answer：Okay, so I have this problem about calculating the cost of first-class postage for a letter that weighs W ounces. The postal rate is 10 cents for every half ounce or any fraction thereof. That means if something weighs, say, 1.1 ounces, it still counts as 2 half-ounces, right? So, it's like rounding up to the next half ounce.But then there's this discount for letters that weigh more than 5 ounces. For every ounce over 5, there's a 2-cent discount per ounce. Hmm, so if a letter is 6 ounces, I guess I get a 2-cent discount for that extra ounce. If it's 7 ounces, I get a 4-cent discount, and so on.Let me try to break this down step by step. First, I need to figure out how to calculate the cost without any discounts. Since it's 10 cents for every half ounce or fraction thereof, I think I need to convert the weight into half-ounce units and then round up to the nearest whole number. So, if W is the weight in ounces, multiplying by 2 gives me the number of half-ounces. Then, I need to use the ceiling function to round up to the next whole number because any fraction of a half ounce still counts as a full half ounce.So, without any discounts, the cost would be 10 cents multiplied by the ceiling of 2W. That makes sense. For example, if W is 4 ounces, 2W is 8, so the cost is 10 * 8 = 80 cents. If W is 4.1 ounces, 2W is 8.2, which we round up to 9, so the cost is 10 * 9 = 90 cents.Now, for the discount part. The discount applies only if the weight is more than 5 ounces. So, if W is greater than 5, I need to subtract 2 cents for each ounce over 5. Let me think about how to express that mathematically. If W is 6 ounces, the discount is 2 cents for the extra ounce. If W is 7 ounces, the discount is 4 cents, and so on. So, the discount is 2 times (W - 5). But this discount only applies if W is greater than 5. Otherwise, the discount is zero.So, putting it all together, the total cost would be the initial cost minus the discount. The initial cost is 10 times the ceiling of 2W, and the discount is the maximum of zero and 2 times (W - 5). Wait, actually, the discount is 2 times (W - 5), but we need to make sure that we don't subtract more than the initial cost. So, using the max function ensures that if W is less than or equal to 5, the discount is zero.Let me test this with an example. Suppose W is 6 ounces. The initial cost is 10 * ceiling(12) = 10 * 12 = 120 cents. The discount is 2 * (6 - 5) = 2 cents. So, the total cost is 120 - 2 = 118 cents. That seems right.Another example: W is 5 ounces. The initial cost is 10 * ceiling(10) = 100 cents. The discount is max(0, 2 * (5 - 5)) = 0. So, total cost is 100 cents. That also makes sense because the discount only starts after 5 ounces.What if W is 4.5 ounces? The initial cost is 10 * ceiling(9) = 10 * 9 = 90 cents. The discount is max(0, 2 * (4.5 - 5)) = max(0, negative number) = 0. So, total cost is 90 cents. Correct.Wait, let me check another one. W is 7 ounces. Initial cost is 10 * ceiling(14) = 10 * 14 = 140 cents. Discount is 2 * (7 - 5) = 4 cents. So, total cost is 140 - 4 = 136 cents. That seems correct.But hold on, what if W is 5.1 ounces? The initial cost is 10 * ceiling(10.2) = 10 * 11 = 110 cents. The discount is 2 * (5.1 - 5) = 0.2 cents. Wait, but the discount is per ounce, so maybe we should only apply the discount for whole ounces over 5. So, if W is 5.1 ounces, it's still only 0.1 ounces over, which is less than an ounce, so no discount. Hmm, that complicates things.Wait, the problem says "for every ounce after the fifth ounce." So, does that mean we only get a discount for whole ounces over 5? So, if it's 5.1 ounces, we don't get any discount because it's not a full ounce over 5. If it's 6 ounces, we get a 2-cent discount for that extra ounce. If it's 6.5 ounces, we get a 2-cent discount for each of the two extra ounces, so 4 cents total.So, in that case, the discount should be 2 times the integer part of (W - 5). But the problem says "for every ounce after the fifth ounce," which might imply that it's per ounce, regardless of fractions. So, maybe we need to take the floor of (W - 5) and multiply by 2.But looking back at the options, option C is 10 * ceiling(2W) - max(0, 2W - 10). Let me see what that gives us.For W = 6, 2W = 12, ceiling(12) = 12, so 10*12 = 120. Then, max(0, 12 - 10) = 2. So, 120 - 2 = 118. That's correct.For W = 5.1, 2W = 10.2, ceiling(10.2) = 11, so 10*11 = 110. Then, max(0, 10.2 - 10) = 0.2, but since we're dealing with cents, 0.2 cents doesn't make sense. Wait, but in the formula, it's 2W - 10, which is 10.2 - 10 = 0.2. But we can't have a fraction of a cent, so maybe the discount is zero in this case.Wait, but the discount is 2 cents per ounce after the fifth ounce. So, for W = 5.1, it's only 0.1 ounces over, which is less than an ounce, so no discount. So, the discount should be zero. But according to option C, it's subtracting 0.2 cents, which is not possible. So, maybe the formula is slightly off.Alternatively, perhaps the discount is calculated based on the number of whole ounces over 5. So, for W = 5.1, it's 0 extra ounces, so no discount. For W = 6, it's 1 extra ounce, so 2 cents discount. For W = 7, it's 2 extra ounces, so 4 cents discount, etc.In that case, the discount would be 2 * floor(W - 5). But the formula in option C is 2W - 10, which is 2*(W - 5). So, if W is not an integer, this would give a fractional discount, which isn't possible. Therefore, maybe the formula should use the floor function for the discount.But looking at the options, none of them use floor. Option C uses max(0, 2W - 10), which could result in a fractional discount if W is not an integer. That seems problematic because we can't have fractions of a cent in postage.Wait, but in reality, postage rates are in whole cents, so maybe the discount is applied per whole ounce over 5, regardless of fractions. So, for any W > 5, even if it's just a little over, you get a discount for each full ounce beyond 5.But then, how does that translate into the formula? If W is 5.1, it's still only 0 extra ounces, so no discount. If W is 6, it's 1 extra ounce, so 2 cents discount. If W is 6.9, it's still 1 extra ounce, so 2 cents discount. If W is 7, it's 2 extra ounces, so 4 cents discount.So, the discount should be 2 * floor(W - 5). But in the formula, it's 2*(W - 5). So, if W is 5.1, it's 2*(0.1) = 0.2, which is not correct. Therefore, maybe the formula is intended to use the integer part of W - 5.But none of the options specify floor or integer part. They just have 2W - 10. So, perhaps in the context of the problem, W is considered in whole ounces. Maybe W is an integer. The problem doesn't specify, but it's possible.Assuming W is an integer, then for W = 5, discount is 0. For W = 6, discount is 2*(6 - 5) = 2. For W = 7, discount is 4, etc. So, in that case, option C would work because 2W - 10 would be 2*(6) - 10 = 12 - 10 = 2, which is correct.But if W can be a non-integer, then the formula might not be accurate because it could result in fractional discounts. However, since postage rates are in whole cents, the discount should also be in whole cents. Therefore, perhaps the problem assumes that W is an integer, or that the discount is applied per whole ounce beyond 5, regardless of fractions.Given that, option C seems to be the correct formula because it correctly applies the discount for each ounce beyond 5, assuming W is an integer or that the discount is applied per whole ounce beyond 5.Let me test it with W = 5.5. If W is 5.5, then 2W = 11, ceiling(11) = 11, so 10*11 = 110 cents. The discount is max(0, 11 - 10) = 1, so total cost is 110 - 1 = 109 cents. But according to the postal rate, for W = 5.5, it's still 5 ounces plus 0.5 ounces, which is less than an ounce, so no discount. Therefore, the cost should be 110 cents, not 109. So, the formula is giving an incorrect discount in this case.Hmm, that's a problem. So, maybe the formula is not accurate for non-integer W. But the problem doesn't specify whether W is an integer or not. If W is allowed to be a non-integer, then the discount calculation in option C might not be correct because it subtracts 2*(W - 5), which could be a fraction.Alternatively, maybe the discount is applied based on the number of whole ounces beyond 5, regardless of fractions. So, for W = 5.5, it's still 0 extra ounces, so no discount. For W = 6, it's 1 extra ounce, so 2 cents discount. For W = 6.9, it's still 1 extra ounce, so 2 cents discount. For W = 7, it's 2 extra ounces, so 4 cents discount.In that case, the discount should be 2 * floor(W - 5). But none of the options have floor. So, perhaps the problem assumes that W is an integer. If that's the case, then option C works.Alternatively, maybe the discount is applied per ounce, regardless of whether it's a whole ounce or not. So, for W = 5.5, the discount is 2*(0.5) = 1 cent. But that would mean that for any fraction over 5 ounces, you get a partial discount, which might not be how the postal service works.Given that, I think the problem assumes that W is an integer, or that the discount is applied per whole ounce beyond 5. Therefore, option C is the correct formula because it correctly subtracts 2*(W - 5) for W > 5, assuming W is an integer.So, after considering all this, I think the correct answer is option C: 10 * ceiling(2W) - max(0, 2W - 10).

answer：Okay, so I have this problem about a function f(x) that's defined on the positive real numbers, (0, +∞). It says that f is an increasing function, which means if a < b, then f(a) < f(b). Also, it has this property that for any a and b in (0, +∞), f(a * b) = f(a) + f(b). They also tell me that f(2) = 1. The first part asks me to find f(1) and f(4). Hmm, okay. Let me think about how to approach this. The function has this multiplicative property, right? So, f(a * b) = f(a) + f(b). That reminds me of logarithms because log(a * b) = log(a) + log(b). So, maybe f(x) is a logarithmic function? But they also say it's increasing, which logarithms are, depending on the base. Since f(2) = 1, if it were a logarithm, it would be log base 2, because log₂(2) = 1. So, maybe f(x) = log₂(x). Let me check that.If f(x) = log₂(x), then f(1) would be log₂(1) = 0, because 2^0 = 1. That makes sense. And f(4) would be log₂(4) = 2, since 2^2 = 4. That also checks out. So, maybe that's the function. But wait, the problem doesn't specify that f is logarithmic, just that it's increasing and has this multiplicative property. So, is f(x) necessarily a logarithm? Or could it be something else?Well, in general, functions that satisfy f(a * b) = f(a) + f(b) are logarithmic functions. So, I think it's safe to assume that f(x) is a logarithm, specifically log base 2, since f(2) = 1. So, f(x) = log₂(x). Therefore, f(1) = 0 and f(4) = 2. That seems straightforward.But maybe I should verify it without assuming it's a logarithm. Let's try to find f(1) first. If I set a = 1 and b = 1 in the given property, then f(1 * 1) = f(1) + f(1). So, f(1) = 2 * f(1). Subtracting f(1) from both sides, I get 0 = f(1). So, f(1) = 0. That's confirmed without assuming it's a logarithm.Now, for f(4). Since 4 is 2 * 2, I can use the property again. f(4) = f(2 * 2) = f(2) + f(2) = 1 + 1 = 2. So, f(4) = 2. That also checks out. So, part (1) is done. f(1) is 0 and f(4) is 2.Moving on to part (2). The inequality is f(x²) < 2f(4). Since we found f(4) = 2, 2f(4) is 4. So, the inequality becomes f(x²) < 4. But wait, f(x) is defined on (0, +∞), so x² must be positive, which it always is except when x = 0, but x = 0 isn't in the domain. So, x can be any real number except 0, but x² is positive.But f is increasing, so if f(x²) < 4, then x² must be less than the value that f maps to 4. Since f is increasing, f(a) < f(b) implies a < b. So, if f(x²) < f(something), then x² < that something. But wait, what's f inverse of 4? Since f(4) = 2, and f(16) would be f(4 * 4) = f(4) + f(4) = 2 + 2 = 4. So, f(16) = 4. Therefore, f(x²) < f(16) implies x² < 16 because f is increasing.So, x² < 16. Solving that inequality, x² < 16 means that x is between -4 and 4, but excluding 0 because x can't be 0. Wait, no, x² is always positive, so x can be any real number except 0, but x² < 16 implies that x is between -4 and 4, excluding 0. So, the solution set is (-4, 0) union (0, 4).But let me double-check. If x² < 16, then x must satisfy -4 < x < 4, but since x can't be 0 (because f is defined on (0, +∞)), we exclude 0. So, the solution set is indeed (-4, 0) union (0, 4). Wait, but hold on. The function f is defined on (0, +∞), so x² is in (0, +∞), but x itself can be any real number except 0. So, when we solve x² < 16, x can be negative or positive, as long as it's not 0. So, yeah, the solution set includes both negative and positive values of x, as long as they are between -4 and 4, excluding 0.So, putting it all together, part (1) gives f(1) = 0 and f(4) = 2, and part (2) gives the solution set as (-4, 0) union (0, 4).I think that's it. I don't see any mistakes in my reasoning. I used the properties of the function given, checked the values step by step, and applied the monotonicity to solve the inequality. It all seems consistent.

answer：Alright, let's try to figure out this problem. So, we have these decompositions of powers of natural numbers, and we need to find the value of ( m ) such that the smallest number in the decomposition of ( m^3 ) is 73. Hmm, okay.First, let me look at the examples given to understand the pattern. For ( 2^2 ), it's decomposed into ( 1 + 3 ). For ( 2^3 ), it's ( 3 + 5 ). Then, ( 3^2 ) is ( 1 + 3 + 5 ), and ( 3^3 ) is ( 7 + 9 + 11 ). Similarly, ( 4^2 = 1 + 3 + 5 + 7 ) and ( 4^3 = 13 + 15 + 17 + 19 ). And so on up to ( 5^3 ).So, it seems like for squares (( m^2 )), the decomposition is the sum of the first ( m ) odd numbers. That makes sense because I remember that the sum of the first ( n ) odd numbers is ( n^2 ). So, ( 2^2 ) is the sum of the first 2 odd numbers, ( 3^2 ) is the sum of the first 3, etc.But what about the cubes (( m^3 ))? Let's see. For ( 2^3 ), it's ( 3 + 5 ). For ( 3^3 ), it's ( 7 + 9 + 11 ). For ( 4^3 ), it's ( 13 + 15 + 17 + 19 ), and ( 5^3 ) is ( 21 + 23 + 25 + 27 + 29 ). Hmm, interesting. It seems like each ( m^3 ) is being decomposed into the sum of ( m ) consecutive odd numbers, but not starting from 1. Instead, they start from a higher odd number.Let me try to find a pattern in the starting numbers. For ( 2^3 ), it starts at 3. For ( 3^3 ), it starts at 7. For ( 4^3 ), it starts at 13. For ( 5^3 ), it starts at 21. Let me list these starting numbers:- ( m = 2 ): 3- ( m = 3 ): 7- ( m = 4 ): 13- ( m = 5 ): 21Now, let's see if there's a formula for the starting number in terms of ( m ). Let's denote the starting number as ( S(m) ).Looking at the values:- ( S(2) = 3 )- ( S(3) = 7 )- ( S(4) = 13 )- ( S(5) = 21 )Let me compute the differences between consecutive ( S(m) ):- ( 7 - 3 = 4 )- ( 13 - 7 = 6 )- ( 21 - 13 = 8 )Hmm, the differences are 4, 6, 8. That seems like an arithmetic sequence with a common difference of 2. So, the difference between ( S(m) ) and ( S(m-1) ) is ( 2(m-1) ). Let me check:For ( m = 3 ), the difference should be ( 2(3-1) = 4 ). Indeed, ( 7 - 3 = 4 ).For ( m = 4 ), the difference should be ( 2(4-1) = 6 ). Indeed, ( 13 - 7 = 6 ).For ( m = 5 ), the difference should be ( 2(5-1) = 8 ). Indeed, ( 21 - 13 = 8 ).So, it seems that ( S(m) = S(m-1) + 2(m-1) ). This is a recursive formula. Let's try to find a closed-form expression for ( S(m) ).Starting from ( S(2) = 3 ), let's compute ( S(m) ) for higher ( m ):- ( S(2) = 3 )- ( S(3) = S(2) + 2(2) = 3 + 4 = 7 )- ( S(4) = S(3) + 2(3) = 7 + 6 = 13 )- ( S(5) = S(4) + 2(4) = 13 + 8 = 21 )- ( S(6) = S(5) + 2(5) = 21 + 10 = 31 )- And so on.This recursive relation can be expressed as:( S(m) = S(m-1) + 2(m-1) )To find a closed-form, let's unroll the recursion:( S(m) = S(m-1) + 2(m-1) )( S(m-1) = S(m-2) + 2(m-2) )...( S(2) = 3 )So, summing all these up:( S(m) = 3 + 2(2) + 2(3) + 2(4) + dots + 2(m-1) )Factor out the 2:( S(m) = 3 + 2[2 + 3 + 4 + dots + (m-1)] )The sum inside the brackets is the sum from ( k=2 ) to ( k=m-1 ) of ( k ). The sum from ( k=1 ) to ( k=n ) is ( frac{n(n+1)}{2} ), so the sum from ( k=2 ) to ( k=m-1 ) is ( frac{(m-1)m}{2} - 1 ).Wait, let me verify that. The sum from ( k=1 ) to ( k=m-1 ) is ( frac{(m-1)m}{2} ). Therefore, the sum from ( k=2 ) to ( k=m-1 ) is ( frac{(m-1)m}{2} - 1 ).So, substituting back:( S(m) = 3 + 2left( frac{(m-1)m}{2} - 1 right) )Simplify:( S(m) = 3 + (m-1)m - 2 )( S(m) = (m-1)m + 1 )( S(m) = m^2 - m + 1 )Okay, so the starting number ( S(m) ) for the decomposition of ( m^3 ) is ( m^2 - m + 1 ). Let me check this with the examples:- For ( m = 2 ): ( 2^2 - 2 + 1 = 4 - 2 + 1 = 3 ). Correct.- For ( m = 3 ): ( 9 - 3 + 1 = 7 ). Correct.- For ( m = 4 ): ( 16 - 4 + 1 = 13 ). Correct.- For ( m = 5 ): ( 25 - 5 + 1 = 21 ). Correct.Perfect, so the formula holds.Now, the problem states that the smallest number in the decomposition of ( m^3 ) is 73. So, according to our formula, ( S(m) = m^2 - m + 1 = 73 ).Let's solve for ( m ):( m^2 - m + 1 = 73 )Subtract 73 from both sides:( m^2 - m + 1 - 73 = 0 )Simplify:( m^2 - m - 72 = 0 )Now, we have a quadratic equation: ( m^2 - m - 72 = 0 ). Let's solve for ( m ).Using the quadratic formula:( m = frac{1 pm sqrt{1 + 4 times 72}}{2} )Calculate the discriminant:( sqrt{1 + 288} = sqrt{289} = 17 )So,( m = frac{1 pm 17}{2} )This gives two solutions:1. ( m = frac{1 + 17}{2} = frac{18}{2} = 9 )2. ( m = frac{1 - 17}{2} = frac{-16}{2} = -8 )Since ( m ) is a natural number (( m in mathbb{N}^* )), we discard the negative solution. Therefore, ( m = 9 ).Let me verify this. If ( m = 9 ), then the starting number should be ( 9^2 - 9 + 1 = 81 - 9 + 1 = 73 ). That's exactly what the problem states. So, the decomposition of ( 9^3 ) starts with 73, which is the smallest number in that decomposition.Therefore, the value of ( m ) is 9.

answer：Okay, so I have this problem here: Given the set ( A = {x in mathbb{R} mid (sqrt{x})^{2} neq a} ), I need to find the range of values for ( a ) such that ( A ) is a non-empty set. Hmm, let me try to break this down step by step.First, let's understand what the set ( A ) represents. It's the set of all real numbers ( x ) such that when you take the square root of ( x ) and then square it, the result is not equal to ( a ). So, ( A ) contains all real numbers ( x ) where ( (sqrt{x})^2 neq a ).Wait a second, ( (sqrt{x})^2 ) is just ( x ), right? Because squaring the square root of ( x ) should give me back ( x ), as long as ( x ) is non-negative. So, ( (sqrt{x})^2 = x ) for ( x geq 0 ). That makes sense because the square root function is only defined for non-negative numbers in the real number system.So, if ( (sqrt{x})^2 = x ), then the condition ( (sqrt{x})^2 neq a ) simplifies to ( x neq a ). Therefore, the set ( A ) is all real numbers ( x ) such that ( x neq a ). Wait, but ( x ) has to be in the domain of ( sqrt{x} ), which is ( x geq 0 ). So actually, ( A ) is all non-negative real numbers ( x ) where ( x neq a ).But the problem states that ( A ) is a non-empty set. So, we need to ensure that there exists at least one ( x ) in ( mathbb{R} ) such that ( x neq a ). But hold on, ( mathbb{R} ) is the set of all real numbers, and ( x ) is restricted to ( x geq 0 ) because of the square root. So, ( A ) is non-empty if there exists at least one ( x geq 0 ) such that ( x neq a ).Now, let's think about what ( a ) can be. If ( a ) is a non-negative real number, then ( x = a ) is a valid element in the domain of ( sqrt{x} ). So, if ( a ) is non-negative, then ( x = a ) would satisfy ( (sqrt{x})^2 = a ), meaning that ( x = a ) is excluded from set ( A ). However, there are still other values of ( x ) (both greater than and less than ( a )) that are non-negative and would satisfy ( x neq a ). Therefore, even if ( a ) is non-negative, ( A ) would still be non-empty because there are infinitely many other ( x ) values that are not equal to ( a ).But wait, the problem is asking for the range of ( a ) such that ( A ) is non-empty. So, if ( a ) is non-negative, ( A ) is non-empty. What if ( a ) is negative? Let's consider that case.If ( a ) is negative, then ( x = a ) is not in the domain of ( sqrt{x} ) because ( a ) is negative, and ( x ) must be non-negative. Therefore, for all ( x geq 0 ), ( x ) cannot be equal to ( a ) because ( a ) is negative. So, in this case, ( A ) would be all non-negative real numbers, since none of them equal ( a ). Therefore, ( A ) is still non-empty.Wait, so whether ( a ) is positive or negative, ( A ) is non-empty? That can't be right because the problem is asking for the range of ( a ) such that ( A ) is non-empty. If ( A ) is always non-empty regardless of ( a ), then the range of ( a ) would be all real numbers. But that doesn't seem right because the problem is specifically asking for the range where ( A ) is non-empty, implying that there might be some values of ( a ) where ( A ) is empty.Let me re-examine the problem. The set ( A ) is defined as ( {x in mathbb{R} mid (sqrt{x})^{2} neq a} ). So, ( x ) must be a real number, but ( sqrt{x} ) is only defined for ( x geq 0 ). Therefore, ( A ) is actually the set of all ( x geq 0 ) such that ( x neq a ).Now, if ( a ) is a non-negative real number, then ( x = a ) is excluded from ( A ), but there are still infinitely many other ( x ) values in ( A ). So, ( A ) is non-empty.If ( a ) is negative, then ( x = a ) is not in the domain of ( sqrt{x} ), so all ( x geq 0 ) satisfy ( x neq a ). Therefore, ( A ) is the entire set of non-negative real numbers, which is certainly non-empty.Wait, so in both cases, whether ( a ) is positive or negative, ( A ) is non-empty. That suggests that ( A ) is always non-empty, regardless of the value of ( a ). But the problem is asking for the range of ( a ) such that ( A ) is non-empty. If ( A ) is always non-empty, then the range of ( a ) would be all real numbers.But that contradicts my initial thought. Maybe I'm misunderstanding the problem. Let me read it again: "Given the set ( A = {x in mathbb{R} mid (sqrt{x})^{2} neq a} ), find the range of values for ( a ) such that ( A ) is a non-empty set."Wait, perhaps I misinterpreted the set ( A ). It says ( x in mathbb{R} ), but ( sqrt{x} ) is only defined for ( x geq 0 ). So, does that mean that ( A ) is actually the set of all ( x geq 0 ) such that ( x neq a )? Because for ( x < 0 ), ( sqrt{x} ) is not a real number, so those ( x ) values are not included in ( A ).Therefore, ( A ) is the set ( {x in mathbb{R} mid x geq 0 text{ and } x neq a} ). So, ( A ) is non-empty if there exists at least one ( x geq 0 ) such that ( x neq a ).Now, if ( a ) is a non-negative real number, then ( x = a ) is excluded, but there are still other ( x geq 0 ) that are not equal to ( a ). So, ( A ) is non-empty.If ( a ) is negative, then ( x = a ) is not in the domain of ( sqrt{x} ), so all ( x geq 0 ) are included in ( A ), making ( A ) non-empty.Wait, so in both cases, whether ( a ) is positive or negative, ( A ) is non-empty. That suggests that ( A ) is always non-empty, regardless of ( a ). Therefore, the range of ( a ) would be all real numbers.But that seems counterintuitive because the problem is specifically asking for the range where ( A ) is non-empty, implying that there might be some values of ( a ) where ( A ) is empty. Maybe I'm missing something.Let me consider the case where ( a ) is such that ( (sqrt{x})^2 = a ) has no solution. If ( a ) is negative, then ( (sqrt{x})^2 = a ) has no solution because ( (sqrt{x})^2 ) is always non-negative. Therefore, ( A ) would be all ( x geq 0 ), which is non-empty.If ( a ) is non-negative, then ( x = a ) is a solution, so ( A ) excludes ( x = a ), but still contains all other ( x geq 0 ), making ( A ) non-empty.Wait, so in both cases, ( A ) is non-empty. Therefore, for all real numbers ( a ), ( A ) is non-empty. So, the range of ( a ) is all real numbers.But the problem is asking for the range of ( a ) such that ( A ) is non-empty. If ( A ) is always non-empty, then the answer would be all real numbers. However, the initial assistant's answer was ( a in (-infty, 0) ), which suggests that only negative ( a ) values make ( A ) non-empty. That seems incorrect based on my reasoning.Let me double-check. If ( a ) is positive, say ( a = 4 ), then ( A ) would be all ( x geq 0 ) except ( x = 4 ). So, ( A ) is still non-empty. If ( a = 0 ), then ( A ) is all ( x > 0 ), which is non-empty. If ( a = -1 ), then ( A ) is all ( x geq 0 ), which is non-empty.Therefore, regardless of whether ( a ) is positive, zero, or negative, ( A ) is non-empty. So, the range of ( a ) should be all real numbers.But the initial assistant's answer was ( a in (-infty, 0) ), which only includes negative numbers. That seems to be a mistake because ( A ) is non-empty for all ( a ).Wait, maybe the problem is interpreted differently. Perhaps the set ( A ) is defined as ( x in mathbb{R} ), but ( (sqrt{x})^2 ) is only defined for ( x geq 0 ). So, for ( x < 0 ), ( (sqrt{x})^2 ) is not a real number, so those ( x ) values are not included in ( A ). Therefore, ( A ) is the set of all ( x geq 0 ) such that ( x neq a ).Now, if ( a ) is a non-negative real number, then ( x = a ) is excluded, but there are still other ( x geq 0 ) that are not equal to ( a ), so ( A ) is non-empty.If ( a ) is negative, then ( x = a ) is not in the domain, so ( A ) includes all ( x geq 0 ), which is non-empty.Therefore, ( A ) is non-empty for all real numbers ( a ). So, the range of ( a ) is all real numbers.But the initial assistant's answer was ( a in (-infty, 0) ), which is incorrect because ( A ) is non-empty for all ( a ).Wait, perhaps the problem is asking for ( A ) to be non-empty in the sense that it's not the entire set of real numbers. But no, the problem just says "non-empty," which means it has at least one element.Given that, ( A ) is non-empty for all ( a ), so the range of ( a ) is all real numbers.But the initial assistant's answer was ( a in (-infty, 0) ), which is incorrect. Therefore, the correct answer should be that ( a ) can be any real number, so ( a in mathbb{R} ).Wait, but let me think again. If ( a ) is such that ( (sqrt{x})^2 = a ) has no solution, then ( A ) would be all ( x geq 0 ). If ( a ) is such that ( (sqrt{x})^2 = a ) has solutions, then ( A ) excludes those solutions but still contains other ( x geq 0 ).Therefore, ( A ) is non-empty regardless of ( a ). So, the range of ( a ) is all real numbers.But the initial assistant's answer was ( a in (-infty, 0) ), which is only considering the case where ( a ) is negative. That seems to be a mistake.Therefore, the correct range of ( a ) is all real numbers, ( a in mathbb{R} ).

answer：First, I need to determine the total number of possible outcomes when drawing two labels from the box without replacement. Since there are 5 labels, the number of ways to choose 2 labels is calculated using the combination formula:[C(5,2) = frac{5!}{2!(5-2)!} = 10]Next, I identify the pairs of labels that represent adjacent integers. These pairs are (1,2), (2,3), (3,4), and (4,5), which gives me 4 favorable outcomes.Finally, I calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:[text{Probability} = frac{4}{10} = frac{2}{5}]

answer：Alright, so I've got this function f(x) = 2022^{x-3} + (x-3)^3 - 2022^{3-x} + 2x, and I need to solve the inequality f(x² - 4) + f(2 - 2x) ≤ 12. Hmm, okay. Let me try to break this down step by step.First, I should probably understand what the function f(x) looks like. It has exponential terms with 2022 as the base, a cubic term, and a linear term. The exponents are x-3 and 3-x, which are negatives of each other. That might be useful. Also, the cubic term is (x-3)^3, which is similar to the exponent terms. Maybe there's some symmetry or property I can exploit here.Let me see if I can rewrite f(x) in a simpler form. If I let y = x - 3, then x = y + 3. Substituting that into f(x), I get:f(x) = 2022^{y} + y^3 - 2022^{-y} + 2(y + 3)Simplifying that, it becomes:f(x) = 2022^{y} - 2022^{-y} + y^3 + 2y + 6So, f(x) can be expressed as g(y) + 6, where g(y) = 2022^{y} - 2022^{-y} + y^3 + 2y.Interesting. So, f(x) is just a shifted version of this new function g(y). Maybe g(y) has some nice properties that I can use.Looking at g(y), it has an exponential term, its negative counterpart, a cubic term, and a linear term. Let me check if g(y) is an odd function. An odd function satisfies g(-y) = -g(y).Let's compute g(-y):g(-y) = 2022^{-y} - 2022^{y} + (-y)^3 + 2(-y) = - (2022^{y} - 2022^{-y}) - y^3 - 2y = - [2022^{y} - 2022^{-y} + y^3 + 2y] = -g(y)Yes! So, g(y) is indeed an odd function. That's a useful property.Also, let's think about the behavior of g(y). The exponential terms 2022^{y} and -2022^{-y} will dominate as y becomes large in magnitude. Since 2022 is a large base, 2022^{y} grows rapidly for positive y and decays rapidly for negative y. Similarly, -2022^{-y} decays for positive y and grows negatively for negative y. The cubic term y^3 will dominate for large |y|, but the exponential terms will dominate for moderate y.But more importantly, since g(y) is odd, it's symmetric about the origin. Also, let's check if g(y) is increasing or decreasing. To do that, I can compute its derivative.g'(y) = d/dy [2022^{y} - 2022^{-y} + y^3 + 2y] = ln(2022) * 2022^{y} + ln(2022) * 2022^{-y} + 3y^2 + 2All the terms here are positive for all real y because 2022^{y} and 2022^{-y} are always positive, ln(2022) is positive, and 3y^2 + 2 is always positive. So, g'(y) is always positive, which means g(y) is strictly increasing over the entire real line.Okay, so g(y) is an odd, strictly increasing function. That's good to know.Now, going back to f(x). Since f(x) = g(y) + 6, where y = x - 3, we can write:f(x) = g(x - 3) + 6So, f is just a shifted version of g, shifted to the right by 3 units and then shifted up by 6 units.Now, the inequality we need to solve is f(x² - 4) + f(2 - 2x) ≤ 12.Let's substitute f(x) with g(x - 3) + 6:f(x² - 4) = g((x² - 4) - 3) + 6 = g(x² - 7) + 6f(2 - 2x) = g((2 - 2x) - 3) + 6 = g(-2x - 1) + 6So, the inequality becomes:g(x² - 7) + 6 + g(-2x - 1) + 6 ≤ 12Simplify this:g(x² - 7) + g(-2x - 1) + 12 ≤ 12Subtract 12 from both sides:g(x² - 7) + g(-2x - 1) ≤ 0Now, since g is an odd function, g(-z) = -g(z). So, g(-2x - 1) = -g(2x + 1). Let's substitute that in:g(x² - 7) - g(2x + 1) ≤ 0Which can be rewritten as:g(x² - 7) ≤ g(2x + 1)Now, since g is strictly increasing, if g(a) ≤ g(b), then a ≤ b. So, we can write:x² - 7 ≤ 2x + 1Let's solve this inequality:x² - 7 ≤ 2x + 1Bring all terms to one side:x² - 2x - 8 ≤ 0Factor the quadratic:x² - 2x - 8 = (x - 4)(x + 2)So, (x - 4)(x + 2) ≤ 0To solve this inequality, we can find the critical points and test intervals.Critical points are x = -2 and x = 4.Test intervals:1. x < -2: Let's pick x = -3: (-3 - 4)(-3 + 2) = (-7)(-1) = 7 > 02. -2 < x < 4: Let's pick x = 0: (0 - 4)(0 + 2) = (-4)(2) = -8 < 03. x > 4: Let's pick x = 5: (5 - 4)(5 + 2) = (1)(7) = 7 > 0Since we're looking for where the expression is ≤ 0, the solution is the interval where the expression is negative or zero, which is between -2 and 4, including the endpoints.So, the solution set is [-2, 4].Wait, let me double-check to make sure I didn't make any mistakes. So, starting from f(x² - 4) + f(2 - 2x) ≤ 12, I transformed f(x) into g(x - 3) + 6, then substituted into the inequality, simplified, used the property that g is odd, and then used the fact that g is increasing to get the inequality x² - 7 ≤ 2x + 1, which led to x ∈ [-2, 4].Does this make sense? Let me plug in x = -2 and x = 4 into the original inequality to verify.First, x = -2:f((-2)^2 - 4) + f(2 - 2*(-2)) = f(4 - 4) + f(2 + 4) = f(0) + f(6)Compute f(0):f(0) = 2022^{0 - 3} + (0 - 3)^3 - 2022^{3 - 0} + 2*0 = 2022^{-3} + (-3)^3 - 2022^{3} + 0 = 1/2022^3 - 27 - 2022^3Compute f(6):f(6) = 2022^{6 - 3} + (6 - 3)^3 - 2022^{3 - 6} + 2*6 = 2022^{3} + 3^3 - 2022^{-3} + 12 = 2022^3 + 27 - 1/2022^3 + 12So, f(0) + f(6) = [1/2022^3 - 27 - 2022^3] + [2022^3 + 27 - 1/2022^3 + 12] = 12So, f(0) + f(6) = 12, which satisfies the inequality f(x² - 4) + f(2 - 2x) ≤ 12.Now, x = 4:f(4² - 4) + f(2 - 2*4) = f(16 - 4) + f(2 - 8) = f(12) + f(-6)Compute f(12):f(12) = 2022^{12 - 3} + (12 - 3)^3 - 2022^{3 - 12} + 2*12 = 2022^{9} + 9^3 - 2022^{-9} + 24 = 2022^9 + 729 - 1/2022^9 + 24Compute f(-6):f(-6) = 2022^{-6 - 3} + (-6 - 3)^3 - 2022^{3 - (-6)} + 2*(-6) = 2022^{-9} + (-9)^3 - 2022^{9} - 12 = 1/2022^9 - 729 - 2022^9 - 12So, f(12) + f(-6) = [2022^9 + 729 - 1/2022^9 + 24] + [1/2022^9 - 729 - 2022^9 - 12] = 12Again, f(12) + f(-6) = 12, which satisfies the inequality.Now, let's pick a value inside the interval, say x = 0:f(0² - 4) + f(2 - 2*0) = f(-4) + f(2)Compute f(-4):f(-4) = 2022^{-4 - 3} + (-4 - 3)^3 - 2022^{3 - (-4)} + 2*(-4) = 2022^{-7} + (-7)^3 - 2022^{7} - 8 = 1/2022^7 - 343 - 2022^7 - 8Compute f(2):f(2) = 2022^{2 - 3} + (2 - 3)^3 - 2022^{3 - 2} + 2*2 = 2022^{-1} + (-1)^3 - 2022^{1} + 4 = 1/2022 - 1 - 2022 + 4So, f(-4) + f(2) = [1/2022^7 - 343 - 2022^7 - 8] + [1/2022 - 1 - 2022 + 4]This sum is definitely less than 12 because the negative terms dominate.Now, let's pick a value outside the interval, say x = 5:f(5² - 4) + f(2 - 2*5) = f(25 - 4) + f(2 - 10) = f(21) + f(-8)Compute f(21):f(21) = 2022^{21 - 3} + (21 - 3)^3 - 2022^{3 - 21} + 2*21 = 2022^{18} + 18^3 - 2022^{-18} + 42 = 2022^18 + 5832 - 1/2022^18 + 42Compute f(-8):f(-8) = 2022^{-8 - 3} + (-8 - 3)^3 - 2022^{3 - (-8)} + 2*(-8) = 2022^{-11} + (-11)^3 - 2022^{11} - 16 = 1/2022^11 - 1331 - 2022^11 - 16So, f(21) + f(-8) = [2022^18 + 5832 - 1/2022^18 + 42] + [1/2022^11 - 1331 - 2022^11 - 16]This sum is way larger than 12 because 2022^18 and 2022^11 are extremely large positive numbers, while the negative terms are negligible in comparison. So, the inequality is not satisfied for x = 5, which is outside our interval.Similarly, if I pick x = -3, which is outside the interval on the left:f((-3)^2 - 4) + f(2 - 2*(-3)) = f(9 - 4) + f(2 + 6) = f(5) + f(8)Compute f(5):f(5) = 2022^{5 - 3} + (5 - 3)^3 - 2022^{3 - 5} + 2*5 = 2022^{2} + 2^3 - 2022^{-2} + 10 = 2022^2 + 8 - 1/2022^2 + 10Compute f(8):f(8) = 2022^{8 - 3} + (8 - 3)^3 - 2022^{3 - 8} + 2*8 = 2022^{5} + 5^3 - 2022^{-5} + 16 = 2022^5 + 125 - 1/2022^5 + 16So, f(5) + f(8) = [2022^2 + 8 - 1/2022^2 + 10] + [2022^5 + 125 - 1/2022^5 + 16]Again, this sum is way larger than 12 because of the large positive terms. So, the inequality is not satisfied for x = -3.Therefore, it seems that the interval [-2, 4] is indeed the correct solution set for the inequality.I think I've covered all the steps carefully and verified the endpoints and some values inside and outside the interval. It all checks out. So, I'm confident that the solution set is [-2, 4].