💡Alright, so I'm trying to figure out whether Chichikov is right in thinking that for any company, he can draw a circle with chords representing each person such that acquaintances' chords intersect and strangers' chords don't. Hmm, that sounds interesting. Let me break it down step by step.First, I need to understand the problem clearly. We have a company with several people, and Chichikov wants to represent each person as a chord on a circle. If two people are acquaintances, their chords should intersect; if they're strangers, their chords shouldn't intersect. Even if the endpoints touch, that's considered an intersection. So, the goal is to see if this is always possible, no matter who knows whom in the company.Okay, so let's think about a simple case first. Suppose there are two people in the company. If they are acquaintances, their chords should intersect. If they are strangers, their chords shouldn't intersect. That seems straightforward. We can draw two chords that intersect or not, depending on their relationship.What about three people? Let's say we have three people: A, B, and C. If A knows B and C, but B and C don't know each other, can we represent this with chords? Let's try. We can draw a circle and place three chords such that A's chord intersects both B's and C's chords, but B's and C's chords don't intersect each other. That seems possible.Wait, but what if all three know each other? Then all three chords should intersect each other. Is that possible? Hmm, if we have three chords in a circle, each intersecting the other two, that would mean each chord crosses the other two. But in a circle, if you have three chords, each intersecting the other two, that would require each chord to cross the other two inside the circle. But in reality, with three chords, you can only have a certain number of intersections. Let me visualize this.Imagine three chords: AB, AC, and BC. If AB and AC intersect, and AB and BC intersect, then AC and BC must also intersect. But in a circle, if AB intersects both AC and BC, then AC and BC would have to intersect as well. So, actually, it's not possible to have three chords where each pair intersects because that would require each chord to cross the other two, which isn't possible in a circle without overlapping endpoints.Wait, but in the problem statement, it says that touching endpoints are considered intersections. So, does that mean that if two chords share an endpoint, that's considered an intersection? If so, then in the case of three people all knowing each other, their chords would all share endpoints, which would mean they all intersect at those endpoints. But in reality, each chord is defined by two distinct endpoints, so if all three chords share endpoints, that would mean all three chords are the same chord, which isn't possible.Hmm, maybe I'm misunderstanding. Perhaps the chords don't necessarily share endpoints, but just intersect somewhere on the circle. So, maybe the chords can be arranged such that each pair intersects inside the circle, not just at the endpoints. But in that case, with three chords, it's still not possible for all three to intersect each other inside the circle without overlapping.Wait, no, actually, it is possible. If you arrange three chords such that each pair crosses inside the circle, forming a triangle-like intersection pattern. So, each chord intersects the other two inside the circle. That would satisfy the condition that all three are acquaintances. So, maybe that works.Okay, so for three people, it's possible. What about four people? Let's say we have four people: A, B, C, and D. Suppose A knows B and C, B knows C and D, C knows D, and D knows A. Can we represent this with chords?Let me try to visualize. We need chords for A, B, C, and D. A's chord should intersect B and C's chords but not D's. B's chord should intersect A and C's chords but not D's. C's chord should intersect A and B's chords and also D's chord. D's chord should intersect C's chord but not A's or B's.Hmm, that seems tricky. Let's try to draw this out mentally. Place A's chord somewhere, then B's chord should intersect A's. Then C's chord should intersect both A's and B's chords. D's chord needs to intersect C's but not A's or B's.Wait, if C's chord intersects both A's and B's chords, then D's chord has to intersect C's chord somewhere. But if D's chord is placed such that it only intersects C's chord, it shouldn't intersect A's or B's. Is that possible?Maybe if we arrange the chords in a way that D's chord is nested inside C's chord, but outside A's and B's chords. But then, would D's chord intersect C's chord without intersecting A's or B's? I'm not sure. It might depend on how the chords are arranged.Alternatively, maybe it's not possible to represent this particular relationship with chords. If that's the case, then Chichikov's assumption might be wrong.Wait, but maybe I'm overcomplicating it. Let's think about the general case. Is there a way to represent any acquaintance relationship with chords on a circle such that acquaintances intersect and strangers don't?I recall something called a "circle graph," which is an intersection graph of chords on a circle. In a circle graph, each vertex represents a chord, and edges represent intersections between chords. So, the problem is essentially asking whether any graph can be represented as a circle graph.But I also remember that not all graphs are circle graphs. There are certain properties that a graph must satisfy to be a circle graph. For example, circle graphs are closely related to Eulerian tours and have specific forbidden induced subgraphs.So, if not all graphs are circle graphs, then Chichikov's assumption that such an arrangement exists for any company is incorrect.But wait, let me make sure. Maybe I'm misremembering. Are there specific types of graphs that cannot be represented as circle graphs? For example, certain types of bipartite graphs or graphs with high treewidth?I think the key point is that circle graphs have a specific structure, and not every graph can be embedded as chords on a circle with the given intersection properties. Therefore, Chichikov's belief is not universally correct.To confirm, let's think about a specific example. Consider the complete graph on four vertices, K4. Can K4 be represented as a circle graph? In K4, every pair of vertices is connected, so every pair of chords must intersect. Is that possible?If we try to draw four chords on a circle where every pair intersects, we run into a problem. Each chord can only intersect so many other chords before they start overlapping or coinciding, which isn't allowed. Specifically, with four chords, each chord can intersect at most two others without overlapping. But in K4, each chord needs to intersect three others, which isn't possible.Therefore, K4 cannot be represented as a circle graph, meaning that Chichikov's method wouldn't work for a company where everyone knows each other.Wait, but earlier I thought that three chords could intersect each other, but now with four, it's impossible. So, that seems to confirm that not all graphs can be represented as circle graphs.Another example is the cycle graph C5, which is a pentagon. Is C5 a circle graph? I think it is, because you can arrange five chords on a circle such that each chord intersects its two neighbors, forming a cycle.But what about the complement of C5? The complement of C5 is another graph where each vertex is connected to the two vertices it wasn't connected to in C5. Is that a circle graph? I'm not sure, but if the complement isn't a circle graph, then that would be another example.Alternatively, consider the graph formed by two triangles sharing a common edge. Is that a circle graph? I'm not sure, but it might not be.In any case, the key takeaway is that not all graphs can be represented as circle graphs. Therefore, Chichikov's method doesn't work for every possible company.So, to sum up, Chichikov's belief is incorrect because there exist companies (graphs) where it's impossible to arrange the chords such that acquaintances intersect and strangers don't. Specifically, certain graphs like K4 cannot be represented as circle graphs, making Chichikov's method inapplicable in those cases.**Final Answer**boxed{text{Not right}}

💡Okay, so I have this expression to factor: (x^{10} - 1024). Hmm, let me think about how to approach this. I remember that when factoring expressions, especially polynomials, it's helpful to look for patterns or known factoring formulas. First, I notice that (x^{10}) is a power of (x), and 1024 is a number I recognize as a power of 2. Let me check: (2^{10} = 1024). Oh, that's right! So, (1024 = 2^{10}). That means I can rewrite the expression as (x^{10} - 2^{10}). Now, this looks like a difference of two terms raised to the 10th power. I remember that a difference of squares can be factored as (a^2 - b^2 = (a - b)(a + b)). But here, we have a difference of 10th powers, not squares. Is there a similar factoring formula for higher powers?I think so. For any exponent (n), the difference of (n)th powers can be factored if (n) is even or odd. Since 10 is even, maybe I can factor it as a difference of squares first. Let me try that.So, (x^{10} - 2^{10}) can be seen as ((x^5)^2 - (2^5)^2). That simplifies to ((x^5)^2 - 32^2), since (2^5 = 32). Now, applying the difference of squares formula, this becomes ((x^5 - 32)(x^5 + 32)). Alright, so now I have factored (x^{10} - 1024) into ((x^5 - 32)(x^5 + 32)). But I wonder if these factors can be factored further. Let me check each one.Starting with (x^5 - 32), this is a difference of fifth powers. I recall that (a^5 - b^5) can be factored as ((a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4)). Let me apply that here.Let (a = x) and (b = 2), since (32 = 2^5). So, (x^5 - 2^5) factors into ((x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)). Now, looking at the quartic polynomial (x^4 + 2x^3 + 4x^2 + 8x + 16), I wonder if it can be factored further. Maybe it can be factored into quadratics or even linear terms. Let me try to factor it.I'll attempt to factor (x^4 + 2x^3 + 4x^2 + 8x + 16) into two quadratics. Suppose it factors as ((x^2 + ax + b)(x^2 + cx + d)). Multiplying these out gives:[x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd]Comparing coefficients with (x^4 + 2x^3 + 4x^2 + 8x + 16), we get the following system of equations:1. (a + c = 2)2. (ac + b + d = 4)3. (ad + bc = 8)4. (bd = 16)Let me try to find integers (a, b, c, d) that satisfy these equations. Since (bd = 16), possible pairs for ((b, d)) are (1,16), (2,8), (4,4), (-1,-16), (-2,-8), (-4,-4). Let's try positive pairs first.Trying (b = 4) and (d = 4):From equation 1: (a + c = 2)From equation 2: (ac + 4 + 4 = 4) ⇒ (ac = -4)From equation 3: (a*4 + c*4 = 8) ⇒ (4(a + c) = 8) ⇒ (a + c = 2), which is consistent with equation 1.So, we have (a + c = 2) and (ac = -4). Let's solve for (a) and (c).The quadratic equation (t^2 - 2t - 4 = 0) has solutions:[t = frac{2 pm sqrt{4 + 16}}{2} = frac{2 pm sqrt{20}}{2} = frac{2 pm 2sqrt{5}}{2} = 1 pm sqrt{5}]Hmm, these are not integers. Maybe (b) and (d) aren't both 4. Let's try another pair.Trying (b = 2) and (d = 8):From equation 1: (a + c = 2)From equation 2: (ac + 2 + 8 = 4) ⇒ (ac = -6)From equation 3: (a*8 + c*2 = 8) ⇒ (8a + 2c = 8). Since (a + c = 2), we can express (c = 2 - a) and substitute into equation 3:(8a + 2(2 - a) = 8) ⇒ (8a + 4 - 2a = 8) ⇒ (6a + 4 = 8) ⇒ (6a = 4) ⇒ (a = frac{2}{3})Not an integer, so discard this pair.Trying (b = 8) and (d = 2):From equation 1: (a + c = 2)From equation 2: (ac + 8 + 2 = 4) ⇒ (ac = -6)From equation 3: (a*2 + c*8 = 8) ⇒ (2a + 8c = 8). Again, (c = 2 - a), substitute:(2a + 8(2 - a) = 8) ⇒ (2a + 16 - 8a = 8) ⇒ (-6a + 16 = 8) ⇒ (-6a = -8) ⇒ (a = frac{4}{3})Still not an integer. Next pair.Trying (b = 1) and (d = 16):From equation 1: (a + c = 2)From equation 2: (ac + 1 + 16 = 4) ⇒ (ac = -13)From equation 3: (a*16 + c*1 = 8) ⇒ (16a + c = 8). With (c = 2 - a):(16a + (2 - a) = 8) ⇒ (15a + 2 = 8) ⇒ (15a = 6) ⇒ (a = frac{2}{5})Not an integer. Similarly, negative pairs might not help since the middle terms are positive. So, it seems this quartic doesn't factor nicely into quadratics with integer coefficients. Maybe it's irreducible over integers.Alternatively, perhaps it can be factored into a product involving the original factors. Wait, I already factored (x^5 - 32) as ((x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)), and the quartic doesn't factor further with integer coefficients. So, I'll leave it as is for now.Now, let's look at the other factor: (x^5 + 32). This is a sum of fifth powers. I recall that (a^5 + b^5) can be factored as ((a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4)). Let's apply that here.Let (a = x) and (b = 2), so (x^5 + 2^5) factors into ((x + 2)(x^4 - 2x^3 + 4x^2 - 8x + 16)).Again, let's check if this quartic (x^4 - 2x^3 + 4x^2 - 8x + 16) can be factored further. I'll try the same approach as before, attempting to factor it into two quadratics.Assume it factors as ((x^2 + ax + b)(x^2 + cx + d)). Multiplying out:[x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd]Comparing coefficients with (x^4 - 2x^3 + 4x^2 - 8x + 16), we have:1. (a + c = -2)2. (ac + b + d = 4)3. (ad + bc = -8)4. (bd = 16)Again, trying integer pairs for (b) and (d) that multiply to 16. Let's try (b = 4) and (d = 4):From equation 1: (a + c = -2)From equation 2: (ac + 4 + 4 = 4) ⇒ (ac = -4)From equation 3: (a*4 + c*4 = -8) ⇒ (4(a + c) = -8) ⇒ (a + c = -2), which is consistent.So, (a + c = -2) and (ac = -4). Solving the quadratic (t^2 + 2t - 4 = 0):[t = frac{-2 pm sqrt{4 + 16}}{2} = frac{-2 pm sqrt{20}}{2} = frac{-2 pm 2sqrt{5}}{2} = -1 pm sqrt{5}]Again, not integers. Trying (b = 2) and (d = 8):From equation 1: (a + c = -2)From equation 2: (ac + 2 + 8 = 4) ⇒ (ac = -6)From equation 3: (a*8 + c*2 = -8) ⇒ (8a + 2c = -8). With (c = -2 - a):(8a + 2(-2 - a) = -8) ⇒ (8a - 4 - 2a = -8) ⇒ (6a - 4 = -8) ⇒ (6a = -4) ⇒ (a = -frac{2}{3})Not an integer. Trying (b = 8) and (d = 2):From equation 1: (a + c = -2)From equation 2: (ac + 8 + 2 = 4) ⇒ (ac = -6)From equation 3: (a*2 + c*8 = -8) ⇒ (2a + 8c = -8). With (c = -2 - a):(2a + 8(-2 - a) = -8) ⇒ (2a - 16 - 8a = -8) ⇒ (-6a - 16 = -8) ⇒ (-6a = 8) ⇒ (a = -frac{4}{3})Still not an integer. Trying (b = 1) and (d = 16):From equation 1: (a + c = -2)From equation 2: (ac + 1 + 16 = 4) ⇒ (ac = -13)From equation 3: (a*16 + c*1 = -8) ⇒ (16a + c = -8). With (c = -2 - a):(16a + (-2 - a) = -8) ⇒ (15a - 2 = -8) ⇒ (15a = -6) ⇒ (a = -frac{2}{5})Not an integer. It seems this quartic also doesn't factor nicely into quadratics with integer coefficients. So, I'll leave it as is.Putting it all together, the original expression (x^{10} - 1024) factors as:[(x - 2)(x + 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)(x^4 - 2x^3 + 4x^2 - 8x + 16)]I think this is as far as I can factor it with integer coefficients. If I were to factor it completely over the complex numbers, I would need to factor each quartic into linear terms, but that involves finding roots which might not be rational or even real. Since the problem just says "completely factor," and without specifying the domain, I think factoring into polynomials with integer coefficients is sufficient.Wait, but I recall that (x^{10} - 1024) can also be factored using cyclotomic polynomials, which factor into minimal polynomials over integers. Let me see if I can factor it further using that approach.The expression (x^{10} - 2^{10}) can be written as (x^{10} - (2)^{10}). The general formula for (x^n - a^n) is:[x^n - a^n = prod_{d | n} Phi_d(x) cdot a^{n/d}]Where (Phi_d(x)) is the dth cyclotomic polynomial. For (n = 10), the divisors are 1, 2, 5, 10. So,[x^{10} - 2^{10} = Phi_1(x) cdot Phi_2(x) cdot Phi_5(x) cdot Phi_{10}(x) cdot 2^{10/1} cdot 2^{10/2} cdot 2^{10/5} cdot 2^{10/10}]Wait, that doesn't seem right. Let me correct that. Actually, the factorization is:[x^{10} - 2^{10} = prod_{d | 10} Phi_d(x) cdot 2^{10/d}]But I think I'm mixing up the formula. Let me recall that:[x^n - a^n = prod_{d | n} Phi_d(x) cdot a^{n/d}]But actually, it's:[x^n - a^n = prod_{d | n} Phi_d(x) cdot a^{n/d}]Wait, no, that's not quite accurate. The correct factorization is:[x^n - a^n = prod_{d | n} Phi_d(x) cdot a^{n/d}]But I think I'm overcomplicating it. Let me look up the cyclotomic polynomial factorization for (x^{10} - 1), which is similar, and then adjust for the 2^{10}.The 10th cyclotomic polynomial is (Phi_{10}(x) = x^4 - x^3 + x^2 - x + 1). But I'm not sure if that helps directly here. Maybe I should stick with the earlier factorization.Alternatively, since (x^{10} - 1024 = (x^5)^2 - (32)^2 = (x^5 - 32)(x^5 + 32)), and each of those can be factored further as I did before, leading to the four factors I have. I think that's the complete factorization over integers.So, to summarize, the complete factorization of (x^{10} - 1024) is:[(x - 2)(x + 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)(x^4 - 2x^3 + 4x^2 - 8x + 16)]I don't think I can factor it any further with integer coefficients. If I were to factor it completely over the complex numbers, I would need to factor each quartic into linear terms, but that would involve complex roots and might not be necessary here unless specified.

💡Alright, so I have this problem here with two propositions, p and q, and I need to find the range of the real number a based on whether p is true and whether the conjunction of p and q is false. Let me try to break this down step by step.First, let's understand what each proposition is saying.Proposition p: "For all x in [1,2], x² - a ≥ 0." So this means that for every x between 1 and 2, the expression x squared minus a is greater than or equal to zero. In other words, x² is always at least a in that interval.Proposition q: "There exists an x in ℝ such that x² + 2a x + 2 - a = 0." This is saying that there is at least one real number x that satisfies this quadratic equation. So, for q to be true, the quadratic equation must have real roots.Okay, so part (I) asks: If proposition p is true, find the range of the real number a.Let me focus on p first. Since p is saying that for all x in [1,2], x² - a ≥ 0, that implies that a must be less than or equal to x² for all x in [1,2]. So, to find the range of a, I need to find the minimum value of x² in the interval [1,2] because a has to be less than or equal to that minimum to satisfy the inequality for all x in [1,2].What's the minimum value of x² on [1,2]? Well, x² is a parabola opening upwards, and on the interval [1,2], it's increasing because the derivative 2x is positive for x > 0. So, the minimum occurs at x=1, which is 1² = 1. Therefore, the minimum value of x² on [1,2] is 1.So, for p to be true, a must be less than or equal to 1. That gives us the range of a as (-∞, 1].Wait, let me double-check that. If a is less than or equal to 1, then for all x in [1,2], x² is at least 1, so x² - a would be at least 1 - a, which is non-negative since a ≤ 1. Yes, that makes sense.So, part (I) is done, a ≤ 1.Now, part (II): If the proposition "p ∧ q" is false, find the range of the real number a.Okay, so "p ∧ q" being false means that it's not the case that both p and q are true. In logic, this is equivalent to saying that either p is false or q is false (or both). So, to find when "p ∧ q" is false, we need to consider when p is false or q is false.From part (I), we know that p is true when a ≤ 1. Therefore, p is false when a > 1.Now, let's figure out when q is false. q is the statement that there exists an x in ℝ such that x² + 2a x + 2 - a = 0. So, q being false would mean that there does not exist any real x satisfying that equation. In other words, the quadratic equation has no real roots.For a quadratic equation ax² + bx + c = 0, the discriminant D is b² - 4ac. If D < 0, there are no real roots.So, let's compute the discriminant for the equation x² + 2a x + (2 - a) = 0.Here, the coefficients are:- A = 1- B = 2a- C = 2 - aSo, discriminant D = B² - 4AC = (2a)² - 4 * 1 * (2 - a) = 4a² - 4(2 - a).Let me compute that:4a² - 4(2 - a) = 4a² - 8 + 4a.So, D = 4a² + 4a - 8.We can factor out a 4: D = 4(a² + a - 2).Now, let's factor the quadratic inside the parentheses: a² + a - 2.Looking for two numbers that multiply to -2 and add to 1. Those numbers are 2 and -1.So, a² + a - 2 = (a + 2)(a - 1).Therefore, D = 4(a + 2)(a - 1).So, q is false when D < 0, which is when 4(a + 2)(a - 1) < 0.Since 4 is positive, we can ignore it for the inequality, so we have (a + 2)(a - 1) < 0.To solve this inequality, let's find the critical points where the expression is zero: a = -2 and a = 1.We can use a sign chart to determine where the product is negative.- For a < -2: Let's pick a = -3. Then (-3 + 2) = -1, (-3 - 1) = -4. Multiplying two negatives gives positive. So, positive.- For -2 < a < 1: Let's pick a = 0. (0 + 2) = 2, (0 - 1) = -1. Multiplying positive and negative gives negative. So, negative.- For a > 1: Let's pick a = 2. (2 + 2) = 4, (2 - 1) = 1. Multiplying two positives gives positive.So, the inequality (a + 2)(a - 1) < 0 holds when -2 < a < 1.Therefore, q is false when -2 < a < 1.So, to recap:- p is false when a > 1.- q is false when -2 < a < 1.Therefore, "p ∧ q" is false when either p is false (a > 1) or q is false (-2 < a < 1). So, combining these two intervals, we get a ∈ (-2, 1) ∪ (1, ∞).Wait, but hold on. Is that correct? Let me think.When we have "p ∧ q" is false, it means that either p is false or q is false. So, the set of a where p is false is (1, ∞), and the set where q is false is (-2, 1). So, combining these two intervals, we get all a such that a > -2 except a = 1? Wait, no, because (1, ∞) and (-2, 1) together make (-2, ∞) except for a = 1? Wait, no, actually, it's (-2, 1) union (1, ∞), which is just (-2, ∞) without the point a=1? Wait, no, because (1, ∞) is separate from (-2,1). So, together, they form (-2, ∞). But wait, is that correct?Wait, no, because when a is between -2 and 1, q is false, and when a is greater than 1, p is false. So, the union is (-2,1) union (1, ∞). So, the total range is (-2, ∞) excluding a=1? Wait, no, because at a=1, let's check.At a=1, p is true because a=1 is the boundary. And for q, let's compute D at a=1.D = 4(a + 2)(a - 1) = 4(3)(0) = 0. So, discriminant is zero, meaning the quadratic has exactly one real root (a repeated root). So, q is true when a=1 because there exists an x (in fact, exactly one x) that satisfies the equation.Therefore, at a=1, p is true and q is true, so "p ∧ q" is true. Therefore, a=1 is not included in the range where "p ∧ q" is false.So, the range where "p ∧ q" is false is (-2,1) union (1, ∞). So, all real numbers a greater than -2 except a=1.Wait, but let me think again. If a is less than or equal to -2, what happens?If a ≤ -2, then q is true because D = 4(a + 2)(a - 1). If a ≤ -2, then (a + 2) ≤ 0 and (a - 1) ≤ -3, so their product is positive (negative times negative). Therefore, D > 0, so q is true.So, for a ≤ -2, q is true, and p is true when a ≤1, which includes a ≤ -2. So, for a ≤ -2, both p and q are true, so "p ∧ q" is true. Therefore, a ≤ -2 is not included in the range where "p ∧ q" is false.Therefore, the range where "p ∧ q" is false is when either p is false (a >1) or q is false (-2 < a <1). So, combining these, it's (-2,1) union (1, ∞).So, the final answer for part (II) is a ∈ (-2,1) ∪ (1, ∞).Let me just verify this with an example.Take a=0, which is in (-2,1). Then p is true because a=0 ≤1, and q is false because D=4(0 +2)(0 -1)=4*2*(-1)=-8 <0. So, "p ∧ q" is false, which is correct.Take a=2, which is in (1, ∞). Then p is false because a=2 >1, and q is true because D=4(2+2)(2-1)=4*4*1=16>0. So, "p ∧ q" is false because p is false, which is correct.Take a=-3, which is not in the range. Then p is true because a=-3 ≤1, and q is true because D=4(-3 +2)(-3 -1)=4*(-1)(-4)=16>0. So, "p ∧ q" is true, which is correct because a=-3 is not in our desired range.Take a=1, which is not in the range. Then p is true, and q is true because D=0, so "p ∧ q" is true, which is correct.Take a=-1, which is in (-2,1). Then p is true, and q is false because D=4(-1 +2)(-1 -1)=4*(1)*(-2)=-8 <0. So, "p ∧ q" is false, which is correct.Take a=1.5, which is in (1, ∞). Then p is false, and q is true because D=4(1.5 +2)(1.5 -1)=4*(3.5)*(0.5)=7>0. So, "p ∧ q" is false because p is false, which is correct.So, all these test cases seem to confirm that the range is indeed (-2,1) union (1, ∞).Therefore, I think I've got it right.

💡First, identify the original point ((-2, 3)).According to the mapping ( f ), the corresponding point is calculated as ((x + y, xy)).Calculate ( x + y ):[-2 + 3 = 1]Calculate ( xy ):[-2 times 3 = -6]Therefore, the corresponding point under the mapping ( f ) is ((1, -6)).

💡Alright, so I have this probability problem about a calligraphy competition with five students: A, B, C, D, and E. The problem says that A and B went to inquire about the results, and the organizing committee gave them some information. Based on that, I need to find the probability that C and D will ultimately win the top two places.First, let me try to understand the information given:1. To A, they said, "Unfortunately, neither you nor B won the championship." So, this means that A is not first, and B is also not first. So, the first place must be either C, D, or E.2. To B, they said, "You certainly are not the worst among the five." So, B is not last, meaning B is not fifth. So, B can be in positions 1, 2, 3, or 4. But from the first statement, we already know B is not first, so B must be in positions 2, 3, or 4.So, summarizing:- A is not first.- B is not first and not last, so B is in positions 2, 3, or 4.- Therefore, the first place must be C, D, or E.Now, the question is about the probability that C and D will win the top two places. So, we need to find the number of possible arrangements where C and D are first and second, divided by the total number of possible arrangements given the constraints.Let me think about how to approach this.First, let's figure out the total number of possible rankings given the constraints.We have five positions: 1st, 2nd, 3rd, 4th, and 5th.Constraints:1. A is not first.2. B is not first and not last.So, let's consider the possible positions for A and B.Since A is not first, A can be in positions 2, 3, 4, or 5.But B is not first and not last, so B can be in positions 2, 3, or 4.So, A and B are both in positions 2, 3, 4, or 5, but B cannot be in position 5.Therefore, A can be in 2, 3, 4, or 5, and B can be in 2, 3, or 4.But A and B are distinct, so we need to consider their positions without overlapping.Let me try to count the total number of possible arrangements.First, let's consider the positions for A and B.Case 1: A is in position 5 (last place).Then, B can be in positions 2, 3, or 4. So, 3 choices for B.Once A and B are placed, the remaining three positions (1, 2, 3, 4 excluding A and B's positions) are to be filled by C, D, and E.So, for each choice of B, the remaining three positions can be filled in 3! = 6 ways.Therefore, total arrangements in this case: 3 * 6 = 18.Case 2: A is not in position 5.So, A can be in positions 2, 3, or 4.Similarly, B can be in positions 2, 3, or 4, but B cannot be in the same position as A.So, for each position of A (2, 3, 4), B has 2 remaining positions.Therefore, number of ways to place A and B: 3 (positions for A) * 2 (positions for B) = 6.Once A and B are placed, the remaining three positions (including position 1) are to be filled by C, D, and E.So, the number of ways to arrange C, D, and E is 3! = 6.Therefore, total arrangements in this case: 6 * 6 = 36.So, total number of possible arrangements is 18 (Case 1) + 36 (Case 2) = 54.Now, we need to find the number of arrangements where C and D are in the top two places.So, C and D must be in positions 1 and 2.But we have constraints:- Position 1 cannot be A or B, so it must be C, D, or E.- Position 2 can be anyone except A or B, but in this case, we want C and D in positions 1 and 2.So, let's consider the favorable cases where C and D are in positions 1 and 2.There are two possibilities:1. C is first, D is second.2. D is first, C is second.So, two possibilities for the top two positions.Now, for each of these, we need to arrange the remaining three students (A, B, E) in the remaining three positions (3, 4, 5), with the constraints:- A cannot be first, but since A is already in positions 3, 4, or 5, that's fine.- B cannot be last (position 5).So, let's consider each case.Case 1: C is first, D is second.Then, we have to arrange A, B, and E in positions 3, 4, and 5, with B not in position 5.So, how many ways can we arrange A, B, E in positions 3, 4, 5 with B not in position 5?Total arrangements without any constraints: 3! = 6.Number of arrangements where B is in position 5: 2! = 2 (since A and E can be in positions 3 and 4).Therefore, number of valid arrangements: 6 - 2 = 4.Similarly, for Case 2: D is first, C is second.Same reasoning applies.We have to arrange A, B, E in positions 3, 4, 5 with B not in position 5.So, again, 4 valid arrangements.Therefore, total favorable arrangements: 2 (for C first and D first) * 4 = 8.Therefore, the probability is favorable / total = 8 / 54.Simplify this fraction: divide numerator and denominator by 2, we get 4 / 27.So, the probability is 4/27.Looking back at the options:A: 2/9B: 4/9C: 8/27D: 4/27So, the correct answer is D: 4/27.

💡Okay, so I have this problem here where I need to find the value of ( k ) such that the infinite series[ 2 + frac{2 + k}{3} + frac{2 + 2k}{3^2} + frac{2 + 3k}{3^3} + dotsb = 7 ]equals 7. Hmm, infinite series can be tricky, but let me see if I can break it down step by step.First, I notice that each term in the series has a numerator that's a linear expression in terms of ( k ), and the denominator is a power of 3. The numerators start at 2 and then each subsequent term adds another ( k ). So, the first term is 2, the second is ( 2 + k ), the third is ( 2 + 2k ), and so on. That seems like a pattern where the coefficient of ( k ) increases by 1 each time, while the constant term remains 2.I also notice that each term is divided by an increasing power of 3. The first term is divided by ( 3^0 = 1 ), the second by ( 3^1 = 3 ), the third by ( 3^2 = 9 ), etc. So, the general term of the series seems to be:[ frac{2 + (n-1)k}{3^{n-1}} ]where ( n ) starts at 1 and goes to infinity. So, the series can be written as:[ sum_{n=1}^{infty} frac{2 + (n-1)k}{3^{n-1}} ]But I need to find ( k ) such that this sum equals 7. Hmm, maybe I can split this series into two separate series: one involving the constant term 2 and the other involving the term with ( k ). Let me try that.So, splitting the series:[ sum_{n=1}^{infty} frac{2}{3^{n-1}} + sum_{n=1}^{infty} frac{(n-1)k}{3^{n-1}} = 7 ]Let me denote the first series as ( S_1 ) and the second series as ( S_2 ):[ S_1 = sum_{n=1}^{infty} frac{2}{3^{n-1}} ][ S_2 = sum_{n=1}^{infty} frac{(n-1)k}{3^{n-1}} ]So, the equation becomes:[ S_1 + S_2 = 7 ]Now, let me compute ( S_1 ) first. ( S_1 ) is a geometric series where each term is ( frac{2}{3^{n-1}} ). The general form of a geometric series is ( sum_{n=0}^{infty} ar^n ), which sums to ( frac{a}{1 - r} ) if ( |r| < 1 ).But in ( S_1 ), the index starts at ( n = 1 ), so let me adjust it to start at ( n = 0 ):[ S_1 = sum_{n=1}^{infty} frac{2}{3^{n-1}} = 2 sum_{n=1}^{infty} left( frac{1}{3} right)^{n-1} ]Let me make a substitution: let ( m = n - 1 ). Then, when ( n = 1 ), ( m = 0 ), and as ( n ) approaches infinity, so does ( m ). So,[ S_1 = 2 sum_{m=0}^{infty} left( frac{1}{3} right)^m ]This is a geometric series with first term ( a = 1 ) and common ratio ( r = frac{1}{3} ). The sum is:[ S_1 = 2 times frac{1}{1 - frac{1}{3}} = 2 times frac{3}{2} = 3 ]Okay, so ( S_1 = 3 ). Now, moving on to ( S_2 ):[ S_2 = sum_{n=1}^{infty} frac{(n-1)k}{3^{n-1}} ]Again, let me adjust the index to start at ( m = 0 ). Let ( m = n - 1 ), so when ( n = 1 ), ( m = 0 ), and as ( n ) approaches infinity, so does ( m ). Then,[ S_2 = sum_{m=0}^{infty} frac{m k}{3^m} ]So, ( S_2 = k sum_{m=0}^{infty} frac{m}{3^m} )Hmm, this looks like a standard series. I remember that the sum ( sum_{m=0}^{infty} m r^m ) is equal to ( frac{r}{(1 - r)^2} ) for ( |r| < 1 ). Let me verify that.Yes, the formula for the sum of ( sum_{m=0}^{infty} m r^m ) is indeed ( frac{r}{(1 - r)^2} ). So, in this case, ( r = frac{1}{3} ), so:[ sum_{m=0}^{infty} frac{m}{3^m} = frac{frac{1}{3}}{(1 - frac{1}{3})^2} ]Let me compute that:First, compute the denominator:( 1 - frac{1}{3} = frac{2}{3} )So, ( (1 - frac{1}{3})^2 = left( frac{2}{3} right)^2 = frac{4}{9} )Then, the numerator is ( frac{1}{3} ), so:[ frac{frac{1}{3}}{frac{4}{9}} = frac{1}{3} times frac{9}{4} = frac{3}{4} ]Therefore,[ sum_{m=0}^{infty} frac{m}{3^m} = frac{3}{4} ]So, ( S_2 = k times frac{3}{4} = frac{3k}{4} )Now, going back to the original equation:[ S_1 + S_2 = 7 ][ 3 + frac{3k}{4} = 7 ]Let me solve for ( k ):Subtract 3 from both sides:[ frac{3k}{4} = 7 - 3 ][ frac{3k}{4} = 4 ]Multiply both sides by ( frac{4}{3} ):[ k = 4 times frac{4}{3} ][ k = frac{16}{3} ]Wait, that can't be right. Wait, 4 times 4/3 is 16/3? Wait, 4 times 4 is 16, divided by 3 is 16/3. Yeah, that's correct.But let me double-check my steps to make sure I didn't make a mistake.First, I split the series into ( S_1 ) and ( S_2 ). ( S_1 ) was the sum of 2 over 3^{n-1}, which I converted into a geometric series starting at m=0, which gave me 3. That seems correct.Then, ( S_2 ) was the sum of (n-1)k over 3^{n-1}, which I converted into a sum starting at m=0, which gave me k times the sum of m over 3^m. Then, using the formula for the sum of m r^m, which is r/(1 - r)^2, I plugged in r = 1/3, which gave me 3/4. So, ( S_2 = (3/4)k ). Then, adding ( S_1 + S_2 = 3 + (3/4)k = 7 ). Solving for k, I subtract 3 to get (3/4)k = 4, then multiply both sides by 4/3 to get k = 16/3.Wait, 16/3 is approximately 5.333... Let me plug that back into the original series to see if it sums to 7.But wait, maybe I should check my steps again.Wait, when I split the series, I had:[ sum_{n=1}^{infty} frac{2 + (n-1)k}{3^{n-1}} = sum_{n=1}^{infty} frac{2}{3^{n-1}} + sum_{n=1}^{infty} frac{(n-1)k}{3^{n-1}} ]Yes, that seems correct.Then, ( S_1 = 2 times sum_{n=1}^{infty} frac{1}{3^{n-1}} ), which is 2 times the sum from m=0 to infinity of (1/3)^m, which is 2 times 1/(1 - 1/3) = 2 times 3/2 = 3. Correct.Then, ( S_2 = k times sum_{m=0}^{infty} frac{m}{3^m} ). The sum of m r^m from m=0 is r/(1 - r)^2, so with r=1/3, it's (1/3)/(1 - 1/3)^2 = (1/3)/(4/9) = (1/3)*(9/4) = 3/4. So, ( S_2 = (3/4)k ). Correct.So, 3 + (3/4)k = 7. Subtract 3: (3/4)k = 4. Multiply both sides by 4/3: k = 16/3. That seems correct.Wait, but let me think again. When I split the series, I assumed that both series converge, which they do because the common ratio is 1/3, which is less than 1. So, the geometric series converges, and the sum involving m r^m also converges because |r| < 1.So, I think my steps are correct. Therefore, k should be 16/3.But just to be thorough, let me consider another approach to verify.Another way to approach this problem is to consider the series as a combination of two separate geometric series. Let me write the original series again:[ 2 + frac{2 + k}{3} + frac{2 + 2k}{3^2} + frac{2 + 3k}{3^3} + dotsb ]Let me factor out the 2 and the k:[ 2 left(1 + frac{1}{3} + frac{1}{3^2} + frac{1}{3^3} + dotsb right) + k left( frac{1}{3} + frac{2}{3^2} + frac{3}{3^3} + dotsb right) ]Yes, that's the same as splitting into ( S_1 ) and ( S_2 ). So, the first part is 2 times the geometric series sum, which is 2*(1/(1 - 1/3)) = 2*(3/2) = 3. The second part is k times the sum of n/(3^n) starting from n=1.Wait, in my previous approach, I had the sum starting from m=0, but here it's starting from n=1. Let me check that.In the original series, the first term is 2, which is 2*(1/3^0). The next term is (2 + k)/3, which is 2*(1/3^1) + k*(1/3^1). The third term is (2 + 2k)/3^2, which is 2*(1/3^2) + 2k*(1/3^2). So, when I factor out the 2 and k, I get:2*(1 + 1/3 + 1/3^2 + 1/3^3 + ...) + k*(1/3 + 2/3^2 + 3/3^3 + ...)So, the first series is 2 times the geometric series starting at n=0, which sums to 2*(3/2) = 3.The second series is k times the sum from n=1 to infinity of n/(3^n). Wait, earlier I had m starting at 0, but here n starts at 1. So, let me compute that sum.I know that the sum from n=1 to infinity of n r^n is r/(1 - r)^2. So, in this case, r = 1/3, so the sum is (1/3)/(1 - 1/3)^2 = (1/3)/(4/9) = 3/4. So, the second series is k*(3/4).Therefore, the total sum is 3 + (3/4)k = 7, leading to k = 16/3, same as before.Wait, so that's consistent with my previous result. So, I think that's correct.Alternatively, another approach is to use the method of multiplying the series by the common ratio and subtracting. Let me try that method as well to confirm.Let me denote the original series as S:[ S = 2 + frac{2 + k}{3} + frac{2 + 2k}{3^2} + frac{2 + 3k}{3^3} + dotsb ]We are told that S = 7.Let me multiply both sides by 3:[ 3S = 6 + (2 + k) + frac{2 + 2k}{3} + frac{2 + 3k}{3^2} + dotsb ]Now, subtract the original series S from this equation:[ 3S - S = 6 + (2 + k) + frac{2 + 2k}{3} + frac{2 + 3k}{3^2} + dotsb - left( 2 + frac{2 + k}{3} + frac{2 + 2k}{3^2} + frac{2 + 3k}{3^3} + dotsb right) ]Simplifying the left side:[ 2S = 6 + (2 + k) + frac{2 + 2k}{3} + frac{2 + 3k}{3^2} + dotsb - 2 - frac{2 + k}{3} - frac{2 + 2k}{3^2} - frac{2 + 3k}{3^3} - dotsb ]Let me look at the right side term by term.First, 6 remains.Then, (2 + k) - 2 = k.Next, (frac{2 + 2k}{3} - frac{2 + k}{3} = frac{(2 + 2k) - (2 + k)}{3} = frac{k}{3}).Similarly, (frac{2 + 3k}{3^2} - frac{2 + 2k}{3^2} = frac{(2 + 3k) - (2 + 2k)}{3^2} = frac{k}{3^2}).This pattern continues, so each subsequent term is (frac{k}{3^n}) for n starting at 1.Therefore, the right side becomes:[ 6 + k + frac{k}{3} + frac{k}{3^2} + frac{k}{3^3} + dotsb ]This is a geometric series with first term k and common ratio 1/3. The sum of this series is:[ 6 + k left( 1 + frac{1}{3} + frac{1}{3^2} + frac{1}{3^3} + dotsb right) ]The series in the parentheses is a geometric series with sum ( frac{1}{1 - 1/3} = frac{3}{2} ).So, the right side becomes:[ 6 + k times frac{3}{2} ]Therefore, we have:[ 2S = 6 + frac{3k}{2} ]But we know that S = 7, so:[ 2 times 7 = 6 + frac{3k}{2} ][ 14 = 6 + frac{3k}{2} ][ 14 - 6 = frac{3k}{2} ][ 8 = frac{3k}{2} ][ k = 8 times frac{2}{3} ][ k = frac{16}{3} ]So, again, we get k = 16/3. This confirms the result from the previous methods.Therefore, I'm confident that the value of k is 16/3.