💡First, I need to determine the total number of possible single-digit numbers and letters that can be used in the password. There are 10 possible single-digit numbers (0-9) and 26 capital letters in the English alphabet.Next, I'll calculate the probability for each part of the password:1. **First Digit (Even Number):** - There are 5 even single-digit numbers: 0, 2, 4, 6, 8. - Probability = Number of favorable outcomes / Total possible outcomes = 5/10 = 1/2.2. **Middle Character (Roman Capital Letter A-E):** - There are 5 specific letters: A, B, C, D, E. - Probability = 5/26.3. **Last Digit (Odd Number Greater Than 5):** - The odd numbers greater than 5 are 7 and 9. - Probability = 2/10 = 1/5.Finally, I'll multiply the probabilities of each independent event to find the overall probability of the password meeting all the specified criteria:Probability = (1/2) × (5/26) × (1/5) = 5/260 = 1/52.

💡Okay, so I have these five propositions about functions, and I need to figure out which ones are correct. Let me go through each one step by step.Starting with proposition ①: "The function ( y = |x| ) and the function ( y = (sqrt{x})^2 ) represent the same function." Hmm, I remember that for two functions to be the same, they need to have the same domain, the same codomain, and the same rule of assignment. Let me think about the domains here. The function ( y = |x| ) is defined for all real numbers because absolute value is defined everywhere. On the other hand, ( y = (sqrt{x})^2 ) is only defined for ( x geq 0 ) because the square root of a negative number isn't a real number. So even though both functions simplify to ( y = x ) when ( x ) is non-negative, their domains are different. Therefore, they can't be the same function. So proposition ① is incorrect.Moving on to proposition ②: "The graph of an odd function must pass through the origin of the Cartesian coordinate system." I recall that an odd function satisfies the condition ( f(-x) = -f(x) ) for all ( x ) in its domain. If I plug in ( x = 0 ), then ( f(-0) = -f(0) ), which simplifies to ( f(0) = -f(0) ). Adding ( f(0) ) to both sides gives ( 2f(0) = 0 ), so ( f(0) = 0 ). That means the point ( (0, 0) ) is on the graph of the function. So the graph must pass through the origin. Wait, but is this always true? Let me think of an example. The function ( f(x) = frac{1}{x} ) is odd because ( f(-x) = frac{1}{-x} = -frac{1}{x} = -f(x) ). However, ( f(0) ) is undefined here because division by zero isn't allowed. So in this case, the function doesn't pass through the origin because it's not defined there. Therefore, proposition ② is not necessarily true because some odd functions might not pass through the origin if they aren't defined at zero. So ② is incorrect.Next is proposition ③: "If the domain of the function ( f(x) ) is ([0, 2]), then the domain of the function ( f(2x) ) is ([0, 4])." Let me recall how function transformations affect the domain. If I have ( f(2x) ), it's a horizontal scaling of the original function. Specifically, replacing ( x ) with ( 2x ) compresses the graph horizontally by a factor of 2. So, if the original domain is ([0, 2]), then for ( f(2x) ) to be defined, ( 2x ) must lie within ([0, 2]). That means ( x ) must satisfy ( 0 leq 2x leq 2 ), which simplifies to ( 0 leq x leq 1 ). So the domain of ( f(2x) ) is actually ([0, 1]), not ([0, 4]). Therefore, proposition ③ is incorrect.Proposition ④: "The graph of the function ( y = 3(x - 1)^2 ) can be obtained by shifting the graph of ( y = 3x^2 ) to the right by one unit." I remember that shifting a function horizontally involves replacing ( x ) with ( x - h ) for a shift to the right by ( h ) units. So in this case, ( y = 3(x - 1)^2 ) is indeed a horizontal shift of ( y = 3x^2 ) to the right by 1 unit. Let me visualize it: the vertex of the parabola ( y = 3x^2 ) is at (0, 0), and shifting it right by 1 unit would move the vertex to (1, 0). That seems correct. So proposition ④ is correct.Finally, proposition ⑤: "Suppose the function ( f(x) ) is continuous on the interval ([a, b]) and ( f(a) cdot f(b) < 0 ), then the equation ( f(x) = 0 ) has at least one real root in the interval ([a, b])." This sounds familiar. I think it's related to the Intermediate Value Theorem. The theorem states that if a function is continuous on ([a, b]) and takes on values ( f(a) ) and ( f(b) ) at each end of the interval, then it also takes on any value between ( f(a) ) and ( f(b) ) at some point within the interval. If ( f(a) cdot f(b) < 0 ), that means ( f(a) ) and ( f(b) ) have opposite signs. Therefore, by the Intermediate Value Theorem, there must be some ( c ) in ([a, b]) where ( f(c) = 0 ). So proposition ⑤ is correct.To recap:- ①: Incorrect (different domains)- ②: Incorrect (some odd functions aren't defined at zero)- ③: Incorrect (domain scales down, not up)- ④: Correct (horizontal shift)- ⑤: Correct (Intermediate Value Theorem)So the correct propositions are ④ and ⑤.**Final Answer**The correct propositions are boxed{④⑤}.

💡To determine the fourth equation in the given pattern, we observe that each equation involves the sum of cubes of the first ( n ) natural numbers being equal to the square of their sum. 1. **Identify the Pattern:** - The first equation involves the sum of cubes from 1 to 2. - The second equation involves the sum of cubes from 1 to 3. - The third equation involves the sum of cubes from 1 to 4. - Following this pattern, the fourth equation should involve the sum of cubes from 1 to 5.2. **Calculate the Sum of Cubes:** [ 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} ]3. **Calculate the Sum of Natural Numbers:** [ 1 + 2 + 3 + 4 + 5 = 15 ]4. **Formulate the Equation:** [ 1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} = (1 + 2 + 3 + 4 + 5)^{2} = 15^{2} ]Therefore, the fourth equation in the pattern is:

💡First, I need to determine the initial number of books in the library. Let's denote this initial number as ( X ).On Monday, 124 books are taken out, so the number of books remaining in the library is ( X - 124 ).On Tuesday, 22 books are brought back, which means the number of books in the library becomes ( (X - 124) + 22 ).According to the problem, after these transactions, there are 234 books in the library. This gives us the equation:[(X - 124) + 22 = 234]Simplifying the equation:[X - 102 = 234]Solving for ( X ):[X = 234 + 102][X = 336]Therefore, the initial number of books in the library was 336.

💡Okay, so I have this problem about a geometric sequence. It says that in the sequence {a_n}, a₄ + a₈ equals -2, and I need to find the value of a₆(a₂ + 2a₆ + a₁₀). The options are A) 4, B) 6, C) 8, D) -9. Hmm, let me try to figure this out step by step.First, I remember that in a geometric sequence, each term is the previous term multiplied by a common ratio, r. So, the general form is a_n = a₁ * r^(n-1). That means I can express any term in the sequence in terms of a₁ and r. But maybe there's a smarter way to do this without having to find a₁ and r explicitly.The problem gives me a₄ + a₈ = -2. Let me write those terms using the general formula. So, a₄ = a₁ * r³ and a₈ = a₁ * r⁷. Adding them together, a₄ + a₈ = a₁ * r³ + a₁ * r⁷ = a₁ r³ (1 + r⁴) = -2. Hmm, that's one equation, but I don't know a₁ or r yet. Maybe I can find another equation or relate it to the expression I need to find.The expression I need to find is a₆(a₂ + 2a₆ + a₁₀). Let me break that down. First, a₆ is a term in the sequence, so a₆ = a₁ * r⁵. Similarly, a₂ = a₁ * r, a₆ is a₁ * r⁵, and a₁₀ = a₁ * r⁹. So, substituting these into the expression:a₆(a₂ + 2a₆ + a₁₀) = a₁ r⁵ (a₁ r + 2a₁ r⁵ + a₁ r⁹)Let me factor out a₁ from each term inside the parentheses:= a₁ r⁵ [a₁ (r + 2r⁵ + r⁹)]= a₁² r⁵ (r + 2r⁵ + r⁹)Hmm, that looks a bit complicated. Maybe I can factor out r from the terms inside the parentheses:= a₁² r⁵ * r (1 + 2r⁴ + r⁸)= a₁² r⁶ (1 + 2r⁴ + r⁸)Wait, 1 + 2r⁴ + r⁸ looks like a perfect square. Let me check: (1 + r⁴)² = 1 + 2r⁴ + r⁸. Yes! So, that simplifies to:= a₁² r⁶ (1 + r⁴)²Now, from the given information, I know that a₁ r³ (1 + r⁴) = -2. Let me write that down:a₁ r³ (1 + r⁴) = -2So, if I square both sides of this equation, I get:(a₁ r³ (1 + r⁴))² = (-2)²Which is:a₁² r⁶ (1 + r⁴)² = 4But wait, that's exactly the expression I have for a₆(a₂ + 2a₆ + a₁₀)! So, a₆(a₂ + 2a₆ + a₁₀) = 4.Let me double-check to make sure I didn't make any mistakes. I expressed each term in the sequence in terms of a₁ and r, substituted them into the expression, factored, recognized the perfect square, and then used the given equation to find the value. It seems to add up.So, the answer should be 4, which is option A.

💡Okay, so I have this problem about a function f(x) that's defined on the positive real numbers, right? It's a decreasing function, which means as x increases, f(x) decreases. The problem gives me two things: first, that for any x and y in (0, +∞), the function satisfies f(x + y) = f(x) + f(y) - 1. Second, it tells me that f(4) = 5. There are two parts to the problem. The first part is to find f(2), and the second part is to solve the inequality f(m - 2) ≤ 3. Let me try to tackle each part step by step.Starting with part (1): Find the value of f(2).Hmm, so I know that f is a function that satisfies this additive property, almost like a linear function but with a twist because of the "-1" term. Maybe I can use the given functional equation to express f(4) in terms of f(2). Let me try that.If I set x = y = 2, then x + y = 4. Plugging into the functional equation:f(4) = f(2) + f(2) - 1Simplify that:f(4) = 2f(2) - 1But I know that f(4) is given as 5. So:5 = 2f(2) - 1Now, solving for f(2):Add 1 to both sides:5 + 1 = 2f(2)6 = 2f(2)Divide both sides by 2:f(2) = 3Okay, that seems straightforward. So f(2) is 3. I think that's the answer for part (1). Let me just double-check my steps to make sure I didn't make a mistake.I used the functional equation with x = y = 2, substituted f(4) = 5, solved for f(2), and got 3. That seems correct. I don't see any errors here.Moving on to part (2): Solve the inequality f(m - 2) ≤ 3.Alright, so I need to find all m such that when I plug m - 2 into the function f, the result is less than or equal to 3. Since f is a decreasing function, that might help me relate m - 2 to some other value where f is known.From part (1), I know that f(2) = 3. So the inequality f(m - 2) ≤ 3 is equivalent to f(m - 2) ≤ f(2). Since f is decreasing, this should imply that m - 2 is greater than or equal to 2. Let me explain why.In general, for a decreasing function, if a ≤ b, then f(a) ≥ f(b). So, if I have f(a) ≤ f(b), that would imply that a ≥ b. Applying this to my inequality:f(m - 2) ≤ f(2) implies that m - 2 ≥ 2.So, solving for m:m - 2 ≥ 2Add 2 to both sides:m ≥ 4Therefore, the solution to the inequality is m ≥ 4.Wait, let me make sure I didn't flip the inequality sign or anything. Since f is decreasing, as x increases, f(x) decreases. So, if f(m - 2) is less than or equal to f(2), that means m - 2 must be greater than or equal to 2 because f is decreasing. Yeah, that makes sense. If m - 2 were less than 2, then f(m - 2) would be greater than f(2), which is 3, but we want f(m - 2) to be less than or equal to 3, so m - 2 has to be at least 2. So, m has to be at least 4.Is there anything else I need to consider? Let me think. The function f is defined on (0, +∞), so m - 2 must also be in (0, +∞). That means m - 2 > 0, so m > 2. But since we have m ≥ 4, which is already greater than 2, that condition is automatically satisfied. So, I don't need to worry about m being too small; the inequality m ≥ 4 already ensures that m - 2 is positive.Let me also check if the function f is linear or has some other form. Maybe I can find a general expression for f(x). The functional equation f(x + y) = f(x) + f(y) - 1 looks similar to Cauchy's functional equation, which is f(x + y) = f(x) + f(y). The solution to Cauchy's equation is linear, f(x) = kx, but here we have an extra constant term. Maybe I can adjust for that.Let me suppose that f(x) = kx + c, where k and c are constants. Let's plug this into the functional equation:f(x + y) = k(x + y) + c = kx + ky + cOn the other hand, f(x) + f(y) - 1 = (kx + c) + (ky + c) - 1 = kx + ky + 2c - 1Set these equal:kx + ky + c = kx + ky + 2c - 1Simplify:c = 2c - 1Subtract c from both sides:0 = c - 1So, c = 1Therefore, the function is f(x) = kx + 1.Now, we can use the given information f(4) = 5 to find k.f(4) = 4k + 1 = 5Subtract 1:4k = 4Divide by 4:k = 1So, the function is f(x) = x + 1.Wait, but hold on. The function f(x) = x + 1 is actually increasing, not decreasing. But the problem states that f is a decreasing function. That's a contradiction. Did I make a mistake somewhere?Hmm, let's see. I assumed f(x) is linear, f(x) = kx + c, and found that c = 1, and k = 1. So f(x) = x + 1, which is increasing. But the problem says f is decreasing. That can't be right. So, maybe my assumption that f is linear is incorrect, or perhaps there's another approach.Wait, maybe f(x) isn't linear. Let me think again. The functional equation is f(x + y) = f(x) + f(y) - 1. Let me rearrange this equation:f(x + y) - 1 = f(x) - 1 + f(y) - 1Let me define a new function g(x) = f(x) - 1. Then, substituting into the equation:g(x + y) = g(x) + g(y)Ah, so now the functional equation becomes Cauchy's equation for g(x). So, g(x + y) = g(x) + g(y). The solutions to Cauchy's equation are linear functions of the form g(x) = kx, provided that g is continuous or monotonic, which in this case, since f is decreasing, g must also be decreasing, hence linear.Therefore, g(x) = kx, so f(x) = kx + 1.But wait, earlier, when I plugged in f(4) = 5, I found k = 1, which would make f(x) = x + 1, which is increasing. But the problem says f is decreasing. So, that suggests that k must be negative. Let me check.If f(x) = kx + 1 is decreasing, then k must be negative. So, let's recast the problem.Given f(x) = kx + 1, and f(4) = 5.So, f(4) = 4k + 1 = 5So, 4k = 4k = 1But that's positive, which contradicts f being decreasing. Hmm, that's a problem.Wait, maybe I made a wrong assumption in defining g(x). Let me double-check.I set g(x) = f(x) - 1, so f(x) = g(x) + 1.Then, the functional equation f(x + y) = f(x) + f(y) - 1 becomes:g(x + y) + 1 = (g(x) + 1) + (g(y) + 1) - 1Simplify the right-hand side:g(x) + 1 + g(y) + 1 - 1 = g(x) + g(y) + 1So, the equation becomes:g(x + y) + 1 = g(x) + g(y) + 1Subtract 1 from both sides:g(x + y) = g(x) + g(y)So, yes, that's correct. So, g satisfies Cauchy's equation. Therefore, g(x) is linear, g(x) = kx.Thus, f(x) = kx + 1.But f(x) is decreasing, so k must be negative. However, when I plug in f(4) = 5, I get k = 1, which is positive. That's a contradiction. So, something is wrong here.Wait, maybe I misapplied the functional equation. Let me check again.Given f(x + y) = f(x) + f(y) - 1.Let me define g(x) = f(x) - c, where c is a constant to be determined.Then, f(x) = g(x) + c.Substitute into the functional equation:g(x + y) + c = (g(x) + c) + (g(y) + c) - 1Simplify:g(x + y) + c = g(x) + g(y) + 2c - 1Subtract c from both sides:g(x + y) = g(x) + g(y) + c - 1We want this to be Cauchy's equation, so we set c - 1 = 0, which gives c = 1.So, g(x + y) = g(x) + g(y). Therefore, g is linear, g(x) = kx.Thus, f(x) = kx + 1.But as before, f(4) = 5 implies k = 1, which is positive, contradicting f being decreasing.Hmm, so maybe there's no such function f(x) that is decreasing and satisfies the given conditions? But the problem says it's given, so perhaps I made a wrong assumption.Wait, maybe f(x) is not linear? But the functional equation f(x + y) = f(x) + f(y) - 1 suggests that f is linear up to a constant. But if f is decreasing, then the slope must be negative, but f(4) = 5 would require the slope to be positive. That seems impossible.Wait, maybe f(x) is not defined for all positive real numbers? But the problem says it's defined on (0, +∞). Hmm.Alternatively, perhaps I made a mistake in assuming f is linear. Maybe f(x) is of the form f(x) = a - kx, where k is positive, making f decreasing.Let me try that. Let f(x) = a - kx.Then, f(x + y) = a - k(x + y) = a - kx - kyOn the other hand, f(x) + f(y) - 1 = (a - kx) + (a - ky) - 1 = 2a - kx - ky - 1Set them equal:a - kx - ky = 2a - kx - ky - 1Simplify:a = 2a - 1Subtract a:0 = a - 1So, a = 1Therefore, f(x) = 1 - kxNow, f(4) = 5:1 - 4k = 5Subtract 1:-4k = 4Divide by -4:k = -1But k is supposed to be positive because f is decreasing (since f(x) = 1 - kx, if k is positive, then f is decreasing). But here, k = -1, which would make f(x) = 1 - (-1)x = 1 + x, which is increasing. Again, a contradiction.Wait, so if I define f(x) = a - kx, then to have f decreasing, k must be positive. But solving f(4) = 5 gives k = -1, which is negative, making f increasing. So, again, a contradiction.This suggests that there is no function f(x) of the form a - kx that satisfies both the functional equation and the condition that f is decreasing with f(4) = 5. But the problem states that such a function exists. So, maybe my approach is wrong.Alternatively, perhaps f(x) is not linear, but exponential? Let me try that.Suppose f(x) = ab^x + c. Let's see if that works.But before I go into that, maybe I should look back at the problem. It says f is decreasing, defined on (0, +∞), satisfies f(x + y) = f(x) + f(y) - 1, and f(4) = 5.From part (1), I found f(2) = 3, which seems correct. Then, using the functional equation, maybe I can find f(1), f(3), etc., to see if a pattern emerges.Let me try to find f(1). Let me set x = y = 1:f(2) = f(1) + f(1) - 1We know f(2) = 3, so:3 = 2f(1) - 1Add 1:4 = 2f(1)So, f(1) = 2Similarly, let's find f(3). Let me set x = 2 and y = 1:f(3) = f(2) + f(1) - 1 = 3 + 2 - 1 = 4Similarly, f(4) = 5, which is given.So, f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5. It seems like f(n) = n + 1 for integer n. But wait, f(x) is defined for all positive real numbers, not just integers. So, maybe f(x) = x + 1 for all x? But that's increasing, which contradicts the given that f is decreasing.Wait, but in the problem, f is decreasing, so f(x) = x + 1 can't be the solution. So, perhaps my initial assumption that f is linear is incorrect, or maybe the function is piecewise defined?Alternatively, maybe f(x) = c - x, but then f(4) = c - 4 = 5, so c = 9, making f(x) = 9 - x, which is decreasing. Let's check if this satisfies the functional equation.f(x + y) = 9 - (x + y) = 9 - x - yf(x) + f(y) - 1 = (9 - x) + (9 - y) - 1 = 18 - x - y - 1 = 17 - x - yBut 9 - x - y ≠ 17 - x - y, so that doesn't work. So, f(x) = 9 - x doesn't satisfy the functional equation.Hmm, this is confusing. Maybe I need to think differently.Wait, from the functional equation, f(x + y) = f(x) + f(y) - 1, and f is decreasing. Let me see if I can find f(x) in terms of f(1).Let me define f(1) = a. Then, f(2) = f(1 + 1) = f(1) + f(1) - 1 = 2a - 1. We know f(2) = 3, so 2a - 1 = 3 ⇒ a = 2. So, f(1) = 2.Similarly, f(3) = f(2 + 1) = f(2) + f(1) - 1 = 3 + 2 - 1 = 4.f(4) = f(3 + 1) = f(3) + f(1) - 1 = 4 + 2 - 1 = 5, which matches the given.So, for integer values, f(n) = n + 1. But since f is defined on all positive reals, maybe f(x) = x + 1 for all x? But that's increasing, which contradicts the given.Wait, but maybe f(x) = -x + c? Let's try that.Suppose f(x) = -x + c.Then, f(x + y) = -(x + y) + c = -x - y + cf(x) + f(y) - 1 = (-x + c) + (-y + c) - 1 = -x - y + 2c - 1Set equal:-x - y + c = -x - y + 2c - 1Simplify:c = 2c - 1 ⇒ c = 1So, f(x) = -x + 1But f(4) = -4 + 1 = -3, which is not equal to 5. So, that doesn't work.Hmm, maybe f(x) is of the form f(x) = k - x, where k is a constant.Then, f(x + y) = k - (x + y) = k - x - yf(x) + f(y) - 1 = (k - x) + (k - y) - 1 = 2k - x - y - 1Set equal:k - x - y = 2k - x - y - 1Simplify:k = 2k - 1 ⇒ k = 1So, f(x) = 1 - xBut f(4) = 1 - 4 = -3 ≠ 5. So, that's not it either.Wait, maybe f(x) = kx + c, but with k negative. Let's try again.Suppose f(x) = kx + c, with k < 0.Then, f(x + y) = k(x + y) + c = kx + ky + cf(x) + f(y) - 1 = (kx + c) + (ky + c) - 1 = kx + ky + 2c - 1Set equal:kx + ky + c = kx + ky + 2c - 1Simplify:c = 2c - 1 ⇒ c = 1So, f(x) = kx + 1, with k < 0.Now, f(4) = 5:4k + 1 = 5 ⇒ 4k = 4 ⇒ k = 1But k must be negative, so this is a contradiction. Therefore, there is no such linear function f(x) = kx + 1 with k < 0 that satisfies f(4) = 5.This suggests that f(x) is not linear, which contradicts the functional equation because f(x + y) = f(x) + f(y) - 1 suggests linearity up to a constant. Maybe the function is affine linear, but with a negative slope, but as we saw, that leads to a contradiction with f(4) = 5.Wait, maybe I'm missing something. Let me try to express f(x) in terms of f(1). Since f is defined for all positive reals, maybe f(x) = 1 + x for x in some interval, but decreasing elsewhere? That seems complicated.Alternatively, perhaps f(x) = 1 + x for x ≤ 4, and then decreases beyond that. But that would make f(x) non-linear and potentially not satisfying the functional equation for all x and y.Wait, let's test the functional equation with f(x) = 1 + x for x ≤ 4 and some other form for x > 4. But that seems too vague.Alternatively, maybe f(x) is a constant function? But if f(x) is constant, say f(x) = c, then f(x + y) = c = c + c - 1 ⇒ c = 2c - 1 ⇒ c = 1. But f(4) = 5, which contradicts c = 1. So, f(x) can't be constant.Hmm, I'm stuck here. Maybe I should accept that f(x) = x + 1 is the solution, even though it's increasing, but the problem says it's decreasing. That doesn't make sense. Maybe the problem has a typo, or I'm misunderstanding something.Wait, let's go back to the problem statement. It says f is decreasing on (0, +∞). So, f(x) must be decreasing. But from the functional equation and f(4) = 5, we seem to get f(x) = x + 1, which is increasing. That's a contradiction. So, perhaps the problem is misstated, or I'm misapplying the functional equation.Alternatively, maybe f(x) is not defined for all positive reals, but only for integers? But the problem says it's defined on (0, +∞), so it's for all positive real numbers.Wait, maybe f(x) is a step function or something else. But without more information, it's hard to say.Alternatively, perhaps the functional equation is f(x + y) = f(x) + f(y) + 1, but the problem says minus 1. Hmm.Wait, let me double-check the functional equation. It says f(x + y) = f(x) + f(y) - 1. So, it's minus 1.Given that, and f is decreasing, f(4) = 5, and f(2) = 3, as found in part (1).But if f(x) = x + 1, which is increasing, but we have f(4) = 5, which fits, but contradicts f being decreasing.Wait, maybe f(x) = 5 - x? Let's test that.f(x + y) = 5 - (x + y) = 5 - x - yf(x) + f(y) - 1 = (5 - x) + (5 - y) - 1 = 10 - x - y - 1 = 9 - x - ySo, 5 - x - y ≠ 9 - x - y. Doesn't work.Wait, maybe f(x) = 6 - x?f(x + y) = 6 - x - yf(x) + f(y) - 1 = (6 - x) + (6 - y) - 1 = 12 - x - y - 1 = 11 - x - yNot equal.Hmm, I'm stuck. Maybe I should accept that f(x) = x + 1 is the solution, even though it's increasing, and perhaps the problem has a typo, or I'm missing something.But since the problem states that f is decreasing, and we have f(4) = 5, which would require f(x) = x + 1, which is increasing, there must be a mistake in my reasoning.Wait, maybe I should consider that f(x) is decreasing, so f(x + y) ≤ f(x) and f(x + y) ≤ f(y). But from the functional equation, f(x + y) = f(x) + f(y) - 1. So, if f is decreasing, then f(x + y) ≤ f(x) and f(x + y) ≤ f(y). Therefore, f(x) + f(y) - 1 ≤ f(x) and f(x) + f(y) - 1 ≤ f(y). Let's see what that implies.From f(x) + f(y) - 1 ≤ f(x), subtract f(x):f(y) - 1 ≤ 0 ⇒ f(y) ≤ 1Similarly, from f(x) + f(y) - 1 ≤ f(y), subtract f(y):f(x) - 1 ≤ 0 ⇒ f(x) ≤ 1So, f(x) ≤ 1 for all x. But f(4) = 5, which is greater than 1. Contradiction.Wait, that can't be. So, if f is decreasing, then f(x + y) ≤ f(x) and f(x + y) ≤ f(y). But from the functional equation, f(x + y) = f(x) + f(y) - 1. So, f(x) + f(y) - 1 ≤ f(x) ⇒ f(y) ≤ 1, and similarly f(x) ≤ 1. But f(4) = 5 > 1, which is a contradiction.Therefore, there is no such function f(x) that is decreasing, defined on (0, +∞), satisfies f(x + y) = f(x) + f(y) - 1, and f(4) = 5. But the problem states that such a function exists, so I must have made a wrong assumption.Wait, maybe f(x) is not defined for all positive reals, but only for x ≥ some value? But the problem says (0, +∞).Alternatively, maybe f(x) is a constant function beyond a certain point. But as I saw earlier, constant functions don't satisfy the functional equation unless they're 1, which contradicts f(4) = 5.Wait, perhaps f(x) is piecewise defined. For example, f(x) = x + 1 for x ≤ 4, and then decreases beyond that. But then, for x + y > 4, the functional equation might not hold.Alternatively, maybe f(x) = 5 - (x - 4) for x ≥ 4, but let's test that.Wait, let me try to define f(x) as follows:For x ≤ 4, f(x) = x + 1.For x > 4, f(x) = 5 - (x - 4) = 9 - x.But let's check the functional equation for x = 2 and y = 3:f(5) = 9 - 5 = 4f(2) + f(3) - 1 = 3 + 4 - 1 = 6But f(5) = 4 ≠ 6. So, that doesn't work.Alternatively, maybe f(x) = 5 - (x/4). Let's test that.f(x + y) = 5 - (x + y)/4f(x) + f(y) - 1 = [5 - x/4] + [5 - y/4] - 1 = 10 - (x + y)/4 - 1 = 9 - (x + y)/4So, f(x + y) = 5 - (x + y)/4 ≠ 9 - (x + y)/4. Doesn't work.Hmm, I'm really stuck here. Maybe I should accept that f(x) = x + 1 is the solution, even though it's increasing, and perhaps the problem has a typo, or I'm misinterpreting the functional equation.Alternatively, maybe the functional equation is f(x + y) = f(x) + f(y) + 1, which would make f(x) = -x + c, but let's see.Wait, if the functional equation were f(x + y) = f(x) + f(y) + 1, then defining g(x) = f(x) + 1, we'd have g(x + y) = g(x) + g(y), so g is linear, g(x) = kx. Then, f(x) = kx - 1. If f is decreasing, k < 0. Then, f(4) = 4k - 1 = 5 ⇒ 4k = 6 ⇒ k = 1.5, which is positive, again a contradiction.Wait, but if the functional equation is f(x + y) = f(x) + f(y) + 1, then f(x) = -x + c. Let's try that.f(x + y) = -(x + y) + c = -x - y + cf(x) + f(y) + 1 = (-x + c) + (-y + c) + 1 = -x - y + 2c + 1Set equal:-x - y + c = -x - y + 2c + 1 ⇒ c = 2c + 1 ⇒ c = -1So, f(x) = -x - 1But f(4) = -4 - 1 = -5 ≠ 5. So, that doesn't work either.I'm really confused now. Maybe the problem is misstated, or I'm missing something obvious.Wait, going back to the problem, it says f is decreasing on (0, +∞), satisfies f(x + y) = f(x) + f(y) - 1, and f(4) = 5.From part (1), I found f(2) = 3, which seems correct.For part (2), I used the fact that f is decreasing to solve f(m - 2) ≤ 3, which led me to m ≥ 4. That seems correct, regardless of the form of f(x).But the issue is that f(x) = x + 1 is increasing, which contradicts the given that f is decreasing. So, perhaps the problem has a typo, or I'm misapplying the functional equation.Alternatively, maybe f(x) is defined differently. Wait, let me try to find f(x) for x = 1/2.Let me set x = y = 1/2:f(1) = f(1/2) + f(1/2) - 1 ⇒ 2 = 2f(1/2) - 1 ⇒ 2f(1/2) = 3 ⇒ f(1/2) = 1.5Similarly, f(1/2 + 1/2) = f(1) = 2 = f(1/2) + f(1/2) - 1 ⇒ 2 = 1.5 + 1.5 - 1 ⇒ 2 = 2, which checks out.Now, f(1/2) = 1.5, f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5.So, it seems like f(x) = x + 1 for x = 1/2, 1, 2, 3, 4, etc. But f(x) is supposed to be decreasing, so f(1/2) = 1.5, f(1) = 2, which is increasing. Contradiction.Therefore, there is no such function f(x) that is decreasing, defined on (0, +∞), satisfies f(x + y) = f(x) + f(y) - 1, and f(4) = 5. The problem is contradictory.But since the problem is given, perhaps I should proceed with the answers I found, assuming that f(x) = x + 1 despite it being increasing, or perhaps the problem meant f is increasing.Alternatively, maybe I made a mistake in assuming f(x) is linear. Maybe f(x) is of the form f(x) = c - kx, but as I saw earlier, that leads to contradictions.Wait, maybe f(x) = 1 + x for x ≤ 4, and then f(x) decreases beyond that. But then, for x + y > 4, the functional equation might not hold.Alternatively, maybe f(x) is a constant function beyond x = 4, but that would contradict the functional equation.I think I've exhausted all possibilities, and the only consistent solution is f(x) = x + 1, which is increasing, contradicting the given that f is decreasing. Therefore, the problem as stated has no solution. But since the problem asks for answers, perhaps I should proceed with the answers I found, assuming that f(x) = x + 1 despite the contradiction.So, for part (1), f(2) = 3, and for part (2), m ≥ 4.But I'm really unsure because of the contradiction. Maybe the problem meant f is increasing, or there's a typo in the functional equation. Alternatively, perhaps I'm overcomplicating things and should just proceed with the answers I found, assuming that f(x) = x + 1 is the solution, even though it's increasing.In conclusion, despite the contradiction, I think the answers are f(2) = 3 and m ≥ 4.