🧠:[<think>Okay, let me try to figure out this problem. So, we have three non-negative rational numbers x, y, z, and they satisfy two equations: 3x + 2y + z = 5 and x + y - z = 2. We need to find the sum of the maximum and minimum values of S, where S = 2x + y - z. Hmm, okay. Let's start by understanding the constraints and what exactly we need to optimize.First, since we have three variables and two equations, we can probably express two variables in terms of the third, which might help in finding the possible values. Let's see. The equations given are:1. 3x + 2y + z = 52. x + y - z = 2We need to find S = 2x + y - z. Maybe if we can express S in terms of one variable, we can then find its maximum and minimum. Also, since all variables are non-negative and rational, we have to consider those constraints as well.Let me try to solve the system of equations first. Let's take the two equations:3x + 2y + z = 5 ...(1)x + y - z = 2 ...(2)Maybe if I add these two equations or subtract them, I can eliminate z. Let's try adding them. If I add equation (1) and equation (2), we get:3x + 2y + z + x + y - z = 5 + 2Which simplifies to 4x + 3y = 7. Okay, that's helpful. So 4x + 3y = 7. Let's keep that in mind.Alternatively, we could subtract the equations. Let's see. If we subtract equation (2) from equation (1):(3x + 2y + z) - (x + y - z) = 5 - 2Which simplifies to 2x + y + 2z = 3. Hmm, not sure if that's helpful yet.Alternatively, maybe solve equation (2) for z. Let's do that. From equation (2):x + y - z = 2 => z = x + y - 2.But since z is non-negative, z >= 0, so x + y - 2 >= 0 => x + y >= 2.Okay, so substituting z = x + y - 2 into equation (1):3x + 2y + (x + y - 2) = 5Simplify: 3x + 2y + x + y - 2 = 5 => 4x + 3y - 2 = 5 => 4x + 3y = 7.Ah, so that's the same as before. So we have 4x + 3y = 7, and z = x + y - 2.Now, since x, y, z are all non-negative, we have constraints:1. x >= 02. y >= 03. z = x + y - 2 >= 0 => x + y >= 24. From 4x + 3y = 7, we can express y in terms of x or vice versa. Let's solve for y.From 4x + 3y = 7 => 3y = 7 - 4x => y = (7 - 4x)/3.Since y >= 0, (7 - 4x)/3 >= 0 => 7 - 4x >= 0 => 4x <= 7 => x <= 7/4.Similarly, since x >= 0, so x is between 0 and 7/4.Also, z = x + y - 2. Let's substitute y into z:z = x + (7 - 4x)/3 - 2 = x + 7/3 - (4x)/3 - 2 = (3x)/3 + 7/3 - (4x)/3 - 6/3 = (-x + 1)/3.So z = (-x + 1)/3. But z >= 0, so (-x + 1)/3 >= 0 => -x + 1 >= 0 => x <= 1.Wait, this is a new constraint. Earlier, we had x <= 7/4 (which is 1.75), but now z >= 0 gives x <= 1. So x must be between 0 and 1.Wait, that's conflicting with previous? Let me check.We had z = x + y - 2. Then we substituted y = (7 - 4x)/3 into z:z = x + (7 - 4x)/3 - 2Calculating that:x is 3x/3, so 3x/3 + (7 -4x)/3 - 6/3 = (3x +7 -4x -6)/3 = (-x +1)/3. So yes, z = (1 - x)/3.Therefore, z >=0 => (1 - x)/3 >=0 => 1 - x >=0 => x <=1.Therefore, even though x could be up to 7/4 from the previous equation, the constraint on z reduces x to be at most 1. So x is in [0,1].So that's a key point. So x is between 0 and 1. Let's verify that.If x =1, then y = (7 -4*1)/3=3/3=1. Then z = (1 -1)/3=0. So x=1, y=1, z=0. All non-negative. That's okay.If x=0, then y = (7 -0)/3=7/3≈2.333. Then z=(1 -0)/3=1/3≈0.333. But z=0 +7/3 -2=7/3 -2=1/3. So z=1/3. So x=0, y=7/3, z=1/3. All non-negative.So the x ranges from 0 to 1. So we can parameterize the problem in terms of x, where x is between 0 and 1, and then y=(7 -4x)/3, z=(1 -x)/3.Then S =2x + y - z. Let's substitute y and z into S.Compute S:2x + y - z =2x + (7 -4x)/3 - (1 -x)/3.Let's combine the terms:First, write 2x as 6x/3 to have a common denominator.6x/3 + (7 -4x)/3 - (1 -x)/3 = [6x +7 -4x -1 +x]/3Combine like terms in the numerator:6x -4x +x = 3x7 -1=6So numerator is 3x +6Therefore, S=(3x +6)/3= x +2.Wait, that simplifies nicely. So S = x +2.Wow, so S is simply x +2. That's straightforward.But x is in [0,1], so then S ranges from 2 +0=2 to 2 +1=3.Therefore, the minimum value of S is 2, and the maximum value is 3. Then the sum of maximum and minimum is 2 +3=5.But wait, this seems too easy. Let me check again.Wait, let's verify with the variables. If x=0, then S=0+2=2. Let's compute S using original variables. If x=0, y=7/3, z=1/3. Then S=2*0 +7/3 -1/3=6/3=2. Correct.If x=1, then y=1, z=0. Then S=2*1 +1 -0=3. Correct. So S=3.But wait, can S be higher than 3? Let's check if there is any other constraints that I might have missed. For example, x can only go up to 1, so S can't be more than 3. Similarly, x can't be less than 0, so S can't be less than 2. Therefore, S ranges between 2 and 3. So sum is 5.But let me check with another point. For example, take x=0.5. Then y=(7 -4*0.5)/3=(7-2)/3=5/3≈1.666. z=(1 -0.5)/3=0.5/3≈0.166. Then S=2*0.5 +5/3 -0.5/3=1 + (5 -0.5)/3=1 +4.5/3=1 +1.5=2.5, which is 0.5 +2=2.5. So that's correct.But just to make sure, maybe there's a different way. Suppose we think in terms of linear programming. Since all the constraints and the objective function are linear, the extrema should occur at the vertices of the feasible region.So, the feasible region is defined by the constraints:1. 4x + 3y =72. z =x + y -2 >=03. x >=04. y >=05. z >=0But since z is determined by x and y, and we already derived x must be between 0 and1, y=(7 -4x)/3, which is non-negative as x <=7/4, but since x <=1, y=(7 -4x)/3. When x=1, y=1; when x=0, y=7/3. So y is always positive in this interval.Therefore, the feasible region is the line segment from (x=0, y=7/3, z=1/3) to (x=1, y=1, z=0). So the vertices are these two points.Therefore, evaluating S at these two points gives the minimum and maximum.At x=0: S=2*0 +7/3 -1/3=6/3=2.At x=1: S=2*1 +1 -0=3.Therefore, the maximum is 3, the minimum is 2. Sum is 5.But the question mentions that x, y, z are non-negative rational numbers. But since we found that even with rational numbers, the extrema occur at x=0 and x=1, which are rational, so even if considering only rational numbers, the maximum and minimum are still 3 and 2.Therefore, the answer should be 5.Wait, but just to be thorough, let me check if there's a possibility that S can be higher or lower when x is not at the endpoints. But since S is linear in x (S =x +2), increasing x increases S, and decreasing x decreases S. So since x is bounded between 0 and1, S is bounded between 2 and3. Therefore, there's no other points where S can be higher or lower.Therefore, the sum is 2 +3=5.Final AnswerThe sum of the maximum and minimum values of ( S ) is boxed{5}.</think>Given the equations (3x + 2y + z = 5) and (x + y - z = 2) with non-negative rational numbers (x, y, z), we need to find the sum of the maximum and minimum values of (S = 2x + y - z).First, we solve the system of equations. By adding the two given equations, we get:[4x + 3y = 7]From the second equation, we express (z) in terms of (x) and (y):[z = x + y - 2]Substituting (z) into the first equation, we get:[3x + 2y + (x + y - 2) = 5 implies 4x + 3y = 7]Solving for (y) in terms of (x):[y = frac{7 - 4x}{3}]Since (z geq 0), we have:[x + y - 2 geq 0 implies x + frac{7 - 4x}{3} geq 2 implies frac{-x + 1}{3} geq 0 implies x leq 1]Thus, (x) must be in the interval ([0, 1]). Substituting (y) and (z) into (S):[S = 2x + frac{7 - 4x}{3} - frac{1 - x}{3}]Simplifying this expression:[S = 2x + frac{7 - 4x - 1 + x}{3} = 2x + frac{6 - 3x}{3} = 2x + 2 - x = x + 2]Since (x) ranges from 0 to 1, (S) ranges from (0 + 2 = 2) to (1 + 2 = 3). Therefore, the minimum value of (S) is 2 and the maximum value is 3. The sum of these values is:[2 + 3 = 5]The sum of the maximum and minimum values of (S) is (boxed{5}).,<think>Okay, so I need to find the sum of the maximum and minimum values of S, where S is defined as 2x + y - z. The constraints are that x, y, z are non-negative rational numbers satisfying two equations: 3x + 2y + z = 5 and x + y - z = 2. Hmm, let's see. First, maybe I can solve the system of equations to express some variables in terms of others. Let me write down the two equations again:1) 3x + 2y + z = 52) x + y - z = 2Since both equations have z, maybe I can eliminate z to find a relationship between x and y. If I add the two equations together, maybe that will help. Let's try:Adding equation 1 and equation 2:3x + 2y + z + x + y - z = 5 + 2Which simplifies to 4x + 3y = 7.So that gives me 4x + 3y = 7. That's one equation with two variables. Maybe I can solve for one variable in terms of the other. Let's solve for y:3y = 7 - 4x=> y = (7 - 4x)/3But y has to be non-negative, so (7 - 4x)/3 ≥ 0. That means 7 - 4x ≥ 0, so 4x ≤ 7, so x ≤ 7/4. Similarly, x must be non-negative, so x ≥ 0. So x is between 0 and 7/4.Similarly, let's find expressions for z. From equation 2: x + y - z = 2 => z = x + y - 2. Since z must be non-negative, x + y - 2 ≥ 0 => x + y ≥ 2. But we already have y in terms of x: y = (7 - 4x)/3. So substituting that into x + y ≥ 2:x + (7 - 4x)/3 ≥ 2Multiply both sides by 3:3x + 7 - 4x ≥ 6Simplify:(-x + 7) ≥ 6 => -x ≥ -1 => x ≤ 1Wait, that's interesting. So from the non-negativity of z, we have x ≤ 1. But earlier, from the non-negativity of y, we had x ≤ 7/4, which is 1.75. So the stricter condition here is x ≤ 1. Therefore, x is between 0 and 1. So summarizing: x ∈ [0, 1], y = (7 - 4x)/3, z = x + y - 2. Let me check if z is non-negative. Since z = x + y - 2, and we already derived that x + y ≥ 2 (to make z non-negative), so substituting y:z = x + (7 - 4x)/3 - 2Let me compute that:z = x + 7/3 - (4x)/3 - 2Combine like terms:x - (4x)/3 = (-x)/37/3 - 2 = 7/3 - 6/3 = 1/3So z = (-x)/3 + 1/3So z = (1 - x)/3Since z ≥ 0, (1 - x)/3 ≥ 0 => 1 - x ≥ 0 => x ≤ 1, which we already have. So that's consistent.So z is expressed as (1 - x)/3. Let's confirm that. From z = x + y - 2. Since y = (7 - 4x)/3, substitute:z = x + (7 - 4x)/3 - 2Convert x to 3x/3 and 2 to 6/3:z = (3x + 7 - 4x - 6)/3 = (-x + 1)/3, which is (1 - x)/3. Yep, that's correct.So now, variables are:x ∈ [0, 1]y = (7 - 4x)/3z = (1 - x)/3All non-negative.Now, we need to find S = 2x + y - z.Let me express S in terms of x. Since y and z are expressed in terms of x, substitute:S = 2x + [(7 - 4x)/3] - [(1 - x)/3]First, combine the fractions:[(7 - 4x) - (1 - x)] / 3 = [7 - 4x -1 + x]/3 = (6 - 3x)/3 = 2 - xSo S = 2x + (2 - x) = 2x + 2 - x = x + 2Wait, that's interesting. So S simplifies to x + 2.Therefore, S is just x + 2.But x is in [0, 1], so then S is maximized when x is maximized (x=1) and minimized when x is minimized (x=0).So maximum S = 1 + 2 = 3, minimum S = 0 + 2 = 2. Therefore, the sum of max and min is 3 + 2 = 5.Wait, but that seems straightforward. Let me double-check my steps because sometimes when variables are connected, even if S simplifies to something linear, there might be hidden constraints. But in this case, since we already accounted for all constraints (x must be between 0 and 1, y and z expressed in terms of x and non-negative). Let me check with x=0:x=0:y=(7 - 0)/3=7/3 ≈2.333z=(1 -0)/3=1/3 ≈0.333Check the original equations:3x +2y +z=0 + 14/3 +1/3=15/3=5, correct.x + y - z=0 +7/3 -1/3=6/3=2, correct.So S=2*0 +7/3 -1/3=6/3=2, which is correct.For x=1:x=1:y=(7 -4)/3=3/3=1z=(1 -1)/3=0Check equations:3*1 +2*1 +0=3+2=5, correct.1 +1 -0=2, correct.S=2*1 +1 -0=3, correct.So S= x +2 with x in [0,1], so max 3, min 2. Sum is 5.But wait, the problem states "the sum of the maximum and minimum values of S". So 3 + 2=5.Is there a possibility that there are other constraints? Let me see. Suppose x is in [0,1], but maybe y and z have some other constraints? For example, y must be non-negative. So when x=1, y=(7-4)/3=1, which is okay. When x=0, y=7/3, which is okay. Similarly, z=(1 -x)/3 is non-negative for x in [0,1], which is already considered.Therefore, S is indeed linear in x, with coefficient 1, so it's increasing as x increases. Therefore, the maximum and minimum are achieved at the endpoints x=1 and x=0.Thus, the answer is 3 + 2 = 5.Wait, but the problem says "non-negative rational numbers". So since x, y, z are rational, but when x is 0 or 1, are y and z rational? For example, x=0: y=7/3, z=1/3, which are rational. x=1: y=1, z=0, which are integers. So yes, those points are valid. So the max and min are achievable.Therefore, the sum of maximum and minimum is 5.But just to make sure there's no other critical point in between. Since S is linear in x, there are no critical points except the endpoints. So yes, it's straightforward.Therefore, the answer should be 5. So boxed answer is boxed{5}.Final Answerboxed{5}</think>Given the problem, we need to find the sum of the maximum and minimum values of ( S = 2x + y - z ) where ( x, y, z ) are non-negative rational numbers satisfying the equations ( 3x + 2y + z = 5 ) and ( x + y - z = 2 ).First, we solve the system of equations. Adding the two equations:[3x + 2y + z + x + y - z = 5 + 2 implies 4x + 3y = 7]Solving for ( y ):[y = frac{7 - 4x}{3}]From the second equation ( x + y - z = 2 ), we express ( z ):[z = x + y - 2]Substituting ( y ) into the expression for ( z ):[z = x + frac{7 - 4x}{3} - 2 = frac{3x + 7 - 4x - 6}{3} = frac{1 - x}{3}]We ensure non-negativity constraints:- ( y geq 0 implies frac{7 - 4x}{3} geq 0 implies x leq frac{7}{4} )- ( z geq 0 implies frac{1 - x}{3} geq 0 implies x leq 1 )Thus, ( x ) is constrained to ( [0, 1] ).Next, we express ( S ) in terms of ( x ):[S = 2x + y - z = 2x + frac{7 - 4x}{3} - frac{1 - x}{3}]Simplifying:[S = 2x + frac{7 - 4x - 1 + x}{3} = 2x + frac{6 - 3x}{3} = 2x + 2 - x = x + 2]Since ( x ) is in the interval ([0, 1]), the maximum value of ( S ) occurs at ( x = 1 ) (yielding ( S = 3 )) and the minimum value occurs at ( x = 0 ) (yielding ( S = 2 )).The sum of the maximum and minimum values of ( S ) is:[3 + 2 = 5]Thus, the final answer is (boxed{5}).]

🔑:9. left{begin{array}{l}3 x+y+z=5, x+y-z=2, 2 x+y-z=S .end{array}right. Solving, we get left{begin{array}{l}x=S-2, y=frac{15-4 S}{3}, z=frac{3-S}{3} .end{array}right. Since x, y, z are all non-negative, we haveleft{begin{array} { l } { S - 2 geqslant 0 , } { frac { 1 5 - 4 S } { 3 } geqslant 0 , } { frac { 3 - S } { 3 } geqslant 0 }end{array} Rightarrow left{begin{array}{l}S geqslant 2, S leqslant frac{15}{4}, S leqslant 3 .end{array}right.right.Therefore, 2 leqslant S leqslant 3. The sum of the maximum and minimum values of S is 5.

🧠:[<think>Okay, so I need to find the greatest number that is the product of some positive integers, and the sum of these numbers is 1976. Hmm, interesting problem. Let me think about how to approach this.First, I remember that when you want to break a number into parts to maximize the product, it's generally best to use as many 3s as possible. Because 3 is the number that gives the highest product when split. For example, if you take a number like 6, splitting it into 3 + 3 gives a product of 9, which is better than 2 + 2 + 2 (product 8) or 4 + 2 (product 8). So the strategy is to divide the number into 3s, and then handle the remainder with 2s if needed. If the remainder is 1, it's better to replace a 3 and a 1 with two 2s, since 2*2 is better than 3*1.Let me verify that again. If I have a number N. Divide N by 3. If the remainder is 0, then it's just all 3s. If the remainder is 1, then you take one of the 3s and make it into two 2s. If the remainder is 2, then you just add a 2. So for example, if N=10. 10 divided by 3 is 3 with remainder 1. So instead of 3*3*3*1, which is 27, we do 3*3*2*2, which is 36. That's better.So applying that logic here. The sum of the numbers needs to be 1976. Let's divide 1976 by 3.1976 ÷ 3. Let me compute that. 3*658 = 1974. So 3 goes into 1976 six hundred fifty-eight times with a remainder of 2. Because 1974 + 2 = 1976. So the remainder is 2.So according to the strategy, since the remainder is 2, we can just have 658 3s and one 2. So the product would be 3^658 * 2. But wait, is that the maximum? Let me check.Alternatively, if the remainder was 1, we would have to adjust by taking away a 3 and adding two 2s, but here the remainder is 2, so we just add the 2. So in this case, the maximum product would be 3^658 * 2.But let me make sure. Let me test this with a smaller number where I know the answer. For example, sum of 5. The maximum product is 3*2=6. Which works. If sum is 6, then 3*3=9. If sum is 7, then 3*2*2=12. Sum 8: 3*3*2=18. Sum 9: 3*3*3=27. So yeah, the pattern holds. So seems like 3s and 2s are the way to go.Therefore, for 1976, which is 3*658 + 2, the maximum product is 3^658 * 2.But wait, let me check if that's correct. Let me see: 3^658 is a massive number. But perhaps writing it in terms of exponents is acceptable? But maybe the problem expects the numerical value? But 3^658 is an astronomically large number, impossible to compute directly. So I think the answer is simply 2 multiplied by 3 raised to the 658th power, written as 2 × 3^658. So the greatest number is 2×3^658.But let me verify once again. Let me check if there's a different combination that could give a higher product. For example, maybe using some 4s? But 4 is equivalent to two 2s, and 2*2=4, so it's the same. 5s? If we split 5 into 2+3, 2*3=6 which is higher than 5. So 5 is worse. 6: 3+3, product 9 vs. 6, so yeah, 3s are better. So any number higher than 3 can be split into 2s and 3s for a higher product. So using only 2s and 3s is optimal.Hence, the maximum product should be achieved by breaking 1976 into as many 3s as possible, and the remainder (which is 2) becomes a 2. So the number of 3s is (1976 - 2)/3 = 1974/3 = 658. So 658 3s and one 2. Therefore, the product is 3^658 * 2.So the answer is 2 × 3^658. I think that's correct. Let me check another example to be safe. Let's take sum = 10. 10 divided by 3 is 3 with remainder 1. So according to the rule, replace one 3 with two 2s. So instead of 3+3+3+1, we have 3+3+2+2. Product is 3*3*2*2 = 36. If we use 3*3*3*1, the product is 27, which is worse. So that works. So similarly, in our case, since the remainder is 2, we don't need to replace any 3s. We just add the 2. So the product is 3^658 * 2.Therefore, I think that's the correct answer.Final AnswerThe greatest product is boxed{2 times 3^{658}}.</think>To determine the greatest number that is the product of some positive integers whose sum is 1976, we use the strategy of breaking the number into as many 3s as possible, and handling the remainder with 2s if needed.1. Divide 1976 by 3: - 1976 ÷ 3 = 658 with a remainder of 2. - This means we can have 658 threes and one two.2. The product of these numbers is (3^{658} times 2).3. Verification with smaller numbers confirms that using 3s and 2s gives the highest product. For example, splitting 10 into 3s and 2s (3+3+2+2) gives a higher product (36) than using 1s (27).Thus, the greatest product is (boxed{2 times 3^{658}}).,<think>Okay, so I need to find the greatest number that is the product of some positive integers, and the sum of these numbers is 1976. Hmm, let's think about this step by step. First, I remember that when you want to break a number into parts to maximize the product, you should use numbers that are as close as possible to the mathematical constant 'e', which is approximately 2.718. So, the numbers closest to 3 would give the highest product. But I also remember that 3s and 2s are typically the best numbers to use for this kind of problem. Let me verify that.Suppose I have a number S that I need to split into numbers such that their sum is S, and the product is maximized. The key idea is to break S into as many 3s as possible, and if the remainder is 1, use a 4 instead of a 3 and 1 because 2*2 is better than 3*1. If the remainder is 2, then just use a 2 along with the 3s. Let me check if this logic applies here.So, 1976 divided by 3. Let's do that division. 3 goes into 1976 how many times? Let me compute 1976 ÷ 3. 3*600 = 1800, which leaves 1976 - 1800 = 176. Then 3*50 = 150, so 176 - 150 = 26. Then 3*8 = 24, so 26 -24 = 2. So total is 600 + 50 + 8 = 658 times with a remainder of 2. So that means 1976 = 3*658 + 2. Therefore, according to the logic, we should use 658 3s and one 2. Then the product would be 3^658 * 2^1. But let me check if that's the case. Wait, if the remainder is 1, then we replace one 3 and the 1 with two 2s. But in this case, the remainder is 2, so we don't need to do that. So, the maximum product would be 3^658 * 2. But let me think again. Is this actually the maximum? Let me test with smaller numbers to see if this approach works. For example, suppose the sum is 6. The maximum product is 3*3 = 9. If the sum is 5, then 3*2=6. If the sum is 4, then 2*2=4. If the sum is 3, then 3. If the sum is 2, then 2. If the sum is 1, then 1. That seems correct. So, the formula seems to hold. If you can divide the number into 3s, that's best. If the remainder is 1, then replace one 3 with two 2s (since 2+2=4, and 3+1=4, but 2*2=4 is better than 3*1=3). If the remainder is 2, then just add a 2. So, in our case, since 1976 divided by 3 gives a quotient of 658 and a remainder of 2, the optimal decomposition is 658 3s and one 2. Therefore, the product is 3^658 * 2. But wait, let's check if that is actually the case. Let's take a slightly bigger remainder. For example, sum=10. According to the rule, 10 divided by 3 is 3 with remainder 1. So, we need to replace one 3 with two 2s. So, the decomposition would be 3*3*2*2=36. Let's see if that's the maximum. Alternatively, 2+2+2+2+2=10, product=32. 3+3+2+2=10, product=36. 3+3+3+1=10, product=27. 5+5=10, product=25. So yes, 36 is the maximum. So the rule works here. Another example: sum=8. Divided by 3 is 2 with remainder 2. So, two 3s and one 2. Product=3*3*2=18. Alternatively, 2*2*2*2=16. 4*4=16. 5*3=15. So 18 is indeed the maximum. So, the rule works here. So, going back, for 1976, which gives remainder 2 when divided by 3. So, the maximum product is 3^658 * 2. But let me check another approach. Suppose instead of using as many 3s as possible, perhaps using some 2s would yield a better product. Let's think about the relationship between 3s and 2s. Since 3 is larger than 2, but 2 is closer to e. Let's see, 3^1 = 3, 2^(1.5) ≈ 2.828. So, 3 is better. But when you have two 2s (sum=4) vs. one 3 and one 1 (sum=4). The product of two 2s is 4, which is better than 3*1=3. So, 2s are better when the alternative is a 3 and a 1. But in the case where we have a remainder of 2, which is not 1, so adding a 2 is better than splitting into 1s. So, in our problem, since 1976 divided by 3 leaves a remainder of 2, we can just use that 2 as part of the product. So, 3^658 * 2. But let me think again. Let's suppose we have a number that's divisible by 3. For example, sum=6. Then, 3*3=9. If we use 2s, 2*2*2=8. So, 3s are better. So, for numbers divisible by 3, use all 3s. For numbers with remainder 1, use 3s and replace one 3 with two 2s. For numbers with remainder 2, use 3s and one 2. Therefore, 1976 divided by 3 is 658 with remainder 2. So, the optimal decomposition is 658 3s and one 2, resulting in a product of 3^658 * 2. But let me compute the product. However, 3^658 is a huge number. But since the question just asks for the greatest number, which is the product, we can express it in terms of exponents. So, the answer is 2 * 3^658. But wait, is there a way to write this in a different form? Let me check if 658 is correct. Let me recalculate 1976 divided by 3. 3*600=1800. 1976-1800=176. 3*50=150. 176-150=26. 3*8=24. 26-24=2. So, yes, 600+50+8=658. So, 3*658=1974. Then 1976-1974=2. So yes, the remainder is 2. So, 658 3s and one 2. Therefore, the maximum product is 3^658 * 2. But let me check another case. Suppose we use 657 3s, which would sum up to 1971, then we have 1976 -1971=5 left. So, breaking 5 into 2 and 3. Then the product would be 3^657 *3*2=3^658*2. Wait, that's the same as before. So, whether we split the remaining 5 into 2 and 3 or not, we get the same product. Hmm. So, that's an alternative way. But actually, in this case, 5 can be split into 2+3, which gives a product of 6, which is better than using 5 as is (product 5). So, even if we have to break down numbers into smaller ones, as long as they are 2s and 3s, the product is maximized. But in our original calculation, we had already broken down the entire number into 3s and a 2. So, maybe there's no problem here. Alternatively, suppose I use 2s instead. Let me see. 1976 divided by 2 is 988. So, 988 2s. Then the product would be 2^988. Which is larger, 2^988 or 3^658 *2? Let's see. Let's compute the ratio of the two products. Let me write 3^658 *2 vs 2^988. Take the ratio: (3^658 *2)/2^988 = 3^658 / 2^987. To compare these, take the natural logarithm:ln(3^658 / 2^987) = 658 ln 3 - 987 ln 2.Compute 658 ln 3 ≈ 658 * 1.098612 ≈ 658 * 1.0986 ≈ Let's calculate 658 * 1 = 658, 658 * 0.0986 ≈ 658 * 0.1 = 65.8, minus 658 * 0.0014 ≈ 0.9212. So approx 65.8 - 0.9212 ≈ 64.8788. So total ≈ 658 + 64.8788 ≈ 722.8788.Then 987 ln 2 ≈ 987 * 0.6931 ≈ Let's compute 1000 * 0.6931 = 693.1, minus 13 * 0.6931 ≈ 8.9863. So approx 693.1 - 8.9863 ≈ 684.1137.Therefore, ln(ratio) ≈ 722.8788 - 684.1137 ≈ 38.7651. Since this is positive, the ratio is greater than 1, meaning 3^658 *2 is greater than 2^988. Therefore, using as many 3s as possible gives a higher product than using all 2s. So, the initial approach is correct. Therefore, the maximum product is 3^658 * 2. But wait, let me check with another example. For sum=10, as before, which we saw gives product 36. Let's compute 3^3 *2=27*2=54? Wait, no. Wait, sum=10. If divided by 3, we get 3*3=9 with remainder 1, so replace one 3 with two 2s, so 3*3*2*2=36. But if we use 3^3*1=27*1=27, which is worse. So, using two 2s is better. But in our problem, since the remainder is 2, we don't have that problem. So, perhaps in our case, the decomposition is correct. Alternatively, suppose I have a remainder of 2, which is what we have here. So, the decomposition is 3^658 *2. Another check: Let me take sum=5. Divided by 3 is 1 with remainder 2. So, decomposition is 3*2=6. If we use 2+3=5, product=6. If we use 2+2+1=5, product=4. So, 3*2=6 is better. So, correct. Similarly, sum=8. 8 divided by 3 is 2 with remainder 2. So, 3*3*2=18. Alternatively, 2*2*2*2=16. So, 18 is better. Therefore, the formula holds. So, given that, 1976 is decomposed into 658 3s and one 2, product is 3^658 *2. But let me check if using more 2s would help. For instance, maybe using some 4s or 5s. Wait, but 4 is 2+2, and 2*2=4 is the same as 4. So, using 4 is equivalent to two 2s. Similarly, 5 can be broken into 2+3, which gives a higher product (6) than 5. So, numbers larger than 3 should be broken into 2s and 3s. Similarly, 1s are bad because multiplying by 1 doesn't change the product. Therefore, we should avoid 1s. So, in our problem, since we have a remainder of 2, we use a single 2. So, no 1s are involved, which is good. Therefore, the conclusion is that the maximum product is 3^658 multiplied by 2. But the problem says "the product of some positive integers", so that would be the answer. However, since 3^658 *2 is an astronomically large number, it's not necessary to compute it explicitly. So, the answer is simply 2 multiplied by 3 raised to the 658th power. Alternatively, if we need to present it in terms of exponents, that's the way. But maybe they just want the expression. Wait, the problem states "the greatest number, who is the product of some positive integers, and the sum of these numbers is 1976." So, the answer should be written as boxed{2 times 3^{658}}. But let me confirm once again. Let's say the maximum product is achieved when we have as many 3s as possible, and then a 2 if the remainder is 2. So, 3^658 *2. Yes, that seems to be the case. Alternatively, maybe the answer is written as boxed{2 cdot 3^{658}} or boxed{3^{658} cdot 2}, but usually, the coefficient comes first. So, I think that's the answer. Final AnswerThe greatest product is boxed{2 times 3^{658}}.</think>To determine the greatest number that is the product of some positive integers whose sum is 1976, we need to break the number 1976 into parts that maximize the product. The optimal strategy is to use as many 3s as possible, and if the remainder is 1, replace one 3 with two 2s. If the remainder is 2, simply add a 2.1. Divide 1976 by 3: [ 1976 div 3 = 658 text{ remainder } 2 ] This means 1976 can be expressed as (3 times 658 + 2).2. According to the strategy, since the remainder is 2, we use 658 instances of 3 and one instance of 2. The product is: [ 3^{658} times 2 ]3. Verifying this strategy with smaller numbers confirms that using 3s and 2s (where necessary) yields the highest product. For example, breaking 10 into 3, 3, 2, 2 gives a product of 36, which is higher than other combinations.Thus, the greatest product of positive integers that sum to 1976 is (boxed{2 times 3^{658}}).]

🔑:1. Initial Assumption and Replacement Strategy: - Suppose we have a positive integer ( k geq 5 ). We can replace ( k ) with ( 2 ) and ( k-2 ). This is because: [ 2(k-2) = 2k - 4 > k quad text{for} quad k geq 5 ] This replacement increases the product, making it more optimal.2. Handling ( k = 4 ): - If ( k = 4 ), we can replace it with two ( 2 )'s: [ 4 = 2 + 2 quad text{and} quad 2 cdot 2 = 4 ] This does not change the product but keeps the sum the same.3. Optimal Numbers: - From the above steps, we see that the optimal numbers to use are either ( 2 ) or ( 3 ). This is because any number ( k geq 5 ) can be broken down into smaller numbers that increase the product.4. Replacing Three 2's with Two 3's: - We can replace three ( 2 )'s with two ( 3 )'s: [ 2^3 = 8 quad text{and} quad 3^2 = 9 quad text{since} quad 9 > 8 ] This replacement increases the product.5. Sum of Numbers: - We need the sum of the numbers to be ( 1976 ). Let ( x ) be the number of ( 2 )'s and ( y ) be the number of ( 3 )'s. Then: [ 2x + 3y = 1976 ]6. Maximizing the Product: - To maximize the product, we should use as many ( 3 )'s as possible because ( 3 ) contributes more to the product than ( 2 ). We solve for ( y ) by maximizing it: [ 3y leq 1976 quad Rightarrow quad y leq leftlfloor frac{1976}{3} rightrfloor = 658 ] Using ( y = 658 ): [ 3 cdot 658 = 1974 ] This leaves: [ 1976 - 1974 = 2 ] So, we have one ( 2 ) left.7. Final Product: - The final product is: [ 2 cdot 3^{658} ](blacksquare)The final answer is ( boxed{ 2 cdot 3^{658} } )

🧠:[<think>Okay, so I need to find the value of 'a' in the binomial expansion of ((frac{a}{x} - sqrt{frac{x}{2}})^9) such that the coefficient of (x^3) is (frac{9}{4}). Hmm, binomial expansions can sometimes be tricky, but let's take it step by step.First, I remember that the binomial expansion of ((x + y)^n) is the sum from k=0 to n of (binom{n}{k} x^{n - k} y^{k}). In this problem, our two terms are (frac{a}{x}) and (-sqrt{frac{x}{2}}). So, substituting into the binomial theorem, each term in the expansion will look like (binom{9}{k} left(frac{a}{x}right)^{9 - k} left(-sqrt{frac{x}{2}}right)^k).Let me write that out more clearly:Each term is (binom{9}{k} cdot left(frac{a}{x}right)^{9 - k} cdot left(-sqrt{frac{x}{2}}right)^k).I need to find the term where the power of x is 3. So, I have to figure out for which value of k the exponent of x in that term is 3.Let's break down the exponents. For the first part, (left(frac{a}{x}right)^{9 - k}) is the same as (a^{9 - k} cdot x^{-(9 - k)}). Then the second part, (left(-sqrt{frac{x}{2}}right)^k), can be rewritten as ((-1)^k cdot left(frac{x}{2}right)^{k/2}), which simplifies to ((-1)^k cdot 2^{-k/2} cdot x^{k/2}).So, combining the exponents of x from both parts: the first part gives (-(9 - k)) and the second part gives (k/2). Therefore, the total exponent of x in each term is (-(9 - k) + frac{k}{2}).We need this exponent to equal 3 because we want the coefficient of (x^3). So let's set up the equation:[-(9 - k) + frac{k}{2} = 3]Let me solve this equation for k.First, distribute the negative sign:[-9 + k + frac{k}{2} = 3]Combine like terms:[-9 + frac{3k}{2} = 3]Add 9 to both sides:[frac{3k}{2} = 12]Multiply both sides by (frac{2}{3}):[k = 8]So, the term where k=8 will have the exponent of x equal to 3. Now, let's find the coefficient of that term.The coefficient is given by:[binom{9}{8} cdot a^{9 - 8} cdot (-1)^8 cdot 2^{-8/2}]Simplify each part:- (binom{9}{8} = 9)- (a^{1} = a)- ((-1)^8 = 1)- (2^{-4} = frac{1}{16})So multiplying these together:[9 cdot a cdot 1 cdot frac{1}{16} = frac{9a}{16}]According to the problem, this coefficient is equal to (frac{9}{4}). So set up the equation:[frac{9a}{16} = frac{9}{4}]To solve for 'a', multiply both sides by 16:[9a = frac{9}{4} cdot 16]Simplify the right side:[9a = 36]Divide both sides by 9:[a = 4]Wait, let me check that again. Let's verify each step to be sure.First, confirming the exponent equation:Total exponent of x is (-(9 - k) + k/2). Setting that equal to 3:-9 + k + k/2 = 3Which becomes:-9 + (3k)/2 = 3Adding 9: (3k)/2 = 12Multiply by 2: 3k = 24Divide by 3: k = 8. Correct.Then, coefficient calculation:Binomial coefficient (binom{9}{8} = 9), correct.Then (a^{9 - 8} = a^1 = a), correct.((-1)^8 = 1), yes.(2^{-8/2} = 2^{-4} = 1/16), yes.So coefficient is 9 * a * 1 * 1/16 = 9a/16. Correct.Set equal to 9/4:9a/16 = 9/4Multiply both sides by 16:9a = (9/4)*16 = 9*4 = 36Divide by 9:a = 4. Correct.Therefore, the value of a is 4. But let me check once more if I considered all signs correctly. The term is (left(frac{a}{x}right)^{9 - k}) times (left(-sqrt{frac{x}{2}}right)^k). The negative sign is included in the second term, which when raised to the 8th power becomes positive, since (-1)^8 = 1. So that's correct. So coefficient is positive 9a/16. So yes, set to 9/4 gives a=4. Hmm. Maybe I should check by expanding a bit more manually to see if I didn't make a mistake.Alternatively, let's verify with a=4. Let's compute the coefficient of x^3 in the expansion of ((frac{4}{x} - sqrt{frac{x}{2}})^9) and see if it's 9/4.Following the same steps: For k=8,Coefficient: (binom{9}{8} * 4^{1} * (-1)^8 * 2^{-4})Which is 9 * 4 * 1 * (1/16) = 9 * 4 / 16 = 9/4. Exactly matches. So that's correct. Therefore, a=4 is the right answer.Final AnswerThe value of (a) is boxed{4}.</think>To find the value of (a) in the binomial expansion of (left(frac{a}{x} - sqrt{frac{x}{2}}right)^9) such that the coefficient of (x^3) is (frac{9}{4}), we start by considering the general term in the expansion:[binom{9}{k} left(frac{a}{x}right)^{9 - k} left(-sqrt{frac{x}{2}}right)^k]We need the exponent of (x) in this term to be 3. The exponent of (x) in (left(frac{a}{x}right)^{9 - k}) is (-(9 - k)) and the exponent of (x) in (left(-sqrt{frac{x}{2}}right)^k) is (frac{k}{2}). Therefore, we set up the equation:[-(9 - k) + frac{k}{2} = 3]Solving for (k):[-9 + k + frac{k}{2} = 3 -9 + frac{3k}{2} = 3 frac{3k}{2} = 12 k = 8]Next, we calculate the coefficient of the term when (k = 8):[binom{9}{8} cdot a^{9 - 8} cdot (-1)^8 cdot 2^{-8/2}]Simplifying each part:[binom{9}{8} = 9, quad a^1 = a, quad (-1)^8 = 1, quad 2^{-4} = frac{1}{16}]Thus, the coefficient is:[9 cdot a cdot 1 cdot frac{1}{16} = frac{9a}{16}]Setting this equal to (frac{9}{4}):[frac{9a}{16} = frac{9}{4}]Solving for (a):[9a = 36 a = 4]Therefore, the value of (a) is (boxed{4}).,<think>Okay, so I need to find the value of 'a' in the binomial expansion of (a/x - sqrt(x/2))^9, given that the coefficient of x^3 is 9/4. Hmm, binomial expansions, right? Let me recall how that works. The binomial theorem states that (x + y)^n = sum from k=0 to n of (n choose k) x^(n-k) y^k. So in this case, the two terms are a/x and -sqrt(x/2), and n is 9. First, let me write down the general term of the expansion. For each term in the expansion, it should be something like C(9, k) * (a/x)^(9 - k) * (-sqrt(x/2))^k. Then, each term will have a coefficient and some power of x. The goal is to find the term where the exponent of x is 3, set its coefficient equal to 9/4, and solve for a.Let me break it down step by step. The first term is (a/x)^(9 - k). That would be a^(9 - k) divided by x^(9 - k). The second term is (-sqrt(x/2))^k. Let's handle the square root first. sqrt(x/2) is x^(1/2) divided by 2^(1/2), so sqrt(x/2) = x^(1/2)/2^(1/2). Therefore, (-sqrt(x/2))^k = (-1)^k * (x^(1/2)/2^(1/2))^k = (-1)^k * x^(k/2) / 2^(k/2).So combining these two parts, the general term is:C(9, k) * a^(9 - k) / x^(9 - k) * (-1)^k * x^(k/2) / 2^(k/2).Let me simplify the exponents of x. The denominator has x^(9 - k), which is x^-(9 - k). Then multiplied by x^(k/2). So the exponent of x is -(9 - k) + (k/2). Let me compute that:-(9 - k) + (k/2) = -9 + k + k/2 = -9 + (3k)/2.We need this exponent to equal 3 because we're looking for the coefficient of x^3. So:-9 + (3k)/2 = 3Let me solve for k:(3k)/2 = 3 + 9(3k)/2 = 12Multiply both sides by 2:3k = 24Divide by 3:k = 8Wait, so k is 8? Let me check that again. If k is 8, then the exponent of x is -9 + (3*8)/2 = -9 + 24/2 = -9 + 12 = 3. Yes, that's correct. So the term where k=8 will have x^3.Therefore, the coefficient of x^3 is:C(9, 8) * a^(9 - 8) * (-1)^8 / 2^(8/2)Simplify each part:C(9, 8) is 9 (since combination of 9 choose 8 is 9).a^(1) = a.(-1)^8 is 1.2^(8/2) is 2^4 which is 16.So putting it all together, the coefficient is:9 * a * 1 / 16 = (9a)/16.But the problem states that this coefficient is 9/4. Therefore:(9a)/16 = 9/4Multiply both sides by 16:9a = (9/4)*16Simplify the right side: (9*16)/4 = 9*4 = 36So 9a = 36Divide both sides by 9:a = 4Wait, that seems straightforward, but let me double-check my steps.First, the exponent calculation: For the term in k=8, exponent of x is -9 + (3*8)/2 = -9 + 12 = 3. That's correct.Then the coefficient: C(9,8) is 9. Then a^(9 - 8) = a^1 = a. Then (-1)^8 is 1, and 2^(8/2) is 2^4 = 16. So coefficient is 9*a*1/16. Then setting equal to 9/4: 9a/16 = 9/4. Multiply both sides by 16: 9a = 36. So a = 4. That seems correct.Wait, but let me check if there's another term with x^3. For example, could there be another k value where the exponent is 3? Let's see. Suppose k is some other integer between 0 and 9. Suppose k=7: exponent is -9 + (3*7)/2 = -9 + 21/2 = (-18 +21)/2 = 3/2, which is 1.5. Not integer 3. k=9: exponent is -9 + (3*9)/2 = -9 + 27/2 = (-18 +27)/2 = 9/2, which is 4.5. k=6: exponent is -9 + 18/2 = -9 + 9 = 0. So no. Similarly, k=5: -9 + 15/2 = (-18 +15)/2 = -3/2. So no. So only k=8 gives exponent 3. Therefore, only one term. So coefficient is indeed (9a)/16 = 9/4. So a=4.Hence, the answer is a=4.But wait, just to be thorough, let me compute the combination coefficient again. C(9,8) is 9, correct. And the term is (a/x)^(9 - 8) = (a/x)^1 = a/x. The second part is (-sqrt(x/2))^8. Let me compute that separately.(-sqrt(x/2))^8 = [(-1)^8]*(sqrt(x/2))^8 = 1*(x/2)^(4) = x^4 / 16. So multiplying that by a/x gives a/x * x^4 /16 = a*x^3 /16. Then multiplied by the combination coefficient 9: 9*(a x^3 /16) = (9a/16)x^3. So yes, the coefficient is 9a/16. Setting equal to 9/4: a=4.So, all steps check out. Therefore, the value of a is 4.Final AnswerThe value of ( a ) is boxed{4}.</think>To find the value of ( a ) in the binomial expansion of ( left( frac{a}{x} - sqrt{frac{x}{2}} right)^9 ) where the coefficient of ( x^3 ) is ( frac{9}{4} ), we use the binomial theorem. The general term in the expansion is given by:[binom{9}{k} left( frac{a}{x} right)^{9-k} left( -sqrt{frac{x}{2}} right)^k]Simplifying the term, we have:[binom{9}{k} cdot a^{9-k} cdot (-1)^k cdot frac{x^{k/2}}{2^{k/2}} cdot frac{1}{x^{9-k}}]Combining the exponents of ( x ):[-(9 - k) + frac{k}{2} = -9 + k + frac{k}{2} = -9 + frac{3k}{2}]We need this exponent to be 3:[-9 + frac{3k}{2} = 3 implies frac{3k}{2} = 12 implies k = 8]For ( k = 8 ), the coefficient is:[binom{9}{8} cdot a^{1} cdot (-1)^8 cdot frac{1}{2^{4}} = 9 cdot a cdot 1 cdot frac{1}{16} = frac{9a}{16}]Setting this equal to ( frac{9}{4} ):[frac{9a}{16} = frac{9}{4} implies 9a = 36 implies a = 4]Thus, the value of ( a ) is (boxed{4}).]

🔑:This problem primarily tests the binomial theorem. Using the general term formula of binomial expansion, we find the r+1 term, let the exponent of x be 3 to find the coefficient of x^3 in the expansion, and then solve for a.The general term of the binomial expansion of ({a over x} - sqrt{{x over 2}})^9 is:T_{r+1}=C_{9}^{r}{({a over x})}^{9-r}({- sqrt{{x over 2}}})^{r}=({- {sqrt{2} over 2}})^{r}a^{9-r}C_{9}^{r}x^{ {3r over 2}-9 }Let {3r over 2}-9=3, we get r=8.So, the coefficient of x^3 in the expansion is {9 over 16}a.Given that the coefficient of x^3 in the expansion is {9 over 4}, we have:{9 over 16}a={9 over 4}Solving for a, we get a=4.Thus, the answer is boxed{4}.

🧠:[<think>Alright, let's see. The problem is asking for the range of real numbers m such that the equation 4^x + m*2^x + 1 = 0 has a real solution. So, proposition p is true if there exists at least one real number x that satisfies this equation for some m. We need to find all such m's.Hmm, okay. Let me start by recalling that 4^x is the same as (2^x)^2. So maybe substituting t = 2^x can simplify this equation. Since 2^x is always positive for any real x, t must be greater than 0. That substitution might turn this into a quadratic equation in terms of t. Let me try that.Let t = 2^x. Then 4^x = (2^x)^2 = t^2. So substituting into the original equation:t^2 + m*t + 1 = 0.So now, the equation becomes a quadratic in t: t² + m*t + 1 = 0. And since t = 2^x > 0, we need this quadratic equation to have at least one positive real root. Therefore, the problem reduces to finding the values of m for which the quadratic t² + m*t + 1 has at least one positive real root.Alright, so quadratic equations and their roots. Let me recall the conditions for a quadratic equation ax² + bx + c = 0 to have real roots. The discriminant D must be non-negative: D = b² - 4ac ≥ 0. Then, depending on the signs of the coefficients, we can determine the nature of the roots.In this case, the quadratic equation is t² + m*t + 1 = 0. So a = 1, b = m, c = 1. The discriminant D = m² - 4*1*1 = m² - 4. For the equation to have real roots, we need D ≥ 0, so m² - 4 ≥ 0. This implies m² ≥ 4, so m ≤ -2 or m ≥ 2.But even if the discriminant is non-negative, we need at least one positive root. However, since the coefficients are a = 1, b = m, c = 1. The product of the roots is c/a = 1, which is positive. That means both roots are either positive or both are negative. Since their product is positive and sum is -b/a = -m. So sum of roots is -m.Wait, the product of the roots is positive (1), so both roots have the same sign. Therefore, either both roots are positive or both are negative. For the equation to have a positive real root, both roots need to be positive. But since the original substitution requires t > 0, even if both roots are positive, then there exists t > 0. Alternatively, if both roots are negative, then there are no solutions.Wait, but hold on. Let me think again. If the quadratic equation in t has two real roots, both with the same sign (since product is positive). If both roots are positive, then we have solutions t > 0. If both roots are negative, then t would have to be negative, which is impossible because t = 2^x > 0. So, in order for the original equation to have a solution, the quadratic must have at least one positive root. Therefore, both roots need to be positive. Wait, but how do we ensure that?Alternatively, maybe not both. If one root is positive and one is negative, but since their product is 1 (positive), they must have the same sign. So both positive or both negative. Therefore, in order for the quadratic equation to have a positive root, both roots must be positive. So the quadratic equation must have two positive real roots. So, in addition to discriminant D ≥ 0, we need the sum of the roots to be positive. Wait, the sum of the roots is -b/a = -m. So if the sum is positive, then -m > 0, which implies m < 0. But the product of the roots is 1, which is positive. Therefore, if sum of roots is positive (i.e., m < 0) and product is positive, then both roots are positive. Alternatively, if sum of roots is negative (i.e., m > 0) and product is positive, both roots are negative.Therefore, in order for the quadratic equation to have two positive roots, we need the discriminant D ≥ 0 (i.e., m ≤ -2 or m ≥ 2) and sum of roots positive, which is equivalent to -m > 0, so m < 0. Combining these two conditions, m ≤ -2. So, if m ≤ -2, then the quadratic equation has two positive real roots. Therefore, the equation 4^x + m*2^x +1 =0 will have real solutions x (since t = 2^x can take the positive roots). On the other hand, if m ≥ 2, the quadratic equation has two negative roots, which are invalid since t must be positive. Therefore, in that case, there are no solutions. Therefore, the range of m is m ≤ -2.Wait, but the problem says "If proposition p is true, then the range of values for m is...". So p is true when there exists an m such that the equation has a real solution. Wait, but m is given as a parameter. Wait, maybe I misinterpret the problem.Wait, actually, the proposition p is: "There exists an m ∈ ℝ such that 4^x + m*2^x +1 =0". But no, wait. Wait the problem states: "Given the proposition p: 'There exists an m ∈ ℝ such that 4^x + m*2^x +1 =0'. If proposition p is true, then the range of values for the real number m is ________."Wait, maybe I need to parse the problem more carefully. The proposition p is: "There exists an m ∈ ℝ such that the equation 4^x + m*2^x +1 =0 holds." Wait, but hold on. Wait, no. The wording is a bit confusing.Wait, the proposition p is: "There exists an m ∈ ℝ such that 4^x + m*2^x +1 =0." So, the proposition p asserts that for some real number m, the equation 4^x + m*2^x +1 =0 is true. But wait, for which variable? Is the equation supposed to hold for all x? Or is the proposition p saying that there exists an m and there exists an x such that the equation holds? The problem is a bit ambiguous in the wording.Wait, in the original problem statement: "Given the proposition p: 'There exists an m ∈ ℝ such that 4^x + m*2^x +1 =0.' If proposition p is true, then the range of values for the real number m is ________."Wait, perhaps the wording is: The proposition p is "There exists an m ∈ ℝ such that the equation 4^x + m*2^x +1 =0 [has a solution?]". But in the problem statement, since it's written as an equation, perhaps it's supposed to hold for all x? But that seems unlikely because 4^x + m*2^x +1 =0 cannot hold for all x unless all coefficients are zero, which is impossible. Therefore, perhaps the correct interpretation is: The proposition p is "There exists an m ∈ ℝ and there exists an x ∈ ℝ such that 4^x + m*2^x +1 =0". So, p is true if there is at least one real number x and a real number m such that the equation is satisfied. But then, the problem says, if proposition p is true, then find the range of m. Wait, but the problem says "If proposition p is true, then the range of values for the real number m is...", which is confusing because p is asserting the existence of m. Wait, perhaps the problem is misstated?Wait, maybe the problem is phrased as: Given the proposition p: "There exists an x ∈ ℝ such that 4^x + m*2^x +1 =0". If p is true, then find the range of m. That would make more sense. So, p is the statement that for a given m, there exists an x such that the equation holds. Then, the problem is asking for all m such that p is true, i.e., the range of m for which the equation has at least one real solution x.Yes, that seems more plausible. The original problem might have a slight wording issue. So, assuming that the proposition p is: "For a real number m, there exists an x ∈ ℝ such that 4^x + m*2^x +1 =0". Then, the problem is asking for the range of m for which p is true.In that case, proceeding as before. Substitute t = 2^x > 0. Then equation becomes t² + m*t +1 = 0. So, to have a real solution x, we need a positive real solution t. Therefore, the quadratic equation in t must have at least one positive root. So, the problem reduces to finding m such that the quadratic equation t² + m*t +1 =0 has at least one positive real root.So, conditions for quadratic equation t² + m*t +1 =0 to have at least one positive real root:Case 1: One positive and one negative root. But the product of roots is 1 (from c/a = 1), which is positive. Therefore, both roots are of the same sign. So, either both positive or both negative. Therefore, there cannot be one positive and one negative root.Case 2: Both roots positive. Then, to have both roots positive, the sum of roots must be positive and the product is positive. The product is already positive (1). The sum of roots is -b/a = -m. Therefore, sum of roots positive implies -m > 0 => m < 0. Also, discriminant D = m² -4 ≥ 0 => m² ≥4 => m ≤ -2 or m ≥2. Therefore, combining with m < 0, we get m ≤ -2. Therefore, if m ≤ -2, the quadratic has two positive real roots, so t is positive, so x exists.Case 3: Both roots negative. Sum of roots is -m, which would be positive if m <0, but product is positive. Wait, if both roots are negative, then their sum is negative. Wait, sum of roots is -m. If both roots are negative, then their sum is negative. So, -m < 0 => m >0. But the product is positive. So, if m >0, then both roots are negative. But since t must be positive, these roots would not yield real solutions x. Therefore, if m >0, the equation has two negative roots, which are invalid for t = 2^x.But what if the quadratic has a repeated root? If discriminant is zero, m² -4 =0 => m= ±2. For m=2, the equation is t² +2t +1=0, which factors as (t+1)^2=0, so t=-1 is a repeated root. But t=-1 is invalid. For m=-2, the equation is t² -2t +1=0, which factors as (t-1)^2=0, so t=1 is a repeated root. t=1 is valid, since t>0, so x=0 is a solution. Therefore, m=-2 is allowed.Therefore, the allowable m are m ≤ -2.But let's verify this with an example. Take m=-3. Then the quadratic equation becomes t² -3t +1=0. The roots are [3 ± sqrt(9 -4)]/2 = [3 ± sqrt(5)]/2. Both roots are positive because sqrt(5) ≈2.236, so 3 - sqrt(5) ≈0.764>0. Therefore, both roots are positive. Therefore, t= [3 + sqrt(5)]/2 or [3 - sqrt(5)]/2. Then x= log2(t) exists. Therefore, m=-3 is valid.If m=-2, quadratic equation is t² -2t +1=0, root t=1, which gives x=0. So valid.If m=-1, discriminant is (-1)^2 -4=1-4=-3<0, so no real roots. So equation has no solution. Similarly, for m=0, equation is t² +1=0, which has no real roots. For m=1, discriminant 1-4=-3<0. For m=2, discriminant 4-4=0, but repeated root at t=-1, invalid. For m=3, discriminant 9-4=5>0, roots [-3 ± sqrt(5)]/2. Both roots negative since sqrt(5)≈2.236, so -3 + sqrt(5)≈-0.764<0 and -3 -sqrt(5)≈-5.236<0. Therefore, both roots negative, invalid.Therefore, only when m ≤ -2 does the quadratic equation t² +mt +1=0 have positive real roots, which correspond to real x via t=2^x.Hence, the range of m is (-∞, -2].But wait, let me just check m=-2. For m=-2, quadratic equation is t² -2t +1=0, which factors as (t-1)^2=0, so t=1. Therefore, x= log2(1)=0, which is real. So m=-2 is included.Therefore, the answer is m ≤ -2. In interval notation, (-∞, -2].So, I think that's the solution.Final AnswerThe range of values for ( m ) is boxed{(-infty, -2]}.</think>Given the proposition ( p ): "There exists an ( m in mathbb{R} ) such that ( 4^x + m cdot 2^x + 1 = 0 )." We need to determine the range of values for ( m ) if proposition ( p ) is true.First, we substitute ( t = 2^x ) (where ( t > 0 ) because ( 2^x ) is always positive). This transforms the given equation into a quadratic equation in ( t ):[ t^2 + mt + 1 = 0 ]For this quadratic equation to have real solutions, the discriminant must be non-negative:[ Delta = m^2 - 4 geq 0 implies m^2 geq 4 implies m leq -2 text{ or } m geq 2 ]Next, we need the quadratic equation to have at least one positive root. Since the product of the roots (given by ( frac{c}{a} = 1 )) is positive, both roots must be of the same sign. For the roots to be positive, the sum of the roots (given by ( -frac{b}{a} = -m )) must be positive:[ -m > 0 implies m < 0 ]Combining the conditions ( m leq -2 ) (from the discriminant) and ( m < 0 ), we get ( m leq -2 ).Thus, the range of values for ( m ) is (boxed{(-infty, -2]}).,<think>Okay, so I have this problem here: I need to find the range of values for the real number m such that the equation 4^x + m * 2^x + 1 = 0 has a real solution. The proposition p is saying that there exists an m in real numbers for which this equation is true, and if p is true, then what is the range of m? Hmm, let me think.First, let me rewrite the equation to see if I can simplify it. 4^x is the same as (2^x)^2, right? So maybe substituting t = 2^x would help. Let's try that. Let t = 2^x. Then 4^x becomes t^2, so the equation becomes t^2 + m*t + 1 = 0. That's a quadratic equation in terms of t. Cool, that seems easier to handle.But wait, 2^x is always positive for any real x. So t must be greater than 0. So the quadratic equation t^2 + m*t + 1 = 0 must have at least one positive real root. So the problem reduces to finding the values of m for which this quadratic has a positive real solution. Got it.Okay, so for a quadratic equation at^2 + bt + c = 0 to have real roots, the discriminant must be non-negative. The discriminant D is b^2 - 4ac. Here, a = 1, b = m, c = 1. So D = m^2 - 4*1*1 = m^2 - 4. So for the quadratic to have real roots, m^2 - 4 ≥ 0. That means m^2 ≥ 4, so m ≤ -2 or m ≥ 2. But we also need the roots to be positive. So even if the discriminant is non-negative, the roots might be negative. So we have to check the conditions for the roots to be positive.Wait, quadratic equations can have two roots. So either one of them is positive, or both. But since t must be positive, we need at least one positive root. But since the product of the roots in the quadratic equation t^2 + m*t + 1 = 0 is c/a = 1/1 = 1. So the product of the roots is 1. Since the product is positive, that means both roots are either positive or both are negative. Because if they have opposite signs, the product would be negative. But here the product is 1, which is positive. So either both roots are positive or both are negative.But t has to be positive, so in order for the equation to have a solution, both roots must be positive. Therefore, not only the discriminant has to be non-negative (so m^2 - 4 ≥ 0), but also the sum of the roots, which is -b/a = -m/1 = -m, must be positive. Wait, sum of the roots is -m. Wait, sum of roots in quadratic equation at^2 + bt + c = 0 is -b/a. So here, sum of roots is -m. But the sum of the roots has to be positive because both roots are positive. Therefore, -m > 0 => m < 0.So combining these two conditions: m^2 - 4 ≥ 0 and m < 0. Since m^2 ≥ 4, then m ≤ -2 or m ≥ 2. But since m must be less than 0, the intersection is m ≤ -2. Therefore, the range of m is (-∞, -2]. Is that correct?Wait, let me check again. Let me go step by step.First, substitution: t = 2^x > 0.Equation becomes t^2 + m*t + 1 = 0.We need this quadratic to have at least one positive real root. However, the product of the roots is 1, which is positive. Therefore, both roots are positive or both are negative. Since t must be positive, we need both roots to be positive. Therefore, the quadratic must have two positive real roots. So the conditions are:1. Discriminant D ≥ 0: m^2 - 4 ≥ 0 => m ≤ -2 or m ≥ 2.2. Sum of the roots positive: sum = -m > 0 => m < 0.3. Product of the roots positive: 1 > 0, which is already satisfied.So the intersection of these conditions is m ≤ -2. Therefore, m must be less than or equal to -2. So the range is (-∞, -2].Wait, let me test m = -2. If m = -2, then the quadratic equation becomes t^2 - 2t + 1 = 0, which factors as (t - 1)^2 = 0, so t = 1 is a repeated root. Since t = 1 > 0, this is acceptable. So m = -2 is included.If m is less than -2, say m = -3, then the equation becomes t^2 - 3t + 1 = 0. The roots are [3 ± sqrt(9 - 4)] / 2 = [3 ± sqrt(5)] / 2. Both roots are positive because sqrt(5) is approximately 2.236, so 3 + sqrt(5) ≈ 5.236, divided by 2 is ≈ 2.618, and 3 - sqrt(5) ≈ 0.764, both positive. So yes, when m is less than -2, both roots are positive.If m were between -2 and 0, say m = -1, then the quadratic becomes t^2 - t + 1 = 0. The discriminant is 1 - 4 = -3 < 0, so no real roots. So m between -2 and 0 doesn't work.If m is greater than or equal to 2, say m = 3, then the quadratic is t^2 + 3t + 1 = 0. The roots would be [-3 ± sqrt(9 - 4)] / 2 = [-3 ± sqrt(5)] / 2. Both roots would be negative since sqrt(5) ≈ 2.236, so -3 + sqrt(5) ≈ -0.764, and -3 - sqrt(5) ≈ -5.236. Both negative, so t can't be negative. Therefore, even though discriminant is satisfied, the roots are negative, which is invalid. So m ≥ 2 doesn't work.Therefore, the only valid m are m ≤ -2.Therefore, the answer should be m ∈ (-∞, -2].But wait, let me check one more example. Let's take m = -4. Then equation is t^2 -4t +1 =0. The roots are [4 ± sqrt(16 -4)] /2 = [4 ± sqrt(12)] /2 = [4 ± 2*sqrt(3)] /2 = 2 ± sqrt(3). Both 2 + sqrt(3) ≈ 3.732 and 2 - sqrt(3) ≈ 0.267, both positive. So yes, works.So conclusion: m must be less than or equal to -2. So the range is all real numbers m with m ≤ -2.Final AnswerThe range of values for ( m ) is boxed{(-infty, -2]}.</think>Given the proposition ( p ): "There exists an ( m in mathbb{R} ) such that ( 4^x + m cdot 2^x + 1 = 0 )." We need to determine the range of values for ( m ) if proposition ( p ) is true.First, we substitute ( t = 2^x ) (noting that ( t > 0 )) which transforms the equation into a quadratic equation:[ t^2 + mt + 1 = 0 ]For this quadratic equation to have real solutions, the discriminant must be non-negative:[ Delta = m^2 - 4 geq 0 ]This gives:[ m^2 geq 4 ][ implies m leq -2 text{ or } m geq 2 ]Next, we need the roots of the quadratic equation to be positive. The product of the roots is 1 (positive), so both roots must be positive or both negative. Since ( t ) must be positive, we need both roots to be positive. The sum of the roots is given by:[ -frac{m}{1} = -m ]For the sum to be positive, we require:[ -m > 0 ][ implies m < 0 ]Combining the conditions ( m leq -2 ) or ( m geq 2 ) with ( m < 0 ), we get:[ m leq -2 ]Testing specific values confirms that for ( m leq -2 ), the quadratic equation has positive real roots. Therefore, the range of values for ( m ) is:[ boxed{(-infty, -2]} ]]

🔑:Since the given proposition p: "There exists an m in mathbb{R} such that 4^x + m cdot 2^x + 1 = 0" is true,then -m = 2^x + frac{1}{2^x} has a solution. Let t = 2^x, then y = t + frac{1}{t} (t > 0), which is a hyperbola function with a range of [2, +infty),thus -m geq 2, or m leq -2.Hence, the answer is boxed{m leq -2}.Given the proposition p: "There exists an m in mathbb{R} such that 4^x + m cdot 2^x + 1 = 0" is true, we separate the parameters and convert it into a problem of finding the minimum value.This question tests the ability to solve problems by separating parameters and is considered a simple problem.

🧠:[<think>Okay, let me try to work through this problem step by step. It's about an ellipse and a parabola, and there are two parts. Let's start with part (1). First, the problem says we have an ellipse C with the equation x²/a² + y²/b² = 1, where a > b > 0. This ellipse passes through the focus F of the parabola M: x² = 4y. Then, F₁ and F₂ are the left and right foci of the ellipse C, respectively, and the dot product of vectors F₁F and F₁F₂ is 6. We need to find the standard equation of the ellipse C.Alright, let's break this down. Let's recall that the standard form of a parabola x² = 4py has its focus at (0, p). In this case, the parabola is x² = 4y, so comparing to x² = 4py, we see that 4p = 4, so p = 1. Therefore, the focus F of the parabola M is at (0, 1).Since the ellipse C passes through this focus F, which is (0, 1), substituting (0, 1) into the ellipse equation gives 0²/a² + 1²/b² = 1, which simplifies to 1/b² = 1, so b² = 1, hence b = 1. Since b > 0, we take the positive root. So that gives us the value of b.Now, for an ellipse, the distance from the center to each focus is c, where c² = a² - b². The foci of the ellipse are located at (±c, 0) since the major axis is along the x-axis (because a > b). So F₁ is (-c, 0) and F₂ is (c, 0). Given that F is (0,1), we need to compute the vectors F₁F and F₁F₂. Let's find their components. Vector F₁F goes from F₁ (-c, 0) to F (0,1), so the components are (0 - (-c), 1 - 0) = (c, 1).Vector F₁F₂ goes from F₁ (-c, 0) to F₂ (c, 0), so the components are (c - (-c), 0 - 0) = (2c, 0).Now, the dot product of these two vectors is given by (c)(2c) + (1)(0) = 2c² + 0 = 2c². According to the problem, this dot product is equal to 6. So 2c² = 6, which implies c² = 3. Therefore, c = sqrt(3). But we know that for an ellipse, c² = a² - b². We already found that b² = 1, so substituting, we have 3 = a² - 1, which leads to a² = 4. Therefore, a = 2 (since a > 0).Therefore, the standard equation of the ellipse C is x²/4 + y²/1 = 1, or simplified, x²/4 + y² = 1.Wait, let me check that again. If c² = a² - b², then a² = c² + b² = 3 + 1 = 4. Yes, so a = 2. So the equation is x²/4 + y² = 1. That seems right.Let me just verify if substituting F (0,1) into the ellipse equation works: 0²/4 + 1² = 0 + 1 = 1, which is correct. Also, the foci of the ellipse are at (±sqrt(3), 0). The vectors F₁F is (sqrt(3), 1) and F₁F₂ is (2sqrt(3), 0). Their dot product is sqrt(3)*2sqrt(3) + 1*0 = 2*3 + 0 = 6, which matches the given condition. So part (1) seems okay.Now moving on to part (2): If a line l is tangent to the parabola M and intersects the ellipse C at points A and B, find the maximum value of the area of triangle OAB.Hmm. So we have a line that is tangent to the parabola x² = 4y (which is M), and this same line intersects the ellipse x²/4 + y² = 1 at two points A and B. We need to find the maximum area of triangle OAB, where O is the origin.First, let's recall that the tangent lines to the parabola x² = 4y can be expressed in a general form. For a parabola x² = 4py, the tangent line at a point (x₁, y₁) on the parabola is given by xx₁ = 2p(y + y₁). Since here p = 1, the tangent line at (x₁, y₁) is xx₁ = 2(y + y₁). But since (x₁, y₁) lies on the parabola, y₁ = x₁²/4. So substituting, the tangent line becomes x x₁ = 2(y + x₁²/4) => x x₁ = 2y + x₁²/2. Rearranged, this is 2y = x x₁ - x₁²/2 => y = (x x₁)/2 - x₁²/4.Alternatively, we can parametrize the tangent line using the slope. Let me think. Alternatively, the equation of the tangent line can also be written as y = tx - t², where t is a parameter. Let me confirm that.For the parabola x² = 4y, the derivative is dy/dx = x/2. So the slope at a point (x₁, y₁) is m = x₁/2. So the tangent line at that point is y - y₁ = m(x - x₁). Substituting y₁ = x₁²/4 and m = x₁/2, we get y - x₁²/4 = (x₁/2)(x - x₁) => y = (x₁/2)x - x₁²/2 + x₁²/4 => y = (x₁/2)x - x₁²/4. Let’s denote t = x₁/2, so x₁ = 2t. Then substituting, we get y = t x - t². So yes, the tangent line can be written as y = tx - t², where t is a parameter. So the tangent lines to the parabola M are given by y = tx - t².So any tangent line to the parabola can be expressed as y = tx - t² for some real number t.Now, this line intersects the ellipse C: x²/4 + y² = 1 at points A and B. We need to find the points of intersection between the line and the ellipse, then compute the area of triangle OAB and find its maximum value.First, substitute y = tx - t² into the ellipse equation.So substituting, we have x²/4 + (tx - t²)^2 = 1.Let me expand that:x²/4 + t²x² - 2t³x + t^4 = 1.Combine like terms:(1/4 + t²)x² - 2t³x + (t^4 - 1) = 0.That's a quadratic equation in x. Let's denote coefficients:A = (1/4 + t²)B = -2t³C = t^4 - 1So the quadratic equation is A x² + B x + C = 0.The solutions are x = [-B ± sqrt(B² - 4AC)]/(2A)But since we need the points of intersection, which are A and B, their x-coordinates are given by this quadratic. The corresponding y-coordinates would be y = tx - t².Alternatively, maybe there's a better way to compute the area.But let's think step by step.First, once we have the x-coordinates of points A and B, we can find the corresponding y-coordinates. Then, the coordinates of A and B are known. Then, the area of triangle OAB can be calculated using the determinant formula: 1/2 |x_A y_B - x_B y_A|.Alternatively, since O is the origin, the area is 1/2 |x_A y_B - x_B y_A|.But to compute this, we need expressions for x_A, x_B, y_A, y_B in terms of t.Alternatively, maybe we can parametrize this area in terms of t and find its maximum.Alternatively, perhaps we can use the fact that the area can be expressed as 1/2 |(x_A, y_A) × (x_B, y_B)|, which is 1/2 |x_A y_B - x_B y_A|.So, let's try to compute x_A and x_B first.From the quadratic equation:x = [2t³ ± sqrt{(2t³)^2 - 4*(1/4 + t²)*(t^4 - 1)}]/(2*(1/4 + t²))Let's compute the discriminant D:D = (2t³)^2 - 4*(1/4 + t²)*(t^4 - 1)Compute each term:(2t³)^2 = 4t^6Then, 4*(1/4 + t²)*(t^4 - 1) = 4*(1/4*(t^4 -1) + t²*(t^4 -1)) = (t^4 -1) + 4t²(t^4 -1) = (t^4 -1)(1 + 4t²)Therefore, D = 4t^6 - (t^4 -1)(1 + 4t²)Let me expand (t^4 -1)(1 + 4t²):= t^4*(1) + t^4*(4t²) -1*(1) -1*(4t²)= t^4 + 4t^6 -1 -4t²Therefore, D = 4t^6 - [4t^6 + t^4 -4t² -1]= 4t^6 -4t^6 - t^4 +4t² +1Simplify:= -t^4 +4t² +1So the discriminant is D = -t^4 +4t² +1Therefore, the roots are:x = [2t³ ± sqrt(-t^4 +4t² +1)] / [2*(1/4 + t²)]Simplify denominator: 2*(1/4 + t²) = 1/2 + 2t²So x = [2t³ ± sqrt(-t^4 +4t² +1)] / (1/2 + 2t²)Alternatively, multiply numerator and denominator by 2 to eliminate the fraction:x = [4t³ ± 2*sqrt(-t^4 +4t² +1)] / (1 + 4t²)That might be a bit simpler.So x_A and x_B are given by these two roots. Let's denote them as x₁ and x₂.So x₁ = [4t³ + 2*sqrt(-t^4 +4t² +1)] / (1 + 4t²)x₂ = [4t³ - 2*sqrt(-t^4 +4t² +1)] / (1 + 4t²)Then, the corresponding y-coordinates are y₁ = t x₁ - t² and y₂ = t x₂ - t².Therefore, the area of triangle OAB is 1/2 |x₁ y₂ - x₂ y₁|Let me compute this:First, compute x₁ y₂ - x₂ y₁:= x₁ (t x₂ - t²) - x₂ (t x₁ - t²)= t x₁ x₂ - t² x₁ - t x₂ x₁ + t² x₂= (t x₁ x₂ - t x₁ x₂) + (- t² x₁ + t² x₂)= 0 + t²(x₂ - x₁)Therefore, x₁ y₂ - x₂ y₁ = t²(x₂ - x₁)But x₂ - x₁ is the negative of (x₁ - x₂). Let's compute x₁ - x₂:x₁ - x₂ = [4t³ + 2*sqrt(D)] / (1 + 4t²) - [4t³ - 2*sqrt(D)] / (1 + 4t²) = [4t³ + 2*sqrt(D) -4t³ + 2*sqrt(D)] / (1 + 4t²) = [4*sqrt(D)] / (1 + 4t²)Therefore, x₁ - x₂ = 4*sqrt(D)/(1 + 4t²)Therefore, x₂ - x₁ = -4*sqrt(D)/(1 + 4t²)Thus, x₁ y₂ - x₂ y₁ = t² * (-4 sqrt(D)/(1 + 4t²)) = -4 t² sqrt(D)/(1 + 4t²)Taking absolute value, we have |x₁ y₂ - x₂ y₁| = 4 t² sqrt(D)/(1 + 4t²)Therefore, the area is 1/2 * 4 t² sqrt(D)/(1 + 4t²) = 2 t² sqrt(D)/(1 + 4t²)But D is the discriminant we found earlier: D = -t^4 +4t² +1Therefore, the area is 2 t² sqrt(-t^4 +4t² +1) / (1 + 4t²)So we need to maximize this expression with respect to t.So the area as a function of t is:A(t) = (2 t² sqrt(-t^4 +4t² +1)) / (1 + 4t²)We need to find the maximum value of A(t).This seems a bit complicated, but maybe we can simplify it by substitution.Let me denote u = t², since t is real, u ≥ 0.Then, express A(t) in terms of u:A(u) = (2 u sqrt(-u² +4u +1)) / (1 + 4u)So we need to maximize A(u) for u ≥ 0, and such that the expression under the square root is non-negative, i.e., -u² +4u +1 ≥ 0.Solve -u² +4u +1 ≥0Multiply both sides by -1 (reversing inequality):u² -4u -1 ≤0Find roots of u² -4u -1 =0:u = [4 ± sqrt(16 +4)]/2 = [4 ± sqrt(20)]/2 = [4 ± 2*sqrt(5)]/2 = 2 ± sqrt(5)Therefore, the inequality u² -4u -1 ≤0 holds for u between 2 - sqrt(5) and 2 + sqrt(5). Since u ≥0, the valid interval is u ∈ [0, 2 + sqrt(5)]Therefore, the domain of A(u) is u ∈ [0, 2 + sqrt(5)]So we need to maximize A(u) = (2u sqrt(-u² +4u +1)) / (1 + 4u) over u ∈ [0, 2 + sqrt(5)]This is a single-variable calculus optimization problem. Let's denote f(u) = A(u). We need to find its maximum.To find critical points, compute f'(u), set it to zero, and solve for u.But differentiating this might be messy. Let's see.First, write f(u) = 2u (-u² +4u +1)^(1/2) / (1 + 4u)Let’s compute f'(u):Using the quotient rule and chain rule:Let’s denote numerator = 2u (-u² +4u +1)^(1/2), denominator = (1 +4u)f'(u) = [denominator * d/du(numerator) - numerator * d/du(denominator)] / denominator²First, compute d/du(numerator):d/du [2u (-u² +4u +1)^(1/2)]Apply product rule:= 2 (-u² +4u +1)^(1/2) + 2u * (1/2)(-u² +4u +1)^(-1/2) * (-2u +4)Simplify:= 2 (-u² +4u +1)^(1/2) + 2u * ( (-2u +4) ) / [2 (-u² +4u +1)^(1/2) ]Simplify the second term:= 2u (-2u +4) / [2 sqrt(-u² +4u +1) ]= u (-2u +4) / sqrt(-u² +4u +1)Therefore, d/du(numerator) = 2 sqrt(-u² +4u +1) + [u (-2u +4)] / sqrt(-u² +4u +1)Combine terms over a common denominator:= [2 (-u² +4u +1) + u(-2u +4)] / sqrt(-u² +4u +1)Compute numerator:2(-u² +4u +1) + u(-2u +4) = -2u² +8u +2 -2u² +4u = (-2u² -2u²) + (8u +4u) +2 = -4u² +12u +2Therefore, d/du(numerator) = (-4u² +12u +2)/sqrt(-u² +4u +1)Now compute d/du(denominator) = d/du(1 +4u) = 4Now, putting it all into f'(u):f'(u) = [ (1 +4u) * (-4u² +12u +2)/sqrt(-u² +4u +1) - 2u sqrt(-u² +4u +1) *4 ] / (1 +4u)^2Simplify numerator term by term:First term: (1 +4u)(-4u² +12u +2)/sqrt(...)Second term: -8u sqrt(-u² +4u +1)So the entire numerator:= (1 +4u)(-4u² +12u +2)/sqrt(...) -8u sqrt(...)To combine these terms, let's factor out 1/sqrt(...):= [ (1 +4u)(-4u² +12u +2) -8u (-u² +4u +1) ] / sqrt(-u² +4u +1)Let me compute the numerator inside the brackets:(1 +4u)(-4u² +12u +2) -8u(-u² +4u +1)First expand (1 +4u)(-4u² +12u +2):= 1*(-4u² +12u +2) +4u*(-4u² +12u +2)= -4u² +12u +2 -16u³ +48u² +8uCombine like terms:-16u³ + (-4u² +48u²) + (12u +8u) +2= -16u³ +44u² +20u +2Now subtract 8u(-u² +4u +1):Wait, actually, the expression is [ (1 +4u)(-4u² +12u +2) -8u(-u² +4u +1) ]Wait, no: The original expression is:[ (1 +4u)(-4u² +12u +2) -8u(-u² +4u +1) ]So expand the second term: -8u*(-u² +4u +1) = 8u³ -32u² -8uTherefore, adding to the previous result:-16u³ +44u² +20u +2 +8u³ -32u² -8uCombine like terms:(-16u³ +8u³) + (44u² -32u²) + (20u -8u) +2= (-8u³) +12u² +12u +2Therefore, the numerator of f'(u) is:(-8u³ +12u² +12u +2)/sqrt(-u² +4u +1)So, f'(u) = [ (-8u³ +12u² +12u +2) / sqrt(-u² +4u +1) ] / (1 +4u)^2= (-8u³ +12u² +12u +2) / [ (1 +4u)^2 sqrt(-u² +4u +1) ]Set f'(u) = 0:The numerator must be zero, so:-8u³ +12u² +12u +2 =0Multiply both sides by -1:8u³ -12u² -12u -2 =0Let me try to factor this cubic equation.Looking for rational roots using Rational Root Theorem. Possible roots are ±1, ±2, ±1/2, ±1/4, ±1/8.Test u=1: 8 -12 -12 -2 = -18 ≠0u= -1: -8 -12 +12 -2 = -10 ≠0u=2: 64 -48 -24 -2 = -10 ≠0u=1/2: 8*(1/8) -12*(1/4) -12*(1/2) -2 =1 -3 -6 -2= -10≠0u= -1/2: 8*(-1/8) -12*(1/4) -12*(-1/2) -2 = -1 -3 +6 -2=0Wait, u=-1/2 is a root.Therefore, factor out (u +1/2):Using polynomial division or synthetic division.Divide 8u³ -12u² -12u -2 by (u +1/2)Using synthetic division:root: -1/2Coefficients: 8 | -12 | -12 | -2Bring down 8Multiply by -1/2: 8*(-1/2) = -4Add to next coefficient: -12 + (-4) = -16Multiply by -1/2: -16*(-1/2)=8Add to next coefficient: -12 +8= -4Multiply by -1/2: -4*(-1/2)=2Add to last term: -2 +2=0Therefore, the cubic factors as (u +1/2)(8u² -16u -4)Simplify quadratic factor:8u² -16u -4 = 4(2u² -4u -1)Therefore, the equation factors as:(u +1/2)*4*(2u² -4u -1)=0Therefore, roots are u=-1/2, and solutions to 2u² -4u -1=0.Solve 2u² -4u -1=0:u = [4 ± sqrt(16 +8)]/4 = [4 ± sqrt(24)]/4 = [4 ± 2*sqrt(6)]/4 = [2 ± sqrt(6)]/2Since u ≥0, the only valid roots are u= [2 + sqrt(6)]/2 and u= [2 - sqrt(6)]/2. But [2 - sqrt(6)]/2 is approximately (2 -2.45)/2≈ (-0.45)/2≈-0.225, which is negative, so invalid. Therefore, the only positive real roots are u= [2 + sqrt(6)]/2 ≈ (2 +2.45)/2≈2.225.But let's check if these roots are in the domain [0, 2 + sqrt(5)].sqrt(5)≈2.236, so 2 + sqrt(5)≈4.236, and [2 + sqrt(6)]/2≈(2 +2.45)/2≈2.225, which is less than 4.236, so it is in the domain.Therefore, critical points are at u= [2 + sqrt(6)]/2 and u=-1/2 (discarded). So the only critical point in the domain is u= [2 + sqrt(6)]/2.Therefore, we need to check the value of A(u) at u=0, u= [2 + sqrt(6)]/2, and u=2 + sqrt(5).Compute A(0):A(0) = 0, since numerator is 0.Compute A(2 + sqrt(5)):First, check if the sqrt(-u² +4u +1) is real at u=2 + sqrt(5). Let's compute:-u² +4u +1 = -( (2 + sqrt(5))² ) +4*(2 + sqrt(5)) +1Compute (2 + sqrt(5))² =4 +4 sqrt(5) +5=9 +4 sqrt(5)Thus:- (9 +4 sqrt(5)) +8 +4 sqrt(5) +1= (-9 -4 sqrt(5)) +9 +4 sqrt(5)=0So sqrt(0)=0, so A(u) at u=2 + sqrt(5) is zero.Therefore, the maximum must occur at the critical point u= [2 + sqrt(6)]/2.Therefore, let's compute A(u) at u= [2 + sqrt(6)]/2.First, denote u= [2 + sqrt(6)]/2.Compute A(u) = (2u sqrt(-u² +4u +1)) / (1 +4u)But first, let's compute -u² +4u +1.Let me compute this expression:Given u = [2 + sqrt(6)]/2,First compute u²:u² = [ (2 + sqrt(6)) /2 ]² = (4 +4 sqrt(6) +6)/4 = (10 +4 sqrt(6))/4 = (5 +2 sqrt(6))/2Then, -u² +4u +1 = - (5 +2 sqrt(6))/2 +4*[ (2 + sqrt(6))/2 ] +1Compute term by term:= - (5 +2 sqrt(6))/2 + [ (8 +4 sqrt(6)) /2 ] +1= [ -5 -2 sqrt(6) +8 +4 sqrt(6) ] /2 +1= [3 +2 sqrt(6)] /2 +1= 3/2 + sqrt(6) +1= 5/2 + sqrt(6)Wait, let's check again:Wait, step by step:First term: - (5 +2 sqrt(6))/2Second term: 4u =4*(2 + sqrt(6))/2=2*(2 + sqrt(6))=4 +2 sqrt(6). So 4u +1=4 +2 sqrt(6) +1=5 +2 sqrt(6)Wait, no:Wait, original expression: -u² +4u +1We have:= -u² +4u +1= - [ (5 +2 sqrt(6))/2 ] +4*[ (2 + sqrt(6))/2 ] +1= - (5 +2 sqrt(6))/2 + (8 +4 sqrt(6))/2 +1Combine fractions:= [ -5 -2 sqrt(6) +8 +4 sqrt(6) ] /2 +1= [3 +2 sqrt(6)] /2 +1= 3/2 + sqrt(6) +1= (3/2 +1) + sqrt(6)= 5/2 + sqrt(6)Therefore, sqrt(-u² +4u +1)=sqrt(5/2 + sqrt(6)).But this seems complicated. Let's see if we can simplify sqrt(5/2 + sqrt(6)).Let me see if 5/2 + sqrt(6) can be written as (sqrt(a) + sqrt(b))².Suppose (sqrt(a) + sqrt(b))² = a + 2 sqrt(ab) + b = (a + b) + 2 sqrt(ab)Set equal to 5/2 + sqrt(6)Therefore, we have:a + b =5/22 sqrt(ab)=sqrt(6) => sqrt(ab)=sqrt(6)/2 => ab=6/4=3/2So solving:a + b=5/2ab=3/2The solutions for a and b are roots of x² - (5/2)x + 3/2=0Multiply by 2: 2x² -5x +3=0Solve:x=(5±sqrt(25-24))/4=(5±1)/4Thus, x= (5+1)/4=6/4=3/2 or x=(5-1)/4=4/4=1Therefore, a=3/2, b=1 or a=1, b=3/2Therefore, sqrt(5/2 + sqrt(6))=sqrt(3/2) + sqrt(1)=sqrt(6)/2 +1Wait, let's check:(sqrt(3/2) +1)^2=3/2 + 2 sqrt(3/2) +1=5/2 + sqrt(12)/2=5/2 + (2 sqrt(3))/2=5/2 + sqrt(3). But we needed 5/2 + sqrt(6). Hmm, that doesn't match.Wait, maybe my approach is wrong. Let me check the calculation again.Wait, if (sqrt(a) + sqrt(b))² = a + b + 2 sqrt(ab). We set this equal to 5/2 + sqrt(6). Therefore:a + b =5/22 sqrt(ab)=sqrt(6) => sqrt(ab)=sqrt(6)/2 => ab=6/4=3/2Thus, solving a + b=5/2 and ab=3/2, we get quadratic equation 2x² -5x +3=0, which factors as (2x-3)(x-1)=0, so x=3/2 or x=1.Therefore, a=3/2, b=1. So sqrt(5/2 + sqrt(6))=sqrt(3/2)+sqrt(1)=sqrt(6)/2 +1.Wait, sqrt(3/2)=sqrt(6)/2. So yes, sqrt(3/2)=sqrt(6)/2. Therefore,sqrt(5/2 + sqrt(6))=sqrt(6)/2 +1.Therefore, sqrt(-u² +4u +1)=sqrt(6)/2 +1Therefore, returning to A(u):A(u)= (2u * [sqrt(6)/2 +1 ]) / (1 +4u)But let's compute 1 +4u:u=(2 + sqrt(6))/2, so 4u=2*(2 + sqrt(6))=4 +2 sqrt(6), thus 1 +4u=5 +2 sqrt(6)Therefore, A(u)= [2u (sqrt(6)/2 +1) ] / (5 +2 sqrt(6))Substitute u=(2 + sqrt(6))/2:= [2*(2 + sqrt(6))/2 * (sqrt(6)/2 +1) ] / (5 +2 sqrt(6))Simplify numerator:= (2 + sqrt(6)) * (sqrt(6)/2 +1 )Multiply out:= (2 + sqrt(6))(sqrt(6)/2 +1 )=2*(sqrt(6)/2) +2*1 + sqrt(6)*(sqrt(6)/2) + sqrt(6)*1= sqrt(6) +2 + (6/2) + sqrt(6)= sqrt(6) +2 +3 + sqrt(6)=2 sqrt(6) +5Therefore, numerator=2 sqrt(6) +5Denominator=5 +2 sqrt(6)Therefore, A(u)= (5 +2 sqrt(6))/(5 +2 sqrt(6))=1Wait, that's interesting. So A(u)=1.Wait, so the maximum area is 1?But that seems too simple. Let me check the steps again.Wait, when computing A(u)= numerator/denominator= [2u*(sqrt(6)/2 +1)] / (5 +2 sqrt(6))We found numerator=5 +2 sqrt(6), denominator=5 +2 sqrt(6). So indeed, A(u)=1.Therefore, the maximum area of triangle OAB is 1.Wait, but let me check units. The area is 1. But given that the ellipse is x²/4 + y²=1, which has semi-axes 2 and 1, the area of the ellipse is π*2*1=2π≈6.28. So the maximum area of triangle OAB being 1 seems plausible.Alternatively, maybe there was a miscalculation. Let's verify.Wait, the calculation of A(u) at the critical point u=(2 + sqrt(6))/2 resulted in:Numerator:2u*sqrt(-u² +4u +1)=2u*(sqrt(6)/2 +1). Then, denominator:1 +4u=5 +2 sqrt(6)But in the step where we computed numerator:We had:A(u)= (2u * [sqrt(6)/2 +1 ]) / (5 +2 sqrt(6))But then substituting u=(2 + sqrt(6))/2:2u=2*(2 + sqrt(6))/2=2 + sqrt(6)Therefore, numerator= (2 + sqrt(6))*(sqrt(6)/2 +1 )Let me compute this:First, expand (2 + sqrt(6))(sqrt(6)/2 +1):=2*(sqrt(6)/2) +2*1 + sqrt(6)*(sqrt(6)/2) + sqrt(6)*1= sqrt(6) +2 + (6/2) + sqrt(6)= sqrt(6) +2 +3 + sqrt(6)=2 sqrt(6) +5Yes, as before.Denominator=5 +2 sqrt(6)Thus, A(u)= (2 sqrt(6) +5)/(5 +2 sqrt(6))=1Yes, because numerator and denominator are the same.Therefore, the maximum area is indeed 1.Therefore, the answer to part (2) is 1.But let me check with an example. Suppose t=0. Then the tangent line is y=0x -0²=0, which is the x-axis. This line intersects the ellipse at x²/4 +0=1 => x=±2, so points A(2,0) and B(-2,0). The area of triangle OAB is 1/2 |(2)(0) - (-2)(0)|=0. Makes sense.Another example, take t=1. The tangent line is y=x -1. Substituting into the ellipse:x²/4 + (x -1)^2 =1x²/4 +x² -2x +1=1(5x²/4) -2x=0x(5x/4 -2)=0 => x=0 or x=8/5.At x=0, y=-1; at x=8/5, y=8/5 -1=3/5. So points A(0,-1) and B(8/5,3/5). The area is 1/2 |0*(3/5) - (8/5)*(-1)|=1/2 |0 +8/5|=4/5=0.8. Less than 1. So the maximum seems plausible at 1.Alternatively, take t such that u=(2 + sqrt(6))/2≈(2+2.45)/2≈2.225. Let's compute t= sqrt(u)=sqrt(2.225)≈1.491. Then the tangent line is y=1.491x - (1.491)^2≈1.491x -2.223. Let's find intersection with the ellipse:x²/4 + (1.491x -2.223)^2=1This would require numerical computation, but assuming the area is indeed 1, that's the maximum.Therefore, the answers are:(1) x²/4 + y² =1(2) The maximum area is 1.Final Answer(1) The standard equation of the ellipse C is boxed{dfrac{x^2}{4} + y^2 = 1}.(2) The maximum value of the area of triangle OAB is boxed{1}.</think>Given an ellipse ( C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) (where ( a > b > 0 )) that passes through the focus ( F ) of the parabola ( M: x^2 = 4y ). The foci ( F_1 ) and ( F_2 ) of the ellipse ( C ) are such that ( overrightarrow{F_1F} cdot overrightarrow{F_1F_2} = 6 ).(1) Finding the standard equation of the ellipse ( C ):- The focus ( F ) of the parabola ( x^2 = 4y ) is ( (0, 1) ).- Substituting ( (0, 1) ) into the ellipse equation gives ( frac{1}{b^2} = 1 ), so ( b^2 = 1 ) and ( b = 1 ).- The foci of the ellipse are ( (pm c, 0) ) where ( c^2 = a^2 - b^2 ).- The vectors ( overrightarrow{F_1F} ) and ( overrightarrow{F_1F_2} ) are ( (c, 1) ) and ( (2c, 0) ) respectively.- The dot product ( overrightarrow{F_1F} cdot overrightarrow{F_1F_2} = 2c^2 = 6 ) gives ( c^2 = 3 ).- Using ( c^2 = a^2 - b^2 ), we get ( 3 = a^2 - 1 ), so ( a^2 = 4 ) and ( a = 2 ).- The standard equation of the ellipse is ( frac{x^2}{4} + y^2 = 1 ).(2) Finding the maximum value of the area of ( triangle OAB ):- The tangent line to the parabola ( x^2 = 4y ) can be written as ( y = tx - t^2 ).- Substituting this into the ellipse equation gives ( frac{x^2}{4} + (tx - t^2)^2 = 1 ).- Solving the quadratic equation in ( x ) and finding the discriminant ( D = -t^4 + 4t^2 + 1 ).- The area of ( triangle OAB ) is derived as ( frac{2t^2 sqrt{-t^4 + 4t^2 + 1}}{1 + 4t^2} ).- By substituting ( u = t^2 ) and optimizing, the maximum area is found to be 1. Final Answer(1) The standard equation of the ellipse ( C ) is (boxed{dfrac{x^2}{4} + y^2 = 1}).(2) The maximum value of the area of (triangle OAB) is (boxed{1}).,<think>Okay, let's try to tackle this problem step by step. First, part (1) asks for the standard equation of the ellipse C. Given that the ellipse passes through the focus F of the parabola M, which is x² = 4y. Let me recall that for a parabola in the form x² = 4py, the focus is at (0, p). Comparing that to the given equation x² = 4y, so 4p = 4, which means p = 1. Therefore, the focus F of the parabola M is at (0, 1). So, the ellipse C passes through the point F(0,1). The standard equation of the ellipse is x²/a² + y²/b² = 1 with a > b > 0. Since the ellipse passes through (0,1), substituting x=0 and y=1 into the ellipse equation gives 0 + 1²/b² = 1, so 1/b² = 1, which implies b² = 1. Therefore, b = 1 (since b > 0). So, we know that b² = 1.Now, the foci of the ellipse are F₁ and F₂. For an ellipse, the distance from the center to each focus is c, where c² = a² - b². Since the ellipse is centered at the origin (as its equation is x²/a² + y²/b² = 1), the foci are located at (-c, 0) and (c, 0), which are F₁ and F₂ respectively.The problem also states that the vector F₁F dotted with the vector F₁F₂ equals 6. Let's parse this. First, let's find the coordinates of F₁, F₂, and F.F is (0,1), F₁ is (-c, 0), and F₂ is (c, 0). The vectors in question are F₁F and F₁F₂.Vector F₁F is from F₁ to F, so it's (0 - (-c), 1 - 0) = (c, 1).Vector F₁F₂ is from F₁ to F₂, so it's (c - (-c), 0 - 0) = (2c, 0).The dot product of these two vectors is (c)(2c) + (1)(0) = 2c² + 0 = 2c². According to the problem, this dot product is 6. So, 2c² = 6 => c² = 3 => c = √3 (since c is positive).But we know that for an ellipse, c² = a² - b². We already found that b² = 1, so c² = a² - 1. From above, c² = 3, so:a² - 1 = 3 => a² = 4 => a = 2 (since a > 0).Therefore, the standard equation of the ellipse is x²/4 + y²/1 = 1, or simplified to x²/4 + y² = 1. That should be part (1)'s answer.Now, part (2) is about finding the maximum area of triangle OAB, where O is the origin, and points A and B are the intersection points of a tangent line l to the parabola M with the ellipse C. The line l is tangent to the parabola x² = 4y and intersects the ellipse x²/4 + y² = 1 at A and B.First, let's recall that the tangent line to the parabola x² = 4y at a point (x₀, y₀) on the parabola has the equation xx₀ = 2(y + y₀). Alternatively, since the parabola is x² = 4y, the general equation of a tangent line can be written as y = tx + t², where t is a parameter. Wait, let me verify that.For a parabola x² = 4py, the tangent line at parameter t is given by x = 2pt + ty. Wait, maybe it's better to use the point-slope form. Alternatively, the derivative of x² = 4y is 2x dx = 4 dy, so dy/dx = x/2. So, at a point (x₀, y₀) on the parabola, the slope is x₀/2. Therefore, the tangent line is y - y₀ = (x₀/2)(x - x₀). Since y₀ = x₀²/4 (from the parabola equation), substituting gives:y - x₀²/4 = (x₀/2)(x - x₀)Multiply through:y = (x₀/2)x - (x₀²/2) + x₀²/4Simplify:y = (x₀/2)x - (x₀²/4)So, the tangent line equation is y = (x₀/2)x - (x₀²/4). Alternatively, this can be written as y = t x - t² if we let t = x₀/2. Because if t = x₀/2, then x₀ = 2t, and substituting back:y = t x - ( (2t)² ) / 4 = t x - (4t²)/4 = t x - t². Yes, so the tangent line can be parametrized as y = tx - t² where t is a parameter. So, any tangent line to the parabola x² = 4y can be written in this form.Therefore, the equation of the tangent line l is y = tx - t². This line intersects the ellipse x²/4 + y² = 1. To find points A and B, we need to solve the system:y = tx - t²x²/4 + y² = 1Substituting y from the first equation into the second:x²/4 + (tx - t²)^2 = 1Let's expand this:x²/4 + t²x² - 2t³x + t^4 = 1Combine like terms:(1/4 + t²)x² - 2t³x + (t^4 - 1) = 0This is a quadratic equation in x. Let's denote the coefficients:A = (1/4 + t²)B = -2t³C = t^4 - 1So, the quadratic is A x² + B x + C = 0. The solutions are given by x = [-B ± √(B² - 4AC)]/(2A). However, since we need the points of intersection A and B, let's first check if the line is tangent to the ellipse. Wait, no, the line is tangent to the parabola but intersects the ellipse at two points (A and B). So, the quadratic equation above should have two real solutions for x, given that the line is not tangent to the ellipse. So, the discriminant D must be positive.Discriminant D = B² - 4AC = ( -2t³ )² - 4*(1/4 + t²)*(t^4 - 1)Calculate:D = 4t^6 - 4*(1/4 + t²)*(t^4 - 1)Let's compute the second term:First, expand (1/4 + t²)(t^4 - 1):= (1/4)(t^4 - 1) + t²(t^4 - 1)= (t^4/4 - 1/4) + t^6 - t²So, the entire term is 4*(t^6 - t² + t^4/4 - 1/4) = 4t^6 - 4t² + t^4 - 1Wait, no. Wait, D = 4t^6 - 4*(1/4 + t²)*(t^4 - 1). So, compute 4*(1/4 + t²)*(t^4 - 1):= [ (1 + 4t²)/4 ] * (t^4 - 1) * 4Wait, no. Wait, D = 4t^6 - 4*(1/4 + t²)*(t^4 - 1). Let me compute step by step:First, compute (1/4 + t²)*(t^4 - 1):Multiply term by term:1/4 * t^4 = t^4/41/4 * (-1) = -1/4t² * t^4 = t^6t² * (-1) = -t²So, altogether:t^6 - t² + t^4/4 - 1/4Therefore, D = 4t^6 - 4*(t^6 - t² + t^4/4 - 1/4 )Compute the multiplication:= 4t^6 - [4t^6 - 4t² + t^4 - 1]= 4t^6 -4t^6 +4t² - t^4 +1Simplify:= 0 +4t² - t^4 +1= -t^4 +4t² +1So, discriminant D = -t^4 +4t² +1. For the quadratic equation to have real solutions, we need D >=0. However, since the line is tangent to the parabola but intersects the ellipse, the quadratic equation should have two real solutions, so D >0. So, we need -t^4 +4t² +1 >0.But perhaps we can proceed without worrying about that for now. Let's move forward.The solutions for x are:x = [2t³ ± √(-t^4 +4t² +1)] / [2*(1/4 + t²)] = [2t³ ± √( -t^4 +4t² +1 ) ] / ( 1/2 + 2t² )Multiply numerator and denominator by 2 to eliminate the denominator in the denominator:x = [4t³ ± 2√( -t^4 +4t² +1 ) ] / (1 + 4t² )Therefore, the x-coordinates of points A and B are [4t³ ± 2√( -t^4 +4t² +1 ) ] / (1 + 4t² )Once we have x, we can find y using y = tx - t². But since we need the area of triangle OAB, which is formed by points O(0,0), A(x₁,y₁), and B(x₂,y₂). The area can be calculated using the determinant formula:Area = (1/2)|x₁ y₂ - x₂ y₁|Since points A and B lie on the line y = tx - t², their y-coordinates are y₁ = tx₁ - t² and y₂ = tx₂ - t². Therefore,Area = (1/2)|x₁(tx₂ - t²) - x₂(tx₁ - t²)|Simplify:= (1/2)| t x₁ x₂ - t² x₁ - t x₁ x₂ + t² x₂ |= (1/2)| -t² x₁ + t² x₂ |= (1/2)| t² (x₂ - x₁ ) |= (1/2) t² |x₂ - x₁|So, the area is (1/2) t² |x₂ - x₁|. Therefore, to maximize the area, we need to maximize (1/2) t² |x₂ - x₁|.But |x₂ - x₁| is the absolute difference between the roots of the quadratic equation. For a quadratic equation Ax² + Bx + C = 0, the difference of roots is √D / |A|. In our case, quadratic equation in x is A x² + B x + C = 0 with A = (1/4 + t²), B = -2t³, C = t^4 -1. The discriminant D was found to be -t^4 +4t² +1. Therefore, the difference of roots is √D / A. Therefore:|x₂ - x₁| = √( -t^4 +4t² +1 ) / A = √( -t^4 +4t² +1 ) / (1/4 + t² )But A is (1/4 + t²). So, substituting back:|x₂ - x₁| = √( -t^4 +4t² +1 ) / (1/4 + t² )Therefore, Area = (1/2) t² * [ √( -t^4 +4t² +1 ) / (1/4 + t² ) ]So, Area = (1/2) t² √( -t^4 +4t² +1 ) / (1/4 + t² )This expression is a bit complicated, but perhaps we can simplify it. Let's denote:Let’s let’s write the area expression as:Area(t) = (1/2) * t² * sqrt( -t^4 + 4t² +1 ) / ( t² + 1/4 )To maximize Area(t), we can consider squaring the area to simplify the expression, since the maximum of Area(t) will occur at the same t as the maximum of Area(t)². Let's compute Area(t)²:Area²(t) = (1/4) * t^4 * ( -t^4 +4t² +1 ) / ( t² + 1/4 )²Let’s denote u = t². Since t is a real number, u >=0.Then, Area²(t) = (1/4) * u² * (-u² +4u +1 ) / ( u + 1/4 )²So, define function f(u) = [ u² (-u² +4u +1) ] / ( u + 1/4 )²We need to find the maximum of f(u) for u >=0. Then, the maximum Area is (1/2)sqrt(f(u)).So, let's analyze f(u):f(u) = [ -u^4 +4u³ + u² ] / ( u + 1/4 )²To find the critical points, take derivative f’(u) and set to zero.First, let me compute f(u) = N(u)/D(u), where N(u) = -u^4 +4u³ + u² and D(u) = (u + 1/4)^2.Using the quotient rule:f’(u) = [ N’(u) D(u) - N(u) D’(u) ] / [D(u)]²First, compute N’(u):N’(u) = -4u³ +12u² +2uD’(u) = 2(u +1/4)So,f’(u) = [ (-4u³ +12u² +2u)(u +1/4)^2 - (-u^4 +4u³ + u²)(2(u +1/4)) ] / (u +1/4)^4This seems complicated. Let me compute numerator step by step.First, compute term1 = (-4u³ +12u² +2u)(u +1/4)^2Expand (u +1/4)^2 = u² + (1/2)u + 1/16Multiply by (-4u³ +12u² +2u):Multiply term by term:First, -4u³*(u² + (1/2)u +1/16) = -4u^5 -2u^4 - (1/4)u³Then, 12u²*(u² + (1/2)u +1/16) =12u^4 +6u³ + (12/16)u² =12u^4 +6u³ + (3/4)u²Then, 2u*(u² + (1/2)u +1/16) =2u³ +u² + (1/8)uAdd all together:-4u^5 -2u^4 -0.25u³ +12u^4 +6u³ +0.75u² +2u³ +u² +0.125uCombine like terms:-4u^5 + (-2u^4 +12u^4) + (-0.25u³ +6u³ +2u³) + (0.75u² +u²) +0.125uWhich is:-4u^5 +10u^4 +7.75u³ +1.75u² +0.125uSimilarly, term2 = - (-u^4 +4u³ +u²)*2(u +1/4)First, compute inside term2:= - [ (-u^4 +4u³ +u²) * 2(u +1/4) ]= - [ -2u^5 -0.5u^4 +8u^4 +2u³ +2u³ +0.5u² ]Wait, let's compute (-u^4 +4u³ +u²)*2(u +1/4):First, multiply (-u^4 +4u³ +u²) by 2:= -2u^4 +8u³ +2u²Then multiply by (u +1/4):= (-2u^4)(u) + (-2u^4)(1/4) +8u³(u) +8u³(1/4) +2u²(u) +2u²(1/4)= -2u^5 -0.5u^4 +8u^4 +2u³ +2u³ +0.5u²Simplify:-2u^5 + ( -0.5u^4 +8u^4 ) + (2u³ +2u³) +0.5u²= -2u^5 +7.5u^4 +4u³ +0.5u²Therefore, term2 = - [ -2u^5 +7.5u^4 +4u³ +0.5u² ] = 2u^5 -7.5u^4 -4u³ -0.5u²Therefore, the numerator of f’(u) is term1 + term2:(-4u^5 +10u^4 +7.75u³ +1.75u² +0.125u) + (2u^5 -7.5u^4 -4u³ -0.5u²)Combine like terms:(-4u^5 +2u^5) + (10u^4 -7.5u^4) + (7.75u³ -4u³) + (1.75u² -0.5u²) +0.125u= (-2u^5) + (2.5u^4) + (3.75u³) + (1.25u²) +0.125uFactor out common terms:= -2u^5 +2.5u^4 +3.75u³ +1.25u² +0.125uThis is the numerator of f’(u). We need to set this equal to zero:-2u^5 +2.5u^4 +3.75u³ +1.25u² +0.125u =0Factor out u:u(-2u^4 +2.5u³ +3.75u² +1.25u +0.125)=0So, solutions are u=0 or solving -2u^4 +2.5u³ +3.75u² +1.25u +0.125=0But u >=0, so u=0 is a critical point. Let's check if u=0 gives a maximum. When u=0, f(u)=0, so the area is zero. Hence, not the maximum. Therefore, the other critical points come from solving:-2u^4 +2.5u³ +3.75u² +1.25u +0.125=0Multiply both sides by -1:2u^4 -2.5u³ -3.75u² -1.25u -0.125=0This is a quartic equation, which is difficult to solve analytically. Maybe factor out 0.125 to see if it's a multiple?Wait, let me write coefficients:2u^4 - 2.5u³ -3.75u² -1.25u -0.125 =0Multiply all terms by 8 to eliminate decimals:16u^4 -20u³ -30u² -10u -1=0Hmm, not obvious. Let me check for rational roots using Rational Root Theorem. Possible rational roots are ±1, ±1/2, ±1/4, ±1/8, ±1/16. Let's test u=1:16(1)^4 -20(1)^3 -30(1)^2 -10(1) -1 =16 -20 -30 -10 -1= -45 ≠0u= -1:16 +20 -30 +10 -1=15≠0u=1/2:16*(1/16) -20*(1/8) -30*(1/4) -10*(1/2) -1=1 -2.5 -7.5 -5 -1= -15 ≠0u=1/4:16*(1/256) -20*(1/64) -30*(1/16) -10*(1/4) -1= (0.0625) -0.3125 -1.875 -2.5 -1= -5.625≠0u= -1/2:16*(1/16) -20*(-1/8) -30*(1/4) -10*(-1/2) -1=1 +2.5 -7.5 +5 -1=0Wait, u=-1/2:16*(-1/2)^4 -20*(-1/2)^3 -30*(-1/2)^2 -10*(-1/2) -1Compute each term:16*(1/16)=1-20*(-1/8)=2.5-30*(1/4)= -7.5-10*(-1/2)=5-1So, sum:1 +2.5 -7.5 +5 -1=0. Yes, u=-1/2 is a root. Therefore, (u +1/2) is a factor.Now, perform polynomial division:Divide 16u^4 -20u³ -30u² -10u -1 by (u +1/2). Alternatively, use synthetic division with u=-1/2.Write coefficients:16, -20, -30, -10, -1Using synthetic division for root -1/2:Bring down 16.Multiply by -1/2: 16*(-1/2)= -8. Add to next coefficient: -20 + (-8)= -28Multiply by -1/2: -28*(-1/2)=14. Add to next coefficient: -30 +14= -16Multiply by -1/2: -16*(-1/2)=8. Add to next coefficient: -10 +8= -2Multiply by -1/2: -2*(-1/2)=1. Add to last coefficient: -1 +1=0. Perfect.So, the quotient polynomial is 16u³ -28u² -16u -2Therefore, 16u^4 -20u³ -30u² -10u -1= (u +1/2)(16u³ -28u² -16u -2)Now, we need to factor 16u³ -28u² -16u -2. Let's attempt to factor further.Factor out 2: 2(8u³ -14u² -8u -1)Check for rational roots of 8u³ -14u² -8u -1. Possible roots: ±1, ±1/2, ±1/4, ±1/8.Test u=1: 8 -14 -8 -1= -15≠0u=-1: -8 -14 +8 -1= -15≠0u=1/2: 8*(1/8) -14*(1/4) -8*(1/2) -1=1 -3.5 -4 -1= -7.5≠0u= -1/2: -8*(1/8) -14*(1/4) -8*(-1/2) -1= -1 -3.5 +4 -1= -1.5≠0u=1/4: 8*(1/64) -14*(1/16) -8*(1/4) -1=0.125 -0.875 -2 -1≈-3.75≠0No rational roots. Maybe factor by grouping:8u³ -14u² -8u -1 = (8u³ -14u²) + (-8u -1) = 2u²(4u -7) - (8u +1). Doesn't help.Alternatively, perhaps use rational root theorem with cubic formula. Alternatively, since this is getting too complicated, maybe we can use substitution.Alternatively, perhaps there was a mistake in calculation? Let me check.Wait, going back, the original quartic equation was:-2u^4 +2.5u³ +3.75u² +1.25u +0.125=0But after multiplying by -8, we get 16u^4 -20u³ -30u² -10u -1=0And found u=-1/2 is a root. Then, after factoring, we had 2(u +1/2)(8u³ -14u² -8u -1)=0Alternatively, perhaps we can use numerical methods to approximate the roots. Let's consider that for u >0, since u = t² >=0.Let’s denote the cubic equation:8u³ -14u² -8u -1=0Let’s define g(u)=8u³ -14u² -8u -1. Looking for real roots u>0.Compute g(2)=8*8 -14*4 -8*2 -1=64 -56 -16 -1= -9g(3)=8*27 -14*9 -8*3 -1=216 -126 -24 -1=65So, there is a root between 2 and 3.g(2.5)=8*(15.625) -14*(6.25) -8*2.5 -1=125 -87.5 -20 -1=16.5>0So, between 2 and 2.5, g(u) crosses from -9 to16.5. Therefore, there is a root between 2 and2.5.Similarly, compute g(2.25):8*(2.25)^3 -14*(2.25)^2 -8*(2.25) -1=8*(11.390625) -14*(5.0625) -18 -1=91.125 -70.875 -18 -1=1.25>0g(2.1):8*(9.261) -14*(4.41) -16.8 -1≈74.088 -61.74 -16.8 -1≈-5.452<0So, between 2.1 and2.25, g(u) crosses zero.Compute g(2.2):8*(10.648) -14*(4.84) -17.6 -1≈85.184 -67.76 -17.6 -1≈-1.176<0g(2.25)≈1.25>0g(2.225):8*(2.225)^3=8*(11.289)=90.312-14*(2.225)^2= -14*(4.9506)= -69.308-8*(2.225)= -17.8-1Total≈90.312 -69.308 -17.8 -1≈2.204>0g(2.2125):Approximate using linear approximation between 2.2 (-1.176) and2.225 (2.204). The difference in u is0.025, difference in g is3.38. To reach zero from -1.176 at u=2.2, need delta u where 3.38/0.025 per delta u. So delta u≈ (1.176)/ (3.38/0.025)=1.176*0.025/3.38≈0.0087. Therefore, root≈2.2 +0.0087≈2.2087So, approximate root at u≈2.2087Similarly, check g(2.2087):Calculate 8*(2.2087)^3 -14*(2.2087)^2 -8*(2.2087) -1First compute 2.2087^2≈4.8782.2087^3≈2.2087*4.878≈10.778*10.77≈86.16-14*4.878≈-68.29-8*2.2087≈-17.67-1Total≈86.16 -68.29 -17.67 -1≈-1.8Wait, that's not correct. Maybe more precise calculation is needed, but perhaps this approach is too time-consuming. Let's instead consider that the cubic equation may have one real root and two complex roots. Given that, and given that for the area function, we need u>0, so there's only one positive real root at around u≈2.2087. However, since this is complicated, maybe there's a smarter way.Alternatively, maybe instead of using calculus, we can parametrize the problem differently. Let's consider that the tangent line to the parabola is y=tx -t², as we did. Then, the area is (1/2) t² |x₂ -x₁|. We expressed |x₂ -x₁|=sqrt(D)/A, where D is the discriminant. So, Area=(1/2)t²*sqrt(D)/A.But perhaps we can parametrize using t and use substitution to simplify.Alternatively, maybe use Lagrange multipliers? Not sure.Alternatively, consider parametrizing the tangent line in terms of slope. Wait, for the parabola x²=4y, the tangent line in terms of slope m is y = mx + c, with c = -m². Because for the parabola x²=4y, the condition for a line y=mx +c to be tangent is c= -m². Wait, let's confirm that.For the parabola x²=4y, substituting y=mx +c into x²=4(mx +c) gives x² -4m x -4c=0. For the line to be tangent, discriminant must be zero. The discriminant is ( -4m )² -4*1*(-4c )=16m² +16c=0 ⇒16m² +16c=0⇒c= -m². Hence, the tangent line is y =mx -m². So, in this case, t=m, so the parametrization is the same as before. Therefore, the line is y=mx -m², with slope m.But in our previous parametrization, we used t such that y=tx -t², so t is the slope. Therefore, m=t.So, perhaps using m instead of t. Anyway, this might not help.Alternatively, let's make substitution in the Area expression.Area = (1/2)*t²*sqrt( -t^4 +4t² +1 ) / ( t² +1/4 )Let’s denote v = t². Then, since t² = v, Area = (1/2)*v*sqrt( -v² +4v +1 ) / (v +1/4 )So, Area squared is (1/4)*v²*(-v² +4v +1)/(v +1/4)^2Let’s denote w = v +1/4, then v = w -1/4. Substitute into the expression:Numerator: (w -1/4)^2*(- (w -1/4)^2 +4(w -1/4) +1 )Denominator: w²This substitution might complicate things further. Alternatively, perhaps set z = v +1/4. Not sure.Alternatively, notice that the expression inside the square root is -v² +4v +1 = -(v² -4v -1). Let’s complete the square for v² -4v -1:v² -4v -1 = (v -2)^2 -5So, -(v² -4v -1) = -[(v -2)^2 -5] = - (v -2)^2 +5 =5 - (v -2)^2Therefore, sqrt(5 - (v -2)^2). Therefore, the expression under the square root is 5 - (v -2)^2.Therefore, Area = (1/2) * v * sqrt(5 - (v -2)^2 ) / (v +1/4 )But maybe this is not helpful. Alternatively, perhaps use trigonometric substitution. Let’s set (v -2)^2 =5 sin²θ, so that 5 - (v -2)^2=5 cos²θ. But this seems complicated.Alternatively, note that for the expression sqrt( -t^4 +4t² +1 ), we can write it as sqrt( -(t^4 -4t² -1) )=sqrt( -(t^4 -4t² +4 -5) )=sqrt( -( (t² -2)^2 -5 ) )=sqrt(5 - (t² -2)^2 )Therefore, sqrt(5 - (t² -2)^2 ). Therefore, the expression inside the square root is 5 - (t² -2)^2. Therefore, the maximum value of this expression is when (t² -2)^2 is minimized. However, since we have other terms in the area expression, this might not directly help.Alternatively, we can consider that the maximum occurs when derivative is zero, but since we couldn't solve the quartic equation analytically, maybe we can consider specific values of t to test.Alternatively, perhaps use substitution s = t² +1/4, but not sure.Alternatively, since this is a competition problem, perhaps the maximum area occurs when the tangent line is symmetric in some way, maybe at a specific angle. Alternatively, consider using parametric equations for the ellipse.Wait, the ellipse is x²/4 + y² =1. So, parametric equations for the ellipse can be x=2cosθ, y=sinθ. Then, points A and B are intersections of the tangent line with the ellipse.But the tangent line is to the parabola x²=4y, which is y=tx -t². Suppose this line is tangent at point (2t, t²) on the parabola, since for parabola x²=4y, the parametric equations are x=2t, y=t². Then, the tangent line at this point is y=tx -t², as we had.So, substituting the parametric equations of the ellipse into the tangent line equation:y = tx -t² => sinθ = t*(2cosθ) -t² => sinθ = 2t cosθ -t²So, rearranged:2t cosθ - sinθ -t² =0This equation relates t and θ. The points of intersection A and B on the ellipse correspond to two θ values for a given t. However, this might not simplify the problem.Alternatively, perhaps consider that the line y=tx -t² intersects the ellipse x²/4 + y² =1 at points A and B. The area of triangle OAB is (1/2)|x1y2 -x2y1|, which we already expressed in terms of t. Given the complexity of the derivative, perhaps we can consider maximizing the expression numerically.Alternatively, observe that the quartic equation we obtained might have a nice root, but since earlier attempts didn't find it, perhaps the problem expects a different approach.Wait, another idea: the area of triangle OAB can also be calculated as (1/2)| OA × OB |, where × denotes the cross product. Which is the same as the determinant formula.Alternatively, since points A and B lie on the ellipse and on the line y=tx -t², perhaps use the parametrization of the line and find the chord length, then compute the area.But the area is (1/2)|x1y2 -x2y1|, and since y1= tx1 - t², y2= tx2 - t², so:x1y2 -x2y1 =x1(tx2 - t²) -x2(tx1 - t²)= t x1x2 -t²x1 - t x1x2 +t²x2= t²(x2 -x1)Therefore, Area=(1/2) t² |x2 -x1|But x2 -x1 is the difference of roots of the quadratic, which we found to be sqrt(D)/A, where D= -t^4 +4t² +1, and A=1/4 +t².Thus, Area=(1/2) t² sqrt(-t^4 +4t² +1)/(1/4 +t²)Let’s denote f(t) = t² sqrt(-t^4 +4t² +1)/(1/4 +t²)We need to maximize f(t). Let’s make substitution u =t². Then,f(u) = u sqrt( -u² +4u +1 )/(u +1/4 )We need to maximize f(u) for u >=0, and -u² +4u +1 >=0.The expression under the square root must be non-negative:-u² +4u +1 >=0 ⇒ u² -4u -1 <=0 ⇒ roots of u² -4u -1=0 are u=(4±sqrt(16 +4))/2=(4±sqrt(20))/2=2±sqrt(5). Since sqrt(5)≈2.236, so the roots are approximately 2+2.236=4.236 and 2-2.236≈-0.236. Therefore, u² -4u -1 <=0 for u between -0.236 and4.236. Since u=t² >=0, the valid interval is u in [0, 2+sqrt(5)]≈[0,4.236].Therefore, u must be between0 and4.236.So, our domain for u is [0,2+sqrt(5)]To find maximum of f(u)=u sqrt(-u² +4u +1)/(u +1/4 )Let’s square f(u) to simplify:f(u)^2= u² (-u² +4u +1 )/(u +1/4 )²Let’s denote this as g(u)= u² (-u² +4u +1 )/(u +1/4 )²We need to maximize g(u) over u in [0,2+sqrt(5)]Taking derivative g’(u):As before, we did this earlier and ended up with a complicated expression. However, maybe we can use substitution.Let’s let’s set v= u +1/4. Then, u= v -1/4.But this might not help. Alternatively, write g(u)= [ -u^4 +4u^3 +u² ] / (u +1/4 )²Let’s perform polynomial division or express as a polynomial plus a remainder.But this might not be helpful. Alternatively, express numerator as related to denominator:Let’s compute the numerator: -u^4 +4u³ +u²Let’s factor out -u^4 +4u³ +u² = -u²(u² -4u -1 )But the denominator is (u +1/4 )². So, g(u)= -u²(u² -4u -1 )/(u +1/4 )²But from earlier, u² -4u -1 = (u -2)^2 -5Therefore, g(u)= -u²[(u -2)^2 -5 ]/(u +1/4 )²But I’m not sure if this helps.Alternatively, consider substituting w = u -2. Then, u= w +2. But this might not help either.Alternatively, let's consider that for maximum g(u), the derivative g’(u)=0.But since we couldn't solve analytically, perhaps we can substitute u=2, which is in the domain.Compute g(2)= [ -16 +32 +4 ] / (2 +0.25 )²= (20)/(2.25)^2=20/(5.0625)=20*(16/81)=320/81≈3.95Alternatively, compute f(2)=2*sqrt(-4 +8 +1)/ (2 +0.25 )=2*sqrt(5)/2.25≈2*2.236/2.25≈4.472/2.25≈1.987If u=1:g(1)= [ -1 +4 +1 ] / (1.25 )²=4/1.5625≈2.56f(1)=1*sqrt(-1 +4 +1)/1.25=sqrt(4)/1.25=2/1.25=1.6If u=0.5:g(0.5)= [ -0.0625 +2 +0.25 ] / (0.75 )²=(2.1875)/0.5625≈3.888f(0.5)=0.5*sqrt(-0.25 +2 +1)/0.75=0.5*sqrt(2.75)/0.75≈0.5*1.658/0.75≈1.105If u=3:g(3)= [ -81 +108 +9 ] / (3.25 )²=36/10.5625≈3.41f(3)=3*sqrt(-9 +12 +1)/3.25=3*sqrt(4)/3.25=3*2/3.25≈6/3.25≈1.846At u=2, we had f(u)≈1.987, which is higher than at u=3. So, maximum might be around u=2.Wait, but we had when u≈2.2087, the root of the quartic, perhaps the maximum occurs there.Alternatively, let’s test u=2.2087. Compute g(u):g(2.2087)= [ - (2.2087)^4 +4*(2.2087)^3 + (2.2087)^2 ] / (2.2087 +0.25 )²Compute numerator:First, approximate 2.2087^2≈4.8782.2087^3≈2.2087*4.878≈10.772.2087^4≈2.2087*10.77≈23.78Numerator≈-23.78 +4*10.77 +4.878≈-23.78 +43.08 +4.878≈24.178Denominator≈(2.4587)^2≈6.04So, g(u)≈24.178 /6.04≈4.003Therefore, f(u)=sqrt(g(u))≈2.0Which is similar to the value at u=2.Wait, but we have approximated g(u)=4 at u≈2.2087, which would give f(u)=2.0. But earlier at u=2, g(u)=20/5.0625≈3.95, so f(u)=sqrt(3.95)≈1.987. So, at u≈2.2087, the area is approximately2.0, which is higher. This suggests that the maximum area is approximately2.0.But the problem asks for the maximum value. Given that this is a math competition problem, the answer is likely a nice number. Maybe the maximum area is2.But need to verify.Wait, let's consider when the line is the x-axis. The x-axis is y=0, which is tangent to the parabola at the vertex (0,0). The intersection with the ellipse x²/4 +0=1 ⇒x=±2. So, points A(2,0) and B(-2,0). The area of triangle OAB is (1/2)|2*0 - (-2)*0|=0. So, zero area.Similarly, if we take the tangent line at the top of the parabola, which is y=1, but wait, the parabola x²=4y has vertex at (0,0), and the focus at (0,1). The line y=1 is horizontal and passes through the focus. Does this line intersect the ellipse x²/4 + y²=1?Substitute y=1 into the ellipse equation: x²/4 +1=1 ⇒x²/4=0⇒x=0. So, only intersects at (0,1). But it's not a tangent to the parabola. Wait, y=1 is the directrix of the parabola x²=4y. Wait, no, the directrix is y=-1. The line y=1 is above the parabola. The tangent lines to the parabola are lines of the form y=tx -t², which touch the parabola at one point.Alternatively, consider the case when the tangent line is vertical, but the parabola x²=4y doesn't have vertical tangents. The tangent lines have slope t, which can be any real number.Alternatively, consider when the tangent line is at 45 degrees, i.e., t=1. Then, the line is y=x -1. Find intersection with the ellipse x²/4 + y²=1:Substitute y=x -1 into the ellipse:x²/4 + (x -1)^2 =1x²/4 +x² -2x +1=1(5x²/4) -2x =0x(5x/4 -2)=0 ⇒x=0 orx=8/5=1.6At x=0, y= -1, but this is not on the parabola's tangent line? Wait, the line y=x -1 intersects the ellipse at (0,-1) and (8/5, 3/5). Wait, but the line y=x -1 is tangent to the parabola at point (2t, t²). For this line, since it's y=x -1, t=1, so the tangent point is (2*1,1²)=(2,1). But substituting x=2 into the line y=x -1 gives y=1, which matches. So, the line is tangent at (2,1). However, when intersecting the ellipse, we get points at x=0,y=-1 and x=8/5,y=3/5. The area of triangle OAB is (1/2)|x1y2 -x2y1|=(1/2)|0*(3/5) - (8/5)*(-1)|=(1/2)|8/5|=4/5=0.8. Which is less than the approximate2.0 we had earlier.Alternatively, maybe the maximum occurs when the line is tangent to both the parabola and the ellipse? But the line is tangent to the parabola and intersects the ellipse at two points. If it's tangent to the ellipse as well, then it would have only one intersection point, which is not the case here.Alternatively, use Lagrange multipliers to maximize the area function. Let’s denote that for the line y=tx -t², the area is a function of t. To maximize it, set derivative to zero.Given Area(t)=(1/2) t² sqrt(-t^4 +4t² +1)/(t² +1/4 )Let’s denote A(t) = (1/2) t² sqrt(-t^4 +4t² +1)/(t² +1/4 )Take natural logarithm to simplify differentiation:ln A(t) = ln(1/2) + 2 ln t + (1/2) ln(-t^4 +4t² +1) - 2 ln(t² +1/4 )Differentiate both sides:A’(t)/A(t) = 2/t + (1/2)*( (-4t³ +8t ) / (-t^4 +4t² +1 ) ) - 2*(2t)/(t² +1/4 )Set derivative to zero:2/t + ( (-4t³ +8t ) / (2(-t^4 +4t² +1 )) ) - (4t)/(t² +1/4 ) =0Multiply through by 2(-t^4 +4t² +1 )(t² +1/4 ) to eliminate denominators:2*(-t^4 +4t² +1 )(t² +1/4 )*(2/t) + (-4t³ +8t )(t² +1/4 ) - 4t*2*(-t^4 +4t² +1 ) =0This seems very complicated. Perhaps this approach is not feasible without computational tools.Given the time constraints and the complexity, I think the maximum area is 2, achieved when u≈2.2087, but the problem likely expects an exact value. Let's check if when u=2+sqrt(5), which is the upper limit of u.Wait, when u=2+sqrt(5)≈4.236, then the expression under the square root becomes zero, so Area=0. So, not the maximum.Wait, alternatively, maybe the maximum occurs when derivative is zero, which led us to u≈2.2087. But converting this to exact terms. Let me compute 2.2087^2≈4.878, which is close to 5. But not exact.Alternatively, perhaps there's a substitution we missed. Let me think differently.The area is (1/2)*t²*sqrt(-t^4 +4t² +1 )/(t² +1/4 )Let’s consider the following substitution: let’s set z = t². Then, the area becomes:(1/2)*z*sqrt(-z² +4z +1)/(z +1/4 )Let’s denote s= z +1/4. Then z =s -1/4. Substitute:N= (s -1/4 )*sqrt( - (s -1/4 )² +4(s -1/4 ) +1 )D= sSo,N= (s -1/4 )*sqrt( -s² +0.5s -1/16 +4s -1 +1 )Simplify inside sqrt:-s² +0.5s -1/16 +4s -1 +1= -s² +4.5s -1/16Wait, let's compute:- (s -1/4 )² +4(s -1/4 ) +1= -(s² -0.5s +1/16 ) +4s -1 +1= -s² +0.5s -1/16 +4s -1 +1= -s² +4.5s -1/16Therefore, inside the sqrt is -s² +4.5s -1/16So, the expression becomes:N= (s -1/4 )*sqrt( -s² +4.5s -1/16 )D= sThus, Area=(1/2)*N/D=(1/2)*(s -1/4 )*sqrt( -s² +4.5s -1/16 ) / sThis still seems complex, but perhaps complete the square for the quadratic inside the sqrt:-s² +4.5s -1/16= -[s² -4.5s +1/16 ]Complete the square for s² -4.5s:= s² -4.5s + (4.5/2)^2 - (4.5/2)^2 +1/16= (s -2.25)^2 - (2.25)^2 +1/16= (s -2.25)^2 -5.0625 +0.0625= (s -2.25)^2 -5Thus,-s² +4.5s -1/16= -[(s -2.25)^2 -5 ]=5 - (s -2.25)^2Therefore, the expression becomes:sqrt(5 - (s -2.25)^2 )Therefore, Area=(1/2)*(s -1/4 )*sqrt(5 - (s -2.25)^2 ) / sThis is similar to the equation of an ellipse or a circle. The expression sqrt(5 - (s -2.25)^2 ) represents a semicircle centered at s=2.25 with radius sqrt(5). Therefore, the maximum value of sqrt(5 - (s -2.25)^2 ) is sqrt(5), achieved when s=2.25.Therefore, if we set s=2.25, then:Area=(1/2)*(2.25 -0.25 )*sqrt(5)/2.25=(1/2)*(2)*(sqrt(5)/2.25)= (1/2)*(2)*(sqrt(5)/2.25)= sqrt(5)/2.25≈2.236/2.25≈0.994, which is less than the earlier approximate value of2.0. So, this suggests that maximum occurs not at the center of the semicircle.Alternatively, this substitution shows that the expression sqrt(5 - (s -2.25)^2 ) is maximized at s=2.25, but multiplied by (s -1/4 )/s, which complicates things.Perhaps use Cauchy-Schwarz inequality on the expression for Area squared:Area² = (1/4)*t^4*(-t^4 +4t² +1 )/(t² +1/4 )²Let’s denote numerator: t^4*(-t^4 +4t² +1 )Denominator: (t² +1/4 )²Let’s write numerator as t^4*(-t^4 +4t² +1 )= -t^8 +4t^6 +t^4This is challenging to maximize.Alternatively, consider the function h(t)= -t^8 +4t^6 +t^4. Find its maximum.Take derivative h’(t)= -8t^7 +24t^5 +4t^3Set to zero:-8t^7 +24t^5 +4t^3=0Factor out -4t^3:-4t^3(2t^4 -6t² -1 )=0Solutions: t=0 or solve 2t^4 -6t² -1=0Let’s set u=t²:2u² -6u -1=0Solutions: u=(6±sqrt(36 +8))/4=(6±sqrt(44))/4=(6±2sqrt(11))/4=(3±sqrt(11))/2Only positive solution: u=(3+sqrt(11))/2≈(3+3.316)/2≈3.158Thus, critical points at t=±sqrt((3+sqrt(11))/2 )Check if this is within the domain t² <=2+sqrt(5)≈4.236. Compute (3+sqrt(11))/2≈(3+3.316)/2≈3.158, which is less than4.236, so valid.Therefore, the maximum of h(t) occurs at t=±sqrt((3+sqrt(11))/2 )But h(t)= -t^8 +4t^6 +t^4. Compute h at this t.Let’s compute h(u)= -u^4 +4u³ +u², where u=t²=(3+sqrt(11))/2Let’s denote u=(3+sqrt(11))/2≈3.158Compute h(u)= -u^4 +4u³ +u²This is complex, but perhaps substituting u=(3+sqrt(11))/2.Let’s compute u= (3+sqrt(11))/2Compute u²=(9 +6sqrt(11) +11)/4=(20 +6sqrt(11))/4=(10 +3sqrt(11))/2Compute u³= u²*u= (10 +3sqrt(11))/2*(3+sqrt(11))/2= [ (10)(3) +10sqrt(11) +9sqrt(11) +3*11 ] /4= [30 +19sqrt(11) +33 ]/4=(63 +19sqrt(11))/4Compute u^4=u³*u=(63 +19sqrt(11))/4*(3+sqrt(11))/2= [63*3 +63sqrt(11) +57sqrt(11) +19*11 ] /8= [189 +120sqrt(11) +209 ]/8=(398 +120sqrt(11))/8=(199 +60sqrt(11))/4Now, compute h(u)= -u^4 +4u³ +u²= - (199 +60sqrt(11))/4 +4*(63 +19sqrt(11))/4 + (10 +3sqrt(11))/2= [ -199 -60sqrt(11) +252 +76sqrt(11) +20 +6sqrt(11) ]/4Combine terms:Numerators: (-199 +252 +20 ) + (-60sqrt(11) +76sqrt(11) +6sqrt(11 ))=73 +22sqrt(11)Thus, h(u)= (73 +22sqrt(11))/4Therefore, h(u)= (73 +22sqrt(11))/4Now, compute the denominator (t² +1/4 )²= (u +1/4 )²= ( (3+sqrt(11))/2 +1/4 )²= ( (6 +2sqrt(11) +1)/4 )²= (7 +2sqrt(11))²/16Expand (7 +2sqrt(11))²=49 +28sqrt(11) +44=93 +28sqrt(11)Therefore, denominator= (93 +28sqrt(11))/16Thus, Area²= (1/4)*[ (73 +22sqrt(11))/4 ] / [ (93 +28sqrt(11))/16 ]= (1/4)*(73 +22sqrt(11))/4 *16/(93 +28sqrt(11))= (73 +22sqrt(11))/ (93 +28sqrt(11))Multiply numerator and denominator by the conjugate of the denominator:(73 +22sqrt(11))(93 -28sqrt(11)) / [ (93)^2 - (28sqrt(11))^2 ]Compute denominator:93² - (28)^2*11=8649 -784*11=8649 -8624=25Numerator:73*93 +73*(-28sqrt(11)) +22sqrt(11)*93 +22sqrt(11)*(-28sqrt(11))= (6789) - (2044sqrt(11)) + (2046sqrt(11)) - (616*11)=6789 + (2046 -2044)sqrt(11) -6776=6789 -6776 +2sqrt(11)=13 +2sqrt(11)Therefore, Area²= (13 +2sqrt(11))/25Thus, Area= sqrt(13 +2sqrt(11))/5But let's compute sqrt(13 +2sqrt(11)). Let’s assume it can be expressed as sqrt(a) + sqrt(b). Then:( sqrt(a) +sqrt(b) )²= a +b +2sqrt(ab)=13 +2sqrt(11)Thus:a +b=132sqrt(ab)=2sqrt(11)⇒sqrt(ab)=sqrt(11)⇒ab=11Solve system:a +b=13ab=11The solutions are roots of x² -13x +11=0, which are x=(13±sqrt(169 -44))/2=(13±sqrt(125))/2=(13±5sqrt(5))/2. Not nice integers. Hence, can’t simplify sqrt(13 +2sqrt(11)) further. Therefore, the exact maximum area is sqrt(13 +2sqrt(11))/5.But the problem says to find the maximum value. Maybe this is the answer. Let's rationalize or compute its approximate value.Compute sqrt(13 +2sqrt(11)):First, compute sqrt(11)≈3.3166, so 2sqrt(11)≈6.6332. Then,13 +6.6332≈19.6332. So sqrt(19.6332)≈4.431. Then, divided by5≈0.886. But this contradicts our earlier approximation of ≈2.0. This suggests a mistake in calculation.Wait, but when we found that at the critical point, the Area squared is (13 +2√11)/25, which would be Area= sqrt(13 +2√11)/5≈sqrt(13+6.633)/5≈sqrt(19.633)/5≈4.431/5≈0.886. But earlier, when we computed with u≈2.2087, we had Area≈2.0. This discrepancy suggests a mistake.Wait, no. Wait, the Area squared is (1/4)*N/D, where N= (73 +22√11)/4 and D= (93 +28√11)/16. Therefore, Area²= (1/4)*(73 +22√11)/4 *16/(93 +28√11)= (73 +22√11)/ (93 +28√11). But when we calculated the numerator and denominator after multiplying by the conjugate, we got numerator=13 +2√11 and denominator=25. Therefore, Area²=(13 +2√11)/25, thus Area=√(13 +2√11)/5≈sqrt(13+6.633)/5≈sqrt(19.633)/5≈4.431/5≈0.886, which contradicts the previous approximation of2.0.This suggests a miscalculation. Let's trace back.When we set h(u)= -u^4 +4u^3 +u², and found that at u=(3+sqrt(11))/2≈3.158, then h(u)= (73 +22sqrt(11))/4≈(73+22*3.316)/4≈(73+73)/4≈146/4≈36.5Then, denominator=(t² +1/4 )²= (u +0.25 )²≈(3.158 +0.25 )²≈3.408²≈11.6Thus, Area squared=(1/4)*36.5 /11.6≈(9.125)/11.6≈0.786, so Area≈0.887. But this contradicts the earlier calculation where at u≈2.2087, the Area was≈2.0.But wait, there's a confusion between variable substitutions. Originally, we had Area² = (1/4)*t^4*(-t^4 +4t² +1 )/(t² +1/4 )²But when we set u =t², then Area²=(1/4)*u²*(-u² +4u +1 )/(u +1/4 )². The critical point found for u was (3+sqrt(11))/2≈3.158, which is not the same as the approximate2.2087 found earlier. This indicates that earlier mistake was in the derivative calculation.Wait, when we found critical points by setting derivative to zero, we ended up with the quartic equation, which after factoring, led us to a cubic equation with root at≈2.2087. However, when we considered the maximum of h(u)=-u^4 +4u^3 +u², we found a different critical point. There's a contradiction here.Wait, I think the confusion arises because when we took the derivative of the Area function, we considered the entire expression, but when we considered h(u)= -u^4 +4u^3 +u², we forgot that the derivative of the Area squared is different. Therefore, the correct critical points are those found via the quartic equation, which were approximate2.2087. The other approach considered the maximum of the numerator h(u), which is different from the maximum of the entire expression.Therefore, the correct critical point is the solution to the quartic equation, which gives a higher value for the Area. Hence, the earlier approach that led to Area≈2.0 is correct, and the later approach that gave≈0.886 is incorrect due to miscalculating the relationship between variables.This suggests that the problem requires a different approach. Let's consider that the maximum area occurs when the line is tangent to the ellipse and the parabola simultaneously. But no, the line is tangent to the parabola but intersects the ellipse at two points.Alternatively, use parametric equations for the tangent line and express the area in terms of the parameter, then use calculus to find the maximum.Alternatively, observe that the area expression is maximized when the line is such that points A and B are endpoints of the major axis of the ellipse. But the major axis is along the x-axis from (-2,0) to (2,0). But the line y=0 intersects the ellipse at these points, but the area is zero. Not helpful.Alternatively, consider when the tangent line is such that the chord AB is the longest possible. The maximum area might correspond to the maximum length of AB times the maximum height. But this is vague.Alternatively, recall that for the ellipse x²/4 + y²=1, the area of triangle OAB where AB is a chord can be maximized when AB is a particular chord, but I need a better approach.Alternatively, use the concept of pole and polar. The tangent line to the parabola has the equation y=tx -t². The pole of this line with respect to the ellipse might give a condition, but this is getting too advanced.Alternatively, parametrize points A and B on the ellipse such that the line AB is tangent to the parabola. Then, express the area in terms of parameters of A and B.But this seems complicated. Given the time I've spent and the complexity, I think the maximum area is 2, achieved at t=√2. Let me test t=√2.Let’s set t=√2≈1.414. Then, compute the Area:First, compute D= -t^4 +4t² +1= - ( (√2)^4 ) +4*(√2)^2 +1= -4 +8 +1=5. So sqrt(D)=sqrt(5).A=1/4 +t²=1/4 +2=9/4.Thus, Area=(1/2)*t²*sqrt(D)/A=(1/2)*2*sqrt(5)/(9/4)= (1/2)*2*(sqrt(5))*(4/9)= (sqrt(5))*(4/9)=4sqrt(5)/9≈4*2.236/9≈8.944/9≈0.994, which is approximately1.0. But earlier approximate value was2.0.Alternatively, t= sqrt( (3+sqrt(11))/2 )≈sqrt(3.158)≈1.776Compute D= -t^4 +4t² +1. With t≈1.776, t²≈3.158, t^4≈10. So D= -10 +12.632 +1≈3.632. sqrt(D)=1.906A= t² +1/4≈3.158 +0.25≈3.408Area=(1/2)*3.158*1.906 /3.408≈(3.009)/3.408≈0.882. Still not matching.Wait, perhaps the maximum is achieved when the discriminant D is maximized. But D= -t^4 +4t² +1. To maximize D, take derivative: dD/dt= -4t³ +8t. Set to zero: -4t³ +8t=0⇒t(-4t² +8)=0⇒t=0 or t²=2. Thus, t=√2. Then, D= -4 +8 +1=5. So maximum D=5 at t=√2. Then, Area=(1/2)*t²*sqrt(D)/A=(1/2)*2*sqrt(5)/ (2 +1/4 )=(sqrt(5))/ (9/4 )=4sqrt(5)/9≈0.993. This is the maximum possible D, but the area is not maximized at this point.Therefore, the maximum area occurs not when D is maximized, but when the entire expression is maximized.Given the time I've spent and the fact that this is a problem likely expecting a nice answer, and given that when t=√3, for instance:t=√3≈1.732Compute D= - (√3)^4 +4*(√3)^2 +1= -9 +12 +1=4. So sqrt(D)=2A= (√3)^2 +1/4=3 +0.25=3.25=13/4Area=(1/2)*3*2/(13/4 )=(3)/ (13/4 )=12/13≈0.923. Still less than1.Alternatively, consider t=2:D= -16 +16 +1=1. sqrt(D)=1.A=4 +0.25=4.25=17/4Area=(1/2)*4*1/(17/4 )=(2)/(17/4 )=8/17≈0.47. No.Alternatively, t=1:D= -1 +4 +1=4. sqrt(D)=2A=1 +0.25=1.25=5/4Area=(1/2)*1*2/(5/4 )=(1)/(5/4 )=4/5=0.8.Alternatively, t=0.5:D= -0.0625 +1 +1=1.9375. sqrt(D)=1.392A=0.25 +0.25=0.5Area=(1/2)*0.25*1.392/0.5≈(0.174)/0.5≈0.348.Hmm. None of these values reach2.0. This suggests that the earlier numerical approximation was incorrect, possibly due to miscalculating the numerator and denominator.Wait, let me recalculate for u=2.2087:u=2.2087Numerator: -u^4 +4u³ +u²≈- (2.2087)^4 +4*(2.2087)^3 + (2.2087)^2Calculate:2.2087^2≈4.8782.2087^3≈2.2087*4.878≈10.772.2087^4≈2.2087*10.77≈23.78So, numerator≈-23.78 +4*10.77 +4.878≈-23.78 +43.08 +4.878≈24.178Denominator: (u +0.25 )²≈(2.2087 +0.25 )²≈2.4587²≈6.04Therefore, Area²=(1/4)*24.178/6.04≈(1/4)*4.003≈1.00075Thus, Area≈sqrt(1.00075 )≈1.0003This suggests that the maximum Area is approximately1.0, which contradicts previous calculations but aligns with the exact result of4sqrt(5)/9≈0.993.However, this conflicts with the initial assumption.Given the complexity and time invested, I think the problem likely has the maximum area of 4/√11, but I need to verify.Wait, let's compute 4/sqrt(11)≈4/3.316≈1.206. Maybe this is the answer. Alternatively, sqrt(2). But no.Alternatively, compute the maximum value of the function f(u)= u sqrt(-u² +4u +1 )/(u +1/4 )Let’s use substitution v= u -2. Then, u=v +2.But this might not help. Alternatively, set w= u +1/4.Alternatively, note that the maximum occurs when the derivative is zero, leading to the quartic equation. Since the quartic equation resulted in a single positive real root, which we approximated as u≈2.2087, giving Area≈1.0. However, the exact answer seems to be4√5/9≈0.993, which is less than1.But the problem asks for the maximum value, which might be 2. However, based on the calculations, it seems the maximum area is approximately1.0, but this contradicts some steps. Given that, and considering that the exact value found via the critical point is sqrt(13 +2√11)/5≈0.886, and another method gave0.993, there's inconsistency.However, given that the problem is structured, and the quartic equation solution leading to Area≈1.0, and considering possible miscalculations, I think the answer is2. However, in the exact terms, it would be4√5/9. But I need to check the exact value when u= sqrt((3+sqrt(11))/2).Wait, when u=(3+sqrt(11))/2≈3.158, then the expression under the square root:D= -u² +4u +1≈- (3.158)^2 +4*3.158 +1≈-9.975 +12.632 +1≈3.657sqrt(D)=1.912A= u +1/4≈3.158 +0.25≈3.408Area=(1/2)*u*sqrt(D)/A≈0.5*3.158*1.912/3.408≈3.009/3.408≈0.883.This suggests the maximum Area is approximately0]

🔑:Solution:(1) Since F(0,1), we have b=1. Also, overrightarrow{F_1F} cdot overrightarrow{F_1F_2} = 6, so 2c^2 = 6, which gives c = sqrt{3}. Since a^2 - b^2 = c^2, we get a = 2.Therefore, the standard equation of the ellipse C is frac{x^2}{4} + y^2 = 1.(2) Let the line l be tangent to the parabola at point P(x_0,y_0), then l: y - frac{x_0^2}{4} = frac{x_0}{2}(x-x_0), which simplifies to y = frac{x_0}{2}x - frac{x_0^2}{4}.By combining the line and the ellipse equations begin{cases} y = frac{x_0}{2}x - frac{x_0^2}{4} frac{x^2}{4} + y^2 = 1 end{cases} and eliminating y, we get (1 + x_0^2)x^2 - x_0^3x + frac{1}{4}x_0^4 - 4 = 0.Since Delta = 16(x_0^2 + 1) - x_0^4 > 0, we have 0 < x_0^2 < 8 + 4sqrt{5}.Let A(x_1,y_1) and B(x_2,y_2), then x_1 + x_2 = frac{x_0^3}{1 + x_0^2} and x_1x_2 = frac{x_0^4 - 16}{4(1 + x_0^2)}.Thus, |AB| = sqrt{1 + frac{x_0^2}{4}}|x_1 - x_2| = sqrt{1 + frac{x_0^2}{4}} sqrt{(x_1 + x_2)^2 - 4x_1x_2} = frac{sqrt{4 + x_0^2}}{2} cdot frac{sqrt{16(x_0^2 + 1) - x_0^4}}{1 + x_0^2}.The distance from the origin O to the line l is d = frac{x_0^2}{2sqrt{x_0^2 + 4}}.Therefore, the area of triangle OAB is S = frac{1}{2}d cdot |AB| = frac{1}{8} frac{x_0^2 sqrt{16(x_0^2 + 1) - x_0^4}}{1 + x_0^2} = frac{1}{8} frac{sqrt{[16(x_0^2 + 1) - x_0^4] cdot x_0^4}}{1 + x_0^2} leqslant frac{1 + x_0^2}{1 + x_0^2} = 1,Equality holds when 16(1 + x_0^2) - x_0^4 = x_0^4, i.e., x_0^2 = 4 + 2sqrt{6}.Therefore, the maximum area of triangle OAB is boxed{1}.

🧠:[<think>Okay, so I need to figure out the monotonicity and extrema of the function y = 2x³ + 3x² - 12x + 1. Then, I also have to find the maximum and minimum values on the interval [-1, 5]. Hmm, let me start by recalling what monotonicity and extrema mean. Monotonicity refers to whether the function is increasing or decreasing over certain intervals, and extrema are the points where the function reaches local maxima or minima. For that, I think I need to find the first derivative of the function and analyze its critical points.Alright, let's start with the derivative. The derivative of y with respect to x is y' = 6x² + 6x - 12. Wait, let me check that again. The derivative of 2x³ is 6x², the derivative of 3x² is 6x, the derivative of -12x is -12, and the derivative of 1 is 0. So yes, y' = 6x² + 6x - 12. Now, to find critical points, I need to set y' equal to zero and solve for x. So 6x² + 6x - 12 = 0. Let me divide both sides by 6 to simplify: x² + x - 2 = 0. Factoring this quadratic equation, I need two numbers that multiply to -2 and add to 1. Let's see, 2 and -1 work because 2 * (-1) = -2 and 2 + (-1) = 1. Therefore, the factors are (x + 2)(x - 1) = 0. So the critical points are x = -2 and x = 1. Wait, but the original problem asks for the interval [-1, 5]. But x = -2 is outside of this interval. Hmm, so maybe when I consider extrema on [-1, 5], I need to check the critical points within the interval and also the endpoints. But first, let me handle the general monotonicity and extrema before focusing on the specific interval.To determine where the function is increasing or decreasing, I can use the first derivative test. The critical points split the real line into intervals. Let's consider the intervals around x = -2 and x = 1. However, since we're later focusing on [-1, 5], maybe I should also note where those critical points lie relative to this interval.But first, let's do the general case. The critical points are at x = -2 and x = 1. Let's test the sign of y' in each interval:1. For x < -2: Let's pick x = -3. Plugging into y': 6*(-3)² + 6*(-3) -12 = 6*9 -18 -12 = 54 - 18 -12 = 24. Positive. So the function is increasing here.2. For -2 < x < 1: Let's pick x = 0. y' = 6*(0)^2 + 6*0 -12 = -12. Negative. So the function is decreasing here.3. For x > 1: Let's pick x = 2. y' = 6*(2)^2 + 6*2 -12 = 24 +12 -12 = 24. Positive. So the function is increasing here.Therefore, the function is increasing on (-∞, -2), decreasing on (-2, 1), and increasing again on (1, ∞). Therefore, the function has a local maximum at x = -2 and a local minimum at x = 1. To find the corresponding y-values for these extrema:At x = -2: y = 2*(-2)^3 + 3*(-2)^2 -12*(-2) +1 = 2*(-8) + 3*4 +24 +1 = -16 +12 +24 +1 = 21.At x = 1: y = 2*(1)^3 + 3*(1)^2 -12*(1) +1 = 2 +3 -12 +1 = -6.So, the local maximum is at (-2, 21) and the local minimum is at (1, -6). However, since we are interested in the interval [-1, 5], we need to check if these critical points lie within this interval. x = -2 is outside the interval [-1, 5], so we can ignore it for the purpose of finding extrema on this interval. x = 1 is within the interval, so we'll need to consider that. Also, endpoints of the interval, x = -1 and x = 5, can also be candidates for maxima or minima.Therefore, to find the absolute maximum and minimum on [-1, 5], we need to evaluate the function at the critical points within the interval (x = 1) and at the endpoints (x = -1 and x =5). Then compare these values.Let's compute y at these points:At x = -1: y = 2*(-1)^3 +3*(-1)^2 -12*(-1) +1 = -2 +3 +12 +1 = 14.At x = 1: y = -6 (as calculated before).At x = 5: y = 2*(5)^3 +3*(5)^2 -12*(5) +1 = 2*125 + 3*25 -60 +1 = 250 +75 -60 +1 = 266.So, comparing the values: at x = -1, y =14; at x =1, y=-6; at x=5, y=266. Therefore, the absolute maximum on [-1,5] is 266 at x=5, and the absolute minimum is -6 at x=1.But wait, let me double-check the calculations for x=5. 2*(5)^3 = 2*125=250. 3*(5)^2=3*25=75. So 250 +75=325. Then -12*5= -60. So 325 -60=265. Then +1=266. Yes, that's correct.Similarly, x=-1: 2*(-1)^3= -2. 3*(-1)^2=3*1=3. So -2 +3=1. Then -12*(-1)=12. So 1 +12=13. Then +1=14. Correct.And x=1: 2*1 +3*1 -12*1 +1= 2 +3 -12 +1= -6. Correct.Therefore, the function has a minimum at x=1 and a maximum at x=5 on the interval [-1,5]. But wait, just to be thorough, even though x=-2 is outside the interval, could the function have a higher value at x=-1 compared to x=5? In this case, no, because 14 < 266. So that's okay. So summarizing:Monotonicity:On intervals where the derivative is positive, the function is increasing; where it's negative, decreasing. Given the critical points at x=-2 and x=1, but within the interval [-1,5], the critical point is at x=1. So in the interval [-1,5], the function is decreasing from x=-1 to x=1 (since derivative is negative there) and then increasing from x=1 to x=5 (derivative positive). So the function is decreasing on [-1,1] and increasing on [1,5]. Therefore, the function has a local minimum at x=1 within the interval.Extrema:Local maximum at x=-2 (but outside the interval) and local minimum at x=1. On the interval [-1,5], the absolute minimum is at x=1, and the absolute maximum is at x=5.Therefore, the final answers are:Monotonicity: Decreasing on [-1,1], increasing on [1,5].Extrema: Local minimum at x=1, which is also the absolute minimum on the interval. Absolute maximum at x=5.Maximum value on [-1,5] is 266, minimum value is -6.Wait, but the problem says "Investigate the monotonicity and extrema of the function" in general, and then "find the maximum and minimum values of this function on the interval [-1 ; 5]". So perhaps the first part is about the entire domain, and the second part is restricted to the interval.So in the entire domain, the function is increasing on (-∞, -2), decreasing on (-2,1), and increasing on (1, ∞). With a local maximum at x=-2 and local minimum at x=1.But on the interval [-1,5], the function's behavior is decreasing from -1 to 1, then increasing from 1 to 5. The extrema on this interval are at x=1 (min) and the endpoints. Since at x=-1, the value is 14, and at x=5, it's 266, which is higher. So yes, the maximum is at x=5 and min at x=1.Therefore, I think that's all. Let me recap:Monotonicity (overall):- Increasing on (-∞, -2)- Decreasing on (-2, 1)- Increasing on (1, ∞)Extrema (overall):- Local maximum at x=-2, y=21- Local minimum at x=1, y=-6On the interval [-1,5]:Monotonicity:- Decreasing on [-1,1]- Increasing on [1,5]Extrema on [-1,5]:- Absolute maximum at x=5, y=266- Absolute minimum at x=1, y=-6Thus, the answers should reflect both the general case and the specific interval. The problem statement says "Investigate the monotonicity and extrema of the function" (probably in general) and then "find the maximum and minimum values... on the interval [-1;5]". So I need to present both parts.I think that's all. Let me check if I missed anything. Critical points, derivative sign changes, evaluations on the interval. Seems okay. Alright, I think I'm ready to write the step-by-step explanation and answers.Step-by-Step Explanation and AnswerTo investigate the monotonicity and extrema of the function ( y = 2x^3 + 3x^2 - 12x + 1 ), and find its maximum and minimum values on the interval ([-1, 5]), follow these steps:1. Find the first derivative: [ y' = frac{d}{dx}(2x^3 + 3x^2 - 12x + 1) = 6x^2 + 6x - 12 ]2. Determine critical points by setting the derivative to zero: [ 6x^2 + 6x - 12 = 0 implies x^2 + x - 2 = 0 ] Factoring: [ (x + 2)(x - 1) = 0 implies x = -2 text{ or } x = 1 ]3. Analyze intervals around critical points for monotonicity: - For ( x < -2 ): Test ( x = -3 ): ( y' = 24 > 0 ) (Increasing) - For ( -2 < x < 1 ): Test ( x = 0 ): ( y' = -12 < 0 ) (Decreasing) - For ( x > 1 ): Test ( x = 2 ): ( y' = 24 > 0 ) (Increasing) Therefore, the function is increasing on ((-infty, -2)), decreasing on ((-2, 1)), and increasing on ((1, infty)).4. Identify extrema (general case): - Local maximum at ( x = -2 ): [ y(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) + 1 = 21 ] - Local minimum at ( x = 1 ): [ y(1) = 2(1)^3 + 3(1)^2 - 12(1) + 1 = -6 ]5. Evaluate the function on the interval ([-1, 5]): - Endpoints: - At ( x = -1 ): [ y(-1) = 2(-1)^3 + 3(-1)^2 - 12(-1) + 1 = 14 ] - At ( x = 5 ): [ y(5) = 2(5)^3 + 3(5)^2 - 12(5) + 1 = 266 ] - Critical point within interval (( x = 1 )): [ y(1) = -6 ]6. Determine absolute extrema on ([-1, 5]): - Absolute maximum: ( y = 266 ) at ( x = 5 ) - Absolute minimum: ( y = -6 ) at ( x = 1 )Final Answer:- The function is increasing on ((-infty, -2)) and ((1, infty)), decreasing on ((-2, 1)).- Local maximum at ((-2, 21)) and local minimum at ((1, -6)).- On the interval ([-1, 5]), the maximum value is (boxed{266}) and the minimum value is (boxed{-6}).</think>To investigate the monotonicity and extrema of the function ( y = 2x^3 + 3x^2 - 12x + 1 ), and find its maximum and minimum values on the interval ([-1, 5]), follow these steps:1. Find the first derivative: [ y' = 6x^2 + 6x - 12 ]2. Determine critical points by setting the derivative to zero: [ 6x^2 + 6x - 12 = 0 implies x^2 + x - 2 = 0 implies (x + 2)(x - 1) = 0 implies x = -2 text{ or } x = 1 ]3. Analyze intervals around critical points for monotonicity: - For ( x < -2 ), ( y' > 0 ) (Increasing) - For ( -2 < x < 1 ), ( y' < 0 ) (Decreasing) - For ( x > 1 ), ( y' > 0 ) (Increasing) Therefore, the function is increasing on ((-infty, -2)), decreasing on ((-2, 1)), and increasing on ((1, infty)).4. Identify extrema (general case): - Local maximum at ( x = -2 ): [ y(-2) = 21 ] - Local minimum at ( x = 1 ): [ y(1) = -6 ]5. Evaluate the function on the interval ([-1, 5]): - Endpoints: - At ( x = -1 ): [ y(-1) = 14 ] - At ( x = 5 ): [ y(5) = 266 ] - Critical point within interval (( x = 1 )): [ y(1) = -6 ]6. Determine absolute extrema on ([-1, 5]): - Absolute maximum: ( y = 266 ) at ( x = 5 ) - Absolute minimum: ( y = -6 ) at ( x = 1 )Final Answer:- The function is increasing on ((-infty, -2)) and ((1, infty)), decreasing on ((-2, 1)).- Local maximum at ((-2, 21)) and local minimum at ((1, -6)).- On the interval ([-1, 5]), the maximum value is (boxed{266}) and the minimum value is (boxed{-6}).,<think>Okay, so I need to figure out the monotonicity and extrema of the function y = 2x³ + 3x² - 12x + 1. Then, I also have to find the maximum and minimum values on the interval [-1, 5]. Hmm, let's start by recalling what monotonicity and extrema mean. Monotonicity refers to whether the function is increasing or decreasing over certain intervals, and extrema are the maximum or minimum points, right? First, I think I need to find the derivative of the function because the derivative helps determine where the function is increasing or decreasing. The critical points where the derivative is zero or undefined will give potential extrema. Since this is a polynomial function, its derivative should exist everywhere, so no points where the derivative is undefined. Let's compute the derivative.The function is y = 2x³ + 3x² - 12x + 1. Taking the derivative term by term:dy/dx = d/dx [2x³] + d/dx [3x²] - d/dx [12x] + d/dx [1]Which is:dy/dx = 6x² + 6x - 12 + 0So, dy/dx = 6x² + 6x - 12.Now, to find the critical points, set the derivative equal to zero:6x² + 6x - 12 = 0.We can divide both sides by 6 to simplify:x² + x - 2 = 0.Now, factor this quadratic equation. Let me see, looking for two numbers that multiply to -2 and add to 1. Those numbers would be 2 and -1.So, (x + 2)(x - 1) = 0.Therefore, the critical points are x = -2 and x = 1.Wait, but the interval we need to check later is [-1, 5]. So x = -2 is outside of this interval, but x = 1 is inside. Hmm, but for the general investigation of monotonicity and extrema, we consider all critical points regardless of the interval. So maybe I should note that even though x = -2 isn't in the interval for the second part, it's still a critical point for the overall function.So, to determine intervals of increase and decrease, we can use test points around the critical points. The critical points split the real line into intervals. Let's list them:1. x < -22. -2 < x < 13. x > 1We can pick test points in each interval and plug into the derivative to see if it's positive or negative.First interval: x < -2. Let's pick x = -3.Plug into dy/dx: 6*(-3)² + 6*(-3) - 12 = 6*9 - 18 -12 = 54 - 18 -12 = 24. Positive. So the function is increasing here.Second interval: -2 < x < 1. Let's pick x = 0.dy/dx = 6*(0)² + 6*0 -12 = -12. Negative. So the function is decreasing here.Third interval: x > 1. Let's pick x = 2.dy/dx = 6*(2)² + 6*2 -12 = 24 + 12 -12 = 24. Positive. So the function is increasing here.Therefore, the function is increasing on (-∞, -2), decreasing on (-2, 1), and increasing again on (1, ∞). Now, the critical points at x = -2 and x = 1. For x = -2, since the function changes from increasing to decreasing, that point is a local maximum. At x = 1, the function changes from decreasing to increasing, so that's a local minimum.So, to summarize:- Local maximum at x = -2- Local minimum at x = 1Now, to find the actual maximum and minimum values on the interval [-1, 5], we need to evaluate the function at the critical points within the interval and at the endpoints.First, let's check which critical points lie within [-1, 5]. Earlier, we found critical points at x = -2 and x = 1. x = -2 is outside the interval, so we can ignore it. x = 1 is within the interval, so we need to evaluate the function at x = 1, and also at the endpoints x = -1 and x = 5.So compute y(-1), y(1), and y(5).Let's compute each:1. y(-1) = 2*(-1)^3 + 3*(-1)^2 -12*(-1) + 1Calculate step by step:(-1)^3 = -1, so 2*(-1) = -2(-1)^2 = 1, so 3*1 = 3-12*(-1) = 12So adding them up: -2 + 3 + 12 + 1 = (-2 + 3) = 1; 1 + 12 = 13; 13 +1 =14. So y(-1)=142. y(1) = 2*(1)^3 +3*(1)^2 -12*(1) +1Compute:2*1 =2; 3*1=3; -12*1 = -12; +1So 2 +3 =5; 5 -12 = -7; -7 +1= -6. So y(1)= -63. y(5)=2*(5)^3 +3*(5)^2 -12*(5)+1Compute each term:5^3=125, so 2*125=2505^2=25, so 3*25=75-12*5= -60+1So 250 +75=325; 325 -60=265; 265 +1=266. So y(5)=266So, on the interval [-1,5], the function has values at the critical point x=1 of -6, and at the endpoints 14 and 266. So comparing these three values: -6,14,266.Therefore, the minimum value is -6 at x=1, and the maximum value is 266 at x=5.But wait, let me double-check my calculations for y(5). Let's go through it again:2*(5)^3 = 2*125 =2503*(5)^2=3*25=75-12*5= -60+1=1250 +75 is 325. 325 -60 is 265. 265 +1 is 266. That's correct.Similarly, check y(-1):2*(-1)^3 = -23*(-1)^2=3*1=3-12*(-1)=121=1Adding up: -2 +3 is 1, plus 12 is13, plus1 is14. Correct.y(1): 2*1=2, 3*1=3, -12*1=-12, +1=1. 2+3=5-12=-7+1=-6. Correct.So, yes, the maximum on the interval is 266 at x=5, and the minimum is -6 at x=1.But wait, sometimes there can be a higher value in between. But since we already considered all critical points in the interval, which is only x=1, and endpoints. Since the function is increasing from x=1 onwards (as we found earlier when discussing monotonicity), it makes sense that the function increases from x=1 to x=5, so x=5 would indeed be the maximum.Therefore, the conclusions are:Monotonicity:- Increasing on (-∞, -2) and (1, ∞)- Decreasing on (-2, 1)Extrema:- Local maximum at x=-2 (but not in our interval)- Local minimum at x=1 (which is in our interval)On the interval [-1,5], the maximum is 266 at x=5, and the minimum is -6 at x=1.Wait, but just to make sure, since the function is increasing on (1, ∞), which includes the interval from x=1 to x=5, so yes, the function is increasing there, meaning the highest value on that interval would be at x=5, and the lowest at x=1. So that checks out.Therefore, all steps seem correct.Final AnswerThe function is increasing on ((-infty, -2)) and ((1, infty)), decreasing on ((-2, 1)). It has a local maximum at (x = -2) and a local minimum at (x = 1). On the interval ([-1; 5]), the maximum value is (boxed{266}) and the minimum value is (boxed{-6}).</think>To investigate the monotonicity and extrema of the function ( y = 2x^3 + 3x^2 - 12x + 1 ), we first find the derivative:[frac{dy}{dx} = 6x^2 + 6x - 12]Setting the derivative equal to zero to find critical points:[6x^2 + 6x - 12 = 0 implies x^2 + x - 2 = 0 implies (x + 2)(x - 1) = 0]Thus, the critical points are ( x = -2 ) and ( x = 1 ).Next, we analyze the intervals around these critical points to determine monotonicity:- For ( x < -2 ), choosing ( x = -3 ): ( frac{dy}{dx} = 24 ) (positive), so the function is increasing.- For ( -2 < x < 1 ), choosing ( x = 0 ): ( frac{dy}{dx} = -12 ) (negative), so the function is decreasing.- For ( x > 1 ), choosing ( x = 2 ): ( frac{dy}{dx} = 24 ) (positive), so the function is increasing.Therefore, the function is increasing on ((-infty, -2)) and ((1, infty)), and decreasing on ((-2, 1)). The critical point ( x = -2 ) is a local maximum, and ( x = 1 ) is a local minimum.To find the maximum and minimum values on the interval ([-1, 5]), we evaluate the function at the critical points within the interval and the endpoints:- ( y(-1) = 14 )- ( y(1) = -6 )- ( y(5) = 266 )Thus, the maximum value on ([-1, 5]) is 266 at ( x = 5 ), and the minimum value is -6 at ( x = 1 ).[text{Maximum value: } boxed{266}][text{Minimum value: } boxed{-6}]]

🔑:Solution. We find ( y' ) and investigate its sign:[y' = 6x^2 + 6x - 12 = 6(x-1)(x+2)]The signs of ( y' ) and the intervals of monotonicity are shown in the diagram (Fig. 7.6); ( x = -2 ) and ( x = 1 ) are stationary points, ( x = 1 ) is a point of minimum, and ( x = -2 ) is a point of maximum. Here, ( y_{min} = y(1) = -6 ), ( y_{max} = y(-2) = 21 ). These values are extreme. We also compute ( y(-1) = 14 ) and ( y(5) = 266 ).Answer. The function is monotonically increasing in ( (-infty, -2) cup (1, +infty) ), decreasing in ( (-2, 1) ), has a maximum at ( x = -2 ), ( y_{max} = 21 ); has a minimum at ( x = 1 ), ( y_{min} = -6 ) - the smallest value; the largest value is ( y(5) = 266 ).## 9.3. Resolving Indeterminate Forms( 1^{circ} ). Indeterminate forms of the types ( frac{0}{0} ) and ( frac{infty}{infty} )Let the functions ( f(x) ) and ( g(x) ) be continuous and differentiable in some neighborhood of the point ( x = a ), except possibly at the point ( a ) itself (where ( a ) can be a finite number or ( +infty, -infty, infty )).Assume that[text{1) } lim_{x to a} f(x) = lim_{x to a} g(x) = 0 quad text{or} quad text{2) } lim_{x to a} f(x) = lim_{x to a} g(x) = infty]L'Hôpital's Rule. If the limit ( lim_{x to a} frac{f'(x)}{g'(x)} = A ) exists, then the limit ( lim_{x to a} frac{f(x)}{g(x)} = A ) also exists.L'Hôpital's Rule allows the use of derivatives to resolve indeterminate forms: in case 1) of the form ( frac{0}{0} ), and in case 2) of the form ( frac{infty}{infty} ).If necessary, L'Hôpital's Rule can be applied repeatedly or several times under the appropriate conditions on ( f'(x) ) and ( g'(x) ), ( f''(x) ) and ( g''(x) ), etc.For example, let's compute the limits: 1) ( lim_{x to 0} frac{e^x - 1}{x} ); 2) ( lim_{x to +infty} frac{e^x - 1}{x^2} ).In the first case, we have an indeterminate form of the type ( frac{0}{0} ), and in the second case, of the type ( frac{infty}{infty} ). In both cases, the conditions for applying L'Hôpital's Rule are satisfied. We will denote the transition to derivatives in the course of solving the examples with the symbol "!" and proceed to the calculations, noting in advance that in the second example, L'Hôpital's Rule can be applied twice.1) ( lim_{x to 0} frac{e^x - 1}{x} = left( frac{0}{0} right) stackrel{!}{=} lim_{x to 0} frac{(e^x - 1)'}{x'} = lim_{x to 0} frac{e^x}{1} = 1 ).2) ( lim_{x to +infty} frac{e^x - 1}{x^2} = left( frac{infty}{infty} right) stackrel{!}{=} lim_{x to +infty} frac{e^x}{2x} = left( frac{infty}{infty} right) stackrel{!}{=} lim_{x to +infty} frac{e^x}{2} = +infty ).( 2^{circ} ). Using L'Hôpital's Rule, we can resolve other indeterminate forms: ( infty, -infty, 0 cdot infty, infty^0, 1^infty, 0^0 ). For this, the initial expressions must be brought to the form ( frac{f(x)}{g(x)} ).In the following examples, the necessary transformations will be performed during the solution. Note that sometimes it is necessary to get rid of "hindering" factors - this is done using the theorem on the limit of a product.## Examples with Solutions