁ ≤2 or x₂ ≤2. But since x₁ ≤x₂, and x₂ is the larger root, if x₂ ≤2, then both roots are ≤2. If x₂ >2, then maybe x₁ ≤2.But actually, we need to check if at least one root is in [0,2]. However, since the parabola opens upwards, if f(0) ≤0 or f(2) ≤0, then there is a root in [0,2]. Wait, but since the product of the roots is 1 (positive), both roots are positive. So f(0)=1, which is positive. So f(0)=1>0. Therefore, if the quadratic crosses the x-axis in [0,2], there must be two roots: one entering from above at x=0 (since f(0)=1>0) and exiting below, but since the product is positive, both roots are positive. Wait, but f(0)=1, so the quadratic is positive at x=0. If it has two positive roots, then the quadratic dips below the x-axis between the two roots. Therefore, if the quadratic is positive at x=0, goes negative between the two roots, and then positive again as x approaches infinity.But in our case, if the quadratic has two positive roots, and since f(0)=1>0, the quadratic is positive at x=0, dips below between the roots, and then becomes positive again. Therefore, if the interval [0,2] includes part of the dip, i.e., if the smaller root x₁ is less than 2, then there is a solution in [0,2].But wait, since f(0)=1>0, and f(x) tends to infinity as x→infty, the quadratic is positive at x=0, negative between x₁ and x₂, positive afterwards.Therefore, if x₁ <2, then there is a root at x₁, but x₁ may be less than 2. Wait, no, x₁ is the smaller root. Wait, actually, the roots are x₁ and x₂, with x₁ <x₂. Since f(0)=1>0, the quadratic is positive at x=0, so crosses the x-axis at x₁ and x₂, where x₁ <x₂. Therefore, in the interval [0,2], if x₁ ≤2, then there is a root at x₁ (if x₁ ≥0) or between 0 and x₁ (but since f(0)=1>0 and f(x₁)=0, if x₁ >0, then there is a root at x₁. If x₁ <0, then the root is negative, but since both roots are positive (as product is positive and sum is positive when m ≤-1), x₁ is positive.Therefore, for m ≤-1, x₁ and x₂ are both positive. Therefore, x₁ is the smaller positive root. So, if x₁ ≤2, then the equation has a solution in [0,2]. But x₁ can be calculated as:x₁ = [ (1 - m) - sqrt( (m -1)^2 -4 ) ] /2We need x₁ ≤2.So set:[ (1 - m) - sqrt( (m -1)^2 -4 ) ] /2 ≤2Multiply both sides by 2:(1 - m) - sqrt( (m -1)^2 -4 ) ≤4Rearrange:(1 - m) -4 ≤ sqrt( (m -1)^2 -4 )Left side:(1 - m) -4 = -m -3Right side:sqrt( (m -1)^2 -4 )Note that sqrt(...) is non-negative, so the left side must be ≤ right side. But the left side is -m -3. Since m ≤-1, -m ≥1, so -m -3 ≥1 -3= -2. But sqrt(...) is positive, so even if the left side is negative, the inequality would hold. Wait, but we have:- m -3 ≤ sqrt( (m -1)^2 -4 )But if left side is negative, then since right side is positive, the inequality holds. Therefore, for m such that -m -3 ≤ sqrt(...), which is always true when m ≤-1, because left side can be negative or positive:Wait, let's check when m ≤-1:- m -3: if m =-1, then -(-1) -3=1 -3=-2.If m =-2: -(-2) -3=2 -3=-1.If m =-3: -(-3)-3=3 -3=0.Wait, when m ≤-3, -m -3 ≥0.So when m ≤-3, -m -3 ≥0. For m between -3 and -1, -m -3 <0.Therefore, split into m ≤-3 and -3 ≤m ≤-1.For m ≤-3:Left side is ≥0, so need:- m -3 ≤ sqrt( (m -1)^2 -4 )Square both sides (since both sides are non-negative):(-m -3)^2 ≤ (m -1)^2 -4Expand left side:m² +6m +9 ≤m² -2m +1 -4Simplify:m² +6m +9 ≤m² -2m -3Subtract m² from both sides:6m +9 ≤-2m -3Bring all terms to left:6m +9 +2m +3 ≤0 ⇒8m +12 ≤0 ⇒8m ≤-12 ⇒m ≤-12/8 ⇒m ≤-1.5But in this case, m ≤-3. Since -3 ≤-1.5, the inequality 8m +12 ≤0 ⇒m ≤-1.5 is automatically satisfied for m ≤-3. Therefore, for m ≤-3, the original inequality holds.For m between -3 and -1:Left side (-m -3) is negative, right side sqrt(...) is positive. So the inequality -m -3 ≤ sqrt(...) is automatically true. Therefore, for all m ≤-1, the inequality holds. Wait, but we need x₁ ≤2, and we translated that into the inequality which holds for all m ≤-1. Therefore, does that mean that for all m ≤-1, x₁ ≤2? Wait, but let's verify with an example.Take m=-4:Equation: x² + (-4 -1)x +1 =x² -5x +1=0.Roots: [5 ±√(25 -4)]/2= [5 ±√21]/2≈[5 ±4.583]/2≈(9.583)/2≈4.791 and (0.417)/2≈0.208. So x₁≈0.208 ≤2. So yes.Another example, m=-3:x² + (-3 -1)x +1=x² -4x +1=0.Roots: [4 ±√(16 -4)]/2= [4 ±√12]/2= [4 ±2√3]/2=2 ±√3≈2 ±1.732. So x₁≈0.268, which is ≤2. So yes.m=-1:Equation: x² + (-1 -1)x +1=x² -2x +1=0. Root x=1, which is in [0,2].So in all these cases, x₁ is in [0,2]. Therefore, for m ≤-1, x₁ is always in [0,2]. Therefore, the equation has a solution in [0,2] if and only if m ≤-1 or m ≥3. But wait, earlier we saw that when m ≥3, the roots are both negative, so no solution in [0,2]. Therefore, the only valid m's are m ≤-1.But wait, let's check m=4 (which is ≥3). Equation: x² + (4 -1)x +1= x² +3x +1=0. Roots: [-3 ±√(9 -4)]/2= [-3 ±√5]/2≈(-3 ±2.236)/2≈-0.382 and -2.618. Both negative, so not in [0,2]. Therefore, m=4 doesn't work.Therefore, the only valid m's are m ≤-1. But wait, we need to check the other way. When m ≤-1, x₁ is in [0,2]. But is that always the case?Wait, take m=-10:Equation: x² + (-10 -1)x +1=x² -11x +1=0.Roots: [11 ±√(121 -4)]/2= [11 ±√117]/2≈[11 ±10.816]/2≈(21.816)/2≈10.908, and (0.184)/2≈0.092. So x₁≈0.092, which is in [0,2]. So even for very large negative m, the smaller root approaches 0 but is still in [0,2].Therefore, it seems that for all m ≤-1, the equation has a solution in [0,2].But wait, need to check if x₁ is in [0,2]. As we saw, for m ≤-1, x₁ is positive and less than 2. So the answer is m ≤-1.But let me check m=-1:As before, x=1, which is in [0,2]. So m=-1 is included.But wait, according to the problem statement, it's given that the equation "has a solution in the range 0 ≤x ≤2". So if there exists at least one x in [0,2] such that the equation holds. As we saw, for m ≤-1, there is always x₁ in [0,2]. For m >-1, there are no real roots (if -1 <m <3) or roots are negative (if m ≥3). Therefore, the range of m is m ≤-1.But let me check m=0. Wait, m=0 is in -1 <m <3. The equation becomes x² + (-1)x +1=0. Discriminant:1 -4= -3. No real roots, so no solution. Correct.Therefore, the answer for part (2) is m ≤-1, or in interval notation, (-∞, -1].But let me confirm with another example. Take m=-2, which is in the solution set. As before, the equation has roots ≈0.38 and 2.618. Wait, 2.618 is greater than 2, but 0.38 is in [0,2]. So even though one root is outside, there's still one root inside. So m=-2 is valid. Similarly, m=-1.5.Equation: x² + (-1.5 -1)x +1=x² -2.5x +1=0.Roots: [2.5 ±√(6.25 -4)]/2= [2.5 ±√2.25]/2= [2.5 ±1.5]/2= (4)/2=2 or (1)/2=0.5. So roots at x=2 and x=0.5. Both in [0,2]. So x=2 is included. So m=-1.5 is valid. But x=2 is the endpoint. Wait, the problem states "has a solution in the range 0 ≤x ≤2". So if the root is at x=2, it's included. So m=-1.5 is valid.At m=-1.5, the roots are 0.5 and 2. So 2 is included. Therefore, even if the root is at 2, it's acceptable.Wait, but let's take m such that x₁=2. Let's solve for m when x=2 is a root.Set x=2 into the equation x² + (m -1)x +1=0:2² + (m -1)*2 +1=0 ⇒4 +2m -2 +1=0 ⇒2m +3=0 ⇒m= -3/2.So m=-1.5. So when m=-1.5, x=2 is a root. So included.Similarly, when m approaches -∞, the smaller root approaches 0 but is still in [0,2].Therefore, all m ≤-1 gives at least one root in [0,2].Hence, the answer to part (2) is m ≤-1, or in interval notation (-∞, -1].Final Answer(1) boxed{[-1, 3]}(2) boxed{(-infty, -1]}</think>Given the quadratic function ( y = x^2 - (m + 2)x + 3 ):1. Finding the range of ( m ) such that the maximum value ( M ) and minimum value ( N ) in the interval ( 1 leq x leq 2 ) satisfy ( M - N leq 2 ): - The vertex of the quadratic function is at ( x = frac{m + 2}{2} ). - Depending on the value of ( m ), the vertex can be inside or outside the interval ( [1, 2] ). - Case 1: ( m < 0 ) - The vertex is left of the interval, the function is increasing, so ( N = y(1) ) and ( M = y(2) ). The difference ( M - N = 1 - m leq 2 ) gives ( -1 leq m < 0 ). - Case 2: ( 0 leq m leq 2 ) - The vertex is within the interval. The minimum ( N ) is at the vertex and the maximum ( M ) is at the endpoints. The difference ( M - N leq 2 ) is always satisfied. - Case 3: ( m > 2 ) - The vertex is right of the interval, the function is decreasing, so ( N = y(2) ) and ( M = y(1) ). The difference ( M - N = -1 + m leq 2 ) gives ( 2 < m leq 3 ). - Combining all cases, the range of ( m ) is ( [-1, 3] ).2. Finding the range of ( m ) such that the equation ( x^2 - (m + 2)x + 3 = -(2m + 1)x + 2 ) has a solution in the interval ( 0 leq x leq 2 ): - Simplify the equation to ( x^2 + (m - 1)x + 1 = 0 ). - The discriminant must be non-negative for real roots: ( (m - 1)^2 - 4 geq 0 ) leading to ( m leq -1 ) or ( m geq 3 ). - For ( m geq 3 ), both roots are negative. For ( m leq -1 ), both roots are positive and the smaller root ( x_1 ) is within ( [0, 2] ). - Therefore, the valid range for ( m ) is ( (-infty, -1] ). Final Answer(1) boxed{[-1, 3]}(2) boxed{(-infty, -1]}]

🔑: Part (1)Given y=x^{2}-(m+2)x+3, the axis of symmetry is x=frac{m+2}{2}. We analyze the function behavior based on the position of the axis of symmetry relative to the interval [1, 2].# Case 1: frac{m+2}{2} leq 1, i.e., m leq 0In this case, the function is increasing in the interval [1, 2].- Maximum value, M = 3 - 2m at x=2,- Minimum value, N = 2 - m at x=1.The condition M - N leq 2 gives:[(3 - 2m) - (2 - m) leq 2 implies -1 leq m leq 0]# Case 2: frac{m+2}{2} geq 2, i.e., m geq 2In this case, the function is decreasing in the interval [1, 2].- Maximum value, M = 2 - m at x=1,- Minimum value, N = 3 - 2m at x=2.The condition M - N leq 2 gives:[(2 - m) - (3 - 2m) leq 2 implies 2 leq m leq 3]# Case 3: 1 < frac{m+2}{2} < frac{3}{2}, i.e., 0 < m < 1- Minimum value, N = 3 - frac{1}{4}(m+2)^{2} at the vertex,- Maximum value, M = 3 - 2m at x=2.The condition M - N leq 2 gives:[3 - 2m - left[3 - frac{1}{4}(m+2)^{2}right] leq 2]Solving gives 2 - 2sqrt{2} leq m leq 2 + 2sqrt{2}, but in this case, 0 < m < 1.# Case 4: frac{3}{2} leq frac{m+2}{2} < 2, i.e., 1 leq m < 2- Minimum value, N = 3 - frac{1}{4}(m+2)^{2} at the vertex,- Maximum value, M = 2 - m at x=1.The condition M - N leq 2 gives:[2 - m - left[3 - frac{1}{4}(m+2)^{2}right] leq 2]Solving gives -2sqrt{2} leq m leq 2sqrt{2}, but in this case, 1 leq m < 2.Combining all cases, the range of m is -1 leq m leq 3.boxed{-1 leq m leq 3} Part (2)The equation simplifies to x^{2} + (m-1)x + 1 = 0. Let y = x^{2} + (m-1)x + 1.# Case 1: One solution in 0 leq x leq 2For x=2, y < 0 gives:[4 + 2(m-1) + 1 < 0 implies m < -frac{3}{2}]# Case 2: Two solutions in 0 leq x leq 2The conditions are:- Delta = (m-1)^{2} - 4 geq 0,- 0 leq -frac{m-1}{2} leq 2,- 4 + 2(m-1) + 1 geq 0.Solving gives:- m geq 3 or m leq -1,- -3 leq m leq 1,- m geq -frac{3}{2}.Combining these, we get -frac{3}{2} leq m leq -1.In conclusion, the range of m is left(-infty, -1right].boxed{left(-infty, -1right]}

🧠:[<think>Okay, so I need to figure out which of these propositions about planes in 3D space is incorrect. Let me go through each option one by one and think about whether they're true or not.Starting with option A: "If plane α is perpendicular to plane β, then there must exist a line in plane α that is parallel to plane β." Hmm. Let me visualize two perpendicular planes. Let's say α is the vertical plane (like a wall) and β is the horizontal plane (like the floor). If they're perpendicular, does there exist a line in α that's parallel to β? Well, in the wall, there are horizontal lines. A horizontal line in the wall (plane α) would be parallel to the floor (plane β). So yes, that seems correct. For example, the line where the wall meets the ceiling is in plane α and is horizontal, hence parallel to plane β. So A seems true.Moving on to option B: "If plane α is not perpendicular to plane β, then there must not exist a line in plane α that is perpendicular to plane β." Wait. If two planes are not perpendicular, does that mean there can't be a line in one plane that's perpendicular to the other? Let me think. Suppose two planes intersect at some angle that's not 90 degrees. Is it possible for there to be a line in one plane that's perpendicular to the other? Hmm. Let's consider a line that's along the intersection of the two planes. The intersection line is in both planes. If the planes are not perpendicular, then this line is not perpendicular to either plane. Wait, but maybe there's another line in α that's perpendicular to β. Let's think about the normal vectors. If plane α has normal vector nα and plane β has normal vector nβ. If the planes are not perpendicular, then the dot product of nα and nβ is not zero. But can there be a line in α that's perpendicular to β? A line perpendicular to β would have to be parallel to nβ. So, is there a line in α that's parallel to nβ? That would require that nβ is orthogonal to nα, because the line is in α, so its direction vector is perpendicular to nα. Wait, but if nβ is not orthogonal to nα (since the planes are not perpendicular), then nβ can't be in the plane α. Wait, this is getting a bit confusing. Let me try a concrete example. Suppose plane α is the floor (horizontal) and plane β is a wall that's not perpendicular to the floor, like slanted at 45 degrees. Is there a line in the floor that is perpendicular to the slanted wall? The line would have to be in the floor and perpendicular to the wall. The normal vector of the slanted wall would be pointing at 45 degrees. A line in the floor (which is horizontal) would need to be in the direction of the normal vector of the wall. But the normal vector of the wall is slanted, not horizontal or vertical. So a line in the floor can't be in the direction of the normal vector of the wall, because that direction has a vertical component. Therefore, there can't be a line in α (floor) that's perpendicular to β (slanted wall). So B's statement is saying that if α is not perpendicular to β, then there must not exist such a line. In my example, that seems to hold. Therefore, B is true? Wait, but maybe there's a case where even if two planes aren't perpendicular, there's still a line in one that's perpendicular to the other. Let me think again. Suppose two planes intersect at some angle θ ≠ 90°. The line of intersection is common to both. If I take a line in plane α that is perpendicular to the line of intersection, would that line be perpendicular to plane β? Hmm. The direction of such a line in plane α would be perpendicular to the line of intersection. To check if it's perpendicular to plane β, its direction must be parallel to the normal vector of β. The normal vector of β is perpendicular to plane β. So, unless the normal vector of β is in plane α, there can't be such a line. But if plane α is not perpendicular to β, then the normal vector of β is not in plane α. Because if it were, then plane α would contain the normal vector of β, which would make them perpendicular. Since they are not, the normal vector of β is not in α, so no line in α can be parallel to it. Therefore, B is correct. So B's statement is true.Option C: "If plane α is perpendicular to plane γ, and plane β is perpendicular to plane γ, and α ∩ β = l, then l is perpendicular to plane γ." Okay, so if both α and β are perpendicular to γ, and they intersect along line l, then l is perpendicular to γ. Let me visualize. Let γ be the horizontal plane. Then α and β are both vertical planes. The intersection of two vertical planes is a vertical line, which is indeed perpendicular to the horizontal plane γ. So yes, l is perpendicular to γ. So this seems correct. Let me check another case. Suppose γ is some arbitrary plane, and α and β are both perpendicular to γ. Then their intersection line l must lie along the normals of γ? Wait, if α is perpendicular to γ, then the normal of α is parallel to γ. Wait, maybe not. Let me recall: If two planes are perpendicular, then their normals are also perpendicular. So if α is perpendicular to γ, then the normal vector of α is perpendicular to the normal vector of γ. Similarly for β. So if both α and β are perpendicular to γ, then their normals are both perpendicular to the normal of γ. That means the normals of α and β lie in a plane that's perpendicular to the normal of γ. Then the intersection line l of α and β would be along a direction that's perpendicular to both normals of α and β. Wait, this is getting a bit complicated. Alternatively, since both α and β are perpendicular to γ, their intersection line l must be perpendicular to γ. Because each plane α and β contains a line that's perpendicular to γ. Actually, if a plane is perpendicular to γ, then it must contain the normal vector of γ. Wait, no. If plane α is perpendicular to plane γ, then the normal vector of α is parallel to plane γ. Because the normal of α would be perpendicular to the normal of γ. Wait, let me clarify: The angle between two planes is equal to the angle between their normals. So if two planes are perpendicular, their normals are also perpendicular. So if α is perpendicular to γ, then the normal vector nα is perpendicular to nγ. Similarly, nβ is perpendicular to nγ. So both nα and nβ are perpendicular to nγ, meaning nα and nβ lie in a plane that's parallel to γ. Then the line l, which is the intersection of α and β, must be along the cross product of nα and nβ. But since nα and nβ are both perpendicular to nγ, their cross product would be in the direction of nγ or opposite. Wait, no. Let me think again. If nα is perpendicular to nγ, and nβ is perpendicular to nγ, then nα and nβ are both in a plane perpendicular to nγ. Therefore, the cross product of nα and nβ would be a vector that's either parallel or antiparallel to nγ. Therefore, the line l, which is the intersection of α and β, is parallel to the cross product of nα and nβ, which is parallel to nγ. Therefore, l is parallel to nγ, which is the normal of γ. Hence, l is perpendicular to plane γ. So yes, l is perpendicular to γ. Therefore, C is correct.Option D: "If plane α is perpendicular to plane β, then all lines in plane α are perpendicular to plane β." Wait, that seems too strong. If α is perpendicular to β, does every line in α have to be perpendicular to β? Let me think. Take α as a vertical plane and β as the horizontal plane. Then a line in α can be vertical or horizontal. The vertical lines in α are perpendicular to β, but the horizontal lines in α are parallel to β. So a horizontal line in α is not perpendicular to β. Therefore, not all lines in α are perpendicular to β. Therefore, D is incorrect. So the answer must be D.But let me verify once more. When two planes are perpendicular, any line in α that is perpendicular to the line of intersection of α and β would be perpendicular to β. But other lines in α, which are not perpendicular to the line of intersection, are not necessarily perpendicular to β. For example, take the line of intersection itself. That line lies in both planes, so it's not perpendicular to β. Therefore, D is indeed incorrect because only some lines in α are perpendicular to β, not all. So the incorrect proposition is D.Final Answerboxed{D}</think>To determine which proposition is incorrect, we analyze each option step by step:Option A: If plane (alpha) is perpendicular to plane (beta), there must exist a line in (alpha) parallel to (beta). This is true because in a vertical plane (alpha) (e.g., a wall) and horizontal plane (beta) (e.g., the floor), a horizontal line in (alpha) is parallel to (beta).Option B: If plane (alpha) is not perpendicular to (beta), there must not exist a line in (alpha) perpendicular to (beta). This is true because if (alpha) and (beta) are not perpendicular, the normal vector of (beta) is not in (alpha), so no line in (alpha) can be parallel to this normal vector.Option C: If planes (alpha) and (beta) are both perpendicular to (gamma) and intersect along line (l), then (l) is perpendicular to (gamma). This is true because the intersection line (l) of (alpha) and (beta) must be parallel to the normal vector of (gamma), making (l) perpendicular to (gamma).Option D: If plane (alpha) is perpendicular to (beta), then all lines in (alpha) are perpendicular to (beta). This is false because while some lines in (alpha) (e.g., vertical lines) are perpendicular to (beta), others (e.g., horizontal lines) are not. For example, the line of intersection of (alpha) and (beta) lies in both planes and is not perpendicular to (beta).Thus, the incorrect proposition is (boxed{D}).,<think>Okay, let's tackle this problem step by step. The question is asking which of the propositions A, B, C, D is incorrect. All of them are about the relationships between planes and lines in three-dimensional space. Let me recall some geometry concepts to analyze each option.Starting with option A: "If plane α is perpendicular to plane β, then there must exist a line in plane α that is parallel to plane β."Hmm, if two planes are perpendicular, their dihedral angle is 90 degrees. So, if α is perpendicular to β, then they intersect along a line, let's say line l. Now, in plane α, can we find a line that's parallel to plane β? I remember that if a line is in one plane and parallel to another plane, it must be parallel to the line of intersection of the two planes or lie in some way that it doesn't intersect the other plane. Wait, but actually, for two perpendicular planes, any line in α that is not the line of intersection (l) would either be parallel to β or intersect β. Wait, but how can a line in α be parallel to β?Wait, maybe if we take a line in α that's perpendicular to the line of intersection l. Since α and β are perpendicular, the line in α that is perpendicular to l would be parallel to β. Because in β, the direction perpendicular to l would be along the normal vector of α, but maybe I need to think more clearly.Alternatively, think of the standard example: α is the xy-plane, β is the xz-plane, intersecting along the x-axis. In α (the xy-plane), the y-axis is a line that is perpendicular to the x-axis (the line of intersection). Then, is the y-axis parallel to β (the xz-plane)? Yes, because the y-axis is perpendicular to the xz-plane's normal vector (which is along the y-axis). Wait, actually, the y-axis is orthogonal to the xz-plane. Wait, no. Wait, the y-axis is in the xy-plane and is perpendicular to the xz-plane. But actually, a line being parallel to a plane means that the line is either lying on the plane or parallel to some line in the plane. Wait, no. The definition of a line being parallel to a plane is that the line does not intersect the plane and is not in the plane. Wait, actually, in three dimensions, a line is parallel to a plane if it is either lying on the plane or doesn't intersect it. But if the line is in another plane that's perpendicular, then maybe it can be parallel.Wait, in the example, the y-axis is in the xy-plane (α), and the xz-plane (β) is perpendicular. The y-axis is orthogonal to the xz-plane, right? Because the direction of the y-axis is perpendicular to the xz-plane. Wait, but does that mean the line (y-axis) is perpendicular to the plane β? If a line is perpendicular to a plane, then it's perpendicular to every line in that plane. So, the y-axis is perpendicular to the xz-plane. Therefore, the y-axis is perpendicular to β, not parallel. So, in this case, there exists a line in α (y-axis) that is perpendicular to β. But the question is about a line in α that is parallel to β. So, maybe in this example, is there a line in α that is parallel to β?Let me think. Take the x-axis, which is the line of intersection. The x-axis is in both α and β. So, if we take a line in α that's parallel to the x-axis, then since the x-axis is in β, that line would be parallel to a line in β. Therefore, such a line in α (parallel to x-axis) would be parallel to β. Wait, but the x-axis is in β, so a line in α parallel to x-axis would be parallel to the line x-axis in β, hence parallel to β? Wait, maybe not. A line parallel to another line in the plane would be parallel to the plane? Wait, no. A line is parallel to a plane if it is parallel to some line in the plane. So, if there's a line in α that's parallel to a line in β, then it's parallel to β. In this case, if we take a line in α that's parallel to the x-axis (which is in β), then that line would be parallel to β. So yes, such a line exists. Therefore, proposition A is correct. So, maybe A is correct.Moving on to option B: "If plane α is not perpendicular to plane β, then there must not exist a line in plane α that is perpendicular to plane β."Hmm. So, if α and β are not perpendicular (i.e., their dihedral angle is not 90 degrees), then can there be a line in α that is perpendicular to β? Let's think.A line perpendicular to plane β would have to be along the normal vector of β. So, if there is a line in α that is along the normal vector of β, then α must contain the normal vector of β. If α contains the normal vector of β, then α is perpendicular to β. Because the normal vector of β is perpendicular to β, so if α contains that normal vector, then α must be perpendicular to β. Therefore, if α is not perpendicular to β, then α cannot contain the normal vector of β, hence there cannot be a line in α that is perpendicular to β. Therefore, proposition B is correct. So B is correct.Option C: "If plane α is perpendicular to plane γ, and plane β is perpendicular to plane γ, and α ∩ β = l, then l is perpendicular to plane γ."So, if both α and β are perpendicular to γ, and their intersection is line l, then l is perpendicular to γ. Let me visualize this. Suppose γ is the xy-plane. Then, planes perpendicular to γ would be vertical planes, like the xz-plane and yz-plane. If α is xz-plane and β is yz-plane, their intersection is the z-axis. The z-axis is perpendicular to the xy-plane (γ). So in this case, l is indeed perpendicular to γ. But is this always the case?Suppose α and β are both perpendicular to γ, so their normals are parallel to γ's normal? Wait, no. If a plane is perpendicular to γ, then its normal vector is parallel to γ. Wait, no. Wait, plane γ has a normal vector n_γ. If α is perpendicular to γ, then the normal vector n_α of α is perpendicular to n_γ. Similarly, n_β is perpendicular to n_γ. So, both n_α and n_β are perpendicular to n_γ. Therefore, n_α and n_β lie in a plane that's perpendicular to n_γ. Now, if α and β intersect along line l, then l is the intersection of two planes whose normals are both perpendicular to n_γ. So, the direction of l would be along the cross product of n_α and n_β. But since both n_α and n_β are perpendicular to n_γ, their cross product would be parallel to n_γ. Therefore, line l is parallel to n_γ, which is the normal vector of γ. Therefore, l is perpendicular to plane γ. Wait, but if l is parallel to the normal vector of γ, then l is perpendicular to γ. Therefore, yes, l is perpendicular to γ. So, in this case, proposition C is correct.Therefore, C is correct.Option D: "If plane α is perpendicular to plane β, then all lines in plane α are perpendicular to plane β."Wait, this seems too strong. If plane α is perpendicular to β, does that mean every line in α is perpendicular to β? Let's recall that if two planes are perpendicular, then a line in one plane is perpendicular to the other plane only if it is perpendicular to the line of intersection of the two planes. Otherwise, the line is only perpendicular to the other plane's normal vector if it's along the normal direction.Wait, take the example where α is the xy-plane and β is the xz-plane, which are perpendicular. The line of intersection is the x-axis. In plane α (xy-plane), the x-axis is in both planes, so it's not perpendicular to β. Similarly, the y-axis in α is perpendicular to β (since it's along the normal vector of β). But other lines in α, like the line y = x, are in α but are not perpendicular to β. Because for a line to be perpendicular to a plane, it must be perpendicular to every line in that plane. The line y = x in the xy-plane is not perpendicular to the xz-plane. For example, the direction vector of y = x is (1,1,0). In the xz-plane, take a direction vector (1,0,1). The dot product is 1*1 + 1*0 + 0*1 = 1 ≠ 0, so they are not perpendicular. Therefore, the line y = x in α is not perpendicular to β. Hence, proposition D is incorrect. Therefore, D is the answer.So, the incorrect proposition is D.Final Answerboxed{D}</think>Let's analyze each proposition step by step:Proposition A: If plane (alpha) is perpendicular to plane (beta), then there must exist a line in plane (alpha) that is parallel to plane (beta).- When two planes are perpendicular, they intersect along a line (l). In plane (alpha), any line parallel to the line of intersection (l) will be parallel to plane (beta). For example, in the xy-plane (α) and xz-plane (β), the x-axis is the line of intersection. A line in α parallel to the x-axis is parallel to β. Therefore, A is correct.Proposition B: If plane (alpha) is not perpendicular to plane (beta), then there must not exist a line in plane (alpha) that is perpendicular to plane (beta).- A line perpendicular to plane (beta) must be along the normal vector of (beta). If (alpha) is not perpendicular to (beta), it cannot contain the normal vector of (beta). Hence, no line in (alpha) can be perpendicular to (beta). Therefore, B is correct.Proposition C: If plane (alpha) is perpendicular to plane (gamma), and plane (beta) is perpendicular to plane (gamma), and (alpha cap beta = l), then (l) is perpendicular to plane (gamma).- If both (alpha) and (beta) are perpendicular to (gamma), their normals are perpendicular to (gamma)'s normal. The intersection line (l) is the cross product of (alpha) and (beta)'s normals, which is parallel to (gamma)'s normal. Hence, (l) is perpendicular to (gamma). Therefore, C is correct.Proposition D: If plane (alpha) is perpendicular to plane (beta), then all lines in plane (alpha) are perpendicular to plane (beta).- A line in (alpha) must be perpendicular to (beta)'s normal to be perpendicular to (beta). Only lines in (alpha) that are perpendicular to the line of intersection are perpendicular to (beta). Other lines are not. For example, in the xy-plane and xz-plane, the line (y = x) in the xy-plane is not perpendicular to the xz-plane. Therefore, D is incorrect.The incorrect proposition is (boxed{D}).]

🔑:From the problem statement, we know:A: By considering real objects: the door surface is perpendicular to the ground, and the upper edge of the door corresponds to a line that is parallel to the ground. Therefore, this proposition is true.B: If there exists a line in plane alpha that is perpendicular to plane beta, according to the theorem of perpendicular planes, the two planes are perpendicular. Therefore, this proposition is true.C: By using the properties of perpendicular planes, we can construct lines in alpha and beta, different from l, that are perpendicular to the intersection line. Then, by the theorem of line-plane perpendicularity, the constructed perpendicular lines are parallel, which leads to the conclusion that the constructed lines are parallel to l. Since if one of two parallel lines is perpendicular to a plane, then the other is also perpendicular to the plane, the proposition is true.D: By counterexample: the inner side wall of a classroom is perpendicular to the ground, but there are many lines within the side wall that are not perpendicular to the ground. Therefore, this proposition is incorrect.Hence, the correct choice is boxed{D}.

🧠:[<think>Okay, so I have this geometry problem here with four statements about a line l and two distinct planes α and β. I need to figure out which of the options A, B, C, or D is correct. Let me take this step by step.First, let me recall some basic concepts about lines and planes in three-dimensional space. When a line is parallel to a plane, it means the line doesn't intersect the plane and lies entirely outside of it, right? Or it could also be that the line is in the plane, but since the problem says "distinct planes," maybe the line can't be in both planes if they're distinct. Wait, but the line could be in one plane and parallel to the other? Hmm, maybe I need to be careful here.Similarly, if a line is perpendicular to a plane, that means it's orthogonal to every line in the plane that it intersects. So, the line forms a 90-degree angle with the plane. Also, if two planes are perpendicular, their dihedral angle is 90 degrees, which means their normal vectors are perpendicular.Now let's look at each option one by one.Option A: If l is parallel to α and l is parallel to β, then α is parallel to β.Hmm. If a line is parallel to two different planes, does that mean the planes must be parallel? Let me visualize this. Suppose we have two planes that are not parallel; they intersect along some line. If there's a line that's parallel to both planes, would that be possible?Imagine two intersecting planes, like the floor and a wall meeting at a corner. If I take a line that's parallel to both the floor and the wall, it would have to be parallel to the line of intersection of the two planes. Wait, but in that case, the line is parallel to both planes. So even though the planes are not parallel, there exists a line that's parallel to both. So this would mean that Option A is not necessarily true. So A is incorrect.Option B: If l is parallel to α and l is perpendicular to β, then α is perpendicular to β.Okay, so if a line is parallel to plane α and perpendicular to plane β, does that imply α and β are perpendicular? Let's think.If line l is parallel to α, then l lies in some plane that's parallel to α or is outside of α but doesn't intersect it. But l is also perpendicular to β. If a line is perpendicular to a plane, it's perpendicular to every line in that plane. So, if l is perpendicular to β, then the direction of l is along the normal vector of β.Now, if l is parallel to α, then the direction of l (which is the normal vector of β) is parallel to α. So, the normal vector of β is parallel to α. That would mean that the normal vector of β is perpendicular to the normal vector of α, right? Because if a vector is parallel to a plane, it's perpendicular to the normal vector of that plane.Wait, let me verify that. The normal vector of α is perpendicular to plane α. If the normal vector of β is parallel to plane α, then the normal vector of β is perpendicular to the normal vector of α. Therefore, the normals are perpendicular, so the planes α and β are perpendicular. Yes, that seems right. So B is correct.But wait, let me think again. Suppose plane α has normal vector n₁, and plane β has normal vector n₂. If l is parallel to α, then the direction of l is perpendicular to n₁. Also, since l is perpendicular to β, the direction of l is parallel to n₂. Therefore, n₂ is perpendicular to n₁, which implies that the planes α and β are perpendicular. So yes, B is correct.Option C: If α is perpendicular to β and l is perpendicular to α, then l is parallel to β.Hmm. If α and β are perpendicular, and l is perpendicular to α, does that mean l is parallel to β?Let me visualize. If α is the xy-plane and β is the xz-plane, which are perpendicular. If l is perpendicular to α (the xy-plane), then l is along the z-axis. But the z-axis is not parallel to the xz-plane (β). Wait, the z-axis lies within the xz-plane. Wait, no. Wait, the z-axis is perpendicular to the xy-plane, but the xz-plane includes the z-axis. Wait, the xz-plane contains the z-axis. So if l is the z-axis, which is perpendicular to α (xy-plane), then l is actually lying in β (xz-plane). But being in a plane isn't the same as being parallel to it. Wait, if a line lies in a plane, it's not parallel, it's intersecting. Wait, no. If a line is in the plane, it's not considered parallel; parallel lines don't intersect and are not in the plane. So in this case, the z-axis is contained within β, so it's not parallel. So l is not parallel to β. Therefore, the conclusion that l is parallel to β is false. So C is incorrect.Alternatively, take another example. Suppose α is the xy-plane, β is the yz-plane, which are perpendicular. If l is perpendicular to α, then l is vertical along the z-axis. But β is the yz-plane. The z-axis is contained in the yz-plane, so again, l is not parallel but lies within β. So C is false.Wait, but maybe if the line is perpendicular to α, then it's parallel to the normal of α. If α and β are perpendicular, then the normal of α is parallel to β? Wait, no. If two planes are perpendicular, their normals are perpendicular. So the normal of α is perpendicular to the normal of β. So if l is along the normal of α, then it is perpendicular to the normal of β, meaning l is parallel to β? Wait, maybe my previous examples are conflicting.Wait, in the example, the normal of α is along the z-axis (for the xy-plane). The normal of β (xz-plane) is along the y-axis. So the normals are perpendicular, which is correct. Then, l is along the z-axis (normal of α). Is the z-axis parallel to the xz-plane (β)? Well, a line is parallel to a plane if it is parallel to any line in the plane or lies in the plane. The z-axis is in the xz-plane, so is a line that lies in a plane considered parallel? Wait, definitions might vary here. In some contexts, a line lying in the plane is considered to be parallel (since it doesn't intersect the plane outside of itself), but in other contexts, parallel might mean not intersecting and not lying within. So I need to clarify.In standard terminology, a line that lies in a plane is not considered parallel; parallelism is for lines that do not intersect and are not skew. But for a line and a plane, if the line is in the plane, it's not parallel—they intersect at every point. However, if the line is outside the plane and doesn't intersect it, then it's parallel. So in the example, the z-axis is in the xz-plane, so it's not parallel. Therefore, the conclusion that l is parallel to β is incorrect.Alternatively, if α and β are two perpendicular planes, and l is perpendicular to α, then l could either lie in β or be parallel to β. Wait, in my previous example, it lies in β. But maybe in another configuration. Suppose α is the xy-plane and β is some other vertical plane that's perpendicular to α but not containing the z-axis. For example, β could be the yz-plane shifted along the x-axis. Then, the z-axis (perpendicular to α) would not lie in β. In that case, would the z-axis be parallel to β?Hmm, the z-axis is a vertical line. The shifted yz-plane is a vertical plane. The z-axis is outside of this shifted plane. Is the z-axis parallel to the shifted yz-plane? If two lines are parallel, they don't intersect and are in the same direction. But a line and a plane: the line is parallel to the plane if it doesn't intersect the plane and is parallel to some line in the plane. So the z-axis doesn't intersect the shifted yz-plane? Wait, if β is the yz-plane shifted along the x-axis, say x=1, then the z-axis (x=0, y=0) doesn't intersect β, right? So the z-axis would be parallel to β? Because it doesn't intersect β and is parallel to the z-axis direction, which is a direction in β. So in this case, yes, the z-axis is parallel to β.Wait, but hold on. If β is the plane x=1, which is parallel to the yz-plane, then the normal vector of β is along the x-axis. Then, if α is the xy-plane with normal along the z-axis. Then α and β are perpendicular because their normals (z-axis and x-axis) are perpendicular.But in this case, l is the z-axis, which is perpendicular to α (xy-plane). Then l is parallel to β (x=1 plane). Because the z-axis doesn't intersect the x=1 plane and is parallel to the z-direction, which is a direction in the x=1 plane. So in this case, l is indeed parallel to β.But earlier, when β was the original yz-plane (x=0), the z-axis lies in β, so it's not parallel. So this seems to depend on the specific orientation. Therefore, in some cases, l is parallel to β, and in others, it's not. So the statement says "if α is perpendicular to β and l is perpendicular to α, then l is parallel to β." But this isn't always true. Because depending on the position of β, l could be either parallel to β or lying in β. Wait, but if planes α and β are perpendicular, and l is perpendicular to α, then l must be parallel to β or lie in β. But the problem states that the two planes are distinct. Wait, but if l is lying in β, does that violate anything? The planes α and β are distinct, but l is just a line.But the statement claims that l must be parallel to β. However, in some cases, l can lie within β, so the conclusion isn't necessarily true. Therefore, the answer C is incorrect because the conclusion doesn't hold in all cases.Option D: If α is perpendicular to β and l is parallel to α, then l is perpendicular to β.Hmm. If two planes α and β are perpendicular, and a line l is parallel to α, does that make l perpendicular to β?Let me take an example. Let α be the xy-plane and β be the xz-plane, which are perpendicular. If l is parallel to α, then l could be any line in the xy-plane or parallel to it. For example, take the x-axis, which is parallel to the xy-plane. Is the x-axis perpendicular to β (xz-plane)? The x-axis is in the xz-plane, actually. Wait, the x-axis is in both the xy-plane and the xz-plane. If l is the x-axis, which is parallel to α (xy-plane), but it's not perpendicular to β (xz-plane); it's lying within β. So in this case, l is not perpendicular to β. So D is incorrect.Alternatively, take a line that's parallel to α but not lying in β. For example, take a line parallel to the xy-plane but at some z-coordinate, say z=1, and running along the x-direction. So this line is parallel to the x-axis, hence parallel to the xy-plane. Is this line perpendicular to the xz-plane (β)? The line is horizontal, and the xz-plane is vertical. The line is not perpendicular to the xz-plane. To be perpendicular to the xz-plane, the line would have to be vertical along the y-axis. But this line is along the x-axis, so it's not perpendicular. Therefore, D is false.Wait another example. Suppose α is the xy-plane, β is the yz-plane, which are perpendicular. If l is a line parallel to α, say the line y=0, z=1 (a horizontal line). Is this line perpendicular to β? The direction of l is along the x-axis. The normal vector of β is along the x-axis. Wait, the yz-plane has a normal vector along the x-axis. So if a line is along the x-axis direction, then it's parallel to the normal vector of β, which would mean the line is perpendicular to β. Wait, but in my previous example, the line at z=1 along the x-axis is indeed perpendicular to β (yz-plane)? Because the direction of the line is along the x-axis, which is the normal of β. So in that case, the line is perpendicular to β.But wait, hold on. If l is parallel to α (xy-plane), then its direction vector lies in the xy-plane. Wait, no. If a line is parallel to a plane, its direction vector is parallel to the plane, meaning it is perpendicular to the plane's normal vector. So the direction vector of l should be perpendicular to the normal vector of α.If α is the xy-plane, its normal is along the z-axis. So a line parallel to α must have a direction vector perpendicular to the z-axis, i.e., lying in the xy-plane. Therefore, the direction vector has components (a, b, 0). But the normal vector of β (if β is the yz-plane) is along the x-axis. So for a line to be perpendicular to β, its direction vector must be parallel to the x-axis. So if a line is parallel to α (xy-plane), its direction vector is in the xy-plane. If it's also perpendicular to β (yz-plane), then its direction vector must be along the x-axis.So in this case, a line that is parallel to α (direction in xy-plane) and perpendicular to β (direction along x-axis) is possible. But does every line parallel to α have to be perpendicular to β?No. For example, take a line in the xy-plane with direction vector (1,1,0). This line is parallel to α (since it's in the xy-plane), but its direction vector is not along the x-axis. Therefore, it's not perpendicular to β (yz-plane), since the normal of β is the x-axis. The line's direction is (1,1,0), which is not parallel to the x-axis. Therefore, the line is not perpendicular to β.Therefore, the statement D says that if α is perpendicular to β and l is parallel to α, then l is perpendicular to β. But in this example, l is parallel to α but not perpendicular to β. Hence, D is incorrect.Wait, but in the case where l is along the x-axis, which is parallel to α (xy-plane) and also perpendicular to β (yz-plane). So sometimes it can be true, but the statement requires it to be always true. Since there exists a line parallel to α that is not perpendicular to β, the statement D is false.So, summarizing:A: FalseB: TrueC: FalseD: FalseTherefore, the correct answer is B.But let me double-check each conclusion.For A: As I considered two intersecting planes, a line parallel to both can exist, so A is false.B: The logic about the normals seems solid. If l is parallel to α, then its direction is perpendicular to α's normal. If l is perpendicular to β, then its direction is parallel to β's normal. Therefore, β's normal is perpendicular to α's normal, so planes are perpendicular. So B is correct.C: Gave counterexample where l is in β, hence not parallel. So C is wrong.D: Gave counterexample where l is parallel to α but not perpendicular to β. So D is wrong.Yep, so B is the correct answer.Final Answerboxed{B}</think>Given a line ( l ) and two distinct planes ( alpha ) and ( beta ), we need to determine which of the given statements is correct.Option A: If ( l parallel alpha ) and ( l parallel beta ), then ( alpha parallel beta ).- If two planes intersect, a line can be parallel to both planes without the planes being parallel. For example, a line parallel to the line of intersection of two intersecting planes is parallel to both planes. Thus, this statement is false.Option B: If ( l parallel alpha ) and ( l perp beta ), then ( alpha perp beta ).- If ( l parallel alpha ), the direction of ( l ) is perpendicular to the normal vector of ( alpha ). If ( l perp beta ), the direction of ( l ) is parallel to the normal vector of ( beta ). Therefore, the normal vectors of ( alpha ) and ( beta ) are perpendicular, implying ( alpha perp beta ). This statement is true.Option C: If ( alpha perp beta ) and ( l perp alpha ), then ( l parallel beta ).- If ( alpha perp beta ) and ( l perp alpha ), ( l ) could lie within ( beta ) or be parallel to ( beta ). For example, the z-axis (perpendicular to the xy-plane) lies in the xz-plane (which is perpendicular to the xy-plane). Since lying in a plane is not considered parallel, this statement is false.Option D: If ( alpha perp beta ) and ( l parallel alpha ), then ( l perp beta ).- If ( alpha perp beta ) and ( l parallel alpha ), ( l ) might not be perpendicular to ( beta ). For example, a line parallel to the xy-plane (with a direction vector in the xy-plane) is not necessarily perpendicular to the xz-plane. This statement is false.Thus, the correct answer is (boxed{B}).,<think>Okay, let's try to figure out which of these statements about lines and planes is correct. I need to go through each option one by one and check their validity. Hmm, I remember some basic concepts from geometry, like if a line is parallel to a plane, it means the line doesn't intersect the plane and is going in the same direction. If a line is perpendicular to a plane, it means it meets the plane at a right angle. Also, two planes can be parallel or intersect, and if they intersect, their intersection is a line. When two planes are perpendicular, they form a dihedral angle of 90 degrees. Let me think about each statement carefully.Starting with option A: If line l is parallel to both planes α and β, then does that mean α and β are parallel? Hmm, I recall that two planes are parallel if every line in one plane is parallel to the other plane. But here, we only know that a single line l is parallel to both. Wait, but a single line can lie in multiple planes. For example, imagine two planes that intersect each other; even if a line is parallel to both, the planes themselves aren't necessarily parallel. Let me visualize this. Suppose we have two planes that intersect along some line m. If line l is parallel to both planes, then l must be parallel to line m as well. But actually, l could be parallel to the direction of the intersection line, but the two planes are still not parallel. So in that case, even though l is parallel to both α and β, the planes themselves can intersect. Therefore, statement A is not necessarily true. So A is incorrect.Moving on to option B: If line l is parallel to plane α and perpendicular to plane β, then α is perpendicular to β. Let's see. If l is perpendicular to β, that means l is orthogonal to every line in β, right? And if l is parallel to α, then l is either lying on α or is parallel to it. Wait, if a line is parallel to a plane, it can be in the plane or outside but parallel. But if l is perpendicular to β, then the direction of l is perpendicular to the normal vector of β. And if l is parallel to α, then the direction of l is parallel to α, which would mean that the normal vector of α is perpendicular to l. So if the normal vector of β is parallel to l, and the normal vector of α is perpendicular to l, then the normal vectors of α and β are perpendicular. Therefore, the planes α and β must be perpendicular. That makes sense. Let me check again. If l is parallel to α, then l's direction is in α. If l is perpendicular to β, then the normal vector of β is parallel to l. Therefore, the normal vector of β is perpendicular to α, hence the planes are perpendicular. So yes, statement B is correct. Maybe B is the answer, but let me check the other options just in case.Option C: If α is perpendicular to β and l is perpendicular to α, then l is parallel to β. Hmm. So if a line is perpendicular to plane α, then it's parallel to the normal vector of α. If α and β are perpendicular, their normal vectors are also perpendicular. So if l is parallel to the normal vector of α, and since α is perpendicular to β, then the normal vector of α is perpendicular to the normal vector of β. Therefore, l would be perpendicular to the normal vector of β. But does that mean l is parallel to β? Wait, if a line is perpendicular to the normal vector of a plane, then the line is parallel to the plane. So yes, because if a line's direction is perpendicular to the normal of the plane, then the line is parallel to the plane. Wait, but hold on. Let's think carefully. If l is perpendicular to α, then l is parallel to the normal vector of α. Since α is perpendicular to β, the normal vector of α is perpendicular to the normal vector of β. Therefore, l is parallel to a vector (normal of α) that is perpendicular to the normal of β. Hence, l is in the plane β? Or parallel to β? Wait, if the direction of l is perpendicular to the normal of β, then l is either lying on β or parallel to β. But the line l could be outside of β but parallel. So, in that case, l is parallel to β. So statement C says l is parallel to β. So that should be correct. Wait, but let me test with an example. Suppose α is the xy-plane, so its normal vector is along the z-axis. If β is the xz-plane, which is perpendicular to α (since their normals are along y and z, which are perpendicular). Now, if l is perpendicular to α, then l is along the z-axis. But the xz-plane β contains the z-axis. So in this case, l is actually lying on β, which is a special case of being parallel (since parallel includes lines that lie on the plane). Wait, but if β is the yz-plane instead, then the normal of β is along the x-axis. If α is the xy-plane (normal along z), then α and β are perpendicular. If l is perpendicular to α, it's along the z-axis. Then l is perpendicular to β (the yz-plane's normal is x-axis, z-axis is not perpendicular to x-axis... Wait, no. If β is the yz-plane, its normal is along the x-axis. Then a line along the z-axis (which is l) is actually lying on β (yz-plane). So l is lying on β, which is parallel. Alternatively, if β is some other perpendicular plane, maybe l is not lying on β but is parallel. Wait, no. If α is the xy-plane, l is the z-axis, then any plane β perpendicular to α must have a normal vector that is in the xy-plane. For example, if β is the xz-plane, normal along y-axis. Wait, no, the xz-plane's normal is along y-axis, which is perpendicular to the normal of α (z-axis). Wait, the normal vectors are perpendicular, so the planes are perpendicular. Then l, which is the z-axis, is lying on β (xz-plane). So in that case, l is in β, which is a subset of being parallel. But if β is another perpendicular plane, say, with normal vector along some direction in the xy-plane, like the x+y direction. Then the plane β would be something like x + y = 0, and its normal is (1,1,0). Then l is the z-axis. Is the z-axis parallel to β? The direction of the z-axis is (0,0,1). The plane β has normal (1,1,0), so the plane is horizontal in some way. The z-axis would intersect β at the origin, but if l is the z-axis, then it actually intersects β at the origin, so if l is supposed to be parallel to β, it shouldn't intersect. But in this case, l is the z-axis, which passes through the origin, which is on β. So in this case, l is not parallel to β, but intersects it. Wait, but in the problem statement, it says "given line l and two distinct planes α and β". So maybe l is not necessarily passing through β? Wait, but in the problem statement, it's not specified whether l is in a particular position. So if α is perpendicular to β, and l is perpendicular to α, then is l necessarily parallel to β?Wait, in the example I just thought of, the z-axis (l) is perpendicular to the xy-plane (α), and β is the plane x + y = 0, which is perpendicular to α. But l (the z-axis) is not parallel to β because it intersects β at the origin. So in this case, l is not parallel to β. Therefore, statement C is not necessarily true. Therefore, C is incorrect. So my initial thought that C was correct was wrong. That's a good catch. So the answer might not be C. Let me check this again.Another example: Let α be the xy-plane, β be the xz-plane, which are perpendicular. If l is perpendicular to α, then l is along the z-axis. Then l is lying on β (the xz-plane), which is a case of being parallel (since being in the plane is a subset of being parallel). So in that case, it works. But in the previous example, if β is the plane x + y = 0, then l (the z-axis) intersects β at the origin, so it's not parallel. Therefore, depending on the orientation of β, l can be either parallel or not. Therefore, the statement C is not necessarily true. So C is incorrect.Now, moving to option D: If α is perpendicular to β and l is parallel to α, then l is perpendicular to β. Hmm. Let's see. If l is parallel to α, then the direction of l lies within α. Since α is perpendicular to β, the normal vector of α is perpendicular to the normal vector of β. But does that mean that any line in α is perpendicular to β? Wait, for a line to be perpendicular to β, its direction must be parallel to the normal vector of β. If α and β are perpendicular, their normal vectors are perpendicular. If l is parallel to α, its direction is perpendicular to the normal vector of α. But the normal vector of β is perpendicular to the normal vector of α. Therefore, the direction of l is parallel to the normal vector of β? Wait, if the normal vectors of α and β are perpendicular, then the direction of l (which is perpendicular to the normal of α) would be parallel to the normal of β? Let's use coordinates to check.Let’s take α as the xy-plane with normal vector (0,0,1), and β as the xz-plane with normal vector (0,1,0). These are perpendicular planes. If a line l is parallel to α, then its direction vector lies in the xy-plane, say (1,1,0). Then is this line perpendicular to β? For a line to be perpendicular to β, its direction must be parallel to the normal vector of β, which is (0,1,0). But (1,1,0) is not parallel to (0,1,0). Therefore, the line l is not perpendicular to β. So statement D is not true. Therefore, D is incorrect.Wait, but what if l is along the x-axis, which is parallel to α (the xy-plane). The x-axis direction is (1,0,0). The normal vector of β (xz-plane) is (0,1,0). So (1,0,0) is perpendicular to (0,1,0)? No, the dot product is zero, so they are perpendicular. Wait, the direction of the line is (1,0,0). To check if the line is perpendicular to the plane β, we need the direction of the line to be parallel to the normal vector of β. Since the normal vector of β is (0,1,0), which is not parallel to (1,0,0). Therefore, the line is not perpendicular to β, even though it's parallel to α. So in this case, statement D is false. Therefore, D is incorrect.So reviewing all options: A is incorrect, B is correct, C is incorrect, D is incorrect. Therefore, the correct answer should be B. Let me just verify once more.Statement B: If l is parallel to α and perpendicular to β, then α is perpendicular to β. Let's use coordinates again. Let l be the z-axis, which is parallel to α (if α is, say, the xz-plane). Wait, no. If l is parallel to α, then α could be any plane that has the direction of l. Let me pick α as the xy-plane. If l is parallel to α, then l is in the xy-plane or parallel to it. Wait, if l is parallel to α, then the direction of l is in α. If l is perpendicular to β, then l is parallel to the normal vector of β. So if l is in the direction of α and also parallel to the normal vector of β, then the normal vector of β is in α. Therefore, the normal vector of β is in α, which is perpendicular to the normal vector of α. Wait, no. Let me think again. Suppose l is a line parallel to α. So the direction of l is parallel to α. If l is perpendicular to β, then the direction of l is parallel to the normal vector of β. Therefore, the normal vector of β is parallel to the direction of l, which is parallel to α. Therefore, the normal vector of β is parallel to α. Therefore, the normal vector of β is perpendicular to the normal vector of α, which would mean that α and β are perpendicular. Therefore, this seems correct. Let me take an example. Let α be the xy-plane (normal vector (0,0,1)), and let l be a line in the direction (1,0,0), which is parallel to α. If l is perpendicular to β, then the normal vector of β must be in the direction (1,0,0). Therefore, β is a plane with normal (1,0,0), like the yz-plane. Then α (xy-plane) and β (yz-plane) are perpendicular. So yes, in this case, α and β are perpendicular. Therefore, statement B is correct. Therefore, the answer is B.Final Answerboxed{B}</think>Given line ( l ) and two distinct planes ( alpha ) and ( beta ), we need to determine which of the given statements is correct.Statement A: If ( l parallel alpha ) and ( l parallel beta ), then ( alpha parallel beta ). - A line being parallel to two planes does not necessarily mean the planes are parallel. They could intersect. Hence, A is incorrect.Statement B: If ( l parallel alpha ) and ( l perp beta ), then ( alpha perp beta ). - If ( l parallel alpha ), the direction of ( l ) is parallel to ( alpha ). If ( l perp beta ), the direction of ( l ) is parallel to the normal vector of ( beta ). This implies the normal vector of ( beta ) is parallel to ( alpha ), making ( alpha ) and ( beta ) perpendicular. Hence, B is correct.Statement C: If ( alpha perp beta ) and ( l perp alpha ), then ( l parallel beta ). - If ( l perp alpha ), ( l ) is parallel to the normal vector of ( alpha ). However, this does not necessarily mean ( l ) is parallel to ( beta ). For example, the z-axis (perpendicular to the xy-plane) intersects the plane ( x + y = 0 ), so it is not parallel. Hence, C is incorrect.Statement D: If ( alpha perp beta ) and ( l parallel alpha ), then ( l perp beta ). - If ( l parallel alpha ), the direction of ( l ) is in ( alpha ). This does not imply ( l ) is perpendicular to ( beta ). For example, a line in the xy-plane is not perpendicular to the xz-plane. Hence, D is incorrect.Thus, the correct answer is (boxed{B}).]

🔑:We analyze each option separately:- For option A: If l parallel alpha and l parallel beta, then alpha and beta may either be parallel (alpha parallel beta) or intersect each other. Therefore, A is incorrect.- For option B: If l parallel alpha and l perp beta, then by the theorem of the properties of line-to-plane parallelism, there exists a plane gamma through l that intersects alpha at line m, where m parallel l and m perp beta. By the theorem for determining perpendicular planes, alpha perp beta is derived. Hence, B is correct.- For option C: If alpha perp beta and l perp alpha, then l may either be parallel to beta (l parallel beta) or contained in beta (l subset beta). Therefore, C is incorrect.- For option D: If alpha perp beta and l parallel alpha, and if l is parallel to the intersection line of alpha and beta, then l parallel beta. Therefore, D is incorrect.The correct answer is boxed{B}.

🧠:[<think>Okay, let's see. So, Xiao Long bought both Package A and Package B. Each Package A has 2 bottles of milk and a souvenir, and each Package B has 3 bottles of cola and a souvenir. He ended up with 50 bottles total (milk and cola combined) and 22 souvenirs. The question is asking how many bottles of milk he bought. First, I need to translate this into equations. Let me define variables. Let's say the number of Package A he bought is 'a', and the number of Package B he bought is 'b'. Then, each Package A gives 2 milk bottles and 1 souvenir, and each Package B gives 3 cola bottles and 1 souvenir. So, total souvenirs would be the sum of souvenirs from both packages, which is a + b. And he received 22 souvenirs, so that gives the equation:a + b = 22Then, the total number of bottles. Each Package A gives 2 milk bottles, so total milk is 2a. Each Package B gives 3 cola bottles, so total cola is 3b. The total number of bottles is 50, so:2a + 3b = 50Now, we have a system of two equations:1) a + b = 222) 2a + 3b = 50We need to solve for 'a' and 'b', and then the total milk is 2a, so once we find 'a', we can compute that.Let me solve the first equation for one variable. Let's solve for 'a':a = 22 - bThen substitute this into the second equation:2(22 - b) + 3b = 50Simplify:44 - 2b + 3b = 50Combine like terms:44 + b = 50Subtract 44 from both sides:b = 6Then, substitute back into a = 22 - b:a = 22 - 6 = 16So, he bought 16 Package A and 6 Package B. Therefore, the total milk bottles are 2a = 2 * 16 = 32.Wait, but let me check. If he bought 16 Package A, each with 2 milk bottles, that's 32 milk. Then 6 Package B, each with 3 cola, that's 18 cola. 32 + 18 = 50, which matches the total. And souvenirs: 16 + 6 = 22, which also matches. So the numbers check out. So the answer should be 32 bottles of milk. But let me just make sure I didn't make any mistakes. Let's go through the steps again.We have two equations:1) a + b = 222) 2a + 3b = 50Solving equation 1 for a: a = 22 - bPlug into equation 2:2*(22 - b) + 3b = 5044 - 2b + 3b = 5044 + b = 50b = 6Then a = 22 - 6 = 16Total milk is 2a = 32. Yes, that seems correct.Alternatively, maybe there's another way to approach this problem, like using substitution or elimination. Let me try elimination.Multiply equation 1 by 2: 2a + 2b = 44Subtract this from equation 2: (2a + 3b) - (2a + 2b) = 50 - 44Which gives: b = 6Same result. Then a = 16. So, same answer. Another way: Let's think about the difference in bottles. Each Package A gives 2 milk, each Package B gives 3 cola. If all 22 souvenirs were from Package A, he would have 22*2 = 44 milk bottles and 0 cola, total 44 bottles. But he has 50, which is 6 more. Each time we replace a Package A with a Package B, the total number of bottles increases by (3 cola - 2 milk) = 1 bottle. Because removing one Package A reduces milk by 2, adding one Package B adds 3 cola. So net change is +1 bottle. To get 6 more bottles, we need to replace 6 Package A with Package B. But wait, since he bought both types, but the initial assumption was all Package A. Wait, maybe that's another method.Wait, let's consider starting with all cola. Wait, but souvenirs are 22. So actually, the number of souvenirs is fixed at 22. So each purchase gives one souvenir, regardless of the package. Therefore, the total number of packages is 22. Each package is either A or B. So the problem is: 22 packages, each contributes either 2 milk or 3 cola, totaling 50 bottles. So, the total bottles from 22 packages is 50. Each Package A is 2, each Package B is 3. Wait, no, because cola is 3 per package. So total bottles per package: Package A gives 2, Package B gives 3. So total bottles are 2a + 3b = 50, and a + b = 22. Which is the same as before. So this is a standard system. So solving gives the same answer. Alternatively, let's think of it as a system where each "unit" (package) gives certain bottles. So the problem is similar to some classic ones, like tickets sold for adults and children. But all methods lead to the same equations, so solving gives 16 Package A, which is 32 milk bottles. So the answer is 32. Seems correct. I think that's solid.Final AnswerThe number of bottles of milk Xiao Long bought is boxed{32}.</think>Let ( a ) be the number of Package A purchased and ( b ) be the number of Package B purchased. Each Package A contains 2 bottles of milk and 1 souvenir, and each Package B contains 3 bottles of cola and 1 souvenir. We know the following:1. The total number of souvenirs received is 22: [ a + b = 22 ]2. The total number of bottles (milk and cola) is 50: [ 2a + 3b = 50 ]Solving the first equation for ( a ):[a = 22 - b]Substitute ( a ) into the second equation:[2(22 - b) + 3b = 50]Simplify:[44 - 2b + 3b = 50][44 + b = 50][b = 6]Substitute ( b = 6 ) back into the equation ( a = 22 - b ):[a = 22 - 6 = 16]The number of bottles of milk is ( 2a ):[2 times 16 = 32]Thus, the number of bottles of milk Xiao Long bought is (boxed{32}).,<think>Okay, so Xiao Long bought both Package A and Package B, right? Let me try to figure out how many bottles of milk he ended up with. The problem says that Package A has 2 bottles of milk and a souvenir, while Package B has 3 bottles of cola and a souvenir. He bought a total of 50 bottles (milk and cola combined) and received 22 souvenirs. The question is asking for the number of milk bottles he bought. First, I need to set up some variables. Let me denote the number of Package A he bought as 'a' and the number of Package B as 'b'. Each Package A gives 2 milk bottles and 1 souvenir, and each Package B gives 3 cola bottles and 1 souvenir. So, the total number of milk bottles he has is 2 times the number of Package A, which is 2a. Similarly, the total cola bottles would be 3 times the number of Package B, which is 3b. The total bottles he bought are milk and cola, so that's 2a + 3b = 50. Now, the souvenirs. Each package, whether A or B, gives 1 souvenir. So the total souvenirs would be the number of Package A plus the number of Package B. That's a + b = 22. So now I have two equations:1. 2a + 3b = 50 (total bottles)2. a + b = 22 (total souvenirs)I need to solve these two equations to find 'a' and 'b', and then use 'a' to find the total milk bottles (which is 2a). Let me solve the second equation for one variable. Let's solve for 'a': a = 22 - b. Then substitute this into the first equation.Substituting into equation 1: 2(22 - b) + 3b = 50Let me compute that:2*22 is 44, and 2*(-b) is -2b. So, 44 - 2b + 3b = 50Combine like terms: (-2b + 3b) is b. So, 44 + b = 50Subtract 44 from both sides: b = 50 - 44 = 6So, b = 6. Then, substituting back into a = 22 - b: a = 22 - 6 = 16So, he bought 16 Package A and 6 Package B.Therefore, the total milk bottles he bought is 2a = 2*16 = 32.Wait, let me check that again. If Package A is 16, then milk is 2*16=32. Package B is 6, so cola is 3*6=18. Total bottles would be 32 + 18 = 50, which matches. Souvenirs would be 16 + 6 = 22, which also matches. So that seems correct.So, the answer should be 32 bottles of milk.Final AnswerThe number of bottles of milk Xiao Long bought is boxed{32}.</think>Let ( a ) be the number of Package A purchased and ( b ) be the number of Package B purchased. Each Package A contains 2 bottles of milk and 1 souvenir, while each Package B contains 3 bottles of cola and 1 souvenir. We know the following:1. The total number of bottles (milk and cola) is 50: [ 2a + 3b = 50 ]2. The total number of souvenirs received is 22: [ a + b = 22 ]Solving the second equation for ( a ):[a = 22 - b]Substituting ( a = 22 - b ) into the first equation:[2(22 - b) + 3b = 50]Simplifying:[44 - 2b + 3b = 50][44 + b = 50][b = 6]Substituting ( b = 6 ) back into ( a = 22 - b ):[a = 22 - 6 = 16]The total number of bottles of milk is ( 2a ):[2 times 16 = 32]Thus, the number of bottles of milk Xiao Long bought is (boxed{32}).]

🔑:匹答案》32Match the answer》32

🧠:[<think>Okay, so I have this problem here with vectors. Let me see... The vectors are given as a = (1, 1, 0) and b = (-1, 0, c). First, I need to find the value of c such that the magnitude of a + b is sqrt(5). Then, there's another part where if k times a plus b is perpendicular to 2a minus b, find the real number k. Alright, let's take this step by step.Starting with the first part: |a + b| = sqrt(5). Hmm. Remember that the magnitude of a vector is calculated using the square root of the sum of the squares of its components. So first, I need to compute the vector a + b.Given a = (1, 1, 0) and b = (-1, 0, c), adding them component-wise gives:a + b = (1 + (-1), 1 + 0, 0 + c) = (0, 1, c).So the resulting vector is (0, 1, c). Now, the magnitude of this vector is sqrt(5). Let me write that out as an equation:|a + b| = sqrt(0^2 + 1^2 + c^2) = sqrt(1 + c^2).And they say this is equal to sqrt(5). So:sqrt(1 + c^2) = sqrt(5).To solve for c, square both sides:1 + c^2 = 5.Subtract 1:c^2 = 4.So c = ±2. Wait, the problem doesn't specify if c is positive or negative, so both solutions are possible? Hmm. Let me check again. The problem says "Find the value of c." But maybe there are two possible values. So c can be 2 or -2. Hmm. Let me verify.If c = 2, then a + b is (0, 1, 2), and its magnitude is sqrt(0 + 1 + 4) = sqrt(5), which is correct. Similarly, if c = -2, then a + b is (0, 1, -2), magnitude is sqrt(0 + 1 + 4) = sqrt(5) as well. So yes, both values are valid. But wait, the problem doesn't mention any constraints on c. So maybe the answer is c = ±2? But the problem says "the value of c," implying a single value. Wait, maybe I misread something. Let me check.Wait, the vectors a and b are given as a = (1,1,0) and b = (-1,0,c). The problem states |a + b| = sqrt(5). So based on calculation, c can be 2 or -2. So unless there's an additional constraint, both are acceptable. Maybe the problem expects both answers? Let me check the problem again. It says "Find the value of c." Hmm, maybe I need to present both possible values. But maybe I made a mistake. Let me double-check the calculation.Compute a + b: (1 + (-1), 1 + 0, 0 + c) = (0, 1, c). Correct. Then the magnitude squared is 0^2 + 1^2 + c^2 = 1 + c^2. Set equal to (sqrt(5))² = 5. So 1 + c² = 5 => c² = 4 => c = ±2. So yes, two solutions. The problem might be expecting both. But let me see if there's any restriction. The problem says "find the value of c". If it's from a math competition or something, sometimes they might expect both. But maybe in the original problem, c is a real number, so both are valid. Let me note that and proceed. Maybe in the answer, they need both values. I'll keep that in mind.Now, the second part: k a + b is perpendicular to 2a - b. So when two vectors are perpendicular, their dot product is zero. So I need to compute the dot product of (k a + b) and (2a - b) and set it equal to zero.First, let's compute k a + b. Let me express a and b as vectors:a = (1, 1, 0), so k a = (k, k, 0).b = (-1, 0, c). So k a + b = (k - 1, k + 0, 0 + c) = (k - 1, k, c).Similarly, 2a - b = 2*(1,1,0) - (-1,0,c) = (2 + 1, 2 - 0, 0 - c) = (3, 2, -c).Now, compute the dot product of (k - 1, k, c) and (3, 2, -c):Dot product = (k - 1)*3 + k*2 + c*(-c) = 3(k - 1) + 2k - c².Simplify:3k - 3 + 2k - c² = (3k + 2k) + (-3 - c²) = 5k - 3 - c².Set this equal to zero:5k - 3 - c² = 0.So solving for k:5k = 3 + c² => k = (3 + c²)/5.But from the first part, we found that c² = 4. Because earlier, c² = 4. Wait, c² is 4 regardless of c being 2 or -2. So c² = 4. Therefore, substitute that into the equation for k:k = (3 + 4)/5 = 7/5.So k is 7/5. So even though c can be ±2, since we square it, c² is 4 in both cases, so k is uniquely determined as 7/5.Therefore, the value of c is ±2, and k is 7/5.Wait, but the problem says "the value of c" and "the value of the real number k". So perhaps they want both values for c? Let me check again.First problem: Find the value of c such that |a + b| = sqrt(5). As we saw, c can be 2 or -2, so both are solutions. Then the second part: If k a + b is perpendicular to 2a - b, find the value of k. Since in this part, the value of c is already determined (from the first part), but wait, does the second part depend on the first part?Wait, the problem is structured as two separate questions: First, given the vectors, find c such that |a + b| = sqrt(5). Then, separately, if k a + b is perpendicular to 2a - b, find k. Wait, but is the value of c determined in the first part used in the second part? Or is the second part independent? Let me check the original problem statement again.The problem says: "Given vectors a = (1,1,0) and b = (-1,0,c), where |a + b| = sqrt(5). Find the value of c. Also, if k a + b is perpendicular to 2a - b, find the value of the real number k."Hmm. So the "where" clause seems to link the condition |a + b| = sqrt(5) to the given vectors. So in other words, first, we are given vectors a and b with a parameter c, and the condition that |a + b| = sqrt(5). Then, find c. Then, with that c, in the same problem, if k a + b is perpendicular to 2a - b, find k. So the two parts are connected. So c is first determined (to be ±2), and then using that c, find k.But in the second part, if we use c = ±2, then c² is 4, so k is (3 + 4)/5 = 7/5, as before. So regardless of whether c is 2 or -2, k is 7/5. Therefore, the answer is c = ±2 and k = 7/5.But let me verify once again if the problem is structured as two separate parts or as two parts depending on each other. The problem says: "Given vectors a = (1,1,0) and b = (-1,0,c), where |a + b| = sqrt(5). Find the value of c. Also, if k a + b is perpendicular to 2a - b, find the value of the real number k."So the "where" seems to define the vectors in the problem. So first, you have vectors a and b with |a + b| = sqrt(5). So in that setup, c is determined. Then, in the same setup (i.e., with that c), you need to find k such that k a + b is perpendicular to 2a - b. Therefore, the c is fixed first, and then k is found based on that c.Therefore, perhaps c has two possible values, and for each c, k is calculated? Wait, but when we calculated k, it only depends on c², which is 4 in either case. So even if c is 2 or -2, the k is the same. So in that case, k is uniquely 7/5 regardless of c's sign. So maybe in this problem, they just want c = ±2 and k = 7/5. So both answers are required.But in some contexts, if the problem expects numerical answers, they might want both values of c. Let me check if the problem allows multiple answers or if there's a unique solution.Alternatively, perhaps I made an error in assuming both c = 2 and c = -2 are valid. Let me check again. For the first part:a + b = (0, 1, c). The magnitude is sqrt(0 + 1 + c²) = sqrt(1 + c²) = sqrt(5). So 1 + c² = 5 => c² = 4 => c = ±2. So yes, that's correct. Both values are valid.In the second part, since k depends on c², which is 4 in both cases, so k is uniquely 7/5. Therefore, the answers are c = ±2 and k = 7/5.But let me check the second part again. If c is 2, then vector b is (-1, 0, 2). If c is -2, then vector b is (-1, 0, -2). Then, when we compute k a + b and 2a - b, does the value of k change? Let me test with c = 2 and c = -2.Case 1: c = 2.Compute k a + b = (k - 1, k, 2). Compute 2a - b = (3, 2, -2). Dot product is 3(k - 1) + 2k + 2*(-2) = 3k - 3 + 2k - 4 = 5k - 7. Set equal to zero: 5k - 7 = 0 => k = 7/5.Case 2: c = -2.Compute k a + b = (k - 1, k, -2). Compute 2a - b = (3, 2, 2). Dot product is 3(k - 1) + 2k + (-2)*2 = 3k - 3 + 2k - 4 = 5k - 7. Same equation: 5k - 7 = 0 => k = 7/5.So in both cases, k = 7/5. Therefore, regardless of the value of c, k is the same. Therefore, the answer is c = ±2 and k = 7/5.Therefore, the problem requires us to find c as ±2 and k as 7/5.But let me check once more if there's a mistake in calculation. Maybe I messed up the components.First part: Adding a and b.a is (1,1,0), b is (-1,0,c). So component-wise addition: (1 -1, 1 + 0, 0 + c) = (0,1,c). Correct. Then the magnitude squared is 0 + 1 + c² = 1 + c². Set equal to 5: 1 + c² = 5 => c² = 4 => c = ±2. Correct.Second part: Vectors k a + b and 2a - b.k a is (k, k, 0). Adding b: (k -1, k, c). 2a is (2, 2, 0). Subtracting b: (2 - (-1), 2 - 0, 0 - c) = (3, 2, -c). Dot product: (k -1)*3 + k*2 + c*(-c) = 3k -3 + 2k -c² = 5k -3 -c². Set to zero: 5k -3 -c² =0. From first part, c² =4. So 5k -3 -4=0 =>5k -7=0 =>k=7/5. All steps correct.So yes, the answer is c=±2 and k=7/5.But the problem says "Find the value of c" and "find the value of the real number k". So maybe they expect two answers: c can be 2 or -2, and k is 7/5. So in the final answer, we need to present c as ±2 and k as 7/5.Therefore, summarizing:First answer: c = ±2 (but how to write that in boxed notation? Maybe boxed{pm 2})Second answer: k = 7/5, which is boxed{dfrac{7}{5}}.But let me check the original problem again. It says "Find the value of c". If they expect both values, then we need to present both. Otherwise, maybe they accept ±2. In any case, since both are correct, I'll go with c = ±2 and k = 7/5.Final AnswerThe value of ( c ) is (boxed{pm 2}) and the value of ( k ) is (boxed{dfrac{7}{5}}).</think>Given vectors (overrightarrow{a} = (1, 1, 0)) and (overrightarrow{b} = (-1, 0, c)), where (|overrightarrow{a} + overrightarrow{b}| = sqrt{5}), we need to find the value of (c). Additionally, if (koverrightarrow{a} + overrightarrow{b}) is perpendicular to (2overrightarrow{a} - overrightarrow{b}), we need to find the value of the real number (k).First, we compute the vector (overrightarrow{a} + overrightarrow{b}):[overrightarrow{a} + overrightarrow{b} = (1 - 1, 1 + 0, 0 + c) = (0, 1, c)]The magnitude of this vector is given by:[|overrightarrow{a} + overrightarrow{b}| = sqrt{0^2 + 1^2 + c^2} = sqrt{1 + c^2}]We are told this magnitude is (sqrt{5}), so:[sqrt{1 + c^2} = sqrt{5} implies 1 + c^2 = 5 implies c^2 = 4 implies c = pm 2]Next, we need to find (k) such that (koverrightarrow{a} + overrightarrow{b}) is perpendicular to (2overrightarrow{a} - overrightarrow{b}). First, we compute the vectors (koverrightarrow{a} + overrightarrow{b}) and (2overrightarrow{a} - overrightarrow{b}):[koverrightarrow{a} + overrightarrow{b} = (k - 1, k, c)][2overrightarrow{a} - overrightarrow{b} = (2 + 1, 2 - 0, 0 - c) = (3, 2, -c)]The dot product of these vectors must be zero:[(k - 1) cdot 3 + k cdot 2 + c cdot (-c) = 0]Simplifying this expression:[3(k - 1) + 2k - c^2 = 0 implies 3k - 3 + 2k - c^2 = 0 implies 5k - 3 - c^2 = 0]From the first part, we know (c^2 = 4), so substituting this value:[5k - 3 - 4 = 0 implies 5k - 7 = 0 implies k = frac{7}{5}]Thus, the value of (c) is (boxed{pm 2}) and the value of (k) is (boxed{dfrac{7}{5}}).,<think>Okay, let's see. I need to find the value of c and then the value of k. Hmm. Let's start with the first part. The problem says that the magnitude of vector a plus vector b is equal to sqrt(5). Vector a is (1,1,0) and vector b is (-1,0,c). So first, I should compute the sum of vectors a and b.Adding vectors a and b component-wise: (1 + (-1), 1 + 0, 0 + c). Let me check that. First component: 1 + (-1) is 0. Second component: 1 + 0 is 1. Third component: 0 + c is c. So vector a + vector b is (0, 1, c). Right?Now, the magnitude of this vector is sqrt(5). The magnitude of a vector (x, y, z) is sqrt(x² + y² + z²). So in this case, the magnitude is sqrt(0² + 1² + c²) = sqrt(0 + 1 + c²) = sqrt(1 + c²). And they say that's equal to sqrt(5). So sqrt(1 + c²) = sqrt(5). If I square both sides, I get 1 + c² = 5. Then subtract 1: c² = 4. So c = ±2. Wait, the problem didn't specify if c is positive or negative. So is there any restriction on c? The problem just says "find the value of c", so maybe both are possible? But maybe in the context, there's something else. Wait, let me check the problem again."Given vectors a=(1,1,0) and b=(-1,0,c), where |a + b| = sqrt(5). Find the value of c."Hmm, it just says "c", so maybe both solutions are acceptable. But in the answer, do they want both values? Wait, but sometimes in these problems, sometimes c is a real number, so maybe both values. But let me think again. Wait, the vector b is given as (-1,0,c). The problem doesn't specify anything else about c. So mathematically, both c=2 and c=-2 satisfy the equation. So would the answer be c=±2? But maybe the problem expects both answers. Wait, the original problem says "Find the value of c." So maybe both? Let me see. Let me check my steps again.First, |a + b| = sqrt(5). Vector a + b is (0,1,c). Then magnitude is sqrt(0 + 1 + c²) = sqrt(1 + c²). Setting that equal to sqrt(5). Squared both sides: 1 + c² = 5 => c²=4 => c=2 or c=-2. Yes, that seems right. So maybe the answer is c=±2. But the problem might have more context. Wait, maybe in the next part, when they ask for k, does the value of c affect k? If c is 2 or -2, does k change? Hmm, maybe. So perhaps I need to consider both possibilities for c when solving for k? But the problem says "Find the value of the real number k". So maybe k has a unique solution regardless of c? Hmm. Let me think.Wait, but the problem is split into two parts: first, find c, then given that k a + b is perpendicular to 2a - b, find k. So maybe first find c, which is ±2, then use that c to find k. But if c can be two values, then k might also have two values. But the problem says "find the value of the real number k", implying maybe a single value. Hmm. This is confusing. Maybe I need to check if both c=2 and c=-2 lead to the same k? Let me see.Wait, perhaps I made a mistake in the first part. Let me double-check.Wait, the vector a is (1,1,0), vector b is (-1,0,c). Adding them gives (1 + (-1), 1 + 0, 0 + c) = (0,1,c). Then the magnitude squared is 0² + 1² + c² = 1 + c². So sqrt(1 + c²) = sqrt(5). Squaring gives 1 + c² = 5 => c² = 4 => c=2 or -2. So that seems correct.Then, for the second part: k a + b is perpendicular to 2a - b. So their dot product must be zero. Let's compute (k a + b) ⋅ (2a - b) = 0.First, let's express k a + b. Vector a is (1,1,0), so k a is (k, k, 0). Vector b is (-1,0,c). So k a + b is (k -1, k + 0, 0 + c) = (k -1, k, c). Similarly, 2a - b: 2a is (2, 2, 0), subtracting vector b: (2 - (-1), 2 - 0, 0 - c) = (3, 2, -c). So 2a - b is (3, 2, -c).Now, the dot product of (k -1, k, c) and (3, 2, -c) must be zero. So:(k -1)*3 + k*2 + c*(-c) = 0Let me compute each term:First term: 3(k -1) = 3k -3Second term: 2kThird term: -c²So overall: 3k -3 + 2k - c² = 0 => Combine like terms: (3k + 2k) + (-3 - c²) = 0 => 5k -3 -c² = 0Therefore: 5k = 3 + c² => k = (3 + c²)/5But from the first part, c² = 4, so c² is 4 regardless of c being 2 or -2. Therefore, k = (3 + 4)/5 = 7/5. So k is 7/5, regardless of the value of c. Therefore, even if c is 2 or -2, k is the same. So the answer for c is ±2 and k is 7/5.But let me verify once again. If I substitute c=2 into k:k = (3 + (2)^2)/5 = (3 +4)/5 =7/5. Similarly, c=-2: k = (3 + (-2)^2)/5 = same thing. So yes, k is uniquely 7/5. Therefore, the value of c is ±2, and k is 7/5. Therefore, the problem is solved. Let me check once again for any mistakes.First part: |a + b| = sqrt(5). Calculated a + b correctly as (0,1,c). Then magnitude squared is 0 +1 +c² =1 +c². Set equal to 5, so c²=4, c=±2. Correct.Second part: perpendicular vectors. Calculated k a + b = (k-1, k, c). 2a -b = (3, 2, -c). Dot product: 3(k-1) +2k + (-c)(c) = 3k -3 +2k -c² =5k -3 -c²=0. Therefore, 5k=3 +c². Since c²=4, 5k=7, so k=7/5. Correct.Therefore, the answers are c=±2 and k=7/5. So in the problem statement, the first answer is c=±2, the second answer is k=7/5.Final AnswerThe value of ( c ) is ( boxed{pm 2} ) and the value of ( k ) is ( boxed{dfrac{7}{5}} ).</think>Given vectors (overrightarrow{a} = (1, 1, 0)) and (overrightarrow{b} = (-1, 0, c)), where (|overrightarrow{a} + overrightarrow{b}| = sqrt{5}), we need to find the value of (c). Additionally, if (koverrightarrow{a} + overrightarrow{b}) is perpendicular to (2overrightarrow{a} - overrightarrow{b}), we need to find the value of (k).First, we compute the sum of the vectors (overrightarrow{a}) and (overrightarrow{b}):[overrightarrow{a} + overrightarrow{b} = (1 - 1, 1 + 0, 0 + c) = (0, 1, c)]The magnitude of this vector is given as (sqrt{5}). Therefore, we have:[sqrt{0^2 + 1^2 + c^2} = sqrt{5} implies sqrt{1 + c^2} = sqrt{5}]Squaring both sides, we get:[1 + c^2 = 5 implies c^2 = 4 implies c = pm 2]Next, we need to find the value of (k) such that (koverrightarrow{a} + overrightarrow{b}) is perpendicular to (2overrightarrow{a} - overrightarrow{b}). First, we express the vectors:[koverrightarrow{a} + overrightarrow{b} = (k - 1, k, c)][2overrightarrow{a} - overrightarrow{b} = (2 + 1, 2 - 0, 0 - c) = (3, 2, -c)]The dot product of these vectors must be zero:[(k - 1) cdot 3 + k cdot 2 + c cdot (-c) = 0]Simplifying, we get:[3(k - 1) + 2k - c^2 = 0 implies 3k - 3 + 2k - c^2 = 0 implies 5k - 3 - c^2 = 0]Since (c^2 = 4), we substitute:[5k - 3 - 4 = 0 implies 5k = 7 implies k = frac{7}{5}]Thus, the value of (c) is (boxed{pm 2}) and the value of (k) is (boxed{dfrac{7}{5}}).]

🔑: Step-by-Step SolutionPart 1: Finding the value of cGiven vectors overrightarrow{a}=(1,1,0) and overrightarrow{b}=(-1,0,c), we first find overrightarrow{a}+overrightarrow{b}:overrightarrow{a}+overrightarrow{b} = (1,1,0) + (-1,0,c) = (1-1, 1+0, 0+c) = (0,1,c).The magnitude of overrightarrow{a}+overrightarrow{b} is given by |{overrightarrow{a}+overrightarrow{b}}|=sqrt{0^2+1^2+c^2}=sqrt{1+c^2}. We know that |{overrightarrow{a}+overrightarrow{b}}|=sqrt{5}, so:sqrt{1+c^2} = sqrt{5} implies 1+c^2 = 5 implies c^2 = 4 implies c = pm 2.Therefore, c = pm 2.Part 2: Finding the value of k1. When c=2:koverrightarrow{a}+overrightarrow{b} = k(1,1,0) + (-1,0,2) = (k-1,k,2).2overrightarrow{a}-overrightarrow{b} = 2(1,1,0) - (-1,0,2) = (3,2,-2).For these vectors to be perpendicular, their dot product must be zero:(k-1,k,2) cdot (3,2,-2) = 3(k-1) + 2k - 2^2 = 0.Solving for k gives:3k - 3 + 2k - 4 = 0 implies 5k = 7 implies k = frac{7}{5}.2. When c=-2:koverrightarrow{a}+overrightarrow{b} = k(1,1,0) + (-1,0,-2) = (k-1,k,-2).2overrightarrow{a}-overrightarrow{b} = 2(1,1,0) - (-1,0,-2) = (3,2,2).Again, for these vectors to be perpendicular, their dot product must be zero:(k-1,k,-2) cdot (3,2,2) = 3(k-1) + 2k - 2^2 = 0.Solving for k in this case also gives:3k - 3 + 2k - 4 = 0 implies 5k = 7 implies k = frac{7}{5}.Conclusion:Regardless of the value of c, we find that k = frac{7}{5}. Therefore, the final answer for k is boxed{frac{7}{5}}.

🧠:[<think>Alright, let's try to figure out this problem step by step. So, we have four cars, A, B, C, and D, starting from the same point on a circular track. A and B are going clockwise, and C and D are going counterclockwise. All of them have different constant speeds. First, some key information given:1. 7 minutes after the start, A meets C for the first time, and at the same time, B meets D for the first time.2. Then, after another 46 minutes (so 53 minutes total from the start), A and B meet for the first time.3. We need to find out after how much time all four cars will meet again at the starting point for the first time.Hmm, okay. Let's break this down.First, since the cars are moving in opposite directions, their relative speed is the sum of their speeds when approaching each other. For cars moving in the same direction, the relative speed is the difference. But since all speeds are different, there might be some overtaking, but the problem mentions meetings between specific pairs at specific times.Let me consider the track length as L. But since the track is circular, maybe we can set L as 1 unit for simplicity? Because actual length might not matter, and we can deal with speeds in terms of laps per minute. Wait, but maybe not. Let's see.Alternatively, let's denote the circumference of the track as C. Then, the time it takes for two cars moving in opposite directions to meet for the first time would be C divided by the sum of their speeds. Similarly, for cars moving in the same direction, the time to meet would be C divided by the difference of their speeds. But in our problem, the first meetings between A-C and B-D happen at 7 minutes. Then A and B meet after 53 minutes.But all cars have different speeds, so we need to figure out their speeds in terms of C per minute.Wait, maybe I should set the track length to 1 unit. Let's do that. Let’s assume the circumference of the track is 1. Then, speeds can be represented as fractions of the track per minute. Let’s denote the speeds of cars A, B, C, D as v_A, v_B, v_C, v_D respectively. All speeds are different, and A, B are moving clockwise (let's take clockwise as positive direction), so their speeds are positive. C and D are moving counterclockwise, so their speeds would be negative if we take clockwise as positive. But maybe it's better to represent their speeds as absolute values and consider direction when calculating relative speed.Wait, actually, when two objects move towards each other on a circular track, their relative speed is the sum of their absolute speeds, regardless of direction. So if A is moving clockwise at speed v_A and C is moving counterclockwise at speed v_C, then their relative speed is v_A + v_C. Therefore, the time until they meet for the first time would be 1/(v_A + v_C) = 7 minutes. Similarly for B and D: 1/(v_B + v_D) = 7 minutes.Similarly, when two cars are moving in the same direction, the time between meetings is 1/|v_A - v_B|. But in our problem, A and B meet for the first time after 53 minutes. Since A and B are moving in the same direction, the time until they meet should be 1/(v_A - v_B) if v_A > v_B or 1/(v_B - v_A) if v_B > v_A. But since they meet after 53 minutes, which is longer than 7 minutes, so the relative speed must be smaller. So, the slower one is ahead, and the faster one catches up. Wait, but they all start at the same point. Hmm.Wait, hold on. If they start at the same point and move in the same direction, then the first time they meet again would be when the faster one has lapped the slower one. But since they start together, the first meeting would be when the faster one has gained a full lap on the slower one. Wait, but the track is circular, so if they go the same direction, the time to meet again would be 1/(v_faster - v_slower). But since they start at the same point, moving in the same direction, the first time they meet again is when the faster one has covered one more lap than the slower one. So the time taken would be 1/(v_faster - v_slower). So if A and B meet after 53 minutes, then |v_A - v_B| = 1/53. But since A and B are both moving clockwise, and all speeds are different, but we don't know which is faster. However, the problem states "after another 46 minutes," meaning 7 + 46 = 53 minutes from start. So the first meeting between A and B is at 53 minutes. So the time between start and first meeting is 53 minutes, so the relative speed is 1/53 per minute. Therefore, |v_A - v_B| = 1/53.But we don't know which one is faster. But since A and B are moving in the same direction, and the first meeting is after 53 minutes, which is a long time, it's possible that one is only slightly faster than the other. Let's assume that A is faster than B. Then, v_A - v_B = 1/53. If B is faster, then v_B - v_A = 1/53. But maybe we can assign variables accordingly.But let me note the equations so far:For A and C meeting at 7 minutes:1/(v_A + v_C) = 7 → v_A + v_C = 1/7For B and D meeting at 7 minutes:1/(v_B + v_D) = 7 → v_B + v_D = 1/7For A and B meeting at 53 minutes:1/|v_A - v_B| = 53 → |v_A - v_B| = 1/53So we have three equations here. But there are four variables: v_A, v_B, v_C, v_D. So we need another equation. But also, all speeds are pairwise different, which might come into play.Additionally, we need to find the time when all four cars meet again at the starting point. For that to happen, the time must be a multiple of the periods of each car. Since they start at the same point, they will meet again when each car has completed an integer number of laps. The first time this happens is the least common multiple (LCM) of their periods. The period of each car is 1/v for clockwise cars (A, B) and 1/|v| for counterclockwise cars (C, D). Since they are moving counterclockwise, their speeds are positive in the counterclockwise direction, but since we took clockwise as positive, their speeds would be negative. Wait, this might complicate things.Alternatively, let's consider the time when all four cars return to the starting point. For each car, the time must be a multiple of their individual periods. So for car A, time t must satisfy t = k_A * (1/v_A), where k_A is a positive integer. Similarly, for car B: t = k_B * (1/v_B), car C: t = k_C * (1/v_C), car D: t = k_D * (1/v_D). The first time all meet is the LCM of their individual periods. But since the periods might not be integers, finding LCM of non-integers is tricky. Instead, we can think in terms of finding the smallest t such that t is a multiple of each car's period. That is, t must satisfy t = n / v_A = m / v_B = p / v_C = q / v_D for some integers n, m, p, q. The smallest such t is the least common multiple of the periods, but expressed in terms of their speeds.Alternatively, t must be such that v_A * t, v_B * t, v_C * t, v_D * t are all integers. Because each car would have completed an integer number of laps. So we need the minimal t > 0 such that v_A * t, v_B * t, v_C * t, v_D * t are integers. This is equivalent to t being a common multiple of 1/v_A, 1/v_B, 1/v_C, 1/v_D. The minimal such t is the LCM of 1/v_A, 1/v_B, 1/v_C, 1/v_D. However, since we don't know the speeds, we need to find relations between them using the given equations.Let me summarize:1. v_A + v_C = 1/72. v_B + v_D = 1/73. |v_A - v_B| = 1/53Additionally, all speeds are different.We need to find t such that t = k / v_A = m / v_B = n / v_C = p / v_D, where k, m, n, p are integers. The minimal such t is the answer.Alternatively, since LCM is involved, perhaps the answer is LCM of the individual times it takes each car to complete a lap, but since they are all different and not necessarily integers, maybe there is a relation in the given data that allows us to find t.Alternatively, perhaps we can find the speeds in terms of fractions with a common denominator, then find LCM of denominators? Not sure. Let me think.First, let's assume the track length is 1 unit. Then, the speed is in laps per minute.From equation 1: v_A + v_C = 1/7From equation 2: v_B + v_D = 1/7From equation 3: |v_A - v_B| = 1/53Assuming v_A > v_B, then v_A - v_B = 1/53. If not, we can adjust later.So let's suppose v_A = v_B + 1/53.Now, from equations 1 and 2, we can express v_C = 1/7 - v_A and v_D = 1/7 - v_B.So, v_C = 1/7 - v_A = 1/7 - (v_B + 1/53) = 1/7 - v_B - 1/53But from equation 2, 1/7 - v_B = v_D, so v_C = v_D - 1/53Similarly, since v_C and v_D are counterclockwise speeds, they must be positive (since speed is a scalar). So, we have:v_C = v_D - 1/53 > 0 → v_D > 1/53But v_D = 1/7 - v_B, so 1/7 - v_B > 1/53 → v_B < 1/7 - 1/53 = (53 - 7)/ (7*53) = 46/371 ≈ 0.124 laps per minute.Similarly, since v_A = v_B + 1/53, and v_A must be less than 1/7 (since v_C = 1/7 - v_A > 0), so v_A < 1/7 → v_B + 1/53 < 1/7 → v_B < 1/7 - 1/53 ≈ 0.1428 - 0.0189 ≈ 0.1239 laps per minute, which matches the above.So, we have:v_A = v_B + 1/53v_D = 1/7 - v_Bv_C = v_D - 1/53 = 1/7 - v_B - 1/53Now, since all speeds must be positive:1. v_B > 02. v_A = v_B + 1/53 < 1/7 (from v_C = 1/7 - v_A > 0)3. v_D = 1/7 - v_B > 0 → v_B < 1/74. v_C = 1/7 - v_B - 1/53 > 0 → v_B < 1/7 - 1/53 = (53 - 7)/ (7*53) = 46/371 ≈ 0.124So, combining these, we have 0 < v_B < 46/371 ≈ 0.124.Similarly, all speeds must be different. Since v_A = v_B + 1/53, and v_D = 1/7 - v_B, and v_C = 1/7 - v_B - 1/53, so all these are determined by v_B. So we need to ensure that all four speeds are different. Let's check:Given that v_A = v_B + 1/53v_C = 1/7 - v_A = 1/7 - v_B - 1/53v_D = 1/7 - v_BSo, v_A, v_B, v_C, v_D are all different if v_B is chosen such that none of these expressions equal each other.Since v_A = v_B + 1/53, and v_D = 1/7 - v_B. So, if v_A = v_D, then:v_B + 1/53 = 1/7 - v_B → 2v_B = 1/7 - 1/53 = (53 - 7)/371 = 46/371 → v_B = 23/371 ≈ 0.062Similarly, if v_A = v_C:v_B + 1/53 = 1/7 - v_B - 1/53 → 2v_B + 2/53 = 1/7 → 2v_B = 1/7 - 2/53 = (53 - 14)/371 = 39/371 → v_B = 39/(2*371) ≈ 0.0525If v_B = 23/371, then v_A = 23/371 + 1/53 = 23/371 + 7/371 = 30/371, and v_D = 1/7 - 23/371 = 53/371 - 23/371 = 30/371. So v_A = v_D in this case, which would make two cars have the same speed, which is not allowed. Therefore, v_B cannot be 23/371.Similarly, if v_B = 39/(2*371) ≈ 0.0525, then v_A = 39/(2*371) + 1/53 = 39/(742) + 14/742 = 53/742 = 1/14, and v_C = 1/7 - 1/14 - 1/53 = 1/14 - 1/53 = (53 - 14)/742 = 39/742. So v_A = 1/14, v_B ≈ 0.0525, v_C = 39/742 ≈ 0.0525, which would make v_B = v_C, which is not allowed. Therefore, v_B cannot be 39/(2*371).Therefore, as long as v_B is chosen such that it's not 23/371 or 39/(2*371), all speeds are different. But since the problem states that all speeds are pairwise different, we can assume that such a v_B exists in the allowed range.So, now, the problem is to find the minimal t such that t is a multiple of 1/v_A, 1/v_B, 1/v_C, 1/v_D. That is, t must satisfy:v_A * t is an integer,v_B * t is an integer,v_C * t is an integer,v_D * t is an integer.So, t must be a common multiple of 1/v_A, 1/v_B, 1/v_C, 1/v_D. The minimal such t is the LCM of these four periods. But since we don't have numerical values, maybe we can express t in terms of the given data.Alternatively, let's try to express all speeds in terms of fractions.Let me note that 1/7 and 1/53 are involved. Let me see:From the equations:v_A + v_C = 1/7v_B + v_D = 1/7v_A - v_B = 1/53Assuming v_A > v_BTherefore, we can set up variables as follows:Let’s denote:v_A = v_B + 1/53Then,v_C = 1/7 - v_A = 1/7 - v_B - 1/53v_D = 1/7 - v_BSo, all speeds are expressed in terms of v_B.Now, to find t such that:v_A * t = integer,v_B * t = integer,v_C * t = integer,v_D * t = integer.So substituting:(v_B + 1/53) * t = integerv_B * t = integer(1/7 - v_B - 1/53) * t = integer(1/7 - v_B) * t = integerLet’s denote k = v_B * t. Then, from the second equation, k is integer.From the first equation: (v_B + 1/53) * t = k + t/53 = integer. Therefore, t/53 must be an integer. Let’s denote m = t/53, so m is integer. Therefore, t = 53m.From the fourth equation: (1/7 - v_B) * t = (1/7 - v_B) * 53m = (53m/7 - 53m v_B) = (53m/7 - 53k). Since this must be integer, 53m/7 must be integer because 53k is integer (k is integer). Therefore, 53m must be divisible by 7. Since 53 and 7 are coprime (53 is prime, 7 is prime, 53 ≠ 7), so m must be a multiple of 7. Let m = 7n. Then, t = 53 * 7n = 371n.Now, substituting back, m = 7n, so t = 371n.Then, k = v_B * t = v_B * 371n must be integer. Similarly, from the third equation: (1/7 - v_B - 1/53) * t = (1/7 - v_B - 1/53) * 371n.Let’s compute each term:First, compute k = v_B * 371n. We need k integer. Since n is integer, we need v_B * 371 to be rational, which it is, as v_B is a speed derived from the problem which likely gives rational numbers. But let's see.From equation 3: v_A - v_B = 1/53. So v_A = v_B + 1/53.From equation 1: v_A + v_C = 1/7. So v_C = 1/7 - v_A = 1/7 - v_B - 1/53.Similarly, v_D = 1/7 - v_B.So, substituting t = 371n into the third equation:(1/7 - v_B - 1/53) * 371n = integer.Compute 1/7 - v_B - 1/53:Convert to common denominator, which is 7*53=371.1/7 = 53/3711/53 = 7/371So, 1/7 - 1/53 = 53/371 - 7/371 = 46/371Therefore, 1/7 - v_B - 1/53 = 46/371 - v_BThus, (46/371 - v_B) * 371n = (46 - 371v_B) * nThis must be integer. Since n is integer, we need (46 - 371v_B) to be integer. Because 46 is integer, 371v_B must be integer. Let’s denote 371v_B = integer. Let’s call that integer p. So, p = 371v_B. Then, v_B = p/371.Similarly, from k = v_B * 371n = p/371 * 371n = pn. Since k must be integer, and n is integer, p must be integer. So, p is an integer, and v_B = p/371.From the earlier constraints on v_B:0 < v_B < 46/371 ≈ 0.124Thus, 0 < p/371 < 46/371 → 0 < p < 46. So p can be 1, 2, ..., 45. Also, since all speeds must be different, p must be chosen such that v_A, v_B, v_C, v_D are all different.But given that v_A = v_B + 1/53 = p/371 + 1/53 = p/371 + 7/371 = (p + 7)/371v_C = 46/371 - p/371 = (46 - p)/371v_D = 1/7 - v_B = 53/371 - p/371 = (53 - p)/371So, the speeds are:v_A = (p + 7)/371v_B = p/371v_C = (46 - p)/371v_D = (53 - p)/371All these must be distinct positive numbers.So:1. (p + 7)/371 > 0 → p + 7 > 0 → p > -7, which is already satisfied since p is positive.2. p/371 > 0 → p > 03. (46 - p)/371 > 0 → 46 - p > 0 → p < 464. (53 - p)/371 > 0 → 53 - p > 0 → p < 53But since p < 46 from above, this is satisfied.Also, all speeds must be distinct:So, (p + 7) ≠ p → 7 ≠ 0, which is true.(p + 7) ≠ (46 - p) → p + 7 ≠ 46 - p → 2p ≠ 39 → p ≠ 19.5. Since p is integer, p ≠ 19.5, so okay.(p + 7) ≠ (53 - p) → p + 7 ≠ 53 - p → 2p ≠ 46 → p ≠ 23.p/371 ≠ (46 - p)/371 → p ≠ 46 - p → 2p ≠ 46 → p ≠ 23.p/371 ≠ (53 - p)/371 → p ≠ 53 - p → 2p ≠ 53 → p ≠ 26.5. Since p is integer, okay.(46 - p)/371 ≠ (53 - p)/371 → 46 - p ≠ 53 - p → 46 ≠ 53, which is true.So, the only possible duplicate would be if p = 23, which would make v_A = 30/371, v_D = 53 - 23 = 30/371. So, v_A = v_D if p = 23, which is not allowed. Therefore, p ≠ 23.Similarly, if p = 19.5, but p must be integer. So, p can be from 1 to 45, excluding 23.Therefore, p is integer, 1 ≤ p ≤ 45, p ≠ 23.Therefore, for some integer p (1 ≤ p ≤45, p≠23), we have the speeds as above.Now, returning to the requirement that (46 - 371v_B) must be integer. Since v_B = p/371, then 46 - 371*(p/371) = 46 - p must be integer, which it is because p is integer.Therefore, (46 - p)*n must be integer. Since n is integer, and p is integer, (46 - p) is integer, so this is satisfied.Therefore, the third equation gives us no new constraints. So, all conditions are satisfied as long as p is integer between 1 and 45, p ≠23.Therefore, t must be 371n, where n is integer, and the minimal t is when n=1, so t=371 minutes. Wait, but that seems very large. 371 minutes is over 6 hours. The problem states "after how much time from the start of the race will all the cars meet for the first time?".But maybe that's the answer? But let's check.Wait, if t = 371 minutes, then each car would have completed:- Car A: v_A * t = (p + 7)/371 * 371 = p + 7 laps- Car B: v_B * t = p/371 * 371 = p laps- Car C: v_C * t = (46 - p)/371 * 371 = 46 - p laps- Car D: v_D * t = (53 - p)/371 * 371 = 53 - p lapsAll integers, so yes, they all meet at the starting point after 371 minutes.But is 371 the minimal such t? Since we set n=1, and t=371n, n=1 is minimal. Because if we take n=1, then t=371. If there were a smaller t, say t= k < 371, then k must be a multiple of 53 and 7, from earlier steps. Because t must be multiple of 53 (from t=53m) and m must be multiple of 7 (from 53m/7 must be integer). Therefore, the minimal such t is LCM(53,7)= 53*7=371. Therefore, 371 is indeed the minimal time.But let me verify with an example. Suppose p=1 (smallest possible):Then,v_A = (1+7)/371 = 8/371 laps per minutev_B = 1/371 laps per minutev_C = (46 -1)/371 = 45/371 laps per minutev_D = (53 -1)/371 = 52/371 laps per minuteThen, t=371 minutes:A completes 8/371 * 371 = 8 lapsB completes 1 lapC completes 45 lapsD completes 52 lapsAll integers, so they meet at the start.Similarly, if p=22 (close to 23, which is excluded):v_A = 29/371v_B =22/371v_C=24/371v_D=31/371t=371 minutes:A:29, B:22, C:24, D:31 laps. All integers.Therefore, regardless of p (as long as it's valid), t=371 minutes is the minimal time when all meet again.But let's check if there is a smaller t. Suppose t is 371/2 minutes. Then:For car A: v_A * t = (p +7)/371 * 371/2 = (p +7)/2 laps. To be integer, (p +7) must be even. But since p can be any integer from 1-45 except 23, if p is odd, then p +7 is even if p is odd. For example, p=1: 1+7=8 even. So t=371/2=185.5 minutes, A completes 8/2=4 laps. But for car B: v_B * t = p/371 *371/2= p/2. If p is even, then p/2 is integer. But p can be even or odd. If p is even and p +7 is even, then t=185.5 would work for A and B. But for C and D:v_C * t = (46 - p)/371 *371/2 = (46 -p)/2. So 46 -p must be even. If p is even, 46 -p is even. Similarly, v_D * t = (53 -p)/2. 53 -p must be even. If p is odd, then 53 -p is even. Therefore, if p is even:- p even- p +7 odd → A would have (p +7)/2 which is not integerContradiction.If p is odd:- p odd- p +7 even → A: (p +7)/2 integer- B: p/2 not integer- C: (46 -p)/2, since p odd, 46 -p is odd → not integer- D: (53 -p)/2, since p odd, 53 -p is even → integerSo, for p odd: only A and D would have integer laps at t=185.5, but B and C would not. Therefore, t=185.5 is invalid.Similarly, t=371/3 ≈123.666 minutes:For car A: (p +7)/3. Needs p +7 ≡0 mod3.Car B: p/3. Needs p ≡0 mod3.Car C: (46 -p)/3. Needs 46 -p ≡0 mod3 → p ≡1 mod3 (since 46 ≡1 mod3).Car D: (53 -p)/3. 53≡2 mod3, so 2 -p ≡0 mod3 → p≡2 mod3.But if p≡0 mod3 (from B), and p≡1 mod3 (from C), contradiction. So impossible.Therefore, no t smaller than 371 satisfies all four cars meeting. Therefore, the minimal time is 371 minutes.But wait, the problem says "after how much time from the start of the race will all the cars meet for the first time?" So 371 minutes is the answer? That's more than 6 hours. But in the problem, the first meetings happen at 7 minutes and 53 minutes. But maybe that's correct because with all different speeds, it takes much longer for all to align again.Alternatively, perhaps there is a mistake in assuming LCM(53,7)=371. Let's check that.Yes, LCM(53,7)=53*7=371, since they are primes. Therefore, if the periods are 53 and 7, LCM is 371. But in our problem, the key equations led us to t=371 minutes as the minimal common multiple. So, unless there is a different interpretation, 371 seems correct.Wait, but in the problem, we have two cars going clockwise and two counterclockwise. Maybe there is a different way they could meet earlier? But the question specifically states "all the cars meet for the first time". So, all four cars must be at the starting point simultaneously. Even if some cars meet each other earlier at other points on the track, the first time they all meet again at the starting point would be LCM of their individual periods.But given that their periods are 1/v_A, 1/v_B, 1/v_C, 1/v_D. But since we expressed their speeds as fractions with denominator 371, their periods are 371/(p+7), 371/p, 371/(46 - p), 371/(53 - p). Therefore, the LCM of these four numbers. However, since p can vary, but LCM must be same for all p?Wait, no. Wait, but earlier, we found that t=371 is the minimal time when t is multiple of 1/v_A, etc., because when t=371, then each car completes integer laps. However, depending on p, the LCM could be larger. Wait, but in our derivation, we found that t=371 works for any p, because:v_A * t = (p +7)/371 *371 = p +7v_B * t = pv_C * t =46 -pv_D * t =53 -pAll integers. So, regardless of p (as long as it's valid), t=371 makes all cars at the starting point. Therefore, t=371 is indeed the minimal such time, because for any p, t=371 works, and any smaller t does not satisfy the divisibility conditions.Therefore, the answer is 371 minutes, which is 371 minutes after the start.But let me check the problem statement again. It says "after how much time from the start of the race will all the cars meet for the first time?"So, 371 minutes? That's 6 hours and 11 minutes. Seems like a lot, but given the different speeds, and the conditions given, it might be the case.Alternatively, perhaps there is a different approach.Let me think differently. Suppose the time when all four meet is t minutes. Then, in t minutes, each car would have traveled a distance equal to their speed multiplied by t. Since the track is circular, they meet at the starting point when the distance is a multiple of the track length (let’s say track length is 1). So, for each car, speed * t must be integer.Therefore, we need t such that:v_A * t ∈ ℤv_B * t ∈ ℤv_C * t ∈ ℤv_D * t ∈ ℤTherefore, t must be a common multiple of 1/v_A, 1/v_B, 1/v_C, 1/v_D. The minimal such t is the LCM of these four numbers.From earlier equations:v_A + v_C = 1/7v_B + v_D = 1/7v_A - v_B = 1/53Assuming v_A > v_B.From these, we can express all speeds in terms of v_B.v_A = v_B + 1/53v_C = 1/7 - v_A = 1/7 - v_B - 1/53v_D = 1/7 - v_BSo, LCM of 1/v_A, 1/v_B, 1/v_C, 1/v_D.But how can we compute LCM without knowing v_B? However, as shown earlier, if we set t=371, then:v_A * t = (v_B + 1/53)*371 = v_B*371 + 371/53 = v_B*371 +7But since v_B*371 = p (from earlier), so p +7, which is integer.Similarly, v_B*t = pv_C*t = (1/7 - v_B -1/53)*371 = (46/371 - v_B)*371 =46 -371v_B =46 -pv_D*t =53 -pSo all are integers. Hence, t=371 is indeed the minimal such time.Therefore, the answer is 371 minutes, which is 6 hours and 11 minutes. So, boxed, that would be boxed{371}Final Answerboxed{371}</think>Given four cars (A, B, C,) and (D) starting from the same point on a circular track, with (A) and (B) driving clockwise and (C) and (D) driving counterclockwise, all at constant but different speeds. The key information provided is:1. (A) meets (C) and (B) meets (D) for the first time 7 minutes after the start.2. (A) and (B) meet for the first time 53 minutes after the start.We need to determine the first time all four cars meet again at the starting point. Key Steps:1. Relative Speeds and Meeting Times: - For cars moving in opposite directions, the relative speed is the sum of their speeds. The first meetings at 7 minutes give: [ v_A + v_C = frac{1}{7} quad text{and} quad v_B + v_D = frac{1}{7} ] - For cars moving in the same direction, the relative speed is the difference of their speeds. The meeting of (A) and (B) at 53 minutes gives: [ |v_A - v_B| = frac{1}{53} ]2. Expressing Speeds: - Assuming (v_A > v_B), we have (v_A = v_B + frac{1}{53}). - Substituting into the equations for (v_C) and (v_D): [ v_C = frac{1}{7} - v_A = frac{1}{7} - v_B - frac{1}{53} ] [ v_D = frac{1}{7} - v_B ]3. Finding the Least Common Multiple (LCM): - The time (t) when all cars meet again must be a multiple of the periods of each car. The periods are (frac{1}{v_A}, frac{1}{v_B}, frac{1}{v_C}, frac{1}{v_D}). - Converting speeds to a common denominator (371), we find that the minimal (t) satisfying all conditions is the LCM of 53 and 7, which are primes.4. Conclusion: - The LCM of 53 and 7 is (53 times 7 = 371). Therefore, the minimal time when all cars meet again at the starting point is 371 minutes.[boxed{371}],<think>Okay, let's try to tackle this problem step by step. So, we have four cars, A, B, C, and D, starting from the same point on a circular track. A and B are going clockwise, while C and D are going counterclockwise. They all have different constant speeds. First, the problem says that exactly 7 minutes after the start, A meets C for the first time, and at the same moment, B meets D for the first time. Then, after another 46 minutes (so 53 minutes total), A and B meet for the first time. The question is: after how much time from the start will all four cars meet again for the first time?Hmm. Alright, let's start by recalling some concepts about circular motion and relative speed. When two objects are moving in opposite directions around a circular track, their relative speed is the sum of their individual speeds. When moving in the same direction, their relative speed is the difference. Since A and B are moving clockwise, and C and D are moving counterclockwise, A and C are moving towards each other, as are B and D. Their first meeting after 7 minutes suggests that the time it takes for them to meet is determined by the length of the track divided by their combined speeds. Similarly, when A and B meet after 53 minutes, since they're moving in the same direction, the time taken to meet would be related to the difference in their speeds.Let me denote the circumference of the track as L. Let's assign speeds to each car: let’s say the speed of car A is v_A, B is v_B, C is v_C, and D is v_D. All speeds are in units per minute, and since they are constant but different, all v's are distinct.First, when A and C meet after 7 minutes, the combined distance they've covered equals one full lap (L). So,(v_A + v_C) * 7 = LSimilarly, when B and D meet after 7 minutes,(v_B + v_D) * 7 = LTherefore, both pairs (A,C) and (B,D) have combined speeds such that their sum multiplied by 7 minutes equals L. Therefore, (v_A + v_C) = (v_B + v_D) = L / 7.Then, after another 46 minutes (so 53 minutes total), A and B meet for the first time. Since A and B are moving in the same direction, the time taken for them to meet is determined by the difference in their speeds. The time it takes for A to catch up to B is when A has covered an extra lap compared to B. So,(v_A - v_B) * 53 = LThus, (v_A - v_B) = L / 53.So, summarizing:1. (v_A + v_C) = L / 72. (v_B + v_D) = L / 73. (v_A - v_B) = L / 53We need to find the time T when all four cars meet again at the starting point. That would be the least common multiple (LCM) of their individual periods to complete a lap around the track. However, since they start at the same point, we need T to be the smallest time such that each car has completed an integer number of laps, i.e., T is a multiple of L / v_A, L / v_B, L / v_C, and L / v_D. Therefore, T is the LCM of the periods L / v_A, L / v_B, L / v_C, L / v_D.But since L is common, we can factor it out. Let me denote the time for each car to complete a lap as T_A = L / v_A, T_B = L / v_B, T_C = L / v_C, T_D = L / v_D. Then, T must be the LCM of T_A, T_B, T_C, T_D.But since L is the circumference, perhaps we can normalize the speeds by letting L = 1 for simplicity. Then, the equations become:1. (v_A + v_C) = 1 / 72. (v_B + v_D) = 1 / 73. (v_A - v_B) = 1 / 53Therefore, we can express v_C = 1/7 - v_A and v_D = 1/7 - v_B.Also, from equation 3, v_A = v_B + 1/53.So, substituting into equation 1: v_A + (1/7 - v_A) = 1/7, which checks out. Similarly for equation 2.But we need to find LCM of 1/v_A, 1/v_B, 1/(1/7 - v_A), 1/(1/7 - v_B). Since all speeds are different, these periods will all be different. Hmm, this seems complicated. Maybe there's another approach.Alternatively, perhaps all four cars meet when the distance each has traveled is an integer multiple of the track length. So, the time T must satisfy that v_A * T is integer, v_B * T is integer, v_C * T is integer, and v_D * T is integer. But since they start at the same point, meeting again at the starting point requires that T is a multiple of each car's period. Therefore, LCM of periods, as before.But since the problem says "after how much time from the start of the race will all the cars meet for the first time?" So, we need the first time T when all four are back at the starting point. So, LCM(T_A, T_B, T_C, T_D).But since we don't have values for the speeds, maybe we can express T in terms of the given data (7 minutes, 46 minutes, 53 minutes total for A and B to meet). Let's see.From the above equations:v_A + v_C = 1/7v_B + v_D = 1/7v_A - v_B = 1/53So, let me solve for the variables. Let's express everything in terms of v_A and v_B.From equation 3, v_A = v_B + 1/53.From equation 1, v_C = 1/7 - v_A = 1/7 - v_B - 1/53.From equation 2, v_D = 1/7 - v_B.Therefore, all speeds can be written in terms of v_B.So, let's write:v_A = v_B + 1/53v_C = (1/7 - 1/53) - v_Bv_D = 1/7 - v_BTherefore, all four speeds are expressed in terms of v_B. Now, we need to find LCM(T_A, T_B, T_C, T_D), where T_X = 1 / v_X (assuming L=1). Wait, but if L=1, then time to complete a lap is 1 / v_X. So, T_A = 1 / v_A, etc.But to compute LCM(T_A, T_B, T_C, T_D), which would be LCM(1 / v_A, 1 / v_B, 1 / v_C, 1 / v_D). However, LCM is generally for integers. For rational numbers, we can express them as fractions and take LCM of the numerators divided by GCD of denominators. But this might get complicated.Alternatively, perhaps we can model the problem in terms of their angular positions. Let’s consider the track as a unit circle (circumference 1). Let’s define the position of each car as a function of time, modulo 1.For car A: position at time t is (v_A * t) mod 1 (clockwise, so increasing position)For car B: (v_B * t) mod 1For car C: since counterclockwise, position can be represented as (-v_C * t) mod 1, or equivalently (1 - v_C * t mod 1)Similarly for car D: (-v_D * t) mod 1They all start at position 0. So, meeting at the starting point means that all their positions modulo 1 are 0. So, v_A * t ≡ 0 mod 1, v_B * t ≡ 0 mod 1, v_C * t ≡ 0 mod 1, v_D * t ≡ 0 mod 1.Which implies that t must be a multiple of 1 / v_A, 1 / v_B, 1 / v_C, and 1 / v_D. So, the minimal such t is the LCM of 1 / v_A, 1 / v_B, 1 / v_C, 1 / v_D.But with the relations we have from the problem, maybe we can find such t.But since all speeds are expressed in terms of v_B, perhaps we can find expressions for 1 / v_A, 1 / v_B, etc., in terms of v_B, then find t such that t is multiple of all four.But this seems a bit abstract. Let's see if we can find more relations.We have:v_A = v_B + 1/53v_C = (1/7 - 1/53) - v_B = (53 - 7)/ (7*53) - v_B = 46 / 371 - v_Bv_D = 1/7 - v_BSo, speeds:v_A = v_B + 1/53v_B = v_Bv_C = 46/371 - v_Bv_D = 1/7 - v_BNote that 1/7 is 53/371, and 1/53 is 7/371, so maybe converting everything to denominator 371.So:v_A = v_B + 7/371v_C = 46/371 - v_Bv_D = 53/371 - v_BSo, in terms of v_B, let's represent v_B as x/371, where x is some integer? Wait, but speeds don't have to be rational numbers. Hmm, but given that all the meeting times are integers (7, 53), maybe the speeds are rational?Assuming that the track length L is 1 unit, then speeds would be in units per minute. The meeting after 7 minutes implies that (v_A + v_C)*7 = 1, so v_A + v_C = 1/7, which is rational. Similarly, v_B + v_D = 1/7, and v_A - v_B = 1/53. Therefore, if we assume that all speeds are rational numbers, then LCM would make sense.Alternatively, maybe L is 371*53 units? To make all the denominators go away. Let me see.But perhaps I can set L = 371*53 to make the arithmetic easier. Wait, 371 is 53*7, right? 53*7 = 371. Yes, because 50*7=350, 3*7=21, so 350+21=371. So 53*7=371. Therefore, 1/7 of L is 53, and 1/53 of L is 7.If we set L=371*53, then 1/7 of L is 53*53=2809, and 1/53 of L is 7*371=2597. Hmm, maybe not helpful. Wait, maybe setting L=371?Wait, if L=371, then 1/7 of L is 53, and 1/53 of L is 7. That might be helpful. Let's try that.Let’s set L=371 units. Then:(v_A + v_C) * 7 = 371 => v_A + v_C = 371 / 7 = 53 units per minute.Similarly, (v_B + v_D) *7 = 371 => v_B + v_D = 53.(v_A - v_B) *53 = 371 => v_A - v_B = 371 /53 = 7.So, in this case, speeds are in units per minute, L=371.So, from the third equation, v_A = v_B + 7.From the first equation, (v_B +7) + v_C =53 => v_C =53 -v_B -7=46 -v_B.From the second equation, v_B + v_D=53 => v_D=53 -v_B.Therefore, we have:v_A = v_B +7v_C =46 -v_Bv_D=53 -v_BNow, all speeds must be positive and distinct. Since all speeds are different, and L=371.Let’s check the constraints:1. v_A = v_B +7 >0. Since v_B must be positive, so v_B >0.2. v_C =46 -v_B >0 => v_B <46.3. v_D=53 -v_B >0 => v_B <53.So, combining these, v_B must be in (0,46) to keep all speeds positive. Also, since all speeds are different:v_A = v_B +7, which is different from v_B. Then, v_C=46 -v_B and v_D=53 -v_B must be different from each other and from v_A, v_B.So, 46 -v_B ≠53 -v_B → which is always true. So, v_C ≠v_D.Also, v_C=46 -v_B ≠v_A= v_B +7 → 46 -v_B ≠v_B +7 →46 -7=2v_B →39=2v_B →v_B=19.5. So, unless v_B=19.5, v_C ≠v_A. Similarly, v_D=53 -v_B ≠v_A= v_B +7 →53 -v_B ≠v_B +7 →53 -7=2v_B →46=2v_B →v_B=23. So, unless v_B=23, v_D≠v_A. Similarly, check if v_C or v_D equal v_B or v_A.v_C=46 -v_B. For v_C to equal v_B: 46 -v_B =v_B →46=2v_B →v_B=23. But if v_B=23, then v_D=53 -23=30, and v_A=23 +7=30. So, in this case, v_A and v_D would both be 30, which violates the "pairwise different" speeds. Hence, v_B cannot be 23.Similarly, if v_C=46 -v_B= v_A= v_B +7 →46 -v_B =v_B +7 → same as before, 39=2v_B →v_B=19.5. If v_B=19.5, then v_A=26.5, v_C=26.5, so v_A and v_C would be the same, which is not allowed. Therefore, v_B cannot be 19.5.Similarly, v_D=53 -v_B. For v_D to equal v_B:53 -v_B =v_B →53=2v_B →v_B=26.5. Then, v_A=26.5 +7=33.5, v_C=46 -26.5=19.5, v_D=53 -26.5=26.5. So, v_D and v_B would both be 26.5, which is again a conflict. So, v_B cannot be 26.5.Therefore, as long as v_B is not 19.5,23,26.5, all speeds are distinct. Given that the problem states that all speeds are pairwise different, we can assume that v_B is chosen such that none of these equalities occur. Therefore, we can proceed.Now, we need to compute the LCM of the periods of each car. The period for each car is L divided by its speed. Since we set L=371, the periods are:T_A = 371 / v_A = 371 / (v_B +7)T_B = 371 / v_BT_C = 371 / v_C = 371 / (46 -v_B)T_D = 371 / v_D = 371 / (53 -v_B)So, LCM(T_A, T_B, T_C, T_D) = LCM(371/(v_B +7), 371/v_B, 371/(46 -v_B), 371/(53 -v_B))Since 371 is common in all terms, we can factor that out:371 * LCM(1/(v_B +7), 1/v_B, 1/(46 -v_B), 1/(53 -v_B))But LCM of reciprocals is the reciprocal of the GCD of the denominators. Wait, LCM of fractions is a bit tricky. Let me recall that LCM(a/b, c/d) = LCM(a,c)/GCD(b,d). But here, all terms are 1 over something. So, LCM(1/x, 1/y, 1/z, 1/w) is 1/GCD(x,y,z,w). Wait, is that true?Wait, for example, LCM(1/2, 1/3) is 1, since 1 is the smallest number where 1/2 and 1/3 divide into integers. Wait, but LCM is typically defined for integers. Maybe in this context, LCM of periods (which are real numbers) is the smallest positive real number T such that T is a multiple of each period. So, for periods T1, T2, T3, T4, LCM(T1, T2, T3, T4) is the smallest T such that T = k1*T1 = k2*T2 = k3*T3 = k4*T4 for integers k1, k2, k3, k4.But in this problem, we need T such that T is a multiple of all four periods. Since periods are 371/(v_B +7), 371/v_B, etc., T must be a multiple of each. Therefore, T must be the least common multiple of these four periods.But how do we compute that? Since we have expressions in terms of v_B, which is a variable here. Wait, but maybe the answer is 371 minutes? Because L=371, if all speeds are 1 unit per minute, they would meet after 371 minutes. But the speeds here are different. Wait, but L=371 was arbitrary. Wait, but we set L=371 to make the numbers work out. Hmm.Alternatively, maybe the LCM is 371 minutes. Let's check.If T = 371 minutes, then each car would have completed:- Car A: v_A * T = (v_B +7)*371 /371 = v_B +7 laps. So, integer.Wait, no. Wait, the number of laps each car completes is speed * time / L. Since speed is in units per minute, time is T, and L is 371. So, number of laps is (v_X * T)/371.So, for T=371, number of laps is (v_X * 371)/371 = v_X. So, unless v_X is integer, it's not an integer number of laps.But from earlier, we have:v_A +v_C =53, which are in units per minute. Since L=371, to get the number of laps per minute, it's v_X /371. Wait, maybe I confused units here.Wait, perhaps setting L=1 is better. Let me try again.Let’s take L=1. Then the equations are:(v_A + v_C) *7 =1 →v_A +v_C=1/7(v_B + v_D)*7=1 →v_B +v_D=1/7(v_A -v_B)*53=1 →v_A -v_B=1/53So, solving:v_A =v_B +1/53v_C =1/7 -v_A=1/7 -v_B -1/53v_D=1/7 -v_BSo, speeds:v_A = v_B +1/53v_C = (1/7 -1/53) -v_B= (53 -7)/ (7*53) -v_B=46/371 -v_Bv_D=1/7 -v_B=53/371 -v_BSo, same as before.Then, the periods for each car (time to complete a lap) are:T_A=1/v_A=1/(v_B +1/53)T_B=1/v_BT_C=1/v_C=1/(46/371 -v_B)T_D=1/v_D=1/(53/371 -v_B)We need to find the minimal T such that T is a multiple of T_A, T_B, T_C, T_D. That is,T = k_A * T_A = k_B * T_B = k_C * T_C = k_D * T_D, where k_A, k_B, k_C, k_D are integers.But since T must be the same for all, we can write:k_A * T_A = k_B * T_B = k_C * T_C = k_D * T_D = TBut this seems too abstract. Maybe another approach.Alternatively, perhaps all cars meet at the starting point when the number of minutes passed is a multiple of each car's period. So, T must satisfy:v_A * T ≡0 mod 1v_B * T ≡0 mod1v_C * T ≡0 mod1v_D * T ≡0 mod1Which means that T must be such that:T * v_A is integer,T * v_B is integer,T * v_C is integer,T * v_D is integer.Therefore, T must be a common multiple of 1/v_A, 1/v_B, 1/v_C, 1/v_D. So, LCM(1/v_A, 1/v_B, 1/v_C, 1/v_D).But 1/v_A, etc., are the periods. So, same as LCM of periods.But how do we compute LCM of these four periods given the relations between the speeds?Alternatively, since we have expressions for each speed in terms of v_B, maybe we can express T in terms of v_B and find minimal T.Let’s denote T as LCM(1/(v_B +1/53), 1/v_B, 1/(46/371 -v_B), 1/(53/371 -v_B)).This looks complicated, but maybe we can make substitutions. Let’s let x = v_B.Then, the periods are:T_A = 1/(x +1/53)T_B =1/xT_C=1/(46/371 -x)T_D=1/(53/371 -x)So, LCM(T_A, T_B, T_C, T_D). To find LCM of these four, we need to find the smallest T such that T is a multiple of each T_A, T_B, T_C, T_D.Alternatively, T must be the minimal positive real number such that there exist integers k_A, k_B, k_C, k_D with:k_A * T_A = Tk_B * T_B = Tk_C * T_C = Tk_D * T_D = TWhich implies:k_A = T / T_A = T * (x +1/53)k_B = T / T_B = T * xk_C = T / T_C = T * (46/371 -x)k_D = T / T_D = T * (53/371 -x)Therefore, T must be such that T*(x +1/53), T*x, T*(46/371 -x), T*(53/371 -x) are all integers.So, T must be chosen such that T multiplied by each of these expressions is integer.Let’s denote T as a rational number, since we have rational coefficients (assuming x is rational). Let's suppose that x is rational. Let me assume x = p/q, where p and q are integers. Then, let's substitute x = p/q into the equations.But this might get too involved. Alternatively, let's note that T must be a multiple of 1/(x +1/53), 1/x, 1/(46/371 -x), 1/(53/371 -x). Let's consider these denominators:Let’s denote:A: x +1/53B: xC:46/371 -xD:53/371 -xSo, LCM(1/A, 1/B, 1/C, 1/D) = 1/GCD(A,B,C,D). Wait, is this a valid approach? Wait, GCD of A,B,C,D would be the greatest common divisor of the denominators, but LCM of 1/A,1/B,1/C,1/D is not necessarily 1/GCD(A,B,C,D). For example, LCM(1/2,1/3)=1, since 1 is the smallest number that is a multiple of both 1/2 and 1/3 (1=2*(1/2)=3*(1/3)). However, GCD(2,3)=1, so 1/GCD(2,3)=1, which matches LCM(1/2,1/3)=1. Similarly, LCM(1/4,1/6)=1/2, since 1/2=2*(1/4)=3*(1/6). GCD(4,6)=2, so 1/GCD(4,6)=1/2. So, in this case, LCM(1/a,1/b)=1/GCD(a,b). This seems to hold. So, generalizing, LCM(1/a,1/b,1/c,1/d)=1/GCD(a,b,c,d).Therefore, if we can find GCD of A,B,C,D, then LCM of their reciprocals is 1 divided by that GCD. Therefore, T=1/GCD(A,B,C,D).But in our case, A,B,C,D are:A =x +1/53B =xC=46/371 -xD=53/371 -xSo, GCD(A,B,C,D) is the GCD of these four expressions. However, since x is a variable, unless we can express them in terms of integers, this is not straightforward.Alternatively, maybe there's a better approach. Let's think differently.Given that cars A and B meet after 53 minutes, which is the same as LCM of their periods. Wait, but they meet when the difference in their distances is equal to L. Since they are moving in the same direction, the time to meet is L/(v_A -v_B). Given that they meet after 53 minutes, so:(v_A -v_B)*53 = L.But we set L=1, so (v_A -v_B)=1/53. Which matches our earlier equation.Similarly, cars moving towards each other meet when their combined distance is L. So, (v_A +v_C)*7=1, etc.But maybe all these times (7 and 53) are related to the LCM we need.Alternatively, note that when all four cars meet again, they must have completed an integer number of laps. Therefore, the time T must satisfy:v_A * T = integerv_B * T = integerv_C * T = integerv_D * T = integerGiven that L=1, these integers represent the number of laps each car has completed.Therefore, T must be such that T is a multiple of 1/v_A, 1/v_B, 1/v_C, 1/v_D. Hence, T is LCM(1/v_A, 1/v_B, 1/v_C, 1/v_D).But how do we compute this LCM given the relationships between the speeds?Let’s see:From earlier, we have:v_A = v_B +1/53v_C =46/371 -v_Bv_D=53/371 -v_BNote that 46/371 = 46/(53*7) = (46/53)/7 ≈0.867/7≈0.124, and 53/371=53/(53*7)=1/7≈0.142.So, let's express all speeds in terms of 1/371 to make the denominators same.Multiply all terms by 371:371*v_A =371*v_B +7371*v_C=46 -371*v_B371*v_D=53 -371*v_BLet’s denote:Let’s set 371*v_B =k, where k is a positive real number (since v_B>0). Then:371*v_A =k +7371*v_C=46 -k371*v_D=53 -kSince all speeds must be positive, then:From 371*v_C =46 -k >0 ⇒k <46From 371*v_D =53 -k >0 ⇒k <53From 371*v_B =k >0 ⇒k >0Thus, 0 <k <46.Therefore, k must be in (0,46). Also, since all speeds must be different:371*v_A =k +7 ≠371*v_B =k ⇒ which holds.371*v_C=46 -k must be different from others.46 -k ≠k ⇒46 ≠2k ⇒k ≠2346 -k ≠k +7 ⇒46 -7 =2k ⇒39=2k ⇒k=19.5Similarly, 53 -k ≠k +7 ⇒53 -7=2k ⇒46=2k ⇒k=2353 -k ≠k ⇒53=2k ⇒k=26.5, which is outside the range since k <46.So, to ensure all speeds are distinct, k must not be 19.5,23.Therefore, k ∈(0,46) excluding 19.5,23.Now, the periods of each car in terms of k:T_A=1/v_A=371/(k +7)T_B=1/v_B=371/kT_C=1/v_C=371/(46 -k)T_D=1/v_D=371/(53 -k)So, LCM(T_A, T_B, T_C, T_D) = LCM(371/(k +7), 371/k, 371/(46 -k), 371/(53 -k))=371 * LCM(1/(k +7), 1/k, 1/(46 -k), 1/(53 -k))=371 / GCD(k +7, k, 46 -k, 53 -k)Wait, based on the previous thought that LCM(1/a,1/b)=1/GCD(a,b). If this holds for multiple terms, then LCM(1/a,1/b,1/c,1/d)=1/GCD(a,b,c,d). So, then LCM of the reciprocals is 1 over the GCD of the original denominators.Therefore, LCM(1/(k +7), 1/k, 1/(46 -k), 1/(53 -k))=1/GCD(k +7, k, 46 -k, 53 -k)Therefore, LCM(T_A, T_B, T_C, T_D)=371 / GCD(k +7, k, 46 -k, 53 -k)Thus, we need to compute GCD(k +7, k, 46 -k, 53 -k).Let’s compute GCD of these four numbers: k +7, k, 46 -k, 53 -k.Let’s denote:Let’s first compute GCD(k, k+7). GCD(k, k+7) = GCD(k,7).Then, GCD(k,7, 46 -k, 53 -k). Let’s compute GCD(7, 46 -k, 53 -k).Note that 46 -k and 53 -k differ by 7. So, 53 -k = (46 -k) +7. Therefore, GCD(7,46 -k, (46 -k)+7) = GCD(7,46 -k). Since GCD(7,46 -k, (46 -k)+7) = GCD(7,46 -k).Therefore, GCD(k,7,46 -k)= GCD(GCD(k,7),46 -k)=GCD(7,46 -k), since GCD(k,7) divides 7, so it's either 1 or7.Thus, if GCD(k,7)=d, where d=1 or7, then GCD(d,46 -k). So,If d=7, then 7 divides k, and GCD(7,46 -k). If 7 divides k, then k=7m, so 46 -k=46 -7m. Then GCD(7,46 -7m)=GCD(7,46)=1, since 46=7*6 +4. Therefore, GCD(7,46 -7m)=GCD(7,46)=1. Therefore, GCD(7,46 -k)=1.If d=1, then GCD(1,46 -k)=1.Thus, overall, GCD(k,7,46 -k)=1.Therefore, GCD(k, k+7,46 -k,53 -k)=GCD(1,53 -k)=1.Wait, but 53 -k= (46 -k) +7. So, GCD(1,53 -k)=1. So, regardless of k, the GCD is 1.Therefore, GCD(k +7, k, 46 -k, 53 -k)=1.Therefore, LCM(T_A, T_B, T_C, T_D)=371 /1=371.Therefore, the minimal time T when all four cars meet again is 371 minutes.But wait, 371 minutes is 7*53, since 7*50=350, 7*3=21, 350+21=371. So, 371=7*53.Therefore, regardless of the value of k (as long as the constraints are satisfied), the GCD is 1, so LCM is 371. Therefore, the answer is 371 minutes.But let's verify this with an example. Let’s pick a value for k that satisfies 0 <k <46, k≠19.5,23.Let’s take k=1 (arbitrary choice within constraints).Then:v_B =1/371v_A=(1+7)/371=8/371v_C=(46 -1)/371=45/371v_D=(53 -1)/371=52/371Periods:T_A=371/8=46.375 minutesT_B=371/1=371 minutesT_C=371/45≈8.244 minutesT_D=371/52≈7.134 minutesLCM of 46.375, 371, 8.244,7.134. Hmm, this seems like it might not be 371. Wait, but according to our previous logic, LCM is 371. But in reality, LCM(46.375,371,8.244,7.134). Let's check if 371 is a multiple of all periods.Check 371 /46.375=371 / (371/8)=8 → integer. Yes.371 /371=1→ integer.371 / (371/45)=45→ integer.371 / (371/52)=52→ integer.Therefore, yes, 371 is a multiple of all periods. Therefore, T=371 minutes is indeed the LCM.Similarly, take k=10.v_B=10/371v_A=17/371v_C=36/371v_D=43/371Periods:T_A=371/17≈21.823T_B=371/10=37.1T_C=371/36≈10.305T_D=371/43≈8.627Check 371 divided by each period:371 /21.823≈371/(371/17)=17→ integer.371 /37.1=371/(371/10)=10→ integer.371 /10.305≈371/(371/36)=36→ integer.371 /8.627≈371/(371/43)=43→ integer.So, again, T=371 minutes is the LCM.Therefore, regardless of k (as long as GCD is 1), the LCM is 371. Therefore, the answer is 371 minutes.But wait, the problem says "after how much time from the start of the race will all the cars meet for the first time?" So, 371 minutes. However, we need to check if 371 is indeed the first time. Is there a smaller T?But according to our reasoning, since the GCD is always 1, LCM is 371, so the first time they meet is 371 minutes. Therefore, the answer is 371.But let's recall that in the problem, we were given L as some track, but we set L=1 (or 371) for calculation. However, since the actual track length was normalized, the answer should be independent of L. Wait, but in the problem, the answer is expected in minutes, which matches our calculation.Alternatively, if we didn't normalize L, but kept it as a variable:From the first meeting after 7 minutes:(v_A +v_C)*7 =LSimilarly, (v_B +v_D)*7=L(v_A -v_B)*53=LTherefore, we have:(v_A +v_C)=L/7(v_B +v_D)=L/7(v_A -v_B)=L/53Then, solving for speeds:v_A = v_B + L/53v_C = L/7 -v_A = L/7 -v_B -L/53v_D = L/7 -v_BTherefore, periods:T_A= L /v_A= L/(v_B +L/53)= 53L/(53v_B +L)T_B= L/v_BT_C= L/v_C= L/(L/7 -v_B -L/53)= L/(L(1/7 -1/53) -v_B)= L/(L*(46/371) -v_B)Similarly for T_D.But in this case, the LCM would involve expressions with L. However, through substitution, we can find that LCM is 371 minutes regardless of L, which seems conflicting. Wait, actually, when we set L=371, LCM became 371 minutes, but if L is different, the LCM would scale accordingly. However, since all given times (7 minutes and 53 minutes) are independent of L, the answer must be in absolute time, so 371 minutes is the answer.Alternatively, perhaps my initial approach by setting L=371 was the key, and since LCM came out as 371 minutes, which is 7*53, which are the given times, that seems plausible.Alternatively, is there a different way to see this?Since A and B start moving clockwise, and C and D counterclockwise. They meet after 7 minutes, then A and B meet after 53 minutes. The total time for all to meet is LCM of 7,53, and the other meeting times. But 7 and 53 are primes, so LCM(7,53)=371. Maybe that's the answer.But why LCM(7,53)? The problem involves multiple cars with different speeds. However, the fact that the two meeting times given are 7 and 53, which are primes, and their LCM is 371, which is the answer we arrived at earlier through the track length normalization method. So, likely 371 is the answer.But to ensure this is correct, let's think again.If all four cars meet at time T, then T must be a multiple of 7 minutes (since A and C meet every 7 minutes, as do B and D). Also, T must be a multiple of 53 minutes (since A and B meet every 53 minutes). However, 7 and 53 are coprime, so LCM(7,53)=371. Therefore, the earliest time when T is a multiple of both 7 and 53 is 371. However, is this sufficient?Wait, but meeting at the starting point is different from meeting somewhere on the track. The problem asks when all four cars meet at the starting point for the first time. The meetings after 7 minutes and 53 minutes are meetings on the track, not necessarily at the starting point.So, perhaps the earlier approach with normalizing the track length is more accurate. But through that method, we arrived at 371 minutes regardless of the variable speed, which seems to align with LCM(7,53).Therefore, considering both methods lead to 371, I think the answer is 371 minutes.But let's verify once more with k=23, even though k=23 causes two cars to have the same speed. Wait, if k=23, which is prohibited, but let's check:If k=23, then v_B=23/371, v_A=23/371 +7/371=30/371, v_C=46/371 -23/371=23/371, v_D=53/371 -23/371=30/371. So, v_A and v_D are both 30/371, and v_C and v_B are both 23/371. So, two pairs have same speed, which is invalid. So, T=371 would still be LCM, but since the problem states all speeds are different, this case is excluded. But in allowed cases where speeds are distinct, LCM remains 371.Therefore, the answer is 371 minutes, which is 7*53.Final Answerboxed{371}</think>Given four cars (A, B, C,) and (D) starting from the same point on a circular track, with (A) and (B) driving clockwise and (C) and (D) driving counterclockwise, all at constant but different speeds. The key events are:- After 7 minutes, (A) meets (C) and (B) meets (D) for the first time.- After another 46 minutes (total 53 minutes), (A) and (B) meet for the first time.We need to determine the first time all four cars meet again at the starting point.1. Meeting Conditions: - (A) and (C) meet after 7 minutes: ((v_A + v_C) cdot 7 = L) - (B) and (D) meet after 7 minutes: ((v_B + v_D) cdot 7 = L) - (A) and (B) meet after 53 minutes: ((v_A - v_B) cdot 53 = L)2. Speed Relationships: - From the meeting conditions, we derive: - (v_A + v_C = frac{L}{7}) - (v_B + v_D = frac{L}{7}) - (v_A - v_B = frac{L}{53})3. Expressing Speeds: - Let (L = 371) (chosen for simplification): - (v_A + v_C = 53) - (v_B + v_D = 53) - (v_A - v_B = 7)4. Periods of Each Car: - The periods (time to complete a lap) are: - (T_A = frac{371}{v_A}) - (T_B = frac{371}{v_B}) - (T_C = frac{371}{v_C}) - (T_D = frac{371}{v_D})5. Least Common Multiple (LCM): - The LCM of the periods must be found. Through analysis, it is determined that the GCD of the denominators of the speeds (when expressed as fractions) is 1, leading to LCM being (371).Thus, the first time all four cars meet again at the starting point is (boxed{371}) minutes.]

🔑:Solution. A and C meet every 7 minutes, while A and B meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is 7 cdot 53 = 371 minutes. At the same time, B and D also meet every 7 minutes, so on the 371st minute, car D will be at the same point as the other three cars.Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.