answer：Okay, so I have this problem here: How many pairs (m, n) of integers satisfy the equation m + n = mn? The options are from A to E, with E being more than 4. Hmm, interesting. Let me try to figure this out step by step.First, I need to understand what the equation is saying. It says that the sum of m and n is equal to their product. So, m + n = mn. That seems a bit tricky because it's not a straightforward linear equation. Maybe I can rearrange it to make it easier to solve.Let me subtract m and n from both sides to get all terms on one side. So, mn - m - n = 0. Hmm, that looks a bit better. Maybe I can factor this somehow. I remember that sometimes adding 1 to both sides can help factor expressions like this. Let me try that.Adding 1 to both sides: mn - m - n + 1 = 1. Now, let's see if I can factor the left side. It looks like it might factor into (m - 1)(n - 1). Let me check:(m - 1)(n - 1) = mn - m - n + 1. Yes! That's exactly what I have on the left side. So, now the equation becomes (m - 1)(n - 1) = 1.Okay, so now I have the product of (m - 1) and (n - 1) equal to 1. Since m and n are integers, both (m - 1) and (n - 1) must be integers as well. So, I need to find all pairs of integers whose product is 1.I know that the integer pairs that multiply to 1 are (1, 1) and (-1, -1). Because 1 * 1 = 1 and (-1) * (-1) = 1. Are there any other integer pairs? Let me think. 1 and -1 are the only integers whose product is 1 because any other integer multiplied by something else wouldn't give 1. For example, 2 * 0.5 = 1, but 0.5 isn't an integer. So, yeah, only (1,1) and (-1,-1).So, now I can set up the equations for each case.Case 1: (m - 1) = 1 and (n - 1) = 1.Solving for m and n:m - 1 = 1 => m = 2n - 1 = 1 => n = 2So, one solution is (2, 2).Case 2: (m - 1) = -1 and (n - 1) = -1.Solving for m and n:m - 1 = -1 => m = 0n - 1 = -1 => n = 0So, another solution is (0, 0).Wait, are these the only solutions? Let me think. Since (m - 1) and (n - 1) must multiply to 1, and 1 only has two factor pairs in integers: (1,1) and (-1,-1). So, yeah, these are the only possible cases.But hold on, let me plug these back into the original equation to make sure they work.First, (2, 2):m + n = 2 + 2 = 4mn = 2 * 2 = 4So, 4 = 4. That works.Next, (0, 0):m + n = 0 + 0 = 0mn = 0 * 0 = 0So, 0 = 0. That also works.Hmm, so that gives me two solutions. But wait, the options go up to E, which is more than 4. So, maybe I missed something. Let me double-check.Wait, another thought: when I factored the equation, I assumed that (m - 1)(n - 1) = 1, but maybe I should consider that 1 can also be expressed as (1,1) and (-1,-1), but are there other possibilities? For example, if one of them is 1 and the other is 1, or both are -1. But in integers, 1 and -1 are the only divisors of 1.Wait, unless... Hmm, actually, in integers, 1 can be factored as 1*1 or (-1)*(-1). So, no, I think that's all. So, only two pairs.But let me think again. Maybe I can approach the problem differently. Let's solve for one variable in terms of the other.Starting from m + n = mn.Let me solve for n in terms of m.mn - n = mn(m - 1) = mSo, n = m / (m - 1)Since n must be an integer, m / (m - 1) must be an integer. So, let's think about when m / (m - 1) is an integer.Let me denote k = m - 1, so m = k + 1.Then, n = (k + 1) / k = 1 + 1/kSo, n must be an integer, which means that 1/k must be an integer. Therefore, k must be a divisor of 1.The divisors of 1 are 1 and -1.So, k = 1 or k = -1.Case 1: k = 1Then, m = k + 1 = 2n = 1 + 1/k = 1 + 1 = 2So, (2, 2) as before.Case 2: k = -1Then, m = k + 1 = 0n = 1 + 1/k = 1 + (-1) = 0So, (0, 0) again.So, same solutions. So, only two solutions.Wait, but the answer choices are A)1, B)2, C)3, D)4, E)more than 4.So, according to this, the answer is 2, which is option B.But hold on, let me think again. Maybe I missed some cases when I factored.Wait, in the equation (m - 1)(n - 1) = 1, I considered only the cases where both factors are 1 or both are -1. But in integers, 1 can also be expressed as 1*1 or (-1)*(-1). There are no other integer pairs because 1 is a prime in integers, right? So, only those two possibilities.Alternatively, maybe I should consider that m and n could be swapped, but in this case, since both are 2 and 0, swapping doesn't give new solutions. So, (2,2) and (0,0) are the only solutions.Wait, but let me test another approach. Let's suppose that m and n are integers, so let's list possible small integers and see.Let me try m = 0:Then, 0 + n = 0 * n => n = 0. So, n = 0. So, (0, 0) is a solution.m = 1:1 + n = 1 * n => 1 + n = n => 1 = 0, which is impossible. So, no solution.m = 2:2 + n = 2n => 2 = n. So, n = 2. So, (2, 2) is a solution.m = 3:3 + n = 3n => 3 = 2n => n = 3/2. Not integer.m = -1:-1 + n = (-1)*n => -1 + n = -n => -1 = -2n => n = 1/2. Not integer.m = -2:-2 + n = (-2)*n => -2 + n = -2n => -2 = -3n => n = 2/3. Not integer.Hmm, so it seems like only m = 0 and m = 2 give integer solutions for n. So, only two solutions: (0, 0) and (2, 2).Wait, but let me check m = n. If m = n, then the equation becomes 2m = m^2 => m^2 - 2m = 0 => m(m - 2) = 0 => m = 0 or m = 2. So, that gives us (0,0) and (2,2). So, same as before.Alternatively, if m ≠ n, is there any solution? Let's see.Suppose m ≠ n. Then, from the equation m + n = mn, let's rearrange it as mn - m - n = 0. Then, as before, adding 1 gives (m - 1)(n - 1) = 1. So, same as before, only two solutions. So, even if m ≠ n, but in integers, the only solutions are when both (m -1) and (n -1) are 1 or -1, which leads to m and n being 2 or 0. But since m and n are both 2 or both 0, they are equal. So, actually, all solutions have m = n.So, that means only two solutions: (0,0) and (2,2). So, the answer is 2, which is option B.But wait, hold on. Let me think again. Maybe I missed some other factor pairs. For example, 1 can be written as (1,1), (-1,-1), but also, in integers, 1 can be written as (1,1), (-1,-1), but also, is 1 considered as (1,1) and (-1,-1) only? Because in integers, 1 has only two divisors: 1 and -1. So, yeah, only two factor pairs.Wait, but in the equation (m - 1)(n - 1) = 1, if we consider that 1 can be expressed as 1*1 or (-1)*(-1), but also, in integers, 1 can be expressed as 1*1, (-1)*(-1), but also, is 1 considered as 1*1, (-1)*(-1), but also, can we have (1,1), (1,1), (-1,-1), (-1,-1)? No, because each factor pair is unique.Wait, maybe I'm overcomplicating. Let me think about the equation (m - 1)(n - 1) = 1. So, in integers, the possible pairs (a, b) such that a*b = 1 are (1,1) and (-1,-1). So, that gives us two possibilities:1. a = 1, b = 12. a = -1, b = -1Which correspond to:1. m - 1 = 1 => m = 2; n - 1 = 1 => n = 22. m - 1 = -1 => m = 0; n - 1 = -1 => n = 0So, only two solutions.Wait, but let me think about negative numbers. Suppose m is negative, does that give any solution?Let me try m = -1:Then, from m + n = mn => -1 + n = (-1)*n => -1 + n = -n => -1 = -2n => n = 1/2. Not integer.m = -2:-2 + n = (-2)*n => -2 + n = -2n => -2 = -3n => n = 2/3. Not integer.m = -3:-3 + n = (-3)*n => -3 + n = -3n => -3 = -4n => n = 3/4. Not integer.So, no solutions for negative m except m = 0, which we already considered.Similarly, trying n as negative:If n = -1:m + (-1) = m*(-1) => m -1 = -m => 2m = 1 => m = 1/2. Not integer.n = -2:m + (-2) = m*(-2) => m - 2 = -2m => 3m = 2 => m = 2/3. Not integer.So, no solutions for negative n except n = 0.Therefore, only two solutions: (0,0) and (2,2).Wait, but let me think again. Maybe I can consider that (m -1) and (n -1) can be 1 or -1, but also, is there a case where one is 1 and the other is -1? Let me see.If (m -1) = 1 and (n -1) = -1, then m = 2 and n = 0.Similarly, if (m -1) = -1 and (n -1) = 1, then m = 0 and n = 2.So, that gives two more solutions: (2, 0) and (0, 2).Wait, hold on, that's different from before. So, in the equation (m -1)(n -1) = 1, the factors can be (1,1), (-1,-1), (1,-1), (-1,1). But wait, 1*1=1, (-1)*(-1)=1, but 1*(-1)=-1, which is not equal to 1. So, actually, only (1,1) and (-1,-1) are valid because their product is 1. The other pairs give -1, which is not equal to 1.So, that means, (m -1) and (n -1) can't be 1 and -1 because their product would be -1, which is not 1. So, only the cases where both are 1 or both are -1.Therefore, (2,2) and (0,0) are the only solutions.Wait, but earlier when I considered m = 2 and n = 0, does that satisfy the original equation?Let me check: m + n = 2 + 0 = 2mn = 2 * 0 = 0So, 2 ≠ 0. So, that doesn't work. So, (2,0) is not a solution.Similarly, (0,2): m + n = 0 + 2 = 2mn = 0 * 2 = 0Again, 2 ≠ 0. So, that doesn't work either.So, my mistake earlier was thinking that (m -1) and (n -1) could be 1 and -1, but that would make their product -1, which is not equal to 1. So, those cases don't actually satisfy the equation. So, only the cases where both are 1 or both are -1.Therefore, only two solutions: (2,2) and (0,0).Wait, but let me think again. Maybe I can consider that (m -1) and (n -1) can be 1 or -1, but also, in the equation (m -1)(n -1) = 1, if one of them is 1 and the other is 1, or both are -1, but also, if one of them is 1 and the other is 1, or both are -1, but also, if one is 1 and the other is 1, or both are -1.Wait, I'm repeating myself. Let me think differently.Let me list all possible integer pairs (a, b) such that a*b = 1.In integers, the pairs are (1,1) and (-1,-1). There are no other integer pairs because 1 is a unit in integers, and its only divisors are 1 and -1.Therefore, only two possibilities for (a, b): (1,1) and (-1,-1). So, only two solutions.Therefore, the number of integer pairs (m, n) is 2.So, the answer is B) 2.Wait, but hold on, let me make sure I didn't miss any other factor pairs. For example, in some cases, people might consider 1 as 1*1, (-1)*(-1), but also, sometimes, people might think of other factorizations, but in integers, 1 only has those two factor pairs.Alternatively, if we consider that 1 can be written as (1,1), (-1,-1), but also, in some contexts, people might think of 1 as 1*1, (-1)*(-1), but also, in the case of zero, but zero can't be because 0 multiplied by anything is 0, not 1.So, yeah, only two solutions.Therefore, I think the answer is 2, which is option B.**Final Answer**boxed{B}

answer：Alright, so I have this problem about reciprocal sets. A reciprocal set is defined as a non-empty set A where 0 is not in A, and for every x in A, 1/x is also in A. I need to check each of the four given sets to see if they are reciprocal sets. The answer is supposed to be 2, but I need to figure out which ones they are.Let me go through each set one by one.**Set (1): {x ∈ ℝ | x² + a x + 1 = 0}**Okay, so this is the set of real solutions to the quadratic equation x² + a x + 1 = 0. First, I should check if this set is non-empty. For a quadratic equation, the discriminant is D = a² - 4*1*1 = a² - 4. So, if a² - 4 ≥ 0, there are real solutions. So, if |a| ≥ 2, the set is non-empty.Now, assuming |a| ≥ 2, the solutions are x = [-a ± sqrt(a² - 4)] / 2. Let me denote them as x₁ and x₂.I need to check if for every x in the set, 1/x is also in the set. So, let's see if 1/x₁ and 1/x₂ are roots of the equation.Let me compute 1/x₁. If x₁ is a root, then x₁² + a x₁ + 1 = 0. Let's solve for 1/x₁:1/x₁ = ?Wait, maybe it's better to see if 1/x is also a root. Let's substitute x = 1/x into the equation:(1/x)² + a*(1/x) + 1 = 0 => 1/x² + a/x + 1 = 0. Multiply both sides by x² to eliminate denominators:1 + a x + x² = 0, which is the same as x² + a x + 1 = 0. So, yes, if x is a root, then 1/x is also a root.Therefore, the set {x | x² + a x + 1 = 0} is a reciprocal set because it satisfies both conditions: it's non-empty (if |a| ≥ 2) and for every x in the set, 1/x is also in the set.Wait, but what if the quadratic has only one root? That is, when D = 0, which happens when a² - 4 = 0, so a = ±2. Then, the equation has a repeated root x = [-a]/2. Let's see, if a = 2, then x = -1. If a = -2, then x = 1.So, in the case of a = 2, the set is {-1}, and 1/(-1) = -1 is also in the set. Similarly, for a = -2, the set is {1}, and 1/1 = 1 is in the set. So, even in the case of a repeated root, it's still a reciprocal set.Therefore, Set (1) is a reciprocal set.**Set (2): {x | x² - 4x + 1 < 0}**This is the set of real numbers x such that x² - 4x + 1 < 0. Let's first find the solution set.The quadratic equation x² - 4x + 1 = 0 has roots at x = [4 ± sqrt(16 - 4)] / 2 = [4 ± sqrt(12)] / 2 = [4 ± 2*sqrt(3)] / 2 = 2 ± sqrt(3). So, the quadratic is positive outside the interval (2 - sqrt(3), 2 + sqrt(3)) and negative inside.Therefore, the set is (2 - sqrt(3), 2 + sqrt(3)). So, it's an open interval centered at 2 with radius sqrt(3).Now, we need to check if this set is a reciprocal set. First, it's non-empty because it's an interval. Second, for every x in the set, 1/x must also be in the set.So, let's see if for every x in (2 - sqrt(3), 2 + sqrt(3)), 1/x is also in that interval.First, note that 2 - sqrt(3) is approximately 2 - 1.732 = 0.2679, and 2 + sqrt(3) is approximately 3.732.So, the interval is (0.2679, 3.732). Now, let's check if 1/x is in the same interval for every x in (0.2679, 3.732).Take x = 0.2679, which is the lower bound. 1/x is approximately 3.732, which is the upper bound. Similarly, x = 3.732, 1/x is approximately 0.2679.But what about x in between? For example, take x = 1. Then 1/x = 1, which is in the interval. Take x = 2, 1/x = 0.5, which is in the interval. Take x = 0.5, 1/x = 2, which is in the interval.Wait, but let's check the endpoints. The interval is open, so x approaches 0.2679 from above, 1/x approaches 3.732 from below, which is still in the interval. Similarly, x approaching 3.732 from below, 1/x approaches 0.2679 from above, which is in the interval.But wait, is 1/x always in the interval for all x in the interval? Let's see.Suppose x is in (0.2679, 3.732). Then, 1/x is in (1/3.732, 1/0.2679) ≈ (0.2679, 3.732). So, 1/x is in the same interval. Therefore, for every x in the set, 1/x is also in the set.But wait, hold on. Let me test x = 0.5. 1/x = 2, which is in the interval. x = 2, 1/x = 0.5, which is in the interval. x = 1, 1/x = 1, which is in the interval. So, seems okay.But wait, is there any x in the interval such that 1/x is not in the interval? Suppose x is 0.3, which is in (0.2679, 3.732). Then 1/x ≈ 3.333, which is less than 3.732, so it's in the interval. Similarly, x = 3, 1/x = 1/3 ≈ 0.333, which is greater than 0.2679, so it's in the interval.Wait, but 0.2679 is approximately 2 - sqrt(3) ≈ 0.2679, and 1/(2 - sqrt(3)) = 2 + sqrt(3) ≈ 3.732. So, 1/(2 - sqrt(3)) = 2 + sqrt(3), which is exactly the upper bound.Similarly, 1/(2 + sqrt(3)) = 2 - sqrt(3). So, the reciprocal of the lower bound is the upper bound and vice versa.Therefore, the interval is reciprocal in the sense that taking reciprocals maps the interval to itself. So, for every x in (2 - sqrt(3), 2 + sqrt(3)), 1/x is also in that interval.Therefore, Set (2) is a reciprocal set.Wait, but hold on. Let me think again. The set is defined by x² - 4x + 1 < 0, which is equivalent to x being between 2 - sqrt(3) and 2 + sqrt(3). So, if x is in that interval, then 1/x is also in that interval because 1/(2 + sqrt(3)) = 2 - sqrt(3) and 1/(2 - sqrt(3)) = 2 + sqrt(3). So, the reciprocals of the endpoints are the other endpoints, and since the interval is open, the reciprocals of the interior points are also in the interior.Therefore, yes, Set (2) is a reciprocal set.Wait, but hold on. Let me test x = 1.5, which is in the interval. 1/x ≈ 0.6667, which is in the interval. Similarly, x = 0.3, 1/x ≈ 3.333, which is in the interval. So, seems okay.But wait, is 1/x always in the interval? Let's suppose x is in (2 - sqrt(3), 2 + sqrt(3)). Then, 1/x is in (1/(2 + sqrt(3)), 1/(2 - sqrt(3))) which is (2 - sqrt(3), 2 + sqrt(3)). So, yes, 1/x is in the same interval.Therefore, Set (2) is a reciprocal set.Wait, but I thought the answer was 2, but so far both (1) and (2) are reciprocal sets. Let me check the other sets.**Set (3): {y | y = (ln x)/x, x ∈ [1/e, 1) ∪ (1, e]}**Okay, so this set is the image of the function y = (ln x)/x over the domain x ∈ [1/e, 1) ∪ (1, e]. I need to find the range of this function and check if it's a reciprocal set.First, let's analyze the function y = (ln x)/x.We know that the function y = (ln x)/x has a maximum at x = e, where y = 1/e. It decreases on (0, e) and increases on (e, ∞). But in our case, x is in [1/e, 1) ∪ (1, e].So, let's compute y at the endpoints:At x = 1/e: y = (ln(1/e))/(1/e) = (-1)/(1/e) = -e.At x = 1: y = (ln 1)/1 = 0/1 = 0.At x = e: y = (ln e)/e = 1/e.Now, let's see the behavior of y in [1/e, 1):As x increases from 1/e to 1, ln x increases from -1 to 0, and x increases from 1/e to 1. So, y = (ln x)/x goes from -e to 0. But since ln x is negative in (1/e, 1), y is negative in this interval.Similarly, in (1, e], as x increases from 1 to e, ln x increases from 0 to 1, and x increases from 1 to e. So, y = (ln x)/x increases from 0 to 1/e.Therefore, the range of y is [-e, 0) ∪ (0, 1/e].Wait, but let me confirm:For x ∈ [1/e, 1), y = (ln x)/x. Since ln x is negative here, y is negative. Let's see the maximum and minimum.At x = 1/e, y = -e.As x approaches 1 from the left, y approaches 0 from below.Similarly, for x ∈ (1, e], y = (ln x)/x is positive, starting from 0 (as x approaches 1 from the right) up to 1/e at x = e.Therefore, the set {y} is [-e, 0) ∪ (0, 1/e].Now, we need to check if this set is a reciprocal set. So, first, it's non-empty. Second, for every y in the set, 1/y must also be in the set.But let's see:Take y = -e. Then 1/y = -1/e. Is -1/e in the set? The set is [-e, 0) ∪ (0, 1/e]. So, -1/e is approximately -0.3679, which is greater than -e (≈ -2.718). So, yes, -1/e is in [-e, 0).Similarly, take y = 1/e. Then 1/y = e. But e is approximately 2.718, which is not in the set {y | y ∈ [-e, 0) ∪ (0, 1/e]}. So, 1/y = e is not in the set.Therefore, for y = 1/e, 1/y is not in the set. Therefore, Set (3) is not a reciprocal set.Wait, but hold on. The set is [-e, 0) ∪ (0, 1/e]. So, 1/y for y = 1/e is e, which is outside the set. Similarly, for y approaching 0 from above, 1/y approaches infinity, which is not in the set. For y approaching 0 from below, 1/y approaches negative infinity, which is not in the set.Therefore, Set (3) is not a reciprocal set.Wait, but let me think again. The set is [-e, 0) ∪ (0, 1/e]. So, it's symmetric around y = 0 in a way, but reciprocals would require that for every y, 1/y is also in the set. But as we saw, for y = 1/e, 1/y = e is not in the set. Similarly, for y = -e, 1/y = -1/e is in the set. So, some reciprocals are in, some are not.Therefore, Set (3) is not a reciprocal set.**Set (4): { y | y = (2x + 2/5)/(x + 1/x), x ∈ [0,1) ∪ [1,2] }**Okay, so this set is the image of the function y = (2x + 2/5)/(x + 1/x) over the domain x ∈ [0,1) ∪ [1,2]. Let me first simplify the function.First, note that x + 1/x is in the denominator. So, x cannot be 0 because 1/x would be undefined. But in the domain, x ∈ [0,1) ∪ [1,2], so x = 0 is included? Wait, x ∈ [0,1) ∪ [1,2], so x can be 0. But if x = 0, then 1/x is undefined. So, actually, the domain should exclude x = 0. Wait, the problem statement says x ∈ [0,1) ∪ [1,2]. Hmm, but x = 0 is problematic because 1/x is undefined. Maybe it's a typo, or perhaps x ∈ (0,1) ∪ [1,2]. But as written, x can be 0, but then 1/x is undefined. So, perhaps the domain is actually (0,1) ∪ [1,2]. Maybe the problem meant that.Alternatively, maybe the function is defined piecewise, but in the problem statement, it's written as:y = (2x + 2/5)/(x + 1/x) for x ∈ [0,1) and x ∈ [1,2]. So, for x ∈ [0,1), it's defined, but at x = 0, it's undefined. So, perhaps the domain is (0,1) ∪ [1,2]. Otherwise, the set would be undefined at x = 0.Assuming that, let's proceed.So, the function is y = (2x + 2/5)/(x + 1/x). Let me simplify this expression.Multiply numerator and denominator by x to eliminate the fraction in the denominator:y = [ (2x + 2/5) * x ] / [x*(x + 1/x)] = [2x² + (2/5)x] / [x² + 1]So, y = (2x² + (2/5)x)/(x² + 1)That's a bit simpler.Now, we can analyze this function over x ∈ (0,1) ∪ [1,2].Let me first consider x ∈ (0,1):As x approaches 0 from the right, y approaches (0 + 0)/(0 + 1) = 0.At x = 1, y = (2*1 + 2/5*1)/(1 + 1) = (2 + 0.4)/2 = 2.4/2 = 1.2.Wait, but x =1 is included in both intervals, but in the problem statement, it's written as x ∈ [0,1) ∪ [1,2], so x=1 is included in both, but in reality, it's just x ∈ [0,2], excluding x=0? Wait, no, x=0 is included but the function is undefined there. So, perhaps the domain is (0,2].But regardless, let's compute y at x=1: y = (2*1 + 2/5)/(1 + 1/1) = (2 + 0.4)/2 = 2.4/2 = 1.2.Now, let's compute y at x=2: y = (2*4 + (2/5)*2)/(4 + 1) = (8 + 0.8)/5 = 8.8/5 = 1.76.Wait, but let me compute that again.Wait, x=2:Numerator: 2*(2)^2 + (2/5)*2 = 2*4 + (4/5) = 8 + 0.8 = 8.8Denominator: (2)^2 + 1 = 4 + 1 = 5So, y = 8.8 / 5 = 1.76.Now, let's see the behavior of y in (0,1):As x approaches 0+, y approaches 0.At x=1, y=1.2.What about the derivative? Let me compute dy/dx to see if the function is increasing or decreasing.But maybe it's easier to analyze the function.Wait, let me compute y at x=0.5:y = (2*(0.5)^2 + (2/5)*0.5)/( (0.5)^2 + 1 ) = (2*0.25 + 0.2)/ (0.25 + 1 ) = (0.5 + 0.2)/1.25 = 0.7 / 1.25 = 0.56.Similarly, at x=0.25:y = (2*(0.0625) + (2/5)*0.25)/(0.0625 + 1) = (0.125 + 0.1)/1.0625 ≈ 0.225 / 1.0625 ≈ 0.2118.So, as x increases from 0 to 1, y increases from 0 to 1.2.Similarly, in the interval [1,2], let's see:At x=1, y=1.2.At x=2, y=1.76.Let me compute y at x=1.5:y = (2*(2.25) + (2/5)*1.5)/(2.25 + 1) = (4.5 + 0.6)/3.25 = 5.1 / 3.25 ≈ 1.57.So, the function seems to be increasing in both intervals.Therefore, the range of y is:For x ∈ (0,1): y ∈ (0, 1.2)For x ∈ [1,2]: y ∈ [1.2, 1.76]Therefore, the entire set is y ∈ (0, 1.76].Wait, but at x=1, y=1.2, which is included in both intervals.So, the set {y} is (0, 1.76].Now, we need to check if this set is a reciprocal set. So, first, it's non-empty. Second, for every y in (0, 1.76], 1/y must also be in the set.But let's see:Take y = 1.76. Then 1/y ≈ 0.568. Is 0.568 in (0, 1.76]? Yes.Take y = 1. Then 1/y = 1, which is in the set.Take y = 0.5. Then 1/y = 2, which is not in the set because the set only goes up to 1.76.Therefore, 1/y = 2 is not in the set. Hence, Set (4) is not a reciprocal set.Wait, but let me confirm:The set is (0, 1.76]. So, for y ∈ (0, 1.76], 1/y must be in (0, 1.76].But 1/y is in [1/1.76, ∞) ≈ [0.568, ∞). So, 1/y is not necessarily in (0, 1.76]. For example, y = 0.5, 1/y = 2, which is outside the set.Therefore, Set (4) is not a reciprocal set.Wait, but let me think again. Maybe I made a mistake in computing the range.Wait, the function y = (2x² + (2/5)x)/(x² + 1). Let me see if this function can take values beyond 1.76.Wait, at x=2, y=1.76. As x increases beyond 2, y would approach 2 as x approaches infinity, but in our case, x is only up to 2. So, the maximum y is 1.76.But let me check the behavior as x approaches infinity, but since x is limited to 2, the maximum is 1.76.Therefore, the range is (0, 1.76]. So, reciprocals of y in (0,1) would be greater than 1, but since the set only goes up to 1.76, reciprocals of y < 1/1.76 ≈ 0.568 would be greater than 1.76, which are not in the set.Therefore, Set (4) is not a reciprocal set.So, summarizing:Set (1): Reciprocal set.Set (2): Reciprocal set.Set (3): Not a reciprocal set.Set (4): Not a reciprocal set.Therefore, the number of reciprocal sets is 2.**Final Answer**boxed{2}

answer：Okay, so I have this problem here about finding the number of common tangent lines between two circles, Q₁ and Q₂. Let me see. Q₁ is given by the equation x² + y² = 9, and Q₂ is (x - 3)² + (y - 4)² = 1. Hmm, I need to figure out how many common tangent lines these two circles have.First, I remember that the number of common tangents between two circles depends on their positions relative to each other. There are different cases: if one circle is entirely inside another, they might have no common tangents. If they intersect, they have two common tangents. If they're separate, they might have four common tangents. Wait, actually, let me recall: when two circles are separate and don't intersect, they have four common tangents—two direct and two transverse. If they intersect at two points, they have two common tangents. If one is inside the other without touching, they have no common tangents. If they're tangent to each other, either externally or internally, they have three common tangents. So, I need to figure out the relationship between these two circles.Let me write down the details of each circle. Q₁ is x² + y² = 9, so that's a circle centered at the origin (0,0) with radius 3. Q₂ is (x - 3)² + (y - 4)² = 1, so its center is at (3,4) and radius is 1. Okay, so centers are at (0,0) and (3,4), radii 3 and 1.Now, I should find the distance between the centers. The distance formula between two points (x₁, y₁) and (x₂, y₂) is √[(x₂ - x₁)² + (y₂ - y₁)²]. So, plugging in the centers, that would be √[(3 - 0)² + (4 - 0)²] = √[9 + 16] = √25 = 5. So the distance between the centers is 5 units.Now, the radii are 3 and 1. Let me see: the sum of the radii is 3 + 1 = 4, and the difference is 3 - 1 = 2. So, the distance between centers is 5, which is greater than the sum of the radii (4). That means the circles are separate from each other, right? Because if the distance between centers is greater than the sum of radii, they don't intersect and are separate. So, in that case, they should have four common tangents.Wait, but let me make sure. If the distance is greater than the sum, they are external to each other, so four tangents. If the distance is equal to the sum, they are externally tangent, so three tangents. If the distance is less than the sum but greater than the difference, they intersect at two points, so two tangents. If the distance is equal to the difference, they are internally tangent, so one tangent. If the distance is less than the difference, one is inside the other, so no common tangents.So in this case, distance is 5, sum is 4. Since 5 > 4, they are separate, so four common tangents. Therefore, the answer should be 4.But wait, let me visualize it. Q₁ is a bigger circle with radius 3, and Q₂ is a smaller circle with radius 1, located at (3,4). The distance between centers is 5. So, the circles are definitely separate because 5 is more than 3 + 1. So, yes, four common tangents.Alternatively, maybe I can calculate using the formula for the number of common tangents. The number of common tangents can be determined by the distance between centers and the radii. The formula is:- If d > r₁ + r₂: 4 common tangents- If d = r₁ + r₂: 3 common tangents- If |r₁ - r₂| < d < r₁ + r₂: 2 common tangents- If d = |r₁ - r₂|: 1 common tangent- If d < |r₁ - r₂|: 0 common tangentsWhere d is the distance between centers, r₁ and r₂ are the radii.In our case, d = 5, r₁ = 3, r₂ = 1. So, |r₁ - r₂| = 2, and r₁ + r₂ = 4. Since 5 > 4, it falls into the first category, so 4 common tangents.Therefore, the answer is 4.Wait, just to make sure, maybe I can think about drawing the circles. The first circle is centered at (0,0) with radius 3, so it touches (3,0), (0,3), (-3,0), (0,-3). The second circle is at (3,4) with radius 1, so it's somewhere in the first quadrant, not overlapping with the first circle because the distance between centers is 5, which is more than 3 + 1 = 4. So, they don't intersect and are separate, so four tangents.Alternatively, maybe I can calculate the equations of the common tangents, but that might be more complicated. But since the question just asks for the number, I think 4 is correct.Wait, but just to be thorough, let me think about the possible common tangents. There are two types: external and internal. External tangents don't cross between the circles, while internal tangents do. Since the circles are separate, both external and internal tangents exist. For external tangents, there are two lines, and for internal tangents, there are also two lines, making a total of four.Yes, that makes sense. So, four common tangent lines.**Final Answer**The number of common tangent lines is boxed{4}.

answer：Alright, so I have this problem here: I need to find how many points the graphs of ( y = f(x^2) ) and ( y = f(x^4) ) intersect. The function ( f ) is defined for all real numbers and is invertible, which means it has an inverse function ( f^{-1} ). First, let me understand what it means for two graphs to intersect. They intersect at points where their ( y )-values are equal for the same ( x )-value. So, I need to solve the equation ( f(x^2) = f(x^4) ) for ( x ). Since ( f ) is invertible, it must be one-to-one, meaning that if ( f(a) = f(b) ), then ( a = b ). That's a crucial property because it allows me to set the arguments of ( f ) equal to each other. So, from ( f(x^2) = f(x^4) ), I can conclude that ( x^2 = x^4 ).Now, I need to solve ( x^2 = x^4 ). Let me rearrange this equation:( x^4 - x^2 = 0 )Factor out an ( x^2 ):( x^2(x^2 - 1) = 0 )So, this gives me two factors: ( x^2 = 0 ) and ( x^2 - 1 = 0 ).Solving ( x^2 = 0 ) gives ( x = 0 ).Solving ( x^2 - 1 = 0 ) gives ( x^2 = 1 ), so ( x = 1 ) or ( x = -1 ).Therefore, the solutions are ( x = -1 ), ( x = 0 ), and ( x = 1 ).Wait, but I should check if these solutions are valid. Since ( f ) is invertible, it's defined for all real numbers, so plugging these ( x )-values into ( f(x^2) ) and ( f(x^4) ) should be fine. Let me verify each solution:1. For ( x = 0 ): ( f(0^2) = f(0) ) and ( f(0^4) = f(0) ), so they are equal.2. For ( x = 1 ): ( f(1^2) = f(1) ) and ( f(1^4) = f(1) ), so they are equal.3. For ( x = -1 ): ( f((-1)^2) = f(1) ) and ( f((-1)^4) = f(1) ), so they are equal.So, all three solutions are valid.But wait, is that all? Let me think if there are any other solutions. The equation ( x^2 = x^4 ) is a quartic equation, which can have up to four real roots. But in this case, we've found three real roots: -1, 0, and 1. Is there another root?Let me consider the equation again:( x^4 - x^2 = 0 )( x^2(x^2 - 1) = 0 )Which factors into ( x^2 = 0 ) and ( x^2 = 1 ). So, the solutions are ( x = 0 ), ( x = 1 ), and ( x = -1 ). There are no other real solutions because ( x^2 ) can't be negative, so ( x^2 - 1 = 0 ) only gives ( x = pm 1 ), and ( x^2 = 0 ) only gives ( x = 0 ).Therefore, there are exactly three real solutions: ( x = -1 ), ( x = 0 ), and ( x = 1 ). But hold on, let me think about the graphs of ( y = f(x^2) ) and ( y = f(x^4) ). Since ( f ) is invertible, it's either strictly increasing or strictly decreasing. Let's assume ( f ) is increasing without loss of generality because if it's decreasing, the reasoning would be similar.So, if ( f ) is increasing, then ( f(x^2) ) and ( f(x^4) ) are both functions that take ( x ) to some positive value through ( x^2 ) and ( x^4 ). But wait, ( x^2 ) and ( x^4 ) are both even functions, so their graphs are symmetric about the y-axis. Therefore, for every positive solution ( x ), there is a corresponding negative solution ( -x ). But in our case, we already have ( x = 1 ) and ( x = -1 ) as solutions, so that symmetry is already accounted for. The third solution is at ( x = 0 ), which is on the y-axis and doesn't have a symmetric counterpart.So, plotting these functions, ( y = f(x^2) ) would have a shape similar to ( f ) composed with a parabola, and ( y = f(x^4) ) would have a shape similar to ( f ) composed with a quartic curve. Since ( x^4 ) grows faster than ( x^2 ) for ( |x| > 1 ), but for ( |x| < 1 ), ( x^4 ) is less than ( x^2 ). But since ( f ) is invertible, it's either always increasing or always decreasing. If it's increasing, then ( f(x^4) ) will be increasing for ( x > 0 ) and decreasing for ( x < 0 ), but since it's an even function, it's symmetric. Similarly, ( f(x^2) ) is also symmetric.Wait, actually, ( f(x^2) ) and ( f(x^4) ) are both even functions because ( x^2 ) and ( x^4 ) are even, so their composition with ( f ) preserves the evenness. Therefore, their graphs are symmetric about the y-axis.So, when I found the solutions at ( x = -1 ), ( x = 0 ), and ( x = 1 ), these are all the intersection points because beyond these points, the functions don't intersect again. Let me think about for ( x > 1 ), ( x^4 > x^2 ), so ( f(x^4) > f(x^2) ) if ( f ) is increasing, and ( f(x^4) < f(x^2) ) if ( f ) is decreasing. Similarly, for ( 0 < x < 1 ), ( x^4 < x^2 ), so ( f(x^4) < f(x^2) ) if ( f ) is increasing, and ( f(x^4) > f(x^2) ) if ( f ) is decreasing.But regardless of whether ( f ) is increasing or decreasing, the functions ( f(x^2) ) and ( f(x^4) ) will only intersect at points where ( x^2 = x^4 ), which are exactly the points we found: ( x = -1 ), ( x = 0 ), and ( x = 1 ). Therefore, the number of intersection points is three.Wait, but hold on a second. Let me visualize the graphs. For ( x = 0 ), both functions equal ( f(0) ). For ( x = 1 ) and ( x = -1 ), both functions equal ( f(1) ). But what about for other values? For example, is there a point where ( f(x^2) = f(x^4) ) without ( x^2 = x^4 )? Since ( f ) is invertible, that can't happen because if ( f(a) = f(b) ), then ( a = b ). So, no, there are no other solutions.Therefore, the total number of intersection points is three.But wait, hold on again. Let me think about the multiplicity of roots. The equation ( x^4 - x^2 = 0 ) can be written as ( x^2(x^2 - 1) = 0 ), which gives roots at ( x = 0 ) (with multiplicity 2) and ( x = pm 1 ) (each with multiplicity 1). But in terms of distinct real solutions, it's still three: -1, 0, and 1.So, even though ( x = 0 ) is a double root, it's still just one point on the graph where they intersect. So, the number of intersection points is three.Therefore, I think the answer is three points.But wait, let me just make sure I didn't miss any other solutions. Let me consider substituting some other values:For example, ( x = 2 ): ( f(4) ) vs ( f(16) ). Since ( f ) is invertible, unless ( f(4) = f(16) ), which would require ( 4 = 16 ), which is not true. So, they don't intersect there.Similarly, ( x = sqrt{2} ): ( f(2) ) vs ( f(4) ). Again, unless ( 2 = 4 ), which isn't, so no intersection.What about ( x = frac{1}{2} ): ( fleft(frac{1}{4}right) ) vs ( fleft(frac{1}{16}right) ). Since ( frac{1}{4} neq frac{1}{16} ), they don't intersect here either.So, indeed, the only points where ( f(x^2) = f(x^4) ) are at ( x = -1 ), ( x = 0 ), and ( x = 1 ).Therefore, the graphs intersect at three points.**Final Answer**The graphs intersect at boxed{3} points.

answer：Alright, so I need to solve this problem where I have two very large numbers, and I have to compute the first number raised to the power of the second number. Then, from that huge result, I only need the last digit. Hmm, okay, let's break this down step by step.First, the numbers are really big. The first number is 954950230952380948328708, and the second is 470128749397540235934750230. Raising such a large number to another large power is going to result in an astronomically huge number. Obviously, I can't compute that directly because it's way beyond what any standard calculator or even a computer can handle without some special algorithms. So, I need a smarter way to find just the last digit without computing the entire number.I remember that when dealing with exponents, the last digit of the result depends only on the last digit of the base and the exponent. So, maybe I can simplify this problem by looking at the last digit of the first number and the exponent.Let me check the last digit of the first number. The first number is 954950230952380948328708. The last digit here is 8. So, the base is effectively 8 for the purpose of finding the last digit of the result.Now, the exponent is 470128749397540235934750230. That's a massive number. But again, for the purpose of finding the last digit of 8 raised to this exponent, I don't need the entire exponent. I just need to know the exponent modulo something, probably 4, because the last digit of powers of 8 cycles every 4.Let me recall the pattern of the last digit of powers of 8:- 8^1 = 8 → last digit 8- 8^2 = 64 → last digit 4- 8^3 = 512 → last digit 2- 8^4 = 4096 → last digit 6- 8^5 = 32768 → last digit 8- And then it repeats every 4 exponents.So, the cycle is 8, 4, 2, 6, and then repeats. Therefore, the last digit cycles every 4. That means if I can find the exponent modulo 4, I can determine where in the cycle the last digit falls.But wait, the exponent is 470128749397540235934750230. Let me find this exponent modulo 4. To find a number modulo 4, I can look at its last two binary digits, but since it's a decimal number, another way is to take the last two digits and compute that modulo 4.Wait, actually, for modulo 4, any number is congruent to its last two digits modulo 4. So, let me take the last two digits of the exponent. The exponent is 470128749397540235934750230. The last two digits are 30.Now, 30 divided by 4 is 7 with a remainder of 2. So, 30 mod 4 is 2. Therefore, the exponent is congruent to 2 modulo 4.So, that means the exponent is of the form 4k + 2. Therefore, when we raise 8 to this exponent, the last digit will be the same as 8^(4k + 2), which is (8^4)^k * 8^2. Since 8^4 ends with 6, and any power of 6 ends with 6, so (8^4)^k ends with 6. Then, multiplying by 8^2, which ends with 4, so 6 * 4 = 24, which ends with 4.Wait, hold on. Let me double-check that. If the exponent is 2 modulo 4, then according to the cycle, the last digit should be 4. Because:- If exponent mod 4 is 1 → last digit 8- If exponent mod 4 is 2 → last digit 4- If exponent mod 4 is 3 → last digit 2- If exponent mod 4 is 0 → last digit 6Yes, that's correct. So, since the exponent mod 4 is 2, the last digit is 4.But wait, let me make sure I didn't make a mistake in calculating the exponent mod 4. The exponent is 470128749397540235934750230. The last two digits are 30. 30 divided by 4 is 7 with a remainder of 2, so 30 mod 4 is indeed 2. Therefore, the exponent mod 4 is 2.Therefore, the last digit of 8 raised to this exponent is 4.So, putting it all together, the last digit of 954950230952380948328708 raised to the power of 470128749397540235934750230 is 4.Wait, just to be thorough, let me test this with smaller numbers to make sure my reasoning is correct.For example, take 8^2 = 64 → last digit 4. That matches.8^6: 8^6 = (8^4)*(8^2) = 4096*64 = ... well, last digit is 6*4=24 → 4. Wait, 8^6 is 262144, which ends with 4. Correct.Another example: 8^10. 10 mod 4 is 2, so last digit should be 4. 8^10 is 1073741824, which indeed ends with 4.Another test: 8^3. 3 mod 4 is 3, so last digit should be 2. 8^3 is 512, ends with 2. Correct.8^4: 4 mod 4 is 0, last digit 6. Correct.8^5: 5 mod 4 is 1, last digit 8. Correct.So, the pattern holds. Therefore, my reasoning seems solid.Therefore, the last digit is 4.

answer：First, I need to determine the number of ways to select one mathematics book from the three available. Since there are three distinct mathematics books, there are 3 possible choices for selecting one.Next, I need to determine the number of ways to select one physics book from the two available. With two distinct physics books, there are 2 possible choices for selecting one.To find the total number of ways to select one mathematics book and one physics book, I multiply the number of choices for each subject. Therefore, the total number of ways is 3 multiplied by 2, which equals 6.