💡First, identify the number of elements in set ( P ) and set ( Q ). Set ( P ) has 3 elements, and set ( Q ) has 4 elements.Next, calculate the Cartesian product of sets ( P ) and ( Q ). The Cartesian product ( P times Q ) consists of all ordered pairs ( (a, b) ) where ( a ) is an element of ( P ) and ( b ) is an element of ( Q ).Since there are 3 elements in ( P ) and 4 elements in ( Q ), the total number of ordered pairs in the Cartesian product is ( 3 times 4 = 12 ).Therefore, the number of elements in ( P times Q ) is 12.

💡Alright, so I have this quadratic function: y = ax² - bx, where a is not zero. It passes through the point P(m, 2). When y is greater than or equal to -2/3, the range of x is x ≤ n - 1 or x ≥ -3 - n. I need to find the axis of symmetry and a possible value of m.First, let me recall that the general form of a quadratic function is y = ax² + bx + c. In this case, the function is given as y = ax² - bx, so c is zero. That simplifies things a bit.Since the function passes through the point P(m, 2), that means when x = m, y = 2. So, substituting m into the equation, we get:2 = a(m)² - b(m)So, 2 = am² - bm. I'll keep that in mind.Now, the problem states that when y ≥ -2/3, the range of x is x ≤ n - 1 or x ≥ -3 - n. This tells me that the quadratic function has a minimum value of -2/3, and the x-values where y is at least -2/3 are outside the interval (n - 1, -3 - n). Wait, actually, it's x ≤ n - 1 or x ≥ -3 - n. So, the quadratic is above or equal to -2/3 outside the interval [n - 1, -3 - n]. Hmm, that seems a bit confusing because usually, for a quadratic that opens upwards, the function is above its minimum between the roots, but here it's saying it's above -2/3 outside of some interval. Maybe I need to think about this differently.Wait, actually, if y ≥ -2/3, then the quadratic is above or equal to -2/3. For a quadratic that opens upwards, the function will be above its minimum value everywhere except between its two roots. But in this case, the function is above -2/3 outside the interval [n - 1, -3 - n]. That suggests that the quadratic is below -2/3 between n - 1 and -3 - n, and above -2/3 outside of that interval. So, the quadratic must open upwards because it's above -2/3 outside the interval.Therefore, the quadratic has a minimum value of -2/3, and the roots of the equation y = -2/3 are at x = n - 1 and x = -3 - n. So, the quadratic can be written in its vertex form, but maybe it's easier to use the standard form.Let me write the quadratic as y = ax² - bx. We know that it passes through (m, 2), so 2 = am² - bm.Also, the quadratic equals -2/3 at x = n - 1 and x = -3 - n. So, substituting these x-values into the equation:For x = n - 1:-2/3 = a(n - 1)² - b(n - 1)For x = -3 - n:-2/3 = a(-3 - n)² - b(-3 - n)So, now I have two equations:1) -2/3 = a(n - 1)² - b(n - 1)2) -2/3 = a(-3 - n)² - b(-3 - n)Let me expand these equations.First, equation 1:-2/3 = a(n² - 2n + 1) - b(n - 1)= a(n² - 2n + 1) - bn + b= a n² - 2a n + a - bn + bSimilarly, equation 2:-2/3 = a(9 + 6n + n²) - b(-3 - n)= a(n² + 6n + 9) + 3b + bn= a n² + 6a n + 9a + 3b + bnSo, now I have:Equation 1:a n² - 2a n + a - bn + b = -2/3Equation 2:a n² + 6a n + 9a + 3b + bn = -2/3Let me write both equations:1) a n² - 2a n + a - bn + b = -2/32) a n² + 6a n + 9a + 3b + bn = -2/3Now, let's subtract equation 1 from equation 2 to eliminate some terms.Equation 2 - Equation 1:(a n² + 6a n + 9a + 3b + bn) - (a n² - 2a n + a - bn + b) = (-2/3) - (-2/3)Simplify:a n² - a n² + 6a n + 2a n + 9a - a + 3b - b + bn + bn = 0So:(0) + (8a n) + (8a) + (2b) + (2bn) = 0Factor terms:8a n + 8a + 2b + 2bn = 0Factor further:8a(n + 1) + 2b(n + 1) = 0Factor out (n + 1):(8a + 2b)(n + 1) = 0So, either (n + 1) = 0 or (8a + 2b) = 0.Case 1: n + 1 = 0 => n = -1Case 2: 8a + 2b = 0 => 4a + b = 0 => b = -4aLet me consider both cases.Case 1: n = -1If n = -1, then the roots are x = n - 1 = -2 and x = -3 - n = -2. Wait, both roots are x = -2? That would mean the quadratic touches the x-axis at x = -2, so it's a perfect square. But in our case, the quadratic equals -2/3 at x = -2, not zero. So, that might not make sense because if both roots are the same, it's a double root, but here the quadratic equals -2/3 at that point, not zero. So, maybe n = -1 is not the right case.Alternatively, perhaps n = -1 is acceptable because it's not necessarily a root of y = 0, but a root of y = -2/3. So, if n = -1, then both x = -2 and x = -2 are the points where y = -2/3. So, that would mean the quadratic has a double root at x = -2 for y = -2/3. So, that would mean the quadratic touches the line y = -2/3 at x = -2, making it the vertex.So, if n = -1, then the quadratic has its vertex at x = -2, which is the axis of symmetry. So, that might be the case.Alternatively, let's consider Case 2: b = -4aSo, if b = -4a, then we can substitute back into the equations.Let me take equation 1:a n² - 2a n + a - bn + b = -2/3Substitute b = -4a:a n² - 2a n + a - (-4a)n + (-4a) = -2/3Simplify:a n² - 2a n + a + 4a n - 4a = -2/3Combine like terms:a n² + ( -2a n + 4a n ) + (a - 4a) = -2/3Which is:a n² + 2a n - 3a = -2/3Factor out a:a(n² + 2n - 3) = -2/3Similarly, equation 2:a n² + 6a n + 9a + 3b + bn = -2/3Substitute b = -4a:a n² + 6a n + 9a + 3(-4a) + (-4a)n = -2/3Simplify:a n² + 6a n + 9a - 12a -4a n = -2/3Combine like terms:a n² + (6a n - 4a n) + (9a - 12a) = -2/3Which is:a n² + 2a n - 3a = -2/3So, both equations reduce to the same equation: a(n² + 2n - 3) = -2/3So, from this, we can solve for a:a = (-2/3) / (n² + 2n - 3)But we also have the point P(m, 2) on the quadratic, so:2 = a m² - b mBut since b = -4a, substitute:2 = a m² - (-4a) m = a m² + 4a m = a(m² + 4m)So, a = 2 / (m² + 4m)Therefore, we have two expressions for a:1) a = (-2/3) / (n² + 2n - 3)2) a = 2 / (m² + 4m)So, setting them equal:(-2/3) / (n² + 2n - 3) = 2 / (m² + 4m)Cross-multiplying:(-2/3)(m² + 4m) = 2(n² + 2n - 3)Multiply both sides by 3 to eliminate the denominator:-2(m² + 4m) = 6(n² + 2n - 3)Expand:-2m² - 8m = 6n² + 12n - 18Bring all terms to one side:-2m² - 8m - 6n² - 12n + 18 = 0Multiply both sides by -1:2m² + 8m + 6n² + 12n - 18 = 0Divide both sides by 2:m² + 4m + 3n² + 6n - 9 = 0Hmm, this seems complicated. Maybe I need another approach.Wait, earlier I considered Case 1 where n = -1, which led to the axis of symmetry being x = -2. Let me see if that works.If n = -1, then the roots are x = n - 1 = -2 and x = -3 - n = -2. So, both roots are x = -2, meaning the quadratic touches y = -2/3 at x = -2. So, the vertex is at x = -2, which is the axis of symmetry.So, the axis of symmetry is x = -2.Now, knowing that, let's find a and b.Since the axis of symmetry is x = -b/(2a) = -2So, -b/(2a) = -2 => b/(2a) = 2 => b = 4aWait, earlier in Case 2, I had b = -4a. But here, from the axis of symmetry, I get b = 4a. There's a discrepancy.Wait, let's double-check. The standard form is y = ax² + bx + c. The axis of symmetry is x = -b/(2a). In our case, the function is y = ax² - bx, so b is replaced by -b. So, the axis of symmetry is x = -(-b)/(2a) = b/(2a). So, if the axis of symmetry is x = -2, then:b/(2a) = -2 => b = -4aAh, okay, so that matches with Case 2 where b = -4a.So, that's consistent.So, b = -4a.Now, going back to the equation for a:From the point P(m, 2):2 = a m² - b m = a m² - (-4a) m = a m² + 4a m = a(m² + 4m)So, a = 2 / (m² + 4m)Also, from the quadratic equaling -2/3 at x = n - 1 and x = -3 - n, we had:a(n² + 2n - 3) = -2/3But since n = -1, let's substitute n = -1:a((-1)² + 2*(-1) - 3) = a(1 - 2 - 3) = a(-4) = -2/3So, -4a = -2/3 => a = (-2/3)/(-4) = (2/3)/4 = 1/6So, a = 1/6Then, b = -4a = -4*(1/6) = -2/3So, the quadratic is y = (1/6)x² - (-2/3)x = (1/6)x² + (2/3)xSimplify:y = (1/6)x² + (2/3)xNow, let's verify that this quadratic passes through P(m, 2):2 = (1/6)m² + (2/3)mMultiply both sides by 6 to eliminate denominators:12 = m² + 4mSo, m² + 4m - 12 = 0Solve for m:m = [-4 ± sqrt(16 + 48)] / 2 = [-4 ± sqrt(64)] / 2 = [-4 ± 8]/2So, m = (-4 + 8)/2 = 4/2 = 2 or m = (-4 - 8)/2 = -12/2 = -6So, possible values for m are 2 and -6.Therefore, the axis of symmetry is x = -2, and possible values for m are 2 and -6.Wait, but earlier I considered n = -1 leading to axis of symmetry x = -2, and that seems consistent. So, that must be the answer.So, to recap:- The axis of symmetry is x = -2.- Possible values for m are 2 and -6.But the problem asks for a possible value of m, so I can choose either 2 or -6.I think that's it.

💡Okay, so I have this problem here: Given the function f(x) = |x² + bx| where b is a real number. When x is in the interval [0, 1], the maximum value of f(x) is M(b). I need to find the minimum value of M(b) from the given options.Hmm, let me try to understand the problem first. The function is the absolute value of a quadratic function. Quadratic functions have a parabola shape, and taking the absolute value will reflect any part of the parabola that's below the x-axis to above it. So, f(x) will always be non-negative.Since we're dealing with the interval [0, 1], I need to figure out how the maximum of |x² + bx| behaves as we vary the parameter b. The goal is to find the minimum possible maximum value, which sounds a bit tricky, but let's break it down.First, maybe I should rewrite the quadratic inside the absolute value to make it easier to analyze. Let's complete the square for x² + bx.x² + bx = x² + bx + (b²/4) - (b²/4) = (x + b/2)² - (b²/4)So, f(x) = |(x + b/2)² - (b²/4)|. That's interesting because it shows the vertex of the parabola is at x = -b/2, and the minimum value of the quadratic (before taking absolute value) is -b²/4.But since we're taking the absolute value, the function f(x) will have a V-shape at the point where the quadratic crosses the x-axis. So, the graph of f(x) will have a "corner" where x² + bx = 0, which is at x = 0 and x = -b. But since x is in [0, 1], the relevant roots are x = 0 and x = -b if -b is in [0, 1].Wait, so if -b is between 0 and 1, that is, if b is between -1 and 0, then the quadratic crosses the x-axis within our interval [0, 1]. Otherwise, if b is positive, the quadratic doesn't cross the x-axis in [0, 1], and if b is less than -1, the quadratic crosses the x-axis at x = -b, which is greater than 1, so outside our interval.So, this suggests that the behavior of f(x) on [0, 1] depends on the value of b. Specifically, whether the quadratic crosses the x-axis within [0, 1] or not.Let me consider different cases for b:1. Case 1: b ≥ 02. Case 2: -1 ≤ b < 03. Case 3: b < -1Maybe analyzing each case will help me find M(b) in each scenario and then determine the minimum M(b).**Case 1: b ≥ 0**If b is non-negative, then the quadratic x² + bx is always non-negative on [0, 1] because both x and b are non-negative. So, f(x) = x² + bx, which is a parabola opening upwards. Since it's increasing on [0, 1] (because the vertex is at x = -b/2, which is negative, so the function is increasing on [0, 1]), the maximum occurs at x = 1.Therefore, M(b) = f(1) = 1 + b.**Case 2: -1 ≤ b < 0**Here, b is negative, but not too negative (since b ≥ -1). So, the quadratic x² + bx can be positive or negative on [0, 1]. Specifically, since the vertex is at x = -b/2, which is positive because b is negative. So, the vertex is in [0, 1] because -b/2 is between 0 and 0.5 (since b is between -1 and 0).Therefore, the quadratic will have a minimum at x = -b/2, which is within [0, 1]. The value at the vertex is (x + b/2)² - (b²/4), so plugging x = -b/2, we get -b²/4. Taking absolute value, f(-b/2) = b²/4.Now, we need to compare this value with the value at the endpoints x = 0 and x = 1.At x = 0: f(0) = |0 + 0| = 0.At x = 1: f(1) = |1 + b|.So, M(b) is the maximum of f(-b/2) and f(1). Let's see when each is larger.We need to compare b²/4 and |1 + b|. Since b is between -1 and 0, 1 + b is between 0 and 1, so |1 + b| = 1 + b.So, set b²/4 = 1 + b.Multiply both sides by 4: b² = 4 + 4b.Bring all terms to one side: b² - 4b - 4 = 0.Solve for b: b = [4 ± sqrt(16 + 16)] / 2 = [4 ± sqrt(32)] / 2 = [4 ± 4*sqrt(2)] / 2 = 2 ± 2*sqrt(2).But since b is between -1 and 0, we take the negative root: 2 - 2*sqrt(2). Let's compute that: sqrt(2) is approximately 1.414, so 2*sqrt(2) is about 2.828. So, 2 - 2.828 ≈ -0.828, which is between -1 and 0.Therefore, for b < 2 - 2*sqrt(2) (approximately -0.828), f(-b/2) = b²/4 is larger than f(1) = 1 + b. For b ≥ 2 - 2*sqrt(2), f(1) is larger.So, in this case, M(b) is:- b²/4 when -1 ≤ b < 2 - 2*sqrt(2)- 1 + b when 2 - 2*sqrt(2) ≤ b < 0**Case 3: b < -1**Here, b is less than -1, so the quadratic x² + bx has a vertex at x = -b/2, which is greater than 0.5 (since b < -1, so -b/2 > 0.5). However, since our interval is [0, 1], the vertex is still within [0, 1] because x = -b/2 is less than 1 only if b > -2. Wait, if b < -1, then -b/2 > 0.5, but if b < -2, then -b/2 > 1, which is outside our interval.So, actually, we need to further subdivide Case 3:- Subcase 3a: -2 ≤ b < -1- Subcase 3b: b < -2In Subcase 3a: -2 ≤ b < -1, the vertex x = -b/2 is in [1, 1.5], but since our interval is [0, 1], the function is decreasing on [0, 1] because the vertex is at x = -b/2 > 1. So, the maximum occurs at x = 0 or x = 1.Wait, at x = 0: f(0) = 0.At x = 1: f(1) = |1 + b|. Since b < -1, 1 + b < 0, so f(1) = -(1 + b) = -1 - b.But also, since the function is decreasing on [0, 1], the maximum is at x = 0, which is 0. But that can't be right because f(1) is positive. Wait, perhaps I made a mistake.Wait, if the vertex is at x = -b/2 > 1, then on [0, 1], the function is decreasing because the parabola opens upwards, and the vertex is to the right of 1. So, the function is decreasing from x = 0 to x = 1. Therefore, the maximum is at x = 0, which is 0, but that contradicts because f(1) is positive.Wait, no, f(0) is |0 + 0| = 0, and f(1) is |1 + b|. Since b < -1, 1 + b < 0, so f(1) = -(1 + b) = -1 - b, which is positive because b < -1.But if the function is decreasing on [0, 1], then f(0) is 0 and f(1) is positive. So, the maximum is at x = 1, which is f(1) = -1 - b.Wait, that makes sense. So, in Subcase 3a: -2 ≤ b < -1, M(b) = -1 - b.In Subcase 3b: b < -2, the vertex x = -b/2 > 1, so on [0, 1], the function is still decreasing, so the maximum is at x = 0, which is 0. But wait, f(0) is 0, but f(1) is |1 + b|, which is positive. So, actually, the maximum is at x = 1, which is f(1) = |1 + b| = -1 - b.Wait, so actually, for all b < 0, M(b) is either b²/4 or 1 + b, depending on whether the vertex is within [0, 1] or not.Wait, maybe I need to correct my earlier analysis.Wait, in Case 3: b < -1, the vertex is at x = -b/2. If b < -2, then x = -b/2 > 1, so on [0, 1], the function is decreasing, so maximum at x = 0, which is 0. But f(1) is |1 + b|, which is positive, so actually, the maximum is at x = 1, which is f(1) = |1 + b|.Wait, but if the function is decreasing on [0, 1], then f(0) is the maximum. But f(0) is 0, which is less than f(1). That can't be. So, perhaps my assumption about the function being decreasing is incorrect.Wait, let's think again. The quadratic x² + bx has its vertex at x = -b/2. If x = -b/2 is greater than 1, then on [0, 1], the function is decreasing because the parabola opens upwards and the vertex is to the right of 1. So, from x = 0 to x = 1, the function decreases from f(0) to f(1).But f(0) is 0, and f(1) is |1 + b|. Since b < -1, 1 + b < 0, so f(1) = -(1 + b) = -1 - b, which is positive. So, the function decreases from 0 to a positive value. Therefore, the maximum is at x = 0, which is 0. But that contradicts because f(1) is positive. Wait, no, if the function is decreasing, it starts at 0 and goes down, but since it's a quadratic, it's actually a U-shape, but shifted.Wait, maybe I'm confusing the direction. Let me plot the function for b < -1.For example, take b = -2. Then f(x) = |x² - 2x|. The quadratic x² - 2x has roots at x = 0 and x = 2. The vertex is at x = 1, which is the midpoint of 0 and 2. So, at x = 1, the quadratic is -1, so f(1) = 1.On [0, 1], the quadratic x² - 2x goes from 0 at x = 0, down to -1 at x = 1, so f(x) = |x² - 2x| goes from 0 up to 1. So, actually, the function is increasing on [0, 1] when b = -2.Wait, that's different from what I thought earlier. So, maybe my earlier conclusion was wrong.Wait, let's compute the derivative of f(x) when x² + bx is negative. So, f(x) = |x² + bx|.If x² + bx < 0, then f(x) = -(x² + bx), so f'(x) = -2x - b.If x² + bx ≥ 0, then f(x) = x² + bx, so f'(x) = 2x + b.So, the derivative depends on whether x² + bx is positive or negative.In the case where b < -1, let's say b = -2, then x² - 2x is negative on (0, 2), so on [0, 1], x² - 2x is negative except at x = 0. So, f(x) = -(x² - 2x) = -x² + 2x.So, f'(x) = -2x + 2. Setting derivative to zero: -2x + 2 = 0 => x = 1.So, the function f(x) = -x² + 2x has a maximum at x = 1, which is f(1) = -1 + 2 = 1.So, in this case, the maximum is at x = 1.Wait, so for b = -2, M(b) = 1.Similarly, for other b < -1, let's see.Take b = -3. Then f(x) = |x² - 3x|. The quadratic x² - 3x has roots at x = 0 and x = 3. The vertex is at x = 1.5, which is outside [0, 1]. So, on [0, 1], x² - 3x is negative, so f(x) = -x² + 3x.f'(x) = -2x + 3. Setting to zero: -2x + 3 = 0 => x = 1.5, which is outside [0, 1]. So, on [0, 1], f(x) is increasing because the derivative is positive (since x < 1.5, so -2x + 3 > 0). Therefore, maximum at x = 1: f(1) = -1 + 3 = 2.So, M(b) = 2 for b = -3.Wait, so for b < -1, f(x) is increasing on [0, 1], so maximum at x = 1, which is |1 + b| = -1 - b.But wait, for b = -2, M(b) = 1, which is -1 - (-2) = 1. For b = -3, M(b) = 2, which is -1 - (-3) = 2. So, in general, for b < -1, M(b) = -1 - b.But wait, earlier when I thought about b < -2, I thought the function was decreasing, but in reality, for b = -2, it's increasing on [0, 1]. So, perhaps for all b < -1, M(b) = -1 - b.Wait, let me check for b = -1.5.f(x) = |x² - 1.5x|. The quadratic x² - 1.5x has roots at x = 0 and x = 1.5. The vertex is at x = 0.75, which is within [0, 1]. So, the function f(x) = |x² - 1.5x| will have a V-shape at x = 0.75.At x = 0.75, f(x) = |(0.75)^2 - 1.5*(0.75)| = |0.5625 - 1.125| = | -0.5625 | = 0.5625.At x = 1, f(1) = |1 - 1.5| = 0.5.So, in this case, the maximum is at x = 0.75, which is 0.5625.Wait, so for b = -1.5, which is between -2 and -1, the maximum is at the vertex, not at x = 1.Hmm, so my earlier conclusion was incorrect. So, perhaps I need to re-examine.Wait, for b between -2 and -1, the vertex x = -b/2 is between 0.5 and 1. So, on [0, 1], the function f(x) = |x² + bx| will have a maximum either at the vertex or at x = 1.So, for b between -2 and -1, we need to compare f(-b/2) and f(1).Similarly, for b < -2, the vertex is at x = -b/2 > 1, so on [0, 1], the function is increasing because the vertex is to the right of 1, so the function is increasing on [0, 1], so maximum at x = 1.Wait, let's test b = -2.5.f(x) = |x² - 2.5x|. The quadratic x² - 2.5x has roots at x = 0 and x = 2.5. The vertex is at x = 1.25, which is outside [0, 1]. So, on [0, 1], the quadratic is negative, so f(x) = -x² + 2.5x.f'(x) = -2x + 2.5. Setting to zero: x = 2.5/2 = 1.25, which is outside [0, 1]. So, on [0, 1], f(x) is increasing because the derivative is positive (since x < 1.25, so -2x + 2.5 > 0). Therefore, maximum at x = 1: f(1) = -1 + 2.5 = 1.5.So, for b = -2.5, M(b) = 1.5.Similarly, for b = -3, M(b) = 2.So, in general, for b ≤ -2, M(b) = -1 - b.But for b between -2 and -1, the maximum is at the vertex x = -b/2, which is within [0, 1].So, let's summarize:- For b ≥ 0: M(b) = 1 + b- For -2 ≤ b < 0: M(b) is the maximum of f(-b/2) and f(1)- For b < -2: M(b) = -1 - bWait, but earlier when I considered b between -1 and 0, I had a different analysis. Maybe I need to correct that.Wait, perhaps the correct breakdown is:- For b ≥ 0: M(b) = 1 + b- For -2 ≤ b < 0: M(b) is the maximum of f(-b/2) and f(1)- For b < -2: M(b) = -1 - bBut when I took b = -1.5, which is between -2 and -1, the maximum was at the vertex. So, perhaps for -2 ≤ b < 0, M(b) is the maximum of f(-b/2) and f(1).Wait, let's formalize this.For b in (-∞, 0):- If b ≥ -2, then the vertex x = -b/2 is in [0, 1] (since b ≥ -2 => x = -b/2 ≤ 1). So, the maximum is either at the vertex or at x = 1.- If b < -2, then the vertex x = -b/2 > 1, so on [0, 1], the function is increasing, so maximum at x = 1.Therefore, the correct cases are:1. b ≥ 0: M(b) = 1 + b2. -2 ≤ b < 0: M(b) = max{f(-b/2), f(1)}3. b < -2: M(b) = f(1) = -1 - bNow, let's compute f(-b/2) and f(1) for -2 ≤ b < 0.f(-b/2) = |(-b/2)^2 + b*(-b/2)| = |b²/4 - b²/2| = | -b²/4 | = b²/4.f(1) = |1 + b| = 1 + b (since b ≥ -2, 1 + b ≥ -1, but since b < 0, 1 + b < 1).Wait, but for b < 0, 1 + b can be positive or negative. Wait, if b ≥ -1, then 1 + b ≥ 0, so f(1) = 1 + b.If b < -1, then 1 + b < 0, so f(1) = -(1 + b) = -1 - b.Wait, so within -2 ≤ b < 0, we have:- For -1 ≤ b < 0: f(1) = 1 + b- For -2 ≤ b < -1: f(1) = -1 - bSo, in the case of -2 ≤ b < 0, we have two subcases:- -1 ≤ b < 0: M(b) = max{b²/4, 1 + b}- -2 ≤ b < -1: M(b) = max{b²/4, -1 - b}Therefore, we need to find where b²/4 is greater than 1 + b or -1 - b.Let's first consider -1 ≤ b < 0:We need to find when b²/4 ≥ 1 + b.Solve b²/4 - b - 1 ≥ 0.Multiply by 4: b² - 4b - 4 ≥ 0.Solve b² - 4b - 4 = 0: b = [4 ± sqrt(16 + 16)] / 2 = [4 ± sqrt(32)] / 2 = [4 ± 4*sqrt(2)] / 2 = 2 ± 2*sqrt(2).Since b is between -1 and 0, we take the smaller root: 2 - 2*sqrt(2) ≈ 2 - 2.828 ≈ -0.828.So, for b ≤ 2 - 2*sqrt(2) ≈ -0.828, b²/4 ≥ 1 + b.Therefore, in -1 ≤ b < 0:- For -1 ≤ b ≤ 2 - 2*sqrt(2): M(b) = b²/4- For 2 - 2*sqrt(2) < b < 0: M(b) = 1 + bSimilarly, for -2 ≤ b < -1:We need to find when b²/4 ≥ -1 - b.Solve b²/4 + b + 1 ≥ 0.Multiply by 4: b² + 4b + 4 ≥ 0.This factors as (b + 2)^2 ≥ 0, which is always true. So, b²/4 ≥ -1 - b for all real b.Therefore, in -2 ≤ b < -1, M(b) = b²/4.Wait, but let's check with b = -1.5:f(-b/2) = f(0.75) = |(0.75)^2 + (-1.5)(0.75)| = |0.5625 - 1.125| = 0.5625.f(1) = |1 + (-1.5)| = 0.5.So, M(b) = 0.5625, which is b²/4 = (2.25)/4 = 0.5625.Similarly, for b = -2:f(-b/2) = f(1) = |1 + (-2)| = 1, but wait, f(-b/2) when b = -2 is f(1) = |1 - 2| = 1, which is equal to f(1).Wait, but for b = -2, the vertex is at x = 1, so f(1) is both the vertex and the endpoint. So, it's consistent.Therefore, in -2 ≤ b < -1, M(b) = b²/4.So, putting it all together:- For b ≥ 0: M(b) = 1 + b- For -2 ≤ b < 0: - If -2 ≤ b ≤ 2 - 2*sqrt(2): M(b) = b²/4 - If 2 - 2*sqrt(2) < b < 0: M(b) = 1 + b- For b < -2: M(b) = -1 - bNow, we need to find the minimum value of M(b) over all real b.So, let's analyze M(b) in each interval.1. For b ≥ 0: M(b) = 1 + b, which is increasing as b increases. So, the minimum in this interval is at b = 0: M(0) = 1.2. For -2 ≤ b < 0: - When -2 ≤ b ≤ 2 - 2*sqrt(2): M(b) = b²/4, which is a parabola opening upwards. The minimum occurs at b = 0, but since b is in [-2, 2 - 2*sqrt(2)], the minimum in this interval is at b = 2 - 2*sqrt(2): M(b) = ( (2 - 2*sqrt(2))² ) / 4. Let's compute that: (2 - 2*sqrt(2))² = 4 - 8*sqrt(2) + 8 = 12 - 8*sqrt(2). So, M(b) = (12 - 8*sqrt(2)) / 4 = 3 - 2*sqrt(2). - When 2 - 2*sqrt(2) < b < 0: M(b) = 1 + b, which is decreasing as b increases. So, the minimum in this interval is at b approaching 0 from the left: M(b) approaches 1.3. For b < -2: M(b) = -1 - b, which is a linear function decreasing as b decreases. So, the minimum in this interval is as b approaches -2 from the left: M(b) approaches -1 - (-2) = 1.Therefore, the minimum value of M(b) occurs at b = 2 - 2*sqrt(2), where M(b) = 3 - 2*sqrt(2).So, the answer is A: 3 - 2*sqrt(2).

💡Alright, so I have this problem about an n x n grid where the first row is filled with numbers from 1 to n in order. Then, each subsequent row is created by shifting some parts of the previous row. Specifically, if a row has numbers a₁, a₂, ..., aₖ, a_{k+1}, ..., aₘ, a_{m+1}, ..., aₙ, the next row becomes a_{m+1}, ..., aₙ, a_{k+1}, ..., aₘ, a₁, ..., aₖ. The conditions given are that m and n - k are coprime, and m > k.I need to prove that after filling the entire table, every column will contain all the numbers from 1 to n. Hmm, okay. So, each column should have each number exactly once, right? That means the table is a Latin square.Let me try to visualize this. The first row is straightforward: 1, 2, 3, ..., n. The second row is constructed by taking the last n - m elements, then the middle m - k elements, and then the first k elements. So, it's like a rotation but not a simple cyclic shift.Since m and n - k are coprime, that might be important for ensuring that the shifts eventually cycle through all positions. I remember that in cyclic groups, if two numbers are coprime, their least common multiple is their product, which might relate to the number of shifts needed to return to the original position.Maybe I can model the movement of each number as it moves through the columns. Let's pick a number, say number 1. In the first row, it's in column 1. In the second row, where does it go? Since the first k elements move to the end, number 1 would be in column n - k + 1 in the second row.Wait, no. Let me think again. The next row is a_{m+1}, ..., aₙ, a_{k+1}, ..., aₘ, a₁, ..., aₖ. So, the first k elements of the previous row become the last k elements of the next row. So, number 1, which was in position 1, moves to position n - k + 1 in the next row.Similarly, in the third row, number 1 would move again by the same shift. So, each time, it's moving n - k positions to the right, wrapping around if necessary.Since n - k and m are coprime, the number of shifts needed to cycle back to the original position would be n, right? Because the shift size and the total number of positions are coprime, so it cycles through all positions before repeating.If that's the case, then each number would appear exactly once in each column after n rows. But wait, the grid is n x n, so after n rows, each number has cycled through all columns. Therefore, each column must contain all numbers from 1 to n.But let me check this with a small example to make sure I'm not making a mistake. Let's take n = 5, k = 1, m = 3. So, m = 3 and n - k = 4, which are coprime. The first row is 1, 2, 3, 4, 5.The second row would be a_{4}, a_{5}, a_{2}, a_{3}, a_{1}, which is 4, 5, 2, 3, 1.The third row would be a_{4}, a_{5}, a_{2}, a_{3}, a_{1} shifted again: 2, 3, 1, 4, 5.Wait, hold on. That doesn't seem right. Let me reconstruct the second row properly. The first row is 1, 2, 3, 4, 5. The next row is a_{m+1}, ..., aₙ, a_{k+1}, ..., aₘ, a₁, ..., aₖ. So, m = 3, so a_{4}, a_{5}, then a_{2}, a_{3}, then a_{1}. So, second row is 4, 5, 2, 3, 1.Third row would be a_{4}, a_{5}, a_{2}, a_{3}, a_{1} shifted again: 2, 3, 1, 4, 5.Fourth row: 3, 1, 4, 5, 2.Fifth row: 1, 4, 5, 2, 3.Wait, now looking at the columns:Column 1: 1, 4, 2, 3, 1 – Wait, 1 appears twice. That's a problem. Did I do something wrong?Hmm, maybe my example is not correct. Let me double-check the shifting. Maybe I misapplied the transformation.Given the first row: 1, 2, 3, 4, 5.Second row: a_{4}, a_{5}, a_{2}, a_{3}, a_{1} = 4, 5, 2, 3, 1.Third row: a_{4}, a_{5}, a_{2}, a_{3}, a_{1} of the second row: 2, 3, 1, 4, 5.Fourth row: a_{4}, a_{5}, a_{2}, a_{3}, a_{1} of the third row: 4, 5, 2, 3, 1.Wait, that's the same as the second row. So, it's cycling between two rows: 4,5,2,3,1 and 2,3,1,4,5. So, in this case, the table doesn't fill properly because it starts repeating before n rows. That contradicts the initial assumption.But in my example, m = 3 and n - k = 4, which are coprime. So why is it not working? Maybe my understanding of the transformation is incorrect.Wait, the problem says "the same numbers are recorded in the following order: a_{m+1}, ..., aₙ, a_{k+1}, ..., aₘ, a₁, ..., aₖ." So, in the second row, it's a_{m+1} to aₙ, then a_{k+1} to aₘ, then a₁ to aₖ.In my example, m = 3, so a_{4}, a_{5}, then a_{2}, a_{3}, then a_{1}. So, second row is 4,5,2,3,1.Third row: Take the second row and apply the same transformation. So, a_{4}, a_{5}, a_{2}, a_{3}, a_{1} of the second row: 2,3,1,4,5.Fourth row: Take the third row and apply the transformation: a_{4}, a_{5}, a_{2}, a_{3}, a_{1} of the third row: 4,5,2,3,1.So, it's cycling every two rows. That means the period is 2, but n = 5, which is larger. So, in this case, the table doesn't fill properly because the cycle repeats before n rows.But according to the problem statement, with m and n - k coprime, the table should have each column containing all numbers. So, either my example is wrong, or my understanding is flawed.Wait, maybe m and n - k being coprime is not sufficient? Or perhaps I made a mistake in choosing m and k. Let me check the conditions again.The problem states m > k and m and n - k are coprime. In my example, m = 3, k = 1, n = 5, so n - k = 4, and gcd(3,4) = 1, which satisfies the condition. So, why isn't the table filling correctly?Hmm, perhaps the issue is that in my example, the shift size is n - k = 4, but m = 3. So, the shift is 4, which is larger than m. Maybe that's causing the cycling.Wait, no. The shift is not n - k, but rather the way the elements are rearranged. Maybe I need to think in terms of permutation cycles.Each number is being shifted by a certain amount each row. The shift is determined by the transformation rule. So, perhaps the shift is equivalent to moving the first k elements to the end, and shifting the rest accordingly.In permutation terms, this is a specific permutation of the columns. If I can show that this permutation is a single cycle of length n, then after n rows, each number would have cycled through all columns, ensuring each column contains all numbers.But in my example, the permutation seems to have a cycle of length 2, not 5. So, that contradicts the idea that it's a single cycle.Wait, maybe I need to analyze the permutation more carefully. Let's consider the permutation as a function f where f(i) = position of the element in the next row.Given the transformation, the first k elements move to positions n - k + 1 to n, the next m - k elements move to positions 1 to m - k, and the last n - m elements move to positions m - k + 1 to n - k.Wait, that might not be accurate. Let me try to map the positions.Suppose we have positions 1 to n. After the transformation, the new position of element at position i is:- If i is in 1 to k, it moves to position n - k + i.- If i is in k + 1 to m, it moves to position i - k.- If i is in m + 1 to n, it moves to position i - m + k.Wait, is that correct? Let me see.Original positions: 1, 2, ..., k, k+1, ..., m, m+1, ..., n.After transformation: m+1, ..., n, k+1, ..., m, 1, ..., k.So, position 1 (which is 1) moves to position n - k + 1.Position 2 moves to n - k + 2....Position k moves to n.Position k + 1 moves to 1.Position k + 2 moves to 2....Position m moves to m - k.Position m + 1 moves to m - k + 1....Position n moves to n - m + k.So, the permutation function f is:f(i) = i + (n - k) if i ≤ k,f(i) = i - k if k < i ≤ m,f(i) = i - m + k if m < i ≤ n.But since we're working modulo n, because it's a permutation, we can think of f(i) as:f(i) ≡ i + s (mod n) where s = n - k for i ≤ k,f(i) ≡ i - k (mod n) for k < i ≤ m,f(i) ≡ i - m + k (mod n) for m < i ≤ n.Wait, but this seems inconsistent. Maybe I need to think of it as a permutation composed of different shifts.Alternatively, perhaps it's better to model the permutation as a combination of shifts and then analyze its cycle structure.Given that m and n - k are coprime, maybe the permutation has a single cycle of length n, ensuring that each number cycles through all columns before repeating.But in my earlier example, it didn't seem to be the case. So, perhaps my example was not chosen correctly or I made a mistake in the transformation.Alternatively, maybe the key is that the shift size s = n - k and m are coprime, which implies that the permutation has a single cycle.Wait, in permutation group theory, if a permutation is composed of a single cycle of length n, then it has order n, meaning that applying it n times brings you back to the identity.So, if the permutation induced by the transformation is a single cycle of length n, then after n transformations, each element returns to its original position, implying that in n rows, each element has been in every column exactly once.Therefore, each column would contain all numbers from 1 to n.But in my example, the permutation seemed to have a cycle of length 2, which contradicts this. So, perhaps my example was not constructed correctly.Wait, in my example, n = 5, k = 1, m = 3. So, s = n - k = 4. The permutation function f is:For i = 1: f(1) = 1 + 4 = 5.For i = 2: f(2) = 2 - 1 = 1.For i = 3: f(3) = 3 - 1 = 2.For i = 4: f(4) = 4 - 3 + 1 = 2.Wait, that doesn't make sense. Let me recast it properly.Wait, perhaps I need to think of the permutation as a mapping from the current position to the next position.Given the transformation rule, the new row is constructed by taking the last n - m elements, then the middle m - k elements, then the first k elements.So, in terms of positions:- The first k positions in the new row correspond to the last k positions of the previous row.- The next m - k positions correspond to the middle m - k positions of the previous row.- The last n - m positions correspond to the first n - m positions of the previous row.Wait, that might not be accurate. Let me try to map it correctly.Suppose we have a row: [a₁, a₂, ..., aₖ, a_{k+1}, ..., aₘ, a_{m+1}, ..., aₙ].The next row is: [a_{m+1}, ..., aₙ, a_{k+1}, ..., aₘ, a₁, ..., aₖ].So, the first position in the new row is a_{m+1}, which was at position m + 1 in the previous row.The second position is a_{m+2}, which was at position m + 2....The (n - m)th position is aₙ, which was at position n.Then, the (n - m + 1)th position is a_{k+1}, which was at position k + 1....The mth position is aₘ, which was at position m.Then, the (m + 1)th position is a₁, which was at position 1....The nth position is aₖ, which was at position k.So, the permutation function f maps position i to:- If i ≤ n - m, then f(i) = m + i.- If n - m < i ≤ m, then f(i) = i - (n - m).- If m < i ≤ n, then f(i) = i - m + k.Wait, that seems complicated. Maybe it's better to think in terms of modular arithmetic.Alternatively, perhaps we can consider the permutation as a combination of shifts. The first k elements are moved to the end, and the rest are shifted accordingly.But I'm getting confused. Maybe I should look for a different approach.Another idea: Since m and n - k are coprime, the number of shifts needed to cycle through all positions is n. Therefore, each number will appear in every column exactly once after n rows.But in my earlier example, it didn't seem to hold. Maybe I need to choose a different example where m and n - k are coprime and see if it works.Let me try n = 7, k = 2, m = 4. So, n - k = 5, and m = 4, which are coprime (gcd(4,5)=1).First row: 1, 2, 3, 4, 5, 6, 7.Second row: a_{5}, a_{6}, a_{7}, a_{3}, a_{4}, a_{1}, a_{2} = 5, 6, 7, 3, 4, 1, 2.Third row: a_{5}, a_{6}, a_{7}, a_{3}, a_{4}, a_{1}, a_{2} of the second row: 3, 4, 1, 2, 5, 6, 7.Fourth row: a_{5}, a_{6}, a_{7}, a_{3}, a_{4}, a_{1}, a_{2} of the third row: 5, 6, 7, 3, 4, 1, 2.Wait, again, it's cycling between two rows. So, in this case, the period is 2, not 7. That's a problem.But according to the problem statement, with m and n - k coprime, the table should have each column containing all numbers. So, why is it not happening in my examples?Maybe I'm misunderstanding the transformation rule. Let me read it again."Suppose that in some row, the numbers are recorded as a₁, ..., aₖ, a_{k+1}, ..., aₘ, a_{m+1}, ..., aₙ. Then, in the next row, the same numbers are recorded in the following order: a_{m+1}, ..., aₙ, a_{k+1}, ..., aₘ, a₁, ..., aₖ."So, the next row is constructed by taking the last n - m elements, then the middle m - k elements, then the first k elements.In my first example with n=5, k=1, m=3:First row: 1,2,3,4,5.Second row: a_{4}, a_{5}, a_{2}, a_{3}, a_{1} = 4,5,2,3,1.Third row: a_{4}, a_{5}, a_{2}, a_{3}, a_{1} of the second row: 2,3,1,4,5.Fourth row: a_{4}, a_{5}, a_{2}, a_{3}, a_{1} of the third row: 4,5,2,3,1.So, it's cycling between two rows. Similarly, in the second example with n=7, k=2, m=4, it's cycling between two rows.This suggests that the period is 2, regardless of n, which contradicts the problem statement.Wait, maybe the period is not necessarily n, but depends on the shift. If the shift size s and n are coprime, then the period is n. But in my examples, s = n - k, and m is involved as well.Wait, perhaps the shift is not just s = n - k, but a combination of shifts. Maybe the permutation is more complex.Alternatively, perhaps the key is that the permutation can be decomposed into cycles whose lengths are divisors of n, and since m and n - k are coprime, the only possible cycle length is n, making it a single cycle.But in my examples, the cycle length was 2, which divides n=5 and n=7. So, that contradicts the idea.Wait, maybe I'm not considering the entire permutation correctly. Let me try to write down the permutation mapping for my first example.n=5, k=1, m=3.Positions: 1,2,3,4,5.Transformation: next row is a_{4}, a_{5}, a_{2}, a_{3}, a_{1}.So, the permutation f is:f(1) = 5 (since a₁ moves to position 5),f(2) = 3 (a₂ moves to position 3),f(3) = 4 (a₃ moves to position 4),f(4) = 1 (a₄ moves to position 1),f(5) = 2 (a₅ moves to position 2).So, the permutation is (1 5 2 3 4). Let's see:1 → 5,5 → 2,2 → 3,3 → 4,4 → 1.So, it's a single cycle of length 5. Therefore, the period is 5, meaning that after 5 transformations, the elements return to their original positions.Wait, but in my earlier calculation, after two transformations, it started repeating. That must be a mistake.Wait, let's apply the permutation step by step.First row: 1,2,3,4,5.Second row: f(1)=5, f(2)=3, f(3)=4, f(4)=1, f(5)=2 → 5,3,4,1,2.Third row: Apply f to the second row: f(5)=2, f(3)=4, f(4)=1, f(1)=5, f(2)=3 → 2,4,1,5,3.Fourth row: Apply f to the third row: f(2)=3, f(4)=1, f(1)=5, f(5)=2, f(3)=4 → 3,1,5,2,4.Fifth row: Apply f to the fourth row: f(3)=4, f(1)=5, f(5)=2, f(2)=3, f(4)=1 → 4,5,2,3,1.Sixth row: Apply f to the fifth row: f(4)=1, f(5)=2, f(2)=3, f(3)=4, f(1)=5 → 1,2,3,4,5.Ah, so it does cycle back to the original after 5 transformations. My earlier mistake was that I misapplied the transformation. I thought the second row was 4,5,2,3,1, but actually, it's 5,3,4,1,2.So, in this case, the permutation is indeed a single cycle of length 5, ensuring that after 5 rows, each number has been in every column exactly once. Therefore, each column contains all numbers from 1 to 5.Similarly, in the second example with n=7, k=2, m=4, let's see:First row: 1,2,3,4,5,6,7.Second row: a_{5}, a_{6}, a_{7}, a_{3}, a_{4}, a_{1}, a_{2} = 5,6,7,3,4,1,2.Third row: Apply f to the second row: f(5)=6, f(6)=7, f(7)=3, f(3)=4, f(4)=1, f(1)=2, f(2)=5 → 6,7,4,1,2,5,3.Fourth row: Apply f to the third row: f(6)=7, f(7)=4, f(4)=1, f(1)=2, f(2)=5, f(5)=3, f(3)=6 → 7,4,2,5,3,6,1.Fifth row: Apply f to the fourth row: f(7)=4, f(4)=2, f(2)=5, f(5)=3, f(3)=6, f(6)=1, f(1)=7 → 4,2,5,3,6,1,7.Sixth row: Apply f to the fifth row: f(4)=2, f(2)=5, f(5)=3, f(3)=6, f(6)=1, f(1)=7, f(7)=4 → 2,5,3,6,1,7,4.Seventh row: Apply f to the sixth row: f(2)=5, f(5)=3, f(3)=6, f(6)=1, f(1)=7, f(7)=4, f(4)=2 → 5,3,6,1,7,4,2.Eighth row: Apply f to the seventh row: f(5)=3, f(3)=6, f(6)=1, f(1)=7, f(7)=4, f(4)=2, f(2)=5 → 3,6,1,7,4,2,5.Ninth row: Apply f to the eighth row: f(3)=6, f(6)=1, f(1)=7, f(7)=4, f(4)=2, f(2)=5, f(5)=3 → 6,1,7,4,2,5,3.Tenth row: Apply f to the ninth row: f(6)=1, f(1)=7, f(7)=4, f(4)=2, f(2)=5, f(5)=3, f(3)=6 → 1,7,4,2,5,3,6.Eleventh row: Apply f to the tenth row: f(1)=7, f(7)=4, f(4)=2, f(2)=5, f(5)=3, f(3)=6, f(6)=1 → 7,4,2,5,3,6,1.Twelfth row: Apply f to the eleventh row: f(7)=4, f(4)=2, f(2)=5, f(5)=3, f(3)=6, f(6)=1, f(1)=7 → 4,2,5,3,6,1,7.Thirteenth row: Apply f to the twelfth row: f(4)=2, f(2)=5, f(5)=3, f(3)=6, f(6)=1, f(1)=7, f(7)=4 → 2,5,3,6,1,7,4.Fourteenth row: Apply f to the thirteenth row: f(2)=5, f(5)=3, f(3)=6, f(6)=1, f(1)=7, f(7)=4, f(4)=2 → 5,3,6,1,7,4,2.Wait, this is getting too long. It seems like it's not cycling back to the original row after 7 transformations. Maybe I made a mistake in the permutation.Alternatively, perhaps the permutation is more complex and has multiple cycles. But given that m and n - k are coprime, it should have a single cycle.Wait, maybe I need to reconsider the permutation mapping. Let me try to write the permutation for n=7, k=2, m=4.The transformation is: next row is a_{5}, a_{6}, a_{7}, a_{3}, a_{4}, a_{1}, a_{2}.So, the permutation f maps:f(1) = 6 (since a₁ moves to position 6),f(2) = 7 (a₂ moves to position 7),f(3) = 4 (a₃ moves to position 4),f(4) = 5 (a₄ moves to position 5),f(5) = 1 (a₅ moves to position 1),f(6) = 2 (a₆ moves to position 2),f(7) = 3 (a₇ moves to position 3).So, the permutation is (1 6 2 7 3 4 5). Let's see:1 → 6,6 → 2,2 → 7,7 → 3,3 → 4,4 → 5,5 → 1.Yes, it's a single cycle of length 7. Therefore, after 7 transformations, the elements return to their original positions.So, in this case, the table would have each column containing all numbers from 1 to 7 after 7 rows. My earlier mistake was not following the permutation correctly.Therefore, the key idea is that the permutation induced by the transformation is a single cycle of length n, which happens because m and n - k are coprime. This ensures that each number cycles through all columns before repeating, resulting in each column containing all numbers from 1 to n.So, to summarize, the transformation defines a permutation of the columns, and because m and n - k are coprime, this permutation is a single cycle of length n. Therefore, after n rows, each number has been in every column exactly once, proving that every column contains all numbers from 1 to n.

💡Okay, so I need to prove these statements about numbers raised to powers. Let me start by understanding each statement clearly.First, the number 1 raised to any power is always 1. That seems straightforward because multiplying 1 by itself any number of times will always result in 1. For example, 1^2 = 1*1 = 1, 1^3 = 1*1*1 = 1, and so on. So, this part seems simple enough.Next, the statement says that a number slightly greater than 1, when multiplied by itself repeatedly, yields larger and larger numbers, and its power increases indefinitely as the exponent increases. Hmm, so if I take a number like 1.1 and raise it to higher and higher powers, it should get bigger and bigger without bound. I think this is related to exponential growth. For example, 1.1^1 = 1.1, 1.1^2 = 1.21, 1.1^3 = 1.331, and so on. It does seem like it's increasing each time. But how do I prove that it will keep increasing indefinitely?Maybe I can use induction or some property of exponents. Let me think about induction. If I assume that (1 + α)^n is greater than some value, then (1 + α)^(n+1) would be (1 + α)^n * (1 + α), which is greater than the previous value times (1 + α). Since α is positive, multiplying by (1 + α) will always make it larger. So, by induction, it's always increasing.But does it increase without bound? I think for any large number N, I can find an exponent n such that (1 + α)^n > N. Maybe using logarithms? If I take the natural log of both sides, ln((1 + α)^n) = n * ln(1 + α). Since ln(1 + α) is positive (because α > 0), n can be chosen large enough so that n * ln(1 + α) > ln(N), which implies (1 + α)^n > N. That makes sense.Now, the converse: a number slightly less than 1, when raised to higher powers, decreases indefinitely and approaches zero. Let's take a number like 0.9. 0.9^1 = 0.9, 0.9^2 = 0.81, 0.9^3 = 0.729, and so on. It does seem to be getting smaller each time. How do I prove that it approaches zero?Maybe similar to the previous case, but in reverse. If I take a number less than 1, say (1 - α) where 0 < α < 1, then (1 - α)^n. I can use logarithms again. Take ln((1 - α)^n) = n * ln(1 - α). Since ln(1 - α) is negative (because 1 - α < 1), as n increases, n * ln(1 - α) becomes more negative, which means (1 - α)^n approaches zero.Alternatively, I can relate it to the previous case by noting that (1 - α) = 1 / (1 + β) where β = α / (1 - α). Then, (1 - α)^n = 1 / (1 + β)^n. Since (1 + β)^n increases without bound, its reciprocal approaches zero. That seems like a solid argument.Wait, let me make sure I didn't make any mistakes. For the number slightly greater than 1, I used induction and logarithms to show it increases without bound. For the number slightly less than 1, I related it to the reciprocal of a number slightly greater than 1, which also makes sense because if one grows without bound, the other shrinks to zero.Is there another way to think about this? Maybe using limits. For the first case, the limit as n approaches infinity of (1 + α)^n is infinity. For the second case, the limit as n approaches infinity of (1 - α)^n is zero. These are standard results in calculus, but since I'm supposed to prove them, I need to show why these limits hold.I think the logarithm approach is sufficient because it allows me to relate the exponent to the logarithm, which is a known increasing function. By showing that the logarithm can be made larger than any given value, it implies that the original expression can be made larger than any given value, hence it tends to infinity. Similarly, for the decreasing case, the logarithm tends to negative infinity, meaning the original expression tends to zero.Let me summarize my thoughts:1. For 1^n = 1: This is straightforward because multiplying 1 by itself any number of times doesn't change its value.2. For (1 + α)^n where α > 0: Use induction to show it's increasing, and logarithms to show it can be made larger than any given number, hence it increases without bound.3. For (1 - α)^n where 0 < α < 1: Express it as the reciprocal of (1 + β)^n, which we already know increases without bound, so its reciprocal must approach zero.I think that covers all the statements. I should make sure each step is logically sound and that I haven't assumed anything without justification. The key idea is using logarithms to handle the exponents and relate them to linear functions, which are easier to analyze for growth and decay.Another thing to consider is whether these proofs hold for any real number exponent or just integers. The original statements don't specify, but I think they're referring to integer exponents since they mention "when multiplied by itself" repeatedly. However, the logarithm approach works for real exponents as well, so it's more general.In conclusion, I believe I have a solid understanding of how to prove each statement using induction and logarithmic properties. I'll proceed to write out the proofs step-by-step, making sure each part is clearly explained.

💡First, I need to identify the highest and lowest scores from the judges. The scores provided are 7.5, 8.8, 9.0, 6.0, and 8.5. The highest score is 9.0, and the lowest score is 6.0.Next, I will exclude these extreme scores and focus on the remaining scores: 7.5, 8.8, and 8.5.Now, I will calculate the sum of these remaining scores:7.5 + 8.8 + 8.5 = 24.8Finally, I will multiply this sum by the degree of difficulty, which is 3.2:24.8 × 3.2 = 79.36Therefore, the point value of the dive is 79.36.