answer：Alright, so I have this problem here about shifting a cosine function and figuring out the range of ω based on the monotonicity in a specific interval. Let me try to break it down step by step.First, the original function is f(x) = cos(ωx). They say it's shifted to the left by π/(3ω) units to get g(x). So, shifting a function left by 'a' units usually means replacing x with x + a in the function. So, g(x) should be cos(ω(x + π/(3ω))). Let me write that out:g(x) = cos(ωx + ω*(π/(3ω))) = cos(ωx + π/3)Okay, so g(x) is cos(ωx + π/3). Got that part.Now, the problem states that y = g(x) is monotonically increasing in the interval (2π/3, 4π/3). So, I need to find the values of ω for which this function is increasing in that interval.I remember that for a function to be increasing, its derivative should be positive in that interval. So, let me compute the derivative of g(x):g'(x) = -ω sin(ωx + π/3)So, for g(x) to be increasing, g'(x) > 0. That means:-ω sin(ωx + π/3) > 0Since ω > 0, I can divide both sides by -ω, which will reverse the inequality:sin(ωx + π/3) < 0So, sin(ωx + π/3) < 0 for all x in (2π/3, 4π/3).Hmm, okay. So, the sine function is negative in the intervals where its argument is between π and 2π, and so on, periodically. So, I need to ensure that for all x in (2π/3, 4π/3), the argument ωx + π/3 is in an interval where sine is negative.Let me denote θ = ωx + π/3. Then, θ must lie in an interval where sinθ < 0, which is (π + 2πk, 2π + 2πk) for some integer k.So, for all x in (2π/3, 4π/3), θ = ωx + π/3 must lie in (π + 2πk, 2π + 2πk).Let me find the range of θ when x is in (2π/3, 4π/3). Let's compute θ at the endpoints:When x = 2π/3:θ = ω*(2π/3) + π/3 = (2ωπ)/3 + π/3 = π/3*(2ω + 1)When x = 4π/3:θ = ω*(4π/3) + π/3 = (4ωπ)/3 + π/3 = π/3*(4ω + 1)So, θ ranges from π/3*(2ω + 1) to π/3*(4ω + 1).We need this entire interval to lie within some interval where sinθ < 0, which is between π and 2π, or 3π and 4π, etc. But since we're dealing with a continuous interval, it's likely the first such interval, between π and 2π.So, let's set up the inequalities:π ≤ π/3*(2ω + 1) < π/3*(4ω + 1) ≤ 2πWait, is that correct? Let me think. The entire interval of θ must lie within (π, 2π). So, the lower bound of θ must be ≥ π, and the upper bound must be ≤ 2π.So:π/3*(2ω + 1) ≥ πandπ/3*(4ω + 1) ≤ 2πLet me solve the first inequality:π/3*(2ω + 1) ≥ πDivide both sides by π:(2ω + 1)/3 ≥ 1Multiply both sides by 3:2ω + 1 ≥ 3Subtract 1:2ω ≥ 2Divide by 2:ω ≥ 1Now, the second inequality:π/3*(4ω + 1) ≤ 2πDivide both sides by π:(4ω + 1)/3 ≤ 2Multiply both sides by 3:4ω + 1 ≤ 6Subtract 1:4ω ≤ 5Divide by 4:ω ≤ 5/4So, combining both inequalities, we have:1 ≤ ω ≤ 5/4Therefore, the range of ω is [1, 5/4], which corresponds to option D.But wait, let me double-check if there could be another interval where sinθ is negative. For example, if θ is between 3π and 4π, but that would require a larger ω. Let me see:If θ is between 3π and 4π, then:3π ≤ π/3*(2ω + 1) < π/3*(4ω + 1) ≤ 4πFollowing similar steps:First inequality:π/3*(2ω + 1) ≥ 3πDivide by π:(2ω + 1)/3 ≥ 3Multiply by 3:2ω + 1 ≥ 92ω ≥ 8ω ≥ 4Second inequality:π/3*(4ω + 1) ≤ 4πDivide by π:(4ω + 1)/3 ≤ 4Multiply by 3:4ω + 1 ≤ 124ω ≤ 11ω ≤ 11/4 = 2.75So, in this case, ω would have to be between 4 and 2.75, which is impossible because 4 > 2.75. So, this interval doesn't yield a valid solution.Therefore, the only valid interval is when θ is between π and 2π, leading to ω between 1 and 5/4.I think that's solid. So, the answer should be D: [1, 5/4].**Final Answer**boxed{D}

answer：Okay, so I have this sequence problem here. Let me try to understand it step by step. The sequence is defined as {a_n} where a₁ = 1, and then there's a recursive formula for a_{n+1} depending on whether n is odd or even. Alright, let me write down what I know:- a₁ = 1- If n is odd, then a_{n+1} = a_n - 2ⁿ- If n is even, then a_{n+1} = a_n + 2^{n+1}And the question has four statements, A to D, and I need to figure out which ones are correct.Let me start by computing the first few terms to see if I can spot a pattern or verify the statements.Starting with a₁ = 1.Now, n=1 is odd, so a₂ = a₁ - 2¹ = 1 - 2 = -1.Next, n=2 is even, so a₃ = a₂ + 2^{2+1} = -1 + 8 = 7.n=3 is odd, so a₄ = a₃ - 2³ = 7 - 8 = -1.n=4 is even, so a₅ = a₄ + 2^{4+1} = -1 + 32 = 31.n=5 is odd, so a₆ = a₅ - 2⁵ = 31 - 32 = -1.n=6 is even, so a₇ = a₆ + 2^{6+1} = -1 + 128 = 127.Hmm, interesting. So it seems like every even index term is -1, and every odd index term after that is 2^{k} -1, where k is the exponent.Wait, let's see:a₁ = 1a₂ = -1a₃ = 7 = 8 -1 = 2³ -1a₄ = -1a₅ = 31 = 32 -1 = 2⁵ -1a₆ = -1a₇ = 127 = 128 -1 = 2⁷ -1So, it seems like for odd n, a_n = 2^{n} -1, and for even n, a_n = -1.Is that the case? Let me test it.Assume for some odd k, a_k = 2^{k} -1.Then, since k is odd, a_{k+1} = a_k - 2^{k} = (2^{k} -1) - 2^{k} = -1.Then, for n = k+1, which is even, a_{k+2} = a_{k+1} + 2^{(k+1)+1} = -1 + 2^{k+2}.Since k is odd, k+2 is odd + 2 = odd + even = odd. Wait, no, k is odd, so k+1 is even, and k+2 is odd.Wait, no, k is odd, so k+1 is even, and k+2 is odd. So, a_{k+2} = -1 + 2^{k+2}.But 2^{k+2} is 4*2^{k}, so a_{k+2} = 4*2^{k} -1.But k is odd, so k+2 is also odd. So, a_{k+2} = 2^{(k+2)} -1, which fits our pattern.Therefore, by induction, it seems that for all odd n, a_n = 2^{n} -1, and for all even n, a_n = -1.So, let's see the statements:A: a₃ = 7From above, a₃ = 7, which is correct.B: a_{2022} = a₂Since 2022 is even, a_{2022} = -1, and a₂ = -1, so yes, they are equal.C: a_{2023} = 2^{2023}Wait, 2023 is odd, so a_{2023} = 2^{2023} -1, not exactly 2^{2023}. So this is incorrect.D: 3S_{2n+1} = 2^{2n+3} -6n -5Hmm, S_{2n+1} is the sum of the first 2n+1 terms.Let me compute S_{2n+1}.Given the pattern, every even term is -1, and every odd term is 2^{k} -1.So, let's write S_{2n+1} as the sum from k=1 to 2n+1 of a_k.Which is a₁ + a₂ + a₃ + ... + a_{2n+1}.Breaking it down:Number of odd terms: n+1 (since from 1 to 2n+1, inclusive, there are n+1 odd indices)Number of even terms: n (from 2 to 2n, inclusive)Each odd term a_{2m-1} = 2^{2m-1} -1Each even term a_{2m} = -1So, S_{2n+1} = sum_{m=1}^{n+1} (2^{2m-1} -1) + sum_{m=1}^{n} (-1)Compute the first sum: sum_{m=1}^{n+1} (2^{2m-1} -1) = sum_{m=1}^{n+1} 2^{2m-1} - sum_{m=1}^{n+1} 1Similarly, the second sum is sum_{m=1}^{n} (-1) = -nCompute sum_{m=1}^{n+1} 2^{2m-1}:This is a geometric series with first term 2^{1} = 2, ratio 4, number of terms n+1.Sum = 2*(4^{n+1} -1)/(4 -1) = (2*(4^{n+1} -1))/3Similarly, sum_{m=1}^{n+1} 1 = n+1So, first sum: (2*(4^{n+1} -1))/3 - (n+1)Second sum: -nTherefore, total sum S_{2n+1} = (2*(4^{n+1} -1))/3 - (n+1) - n = (2*(4^{n+1} -1))/3 - 2n -1Simplify:Multiply numerator: 2*4^{n+1} -2So, S_{2n+1} = (2*4^{n+1} -2)/3 -2n -1Let me write 4^{n+1} as 2^{2n+2}, so:S_{2n+1} = (2*2^{2n+2} -2)/3 -2n -1 = (2^{2n+3} -2)/3 -2n -1Combine terms:= (2^{2n+3} -2 -6n -3)/3 = (2^{2n+3} -6n -5)/3Therefore, 3S_{2n+1} = 2^{2n+3} -6n -5, which matches statement D.So, statement D is correct.Thus, the correct statements are A, B, and D.**Final Answer**boxed{ABD}

answer：Okay, so I have this equation: 2x|x| + y² = 1. I need to figure out which of the properties A, B, C, or D are correct. Let me go through each option one by one.First, let me understand the equation better. The term 2x|x| is interesting because it involves an absolute value. I know that |x| is equal to x when x is positive and -x when x is negative. So, 2x|x| can be rewritten depending on whether x is positive or negative.If x is positive, then |x| = x, so 2x|x| becomes 2x². If x is negative, |x| = -x, so 2x|x| becomes 2x*(-x) = -2x². So, the equation can be split into two cases:1. When x ≥ 0: 2x² + y² = 12. When x < 0: -2x² + y² = 1Hmm, okay. So for positive x, it's a standard ellipse equation, and for negative x, it's a hyperbola? Wait, let me check that.For x ≥ 0: 2x² + y² = 1. If I rewrite this, it's (x²)/(1/√2)² + y²/1² = 1, which is an ellipse centered at the origin, stretched along the y-axis.For x < 0: -2x² + y² = 1. Let me rearrange this: y² - 2x² = 1. That looks like a hyperbola opening upwards and downwards, centered at the origin.So the curve consists of an ellipse in the right half-plane (x ≥ 0) and a hyperbola in the left half-plane (x < 0). Interesting.Now, let's look at the options:A: The corresponding curve is symmetric about the y-axis.Symmetry about the y-axis means that if (x, y) is on the curve, then (-x, y) is also on the curve. Let me test this.Take a point (a, b) on the curve. If a is positive, then 2a² + b² = 1. Now, if I consider (-a, b), since a is positive, -a is negative. So, plugging into the equation for x < 0: -2(-a)² + b² = -2a² + b². Is this equal to 1? Well, originally, 2a² + b² = 1, so -2a² + b² would be (b² - 2a²) = 1 - 4a². Unless a = 0, this won't be equal to 1. So, unless a is 0, (-a, b) is not on the curve. Therefore, the curve is not symmetric about the y-axis. So, option A is incorrect.B: The corresponding curve is centrally symmetric about the origin.Central symmetry about the origin means that if (x, y) is on the curve, then (-x, -y) is also on the curve. Let's test this.Take a point (a, b) on the curve. If a is positive, then 2a² + b² = 1. Now, consider (-a, -b). Since a is positive, -a is negative. Plugging into the equation for x < 0: -2(-a)² + (-b)² = -2a² + b². Again, this is equal to (b² - 2a²) = 1 - 4a², which isn't necessarily 1. So, unless a = 0, (-a, -b) isn't on the curve. Therefore, the curve isn't centrally symmetric about the origin. So, option B is incorrect.C: x can take any real number.Let me see. For x ≥ 0: 2x² + y² = 1. Since both terms are non-negative, x² ≤ 1/2, so x ≤ 1/√2 ≈ 0.707. So, x can't be greater than that. For x < 0: -2x² + y² = 1. Here, y² = 1 + 2x². Since y² must be non-negative, this is always true for any real x. But wait, x can be any negative number, but for x positive, it's limited. So, x can't be any real number because for positive x, it's limited to x ≤ 1/√2. Therefore, option C is incorrect.D: y can take any real number.Looking at the equation, y² = 1 - 2x|x|. Let's see what values y² can take. For x ≥ 0: y² = 1 - 2x². Since 2x² ≤ 1, y² ≥ 0, so y can be between -√(1 - 2x²) and √(1 - 2x²). So, y is limited here. For x < 0: y² = 1 + 2x². Since x² is always non-negative, y² can be any value greater than or equal to 1. So, y can be any real number with |y| ≥ 1. But wait, for x ≥ 0, y is limited between -√(1 - 2x²) and √(1 - 2x²), which is a finite range. So, y can't take all real numbers because in the positive x region, y is limited. Therefore, option D is incorrect.Wait, that contradicts my initial thought. Let me double-check.For x < 0: y² = 1 + 2x². Since x² can be any non-negative number, y² can be any number greater than or equal to 1. So, y can be any real number with |y| ≥ 1. But for x ≥ 0: y² = 1 - 2x². Since 2x² ≤ 1, y² ≥ 0, so y is between -√(1 - 2x²) and √(1 - 2x²). So, y can't be greater than 1 or less than -1 in this region. But for x < 0, y can be any real number with |y| ≥ 1. So, overall, y can take any real number because in the left half-plane, y can be any real number beyond 1 or -1, and in the right half-plane, y is limited between -1 and 1. So, combining both, y can take any real number because for any y, there exists an x (either positive or negative) such that the equation holds. Wait, is that true?Let me think. If y is between -1 and 1, then x must be positive because for y² ≤ 1, 1 - y² ≥ 0, so x² = (1 - y²)/2, which is non-negative. So, x can be positive or negative? Wait, no. For y² ≤ 1, x must be positive because if x were negative, y² = 1 + 2x², which would be greater than 1. So, for y between -1 and 1, x must be positive. For y outside that range, x must be negative.So, for any real y, there exists an x such that the equation holds. If |y| ≤ 1, x is positive; if |y| > 1, x is negative. Therefore, y can indeed take any real number. So, option D is correct.Wait, but earlier I thought y was limited in the positive x region, but since for any y, whether inside or outside the range, there's an x that satisfies the equation, y can take any real number. So, D is correct.So, after going through all options, only D is correct.

answer：Okay, so I've got this problem here where I need to choose five numbers from a 5x5 grid, and the catch is that no two numbers can be from the same row or the same column. And I need to prove that no matter how I pick these numbers, their sum will always be the same. Hmm, interesting. Let me see how I can approach this.First, let me look at the grid they've given me:[begin{array}{|c|c|c|c|c|}hline1&4&7&10&13 hline16&19&22&25&28 hline31&34&37&40&43 hline46&49&52&55&58 hline61&64&67&70&73 hlineend{array}]Alright, so it's a 5x5 grid with numbers increasing from left to right and top to bottom. I notice that each row starts with a number that's 15 more than the first number of the previous row. For example, the first row starts with 1, the second with 16 (which is 1 + 15), the third with 31 (16 + 15), and so on. Similarly, within each row, the numbers increase by 3 each time. So, in the first row, it's 1, 4, 7, 10, 13; each number is 3 more than the one before it.This seems like a pattern. Maybe I can generalize the numbers in the grid. Let me try to come up with a formula for the number in the (i)-th row and (j)-th column. Looking at the first row ((i=1)):- The first element is 1.- The second is 4, which is 1 + 3.- The third is 7, which is 1 + 3*2.- And so on.So, for the first row, the (j)-th element can be written as (1 + 3(j - 1)).Now, looking at the second row ((i=2)):- The first element is 16, which is 1 + 15.- The second is 19, which is 16 + 3.- The third is 22, which is 19 + 3.- So, the (j)-th element is (16 + 3(j - 1)).Similarly, the third row starts with 31, which is 1 + 15*2, and each subsequent element increases by 3. So, the (j)-th element is (31 + 3(j - 1)).Putting this together, it seems like the number in the (i)-th row and (j)-th column is given by:[a_{i,j} = 1 + 15(i - 1) + 3(j - 1)]Let me check this formula with a few examples to make sure it's correct.For (i=1, j=1):[a_{1,1} = 1 + 15(0) + 3(0) = 1]That's correct.For (i=2, j=1):[a_{2,1} = 1 + 15(1) + 3(0) = 16]Also correct.For (i=3, j=3):[a_{3,3} = 1 + 15(2) + 3(2) = 1 + 30 + 6 = 37]Yep, that's in the grid.Okay, so the formula seems to hold. Now, the problem is to choose five numbers such that no two are from the same row or column, and show that their sum is always the same.So, essentially, I'm selecting one number from each row and each column. This is similar to selecting a permutation of the columns for each row. For example, in the first row, I pick column 1, in the second row, column 3, and so on, making sure that each column is picked exactly once.Let me denote the permutation of columns as (sigma), where (sigma(i)) gives the column index for the (i)-th row. So, the sum (S) of the selected numbers would be:[S = sum_{i=1}^5 a_{i, sigma(i)} = sum_{i=1}^5 left[1 + 15(i - 1) + 3(sigma(i) - 1)right]]Let me break this sum down into three separate sums:[S = sum_{i=1}^5 1 + sum_{i=1}^5 15(i - 1) + sum_{i=1}^5 3(sigma(i) - 1)]Calculating each part individually:1. The first sum is straightforward:[sum_{i=1}^5 1 = 5]2. The second sum involves the row indices:[sum_{i=1}^5 15(i - 1) = 15 sum_{i=0}^4 i = 15 times frac{4 times 5}{2} = 15 times 10 = 150]Wait, hold on. The sum of (i) from 0 to 4 is (0 + 1 + 2 + 3 + 4 = 10), so multiplying by 15 gives 150. That seems right.3. The third sum involves the permutation (sigma):[sum_{i=1}^5 3(sigma(i) - 1) = 3 sum_{i=1}^5 (sigma(i) - 1) = 3 left( sum_{i=1}^5 sigma(i) - 5 right)]Since (sigma) is a permutation of the columns, the sum of (sigma(i)) for (i = 1) to (5) is just the sum of the numbers 1 through 5:[sum_{i=1}^5 sigma(i) = 1 + 2 + 3 + 4 + 5 = 15]Therefore:[3 left( 15 - 5 right) = 3 times 10 = 30]Now, adding all three parts together:[S = 5 + 150 + 30 = 185]So, no matter what permutation (sigma) I choose, the sum (S) will always be 185. That means, regardless of how I pick one number from each row and column, their total will always be the same.Let me double-check this with an example. Suppose I pick the main diagonal elements: 1, 19, 37, 55, 73.Adding them up:[1 + 19 = 20][20 + 37 = 57][57 + 55 = 112][112 + 73 = 185]Yep, that adds up to 185.Let me try another set. How about picking the first element of each row but shifting columns each time: 1 (row1, col1), 22 (row2, col3), 40 (row3, col4), 58 (row4, col5), and 64 (row5, col2).Adding these:[1 + 22 = 23][23 + 40 = 63][63 + 58 = 121][121 + 64 = 185]Again, 185. Seems consistent.One more for good measure. Let's pick randomly: 4 (row1, col2), 25 (row2, col4), 37 (row3, col3), 52 (row4, col3) – wait, hold on, I already picked column 3 in row3, so I can't pick column3 again in row4. Oops, that's a mistake. Let me correct that.Pick 4 (row1, col2), 25 (row2, col4), 37 (row3, col3), 58 (row4, col5), and 61 (row5, col1).Adding these:[4 + 25 = 29][29 + 37 = 66][66 + 58 = 124][124 + 61 = 185]Still 185. Okay, so it seems like regardless of how I pick the numbers, as long as I follow the rule of one per row and column, the sum is always 185.This makes sense because the way the grid is constructed, each row and column has a consistent increment, which when summed across all rows and columns, results in a fixed total. The permutation of columns doesn't affect the overall sum because each column's contribution is accounted for exactly once in the total sum.So, in conclusion, no matter which five numbers you pick following the given rules, their sum will always be 185.

answer：Okay, so I have this equation here: 50 squared equals 10 to the power of x. I need to solve for x. Hmm, let me think about how to approach this. First, I know that 50 squared is 2500. So, 50^2 is 2500, and that's equal to 10^x. So, I need to find the exponent x such that 10 raised to that power equals 2500. Wait, 10 raised to what power gives me 2500? I know that 10^3 is 1000, and 10^4 is 10,000. So, 2500 is somewhere between 10^3 and 10^4. That means x is between 3 and 4. But how do I find the exact value of x? Maybe I can use logarithms because logarithms are useful for solving equations where the variable is in the exponent. I remember that if I have an equation like a^b = c, then taking the logarithm of both sides can help me solve for b. Specifically, log(a^b) = log(c), and using the logarithm power rule, that becomes b * log(a) = log(c). Then, I can solve for b by dividing both sides by log(a). Applying that to my equation, 10^x = 2500, I can take the logarithm of both sides. Let me write that down: log(10^x) = log(2500). Using the power rule, this simplifies to x * log(10) = log(2500). I know that log(10) is 1 because we're using base 10 logarithms. So, this simplifies further to x = log(2500). Now, I need to calculate log(2500). I don't remember the exact value, but I can approximate it. I know that log(1000) is 3 and log(10,000) is 4, as I thought earlier. Since 2500 is halfway between 1000 and 10,000 in terms of multiplication by 2.5, maybe the logarithm is somewhere around 3.4? Wait, let me think about it more carefully. Logarithms are exponents, so log(2500) is the exponent that you raise 10 to in order to get 2500. Since 2500 is 2.5 times 1000, and 1000 is 10^3, then 2500 is 2.5 * 10^3. So, log(2500) is log(2.5 * 10^3). Using the logarithm product rule, that's log(2.5) + log(10^3). Log(10^3) is 3, so that part is straightforward. Now, I need to find log(2.5). I remember that log(2) is approximately 0.3010 and log(3) is approximately 0.4771. Since 2.5 is between 2 and 3, log(2.5) should be between 0.3010 and 0.4771. Maybe around 0.3979? Let me check that. If I use a calculator, log(2.5) is approximately 0.39794. So, adding that to 3 gives me log(2500) ≈ 3 + 0.39794 = 3.39794. Therefore, x is approximately 3.39794. Wait, let me double-check my steps to make sure I didn't make any mistakes. I started with 50^2 = 10^x, which is 2500 = 10^x. Then, I took the logarithm of both sides, which gave me x = log(2500). I broke down 2500 into 2.5 * 10^3, so log(2500) = log(2.5) + 3. I approximated log(2.5) as 0.39794, so x ≈ 3.39794. That seems correct. I don't think I made any errors in my calculations. So, the solution is x ≈ 3.39794.

answer：To determine the number of ways the books can be arranged with the given constraints, consider the following steps:1. **Total Copies**: There are 10 identical copies of the book.2. **Constraints**: - At least 2 books must be in the library. - At least 2 books must be checked out.3. **Possible Configurations**: - The number of books in the library can range from 2 to 8 (inclusive). This ensures that at least 2 books are checked out (since 10 - 8 = 2) and at least 2 books are in the library.4. **Counting the Configurations**: - For each valid number of books in the library (2 through 8), there is exactly one corresponding configuration since the books are identical.5. **Total Number of Ways**: - There are 7 possible configurations (2, 3, 4, 5, 6, 7, 8 books in the library).Therefore, the number of different ways is 7.