answer：Okay, so I have this problem where I'm given a function f(x) = 5 sin(x - π/6), and I know that f(α) = f(β) = 1, with 0 < α < β < 2π. I need to find cos(β - α). Hmm, let me think about how to approach this step by step.First, since f(α) = 1, I can write the equation:5 sin(α - π/6) = 1Similarly, for β:5 sin(β - π/6) = 1So, if I divide both sides by 5, I get:sin(α - π/6) = 1/5andsin(β - π/6) = 1/5Alright, so both α - π/6 and β - π/6 have the same sine value, which is 1/5. Since sine is positive in the first and second quadrants, that means α - π/6 is in the first quadrant (acute angle) and β - π/6 is in the second quadrant (obtuse angle). That makes sense because α is less than β, so their adjusted angles would be in different quadrants.Now, I need to find cos(β - α). Hmm, maybe I can use the cosine of difference identity. Let me recall that:cos(β - α) = cos[(β - π/6) - (α - π/6)]Which simplifies to:cos[(β - π/6) - (α - π/6)] = cos(β - π/6)cos(α - π/6) + sin(β - π/6)sin(α - π/6)Okay, so I need to find cos(β - π/6) and cos(α - π/6). I already know sin(α - π/6) = 1/5 and sin(β - π/6) = 1/5. Let me find the cosines using the Pythagorean identity.For α - π/6, which is in the first quadrant:cos^2(α - π/6) = 1 - sin^2(α - π/6) = 1 - (1/5)^2 = 1 - 1/25 = 24/25So, cos(α - π/6) = sqrt(24/25) = (2 sqrt(6))/5Since it's in the first quadrant, cosine is positive.For β - π/6, which is in the second quadrant:cos^2(β - π/6) = 1 - sin^2(β - π/6) = 24/25 as wellBut since it's in the second quadrant, cosine is negative, so:cos(β - π/6) = - (2 sqrt(6))/5Alright, now I can plug these into the cosine of difference formula:cos(β - α) = cos(β - π/6) cos(α - π/6) + sin(β - π/6) sin(α - π/6)Let's compute each term:First term: cos(β - π/6) cos(α - π/6) = (-2 sqrt(6)/5) * (2 sqrt(6)/5) = (-4 * 6)/25 = (-24)/25Second term: sin(β - π/6) sin(α - π/6) = (1/5) * (1/5) = 1/25So, adding them together:cos(β - α) = (-24/25) + (1/25) = (-23)/25Hmm, so the answer should be -23/25. Let me double-check my steps to make sure I didn't make any mistakes.1. Started with f(α) = f(β) = 1, leading to sin(α - π/6) = sin(β - π/6) = 1/5. That seems right.2. Determined that α - π/6 is in the first quadrant and β - π/6 is in the second quadrant because α < β. That makes sense since sine is positive in both quadrants, but angles would be in different quadrants to have different solutions.3. Calculated cos(α - π/6) as positive 2 sqrt(6)/5 and cos(β - π/6) as negative 2 sqrt(6)/5. That seems correct using the Pythagorean identity.4. Applied the cosine of difference identity correctly, substituting the values. The multiplication of the cosines gave -24/25, and the multiplication of the sines gave 1/25. Adding them together gives -23/25.I think that's all correct. So, the answer should be -23/25, which is option B.**Final Answer**boxed{B}

answer：Alright, so I've got this problem here: real numbers a, b, c are all greater than 0, and the product (a + c) multiplied by (a + b) equals 6 minus 2 times the square root of 5. I need to find the minimum value of 2a + b + c. The options are given as A: sqrt(5) - 1, B: sqrt(5) + 1, C: 2sqrt(5) + 2, and D: 2sqrt(5) - 2.Hmm, okay. Let me try to unpack this step by step. First, I know that a, b, c are all positive real numbers. So, all these expressions like (a + c) and (a + b) are going to be positive as well.The equation given is (a + c)(a + b) = 6 - 2sqrt(5). I need to find the minimum value of 2a + b + c. Hmm, 2a + b + c. Let me see if I can express this in terms of (a + c) and (a + b) because that's what the equation involves.Wait, 2a + b + c can be rewritten as (a + c) + (a + b). Let me check that: (a + c) + (a + b) = a + c + a + b = 2a + b + c. Yes, that works. So, 2a + b + c is equal to (a + c) + (a + b). That seems useful because the given equation is about the product of these two terms.So, if I let x = a + c and y = a + b, then the given equation becomes x * y = 6 - 2sqrt(5), and I need to minimize x + y. So, the problem reduces to minimizing x + y given that x * y = 6 - 2sqrt(5) and x, y > 0.This seems like a classic optimization problem where I can apply the AM-GM inequality. The Arithmetic Mean - Geometric Mean inequality states that for positive real numbers, the arithmetic mean is always greater than or equal to the geometric mean. So, for x and y, we have:(x + y)/2 >= sqrt(x * y)Multiplying both sides by 2 gives:x + y >= 2 * sqrt(x * y)Since x * y is given as 6 - 2sqrt(5), substituting that in:x + y >= 2 * sqrt(6 - 2sqrt(5))So, the minimum value of x + y is 2 * sqrt(6 - 2sqrt(5)). Therefore, the minimum value of 2a + b + c is 2 * sqrt(6 - 2sqrt(5)).Now, I need to compute sqrt(6 - 2sqrt(5)) and simplify it if possible. Let me see if sqrt(6 - 2sqrt(5)) can be expressed in a simpler form.Suppose sqrt(6 - 2sqrt(5)) can be written as sqrt(a) - sqrt(b), where a and b are positive real numbers. Let's square both sides:(sqrt(a) - sqrt(b))^2 = a + b - 2sqrt(ab) = 6 - 2sqrt(5)So, we have:a + b = 6and2sqrt(ab) = 2sqrt(5)Dividing both sides of the second equation by 2:sqrt(ab) = sqrt(5)Squaring both sides:ab = 5So, now we have a system of equations:a + b = 6ab = 5This is a system that can be solved for a and b. Let me set up the quadratic equation:Let a and b be the roots of x^2 - (a + b)x + ab = 0, which becomes x^2 - 6x + 5 = 0.Solving this quadratic equation:x = [6 ± sqrt(36 - 20)] / 2 = [6 ± sqrt(16)] / 2 = [6 ± 4] / 2So, the solutions are:x = (6 + 4)/2 = 10/2 = 5andx = (6 - 4)/2 = 2/2 = 1So, a and b are 5 and 1. Therefore, sqrt(6 - 2sqrt(5)) = sqrt(5) - sqrt(1) = sqrt(5) - 1.Therefore, sqrt(6 - 2sqrt(5)) = sqrt(5) - 1.So, going back to the expression for the minimum value:2 * sqrt(6 - 2sqrt(5)) = 2 * (sqrt(5) - 1) = 2sqrt(5) - 2.So, the minimum value of 2a + b + c is 2sqrt(5) - 2, which is option D.Let me just double-check my steps to make sure I didn't make any mistakes.1. I expressed 2a + b + c as (a + c) + (a + b), which seems correct.2. I set x = a + c and y = a + b, leading to x * y = 6 - 2sqrt(5) and needing to minimize x + y.3. Applied AM-GM inequality correctly, leading to x + y >= 2sqrt(x * y).4. Calculated sqrt(6 - 2sqrt(5)) by assuming it's sqrt(a) - sqrt(b), squared it, and solved for a and b, which gave me a = 5 and b = 1.5. Substituted back to find sqrt(6 - 2sqrt(5)) = sqrt(5) - 1, which seems correct.6. Multiplied by 2 to get 2sqrt(5) - 2, which is option D.Everything seems to check out. I don't see any errors in my reasoning, so I'm confident that the answer is D: 2sqrt(5) - 2.**Final Answer**boxed{D}

answer：To determine how much the graph of ( y = 2sin(2x) ) needs to be shifted to obtain the graph of ( f(x) = 2sinleft(2x + frac{pi}{6}right) ), I can rewrite the function ( f(x) ) in a form that makes the shift more apparent.Starting with the given function:[f(x) = 2sinleft(2x + frac{pi}{6}right)]I can factor out the coefficient of ( x ) inside the sine function:[f(x) = 2sinleft(2left(x + frac{pi}{12}right)right)]This shows that the graph of ( y = 2sin(2x) ) is shifted to the left by ( frac{pi}{12} ) units.Therefore, the correct answer is:[boxed{frac{pi}{12}}]

answer：Alright, so I'm trying to figure out how many prime numbers are in this sequence that Xiaoming created. The sequence starts with the digit 2, and then keeps adding the digits 0, 1, and 6 in that order repeatedly. So the sequence goes like this: 2, 20, 201, 2016, 20162, 201620, 2016201, 20162016, 201620162, and so on. First off, I know that a prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So, I need to check each number in this sequence to see if it's prime.Starting with the first number, which is 2. That's definitely a prime number because it's only divisible by 1 and itself. So, that's one prime number right there.Next up is 20. Hmm, 20. I know that 20 is an even number, so it's divisible by 2. That means it's not a prime number. So, 20 is out.Moving on to 201. Okay, 201. I'm not sure off the top of my head if this is prime. Maybe I can check by dividing it by some small primes. Let's see, 201 divided by 3. Adding the digits: 2 + 0 + 1 = 3, which is divisible by 3. So, 201 divided by 3 is 67. So, 201 is 3 times 67, which means it's not a prime number.Next is 2016. That's a pretty big number. I know that 2016 is even, so it's divisible by 2. Therefore, it's not a prime number. Also, the sum of its digits is 2 + 0 + 1 + 6 = 9, which is divisible by 3, so it's also divisible by 3. So, definitely not prime.Then we have 20162. Again, this is an even number because it ends with a 2, so it's divisible by 2. Not prime.Next is 201620. This one ends with a 0, so it's divisible by 10, which means it's not prime. Also, the sum of the digits is 2 + 0 + 1 + 6 + 2 + 0 = 11, which isn't divisible by 3, but since it's already divisible by 10, it's not prime.Moving on to 2016201. This one is a bit trickier. Let me see if it's divisible by any small primes. It doesn't end with an even number or a 5, so it's not divisible by 2 or 5. Let's check divisibility by 3: 2 + 0 + 1 + 6 + 2 + 0 + 1 = 12, which is divisible by 3. So, 2016201 divided by 3 is 672067. Therefore, it's not a prime number.Next is 20162016. This is another even number, so it's divisible by 2. Not prime. Also, the sum of the digits is 2 + 0 + 1 + 6 + 2 + 0 + 1 + 6 = 18, which is divisible by 9, so it's also divisible by 9. Definitely not prime.Then we have 201620162. Again, this ends with a 2, so it's even and divisible by 2. Not prime.At this point, I notice a pattern. After the first number, which is 2, all the subsequent numbers are even or divisible by 3 or 5. Since the sequence keeps appending the digits 0, 1, 6, which are all even or 0, the numbers keep getting larger and are consistently even or have other divisors. I also recall that as numbers get larger, especially in a sequence like this, the likelihood of them being prime decreases because there are more potential divisors. So, it's unlikely that any of these larger numbers are prime.To be thorough, I could check a few more numbers in the sequence, but I suspect they'll all follow the same pattern. For example, the next number after 201620162 would be 2016201620, which ends with a 0, making it divisible by 10. The one after that would be 20162016201, which I can check for divisibility by 3: 2 + 0 + 1 + 6 + 2 + 0 + 1 + 6 + 2 + 0 + 1 = 15, which is divisible by 3, so it's not prime.Given all this, it seems like the only prime number in the sequence is the very first one, which is 2. All the other numbers are composite because they're either even, divisible by 3, or have other factors.So, after checking each number step by step and considering the patterns in divisibility, I conclude that there's only one prime number in Xiaoming's sequence.

answer：Okay, so I have this function f(x) defined piecewise. It's 2^{x+1} when x is less than or equal to 0, and it's -log₂(x+1) + 2 when x is greater than 0. I need to find f(6 - a) given that f(a) = -1. Hmm, let me break this down step by step.First, let's understand what f(a) = -1 means. Since f(x) is defined differently for x ≤ 0 and x > 0, I need to figure out which part of the function applies to a. So, I should check both cases.Case 1: If a ≤ 0, then f(a) = 2^{a+1}. I need to see if 2^{a+1} can equal -1. But wait, 2 raised to any power is always positive, right? So 2^{a+1} can never be negative. That means f(a) can't be -1 if a is less than or equal to 0. So, this case doesn't work.Case 2: If a > 0, then f(a) = -log₂(a+1) + 2. Okay, so let's set that equal to -1 and solve for a.So, -log₂(a+1) + 2 = -1.Let me solve this equation step by step. First, subtract 2 from both sides:-log₂(a+1) = -1 - 2-log₂(a+1) = -3Now, multiply both sides by -1 to make it positive:log₂(a+1) = 3Alright, so log base 2 of (a+1) equals 3. To solve for a, I can rewrite this in exponential form. Remember that log_b(c) = d means b^d = c.So, 2^3 = a + 18 = a + 1Subtract 1 from both sides:a = 8 - 1a = 7Okay, so a is 7. Now, I need to find f(6 - a). Since a is 7, 6 - a is 6 - 7, which is -1.So, f(-1). Now, since -1 is less than or equal to 0, I use the first part of the piecewise function: f(x) = 2^{x+1}.Plugging in x = -1:f(-1) = 2^{-1 + 1} = 2^{0} = 1.So, f(6 - a) is 1. Looking at the options, that's option A.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. f(a) = -1. Since f(x) for x ≤ 0 is always positive, a must be greater than 0. So, I used the second part of the function correctly.2. Solved -log₂(a+1) + 2 = -1. Subtracted 2, got -log₂(a+1) = -3. Multiplied by -1, log₂(a+1) = 3. Converted to exponential form, a+1 = 8, so a=7. That seems right.3. Then, 6 - a = 6 -7 = -1. Since -1 ≤0, used the first part of the function: 2^{-1 +1}=2^0=1. Yep, that's correct.So, I think I did everything correctly. The answer should be 1, which is option A.**Final Answer**boxed{A}

answer：Okay, so I have this problem: "What is the smallest positive integer with exactly 12 positive factors?" Hmm, interesting. I remember that the number of positive factors a number has is related to its prime factorization. Let me try to recall how that works.First, if a number is expressed as a product of primes raised to exponents, like ( n = p_1^{a_1} cdot p_2^{a_2} cdot ldots cdot p_k^{a_k} ), then the total number of positive factors is given by multiplying one more than each of the exponents. So, the formula is ( (a_1 + 1)(a_2 + 1) ldots (a_k + 1) ). That makes sense because for each prime, you can choose how many times it appears in a factor, from 0 up to its exponent in the prime factorization.So, in this case, we want the number of positive factors to be exactly 12. That means we need to find exponents ( a_1, a_2, ldots, a_k ) such that when we add 1 to each and multiply them together, we get 12. Let me write down the possible ways to express 12 as a product of integers greater than 1, since each ( (a_i + 1) ) must be at least 2.12 can be factored in several ways:1. 12 = 12 × 12. 12 = 6 × 23. 12 = 4 × 34. 12 = 3 × 2 × 2These correspond to different forms of the prime factorization of the number we're looking for.Let's analyze each case:1. **Case 1: 12 = 12 × 1** This would mean that the number has a single prime factor raised to the 11th power because ( a_1 + 1 = 12 ) implies ( a_1 = 11 ). So, the number would be ( p^{11} ). The smallest such number would be ( 2^{11} = 2048 ). That's pretty large, so maybe there's a smaller number in another case.2. **Case 2: 12 = 6 × 2** Here, we have two prime factors. The exponents would be ( a_1 = 5 ) and ( a_2 = 1 ) because ( (5 + 1)(1 + 1) = 6 × 2 = 12 ). So, the number would be ( p_1^5 cdot p_2^1 ). To get the smallest number, we should use the smallest primes, which are 2 and 3. So, ( 2^5 cdot 3^1 = 32 × 3 = 96 ). That's smaller than 2048, but let's see if we can get even smaller.3. **Case 3: 12 = 4 × 3** This case involves two prime factors as well, but with exponents ( a_1 = 3 ) and ( a_2 = 2 ) because ( (3 + 1)(2 + 1) = 4 × 3 = 12 ). So, the number would be ( p_1^3 cdot p_2^2 ). Again, using the smallest primes, 2 and 3, we get ( 2^3 cdot 3^2 = 8 × 9 = 72 ). That's even smaller than 96.4. **Case 4: 12 = 3 × 2 × 2** This case involves three prime factors. The exponents would be ( a_1 = 2 ), ( a_2 = 1 ), and ( a_3 = 1 ) because ( (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12 ). So, the number would be ( p_1^2 cdot p_2^1 cdot p_3^1 ). Using the smallest primes, 2, 3, and 5, we get ( 2^2 cdot 3 cdot 5 = 4 × 3 × 5 = 60 ). That's smaller than 72.So, comparing all the cases:- Case 1: 2048- Case 2: 96- Case 3: 72- Case 4: 60The smallest number is 60. Let me double-check to make sure 60 indeed has exactly 12 factors.The prime factorization of 60 is ( 2^2 cdot 3^1 cdot 5^1 ). Using the formula for the number of factors: ( (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12 ). Yep, that's correct.Just to be thorough, let me list all the factors of 60 to ensure there are exactly 12:1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. That's 12 factors. Perfect.I think I've considered all possible cases and confirmed that 60 is indeed the smallest positive integer with exactly 12 positive factors.