💡First, I need to find the value of ( g(1) ). Given that ( g(x) = Bx^2 ), substituting ( x = 1 ) gives:[g(1) = B times 1^2 = B]Next, I substitute ( g(1) = B ) into the function ( f(x) ). The function ( f(x) ) is defined as ( f(x) = Ax^2 - 3B^2 ), so:[f(g(1)) = f(B) = A times B^2 - 3B^2]According to the problem, ( f(g(1)) = 0 ). Therefore, I set up the equation:[AB^2 - 3B^2 = 0]I can factor out ( B^2 ) from the equation:[B^2(A - 3) = 0]Since ( B neq 0 ), ( B^2 ) is also not zero. This implies that the factor ( A - 3 ) must be zero:[A - 3 = 0][A = 3]Thus, the value of ( A ) in terms of ( B ) is 3.

💡First, I need to determine how often James takes a water break. I know that James works for a total of 240 minutes and takes a sitting break every 120 minutes. To find out how many sitting breaks James takes, I divide the total working time by the duration between sitting breaks:[text{Number of sitting breaks} = frac{240}{120} = 2]James takes 10 more water breaks than sitting breaks. Therefore, the number of water breaks he takes is:[text{Number of water breaks} = 2 + 10 = 12]Finally, to find out how often James takes a water break, I divide the total working time by the number of water breaks:[text{Duration between water breaks} = frac{240}{12} = 20 text{ minutes}]

💡Okay, so I have this problem here: If (7 = k cdot 3^r) and (49 = k cdot 9^r), then what is (r)? The options are A. (log_5 7), B. (log_7 3), C. (log_3 7), D. (log_2 7), and E. (7).Alright, let me try to figure this out step by step. I remember that when you have equations with exponents, sometimes you can manipulate them to solve for the variable. Here, we have two equations with the same (k) and (r), so maybe I can eliminate (k) by dividing or something.First, let me write down the equations again:1. (7 = k cdot 3^r)2. (49 = k cdot 9^r)Hmm, okay. I notice that 49 is 7 squared, and 9 is 3 squared. Maybe that can help me relate the two equations.Let me try to express both equations in terms of (k). From the first equation, I can solve for (k):(k = frac{7}{3^r})Similarly, from the second equation:(k = frac{49}{9^r})Now, since both expressions equal (k), I can set them equal to each other:(frac{7}{3^r} = frac{49}{9^r})Alright, now I need to solve for (r). Let me simplify this equation. I know that (9^r) is the same as ((3^2)^r), which is (3^{2r}). So, I can rewrite the equation as:(frac{7}{3^r} = frac{49}{3^{2r}})Now, to make it easier, I can cross-multiply to eliminate the denominators:(7 cdot 3^{2r} = 49 cdot 3^r)Hmm, okay. Let me simplify this. I can divide both sides by 7 to make the numbers smaller:(3^{2r} = 7 cdot 3^r)Now, this looks a bit tricky, but maybe I can factor out (3^r) from both sides. Let me try that:(3^{2r} = 7 cdot 3^r)Dividing both sides by (3^r) (assuming (3^r neq 0), which it isn't because exponential functions are always positive):(3^{2r} / 3^r = 7)Simplifying the left side using the properties of exponents ((a^{m}/a^{n} = a^{m-n})):(3^{2r - r} = 7)Which simplifies to:(3^{r} = 7)Okay, so now I have (3^r = 7). I need to solve for (r). To do this, I can take the logarithm of both sides. Since the base is 3, I'll use logarithm base 3.Taking (log_3) of both sides:(log_3(3^r) = log_3(7))Simplifying the left side, since (log_b(b^x) = x):(r = log_3(7))So, (r) is equal to (log_3 7). Let me check the options: option C is (log_3 7), so that must be the answer.Wait, just to make sure I didn't make any mistakes, let me go through the steps again.Starting with the two equations:1. (7 = k cdot 3^r)2. (49 = k cdot 9^r)Express (k) from both:1. (k = 7 / 3^r)2. (k = 49 / 9^r)Set equal:(7 / 3^r = 49 / 9^r)Simplify (9^r) as (3^{2r}):(7 / 3^r = 49 / 3^{2r})Cross-multiply:(7 cdot 3^{2r} = 49 cdot 3^r)Divide both sides by 7:(3^{2r} = 7 cdot 3^r)Divide both sides by (3^r):(3^{r} = 7)Take (log_3) of both sides:(r = log_3 7)Yep, that seems correct. So, the answer is C.

💡Alright, so I've got this problem here about sets, and I need to figure out how many sets X satisfy the condition that A is a subset of X, and X is a subset of M. Let me try to break this down step by step.First, let's parse the given information:- The set M has 10 elements, so card(M) = 10.- A and B are both subsets of M.- A and B are disjoint, meaning they have no elements in common (A ∩ B = ∅).- The number of elements in A is 2, so card(A) = 2.- The number of elements in B is 3, so card(B) = 3.We need to find the number of sets X such that A is a subset of X and X is a subset of M. So, A ⊆ X ⊆ M.Okay, so X has to include all elements of A, and it can't include any elements outside of M. But how many such X are there?Let me think about this. Since A is a subset of X, X must contain all the elements of A. Additionally, X can contain any combination of the remaining elements in M that are not in A. So, the number of possible X's depends on how many elements are left in M after removing A.Given that M has 10 elements and A has 2 elements, the number of elements not in A is 10 - 2 = 8. These 8 elements can either be included in X or not. So, for each of these 8 elements, there are two choices: either include it in X or exclude it.This sounds like a classic combinatorial problem where the number of subsets is 2 raised to the number of elements. So, for 8 elements, the number of subsets is 2^8.Calculating that, 2^8 is 256. So, does that mean there are 256 possible sets X?Wait, hold on. Let me make sure I'm not missing something. The problem mentions set B as well, with card(B) = 3 and B is a subset of M, and A and B are disjoint. Does this affect the number of possible X's?Hmm, since A and B are disjoint, B is entirely separate from A. So, B is a subset of M, but it doesn't overlap with A. So, when considering the elements not in A, which are 8 in total, B is part of that. But does B have any restrictions on whether its elements can be in X or not?The problem only specifies that A is a subset of X, but it doesn't say anything about B. So, B's elements can be either included or excluded from X freely, just like the other elements not in A.Therefore, the presence of B doesn't impose any additional constraints on X. So, my initial thought was correct: the number of possible X's is 2^8, which is 256.But just to double-check, let's think about it another way. If we fix A as a subset of X, then X can be thought of as A union some subset of the remaining elements in M. The remaining elements are M A, which has 8 elements. The number of subsets of M A is 2^8, so the number of possible X's is indeed 256.Another way to think about it is to consider each element in M. For each element, we decide whether it's in X or not. However, since A must be a subset of X, all elements of A must be included. So, for the 2 elements in A, their inclusion is fixed. For the remaining 8 elements, each can be independently included or excluded. So, the total number of possibilities is 2^8 = 256.I think that's solid. The mention of set B might have thrown me off initially, but since there's no restriction on B, it doesn't affect the count. So, yeah, 256 is the number of such sets X.

💡Okay, so I have this problem about matrices and determinants. Let me try to understand what it's asking. We have a matrix ( A ) in ( M_n(mathbb{C}) ), which means it's an ( n times n ) complex matrix. There's also a complex number ( a ) such that ( A - A^* = 2aI_n ). Here, ( A^* ) is the conjugate transpose of ( A ), and ( I_n ) is the identity matrix. Part (i) asks me to show that the absolute value of the determinant of ( A ) is at least ( |a|^n ). Part (ii) is about showing that if the absolute value of the determinant equals ( |a|^n ), then ( A ) must be equal to ( aI_n ).Alright, let's start with part (i). First, I know that ( A^* ) is the conjugate transpose, so ( A - A^* ) is a skew-Hermitian matrix because ( (A - A^*)^* = A^* - A = -(A - A^*) ). Skew-Hermitian matrices have eigenvalues that are purely imaginary. Given that ( A - A^* = 2aI_n ), this tells me that ( 2aI_n ) is skew-Hermitian. So, ( (2aI_n)^* = -2aI_n ). Since the conjugate transpose of ( 2aI_n ) is ( 2overline{a}I_n ), we have ( 2overline{a}I_n = -2aI_n ). Dividing both sides by 2, we get ( overline{a}I_n = -aI_n ). This implies ( overline{a} = -a ), so ( a ) must be purely imaginary. Let me write ( a = ibeta ) where ( beta ) is a real number. So, ( A - A^* = 2ibeta I_n ).Now, since ( A ) is a complex matrix, I wonder if it's normal. A matrix is normal if it commutes with its conjugate transpose, i.e., ( AA^* = A^*A ). Let me check:( AA^* = A(A^*) ) and ( A^*A = (A^*)A ). If ( A - A^* ) is a multiple of the identity matrix, does that imply ( A ) is normal?Wait, if ( A - A^* = 2ibeta I_n ), then ( A = A^* + 2ibeta I_n ). Let's compute ( AA^* ) and ( A^*A ):( AA^* = (A^* + 2ibeta I_n)A^* = A^*A^* + 2ibeta A^* )( A^*A = A^*(A^* + 2ibeta I_n) = A^*A^* + 2ibeta A^* )So, ( AA^* = A^*A ). Therefore, ( A ) is normal. That's good because normal matrices are diagonalizable by a unitary matrix.So, there exists a unitary matrix ( U ) such that ( U^*AU = Lambda ), where ( Lambda ) is a diagonal matrix with the eigenvalues of ( A ). Since ( A ) is normal, its eigenvalues are well-behaved.Given that ( A ) is diagonalizable, let's write ( A = U Lambda U^* ). Then, ( A^* = (U Lambda U^*)^* = U Lambda^* U^* ), where ( Lambda^* ) is the conjugate transpose of ( Lambda ). Since ( Lambda ) is diagonal, ( Lambda^* ) is just the conjugate of each diagonal entry.From ( A - A^* = 2ibeta I_n ), substituting the diagonalized forms:( U Lambda U^* - U Lambda^* U^* = 2ibeta U U^* )Simplify:( U (Lambda - Lambda^*) U^* = 2ibeta I_n )Since ( U ) is unitary, ( U^*U = I_n ). So, multiplying both sides by ( U^* ) on the left and ( U ) on the right:( Lambda - Lambda^* = 2ibeta I_n )This means that each diagonal entry of ( Lambda - Lambda^* ) is ( 2ibeta ). Let me denote the eigenvalues of ( A ) as ( lambda_k ). Then, ( lambda_k - overline{lambda_k} = 2ibeta ).But ( lambda_k - overline{lambda_k} = 2i text{Im}(lambda_k) ). So, ( 2i text{Im}(lambda_k) = 2ibeta ), which implies ( text{Im}(lambda_k) = beta ) for all ( k ).Therefore, each eigenvalue ( lambda_k ) of ( A ) has an imaginary part equal to ( beta ). So, we can write each eigenvalue as ( lambda_k = alpha_k + ibeta ), where ( alpha_k ) is real.Now, the determinant of ( A ) is the product of its eigenvalues:( det(A) = prod_{k=1}^n lambda_k = prod_{k=1}^n (alpha_k + ibeta) )Taking the absolute value:( |det(A)| = prod_{k=1}^n |alpha_k + ibeta| )The modulus of each term ( |alpha_k + ibeta| = sqrt{alpha_k^2 + beta^2} ). So,( |det(A)| = prod_{k=1}^n sqrt{alpha_k^2 + beta^2} )Now, I need to show that this product is at least ( |a|^n ). Since ( a = ibeta ), ( |a| = |beta| ). So, ( |a|^n = |beta|^n ).Therefore, I need to show:( prod_{k=1}^n sqrt{alpha_k^2 + beta^2} geq |beta|^n )This seems like applying the AM-GM inequality or recognizing that each ( sqrt{alpha_k^2 + beta^2} geq |beta| ).Indeed, for each ( k ), ( sqrt{alpha_k^2 + beta^2} geq |beta| ) because ( alpha_k^2 geq 0 ). So, the product of these terms is at least the product of ( |beta| ) n times, which is ( |beta|^n ).Therefore, ( |det(A)| geq |a|^n ).Alright, that seems to work for part (i). Now, moving on to part (ii).Part (ii) says that if ( |det(A)| = |a|^n ), then ( A = aI_n ).From part (i), we have ( |det(A)| = prod_{k=1}^n sqrt{alpha_k^2 + beta^2} ). If this equals ( |beta|^n ), then each term in the product must be equal to ( |beta| ). Because if any ( sqrt{alpha_k^2 + beta^2} > |beta| ), then the product would be greater than ( |beta|^n ). So, equality holds only if each ( sqrt{alpha_k^2 + beta^2} = |beta| ).Which implies ( alpha_k^2 + beta^2 = beta^2 ) for each ( k ). Therefore, ( alpha_k^2 = 0 ), so ( alpha_k = 0 ) for all ( k ).Thus, each eigenvalue ( lambda_k = 0 + ibeta = ibeta = a ) (since ( a = ibeta )).Therefore, all eigenvalues of ( A ) are equal to ( a ). Since ( A ) is diagonalizable (as it's normal) and all its eigenvalues are equal, ( A ) must be a scalar multiple of the identity matrix. Specifically, ( A = aI_n ).So, that completes part (ii).Wait, let me double-check. If all eigenvalues are equal and the matrix is diagonalizable, then it's similar to a diagonal matrix with all entries equal, which is just ( aI_n ). Since ( A ) is diagonalizable by a unitary matrix, it's actually unitarily similar to ( aI_n ), which means ( A = aI_n ).Yes, that makes sense. So, if the determinant's absolute value is exactly ( |a|^n ), then all the ( alpha_k ) must be zero, forcing all eigenvalues to be ( a ), hence ( A ) is ( aI_n ).I think that covers both parts. I should probably write this up more formally, but the reasoning seems solid.**Final Answer**(i) boxed{|det A| geq |a|^n}(ii) boxed{A = aI_n}

💡Okay, so I have this function f(x) defined piecewise. It's 1 minus x when x is less than or equal to 1, and it's negative x squared plus a when x is greater than 1. The question is asking for the range of the real number a that makes this function monotonically decreasing on all real numbers. The options are A: (-∞, 0], B: [0, ∞), C: (-∞, 1], and D: [1, ∞).Alright, first, I need to recall what it means for a function to be monotonically decreasing. A function is monotonically decreasing if, as x increases, f(x) does not increase. In other words, for any two points x1 and x2 where x1 < x2, f(x1) ≥ f(x2). So, the function should never go up as x increases.Now, looking at the function f(x), it's defined differently for x ≤ 1 and x > 1. So, I should check the behavior of each piece separately and then see how they connect at x = 1.Starting with the first piece: f(x) = 1 - x for x ≤ 1. Let's find its derivative to see if it's decreasing. The derivative f'(x) is -1. Since the derivative is negative, this part of the function is indeed decreasing for all x ≤ 1.Next, the second piece: f(x) = -x² + a for x > 1. Let's compute its derivative as well. The derivative f'(x) is -2x. Now, for x > 1, -2x is negative because x is positive and multiplied by -2. So, this part is also decreasing for all x > 1.So, both pieces of the function are decreasing on their respective intervals. But just because each piece is decreasing doesn't necessarily mean the entire function is decreasing on all real numbers. We need to ensure that the function doesn't have a jump or a point where it starts increasing at x = 1.To check this, we need to ensure that the function is continuous at x = 1 and that the value of the function just to the left of x = 1 is greater than or equal to the value just to the right of x = 1. This way, there's no increase at the point where the two pieces meet.First, let's find the value of the function at x = 1 from the left side. Using the first piece, f(1) = 1 - 1 = 0.Now, let's find the value of the function at x = 1 from the right side. Using the second piece, f(1) = -(1)² + a = -1 + a.For the function to be monotonically decreasing, the value from the left (which is 0) must be greater than or equal to the value from the right (which is -1 + a). So, we set up the inequality:0 ≥ -1 + aNow, solving for a:0 ≥ -1 + a Add 1 to both sides: 1 ≥ a Which can be rewritten as: a ≤ 1So, a must be less than or equal to 1 for the function to be monotonically decreasing on all real numbers.Looking back at the options, option C is (-∞, 1], which matches our result. So, the correct answer should be option C.But wait, just to make sure I didn't miss anything, let me think again. The function is decreasing on both sides, but the key was ensuring that at the point where they meet, the function doesn't jump up. So, the left limit at x=1 is 0, and the right limit is -1 + a. To prevent an upward jump, 0 must be greater than or equal to -1 + a, which simplifies to a ≤ 1. Yep, that seems right.I don't think I made any mistakes here. The derivative checks were straightforward, and the continuity condition at x=1 was correctly applied. So, I'm confident that the range of a is (-∞, 1], which is option C.**Final Answer**boxed{C}