🧠:<think>Okay, so I need to figure out how many vertical asymptotes are in the function ( y = frac{x^2 - 14x + 49}{x^3 - 5x^2 - 14x + 70} ). Hmm, vertical asymptotes occur where the denominator is zero but the numerator isn't zero at those points, right? First, I think I should factor both the numerator and the denominator to simplify the function. Let me start with the numerator: ( x^2 - 14x + 49 ). Hmm, can I factor this? It looks like a quadratic. Let me check if it's a perfect square. The first term is ( x^2 ), which is ( x ) squared, and the last term is 49, which is 7 squared. The middle term is -14x, which is -2 times 7 times x. So yeah, it factors into ( (x - 7)^2 ). Okay, that's straightforward.Now, the denominator is ( x^3 - 5x^2 - 14x + 70 ). Factoring a cubic can be a bit trickier. Maybe I can factor by grouping. Let me see if that works here. Let me group the first two terms and the last two terms:( x^3 - 5x^2 ) and ( -14x + 70 ). Factoring out an ( x^2 ) from the first group gives ( x^2(x - 5) ). From the second group, I can factor out a -14, which gives ( -14(x - 5) ). So now the denominator looks like ( x^2(x - 5) - 14(x - 5) ). Oh, I see a common factor of ( (x - 5) ). So I can factor that out, which gives ( (x - 5)(x^2 - 14) ). Wait, is that right? Let me check my factoring steps again. Wait, no, I think I made a mistake there. Let me try again. So the denominator is ( x^3 - 5x^2 - 14x + 70 ). Let me try to factor by grouping again. Group the first two terms: ( x^3 - 5x^2 ) and the last two terms: ( -14x + 70 ). Factoring ( x^2 ) from the first group: ( x^2(x - 5) ). Factoring -14 from the second group: ( -14(x - 5) ). So now, I have ( x^2(x - 5) - 14(x - 5) ). Ah, yes, so now I can factor out ( (x - 5) ), which gives ( (x - 5)(x^2 - 14) ). So, the denominator factors into ( (x - 5)(x^2 - 14) ). Wait, but ( x^2 - 14 ) isn't factorable over the integers, right? So that's as far as I can go. So, putting it all together, the function becomes ( y = frac{(x - 7)^2}{(x - 5)(x^2 - 14)} ). Now, I need to find where the denominator is zero, which would give me the potential vertical asymptotes. So setting the denominator equal to zero: ( (x - 5)(x^2 - 14) = 0 ). This gives me ( x = 5 ) and ( x^2 - 14 = 0 ). Solving ( x^2 - 14 = 0 ) gives ( x = sqrt{14} ) and ( x = -sqrt{14} ). But before I conclude that these are vertical asymptotes, I need to check if the numerator is also zero at these points. The numerator is ( (x - 7)^2 ), which is zero only when ( x = 7 ). So, for ( x = 5 ), the numerator is ( (5 - 7)^2 = (-2)^2 = 4 ), which is not zero. So, ( x = 5 ) is a vertical asymptote. For ( x = sqrt{14} ), the numerator is ( (sqrt{14} - 7)^2 ), which is definitely not zero because ( sqrt{14} ) is approximately 3.7417, so ( 3.7417 - 7 ) is about -3.2583, and squaring that gives a positive number, not zero. So, ( x = sqrt{14} ) is a vertical asymptote. Similarly, for ( x = -sqrt{14} ), the numerator is ( (-sqrt{14} - 7)^2 ). Again, this is not zero because ( -sqrt{14} ) is about -3.7417, so ( -3.7417 - 7 ) is about -10.7417, and squaring that also gives a positive number, not zero. So, ( x = -sqrt{14} ) is also a vertical asymptote. Wait a minute, but earlier I thought ( x = 7 ) might be a problem. Let me check that. At ( x = 7 ), the numerator is ( (7 - 7)^2 = 0 ), and the denominator is ( (7 - 5)(7^2 - 14) = (2)(49 - 14) = 2*35 = 70 ), which is not zero. So, at ( x = 7 ), the function is defined because both numerator and denominator are not zero, but the numerator is zero, so it's just a hole or a point discontinuity. So, ( x = 7 ) is not a vertical asymptote, just a point where the function is zero. So, putting it all together, the function has vertical asymptotes at ( x = 5 ), ( x = sqrt{14} ), and ( x = -sqrt{14} ). That means there are three vertical asymptotes. Wait, but I thought earlier that the denominator factored into ( (x - 5)(x^2 - 14) ), which would give three roots: 5, ( sqrt{14} ), and ( -sqrt{14} ). And since none of these make the numerator zero, all three should be vertical asymptotes. But hold on, let me double-check my factoring of the denominator. Maybe I made a mistake there. The denominator is ( x^3 - 5x^2 - 14x + 70 ). Let me try synthetic division to see if I can factor it correctly. Let me test ( x = 5 ) as a root. Plugging into the denominator: ( 5^3 - 5*5^2 -14*5 +70 = 125 - 125 -70 +70 = 0. Yes, so 5 is a root. So, dividing the denominator by ( x - 5 ), let's perform polynomial long division or synthetic division. Let me use synthetic division with x = 5. Coefficients: 1 (x^3), -5 (x^2), -14 (x), 70 (constant). Set up synthetic division:5 | 1 -5 -14 70Bring down the 1.Multiply 1 by 5: 5. Add to -5: 0.Multiply 0 by 5: 0. Add to -14: -14.Multiply -14 by 5: -70. Add to 70: 0. So, the result is ( x^2 + 0x -14 ), which simplifies to ( x^2 -14 ). So, denominator factors into ( (x - 5)(x^2 -14) ), which is correct. So, no mistake there. Therefore, the denominator has three roots: 5, ( sqrt{14} ), and ( -sqrt{14} ). Thus, the function has vertical asymptotes at all three of these points because the numerator doesn't cancel them out. But wait, earlier I thought that ( x = 7 ) was a point where the numerator and denominator both have a factor, but in reality, in the denominator, after factoring, I only have ( (x -5)(x^2 -14) ), so ( x =7 ) isn't a root of the denominator. Wait, hold on, in the numerator, after factoring, I have ( (x -7)^2 ), and in the denominator, after factoring, I have ( (x -5)(x^2 -14) ). So, there is no common factor between numerator and denominator except perhaps at ( x =7 ), but in the denominator, ( x =7 ) is not a root because ( 7^2 -14 =49 -14=35 neq0 ). So, ( x =7 ) is not a vertical asymptote, but just a regular point where the function is zero. Therefore, the vertical asymptotes are at ( x =5 ), ( x =sqrt{14} ), and ( x = -sqrt{14} ). So, three vertical asymptotes. Wait, but the original function is ( y = frac{x^2 -14x +49}{x^3 -5x^2 -14x +70} ). Let me plug in ( x =5 ) into the denominator: ( 125 - 125 -70 +70 =0), so yes, 5 is a root. Similarly, ( x = sqrt{14} ) and ( x = -sqrt{14} ) are roots because ( x^2 -14 =0 ) when ( x = sqrt{14} ) or ( x = -sqrt{14} ). And as for the numerator, at these points, it's ( (5 -7)^2=4 ), ( (sqrt{14}-7)^2 ) which is positive, and ( (-sqrt{14}-7)^2 ) which is also positive. So, all three are vertical asymptotes. Therefore, the number of vertical asymptotes is three. Wait, but in my initial factoring, I thought the denominator was ( (x -7)(x -5)(x +2) ). Wait, that can't be right because earlier I factored it as ( (x -5)(x^2 -14) ). So, which one is correct? Let me check the denominator again: ( x^3 -5x^2 -14x +70 ). If I factor by grouping, as I did earlier, it becomes ( (x -5)(x^2 -14) ). So, the denominator is ( (x -5)(x^2 -14) ), not ( (x -7)(x -5)(x +2) ). Wait, where did that ( (x -7)(x -5)(x +2) ) come from? Maybe I confused it with another problem. Let me double-check. If I factor ( x^3 -5x^2 -14x +70 ), by synthetic division, we saw that ( x =5 ) is a root, and then the quadratic is ( x^2 -14 ), which doesn't factor further over integers. So, the denominator is ( (x -5)(x^2 -14) ), not involving ( x -7 ) or ( x +2 ). Therefore, my initial thought about factoring was incorrect, and the denominator doesn't have ( x -7 ) or ( x +2 ) as factors. So, the function simplifies to ( y = frac{(x -7)^2}{(x -5)(x^2 -14)} ), and there's no cancellation between numerator and denominator except at ( x =7 ), which doesn't affect the denominator. Therefore, the vertical asymptotes are at ( x =5 ), ( x =sqrt{14} ), and ( x = -sqrt{14} ), totaling three vertical asymptotes. Wait, but in the initial problem, the denominator is ( x^3 -5x^2 -14x +70 ). Let me try plugging in ( x =7 ) into the denominator: ( 343 - 245 -98 +70 = (343 -245) + (-98 +70) =98 -28=70 neq0 ). So, ( x =7 ) is not a vertical asymptote, just a hole or a point where the function is zero. So, to summarize, the vertical asymptotes occur at ( x =5 ), ( x =sqrt{14} ), and ( x = -sqrt{14} ), so three vertical asymptotes. Wait, but earlier I thought the denominator was factored as ( (x -7)(x -5)(x +2) ), but that must have been a mistake. Let me check that again. If I factor ( x^3 -5x^2 -14x +70 ), I can try to factor it as ( (x - a)(x^2 + bx + c) ). But since ( x =5 ) is a root, as we saw, the denominator factors into ( (x -5)(x^2 -14) ), which doesn't factor further over integers. So, no, it's not ( (x -7)(x -5)(x +2) ). That must have been a confusion with another problem. Therefore, the correct factoring is ( (x -5)(x^2 -14) ), leading to vertical asymptotes at ( x =5 ), ( x =sqrt{14} ), and ( x = -sqrt{14} ). Hence, the number of vertical asymptotes is three.</think>

🧠:<think>Okay, so I have this problem where I need to express the number 2021 in the form of a fraction involving factorials. The expression is (frac{a_1!a_2!...a_m!}{b_1!b_2!...b_n!}), where the (a_i)s and (b_i)s are positive integers arranged in non-increasing order. The goal is to make (a_1 + b_1) as small as possible and then find the absolute difference (|a_1 - b_1|).First, I should probably factorize 2021 to understand its prime components. Let me try dividing by small primes. 2021 divided by 43 is... let's see, 43 times 47 is 2021. So, 2021 = 43 × 47. That means 2021 is a product of two primes, 43 and 47. Now, I need to express this as a ratio of products of factorials. Factorials involve multiplying sequences of integers, so I need to find factorials that include 43 and 47 in their products but can also cancel out other terms in the numerator and denominator.Since 47 is a larger prime, it's going to be in the numerator's factorial because if it were in the denominator, we wouldn't be able to get rid of it. So, let's say (a_1) is 47 because that's the larger prime, and it needs to be in the numerator to include 47. Then, (b_1) is going to be the next prime down, which is 43, because we need to cancel out 43 in the denominator. So, if (a_1 = 47) and (b_1 = 43), then (a_1 + b_1 = 47 + 43 = 90). I need to make sure this is the smallest possible sum. If I try a smaller (a_1), say 46, but 46! doesn't include 47, so that won't work. Similarly, if I make (b_1) larger, say 44, but 44 isn't a prime, and 44! would include 43 and 47 if 44 is larger than both. Wait, no, 44 is less than 47, so 44! doesn't include 47. So, 44! would include 43 but not 47, which means we still need 47 in the numerator. Therefore, (a_1) must be at least 47 to include 47 in the numerator, and (b_1) must be at least 43 to include 43 in the denominator. So, 47 and 43 seem to be the minimal values for (a_1) and (b_1) respectively. Now, calculating the absolute difference, (|47 - 43| = 4). That would be the answer. But let me double-check to make sure there isn't a smaller sum. Suppose I tried (a_1 = 47) and (b_1 = 44). Then (47! / 44!) would give me (45×46×47), which is way larger than 2021. So, that's not helpful. Similarly, if I tried (b_1 = 42), then (47! / 42!) would be even larger. So, 43 seems necessary for (b_1) to cancel out the 43 in the numerator. Alternatively, maybe there's a way to have multiple factorials in the numerator and denominator that could somehow combine to give 2021 without needing such large factorials. For example, maybe using 47! in the numerator and 46! and something else in the denominator? But 47! / (46! × something) would give me 47 divided by something, which needs to be 2021. That doesn't seem straightforward because 47 is a prime and 2021 is 47×43. So, to get 43 in the numerator, we need 43! in the numerator, but 43! is already in the denominator as part of (b_1!). Hmm, this is getting a bit tangled.Wait, maybe another approach: Since 2021 = 43×47, perhaps I can write it as (frac{47!}{46!}) multiplied by (frac{43!}{42!}), but that would be 47×43, which is 2021. So, the expression would be (frac{47! times 43!}{46! times 42!}). Let's check: 47! / 46! = 47 and 43! / 42! = 43, so 47×43 = 2021. That works.In this case, (a_1 = 47) and (a_2 = 43), while (b_1 = 46) and (b_2 = 42). But wait, the problem specifies that (a_1 geq a_2 geq ... geq a_m) and (b_1 geq b_2 geq ... geq b_n). So, in this case, (a_1 = 47), (a_2 = 43), which is okay because 47 ≥ 43. Similarly, (b_1 = 46), (b_2 = 42), which is also okay because 46 ≥ 42.But now, (a_1 + b_1 = 47 + 46 = 93), which is actually larger than the previous sum of 90. So, this approach doesn't help us minimize (a_1 + b_1). Therefore, the initial approach where (a_1 = 47) and (b_1 = 43) is better because their sum is smaller.Another thought: could I have (a_1 = 47) and (b_1 = 43) with just single factorials? That would mean (frac{47!}{43!}). Let's compute that: 47! / 43! = 44×45×46×47. Wait, that's 44×45×46×47, which is way larger than 2021. So, that won't work. So, I need something else.Perhaps, instead of just 47! / 43!, I can have multiple factorials in the numerator and denominator that multiply to give the necessary product. For example, maybe 47! / (46! × something) = 47 / something = 2021. But 2021 is 47×43, so that something would have to be 1/43, which isn't an integer factorial. Hmm, that doesn't make sense.Alternatively, maybe (frac{47! times 43!}{46! times 42!}), as I thought earlier, but that gives us 47×43, which is 2021. So, that works, but as I saw before, (a_1 + b_1) is 47 + 46 = 93, which is larger than 47 + 43 = 90. So, 93 is bigger, so we don't want that.Wait, maybe if I have (a_1 = 47) and (b_1 = 43), but with more factorials, maybe I can get a smaller sum. Let me think. If I have (frac{47!}{43!}), that's too big, as it's 44×45×46×47. But if I can somehow divide that by other factorials to get down to 2021.Alternatively, maybe use smaller factorials in the numerator and denominator to get the product 2021. For example, if I have 47! in the numerator and 46! in the denominator, that gives me 47. Then, if I have 43! in the numerator and 42! in the denominator, that gives me 43. Multiplying those together gives 47×43=2021. So, that's the same as before, but with (a_1 + b_1 = 47 + 46 = 93). So, that's worse.Alternatively, maybe I can have a single factorial in the numerator and denominator such that their ratio is 2021. But 2021 is 43×47, so unless I have 47! / (something) = 43×47, which would require that something is 46! / 43!, but that would give 44×45×46, which is not 1. So, that doesn't help.Wait, maybe another approach: Let me think about the exponents of primes in 2021. Since 2021 is 43×47, both primes, so in terms of factorial exponents, each prime will only appear once in the factorization. So, in the numerator, I need to have at least one 43 and one 47, and in the denominator, I shouldn't have any 43 or 47, because 2021 doesn't have them in the denominator. So, that means 43 and 47 must be in the numerator's factorials, but not canceled out by the denominator's factorials.But if I have 47! in the numerator, that includes 47, and 43! in the numerator includes 43, but if I have 43! in the denominator, that would cancel out the 43 in the numerator, which we don't want. So, maybe we need to have 47! in the numerator and 46! in the denominator, which gives us 47, and then 43! in the numerator and 42! in the denominator, which gives us 43. Then, multiplying those together gives 47×43=2021.But as I saw earlier, that leads to (a_1 = 47), (a_2 = 43), (b_1 = 46), (b_2 = 42), and (a_1 + b_1 = 47 + 46 = 93). That's still higher than 47 + 43 = 90.Wait, but if I have (a_1 = 47) and (b_1 = 43), can I write 2021 as (frac{47!}{43!}) divided by something else? Let me see: (frac{47!}{43!} = 44×45×46×47). If I divide that by (44×45×46), I get 47, which is not 2021. Alternatively, if I divide by (44×45×46 / 43), that's complicated. Maybe I need to include another factorial term.Alternatively, maybe (frac{47! times 43!}{46! times 42!}), which is 47×43, as we saw. So, that works, but (a_1 + b_1 = 47 + 46 = 93). Hmm.Wait, maybe there's a way to have (a_1 = 47) and (b_1 = 43) without having to increase (b_1) to 46. Let me think: If I have (frac{47!}{43!}), that's 44×45×46×47. I need this to equal 2021, which is 43×47. So, I need to divide 44×45×46×47 by something to get 43×47. That something would be 44×45×46 / 43. But 44×45×46 is 44×45×46 = 91140, and 91140 / 43 is approximately 2120, which is not an integer, so that doesn't work.Alternatively, maybe I can have more factorials in the numerator and denominator to cancel out the extra terms. For example, if I have (frac{47! times 43!}{46! times 42!}), as before, that gives 47×43, which is 2021. So, that's the expression, but (a_1 + b_1 = 47 + 46 = 93). I need to see if there's a way to get (a_1 + b_1) smaller than 90.Wait, if (a_1 = 47) and (b_1 = 43), their sum is 90. Is there a way to have (a_1) and (b_1) both smaller than 47 and 43 respectively? Well, 47 is a prime, so the smallest factorial that includes 47 is 47!. Similarly, 43 is a prime, so the smallest factorial that includes 43 is 43!. Therefore, I can't have a smaller (a_1) than 47 or a smaller (b_1) than 43. So, 47 and 43 are the minimal values for (a_1) and (b_1).Therefore, (a_1 + b_1 = 47 + 43 = 90), and (|a_1 - b_1| = |47 - 43| = 4). I think that's the minimal possible sum, so the answer should be 4, which is option D.</think>

🧠:<think>Alright, so I have this problem here with two functions: ( f(x) = x^2 - 4x + 3 ) and ( g(x) = m(x - 1) + 2 ), where ( m > 0 ). The question is asking about the range of the real number ( m ) such that there exists an ( x_1 ) in the interval [0, 3] where for any ( x_2 ) in [0, 3], ( f(x_1) = g(x_2) ) holds true.Hmm, okay. Let me try to break this down.First, I need to understand what it means for there to exist an ( x_1 ) such that for any ( x_2 ), ( f(x_1) = g(x_2) ). That sounds like ( f(x_1) ) is a constant value that equals ( g(x_2) ) for every ( x_2 ) in [0, 3]. So, ( g(x_2) ) must be a constant function over [0, 3], right? But wait, ( g(x) ) is a linear function, so unless its slope ( m ) is zero, it won't be constant. But ( m > 0 ), so ( m ) can't be zero. That seems conflicting.Wait, maybe I'm misinterpreting. The problem says there exists an ( x_1 ) such that for any ( x_2 ), ( f(x_1) = g(x_2) ). So, it's not that ( g(x) ) is constant, but rather that there's a specific value ( f(x_1) ) that equals ( g(x_2) ) for all ( x_2 ). That would mean that ( g(x_2) ) is constant, because it's equal to the same value ( f(x_1) ) regardless of ( x_2 ). Therefore, ( g(x) ) must be a constant function on [0, 3], which implies that its slope ( m ) must be zero. But the problem states that ( m > 0 ), so that's a contradiction.Wait, that can't be right. Maybe I need to approach this differently.Perhaps instead of ( g(x) ) being constant, it's that for some fixed ( x_1 ), ( f(x_1) ) is equal to ( g(x_2) ) for all ( x_2 ). That would mean ( g(x_2) ) is constant, but as ( g(x) ) is linear, the only way ( g(x_2) ) is constant is if ( m = 0 ). But since ( m > 0 ), this can't happen. Therefore, maybe my initial interpretation is wrong.Let me read the problem again carefully: "if there exists ( x_1 in [0,3] ) such that for any ( x_2 in [0,3] ), ( f(x_1) = g(x_2) ) holds". So, it's saying that there is a particular ( x_1 ) where ( f(x_1) ) is equal to ( g(x_2) ) for every ( x_2 ) in [0,3]. So, ( g(x_2) ) must be equal to ( f(x_1) ) for all ( x_2 ), meaning ( g(x) ) is a constant function. But since ( g(x) ) is linear and ( m > 0 ), it can't be constant. Therefore, this seems impossible unless ( m = 0 ), but ( m > 0 ).This is confusing. Maybe I need to visualize the functions.First, let's analyze ( f(x) = x^2 - 4x + 3 ). This is a quadratic function opening upwards. Let me rewrite it in vertex form to find its minimum. Completing the square:( f(x) = x^2 - 4x + 3 = (x^2 - 4x + 4) - 1 = (x - 2)^2 - 1 ).So, the vertex is at (2, -1). Since the parabola opens upwards, the minimum value is -1 at x=2, and it goes up from there. Let's check the values at the endpoints of the interval [0, 3]:At x=0: ( f(0) = 0 - 0 + 3 = 3 ).At x=3: ( f(3) = 9 - 12 + 3 = 0 ).So, the range of ( f(x) ) on [0, 3] is from -1 to 3, that is, [-1, 3].Now, ( g(x) = m(x - 1) + 2 ). Since ( m > 0 ), this is a straight line with a positive slope. Let's see its behavior on [0, 3]. Let's compute its values at the endpoints:At x=0: ( g(0) = m(-1) + 2 = -m + 2 ).At x=3: ( g(3) = m(2) + 2 = 2m + 2 ).So, ( g(x) ) increases from ( 2 - m ) to ( 2 + 2m ) as x goes from 0 to 3.Now, the problem states that there exists an ( x_1 ) in [0, 3] such that for any ( x_2 ) in [0, 3], ( f(x_1) = g(x_2) ). So, ( f(x_1) ) must equal ( g(x_2) ) for all ( x_2 ) in [0, 3]. That would mean that ( g(x) ) is a constant function, but as we saw, ( g(x) ) is linear with slope ( m > 0 ), so it's not constant. Therefore, this is only possible if ( g(x) ) is constant, which requires ( m = 0 ), but ( m > 0 ). So, that seems contradictory.Wait, maybe another interpretation: Perhaps the problem is saying that for some ( x_1 ), ( f(x_1) ) is equal to ( g(x_2) ) for all ( x_2 ), which would imply that ( g(x) ) must be constant, but since it's not, the only way this can happen is if ( g(x) ) is equal to ( f(x_1) ) for all ( x ), which would mean ( g(x) ) is constant. But since ( g(x) ) is not constant, the only way is if ( g(x) ) is a constant function, which is not possible here.Hmm, maybe I'm misunderstanding the problem. Perhaps it's not that ( g(x) ) is constant, but that ( f(x_1) ) equals ( g(x_2) ) for some ( x_1 ) and for any ( x_2 ). So, for each ( x_2 ), there exists an ( x_1 ) such that ( f(x_1) = g(x_2) ). But the wording is "there exists ( x_1 ) such that for any ( x_2 ), ( f(x_1) = g(x_2) )". So, it's a single ( x_1 ) that works for all ( x_2 ).Wait, that would mean that ( g(x) ) is equal to ( f(x_1) ) for all ( x ) in [0, 3], so ( g(x) ) must be a constant function. But ( g(x) ) is linear with slope ( m > 0 ), so it's not constant unless ( m = 0 ), which is not allowed.So, this seems impossible. But the problem is giving options, so perhaps I need to think differently.Wait, maybe the problem is not requiring ( g(x) ) to be equal to ( f(x_1) ) for all ( x ) in [0, 3], but rather that for some ( x_1 ), ( f(x_1) ) is equal to ( g(x_2) ) for any ( x_2 ). So, perhaps ( f(x_1) ) is a value that ( g(x) ) can take for any ( x_2 ). But that doesn't make much sense.Alternatively, maybe the problem is saying that there exists an ( x_1 ) such that ( f(x_1) ) is equal to ( g(x_2) ) for all ( x_2 ). So, ( f(x_1) ) must be equal to every value that ( g(x) ) takes on [0, 3]. But the only way a single value can equal every value of another function is if the other function is constant. So, again, this would require ( g(x) ) to be constant, which is not possible unless ( m = 0 ), but ( m > 0 ). Therefore, perhaps the problem is misinterpreted.Wait, maybe the problem is saying that for the specific ( x_1 ), ( f(x_1) ) is equal to ( g(x_2) ) for any ( x_2 ). So, ( f(x_1) ) is a fixed value, and ( g(x_2) ) must equal that fixed value for all ( x_2 ). But as we saw, ( g(x) ) is linear, so unless it's constant, it can't take the same value for all ( x_2 ).Therefore, unless ( m = 0 ), which is not allowed, this is impossible. So, perhaps the answer is that no such ( m ) exists, but since the options are given, maybe I need to think differently.Wait, perhaps the problem is intended to mean that for each ( x_2 ), there exists an ( x_1 ) such that ( f(x_1) = g(x_2) ). That would mean that the range of ( g(x) ) is a subset of the range of ( f(x) ). But that's a different interpretation.Wait, the problem says: "there exists ( x_1 in [0,3] ) such that for any ( x_2 in [0,3] ), ( f(x_1) = g(x_2) )". So, it's a single ( x_1 ) such that for every ( x_2 ), ( f(x_1) = g(x_2) ). So, ( f(x_1) ) must equal ( g(x_2) ) for all ( x_2 ), which as we saw, implies ( g(x) ) is constant, which is not possible because ( m > 0 ).Therefore, perhaps the correct interpretation is that the range of ( g(x) ) must be a subset of the range of ( f(x) ), and also ( g(x) ) must achieve the same value for all ( x_2 ). But that again leads to ( g(x) ) being constant.Alternatively, maybe the problem is that ( f(x_1) ) must be equal to ( g(x_2) ) for all ( x_2 ), meaning that ( f(x_1) ) is a value that ( g(x) ) must take for every ( x_2 ). But since ( g(x) ) is linear, it can only take each value once, unless it's constant. So, again, it's not possible unless ( m = 0 ).Wait, perhaps the problem is intended to mean that ( f(x_1) ) is equal to ( g(x_2) ) for some ( x_1 ) and for any ( x_2 ). So, for any ( x_2 ), there exists an ( x_1 ) such that ( f(x_1) = g(x_2) ). That would mean that the range of ( g(x) ) is a subset of the range of ( f(x) ). But then, ( g(x) ) is linear, so its range is an interval, and ( f(x) ) has a range of [-1, 3]. So, for the range of ( g(x) ) to be a subset of [-1, 3], we need ( g(0) geq -1 ) and ( g(3) leq 3 ).Let me compute ( g(0) = 2 - m ) and ( g(3) = 2 + 2m ).So, to have ( g(0) geq -1 ):( 2 - m geq -1 )( -m geq -3 )Multiply both sides by -1 (remembering to flip the inequality):( m leq 3 )And for ( g(3) leq 3 ):( 2 + 2m leq 3 )( 2m leq 1 )( m leq 1/2 )Also, since ( m > 0 ), combining these, we have ( 0 < m leq 1/2 ).So, the range of ( m ) is (0, 1/2], which is option A.Wait, that makes sense. So, perhaps my initial confusion was due to misinterpreting the problem. The key was to realize that for the range of ( g(x) ) to be a subset of the range of ( f(x) ), we need ( g(0) geq -1 ) and ( g(3) leq 3 ), leading to ( m leq 1/2 ). Therefore, the correct answer is A.</think>

🧠:<think>Alright, so I have this probability problem here, and I need to figure out the ratio q/p, where p is the probability of drawing five cards all with the same number, and q is the probability of drawing four cards with one number and the fifth card with a different number. Let me try to break this down step by step.First, let me understand the setup. There are 52 cards in a box, each labeled with a number from 1 to 13. Each number appears on four cards. So, four cards of each number 1 through 13. That makes sense because 13 numbers times four cards each gives us 52 cards total.Now, the problem says we're drawing five cards from the box without replacement. So, once a card is drawn, it can't be drawn again. I need to find two probabilities: p, which is the probability that all five cards have the same number, and q, which is the probability that four of the cards have a number 'a' and the fifth has a different number 'b' where 'b' is not equal to 'a'. Then, I need to find the ratio q/p.Okay, let's tackle p first. The probability p is the chance that all five cards have the same number. So, how can this happen? Well, since each number only has four cards, it's impossible to have five cards of the same number because there are only four of each. Wait, that can't be right because the problem mentions p as the probability of all five cards having the same number. Hmm, maybe I'm misunderstanding something.Wait, hold on. Let me read the problem again. It says, "Fifty-two cards are placed into a box, each bearing a number from 1 to 13, with each number entered on four cards. Five cards are drawn from the box at random and without replacement. Let p be the probability that all five cards bear the same number."Hmm, so if each number is only on four cards, how can five cards have the same number? That seems impossible. So is the probability p zero? But the problem then goes on to define q and asks for the ratio q/p, so maybe p isn't zero. Maybe I'm misinterpreting something.Wait, no. Maybe the problem is a standard one, and perhaps they mean five cards with the same rank, but in a deck, each rank has four suits. So, perhaps it's considering the number as the rank and the four suits as the different cards. So, in reality, you can't have five cards of the same rank because there are only four suits. Therefore, the probability p should be zero. But then q/p would be undefined. That can't be the case because the problem is asking for the ratio.Wait, maybe I'm overcomplicating. Let me check the original problem again. It says, "Fifty-two cards are placed into a box, each bearing a number from 1 to 13, with each number entered on four cards." So, four cards per number, 13 numbers, 52 cards. Then, five cards are drawn. So, p is the probability that all five have the same number. But since each number only has four cards, five cards can't all have the same number. So p is zero. Hmm, but the problem is asking for q/p, which would be undefined or maybe infinity? But the initial solution provided in the problem states that p is 13 divided by the combination of 52 choose 5, which is non-zero. So, there must be something wrong with my reasoning.Wait, maybe the problem is not considering that each number has exactly four cards, but that each number is written on four cards, but perhaps the numbers can be duplicated beyond four? No, the problem clearly states, "each number entered on four cards." So, there are four cards of each number.Therefore, drawing five cards all with the same number is impossible because there are only four of each number. So, how is p equal to 13 divided by the combination? That must be a mistake.Wait, maybe I'm misunderstanding the problem. Perhaps it's considering that each number is on four cards, but when it says "bearing the same number," it's referring to something else? No, that doesn't make sense.Wait, hold on. Let me try to compute p as if it's possible. Let's see. The number of ways to choose five cards all with the same number would be zero because there are only four of each number. So, the number of favorable outcomes is zero, hence p is zero. Then, q is the number of ways to choose four cards of one number and one card of another number.So, the number of ways to choose four cards of one number is 13 choices for the number, and then for each number, there's only one way to choose all four cards since there are exactly four. Then, for the fifth card, which has to be a different number, we have 12 remaining numbers, each with four cards, so 12 times 4. So, the total number of favorable outcomes for q would be 13 times 1 times (12 times 4) which is 13*1*48 = 624.Then, the total number of possible five-card hands is 52 choose 5, which is 2,598,960. So, p is zero, q is 624 divided by 2,598,960. But then, q/p would be undefined because division by zero is not allowed.But in the initial solution provided, they say p is 13 divided by 52 choose 5, which would be 13/2,598,960. So, where does this discrepancy come from?Wait, maybe the problem is mistyped, or I'm misinterpreting it. Let me check the problem again: "Fifty-two cards are placed into a box, each bearing a number from 1 to 13, with each number entered on four cards." So, four cards per number, 13 numbers, 52 cards. So, you can't have five cards of the same number because each number only has four cards. Therefore, p is zero.But in the initial solution, they say that there are 13 possibilities where all five cards have the same number, implying that they can choose any of the 13 numbers, but wait, each number only has four cards, so you can't have five. So, that initial solution must be wrong.Alternatively, maybe the problem intended that each number is on five cards instead of four? Because if each number was on five cards, then five of a kind would be possible. But the problem clearly states four cards per number.Alternatively, is the problem considering that the four cards are of different suits or something else? Wait, no, the problem is abstracted away from suits, just numbers from 1 to 13, each on four cards.Wait, perhaps the problem is not about five of a kind, but about five cards with the same number, but since each number only has four, it's impossible. So, p is zero, and q is non-zero, making q/p undefined.But the initial solution says p is 13 over 52 choose 5, which is not zero. So, maybe I need to re-examine the initial problem.Wait, perhaps I misread the problem. Let me read it again: "Fifty-two cards are placed into a box, each bearing a number from 1 to 13, with each number entered on four cards. Five cards are drawn from the box at random and without replacement. Let p be the probability that all five cards bear the same number. Let q be the probability that four of the cards bear a number a and the other bears a number b that is not equal to a. What is the value of q/p?"Wait, so perhaps the problem is considering that there are four cards of each number, so five cards cannot all be the same number, so p is zero, but maybe the problem is intended to have p as if it's possible, even though in reality, it's not.Alternatively, perhaps the problem is written incorrectly, and it should be four cards drawn, not five? Because with four cards, p would be 13 times 1 over 52 choose 4, which is 13 / 270725, and q would be 13*12*4 over 270725, giving q/p as 48. That would make sense.Alternatively, maybe the problem is correct, and I'm misinterpreting something. Wait, perhaps the five cards can have the same number if we consider something else, but no, each number only has four cards.Alternatively, maybe the numbers are not unique per four cards, but each number is on five cards, but the problem says four. So, I'm confused.Wait, maybe the initial solution is wrong, and the correct answer is that q/p is undefined because p is zero. But the problem is asking for a numerical value, so maybe I'm missing something.Wait, let me think differently. Maybe the problem is not considering that five of a kind is impossible, but perhaps the person who wrote the problem thought that five of a kind is possible, hence calculated p as 13 over 52 choose 5, which is incorrect because you can't have five of a kind with only four cards per number.Alternatively, maybe the problem is considering that the numbers can be repeated beyond four, but the problem explicitly states that each number is on four cards, so that's not the case.Alternatively, maybe I'm overcomplicating. Let me just calculate p and q as if p is non-zero, even though in reality, it's impossible.So, according to the initial solution, p is 13 / (52 choose 5) because there are 13 possible numbers, each with four cards, but wait, you can't choose five cards of the same number because there are only four. So, that initial solution is wrong.Wait, so maybe the problem is intended to have five cards, but in reality, five of a kind is impossible, so p is zero, making q/p undefined. But the problem is asking for a numerical value, so perhaps I need to proceed differently.Wait, maybe the problem is considering that the numbers can be anything, but each number appears four times, so five of a kind is impossible, but perhaps the problem is considering that you have five cards, and all five have the same number, even though it's impossible, so p is zero, and q is some positive number, so q/p is infinity. But in the initial solution, they calculate q/p as 48, which is finite.Wait, perhaps I need to think of p as the number of ways to choose five cards of the same number, which is zero, but the initial solution incorrectly counts it as 13. So, maybe I need to proceed as if p is 13, even though it's impossible.Alternatively, perhaps the problem is considering that each number is on five cards, not four, making five of a kind possible. So, then, p would be 13 times (5 choose 5) over (52 choose 5), which is 13 / 2598960, and q would be 13*12*(5 choose 4)*(5 choose 1) over (52 choose 5), which would be 13*12*5*5 = 3900, so q/p would be 3900 / 13 = 300. But that's not matching the initial solution.Wait, no, in the initial solution, they have four cards per number, so for p, it's 13*(4 choose 5), which is zero, but they incorrectly write it as 13*(4 choose 4) = 13*1 = 13, which is wrong because (4 choose 5) is zero. So, that initial solution is incorrect.Alternatively, maybe the problem is considering that you can have five of a kind by considering that the four cards are of the same number, and the fifth is also of the same number, but that's impossible. So, maybe the problem is intended to have p as the probability of getting all four of a number and one more, but that's actually q. So, perhaps p is zero, and q is 624 / (52 choose 5), making q/p undefined.Wait, this is getting confusing. Maybe I need to approach this differently. Let me calculate p and q correctly.First, p is the probability that all five cards have the same number. Since each number only has four cards, it's impossible to have five cards of the same number. Therefore, the number of ways to choose five cards of the same number is zero. Hence, p = 0.Now, q is the probability that four cards have a number 'a' and the fifth has a different number 'b'. So, how many ways can this happen? Well, first, choose the number 'a' which will have four cards. There are 13 choices for 'a'. Then, for each 'a', we need to choose all four cards of that number, which is only 1 way because there are exactly four cards. Then, choose a different number 'b', which can be any of the remaining 12 numbers, and for each 'b', choose one card out of the four available. So, for each 'a', there are 12*4 = 48 ways to choose 'b' and the card.Therefore, the total number of favorable outcomes for q is 13 * 1 * 48 = 624. The total number of possible five-card hands is 52 choose 5, which is 2,598,960. Therefore, q = 624 / 2,598,960.But since p is zero, q/p is undefined because you can't divide by zero. However, the problem is asking for a numerical value, so maybe there's a mistake in the problem statement.Alternatively, perhaps the problem intended to have five cards where at least four are the same number, which would include the case of five of a kind, but since five of a kind is impossible, p is zero, making q/p undefined.Alternatively, maybe the problem is considering that the four cards are of the same number, and the fifth can be any number, including the same, but that's not the case because q specifically requires the fifth card to be a different number.Wait, perhaps I need to think of it differently. Maybe the problem is considering that the five cards can have the same number, even though it's impossible, and just calculating the ratio q/p as if p were non-zero. So, if p is 13 / (52 choose 5), and q is 624 / (52 choose 5), then q/p is 624 / 13 = 48.But that would be incorrect because p is actually zero, making the ratio undefined. However, if we proceed under the assumption that p is non-zero, then q/p is 48.Alternatively, maybe the problem is intended to have four cards of a number and one different, and p is the probability of four of a kind, and q is the probability of a full house or something else. But no, the problem clearly defines p as all five cards the same number, which is impossible, and q as four of one number and one of another.So, in conclusion, I think the problem has a mistake because p is zero, making q/p undefined. However, if we ignore that and proceed with the initial solution, the ratio q/p is 48. Maybe the problem intended to have four cards drawn instead of five, making p possible. In that case, p would be 13*(4 choose 4) = 13, and q would be 13*12*(4 choose 4)*(4 choose 1) = 624, so q/p would be 48.Therefore, I think the answer is 48, considering that the problem may have a typo, and we proceed accordingly.</think>