answer：Okay, so I have this problem about 100 students and 25 jury members. Each student is checked by one jury member, and each student likes exactly 10 jury members. There are two parts to this problem: part (a) asks to prove that we can select 7 jury members such that any student likes at least one of them. Part (b) is about proving that we can assign each student to a jury member they like, with each jury member checking at most 10 students.Starting with part (a). I need to show that 7 jury members are enough so that every student likes at least one of them. Hmm, so essentially, I need to cover all 100 students with just 7 jury members, where each student has at least one of these 7 in their list of 10 liked jury members.I remember something about the Pigeonhole Principle. Maybe that can help here. There are 25 jury members, and each student likes 10 of them. So, in total, the number of "likes" is 100 students * 10 likes = 1000 likes. Divided by 25 jury members, that's 40 likes per jury member on average. So, on average, each jury member is liked by 40 students.But how does that help me? Maybe I can use some kind of iterative selection. If I pick a jury member who is liked by the most students, say 40, then exclude those 40 students. Then, with the remaining 60 students, I can pick another jury member who is liked by the most remaining students, which would be 25 on average. Then exclude those 25, leaving 35 students. Then pick another jury member liked by 14 students, and so on.Wait, but 40 + 25 + 14 + ... might not add up to 100. Let me see. If I pick 7 jury members, each liked by roughly 40, 25, 14, etc., would that cover all students? Maybe, but I'm not sure. Perhaps there's a more systematic way.I think this might relate to something called the hitting set problem, where we want the smallest set that intersects all the sets in a collection. In this case, each student's set of liked jury members is a 10-element set, and we want a hitting set of size 7 that intersects all these sets.But hitting set is generally NP-hard, so maybe there's a specific property here because of the uniform size of the liked sets. Each student has exactly 10 liked jury members, so maybe there's a way to bound the size of the hitting set.Alternatively, maybe using probability. If I randomly select 7 jury members, what's the chance that a particular student doesn't like any of them? Each student has 10 liked out of 25, so the probability that a random jury member is disliked by a student is 15/25 = 3/5. The probability that all 7 are disliked would be (3/5)^7, which is approximately 0.028. So, the expected number of students not covered would be 100 * 0.028 = 2.8. So, on average, we'd miss about 3 students. That's not enough, but maybe with some adjustments, we can cover all.But this is just an expectation; it doesn't guarantee coverage. Maybe there's a way to derandomize this. Or perhaps use the probabilistic method to show that such a set exists.Wait, another approach: maybe using the concept of covering codes or something from combinatorics. If each student has 10 liked jury members, then the union of 7 such sets should cover all 100 students.But how to ensure that? Maybe by considering the maximum number of students that can be left uncovered by 7 jury members.Alternatively, think about it in terms of the complement: each jury member not selected can cover at most 10 students. So, if we have 25 - 7 = 18 jury members not selected, they can cover at most 18 * 10 = 180 students. But since there are only 100 students, this doesn't directly help.Wait, no, each student is liked by 10 jury members, so the number of dislikes per student is 15. So, if we have 18 jury members not selected, each student could potentially dislike up to 15 of them. But 18 is more than 15, so actually, each student must like at least 18 - 15 = 3 of the selected jury members. Wait, that doesn't make sense.Wait, maybe it's the other way around. Each student dislikes 15 jury members. If we have 18 jury members not selected, the number of disliked jury members among the not selected is at most 15. Therefore, the number of liked jury members among the selected is at least 18 - 15 = 3. So, each student likes at least 3 of the selected jury members. Therefore, selecting 7 jury members would ensure that each student likes at least 3 of them, which is more than enough to cover all students.Wait, that seems promising. Let me formalize it.Each student dislikes 15 jury members. If we have 25 - 7 = 18 jury members not selected, then the number of disliked jury members among the not selected is at most 15. Therefore, the number of liked jury members among the selected is at least 18 - 15 = 3. So, each student likes at least 3 of the selected 7 jury members. Therefore, selecting any 7 jury members would ensure that each student likes at least 3 of them, which is more than enough to cover all students.But wait, does this guarantee that all students are covered? Yes, because each student has at least one liked jury member in the selected set. In fact, they have at least 3. So, 7 jury members are sufficient to cover all 100 students.Okay, that seems to work. So, for part (a), selecting 7 jury members ensures that every student likes at least one of them.Now, moving on to part (b). We need to assign each student to a jury member they like, such that each jury member checks at most 10 students. So, essentially, we need a matching where each student is matched to one of their liked jury members, and no jury member is matched to more than 10 students.This sounds like a bipartite graph matching problem. We have a bipartite graph with students on one side and jury members on the other, with edges representing "likes." We need to find a matching that covers all students, with each jury member having degree at most 10.To prove that such a matching exists, maybe we can use Hall's Marriage Theorem. Hall's condition states that for any subset of students, the number of liked jury members must be at least as large as the subset.But in our case, each student has 10 liked jury members, and we have 25 jury members. Let's see. For any subset S of students, the number of liked jury members N(S) must satisfy |N(S)| ≥ |S| / 10, but I'm not sure.Wait, Hall's condition requires that for every subset S of students, the number of neighbors N(S) is at least |S|. But in our case, each student has 10 neighbors, so for any subset S, the number of neighbors is at least |S| / something.Wait, maybe not directly applicable. Alternatively, since each student has 10 options, and we need to assign each to one of their options, with each option being used at most 10 times.This is similar to a flow problem, where we can model it as a bipartite graph and find a flow where each jury member can handle up to 10 units.Alternatively, maybe use the concept of fractional matching and then round it. But perhaps a simpler approach is to use the fact that the graph is regular or something.Wait, each student has degree 10, and each jury member can have degree up to 10. So, the total number of edges is 100 * 10 = 1000. The total capacity on the jury side is 25 * 10 = 250. Wait, 250 is less than 1000, so that doesn't make sense. Wait, no, each edge is from a student to a jury member, so the total capacity is 25 * 10 = 250, but we have 100 students, each needing to send 1 unit. So, 100 units need to be sent, but the total capacity is 250, which is more than enough.But that doesn't directly help. Maybe think about it as a hypergraph matching problem.Alternatively, use the probabilistic method again. If we randomly assign each student to one of their 10 liked jury members, what's the expected number of students assigned to each jury member? It would be 100 / 25 = 4. So, on average, each jury member gets 4 students. But we need to ensure that no jury member gets more than 10.But this is just expectation; there could be variance. Maybe use Chernoff bounds to show that the probability that any jury member gets more than 10 is negligible, and then use the Lovász Local Lemma to show that there's a positive probability of a valid assignment.But that might be too involved. Maybe there's a simpler combinatorial argument.Alternatively, use the fact that each student has 10 options, and each jury member can take up to 10 students. So, we can construct the assignment step by step.Start by selecting a jury member who has the most students liking them. Assign 10 students to them, then remove those students and that jury member from the graph. Repeat this process until all students are assigned.But wait, if we do this, we might end up with some jury members who have fewer than 10 students left, but we can still assign the remaining students.Wait, let's formalize this. Let’s denote the number of students liking each jury member. Initially, each jury member is liked by 40 students on average, but some might be liked by more.If we pick the jury member with the most students, say 40, assign 10 of them, and remove those 10 students and the jury member. Now, we have 90 students left and 24 jury members. The total number of likes is now 90 * 10 = 900. The average per jury member is 900 / 24 = 37.5. So, pick the next jury member with the most students, say 38, assign 10, remove them, and so on.Continuing this way, after 7 steps, we would have assigned 70 students, leaving 30 students and 18 jury members. The total likes would be 30 * 10 = 300, so average per jury member is 16.67. Continue assigning 10 students at a time until all are assigned.But wait, this might not always work because as we remove students, the number of likes per jury member decreases, but we might end up with some jury members who have fewer than 10 students left, making it impossible to assign 10 each time.Alternatively, maybe use a more careful selection. Since each student has 10 options, and each jury member can take up to 10 students, we can model this as a bipartite graph where we need to find a matching saturating all students with the degree constraints on the jury side.This is similar to a transportation problem in operations research, where we have supply and demand constraints. Here, each student has a supply of 1, and each jury member has a demand of 10. We need to find a feasible flow.By the max-flow min-cut theorem, such a flow exists if and only if for every subset of students, the number of liked jury members is sufficient to cover their demand.But I'm not sure how to apply this directly. Maybe use Konig's theorem or something else.Wait, another idea: since each student has 10 options, and each jury member can take 10 students, the problem is similar to edge coloring in bipartite graphs. But I'm not sure.Alternatively, think about it as a hypergraph where each hyperedge connects a student to their 10 liked jury members. We need to select a set of hyperedges such that each student is covered exactly once, and each jury member is covered at most 10 times.This is similar to a matching in hypergraphs, but I don't know much about that.Wait, maybe use the fact that the graph is 10-regular on the student side and 10-regular on the jury side. So, it's a biregular bipartite graph with degrees 10 and 10. Then, by some theorem, such a graph has a perfect matching.But I think in bipartite graphs, if both partitions are regular, then a perfect matching exists. But in our case, the graph is not necessarily regular, just that each student has degree 10, and each jury member can have varying degrees, but we need to find a matching where each jury member is matched at most 10 times.Wait, actually, the graph is not regular on the jury side. Each jury member is liked by some number of students, but we need to find a matching where each jury member is matched at most 10 times.But since each student has 10 options, and each jury member can take up to 10 students, it's possible to find such a matching.Alternatively, use the following approach: construct a bipartite graph where students are connected to their liked jury members. We need to find a matching that covers all students, with each jury member matched at most 10 times.To apply Hall's theorem, we need to check that for any subset S of students, the number of neighbors N(S) is at least |S| / 10. Wait, no, Hall's condition is that |N(S)| ≥ |S| for any S. But in our case, each student has 10 neighbors, so for any S, |N(S)| ≥ |S| / something.Wait, maybe not directly applicable. Alternatively, use the concept of fractional matching. If we can show that the fractional matching number is at least 100, then we can round it to an integral matching.But I'm not sure. Maybe another approach: since each student has 10 options, and each jury member can take 10 students, we can use a greedy algorithm. Assign each student one by one to any of their liked jury members who haven't reached their limit yet.But this might fail because a student might only have liked jury members who are already full. So, we need a more sophisticated approach.Wait, maybe use the following theorem: In a bipartite graph where each node on the left has degree at least d, and each node on the right has capacity at least d, then a matching exists that covers all left nodes with the right nodes not exceeding their capacities.But I don't recall the exact theorem. Alternatively, use the fact that the graph is 10-regular on the left and each right node can take up to 10.Wait, maybe use the following: Since each student has 10 options, and each jury member can take 10 students, the total number of possible assignments is 1000, and we need to assign 100, so it's feasible.But that's too vague. Maybe think about it as a matrix where rows are students and columns are jury members, with a 1 if the student likes the jury member. We need to select 100 ones such that each row has exactly one, and each column has at most 10.This is similar to a binary matrix decomposition problem. Maybe use some combinatorial design.Alternatively, use the following approach: since each student has 10 options, and each jury member can take 10 students, we can partition the students into 10 groups, each group assigned to a different set of 10 jury members. But I'm not sure.Wait, another idea: use the fact that the graph is 10-regular on the student side, and we need a matching where each jury member is matched at most 10 times. Since the total number of edges is 1000, and we need to select 100 edges, it's possible.But I need a more concrete argument.Wait, maybe use the following lemma: In a bipartite graph where each left node has degree at least d, and the right side has n nodes each with capacity at least d, then a matching exists that covers all left nodes with capacities not exceeded.But I'm not sure about the exact statement.Alternatively, use the following approach: construct a flow network where each student is connected to their liked jury members, with edges of capacity 1. Each jury member is connected to a sink with capacity 10. Then, find a flow of value 100. If such a flow exists, then the matching exists.By the max-flow min-cut theorem, such a flow exists if and only if for every subset S of students, the number of liked jury members N(S) is at least |S| / 10. Wait, no, the min-cut condition is that the capacity of the cut is at least the demand.Wait, actually, the min-cut would be the sum of capacities on the right side that need to be cut to separate the source from the sink. So, for any subset S of students, the number of liked jury members N(S) must satisfy that the sum of capacities on the right side minus the edges leaving N(S) is at least |S|.But I'm getting confused. Maybe it's better to think in terms of the integral flow.Alternatively, use the following theorem: In a bipartite graph where each node on the left has degree at least d, and each node on the right has capacity at least d, then a matching exists that covers all left nodes with the right nodes not exceeding their capacities.But I think this is similar to what I thought earlier. Since each student has 10 options, and each jury member can take 10 students, such a matching exists.Alternatively, use the following argument: Since each student has 10 options, and each jury member can take 10 students, we can construct the matching incrementally. Start by assigning each student to one of their liked jury members, ensuring that no jury member exceeds their limit.But to formalize this, maybe use induction. Suppose we have k students assigned, and we need to assign the (k+1)-th student. They have 10 options, and each of these options has at most 10 assignments. Since we've only assigned k students, each jury member has at most 10 assignments, so at least one of the student's liked jury members has fewer than 10 assignments. Therefore, we can assign the student to that jury member.Wait, that seems promising. Let's see.Base case: k=0, trivially true.Inductive step: Assume that we've assigned k students without exceeding any jury member's limit. Now, consider the (k+1)-th student. They have 10 liked jury members. Each of these jury members has been assigned at most 10 students so far. Since we've only assigned k students, and k < 100, the number of assignments per jury member is less than 10 (since 10*25=250 > 100). Wait, no, that's not necessarily true. Some jury members could have been assigned up to 10 already.Wait, no, because we're assigning incrementally, ensuring that we don't exceed 10 per jury member. So, when assigning the (k+1)-th student, their 10 liked jury members have each been assigned at most 10 students. But since we've only assigned k students, and k < 100, the total number of assignments is k, so the average number of assignments per jury member is k/25. Since k < 100, k/25 < 4. So, each jury member has been assigned at most 4 students on average, but some could have more.Wait, but we need to ensure that at least one of the student's liked jury members has been assigned fewer than 10 students. Since each student has 10 liked jury members, and each can have up to 10 assignments, but we've only assigned k < 100 students, so the total number of assignments is k. The maximum number of assignments any jury member can have is 10, but since k < 100, the number of jury members that have reached 10 assignments is at most k/10 < 10. So, there are at least 25 - 10 = 15 jury members who have not reached their limit.But the student has 10 liked jury members. If all 10 of them have reached their limit of 10, then we have a problem. But since only k < 100 students have been assigned, the number of jury members that have reached their limit is at most k/10 < 10. Therefore, the student's 10 liked jury members cannot all have reached their limit, because there are only 9 jury members that could have reached their limit (since k < 100 implies k/10 < 10). Therefore, at least one of the student's liked jury members has not reached their limit, so we can assign the student to that jury member.Thus, by induction, we can assign all 100 students to jury members they like, with each jury member checking at most 10 students.Okay, that seems to work. So, for part (b), we can indeed make such an assignment.

answer：Okay, so I have this problem where I need to find the value of ( a + b + c + d ) for a cubic function ( f(x) = ax^3 + bx^2 + cx + d ). The information given is that ( f(x + 2) = 2x^3 - x^2 + 5x + 3 ). Hmm, that seems a bit tricky, but let me try to break it down step by step.First, I know that ( f(x + 2) ) means I substitute ( x + 2 ) into the function ( f(x) ). So, if ( f(x) = ax^3 + bx^2 + cx + d ), then ( f(x + 2) ) would be ( a(x + 2)^3 + b(x + 2)^2 + c(x + 2) + d ). I need to expand this expression and then compare it to the given polynomial ( 2x^3 - x^2 + 5x + 3 ) to find the coefficients ( a, b, c, ) and ( d ).Alright, let's start by expanding each term in ( f(x + 2) ).1. **Expanding ( (x + 2)^3 ):** [ (x + 2)^3 = x^3 + 3 cdot x^2 cdot 2 + 3 cdot x cdot 2^2 + 2^3 = x^3 + 6x^2 + 12x + 8 ] So, ( a(x + 2)^3 = a(x^3 + 6x^2 + 12x + 8) = ax^3 + 6ax^2 + 12ax + 8a ).2. **Expanding ( (x + 2)^2 ):** [ (x + 2)^2 = x^2 + 4x + 4 ] So, ( b(x + 2)^2 = b(x^2 + 4x + 4) = bx^2 + 4bx + 4b ).3. **Expanding ( c(x + 2) ):** [ c(x + 2) = cx + 2c ]4. **The constant term ( d ) remains as it is.**Now, let's combine all these expanded terms together:[f(x + 2) = ax^3 + 6ax^2 + 12ax + 8a + bx^2 + 4bx + 4b + cx + 2c + d]Next, I'll combine like terms:- The ( x^3 ) term: ( ax^3 )- The ( x^2 ) terms: ( 6ax^2 + bx^2 = (6a + b)x^2 )- The ( x ) terms: ( 12ax + 4bx + cx = (12a + 4b + c)x )- The constant terms: ( 8a + 4b + 2c + d )So, putting it all together:[f(x + 2) = ax^3 + (6a + b)x^2 + (12a + 4b + c)x + (8a + 4b + 2c + d)]We know that this expression equals ( 2x^3 - x^2 + 5x + 3 ). Therefore, we can set the coefficients of corresponding powers of ( x ) equal to each other. This gives us a system of equations:1. For ( x^3 ): ( a = 2 )2. For ( x^2 ): ( 6a + b = -1 )3. For ( x ): ( 12a + 4b + c = 5 )4. For the constant term: ( 8a + 4b + 2c + d = 3 )Now, let's solve this system step by step.**Step 1: Solve for ( a )**From the first equation, we directly get:[a = 2]**Step 2: Solve for ( b )**Substitute ( a = 2 ) into the second equation:[6(2) + b = -1 12 + b = -1 b = -1 - 12 b = -13]**Step 3: Solve for ( c )**Substitute ( a = 2 ) and ( b = -13 ) into the third equation:[12(2) + 4(-13) + c = 5 24 - 52 + c = 5 -28 + c = 5 c = 5 + 28 c = 33]**Step 4: Solve for ( d )**Substitute ( a = 2 ), ( b = -13 ), and ( c = 33 ) into the fourth equation:[8(2) + 4(-13) + 2(33) + d = 3 16 - 52 + 66 + d = 3 (16 - 52) + 66 + d = 3 -36 + 66 + d = 3 30 + d = 3 d = 3 - 30 d = -27]Now, we have all the coefficients:- ( a = 2 )- ( b = -13 )- ( c = 33 )- ( d = -27 )The question asks for ( a + b + c + d ). Let's compute that:[a + b + c + d = 2 + (-13) + 33 + (-27)]Calculating step by step:[2 - 13 = -11 -11 + 33 = 22 22 - 27 = -5]So, the value of ( a + b + c + d ) is ( -5 ).**Final Answer**boxed{-5}

answer：First, I need to find the vector resulting from subtracting vector (overrightarrow{b}) from vector (overrightarrow{a}). This involves subtracting the corresponding components of (overrightarrow{b}) from (overrightarrow{a}).Next, I will calculate the dot product of the resulting vector with vector (overrightarrow{c}). The dot product is obtained by multiplying the corresponding components of the two vectors and then summing those products.Finally, I will perform the arithmetic operations to arrive at the final result.

answer：Okay, so I have this problem about a sports lottery where you pick 7 numbers from 01 to 36, and each bet costs 2 yuan. The person wants to include the number 18, and then choose 3 consecutive numbers from 01 to 17, 2 consecutive numbers from 19 to 29, and 1 number from 30 to 36. I need to figure out how much it would cost for this person to buy all possible combinations that meet these requirements.Alright, let me break this down step by step. First, the person is fixing one number, which is 18. So, out of the 7 numbers, one is already decided. That leaves 6 more numbers to choose. But these 6 numbers have specific constraints: 3 consecutive from 01-17, 2 consecutive from 19-29, and 1 from 30-36.So, I think the way to approach this is to calculate how many ways there are to choose each of these sets and then multiply them together to get the total number of possible bets. Then, since each bet costs 2 yuan, I can multiply the total number of bets by 2 to find the total cost.Starting with the 3 consecutive numbers from 01 to 17. How many sets of 3 consecutive numbers are there in this range? Well, if I think about it, the first set would be 01,02,03, the next would be 02,03,04, and so on. The last possible set would be 15,16,17 because after that, the next set would go beyond 17. So, how many sets is that? From 01 to 15 as the starting number, because 15+2=17. So that's 15 sets.Next, the 2 consecutive numbers from 19 to 29. Similarly, the first set is 19,20, the next is 20,21, and so on. The last set would be 28,29. So, how many sets is that? Starting from 19 up to 28, that's 10 sets because 29-19+1=11, but since we're choosing 2 consecutive numbers, the number of sets is 10. Wait, actually, let me think again. If you start at 19, the next number is 20, so that's one set. Then starting at 20, next is 21, that's another set. So, the number of sets is 29 - 19 = 10, but since we're choosing 2 numbers, it's 10 sets. Yeah, that makes sense.Then, the last number is from 30 to 36. That's straightforward. There are 7 numbers in this range: 30,31,32,33,34,35,36. So, 7 choices here.Now, to find the total number of possible bets, I need to multiply the number of choices for each set together. So, that's 15 (for the 3 consecutive numbers) multiplied by 10 (for the 2 consecutive numbers) multiplied by 7 (for the single number). Let me calculate that: 15 * 10 = 150, and 150 * 7 = 1050. So, there are 1050 different possible bets that meet the requirements.Since each bet costs 2 yuan, the total cost would be 1050 * 2. Let me do that multiplication: 1000 * 2 = 2000, and 50 * 2 = 100, so 2000 + 100 = 2100 yuan.Wait, let me double-check my calculations to make sure I didn't make a mistake. For the 3 consecutive numbers, starting from 01 to 17, the number of sets is 15 because 17 - 3 + 1 = 15. That seems right. For the 2 consecutive numbers from 19 to 29, it's 29 - 19 + 1 - 2 + 1 = 10. Wait, is that the right formula? Actually, the formula for the number of consecutive sequences of length k in a range from a to b is b - a - k + 2. So, for 2 consecutive numbers from 19 to 29, it's 29 - 19 - 2 + 2 = 10. Yeah, that matches what I thought earlier.And for the single number from 30 to 36, it's just 7 numbers, so 7 choices. So, 15 * 10 * 7 = 1050. That seems correct.Multiplying 1050 by 2 yuan per bet: 1050 * 2 = 2100 yuan. So, the total cost would be 2100 yuan.I think that's it. I don't see any mistakes in my reasoning. Each step seems logical, and the calculations check out.

answer：Alright, so I've got this table here that shows the relationship between x and y. Let me see if I can figure out what formula connects them. The table looks like this:x | 0 | 1 | 2 | 3 | 4---|---|---|---|---|---y | 200 | 150 | 80 | 0 | -140Hmm, okay. So when x is 0, y is 200; when x is 1, y drops to 150; then it goes down to 80 when x is 2, and then it hits zero at x=3, and even becomes negative at x=4. That seems like a pretty steep decline, especially from x=3 to x=4. I think maybe this is a polynomial relationship because the y-values are changing in a non-linear way. Let me see if I can figure out the degree of the polynomial. If I look at the differences between the y-values, maybe that can help.First, let's list the y-values: 200, 150, 80, 0, -140.Calculating the first differences (the change from one y to the next):150 - 200 = -5080 - 150 = -700 - 80 = -80-140 - 0 = -140So the first differences are: -50, -70, -80, -140.Hmm, these differences are not constant, so it's not a linear relationship. Let's try the second differences.Second differences are the differences of the first differences:-70 - (-50) = -20-80 - (-70) = -10-140 - (-80) = -60So the second differences are: -20, -10, -60.Still not constant. Let's try the third differences.Third differences are the differences of the second differences:-10 - (-20) = 10-60 - (-10) = -50So the third differences are: 10, -50.Still not constant. Wait, but maybe I made a mistake here. Let me double-check.First differences: -50, -70, -80, -140.Second differences: (-70 - (-50)) = -20, (-80 - (-70)) = -10, (-140 - (-80)) = -60.Third differences: (-10 - (-20)) = 10, (-60 - (-10)) = -50.Hmm, that's correct. So the third differences aren't constant either. Maybe I need to go to the fourth differences?Fourth differences would be the differences of the third differences:-50 - 10 = -60.So the fourth difference is -60.Wait, but with only five points, I can only get up to the fourth difference. Since the fourth difference is constant, that suggests that the relationship is a fourth-degree polynomial. But the options given are all cubic polynomials (degree 3). Maybe I'm overcomplicating it.Alternatively, perhaps the relationship is cubic, and the fourth difference being constant is just a coincidence or maybe not necessary. Let me think.Given that the options are all cubic, maybe I should try to fit a cubic polynomial to the data. A cubic polynomial has the form:y = ax³ + bx² + cx + dI need to find the coefficients a, b, c, d that satisfy the given points.Let's set up equations based on the given points.When x=0, y=200:a*(0)³ + b*(0)² + c*(0) + d = 200So, d = 200.That's straightforward.Next, when x=1, y=150:a*(1)³ + b*(1)² + c*(1) + 200 = 150Simplify:a + b + c + 200 = 150So, a + b + c = 150 - 200 = -50Equation 1: a + b + c = -50Next, when x=2, y=80:a*(2)³ + b*(2)² + c*(2) + 200 = 80Simplify:8a + 4b + 2c + 200 = 80So, 8a + 4b + 2c = 80 - 200 = -120Equation 2: 8a + 4b + 2c = -120Next, when x=3, y=0:a*(3)³ + b*(3)² + c*(3) + 200 = 0Simplify:27a + 9b + 3c + 200 = 0So, 27a + 9b + 3c = -200Equation 3: 27a + 9b + 3c = -200Finally, when x=4, y=-140:a*(4)³ + b*(4)² + c*(4) + 200 = -140Simplify:64a + 16b + 4c + 200 = -140So, 64a + 16b + 4c = -140 - 200 = -340Equation 4: 64a + 16b + 4c = -340Now, I have four equations:1. a + b + c = -502. 8a + 4b + 2c = -1203. 27a + 9b + 3c = -2004. 64a + 16b + 4c = -340I need to solve this system of equations to find a, b, c.Let me try to solve equations 1, 2, and 3 first, and then check if they satisfy equation 4.From equation 1: a + b + c = -50From equation 2: 8a + 4b + 2c = -120From equation 3: 27a + 9b + 3c = -200Let me try to eliminate variables step by step.First, let's multiply equation 1 by 2 to make the coefficients of c the same as in equation 2:Equation 1 * 2: 2a + 2b + 2c = -100Now, subtract this from equation 2:Equation 2 - (Equation 1 * 2):(8a + 4b + 2c) - (2a + 2b + 2c) = -120 - (-100)Simplify:6a + 2b = -20Let's call this Equation 5: 6a + 2b = -20Similarly, let's multiply equation 1 by 3 to align with equation 3:Equation 1 * 3: 3a + 3b + 3c = -150Subtract this from equation 3:Equation 3 - (Equation 1 * 3):(27a + 9b + 3c) - (3a + 3b + 3c) = -200 - (-150)Simplify:24a + 6b = -50Let's call this Equation 6: 24a + 6b = -50Now, we have:Equation 5: 6a + 2b = -20Equation 6: 24a + 6b = -50Let's solve these two equations.First, let's multiply Equation 5 by 3 to align the coefficients of b:Equation 5 * 3: 18a + 6b = -60Now, subtract Equation 6 from this:(18a + 6b) - (24a + 6b) = -60 - (-50)Simplify:-6a = -10So, a = (-10)/(-6) = 10/6 = 5/3 ≈ 1.6667Wait, that doesn't seem right because the options have integer coefficients. Maybe I made a mistake somewhere.Let me double-check my calculations.Equation 5: 6a + 2b = -20Equation 6: 24a + 6b = -50If I multiply Equation 5 by 3:18a + 6b = -60Then subtract Equation 6:(18a + 6b) - (24a + 6b) = -60 - (-50)Which simplifies to:-6a = -10So, a = (-10)/(-6) = 5/3Hmm, 5/3 is approximately 1.6667, which is not an integer. But the answer choices have integer coefficients. Maybe I made a mistake earlier.Let me go back to the equations.Equation 1: a + b + c = -50Equation 2: 8a + 4b + 2c = -120Equation 3: 27a + 9b + 3c = -200Equation 4: 64a + 16b + 4c = -340Perhaps I should try a different approach. Let's use equations 1, 2, and 4 instead.From equation 1: a + b + c = -50From equation 2: 8a + 4b + 2c = -120From equation 4: 64a + 16b + 4c = -340Let me try to eliminate c first.From equation 1: c = -50 - a - bSubstitute c into equation 2:8a + 4b + 2*(-50 - a - b) = -120Simplify:8a + 4b - 100 - 2a - 2b = -120Combine like terms:(8a - 2a) + (4b - 2b) - 100 = -1206a + 2b - 100 = -120So, 6a + 2b = -120 + 100 = -20Which is the same as Equation 5: 6a + 2b = -20Now, substitute c into equation 4:64a + 16b + 4*(-50 - a - b) = -340Simplify:64a + 16b - 200 - 4a - 4b = -340Combine like terms:(64a - 4a) + (16b - 4b) - 200 = -34060a + 12b - 200 = -340So, 60a + 12b = -340 + 200 = -140Let's call this Equation 7: 60a + 12b = -140Now, we have:Equation 5: 6a + 2b = -20Equation 7: 60a + 12b = -140Let's simplify Equation 7 by dividing by 6:10a + 2b = -140/6 ≈ -23.333Wait, that's not helpful. Maybe I should express Equation 5 in terms of b.From Equation 5: 6a + 2b = -20Divide both sides by 2:3a + b = -10So, b = -10 - 3aNow, substitute b into Equation 7:60a + 12*(-10 - 3a) = -140Simplify:60a - 120 - 36a = -140Combine like terms:(60a - 36a) - 120 = -14024a - 120 = -14024a = -140 + 120 = -20So, a = -20/24 = -5/6 ≈ -0.8333Again, not an integer. This is confusing because the answer choices have integer coefficients. Maybe I'm missing something.Wait, perhaps I should try to use all four equations to see if the system is consistent.Let me try another approach. Let's use equations 1, 2, and 3 to find a, b, c, and then check if they satisfy equation 4.From equation 1: a + b + c = -50From equation 2: 8a + 4b + 2c = -120From equation 3: 27a + 9b + 3c = -200Let me try to eliminate c.From equation 1: c = -50 - a - bSubstitute into equation 2:8a + 4b + 2*(-50 - a - b) = -120Simplify:8a + 4b - 100 - 2a - 2b = -120Combine like terms:6a + 2b - 100 = -120So, 6a + 2b = -20Which is Equation 5 again.Now, substitute c into equation 3:27a + 9b + 3*(-50 - a - b) = -200Simplify:27a + 9b - 150 - 3a - 3b = -200Combine like terms:24a + 6b - 150 = -200So, 24a + 6b = -50Which is Equation 6.Now, we have:Equation 5: 6a + 2b = -20Equation 6: 24a + 6b = -50Let me solve these two equations.From Equation 5: 6a + 2b = -20Multiply by 3: 18a + 6b = -60Subtract Equation 6:(18a + 6b) - (24a + 6b) = -60 - (-50)Simplify:-6a = -10So, a = (-10)/(-6) = 5/3 ≈ 1.6667Again, not an integer. This suggests that maybe the relationship isn't a cubic polynomial, or perhaps I made a mistake in setting up the equations.Wait, let me check the calculations again.Equation 1: a + b + c = -50Equation 2: 8a + 4b + 2c = -120Equation 3: 27a + 9b + 3c = -200Equation 4: 64a + 16b + 4c = -340Let me try to solve equations 1, 2, and 4.From equation 1: c = -50 - a - bSubstitute into equation 2:8a + 4b + 2*(-50 - a - b) = -120Simplify:8a + 4b - 100 - 2a - 2b = -1206a + 2b - 100 = -1206a + 2b = -20Equation 5: 6a + 2b = -20Now, substitute c into equation 4:64a + 16b + 4*(-50 - a - b) = -340Simplify:64a + 16b - 200 - 4a - 4b = -34060a + 12b - 200 = -34060a + 12b = -140Equation 7: 60a + 12b = -140Now, let's solve Equations 5 and 7.Equation 5: 6a + 2b = -20Equation 7: 60a + 12b = -140Let's multiply Equation 5 by 6:36a + 12b = -120Now, subtract Equation 7:(36a + 12b) - (60a + 12b) = -120 - (-140)Simplify:-24a = 20So, a = 20/(-24) = -5/6 ≈ -0.8333Again, not an integer. This is perplexing. Maybe the relationship isn't cubic after all, or perhaps there's a typo in the problem.Alternatively, maybe I should try to see if any of the given options satisfy all the points.Let's check option A: y = -10x³ + 20x² - 60x + 200Let's plug in x=0:y = -10*(0) + 20*(0) - 60*(0) + 200 = 200. Correct.x=1:y = -10*(1) + 20*(1) - 60*(1) + 200 = -10 + 20 - 60 + 200 = 150. Correct.x=2:y = -10*(8) + 20*(4) - 60*(2) + 200 = -80 + 80 - 120 + 200 = 80. Correct.x=3:y = -10*(27) + 20*(9) - 60*(3) + 200 = -270 + 180 - 180 + 200 = 0. Correct.x=4:y = -10*(64) + 20*(16) - 60*(4) + 200 = -640 + 320 - 240 + 200 = -640 + 320 = -320; -320 -240 = -560; -560 + 200 = -360. Wait, but the table says y=-140. That's not matching.Hmm, so option A gives y=-360 at x=4, but the table says y=-140. So option A is incorrect.Wait, did I calculate correctly?Let me recalculate y at x=4 for option A:y = -10*(4)^3 + 20*(4)^2 - 60*(4) + 200= -10*64 + 20*16 - 60*4 + 200= -640 + 320 - 240 + 200= (-640 + 320) = -320; (-320 - 240) = -560; (-560 + 200) = -360Yes, that's correct. So option A is wrong.Let's try option B: y = 10x³ - 20x² + 60x - 200x=0:y = 0 - 0 + 0 - 200 = -200. But the table says y=200. So option B is incorrect.Option C: y = -5x³ + 15x² - 40x + 200x=0:y=200. Correct.x=1:y = -5 + 15 - 40 + 200 = (-5 +15)=10; (10 -40)=-30; (-30 +200)=170. But table says 150. So incorrect.Option D: y = 20x³ - 10x² + 30x - 100x=0:y=-100. Table says 200. Incorrect.So none of the options A to D fit all the points. Therefore, the answer must be E) None of these.Wait, but I thought option A was correct except for x=4. Maybe I made a mistake in calculating option A.Let me double-check option A at x=4:y = -10*(4)^3 + 20*(4)^2 - 60*(4) + 200= -10*64 + 20*16 - 60*4 + 200= -640 + 320 - 240 + 200= (-640 + 320) = -320; (-320 -240) = -560; (-560 +200) = -360Yes, still -360, which doesn't match the table's -140.So, indeed, none of the options fit all the points. Therefore, the correct answer is E) None of these.

answer：Okay, so I have this problem where the function ( g(x) ) satisfies the equation:[ g(2^x) + xg(2^{-x}) = 2 ]for all real numbers ( x ). I need to find ( g(2) ).Hmm, let me think about how to approach this. Since the equation holds for all real numbers ( x ), maybe I can choose specific values of ( x ) that will help me set up equations involving ( g(2) ) and other related terms. That way, I can solve for ( g(2) ).First, let me try plugging in ( x = 1 ). That should give me an equation involving ( g(2) ) because ( 2^1 = 2 ). So, substituting ( x = 1 ) into the equation:[ g(2^1) + 1 cdot g(2^{-1}) = 2 ]Simplifying that:[ g(2) + gleft(frac{1}{2}right) = 2 ]Okay, so that's one equation. Now, I need another equation to solve for ( g(2) ). Maybe I can plug in another value of ( x ) that relates to ( frac{1}{2} ). Let me try ( x = -1 ). That would give me ( 2^{-1} = frac{1}{2} ), which is the same as the previous term. Let's see:[ g(2^{-1}) + (-1) cdot g(2^{1}) = 2 ]Simplifying:[ gleft(frac{1}{2}right) - g(2) = 2 ]Alright, now I have two equations:1. ( g(2) + gleft(frac{1}{2}right) = 2 )2. ( gleft(frac{1}{2}right) - g(2) = 2 )Hmm, so I have a system of two equations with two variables: ( g(2) ) and ( gleft(frac{1}{2}right) ). Maybe I can solve this system to find ( g(2) ).Let me denote ( a = g(2) ) and ( b = gleft(frac{1}{2}right) ) for simplicity. Then the equations become:1. ( a + b = 2 )2. ( b - a = 2 )Now, I can solve this system. Let's add the two equations together:[ (a + b) + (b - a) = 2 + 2 ]Simplifying the left side:[ a + b + b - a = 2b ]And the right side:[ 4 ]So, ( 2b = 4 ), which means ( b = 2 ).Now that I know ( b = 2 ), I can substitute back into the first equation:[ a + 2 = 2 ]Subtracting 2 from both sides:[ a = 0 ]So, ( a = g(2) = 0 ).Wait, let me double-check that. If ( g(2) = 0 ) and ( gleft(frac{1}{2}right) = 2 ), then plugging back into the original equations:1. ( 0 + 2 = 2 ) which is correct.2. ( 2 - 0 = 2 ) which is also correct.Okay, that seems consistent. So, ( g(2) = 0 ).I wonder if there's another way to approach this problem, maybe by expressing ( g(2^x) ) in terms of ( g(2^{-x}) ) or something like that. But since I already have a solution and it checks out, maybe that's sufficient.Alternatively, I could consider expressing the original functional equation in terms of another variable substitution. Let me try letting ( y = 2^x ). Then, ( 2^{-x} = frac{1}{y} ). So, the original equation becomes:[ g(y) + x gleft(frac{1}{y}right) = 2 ]But since ( y = 2^x ), taking the logarithm base 2 of both sides gives ( x = log_2 y ). So, substituting back:[ g(y) + log_2 y cdot gleft(frac{1}{y}right) = 2 ]Hmm, that might not directly help me unless I can find another equation involving ( g(y) ) and ( gleft(frac{1}{y}right) ). Maybe if I substitute ( y ) with ( frac{1}{y} ), I can get another equation.Let me try that. If I substitute ( y ) with ( frac{1}{y} ), then the equation becomes:[ gleft(frac{1}{y}right) + log_2 left(frac{1}{y}right) cdot g(y) = 2 ]Simplifying ( log_2 left(frac{1}{y}right) ) gives ( -log_2 y ). So, the equation is:[ gleft(frac{1}{y}right) - log_2 y cdot g(y) = 2 ]Now, I have two equations:1. ( g(y) + log_2 y cdot gleft(frac{1}{y}right) = 2 )2. ( gleft(frac{1}{y}right) - log_2 y cdot g(y) = 2 )This is a system of equations in terms of ( g(y) ) and ( gleft(frac{1}{y}right) ). Maybe I can solve for ( g(y) ) in terms of ( y ).Let me denote ( u = g(y) ) and ( v = gleft(frac{1}{y}right) ). Then, the equations become:1. ( u + (log_2 y) v = 2 )2. ( v - (log_2 y) u = 2 )Now, I can solve this system for ( u ) and ( v ). Let's write it as:1. ( u + (log_2 y) v = 2 )2. ( -(log_2 y) u + v = 2 )This is a linear system, so I can use substitution or elimination. Let's use elimination. Multiply the first equation by ( log_2 y ):1. ( (log_2 y) u + (log_2 y)^2 v = 2 log_2 y )2. ( -(log_2 y) u + v = 2 )Now, add the two equations together:[ (log_2 y) u + (log_2 y)^2 v - (log_2 y) u + v = 2 log_2 y + 2 ]Simplifying the left side:[ (log_2 y)^2 v + v = 2 log_2 y + 2 ]Factor out ( v ):[ v left( (log_2 y)^2 + 1 right) = 2 (log_2 y + 1) ]So, solving for ( v ):[ v = frac{2 (log_2 y + 1)}{(log_2 y)^2 + 1} ]Now, substitute ( v ) back into one of the original equations to solve for ( u ). Let's use the second equation:[ -(log_2 y) u + v = 2 ]Substituting ( v ):[ -(log_2 y) u + frac{2 (log_2 y + 1)}{(log_2 y)^2 + 1} = 2 ]Let me solve for ( u ):[ -(log_2 y) u = 2 - frac{2 (log_2 y + 1)}{(log_2 y)^2 + 1} ]Factor out 2 on the right side:[ -(log_2 y) u = 2 left( 1 - frac{log_2 y + 1}{(log_2 y)^2 + 1} right) ]Simplify the expression inside the parentheses:First, write 1 as ( frac{(log_2 y)^2 + 1}{(log_2 y)^2 + 1} ):[ 1 - frac{log_2 y + 1}{(log_2 y)^2 + 1} = frac{(log_2 y)^2 + 1 - (log_2 y + 1)}{(log_2 y)^2 + 1} ]Simplify the numerator:[ (log_2 y)^2 + 1 - log_2 y - 1 = (log_2 y)^2 - log_2 y ]So, the expression becomes:[ frac{(log_2 y)^2 - log_2 y}{(log_2 y)^2 + 1} ]Therefore, going back to the equation:[ -(log_2 y) u = 2 cdot frac{(log_2 y)^2 - log_2 y}{(log_2 y)^2 + 1} ]Simplify the right side:Factor out ( log_2 y ) in the numerator:[ 2 cdot frac{log_2 y (log_2 y - 1)}{(log_2 y)^2 + 1} ]So, the equation is:[ -(log_2 y) u = frac{2 log_2 y (log_2 y - 1)}{(log_2 y)^2 + 1} ]Divide both sides by ( -log_2 y ) (assuming ( log_2 y neq 0 )):[ u = frac{2 (log_2 y - 1)}{-(log_2 y)^2 - 1} ]Simplify the denominator:[ u = frac{2 (log_2 y - 1)}{-(log_2 y)^2 - 1} = frac{2 (1 - log_2 y)}{(log_2 y)^2 + 1} ]So, ( u = g(y) = frac{2 (1 - log_2 y)}{(log_2 y)^2 + 1} )Now, I can express ( g(y) ) in terms of ( y ). Let me write that:[ g(y) = frac{2 (1 - log_2 y)}{(log_2 y)^2 + 1} ]Now, to find ( g(2) ), I substitute ( y = 2 ) into this expression:[ g(2) = frac{2 (1 - log_2 2)}{(log_2 2)^2 + 1} ]Simplify ( log_2 2 = 1 ):[ g(2) = frac{2 (1 - 1)}{1^2 + 1} = frac{2 cdot 0}{2} = 0 ]So, ( g(2) = 0 ). That matches the result I got earlier by plugging in specific values of ( x ). It seems consistent.Just to make sure, let me verify this result by plugging ( g(2) = 0 ) and ( gleft(frac{1}{2}right) = 2 ) back into the original functional equation for ( x = 1 ) and ( x = -1 ).For ( x = 1 ):Left side: ( g(2^1) + 1 cdot g(2^{-1}) = g(2) + gleft(frac{1}{2}right) = 0 + 2 = 2 )Right side: 2So, it holds.For ( x = -1 ):Left side: ( g(2^{-1}) + (-1) cdot g(2^{1}) = gleft(frac{1}{2}right) - g(2) = 2 - 0 = 2 )Right side: 2So, it also holds.Therefore, my solution seems correct. ( g(2) = 0 ).I think I've covered all the steps and checked my work. It makes sense, and both methods—substituting specific ( x ) values and solving the functional equation generally—lead to the same conclusion. I don't see any mistakes in my reasoning, so I'm confident that ( g(2) = 0 ) is the correct answer.**Final Answer**The value of ( g(2) ) is boxed{0}.