💡Okay, so I have this problem here where I need to find the minimum value of ( a + 2b ) given that ( a ) and ( b ) are positive real numbers and they satisfy the equation ( frac{1}{a} + frac{2}{b} = 4 ). The options are A: 2, B: ( frac{9}{4} ), C: ( frac{5}{2} ), and D: 3.Hmm, let me think about how to approach this. I remember that when dealing with optimization problems like finding minima or maxima, methods like calculus or inequalities like AM-GM can be useful. Since the problem gives me a constraint and asks for the minimum of a linear expression, maybe I can use the method of Lagrange multipliers or perhaps apply the AM-GM inequality directly.Wait, the AM-GM inequality might be simpler here. Let me recall what it says. The Arithmetic Mean is always greater than or equal to the Geometric Mean for non-negative real numbers. So, if I can express ( a + 2b ) in terms that can be compared using AM-GM, that might help.But first, let me write down the given equation again:[frac{1}{a} + frac{2}{b} = 4]I need to find the minimum of ( a + 2b ). Maybe I can express ( a ) in terms of ( b ) or vice versa from the given equation and substitute it into the expression I need to minimize.Let me try solving for ( frac{1}{a} ):[frac{1}{a} = 4 - frac{2}{b}]So,[a = frac{1}{4 - frac{2}{b}} = frac{b}{4b - 2}]Hmm, that seems a bit messy. Maybe I can express ( b ) in terms of ( a ) instead.Starting again from the given equation:[frac{1}{a} + frac{2}{b} = 4]Let me solve for ( frac{2}{b} ):[frac{2}{b} = 4 - frac{1}{a}]So,[frac{2}{b} = frac{4a - 1}{a}]Therefore,[b = frac{2a}{4a - 1}]Okay, so now I have ( b ) expressed in terms of ( a ). Let me substitute this into ( a + 2b ):[a + 2b = a + 2 left( frac{2a}{4a - 1} right) = a + frac{4a}{4a - 1}]Now, I have ( a + 2b ) as a function of ( a ). Let me denote this function as ( f(a) ):[f(a) = a + frac{4a}{4a - 1}]I need to find the minimum value of ( f(a) ) for ( a > 0 ). Since ( b ) must also be positive, the denominator ( 4a - 1 ) must be positive as well. So,[4a - 1 > 0 implies a > frac{1}{4}]So, ( a ) must be greater than ( frac{1}{4} ).Now, to find the minimum of ( f(a) ), I can take the derivative of ( f(a) ) with respect to ( a ) and set it equal to zero.First, let's compute the derivative ( f'(a) ):[f(a) = a + frac{4a}{4a - 1}]Let me compute the derivative term by term.The derivative of ( a ) with respect to ( a ) is 1.Now, for the second term ( frac{4a}{4a - 1} ), I can use the quotient rule. The quotient rule states that if I have a function ( frac{u}{v} ), its derivative is ( frac{u'v - uv'}{v^2} ).Here, ( u = 4a ) and ( v = 4a - 1 ).So,[u' = 4][v' = 4]Therefore, the derivative of ( frac{4a}{4a - 1} ) is:[frac{4(4a - 1) - 4a(4)}{(4a - 1)^2} = frac{16a - 4 - 16a}{(4a - 1)^2} = frac{-4}{(4a - 1)^2}]So, putting it all together, the derivative of ( f(a) ) is:[f'(a) = 1 + left( frac{-4}{(4a - 1)^2} right) = 1 - frac{4}{(4a - 1)^2}]To find the critical points, set ( f'(a) = 0 ):[1 - frac{4}{(4a - 1)^2} = 0]Solving for ( a ):[frac{4}{(4a - 1)^2} = 1][(4a - 1)^2 = 4][4a - 1 = pm 2]Since ( a > frac{1}{4} ), ( 4a - 1 ) must be positive. Therefore, we take the positive square root:[4a - 1 = 2][4a = 3][a = frac{3}{4}]Okay, so the critical point is at ( a = frac{3}{4} ). Now, I need to check if this is a minimum.To confirm whether this critical point is a minimum, I can use the second derivative test.First, let's compute the second derivative ( f''(a) ).We have:[f'(a) = 1 - frac{4}{(4a - 1)^2}]Differentiating ( f'(a) ) with respect to ( a ):The derivative of 1 is 0.For the second term, ( -4(4a - 1)^{-2} ), using the chain rule:[frac{d}{da} left( -4(4a - 1)^{-2} right) = -4 times (-2)(4a - 1)^{-3} times 4 = 32(4a - 1)^{-3}]So,[f''(a) = 32(4a - 1)^{-3} = frac{32}{(4a - 1)^3}]Now, evaluate ( f''(a) ) at ( a = frac{3}{4} ):First, compute ( 4a - 1 ):[4 times frac{3}{4} - 1 = 3 - 1 = 2]So,[f''left( frac{3}{4} right) = frac{32}{2^3} = frac{32}{8} = 4]Since ( f''left( frac{3}{4} right) > 0 ), the function is concave upward at this point, which means it's a local minimum.Therefore, the minimum occurs at ( a = frac{3}{4} ). Now, let's find the corresponding ( b ) value.From earlier, we had:[b = frac{2a}{4a - 1}]Substituting ( a = frac{3}{4} ):[b = frac{2 times frac{3}{4}}{4 times frac{3}{4} - 1} = frac{frac{3}{2}}{3 - 1} = frac{frac{3}{2}}{2} = frac{3}{4}]So, ( b = frac{3}{4} ).Now, let's compute ( a + 2b ):[a + 2b = frac{3}{4} + 2 times frac{3}{4} = frac{3}{4} + frac{6}{4} = frac{9}{4}]So, the minimum value of ( a + 2b ) is ( frac{9}{4} ).Wait, but let me double-check my calculations to make sure I didn't make any mistakes.Starting from the given equation:[frac{1}{a} + frac{2}{b} = 4]With ( a = frac{3}{4} ) and ( b = frac{3}{4} ), let's plug these values back into the equation:[frac{1}{frac{3}{4}} + frac{2}{frac{3}{4}} = frac{4}{3} + frac{8}{3} = frac{12}{3} = 4]Yes, that satisfies the equation. So, the calculations seem correct.Alternatively, I could have used the AM-GM inequality directly without calculus. Let me try that approach as a verification.We have:[frac{1}{a} + frac{2}{b} = 4]We need to minimize ( a + 2b ).Let me denote ( x = a ) and ( y = 2b ). Then, the expression to minimize becomes ( x + y ).From the given equation:[frac{1}{x} + frac{2}{b} = 4]But since ( y = 2b ), then ( b = frac{y}{2} ). Substituting into the equation:[frac{1}{x} + frac{2}{frac{y}{2}} = frac{1}{x} + frac{4}{y} = 4]So, we have:[frac{1}{x} + frac{4}{y} = 4]And we need to minimize ( x + y ).Now, let's apply the Cauchy-Schwarz inequality or perhaps the AM-GM inequality.Alternatively, using the method of Lagrange multipliers, but I think using the Cauchy-Schwarz inequality might be more straightforward here.Recall that for positive real numbers, the Cauchy-Schwarz inequality states that:[left( frac{1}{x} + frac{4}{y} right)(x + y) geq left( sqrt{frac{1}{x} cdot x} + sqrt{frac{4}{y} cdot y} right)^2 = (1 + 2)^2 = 9]So,[left( frac{1}{x} + frac{4}{y} right)(x + y) geq 9]Given that ( frac{1}{x} + frac{4}{y} = 4 ), we substitute:[4(x + y) geq 9][x + y geq frac{9}{4}]So, the minimum value of ( x + y ) is ( frac{9}{4} ), which occurs when equality holds in the Cauchy-Schwarz inequality. Equality holds when:[frac{frac{1}{x}}{x} = frac{frac{4}{y}}{y}][frac{1}{x^2} = frac{4}{y^2}][frac{1}{x} = frac{2}{y}][y = 2x]So, ( y = 2x ). Since ( y = 2b ) and ( x = a ), this implies ( 2b = 2a ), so ( b = a ).Therefore, when ( a = b ), the minimum is achieved. Substituting back into the given equation:[frac{1}{a} + frac{2}{a} = 4][frac{3}{a} = 4][a = frac{3}{4}]So, ( a = frac{3}{4} ) and ( b = frac{3}{4} ), which gives ( a + 2b = frac{3}{4} + frac{6}{4} = frac{9}{4} ).This confirms the result I obtained earlier using calculus.Alternatively, another approach is to use substitution and then apply AM-GM.Let me try that as well.We have:[frac{1}{a} + frac{2}{b} = 4]Let me denote ( u = frac{1}{a} ) and ( v = frac{2}{b} ). Then, the equation becomes:[u + v = 4]We need to express ( a + 2b ) in terms of ( u ) and ( v ).Since ( u = frac{1}{a} ), ( a = frac{1}{u} ).Similarly, ( v = frac{2}{b} ), so ( b = frac{2}{v} ).Therefore,[a + 2b = frac{1}{u} + 2 times frac{2}{v} = frac{1}{u} + frac{4}{v}]So, we need to minimize ( frac{1}{u} + frac{4}{v} ) subject to ( u + v = 4 ), where ( u > 0 ) and ( v > 0 ).This seems similar to the earlier problem. Let me apply the AM-GM inequality here.Consider the terms ( frac{1}{u} ) and ( frac{4}{v} ). Let me see if I can relate them to ( u + v ).Alternatively, using the method of Lagrange multipliers, but perhaps using the Cauchy-Schwarz inequality again.Using Cauchy-Schwarz:[left( frac{1}{u} + frac{4}{v} right)(u + v) geq left( sqrt{frac{1}{u} cdot u} + sqrt{frac{4}{v} cdot v} right)^2 = (1 + 2)^2 = 9]So,[left( frac{1}{u} + frac{4}{v} right)(u + v) geq 9]Given that ( u + v = 4 ), we have:[left( frac{1}{u} + frac{4}{v} right) times 4 geq 9][frac{1}{u} + frac{4}{v} geq frac{9}{4}]Thus, the minimum value of ( a + 2b ) is ( frac{9}{4} ), achieved when equality holds in Cauchy-Schwarz, which, as before, occurs when ( frac{frac{1}{u}}{u} = frac{frac{4}{v}}{v} ), leading to ( u = frac{1}{2}v ). Substituting back into ( u + v = 4 ):[frac{1}{2}v + v = 4][frac{3}{2}v = 4][v = frac{8}{3}][u = frac{1}{2} times frac{8}{3} = frac{4}{3}]Then, ( a = frac{1}{u} = frac{3}{4} ) and ( b = frac{2}{v} = frac{2}{frac{8}{3}} = frac{3}{4} ), which again gives ( a + 2b = frac{9}{4} ).So, through multiple approaches—calculus, Cauchy-Schwarz inequality, substitution and AM-GM—I consistently arrive at the conclusion that the minimum value of ( a + 2b ) is ( frac{9}{4} ).Just to ensure there are no mistakes, let me consider if there's another method, perhaps using substitution and completing the square.Starting from the expression ( a + 2b ) and the constraint ( frac{1}{a} + frac{2}{b} = 4 ).Let me express ( b ) in terms of ( a ) as I did earlier:[b = frac{2a}{4a - 1}]Then,[a + 2b = a + 2 times frac{2a}{4a - 1} = a + frac{4a}{4a - 1}]Let me denote ( t = 4a - 1 ). Then, ( a = frac{t + 1}{4} ).Substituting into ( a + 2b ):[a + 2b = frac{t + 1}{4} + frac{4 times frac{t + 1}{4}}{t} = frac{t + 1}{4} + frac{t + 1}{t}]Simplify:[= frac{t + 1}{4} + frac{t + 1}{t}][= frac{t + 1}{4} + 1 + frac{1}{t}][= frac{t}{4} + frac{1}{4} + 1 + frac{1}{t}][= frac{t}{4} + frac{5}{4} + frac{1}{t}]Now, let me denote ( f(t) = frac{t}{4} + frac{5}{4} + frac{1}{t} ). I need to find the minimum of ( f(t) ) for ( t > 0 ).Taking the derivative of ( f(t) ):[f'(t) = frac{1}{4} - frac{1}{t^2}]Setting ( f'(t) = 0 ):[frac{1}{4} - frac{1}{t^2} = 0][frac{1}{4} = frac{1}{t^2}][t^2 = 4][t = 2 quad (text{since } t > 0)]So, the critical point is at ( t = 2 ). Let's check the second derivative to confirm it's a minimum.Second derivative:[f''(t) = frac{2}{t^3}]At ( t = 2 ):[f''(2) = frac{2}{8} = frac{1}{4} > 0]So, it's a local minimum. Therefore, the minimum occurs at ( t = 2 ).Now, compute ( f(2) ):[f(2) = frac{2}{4} + frac{5}{4} + frac{1}{2} = frac{1}{2} + frac{5}{4} + frac{1}{2} = frac{1}{2} + frac{1}{2} + frac{5}{4} = 1 + frac{5}{4} = frac{9}{4}]Again, the minimum value is ( frac{9}{4} ).This method also confirms the previous results. It seems that no matter which approach I take—calculus, inequalities, substitution—the minimum value consistently comes out to be ( frac{9}{4} ).Just to be thorough, let me consider if there's any possibility that the minimum could be lower. For example, if I take ( a ) approaching ( frac{1}{4} ) from above, what happens to ( a + 2b )?As ( a ) approaches ( frac{1}{4} ), ( b ) approaches infinity because ( b = frac{2a}{4a - 1} ), and the denominator approaches zero. So, ( a + 2b ) would approach infinity.On the other hand, as ( a ) increases beyond ( frac{3}{4} ), ( b ) decreases, but ( a + 2b ) might increase again. Let me test with ( a = 1 ):[b = frac{2 times 1}{4 times 1 - 1} = frac{2}{3}][a + 2b = 1 + 2 times frac{2}{3} = 1 + frac{4}{3} = frac{7}{3} approx 2.333]Which is greater than ( frac{9}{4} = 2.25 ).Similarly, if I take ( a = frac{1}{2} ):[b = frac{2 times frac{1}{2}}{4 times frac{1}{2} - 1} = frac{1}{2 - 1} = 1][a + 2b = frac{1}{2} + 2 times 1 = frac{1}{2} + 2 = frac{5}{2} = 2.5]Which is also greater than ( frac{9}{4} ).Therefore, it seems that ( frac{9}{4} ) is indeed the minimum value.In conclusion, after exploring multiple methods—calculus, substitution, inequalities, and testing boundary cases—it is clear that the minimum value of ( a + 2b ) under the given constraint is ( frac{9}{4} ).

💡First, I need to determine the total number of games Henry played by adding up his wins, losses, and draws.He won 2 games, lost 2 games, and there were 10 draws.Adding these together: 2 (wins) + 2 (losses) + 10 (draws) equals 14 games.Therefore, Henry played a total of 14 games.

💡First, I need to determine the circumference of the circular field to find out the total length of the fence required. The circumference of a circle is calculated using the formula:[ C = pi times d ]where ( d ) is the diameter of the circle. Given that the diameter is 16 meters, I can substitute this value into the formula:[ C = pi times 16 , text{m} ]Using the approximation ( pi approx 3.14159 ), the circumference becomes:[ C approx 3.14159 times 16 , text{m} ][ C approx 50.26544 , text{m} ]Next, I need to calculate the total cost of fencing. The cost per meter is Rs. 3, so the total cost (( TC )) is:[ TC = C times text{Rate per meter} ][ TC approx 50.26544 , text{m} times 3 , text{Rs/m} ][ TC approx 150.79632 , text{Rs} ]Finally, rounding the total cost to the nearest whole number, the total cost of fencing around the circular field is approximately Rs. 151.

💡Alright, so I've got this probability problem here, and I need to figure out the probability of a specific outcome when drawing slips from a hat. Let me try to break it down step by step.First, the setup: There are 60 slips in a hat, each labeled with a number from 1 to 20. Each number appears on three slips. So, that makes sense because 20 numbers times three slips each equals 60 slips total. Got that.Now, the problem says that five slips are drawn from the hat at random, without replacement. I need to find the probability, which they're calling 'r', that among these five slips, three of them have the same number, and the other two slips each have different numbers. Moreover, these two different numbers should be distinct from each other and also different from the number that appears on the three slips. Okay, that's a mouthful, but let me try to parse it.So, to rephrase, I need to calculate the probability that when I draw five slips, I get exactly one triplet (three slips with the same number) and two singletons (each with a unique number, different from the triplet and from each other). Got it.To find this probability, I know that probability is generally the number of favorable outcomes divided by the total number of possible outcomes. So, I need to figure out both the total number of ways to draw five slips and the number of ways to draw five slips that meet the specified condition.Let's start with the total number of possible outcomes. Since we're drawing five slips from 60 without replacement, the total number of possible combinations is given by the combination formula:Total outcomes = C(60, 5)Where C(n, k) is the combination of n items taken k at a time. I can calculate this value, but maybe I'll leave it as C(60, 5) for now and compute it later.Now, moving on to the number of favorable outcomes. This is the tricky part. I need to count how many ways I can draw five slips such that three are the same number, and the other two are different numbers, each appearing only once, and all distinct from each other and from the triplet.Let me break this down into steps:1. **Choose the number for the triplet:** There are 20 different numbers, each appearing on three slips. So, I need to choose one of these 20 numbers to be the triplet. The number of ways to do this is C(20, 1).2. **Choose the three slips for the triplet:** Once I've chosen the number, I need to select all three slips of that number. Since each number has exactly three slips, the number of ways to choose all three slips is C(3, 3), which is 1.3. **Choose the numbers for the two singletons:** Now, the remaining two slips must each have different numbers, and these numbers must be different from the triplet number. Since we've already chosen one number for the triplet, there are 19 numbers left. We need to choose two distinct numbers from these 19. The number of ways to do this is C(19, 2).4. **Choose the slips for the singletons:** For each of the two numbers chosen in the previous step, we need to select one slip from the three available for each number. So, for each number, the number of ways to choose one slip is C(3, 1), and since we have two such numbers, the total number of ways is [C(3, 1)]^2.Putting it all together, the number of favorable outcomes is:Number of favorable outcomes = C(20, 1) × C(3, 3) × C(19, 2) × [C(3, 1)]^2Let me compute each part step by step.First, C(20, 1) is simply 20.Next, C(3, 3) is 1, as there's only one way to choose all three slips.Then, C(19, 2) is the number of ways to choose two distinct numbers from the remaining 19. The formula for combinations is C(n, k) = n! / (k! (n - k)!). So, plugging in the numbers:C(19, 2) = 19! / (2! × 17!) = (19 × 18) / (2 × 1) = 171.Okay, so that's 171 ways.Now, for the slips of the two singletons, each has three slips, and we need to choose one from each. So, for each number, it's C(3, 1) = 3. Since we have two numbers, it's 3 × 3 = 9.So, putting it all together:Number of favorable outcomes = 20 × 1 × 171 × 9.Let me compute that:20 × 171 = 3420.3420 × 9 = 30,780.Wait, hold on, 3420 × 9. Let me compute that again to make sure.3420 × 9:3000 × 9 = 27,000420 × 9 = 3,780Adding them together: 27,000 + 3,780 = 30,780.Yes, that's correct.So, the number of favorable outcomes is 30,780.Now, let's compute the total number of possible outcomes, which is C(60, 5).C(60, 5) = 60! / (5! × (60 - 5)!) = (60 × 59 × 58 × 57 × 56) / (5 × 4 × 3 × 2 × 1).Let me compute that step by step.First, compute the numerator: 60 × 59 × 58 × 57 × 56.Let me compute this in parts.60 × 59 = 3,5403,540 × 58 = Let's compute 3,540 × 50 = 177,000 and 3,540 × 8 = 28,320. Adding them together: 177,000 + 28,320 = 205,320.205,320 × 57 = Hmm, this is getting big. Let me break it down.205,320 × 50 = 10,266,000205,320 × 7 = 1,437,240Adding them: 10,266,000 + 1,437,240 = 11,703,24011,703,240 × 56 = Okay, this is getting really large. Maybe I should compute it differently.Wait, perhaps I made a miscalculation earlier. Let me try another approach.Alternatively, I can compute C(60, 5) using a calculator or look up the value, but since I need to compute it manually, let me see.Alternatively, I can compute it step by step:C(60, 5) = (60 × 59 × 58 × 57 × 56) / (5 × 4 × 3 × 2 × 1)First, compute the numerator:60 × 59 = 3,5403,540 × 58 = Let's compute 3,540 × 50 = 177,000 and 3,540 × 8 = 28,320. So, 177,000 + 28,320 = 205,320.205,320 × 57 = Let's compute 205,320 × 50 = 10,266,000 and 205,320 × 7 = 1,437,240. Adding them: 10,266,000 + 1,437,240 = 11,703,240.11,703,240 × 56 = Hmm, this is getting too big. Maybe I should compute the denominator first and see if I can simplify.Denominator: 5 × 4 × 3 × 2 × 1 = 120.So, C(60, 5) = (60 × 59 × 58 × 57 × 56) / 120.Alternatively, I can factor out the 120 from the numerator to simplify.Let me see:60 / 120 = 0.5, so 60 / 120 = 1/2.So, we can write C(60, 5) as (1/2) × 59 × 58 × 57 × 56.Wait, that might not be the most efficient way. Alternatively, let's factor out 5 from 60, which is 12 × 5, so 60 = 12 × 5.So, 60 / 5 = 12.So, now, C(60, 5) = (12 × 59 × 58 × 57 × 56) / (4 × 3 × 2 × 1).Wait, that might not help much either. Maybe I should compute the numerator first and then divide by 120.Let me compute the numerator:60 × 59 = 3,5403,540 × 58 = 205,320205,320 × 57 = Let's compute 205,320 × 50 = 10,266,000 and 205,320 × 7 = 1,437,240. Adding them: 10,266,000 + 1,437,240 = 11,703,240.11,703,240 × 56 = Hmm, 11,703,240 × 50 = 585,162,000 and 11,703,240 × 6 = 70,219,440. Adding them: 585,162,000 + 70,219,440 = 655,381,440.So, numerator is 655,381,440.Now, divide by 120:655,381,440 / 120.Let me compute that.Divide numerator and denominator by 10: 65,538,144 / 12.Now, 65,538,144 divided by 12.12 × 5,461,512 = 65,538,144.So, C(60, 5) = 5,461,512.Wait, let me verify that:12 × 5,461,512 = 12 × 5,000,000 = 60,000,00012 × 461,512 = Let's compute 12 × 400,000 = 4,800,00012 × 61,512 = 738,144So, 4,800,000 + 738,144 = 5,538,144Adding to 60,000,000: 60,000,000 + 5,538,144 = 65,538,144.Yes, that's correct.So, C(60, 5) = 5,461,512.Therefore, the total number of possible outcomes is 5,461,512.Earlier, I computed the number of favorable outcomes as 30,780.So, the probability 'r' is:r = 30,780 / 5,461,512Now, let's simplify this fraction.First, let's see if both numerator and denominator can be divided by 12.30,780 ÷ 12 = 2,5655,461,512 ÷ 12 = 455,126Wait, let me check:30,780 ÷ 12: 12 × 2,500 = 30,000, so 30,780 - 30,000 = 780. 780 ÷ 12 = 65. So, total is 2,500 + 65 = 2,565.Similarly, 5,461,512 ÷ 12: 12 × 455,000 = 5,460,000. 5,461,512 - 5,460,000 = 1,512. 1,512 ÷ 12 = 126. So, total is 455,000 + 126 = 455,126.So, now the fraction is 2,565 / 455,126.Let me see if this can be simplified further.Let's check the greatest common divisor (GCD) of 2,565 and 455,126.First, factorize 2,565:2,565 ÷ 5 = 513513 ÷ 3 = 171171 ÷ 3 = 5757 ÷ 3 = 19So, 2,565 = 5 × 3 × 3 × 3 × 19 = 5 × 3³ × 19.Now, factorize 455,126.Let's see if 455,126 is divisible by 2: Yes, it's even.455,126 ÷ 2 = 227,563.Now, check if 227,563 is divisible by 3: 2 + 2 + 7 + 5 + 6 + 3 = 25, which is not divisible by 3, so no.Check divisibility by 5: Ends with 3, so no.Check divisibility by 7: Let's see, 227,563 ÷ 7. 7 × 32,500 = 227,500. 227,563 - 227,500 = 63. 63 ÷ 7 = 9. So, 227,563 = 7 × 32,509.Now, check 32,509.Check divisibility by 7: 7 × 4,644 = 32,508. 32,509 - 32,508 = 1. Not divisible by 7.Check divisibility by 11: 3 - 2 + 5 - 0 + 9 = 15, which is not divisible by 11.Check divisibility by 13: Let's see, 13 × 2,500 = 32,500. 32,509 - 32,500 = 9. Not divisible by 13.Check divisibility by 17: 17 × 1,912 = 32,504. 32,509 - 32,504 = 5. Not divisible by 17.Check divisibility by 19: 19 × 1,711 = 32,509? Let's see, 19 × 1,700 = 32,300. 32,509 - 32,300 = 209. 209 ÷ 19 = 11. So, yes, 19 × 11 = 209. So, 32,509 = 19 × (1,700 + 11) = 19 × 1,711.So, 32,509 = 19 × 1,711.Now, check 1,711.Check divisibility by 19: 19 × 90 = 1,710. 1,711 - 1,710 = 1. Not divisible by 19.Check divisibility by 23: 23 × 74 = 1,702. 1,711 - 1,702 = 9. Not divisible by 23.Check divisibility by 29: 29 × 59 = 1,711? Let's see, 29 × 60 = 1,740. 1,740 - 29 = 1,711. Yes, so 29 × 59 = 1,711.So, putting it all together:455,126 = 2 × 7 × 19 × 29 × 59.And 2,565 = 5 × 3³ × 19.So, the common factor is 19.Therefore, we can divide numerator and denominator by 19.2,565 ÷ 19 = 135.455,126 ÷ 19 = Let's compute 455,126 ÷ 19.19 × 24,000 = 456,000, which is more than 455,126. So, 19 × 24,000 = 456,000.Subtract: 455,126 - 456,000 = -874. Hmm, that's negative, so maybe 19 × 23,950 = ?Wait, perhaps a better way is to compute 455,126 ÷ 19.19 × 20,000 = 380,000455,126 - 380,000 = 75,12619 × 3,000 = 57,00075,126 - 57,000 = 18,12619 × 900 = 17,10018,126 - 17,100 = 1,02619 × 54 = 1,026So, total is 20,000 + 3,000 + 900 + 54 = 23,954.So, 455,126 ÷ 19 = 23,954.Therefore, the simplified fraction is 135 / 23,954.Now, let's see if this can be simplified further.Factorize 135: 135 = 5 × 27 = 5 × 3³.Factorize 23,954: Let's check divisibility.23,954 is even, so divide by 2: 23,954 ÷ 2 = 11,977.Check if 11,977 is prime.Check divisibility by 3: 1 + 1 + 9 + 7 + 7 = 25, not divisible by 3.Divisible by 5? Ends with 7, no.Check divisibility by 7: 7 × 1,711 = 11,977? Let's see, 7 × 1,700 = 11,900. 11,977 - 11,900 = 77. 77 ÷ 7 = 11. So, yes, 7 × (1,700 + 11) = 7 × 1,711 = 11,977.So, 11,977 = 7 × 1,711.Now, check 1,711.Earlier, we saw that 1,711 = 29 × 59.So, putting it all together:23,954 = 2 × 7 × 29 × 59.And 135 = 5 × 3³.No common factors between 135 and 23,954, since 135 has factors 3 and 5, while 23,954 has factors 2, 7, 29, 59.Therefore, the fraction 135 / 23,954 is in its simplest form.So, the probability 'r' simplifies to 135 / 23,954.But let me double-check my calculations to make sure I didn't make any errors.First, number of favorable outcomes: 20 × 1 × 171 × 9 = 30,780. That seems correct.Total outcomes: C(60, 5) = 5,461,512. That also seems correct.Then, 30,780 / 5,461,512 = 2,565 / 455,126 after dividing numerator and denominator by 12.Then, factoring both, found GCD of 19, so divided by 19 to get 135 / 23,954.Yes, that seems correct.Alternatively, I can express this as a decimal to check the probability.Compute 135 ÷ 23,954.Let me compute that.23,954 goes into 135 zero times. So, 0.Add a decimal point and a zero: 1350 ÷ 23,954 = 0.056...Wait, 23,954 × 0.005 = 119.77So, 0.005 is about 119.77, which is less than 135.So, 0.005 × 23,954 = 119.77Subtract from 135: 135 - 119.77 = 15.23Bring down another zero: 152.3 ÷ 23,954 ≈ 0.00635So, total is approximately 0.005 + 0.00635 ≈ 0.01135.Wait, that seems low. Let me check with another method.Alternatively, 135 / 23,954 ≈ 0.00563.Wait, 23,954 × 0.005 = 119.7723,954 × 0.0006 = approximately 14.37So, 0.005 + 0.0006 = 0.0056, which gives 119.77 + 14.37 ≈ 134.14, which is close to 135.So, approximately 0.00563.So, about 0.563%.That seems reasonable for such a probability.Alternatively, I can leave it as the fraction 135/23,954, but perhaps we can write it in a simpler form or check if it can be reduced further.But as we saw earlier, 135 and 23,954 have no common factors besides 1, so that's the simplest form.Alternatively, I can write it as 135/23,954, but maybe the original fraction 30,780/5,461,512 can be simplified differently.Wait, let me check if 30,780 and 5,461,512 have any other common factors.We already divided by 12 and then by 19, but let me see if there's another common factor.30,780 = 2² × 3³ × 5 × 195,461,512 = 2³ × 3 × 7 × 19 × 29 × 59So, common factors are 2², 3, and 19.We already divided by 12 (which is 2² × 3) and then by 19, so that's all the common factors.Therefore, 135/23,954 is indeed the simplest form.Alternatively, I can write the probability as 135/23,954 or approximately 0.00563, which is about 0.563%.So, to recap, the probability 'r' is 135/23,954, which simplifies from 30,780/5,461,512 after dividing both numerator and denominator by their greatest common divisor, which was 19 after initial division by 12.Therefore, the final answer is 135/23,954.But wait, let me check if 135 and 23,954 have any other common factors.135 is 5 × 3³.23,954 is 2 × 7 × 29 × 59.No common factors, so yes, 135/23,954 is the simplest form.Alternatively, if I want to write it in the original terms without simplifying, it's 30,780/5,461,512, but it's better to simplify it.So, I think that's the answer.

💡Okay, so I have this problem here: I need to find the sum of all integer values ( n ) for which ( binom{25}{n} + binom{25}{12} = binom{26}{13} ). Hmm, binomial coefficients, okay. I remember that binomial coefficients have some identities and properties, so maybe I can use those to solve this.First, let me recall Pascal's identity. It says that ( binom{n}{k} = binom{n-1}{k-1} + binom{n-1}{k} ). So, if I have two binomial coefficients with the same upper index and consecutive lower indices, their sum is another binomial coefficient with an upper index increased by one. Maybe this can help me here.Looking at the equation ( binom{25}{n} + binom{25}{12} = binom{26}{13} ), I notice that on the left side, both terms have the upper index 25, and the right side has 26. So perhaps I can apply Pascal's identity here.If I let ( n = 13 ), then the left side becomes ( binom{25}{13} + binom{25}{12} ). According to Pascal's identity, this should equal ( binom{26}{13} ). Let me check that: yes, ( binom{25}{12} + binom{25}{13} = binom{26}{13} ). So, ( n = 13 ) is definitely a solution.But the problem asks for all integer values of ( n ). So, is there another value of ( n ) that could satisfy this equation? Let me think.I know that binomial coefficients are symmetric, meaning ( binom{n}{k} = binom{n}{n - k} ). So, ( binom{25}{n} = binom{25}{25 - n} ). Maybe this symmetry can give another solution.If ( n = 12 ), then ( binom{25}{12} + binom{25}{12} = 2 times binom{25}{12} ). But does this equal ( binom{26}{13} )? Let me calculate both sides.First, ( binom{25}{12} ) is a specific number, and ( binom{26}{13} ) is another. I know that ( binom{26}{13} = frac{26}{13} times binom{25}{12} = 2 times binom{25}{12} ). Wait, so ( 2 times binom{25}{12} = binom{26}{13} ). That means if ( n = 12 ), the left side becomes ( binom{25}{12} + binom{25}{12} = 2 times binom{25}{12} = binom{26}{13} ). So, ( n = 12 ) is also a solution.Wait, hold on. Earlier, I thought ( n = 13 ) was a solution, and now ( n = 12 ) is also a solution. So, are both 12 and 13 solutions? Let me verify.For ( n = 13 ): ( binom{25}{13} + binom{25}{12} = binom{26}{13} ). That's correct by Pascal's identity.For ( n = 12 ): ( binom{25}{12} + binom{25}{12} = 2 times binom{25}{12} ). And since ( binom{26}{13} = 2 times binom{25}{12} ), that's also correct.So, both ( n = 12 ) and ( n = 13 ) satisfy the equation. Therefore, the integer values of ( n ) are 12 and 13.But wait, could there be more solutions? Let me check ( n = 11 ).If ( n = 11 ), then the left side is ( binom{25}{11} + binom{25}{12} ). Using Pascal's identity again, this equals ( binom{26}{12} ). But ( binom{26}{12} ) is not equal to ( binom{26}{13} ) because ( binom{26}{12} = binom{26}{14} ) and ( binom{26}{13} ) is a different value. So, ( n = 11 ) doesn't work.What about ( n = 14 )? Let's see: ( binom{25}{14} + binom{25}{12} ). But ( binom{25}{14} = binom{25}{11} ), so this becomes ( binom{25}{11} + binom{25}{12} = binom{26}{12} ), which again isn't equal to ( binom{26}{13} ). So, ( n = 14 ) isn't a solution either.How about ( n = 25 )? Then, ( binom{25}{25} + binom{25}{12} = 1 + binom{25}{12} ). That's definitely not equal to ( binom{26}{13} ), which is a much larger number.Similarly, ( n = 0 ): ( binom{25}{0} + binom{25}{12} = 1 + binom{25}{12} ), which is also not equal to ( binom{26}{13} ).So, it seems that only ( n = 12 ) and ( n = 13 ) satisfy the equation. Therefore, the integer values of ( n ) are 12 and 13.The problem asks for the sum of all such integer values. So, I need to add 12 and 13 together.12 + 13 = 25.Therefore, the sum of all integer values ( n ) that satisfy the equation is 25.Wait, but just to make sure I haven't missed any other possible values of ( n ). Let me think again.The equation is ( binom{25}{n} + binom{25}{12} = binom{26}{13} ). So, ( binom{25}{n} = binom{26}{13} - binom{25}{12} ).I can compute ( binom{26}{13} - binom{25}{12} ) to see what it equals.We know that ( binom{26}{13} = 2 times binom{25}{12} ). So, subtracting ( binom{25}{12} ) from that gives ( binom{25}{12} ). Therefore, ( binom{25}{n} = binom{25}{12} ).Since binomial coefficients are symmetric, ( binom{25}{n} = binom{25}{12} ) implies that ( n = 12 ) or ( n = 25 - 12 = 13 ). So, indeed, the only solutions are ( n = 12 ) and ( n = 13 ).Therefore, adding them together gives 25. So, I'm confident that 25 is the correct answer.**Final Answer**The sum of all integer values ( n ) is boxed{25}.

💡Alright, so I have this problem where I'm given a function ( f(x) = |x^2 + ax| ) and I need to do two things: first, find the range of values for ( a ) such that ( f(x) ) is monotonically increasing on the interval ([0, 1]), and second, find the minimum value of the maximum of ( f(x) ) on the same interval. Hmm, okay, let's break this down step by step.Starting with part (I): I need to figure out for which values of ( a ) the function ( f(x) ) is monotonically increasing on ([0, 1]). Monotonically increasing means that as ( x ) increases, ( f(x) ) doesn't decrease. So, the function should either stay the same or increase as ( x ) moves from 0 to 1.First, let's consider the function without the absolute value: ( g(x) = x^2 + ax ). This is a quadratic function, and its graph is a parabola. The coefficient of ( x^2 ) is positive, so the parabola opens upwards. The vertex of this parabola is at ( x = -frac{a}{2} ). So, depending on the value of ( a ), the vertex can be to the left or right of the y-axis.Now, since we're dealing with the absolute value of ( g(x) ), ( f(x) = |g(x)| ), the graph of ( f(x) ) will reflect any part of ( g(x) ) that's below the x-axis to above it. So, if the parabola ( g(x) ) crosses the x-axis within the interval ([0, 1]), the function ( f(x) ) will have a "V" shape there, which could affect its monotonicity.To ensure ( f(x) ) is monotonically increasing on ([0, 1]), we need to make sure that ( f(x) ) doesn't have any decreasing parts in that interval. That could happen if the original function ( g(x) ) doesn't cross the x-axis in ([0, 1]) or if it does, the reflection doesn't cause a decrease.Let me think about when ( g(x) ) crosses the x-axis. The roots of ( g(x) = 0 ) are at ( x = 0 ) and ( x = -a ). So, if ( -a ) is within ([0, 1]), that is, if ( a ) is between (-1) and (0), then ( g(x) ) crosses the x-axis within ([0, 1]). Otherwise, it doesn't.So, if ( a geq 0 ), then ( g(x) ) doesn't cross the x-axis in ([0, 1]) because the other root is at ( x = -a leq 0 ). Therefore, ( f(x) = g(x) ) in this case, which is a parabola opening upwards, and since it's increasing on ([0, 1]) because the vertex is at ( x = -frac{a}{2} leq 0 ), which is outside the interval. So, ( f(x) ) is increasing on ([0, 1]) for ( a geq 0 ).Now, if ( a < 0 ), the situation is different. The root ( x = -a ) is positive, so it could be inside or outside the interval ([0, 1]). If ( -a geq 1 ), that is, ( a leq -1 ), then the root is at ( x = -a geq 1 ), so within ([0, 1]), ( g(x) ) doesn't cross the x-axis. Therefore, ( f(x) = |g(x)| ) would be equal to ( g(x) ) if ( g(x) ) is positive, but since ( g(x) ) is a parabola opening upwards, and if ( a leq -1 ), ( g(x) ) is negative on ([0, -a)) and positive on ((-a, infty)). But since ( -a geq 1 ), on ([0, 1]), ( g(x) ) is negative, so ( f(x) = -g(x) ).So, ( f(x) = -x^2 - ax ) on ([0, 1]). Is this function increasing? Let's compute its derivative: ( f'(x) = -2x - a ). For ( f(x) ) to be increasing, ( f'(x) geq 0 ) on ([0, 1]). So, ( -2x - a geq 0 ) for all ( x in [0, 1] ).The minimum of ( f'(x) ) occurs at ( x = 1 ), so ( -2(1) - a geq 0 ) implies ( -2 - a geq 0 ) which gives ( a leq -2 ). So, if ( a leq -2 ), then ( f'(x) geq 0 ) on ([0, 1]), meaning ( f(x) ) is increasing.But wait, earlier I thought if ( a leq -1 ), ( g(x) ) is negative on ([0, 1]), but actually, for ( a leq -2 ), ( f(x) = -g(x) ) is increasing, but if ( -2 < a < 0 ), then ( f(x) ) might not be increasing on the entire interval.Let me check for ( a = -1 ). Then ( g(x) = x^2 - x ), which has roots at 0 and 1. So, on ([0, 1]), ( g(x) ) is negative between 0 and 1, so ( f(x) = -g(x) = -x^2 + x ). The derivative is ( f'(x) = -2x + 1 ). At ( x = 0 ), it's 1, which is positive, and at ( x = 1 ), it's ( -2 + 1 = -1 ), which is negative. So, ( f(x) ) increases initially but then decreases, meaning it's not monotonically increasing on the entire interval. So, ( a = -1 ) doesn't work.Similarly, for ( a = -1.5 ), which is between -2 and 0, let's see. ( g(x) = x^2 - 1.5x ), which has roots at 0 and 1.5. So, on ([0, 1]), ( g(x) ) is negative, so ( f(x) = -x^2 + 1.5x ). The derivative is ( f'(x) = -2x + 1.5 ). At ( x = 0 ), it's 1.5, positive. At ( x = 1 ), it's ( -2 + 1.5 = -0.5 ), negative. So again, the function increases then decreases, not monotonic.Therefore, only when ( a leq -2 ), the function ( f(x) = -g(x) = -x^2 - ax ) has a derivative ( f'(x) = -2x - a ). For ( a leq -2 ), at ( x = 1 ), ( f'(1) = -2 - a geq 0 ) because ( a leq -2 ), so ( -a geq 2 ), hence ( -2 - a geq 0 ). Also, since ( f'(x) ) is linear decreasing, the minimum derivative is at ( x = 1 ), which is non-negative, so the derivative is non-negative throughout ([0, 1]). Therefore, ( f(x) ) is increasing on ([0, 1]) for ( a leq -2 ).So, combining both cases, when ( a geq 0 ), ( f(x) ) is increasing on ([0, 1]), and when ( a leq -2 ), ( f(x) ) is also increasing on ([0, 1]). Therefore, the range of ( a ) is ( a leq -2 ) or ( a geq 0 ).Okay, that seems solid for part (I). Now, moving on to part (II): We need to find the minimum value of ( M(a) ), where ( M(a) ) is the maximum of ( f(x) ) on ([0, 1]).So, ( M(a) ) is the maximum value of ( |x^2 + ax| ) on ([0, 1]). To find the minimum of this maximum, we need to analyze how ( M(a) ) behaves as ( a ) varies and find the value of ( a ) that minimizes this maximum.From part (I), we know that when ( a leq -2 ) or ( a geq 0 ), ( f(x) ) is monotonically increasing on ([0, 1]). Therefore, in these cases, the maximum of ( f(x) ) occurs at ( x = 1 ), so ( M(a) = f(1) = |1 + a| ).However, when ( -2 < a < 0 ), ( f(x) ) is not monotonically increasing on ([0, 1]). In this case, ( f(x) ) will have a maximum either at ( x = 1 ) or at the critical point where the derivative is zero. Since ( f(x) ) is the absolute value of a quadratic, its maximum could occur either at the endpoints or at the vertex of the parabola.Wait, actually, since ( f(x) = |x^2 + ax| ), the maximum could be at ( x = 0 ), ( x = 1 ), or at the point where ( x^2 + ax = 0 ), which is at ( x = 0 ) and ( x = -a ). But ( x = -a ) is in ([0, 1]) only if ( -a in [0, 1] ), which is when ( a in [-1, 0) ). However, in the interval ( -2 < a < 0 ), ( -a ) can be greater than 1 when ( a < -1 ). Wait, no, if ( a ) is between -2 and 0, ( -a ) is between 0 and 2. So, when ( a ) is between -2 and -1, ( -a ) is between 1 and 2, which is outside ([0, 1]). When ( a ) is between -1 and 0, ( -a ) is between 0 and 1, so within the interval.Therefore, for ( -2 < a < -1 ), the function ( f(x) ) is ( |x^2 + ax| ), which is equal to ( x^2 + ax ) when ( x geq -a ), but since ( -a > 1 ), on ([0, 1]), ( f(x) = |x^2 + ax| ). But since ( a ) is negative, ( x^2 + ax ) is a quadratic that opens upwards, but with a negative linear term. So, it might dip below zero somewhere in ([0, 1]).Wait, actually, when ( a ) is between -2 and 0, the function ( g(x) = x^2 + ax ) has its vertex at ( x = -a/2 ). So, for ( a ) between -2 and 0, the vertex is at ( x = -a/2 ), which is between 0 and 1 when ( a ) is between -2 and 0. Specifically, when ( a = -2 ), the vertex is at ( x = 1 ), and when ( a = 0 ), the vertex is at ( x = 0 ).So, for ( -2 < a < 0 ), the vertex is inside ([0, 1]), meaning that ( g(x) ) attains its minimum at ( x = -a/2 ). Therefore, ( f(x) = |g(x)| ) will have a maximum either at ( x = 0 ), ( x = 1 ), or at the point where ( g(x) ) is minimized (since the absolute value could make that point a maximum if ( g(x) ) is negative there).So, let's compute ( f(0) = |0 + 0| = 0 ), ( f(1) = |1 + a| ), and ( f(-a/2) = |(-a/2)^2 + a*(-a/2)| = |a^2/4 - a^2/2| = | -a^2/4 | = a^2/4 ).Therefore, for ( -2 < a < 0 ), ( M(a) = max { |1 + a|, a^2/4 } ).Now, we need to determine when ( |1 + a| ) is greater than ( a^2/4 ) and vice versa.Let's solve the inequality ( |1 + a| geq a^2/4 ).Since ( a ) is between -2 and 0, ( 1 + a ) is between -1 and 1. Specifically, when ( a ) is between -2 and -1, ( 1 + a ) is between -1 and 0, so ( |1 + a| = -(1 + a) ). When ( a ) is between -1 and 0, ( |1 + a| = 1 + a ).So, let's split into two cases:1. When ( -2 < a < -1 ): ( |1 + a| = -(1 + a) = -1 - a ). So, we need to compare ( -1 - a ) and ( a^2/4 ). Let's set ( -1 - a = a^2/4 ): ( a^2/4 + a + 1 = 0 ) Multiply both sides by 4: ( a^2 + 4a + 4 = 0 ) ( (a + 2)^2 = 0 ) So, ( a = -2 ). But ( a = -2 ) is the boundary, not in the interval ( -2 < a < -1 ). Therefore, in this interval, ( -1 - a ) is greater than ( a^2/4 ) because at ( a = -2 ), ( -1 - (-2) = 1 ) and ( (-2)^2/4 = 1 ), so they are equal. As ( a ) increases from -2 to -1, ( -1 - a ) decreases from 1 to 0, while ( a^2/4 ) decreases from 1 to ( 1/4 ). Wait, actually, at ( a = -1.5 ), ( -1 - (-1.5) = 0.5 ) and ( (-1.5)^2 /4 = 2.25 /4 = 0.5625 ). So, ( a^2/4 ) is greater than ( -1 - a ) at ( a = -1.5 ). Hmm, so maybe my earlier conclusion was wrong.Wait, let's test at ( a = -1.5 ):( |1 + (-1.5)| = 0.5 )( a^2/4 = (2.25)/4 = 0.5625 )So, ( a^2/4 > |1 + a| ) at ( a = -1.5 ).At ( a = -2 ):( |1 + (-2)| = 1 )( a^2/4 = 4/4 = 1 )They are equal.At ( a = -1.9 ):( |1 + (-1.9)| = 0.9 )( a^2/4 = (3.61)/4 ≈ 0.9025 )So, ( a^2/4 ≈ 0.9025 > 0.9 )At ( a = -1.1 ):( |1 + (-1.1)| = 0.1 )( a^2/4 = (1.21)/4 ≈ 0.3025 )So, ( a^2/4 > |1 + a| )Wait, so actually, for all ( a ) in ( (-2, -1) ), ( a^2/4 geq |1 + a| ). Because at ( a = -2 ), they are equal, and as ( a ) increases towards -1, ( a^2/4 ) decreases but ( |1 + a| ) also decreases, but ( a^2/4 ) is always slightly larger.Wait, let me solve the equation ( a^2/4 = |1 + a| ) for ( -2 < a < 0 ).Case 1: ( -2 < a < -1 ), so ( |1 + a| = -1 - a ).Set ( a^2/4 = -1 - a ):( a^2/4 + a + 1 = 0 )Multiply by 4:( a^2 + 4a + 4 = 0 )( (a + 2)^2 = 0 )So, ( a = -2 ). But this is the boundary.Case 2: ( -1 leq a < 0 ), so ( |1 + a| = 1 + a ).Set ( a^2/4 = 1 + a ):( a^2/4 - a - 1 = 0 )Multiply by 4:( a^2 - 4a - 4 = 0 )Solutions:( a = [4 ± sqrt(16 + 16)] / 2 = [4 ± sqrt(32)] / 2 = [4 ± 4*sqrt(2)] / 2 = 2 ± 2*sqrt(2) )Since ( a ) is between -1 and 0, we take the negative root:( a = 2 - 2*sqrt(2) ≈ 2 - 2.828 ≈ -0.828 )So, at ( a = 2 - 2*sqrt(2) ≈ -0.828 ), ( a^2/4 = |1 + a| ).Therefore, for ( -2 < a < 2 - 2*sqrt(2) ), ( a^2/4 geq |1 + a| ), and for ( 2 - 2*sqrt(2) leq a < 0 ), ( |1 + a| geq a^2/4 ).Therefore, ( M(a) ) is:- ( a^2/4 ) when ( -2 < a < 2 - 2*sqrt(2) )- ( |1 + a| ) when ( 2 - 2*sqrt(2) leq a < 0 )- ( |1 + a| ) when ( a geq 0 ) or ( a leq -2 )So, to find the minimum of ( M(a) ), we need to analyze both expressions ( a^2/4 ) and ( |1 + a| ) in their respective intervals.First, consider ( a leq -2 ) or ( a geq 0 ): ( M(a) = |1 + a| ).For ( a geq 0 ), ( |1 + a| = 1 + a ), which is minimized at ( a = 0 ), giving ( M(a) = 1 ).For ( a leq -2 ), ( |1 + a| = -(1 + a) = -1 - a ), which is minimized when ( a ) is as large as possible, i.e., ( a = -2 ), giving ( M(a) = -1 - (-2) = 1 ).So, in these regions, the minimum ( M(a) ) is 1.Now, consider the interval ( -2 < a < 2 - 2*sqrt(2) ): ( M(a) = a^2/4 ). This is a parabola opening upwards, so it attains its minimum at the vertex. The vertex is at ( a = 0 ), but since we're considering ( a ) in ( (-2, 2 - 2*sqrt(2)) ), the minimum occurs at the smallest ( a ) where ( a^2/4 ) is minimized. Wait, actually, since it's a parabola, the minimum is at ( a = 0 ), but ( a = 0 ) is not in this interval. So, the minimum in this interval would be at the left endpoint, ( a = -2 ), but ( a = -2 ) is already covered in the previous case.Wait, no, actually, in the interval ( -2 < a < 2 - 2*sqrt(2) ), the function ( a^2/4 ) is decreasing as ( a ) increases from -2 to 0. So, the minimum value in this interval would be at ( a = 2 - 2*sqrt(2) ), because beyond that point, ( M(a) ) switches to ( |1 + a| ).Wait, let me think again. For ( -2 < a < 2 - 2*sqrt(2) ), ( M(a) = a^2/4 ). The function ( a^2/4 ) is decreasing as ( a ) increases from -2 to 0. So, the minimum value in this interval would be at the right endpoint, which is ( a = 2 - 2*sqrt(2) ). At this point, ( M(a) = ( (2 - 2*sqrt(2))^2 ) / 4 ).Let's compute that:( (2 - 2*sqrt(2))^2 = 4 - 8*sqrt(2) + 8 = 12 - 8*sqrt(2) )Divide by 4:( (12 - 8*sqrt(2))/4 = 3 - 2*sqrt(2) ≈ 3 - 2.828 ≈ 0.172 )So, ( M(a) ) at ( a = 2 - 2*sqrt(2) ) is ( 3 - 2*sqrt(2) ).Now, comparing this with the minimum in the other intervals, which was 1, clearly ( 3 - 2*sqrt(2) ) is smaller. Therefore, the overall minimum of ( M(a) ) is ( 3 - 2*sqrt(2) ).To confirm, let's check the behavior around ( a = 2 - 2*sqrt(2) ). For ( a ) slightly less than ( 2 - 2*sqrt(2) ), ( M(a) = a^2/4 ), which is slightly larger than ( 3 - 2*sqrt(2) ). For ( a ) slightly more than ( 2 - 2*sqrt(2) ), ( M(a) = |1 + a| ), which is slightly larger than ( 3 - 2*sqrt(2) ) as well because ( |1 + a| = 1 + a ) (since ( a > -1 )) and ( a = 2 - 2*sqrt(2) ≈ -0.828 ), so ( 1 + a ≈ 0.172 ), which is ( 3 - 2*sqrt(2) ≈ 0.172 ). So, it's continuous at that point.Therefore, the minimum value of ( M(a) ) is ( 3 - 2*sqrt(2) ), achieved at ( a = 2 - 2*sqrt(2) ).So, summarizing:(I) The range of ( a ) for which ( f(x) ) is monotonically increasing on ([0, 1]) is ( a leq -2 ) or ( a geq 0 ).(II) The minimum value of ( M(a) ) is ( 3 - 2sqrt{2} ).**Final Answer**(I) The range of ( a ) is ( boxed{(-infty, -2] cup [0, infty)} ).(II) The minimum value of ( M(a) ) is ( boxed{3 - 2sqrt{2}} ).