answer：Alright, so I've got this problem here with two propositions, p and q, and I need to figure out the range of values for 'a' based on the conditions given. Let me try to break it down step by step.First, let's understand what each proposition is saying.**Proposition p**: The solution set of the inequality ( x^2 - (a - 1)x + 1 leq 0 ) is empty. That means there are no real numbers x that satisfy this inequality. So, for this quadratic inequality to have no solution, the quadratic equation must never be less than or equal to zero. In other words, the quadratic must always be positive. To find when a quadratic is always positive, I remember that the discriminant must be negative. The discriminant formula is ( Delta = b^2 - 4ac ). If ( Delta < 0 ), the quadratic doesn't cross the x-axis and is always positive (if the coefficient of ( x^2 ) is positive, which it is here since it's 1).So, let's compute the discriminant for the quadratic ( x^2 - (a - 1)x + 1 ).Here, ( a = 1 ), ( b = -(a - 1) ), and ( c = 1 ). Plugging into the discriminant:( Delta = [-(a - 1)]^2 - 4(1)(1) = (a - 1)^2 - 4 )For the quadratic to have no real roots (and thus the inequality ( leq 0 ) to have no solution), we need:( (a - 1)^2 - 4 < 0 )Let me solve this inequality:( (a - 1)^2 < 4 )Taking square roots on both sides:( |a - 1| < 2 )Which means:( -2 < a - 1 < 2 )Adding 1 to all parts:( -1 < a < 3 )So, proposition p is true when ( -1 < a < 3 ).**Proposition q**: The function ( f(x) = (a + 1)^x ) is increasing in its domain. I know that an exponential function ( b^x ) is increasing if the base b is greater than 1. If 0 < b < 1, it's decreasing. So, for ( f(x) ) to be increasing, we need:( a + 1 > 1 )Solving for a:( a > 0 )So, proposition q is true when ( a > 0 ).Now, the problem states that ( p land q ) is false, and ( p lor q ) is true. Let me recall what these logical connectives mean:- ( p land q ) is true only when both p and q are true.- ( p lor q ) is true when at least one of p or q is true.Given that ( p land q ) is false, it means that it's not the case that both p and q are true simultaneously. So, at least one of them must be false.And since ( p lor q ) is true, it means that at least one of p or q is true.Putting these together, the only possibilities are:1. p is true and q is false.2. p is false and q is true.Because if both were false, ( p lor q ) would be false, which contradicts the given condition. Similarly, if both were true, ( p land q ) would be true, which also contradicts the given condition.So, we have two cases to consider:**Case 1: p is true and q is false**From earlier, p is true when ( -1 < a < 3 ).q is false when ( a leq 0 ).So, the overlap of these two conditions is:( -1 < a leq 0 )**Case 2: p is false and q is true**p is false when ( a leq -1 ) or ( a geq 3 ).q is true when ( a > 0 ).So, the overlap here is:( a geq 3 )Because ( a > 0 ) combined with ( a leq -1 ) or ( a geq 3 ) only gives ( a geq 3 ).So, combining both cases, the possible values of a are:( -1 < a leq 0 ) or ( a geq 3 )Let me just double-check to make sure I haven't missed anything.For p being true, we have ( -1 < a < 3 ). For q being false, ( a leq 0 ). So, their intersection is ( -1 < a leq 0 ). That seems right.For p being false, it's ( a leq -1 ) or ( a geq 3 ). For q being true, it's ( a > 0 ). So, the intersection is ( a geq 3 ). That makes sense because ( a > 0 ) doesn't overlap with ( a leq -1 ).Therefore, the range of a is ( -1 < a leq 0 ) or ( a geq 3 ).I think that's all. I don't see any mistakes in the logic.**Final Answer**The range of values for ( a ) is boxed{(-1, 0]} cup boxed{[3, infty)}.Wait, but the original problem didn't specify whether to present the answer as two separate intervals or combine them. Since they are separate, I think it's better to present both intervals clearly.But in the initial problem, the user wrote two separate boxed answers. Maybe I should follow that format.So, the final answer is:boxed{-1 < a leq 0} quad text{or} quad boxed{a geq 3}Yeah, that seems consistent with the initial problem's presentation.**Final Answer**The range of values for ( a ) is boxed{-1 < a leq 0} or boxed{a geq 3}.

answer：First, I need to determine how many pieces of pizza each person eats and how many are left uneaten.Ann and Bill each eat 40% of their 6-piece pizzas. To find out how many pieces they eat:- Ann: 6 pieces * 40% = 2.4 pieces eaten- Bill: 6 pieces * 40% = 2.4 pieces eatenNext, I'll calculate the uneaten pieces for Ann and Bill:- Ann: 6 - 2.4 = 3.6 pieces uneaten- Bill: 6 - 2.4 = 3.6 pieces uneatenCate and Dale each eat 70% of their 6-piece pizzas. To find out how many pieces they eat:- Cate: 6 pieces * 70% = 4.2 pieces eaten- Dale: 6 pieces * 70% = 4.2 pieces eatenNow, I'll calculate the uneaten pieces for Cate and Dale:- Cate: 6 - 4.2 = 1.8 pieces uneaten- Dale: 6 - 4.2 = 1.8 pieces uneatenEddie eats 60% of his 6-piece pizza:- Eddie: 6 pieces * 60% = 3.6 pieces eatenCalculating the uneaten pieces for Eddie:- Eddie: 6 - 3.6 = 2.4 pieces uneatenFinally, I'll add up all the uneaten pieces:- Ann: 3.6 pieces- Bill: 3.6 pieces- Cate: 1.8 pieces- Dale: 1.8 pieces- Eddie: 2.4 piecesTotal uneaten pieces = 3.6 + 3.6 + 1.8 + 1.8 + 2.4 = 13.2 piecesSince you can't have a fraction of a pizza piece, we'll round down to the nearest whole number. Therefore, there are 13 uneaten pizza pieces in total.

answer：Okay, so I have this problem where I need to find Q(7) given that Q(x) is the product of two quartic polynomials, and the set of all complex roots of Q(x) is {1, 2, 3, 4, 6}. Hmm, let me try to unpack this step by step.First, Q(x) is given as the product of two quartic polynomials: [ Q(x) = (3x^4 - 33x^3 + gx^2 + hx + i)(4x^4 - 88x^3 + jx^2 + kx + l) ]So, each of these factors is a fourth-degree polynomial. That means Q(x) is an eighth-degree polynomial because when you multiply two fourth-degree polynomials, you get a degree 8 polynomial.Now, the problem says that the set of all complex roots of Q(x) is {1, 2, 3, 4, 6}. That means that these are the roots, but it doesn't specify their multiplicities. So, I need to figure out how many times each of these roots appears in the factorization of Q(x).Since Q(x) is the product of two quartic polynomials, each quartic can have up to four roots. But the given roots are only five distinct numbers: 1, 2, 3, 4, and 6. So, these roots must be distributed between the two quartic polynomials, possibly with some multiplicities.Let me denote the first quartic polynomial as S(x) and the second as T(x). So, Q(x) = S(x) * T(x). Each quartic polynomial will have four roots (counting multiplicities). So, altogether, Q(x) has eight roots (counting multiplicities). The given roots are five distinct numbers, so some of them must be repeated.Let me think about the sum of the roots for each quartic polynomial. For a quartic polynomial ax^4 + bx^3 + cx^2 + dx + e, the sum of the roots is -b/a.For S(x) = 3x^4 - 33x^3 + gx^2 + hx + i, the sum of the roots is 33/3 = 11.For T(x) = 4x^4 - 88x^3 + jx^2 + kx + l, the sum of the roots is 88/4 = 22.So, the sum of all roots of Q(x) is 11 + 22 = 33.Now, the given roots are 1, 2, 3, 4, 6. Their sum is 1 + 2 + 3 + 4 + 6 = 16. So, the remaining three roots (since Q(x) has eight roots in total) must sum up to 33 - 16 = 17.These remaining three roots must also be from the set {1, 2, 3, 4, 6} because the problem states that the set of all complex roots is exactly {1, 2, 3, 4, 6}. So, we need to find three numbers from this set that add up to 17.Let me list all possible combinations of three numbers from {1, 2, 3, 4, 6} and see which ones add up to 17.- 6 + 6 + 5: But 5 isn't in the set.- 6 + 6 + 4: That's 16, which is too low.- 6 + 6 + 6: That's 18, which is too high.- 6 + 4 + 7: 7 isn't in the set.- 6 + 4 + 6: That's 16 again.- 4 + 4 + 9: 9 isn't in the set.- 3 + 6 + 8: 8 isn't in the set.Wait, maybe I need to consider that the roots can be repeated. So, perhaps the three extra roots are 4, 6, and 7? But 7 isn't in the given set. Hmm, that's a problem.Wait, maybe I made a mistake. Let me think again. The sum of the given roots is 16, and we need three more roots that sum to 17. But all roots must be from {1, 2, 3, 4, 6}.Let me try different combinations:- 6 + 6 + 5: Still 5 isn't in the set.- 6 + 6 + 4: 16, too low.- 6 + 6 + 6: 18, too high.- 6 + 4 + 6: 16, same as above.- 4 + 4 + 9: 9 not in set.- 3 + 6 + 8: 8 not in set.Wait, maybe I need to include 4 twice and 6 once? 4 + 4 + 9? No, 9 isn't in the set. Hmm.Wait, maybe 4 + 6 + 7? But 7 isn't in the set. So, that's not possible.Wait, perhaps I need to include 6 three times? 6 + 6 + 5? No, 5 isn't in the set.Wait, maybe the roots are 4, 6, and 7, but since 7 isn't in the set, that can't be. So, perhaps I need to adjust my approach.Wait, maybe the roots are 4, 6, and 6? Let's check: 4 + 6 + 6 = 16. But we need 17. Hmm, that's still one short.Wait, maybe 4, 6, and 7, but 7 isn't allowed. Hmm.Wait, perhaps I made a mistake in the sum. Let me recalculate. The sum of the given roots is 1 + 2 + 3 + 4 + 6 = 16. The total sum needed is 33, so the remaining three roots must sum to 17.Wait, so 17 is needed. Let me think of three numbers from {1, 2, 3, 4, 6} that add up to 17.Wait, 6 + 6 + 5 is 17, but 5 isn't in the set. 6 + 4 + 7 is 17, but 7 isn't in the set. 4 + 4 + 9 is 17, but 9 isn't in the set. 3 + 6 + 8 is 17, but 8 isn't in the set. 2 + 6 + 9 is 17, but 9 isn't in the set. 1 + 6 + 10 is 17, but 10 isn't in the set.Hmm, this is confusing. Maybe I need to consider that some roots are repeated more than once. Let me think about the multiplicities.Wait, perhaps the roots are 4, 6, and 6, which sum to 16, but we need 17. So, that's one short. Maybe 4, 6, and 7, but 7 isn't allowed. Hmm.Wait, maybe I made a mistake in the initial assumption. Let me check the sum again. The sum of the roots of S(x) is 11, and the sum of the roots of T(x) is 22. So, the total sum is 33. The given roots sum to 16, so the remaining three roots must sum to 17.Wait, maybe the roots are 4, 6, and 7, but since 7 isn't in the set, perhaps the problem allows for multiplicities beyond what's given? But the problem says the set of all complex roots is {1, 2, 3, 4, 6}, so 7 can't be a root.Wait, maybe I need to consider that some roots are repeated more times. For example, maybe 6 is repeated three times, and 4 is repeated once, but that would be 6+6+6+4=22, which is too much.Wait, no, because the remaining three roots need to sum to 17, not four.Wait, perhaps the remaining three roots are 4, 6, and 7, but since 7 isn't allowed, maybe the problem is that I need to adjust the multiplicities differently.Wait, maybe the roots are 4, 6, and 6, which is 16, but we need 17. So, perhaps one of the roots is 4, and the other two are 6 and 7, but 7 isn't allowed.Wait, maybe I'm overcomplicating this. Let me think differently.Since the total sum is 33, and the given roots sum to 16, the remaining three roots must sum to 17. Since all roots must be from {1, 2, 3, 4, 6}, perhaps the only way is to have 4, 6, and 7, but since 7 isn't in the set, maybe the problem is that the multiplicities are such that 6 is repeated three times, and 4 is repeated once, but that would be four roots, not three.Wait, maybe I need to consider that the roots are 4, 6, and 6, which is 16, but we need 17. So, perhaps one of the roots is 4, and the other two are 6 and 7, but 7 isn't allowed.Wait, maybe the problem is that I'm miscounting the multiplicities. Let me try to list all possible roots with their multiplicities.Given that Q(x) has eight roots, and the set is {1, 2, 3, 4, 6}, so each of these can appear multiple times.Let me denote the multiplicities as follows:Let m1 be the multiplicity of 1,m2 be the multiplicity of 2,m3 be the multiplicity of 3,m4 be the multiplicity of 4,m6 be the multiplicity of 6.So, m1 + m2 + m3 + m4 + m6 = 8.And the sum of the roots is:1*m1 + 2*m2 + 3*m3 + 4*m4 + 6*m6 = 33.We know that m1 + m2 + m3 + m4 + m6 = 8,and 1*m1 + 2*m2 + 3*m3 + 4*m4 + 6*m6 = 33.We also know that the given roots are {1, 2, 3, 4, 6}, so each of these must appear at least once. So, m1, m2, m3, m4, m6 are all at least 1.So, let me subtract the minimum multiplicities:Let me set m1 = 1, m2 = 1, m3 = 1, m4 = 1, m6 = 1.Then, the remaining multiplicities are:m1' = m1 - 1 = 0,m2' = m2 - 1 = 0,m3' = m3 - 1 = 0,m4' = m4 - 1 = 0,m6' = m6 - 1 = 0.Wait, no, that's not helpful. Let me think differently.Let me subtract the minimum sum:The minimum sum is 1 + 2 + 3 + 4 + 6 = 16.The remaining sum is 33 - 16 = 17, which must come from the extra multiplicities.So, the extra multiplicities contribute 17 to the sum.Each extra multiplicity of 1 adds 1 to the sum,each extra multiplicity of 2 adds 2,and so on.So, we need to distribute 17 extra among the roots, but each extra multiplicity adds its root's value.So, we need to find non-negative integers a, b, c, d, e such that:a*1 + b*2 + c*3 + d*4 + e*6 = 17,and a + b + c + d + e = 3,since we have 8 total roots, and 5 are already accounted for, so 3 extra multiplicities.So, we need to solve:a + b + c + d + e = 3,anda*1 + b*2 + c*3 + d*4 + e*6 = 17.Hmm, this seems tricky because 3 extra multiplicities can only add up to a maximum of 3*6=18, which is more than 17, but let's see.Let me try to find possible combinations.Let me start by considering the largest possible contribution, which is 6.If e=2, then the contribution is 12, and we have 17-12=5 left, with 3-2=1 multiplicity left.So, we need a + b + c + d =1,and a + 2b + 3c + 4d =5.Possible solutions:If d=1, then 4*1=4, so remaining is 1, which can be a=1.So, a=1, d=1, others zero.So, e=2, a=1, d=1.Thus, the extra multiplicities are:1 extra 1,1 extra 4,and 2 extra 6s.So, total multiplicities:m1=1+1=2,m2=1,m3=1,m4=1+1=2,m6=1+2=3.Let me check the sum:2*1 + 1*2 + 1*3 + 2*4 + 3*6 =2 + 2 + 3 + 8 + 18 = 33. Yes, that works.So, the multiplicities are:1: 2,2:1,3:1,4:2,6:3.So, the roots are:1 (twice),2,3,4 (twice),6 (three times).So, now, we can write Q(x) as:Q(x) = 12*(x-1)^2*(x-2)*(x-3)*(x-4)^2*(x-6)^3.Wait, why 12? Because the leading coefficient of Q(x) is 3*4=12.Yes, because Q(x) is the product of two quartic polynomials with leading coefficients 3 and 4, so the leading coefficient of Q(x) is 3*4=12.So, Q(x) = 12*(x-1)^2*(x-2)*(x-3)*(x-4)^2*(x-6)^3.Now, we need to find Q(7).So, let's compute Q(7):Q(7) = 12*(7-1)^2*(7-2)*(7-3)*(7-4)^2*(7-6)^3.Compute each term:7-1=6,7-2=5,7-3=4,7-4=3,7-6=1.So,Q(7) = 12*(6)^2*(5)*(4)*(3)^2*(1)^3.Compute step by step:6^2 = 36,3^2 = 9,1^3 =1.So,Q(7) = 12 * 36 * 5 * 4 * 9 * 1.Now, compute the product:First, 12 * 36 = 432,432 *5 = 2160,2160 *4 = 8640,8640 *9 = 77760,77760 *1 =77760.Wait, that's 77760, but in the initial problem, the user wrote 12960. Did I make a mistake?Wait, let me check my calculations again.Wait, 12 * 36 is indeed 432.432 *5 is 2160.2160 *4 is 8640.8640 *9 is 77760.Hmm, but the initial thought process concluded 12960. So, where did I go wrong?Wait, perhaps I made a mistake in the multiplicities.Wait, let me go back.Earlier, I concluded that the multiplicities are:1:2,2:1,3:1,4:2,6:3.So, Q(x) =12*(x-1)^2*(x-2)*(x-3)*(x-4)^2*(x-6)^3.But when I plug in x=7, I get:12*(6)^2*(5)*(4)*(3)^2*(1)^3.Which is 12*36*5*4*9*1.Wait, 36*5=180,180*4=720,720*9=6480,6480*12=77760.Hmm, so why did the initial thought process say 12960?Wait, perhaps I made a mistake in the multiplicities.Wait, let me check the sum again.We had:a + b + c + d + e =3,and a + 2b +3c +4d +6e=17.We found e=2, a=1, d=1.So, e=2, a=1, d=1, and b=c=0.Thus, the extra multiplicities are:1 extra 1,1 extra 4,and 2 extra 6s.So, total multiplicities:m1=2,m2=1,m3=1,m4=2,m6=3.So, that's correct.Wait, but when I compute Q(7), I get 77760, but the initial thought process got 12960. So, perhaps I made a mistake in the initial assumption.Wait, perhaps the multiplicities are different.Wait, let me try another approach.Wait, maybe the extra multiplicities are different.Suppose instead of e=2, a=1, d=1, maybe another combination.Let me try e=1, then the contribution is 6, so remaining is 17-6=11, with 3-1=2 multiplicities left.So, a + b + c + d =2,and a + 2b +3c +4d=11.Looking for non-negative integers a,b,c,d.Let me try d=2, then 4*2=8, remaining is 11-8=3, with a + b + c=0, which is not possible.d=1, then 4*1=4, remaining is 11-4=7, with a + b + c=1.Looking for a + 2b +3c=7, with a + b + c=1.This is impossible because a + b + c=1, so maximum a +2b +3c=3*1=3 <7.So, no solution here.d=0, then a +2b +3c=11, with a + b + c=2.So, 11= a +2b +3c,and 2= a + b + c.Subtracting, we get 9= b + 2c.Looking for non-negative integers b,c such that b +2c=9, and a=2 -b -c.But since a must be non-negative, 2 -b -c >=0.So, b + c <=2.But b +2c=9,and b + c <=2.This is impossible because 9 >2.So, no solution for e=1.Next, try e=3, which would contribute 18, but we only need 17, so that's too much.So, e=2 is the only possibility.Thus, the multiplicities are as before.So, Q(7)=77760.But the initial thought process got 12960, so perhaps I made a mistake in the initial assumption.Wait, maybe the multiplicities are different.Wait, let me think again.Wait, the sum of the roots is 33.The given roots are 1,2,3,4,6, summing to 16.So, the remaining three roots must sum to 17.Wait, but if I have three roots, each from {1,2,3,4,6}, summing to 17.Wait, 6 + 6 +5=17, but 5 isn't in the set.Wait, 6 + 6 +4=16, which is too low.Wait, 6 + 4 +7=17, but 7 isn't in the set.Wait, 4 +4 +9=17, but 9 isn't in the set.Wait, 3 +6 +8=17, but 8 isn't in the set.Wait, 2 +6 +9=17, but 9 isn't in the set.Wait, 1 +6 +10=17, but 10 isn't in the set.Wait, maybe the roots are 4, 6, and 6, which sum to 16, but we need 17.Wait, perhaps I made a mistake in the initial assumption that the extra multiplicities are 1,4,6,6.Wait, maybe the extra multiplicities are 4,6,7, but 7 isn't allowed.Wait, perhaps the problem is that the initial thought process assumed the multiplicities differently.Wait, in the initial thought process, the user wrote:"Thus, the roots of Q(x) are 1, 2, 3, 4, 4, 6, 6, 6 (with multiplicity)."So, multiplicities:1:1,2:1,3:1,4:2,6:3.Which is the same as I have.So, why did the initial thought process compute Q(7)=12960, while I get 77760?Wait, let me check the initial thought process:"Q(7) = 12 cdot 6 cdot 5 cdot 4 cdot 3^2 cdot 1^3 = 12 cdot 6 cdot 5 cdot 4 cdot 9 cdot 1 = 12 cdot 6 cdot 5 cdot 4 cdot 9 = 12960."Wait, but 12*6=72,72*5=360,360*4=1440,1440*9=12960.Wait, but in my calculation, I have:12*(6^2)*(5)*(4)*(3^2)*(1^3)=12*36*5*4*9*1=12*36=432,432*5=2160,2160*4=8640,8640*9=77760.Wait, so the discrepancy is because in the initial thought process, they wrote 6,5,4,3^2,1^3, but in reality, it's (6)^2,5,4,(3)^2,1^3.Wait, so in the initial thought process, they might have miscalculated the exponents.Wait, let me check:Q(7)=12*(7-1)^2*(7-2)*(7-3)*(7-4)^2*(7-6)^3.Which is 12*(6)^2*(5)*(4)*(3)^2*(1)^3.So, that's 12*36*5*4*9*1.Which is 12*36=432,432*5=2160,2160*4=8640,8640*9=77760.So, the correct answer should be 77760.But the initial thought process got 12960, which is incorrect.Wait, perhaps the initial thought process made a mistake in the multiplicities.Wait, in the initial thought process, they wrote:"Thus, the roots of Q(x) are 1, 2, 3, 4, 4, 6, 6, 6 (with multiplicity)."So, multiplicities:1:1,2:1,3:1,4:2,6:3.Which is correct.But when computing Q(7), they wrote:"Q(7) = 12 cdot 6 cdot 5 cdot 4 cdot 3^2 cdot 1^3 = 12 cdot 6 cdot 5 cdot 4 cdot 9 cdot 1 = 12 cdot 6 cdot 5 cdot 4 cdot 9 = 12960."Wait, but (7-1)^2 is 6^2=36, not 6.Similarly, (7-4)^2 is 3^2=9, not 3.So, the initial thought process incorrectly took (7-1) as 6, not squared, and (7-4) as 3, not squared.So, that's where the mistake is.Thus, the correct calculation is:Q(7)=12*(6)^2*(5)*(4)*(3)^2*(1)^3=12*36*5*4*9*1=77760.Therefore, the correct answer is 77760, not 12960.But the initial thought process concluded 12960, which is wrong.So, in conclusion, the correct value of Q(7) is 77760.

answer：Okay, so I have this function f(x) = sin²(x/2), and I need to find the value of f(9π/4). Hmm, let me think about how to approach this. I remember that there are some trigonometric identities that might help simplify this expression. Maybe the double angle formula or something like that?Alright, let me recall. The double angle formula for sine is sin²θ = (1 - cos(2θ))/2. Wait, that seems useful here because I have sin²(x/2). If I let θ = x/2, then sin²(x/2) would be equal to (1 - cos(x))/2. So, substituting that into the function, f(x) becomes (1 - cos(x))/2. That seems simpler to work with.So now, f(x) = (1 - cos(x))/2. Now, I need to find f(9π/4). That means I should substitute x with 9π/4 in the function. Let me write that down: f(9π/4) = (1 - cos(9π/4))/2.Now, I need to figure out what cos(9π/4) is. Cosine has a period of 2π, right? So, 9π/4 is more than 2π. Let me subtract 2π from 9π/4 to find an equivalent angle within the first rotation. 2π is equal to 8π/4, so 9π/4 - 8π/4 = π/4. Therefore, cos(9π/4) is the same as cos(π/4).I remember that cos(π/4) is √2/2. So, cos(9π/4) = √2/2. Now, plugging that back into the function: f(9π/4) = (1 - √2/2)/2.Let me simplify that. First, I'll write 1 as 2/2 to have a common denominator: (2/2 - √2/2)/2. That simplifies to (2 - √2)/4. Wait, but looking at the answer choices, I don't see that exact form. Let me check my steps again.Wait, maybe I made a mistake in simplifying. Let's go back. f(9π/4) = (1 - √2/2)/2. To simplify this, I can distribute the division by 2: 1/2 - (√2/2)/2. That becomes 1/2 - √2/4. Hmm, that's one of the answer choices. Let me check the options again.Option A is 1 - √2/2, which is larger than 1/2, but my result is 1/2 - √2/4, which is smaller. Option B is 1/2 - √2/2, which is more negative than my result. Option C is 1/2 - √2/4, which matches my calculation. Option D is √2/2, which is positive and doesn't match.So, it looks like option C is the correct answer. But just to be thorough, let me double-check my steps. I used the double angle identity correctly, right? sin²(x/2) = (1 - cos(x))/2. Yes, that seems right. Then I substituted x = 9π/4 and found that cos(9π/4) = cos(π/4) = √2/2. That part also seems correct.Then, plugging back in: (1 - √2/2)/2. Distribute the division: 1/2 - √2/4. Yep, that's exactly what I did. So, I think I did everything correctly. Therefore, the answer should be option C: 1/2 - √2/4.I guess I was just a bit confused when I saw the options because I initially thought of simplifying differently, but breaking it down step by step helped me see that option C is indeed the right choice. It's important to carefully go through each step and not make any assumptions that could lead to mistakes.

answer：Alright, so I've got this problem: I need to find all two-digit numbers where the sum of their digits doesn't change when the number is multiplied by 2, 3, 4, 5, 6, 7, 8, or 9. Hmm, that sounds a bit tricky, but let's break it down step by step.First off, let's understand what it means for the digit sum to remain unchanged when multiplied by these numbers. If I have a two-digit number, say 18, and I multiply it by 2, I get 36. The sum of the digits of 18 is 1 + 8 = 9, and the sum of the digits of 36 is 3 + 6 = 9. So, in this case, the digit sum remains the same after multiplying by 2. That's interesting.But I need this to hold true for all multipliers from 2 to 9. That means, for a number to be a solution, when I multiply it by any of these numbers (2 through 9), the sum of the digits of the result should still be the same as the original number's digit sum.Okay, so maybe I should start by listing all two-digit numbers and check each one against these conditions. But that seems time-consuming. Maybe there's a smarter way to approach this.I remember that the sum of the digits of a number is related to its divisibility by 9. Specifically, if a number is divisible by 9, the sum of its digits is also divisible by 9. So, perhaps the numbers we're looking for are divisible by 9. Let me test this idea.Looking back at 18, which is divisible by 9, we saw that multiplying it by 2 gives 36, which is also divisible by 9, and the digit sum remains 9. That seems to support the idea. Let's try another number divisible by 9, say 27.Multiplying 27 by 2 gives 54. The sum of the digits of 27 is 2 + 7 = 9, and the sum of the digits of 54 is 5 + 4 = 9. So far, so good. Let's try multiplying 27 by 3: 27 * 3 = 81. The sum of the digits is 8 + 1 = 9. Still the same. What about 27 * 4 = 108? The sum of the digits is 1 + 0 + 8 = 9. Okay, this is working.But wait, does this hold for all multipliers up to 9? Let's check 27 * 5 = 135. Sum of digits: 1 + 3 + 5 = 9. 27 * 6 = 162. Sum: 1 + 6 + 2 = 9. 27 * 7 = 189. Sum: 1 + 8 + 9 = 18, which reduces to 9. 27 * 8 = 216. Sum: 2 + 1 + 6 = 9. 27 * 9 = 243. Sum: 2 + 4 + 3 = 9. Wow, 27 seems to work for all multipliers from 2 to 9.But I need to confirm if all numbers divisible by 9 have this property or if it's just some of them. Let's try another one: 36.36 * 2 = 72. Sum: 7 + 2 = 9. 36 * 3 = 108. Sum: 1 + 0 + 8 = 9. 36 * 4 = 144. Sum: 1 + 4 + 4 = 9. 36 * 5 = 180. Sum: 1 + 8 + 0 = 9. 36 * 6 = 216. Sum: 2 + 1 + 6 = 9. 36 * 7 = 252. Sum: 2 + 5 + 2 = 9. 36 * 8 = 288. Sum: 2 + 8 + 8 = 18, which reduces to 9. 36 * 9 = 324. Sum: 3 + 2 + 4 = 9. Okay, 36 also works.Let's try 45.45 * 2 = 90. Sum: 9 + 0 = 9. 45 * 3 = 135. Sum: 1 + 3 + 5 = 9. 45 * 4 = 180. Sum: 1 + 8 + 0 = 9. 45 * 5 = 225. Sum: 2 + 2 + 5 = 9. 45 * 6 = 270. Sum: 2 + 7 + 0 = 9. 45 * 7 = 315. Sum: 3 + 1 + 5 = 9. 45 * 8 = 360. Sum: 3 + 6 + 0 = 9. 45 * 9 = 405. Sum: 4 + 0 + 5 = 9. So, 45 works too.Continuing this pattern, let's check 54.54 * 2 = 108. Sum: 1 + 0 + 8 = 9. 54 * 3 = 162. Sum: 1 + 6 + 2 = 9. 54 * 4 = 216. Sum: 2 + 1 + 6 = 9. 54 * 5 = 270. Sum: 2 + 7 + 0 = 9. 54 * 6 = 324. Sum: 3 + 2 + 4 = 9. 54 * 7 = 378. Sum: 3 + 7 + 8 = 18, which reduces to 9. 54 * 8 = 432. Sum: 4 + 3 + 2 = 9. 54 * 9 = 486. Sum: 4 + 8 + 6 = 18, which reduces to 9. So, 54 also works.Next, let's check 63.63 * 2 = 126. Sum: 1 + 2 + 6 = 9. 63 * 3 = 189. Sum: 1 + 8 + 9 = 18, which reduces to 9. 63 * 4 = 252. Sum: 2 + 5 + 2 = 9. 63 * 5 = 315. Sum: 3 + 1 + 5 = 9. 63 * 6 = 378. Sum: 3 + 7 + 8 = 18, which reduces to 9. 63 * 7 = 441. Sum: 4 + 4 + 1 = 9. 63 * 8 = 504. Sum: 5 + 0 + 4 = 9. 63 * 9 = 567. Sum: 5 + 6 + 7 = 18, which reduces to 9. So, 63 works as well.Moving on to 72.72 * 2 = 144. Sum: 1 + 4 + 4 = 9. 72 * 3 = 216. Sum: 2 + 1 + 6 = 9. 72 * 4 = 288. Sum: 2 + 8 + 8 = 18, which reduces to 9. 72 * 5 = 360. Sum: 3 + 6 + 0 = 9. 72 * 6 = 432. Sum: 4 + 3 + 2 = 9. 72 * 7 = 504. Sum: 5 + 0 + 4 = 9. 72 * 8 = 576. Sum: 5 + 7 + 6 = 18, which reduces to 9. 72 * 9 = 648. Sum: 6 + 4 + 8 = 18, which reduces to 9. So, 72 works.Next, let's check 81.81 * 2 = 162. Sum: 1 + 6 + 2 = 9. 81 * 3 = 243. Sum: 2 + 4 + 3 = 9. 81 * 4 = 324. Sum: 3 + 2 + 4 = 9. 81 * 5 = 405. Sum: 4 + 0 + 5 = 9. 81 * 6 = 486. Sum: 4 + 8 + 6 = 18, which reduces to 9. 81 * 7 = 567. Sum: 5 + 6 + 7 = 18, which reduces to 9. 81 * 8 = 648. Sum: 6 + 4 + 8 = 18, which reduces to 9. 81 * 9 = 729. Sum: 7 + 2 + 9 = 18, which reduces to 9. So, 81 works.Finally, let's check 90.90 * 2 = 180. Sum: 1 + 8 + 0 = 9. 90 * 3 = 270. Sum: 2 + 7 + 0 = 9. 90 * 4 = 360. Sum: 3 + 6 + 0 = 9. 90 * 5 = 450. Sum: 4 + 5 + 0 = 9. 90 * 6 = 540. Sum: 5 + 4 + 0 = 9. 90 * 7 = 630. Sum: 6 + 3 + 0 = 9. 90 * 8 = 720. Sum: 7 + 2 + 0 = 9. 90 * 9 = 810. Sum: 8 + 1 + 0 = 9. So, 90 works.And finally, 99.99 * 2 = 198. Sum: 1 + 9 + 8 = 18, which reduces to 9. 99 * 3 = 297. Sum: 2 + 9 + 7 = 18, which reduces to 9. 99 * 4 = 396. Sum: 3 + 9 + 6 = 18, which reduces to 9. 99 * 5 = 495. Sum: 4 + 9 + 5 = 18, which reduces to 9. 99 * 6 = 594. Sum: 5 + 9 + 4 = 18, which reduces to 9. 99 * 7 = 693. Sum: 6 + 9 + 3 = 18, which reduces to 9. 99 * 8 = 792. Sum: 7 + 9 + 2 = 18, which reduces to 9. 99 * 9 = 891. Sum: 8 + 9 + 1 = 18, which reduces to 9. So, 99 works.Wait a minute, so all the two-digit numbers divisible by 9 seem to satisfy this condition. That makes sense because multiplying a number divisible by 9 by any integer will still result in a number divisible by 9, and thus the sum of its digits will also be divisible by 9. Since we're dealing with two-digit numbers, the sum of the digits will either be 9 or a multiple of 9 that reduces to 9 when summed again.But let's double-check to make sure there aren't any exceptions. For example, take 18:18 * 2 = 36 (sum = 9)18 * 3 = 54 (sum = 9)18 * 4 = 72 (sum = 9)18 * 5 = 90 (sum = 9)18 * 6 = 108 (sum = 9)18 * 7 = 126 (sum = 9)18 * 8 = 144 (sum = 9)18 * 9 = 162 (sum = 9)Yep, all good. What about 99:99 * 2 = 198 (sum = 18 → 9)99 * 3 = 297 (sum = 18 → 9)99 * 4 = 396 (sum = 18 → 9)99 * 5 = 495 (sum = 18 → 9)99 * 6 = 594 (sum = 18 → 9)99 * 7 = 693 (sum = 18 → 9)99 * 8 = 792 (sum = 18 → 9)99 * 9 = 891 (sum = 18 → 9)Still consistent. It seems that all two-digit numbers divisible by 9 meet the criteria. Therefore, the numbers we're looking for are:18, 27, 36, 45, 54, 63, 72, 81, 90, 99.But wait, the problem specifies two-digit numbers, so 99 is included as it's a two-digit number.Just to be thorough, let's check a number that's not divisible by 9 to see if it could be a solution. Let's take 10.10 * 2 = 20 (sum = 2)10 * 3 = 30 (sum = 3)10 * 4 = 40 (sum = 4)10 * 5 = 50 (sum = 5)10 * 6 = 60 (sum = 6)10 * 7 = 70 (sum = 7)10 * 8 = 80 (sum = 8)10 * 9 = 90 (sum = 9)Clearly, the digit sums are changing, so 10 doesn't work. How about 11?11 * 2 = 22 (sum = 4)11 * 3 = 33 (sum = 6)11 * 4 = 44 (sum = 8)11 * 5 = 55 (sum = 10)11 * 6 = 66 (sum = 12)11 * 7 = 77 (sum = 14)11 * 8 = 88 (sum = 16)11 * 9 = 99 (sum = 18)Again, the sums are changing, so 11 doesn't work either. It seems that only numbers divisible by 9 maintain the same digit sum when multiplied by these numbers.Therefore, the solution set consists of all two-digit numbers divisible by 9.

answer：Alright, so I've got this problem: Given that f(x + 1) equals x squared minus 2x, I need to find f(2). Hmm, okay. Let's see how I can approach this.First off, I know that f is a function, and it's defined in terms of x + 1. So, f(x + 1) = x² - 2x. That means if I plug in x + 1 into the function f, I get this quadratic expression. But I need to find f(2). So, how do I relate f(2) to this expression?Maybe I can think about what value of x would make x + 1 equal to 2. If x + 1 = 2, then x must be 1, right? Because 1 + 1 is 2. So, if I substitute x = 1 into the equation f(x + 1) = x² - 2x, I should get f(2).Let me write that down:f(2) = f(1 + 1) = (1)² - 2*(1) = 1 - 2 = -1.Wait, so is it that simple? Did I just find f(2) by plugging in x = 1? That seems straightforward. But maybe I should double-check to make sure I didn't miss anything.Alternatively, I could try to find the general form of f(x) and then plug in 2. Let's see if that works. If f(x + 1) = x² - 2x, maybe I can express f in terms of another variable. Let me set t = x + 1. Then, x = t - 1.Substituting back into the equation, f(t) = (t - 1)² - 2*(t - 1). Let's expand that:(t - 1)² is t² - 2t + 1, and -2*(t - 1) is -2t + 2. So, putting it all together:f(t) = t² - 2t + 1 - 2t + 2 = t² - 4t + 3.So, f(t) = t² - 4t + 3. That means f(x) = x² - 4x + 3. Now, plugging in x = 2:f(2) = (2)² - 4*(2) + 3 = 4 - 8 + 3 = -1.Okay, so both methods give me the same answer, -1. That makes me more confident that f(2) is indeed -1.Just to be thorough, let me think about what f(x) represents here. It seems like f is a quadratic function because when I expressed it in terms of t, it became a quadratic equation. Quadratic functions have the form ax² + bx + c, and in this case, a is 1, b is -4, and c is 3.So, f(x) = x² - 4x + 3. If I graph this function, it would be a parabola opening upwards with its vertex somewhere. But since the problem only asks for f(2), I don't need to worry about the graph or the vertex. I just need to evaluate the function at x = 2.Another way to think about it is to consider the transformation of the function. Since f(x + 1) is given, it's like the function f is shifted one unit to the left. So, to get back to f(x), I have to shift it one unit to the right, which is what I did by substituting t = x + 1.Let me recap:1. Given f(x + 1) = x² - 2x.2. Let t = x + 1, so x = t - 1.3. Substitute into the equation: f(t) = (t - 1)² - 2*(t - 1).4. Expand and simplify: f(t) = t² - 4t + 3.5. Therefore, f(x) = x² - 4x + 3.6. Evaluate at x = 2: f(2) = 4 - 8 + 3 = -1.Everything checks out. I don't see any mistakes in my calculations or logic. It seems like a solid approach.Just to make sure, I'll plug x = 2 back into the original expression f(x + 1) and see if it aligns. Wait, f(x + 1) is defined in terms of x, so if I set x + 1 = 2, then x = 1, and f(2) = f(1 + 1) = (1)² - 2*(1) = 1 - 2 = -1. Yep, that's consistent with what I found earlier.So, both methods—direct substitution and finding the general form of f(x)—lead me to the same conclusion. I think I'm confident that f(2) is -1.I guess the key takeaway here is understanding how function transformations work, especially horizontal shifts. Knowing that f(x + 1) shifts the graph of f(x) one unit to the left helps in manipulating the equation to find the original function f(x).Also, being comfortable with algebraic manipulations, like expanding and simplifying expressions, is crucial in problems like this. It's all about expressing the function in terms that are useful for the problem at hand.In summary, by either directly substituting the value that makes x + 1 equal to 2 or by finding the general form of f(x) through substitution and simplification, I arrived at the answer f(2) = -1. Both methods reinforce each other, making me sure that the solution is correct.